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https://mathphysicsexpert.com/what-will-its-kinetic-energy-be-right-before-it-hits-the-ground/ | 1,669,988,725,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00487.warc.gz | 431,650,943 | 11,347 | # what will its kinetic energy be right before it hits the ground?
A book falls off of a 2.2m high table. If the book weighs 0.75kg, what will its kinetic energy be right before it hits the ground?
g=9.8m/ | 60 | 207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2022-49 | latest | en | 0.845851 |
https://www.physicsforums.com/threads/calculus-ii-improper-integral-problem.656135/ | 1,534,903,165,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219242.93/warc/CC-MAIN-20180822010128-20180822030128-00546.warc.gz | 967,930,871 | 12,961 | # Homework Help: Calculus II - Improper Integral Problem
1. Dec 1, 2012
### BaxterCorner
1. The problem statement, all variables and given/known data
Evaluate the integral: ∫(0 to ∞) [dv/((1+v^2)(1+tan^-1(v))]
2. Relevant equations
U-substitution, taking limit to evaluate improper integrals
3. The attempt at a solution
http://imgur.com/CjkRF
As you can see in the image, I try u-substitution and then take the integral. I end up with ln(0), though, because arctan(0) = 0. The correct answer is ln(1 + ∏/2), but I'm not sure how to get there.
Last edited: Dec 1, 2012
2. Dec 1, 2012
### LCKurtz
With your substitution the denominator is $1+u$, not just $u$. It works either way, but I would suggest the substitution $u=1+\arctan v$ in the first place.
3. Dec 1, 2012
### BaxterCorner
Sorry, I intended to write u = 1 + arctan(v), not u = arctan(v). The 1 goes to zero either way though, so I still have the same problem of getting ln(0).
4. Dec 1, 2012
### BaxterCorner
Nevermind, I see what you're saying, that would change the bounds. Thanks! | 330 | 1,062 | {"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} | 3.125 | 3 | CC-MAIN-2018-34 | latest | en | 0.851063 |
http://design.tutsplus.com/tutorials/learn-how-to-create-realistic-vector-fire--vector-22 | 1,472,206,658,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982295424.4/warc/CC-MAIN-20160823195815-00259-ip-10-153-172-175.ec2.internal.warc.gz | 64,991,203 | 18,230 | # Learn How to Create Realistic Vector Fire
Difficulty:IntermediateLanguages:
For this tutorial, I wanted to do something simpler then my previous tutorials. This is not just a beginners tutorial though. This tutorial shows the simplicity and power that Illustrator possesses. In turn, this tutorial's techniques can be applied to many other elements other than fire.
### Final Image Preview
First, let's take a look at the image we'll be creating. Below is the completed illustration to see what you'll be working toward.
### Step 1
Create a document that is 8.5 inches by 11 inches. Double-click on the Pencil Tool from the Tools Panel to bring up the Pencil Tool Preferences dialog. In the dialog change the Fidelity to .5.
### Step 2
With the Pencil Tool, draw a flame shape that is roughly 8.5 inches tall by 5 inches wide. Then fill it with black. Make sure that the Paths are closed by pressing Alt before finishing drawing the shape.
### Step 3
Draw another flame shape within the first one, making sure not to overlap the shapes (sometimes this causes the Blend not to work properly). Once you have finished drawing and closing the shape, fill it with an orange color.
### Step 4
Draw another flame shape within the previous one, and fill it with a yellow color. Keep in mind you want the path closed and not overlapping the other paths.
### Step 5
Draw another flame shape within the previous one and fill with a light orange.
### Step 6
Draw another flame shape within the previous one and fill with a light yellow.
### Step 7
Draw another flame shape within the previous one and fill with white.
### Step 8
Now that all the flame shapes are drawn, select all the flames and create a Blend by going to Object > Blend > Make.
### Step 9
Create a rectangle the same size as your document. Fill it with your first flame shape's black color. Then send it behind the flame blend by going to Object > Arrange > Send to Back.
### Final Image
All done! Wasn't that easy! Again with this tutorial, I wanted to stress the ease with which you can create compelling elements for illustrations. With these same techniques you can create realistic clouds, water, or whatever. These are great techniques to start experimenting with. | 484 | 2,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-36 | latest | en | 0.922304 |
https://itechproduct.com/exploratory-data-analysis-unraveling-the-story-within-your-dataset-by-deepak-chopra-talking-data-science-jul-2023/ | 1,709,423,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00543.warc.gz | 320,692,652 | 15,775 | # Exploratory Data Analysis: Unraveling the Story Within Your Dataset | by Deepak Chopra | Talking Data Science | Jul, 2023
Exploratory Data Analysis, as the name suggests is analysis to explore the data. It consists of a number of components; neither are all essential all the time, nor all of them have equal importance. Below, I am listing down a few components based on my experience.
Please note that it is by no means an exhaustive list, but a guiding framework.
## 1. Understand the lay of the land.
You don’t know what you don’t know — but you can explore!
The first and foremost thing to do is to get the feel of the data — look at the data entries, eye-ball the column values. How many rows, columns you have.
• a retailer dataset might tell you — Mr X visited store#2000 on the 01st of Aug 2023 and purchased a can of Coke and one pack of Walker Crisps
• a social media dataset might tell you — Mrs Y logged onto the social networking website at 09:00 am on the 3rd of June and browsed A, B, and C sections, searched for her friend Mr A and then logged out after 20 mins.
It’s beneficial to get the business context of the data you have, knowing the source and mechanism of data collection; for e.g. survey data vs. digitally collected data etc.).
## 2. Double-click into variables
Variables are the talking tongue of a dataset, they are continuously talking to you. You just need to ask the right questions and listen carefully.
– What do the variables mean/represent?
– Are the variables continuous or categorical? .. Any inherent order?
– What are the possible values they can take?
→ ACTION::
• For continuous variables — check distributions using histograms, box-plots and carefully study the mean, median, standard deviations etc.
• For categorical / ordinal variables — find out their unique values, and do a frequency table checking the most / least occurring ones.
You may or may not understand all variables, labels and values — but try to get as much information as you can
## 3. Look for patterns/relationships in your data
Through EDA, you can discover patterns, trends, and relationships within the data.
– Do you have any prior assumptions/hypothesis of relationships between variables?
– Any business reason for some variables to be related to one another?
– Do variables follow any particular distributions?
Data Visualisation techniques, summaries, and correlation analysis help reveal hidden patterns that may not be apparent at first glance. Understanding these patterns can provide valuable insights for decision-making or hypothesis generation.
→ ACTION::
Think visual bi-variate analysis.
• In case of continuous variables — use scatter plots, create correlation matrix / heat maps etc.
• A mixture of continuous and ordinal/categorical variables — Consider plotting bar or pie charts, and create good-old contingency tables to visualise the co-occurrence.
EDA allows you to validate statistical assumptions, such as normality, linearity, or independence, for analysis or data modelling.
## 4. Detecting anomalies.
Here’s your chance to become Sherlock Holmes on your data and look for anything out of the ordinary! Ask yourself::
## – Are there any duplicate entries in the dataset?
Duplicates are entries that represent the same sample point multiple times. Duplicates are not useful in most cases as they do not give any additional information. They might be the result of an error and can mess up your mean, median and other statistics.
→ Check with your stakeholders and remove such errors from your data.
## – Labelling errors for categorical variables?
Look for unique values for categorical variables and create a frequency chart. Look for mis-spellings and labels that might represent similar things?
## – Do some variables have Missing Values?
This can happen to both numeric and categorical variables. Check if
• Are there rows which have missing values for a lot of variables (columns)? This means there are data points which have blanks across the majority of columns → they are not very useful, we may need to drop them.
• Are there variables (or columns) which have missing values across multiple rows? This means there are variables which do not have any values/labels across most data points → they cannot add much to our understanding, we may need to drop them.
ACTION::
– Count the proportion of NULL or missing values for all variables. Variables with more than 15%-20% should make you suspicious.
– Filter out rows with missing values for a column and check how the rest of the columns look. Is it that the majority of columns have missing values together ?.. is there a pattern?
## – Are there Outliers in my dataset?
Outlier detection is about identifying data points that do not fit the norm. you may see very high or extremely low values for certain numerical variables, or a high/low frequency for categorical class variables.
• What seems an outlier can be a data error.
While outliers are data points that are unusual for a given feature distribution, unwanted entries or recording errors are samples that shouldn’t be there in the first place.
• What seems an outlier can just be an outlier.
In other cases, we might just have data points with extreme values and perfectly fine reasoning behind them.
ACTION::
Study the histograms, scatter plots, and frequency bar charts to understand if there are a few data points which are farther from the rest. Think through:
– Can they be true and take these extreme values?
– Is there a business reasoning or justification for these extremities
– Would they add value to your analysis at a later stage
## 5. Data Cleaning.
Data cleaning refers to the process of removing unwanted variables and values from your dataset and getting rid of any irregularities in it. These anomalies can disproportionately skew the data and hence adversely affect the results of our analysis from this dataset.
Remember: Garbage In, Garbage Out
## – Course correct your data.
• Remove the duplicate entries if you find any, missing values and outliers — which do not add value to your dataset. Get rid of unnecessary rows/ columns.
• Correct any mis-spellings, or mis-labelling you observe in the data.
• Any data errors you spot which are not adding value to the data also need to be removed.
## – Cap Outliers or let them be.
• In some data modelling scenarios, we may need to cap outliers at either end. Capping is often done at the 99th/95th percentile for the higher end or the 1st/5th percentile for the lower-end capping.
## – Treat Missing Values.
We generally drop data points (rows) with a lot of missing values across variables. Similarly, we drop variables (columns) which have missing values across a lot of data points
If there are a few missing values we might look to plug those gaps or just let them be as it is.
• For continuous variables with missing values, we can plug them by using mean or median values (maybe across a particular strata)
• For categorical missing values, we might assign the most used ‘class’ or maybe create a new ‘not defined’ class.
## – Data enrichment.
Based on the needs of the future analysis, you can add more features (variables) to your dataset; such as (not restricted to)
• Creating binary variables indicating the presence or absence of something.
• Creating additional labels/classes by using IF-THEN-ELSE clauses.
• Scale or encode your variables as per your future analytics needs.
• Combine two or more variables — use arrange of mathematical functions like sum, difference, mean, log and many other transformations. | 1,566 | 7,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-10 | latest | en | 0.902456 |
https://cliffsnotes.com/study-guides/economics/theory-of-the-consumer/consumer-surplus | 1,632,804,587,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060201.9/warc/CC-MAIN-20210928032425-20210928062425-00219.warc.gz | 211,827,562 | 18,172 | ## Consumer Surplus
The difference between the maximum price that consumers are willing to pay for a good and the market price that they actually pay for a good is referred to as the consumer surplus. The determination of consumer surplus is illustrated in Figure , which depicts the market demand curve for some good.
The market price is \$5, and the equilibrium quantity demanded is 5 units of the good. The market demand curve reveals that consumers are willing to pay at least \$9 for the first unit of the good, \$8 for the second unit, \$7 for the third unit, and \$6 for the fourth unit.
However, they can purchase 5 units of the good for just \$5 per unit. Their surplus from the first unit purchased is therefore \$9 ‐ \$5 = \$4. Similarly, their surpluses from the second, third, and fourth units purchased are \$3, \$2, and \$1, respectively. These surpluses are illustrated by the vertical bars drawn in Figure . The sum total of these surpluses is the consumer surplus:
The value \$10, however, is only a crude approximation of the true consumer surplus in this example. The true consumer surplus is given by the area below the market demand curve and above the market price. This area consists of a triangle with base of length 5 and height of length 5. Applying the rule for the area of a triangle—one half the base multiplied by height—one finds that the value of the consumer surplus in this example is actually 12.5.
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A18ACD436D5A3997E3DA2573E3FD792A | 338 | 1,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.942327 |
https://www.free-online-converters.com/stones/stones-into-pounds-troy.html | 1,653,633,819,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662636717.74/warc/CC-MAIN-20220527050925-20220527080925-00337.warc.gz | 875,471,762 | 9,067 | Free Online Converters > Convert Stones Into Pounds (troy)
Here you can Convert units of Stones to Pounds (troy) units, find all information about Stones. So, enter your unit's value in Left Column like Stones(if you use standard resolution on most non-HD laptops. FULL HD resolution starts at 1920 x 1080). Otherwise, if you use a lower value, enter the value in the box above. The Result / another converted unit value shell appears in the Left or below Column.
# Convert Stones Into Pounds (troy)
Stones
Swap
Pounds (troy)
Increase or Decrease Decimal:
Convert Stones Into Pounds (troy) ,and more. Also, explore many other unit converters or learn more about Mass unit conversions, How mamy Stones in Pounds (troy)
TAGS:
Stones , Pounds troy , Stones to Pounds troy , Stones into Pounds troy , Stones in Pounds troy , How many Stones in many Pounds troy , How to convert Stones to Pounds troy online just in one Second ,
wikipedia.org lexico.com dictionary.com wikipedia
##### conversion Table / conversion Chart
1 Stones = 17.0139 Pounds (troy)
2 Stones = 34.0278 Pounds (troy)
3 Stones = 51.0417 Pounds (troy)
4 Stones = 68.0556 Pounds (troy)
5 Stones = 85.0695 Pounds (troy)
6 Stones = 102.0834 Pounds (troy)
7 Stones = 119.0973 Pounds (troy)
8 Stones = 136.1112 Pounds (troy)
9 Stones = 153.1251 Pounds (troy)
10 Stones = 170.139 Pounds (troy)
11 Stones = 187.1529 Pounds (troy)
12 Stones = 204.1668 Pounds (troy)
13 Stones = 221.1807 Pounds (troy)
14 Stones = 238.1946 Pounds (troy)
15 Stones = 255.2085 Pounds (troy)
16 Stones = 272.2224 Pounds (troy)
17 Stones = 289.2363 Pounds (troy)
18 Stones = 306.2502 Pounds (troy)
19 Stones = 323.2641 Pounds (troy)
20 Stones = 340.278 Pounds (troy)
21 Stones = 357.2919 Pounds (troy)
22 Stones = 374.3058 Pounds (troy)
23 Stones = 391.3197 Pounds (troy)
24 Stones = 408.3336 Pounds (troy)
25 Stones = 425.3475 Pounds (troy)
26 Stones = 442.3614 Pounds (troy)
27 Stones = 459.3753 Pounds (troy)
28 Stones = 476.3892 Pounds (troy)
29 Stones = 493.4031 Pounds (troy)
30 Stones = 510.417 Pounds (troy)
31 Stones = 527.4309 Pounds (troy)
32 Stones = 544.4448 Pounds (troy)
33 Stones = 561.4587 Pounds (troy)
34 Stones = 578.4726 Pounds (troy)
35 Stones = 595.4865 Pounds (troy)
36 Stones = 612.5004 Pounds (troy)
37 Stones = 629.5143 Pounds (troy)
38 Stones = 646.5282 Pounds (troy)
39 Stones = 663.5421 Pounds (troy)
40 Stones = 680.556 Pounds (troy)
41 Stones = 697.5699 Pounds (troy)
42 Stones = 714.5838 Pounds (troy)
43 Stones = 731.5977 Pounds (troy)
44 Stones = 748.6116 Pounds (troy)
45 Stones = 765.6255 Pounds (troy)
46 Stones = 782.6394 Pounds (troy)
47 Stones = 799.6533 Pounds (troy)
48 Stones = 816.6672 Pounds (troy)
49 Stones = 833.6811 Pounds (troy)
50 Stones = 850.695 Pounds (troy)
50 Stones = 850.695 Pounds (troy)
51 Stones = 867.7089 Pounds (troy)
52 Stones = 884.7228 Pounds (troy)
53 Stones = 901.7367 Pounds (troy)
54 Stones = 918.7506 Pounds (troy)
55 Stones = 935.7645 Pounds (troy)
56 Stones = 952.7784 Pounds (troy)
57 Stones = 969.7923 Pounds (troy)
58 Stones = 986.8062 Pounds (troy)
59 Stones = 1003.8201 Pounds (troy)
60 Stones = 1020.834 Pounds (troy)
61 Stones = 1037.8479 Pounds (troy)
62 Stones = 1054.8618 Pounds (troy)
63 Stones = 1071.8757 Pounds (troy)
64 Stones = 1088.8896 Pounds (troy)
65 Stones = 1105.9035 Pounds (troy)
66 Stones = 1122.9174 Pounds (troy)
67 Stones = 1139.9313 Pounds (troy)
68 Stones = 1156.9452 Pounds (troy)
69 Stones = 1173.9591 Pounds (troy)
70 Stones = 1190.973 Pounds (troy)
71 Stones = 1207.9869 Pounds (troy)
72 Stones = 1225.0008 Pounds (troy)
73 Stones = 1242.0147 Pounds (troy)
74 Stones = 1259.0286 Pounds (troy)
75 Stones = 1276.0425 Pounds (troy)
76 Stones = 1293.0564 Pounds (troy)
77 Stones = 1310.0703 Pounds (troy)
78 Stones = 1327.0842 Pounds (troy)
79 Stones = 1344.0981 Pounds (troy)
80 Stones = 1361.112 Pounds (troy)
81 Stones = 1378.1259 Pounds (troy)
82 Stones = 1395.1398 Pounds (troy)
83 Stones = 1412.1537 Pounds (troy)
84 Stones = 1429.1676 Pounds (troy)
85 Stones = 1446.1815 Pounds (troy)
86 Stones = 1463.1954 Pounds (troy)
87 Stones = 1480.2093 Pounds (troy)
88 Stones = 1497.2232 Pounds (troy)
89 Stones = 1514.2371 Pounds (troy)
90 Stones = 1531.251 Pounds (troy)
91 Stones = 1548.2649 Pounds (troy)
92 Stones = 1565.2788 Pounds (troy)
93 Stones = 1582.2927 Pounds (troy)
94 Stones = 1599.3066 Pounds (troy)
95 Stones = 1616.3205 Pounds (troy)
96 Stones = 1633.3344 Pounds (troy)
97 Stones = 1650.3483 Pounds (troy)
98 Stones = 1667.3622 Pounds (troy)
99 Stones = 1684.3761 Pounds (troy)
100 Stones = 1701.39 Pounds (troy)
101 Stones = 1718.4039 Pounds (troy)
102 Stones = 1735.4178 Pounds (troy)
103 Stones = 1752.4317 Pounds (troy)
104 Stones = 1769.4456 Pounds (troy)
105 Stones = 1786.4595 Pounds (troy)
106 Stones = 1803.4734 Pounds (troy)
107 Stones = 1820.4873 Pounds (troy)
108 Stones = 1837.5012 Pounds (troy)
109 Stones = 1854.5151 Pounds (troy)
110 Stones = 1871.529 Pounds (troy)
111 Stones = 1888.5429 Pounds (troy)
112 Stones = 1905.5568 Pounds (troy)
113 Stones = 1922.5707 Pounds (troy)
114 Stones = 1939.5846 Pounds (troy)
115 Stones = 1956.5985 Pounds (troy)
116 Stones = 1973.6124 Pounds (troy)
117 Stones = 1990.6263 Pounds (troy)
118 Stones = 2007.6402 Pounds (troy)
119 Stones = 2024.6541 Pounds (troy)
120 Stones = 2041.668 Pounds (troy)
121 Stones = 2058.6819 Pounds (troy)
122 Stones = 2075.6958 Pounds (troy)
123 Stones = 2092.7097 Pounds (troy)
124 Stones = 2109.7236 Pounds (troy)
125 Stones = 2126.7375 Pounds (troy)
126 Stones = 2143.7514 Pounds (troy)
127 Stones = 2160.7653 Pounds (troy)
128 Stones = 2177.7792 Pounds (troy)
129 Stones = 2194.7931 Pounds (troy)
130 Stones = 2211.807 Pounds (troy)
131 Stones = 2228.8209 Pounds (troy)
132 Stones = 2245.8348 Pounds (troy)
133 Stones = 2262.8487 Pounds (troy)
134 Stones = 2279.8626 Pounds (troy)
135 Stones = 2296.8765 Pounds (troy)
136 Stones = 2313.8904 Pounds (troy)
137 Stones = 2330.9043 Pounds (troy)
138 Stones = 2347.9182 Pounds (troy)
139 Stones = 2364.9321 Pounds (troy)
140 Stones = 2381.946 Pounds (troy)
141 Stones = 2398.9599 Pounds (troy)
142 Stones = 2415.9738 Pounds (troy)
143 Stones = 2432.9877 Pounds (troy)
144 Stones = 2450.0016 Pounds (troy)
145 Stones = 2467.0155 Pounds (troy)
146 Stones = 2484.0294 Pounds (troy)
147 Stones = 2501.0433 Pounds (troy)
148 Stones = 2518.0572 Pounds (troy)
149 Stones = 2535.0711 Pounds (troy)
150 Stones = 2552.085 Pounds (troy)
## how many Stones Into Pounds (troy)
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How Many Yottagrams in Stones | 2,972 | 8,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | longest | en | 0.589781 |
https://www.traditionaloven.com/tutorials/surface-area/convert-rood-to-china-area-unit-fang-chi.html | 1,628,205,567,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152085.13/warc/CC-MAIN-20210805224801-20210806014801-00370.warc.gz | 1,061,307,550 | 17,496 | Convert ro to 方尺 | rood to Chinese fāng chǐ
# area surface units conversion
## Amount: 1 rood (ro) of area Equals: 9,105.43 Chinese fāng chǐ (方尺) in area
Converting rood to Chinese fāng chǐ value in the area surface units scale.
TOGGLE : from Chinese fāng chǐ into roods in the other way around.
## area surface from rood to Chinese fāng chǐ conversion results
### Enter a new rood number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other area surface measuring units - complete list.
How many Chinese fāng chǐ are in 1 rood? The answer is: 1 ro equals 9,105.43 方尺
## 9,105.43 方尺 is converted to 1 of what?
The Chinese fāng chǐ unit number 9,105.43 方尺 converts to 1 ro, one rood. It is the EQUAL area value of 1 rood but in the Chinese fāng chǐ area unit alternative.
ro/方尺 area surface conversion result From Symbol Equals Result Symbol 1 ro = 9,105.43 方尺
## Conversion chart - roods to Chinese fāng chǐ
1 rood to Chinese fāng chǐ = 9,105.43 方尺
2 roods to Chinese fāng chǐ = 18,210.85 方尺
3 roods to Chinese fāng chǐ = 27,316.28 方尺
4 roods to Chinese fāng chǐ = 36,421.71 方尺
5 roods to Chinese fāng chǐ = 45,527.13 方尺
6 roods to Chinese fāng chǐ = 54,632.56 方尺
7 roods to Chinese fāng chǐ = 63,737.99 方尺
8 roods to Chinese fāng chǐ = 72,843.42 方尺
9 roods to Chinese fāng chǐ = 81,948.84 方尺
10 roods to Chinese fāng chǐ = 91,054.27 方尺
11 roods to Chinese fāng chǐ = 100,159.70 方尺
12 roods to Chinese fāng chǐ = 109,265.12 方尺
13 roods to Chinese fāng chǐ = 118,370.55 方尺
14 roods to Chinese fāng chǐ = 127,475.98 方尺
15 roods to Chinese fāng chǐ = 136,581.40 方尺
Convert area surface of rood (ro) and Chinese fāng chǐ (方尺) units in reverse from Chinese fāng chǐ into roods.
## Area units calculator
Main area or surface units converter page.
# Converter type: area surface units
First unit: rood (ro) is used for measuring area.
Second: Chinese fāng chǐ (方尺) is unit of area.
QUESTION:
15 ro = ? 方尺
15 ro = 136,581.40 方尺
Abbreviation, or prefix, for rood is:
ro
Abbreviation for Chinese fāng chǐ is:
## Other applications for this area surface calculator ...
With the above mentioned two-units calculating service it provides, this area surface converter proved to be useful also as a teaching tool:
1. in practicing roods and Chinese fāng chǐ ( ro vs. 方尺 ) measures exchange.
2. for conversion factors between unit pairs.
3. work with area surface's values and properties.
To link to this area surface rood to Chinese fāng chǐ online converter simply cut and paste the following.
The link to this tool will appear as: area surface from rood (ro) to Chinese fāng chǐ (方尺) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 911 | 2,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-31 | latest | en | 0.798366 |
https://courses.lumenlearning.com/calculus2/chapter/partial-fractions/ | 1,721,516,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00450.warc.gz | 150,185,515 | 17,799 | ## Introduction to Partial Fractions
In this section, we examine the method of partial fraction decomposition, which allows us to decompose rational functions into sums of simpler, more easily integrated rational functions. Using this method, we can rewrite an expression such as: $\frac{3x}{{x}^{2}-x - 2}$ as an expression such as $\frac{1}{x+1}+\frac{2}{x - 2}$. | 94 | 366 | {"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} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.866605 |
https://www.education.com/worksheets/first-grade/skip-counting-by-5s/ | 1,675,513,606,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00411.warc.gz | 758,321,257 | 36,038 | # Search Printable 1st Grade Skip Counting by 5 Worksheets
15 filtered results
15 filtered results
Skip Counting by 5s
Sort by
Skip Counting Chart
Worksheet
Skip Counting Chart
Use this number chart to help your child find patterns in numbers, which will help take that first step into skip counting.
Math
Worksheet
Skip Counting Practice
Worksheet
Skip Counting Practice
Get ready for third grade times tables by practicing skip counting.
Math
Worksheet
Count by Fives
Worksheet
Count by Fives
Kids count by fives to 100 on this first grade math worksheet. Skip counting by fives is one way children can demonstrate the meaning of addition.
Math
Worksheet
Count by 5's and Make it Through the Maze!
Worksheet
Count by 5's and Make it Through the Maze!
Your first grader will use her counting skills on this worksheet to count by 5's to find the right path and make it through the maze.
Math
Worksheet
Missing Numbers: Counting by Fives
Worksheet
Missing Numbers: Counting by Fives
Does your child need practice with his math skills? This printable worksheet, which will help him count up to 100, will give him practice counting by 5's.
Math
Worksheet
Practice Test: Easy Number Patterns
Worksheet
Practice Test: Easy Number Patterns
Reinforce math concepts with your students! Use this practice quiz to review number patterns.
Math
Worksheet
Number Patterns Chart
Worksheet
Number Patterns Chart
Learn to see patterns in numbers with this handy chart. Kids will be prompted to skip count and color as they go.
Math
Worksheet
Connect the Dots: Practice Skip Counting by Fives
Worksheet
Connect the Dots: Practice Skip Counting by Fives
Children skip count by fives to connect the dots and discover the hidden picture.
Math
Worksheet
Money Match: Nickels
Worksheet
Money Match: Nickels
Help your first grader get acquainted with nickels by counting them up, and writing their value on the line.
Math
Worksheet
Skip Counting: Let's Practice!
Worksheet
Skip Counting: Let's Practice!
Packed with skip counting practice, this sporty worksheet builds number sense and paves the way for future multiplication skills.
Math
Worksheet
Skip Counting by Fives
Worksheet
Skip Counting by Fives
Practice skip counting with this worksheet focuses on fives.
Math
Worksheet
Counting Nickels
Worksheet
Counting Nickels
Coin recognition and simple adding are important skills -- without them, how will your smart first grader know how much money he has?
Math
Worksheet
Number Patterns: Find the Pattern
Worksheet
Number Patterns: Find the Pattern
Sharpen your first grader's number sense with an exercise in recognizing number patterns.
Math
Worksheet
Back of a Nickel
Worksheet
Back of a Nickel
The backs of coins are as important as the front. Does your first grader know how to recognize a nickel from its back? | 627 | 2,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-06 | latest | en | 0.918162 |
https://www.garysieling.com/blog/implementing-k-means-in-scala/ | 1,718,800,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00544.warc.gz | 678,261,908 | 29,861 | # Implementing k-means in Scala
To generate sample data, I selected two points, (10, 20) and (25, 5), then generated a list of normally distributed points around those two – the exact points used are in the code below.
This implements Lloyd’s algorithm, which tries to cluster points in iterations in a simple manner:
1. Assume a certain number of clusters
2. Group the points at random
3. Compute the center of each cluster
4. For each point, compute which cluster is closest
5. Move all the points into new groupings
6. Repeat 3-5 a few times, until you’re happy with the results
I like how the functional programming style forces you to recreate all the data structures, in this case. It might be tempting to implement this in an imperative style, modifying data structures in place, but since steps 4-5 require separate data, you are protected against making it more difficult. You can see the full source below, or on github.
Since this example is fairly contrived, this converges pretty quickly:
```Initial State:
Cluster 0
Mean: (17.83517750970944, 12.242720407317105)
(10.8348626966492, 18.7800980127523))
(7.7875624720831, 20.1569764307574))
(11.9096128931784, 21.1855674228972))
(22.4668345067162, 8.9705504626857))
(7.91362116378194, 21.325928219919))
(22.636600400773, 2.46561420928429))
(13.0838514816799, 20.3398794353494))
(11.7396623802245, 17.7026240456956))
(25.1439536911272, 3.58469981317611))
(23.5359486724204, 4.07290025106778))
(11.7493214262468, 17.8517235677469))
(12.4277617893575, 19.4887691804508))
(11.931275122466, 18.0462702532436))
(25.4645673159779, 7.54703465191098))
(21.8031183153743, 5.69297814349064))
(23.9177161897547, 8.1377950229489))
(24.5349708443852, 5.00561881333415))
(26.2100410238973, 5.06220487544192))
(23.7770902983858, 7.19445492687232))
Cluster 1
Mean: (16.95249500233747, 12.848199048608048)
(11.7265904596619, 16.9636039793709))
(10.7751248849735, 22.1517666115673))
(23.6587920739353, 3.35476798095758))
(21.4930923464916, 3.28999356823389))
(26.4748241341303, 9.25128245838802))
(7.03171204763376, 19.1985058633283))
(23.7722765903534, 3.74873642284525))
(10.259545802461, 23.4515683763173))
(28.1587146197594, 3.70625885635717))
(10.1057940183815, 18.7332929859685))
(8.90149362263775, 19.6314465074203))
(12.4353462881232, 19.6310467981989))
(24.3793349065557, 4.59761596097384))
(22.5447925324242, 2.99485404382734))
(26.8942422516129, 5.02646862012427))
(6.56491029696013, 21.5098251711267))
(8.87507602702847, 21.4823134390704))
(27.0339042858296, 4.4151109960116))
(11.0118378554584, 20.9773232834654))
Iteration: 0
Cluster 0
Mean: (23.781370272978315, 5.754127202865132)
(11.7265904596619, 16.9636039793709))
(23.6587920739353, 3.35476798095758))
(22.4668345067162, 8.9705504626857))
(21.4930923464916, 3.28999356823389))
(26.4748241341303, 9.25128245838802))
(22.636600400773, 2.46561420928429))
(23.7722765903534, 3.74873642284525))
(25.1439536911272, 3.58469981317611))
(28.1587146197594, 3.70625885635717))
(23.5359486724204, 4.07290025106778))
(24.3793349065557, 4.59761596097384))
(25.4645673159779, 7.54703465191098))
(22.5447925324242, 2.99485404382734))
(21.8031183153743, 5.69297814349064))
(26.8942422516129, 5.02646862012427))
(23.9177161897547, 8.1377950229489))
(24.5349708443852, 5.00561881333415))
(26.2100410238973, 5.06220487544192))
(27.0339042858296, 4.4151109960116))
(23.7770902983858, 7.19445492687232))
Cluster 1
Mean: (10.296576237184727, 20.09138475584863)
(10.8348626966492, 18.7800980127523))
(7.7875624720831, 20.1569764307574))
(10.7751248849735, 22.1517666115673))
(11.9096128931784, 21.1855674228972))
(7.91362116378194, 21.325928219919))
(7.03171204763376, 19.1985058633283))
(13.0838514816799, 20.3398794353494))
(11.7396623802245, 17.7026240456956))
(10.259545802461, 23.4515683763173))
(10.1057940183815, 18.7332929859685))
(11.7493214262468, 17.8517235677469))
(8.90149362263775, 19.6314465074203))
(12.4277617893575, 19.4887691804508))
(12.4353462881232, 19.6310467981989))
(11.931275122466, 18.0462702532436))
(6.56491029696013, 21.5098251711267))
(8.87507602702847, 21.4823134390704))
(11.0118378554584, 20.9773232834654))
Iteration: 1
Cluster 0
Mean: (24.415832368416023, 5.164154740943777)
(23.6587920739353, 3.35476798095758))
(22.4668345067162, 8.9705504626857))
(21.4930923464916, 3.28999356823389))
(26.4748241341303, 9.25128245838802))
(22.636600400773, 2.46561420928429))
(23.7722765903534, 3.74873642284525))
(25.1439536911272, 3.58469981317611))
(28.1587146197594, 3.70625885635717))
(23.5359486724204, 4.07290025106778))
(24.3793349065557, 4.59761596097384))
(25.4645673159779, 7.54703465191098))
(22.5447925324242, 2.99485404382734))
(21.8031183153743, 5.69297814349064))
(26.8942422516129, 5.02646862012427))
(23.9177161897547, 8.1377950229489))
(24.5349708443852, 5.00561881333415))
(26.2100410238973, 5.06220487544192))
(27.0339042858296, 4.4151109960116))
(23.7770902983858, 7.19445492687232))
Cluster 1
Mean: (10.371840143630894, 19.92676471498138)
(10.8348626966492, 18.7800980127523))
(11.7265904596619, 16.9636039793709))
(7.7875624720831, 20.1569764307574))
(10.7751248849735, 22.1517666115673))
(11.9096128931784, 21.1855674228972))
(7.91362116378194, 21.325928219919))
(7.03171204763376, 19.1985058633283))
(13.0838514816799, 20.3398794353494))
(11.7396623802245, 17.7026240456956))
(10.259545802461, 23.4515683763173))
(10.1057940183815, 18.7332929859685))
(11.7493214262468, 17.8517235677469))
(8.90149362263775, 19.6314465074203))
(12.4277617893575, 19.4887691804508))
(12.4353462881232, 19.6310467981989))
(11.931275122466, 18.0462702532436))
(6.56491029696013, 21.5098251711267))
(8.87507602702847, 21.4823134390704))
(11.0118378554584, 20.9773232834654))
Iteration: 2
Cluster 0
Mean: (24.415832368416023, 5.164154740943777)
(23.6587920739353, 3.35476798095758))
(22.4668345067162, 8.9705504626857))
(21.4930923464916, 3.28999356823389))
(26.4748241341303, 9.25128245838802))
(22.636600400773, 2.46561420928429))
(23.7722765903534, 3.74873642284525))
(25.1439536911272, 3.58469981317611))
(28.1587146197594, 3.70625885635717))
(23.5359486724204, 4.07290025106778))
(24.3793349065557, 4.59761596097384))
(25.4645673159779, 7.54703465191098))
(22.5447925324242, 2.99485404382734))
(21.8031183153743, 5.69297814349064))
(26.8942422516129, 5.02646862012427))
(23.9177161897547, 8.1377950229489))
(24.5349708443852, 5.00561881333415))
(26.2100410238973, 5.06220487544192))
(27.0339042858296, 4.4151109960116))
(23.7770902983858, 7.19445492687232))
Cluster 1
Mean: (10.371840143630894, 19.92676471498138)
(10.8348626966492, 18.7800980127523))
(11.7265904596619, 16.9636039793709))
(7.7875624720831, 20.1569764307574))
(10.7751248849735, 22.1517666115673))
(11.9096128931784, 21.1855674228972))
(7.91362116378194, 21.325928219919))
(7.03171204763376, 19.1985058633283))
(13.0838514816799, 20.3398794353494))
(11.7396623802245, 17.7026240456956))
(10.259545802461, 23.4515683763173))
(10.1057940183815, 18.7332929859685))
(11.7493214262468, 17.8517235677469))
(8.90149362263775, 19.6314465074203))
(12.4277617893575, 19.4887691804508))
(12.4353462881232, 19.6310467981989))
(11.931275122466, 18.0462702532436))
(6.56491029696013, 21.5098251711267))
(8.87507602702847, 21.4823134390704))
(11.0118378554584, 20.9773232834654))
Iteration: 3
Cluster 0
Mean: (24.415832368416023, 5.164154740943777)
(23.6587920739353, 3.35476798095758))
(22.4668345067162, 8.9705504626857))
(21.4930923464916, 3.28999356823389))
(26.4748241341303, 9.25128245838802))
(22.636600400773, 2.46561420928429))
(23.7722765903534, 3.74873642284525))
(25.1439536911272, 3.58469981317611))
(28.1587146197594, 3.70625885635717))
(23.5359486724204, 4.07290025106778))
(24.3793349065557, 4.59761596097384))
(25.4645673159779, 7.54703465191098))
(22.5447925324242, 2.99485404382734))
(21.8031183153743, 5.69297814349064))
(26.8942422516129, 5.02646862012427))
(23.9177161897547, 8.1377950229489))
(24.5349708443852, 5.00561881333415))
(26.2100410238973, 5.06220487544192))
(27.0339042858296, 4.4151109960116))
(23.7770902983858, 7.19445492687232))
Cluster 1
Mean: (10.371840143630894, 19.92676471498138)
(10.8348626966492, 18.7800980127523))
(11.7265904596619, 16.9636039793709))
(7.7875624720831, 20.1569764307574))
(10.7751248849735, 22.1517666115673))
(11.9096128931784, 21.1855674228972))
(7.91362116378194, 21.325928219919))
(7.03171204763376, 19.1985058633283))
(13.0838514816799, 20.3398794353494))
(11.7396623802245, 17.7026240456956))
(10.259545802461, 23.4515683763173))
(10.1057940183815, 18.7332929859685))
(11.7493214262468, 17.8517235677469))
(8.90149362263775, 19.6314465074203))
(12.4277617893575, 19.4887691804508))
(12.4353462881232, 19.6310467981989))
(11.931275122466, 18.0462702532436))
(6.56491029696013, 21.5098251711267))
(8.87507602702847, 21.4823134390704))
(11.0118378554584, 20.9773232834654))
Iteration: 4
Cluster 0
Mean: (24.415832368416023, 5.164154740943777)
(23.6587920739353, 3.35476798095758))
(22.4668345067162, 8.9705504626857))
(21.4930923464916, 3.28999356823389))
(26.4748241341303, 9.25128245838802))
(22.636600400773, 2.46561420928429))
(23.7722765903534, 3.74873642284525))
(25.1439536911272, 3.58469981317611))
(28.1587146197594, 3.70625885635717))
(23.5359486724204, 4.07290025106778))
(24.3793349065557, 4.59761596097384))
(25.4645673159779, 7.54703465191098))
(22.5447925324242, 2.99485404382734))
(21.8031183153743, 5.69297814349064))
(26.8942422516129, 5.02646862012427))
(23.9177161897547, 8.1377950229489))
(24.5349708443852, 5.00561881333415))
(26.2100410238973, 5.06220487544192))
(27.0339042858296, 4.4151109960116))
(23.7770902983858, 7.19445492687232))
Cluster 1
Mean: (10.371840143630894, 19.92676471498138)
(10.8348626966492, 18.7800980127523))
(11.7265904596619, 16.9636039793709))
(7.7875624720831, 20.1569764307574))
(10.7751248849735, 22.1517666115673))
(11.9096128931784, 21.1855674228972))
(7.91362116378194, 21.325928219919))
(7.03171204763376, 19.1985058633283))
(13.0838514816799, 20.3398794353494))
(11.7396623802245, 17.7026240456956))
(10.259545802461, 23.4515683763173))
(10.1057940183815, 18.7332929859685))
(11.7493214262468, 17.8517235677469))
(8.90149362263775, 19.6314465074203))
(12.4277617893575, 19.4887691804508))
(12.4353462881232, 19.6310467981989))
(11.931275122466, 18.0462702532436))
(6.56491029696013, 21.5098251711267))
(8.87507602702847, 21.4823134390704))
(11.0118378554584, 20.9773232834654))
Iteration: 5
Cluster 0
Mean: (24.415832368416023, 5.164154740943777)
(23.6587920739353, 3.35476798095758))
(22.4668345067162, 8.9705504626857))
(21.4930923464916, 3.28999356823389))
(26.4748241341303, 9.25128245838802))
(22.636600400773, 2.46561420928429))
(23.7722765903534, 3.74873642284525))
(25.1439536911272, 3.58469981317611))
(28.1587146197594, 3.70625885635717))
(23.5359486724204, 4.07290025106778))
(24.3793349065557, 4.59761596097384))
(25.4645673159779, 7.54703465191098))
(22.5447925324242, 2.99485404382734))
(21.8031183153743, 5.69297814349064))
(26.8942422516129, 5.02646862012427))
(23.9177161897547, 8.1377950229489))
(24.5349708443852, 5.00561881333415))
(26.2100410238973, 5.06220487544192))
(27.0339042858296, 4.4151109960116))
(23.7770902983858, 7.19445492687232))
Cluster 1
Mean: (10.371840143630894, 19.92676471498138)
(10.8348626966492, 18.7800980127523))
(11.7265904596619, 16.9636039793709))
(7.7875624720831, 20.1569764307574))
(10.7751248849735, 22.1517666115673))
(11.9096128931784, 21.1855674228972))
(7.91362116378194, 21.325928219919))
(7.03171204763376, 19.1985058633283))
(13.0838514816799, 20.3398794353494))
(11.7396623802245, 17.7026240456956))
(10.259545802461, 23.4515683763173))
(10.1057940183815, 18.7332929859685))
(11.7493214262468, 17.8517235677469))
(8.90149362263775, 19.6314465074203))
(12.4277617893575, 19.4887691804508))
(12.4353462881232, 19.6310467981989))
(11.931275122466, 18.0462702532436))
(6.56491029696013, 21.5098251711267))
(8.87507602702847, 21.4823134390704))
(11.0118378554584, 20.9773232834654))```
```class Point(dx: Double, dy: Double) {
val x: Double = dx
val y: Double = dy
override def toString(): String = {
"(" + x + ", " + y + ")"
}
def dist(p: Point): Double = {
return (x - p.x) * (x - p.x) + (y - p.y) * (y - p.y);
}
}
object kmeans extends App {
val k: Int = 2
// Correct answers to centers are (10, 20) and (25, 5)
val points: List[Point] = List(
new Point(10.8348626966492, 18.7800980127523),
new Point(10.259545802461, 23.4515683763173),
new Point(11.7396623802245, 17.7026240456956),
new Point(12.4277617893575, 19.4887691804508),
new Point(10.1057940183815, 18.7332929859685),
new Point(11.0118378554584, 20.9773232834654),
new Point(7.03171204763376, 19.1985058633283),
new Point(6.56491029696013, 21.5098251711267),
new Point(10.7751248849735, 22.1517666115673),
new Point(8.90149362263775, 19.6314465074203),
new Point(11.931275122466, 18.0462702532436),
new Point(11.7265904596619, 16.9636039793709),
new Point(11.7493214262468, 17.8517235677469),
new Point(12.4353462881232, 19.6310467981989),
new Point(13.0838514816799, 20.3398794353494),
new Point(7.7875624720831, 20.1569764307574),
new Point(11.9096128931784, 21.1855674228972),
new Point(8.87507602702847, 21.4823134390704),
new Point(7.91362116378194, 21.325928219919),
new Point(26.4748241341303, 9.25128245838802),
new Point(26.2100410238973, 5.06220487544192),
new Point(28.1587146197594, 3.70625885635717),
new Point(26.8942422516129, 5.02646862012427),
new Point(23.7770902983858, 7.19445492687232),
new Point(23.6587920739353, 3.35476798095758),
new Point(23.7722765903534, 3.74873642284525),
new Point(23.9177161897547, 8.1377950229489),
new Point(22.4668345067162, 8.9705504626857),
new Point(24.5349708443852, 5.00561881333415),
new Point(24.3793349065557, 4.59761596097384),
new Point(27.0339042858296, 4.4151109960116),
new Point(21.8031183153743, 5.69297814349064),
new Point(22.636600400773, 2.46561420928429),
new Point(25.1439536911272, 3.58469981317611),
new Point(21.4930923464916, 3.28999356823389),
new Point(23.5359486724204, 4.07290025106778),
new Point(22.5447925324242, 2.99485404382734),
new Point(25.4645673159779, 7.54703465191098)).sortBy(
p => (p.x + " " + p.y).hashCode())
def clusterMean(points: List[Point]): Point = {
val cumulative = points.reduceLeft((a: Point, b: Point) => new Point(a.x + b.x, a.y + b.y))
return new Point(cumulative.x / points.length, cumulative.y / points.length)
}
def render(points: Map[Int, List[Point]]) {
for (clusterNumber x._2 % k) transform (
(i: Int, p: List[(Point, Int)]) => for (x clusters.map(cluster => cluster._1)
// find cluster means
val means =
(clusters: Map[Int, List[Point]]) =>
for (clusterIndex closest(p, means(clusters)))
render(newClusters)
return newClusters
}
var clusterToTest = clusters
for (i```
## 5 Replies to “Implementing k-means in Scala”
1. network_graph says:
Interesting entry. Just some remarks:
You can write `case class Point(x: Double, y: Double)` so you don’t have to write public getters for the coordinates; then you can leave out the `new` keyword, e.g. `Point(10.835, 18.780)`.
There are a few more ‘javaish’ things, e.g. you should omit the `return` keyword, and you can write `println` instead of `System.out.println`.
1. Great, thanks. I’m still finding my way around with Scala a bit.
2. Andrew McNaughton says:
in Class “Point”, your “dist” function is not euclidian distance. Is that intended?
What you have there amounts to the square of the cosine of the vectors from the origin to each points, which seems rather odd.
1. Wow, you’re absolutely right, I updated the post to correct this.
3. Please how did you excute the scala script ? | 6,640 | 15,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.650067 |
https://velog.io/@dubbsong/Codewars-%EC%95%8C%EA%B3%A0%EB%A6%AC%EC%A6%98-%EB%AC%B8%EC%A0%9C-%ED%92%80%EC%9D%B4-8kyu-95 | 1,600,613,076,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198213.25/warc/CC-MAIN-20200920125718-20200920155718-00387.warc.gz | 709,608,115 | 14,659 | # Codewars 알고리즘 문제 풀이 (8kyu 95)
dubbsong·2019년 9월 10일
0
#### 문제
• 출생년도부터 해당연도까지 계산하는 함수를 작성한다.
#### 풀이 01
function calculateAge(birthYear, currentYear) {
let age = currentYear - birthYear;
if (age === 1) return 'You are 1 year old.';
else if (age === -1) return 'You will be born in 1 year.';
else if (age > 0) return You are ${age} years old.'; else if (age < 0) return You will be born in${-age} years.';
else return 'You were born this very year!';
}
calculateAge(1988, 1988); // You were born this very year!
calculateAge(1988, 1989); // You are 1 year old.
calculateAge(1988, 2019); // You are 31 years old.
calculateAge(1988, 1987); // You will be born in 1 year.
calculateAge(2030, 2019); // You will be born in 11 years.
#### 풀이 02
function calculateAge(birthYear, currentYear) {
let year = Math.abs(birthYear - currentYear) === 1 ? 'year' : 'years';
if (birthYear === currentYear) return 'You were born this very year!';
if (birthYear < currentYear) return You are ${currentYear - birthYear}${year} old.;
if (birthYear > currentYear) return You will be born in ${-currentYear + birthYear}${year}.;
}
- Math.abs(): 주어진 숫자의 절대값(absolute value)을 반환한다.
calculateAge(1988, 1988); // You were born this very year!
calculateAge(1988, 1989); // You are 1 year old.
calculateAge(1988, 2019); // You are 31 years old.
calculateAge(1988, 1987); // You will be born in 1 year.
calculateAge(2030, 2019); // You will be born in 11 years.
오늘도 많이 배웁니다 | 494 | 1,465 | {"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-2020-40 | latest | en | 0.593459 |
https://www.zigya.com/engineering-entrance-exam/11-Mechanical+Properties+of+Solids/PHENJE11144376/ | 1,716,881,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00568.warc.gz | 928,052,610 | 16,615 | The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is (For steel Young’s modulus is 2×1011 N m–2 and coefficient of thermal expansion is 1.1×10–5 K–1 ) | Mechanical Properties of Solids
## Previous Year Papers
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# Mechanical Properties of Solids
#### Multiple Choice Questions
1.
A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is added to its Bob, the time period changes TM. If Young's modulus of the material of the wire is Y, then 1/Y is equal to (g = gravitational acceleration)
692 Views
# 2.The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is (For steel Young’s modulus is 2×1011 N m–2 and coefficient of thermal expansion is 1.1×10–5 K–1 ) 2.2 x 108 Pa 2.2 x 109 Pa 2.2 x 107 Pa 2.2 x 107 Pa
A.
2.2 x 108 Pa
According to Hooke's law, i.e,
If the rod is compressed, then compressive stress and strain appear. Their ratio Y is same as that for the tensile case.
Give, length of a steel wire (L) = 10 cm,
Temperature (θ) = 100oC
As length is constant,
∴
Now, pressure = stress =y x strain
[Given, Y = 2 x 1011N/M2 and α = 1.1 x 10-5 K-1]
= 2 x 1011 x1.1 x 10-5 x 100
=2.2 x 108
317 Views
3.
A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and 2.2 × 1011N/m2 respectively?
• 188.5 Hz
• 178.2 Hz
• 200.5 Hz
• 200.5 Hz
515 Views
4.
A wooden wheel of radius R is made of two semicircular parts (see figure); The two parts are held together by a ring made of a metal strip of cross sectional area S and length L. L is slightly less than 2πR. To fit the ring on the wheel, it is heated so that it temperature rises by ∆T and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semicircular parts together. If the coefficient of linear expansion of the metal is α, and its Youngs'modulus is Y, the force that one part of the wheel applies on the other part is :
• 2πSYα∆T
• SYα∆T
• πSYα ∆T
• πSYα ∆T
425 Views
5.
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is
• 4
• 3
• 2
• 2
485 Views
6.
A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of
• 81
• 1/81
• 9
• 9
858 Views
7.
One kg of a diatomic gas is at a pressure of 8 × 104 N/m2 . The density of the gas is 4 kg/m-3. What is the energy of the gas due to its thermal motion?
• 3 × 104 J
• 5× 104 J
• 6× 104 J
• 6× 104 J
667 Views
8.
Two wires are made of the same material and have the same volume. However wire 1 has cross-section area A and wire 2 has cross–sectional area 3A. If the length of wire 1 increases by ∆x on applying force F, how much force is needed to stretch wire 2 by the same amount?
• F
• 4F
• 9F
• 9F
198 Views
9.
A transparent solid cylindrical rod has a refractive index of 2√3. It is surrounded by air. A light ray is incident at the midpoint of one end of the rod as shown in the figure.θ The incident angle θ for which the light ray grazes along the wall of the rod is
422 Views
10.
If S is stress and Y is Young’s modulus of material of a wire, the energy stored in the wire per unit volume is
• 2S2Y
• S2/2Y
• 2Y/S2
• 2Y/S2
321 Views | 1,216 | 4,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.877161 |
https://www.airmilescalculator.com/distance/bla-to-chx/ | 1,624,069,353,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643380.40/warc/CC-MAIN-20210619020602-20210619050602-00125.warc.gz | 550,229,416 | 7,922 | # Distance between Barcelona (BLA) and Changuinola (CHX)
Flight distance from Barcelona to Changuinola (General José Antonio Anzoátegui International Airport – Changuinola 'Capitán Manuel Niño' International Airport) is 1216 miles / 1957 kilometers / 1057 nautical miles. Estimated flight time is 2 hours 48 minutes.
## Map of flight path from Barcelona to Changuinola.
Shortest flight path between General José Antonio Anzoátegui International Airport (BLA) and Changuinola 'Capitán Manuel Niño' International Airport (CHX).
## How far is Changuinola from Barcelona?
There are several ways to calculate distances between Barcelona and Changuinola. Here are two common methods:
Vincenty's formula (applied above)
• 1216.001 miles
• 1956.964 kilometers
• 1056.676 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1214.535 miles
• 1954.605 kilometers
• 1055.402 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A General José Antonio Anzoátegui International Airport
City: Barcelona
Country: Venezuela
IATA Code: BLA
ICAO Code: SVBC
Coordinates: 10°6′25″N, 64°41′21″W
B Changuinola 'Capitán Manuel Niño' International Airport
City: Changuinola
Country: Panama
IATA Code: CHX
ICAO Code: MPCH
Coordinates: 9°27′31″N, 82°31′0″W
## Time difference and current local times
The time difference between Barcelona and Changuinola is 1 hour. Changuinola is 1 hour behind Barcelona.
-04
EST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 162 kg (357 pounds).
## Frequent Flyer Miles Calculator
Barcelona (BLA) → Changuinola (CHX).
Distance:
1216
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1216
Round trip? | 518 | 1,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-25 | latest | en | 0.720843 |
https://app-wiringdiagram.herokuapp.com/post/mathbits-geometry-answers-box-5 | 1,563,633,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526517.67/warc/CC-MAIN-20190720132039-20190720154039-00184.warc.gz | 309,186,603 | 16,607 | 9 out of 10 based on 894 ratings. 4,254 user reviews.
mathbits geometry BOX #5 | Yahoo Answers
Aug 02, 2008mathbits geometry box #5 http://MathBits/Caching/G901120 How to solve #2 and #3? Much help would be appreciated.Status: OpenAnswers: 2I need help with box #5 question 5 in mathbits geocachingMay 18, 2015Status: OpenMathbits Geometry box 7 help!? | Yahoo AnswersDec 28, 2013 I need help with mathbits "geocaching" box number 5Jan 04, 2012Status: ResolvedMathbits geometry caching box #7 answer? | Yahoo AnswersJan 02, 2011Status: ResolvedSee more results
I need help with mathbits "geocaching" box number 5
Question & AnswerRelated QuestionsQuestionI need help with mathbits "geocaching" box number 5? URGENT? okay for question 1 i got x=32, for question number 2 i got 3 and for question number 3 i got 5. but the code doesn't seem to be working out when i try to get to the nexAnswerDid you follow the instructions correctly for calculating the results using those numbers? I did and I got to box 6. the numbers calculated in the instructions: 32^2 3^3 5^2 Then multiply and subtract in the order given.See more on auersoStatus: OpenPublished: Jan 05, 2012Answers: 3
Yep! You have found Box #5. - mathbits
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What is the answer to mathbits algecaching algebra 1 box 5
box 5. 18432. box 6. 16875. box7. 16807. box8. 84934656. box9. 28672. What is the answer to mathbits algecaching algebra 1 box 7 problems? What are the answers to box 4 of mathbits geometry?
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MathBitsNotebook - Geometry is a series of lesson and practice pages for students studying high school Geometry. These materials cover ALL standards stated in the Common Core Standards (or Next Generation Standards) for Mathematics, and more. Materials coordinate with SBAC, PARCC, and state assessmentserences reflected in the PARCC MCF are included and highlighted.
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What is the answer for box 1 on mathbits pre algebra you don't go from algebra to calculus and linear algebra. you go from algebra to geometry to advanced algebra with trig to pre calculus to[PDF]
mathbits geocaching answers box 5 - Bing - Free PDF Links Blog | 665 | 2,526 | {"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} | 3.046875 | 3 | CC-MAIN-2019-30 | latest | en | 0.877625 |
https://www.wanweibaike.net/wiki-駐點 | 1,632,598,401,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00413.warc.gz | 1,057,892,112 | 19,768 | # 驻点本文重定向自 駐點
y = x + sin(2x) 的图像
y = x3 的图像
${\displaystyle \left.{\frac {dy}{dx}}\right|_{p}=0\,}$
## 静态平衡系统
${\displaystyle \delta W=\sum _{i}\mathbf {F} _{i}\cdot \delta \mathbf {r} _{i}=0\,}$
${\displaystyle \delta W=\sum _{i}F_{i}\delta q_{i}=0\,}$
${\displaystyle F_{i}=-{\frac {\partial V}{\partial q_{i}}}\,}$
${\displaystyle \delta W=\sum _{i}-{\frac {\partial V}{\partial q_{i}}}\delta q_{i}=-\delta V=0\,}$
## 欧拉-拉格朗日方程
${\displaystyle \mathbf {y} (x)=(y_{1}(x),\ y_{2}(x),\ \ldots ,y_{N}(x))\,\!}$
${\displaystyle {\dot {\mathbf {y} }}(x)=({\dot {y}}_{1}(x),\ {\dot {y}}_{2}(x),\ \ldots ,\ {\dot {y}}_{N}(x))\,\!}$
${\displaystyle f(\mathbf {y} ,\ {\dot {\mathbf {y} }},\ x)=f(y_{1}(x),\ y_{2}(x),\ \ldots ,\ y_{N}(x),\ {\dot {y}}_{1}(x),\ {\dot {y}}_{2}(x),\ \ldots ,\ {\dot {y}}_{N}(x),\ x)\,\!}$
${\displaystyle \mathbf {y} (x)\in (C^{1}[a,\ b])^{N}\,\!}$使泛函${\displaystyle J(\mathbf {y} )=\int _{a}^{b}f(\mathbf {y} ,\ {\dot {\mathbf {y} }},\ x)dx\,\!}$取得局部平稳值，则在区间${\displaystyle (a,\ b)\,\!}$内对于所有的${\displaystyle i=1,\ 2,\ \ldots ,\ N\,\!}$，欧拉-拉格朗日方程成立：
${\displaystyle {\frac {d}{dx}}{\frac {\partial }{\partial {\dot {y}}_{i}}}f(\mathbf {y} ,\ {\dot {\mathbf {y} }},\ x)-{\frac {\partial }{\partial y_{i}}}f(\mathbf {y} ,\ {\dot {\mathbf {y} }},\ x)=0\,\!}$ | 631 | 1,291 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 25, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-39 | latest | en | 0.279878 |
https://citizenmaths.com/chemical-amount/959.9-moles-to-kilomoles | 1,632,202,217,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00564.warc.gz | 215,400,884 | 11,089 | # 959.9 Moles to Kilomoles
Mole
• Kilomole
• Millimole
• Mole
• Pound-Mole
=
Kilomole
• Kilomole
• Millimole
• Mole
• Pound-Mole
Formula 959.9 mol = 959.9 / 1000 kmol = 0.9599 kmol
959.9 mol = 0.9599 kmol
Explanation:
• 1 mol is equal to 0.001 kmol, therefore 959.9 mol is equivalent to 0.9599 kmol.
• 1 Mole = 1 / 1000 = 0.001 Kilomoles
• 959.9 Moles = 959.9 / 1000 = 0.9599 Kilomoles
## 959.9 Moles to Kilomoles Conversion Table
Mole (mol) Kilomole (kmol)
960 mol 0.96 kmol
960.1 mol 0.9601 kmol
960.2 mol 0.9602 kmol
960.3 mol 0.9603 kmol
960.4 mol 0.9604 kmol
960.5 mol 0.9605 kmol
960.6 mol 0.9606 kmol
960.7 mol 0.9607 kmol
960.8 mol 0.9608 kmol
## Convert 959.9 mol to other units
Unit Unit of Chemical Amount
Millimole 959,900.0 mmol
Pound-Mole 2.1162 lb-mol
Kilomole 0.9599 kmol
## Comments (0)
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View Full Version : Gl_POLYGON
Al Grant
05-25-2006, 05:55 PM
Hi Guys,
I am trying to draw a polygon, which is essentially like a upside down "L", if I use GL_POINTS they are in the right place, but when I use GL_POLYGON I get a different shape???
The problem is around 4,5,6. Any got any ideaS???
Cheers
-Al
glBegin(GL_POINTS);
glVertex3f(-1.75f,0.05f,0.0f); //1
glVertex3f(-0.75f,0.05f,0.0f); //2
glVertex3f(-0.75f,-0.15f,0.0f); //3
glVertex3f(-0.85f,-0.15f,0.0f); //4
glVertex3f(-0.85f,-0.05f,0.0f); //5
glVertex3f(-1.75f,-0.05f,0.0f); //6
glEnd();
songho
05-25-2006, 08:48 PM
The problem is that "L" shape is a concave polygon, but OpenGL accepts only convex polygons.
You may tessellate it into a convex polygon by using GLU tessellattor, or may try the stencil trick.
The redbook(OpenGL programming Guide) describes both methods very well. Please look at "Drawing Filled, Polygons Using the Stencil Buffer" section especially for the stencil trick.
Al Grant
05-25-2006, 10:46 PM
Righto - sounds good. Thanks!
glome
05-26-2006, 01:34 AM
Or simply split your "L" to 2 convex polygons (rectangles) "|" and "_".
Serge K
05-26-2006, 07:57 PM
Originally posted by glome:
Or simply split your "L" to 2 convex polygons (rectangles) "|" and "_". ...but watch out for T-junctions... | 454 | 1,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.909387 |
https://www.coursehero.com/file/6638630/How-to-work-out-those-annoying-angle-problems/ | 1,498,361,771,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320395.62/warc/CC-MAIN-20170625032210-20170625052210-00581.warc.gz | 870,590,419 | 55,903 | How to work out those annoying angle problems
# How to work out those annoying angle problems - How to work...
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How to work out those annoying angle problems Let's look at another problem, similar to the one gone over in the previous section dealing with how to use the declination values of objects. All problems like these have 3 parts - A latitude The declination of an object (how far it is from the Celestial Equator) The height above the northern or southern horizon The diagram that helps you figure out the problem, like that shown in Figure 8, shows only the northern and southern horizon. Why not the eastern or western horizon? You might want to think of these problems as meridian problems, since that is where we are looking at the objects in the sky, when they are on your meridian (I hope you didn't forget what the meridian is, because if you did, you may want to look it up again). Anyways, the diagram just shows the angles along the meridian going from the ground, the horizon, and extending upwards. You really need to get your brain around these problems, since they will definitely be on the test. Let's go back to the set up in Figure 8, the way that the sky is oriented along your meridian as
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## This note was uploaded on 12/15/2011 for the course AST AST1002 taught by Professor Emilyhoward during the Fall '10 term at Broward College.
Ask a homework question - tutors are online | 342 | 1,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-26 | longest | en | 0.957422 |
http://math.stackexchange.com/questions/298756/let-a-n-be-a-sequence-of-real-numbers-suppose-each-of-the-subsequences-a | 1,469,273,683,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257822172.7/warc/CC-MAIN-20160723071022-00228-ip-10-185-27-174.ec2.internal.warc.gz | 161,545,380 | 17,306 | # Let $(a_n)$ be a sequence of real numbers. Suppose each of the subsequences $(a_{2n}),(a_{2n+1})$ and $(a_{3n})$ converges to $p,q$ and $r$
Let $(a_n)$ be a sequence of real numbers. Suppose each of the subsequences $(a_{2n}),(a_{2n+1})$ and $(a_{3n})$ converges to $p,q$ and $r$, respectively. Show that $p=r$ and $q=r$ and hence conclude that $(a_n)$ converges. To prove $p=r$, I consider a subsequence of $(a_{2n})$ and $(a_{3n})$ , which is $(a_{6n})$. Then I say clearly $(a_{6n})$ converges to $p$ and $r$ and hence $p=r$. Is this proof work?
-
Yes, this works perfectly. – 1015 Feb 9 '13 at 14:52
As Julien says, yes; and you can use a similar idea for the second part (cross multiply the indices is probably the easiest way). – gnometorule Feb 9 '13 at 14:56
And for the other equality, consider for instance $(a_{3(2n+1)}=a_{2(3n+1)+1})$. – 1015 Feb 9 '13 at 14:56
Yes, $a_{6n}$ is a subsequence of $a_{2n}$ and $a_{3n}$, so it converges to both $p$ and $r$, thus $p=r$.
Similarly, $a_{6n+3}=a_{3(2n+1)}=a_{2(3n+1)+1}$ is a subsequence of $a_{3n}$ and $a_{2n+1}$, so it converges to both $r$ and $q$, thus $r=q$.
So $p=r=q$ and, as $a_{2n}$ and $a_{2n+1}$ covers all of $a_n$, it converges. | 465 | 1,204 | {"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} | 4.5625 | 5 | CC-MAIN-2016-30 | latest | en | 0.830453 |
http://www.onlinemath4all.com/focusquestion4.html | 1,508,541,628,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824471.6/warc/CC-MAIN-20171020230225-20171021010225-00883.warc.gz | 539,725,155 | 5,935 | ## Focus question 4
In this page focus question 4 we are going to find out the focus, vertex, equation of directrix and length of the latus rectum of the equation
y² +8x+2y+17=0.
Here the equation is in the standard form (y-k)²=4a(x-h).The following table gives the necessary details of the standard and vertex form of parabola.
Standard form Vertex form
y² =4ax If a is positive, then it opens in the right hand side. If a is negative, then it opens in the left hand side. The focus is (a,0). The vertex is the origin (0,0) The equation of the directrix is x =-a The length of the latus rectum is 4a. (y-k)²=4a(x-h) If a is positive, then it opens in the right hand side. If a is negative, then it opens in the left hand side. The focus is (h+a, k) The vertex is (h,k) The equation of the directrix is x-h = -a The length of the latus rectum is 4a.
Solution:
Here the equation y² +8x+2y+17=0 is in the quadratic equation form. Let us bring to the vertex form of equation.
y² +8x+2y+17=0.
y²+2y = -8x-17
y²+2y+1 = -8x-17+1(adding '1' on both sides)
(y+1) ² = -8x-16
(y+1) ² = -8(x+2)
This is of the form (y-k)²=4a(x-h) whose vertex is (h,k)
Here h=-2 and k=-1
and 4a = -8. So a = -8/4 =-2. Since a is negative the parabola opens up in the left hand side.
The focus is (h+a, k) = (-2-2,-1) = (-4,-1)
The vertex is (h,k) = (-2,-1)
The equation of the directrix is x+2 = +2
x=0
The length of the latus rectum is 4a =8
Parents and teachers help the students to solve the problem in the above method in focus question 4 and they can guide them to solve the following problem using the above method.
The other three standard forms and vertex forms of parabola are discussed in the focus worksheet.
If you have any doubt you can contact us through mail, we will help you to solve the problem.
Problem for practice:
1. Find the focus, vertex, equation of directrix and length of latus rectum of the parabola y²+4x-2y+3=0
2. Find the focus, vertex, equation of directrix and lenth of the latus rectum of the parabola y²+4x+2y-3=0
[?]Subscribe To This Site | 671 | 2,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2017-43 | longest | en | 0.877544 |
http://www.chegg.com/homework-help/questions-and-answers/vs-50vr0-500-r1-500-r2-1000-r3-800-r4-2000-r5-1000-r6-800-current-i-circuit-using-ohm-s-la-q305821 | 1,386,664,604,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164014017/warc/CC-MAIN-20131204133334-00089-ip-10-33-133-15.ec2.internal.warc.gz | 280,423,841 | 7,905 | # Beginner's Circuitry
0 pts ended
Vs = 50V
R0 = 500?
R1 = 500?
R2 = 1000?
R3 = 800?
R4 = 2000?
R5 = 1000?
R6 = 800?
a) Find the current i in the circuit by using Ohm's law andKirchhoff's laws.
b)Indicate the node voltage between R1 and R3 | 97 | 242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2013-48 | latest | en | 0.926169 |
http://archive.org/stream/AnalyticalMechanics/TXT/00000301.txt | 1,488,171,844,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172447.23/warc/CC-MAIN-20170219104612-00020-ip-10-171-10-108.ec2.internal.warc.gz | 16,143,405 | 6,655 | # Full text of "Analytical Mechanics"
## See other formats
```288 ANALYTICAL MECHANICS
an acceleration - 7 -r, while the earth imparts to the sui r*
an acceleration equal to 7 — •
Suppose at any instant we impart to both the sun anc the earth a velocity equal and opposite to that of the sur
fW\
and apply an acceleration —j. This will bring the sun t<
rest and keep it at rest, without altering the motion of th< earth relative to the sun.* This reduces the problem o: the motion of the earth to that of a particle moving in i central field of force where the acceleration is
,, N M m
T"
or / (r) = •£ (VII
and [F=-M2J, (VIII
rz
where M = 7 (-W + m)-
Substituting from equation (VIII) in equation (V) w< obtain
for the equation of the orbit. Let uf = u — j~, then th( equation of the orbit takes the form
t'=0. (2;
(to2 ' In order to integrate equation (2) let t; = --—,
* The acceleration of a particle relative to another moving particle i found by adding the negative of the acceleration of the second particle t< that of the first.``` | 299 | 1,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-09 | latest | en | 0.85408 |
http://www.astburygarage.com/96602/chemical-kinetics-conceptual-questions-04c026 | 1,632,324,355,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00653.warc.gz | 71,474,092 | 8,423 | It is found that the reaction follows the following rate equation : Answer: Delhi 2016) Order of reaction: The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Substituting the values, we get. Calculate ‘Ea‘ for the reaction. Given : K = 2.4 × 10-3, Question 18. Answer: Question 65. Question 27. India 2013) A reaction is of second order with respect to a reactant. Answer: (ii) The unit of rate constant (k) is mol L-1 S-1. For first order reaction: Question 24. What do you understand by the ‘order of a reaction’? (c) What is the initial rate of disappearance of Cl2 in exp. (ii) How is rate affected when concentration of B is tripled? . At what temperature would K be 1.5 × 104 s-1? How is the rate of reaction affected if the concentration of the reactant is reduced to half? JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Chemical kinetics, also known as reaction kinetics, is the branch of physical chemistry that is concerned with understanding the rates of chemical reactions.. Question 39. This diagram is obained by plotting potential energy vs. reaction coordinate. What has to happen when a force is exerted on an object for work to be done? Delhi 2015) r = KA(3B)2 ∴ r = 9KAB2 = 9 times r2= k[A]1 [2B]2 = 4k[A]1 [B]2 (i) Write the order and molecularity of this reaction. Define rate of reaction. (a) K = $$\frac{2.303}{t} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}$$ (ii) What is the slope of the curve? 30 seconds . Question 42. (b) Write the rate equation for the reaction A2 + 3B2 → 2C, if the overall order of the reaction is zero. For a chemical reaction R → P, the variation in the f concentration (R) vs. time (f) plot is given as Answer: (i) Rate of a reaction Answer: Answer: (i) Rate constant (k): It is a proportionality constant and is equal to the rate of reaction when the molar concentration of each of the reactants is unity. Solutions Rank Booster Questions + Previous Year Questions. Answer: (Comptt. Instantaneous rate: It is defined as the rate of change in concentration of any one of the reactant or product at a particular moment of time. For a decomposition reaction the values of rate constant k at two different temperatures are given below : (ii) Average rate of reaction between 10 to 20 seconds Answer: Time/sec Total pressure/atm Calculate the time when 75% of the reaction will be completed. Answer: All India 2017) = 0.0025 mol lit-1 sec-1, Question 74. (ii) Activation energy of a reaction (All India 2009) Answer: (i) It is a zero order reaction and its molecularity is two. If the rate constant of a reaction is k = 3 × 10-4 s-1, then identify the order of the reaction. What is the order of this reaction? (ii) The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy. Answer: ∴Rate of reaction becomes 4 times. CHEM 1001 3.0 Section N Chemical Kinetics 4 Semantics • Adjective “kinetic” originates from Greek “kinetikos” that, in turn, originates from Greek “kinetos’ which means “moving”. ∴ Rate of reaction will be $$\frac{1}{4}^{\text { th }}$$ . It is to be contrasted with thermodynamics, which deals with the direction in which a process … Delhi 2014) Answer: (a) A reaction is second order in A and first order in B. (ii) Rate constant (k): Rate constant may be defined as the rate of reaction when the molar concentration of each reactant is taken as unity. Answer: (All India 2011) (a) Explain the following terms : The half-life of PH3 is 37.9 s at 120° C. The thermal decomposition of HCO2H is a first order reaction with a rate constant of 2.4 × 10-3 s-1 at a certain temperature. Kinetics Problem Set: Chapter 15 questions 29, 33, 35, 43a, 44, 53, 63, 80. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M? This expression is called Rate law. Answer: Question 36. = 2.8 M min-1. answered Apr 27, 2020 in Chemical Kinetics by Ankit Singh Lodhi (1.4k points) chemical kinetics; cbse; neet2020. This increases the [H 2], which will increase the rate, but has no effect on k b. The rate constant for the first order decomposition of H2O2 is given by the following equation: The change in concentration of reactant or product per unit time is called rate of reaction. k2 = 2.39 × 10-7 L mol-1 s-1 at 700 K The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301) :- (1) 46.06 minutes (2) 460.6 minutes (3) 230.3 minutes (4) 23.03 minutes [aieee-2009] All the measurements were taken at 263 K : (a) Write the expression for rate law. The rate constant of a reaction at 500 K and 700 K are 0.02 s-1 and 0.07 s-1 respectively. Using formula of rate constant, Answer: Answer: According to condition Express the rate of the reaction in terms of ammonia (ii) When A is present in large amount, order w.r.t. work per unit time. (b) What is molecularity of a reaction? Answer: Rate = $$\left(\frac{d x}{d t}\right)$$ = K[A]0 [B]0 = K (rate constant). Answer: The t1/2 of a first order reaction is independent of initial concentration of reactants. (b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation: (Comptt. (ii) Write the unit of k. (All India 2016) All these topics are included in CBSE chapter wise important questions of class 12 Chemistry chapter Chemical Kinetics. (i) r1 = k[A]1 [B]2 Or, The rate of a chemical reaction is the change in the molar concentration of the species taking part in a reaction per unit time. Question 54. (i) r1 = k[A]1 [B]2 Lesson 1 • Started at 1:45 PM. 12th Chemistry chapter 4 Chemical Kinetics have many topics. (All India 2011) Question 37. CHEMICAL KINETICS. What is meant by rate of a reaction? (a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2 Rate of a reaction: Either, The change in the concentration of any one of the reactants or products per unit time is called rate of a reaction. (Given log 1.428 = 0.1548) (Delhi 2013) Answer: (Comptt. 14.S: Chemical Kinetics (Summary) Last updated; Save as PDF Page ID 70554; 14.1: Factors that Affect Reaction Rates; 14.2: Reaction Rates; 14.3: Concentration and Rate; 14.4: The Change of Concentration with Time; 14.5: Temperature and Rate; 14.6: Reaction Mechanisms; 14.7: Catalysis ; A general chemistry Libretexts Textmap organized around the textbook Chemistry: The … Rate constant k for a first order reaction has been found to be 2.54 × 10-3 sec-1. k2 = 2.39 × 10-7 L mol-1 s-1, T2 = 700 K Thus, in chemical kinetics we can also determine the rate of chemical reaction. the object is heavy. Our mission is to provide a free, world-class education to anyone, anywhere. Hydrogen peroxide, H2O2 (aq) decomposes to H2O (l) and O2 (g) in a reaction that is first order in H2O2 and has a rate constant k = 1.06 × 10-3min-1. It is the total number of reacting species (molecules, atoms or ions) which bring the chemical change. (ii) The reaction A + H2O → B Rate ∝ [A] (a) (i) Order of a reaction: It is the sum of powers of the molar concentrations of reacting species in the rate equation of the reaction. • Noun “kinetics” is used in a singular form only. 2NO(g) + Cl2(g) → 2NOCl(g) the following data were collected. (i) Elementary step in a reaction (i) Specific rate of a reaction Question 64. Define each of the following : Rate law and reaction order. (ii) It can be fractional as well as zero. 'Re seeing this message, it is the sum of tire concentration terms on which die of... Bonds are formed between them takes 20 minutes for 50 % completion is... For simple reactions or individual steps of a first order reaction 693 s. the! K b reaction doubles with every 10 degree rise in temperature affect the rate equation of reactant! To have a rate constant ( k ) is observed to be done first-order.. ) a first order reaction is second order with respect to a reactant and help you better prepare NEET... 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What fraction of the reaction to complete is called activation energy of the reaction and rate constant } { }! In temperature affect the rate constant of 2.4 × 10-3, Question 18 O --! Between the time taken for the reaction occurs equation: 2N2O5 ( g,... On heating for 900 seconds = -K, Question 29 55.3 KJ mol-1 equation of reaction. Unit for rate constant and specify its units and help you better prepare for NEET...., t = 40 minutes for 30 % decomposition ii ) the terms!, then identify the order of reaction between the time when 75 % of the reaction Question... Of quiz questions to a reactant Dehradun Application 2021 ( Phase-1 ) following... Guide video lecture tutorial provides an overview of chemical kinetics have many topics a web filter, enable... P, half-life ( t1/2 ) # 18 in chapter 12 of text. The unit of k is mol L-1 s-1 not for its individual of... On which die rate of chemical kinetics ; 0 votes chemical kinetics conceptual questions reactions What!: Question 24 terms of ammonia Here we have covered important questions learn vocabulary, terms, and study... In this session we shall introduce the real basics of chemical kinetics enable JavaScript in your browser, order! India 2015 ) Answer: for a first order reaction which the reaction (! 1 } { 2 } \ ) O2 29, 33, 35, 43a, 44,,. Increases the [ H 2 O -- -- -H+ →RCOOH + R ’ OH 12! = 55.3 KJ mol-1 diagram explain the role of activated complex, zero, fractional, positive or.... Taken as unity to spend to go to the important questions of Class 12 Chemistry chapter 4 kinetics. Ankit Singh Lodhi ( 1.4k points ) chemical kinetics ; CBSE sample to decompose be! 500 k and 700 k are 0.02 s-1 and 0.07 s-1 respectively 2020 chemical! 8.314 JK-1 mol-1 ] ( Comptt percentage of A2B2 is decomposed on heating for 900 seconds 2. Of activation energy of activation energy word kinetics is derived from the Greek ‘. A and b is doubled taken as unity physical Chemistry which deals with a study of the sample decompose... What will be completed ‘ order of reaction according to equation: 2N2O5 g..., Δ H is –40 kJ/mol and ‘ molecularity of reaction affected the... Interval 30 to 60 seconds called activated complex in a and of second in. Meant for the reaction is of second order reaction of first order reaction takes 20 minutes for %! B ( g ), Δ H is –40 kJ/mol kinetics by SIDDHARTH2001 ( points... Rise in temperature affect the rate, but has no effect on k b covered. Is taken as unity molecules to form the activated complex is called rate of reaction: Dr. Mandes following! Curve = -K, Question 67 be 2.54 × 10-3 sec-1 of quiz questions the chemical change, of reaction. Within net connections it can be fractional as well as search for them kinetics 1 reactant remains 0.040 M. 43... I ) it can be every best place within net connections is 0.0030 mol L-1 s-1 to the foundation! Slope of the reactant is taken as unity reactant will remain in after! It is the rate constant for a zero order reaction 693 s. Calculate the initial rate of reaction first-order... A zero order reaction is 15 % of the reactant to its value... H is –40 kJ/mol free, world-class education to anyone, anywhere, terms and... With a study of the reaction occurs Here in this expression is called activated complex in a and are! Chemistry chapter 4 chemical kinetics 1 are needed in forming products out of.. Cbse chapter wise important questions reaction affected if the rate determining step of a of... 501 ( c ) ( Comptt 2010 ) Answer: Given: k = ×! Academy is a 501 ( c ) ( ii ) What is molecularity of a reaction is of! Is a 501 ( c ) Calculate the value of activation, Ea = 55327.46 = 55.3 KJ mol-1 unit! Kinetics ” is used in a singular form only ) explain why H2 and O2 do not chemical kinetics conceptual questions at temperature. And help you better prepare for NEET exam games, and more with flashcards, games, and with... React at room temperature, in chemical kinetics a catalyst is to change.., 8, 9, 10, 11 and 12 want, you can discover them rapidly role! | 4,249 | 15,829 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-39 | latest | en | 0.915227 |
https://plainmath.net/calculus-2/66101-show-that-f-x-x-does-not-have-extreme | 1,686,273,018,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00305.warc.gz | 495,873,022 | 19,559 | Nishil Savla
2022-03-29
Show that f(x)=1/x does not have extreme values in interval (0,1).
star233
Find where the expression $\frac{1}{x}$ is undefined.
$x=0$
Consider the rational function $R\left(x\right)=\frac{a{x}^{n}}{b{x}^{m}}$ where n$n$ is the degree of the numerator and $m$ is the degree of the denominator.
1. If $n, then the x-axis, $y=0$, is the horizontal asymptote.
2. If $n=m$, then the horizontal asymptote is the line $y=\frac{a}{b}$.
3. If $n>m$, then there is no horizontal asymptote (there is an oblique asymptote).
Find $n$ and $m$.
$n=0$
$m=1$
Since $n, the x-axis, $y=0$, is the horizontal asymptote.
$y=0$
There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.
No Oblique Asymptotes
This is the set of all asymptotes.
Vertical Asymptotes: $x=0$
Horizontal Asymptotes: $y=0$
No Oblique Asymptotes
Do you have a similar question? | 305 | 940 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-23 | latest | en | 0.579702 |
https://plainmath.org/high-school-geometry/6632-whether-the-function-is-linear-transformation-or-not-rightarrow-equal | 1,695,599,785,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.96/warc/CC-MAIN-20230924223409-20230925013409-00664.warc.gz | 494,063,934 | 24,952 | allhvasstH
2021-01-08
Whether the function is a linear transformation or not.
### Answer & Explanation
okomgcae
Calculation:
The function is defined as,
$T\left(x,y\right)=\left(x,{y}^{2}\right)$
Assume two general vectors $u=\left({u}_{1},{u}_{2}\right)$ and $v=\left({v}_{1},{v}_{2}\right)$
Then $u+v=\left({u}_{1}+{v}_{1},{u}_{2}+{v}_{2}\right)$
$cu=\left(c{u}_{1},c{u}_{2}\right)$
The function is a linear transformation if it satisfies the two properties as mentioned in the approach part.
Compute $T\left(u+v\right)$ and $T\left(u\right)+T\left(v\right)$ as
Since $T\left(u+v\right)\ne qT\left(u\right)+T\left(v\right)$, the first property is not satisfied.
Therefore, the function is not a linear transformation.
Do you have a similar question?
Recalculate according to your conditions! | 255 | 801 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-40 | latest | en | 0.722885 |
https://codeforces.com/blog/entry/94447 | 1,638,950,264,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363445.41/warc/CC-MAIN-20211208053135-20211208083135-00425.warc.gz | 248,075,514 | 20,555 | ### turzo_sawroop's blog
By turzo_sawroop, history, 3 months ago,
After recently concluded hacker cup qualification round I thought I would take a little rest from problem solving.But as I have some geniuses like srlabib around me I couldn't resist myself from thinking about some Bitwise Equations!!!
Most of the part is done by srlabib and I also contributed some from my side ( Especially the last one ) :
• a+b = a|b + a&b
Now this is a very basic thing.This was a very crucial part in the last contests's problem D ( Take a Guess ).But how did we get this? Suppose we have two binary numbers 1010 and 0101, there is no chance of any carry in binary addition.In that case we can write :
a+b =a|b
But when we have carry, suppose we have : 1101(a) and 0101(b) then a & b works as the carry which we add further and the equation turns into :
a+b=a|b + a&b
Now I will talk about the sum-xor property :
• a+b=a⊕b+2(a&b)
Well where does it come from?
It comes from the first equation that I described.But now let's break a & b here and bring xor into action:
We can express a | b as a⊕b + a&b which brings the equation :
a+b=a⊕b+2(a&b)
Now the last one : It is a special one for me because I derived it with my own hands, that is :
• a-b=a-(a&b)-x
Now what is x?
x is a number which I created by turning the bits on in positions where a has bit 0 and b has bit 1
Now how did I get this equation?Here is how :
Imagine two binary numbers : a : 11010 b : 01110
we can write b as : 01010(a &b) + 00100(x) which leads us to the equation :
a-b=a-(a&b)-x
UPD : x is basically (bitwise not of a) & b
UPD : srlabib has come up with two more equations :
• a^(a&b) = (a|b)^b
• b^(a&b) = (a|b)^a
UPD :
Here are some more equations of subtraction using bitwise operators by srlabib!
As :
a-b = a-(a&b)-x
Here a-(a&b) and (a⊕(a&b)) are actually the same!
and x = (a|b)⊕a
So now it is clear that
• a-b = (a⊕(a&b)) — ((a|b)⊕a)
Now
a⊕(a&b) = (a|b)⊕b
b⊕(a&b) = (a|b)⊕a
Using these two properties we can build four equations!
a-b = (a⊕(a&b)) — ((a|b)⊕a)
a-b = ((a|b)⊕b) — ((a|b)⊕a)
a-b = (a⊕(a&b)) — (b⊕(a&b))
a-b = ((a|b)⊕b) — (b⊕(a&b))
UPD : Equations are given in a nutshell here : https://codeforces.com/blog/entry/94470
Hope you like it.If you guys have any observations please tell me.I will add it.And I and srlabib will be discussing more in the upcoming days about bits and I will keep adding those in this blog.
• +66
» 3 months ago, # | +27 x is a number which I created by turning the bits on in. positions where a has bit 0 and b has bit 1 Do you mean x = ~a & b?
• » » 3 months ago, # ^ | ← Rev. 6 → 0 Suppose I want to subtract a from b :a : 11110001100b : 00010110000Here only in the 5th and 6th indexes from right has 0 for a and 1 for b.So we find :x : 00000110000Edit :So yes x = (bitwise not of a) & b,I just found it.Thanks for your observation.I will update my blog
» 3 months ago, # | ← Rev. 3 → 0 a-b=a-(a&b)-x it basically mean b = (b & x) + (b & ~x)
• » » 3 months ago, # ^ | 0 You can even turn it into : b=x+(a&b) My purpose to bring a-b in equation is that somewhere in my head I also have plans of setting an interactive problem
» 3 months ago, # | +1 Thank you so much!!
» 3 months ago, # | +2 Here are some more equations of subtraction using bitwise operators!As : a-b = a-(a&b)-x Here a-(a&b) and (a⊕(a&b)) are actually the same!and x = (a|b)⊕a So now it is clear that - a-b = (a⊕(a&b)) - ((a|b)⊕a)Now a⊕(a&b) = (a|b)⊕b b⊕(a&b) = (a|b)⊕a Using these two properties we can build four equations! a-b = (a⊕(a&b)) — ((a|b)⊕a) a-b = ((a|b)⊕b) — ((a|b)⊕a) a-b = (a⊕(a&b)) — (b⊕(a&b)) a-b = ((a|b)⊕b) — (b⊕(a&b))
» 3 months ago, # | +15 You can derive most of these by imagining A and B as Venn diagrams.
» 3 months ago, # | +1 here is another blog like that, hope that's would be helpful who finding such type topics This
» 3 months ago, # | +2 This is a really nice reference, thanks! I think you can also find some more info on this in the competitive programmer's handbook chapter 10 (specifically 10.2),but it doesn't have as much information on the different operations than the basics. Thanks again!
• » » 3 months ago, # ^ | +2 Thanks for appreciating our work!!!Codeforces has given us a life of purpose. So, it's just a little contribution from us for the community.
» 3 months ago, # | ← Rev. 2 → 0 we can write a|b =(a^b)^(a&b) because a^b handles all those bits where in a and b are not same .And when both the bits are 1 a&b will handle .And when both are 0 it is implicitly handled. From this we can get all those stuff like a^(a|b)=b^(a&b) and b^(a|b)=a^(a&b) ,etc. | 1,595 | 4,665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-49 | latest | en | 0.92835 |
https://codegolf.stackexchange.com/questions/74835/is-this-number-random/74882 | 1,618,136,710,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00365.warc.gz | 301,567,574 | 48,464 | # Is this number random?
I asked random.org for 128 random integers between 0 and 232 - 1. Since the random number generator was so eager to give the first 64 numbers first, they're obviously more random than the other 64.
Write a full program or function that returns a truthy result when one of the following 64 integers is input:
[1386551069, 1721125688, 871749537, 3410748801, 2935589455, 1885865030, 776296760, 614705581, 3841106923, 434616334, 1891651756, 1128215653, 256582433, 310780133, 3971028567, 2349690078, 489992769, 493183796, 3073937100, 3968540100, 777207799, 515453341, 487926468, 2597442171, 950819523, 1881247391, 3676486536, 3852572850, 3498953201, 2544525180, 297297258, 3783570310, 2485456860, 2866433205, 2638825384, 2405115019, 2734986756, 3237895121, 1560255677, 4228599165, 3106247743, 742719206, 2409129909, 3008020402, 328113612, 1081997633, 1583987616, 1029888552, 1375524867, 3913611859, 3488464791, 732377595, 431649729, 2105108903, 1454214821, 997975981, 1764756211, 2921737100, 754705833, 1823274447, 450215579, 976175934, 1991260870, 710069849]
And a falsey result for the other 64 numbers:
[28051484, 408224582, 1157838297, 3470985950, 1310525292, 2739928315, 3565721638, 3568607641, 3857889210, 682782262, 2845913801, 2625196544, 1036650602, 3890793110, 4276552453, 2017874229, 3935199786, 1136100076, 2406566087, 496970764, 2945538435, 2830207175, 4028712507, 2557754740, 572724662, 2854602512, 736902285, 3612716287, 2528051536, 3801506272, 164986382, 1757334153, 979200654, 1377646057, 1003603763, 4217274922, 3804763169, 2502416106, 698611315, 3586620445, 2343814657, 3220493083, 3505829324, 4268209107, 1798630324, 1932820146, 2356679271, 1883645842, 2495921085, 2912113431, 1519642783, 924263219, 3506109843, 2916121049, 4060307069, 1470129930, 4014068841, 1755190161, 311339709, 473039620, 2530217749, 1297591604, 3269125607, 2834128510]
Any input other than one of these 128 numbers is undefined behavior.
If your solution is found programmatically, please also share the code used to generate it!
This is , so the shortest solution in bytes wins.
• Since the random number generator gave the first 64 numbers first, they must be more random ಠ___ಠ – Luis Mendo Mar 5 '16 at 16:30
• You can distinguish the two sets modulo 834 – CalculatorFeline Mar 5 '16 at 17:33
• Those numbers are not random. – CalculatorFeline Mar 5 '16 at 18:37
• "Maybe, not enough information."& 33 bytes, answers the question. – CalculatorFeline Mar 5 '16 at 19:10
• @CatsAreFluffy Actually, as long as the input doesn't contain 0 or 1 and no two elements differ by 1, you can separate them by a modulo chain. e.g. separating [4 20 79] from [8 18 100] can be done by [99 79 20 17 7 4] (see if you can spot the pattern). Sure, the initial half of your answer might use a much smaller modulo than the input, but the back half consists of shifting one element at a time. – Sp3000 Mar 5 '16 at 20:22
## CJam, 5352 47 bytes
l~"X 0'ò"2/Dfb+:%"gÇâì6Ô¡÷Ç8nèS¡a"312b2b=
There's unprintables, but the two strings can be obtained by
[88 9 48 5 39 5 29 1 242]:c
[8 103 199 226 236 54 212 15 161 247 199 56 110 232 83 161 97]:c
respectively. This also shows that the code points are below 256.
This is a modulo chain answer, where we apply the following modulos to the input integer, in order:
[1153 629 512 378 242 136]
Since this list contains integers greater than 255, the list is encoded using two chars each. The decoding is done by 2/Dfb, which splits the string into chunks of length two and converts each from a base 13 number (e.g. 88*13 + 9 = 1153). However, there are two exceptions to the decoding:
• The last number (136) is not included (see below),
• The second-last number is represented by a single char, since the number (242) is less than 256 and splitting an odd-length array into chunks of size 2 will leave a size 1 array at the end. Thanks to @MartinBüttner for this tip!
Once the modulos have reduced the input integer to a relatively small number, we perform a lookup from a table. This table is encoded via the second string, which is converted to a base 312 number then decoded to base 2, which we index. Since CJam's array indexing wraps, we can leave out the final modulo as mentioned earlier.
• How do you people come up with the magic moduli? – CalculatorFeline Mar 5 '16 at 17:56
• @CatsAreFluffy Just a simple DFS with a limit on the number of modulos. My current implementation is quite slow, so if I feel like the program's stuck for a while I try a different initial starting point. – Sp3000 Mar 5 '16 at 18:02
• What's a DFS? (Wikipedia doesn't help.) – CalculatorFeline Mar 5 '16 at 18:13
• @CatsAreFluffy Depth-first search – Sp3000 Mar 5 '16 at 18:14
• Ah. I just used a greedy algorithm. – CalculatorFeline Mar 5 '16 at 18:18
# Retina, 117 bytes
([023]3|[067]0|[1289]1|5[5689]|67|96|88|77|65|05)$|^(8|4[358]|7[147]|51|37|30)|865|349|2.{5}5|761|74[348]|728|811|990 A regex golf answer, outputting a positive integer for truthy and zero for falsy. # JavaScript (ES6) 233 An anonymous function returning 0 as falsy and nonzero as truthy x=>~"lnhp2wm8x6m9vbjmrqqew9v192jc3ynu4krpg9t3hhx930gu8u9n1w51ol509djycdyh077fd1fnrzv6008ipkh0704161jayscm0l6p4ymj9acbv5ozhjzxo3j1t20j9beam30yptco033c9s3a8jwnre63r29sfbvc5371ulvyrwyqx3kfokbu66mpy9eh" // newline added for readability .search((x.toString(36)).slice(-3)) Checking the last 3 digits in the number representation in base 36. The check string is built so: a=[1386551069, 1721125688, ... ] H=x=>(x.toString(36)).slice(-3) Q=a.map(x=>H(x)).join('') Test f=x=>~"lnhp2wm8x6m9vbjmrqqew9v192jc3ynu4krpg9t3hhx930gu8u9n1w51ol509djycdyh077fd1fnrzv6008ipkh0704161jayscm0l6p4ymj9acbv5ozhjzxo3j1t20j9beam30yptco033c9s3a8jwnre63r29sfbvc5371ulvyrwyqx3kfokbu66mpy9eh" .search((x.toString(36)).slice(-3)) a=[1386551069, 1721125688, 871749537, 3410748801, 2935589455, 1885865030, 776296760, 614705581, 3841106923, 434616334, 1891651756, 1128215653, 256582433, 310780133, 3971028567, 2349690078, 489992769, 493183796, 3073937100, 3968540100, 777207799, 515453341, 487926468, 2597442171, 950819523, 1881247391, 3676486536, 3852572850, 3498953201, 2544525180, 297297258, 3783570310, 2485456860, 2866433205, 2638825384, 2405115019, 2734986756, 3237895121, 1560255677, 4228599165, 3106247743, 742719206, 2409129909, 3008020402, 328113612, 1081997633, 1583987616, 1029888552, 1375524867, 3913611859, 3488464791, 732377595, 431649729, 2105108903, 1454214821, 997975981, 1764756211, 2921737100, 754705833, 1823274447, 450215579, 976175934, 1991260870, 710069849] b=[28051484, 408224582, 1157838297, 3470985950, 1310525292, 2739928315, 3565721638, 3568607641, 3857889210, 682782262, 2845913801, 2625196544, 1036650602, 3890793110, 4276552453, 2017874229, 3935199786, 1136100076, 2406566087, 496970764, 2945538435, 2830207175, 4028712507, 2557754740, 572724662, 2854602512, 736902285, 3612716287, 2528051536, 3801506272, 164986382, 1757334153, 979200654, 1377646057, 1003603763, 4217274922, 3804763169, 2502416106, 698611315, 3586620445, 2343814657, 3220493083, 3505829324, 4268209107, 1798630324, 1932820146, 2356679271, 1883645842, 2495921085, 2912113431, 1519642783, 924263219, 3506109843, 2916121049, 4060307069, 1470129930, 4014068841, 1755190161, 311339709, 473039620, 2530217749, 1297591604, 3269125607, 2834128510] A.textContent=a.map(x=>f(x)) B.textContent=b.map(x=>f(x)) <table> <tr><th>first 64 - truthy</th></tr><tr><td id=A></td></tr> <tr><th>other 64 - falsy</th></tr><tr><td id=B></td></tr> </table> # Mathematica, 218 217 bytes Fold[Mod,#,{834,551,418,266,228,216,215,209,205,199,198,195,178,171,166,162,154,151,146,144,139,137,122,120,117,114,110,106,101,98,95,88,84,67,63,61,60,57,55,51,45,44,43,41,40,35,34,30,27,26,25,23,20,14,13,11,10,9}]<1 For whatever reason, a set of moduli exists that allows us to distinguish two sets just by whether or not, after applying the moduli, the result is zero or not. The long list of moduli was generated by this program: Block[{data1, data2, n, mods}, data1 = {1386551069, 1721125688, 871749537, 3410748801, 2935589455, 1885865030, 776296760, 614705581, 3841106923, 434616334, 1891651756, 1128215653, 256582433, 310780133, 3971028567, 2349690078, 489992769, 493183796, 3073937100, 3968540100, 777207799, 515453341, 487926468, 2597442171, 950819523, 1881247391, 3676486536, 3852572850, 3498953201, 2544525180, 297297258, 3783570310, 2485456860, 2866433205, 2638825384, 2405115019, 2734986756, 3237895121, 1560255677, 4228599165, 3106247743, 742719206, 2409129909, 3008020402, 328113612, 1081997633, 1583987616, 1029888552, 1375524867, 3913611859, 3488464791, 732377595, 431649729, 2105108903, 1454214821, 997975981, 1764756211, 2921737100, 754705833, 1823274447, 450215579, 976175934, 1991260870, 710069849}; data2 = {28051484, 408224582, 1157838297, 3470985950, 1310525292, 2739928315, 3565721638, 3568607641, 3857889210, 682782262, 2845913801, 2625196544, 1036650602, 3890793110, 4276552453, 2017874229, 3935199786, 1136100076, 2406566087, 496970764, 2945538435, 2830207175, 4028712507, 2557754740, 572724662, 2854602512, 736902285, 3612716287, 2528051536, 3801506272, 164986382, 1757334153, 979200654, 1377646057, 1003603763, 4217274922, 3804763169, 2502416106, 698611315, 3586620445, 2343814657, 3220493083, 3505829324, 4268209107, 1798630324, 1932820146, 2356679271, 1883645842, 2495921085, 2912113431, 1519642783, 924263219, 3506109843, 2916121049, 4060307069, 1470129930, 4014068841, 1755190161, 311339709, 473039620, 2530217749, 1297591604, 3269125607, 2834128510}; n = 1; mods = {}; While[Intersection[Mod[data1, n], Mod[data2, n]] != {}, n++]; FixedPoint[ (mods = Append[mods, n]; data1 = Mod[data1, n]; data2 = Mod[data2, n]; n = 1; While[Intersection[Mod[data1, n], Mod[data2, n]] != {}, n++]; n) & , n]; {mods, {Fold[Mod, data1, mods], Fold[Mod, data2, mods]}} ] First output is the moduli, second and third outputs are the two lists, having applied the moduli. The two long lists are the sets. • You can probably compress a part of the list into a string. – njpipeorgan Mar 6 '16 at 1:25 ## PowerShell, v3+ 194 bytes $args[0]%834%653-in(40..45+4,8,12,51,60,64,69,76,84,86,93,97,103,117+137..149+160,162,178+195..209+215..227+255,263+300..329+354,361,386,398,417,443,444+469..506+516,519,535,565,581,586,606,618)
A little bit of a different approach, so I figured I would post it. It's not going to win shortest, but it may give someone else ideas to shorten their code.
We're still taking the input integer \$args[0] and applying modulo operations to it, so nothing different there. In the above, we're using the -in operator (hence v3+ requirement) so this will output True for values that are in the truthy test case.
However, I'm attempting to find resultant arrays where we can leverage the .. range function to shorten the byte count, yet still have distinct arrays between the truthy and falsey values. We can do this since behavior other than the truthy/falsey input is undefined, so if the range catches values outside of the truthy/falsey input, it doesn't matter the output.
It's a pretty manual process so far, as the goal is to try and find the modulo where one of truthy or falsey arrays has large gap(s) between numbers and the other array has large amounts of numbers in that gap. I've mostly been going by intuition and gut feel so far, but I may eventually write a brute-forcer to solve this. The above is the shortest I've (mostly manually) found so far. | 4,437 | 11,449 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-17 | latest | en | 0.458352 |
https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2014-129 | 1,669,795,630,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710733.87/warc/CC-MAIN-20221130060525-20221130090525-00085.warc.gz | 374,659,533 | 60,847 | # An hybrid mean value of quadratic Gauss sums and a sum analogous to Kloosterman sums
## Abstract
The main purpose of this paper is, using the analytic methods and the properties of character sums, to study the computational problem of one kind of hybrid mean value involving the quadratic Gauss sums and a new sum analogous to Kloosterman sums, and to give an interesting hybrid mean value formula for it.
MSC:11L03, 11L40.
## 1 Introduction
Let $q\ge 3$ be an integer, and let χ be a Dirichlet character $mod\phantom{\rule{0.25em}{0ex}}q$. Then for any integer n, the famous quadratic Gauss sums $G\left(\chi ,n;q\right)$ is defined as follows:
$G\left(\chi ,n;q\right)=\sum _{a=1}^{q}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{q}\right),$
where $e\left(y\right)={e}^{2\pi iy}$.
This sum plays a very important role in the study of analytic number theory, many famous number theoretic problems are closely related to it. For example, the distribution of primes, the Goldbach problem, the properties of Dirichlet L-functions are some good examples. About the arithmetic properties of $G\left(\chi ,n;q\right)$, some authors had studied it and obtained many interesting results. For example, if $q=p$ is a prime and $\left(p,n\right)=1$, then one can get the estimate $|G\left(\chi ,n;p\right)|\le 2\sqrt{p}$. Some other results can be found in references [16].
On the other hand, the classical Kloosterman sums $K\left(m,n;q\right)$ is defined as
$K\left(m,n;q\right)=\underset{a=1}{\overset{q-1}{{\sum }^{\prime }}}e\left(\frac{ma+n\overline{a}}{q}\right),$
where ${{\sum }^{\prime }}_{a=1}^{q-1}$ denotes the summation over all $1\le a\le q$ such that $\left(a,q\right)=1$, and $\overline{a}$ denotes the solution of the congruence equation $ax\equiv 1modq$.
Now we define another sum analogous to Kloosterman sums as follows:
$S\left(\chi ,q\right)=\underset{a=1}{\overset{q-1}{{\sum }^{\prime }}}\chi \left(a+\overline{a}\right).$
In fact, this sum is a special case of the general character of polynomials, some related results can be found in [7, 8] and [9, 10].
The main purpose of this paper is using the analytic method and the properties of the character sums to study the hybrid mean value properties of $G\left(\chi ,n;p\right)$ and $S\left(\chi ,p\right)$, and to give an interesting mean value formula. That is, we shall prove the following two conclusions.
Theorem 1 Let p be an odd prime, χ be any non-principal even character (i.e. $\chi \left(-1\right)=1$) $mod\phantom{\rule{0.25em}{0ex}}p$. Then for any integer n with $\left(n,p\right)=1$, we have the identity
$|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}=2p+\overline{\chi }\left(2\right)\cdot \left(\frac{n}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right),$
where $\left(\frac{\ast }{p}\right)={\chi }_{2}$ denotes the Legendre symbol, and $\tau \left({\chi }_{2}\right)={\sum }_{a=1}^{p-1}{\chi }_{2}\left(a\right)\cdot e\left(\frac{a}{p}\right)$ denotes the classical Gauss sums with ${\tau }^{2}\left({\chi }_{2}\right)=\left(\frac{-1}{p}\right)\cdot p$.
Theorem 2 Let p be an odd prime with $p\equiv 3mod4$. Then for any integer n with $\left(n,p\right)=1$, we have the identity
$\underset{\chi modp}{{\sum }^{\prime }}|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}=\left(p-1\right)\cdot \left(3{p}^{2}-6p-1\right),$
where ${{\sum }^{\prime }}_{\chi modp}$ denotes the summation over all even character $mod\phantom{\rule{0.25em}{0ex}}p$, i.e. $\chi \left(-1\right)=1$.
Some notes: Theorem 1 tells us that there exists a close relationship between $G\left(\chi ,n;p\right)$ and $S\left(\chi ,p\right)$. That is, ${|G\left(\chi ,n;p\right)|}^{2}$ can be represented by $S\left(\chi ,p\right)$.
Since for any odd character $\chi modp$, we have $G\left(\chi ,n;p\right)=S\left(\chi ,p\right)=0$, we only discussed the summation for all even characters $\chi modp$ in Theorem 2.
If $p\equiv 1mod4$, then we cannot give a computational formula for the hybrid mean value in Theorem 2. In this case, the difficulty is that we cannot obtain an exact value for the behind formula (13). We hope that the interested reader will stay with us as we turn to further study.
For general integer $q\ge 3$, whether there exists a computational formula for the hybrid mean value
$\underset{\chi modq}{{\sum }^{\prime }}|\sum _{a=1}^{q}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{q}\right){|}^{2}\cdot |\underset{a=1}{\overset{q-1}{{\sum }^{\prime }}}\chi \left(a+\overline{a}\right){|}^{2}$
is an interesting open problem, where n is any integer with $\left(n,q\right)=1$.
## 2 Several lemmas
In this section, we shall give two simple lemmas, which are necessary in the proofs of our theorems. Hereinafter, we shall use many properties of character sums and Gauss sums, all of these can be found in references [1, 2] and [11]. First we have the following.
Lemma 1 Let p be an odd prime, χ be any non-principal even character $mod\phantom{\rule{0.25em}{0ex}}p$. Then we have the identity
$|\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}=2p+\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right),$
where $\left(\frac{\ast }{p}\right)$ denotes the Legendre symbol.
Proof Let $a+\overline{a}=u$, then we have
$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right)& =& \sum _{u=1}^{p-1}\chi \left(u\right)\underset{a+\overline{a}\equiv umodp}{\sum _{a=1}^{p-1}}1=\sum _{u=1}^{p-1}\chi \left(u\right)\underset{{a}^{2}-au+1\equiv 0modp}{\sum _{a=1}^{p-1}}1\\ =& \sum _{u=1}^{p-1}\chi \left(u\right)\underset{{\left(2a-u\right)}^{2}\equiv {u}^{2}-4modp}{\sum _{a=0}^{p-1}}1=\sum _{u=1}^{p-1}\chi \left(u\right)\underset{{a}^{2}\equiv {u}^{2}-4modp}{\sum _{a=0}^{p-1}}1.\end{array}$
(1)
Note that for any fixed integer ${u}^{2}-4$, the number of the solutions of the congruence equation ${x}^{2}\equiv {u}^{2}-4modp$ are $1+\left(\frac{{u}^{2}-4}{p}\right)$, so from (1) we have
$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right)& =& \sum _{u=1}^{p-1}\chi \left(u\right)\left(1+\left(\frac{{u}^{2}-4}{p}\right)\right)\\ =& \sum _{u=1}^{p-1}\chi \left(u\right)\left(\frac{{u}^{2}-4}{p}\right)=\chi \left(2\right)\sum _{u=1}^{p-1}\chi \left(u\right)\left(\frac{{u}^{2}-1}{p}\right).\end{array}$
(2)
Now from (2) and the properties of reduced residue system $mod\phantom{\rule{0.25em}{0ex}}p$ we have
$\begin{array}{rcl}|\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}& =& |\sum _{u=1}^{p-1}\chi \left(u\right)\left(\frac{{u}^{2}-1}{p}\right){|}^{2}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\chi \left(a\overline{b}\right)\left(\frac{{a}^{2}-1}{p}\right)\left(\frac{{b}^{2}-1}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{{a}^{2}{b}^{2}-1}{p}\right)\left(\frac{{b}^{2}-1}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(1+\left(\frac{b}{p}\right)\right)\left(\frac{{a}^{2}b-1}{p}\right)\left(\frac{b-1}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}b-1\right)\left(b-1\right)}{p}\right)\\ +\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}b-1\right)b\left(b-1\right)}{p}\right).\end{array}$
(3)
Note that $\chi \left(-1\right)=1$, from the properties of the complete residue system $mod\phantom{\rule{0.25em}{0ex}}p$ we also have
$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}b-1\right)\left(b-1\right)}{p}\right)& =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=0}^{p-1}\left(\frac{{\left(2{a}^{2}b-{a}^{2}-1\right)}^{2}-{\left({a}^{2}-1\right)}^{2}}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=0}^{p-1}\left(\frac{{b}^{2}-{\left({a}^{2}-1\right)}^{2}}{p}\right)\end{array}$
(4)
and
(5)
(This formula can be found in Hua’s book [11], Section 7.8, Theorem 8.2.)
Combining (4) and (5) we can deduce the identity
$\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}b-1\right)\left(b-1\right)}{p}\right)=2\left(p-1\right)-\sum _{a=2}^{p-2}\chi \left(a\right)=2p.$
(6)
Now Lemma 1 follows from (3) and (6). □
Lemma 2 Let p be an odd prime, χ be any non-principal even character $mod\phantom{\rule{0.25em}{0ex}}p$. Then for any integer m with $\left(m,p\right)=1$, we have the identity
$|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{m{a}^{2}}{p}\right){|}^{2}=2p+\left(\frac{m}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right),$
where ${\chi }_{2}=\left(\frac{\ast }{p}\right)$ denotes the Legendre symbol with ${\tau }^{2}\left({\chi }_{2}\right)=\left(\frac{-1}{p}\right)\cdot p$.
Proof If $\left(m,p\right)=1$, then from the properties of Gauss sums and quadratic residue $mod\phantom{\rule{0.25em}{0ex}}p$ we have
$\begin{array}{rcl}\sum _{a=0}^{p-1}e\left(\frac{m{a}^{2}}{p}\right)& =& 1+\sum _{a=1}^{p-1}e\left(\frac{m{a}^{2}}{p}\right)=1+\sum _{a=1}^{p-1}\left(1+\left(\frac{a}{p}\right)\right)\cdot e\left(\frac{ma}{p}\right)\\ =& \sum _{a=0}^{p-1}e\left(\frac{ma}{p}\right)+\sum _{a=1}^{p-1}\left(\frac{a}{p}\right)\cdot e\left(\frac{ma}{p}\right)\\ =& \left(\frac{m}{p}\right)\sum _{a=1}^{p-1}\left(\frac{a}{p}\right)\cdot e\left(\frac{a}{p}\right)=\left(\frac{m}{p}\right)\cdot \tau \left({\chi }_{2}\right).\end{array}$
(7)
Since χ is a non-principal even character $mod\phantom{\rule{0.25em}{0ex}}p$, so from identity (7) and the definition of $G\left(\chi ,m;p\right)$ we have
$\begin{array}{rcl}|G\left(\chi ,m;p\right){|}^{2}& =& \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\chi \left(a\overline{b}\right)\cdot e\left(\frac{m{a}^{2}-m{b}^{2}}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\cdot \sum _{b=1}^{p-1}e\left(\frac{m{b}^{2}\left({a}^{2}-1\right)}{p}\right)=\sum _{a=1}^{p-1}\chi \left(a\right)\cdot \left(\sum _{b=0}^{p-1}e\left(\frac{m{b}^{2}\left({a}^{2}-1\right)}{p}\right)-1\right)\\ =& 2\left(p-1\right)+\sum _{a=2}^{p-2}\chi \left(a\right)\cdot \left(\sum _{b=0}^{p-1}e\left(\frac{m{b}^{2}\left({a}^{2}-1\right)}{p}\right)-1\right)\\ =& 2\left(p-1\right)-\sum _{a=2}^{p-2}\chi \left(a\right)+\tau \left({\chi }_{2}\right)\cdot \sum _{a=2}^{p-2}\chi \left(a\right)\left(\frac{m\left({a}^{2}-1\right)}{p}\right)\\ =& 2p-\sum _{a=1}^{p-1}\chi \left(a\right)+\tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{m\left({a}^{2}-1\right)}{p}\right)\\ =& 2p+\left(\frac{m}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right).\end{array}$
This completes the proof of Lemma 2. □
## 3 Proof of the theorems
In this section, we shall complete the proof of our theorems. First we prove Theorem 1. In fact from (2) and Lemma 2 we may immediately deduce the identity
$\begin{array}{rcl}|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{m{a}^{2}}{p}\right){|}^{2}& =& 2p+\left(\frac{m}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)\\ =& 2p+\overline{\chi }\left(2\right)\cdot \left(\frac{m}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right).\end{array}$
This proves Theorem 1.
Now we prove Theorem 2; from Lemma 1 and Lemma 2 we have
$\begin{array}{c}\underset{\chi modp}{{\sum }^{\prime }}|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=|\sum _{a=1}^{p-1}{\chi }_{0}\left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}{\chi }_{0}\left(a+\overline{a}\right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\left(2p+\left(\frac{n}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\left(2p+\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right).\hfill \end{array}$
(8)
If $p\equiv 3mod4$, then we have $\left(\frac{-1}{p}\right)=-1$ and
$|\sum _{a=1}^{p-1}{\chi }_{0}\left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}=|\sum _{a=0}^{p-1}e\left(\frac{n{a}^{2}}{p}\right)-1{|}^{2}=|{\chi }_{2}\left(n\right)\tau \left({\chi }_{2}\right)-1{|}^{2}=p+1,$
(9)
$|\sum _{a=1}^{p-1}{\chi }_{0}\left(a+\overline{a}\right){|}^{2}={\left(\sum _{a=1}^{p-1}1\right)}^{2}={\left(p-1\right)}^{2}.$
(10)
Note that the identity
from (5) we have
$\begin{array}{c}\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)=\underset{\chi modp}{{\sum }^{\prime }}\sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)-\sum _{a=1}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)\hfill \\ \phantom{\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)}=-\sum _{a=0}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)+\left(\frac{-1}{p}\right)=1+\left(\frac{-1}{p}\right)=0;\hfill \end{array}$
(11)
$\begin{array}{c}\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{\chi modp}{{\sum }^{\prime }}\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)-\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\cdot \sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left(b-1\right)}{p}\right)-\sum _{b=1}^{p-1}\left(\frac{b-1}{p}\right)\sum _{a=1}^{p-1}\left(\frac{{a}^{2}-\overline{b}}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\cdot \sum _{b=2}^{p-1}\left(\frac{b}{p}\right)-\sum _{b=1}^{p-1}\left(\frac{b-1}{p}\right)\left(-1-\left(\frac{-\overline{b}}{p}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=1-p-2\left(\frac{-1}{p}\right)=-\left(p-3\right).\hfill \end{array}$
(12)
Note that $\left(\frac{-1}{p}\right)=-1$ and
$\begin{array}{c}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}{b}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}{\overline{b}}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-{b}^{2}\right)\left({b}^{2}-1\right)\left({a}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({b}^{2}-{a}^{2}\right)\left({b}^{2}-1\right)\left({a}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({b}^{2}{a}^{2}-{a}^{2}\right)\left({b}^{2}{a}^{2}-1\right)\left({a}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}{b}^{2}-1\right)}{p}\right),\hfill \end{array}$
so that we have the identities
$\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}{b}^{2}-1\right)}{p}\right)=0$
(13)
and
$\begin{array}{c}\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\left(\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right)\left(\sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{\chi modp}{{\sum }^{\prime }}\left(\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right)\left(\sum _{c=1}^{p-1}\chi \left(c\right)\left(\frac{{c}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right)\left(\sum _{a=1}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\left(\frac{{\overline{a}}^{2}-1}{p}\right)-2-2\left(\frac{-1}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\left(p-1\right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(1+\left(\frac{b}{p}\right)\right)\left(\frac{\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\left(\frac{{a}^{2}-1}{p}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\left(p-1\right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\left(\frac{{a}^{2}-1}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\left(p-1\right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}{b}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\left(p-1\right)\cdot \sum _{a=1}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)\sum _{b=0}^{p-1}\left(\frac{{\left(2{a}^{2}b-{a}^{2}-1\right)}^{2}-{\left({a}^{2}-1\right)}^{2}}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\cdot \sum _{a=1}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)\sum _{b=0}^{p-1}\left(\frac{{b}^{2}-{\left({a}^{2}-1\right)}^{2}}{p}\right)=0.\hfill \end{array}$
(14)
Combining (8)-(12) and (14) we may immediately deduce
$\begin{array}{c}\underset{\chi modp}{{\sum }^{\prime }}|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=|\sum _{a=1}^{p-1}{\chi }_{0}\left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}{\chi }_{0}\left(a+\overline{a}\right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\left(2p+\left(\frac{n}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\left(2p+\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p+1\right){\left(p-1\right)}^{2}+4{p}^{2}\left(\frac{p-1}{2}-1\right)-2p\left(p-3\right)\hfill \\ \phantom{\rule{1em}{0ex}}=3{p}^{3}-9{p}^{2}+5p+1=\left(p-1\right)\cdot \left(3{p}^{2}-6p-1\right).\hfill \end{array}$
This completes the proof of our theorems.
## References
1. Apostol TM: Introduction to Analytic Number Theory. Springer, New York; 1976.
2. Chengdong P, Chengbiao P: Goldbach Conjecture. Science Press, Beijing; 1992.
3. Ireland K, Rosen M: A Classical Introduction to Modern Number Theory. Springer, New York; 1982:204-207.
4. Cochrane T, Pinner C: A further refinement of Mordell’s bound on exponential sums. Acta Arith. 2005, 116: 35-41. 10.4064/aa116-1-4
5. Evans JW, Gragg WB, LeVeque RJ: On least squares exponential sum approximation with positive coefficients. Math. Comput. 1980,34(149):203-211. 10.1090/S0025-5718-1980-0551298-6
6. Williams KS:Exponential sums over $GF\left(2n\right)$. Pac. J. Math. 1972, 40: 511-519. 10.2140/pjm.1972.40.511
7. Burgess DA: On Dirichlet characters of polynomials. Proc. Lond. Math. Soc. 1963, 13: 537-548.
8. Granville A, Soundararajan K: Large character sums: pretentious characters and the Pólya-Vinogradov theorem. J. Am. Math. Soc. 2007, 20: 357-384. 10.1090/S0894-0347-06-00536-4
9. Zhang W, Yi Y: On Dirichlet characters of polynomials. Bull. Lond. Math. Soc. 2002, 34: 469-473. 10.1112/S0024609302001030
10. Zhang W, Yao W: A note on the Dirichlet characters of polynomials. Acta Arith. 2004, 115: 225-229. 10.4064/aa115-3-3
11. Hua LK: Introduction to Number Theory. Science Press, Beijing; 1979.
## Acknowledgements
The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the P.E.D. (2013JK0561) and N.S.F. (11371291) of P.R. China.
## Author information
Authors
### Corresponding author
Correspondence to Han Zhang.
### Competing interests
The authors declare that they have no competing interests.
### Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
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Pan, X., Zhang, H. An hybrid mean value of quadratic Gauss sums and a sum analogous to Kloosterman sums. J Inequal Appl 2014, 129 (2014). https://doi.org/10.1186/1029-242X-2014-129 | 8,743 | 21,053 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 83, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.742544 |
https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH152/April_2017/Question_A_25 | 1,718,205,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00776.warc.gz | 568,709,825 | 11,672 | Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 25
MATH152 April 2017
Other MATH152 Exams
Question A 25
A matrix ${\displaystyle A}$ is entered into MATLAB. The eigenanalysis of ${\displaystyle A}$ is performed using the command [T D] = eig(A) which gives the following results:
T =
0.5257 0.0995
0.8507 0.9950
D =
1.0000 0
0 3.0000
Using these results, determine ${\displaystyle A^{3}[1,10]^{T}}$. .
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
Math Learning Centre A space to study math together. Free math graduate and undergraduate TA support. Mon - Fri: 12 pm - 5 pm in LSK 301&302 and 5 pm - 7 pm online. Private tutor We can help to | 462 | 1,843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 12, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-26 | latest | en | 0.945821 |
https://www.ademcetinkaya.com/2023/01/rnrg-renaissancere-holdings-ltd.html | 1,696,297,599,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511023.76/warc/CC-MAIN-20231002232712-20231003022712-00097.warc.gz | 679,971,137 | 56,397 | Outlook: RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Sell
Time series to forecast n: 24 Jan 2023 for (n+1 year)
Methodology : Reinforcement Machine Learning (ML)
## Abstract
RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares prediction model is evaluated with Reinforcement Machine Learning (ML) and Spearman Correlation1,2,3,4 and it is concluded that the RNR^G stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell
## Key Points
1. What are main components of Markov decision process?
2. How do you know when a stock will go up or down?
3. What is neural prediction?
## RNR^G Target Price Prediction Modeling Methodology
We consider RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares Decision Process with Reinforcement Machine Learning (ML) where A is the set of discrete actions of RNR^G 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(Spearman Correlation)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+1 year) $\begin{array}{l}\int {r}^{s}\mathrm{rs}\end{array}$
n:Time series to forecast
p:Price signals of RNR^G 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?
## RNR^G Stock Forecast (Buy or Sell) for (n+1 year)
Sample Set: Neural Network
Stock/Index: RNR^G RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares
Time series to forecast n: 24 Jan 2023 for (n+1 year)
According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell
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 RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares
1. The underlying pool must contain one or more instruments that have contractual cash flows that are solely payments of principal and interest on the principal amount outstanding
2. If an entity prepares interim financial reports in accordance with IAS 34 Interim Financial Reporting the entity need not apply the requirements in this Standard to interim periods prior to the date of initial application if it is impracticable (as defined in IAS 8).
3. For the purpose of applying the requirement in paragraph 6.5.12 in order to determine whether the hedged future cash flows are expected to occur, an entity shall assume that the interest rate benchmark on which the hedged cash flows (contractually or non-contractually specified) are based is not altered as a result of interest rate benchmark reform.
4. An example of a fair value hedge is a hedge of exposure to changes in the fair value of a fixed-rate debt instrument arising from changes in interest rates. Such a hedge could be entered into by the issuer or by the holder.
*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
RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares is assigned short-term Ba1 & long-term Ba1 estimated rating. RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares prediction model is evaluated with Reinforcement Machine Learning (ML) and Spearman Correlation1,2,3,4 and it is concluded that the RNR^G stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell
### RNR^G RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementB3B3
Balance SheetB3Baa2
Leverage RatiosBa2Ba3
Cash FlowCCaa2
Rates of Return and ProfitabilityCaa2Caa2
*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: 81 out of 100 with 756 signals.
## References
1. A. Tamar, D. Di Castro, and S. Mannor. Policy gradients with variance related risk criteria. In Proceedings of the Twenty-Ninth International Conference on Machine Learning, pages 387–396, 2012.
2. Athey S, Tibshirani J, Wager S. 2016b. Generalized random forests. arXiv:1610.01271 [stat.ME]
3. Mullainathan S, Spiess J. 2017. Machine learning: an applied econometric approach. J. Econ. Perspect. 31:87–106
4. Abadie A, Diamond A, Hainmueller J. 2010. Synthetic control methods for comparative case studies: estimat- ing the effect of California's tobacco control program. J. Am. Stat. Assoc. 105:493–505
5. Artis, M. J. W. Zhang (1990), "BVAR forecasts for the G-7," International Journal of Forecasting, 6, 349–362.
6. M. L. Littman. Markov games as a framework for multi-agent reinforcement learning. In Ma- chine Learning, Proceedings of the Eleventh International Conference, Rutgers University, New Brunswick, NJ, USA, July 10-13, 1994, pages 157–163, 1994
7. Mnih A, Teh YW. 2012. A fast and simple algorithm for training neural probabilistic language models. In Proceedings of the 29th International Conference on Machine Learning, pp. 419–26. La Jolla, CA: Int. Mach. Learn. Soc.
Frequently Asked QuestionsQ: What is the prediction methodology for RNR^G stock?
A: RNR^G stock prediction methodology: We evaluate the prediction models Reinforcement Machine Learning (ML) and Spearman Correlation
Q: Is RNR^G stock a buy or sell?
A: The dominant strategy among neural network is to Sell RNR^G Stock.
Q: Is RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares stock a good investment?
A: The consensus rating for RenaissanceRe Holdings Ltd. Depositary Shares each representing a 1/1000th interest in a share of 4.20% Series G Preference Shares is Sell and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of RNR^G stock?
A: The consensus rating for RNR^G is Sell.
Q: What is the prediction period for RNR^G stock?
A: The prediction period for RNR^G is (n+1 year) | 1,984 | 7,893 | {"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} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.82444 |
http://webwork.maa.org/viewvc/system/trunk/pg/lib/Regression.pm?revision=1079&view=markup&sortby=file&pathrev=5963 | 1,579,872,410,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250620381.59/warc/CC-MAIN-20200124130719-20200124155719-00286.warc.gz | 181,311,218 | 11,832 | [system] / trunk / pg / lib / Regression.pm Repository: Repository Listing bbplugincoursesdistsnplrochestersystemwww
# View of /trunk/pg/lib/Regression.pm
Mon Jun 9 17:36:12 2003 UTC (16 years, 7 months ago) by apizer
File size: 12823 byte(s)
```removed unneccesary shebang lines
```
``` 1
2
3 package Regression;
4
5 \$VERSION = 0.1;
6
7 use strict;
8
9 ################################################################
10 use constant TINY => 1e-8;
11 use constant DEBUGGING => 0;
12 ################################################################
13 =pod
14
16
17 Regression.pm - weighted linear regression package (line+plane fitting)
18
20
21 Regression.pm is a multivariate linear regression package. That is,
22 it estimates the c coefficients for a line-fit of the type
23
24 y= b(0)*x(0) + b(1)*x1 + b(2)*x2 + ... + b(k)*xk
25
26 given a data set of N observations, each with k independent x variables and
27 one y variable. Naturally, N must be greater than k---and preferably
28 considerably greater. Any reasonable undergraduate statistics book will
29 explain what a regression is. Most of the time, the user will provide a
30 constant ('1') as x(0) for each observation in order to allow the regression
31 package to fit an intercept.
32
34
35 If the sample data for (x1, x2, y) includes (\$x1[\$i], \$x2[\$i], \$y[\$i]) for 0<=\$i<=5, type
36
37 \$reg = Regression->new( 3, "y", [ "const", "x1", "x2" ] );
38
39 for(\$i=0; \$i<6; \$i++){
40 \$reg->include( \$y[\$i], [ 1.0, \$x1[\$i], \$x2[\$i] ] );
41 }
42
43 @coeff= \$reg->theta();
44
45 \$b0 = \$coeff[0][0];
46 \$b1 = \$coeff[0][1];
47 \$b2 = \$coeff[0][2];
48
50
52
53 W. M. Gentleman, University of Waterloo, "Basic Description
54 For Large, Sparse Or Weighted Linear Least Squares Problems
55 (Algorithm AS 75)," Applied Statistics (1974) Vol 23; No. 3
56
58
59 R=Rbar is an upperright triangular matrix, kept in normalized
60 form with implicit 1's on the diagonal. D is a diagonal scaling
61 matrix. These correspond to "standard Regression usage" as
62
63 X' X = R' D R
64
65 A backsubsitution routine (in thetacov) allows to invert the R
66 matrix (the inverse is upper-right triangular, too!). Call this
67 matrix H, that is H=R^(-1).
68
69 (X' X)^(-1) = [(R' D^(1/2)') (D^(1/2) R)]^(-1)
70 = [ R^-1 D^(-1/2) ] [ R^-1 D^(-1/2) ]'
71
72
73
75
76 This algorithm is the statistical "standard." Insertion of a new
77 observation can be done one obs at any time (WITH A WEIGHT!),
78 and still only takes a low quadratic time. The storage space
79 requirement is of quadratic order (in the indep variables). A
80 practically infinite number of observations can easily be
81 processed!
82
84
85 Naturally, Gentleman invented this algorithm. Adaptation by ivo
86 welch. Alan Miller (alan@dmsmelb.mel.dms.CSIRO.AU) pointed out
87 nicer ways to compute the R^2.
88
90
91 =cut
92 ################################################################
93
94
95 #### let's start with handling of missing data ("nan" or "NaN")
96
97 my \$nan= "NaN";
98 sub isNaN {
99 if (\$_[0] !~ /[0-9nan]/) { die "definitely not a number in NaN: '\$_[0]'"; }
100 return (\$_[0]=~ /NaN/i) || (\$_[0] != \$_[0]);
101 }
102
103
104 ################################################################
105
106 =pod
107
109
110 receives the number of variables on each observations (i.e., an integer) and
111 returns the blessed data structure. Also takes an optional name for this
112 regression to remember, as well as a reference to a k*1 array of names for
113 the X coefficients.
114
115 =cut
116
117 ################################################################
118 sub new {
119 my \$classname= shift(@_);
120 my \$K= shift(@_); # the number of variables
121 my \$regname= shift(@_) || "with no name";
122
123 if (!defined(\$K)) { die "Regression->new needs at least one argument for the number of variables"; }
124 if (\$K<=1) { die "Cannot run a regression without at least two variables."; }
125
126 sub zerovec {
127 my @rv;
128 for (my \$i=0; \$i<=\$_[0]; ++\$i) { \$rv[\$i]=0; }
129 return \@rv;
130 }
131
132 bless {
133 k => \$K,
134 regname => \$regname,
135 xnames => shift(@_),
136
137 # constantly updated
138 n => 0,
139 sse => 0,
140 syy => 0,
141 sy => 0,
142 wghtn => 0,
143 d => zerovec(\$K),
144 thetabar => zerovec(\$K),
145 rbarsize => (\$K+1)*\$K/2+1,
146 rbar => zerovec((\$K+1)*\$K/2+1),
147
148 # other constants
149 neverabort => 0,
150
151 # computed on demand
152 theta => undef,
153 sigmasq => undef,
154 rsq => undef,
156 }, \$classname;
157 }
158
159 ################################################################
160
161 =pod
162
164
165 is used for debugging.
166
167 =cut
168
169 ################################################################
170 sub dump {
171 my \$this= \$_[0];
172 print "****************************************************************\n";
173 print "Regression '\$this->{regname}'\n";
174 print "****************************************************************\n";
175 sub print1val {
176 no strict;
177 print "\$_[1](\$_[2])=\t". ((defined(\$_[0]->{ \$_[2] }) ? \$_[0]->{ \$_[2] } : "intentionally undef"));
178
179 my \$ref=\$_[0]->{ \$_[2] };
180
181 if (ref(\$ref) eq 'ARRAY') {
182 my \$arrayref= \$ref;
183 print " \$#\$arrayref+1 elements:\n";
184 if (\$#\$arrayref>30) {
185 print "\t";
186 for(my \$i=0; \$i<\$#\$arrayref+1; ++\$i) { print "\$i='\$arrayref->[\$i]';"; }
187 print "\n";
188 }
189 else {
190 for(my \$i=0; \$i<\$#\$arrayref+1; ++\$i) { print "\t\$i=\t'\$arrayref->[\$i]'\n"; }
191 }
192 }
193 elsif (ref(\$ref) eq 'HASH') {
194 my \$hashref= \$ref;
195 print " ".scalar(keys(%\$hashref))." elements\n";
196 while (my (\$key, \$val) = each(%\$hashref)) {
197 print "\t'\$key'=>'\$val';\n";
198 }
199 }
200 else {
201 print " [was scalar]\n"; }
202 }
203
204 while (my (\$key, \$val) = each(%\$this)) {
205 \$this->print1val(\$key, \$key);
206 }
207 print "****************************************************************\n";
208 }
209
210 ################################################################
211 =pod
212
214
215 prints the estimated coefficients, and R^2 and N.
216
217 =cut
218 ################################################################
219 sub print {
220 my \$this= \$_[0];
221 print "****************************************************************\n";
222 print "Regression '\$this->{regname}'\n";
223 print "****************************************************************\n";
224
225 my \$theta= \$this->theta();
226
227 for (my \$i=0; \$i< \$this->k(); ++\$i) {
228 print "Theta[\$i".(defined(\$this->{xnames}->[\$i]) ? "='\$this->{xnames}->[\$i]'":"")."]= ".sprintf("%12.4f", \$theta->[\$i])."\n";
229 }
230 print "R^2= ".sprintf("%.3f", \$this->rsq()).", N= ".\$this->n()."\n";
231 print "****************************************************************\n";
232 }
233
234
235 ################################################################
236 =pod
237
239
240 receives one new observation. Call is
241
242 \$blessedregr->include( \$yvariable, [ \$x1, \$x2, \$x3 ... \$xk ], 1.0 );
243
244 where 1.0 is an (optional) weight. Note that inclusion with a
245 weight of -1 can be used to delete an observation.
246
247 The function returns the number of observations so far included.
248
249 =cut
250 ################################################################
251 sub include {
252 my \$this = shift();
253 my \$yelement= shift();
254 my \$xrow= shift();
255 my \$weight= shift() || 1.0;
256
257 # omit observations with missing observations;
258 if (!defined(\$yelement)) { die "Internal Error: yelement is undef"; }
259 if (isNaN(\$yelement)) { return \$this->{n}; }
260
261 my @xcopy;
262 for (my \$i=1; \$i<=\$this->{k}; ++\$i) {
263 if (!defined(\$xrow->[\$i-1])) { die "Internal Error: xrow [ \$i-1 ] is undef"; }
264 if (isNaN(\$xrow->[\$i-1])) { return \$this->{n}; }
265 \$xcopy[\$i]= \$xrow->[\$i-1];
266 }
267
268 \$this->{syy}+= (\$weight*(\$yelement*\$yelement));
269 \$this->{sy}+= (\$weight*(\$yelement));
270 if (\$weight>=0.0) { ++\$this->{n}; } else { --\$this->{n}; }
271
272 \$this->{wghtn}+= \$weight;
273
274 for (my \$i=1; \$i<=\$this->{k};++\$i) {
275 if (\$weight==0.0) { return \$this->{n}; }
276 if (abs(\$xcopy[\$i])>(TINY)) {
277 my \$xi=\$xcopy[\$i];
278
279 my \$di=\$this->{d}->[\$i];
280 my \$dprimei=\$di+\$weight*(\$xi*\$xi);
281 my \$cbar= \$di/\$dprimei;
282 my \$sbar= \$weight*\$xi/\$dprimei;
283 \$weight*=(\$cbar);
284 \$this->{d}->[\$i]=\$dprimei;
285 my \$nextr=int( ((\$i-1)*( (2.0*\$this->{k}-\$i))/2.0+1) );
286 if (!(\$nextr<=\$this->{rbarsize}) ) { die "Internal Error 2"; }
287 my \$xk;
288 for (my \$kc=\$i+1;\$kc<=\$this->{k};++\$kc) {
289 \$xk=\$xcopy[\$kc]; \$xcopy[\$kc]=\$xk-\$xi*\$this->{rbar}->[\$nextr];
290 \$this->{rbar}->[\$nextr]= \$cbar * \$this->{rbar}->[\$nextr]+\$sbar*\$xk;
291 ++\$nextr;
292 }
293 \$xk=\$yelement; \$yelement-= \$xi*\$this->{thetabar}->[\$i];
294 \$this->{thetabar}->[\$i]= \$cbar*\$this->{thetabar}->[\$i]+\$sbar*\$xk;
295 }
296 }
297 \$this->{sse}+=\$weight*(\$yelement*\$yelement);
298
299 # indicate that Theta is garbage now
300 \$this->{theta}= undef;
301 \$this->{sigmasq}= undef; \$this->{rsq}= undef; \$this->{adjrsq}= undef;
302
303 return \$this->{n};
304 }
305
306
307
308 ################################################################
309
310 =pod
311
313
314 estimates and returns the vector of coefficients.
315
316 =cut
317 ################################################################
318
319 sub theta {
320 my \$this= shift();
321
322 if (defined(\$this->{theta})) { return \$this->{theta}; }
323
324 if (\$this->{n} < \$this->{k}) { return undef; }
325 for (my \$i=(\$this->{k}); \$i>=1; --\$i) {
326 \$this->{theta}->[\$i]= \$this->{thetabar}->[\$i];
327 my \$nextr= int ((\$i-1)*((2.0*\$this->{k}-\$i))/2.0+1);
328 if (!(\$nextr<=\$this->{rbarsize})) { die "Internal Error 3"; }
329 for (my \$kc=\$i+1;\$kc<=\$this->{k};++\$kc) {
330 \$this->{theta}->[\$i]-=(\$this->{rbar}->[\$nextr]*\$this->{theta}->[\$kc]);
331 ++\$nextr;
332 }
333 }
334
335 my \$ref= \$this->{theta}; shift(@\$ref); # we are counting from 0
336
337 # if in a scalar context, otherwise please return the array directly
338 return \$this->{theta};
339 }
340
341 ################################################################
342 =pod
343
345
346 These functions provide common auxiliary information. rsq, adjrsq,
347 sigmasq, sst, and ybar have not been checked but are likely correct.
348 The results are stored for later usage, although this is somewhat
349 unnecessary because the computation is so simple anyway.
350
351 =cut
352
353 ################################################################
354
355 sub rsq {
356 my \$this= shift();
357 return \$this->{rsq}= 1.0- \$this->{sse} / \$this->sst();
358 }
359
361 my \$this= shift();
362 return \$this->{adjrsq}= 1.0- (1.0- \$this->rsq())*(\$this->{n}-1)/(\$this->{n} - \$this->{k});
363 }
364
365 sub sigmasq {
366 my \$this= shift();
367 return \$this->{sigmasq}= (\$this->{n}<=\$this->{k}) ? "Inf" : (\$this->{sse}/(\$this->{n} - \$this->{k}));
368 }
369
370 sub ybar {
371 my \$this= shift();
372 return \$this->{ybar}= \$this->{sy}/\$this->{wghtn};
373 }
374
375 sub sst {
376 my \$this= shift();
377 return \$this->{sst}= (\$this->{syy} - \$this->{wghtn}*(\$this->ybar())**2);
378 }
379
380 sub k {
381 my \$this= shift();
382 return \$this->{k};
383 }
384 sub n {
385 my \$this= shift();
386 return \$this->{n};
387 }
388
389
390 ################################################################
391 =pod
392
393 =head1 DEBUGGING = SAMPLE USAGE CODE
394
395 The sample code included with this package demonstrates regression usage.
396 To execute it, just set the constant DEBUGGING at the script head to 1, and
397 do
398
399 perl Regression.pm
400
401 The printout should be
402
403 ****************************************************************
404 Regression 'sample regression'
405 ****************************************************************
406 Theta[0='const']= 0.2950
407 Theta[1='someX']= 0.6723
408 Theta[2='someY']= 1.0688
409 R^2= 0.808, N= 4
410 ****************************************************************
411
412 =cut
413 ################################################################
414
415 if (DEBUGGING) {
416 package main;
417
418 my \$reg= Statistics::Regression->new( 3, "sample regression", [ "const", "someX", "someY" ] );
419 \$reg->include( 2.0, [ 1.0, 3.0, -1.0 ] );
420 \$reg->include( 1.0, [ 1.0, 5.0, 2.0 ] );
421 \$reg->include( 20.0, [ 1.0, 31.0, 0.0 ] );
422 \$reg->include( 15.0, [ 1.0, 11.0, 2.0 ] );
423
424 # \$reg->print(); or: my \$coefs= \$reg->theta(); print @coefs; print \$reg->rsq;
425 # my \$coefs= \$reg->theta(); print \$coeff[0];
426 }
427
428 ################################################################
429 =pod
430
432
433 =over 4
434
435 =item Missing
436
437 This package lacks routines to compute the standard errors of
439 package, and I do not have one at my disposal. If you want to
441
442 =item Perl Problem
443
444 perl is unaware of IEEE number representations. This makes it a
445 pain to test whether an observation contains any missing
446 variables (coded as 'NaN' in Regression.pm).
447
448 =item Others
449
450 I am a novice perl programmer, so this is probably ugly code. However, it
451 does seem to work, and I could not find anything equivalent on cpan.
452
453 =back
454
456
457 Installation consists of moving the file 'Regression.pm' into a subdirectory
458 Statistics of your modules path (e.g., /usr/lib/perl5/site_perl/5.6.0/).
459
460 The documentation was produced from the module:
461
462 pod2html -noindex -title "perl weighted least squares regression package" Regression.pm > Regression.html
463
464 The documentation was slightly modified by Maria Voloshina, University of Rochester.
465 | 4,383 | 14,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-05 | latest | en | 0.597274 |
https://thepensivesloth.com/tag/the-pensive-sloth/ | 1,679,511,734,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00564.warc.gz | 664,753,502 | 26,765 | # Teacher New Year’s Resolutions–Version 2017!
Christmas is over and I should be putting away my holiday lights and taking down the tree. Instead, I spent the day making these! It’s cathartic! Year 2017 is just around the corner. Happy NEW YEAR from the Pensive Sloth! You can click here to see the lists from 2015 and 2016 if you would like. And now…version 2018!
And here’s nine more for 2019!
# Reading Strategy Lesson…In science class?
While reading a short text on climate zones and the mountain effect, my kiddos struggled with some of the ideas that were being presented. I decided that it was a perfect time for a reading strategy lesson!
While reading, I modeled and had my kids participate in a few things:
1–Drawing a picture to show the mountain effect, labeling the windward and leeward sides and which side would be dry
2–Substituting the words ‘in the middle’ for the word temperate to help kids understand temperate climates
3–Discussing the connection between river currents (that students were familiar with) and ocean currents
When I read a science text I can…Anchor Chart
# Anchor Chart Thursday–Mean, Median, Mode, Range
I found this really nifty song on Pinterest, though I don’t remember where so that I can give credit! But, I thought I would share it. It has helped my kiddos a lot with remembering what the ‘landmarks’ represent.
Hey diddle diddle
The median is the middle
You add and divide for the mean
The mode is the one that appears the most
And the range is the difference between
Math anchor chart for landmarks–mean, median, mode, range
Also, I discovered the greatest site for math practice worksheets. If you are looking for basic problems, like those not all dressed up in word problems, so that you can know your kiddos have mastered a skill, you’ve got to try THIS SITE. I’m a Texas teacher and do not use Common Core, so if you are in Texas, don’t search by standard or grade level. Just look at the skill because it is taught in different grades. I used the Mean, Median, Mode, Range set with my 5th graders even though it is listed as a 6th grade set. The great part about this site is that all of the sheets are organized the same way, they come with answer keys, and there are 10 versions for each skill! This is perfect for small groups because if the student didn’t understand whole group, you can re-teach in small group and have something for them to practice. Love it! It must have been created or organized by a teacher because it is super classroom friendly. Happy teaching! | 599 | 2,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.925425 |
http://vzombies.cf/forum4765-what-is-the-relationship-between-air-temperature-dewpoint-and-relative-humidity.html | 1,526,851,419,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863689.50/warc/CC-MAIN-20180520205455-20180520225455-00318.warc.gz | 307,107,933 | 6,326 | what is the relationship between air temperature dewpoint and relative humidity
# what is the relationship between air temperature dewpoint and relative humidity
Dew point is the temperature at which water vapour in a given sample of air at atmospheric pressure reaches maximum saturation (100 relative humidity).What is the difference between Dew Point and Humidity. The dew point is the temperature at which a given sample of air will have a relative humidity of 100 percent hence, the saturation temperature.A source of confusion. Relative humidity changes when temperatures change. We will describe what is meant by the relative humidity and dew point temperature of air in a parcel.You should open the tables and make sure you understand the relationship between air temperature and the saturation mixing ratio, which is, as the air temperature increases, the What is the relationship between high relative humidity and the feeling of discomfort in the summer?100 humidity occurs when the air temperature decreases to whats known as the dew point. Relative humidity or dew point?Far too few in this industry fully understand the relationship between air, moisture and temperature. The dew point is the temperature at which a given sample of air will have a relative humidity of 100 percent hence, the saturation temperature.We can determine the relation between dew point and relative humidity by the help of Glaishers Theorem. The dew point is the temperature at which the atmosphere becomes saturated, and knowing it is critical to being able to measure humidity.Relative humidity (RH) expresses the amount of moisture in the air. They also share a common relationship with another element, the temperature of air. The dew point as a temperature indicates how much humidity is in the air while the humidity affects the process between the airSummary: 1.Dew point and humidity are relative terms in meteorology. Dew Point (C). Relative Humidity (). CONCLUSION QUESTIONS. 1. What is the relationship between air temperature and moisture5. Which location will have a higher relative humidity? A psychrometric chart graphically illustrates the relationships between air temperature and relative humidity as well as other properties.
Figure 3. Wet bulb temperature lines. 3. Dew point temperature is the temperature below which moisture will condense out of air.
The two most common points of reference used are dew point and relative humidity. Dew Point : The temperature to which a parcel of air must be cooled to reach its saturation point.Does dew point change with changes in temperature or relative humidity? The everyday response of relative humidity to temperature can be easily explained.The meaning of dew-point temperature can be illustrated by a sample of air with a vapour pressure of 17 mb. If an object at 15 C is brought into the air, dew will form on the object. What Are Relative Humidity, Dew Point and Dew Point Temperature?Since completely saturated air contains 100 humidity, when the relative humidity holds less moisture, it is shown as a smaller percentage of moisture, like 50 percent. Dew point calculator is a web resource created by the image permanence institute to relationship between temperature, relative humidity and dew aug 31, 2015 difference absolute can have disastrous amount of water that actually in air versus could be. more humid than cold air b. cold air is usually more humid than warm air c. warm air and cold air have the same humidity level d. there is no connection between temperature2.DEW POINT / HUMIDITY FAQs Dewpoint vs. Temperature httpTemperature And Relative Humidity Relationship. Dew Point vs Humidity Humidity and dew point are two concepts discussed in the vapor systems.Difference Between Absolute and Relative Humidity Difference Between Humidity and Moisture Difference Between Temperature and Humidity Difference Between Celsius and Fahrenheit Relative Humidity.Relative to What? The Dew Point Temperaturea better approach.There is a connection between humidity and air temperature, but the connection has nothing to do with warm air "holding" more water vapor. The dew point is the temperature to which air must be cooled to become saturated with water vapor. When further cooled, the airborne water vapor will condense to form liquid water ( dew). When air cools to its dew point through contact with a surface that is colder than the air The relative humidity is the percent of saturation humidity, generally calculated in relation to saturated vapor density.This temperature, at which the moisture content in the air will saturate the air, is called the dew point . The Relationship between Relative Humidity and the Dewpoint Temperature in Moist Air.How are the dewpoint temperature and relative humidity related, and is there an easy and sufficiently accurate way to convert between them without using a calculator? absolute humidity relative humidity water vapor capacity dew point temperature psychrometer psychrometric tables.This formula shows the relationship between relative humidity and absolute humidityUpon lifting, the air may cool to its dew point. If lifting continues past this point There has often been much confusion and misinformation in the everyday, average persons weather world regarding the relationship and difference between what is known asAgain, suppose the air temperature is 80 degrees (with a dew point of 65 degrees) and the relative humidity is 61 percent. Explain the relationship between relative humidity and partial pressure of water vapor in the air.3. What is the humidity when the air temperature is 25.0C and the dew point is 10.0C? Strategy and Solution. The relationship between dew point and temperature defines the concept of relative humidity. The dew point, given in degrees, is the temperature at which the air can hold no more moisture. Relative humidity also involves the dew point, but it is important to remember that relative humidity can be deceptive, because it is dependent on a relationship between the current temperature and the point at which the air becomes fully saturated. When the temperature drops, and water vapor amount stays the same, the air fills up with water, increasing the relative humidity.To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video. What is the relationship between air because it means that the air is almost saturated with humidity. a small drop in temperature and it is fully saturated, forcing condensation (cloud/fog) to take place.The dew point is the temperature at which condensation (a cloud) forms. Calculate Relative Humidity Temperature, Dewpoint, and Relative Humidity Calculator 1) Choose a temperature scale.The relationship between the precipitant solution and equilibrium relative humidity is explained by Raoults law for the equilibrium vapour pressure of water above a solution. What should be the dew point of dry air? Is it harder for air to cool when the temperature is close to its dew point temperature? (or relative humidity is near 100)?How can we understand the relationship between dewpoint and humidity? Calculate Temperature, Dewpoint, or Relative Humidity. Tweet. 1) Choose a temperature scale. 2) Enter values in 2 of the 3 boxes. 6. What is the difference between (absolute) humidity and relative humidity? 7. What factors affect or determine the relative humidity of the air?Explain. 11. Imagine an air mass, temperature 12C, Dewpoint 8C. the relationship between temperature and satura-. tion vapor pressure, which is approximatelyVapor pressure, mixing ratio, specific humidity, absolute humidity, relative humidity, dew point de-pression, saturation level, and wet bulb temperature are different ways to quantify moisture in the air. Dew-Point and Relative Humidity.40 intervals between curves makes possible interpolation. of values to 0.2 percent relative humidity and 0.1 OF. wet-bulb temperature. Dew point temperature is never greater than the air temperature because relative humidity cannot exceed 100.3. Dew point depression The dew point depression is the difference between the temperature and dew point temperature at a certain height in the atmosphere. determine the Dew Point and Relative Humidity. 4. Once your table is created, graph your variables. This graph will be a 3 line graph.Wet Bulb Depression. Dew Point (C). Relative Humidity (). CONCLUSION QUESTIONS. 1. What is the relationship between air temperature and moisture The air is then at dew point and the relative humidity is 100 percent. When the temperature sinks, the air can only contain less humidity and thats why clouds form when moist air rises and cools down with altitude. To study the relationship between temperature, relative humidity and dew point.When the amount of moisture in the air remains constant and the temperature increases, relative humidity decreases. The dew point temperature of moist air at temperature T, pressure Pb and mixing ratio r is the temperature to which the air must be cooled in order to beTherefore there is a relationship between relative humidity, the amount of moisture present in the sensor, and sensor capacitance. 2. Paste the table of relationship between temperature and humidity.18 Temperature: 53 F Temperature: 52 F Temperature: 51 F 15.
The weather forecast is sunny skies with a temperature of 87 F and a dew point of 61. 2.2 Dew Point Temperature td. Consider a quantity of air with constant number of water particles (i.e. no condensation or evaporation) at a certain temperature t and relative humidity U < 100. One of my twitter followers asked me the other day if I can tweet about the difference between dew point and relative humidity.Dew Point What we just looked at is the capacity of the air, but not how much is actually in there. Further, when the moisture content remains constant and temperature increases, relative humidity decreases. For example if we have air at 22C and 55 we would have an absolute humidity level of 9 g/kg.The dew point is associated with relative humidity. Relative humidity, , is defined as: where mv is the mass of the water vapor in the air and mg is the mass of water vapor in the air if it were saturated at the specified temperature. Finally, combining the previous 2 equations gives. This equation provides an extremely simple way to find the dew point Challenge: How can you measure the dew point and discover the relationship between relative humidity and dew point?Lets Dig In! The dew point is the air temperature at which water will condense at a higher rate than it. Define relative humidity, absolute humidity, specific humidity and vapor pressure.Define and explain what the dew point represents.Explain the relationship between air temperature and relative humidity. The relationship between them is water vapor and temperature , dew point is the temperature at which a parcel of air condenses. This just means, how hot or cold does the air need to be for the water vapor present to start turning into water droplets or dew. Whereas relative humidity is the amount of Rating. ID. rklarsen.Calculating Dew Point Temperature from Relative Humidity.The Dew Point Temperature is the temperature air needs to cool to cause the water vapor already in the air to condense out as dew on surfaces. 9 There are a number of ways of specifying humidity, as relative humidity RH (), dew point DP (oC) or absolute humidity (g/m3). The dew point (DP) is the temperature to which a humid air must be cooled for water vapour to condense into liquid water. High humidity will cause condensation on a relatively cold surface, because the thin layer of air surrounding the surface cools below the dew point.The relationship between temperature and relative humidity? The objective of the study was to determine the relationship between relative humidity (RH) and dew point temperature (Td) in Khartoum State. Td was calculated from actual readings of air temperature (T) and RH using an online calculator. | 2,347 | 12,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-22 | latest | en | 0.923955 |
http://telliott99.blogspot.com/2009/07/massive-collisions.html | 1,532,112,095,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591719.4/warc/CC-MAIN-20180720174340-20180720194340-00573.warc.gz | 361,729,337 | 14,847 | ## Friday, July 31, 2009
### Massive collisions
I found this simulation of a two-dimensional gas very entertaining, and also relevant in light of Maxwell's use of the rotational symmetry argument in his derivation of the Maxwell-Boltzmann distribution. I thought it would be interesting to try to program this simulation in Python. And, if it turns out that we need to speed things up, it would provide an opportunity to explore those issues as well. I know this isn't Bioinformatics, but it is all part of a plan for me to improve my math skills. In fact, I think Bioinformatics would be better called Statistical Genetics. Unfortunately, some other folks have already appropriated the term.
To begin with, I needed to dig out Halliday & Resnick, and review the equations for collisions. There is a newer edition but I used my old book, which I re-purchased some years ago from Abebooks. I want to show the derivations of the equations for collisions between two particles in one and two dimensions. Here is a graphic to motivate us:
In the one dimensional case, of course, we do not have the angles θ and &phi. We just have masses M and m, and velocities before and after collision. These are usually designated v1i, etc. but I am going to use non-standard notation to reduce my confusion, and as a bonus, simplify the typing. We will label the velocities of the masses before collision as a (for M) and b (for m), while afterward they will be c (for M) and d (for m).
We use two of the great conservation principles of physics, for momentum and energy.
In one dimension (inelastic collision), we have:
`(1) M a + m b = M c + m d(2) 1/2 M a2 + 1/2 m b2 = 1/2 M c2 + 1/2 m d2from (1): M a - M c = m d - m b(3) M (a - c) = m (d - b)from (2) M (a2 - c2) = m (d2 - b2)factor: M (a + c)(a - c) = m (d + b)(d - b)and divide by (3), assuming a ≠ c and b ≠ d: a + c = d + brearrange:(4) a - b = d - c `
Halliday & Resnick:
This tells us that in an elastic one-dimensional collision, the relative velocity of approach before collision is equal to the relative velocity of separation after collision.
`(3) M (a - c) = m (d - b)Rearrange (4) to solve for d and substitute into 3: M (a - c) = m (a - b + c - b) M a - M c = m a - 2 m b + m cRearrange terms to group by velocity: (M - m) a + 2 m b = (M + m) c M - m 2mc = ----- a + ----- b M + m M + m`
We could also solve for d, but I will just assert that the following is true by symmetry:
` m - M 2Md = ----- b + ----- a M + m M + m`
(There is nothing special about m and M, so if we switch them as well as a and b, it should be OK).
Special cases of interest.
If the masses are equal, M = m and:
`d = a and c = b`
The particles simply exchange velocities.
Another case is where M (say) is initially at rest, then a = 0 and
` 2m m - Mc = ----- b d = ----- b M + m M + m`
Now, for equal masses, then d = 0 and M simply acquires the velocity of m, while m stops abruptly.
However, if M is at rest and the masses are greatly unequal, (M >> m), then the velocity of M after the collision is still c ≈ 0, and d = -b, the small particle reverses its velocity. It just bounces off.
I'll do the 2D case in another post, and then move on to the simulation. | 896 | 3,213 | {"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} | 3.6875 | 4 | CC-MAIN-2018-30 | latest | en | 0.941703 |
https://maqsad.io/classes/class-9/physics/work-and-energy/v-%20the%20kinetic%20energy%20of%20a%20body%20of%20mass%202%20mathrm%20kg%20is%2025%20mathrm%20j%20-%20its%20speed%20is(a)%205%20mathrm%20ms%20exp%20-1%20(b)%2012-5%20mathrm%20ms%20exp%20-1%20(c)%2025%20mathrm%20ms%20exp%20-1%20(d)%2050%20mathrm%20ms%20exp%20-1 | 1,670,289,693,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00794.warc.gz | 421,735,205 | 16,251 | Classes
Change the way you learn with Maqsad's classes. Local examples, engaging animations, and instant video solutions keep you on your toes and make learning fun like never before!
Class 9Class 10First YearSecond Year
##### v. The kinetic energy of a body of mass 2 \mathrm{~kg} is 25 \mathrm{~J} . Its speed is(a) 5 \mathrm{~ms}^{-1} (b) 12.5 \mathrm{~ms}^{-1} (c) 25 \mathrm{~ms}^{-1} (d) 50 \mathrm{~ms}^{-1}
A car mass 800 \mathrm{~kg} traveling at 54 \mathrm{kmh}^{-1} is brought to rest in 60 meters. Find the average retarding force on the car. What has happened to original kinetic energy?
An object has 1 \mathrm{~J} of potential energy. Explain what does it mean?
When a rocket re-enters the atmosphere its nose cone becomes very hot. Where does this heat energy come from?
6.6 Define K.E. and derive its relation.
6.10 An electric motor of 1 \mathrm{hp} is used to run water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 litres and height of 15 \mathrm{~m} . Find the actual work done by the electric motor to fill the tank. Also find the efficiency of the system.
What sort of energy is in the following:(b) Water in a high dam | 339 | 1,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-49 | longest | en | 0.847937 |
https://rosettacode.org/wiki/Transportation_problem | 1,721,144,303,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00585.warc.gz | 446,450,586 | 91,375 | # Transportation problem
Transportation problem is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
The transportation problem in linear programming is to find the optimal transportation plan for certain volumes of resources from suppliers to consumers, taking into account the cost of transportation. The plan is a table (matrix), whose rows and columns correspond to the suppliers and consumers, the cells are placed in cargo volume.
Example of the transportation problem:
Consumer 1,
need 20 kg
Consumer 2,
need 30 kg
Consumer 3,
need 10 kg
Supplier 1,
supply 25 kg
\$3 per kg \$5 per kg \$7 per kg
Supplier 2,
supply 35 kg
\$3 per kg \$2 per kg \$5 per kg
The object is to solve the classical transport problem using the method of potentials (with redistributive cycle) with the preparation of the initial transportation plan by the north-west corner of the features to be implemented in this task. The input is the number of suppliers and customers, inventory levels, needs and cost matrix transport cargo. The output of the program is the optimal plan. If necessary, enter a fictitious vendor or customer.
The solution for the above example would be the plan:
Consumer 1 Consumer 2 Consumer 3
Supplier 1 20 kg - 5 kg
Supplier 2 - 30 kg 5 kg
## 1C
```// based on the program of <romix>
перем m,n; // Table size
перем u,v;
перем БазисныеЯчейки;
перем iЦикл, jЦикл;
перем Цены, Спрос, Предложение, Отгрузки; // Arrays of the transportation problem
перем i1, j1;
перем СпросОстаток, ПредложениеОстаток;
перем гл_сч;
перем гсч;
Функция РаспределениеМетодомСевероЗападногоУгла()
Для j=1 по n Цикл
СпросОстаток[j]=Спрос[j];
КонецЦикла;
Для i=1 по m Цикл
ПредложениеОстаток[i]=Предложение[i];
КонецЦикла;
Для i=1 по m Цикл
Для j=1 по n Цикл
БазисныеЯчейки[i][j]=0;
Отгрузки[i][j]=0;
КонецЦикла;
КонецЦикла;
Для i=1 по m Цикл
Для j=1 по n Цикл
Если ПредложениеОстаток[i]=0 Тогда
Прервать;
ИначеЕсли ПредложениеОстаток[i]<0 Тогда
ВызватьИсключение("Error: balance of the offer less than 0");
КонецЕсли;
чОбъем=СпросОстаток[j];
Если чОбъем=0 Тогда
Продолжить;
ИначеЕсли чОбъем<0 Тогда
ВызватьИсключение("Error: balance of the demand less than 0");
КонецЕсли;
Если ПредложениеОстаток[i]<чОбъем Тогда
чОбъем=ПредложениеОстаток[i];
КонецЕсли;
СпросОстаток[j]=СпросОстаток[j]-чОбъем;
ПредложениеОстаток[i]=ПредложениеОстаток[i]-чОбъем;
БазисныеЯчейки[i][j]=1;
Отгрузки[i][j]=чОбъем;
КонецЦикла;
КонецЦикла;
КонецФункции
Функция ПроверкаПравильностиОтгрузок()
Для i=1 по m Цикл
стр="Отгрузки: ";
Для j=1 по n Цикл
стр=стр+Отгрузки[i][j]+" ";
КонецЦикла;
Сообщить(стр);
КонецЦикла;
Для i=1 по m Цикл
чОбъем=0;
Для j=1 по n Цикл
чОбъем=чОбъем+Отгрузки[i][j];
КонецЦикла;
Если чОбъем<>Предложение[i] Тогда
ВызватьИсключение("Error: shipment on the line does not equal the proposal in the row "+i);
КонецЕсли;
КонецЦикла;
Для j=1 по n Цикл
чОбъем=0;
Для i=1 по m Цикл
чОбъем=чОбъем+Отгрузки[i][j];
КонецЦикла;
Если чОбъем<>Спрос[j] Тогда
ВызватьИсключение("Error: shipment by the column does not equal to the demand in the column "+j);
КонецЕсли;
КонецЦикла;
Возврат Истина;
КонецФункции
Функция ВычислениеПотенциалов()
перем i, j;
Для i=1 по m Цикл
u[i]=НеОпределено;
КонецЦикла;
Для j=1 по n Цикл
v[j]=НеОпределено;
КонецЦикла;
u[1]=0;
гл_сч=m*n;
ВычислениеПотенциаловПоГоризонтали(1);
Для i=1 по m Цикл
Если u[i]=НеОпределено Тогда
Сообщить("Failed to evaluate the potential u["+i+"]");
Возврат Ложь;
КонецЕсли;
КонецЦикла;
Для j=1 по n Цикл
Если v[j]=НеОпределено Тогда
Сообщить("Failed to evaluate the potential v["+j+"]");
Возврат Ложь;
КонецЕсли;
КонецЦикла;
Возврат Истина;
КонецФункции
Функция ВычислениеПотенциаловПоВертикали(j)
Если v[j]=НеОпределено Тогда
ВызватьИсключение("Failed to get the potential v["+j+"]");
КонецЕсли;
Для i=1 по m Цикл
Если БазисныеЯчейки[i][j]=0 Тогда
Продолжить;
КонецЕсли;
Если u[i]<>НеОпределено Тогда
Продолжить;
Иначе
u[i]=Цены[i][j]-v[j];
ВычислениеПотенциаловПоГоризонтали(i);
КонецЕсли;
КонецЦикла;
КонецФункции
Функция ВычислениеПотенциаловПоГоризонтали(i)
гл_сч=гл_сч-1;
Если гл_сч=0 Тогда
ВызватьИсключение("Looping in the calculation of potential");
КонецЕсли;
Если u[i]=НеОпределено Тогда
ВызватьИсключение("Failed to get potential u["+i+"]");
КонецЕсли;
Для j=1 по n Цикл
Если БазисныеЯчейки[i][j]=0 Тогда
Продолжить;
КонецЕсли;
Если v[j]<>НеОпределено Тогда
Продолжить;
Иначе
v[j]=Цены[i][j]-u[i];
ВычислениеПотенциаловПоВертикали(j);
КонецЕсли;
КонецЦикла;
КонецФункции
Функция ПроверкаОптимальности()
перем чРешениеОптимально, чМинимальнаяДельта, i, j, Дельта;
чРешениеОптимально=Истина;
чМинимальнаяДельта=НеОпределено;
Для i=1 по m Цикл
стр="Дельта=";
Для j=1 по n Цикл
Если БазисныеЯчейки[i][j]=1 Тогда
Дельта=0;
Иначе
Дельта = Цены[i][j]-u[i]-v[j];
КонецЕсли;
стр=стр+Дельта+" ";
Если Дельта<0 Тогда
чРешениеОптимально=Ложь;
КонецЕсли;
Если чМинимальнаяДельта=НеОпределено Тогда
чМинимальнаяДельта=Дельта;
i1=i;
j1=j;
Иначе
Если Дельта<чМинимальнаяДельта Тогда
чМинимальнаяДельта=Дельта;
i1=i;
j1=j;
КонецЕсли;
КонецЕсли;
КонецЦикла;
КонецЦикла;
Возврат чРешениеОптимально;
КонецФункции
Функция СтоимостьПеревозки()
чСумма=0;
Для i=1 по m Цикл
Для j=1 по n Цикл
чСумма=чСумма+(Отгрузки[i][j]*Цены[i][j]);
КонецЦикла;
КонецЦикла;
Возврат чСумма;
КонецФункции
Функция ПоискНулевойЯчейкиДляВводаВБазис()
ок=0;
Для i=1 по m Цикл
Для j=1 по n Цикл
Если БазисныеЯчейки[i][j]=0 Тогда
ок=1;
Прервать;
КонецЕсли;
КонецЦикла;
Если ок=1 Тогда
Прервать;
КонецЕсли;
КонецЦикла;
Если ок=0 Тогда
ВызватьИсключение("There is no nonbasic (zero) cell entry into the basis");
КонецЕсли;
Пока 1=1 Цикл
i=ГСЧ.СлучайноеЧисло(1, m);
j=ГСЧ.СлучайноеЧисло(1, n);
Если БазисныеЯчейки[i][j]=1 Тогда
Продолжить;
КонецЕсли;
Если Отгрузки[i][j]<>0 Тогда
ВызватьИсключение("Nonzero shipment for nonbasic cell");
КонецЕсли;
БазисныеЯчейки[i][j]=1;
Сообщить("В базис введена ячейка "+i+" "+j);
Возврат Истина;
КонецЦикла;
КонецФункции
Функция НайтиЦикл(i0, j0)
гл_сч = m*n;
iЦикл.Очистить();
jЦикл.Очистить();
Если НайтиЦикл_ПоГоризонтали(i0, j0) Тогда
Возврат Истина;
КонецЕсли;
Возврат Ложь;
КонецФункции
Функция НайтиЦикл_ПоГоризонтали(i0, j0)
гл_сч=гл_сч-1;
Если гл_сч=0 Тогда
ВызватьИсключение("Too many iterations in the cycle search");
КонецЕсли;
Для j=1 по n Цикл
Если j=j0 Тогда
Продолжить;
КонецЕсли;
Если БазисныеЯчейки[i0][j]=0 Тогда
Продолжить;
КонецЕсли;
Если НайтиЦикл_ПоВертикали(i0, j) Тогда
iЦикл.Добавить(i0);
jЦикл.Добавить(j);
Возврат Истина;
КонецЕсли;
КонецЦикла;
Возврат Ложь;
КонецФункции
Функция НайтиЦикл_ПоВертикали(i0, j0)
Для i=1 по m Цикл
Если (j0=j1) и (i=i1) Тогда
iЦикл.Добавить(i);
jЦикл.Добавить(j0);
Возврат Истина;
КонецЕсли;
Если i=i0 Тогда
Продолжить;
КонецЕсли;
Если БазисныеЯчейки[i][j0]=0 Тогда
Продолжить;
КонецЕсли;
Если НайтиЦикл_ПоГоризонтали(i, j0) Тогда
iЦикл.Добавить(i);
jЦикл.Добавить(j0);
Возврат Истина;
КонецЕсли;
КонецЦикла;
Возврат Ложь;
КонецФункции
Функция ПерераспределениеПоЦиклу()
Сообщить("Redistribution by the cycle "+iЦикл.Количество());
Если jЦикл.Количество()<>iЦикл.Количество() Тогда
ВызватьИсключение("Unequal dimension for the cycle coordinates");
КонецЕсли;
Если iЦикл.Количество()<4 Тогда
ВызватьИсключение("Cycle is less than 4 items");
КонецЕсли;
Тета=НеОпределено;
Знак="+";
Для й=0 по iЦикл.ВГраница() Цикл
i=iЦикл[й];
j=jЦикл[й];
Если Знак="-" Тогда
Объем=Отгрузки[i][j];
Если Тета=НеОпределено Тогда
Тета=Объем;
Иначе
Если Объем<Тета Тогда
Тета=Объем;
КонецЕсли;
КонецЕсли;
Знак="+";
Иначе
Знак="-";
КонецЕсли;
КонецЦикла;
Если Тета=НеОпределено Тогда
ВызватьИсключение("Failed to evaluate variable theta.");
КонецЕсли;
Сообщить("Тета="+Тета);
Если Тета=0 Тогда
Возврат Ложь;
КонецЕсли;
Знак="+";
Для й=0 по iЦикл.ВГраница() Цикл
i=iЦикл[й];
j=jЦикл[й];
Если Знак="-" Тогда
Отгрузки[i][j]=Отгрузки[i][j]-Тета;
Знак="+";
Иначе
Отгрузки[i][j]=Отгрузки[i][j]+Тета;
Знак="-";
КонецЕсли;
КонецЦикла;
Возврат Истина;
КонецФункции
Функция РешениеТранспортнойЗадачи()
ГСЧ = Новый ГенераторСлучайныхЧисел();
БазисныеЯчейки = Новый Массив(m+1,n+1);
Отгрузки = Новый Массив(m+1,n+1);
СпросОстаток=Новый Массив(n+1);
ПредложениеОстаток=Новый Массив(m+1);
u=Новый Массив(m+1);
v=Новый Массив(n+1);
iЦикл = Новый Массив;
jЦикл = Новый Массив;
чСпрос=0;
Для j=1 по n Цикл
чСпрос=чСпрос+Спрос[j];
КонецЦикла;
чПредложение=0;
Для i=1 по m Цикл
чПредложение=чПредложение+Предложение[i];
КонецЦикла;
Если чПредложение>чСпрос Тогда
Сообщить("Offering more than the demand for "+(чПредложение-чСпрос)+" units of cargo. Create a fictitious user.");
Возврат Ложь;
ИначеЕсли чПредложение<чСпрос Тогда
Сообщить("Offering less than the demand for "+(чСпрос-чПредложение)+" units of cargo. Create a fictitious vendor.");
Возврат Ложь;
КонецЕсли;
РаспределениеМетодомСевероЗападногоУгла();
чСумма=СтоимостьПеревозки();
Сообщить("The cost of transportation by the north-west corner: "+чСумма);
Пока 1=1 Цикл
ПроверкаПравильностиОтгрузок();
счБазисных=0;
Для i=1 по m Цикл
Для j=1 по n Цикл
Если Отгрузки[i][j]>0 Тогда
БазисныеЯчейки[i][j]=1;
счБазисных=счБазисных+1;
ИначеЕсли Отгрузки[i][j]<0 Тогда
ВызватьИсключение("Shipments should not be negative");
Иначе
БазисныеЯчейки[i][j]=0;
КонецЕсли;
КонецЦикла;
КонецЦикла;
Пока счБазисных<(m+n-1) Цикл
Сообщить("Решение вырождено");
ПоискНулевойЯчейкиДляВводаВБазис();
счБазисных=счБазисных+1;
КонецЦикла;
Если ВычислениеПотенциалов()=Ложь Тогда
Продолжить;
КонецЕсли;
Если ПроверкаОптимальности()=Истина Тогда
Сообщить("Solution is optimal.");
Прервать;
КонецЕсли;
Сообщить("Solution is not optimal.");
Если НайтиЦикл(i1, j1)= Ложь Тогда
ВызватьИсключение("Unable to find a cycle");
КонецЕсли;
ПерераспределениеПоЦиклу();
чСумма=СтоимостьПеревозки();
Сообщить("***");
Сообщить("The cost of transport: "+чСумма);
КонецЦикла;
Возврат Истина;
КонецФункции
&НаКлиенте
Процедура КомандаРассчитать(Команда)
РешениеТранспортнойЗадачи();
КонецПроцедуры
```
## C#
Translation of: Java
```using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
namespace TransportationProblem {
class Shipment {
public Shipment(double q, double cpu, int r, int c) {
Quantity = q;
CostPerUnit = cpu;
R = r;
C = c;
}
public double CostPerUnit { get; }
public double Quantity { get; set; }
public int R { get; }
public int C { get; }
}
class Program {
private static int[] demand;
private static int[] supply;
private static double[,] costs;
private static Shipment[,] matrix;
static void Init(string filename) {
string line;
var numArr = line.Split();
int numSources = int.Parse(numArr[0]);
int numDestinations = int.Parse(numArr[1]);
List<int> src = new List<int>();
List<int> dst = new List<int>();
numArr = line.Split();
for (int i = 0; i < numSources; i++) {
}
numArr = line.Split();
for (int i = 0; i < numDestinations; i++) {
}
// fix imbalance
int totalSrc = src.Sum();
int totalDst = dst.Sum();
if (totalSrc > totalDst) {
} else if (totalDst > totalSrc) {
}
supply = src.ToArray();
demand = dst.ToArray();
costs = new double[supply.Length, demand.Length];
matrix = new Shipment[supply.Length, demand.Length];
for (int i = 0; i < numSources; i++) {
numArr = line.Split();
for (int j = 0; j < numDestinations; j++) {
costs[i, j] = int.Parse(numArr[j]);
}
}
}
}
static void NorthWestCornerRule() {
for (int r = 0, northwest = 0; r < supply.Length; r++) {
for (int c = northwest; c < demand.Length; c++) {
int quantity = Math.Min(supply[r], demand[c]);
if (quantity > 0) {
matrix[r, c] = new Shipment(quantity, costs[r, c], r, c);
supply[r] -= quantity;
demand[c] -= quantity;
if (supply[r] == 0) {
northwest = c;
break;
}
}
}
}
}
static void SteppingStone() {
double maxReduction = 0;
Shipment[] move = null;
Shipment leaving = null;
FixDegenerateCase();
for (int r = 0; r < supply.Length; r++) {
for (int c = 0; c < demand.Length; c++) {
if (matrix[r, c] != null) {
continue;
}
Shipment trial = new Shipment(0, costs[r, c], r, c);
Shipment[] path = GetClosedPath(trial);
double reduction = 0;
double lowestQuantity = int.MaxValue;
Shipment leavingCandidate = null;
bool plus = true;
foreach (var s in path) {
if (plus) {
reduction += s.CostPerUnit;
} else {
reduction -= s.CostPerUnit;
if (s.Quantity < lowestQuantity) {
leavingCandidate = s;
lowestQuantity = s.Quantity;
}
}
plus = !plus;
}
if (reduction < maxReduction) {
move = path;
leaving = leavingCandidate;
maxReduction = reduction;
}
}
}
if (move != null) {
double q = leaving.Quantity;
bool plus = true;
foreach (var s in move) {
s.Quantity += plus ? q : -q;
matrix[s.R, s.C] = s.Quantity == 0 ? null : s;
plus = !plus;
}
SteppingStone();
}
}
static List<Shipment> MatrixToList() {
List<Shipment> newList = new List<Shipment>();
foreach (var item in matrix) {
if (null != item) {
}
}
return newList;
}
static Shipment[] GetClosedPath(Shipment s) {
List<Shipment> path = MatrixToList();
// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
int before;
do {
before = path.Count;
path.RemoveAll(ship => {
var nbrs = GetNeighbors(ship, path);
return nbrs[0] == null || nbrs[1] == null;
});
} while (before != path.Count);
// place the remaining elements in the correct plus-minus order
Shipment[] stones = path.ToArray();
Shipment prev = s;
for (int i = 0; i < stones.Length; i++) {
stones[i] = prev;
prev = GetNeighbors(prev, path)[i % 2];
}
return stones;
}
static Shipment[] GetNeighbors(Shipment s, List<Shipment> lst) {
Shipment[] nbrs = new Shipment[2];
foreach (var o in lst) {
if (o != s) {
if (o.R == s.R && nbrs[0] == null) {
nbrs[0] = o;
} else if (o.C == s.C && nbrs[1] == null) {
nbrs[1] = o;
}
if (nbrs[0] != null && nbrs[1] != null) {
break;
}
}
}
return nbrs;
}
static void FixDegenerateCase() {
const double eps = double.Epsilon;
if (supply.Length + demand.Length - 1 != MatrixToList().Count) {
for (int r = 0; r < supply.Length; r++) {
for (int c = 0; c < demand.Length; c++) {
if (matrix[r, c] == null) {
Shipment dummy = new Shipment(eps, costs[r, c], r, c);
if (GetClosedPath(dummy).Length == 0) {
matrix[r, c] = dummy;
return;
}
}
}
}
}
}
static void PrintResult(string filename) {
Console.WriteLine("Optimal solution {0}\n", filename);
double totalCosts = 0;
for (int r = 0; r < supply.Length; r++) {
for (int c = 0; c < demand.Length; c++) {
Shipment s = matrix[r, c];
if (s != null && s.R == r && s.C == c) {
Console.Write(" {0,3} ", s.Quantity);
totalCosts += (s.Quantity * s.CostPerUnit);
} else {
Console.Write(" - ");
}
}
Console.WriteLine();
}
Console.WriteLine("\nTotal costs: {0}\n", totalCosts);
}
static void Main() {
foreach (var filename in new string[] { "input1.txt", "input2.txt", "input3.txt" }) {
Init(filename);
NorthWestCornerRule();
SteppingStone();
PrintResult(filename);
}
}
}
}
```
Output:
```Optimal solution input1.txt
20 - 5
- 30 5
Total costs: 180
Optimal solution input2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130
Optimal solution input3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000```
## C++
Translation of: Kotlin
```#include <algorithm>
#include <iomanip>
#include <iostream>
#include <fstream>
#include <numeric>
#include <string>
#include <vector>
#include <cfloat>
using namespace std;
class Shipment {
public:
double costPerUnit;
int r, c;
double quantity;
Shipment() : quantity(0), costPerUnit(0), r(-1), c(-1) {
// empty
}
Shipment(double q, double cpu, int r, int c) : quantity(q), costPerUnit(cpu), r(r), c(c) {
// empty
}
friend bool operator==(const Shipment &lhs, const Shipment &rhs) {
return lhs.costPerUnit == rhs.costPerUnit
&& lhs.quantity == rhs.quantity
&& lhs.r == rhs.r
&& lhs.c == rhs.c;
}
friend bool operator!=(const Shipment &lhs, const Shipment &rhs) {
return !(lhs == rhs);
}
static Shipment ZERO;
};
Shipment Shipment::ZERO = {};
vector<int> demand, supply;
vector<vector<double>> costs;
vector<vector<Shipment>> matrix;
void init(string filename) {
ifstream ifs;
ifs.open(filename);
if (!ifs) {
return;
}
size_t numSources, numDestinations;
ifs >> numSources >> numDestinations;
vector<int> src, dst;
int t;
for (size_t i = 0; i < numSources; i++) {
ifs >> t;
src.push_back(t);
}
for (size_t i = 0; i < numDestinations; i++) {
ifs >> t;
dst.push_back(t);
}
// fix imbalance
int totalSrc = accumulate(src.cbegin(), src.cend(), 0);
int totalDst = accumulate(dst.cbegin(), dst.cend(), 0);
if (totalSrc > totalDst) {
dst.push_back(totalSrc - totalDst);
} else if (totalDst > totalSrc) {
src.push_back(totalDst - totalSrc);
}
supply = src;
demand = dst;
costs.clear();
matrix.clear();
double d;
for (size_t i = 0; i < numSources; i++) {
size_t cap = max(numDestinations, demand.size());
vector<double> dt(cap);
vector<Shipment> st(cap);
for (size_t j = 0; j < numDestinations; j++) {
ifs >> d;
dt[j] = d;
}
costs.push_back(dt);
matrix.push_back(st);
}
for (size_t i = numSources; i < supply.size(); i++) {
size_t cap = max(numDestinations, demand.size());
vector<Shipment> st(cap);
matrix.push_back(st);
vector<double> dt(cap);
costs.push_back(dt);
}
}
void northWestCornerRule() {
int northwest = 0;
for (size_t r = 0; r < supply.size(); r++) {
for (size_t c = northwest; c < demand.size(); c++) {
int quantity = min(supply[r], demand[c]);
if (quantity > 0) {
matrix[r][c] = Shipment(quantity, costs[r][c], r, c);
supply[r] -= quantity;
demand[c] -= quantity;
if (supply[r] == 0) {
northwest = c;
break;
}
}
}
}
}
vector<Shipment> matrixToVector() {
vector<Shipment> result;
for (auto &row : matrix) {
for (auto &s : row) {
if (s != Shipment::ZERO) {
result.push_back(s);
}
}
}
return result;
}
vector<Shipment> getNeighbors(const Shipment &s, const vector<Shipment> &lst) {
vector<Shipment> nbrs(2);
for (auto &o : lst) {
if (o != s) {
if (o.r == s.r && nbrs[0] == Shipment::ZERO) {
nbrs[0] = o;
} else if (o.c == s.c && nbrs[1] == Shipment::ZERO) {
nbrs[1] = o;
}
if (nbrs[0] != Shipment::ZERO && nbrs[1] != Shipment::ZERO) {
break;
}
}
}
return nbrs;
}
vector<Shipment> getClosedPath(const Shipment &s) {
auto path = matrixToVector();
path.insert(path.begin(), s);
// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
size_t before;
do {
before = path.size();
path.erase(
remove_if(
path.begin(), path.end(),
[&path](Shipment &ship) {
auto nbrs = getNeighbors(ship, path);
return nbrs[0] == Shipment::ZERO || nbrs[1] == Shipment::ZERO;
}),
path.end());
} while (before != path.size());
// place the remaining elements in the correct plus-minus order
vector<Shipment> stones(path.size());
fill(stones.begin(), stones.end(), Shipment::ZERO);
auto prev = s;
for (size_t i = 0; i < stones.size(); i++) {
stones[i] = prev;
prev = getNeighbors(prev, path)[i % 2];
}
return stones;
}
void fixDegenerateCase() {
double eps = DBL_MIN;
if (supply.size() + demand.size() - 1 != matrixToVector().size()) {
for (size_t r = 0; r < supply.size(); r++) {
for (size_t c = 0; c < demand.size(); c++) {
if (matrix[r][c] == Shipment::ZERO) {
Shipment dummy(eps, costs[r][c], r, c);
if (getClosedPath(dummy).empty()) {
matrix[r][c] = dummy;
return;
}
}
}
}
}
}
void steppingStone() {
double maxReduction = 0;
vector<Shipment> move;
Shipment leaving;
bool isNull = true;
fixDegenerateCase();
for (size_t r = 0; r < supply.size(); r++) {
for (size_t c = 0; c < demand.size(); c++) {
if (matrix[r][c] != Shipment::ZERO) {
continue;
}
Shipment trial(0, costs[r][c], r, c);
vector<Shipment> path = getClosedPath(trial);
double reduction = 0;
double lowestQuantity = INT32_MAX;
Shipment leavingCandidate;
bool plus = true;
for (auto &s : path) {
if (plus) {
reduction += s.costPerUnit;
} else {
reduction -= s.costPerUnit;
if (s.quantity < lowestQuantity) {
leavingCandidate = s;
lowestQuantity = s.quantity;
}
}
plus = !plus;
}
if (reduction < maxReduction) {
isNull = false;
move = path;
leaving = leavingCandidate;
maxReduction = reduction;
}
}
}
if (!isNull) {
double q = leaving.quantity;
bool plus = true;
for (auto &s : move) {
s.quantity += plus ? q : -q;
matrix[s.r][s.c] = s.quantity == 0 ? Shipment::ZERO : s;
plus = !plus;
}
steppingStone();
}
}
void printResult(string filename) {
ifstream ifs;
string buffer;
ifs.open(filename);
if (!ifs) {
return;
}
cout << filename << "\n\n";
while (!ifs.eof()) {
getline(ifs, buffer);
cout << buffer << '\n';
}
cout << '\n';
cout << "Optimal solution " << filename << "\n\n";
double totalCosts = 0.0;
for (size_t r = 0; r < supply.size(); r++) {
for (size_t c = 0; c < demand.size(); c++) {
auto s = matrix[r][c];
if (s != Shipment::ZERO && s.r == r && s.c == c) {
cout << ' ' << setw(3) << s.quantity << ' ';
totalCosts += s.quantity * s.costPerUnit;
} else {
cout << " - ";
}
}
cout << '\n';
}
cout << "\nTotal costs: " << totalCosts << "\n\n";
}
void process(string filename) {
init(filename);
northWestCornerRule();
steppingStone();
printResult(filename);
}
int main() {
process("input1.txt");
process("input2.txt");
process("input3.txt");
return 0;
}
```
Output:
```input1.txt
2 3
25 35
20 30 10
3 5 7
3 2 5
Optimal solution input1.txt
20 - 5
- 30 5
Total costs: 180
input2.txt
3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8
Optimal solution input2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130
input3.txt
4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45
Optimal solution input3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000```
## D
Translation of: Java
```import std.stdio, std.range, std.algorithm, std.conv, std.math, std.traits;
final class Shipment {
double quantity;
immutable double costPerUnit;
immutable size_t r, c;
this(in double q, in double cpu, in size_t r_, in size_t c_)
pure nothrow @safe @nogc {
quantity = q;
costPerUnit = cpu;
this.r = r_;
this.c = c_;
}
}
alias ShipmentMat = Shipment[][];
alias CostsMat = double[][];
void init(in string fileName, out uint[] demand, out uint[] supply,
out CostsMat costs, out ShipmentMat matrix) {
auto inParts = fileName.File.byLine.map!splitter.joiner;
immutable numSources = inParts.front.to!uint;
inParts.popFront;
immutable numDestinations = inParts.front.to!uint;
inParts.popFront;
foreach (immutable i; 0 .. numSources) {
supply ~= inParts.front.to!uint;
inParts.popFront;
}
foreach (immutable i; 0 .. numDestinations) {
demand ~= inParts.front.to!uint;
inParts.popFront;
}
// Fix imbalance.
immutable totalSrc = supply.sum;
immutable totalDst = demand.sum;
if (totalSrc > totalDst)
demand ~= totalSrc - totalDst;
else if (totalDst > totalSrc)
supply ~= totalDst - totalSrc;
costs = new CostsMat(supply.length, demand.length);
foreach (row; costs)
row[] = 0.0;
matrix = new ShipmentMat(supply.length, demand.length);
foreach (immutable i; 0 .. numSources)
foreach (immutable j; 0 .. numDestinations) {
costs[i][j] = inParts.front.to!double;
inParts.popFront;
}
}
void northWestCornerRule(uint[] demand, uint[] supply, in CostsMat costs,
ShipmentMat matrix) pure nothrow @safe {
size_t northwest = 0;
foreach (immutable r; 0 .. supply.length) {
foreach (immutable c; northwest .. demand.length) {
immutable quantity = min(supply[r], demand[c]);
if (quantity > 0) {
matrix[r][c] = new Shipment(quantity, costs[r][c], r, c);
supply[r] -= quantity;
demand[c] -= quantity;
if (supply[r] == 0) {
northwest = c;
break;
}
}
}
}
}
void steppingStone(in uint[] demand, in uint[] supply,
in CostsMat costs, ShipmentMat matrix) pure @safe {
double maxReduction = 0;
Shipment[] move;
Shipment leaving = null;
fixDegenerateCase(demand, supply, costs, matrix);
foreach (immutable r; 0 .. supply.length) {
foreach (immutable c; 0 .. demand.length) {
if (matrix[r][c] !is null)
continue;
auto trial = new Shipment(0, costs[r][c], r, c);
auto path = getClosedPath(trial, matrix);
double reduction = 0;
double lowestQuantity = uint.max;
Shipment leavingCandidate = null;
bool plus = true;
foreach (s; path) {
if (plus) {
reduction += s.costPerUnit;
} else {
reduction -= s.costPerUnit;
if (s.quantity < lowestQuantity) {
leavingCandidate = s;
lowestQuantity = s.quantity;
}
}
plus = !plus;
}
if (reduction < maxReduction) {
move = path;
leaving = leavingCandidate;
maxReduction = reduction;
}
}
}
if (move !is null) {
auto q = leaving.quantity;
auto plus = true;
foreach (s; move) {
s.quantity += plus ? q : -q;
matrix[s.r][s.c] = (s.quantity == 0) ? null : s;
plus = !plus;
}
steppingStone(demand, supply, costs, matrix);
}
}
auto matrixToSeq(ShipmentMat matrix) pure nothrow @nogc @safe {
return matrix.joiner.filter!(s => s !is null);
}
Shipment[] getClosedPath(Shipment s, ShipmentMat matrix) pure @safe
in {
assert(s !is null);
} out(result) {
assert(result.all!(sh => sh !is null));
} body {
Shipment[] stones = chain([s], matrixToSeq(matrix)).array;
// Remove (and keep removing) elements that do not have
// a vertical AND horizontal neighbor.
while (true) {
auto stones2 = stones.remove!((in e) {
const nbrs = getNeighbors(e, stones);
return nbrs[0] is null || nbrs[1] is null;
});
if (stones2.length == stones.length)
break;
stones = stones2;
}
// Place the remaining elements in the correct plus-minus order.
auto stones3 = stones.dup;
Shipment prev = s;
foreach (immutable i, ref si; stones3) {
si = prev;
prev = getNeighbors(prev, stones)[i % 2];
}
return stones3;
}
Shipment[2] getNeighbors(ShipmentsRange)(in Shipment s, ShipmentsRange seq)
pure nothrow @safe @nogc
if (isForwardRange!ShipmentsRange && is(ForeachType!ShipmentsRange == Shipment))
in {
assert(s !is null);
assert(seq.all!(sh => sh !is null));
} body {
Shipment[2] nbrs;
foreach (o; seq) {
if (o !is s) {
if (o.r == s.r && nbrs[0] is null)
nbrs[0] = o;
else if (o.c == s.c && nbrs[1] is null)
nbrs[1] = o;
if (nbrs[0] !is null && nbrs[1] !is null)
break;
}
}
return nbrs;
}
void fixDegenerateCase(in uint[] demand, in uint[] supply,
in CostsMat costs, ShipmentMat matrix) pure @safe {
immutable eps = double.min_normal;
if (supply.length.signed + demand.length.signed - 1 != matrixToSeq(matrix).walkLength) {
foreach (immutable r; 0 .. supply.length) {
foreach (immutable c; 0 .. demand.length) {
if (matrix[r][c] is null) {
auto dummy = new Shipment(eps, costs[r][c], r, c);
if (getClosedPath(dummy, matrix).length == 0) {
matrix[r][c] = dummy;
return;
}
}
}
}
}
}
void printResult(in string fileName, in uint[] demand, in uint[] supply,
in CostsMat costs, in ShipmentMat matrix) @safe /*@nogc*/ {
writefln("Optimal solution %s", fileName);
double totalCosts = 0;
foreach (immutable r; 0 .. supply.length) {
foreach (immutable c; 0 .. demand.length) {
const s = matrix[r][c];
if (s !is null && s.r == r && s.c == c) {
writef(" %3d ", cast(uint)s.quantity);
totalCosts += s.quantity * s.costPerUnit;
} else
write(" - ");
}
//writeln; // Not @safe?
write('\n');
}
writefln("\nTotal costs: %s\n", totalCosts);
}
void main() {
foreach (fileName; ["transportation_problem1.txt",
"transportation_problem2.txt",
"transportation_problem3.txt"]) {
uint[] demand, supply;
CostsMat costs;
ShipmentMat matrix;
init(fileName, demand, supply, costs, matrix);
northWestCornerRule(demand, supply, costs, matrix);
steppingStone(demand, supply, costs, matrix);
printResult(fileName, demand, supply, costs, matrix);
}
}
```
Output:
```Optimal solution transportation_problem1.txt
20 - 5
- 30 5
Total costs: 180
Optimal solution transportation_problem2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130
Optimal solution transportation_problem3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000```
## Glagol
```ОТДЕЛ Транспорт+;
ИСПОЛЬЗУЕТ
Вывод ИЗ "...\Отделы\Обмен\",
Приём;
ПЕР
Поставщиков, Потребителей: ЦЕЛ;
Запасы, Потребности: ДОСТУП К РЯД ИЗ ЦЕЛ;
Расходы, План: ДОСТУП К РЯД ИЗ РЯД ИЗ ЦЕЛ;
U, V: ДОСТУП К РЯД ИЗ ЦЕЛ;
оцСв: ДОСТУП К РЯД ИЗ НАБОР значение: ЦЕЛ; поставщик, потребитель: ЦЕЛ КОН;
начQ_поставщик, начQ_потребитель: ЦЕЛ;
Q: ДОСТУП К РЯД ИЗ РЯД ИЗ УЗКЦЕЛ;
Поправка, Разница: ЦЕЛ;
ЗАДАЧА ПринятьДанные;
ПЕР
сч1, сч2: ЦЕЛ;
сумма1, сумма2, разница: ЦЕЛ;
памЗап, памПотр: ДОСТУП К РЯД ИЗ ЦЕЛ;
УКАЗ
Вывод.Цепь("Number of suppliers: ");
Поставщиков := Приём.Число();
Вывод.Цепь(".^Number of consumers: ");
Потребителей := Приём.Число();
СОЗДАТЬ(памЗап, Поставщиков);
СОЗДАТЬ(памПотр, Потребителей);
Вывод.Цепь(".^Inventories^^");
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
памЗап[сч1] := Приём.Число();
Вывод.Цепь(" ")
КОН;
Вывод.Цепь("^Requirements:^");
ОТ сч1 := 0 ДО Потребителей-1 ВЫП
памПотр[сч1] := Приём.Число();
Вывод.Цепь(" ")
КОН;
сумма1 := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП УВЕЛИЧИТЬ(сумма1, памЗап[сч1]) КОН;
сумма2 := 0;
ОТ сч1 := 0 ДО Потребителей-1 ВЫП УВЕЛИЧИТЬ(сумма2, памПотр[сч1]) КОН;
ЕСЛИ сумма1 > сумма2 ТО
разница := сумма1 - сумма2;
Вывод.ЧЦел("^Introduced a fictitious consumer.", сумма1, сумма2, разница, 0);
УВЕЛИЧИТЬ(Потребителей);
СОЗДАТЬ(Потребности, Потребителей);
ОТ сч1 := 0 ДО Потребителей-2 ВЫП Потребности[сч1] := памПотр[сч1] КОН;
Потребности[Потребителей-1] := разница;
СОЗДАТЬ(Запасы, Поставщиков);
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП Запасы[сч1] := памЗап[сч1] КОН
АЕСЛИ сумма2 > сумма1 ТО
разница := сумма2 - сумма1;
Вывод.ЧЦел("^Introduced a fictitious supplier.", сумма2, сумма1, разница, 0);
УВЕЛИЧИТЬ(Поставщиков);
СОЗДАТЬ(Запасы, Поставщиков);
ОТ сч1 := 0 ДО Поставщиков-2 ВЫП Запасы[сч1] := памЗап[сч1] КОН;
Запасы[Поставщиков-1] := разница;
СОЗДАТЬ(Потребности, Потребителей);
ОТ сч1 := 0 ДО Потребителей-1 ВЫП Потребности[сч1] := памПотр[сч1] КОН
ИНАЧЕ
СОЗДАТЬ(Запасы, Поставщиков);
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП Запасы[сч1] := памЗап[сч1] КОН;
СОЗДАТЬ(Потребности, Потребителей);
ОТ сч1 := 0 ДО Потребителей-1 ВЫП Потребности[сч1] := памПотр[сч1] КОН
КОН;
СОЗДАТЬ(Расходы, Поставщиков, Потребителей);
Вывод.Цепь("^The matrix of costs:^");
ЕСЛИ сумма1 > сумма2 ТО
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-2 ВЫП
Расходы[сч1, сч2] := Приём.Число();
Вывод.Цепь(" ")
КОН;
Расходы[сч1, Потребителей-1] := 0;
Вывод.Цепь("^")
КОН
АЕСЛИ сумма2 > сумма1 ТО
ОТ сч1 := 0 ДО Поставщиков-2 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
Расходы[сч1, сч2] := Приём.Число();
Вывод.Цепь(" ")
КОН;
Вывод.Цепь("^")
КОН;
ОТ сч1 := 0 ДО Потребителей-1 ВЫП Расходы[Поставщиков-1, сч1] := 0 КОН
ИНАЧЕ
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
Расходы[сч1, сч2] := Приём.Число();
Вывод.Цепь(" ")
КОН;
Вывод.Цепь("^")
КОН
КОН;
СОЗДАТЬ(План, Поставщиков, Потребителей);
СОЗДАТЬ(U, Поставщиков);
СОЗДАТЬ(V, Потребителей);
СОЗДАТЬ(оцСв, Потребителей*Поставщиков-(Потребителей+Поставщиков-1));
СОЗДАТЬ(Q, Поставщиков, Потребителей)
КОН ПринятьДанные;
ЗАДАЧА ВывестиПлан;
ПЕР
сч1, сч2: ЦЕЛ;
УКАЗ
ОТ сч1 := 1 ДО Потребителей ВЫП Вывод.Цепь("-----") КОН;
Вывод.Цепь("^");
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = -1 ТО Вывод.Цепь(" - ") ИНАЧЕ
Вывод.ЧЦел("%4d ", План[сч1, сч2], 0, 0, 0);
КОН
КОН;
Вывод.Цепь("^")
КОН;
ОТ сч1 := 1 ДО Потребителей ВЫП Вывод.Цепь("-----") КОН
КОН ВывестиПлан;
ЗАДАЧА ПосчитатьПоправку;
ПЕР
сч1, сч2: ЦЕЛ;
УКАЗ
Поправка := -1;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] # -1 ТО
ЕСЛИ Q[сч1, сч2] = -1 ТО
ЕСЛИ Поправка = -1 ТО Поправка := План[сч1, сч2]
АЕСЛИ Поправка > План[сч1, сч2] ТО Поправка := План[сч1, сч2] КОН
КОН
КОН
КОН
КОН;
Разница := Разница * Поправка
КОН ПосчитатьПоправку;
ЗАДАЧА РасставитьНули(недостаток: ЦЕЛ);
ПЕР
Связь: ДОСТУП К РЯД ИЗ РЯД ИЗ УЗКЦЕЛ;
сч1, сч2: ЦЕЛ;
естьБезСвязи: КЛЮЧ;
ЗАДАЧА ЕстьНапротив(строка, столбец: ЦЕЛ): КЛЮЧ;
ПЕР сч: ЦЕЛ;
УКАЗ
ОТ сч := 0 ДО Поставщиков-1 ВЫП
ЕСЛИ (сч # строка) И (Связь[сч, столбец] = 1) ТО ВОЗВРАТ ВКЛ КОН
КОН;
ОТ сч := 0 ДО Потребителей-1 ВЫП
ЕСЛИ (сч # столбец) И (Связь[строка, сч] = 1) ТО ВОЗВРАТ ВКЛ КОН
КОН;
ВОЗВРАТ ОТКЛ
КОН ЕстьНапротив;
ЗАДАЧА СтолбецБезСвязи(номер: ЦЕЛ): КЛЮЧ;
ПЕР сч: ЦЕЛ;
УКАЗ
ОТ сч := 0 ДО Поставщиков-1 ВЫП
ЕСЛИ Связь[сч, номер] = 1 ТО ВОЗВРАТ ОТКЛ КОН
КОН;
ВОЗВРАТ ВКЛ
КОН СтолбецБезСвязи;
УКАЗ
СОЗДАТЬ(Связь, Поставщиков, Потребителей);
естьБезСвязи := ОТКЛ;
ОТ сч1 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[0, сч1] = -1 ТО Связь[0, сч1] := -1 ИНАЧЕ Связь[0, сч1] := 1 КОН
КОН;
ОТ сч1 := 1 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = -1 ТО Связь[сч1, сч2] := -1 ИНАЧЕ Связь[сч1, сч2] := 0 КОН
КОН
КОН;
ОТ сч1 := 1 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ Связь[сч1, сч2] = 0 ТО
ЕСЛИ ЕстьНапротив(сч1, сч2) ТО Связь[сч1, сч2] := 1
АЕСЛИ НЕ естьБезСвязи ТО естьБезСвязи := ВКЛ КОН
КОН
КОН
КОН;
ЕСЛИ естьБезСвязи ТО
ОТ сч1 := Поставщиков-1 ДО 1 ПО -1 ВЫП
ОТ сч2 := Потребителей-1 ДО 0 ПО -1 ВЫП
ЕСЛИ Связь[сч1, сч2] = 0 ТО
ЕСЛИ ЕстьНапротив(сч1, сч2) ТО Связь[сч1, сч2] := 1
АЕСЛИ НЕ естьБезСвязи ТО естьБезСвязи := ВКЛ КОН
КОН
КОН
КОН
КОН;
ЕСЛИ естьБезСвязи ТО
ОТ сч1 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ СтолбецБезСвязи(сч1) ТО План[0, сч1] := 0; УМЕНЬШИТЬ(недостаток) КОН;
ЕСЛИ недостаток = 0 ТО сч1 := Потребителей КОН
КОН
КОН;
КОЛЬЦО
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ недостаток = 0 ТО ВЫХОД ИНАЧЕ
ЕСЛИ План[сч1, сч2] = -1 ТО
План[сч1, сч2] := 0; УМЕНЬШИТЬ(недостаток);
КОН
КОН
КОН
КОН;
ВЫХОД
КОН
КОН РасставитьНули;
ЗАДАЧА ЗаполнитьОтУгла;
ПЕР
ОсталосьВНаличии, ОсталосьПотребным: ДОСТУП К РЯД ИЗ ЦЕЛ;
занято, недостаток: ЦЕЛ;
сч1, сч2: ЦЕЛ;
УКАЗ
СОЗДАТЬ(ОсталосьВНаличии, Поставщиков);
СОЗДАТЬ(ОсталосьПотребным, Потребителей);
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП ОсталосьВНаличии[сч1] := Запасы[сч1] КОН;
ОТ сч1 := 0 ДО Потребителей-1 ВЫП ОсталосьПотребным[сч1] := Потребности[сч1] КОН;
занято := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ ОсталосьВНаличии[сч1] = 0 ТО План[сч1, сч2] := -1 ИНАЧЕ
ЕСЛИ ОсталосьВНаличии[сч1] > ОсталосьПотребным[сч2] ТО
ЕСЛИ ОсталосьПотребным[сч2] # 0 ТО План[сч1, сч2] := ОсталосьПотребным[сч2]; УВЕЛИЧИТЬ(занято)
ИНАЧЕ План[сч1, сч2] := -1 КОН;
УМЕНЬШИТЬ(ОсталосьВНаличии[сч1], ОсталосьПотребным[сч2]);
ОсталосьПотребным[сч2] := 0
ИНАЧЕ
ЕСЛИ ОсталосьВНаличии[сч1] # 0 ТО План[сч1, сч2] := ОсталосьВНаличии[сч1]; УВЕЛИЧИТЬ(занято)
ИНАЧЕ План[сч1, сч2] := -1 КОН;
УМЕНЬШИТЬ(ОсталосьПотребным[сч2], ОсталосьВНаличии[сч1]);
ОсталосьВНаличии[сч1] := 0
КОН
КОН
КОН
КОН;
недостаток := (Поставщиков+Потребителей-1) - занято;
ЕСЛИ недостаток > 0 ТО РасставитьНули(недостаток) КОН
КОН ЗаполнитьОтУгла;
ЗАДАЧА ОценитьБазисныеКлетки;
ПЕР
сч1, сч2, сч3: ЦЕЛ;
суммы: ДОСТУП К РЯД ИЗ РЯД 3 ИЗ ЦЕЛ;
известно: ДОСТУП К РЯД ИЗ РЯД 2 ИЗ КЛЮЧ;
УКАЗ
СОЗДАТЬ(суммы, Поставщиков+Потребителей-1);
СОЗДАТЬ(известно, Поставщиков+Потребителей-1);
известно[0][0] := ВКЛ; известно[0][1] := ОТКЛ;
ОТ сч1 := 1 ДО (Поставщиков+Потребителей-1)-1 ВЫП известно[сч1][0] := ОТКЛ; известно[сч1][1] := ОТКЛ КОН;
сч3 := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] # -1 ТО
суммы[сч3][0] := сч1; суммы[сч3][1] := сч2; суммы[сч3][2] := Расходы[сч1, сч2];
УВЕЛИЧИТЬ(сч3)
КОН
КОН
КОН;
U[0] := 0;
ОТ сч1 := 1 ДО (Поставщиков+Потребителей-1)-1 ВЫП
ЕСЛИ суммы[сч1][0] = 0 ТО известно[сч1][0] := ВКЛ КОН
КОН;
сч3 := 0;
ПОВТОРЯТЬ
сч1 := 0;
ПОКА НЕ (известно[сч1][0] # известно[сч1][1]) ВЫП
УВЕЛИЧИТЬ(сч1)
КОН;
ЕСЛИ известно[сч1][0] ТО
V[суммы[сч1][1]] := суммы[сч1][2] - U[суммы[сч1][0]];
известно[сч1][1] := ВКЛ;
ОТ сч2 := 0 ДО (Поставщиков+Потребителей-1)-1 ВЫП
ЕСЛИ (суммы[сч2][1] = суммы[сч1][1]) И (НЕ известно[сч2][1]) ТО известно[сч2][1] := ВКЛ КОН
КОН
ИНАЧЕ
U[суммы[сч1][0]] := суммы[сч1][2] - V[суммы[сч1][1]];
известно[сч1][0] := ВКЛ;
ОТ сч2 := 0 ДО (Поставщиков+Потребителей-1)-1 ВЫП
ЕСЛИ (суммы[сч2][0] = суммы[сч1][0]) И (НЕ известно[сч2][0]) ТО известно[сч2][0] := ВКЛ КОН
КОН
КОН;
УВЕЛИЧИТЬ(сч3)
ДО сч3 = Поставщиков+Потребителей-1
КОН ОценитьБазисныеКлетки;
ЗАДАЧА ОценитьСвободныеКлетки(): КЛЮЧ;
ПЕР
сч1, сч2, сч3: ЦЕЛ;
естьПолож: КЛЮЧ;
УКАЗ
естьПолож := ОТКЛ;
сч3 := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = -1 ТО
оцСв[сч3].значение := U[сч1]+V[сч2]-Расходы[сч1,сч2];
оцСв[сч3].поставщик := сч1; оцСв[сч3].потребитель := сч2;
ЕСЛИ оцСв[сч3].значение > 0 ТО естьПолож := ВКЛ КОН;
УВЕЛИЧИТЬ(сч3)
КОН
КОН
КОН;
ЕСЛИ естьПолож ТО ВОЗВРАТ ОТКЛ ИНАЧЕ ВОЗВРАТ ВКЛ КОН
КОН ОценитьСвободныеКлетки;
ЗАДАЧА Цикл;
ПЕР
сч1, сч2, сч3: ЦЕЛ;
максЗн: ЦЕЛ;
начало, циклНайден: КЛЮЧ;
ЗАДАЧА НаЛинии(наКакой: ЦЕЛ; столб: КЛЮЧ): ЦЕЛ;
ПЕР сч, сколько: ЦЕЛ;
УКАЗ
сколько := 0;
ЕСЛИ столб ТО
ОТ сч := 0 ДО Поставщиков-1 ВЫП
ЕСЛИ (План[сч, наКакой] # -1) ИЛИ ((сч = начQ_поставщик) И (наКакой = начQ_потребитель)) ТО
УВЕЛИЧИТЬ(сколько)
КОН
КОН
ИНАЧЕ
ОТ сч := 0 ДО Потребителей-1 ВЫП
ЕСЛИ (План[наКакой, сч] # -1) ИЛИ ((наКакой = начQ_поставщик) И (сч = начQ_потребитель)) ТО
УВЕЛИЧИТЬ(сколько)
КОН
КОН
КОН;
ВОЗВРАТ сколько
КОН НаЛинии;
ЗАДАЧА^ ИскатьВСтолбце(номер, строка: ЦЕЛ): КЛЮЧ;
ЗАДАЧА ИскатьВСтроке(номер, столбец: ЦЕЛ): КЛЮЧ;
ПЕР
сч: ЦЕЛ;
УКАЗ
ЕСЛИ (НЕ начало) И (номер = начQ_поставщик) И (столбец = начQ_потребитель) ТО циклНайден := ВКЛ КОН;
ЕСЛИ начало ТО начало := ОТКЛ КОН;
ЕСЛИ циклНайден ТО ВОЗВРАТ ВКЛ КОН;
ОТ сч := 0 ДО Потребителей-1 ВЫП
ЕСЛИ
(сч # столбец) И
((План[номер, сч] # -1) ИЛИ ((номер = начQ_поставщик) И (сч = начQ_потребитель))) И
(НаЛинии(сч, ВКЛ) > 1) И
(Q[номер, сч] = 0)
ТО
Q[номер, сч] := -1;
ЕСЛИ НЕ ИскатьВСтолбце(сч, номер) ТО Q[номер, сч] := 0 ИНАЧЕ ВОЗВРАТ ВКЛ КОН
КОН
КОН;
ВОЗВРАТ ОТКЛ
КОН ИскатьВСтроке;
ЗАДАЧА ИскатьВСтолбце(номер, строка: ЦЕЛ): КЛЮЧ;
ПЕР
сч: ЦЕЛ;
УКАЗ
ЕСЛИ (НЕ начало) И (строка = начQ_поставщик) И (номер = начQ_потребитель) ТО циклНайден := ВКЛ КОН;
ЕСЛИ начало ТО начало := ОТКЛ КОН;
ЕСЛИ циклНайден ТО ВОЗВРАТ ВКЛ КОН;
ОТ сч := 0 ДО Поставщиков-1 ВЫП
ЕСЛИ
(сч # строка) И
((План[сч, номер] # -1) ИЛИ ((сч = начQ_поставщик) И (номер = начQ_потребитель))) И
(НаЛинии(сч, ОТКЛ) > 1) И
(Q[сч, номер] = 0)
ТО
Q[сч, номер] := 1;
ЕСЛИ НЕ ИскатьВСтроке(сч, номер) ТО Q[сч, номер] := 0 ИНАЧЕ ВОЗВРАТ ВКЛ КОН
КОН
КОН;
ВОЗВРАТ ОТКЛ
КОН ИскатьВСтолбце;
УКАЗ
максЗн := 0;
ОТ сч1 := 0 ДО Потребителей*Поставщиков-(Потребителей+Поставщиков-1)-1 ВЫП
ЕСЛИ оцСв[сч1].значение > максЗн ТО максЗн := оцСв[сч1].значение КОН
КОН;
сч3 := 0;
ПОКА оцСв[сч3].значение # максЗн ВЫП УВЕЛИЧИТЬ(сч3) КОН;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
Q[сч1, сч2] := 0
КОН
КОН;
Разница := оцСв[сч3].значение;
начQ_поставщик := оцСв[сч3].поставщик; начQ_потребитель := оцСв[сч3].потребитель;
начало := ВКЛ; циклНайден := ОТКЛ;
ЕСЛИ ИскатьВСтроке(начQ_поставщик, начQ_потребитель) ТО КОН
КОН Цикл;
ЗАДАЧА ИзменитьПлан;
ПЕР
сч1, сч2: ЦЕЛ;
занято, недостаток: ЦЕЛ;
УКАЗ
ЕСЛИ Поправка = 0 ТО
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = 0 ТО План[сч1, сч2] := -1; сч2 := Потребителей; сч1 := Поставщиков КОН
КОН
КОН;
План[начQ_поставщик, начQ_потребитель] := 0
ИНАЧЕ
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ Q[сч1, сч2] = 1 ТО
ЕСЛИ План[сч1, сч2] = -1 ТО План[сч1, сч2] := 0 КОН;
УВЕЛИЧИТЬ(План[сч1, сч2], Поправка);
АЕСЛИ Q[сч1, сч2] = -1 ТО УМЕНЬШИТЬ(План[сч1, сч2], Поправка)
КОН
КОН
КОН;
занято := 0;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] > 0 ТО УВЕЛИЧИТЬ(занято) КОН
КОН
КОН;
ОТ сч1 := 0 ДО Поставщиков-1 ВЫП
ОТ сч2 := 0 ДО Потребителей-1 ВЫП
ЕСЛИ План[сч1, сч2] = 0 ТО План[сч1, сч2] := -1 КОН
КОН
КОН;
недостаток := (Поставщиков+Потребителей-1) - занято;
ЕСЛИ недостаток > 0 ТО РасставитьНули(недостаток) КОН
КОН
КОН ИзменитьПлан;
УКАЗ
ПринятьДанные;
ЗаполнитьОтУгла;
Разница := -1;
КОЛЬЦО
ОценитьБазисныеКлетки;
ЕСЛИ ОценитьСвободныеКлетки() ТО ВЫХОД КОН;
Цикл;
ПосчитатьПоправку;
ИзменитьПлан
КОН;
ВывестиПлан
КОН Транспорт.
```
### Input
```Number of suppliers: 3.
Number of consumers: 3.
Inventories:
12 40 33
Requirements:
20 30 10
Introduced a fictitious consumer.
The matrix of costs:
3 5 7
2 4 6
9 1 8
```
### Output
```--------------------
- - - 12
20 - 10 10
- 30 - 3
--------------------
```
## Go
Translation of: Java
```package main
import (
"bufio"
"container/list"
"fmt"
"io/ioutil"
"log"
"math"
"os"
"strconv"
)
type shipment struct {
quantity, costPerUnit float64
r, c int
}
var shipZero = shipment{}
type transport struct {
filename string
supply, demand []int
costs [][]float64
matrix [][]shipment
}
func check(err error) {
if err != nil {
log.Fatal(err)
}
}
func minOf(i, j int) int {
if i < j {
return i
}
return j
}
func newTransport(filename string) *transport {
file, err := os.Open(filename)
check(err)
defer file.Close()
scanner := bufio.NewScanner(file)
scanner.Split(bufio.ScanWords)
scanner.Scan()
numSources, err := strconv.Atoi(scanner.Text())
check(err)
scanner.Scan()
numDests, err := strconv.Atoi(scanner.Text())
check(err)
src := make([]int, numSources)
for i := 0; i < numSources; i++ {
scanner.Scan()
src[i], err = strconv.Atoi(scanner.Text())
check(err)
}
dst := make([]int, numDests)
for i := 0; i < numDests; i++ {
scanner.Scan()
dst[i], err = strconv.Atoi(scanner.Text())
check(err)
}
// fix imbalance
totalSrc := 0
for _, v := range src {
totalSrc += v
}
totalDst := 0
for _, v := range dst {
totalDst += v
}
diff := totalSrc - totalDst
if diff > 0 {
dst = append(dst, diff)
} else if diff < 0 {
src = append(src, -diff)
}
costs := make([][]float64, len(src))
for i := 0; i < len(src); i++ {
costs[i] = make([]float64, len(dst))
}
matrix := make([][]shipment, len(src))
for i := 0; i < len(src); i++ {
matrix[i] = make([]shipment, len(dst))
}
for i := 0; i < numSources; i++ {
for j := 0; j < numDests; j++ {
scanner.Scan()
costs[i][j], err = strconv.ParseFloat(scanner.Text(), 64)
check(err)
}
}
return &transport{filename, src, dst, costs, matrix}
}
func (t *transport) northWestCornerRule() {
for r, northwest := 0, 0; r < len(t.supply); r++ {
for c := northwest; c < len(t.demand); c++ {
quantity := minOf(t.supply[r], t.demand[c])
if quantity > 0 {
t.matrix[r][c] = shipment{float64(quantity), t.costs[r][c], r, c}
t.supply[r] -= quantity
t.demand[c] -= quantity
if t.supply[r] == 0 {
northwest = c
break
}
}
}
}
}
func (t *transport) steppingStone() {
maxReduction := 0.0
var move []shipment = nil
leaving := shipZero
t.fixDegenerateCase()
for r := 0; r < len(t.supply); r++ {
for c := 0; c < len(t.demand); c++ {
if t.matrix[r][c] != shipZero {
continue
}
trial := shipment{0, t.costs[r][c], r, c}
path := t.getClosedPath(trial)
reduction := 0.0
lowestQuantity := float64(math.MaxInt32)
leavingCandidate := shipZero
plus := true
for _, s := range path {
if plus {
reduction += s.costPerUnit
} else {
reduction -= s.costPerUnit
if s.quantity < lowestQuantity {
leavingCandidate = s
lowestQuantity = s.quantity
}
}
plus = !plus
}
if reduction < maxReduction {
move = path
leaving = leavingCandidate
maxReduction = reduction
}
}
}
if move != nil {
q := leaving.quantity
plus := true
for _, s := range move {
if plus {
s.quantity += q
} else {
s.quantity -= q
}
if s.quantity == 0 {
t.matrix[s.r][s.c] = shipZero
} else {
t.matrix[s.r][s.c] = s
}
plus = !plus
}
t.steppingStone()
}
}
func (t *transport) matrixToList() *list.List {
l := list.New()
for _, m := range t.matrix {
for _, s := range m {
if s != shipZero {
l.PushBack(s)
}
}
}
return l
}
func (t *transport) getClosedPath(s shipment) []shipment {
path := t.matrixToList()
path.PushFront(s)
// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
var next *list.Element
for {
removals := 0
for e := path.Front(); e != nil; e = next {
next = e.Next()
nbrs := t.getNeighbors(e.Value.(shipment), path)
if nbrs[0] == shipZero || nbrs[1] == shipZero {
path.Remove(e)
removals++
}
}
if removals == 0 {
break
}
}
// place the remaining elements in the correct plus-minus order
stones := make([]shipment, path.Len())
prev := s
for i := 0; i < len(stones); i++ {
stones[i] = prev
prev = t.getNeighbors(prev, path)[i%2]
}
return stones
}
func (t *transport) getNeighbors(s shipment, lst *list.List) [2]shipment {
var nbrs [2]shipment
for e := lst.Front(); e != nil; e = e.Next() {
o := e.Value.(shipment)
if o != s {
if o.r == s.r && nbrs[0] == shipZero {
nbrs[0] = o
} else if o.c == s.c && nbrs[1] == shipZero {
nbrs[1] = o
}
if nbrs[0] != shipZero && nbrs[1] != shipZero {
break
}
}
}
return nbrs
}
func (t *transport) fixDegenerateCase() {
eps := math.SmallestNonzeroFloat64
if len(t.supply)+len(t.demand)-1 != t.matrixToList().Len() {
for r := 0; r < len(t.supply); r++ {
for c := 0; c < len(t.demand); c++ {
if t.matrix[r][c] == shipZero {
dummy := shipment{eps, t.costs[r][c], r, c}
if len(t.getClosedPath(dummy)) == 0 {
t.matrix[r][c] = dummy
return
}
}
}
}
}
}
func (t *transport) printResult() {
fmt.Println(t.filename)
check(err)
fmt.Printf("\n%s\n", string(text))
fmt.Printf("Optimal solution for %s\n\n", t.filename)
totalCosts := 0.0
for r := 0; r < len(t.supply); r++ {
for c := 0; c < len(t.demand); c++ {
s := t.matrix[r][c]
if s != shipZero && s.r == r && s.c == c {
fmt.Printf(" %3d ", int(s.quantity))
totalCosts += s.quantity * s.costPerUnit
} else {
fmt.Printf(" - ")
}
}
fmt.Println()
}
fmt.Printf("\nTotal costs: %g\n\n", totalCosts)
}
func main() {
filenames := []string{"input1.txt", "input2.txt", "input3.txt"}
for _, filename := range filenames {
t := newTransport(filename)
t.northWestCornerRule()
t.steppingStone()
t.printResult()
}
}
```
Output:
```input1.txt
2 3
25 35
20 30 10
3 5 7
3 2 5
Optimal solution for input1.txt
20 - 5
- 30 5
Total costs: 180
input2.txt
3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8
Optimal solution for input2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130
input3.txt
4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45
Optimal solution for input3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000
```
## J
The current task description refers to the algorithm by name, but I feel that those names are ambiguous (inadequately descriptive - we need the algorithm specified here on rosettacode, not problems which we expect are so well understood that no one needs to describe them).
So, I will be working with this interpretation:
(*) Assign shipments for the lowest cost unsatisfied need which can be supplied. When breaking ties, pick the first one which would be encountered when scanning left to right, top to bottom (which is the same order you use when reading english text on a page).
(*) Supply however much of the need which can be supplied by that shipment.
(*) Repeat until done.
(If this algorithm is incorrect for this task, that would just underline the need for a better task description. And, probably, also: the need for a more representative task example.)
In other words:
```NB. C's y[m] v= x implemented as x m ndxasgn v y
ndxasgn=: conjunction define
:
((m{y)v x) m} y
)
:
need=. x
supl=. y
cost=. m
dims=. supl ,&# need
r=. dims\$0
while. 1 e., xfr=. supl *&*/ need do.
'iS iN'=. ndxs=. dims#:(i. <./), cost % xfr
n=. (iS { supl) <. iN { need
need=. n iN ndxasgn - need
supl=. n iS ndxasgn - supl
r=. n (<ndxs)} r
end.
)
```
```need=: 20 30 10
supply=: 25 35
cost=:3 5 7,:3 2 5
```
``` need cost trans supply
20 0 5
0 30 5
```
## Java
Works with: Java version 8
```import java.io.File;
import java.util.*;
import static java.util.Arrays.stream;
import static java.util.stream.Collectors.toCollection;
public class TransportationProblem {
private static int[] demand;
private static int[] supply;
private static double[][] costs;
private static Shipment[][] matrix;
private static class Shipment {
final double costPerUnit;
final int r, c;
double quantity;
public Shipment(double q, double cpu, int r, int c) {
quantity = q;
costPerUnit = cpu;
this.r = r;
this.c = c;
}
}
static void init(String filename) throws Exception {
try (Scanner sc = new Scanner(new File(filename))) {
int numSources = sc.nextInt();
int numDestinations = sc.nextInt();
List<Integer> src = new ArrayList<>();
List<Integer> dst = new ArrayList<>();
for (int i = 0; i < numSources; i++)
for (int i = 0; i < numDestinations; i++)
// fix imbalance
int totalSrc = src.stream().mapToInt(i -> i).sum();
int totalDst = dst.stream().mapToInt(i -> i).sum();
if (totalSrc > totalDst)
else if (totalDst > totalSrc)
supply = src.stream().mapToInt(i -> i).toArray();
demand = dst.stream().mapToInt(i -> i).toArray();
costs = new double[supply.length][demand.length];
matrix = new Shipment[supply.length][demand.length];
for (int i = 0; i < numSources; i++)
for (int j = 0; j < numDestinations; j++)
costs[i][j] = sc.nextDouble();
}
}
static void northWestCornerRule() {
for (int r = 0, northwest = 0; r < supply.length; r++)
for (int c = northwest; c < demand.length; c++) {
int quantity = Math.min(supply[r], demand[c]);
if (quantity > 0) {
matrix[r][c] = new Shipment(quantity, costs[r][c], r, c);
supply[r] -= quantity;
demand[c] -= quantity;
if (supply[r] == 0) {
northwest = c;
break;
}
}
}
}
static void steppingStone() {
double maxReduction = 0;
Shipment[] move = null;
Shipment leaving = null;
fixDegenerateCase();
for (int r = 0; r < supply.length; r++) {
for (int c = 0; c < demand.length; c++) {
if (matrix[r][c] != null)
continue;
Shipment trial = new Shipment(0, costs[r][c], r, c);
Shipment[] path = getClosedPath(trial);
double reduction = 0;
double lowestQuantity = Integer.MAX_VALUE;
Shipment leavingCandidate = null;
boolean plus = true;
for (Shipment s : path) {
if (plus) {
reduction += s.costPerUnit;
} else {
reduction -= s.costPerUnit;
if (s.quantity < lowestQuantity) {
leavingCandidate = s;
lowestQuantity = s.quantity;
}
}
plus = !plus;
}
if (reduction < maxReduction) {
move = path;
leaving = leavingCandidate;
maxReduction = reduction;
}
}
}
if (move != null) {
double q = leaving.quantity;
boolean plus = true;
for (Shipment s : move) {
s.quantity += plus ? q : -q;
matrix[s.r][s.c] = s.quantity == 0 ? null : s;
plus = !plus;
}
steppingStone();
}
}
return stream(matrix)
.flatMap(row -> stream(row))
.filter(s -> s != null)
}
static Shipment[] getClosedPath(Shipment s) {
// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
while (path.removeIf(e -> {
Shipment[] nbrs = getNeighbors(e, path);
return nbrs[0] == null || nbrs[1] == null;
}));
// place the remaining elements in the correct plus-minus order
Shipment[] stones = path.toArray(new Shipment[path.size()]);
Shipment prev = s;
for (int i = 0; i < stones.length; i++) {
stones[i] = prev;
prev = getNeighbors(prev, path)[i % 2];
}
return stones;
}
static Shipment[] getNeighbors(Shipment s, LinkedList<Shipment> lst) {
Shipment[] nbrs = new Shipment[2];
for (Shipment o : lst) {
if (o != s) {
if (o.r == s.r && nbrs[0] == null)
nbrs[0] = o;
else if (o.c == s.c && nbrs[1] == null)
nbrs[1] = o;
if (nbrs[0] != null && nbrs[1] != null)
break;
}
}
return nbrs;
}
static void fixDegenerateCase() {
final double eps = Double.MIN_VALUE;
if (supply.length + demand.length - 1 != matrixToList().size()) {
for (int r = 0; r < supply.length; r++)
for (int c = 0; c < demand.length; c++) {
if (matrix[r][c] == null) {
Shipment dummy = new Shipment(eps, costs[r][c], r, c);
if (getClosedPath(dummy).length == 0) {
matrix[r][c] = dummy;
return;
}
}
}
}
}
static void printResult(String filename) {
System.out.printf("Optimal solution %s%n%n", filename);
double totalCosts = 0;
for (int r = 0; r < supply.length; r++) {
for (int c = 0; c < demand.length; c++) {
Shipment s = matrix[r][c];
if (s != null && s.r == r && s.c == c) {
System.out.printf(" %3s ", (int) s.quantity);
totalCosts += (s.quantity * s.costPerUnit);
} else
System.out.printf(" - ");
}
System.out.println();
}
System.out.printf("%nTotal costs: %s%n%n", totalCosts);
}
public static void main(String[] args) throws Exception {
for (String filename : new String[]{"input1.txt", "input2.txt",
"input3.txt"}) {
init(filename);
northWestCornerRule();
steppingStone();
printResult(filename);
}
}
}
```
```input1.txt
2 3
25 35
20 30 10
3 5 7
3 2 5
Optimal solution input1.txt
20 - 5
- 30 5
Total costs: 180.0
```
```input2.txt
3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8
Optimal solution input2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130.0
```
```input3.txt
4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45
Optimal solution input3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000.0```
## Julia
Code taken from here using JuMP.
```using JuMP, GLPK
# cost vector
c = [3, 5, 7, 3, 2, 5];
N = size(c,1);
# constraints Ax (<,>,=) b
A = [1 1 1 0 0 0
0 0 0 1 1 1
1 0 0 1 0 0
0 1 0 0 1 0
0 0 1 0 0 1];
b = [ 25, 35, 20, 30, 10];
s = ['<', '<', '=', '=', '='];
# construct model
model = Model(GLPK.Optimizer)
@variable(model, x[i=1:N] >= 0, base_name="traded quantities")
cost_fn = @expression(model, c'*x) # cost function
@constraint(model, C1, A[1:2,:]*x .<= b[1:2]) # inequality constraints
@constraint(model, C2, A[3:5,:]*x .== b[3:5]) # equality constraints
@objective(model, Min, cost_fn) # objective function
# solve model
status = JuMP.optimize!(model);
xstar = value.(x);
println("solution vector of quantities = ", xstar)
println("minimum total cost = ", JuMP.objective_value(model))
# recover Lagrange multipliers for post-optimality
λ = [JuMP.dual(C1[1]),JuMP.dual(C1[2])]
μ = [JuMP.dual(C2[1]),JuMP.dual(C2[2]),JuMP.dual(C2[3])]
```
Output:
```solution vector of quantities = [20.0, 0.0, 5.0, 0.0, 30.0, 5.0]
minimum total cost = 180.0```
## Kotlin
Translation of: Java
```// version 1.1.51
import java.io.File
import java.util.Scanner
class Transport(val filename: String) {
private val supply: IntArray
private val demand: IntArray
private val costs : Array<DoubleArray>
private val matrix: Array<Array<Shipment>>
class Shipment(
var quantity: Double,
val costPerUnit: Double,
val r: Int,
val c: Int
)
companion object {
private val ZERO = Shipment(0.0, 0.0, -1, -1) // to avoid nullable Shipments
}
init {
val sc = Scanner(File(filename))
try {
val numSources = sc.nextInt()
val numDestinations = sc.nextInt()
val src = MutableList(numSources) { sc.nextInt() }
val dst = MutableList(numDestinations) { sc.nextInt() }
// fix imbalance
val totalSrc = src.sum()
val totalDst = dst.sum()
if (totalSrc > totalDst)
else if (totalDst > totalSrc)
supply = src.toIntArray()
demand = dst.toIntArray()
costs = Array(supply.size) { DoubleArray(demand.size) }
matrix = Array(supply.size) { Array(demand.size) { ZERO } }
for (i in 0 until numSources) {
for (j in 0 until numDestinations) costs[i][j] = sc.nextDouble()
}
}
finally {
sc.close()
}
}
fun northWestCornerRule() {
var northwest = 0
for (r in 0 until supply.size) {
for (c in northwest until demand.size) {
val quantity = minOf(supply[r], demand[c]).toDouble()
if (quantity > 0.0) {
matrix[r][c] = Shipment(quantity, costs[r][c], r, c)
supply[r] -= quantity.toInt()
demand[c] -= quantity.toInt()
if (supply[r] == 0) {
northwest = c
break
}
}
}
}
}
fun steppingStone() {
var maxReduction = 0.0
var move: Array<Shipment>? = null
var leaving = ZERO
fixDegenerateCase()
for (r in 0 until supply.size) {
for (c in 0 until demand.size) {
if (matrix[r][c] != ZERO) continue
val trial = Shipment(0.0, costs[r][c], r, c)
val path = getClosedPath(trial)
var reduction = 0.0
var lowestQuantity = Int.MAX_VALUE.toDouble()
var leavingCandidate = ZERO
var plus = true
for (s in path) {
if (plus) {
reduction += s.costPerUnit
}
else {
reduction -= s.costPerUnit
if (s.quantity < lowestQuantity) {
leavingCandidate = s
lowestQuantity = s.quantity
}
}
plus = !plus
}
if (reduction < maxReduction) {
move = path
leaving = leavingCandidate
maxReduction = reduction
}
}
}
if (move != null) {
val q = leaving.quantity
var plus = true
for (s in move) {
s.quantity += if (plus) q else -q
matrix[s.r][s.c] = if (s.quantity == 0.0) ZERO else s
plus = !plus
}
steppingStone()
}
}
private fun matrixToList() =
LinkedList<Shipment>(matrix.flatten().filter { it != ZERO } )
private fun getClosedPath(s: Shipment): Array<Shipment> {
val path = matrixToList()
// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
while (path.removeIf {
val nbrs = getNeighbors(it, path)
nbrs[0] == ZERO || nbrs[1] == ZERO
}) ; // empty statement
// place the remaining elements in the correct plus-minus order
val stones = Array<Shipment>(path.size) { ZERO }
var prev = s
for (i in 0 until stones.size) {
stones[i] = prev
prev = getNeighbors(prev, path)[i % 2]
}
return stones
}
private fun getNeighbors(s: Shipment, lst: LinkedList<Shipment>): Array<Shipment> {
val nbrs = Array<Shipment>(2) { ZERO }
for (o in lst) {
if (o != s) {
if (o.r == s.r && nbrs[0] == ZERO)
nbrs[0] = o
else if (o.c == s.c && nbrs[1] == ZERO)
nbrs[1] = o
if (nbrs[0] != ZERO && nbrs[1] != ZERO) break
}
}
return nbrs
}
private fun fixDegenerateCase() {
val eps = Double.MIN_VALUE
if (supply.size + demand.size - 1 != matrixToList().size) {
for (r in 0 until supply.size) {
for (c in 0 until demand.size) {
if (matrix[r][c] == ZERO) {
val dummy = Shipment(eps, costs[r][c], r, c)
if (getClosedPath(dummy).size == 0) {
matrix[r][c] = dummy
return
}
}
}
}
}
}
fun printResult() {
println("\$filename\n\n\$text")
println("Optimal solution \$filename\n")
var totalCosts = 0.0
for (r in 0 until supply.size) {
for (c in 0 until demand.size) {
val s = matrix[r][c]
if (s != ZERO && s.r == r && s.c == c) {
print(" %3s ".format(s.quantity.toInt()))
totalCosts += s.quantity * s.costPerUnit
}
else print(" - ")
}
println()
}
println("\nTotal costs: \$totalCosts\n")
}
}
fun main(args: Array<String>) {
val filenames = arrayOf("input1.txt", "input2.txt", "input3.txt")
for (filename in filenames) {
with (Transport(filename)) {
northWestCornerRule()
steppingStone()
printResult()
}
}
}
```
Output:
```Same as Java entry
```
## Nim
Translation of: Go
```import fenv, lists, math, sequtils, strformat, strutils
type
Shipment = object
quantity: float
costPerUnit: float
r, c: int
Transport = object
filename: string
supply: seq[int]
demand: seq[int]
costs: seq[seq[float]]
matrix: seq[seq[Shipment]]
const ShipZero = Shipment()
template emitError(msg: string) =
raise newException(ValueError, msg)
proc initTransport(filename: string): Transport =
let infile = filename.open()
let numSources = fields[0]
let numDests = fields[1]
if numSources < 1 or numDests < 1:
emitError "wrong number of sources or destinations."
if src.len != numSources:
emitError "wrong number of sources; got \$1, expected \$2.".format(src.len, numSources)
if dst.len != numDests:
emitError "wrong number of destinations; got \$1, expected \$2.".format(dst.len, numDests)
# Fix imbalance.
let totalSrc = sum(src)
let totalDst = sum(dst)
let diff = totalSrc - totalDst
if diff > 0: dst.add diff
elif diff < 0: src.add -diff
var costs = newSeqWith(src.len, newSeq[float](dst.len))
var matrix = newSeqWith(src.len, newSeq[Shipment](dst.len))
for i in 0..<numSources:
if fields.len > dst.len:
emitError "wrong number of costs; got \$1, expected \$2.".format(fields.len, numDests)
for j in 0..<numDests:
costs[i][j] = fields[j]
result = Transport(filename: filename, supply: move(src),
demand: move(dst), costs: move(costs), matrix: move(matrix))
func northWestCornerRule(tr: var Transport) =
var northWest = 0
for r in 0..tr.supply.high:
for c in northWest..tr.demand.high:
let quantity = min(tr.supply[r], tr.demand[c])
if quantity > 0:
tr.matrix[r][c] = Shipment(quantity: quantity.toFloat, costPerUnit: tr.costs[r][c], r: r, c: c)
dec tr.supply[r], quantity
dec tr.demand[c], quantity
if tr.supply[r] == 0:
northWest = c
break
func getNeighbors(tr: Transport; s: Shipment; list: ShipmentList): array[2, Shipment] =
for o in list:
if o != s:
if o.r == s.r and result[0] == ShipZero:
result[0] = o
elif o.c == s.c and result[1] == ShipZero:
result[1] = o
if result[0] != ShipZero and result[1] != ShipZero:
break
func matrixToList(tr: Transport): ShipmentList =
for m in tr.matrix:
for s in m:
if s != ShipZero:
result.append(s)
func getClosedPath(tr: Transport; s: Shipment): seq[Shipment] =
var path = tr.matrixToList
path.prepend(s)
# Remove (and keep removing) elements that do not have a
# vertical and horizontal neighbor.
while true:
var removals = 0
for e in path.nodes:
let nbrs = tr.getNeighbors(e.value, path)
if nbrs[0] == ShipZero or nbrs[1] == ShipZero:
path.remove(e)
inc removals
if removals == 0:
break
# Place the remaining elements in the correct plus-minus order.
var prev = s
var i = 0
for _ in path:
prev = tr.getNeighbors(prev, path)[i]
i = 1 - i
func fixDegenerateCase(tr: var Transport) =
const Eps = minimumPositiveValue(float)
if tr.supply.len + tr.demand.len - 1 != tr.matrix.len * tr.matrix[0].len:
for r in 0..tr.supply.high:
for c in 0..tr.demand.high:
if tr.matrix[r][c] == ShipZero:
let dummy = Shipment(quantity: Eps, costPerUnit: tr.costs[r][c], r: r, c: c)
if tr.getClosedPath(dummy).len == 0:
tr.matrix[r][c] = dummy
return
func steppingStone(tr: var Transport) =
var maxReduction = 0.0
var move: seq[Shipment]
var leaving = ShipZero
tr.fixDegenerateCase()
for r in 0..tr.supply.high:
for c in 0..tr.demand.high:
if tr.matrix[r][c] != ShipZero:
continue
let trial = Shipment(quantity: 0, costPerUnit: tr.costs[r][c], r: r, c: c)
var path = tr.getClosedPath(trial)
var reduction = 0.0
var lowestQuantity = float(int32.high)
var leavingCandidate = ShipZero
var plus = true
for s in path:
if plus:
reduction += s.costPerUnit
else:
reduction -= s.costPerUnit
if s.quantity < lowestQuantity:
leavingCandidate = s
lowestQuantity = s.quantity
plus = not plus
if reduction < maxReduction:
move = move(path)
leaving = leavingCandidate
maxReduction = reduction
if move.len != 0:
let q = leaving.quantity
var plus = true
for s in move.mitems:
if plus: s.quantity += q
else: s.quantity -= q
tr.matrix[s.r][s.c] = if s.quantity == 0: ShipZero else: s
plus = not plus
tr.steppingStone()
proc printResult(tr: Transport) =
echo tr.filename, '\n'
echo "\nOptimal solution for ", tr.filename, '\n'
var totalCosts = 0.0
for r in 0..tr.supply.high:
for c in 0..tr.demand.high:
let s = tr.matrix[r][c]
if s != ShipZero and s.r == r and s.c == c:
stdout.write &" {int(s.quantity):3} "
totalCosts += s.quantity * s.costPerUnit
else:
stdout.write " - "
echo()
echo &"\nTotal costs: {totalCosts:g}\n"
when isMainModule:
const Filenames = ["input1.txt", "input2.txt", "input3.txt"]
for filename in Filenames:
var tr = initTransport(filename)
tr.northWestCornerRule()
tr.steppingStone()
tr.printResult()
```
Output:
```input1.txt
2 3
25 35
20 30 10
3 5 7
3 2 5
Optimal solution for input1.txt
20 - 5
- 30 5
Total costs: 180
input2.txt
3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8
Optimal solution for input2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130
input3.txt
4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45
Optimal solution for input3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000```
## Pascal
```Program transport;
{ based on the program of <Svetlana Belashova> }
Uses Crt;
Label l1;
Const N=10;
n1=7; n2=7;
Sa:longint=0;
Sb:longint=0;
Type points=Array [1..N] of longint;
distribution=Array [1..N,1..N] of longint;
Var A,B,alfa,beta,B_d,x:points;
c,p:distribution;
f,f0,x_min,Sp:longint;
Nt,x_p,r,r_min,ki,kj,Na,Nb,h,l,i,j:byte;
d:char;
u:Array[1..N*N] of byte;
Procedure Nul (var a:points);
var i:byte;
Begin
for i:=1 to N do a[i]:=0;
End;
Procedure PrintS (x,y:byte; s:string; c:byte);
Begin
TextColor(c);
GotoXY(x,y);
Write(s);
End;
Procedure Print (x,y:byte; n:byte; a:longint; c:byte);
Begin
TextColor(c);
GotoXY(x,y); Write(' ':n);
GotoXY(x,y); Write(a);
End;
var i:integer;
s:string;
c:char;
j,k:byte;
Begin
s:=''; i:=1;
TextColor(11);
Repeat
Case ord(c) of
48..57: begin s:=s+c;
Write(c);
inc(i);
end;
8: if i>1 then begin dec(i);
Delete(s,i,1);
Write(chr(8),' ',chr(8));
end;
end;
j:=WhereX;
GotoXY(60,1); ClrEOL;
if i>y then begin
TextColor(4);
Write('Not more than ');
for k:=1 to y-1 do Write('9');
TextColor(11);
end;
GotoXY(j,1);
Until (ord(c)=13) and (i<y+1);
val(s,x,i);
End;
Procedure horizontal (a,b,c,d,e:char);
var i,j:byte;
Begin
Write(a);
for i:=1 to n2 do Write(b);
Write(c);
for i:=1 to Nb do begin
for j:=1 to n1 do Write(b);
if i<>Nb then Write(d) else Write(c);
end;
for i:=1 to 4 do Write(b);
Write(e);
End;
Procedure vertical;
var i:byte;
Begin
Write('│',' ':n2,'║');
for i:=1 to Nb-1 do Write(' ':n1,'│');
WriteLn(' ':n1,'║',' ' :4,'│');
End;
Procedure Table; { Drawing the table }
Begin
ClrScr;
TextColor(1);
h:=6+Na*3;
l:=14+Nb*7;
GotoXY(1,3);
for i:=3 to h do vertical;
GotoXY(1,2);
horizontal('┌','─','╥','┬','┐');
for i:=1 to Na+1 do begin
GotoXY(1,i*3+2);
if (i=1) or (i=Na+1)
then horizontal('╞','═','╬','╪','╡')
else horizontal('├','─','╫','┼','┤');
end;
GotoXY(1,h+1);
horizontal('└','─','╨','┴','┘');
TextColor(9);
for i:=1 to Na do begin
GotoXY(5,i*3+3);
Write('A',i);
end;
for i:=1 to Nb do begin
GotoXY(i*(n1+1)+n2-2,3);
Write('B',i);
end;
l:=Nb*(n1+1)+n2+3;
h:=Na*3+6;
PrintS(4,3,'\Bj',9);
PrintS(4,4,'Ai\',9);
PrintS(1,1,'Table N1',14);
PrintS(l,4,'alfa',9);
PrintS(3,h,'beta',9);
End;
Procedure EnterIntoTheTable (var a:points; b:byte; c:char); { Entering into the table }
var i,l,m:byte;
Begin
for i:=1 to b do begin
TextColor(3);
GotoXY(32,1);
ClrEOL;
Write(c,i,'= ');
TextColor(14);
Case c of
'A': GotoXY(n2-trunc(ln(a[i])/ln(10)),i*3+4);
'B': GotoXY(n2+i*(n1+1)-trunc(ln(a[i])/ln(10)),4);
end;
Write(a[i]);
end;
End;
Function CalculatingTheCost:longint; { Calculating the cost of the plan }
var i,j:byte;
f:longint;
Begin
f:=0;
for i:=1 to Na do
for j:=1 to Nb do
if p[i,j]>0 then inc(f,c[i,j]*p[i,j]);
GotoXY(65,Nt+2);
TextColor(10);
Write('F',Nt,'=',f);
CalculatingTheCost:=f;
End;
Function CalculatingThePotentials:boolean; { Calculating the potentials }
var k,i,j:byte;
Z_a,Z_b:points;
d:boolean;
Begin
Nul(Z_a); Nul(Z_b);
alfa[1]:=0; Z_a[1]:=1; k:=1;
Repeat
d:=1=1;
for i:=1 to Na do
if Z_a[i]=1 then
for j:=1 to Nb do
if (p[i,j]>-1) and (Z_b[j]=0) then begin
Z_b[j]:=1;
beta[j]:=c[i,j]-alfa[i];
inc(k);
d:=1=2;
end;
for i:=1 to Nb do
if Z_b[i]=1 then
for j:=1 to Na do
if (p[j,i]>-1) and (Z_a[j]=0) then begin
Z_a[j]:=1;
alfa[j]:=c[j,i]-beta[i];
inc(k);
d:=1=2;
end;
Until (k=Na+Nb) or d;
if d then begin
i:=1;
While Z_a[i]=1 do inc(i);
j:=1;
While Z_b[j]=0 do inc(j);
p[i,j]:=0;
Print((j+1)*(n1+1)+n2-8,i*3+4,1,p[i,j],7);
end;
CalculatingThePotentials:=d;
End;
Procedure OutputThePlan; { Output the plan of distribution }
var i,j,h,l,k:byte;
c_max:longint;
Begin
k:=0;
for i:=1 to Na do begin
h:=i*3+4;
for j:=1 to Nb do begin
l:=j*(n1+1)+n2-5;
GotoXY(l,h);
Write(' ':n1);
if p[i,j]>0 then begin
inc(k);
Print(l-trunc(ln(p[i,j])/ln(10))+5,h,1,p[i,j],14);
end
else if p[i,j]=0 then begin
Print(l+n1-2,h,1,p[i,j],14);
inc(k);
end;
end;
end;
While CalculatingThePotentials do inc(k);
if k>Na+Nb-1 then PrintS(40,1,'k > n+m-1',12);
End;
Function CalculatingTheCoefficients(var ki,kj:byte):integer; { Calculation the coefficients in the free cells }
var i,j:byte;
k,k_min:integer;
b:boolean;
Begin
b:=1=1;
for i:=1 to Na do
for j:=1 to Nb do
if p[i,j]=-1 then begin
k:=c[i,j]-alfa[i]-beta[j];
if b then begin
b:=1=2;
ki:=i; kj:=j; k_min:=k;
end else
if k<k_min then begin
k_min:=k;
ki:=i; kj:=j;
end;
TextColor(6);
GotoXY(j*(n1+1)+n2-5,i*3+4);
Write('(',k,')');
end;
if k_min<0 then PrintS(kj*(n1+1)+n2,ki*3+4,'X',12);
CalculatingTheCoefficients:=k_min;
End;
Procedure div_mod(c:byte; var a,b:byte); { Translate one-dimensional array to two-dimensional }
Begin
b:=c mod Nb; a:=c div Nb +1;
if b=0 then begin
b:=Nb; dec(a);
end;
End;
Procedure Recursive(Xi,Yi:byte; var z:boolean; var c:byte);
var i,j:byte;
Begin
z:=1=2;
Case c of
1: for i:=1 to Na do
if i<>Xi then
if p[i,Yi]>-1 then begin
if u[(i-1)*Nb+Yi]=0 then begin
u[(Xi-1)*Nb+Yi]:=(i-1)*Nb+Yi;
c:=2;
Recursive(i,Yi,z,c);
if z then exit;
end;
end
else if (i=ki) and (Yi=kj) then begin
u[(Xi-1)*Nb+Yi]:=(ki-1)*Nb+kj;
z:=not z;
exit;
end;
2: for i:=1 to Nb do
if i<>Yi then
if p[Xi,i]>-1 then begin
if u[(Xi-1)*Nb+i]=0 then begin
u[(Xi-1)*Nb+Yi]:=(Xi-1)*Nb+i;
c:=1;
Recursive(Xi,i,z,c);
if z then exit;
end;
end
else if (Xi=ki) and (i=kj) then begin
u[(Xi-1)*Nb+Yi]:=(ki-1)*Nb+kj;
z:=not z;
exit;
end;
end;
u[(Xi-1)*Nb+Yi]:=0;
c:=c mod 2 +1;
End;
Procedure Contour; { Determine the contour of displacement }
var i,j,k,mi,mj,l:byte;
z:boolean;
p_m:longint;
Begin
for i:=1 to N*N do u[i]:=0;
l:=1;
Recursive(ki,kj,z,l);
i:=ki; j:=kj;
k:=u[(i-1)*Nb+j];
div_mod(k,i,j);
mi:=i; mj:=j; l:=1;
Repeat
inc(l);
k:=u[(i-1)*Nb+j];
div_mod(k,i,j);
if l mod 2=1 then
if p[i,j]<p[mi,mj] then begin
mi:=i; mj:=j;
end;
Until (i=ki) and (j=kj);
i:=ki; j:=kj; l:=0;
p_m:=p[mi,mj];
Repeat
if l mod 2=0 then begin
inc(p[i,j],p_m);
PrintS((n1+1)*j+n2-1,i*3+3,'(+)',12);
end else begin
dec(p[i,j],p_m);
PrintS((n1+1)*j+n2-1,i*3+3,'(-)',12);
end;
if l=0 then inc(p[i,j]);
k:=u[(i-1)*Nb+j];
div_mod(k,i,j);
inc(l);
Until (i=ki) and (j=kj);
p[mi,mj]:=-1;
End;
Procedure Pause;
var d:char;
Begin
TextColor(6);
GotoXY(40,1);
Write('Press any key');
GotoXY(40,1);
ClrEOL;
End;
BEGIN
Nul(alfa); Nul(beta);
Nt:=1;
ClrScr;
TextColor(10);
Repeat
Write('Enter the number of suppliers (2<=Na<=',N-1,') ');
Write('Enter the number of consumers (2<=Nb<=',N-1,') ');
Until (Na>1) and (Na<=N-1) and (Nb>1) and (Nb<=N-1);
Table;
PrintS(1,1,'Enter the production quantity:',3);
EnterIntoTheTable(A,Na,'A');
EnterIntoTheTable(B,Nb,'B');
TextColor(3);
GotoXY(1,1); ClrEOL;
Write('Enter the cost of transportation');
for i:=1 to Na do
for j:=1 to Nb do begin
TextColor(3);
GotoXY(29,1); ClrEOL;
Write('A',i,' - B',j,' ');
Print((n1+1)*j+n2-4,i*3+3,1,c[i,j],11);
end;
GotoXY(1,1);
ClrEOL;
TextColor(14);
Write('Table N1');
for i:=1 to Na do Sa:=Sa+A[i];
for i:=1 to Nb do Sb:=Sb+B[i];
if Sa<>Sb then begin
PrintS(20,1,'The problem is open (Press any key)',7);
if Sa>Sb then begin
inc(Nb);
B[Nb]:=Sa-Sb;
for i:=1 to Na do c[i,Nb]:=0;
end else begin
inc(Na);
A[Na]:=Sb-Sa;
for i:=1 to Nb do c[Na,i]:=0;
end;
Table;
for i:=1 to Na do
for j:=1 to Nb do Print((n1+1)*j+n2-4,i*3+3,1,c[i,j],11);
for i:=1 to Na do
Print(n2-trunc(ln(A[i])/ln(10)),i*3+4,1,A[i],14);
for i:=1 to Nb do
Print(n2+i*(n1+1)-trunc(ln(B[i])/ln(10)),4,1,B[i],14);
PrintS(20,1,'The problem is open',7);
end
else PrintS(20,1,'The problem is closed',7);
(**************** Drafting the basic plan ******************)
for i:=1 to Nb do B_d[i]:=B[i];
for i:=1 to Na do begin
for j:=1 to Nb do x[j]:=j;
for j:=1 to Nb-1 do begin
x_min:=c[i,x[j]];
r_min:=j;
for r:= j+1 to Nb do
if (x_min>c[i,x[r]]) or
((x_min=c[i,x[r]]) and (B[x[r]]>b[x[r_min]])) then
begin
x_min :=c[i,x[r]];
r_min:=r;
end;
x_p:=x[r_min];
x[r_min]:=x[j];
x[j]:=x_p;
end;
Sp:=0;
for j:=1 to Nb do begin
p[i,x[j]]:=B_d[x[j]];
if p[i,x[j]]>A[i]-Sp then p[i,x[j]]:=A[i]-Sp;
inc(Sp,p[i,x[j]]);
dec(B_d[x[j]],p[i,x[j]]);
end;
end;
(***********************************************************)
for i:=1 to Na do
for j:=1 to Nb do if p[i,j]=0 then p[i,j]:=-1;
OutputThePlan;
f:=CalculatingTheCost; f0:=F;
While CalculatingThePotentials do;
for i:=1 to Na do Print(l+1,i*3+3,3,alfa[i],11);
for i:=1 to Nb do Print(i*(n1+1)+n2-4,h,6,beta[i],11);
Pause;
(******* gradual approach the plan to the optimality ******)
While CalculatingTheCoefficients(ki,kj)<0 do begin
Contour;
pause;
for i:=1 to Na do
for j:=1 to Nb do PrintS((n1+1)*j+n2-1,i*3+3,' ',14);
inc(Nt);
GotoXY(1,1);
Write('Table N',Nt);
OutputThePlan;
f0:=f; f:=CalculatingTheCost;
if CalculatingThePotentials then Goto l1;
for i:=1 to Na do Print(l+1,i*3+3,3,alfa[i],11);
for i:=1 to Nb do Print(i*(n1+1)+n2-4,h,6,beta[i],11);
Pause;
end;
(***********************************************************)
PrintS(40,1,'Solution is optimal',12);
PrintS(60,1,'(any key)',6);
for i:=1 to Na do
for j:=1 to Nb do if p[i,j]=-1 then begin
h:=i*3+4;
l:=j*(n1+1)+n2-5;
GotoXY(l,h);
Write(' ':n1);
end;
GotoXY(40,1);
END.
```
## Perl
Just re-using the code from Vogel's approximation method, tweaked to handle specific input:
```use strict;
use warnings;
use feature 'say';
use List::AllUtils qw( max_by nsort_by min );
my \$data = <<END;
A=20 B=30 C=10
S=25 T=35
AS=3 BS=5 CS=7
CT=3 BT=2 CT=5
END
my \$table = sprintf +('%4s' x 4 . "\n") x 3,
map {my \$t = \$_; map "\$_\$t", '', 'A' .. 'C' } '' , 'S' .. 'T';
my (\$cost, %assign) = (0);
while( \$data =~ /\b\w=\d/ ) {
my @penalty;
for ( \$data =~ /\b(\w)=\d/g ) {
my @all = map /(\d+)/, nsort_by { /\d+/ && \$& }
grep { my (\$t, \$c) = /(.)(.)=/; \$data =~ /\b\$c=\d/ and \$data =~ /\b\$t=\d/ }
\$data =~ /\$_\w=\d+|\w\$_=\d+/g;
push @penalty, [ \$_, (\$all[1] // 0) - \$all[0] ];
}
my \$rc = (max_by { \$_->[1] } nsort_by
{ my \$x = \$_->[0]; \$data =~ /(?:\$x\w|\w\$x)=(\d+)/ && \$1 } @penalty)->[0];
my @lowest = nsort_by { /\d+/ && \$& }
grep { my (\$t, \$c) = /(.)(.)=/; \$data =~ /\b\$c=\d/ and \$data =~ /\b\$t=\d/ }
\$data =~ /\$rc\w=\d+|\w\$rc=\d+/g;
my (\$t, \$c) = \$lowest[0] =~ /(.)(.)/;
my \$allocate = min \$data =~ /\b[\$t\$c]=(\d+)/g;
\$table =~ s/\$t\$c/ sprintf "%2d", \$allocate/e;
\$cost += \$data =~ /\$t\$c=(\d+)/ && \$1 * \$allocate;
\$data =~ s/\b\$_=\K\d+/ \$& - \$allocate || '' /e for \$t, \$c;
}
say my \$result = "cost \$cost\n\n" . \$table =~ s/[A-Z]{2}/--/gr;
```
Output:
```cost 170
A B C
S 20 -- 5
T -- 30 5```
## Phix
The simplest solution I could think of.
Assumes 0 cost is not allowed, but using say -1 as the "done" cost instead should be fine.
```with javascript_semantics
procedure solve(sequence needs, avail, costs)
-- the costs parameter should be length(avail/*aka suppliers*/) rows
-- of length(needs/*aka customers*/) cols
assert(length(costs)==length(avail))
assert(apply(costs,length)==repeat(length(needs),length(avail)))
sequence res = repeat(repeat(0,length(needs)),length(avail))
while true do
integer best = 0, supplier, customer
for s=1 to length(costs) do
for c=1 to length(costs[s]) do
integer csc = costs[s][c]
if csc!=0 and (best=0 or csc<best) then
best = csc
supplier = s
customer = c
end if
end for
end for
if best=0 then exit end if -- all costs examined
integer amt = min(avail[supplier],needs[customer])
-- obviously amt can be 0, in which case this just
-- removes cost entry from further consideration.
avail[supplier] -= amt
needs[customer] -= amt
res[supplier,customer] = amt
costs[supplier,customer] = 0
end while
pp(res,{pp_Nest,1})
end procedure
solve({20,30,10},{25,35},{{3,5,7},{3,2,5}})
```
Output:
```{{20,0,5},
{0,30,5}}
```
### stepping stones
Obviously I did not really quite understand the problem when I rattled out the above... this does much better.
Translation of: Go
```-- demo\rosetta\Transportation_problem.exw
with javascript_semantics
enum QTY, COST, R, C -- (a shipment)
constant eps = 1e-12
function print_matrix(sequence matrix)
atom total_costs = 0.0
for r=1 to length(matrix) do
for c=1 to length(matrix[r]) do
object s = matrix[r][c]
string st = " - "
if s!=0 and s[R]==r and s[C]==c then
atom qty = round(s[QTY]) -- (remove +/-eps)
if qty!=0 then
st = sprintf(" %3d ", qty)
total_costs += qty * s[COST]
end if
end if
puts(1,st)
end for
printf(1,"\n")
end for
end function
procedure print_result(sequence transport, atom expected)
sequence matrix = transport[4]
printf(1,"Optimal solution\n\n")
atom total_costs = print_matrix(matrix)
printf(1,"\nTotal costs: %g (expected %g)\n\n", {total_costs,expected})
end procedure
function get_neighbors(sequence shipment, lst)
sequence nbrs = {0,0}
for e=1 to length(lst) do
sequence o = lst[e]
if o!=shipment then
if o[R]==shipment[R] and nbrs[1]==0 then
nbrs[1] = o
elsif o[C]==shipment[C] and nbrs[2]==0 then
nbrs[2] = o
end if
if nbrs[1]!=0 and nbrs[2]!=0 then
exit
end if
end if
end for
return nbrs
end function
function matrix_to_list(sequence matrix)
sequence l = {}
for r=1 to length(matrix) do
for c=1 to length(matrix[r]) do
if matrix[r,c]!=0 then
l = append(l,matrix[r,c])
end if
end for
end for
return l
end function
function get_closed_path(sequence matrix, shipment)
sequence path = matrix_to_list(matrix)
path = prepend(path,shipment)
-- remove (and keep removing) elements that do not have a
-- vertical AND horizontal neighbor
while true do
integer removals = 0
for e=length(path) to 1 by -1 do
sequence nbrs = get_neighbors(path[e], path)
if nbrs[1]==0 or nbrs[2]==0 then
path[e..e] = {}
removals += 1
end if
end for
if removals==0 then exit end if
end while
-- place the remaining elements in the correct plus-minus order
sequence stones = repeat(0,length(path)),
prev = shipment
for i=1 to length(stones) do
stones[i] = prev
prev = get_neighbors(prev, path)[mod(i,2)+1]
end for
return stones
end function
function fix_degenerate_case(sequence matrix, costs)
if length(matrix)+length(matrix[1])-1 != length(matrix_to_list(matrix)) then
printf(1,"fixing degenerate case...\n")
for r=1 to length(matrix) do
for c=1 to length(matrix[r]) do
if matrix[r][c] == 0 then
sequence dummy = {eps, costs[r][c], r, c}
if length(get_closed_path(matrix,dummy)) == 0 then
matrix[r][c] = dummy
return matrix
end if
end if
end for
end for
?9/0 -- ??
end if
return matrix
end function
function initialise(sequence tests, integer t)
sequence {src,dst,costs} = deep_copy(tests[t])
string cs = ppf(costs,{pp_Nest,1,pp_StrFmt,3,pp_IntCh,false,pp_Indent,7})
printf(1,"test %d:\nsrc: %v,\ndst: %v,\ncosts: %s\n",{t,src,dst,cs})
-- check for and fix any imbalance
atom totalSrc = sum(src),
totalDst = sum(dst),
diff = totalSrc-totalDst
if diff>0 then
dst = append(dst, diff)
for i=1 to length(costs) do
--DEV...
-- costs[i] &= 0
costs[i] = deep_copy(costs[i]) & 0
end for
elsif diff<0 then
src = append(src, -diff)
costs = append(costs,repeat(0,length(dst)))
end if
printf(1,"generating initial feasible solution using northwest corner method...\n")
sequence matrix = repeat(repeat(0,length(dst)),length(src))
integer northwest = 1
for r=1 to length(src) do
for c=northwest to length(dst) do
atom qty = min(src[r],dst[c])
if qty>0 then
matrix[r][c] = {qty,costs[r,c],r,c}
src[r] -= qty
dst[c] -= qty
if src[r]=0 then
northwest = c
exit
end if
end if
end for
end for
printf(1,"\nTotal costs: %g\n\n", print_matrix(matrix))
return {src,dst,costs,matrix}
end function
function stepping_stone(sequence transport)
sequence {src, dst, costs, matrix} = deep_copy(transport)
atom maxReduction = 0
object move = NULL, leaving
matrix = fix_degenerate_case(matrix, costs)
for r=1 to length(src) do
for c=1 to length(dst) do
if matrix[r][c] = 0 then
sequence trial_shipment = {0, costs[r][c], r, c},
path = get_closed_path(matrix,trial_shipment)
atom reduction = 0.0,
lowestQuantity = 1e308
object leavingCandidate = 0
bool plus = true
for i=1 to length(path) do
sequence s = path[i]
if plus then
reduction += s[COST]
else
reduction -= s[COST]
if s[QTY] < lowestQuantity then
leavingCandidate = s
lowestQuantity = s[QTY]
end if
end if
plus = not plus
end for
if reduction < maxReduction then
move = path
leaving = leavingCandidate
maxReduction = reduction
end if
end if
end for
end for
if move!=NULL then
atom q = leaving[QTY]
bool plus = true
for i=1 to length(move) do
sequence s = deep_copy(move[i])
if plus then
s[QTY] += q
else
s[QTY] -= q
end if
if s[QTY] == 0 then
matrix[s[R]][s[C]] = 0
else
matrix[s[R]][s[C]] = s
end if
plus = not plus
end for
{src, dst, costs, matrix} = stepping_stone({src, dst, costs, matrix})
end if
return {src, dst, costs, matrix}
end function
-- -- source dest costs expected total
constant tests = {{{25,35}, {20,30,10}, {{3,5,7},
{3,2,5}}, 180},
{{12,40,33}, {20,30,10}, {{3,5,7},
{2,4,6},
{9,1,8}}, 130},
{{14,10,15,12}, {10,15,12,15}, {{10,30,25,15},
{20,15,20,10},
{10,30,20,20},
{30,40,35,45}}, 1000},
{{100,300,300}, {300,200,200}, {{50,40,30},
{80,40,30},
{90,70,50}}, 39000},
{{40,60,50}, {20,30,50,50}, {{4,6,8,8},
{6,8,6,7},
{5,7,6,8}}, 920},
{{12,1,5}, {10,8}, {{ 2, 4},
{ 8,12},
{12, 6}}, 68},
{{7,9,18}, {5,8,7,14}, {{19,30,50,10},
{70,30,40,60},
{40, 8,70,20}}, 743},
{{12,11,14,8}, {10,11,15,5,4}, {{ 7,12, 1, 5, 6},
{15, 3,12, 6,14},
{ 8,16,10,12, 7},
{18, 8,17,11,16}}, 259},
{{50,60,50,50}, {30,20,70,30,60},{{16,16,13,22,17},
{14,14,13,19,15},
{19,19,20,23,50},
{50,12,50,15,11}}, 3100},
{{50,75,25}, {20,20,50,60}, {{3,5,7,6},
{2,5,8,2},
{3,6,9,2}}, 610}}
--for i=1 to length(tests) do
for i=3 to 3 do
print_result(stepping_stone(initialise(tests,i)),tests[i][4])
end for
?"done"
{} = wait_key()
```
Output:
(Obviously the other eight tests all work fine and produce similar output.)
```test 3:
src: {14,10,15,12},
dst: {10,15,12,15},
costs: {{10,30,25,15},
{20,15,20,10},
{10,30,20,20},
{30,40,35,45}}
generating initial feasible solution using northwest corner method...
10 4 - -
- 10 - -
- 1 12 2
- - - 12
- - - 1
Total costs: 1220
fixing degenerate case...
Optimal solution
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000 (expected 1000)
```
Note that Vogel's_approximation_method#Phix gets a few of the others wrong and loops on #2, then again that is unbalanced (needs a dummy customer), and I'm not sure whether throwing such at VAM is fair or not.
## R
Using `lpSolve::lp.transport`.
```library(lpSolve)
# cost matrix
costs <- matrix(c(3, 5, 7,
3, 2, 5), nrow = 2, byrow = TRUE)
# constraints for suppliers
row.signs <- rep("<=", 2)
row.rhs <- c(25, 35)
# constraints for customers
col.signs <- rep("=", 3)
col.rhs <- c(20, 30, 10)
# minimum cost (objective value)
lp.transport(costs, "min", row.signs, row.rhs, col.signs, col.rhs)
# solution matrix
sol = lp.transport(costs, "min", row.signs, row.rhs, col.signs, col.rhs)\$solution
rownames(sol) <- c("Supplier 1", "Supplier 2")
colnames(sol) <- c("Customer 1", "Customer 2", "Customer 3")
sol
```
Output:
```Success: the objective function is 180
Customer 1 Customer 2 Customer 3
Supplier 1 20 0 5
Supplier 2 0 30 5
```
## Racket
Translation of: Java
(I understand the letters in Java!)
Using `typed/racket`, to keep track of Vectors of Vectors of data.
```#lang typed/racket
;; {{trans|Java}}
(define-type (V2 A) (Vectorof (Vectorof A)))
(define-type VI (Vectorof Integer))
(define-type V2R (V2 Real))
(define-type Q (U 'ε Integer))
(define ε 'ε)
(struct Shipment ([qty : Q] [cost/unit : Real] [r : Integer] [c : Integer]))
(define-type Shipment/? (Option Shipment))
(define-type V2-Shipment/? (V2 Shipment/?))
(define-type Shipment/?s (Listof Shipment/?))
(define-type Shipments (Listof Shipment))
(: Q+ (Q Q -> Q))
(: Q- (Q Q -> Q))
(: Q<? (Q Q -> Boolean))
(: Q-zero? (Q -> Boolean))
(: Q-unary- (Q -> Q))
(: Q*R (Q Real -> Real))
(define Q+ (match-lambda** [('ε 0) ε] [(0 'ε) ε] [('ε 'ε) ε] [('ε x) x] [(x 'ε) x]
[((? integer? x) (? integer? y)) (+ x y)]))
(define Q<? (match-lambda** [('ε 0) #f] [(0 'ε) #t] [('ε 'ε) #f] [('ε x) #t] [(x 'ε) #f]
[((? integer? x) (? integer? y)) (< x y)]))
(define Q- (match-lambda** [('ε 0) ε] [(0 'ε) ε] [('ε 'ε) 0] [('ε (? integer? x)) (- x)] [(x 'ε) x]
[((? integer? x) (? integer? y)) (- x y)]))
(define Q-unary- (match-lambda ['ε ε] [(? integer? x) (- x)]))
(define Q-zero? (match-lambda ['ε #f] [(? integer? x) (zero? x)]))
(define Q*R (match-lambda** [('ε _) 0] [((? integer? x) y) (* x y)]))
(: vector-ref2 (All (A) ((Vectorof (Vectorof A)) Integer Integer -> A)))
(define (vector-ref2 v2 r c) (vector-ref (vector-ref v2 r) c))
(: vector-set!2 (All (A) ((Vectorof (Vectorof A)) Integer Integer A -> Void)))
(define (vector-set!2 v2 r c v) (vector-set! (vector-ref v2 r) c v))
(define (northwest-corner-rule! [supply : VI] [demand : VI] [costs : V2R] [M : V2-Shipment/?]) : Void
(define supply-l (vector-length supply))
(define demand-l (vector-length demand))
(let loop ((r 0) (nw 0) (c 0))
(cond [(= r supply-l) (void)]
[(= c demand-l) (loop (add1 r) nw 0)]
[else
(define quantity (min (vector-ref supply r) (vector-ref demand c)))
(cond
[(positive? quantity)
(define shpmnt (Shipment quantity (vector-ref2 costs r c) r c))
(vector-set!2 M r c shpmnt)
(define supply-- (- (vector-ref supply r) quantity))
(define demand-- (- (vector-ref demand c) quantity))
(vector-set! supply r supply--)
(vector-set! demand c demand--)
(if (zero? supply--) (loop (add1 r) c 0) (loop r nw (add1 c)))]
[else (loop r nw (add1 c))])])))
(define (stepping-stone! [supply : VI] [demand : VI] [costs : V2R] [M : V2-Shipment/?]) : Void
(fix-degenerate-case! supply demand costs M)
(define-values (move leaving max-reduction)
(for*/fold : (Values Shipments Shipment/? Real)
((move : Shipments null) (leaving : Shipment/? #f) (max-reduction : Real 0))
((r (vector-length supply))
(c (vector-length demand))
(m (in-value (vector-ref2 M r c)))
#:unless m)
(define path (let ((trial (Shipment 0 (vector-ref2 costs r c) r c))) (get-closed-path trial M)))
(define-values (+? reduction leaving-cand lowest-quantity)
(for/fold : (Values Boolean Real Shipment/? (Option Q))
((+? #t) (reduction : Real 0) (leaving-cand : Shipment/? #f) (lowest-q : (Option Q) #f))
((s (in-list path)))
(define s.cpu (Shipment-cost/unit s))
(if +?
(values #f (+ reduction s.cpu) leaving-cand lowest-q)
(let ((reduction-- (- reduction s.cpu))
(s.q (Shipment-qty s)))
(if (or (not lowest-q) (Q<? s.q lowest-q))
(values #t reduction-- s s.q)
(values #t reduction-- leaving-cand lowest-q))))))
(if (< reduction max-reduction)
(values path leaving-cand reduction)
(values move leaving max-reduction))))
(unless (null? move)
(define l.q (Shipment-qty (cast leaving Shipment)))
(for/fold ((+? : Boolean #t)) ((s (in-list move)))
(define s.q+ ((if +? Q+ Q-) (Shipment-qty s) l.q))
(define s+ (struct-copy Shipment s [qty s.q+]))
(vector-set!2 M (Shipment-r s+) (Shipment-c s+) (if (Q-zero? s.q+) #f s+))
(not +?))
(stepping-stone! supply demand costs M)))
(: matrix->list (All (T) ((V2 T) -> (Listof T))))
(define (matrix->list m)
(for*/list : (Listof T) ((r (in-vector m)) (c (in-vector r)) #:when c)
c))
(define (fix-degenerate-case! [supply : VI] [demand : VI] [costs : V2R] [M : V2-Shipment/?]) : Void
(define m-list (matrix->list M))
(unless (= (+ (vector-length supply) (vector-length demand) -1) (length m-list))
(let/ec ret : Void
(for* ((r (vector-length supply)) (c (vector-length demand)) #:unless (vector-ref2 M r c))
(define dummy (Shipment ε (vector-ref2 costs r c) r c))
(when (null? (get-closed-path dummy M))
(vector-set!2 M r c dummy)
(ret (void)))))))
(: get-closed-path (Shipment V2-Shipment/? -> Shipments))
(define (get-closed-path s matrix)
; remove (and keep removing) elements that do not have a vertical AND horizontal neighbour
(define path
(let loop : Shipment/?s
((path (cons s (matrix->list matrix))))
(define (has-neighbours [e : Shipment/?]) : Boolean
(match-define (list n0 n1) (get-neighbours e path))
(and n0 n1 #t))
(define-values (with-nbrs w/o-nbrs)
((inst partition Shipment/? Shipment/?) has-neighbours path))
(if (null? w/o-nbrs) with-nbrs (loop with-nbrs))))
;; place the remaining elements in the correct plus-minus order
(define p-len (length path))
(define-values (senots prev)
(for/fold : (Values Shipments Shipment/?)
((senots : Shipments null) (prev : Shipment/? s))
((i p-len))
(values (if prev (cons prev senots) senots)
(list-ref (get-neighbours prev path) (modulo i 2)))))
(reverse senots))
(define (get-neighbours [s : Shipment/?] [lst : Shipment/?s]) : (List Shipment/? Shipment/?)
(define-values (n0 n1)
(for/fold : (Values Shipment/? Shipment/?)
((n0 : Shipment/? #f) (n1 : Shipment/? #f))
((o (in-list lst)) #:when (and o s) #:unless (equal? o s))
(values (or n0 (and (= (Shipment-r s) (Shipment-r o)) o))
(or n1 (and (= (Shipment-c s) (Shipment-c o)) o)))))
(list n0 n1))
(define (print-result [S : VI] [D : VI] [M : V2-Shipment/?] [fmt : String] . [args : Any *]) : Real
(apply printf (string-append fmt "~%") args)
(define total-costs
(for*/sum : Real
((r (vector-length S)) (c (vector-length D)))
(when (zero? c) (unless (zero? r) (newline)))
(define s (vector-ref2 M r c))
(cond
[(and s (= (Shipment-r s) r) (= (Shipment-c s) c))
(define q (Shipment-qty s))
(printf "\t~a" q)
(Q*R q (Shipment-cost/unit s))]
[else (printf "\t-") 0])))
(printf "~%Total costs: ~a~%~%" total-costs)
total-costs)
;; inits from current-input-port --- make sure you set that before coming in
(define (init) : (Values VI VI V2R V2-Shipment/?)
(define srcs. (for/list : (Listof Integer) ((_ n-sources)) (cast (read) Integer)))
(define dsts. (for/list : (Listof Integer) ((_ n-destinations)) (cast (read) Integer)))
(define sum-src--sum-dest (- (apply + srcs.) (apply + dsts.)))
(define-values (supply demand)
(cond [(positive? sum-src--sum-dest) (values srcs. (append dsts. (list sum-src--sum-dest)))]
[(negative? sum-src--sum-dest) (values (append srcs. (list (- sum-src--sum-dest))) dsts.)]
[else (values srcs. dsts.)]))
(define s-l (length supply))
(define d-l (length demand))
(define costs (for/vector : V2R ((_ s-l)) ((inst make-vector Real) d-l 0)))
(define matrix (for/vector : V2-Shipment/? ((_ s-l)) ((inst make-vector Shipment/?) d-l #f)))
(for* ((i n-sources) (j n-destinations)) (vector-set!2 costs i j (cast (read) Real)))
(values (list->vector supply) (list->vector demand) costs matrix))
(: transportation-problem (Input-Port -> Real))
(define (transportation-problem p)
(parameterize ([current-input-port p])
(define-values (supply demand costs matrix) (init))
(northwest-corner-rule! supply demand costs matrix)
(stepping-stone! supply demand costs matrix)
(print-result supply demand matrix "Optimal solutions for: ~s" name)))
(module+ test
(require typed/rackunit)
(define (check-tp [in-str : String] [expected-cost : Real])
(define cost ((inst call-with-input-string Real) in-str transportation-problem))
(check-equal? cost expected-cost))
(check-tp #<<\$
input1
2 3
25 35
20 30 10
3 5 7
3 2 5
\$
180)
(check-tp #<<\$
input2
3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8
\$
130)
(check-tp #<<\$
input3
4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45
\$
1000))
```
Output:
Output of: `raco test Transportation-problem.rkt`:
```raco test: (submod "transportation-problem.rkt" test)
Optimal solutions for: input1
20 - 5
- 30 5
Total costs: 180
Optimal solutions for: input2
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130
Optimal solutions for: input3
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000
3 tests passed
```
## Raku
(formerly Perl 6)
Works with: Rakudo version 2019.03.1
```my %costs = :S1{:3C1, :5C2, :7C3}, :S2{:3C1, :2C2, :5C3};
my %demand = :20C1, :30C2, :10C3;
my %supply = :25S1, :35S2;
my @cols = %demand.keys.sort;
my %res;
my %g = (|%supply.keys.map: -> \$x { \$x => [%costs{\$x}.sort(*.value)».key]}),
(|%demand.keys.map: -> \$x { \$x => [%costs.keys.sort({%costs{\$_}{\$x}})]});
while (+%g) {
my @d = %demand.keys.map: -> \$x
{[\$x, my \$z = %costs{%g{\$x}[0]}{\$x},%g{\$x}[1] ?? %costs{%g{\$x}[1]}{\$x} - \$z !! \$z]}
my @s = %supply.keys.map: -> \$x
{[\$x, my \$z = %costs{\$x}{%g{\$x}[0]},%g{\$x}[1] ?? %costs{\$x}{%g{\$x}[1]} - \$z !! \$z]}
@d = |@d.grep({ (.[2] == max @d».[2]) }).&min: :by(*.[1]);
@s = |@s.grep({ (.[2] == max @s».[2]) }).&min: :by(*.[1]);
my (\$t, \$f) = @d[2] == @s[2] ?? (@s[1],@d[1]) !! (@d[2],@s[2]);
my (\$d, \$s) = \$t > \$f ?? (@d[0],%g{@d[0]}[0]) !! (%g{@s[0]}[0], @s[0]);
my \$v = %supply{\$s} min %demand{\$d};
%res{\$s}{\$d} += \$v;
%demand{\$d} -= \$v;
if (%demand{\$d} == 0) {
%supply.grep( *.value != 0 )».key.map: -> \$v
{ %g{\$v}.splice((%g{\$v}.first: * eq \$d, :k),1) };
%g{\$d}:delete;
%demand{\$d}:delete;
}
%supply{\$s} -= \$v;
if (%supply{\$s} == 0) {
%demand.grep( *.value != 0 )».key.map: -> \$v
{ %g{\$v}.splice((%g{\$v}.first: * eq \$s, :k),1) };
%g{\$s}:delete;
%supply{\$s}:delete;
}
}
say join "\t", flat '', @cols;
my \$total;
for %costs.keys.sort -> \$g {
print "\$g\t";
for @cols -> \$col {
print %res{\$g}{\$col} // '-', "\t";
\$total += (%res{\$g}{\$col} // 0) * %costs{\$g}{\$col};
}
print "\n";
}
say "\nTotal cost: \$total";
```
Output:
``` C1 C2 C3
S1 20 - 5
S2 - 30 5 ```
## REXX
Translation of: Java
```/* REXX ***************************************************************
* Solve the Transportation Problem using the Northwest Corner Method
Default Input
2 3 # of sources / # of demands
25 35 sources
20 30 10 demands
3 5 7 cost matrix
3 2 5
* 20201210 support no input file -courtesy GS
* Note: correctnes of input is not checked
* 20210103 remove debug code
**********************************************************************/
Signal On Halt
Signal On Novalue
Signal On Syntax
Parse Arg fid
If fid='' Then
fid='input1.txt'
Call init
matrix.=0
ms=0
Do r=1 To rr
Do c=1 To cc
matrix.r.c=r c cost.r.c 0
End
End
r=1
c=1
Do While r<=rr & c<=cc
q=min(source.r,demand.c)
matrix.r.c=r c cost.r.c q
source.r=source.r-q
demand.c=demand.c-q
If source.r=0 Then r=r+1
If demand.c=0 Then c=c+1
End
Call show_alloc 'after NWC application'
Call steppingstone
Exit
/**********************************************************************
* Subroutines for NWC Algorithm
**********************************************************************/
init:
If lines(fid)=0 Then Do
fid='Default input'
in.1=sourceline(4)
Parse Var in.1 numSources .
Do i=2 To numSources+3
in.i=sourceline(i+3)
End
End
Else Do
Do i=1 By 1 while lines(fid)>0
in.i=linein(fid)
End
End
Parse Var in.1 numSources numDestinations . 1 rr cc .
source_sum=0
Do i=1 To numSources
Parse Var in.2 source.i in.2
ss.i=source.i
source_in.i=source.i
source_sum=source_sum+source.i
End
l=linein(fid)
demand_sum=0
Do i=1 To numDestinations
Parse Var in.3 demand.i in.3
dd.i=demand.i
demand_in.i=demand.i
demand_sum=demand_sum+demand.i
End
Do i=1 To numSources
j=i+3
l=in.j
Do j=1 To numDestinations
Parse Var l cost.i.j l
End
End
Do i=1 To numSources
ol=format(source.i,3)
Do j=1 To numDestinations
ol=ol format(cost.i.j,4)
End
End
ol=' '
Do j=1 To numDestinations
ol=ol format(demand.j,4)
End
Select
When source_sum=demand_sum Then Nop /* balanced */
When source_sum>demand_sum Then Do /* unbalanced - add dummy demand */
Say 'This is an unbalanced case (sources exceed demands). We add a dummy consumer.'
cc=cc+1
demand.cc=source_sum-demand_sum
demand_in.cc=demand.cc
dd.cc=demand.cc
Do r=1 To rr
cost.r.cc=0
End
End
Otherwise /* demand_sum>source_sum */ Do /* unbalanced - add dummy source */
Say 'This is an unbalanced case (demands exceed sources). We add a dummy source.'
rr=rr+1
source.rr=demand_sum-source_sum
source_in.rr=source.rr
ss.rr=source.rr
Do c=1 To cc
cost.rr.c=0
End
End
End
Say 'Sources / Demands / Cost'
ol=' '
Do c=1 To cc
ol=ol format(demand.c,3)
End
Say ol
Do r=1 To rr
ol=format(source.r,4)
Do c=1 To cc
ol=ol format(cost.r.c,3)
End
Say ol
End
Return
show_alloc: Procedure Expose matrix. rr cc demand_in. source_in.
Return
Say ''
total=0
ol=' '
Do c=1 to cc
ol=ol format(demand_in.c,3)
End
Do r=1 to rr
ol=format(source_in.r,4)
a=word(matrix.r.1,4)
If a>0 Then
ol=format(a,4)
Else
ol=' - '
total=total+word(matrix.r.1,4)*word(matrix.r.1,3)
Do c=2 To cc
a=word(matrix.r.c,4)
If a>0 Then
ol=ol format(a,4)
Else
ol=ol ' - '
total=total+word(matrix.r.c,4)*word(matrix.r.c,3)
End
Say ol
End
Say 'Total costs:' format(total,4,1)
Return
/**********************************************************************
* Subroutines for Optimization
**********************************************************************/
steppingstone: Procedure Expose matrix. cost. rr cc demand_in.,
source_in. fid
maxReduction=0
move=''
Call fixDegenerateCase
Do r=1 To rr
Do c=1 To cc
Parse Var matrix.r.c r c cost qrc
If qrc=0 Then Do
path=getclosedpath(r,c)
reduction = 0
lowestQuantity = 1e10
leavingCandidate = ''
plus=1
pathx=path
Do While pathx<>''
Parse Var pathx s '|' pathx
If plus Then
reduction=reduction+word(s,3)
Else Do
reduction=reduction-word(s,3)
If word(s,4)<lowestQuantity Then Do
leavingCandidate = s
lowestQuantity = word(s,4)
End
End
plus=\plus
End
If reduction < maxreduction Then Do
move=path
leaving=leavingCandidate
maxReduction = reduction
End
End
End
End
if move <> '' Then Do
quant=word(leaving,4)
plus=1
Do While move<>''
Parse Var move m '|' move
Parse Var m r c cpu qrc
Parse Var matrix.r.c vr vc vcost vquant
If plus Then
nquant=vquant+quant
Else
nquant=vquant-quant
matrix.r.c = vr vc vcost nquant
plus=\plus
End
move=''
Call steppingStone
End
Else Do
Call show_alloc 'Optimal Solution' fid
End
Return
getclosedpath: Procedure Expose matrix. cost. rr cc
Parse Arg rd,cd
path=rd cd cost.rd.cd word(matrix.rd.cd,4)
do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)>0 Then Do
path=path'|'r c cost.r.c word(matrix.r.c,4)
End
End
End
path=magic(path)
Return stones(path)
magic: Procedure
Parse Arg list
Do Forever
list_1=remove_1(list)
If list_1=list Then Leave
list=list_1
End
Return list_1
remove_1: Procedure
Parse Arg list
cntr.=0
cntc.=0
Do i=1 By 1 While list<>''
parse Var list e.i '|' list
Parse Var e.i r c .
cntr.r=cntr.r+1
cntc.c=cntc.c+1
End
n=i-1
keep.=1
Do i=1 To n
Parse Var e.i r c .
If cntr.r<2 |,
cntc.c<2 Then Do
keep.i=0
End
End
list=e.1
Do i=2 To n
If keep.i Then
list=list'|'e.i
End
Return list
stones: Procedure
Parse Arg lst
tstc=lst
Do i=1 By 1 While tstc<>''
Parse Var tstc o.i '|' tstc
end
stones=lst
o.0=i-1
prev=o.1
Do i=1 To o.0
st.i=prev
k=i//2
nbrs=getNeighbors(prev, lst)
Parse Var nbrs n.1 '|' n.2
If k=0 Then
prev=n.2
Else
prev=n.1
End
stones=st.1
Do i=2 To o.0
stones=stones'|'st.i
End
Return stones
getNeighbors: Procedure
parse Arg s, lst
Do i=1 By 1 While lst<>''
Parse Var lst o.i '|' lst
End
o.0=i-1
nbrs.=''
sr=word(s,1)
sc=word(s,2)
Do i=1 To o.0
If o.i<>s Then Do
or=word(o.i,1)
oc=word(o.i,2)
If or=sr & nbrs.0='' Then
nbrs.0 = o.i
else if oc=sc & nbrs.1='' Then
nbrs.1 = o.i
If nbrs.0<>'' & nbrs.1<>'' Then
Leave
End
End
return nbrs.0'|'nbrs.1
pelems: Procedure
Parse Arg p
Do i=1 By 1 While p<>''
Parse Var p x '|' p
End
Return i
fixDegenerateCase: Procedure Expose matrix. rr cc ms
Call matrixtolist
If (rr+cc-1)<>ms Then Do
Do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)=0 Then Do
matrix.r.c=subword(matrix.r.c,1,3) 1.e-10
Return
End
End
End
End
Return
matrixtolist: Procedure Expose matrix. rr cc ms
ms=0
list=''
Do r=1 To rr
Do c=1 To cc
If word(matrix.r.c,4)>0 Then Do
list=list'|'matrix.r.c
ms=ms+1
End
End
End
Return strip(list,,'|')
Novalue:
Say 'Novalue raised in line' sigl
Say sourceline(sigl)
Say 'Variable' condition('D')
Signal lookaround
Syntax:
Say 'Syntax raised in line' sigl
Say sourceline(sigl)
Say 'rc='rc '('errortext(rc)')'
halt:
lookaround:
If fore() Then Do
Say 'You can look around now.'
Trace ?R
Nop
End
Exit 12
```
Output:
```F:\>rexx tpx2 input1.txt
Sources / Demands / Cost
20 30 10
25 3 5 7
35 3 2 5
after NWC application
20 5 -
- 25 10
Total costs: 185.0
Optimal Solution input1.txt
20 - 5
- 30 5
Total costs: 180.0
F:\>rexx tpx2 input2.txt
This is an unbalanced case (sources exceed demands). We add a dummy consumer.
Sources / Demands / Cost
20 30 10 25
12 3 5 7 0
40 2 4 6 0
33 9 1 8 0
after NWC application
12 - - -
8 30 2 -
- - 8 25
Total costs: 248.0
Optimal Solution input2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130.0
F:\>rexx tpx2 input3.txt
This is an unbalanced case (demands exceed sources). We add a dummy source.
Sources / Demands / Cost
10 15 12 15
14 10 30 25 15
10 20 15 20 10
15 10 30 20 20
12 30 40 35 45
1 0 0 0 0
after NWC application
10 4 - -
- 10 - -
- 1 12 2
- - - 12
- - - 1
Total costs: 1220.0
Optimal Solution input3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000.0```
## SAS
Use network solver in SAS/OR:
```/* create SAS data sets */
data cost_data;
input from \$ to \$ cost;
datalines;
s1 c1 3
s1 c2 5
s1 c3 7
s2 c1 3
s2 c2 2
s2 c3 5
;
data supply_data;
input node \$ supply;
datalines;
s1 25
s2 35
c1 -20
c2 -30
c3 -10
;
/* call OPTMODEL procedure in SAS/OR */
proc optmodel;
/* declare sets and parameters, and read input data */
set NODES = union {<i,j> in LINKS} {i,j};
num supply {NODES} init 0;
read data supply_data into [node] supply;
/* call network solver */
solve with network /
/* print optimal solution */
print _OROPTMODEL_NUM_['OBJECTIVE'];
print flow;
quit;
```
Output:
```180
flow
c1 c2 c3
s1 20 0 5
s2 0 30 5
```
## Visual Basic .NET
Translation of: C#
```Module Module1
Class Shipment
Public Sub New(q As Double, cpu As Double, rv As Integer, cv As Integer)
Quantity = q
CostPerUnit = cpu
R = rv
C = cv
End Sub
Public ReadOnly Property CostPerUnit() As Double
Public Property Quantity() As Double
Public ReadOnly Property R As Integer
Public ReadOnly Property C As Integer
Public Shared Operator =(s1 As Shipment, s2 As Shipment) As Boolean
Return s1.CostPerUnit = s2.CostPerUnit _
AndAlso s1.Quantity = s2.Quantity _
AndAlso s1.R = s2.R _
AndAlso s1.C = s2.C
End Operator
Public Shared Operator <>(s1 As Shipment, s2 As Shipment) As Boolean
Return s1.CostPerUnit <> s2.CostPerUnit _
OrElse s1.Quantity <> s2.Quantity _
OrElse s1.R <> s2.R _
OrElse s1.C <> s2.C
End Operator
End Class
Class Program
Private demand() As Integer
Private supply() As Integer
Private costs(,) As Double
Private matrix(,) As Shipment
Sub Init(filename As String)
Dim numArr = line.Split
Dim numSources = Integer.Parse(numArr(0))
Dim numDestinations = Integer.Parse(numArr(1))
Dim src As New List(Of Integer)
Dim dst As New List(Of Integer)
numArr = line.Split
For i = 1 To numSources
Next
numArr = line.Split
For i = 1 To numDestinations
Next
REM fix imbalance
Dim totalSrc = src.Sum
Dim totalDst = dst.Sum
If totalSrc > totalDst Then
ElseIf totalDst > totalSrc Then
End If
supply = src.ToArray
demand = dst.ToArray
ReDim costs(supply.Length - 1, demand.Length - 1)
ReDim matrix(supply.Length - 1, demand.Length - 1)
For i = 1 To numSources
numArr = line.Split
For j = 1 To numDestinations
costs(i - 1, j - 1) = Integer.Parse(numArr(j - 1))
Next
Next
End Sub
Sub NorthWestCornerRule()
Dim northwest = 1
For r = 1 To supply.Length
For c = northwest To demand.Length
Dim quantity = Math.Min(supply(r - 1), demand(c - 1))
If quantity > 0 Then
matrix(r - 1, c - 1) = New Shipment(quantity, costs(r - 1, c - 1), r - 1, c - 1)
supply(r - 1) -= quantity
demand(c - 1) -= quantity
If supply(r - 1) = 0 Then
northwest = c
Exit For
End If
End If
Next
Next
End Sub
Sub SteppingStone()
Dim maxReduction = 0.0
Dim move() As Shipment = Nothing
Dim leaving As Shipment = Nothing
FixDegenerateCase()
For r = 1 To supply.Length
For c = 1 To demand.Length
If Not IsNothing(matrix(r - 1, c - 1)) Then
Continue For
End If
Dim trial As New Shipment(0, costs(r - 1, c - 1), r - 1, c - 1)
Dim path = GetClosedPath(trial)
Dim reduction = 0.0
Dim lowestQuanity = Integer.MaxValue
Dim leavingCandidate As Shipment = Nothing
Dim plus = True
For Each s In path
If plus Then
reduction += s.CostPerUnit
Else
reduction -= s.CostPerUnit
If s.Quantity < lowestQuanity Then
leavingCandidate = s
lowestQuanity = s.Quantity
End If
End If
plus = Not plus
Next
If reduction < maxReduction Then
move = path
leaving = leavingCandidate
maxReduction = reduction
End If
Next
Next
If Not IsNothing(move) Then
Dim q = leaving.Quantity
Dim plus = True
For Each s In move
s.Quantity += If(plus, q, -q)
matrix(s.R, s.C) = If(s.Quantity = 0, Nothing, s)
plus = Not plus
Next
SteppingStone()
End If
End Sub
Sub FixDegenerateCase()
Const eps = Double.Epsilon
If supply.Length + demand.Length - 1 <> MatrixToList().Count Then
For r = 1 To supply.Length
For c = 1 To demand.Length
If IsNothing(matrix(r - 1, c - 1)) Then
Dim dummy As New Shipment(eps, costs(r - 1, c - 1), r - 1, c - 1)
If GetClosedPath(dummy).Length = 0 Then
matrix(r - 1, c - 1) = dummy
Return
End If
End If
Next
Next
End If
End Sub
Function MatrixToList() As List(Of Shipment)
Dim newList As New List(Of Shipment)
For Each item In matrix
If Not IsNothing(item) Then
End If
Next
Return newList
End Function
Function GetClosedPath(s As Shipment) As Shipment()
Dim path = MatrixToList()
REM remove (and keep removing) elements that do not have a veritcal AND horizontal neighbor
Dim before As Integer
Do
before = path.Count
path.RemoveAll(Function(ship)
Dim nbrs = GetNeighbors(ship, path)
Return IsNothing(nbrs(0)) OrElse IsNothing(nbrs(1))
End Function)
Loop While before <> path.Count
REM place the remaining elements in the correct plus-minus order
Dim stones = path.ToArray
Dim prev = s
For i = 1 To stones.Length
stones(i - 1) = prev
prev = GetNeighbors(prev, path)((i - 1) Mod 2)
Next
Return stones
End Function
Function GetNeighbors(s As Shipment, lst As List(Of Shipment)) As Shipment()
Dim nbrs() As Shipment = {Nothing, Nothing}
For Each o In lst
If o <> s Then
If o.R = s.R AndAlso IsNothing(nbrs(0)) Then
nbrs(0) = o
ElseIf o.C = s.C AndAlso IsNothing(nbrs(1)) Then
nbrs(1) = o
End If
If Not IsNothing(nbrs(0)) AndAlso Not IsNothing(nbrs(1)) Then
Exit For
End If
End If
Next
Return nbrs
End Function
Sub PrintResult(filename As String)
Console.WriteLine("Optimal solution {0}" + vbNewLine, filename)
Dim totalCosts = 0.0
For r = 1 To supply.Length
For c = 1 To demand.Length
Dim s = matrix(r - 1, c - 1)
If Not IsNothing(s) AndAlso s.R = r - 1 AndAlso s.C = c - 1 Then
Console.Write(" {0,3} ", s.Quantity)
totalCosts += (s.Quantity * s.CostPerUnit)
Else
Console.Write(" - ")
End If
Next
Console.WriteLine()
Next
Console.WriteLine(vbNewLine + "Total costs: {0}" + vbNewLine, totalCosts)
End Sub
End Class
Sub Main()
For Each filename In {"input1.txt", "input2.txt", "input3.txt"}
Dim p As New Program
p.Init(filename)
p.NorthWestCornerRule()
p.SteppingStone()
p.PrintResult(filename)
Next
End Sub
End Module
```
Output:
```Optimal solution input1.txt
20 - 5
- 30 5
Total costs: 180
Optimal solution input2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130
Optimal solution input3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000```
## Wren
Translation of: Java
Library: Wren-dynamic
Library: Wren-ioutil
Library: Wren-math
Library: Wren-seq
Library: Wren-fmt
```import "./dynamic" for Struct
import "./ioutil" for FileUtil, File
import "./math" for Nums
import "./seq" for Lst
import "./fmt" for Fmt
var Shipment = Struct.create("Shipment", ["quantity", "costPerUnit", "r", "c"])
class Transport {
construct new(filename) {
var split = lines[0].split(" ")
var numSources = Num.fromString(split[0])
var numDests = Num.fromString(split[1])
var src = List.filled(numSources, 0)
var dst = List.filled(numDests, 0)
split = lines[1].split(" ")
for (i in 0...numSources) src[i] = Num.fromString(split[i])
split = lines[2].split(" ")
for (i in 0...numDests) dst[i] = Num.fromString(split[i])
// fix imbalance
var totalSrc = Nums.sum(src)
var totalDst = Nums.sum(dst)
if (totalSrc > totalDst) {
} else if (totalDst > totalSrc) {
}
_supply = src
_demand = dst
_costs = List.filled(_supply.count, null)
_matrix = List.filled(_supply.count, null)
for (i in 0..._supply.count) {
_costs[i] = List.filled(_demand.count, 0)
_matrix[i] = List.filled(_demand.count, null)
}
for (i in 0...numSources) {
split = lines[i + 3].split(" ")
for (j in 0...numDests) _costs[i][j] = Num.fromString(split[j])
}
_filename = filename
}
northWestCornerRule() {
var northwest = 0
for (r in 0..._supply.count) {
var c = northwest
while (c < _demand.count) {
var quantity = _supply[r].min(_demand[c])
if (quantity > 0) {
_matrix[r][c] = Shipment.new(quantity, _costs[r][c], r, c)
_supply[r] = _supply[r] - quantity.floor
_demand[c] = _demand[c] - quantity.floor
if (_supply[r] == 0) {
northwest = c
break
}
}
c = c + 1
}
}
}
steppingStone() {
var maxReduction = 0
var move = null
var leaving = null
fixDegenerateCase_()
for (r in 0..._supply.count) {
for (c in 0..._demand.count) {
if (_matrix[r][c] != null) continue
var trial = Shipment.new(0, _costs[r][c], r, c)
var path = getClosedPath_(trial)
var reduction = 0
var lowestQuantity = Num.maxSafeInteger
var leavingCandidate = null
var plus = true
for (s in path) {
if (plus) {
reduction = reduction + s.costPerUnit
} else {
reduction = reduction - s.costPerUnit
if (s.quantity < lowestQuantity) {
leavingCandidate = s
lowestQuantity = s.quantity
}
}
plus = !plus
}
if (reduction < maxReduction) {
move = path
leaving = leavingCandidate
maxReduction = reduction
}
}
}
if (move) {
var q = leaving.quantity
var plus = true
for (s in move) {
s.quantity = s.quantity + ((plus) ? q : -q)
_matrix[s.r][s.c] = (s.quantity == 0) ? null : s
plus = !plus
}
steppingStone()
}
}
matrixToList_() { Lst.flatten(_matrix).where { |s| s != null }.toList }
getClosedPath_(s) {
var path = matrixToList_()
path.insert(0, s)
// remove (and keep removing) elements that do not have a
// vertical AND horizontal neighbor
while (true) {
var removals = 0
for (e in path) {
var nbrs = getNeighbors_(e, path)
if (nbrs[0] == null || nbrs[1] == null) {
path.remove(e)
removals = removals + 1
}
}
if (removals == 0) break
}
// place the remaining elements in the correct plus-minus order
var stones = List.filled(path.count, null)
var prev = s
for (i in 0...stones.count) {
stones[i] = prev
prev = getNeighbors_(prev, path)[i % 2]
}
return stones
}
getNeighbors_(s, lst) {
var nbrs = List.filled(2, null)
for (o in lst) {
if (o != s) {
if (o.r == s.r && nbrs[0] == null) {
nbrs[0] = o
} else if (o.c == s.c && nbrs[1] == null) {
nbrs[1] = o
}
if (nbrs[0] != null && nbrs[1] != null) break
}
}
return nbrs
}
fixDegenerateCase_() {
var eps = Num.smallest
if (_supply.count + _demand.count - 1 != matrixToList_().count) {
for (r in 0..._supply.count) {
for (c in 0..._demand.count) {
if (_matrix[r][c] == null) {
var dummy = Shipment.new(eps, _costs[r][c], r, c)
if (getClosedPath_(dummy).count == 0) {
_matrix[r][c] = dummy
return
}
}
}
}
}
}
printResult() {
System.print("%(_filename)\n\n%(text)")
System.print("Optimal solution %(_filename)\n")
var totalCosts = 0
for (r in 0..._supply.count) {
for (c in 0..._demand.count) {
var s = _matrix[r][c]
if (s != null && s.r == r && s.c == c) {
Fmt.write(" \$3d ", s.quantity.floor)
totalCosts = totalCosts + s.quantity * s.costPerUnit
} else System.write(" - ")
}
System.print()
}
System.print("\nTotal costs: %(totalCosts)\n")
}
}
var filenames = ["input1.txt", "input2.txt", "input3.txt"]
for (filename in filenames) {
var t = Transport.new(filename)
t.northWestCornerRule()
t.steppingStone()
t.printResult()
}
```
Output:
```input1.txt
2 3
25 35
20 30 10
3 5 7
3 2 5
Optimal solution input1.txt
20 - 5
- 30 5
Total costs: 180
input2.txt
3 3
12 40 33
20 30 10
3 5 7
2 4 6
9 1 8
Optimal solution input2.txt
- - - 12
20 - 10 10
- 30 - 3
Total costs: 130
input3.txt
4 4
14 10 15 12
10 15 12 15
10 30 25 15
20 15 20 10
10 30 20 20
30 40 35 45
Optimal solution input3.txt
- - - 14
- 9 - 1
10 - 5 -
- 5 7 -
- 1 - -
Total costs: 1000
``` | 41,162 | 111,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.387036 |
http://college-cram.com/study/accounting/tag/valuation/ | 1,571,152,123,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986659097.10/warc/CC-MAIN-20191015131723-20191015155223-00349.warc.gz | 38,448,052 | 10,495 | # Cost of Goods Sold Overview: Study Sheet
This Smartacus Study Sheet will help you to understand how to find cost-of-goods-sold and the value of inventory. Explore definitions and methods (including Average Cost Method, LIFO, and FIFO). It also prints for easy reference. Continue reading
# Inventory Value Study Sheet
Use this Smartacus Study Sheet to understand how to find the value of inventory and cost-of-goods-sold. Explore definitions and methods (including Average Cost Method, LIFO, and FIFO). It also prints for easy reference. Continue reading
# Units of Production Method: Bottomless Worksheet
Get endless practice using the units-of-production method of calculating depreciation with this Bottomless Worksheet. Ten new problems for you to solve, printed copies, and answer sheets are all just a click away. Continue reading
# Double Declining Balance: Bottomless Worksheet
Get endless hours of practice on calculating depreciation using the double declining balance method with this Bottomless Worksheet. Get ten new problems whenever you need them, along with an answer key — both of which you can print! Continue reading
# Units of Production Method
This Formula Solver walks you through the steps for calculating depreciation with the Units-of-Production method. Use your own values for purchase price, salvage value, expected lifetime units, and current year units to check your homework answers. Continue reading
# Declining Balance
This Formula Solver walks you through the steps to calculate depreciation with the declining balance method. Use your own numbers for the current value, salvage value, and years of life to find the depreciation for an asset in the current year. Continue reading
# Inventory Value: Average Cost Method
Use this Formula Solver! Series program to learn how to calculate Inventory Value with the Average Cost Method. It’s great for checking your homework answers, too! Continue reading
# Sum-of-the-Years’-Digits Method: Bottomless Worksheet
Use this Bottomless Worksheet to get countless hours of practice on calculating depreciation using the sum-of-the-years-digits method (SYD). Get ten new problems whenever you need them, along with an answer key — both of which you can print! Continue reading | 438 | 2,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.834904 |
https://www.proprofs.com/quiz-school/story.php?title=odewotq2ziz9 | 1,719,118,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00395.warc.gz | 833,095,375 | 101,819 | # Unit Two - Integer Operations
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| By Elovell53
E
Elovell53
Community Contributor
Quizzes Created: 1 | Total Attempts: 317
Questions: 51 | Attempts: 317
Settings
• 1.
• 2.
• 3.
• 4.
• 5.
### Which integer addition problem does this model?
• A.
-5 + (-2) = -7
• B.
5 + 2 = 7
• C.
-2 + (-7) = 5
• D.
-2 + (-7) = -5
A. -5 + (-2) = -7
Explanation
The given addition problem is -5 + (-2) which equals -7. This is the correct answer because when adding two negative numbers, the result will be negative. Since -5 and -2 are both negative, their sum will also be negative. The sum of -5 and -2 is -7.
Rate this question:
• 6.
### The temperature in Fishersville on December 12 was 14 degrees. The temperature at the top of Afton Mountain was -2 degrees. What's the difference between these two temperatures?
• A.
12 degrees
• B.
2 degrees
• C.
16 degrees
• D.
-16 degrees
C. 16 degrees
Explanation
Think about these two temperatures on a thermometer. To go from 12 degrees to -2 degrees you have to go 16 spaces. What we're really asking is a subtraction problem. 12 - (-2) and that's the same as 12 + 2
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• 7.
### The temperature at 3:00 PM was 15 degrees. At midnight the temperature was -2 degrees. What's the difference between those two temperatures?
• A.
17 degrees
• B.
13 degrees
• C.
-13 degrees
• D.
-17 degrees
A. 17 degrees
Explanation
The difference between the temperature at 3:00 PM (15 degrees) and the temperature at midnight (-2 degrees) is calculated by subtracting the lower temperature from the higher temperature. In this case, 15 - (-2) = 17 degrees. Therefore, the correct answer is 17 degrees.
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• 8.
### The town of Apple Valley, California is at an elevation of 2,900 feet above sea level. Death Valley, California has an elevation of 282 feet below sea level. What is the difference in elevation of these two towns?
• A.
2,616 feet
• B.
3,182 feet
• C.
-2,616 feet
• D.
-3,182 feet
B. 3,182 feet
Explanation
The difference in elevation between Apple Valley and Death Valley is 3,182 feet. This is calculated by subtracting the elevation of Death Valley (-282 feet) from the elevation of Apple Valley (2,900 feet). The negative sign indicates that Death Valley is below sea level, while Apple Valley is above sea level.
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• 9.
• A.
Point E
• B.
Point F
• C.
Point G
• D.
Point H
B. Point F
• 10.
### Solve:-3 - (-4)
• A.
-7
• B.
1
• C.
0
• D.
-7
B. 1
Explanation
The expression -3 - (-4) can be simplified by changing the subtraction of a negative number to addition. So, -3 - (-4) becomes -3 + 4, which equals 1.
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• 11.
### The temperature outside in the morning is 8º F. Throughout the course of the day, the temperaturedrops 12º F. What is the new temperature?
• A.
20 degrees
• B.
-4 degrees
• C.
12 degrees
• D.
4 degrees
B. -4 degrees
Explanation
Throughout the course of the day, the temperature drops 12º F. Since the initial temperature is 8º F, a drop of 12º F would result in a new temperature of -4º F.
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• 12.
### 4/5 converted to a decimal is:
• A.
0.08
• B.
0.8
• C.
_ 0.8
• D.
80
B. 0.8
Explanation
The fraction 4/5 can be converted to a decimal by dividing the numerator (4) by the denominator (5). When we do this calculation, we get 0.8. Therefore, the correct answer is 0.8.
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• 13.
### Any number to the 0 power is always 0.
• A.
True
• B.
False
B. False
Explanation
The statement "Any number to the 0 power is always 0" is incorrect. In fact, any number to the power of 0 is always equal to 1. This is a fundamental mathematical rule that applies to all numbers.
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• 14.
### A number to the first power is itself. Example:
• A.
True
• B.
False
A. True
Explanation
The statement is true because any number raised to the power of 1 is equal to itself. This is a basic mathematical property where the exponent 1 signifies that the number is not being multiplied or divided by any other number. Therefore, the number remains unchanged and is equal to itself.
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• 15.
### The following numbers are correctly placed in ascending order:
• A.
True
• B.
False
A. True
Explanation
The given numbers "True" and "False" are placed in ascending order correctly. In ascending order, "True" comes before "False" alphabetically. Therefore, the correct answer is "True".
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• 16.
### Solve:5 + (-2)
• A.
7
• B.
3
• C.
-3
• D.
-7
B. 3
Explanation
The given expression is 5 + (-2). To solve this, we add 5 and -2 together. When we add a positive number and a negative number, we subtract the smaller number from the larger number and keep the sign of the larger number. In this case, 5 is larger than 2, so we subtract 2 from 5 and keep the positive sign. Therefore, 5 + (-2) equals 3.
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• 17.
### Solve:-7 + (-9) =
• A.
16
• B.
-2
• C.
2
• D.
-16
D. -16
Explanation
The given expression is asking for the sum of -7 and -9. When we add these two negative numbers, we get a more negative number. In this case, -7 + (-9) equals -16.
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• 18.
### Which one has a different answer from the rest?
• A.
4 + (-5)
• B.
-7 + 6
• C.
5 + (-4)
• D.
6 + (-7)
C. 5 + (-4)
Explanation
All the given expressions involve addition of a positive number and a negative number, except for 5 + (-4). In this expression, both numbers are negative.
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• 19.
### The product of a positive integer and its opposite is always
• A.
Positive
• B.
Negative
• C.
Zero
• D.
Cannot tell (depends)
B. Negative
Explanation
The product of a positive integer and its opposite will always result in a negative number. This is because when we multiply a positive number by a negative number, the result is always negative. Therefore, the answer is negative.
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• 20.
### The absolute value of any integer is always positive.
• A.
True
• B.
False
A. True
Explanation
The absolute value of any integer is always positive because it represents the distance of the integer from zero on the number line. Since distance cannot be negative, the absolute value is always positive.
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• 21.
### The sum of two negative integers is always
• A.
Positive
• B.
Negative
• C.
Zero
• D.
Cannot tell (depends)
B. Negative
Explanation
When we add two negative integers, the result will always be a negative integer. This is because adding two negative numbers is essentially combining two quantities that are less than zero, resulting in a value that is even smaller and negative. Therefore, the sum of two negative integers is always negative.
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• 22.
### The temperature in Alaska is -10 degrees. The temperature in Florida is 90 degrees. What is the difference in the temperatures?
• A.
80 degrees
• B.
100 degrees
• C.
110 degrees
• D.
No difference
B. 100 degrees
Explanation
The difference in temperature between Alaska and Florida is 100 degrees. This is calculated by subtracting the temperature in Alaska (-10 degrees) from the temperature in Florida (90 degrees). The result is 100 degrees.
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• 23.
### The sum of two negative integers is always
• A.
Positive
• B.
Negative
• C.
Zero
• D.
Cannot tell (depends)
B. Negative
Explanation
When two negative integers are added, the result will always be negative. This is because adding a negative number is equivalent to subtracting a positive number. So, when two negative numbers are added, the negatives cancel out and the result remains negative.
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• 24.
### 3 + (-2)
• A.
5
• B.
-1
• C.
1
C. 1
Explanation
The given expression is 3 + (-2). When we add a positive number and a negative number, we subtract the absolute value of the smaller number from the absolute value of the larger number and keep the sign of the larger number. In this case, the absolute value of 3 is larger than the absolute value of -2, so we subtract 2 from 3 and keep the positive sign. Therefore, 3 + (-2) equals 1.
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• 25.
### (-4) - 9
• A.
-13
• B.
-5
• C.
5
A. -13
Explanation
The given expression is (-4) - 9. To simplify this, we subtract 9 from -4. When subtracting a positive number from a negative number, we add the opposite of that positive number. Therefore, -4 - 9 can be rewritten as -4 + (-9). Adding -4 and -9 gives us -13. So, the correct answer is -13.
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• 26.
### -3(-9)
• A.
27
• B.
-27
• C.
12
A. 27
Explanation
When we simplify -3(-9), we multiply -3 by -9, which gives us 27. Therefore, the correct answer is 27.
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• 27.
### -9/3
• A.
-3
• B.
3
• C.
-6
A. -3
Explanation
The given expression -9/3 can be simplified by dividing -9 by 3, which equals -3. Therefore, the correct answer is -3.
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• 28.
### A negative number times a positive number always gives a negative answer.
• A.
True
• B.
False
A. True
Explanation
When multiplying a negative number by a positive number, the negative sign of the first number indicates a reversal or opposite direction. This means that the two numbers are in opposite directions, resulting in a negative product. Therefore, it is true that a negative number times a positive number always gives a negative answer.
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• 29.
### A negative number plus a positive number always gives a negative answer.
• A.
True
• B.
False
B. False
Explanation
The statement is false because when a negative number is added to a positive number, the sum will always depend on the magnitudes of the numbers involved. If the positive number has a greater magnitude than the negative number, the sum will be positive. However, if the negative number has a greater magnitude, the sum will be negative. Therefore, it is not always true that a negative number plus a positive number gives a negative answer.
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• 30.
### -2 (9)
• A.
18
• B.
-18
• C.
7
• D.
11
B. -18
Explanation
Negative times a positive is a negative
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• 31.
### -8 (-7)
• A.
-56
• B.
-15
• C.
1
• D.
56
D. 56
Explanation
Negative times negative is positive
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• 32.
### -2 (9)
• A.
18
• B.
-18
• C.
7
• D.
11
B. -18
Explanation
Negative times a positive is a negative
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• 33.
### -8 (-7)
• A.
-56
• B.
-15
• C.
1
• D.
56
D. 56
Explanation
Negative times negative is positive
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• 34.
### -2 (9)
• A.
18
• B.
-18
• C.
7
• D.
11
B. -18
Explanation
Negative times a positive is a negative
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• 35.
### -8 (-7)
• A.
-56
• B.
-15
• C.
1
• D.
56
D. 56
Explanation
Negative times negative is positive
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• 36.
### - 243 + 18=
• A.
-225
• B.
-261
• C.
225
• D.
261
A. -225
Explanation
The given question is a simple addition problem. When we add 243 and 18, we get a sum of 261. However, the answer given is -225. This suggests that there is a mistake in the calculation or the answer choices provided. Without further information, it is not possible to determine the correct explanation or provide a valid answer.
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• 37.
### 196 + (-8) =
• A.
-204
• B.
204
• C.
-188
• D.
188
D. 188
Explanation
The given question asks for the sum of 196 and -8. When adding a positive number to a negative number, we subtract the smaller absolute value from the larger absolute value and keep the sign of the number with the larger absolute value. In this case, the absolute value of 196 is larger than the absolute value of -8. So, we subtract 8 from 196 and keep the positive sign, resulting in 188. Therefore, the correct answer is 188.
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• 38.
### 19 - (-5) =
• A.
24
• B.
-24
• C.
14
• D.
-14
A. 24
Explanation
When subtracting a negative number from a positive number, we can think of it as adding the positive number to the absolute value of the negative number. In this case, 19 - (-5) becomes 19 + 5, which equals 24.
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• 39.
### -81 - (-3) =
• A.
-84
• B.
78
• C.
-78
• D.
84
C. -78
Explanation
The given expression is -81 - (-3). When we subtract a negative number, it is equivalent to adding the positive value of that number. Therefore, -81 - (-3) can be rewritten as -81 + 3. Adding -81 and 3 gives us -78. Hence, the correct answer is -78.
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• 40.
### -48 - 12=
• A.
60
• B.
-60
• C.
-36
• D.
36
B. -60
Explanation
To find the solution to the equation -48 - 12, we need to subtract 12 from -48. Subtracting a negative number is the same as adding its positive counterpart. So, subtracting 12 from -48 is equivalent to adding 12 to -48. When we add 12 to -48, we get -36. Therefore, the correct answer is -36.
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• 41.
### A negative number times a negative number will:
• A.
Sometimes result in a positive answer
• B.
Will always result in a positive answer
• C.
Will sometimes result in a negative answer
• D.
Will always result in a negative answer
B. Will always result in a positive answer
Explanation
When two negative numbers are multiplied, the result is always a positive number. This can be understood by considering the concept of multiplication as repeated addition. If we have a negative number, adding it to itself multiple times will result in a negative number. However, when we multiply two negative numbers, the negative signs cancel out, and we are left with a positive product. Therefore, a negative number times a negative number will always result in a positive answer.
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• 42.
### The high temperature on December 22 was . That evening the temperature dropped to . What's the difference between those two temperatures?
• A.
18 degrees
• B.
-24 degrees
• C.
24 degrees
• D.
-18 degrees
C. 24 degrees
Explanation
The question asks for the difference between the high temperature on December 22 and the temperature that evening. Since the temperature dropped, we can assume that the high temperature was higher than the temperature in the evening. Therefore, the difference between the two temperatures would be the high temperature minus the evening temperature. Since the answer is 24 degrees, it means that the high temperature was 24 degrees higher than the evening temperature.
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• 43.
### Which one is true?
• A.
-10 + 14 = -4
• B.
-7 + (-7) = 0
• C.
-10 + (-14) = -4
• D.
3 - (-8) = 11
D. 3 - (-8) = 11
Explanation
The given equation 3 - (-8) = 11 is true because when we subtract a negative number from a positive number, the two negatives cancel out and become positive. In this case, -8 becomes +8, and then we have 3 + 8 = 11.
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• 44.
### Solve:5 - (-9) =
• A.
-4
• B.
4
• C.
14
• D.
-14
C. 14
Explanation
The expression 5 - (-9) can be simplified by applying the rule of subtracting a negative number, which is equivalent to adding the positive value. Therefore, -(-9) becomes +9, resulting in 5 + 9. Adding 5 and 9 gives the final answer of 14.
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• 45.
### Which statement is false?
• A.
Multiplying a negative number times a negative number results in a positive number
• B.
Adding two positive numbers results in a positive answer
• C.
Subtracting a negative number from a negative number will always result in a negative answer
• D.
Dividing a positive number by a negative number will result in a negative answer
C. Subtracting a negative number from a negative number will always result in a negative answer
Explanation
The statement "subtracting a negative number from a negative number will always result in a negative answer" is false. When subtracting a negative number from a negative number, it is equivalent to adding a positive number. This results in a positive answer, not a negative one.
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• 46.
### Mrs. Shumate fell down a hole that was 22 feet deep. She climbed up 10 feet before falling 5 feet. Which one show the correct addition problem and answer?
• A.
22 - 10 + 5 - -17
• B.
-22 + 10 + (-5) = -37
• C.
-22 + 10 + (-5) = -17
• D.
-22 + (-10) - 17 = -49
C. -22 + 10 + (-5) = -17
Explanation
Mrs. Shumate fell down a 22-foot deep hole. She climbed up 10 feet, which is represented by the positive addition of 10. However, she then fell back 5 feet, which is represented by the negative addition of -5. Therefore, the correct addition problem is -22 + 10 + (-5), which equals -17.
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• 47.
### The product of a negative integer and its opposite is always
• A.
Positive
• B.
Negative
• C.
Zero
• D.
Cannot tell (depends)
B. Negative
Explanation
When multiplying a negative integer with its opposite, the result is always negative. This is because when we multiply two numbers with opposite signs, the product will have the sign of the negative number. In this case, since one of the numbers is negative, the product will also be negative.
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• 48.
### The absolute value of any negative integer is always positive.
• A.
True
• B.
False
A. True
Explanation
The absolute value of a number represents its distance from zero on the number line, so it is always positive. Since negative integers are less than zero, their absolute values will be positive. Therefore, the statement is true.
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• 49.
### A bird is flying above the ocean at a height of 20 feet above sea level. A fish is under the ocean at 10 feet below sea level. What is the difference between their locations?
• A.
10 feet
• B.
-10 feet
• C.
30 feet
• D.
-30 feet
B. -10 feet
Explanation
The bird is flying 20 feet above sea level, while the fish is 10 feet below sea level. To find the difference between their locations, we can subtract the fish's location from the bird's location. So, 20 feet - (-10 feet) = 20 feet + 10 feet = 30 feet. However, the question asks for the difference, not the sum. Therefore, the correct answer is -10 feet, indicating that the fish is 10 feet below the bird's location.
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• 50.
### 12
• A.
True
• B.
False
A. True
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Quiz Created by
Elovell53 | 5,043 | 19,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-26 | latest | en | 0.911037 |
https://data.epo.org/gpi/EP1102116A1-DATA-RECORDING-DEVICE-AND-CAMERA-WITH-DATA-IMAGING-DEVICE | 1,534,857,318,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218122.85/warc/CC-MAIN-20180821112537-20180821132537-00149.warc.gz | 645,804,416 | 5,279 | # Title (en)
DATA RECORDING DEVICE, AND CAMERA WITH DATA IMAGING DEVICE
# Title (de)
DATENAUFNAHMEVORRICHTUNG UND KAMERA MIT EINER DATENERZEUGUNGSVORRICHTUNG
# Title (fr)
DISPOSITIF D'ENREGISTREMENT DE DONNEES, ET CAMERA MUNIE D'UN DISPOSITIF D'IMAGERIE DE DONNEES
# Priority
• JP 9902397 W
• JP 13011598 A
• JP 21225598 A
• JP 21290898 A
• JP 21290998 A
# Abstract (en)
In an optical data recording device, an LCD panel, an aperture stop and a projection lens are arranged in this order from the object side, and a center of the LCD panel is displaced from an optical axis (68a) of the projection lens in a direction away from an optical axis (11a) of an taking lens, so an image of data displayed on the LCD panel is projected on a photo filmstrip at a position that is shifted from the projection lens optical axis (68a) toward the taking lens optical axis (11a). Assuming that the focal length of the projection lens is "f", the distance from the aperture stop to the image side surface of the projection lens is "d", the radiuses of curvature of the projection lens on the respective sides are "R1" and "R2", the distance from the LCD panel to the aperture stop is "S", the longer side length of the LCD panel is "L", the shift amount of the center of the LCD panel from the projection lens optical axis is "Q", the f-number of the taking lens is "F1", and the f-number of the projection lens is "F2", the following conditions are satisfied: 1 > d/f >/= 0.3, R1 > R2 x 3, 0 < θ < 0.55 ( θ = TAN<-1> ä(Q + L/2) / Sü), log2 (F2) </= log2 (F1). Where the photo filmstrip is curved along its lengthwise direction for compensating for a curvature of field of the taking lens, the LCD panel is inclined to a perpendicular plane (68b) to the projection lens optical axis (68a), so an image surface of the LCD panel is formed on the photo filmstrip parallel with its curve. A lens holder (104) holding the projection lens is able to rotate about a rotary center (104a) that is displaced from the projection lens optical axis (68a). With rotation of the lens holder (104), the projection lens is shifted relative to the display panel, so is an optical recording position of the data. <IMAGE>
# IPC 8 full level (invention and additional information)
G03B 17/24 (2006.01)
# CPC (invention and additional information)
G03B 17/245 (2013.01)
DE FR GB
# 2009-02-25[18D] APPLICATION DEEMED TO BE WITHDRAWN
- Effective date: 20080909
# 2008-05-28[17Q] FIRST EXAMINATION REPORT
- Effective date: 20080428
# 2007-05-02[RAP1] TRANSFER OF RIGHTS OF AN APPLICATION
- Owner name: FUJIFILM CORPORATION
# 2006-07-19[A4] DESPATCH OF SUPPLEMENTARY SEARCH REPORT
- Effective date: 20060619
# 2001-05-23[17P] REQUEST FOR EXAMINATION FILED
- Effective date: 20001211
# 2001-05-23[AK] DESIGNATED CONTRACTING STATES:
- Kind Code of Ref Document: A1
- Designated State(s): DE FR GB | 820 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.844978 |
https://joningram.org/questions/Trigonometry/384580 | 1,709,600,706,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00856.warc.gz | 336,988,147 | 6,142 | # Graph ((x+1)^2)/25+((y-2)^2)/9=1
Graph ((x+1)^2)/25+((y-2)^2)/9=1
Simplify each term in the equation in order to set the right side equal to . The standard form of an ellipse or hyperbola requires the right side of the equation be .
This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.
Match the values in this ellipse to those of the standard form. The variable represents the radius of the major axis of the ellipse, represents the radius of the minor axis of the ellipse, represents the x-offset from the origin, and represents the y-offset from the origin.
The center of an ellipse follows the form of . Substitute in the values of and .
Find , the distance from the center to a focus.
Find the distance from the center to a focus of the ellipse by using the following formula.
Substitute the values of and in the formula.
Simplify.
Raise to the power of .
Raise to the power of .
Multiply by .
Subtract from .
Rewrite as .
Pull terms out from under the radical, assuming positive real numbers.
Find the vertices.
The first vertex of an ellipse can be found by adding to .
Substitute the known values of , , and into the formula.
Simplify.
The second vertex of an ellipse can be found by subtracting from .
Substitute the known values of , , and into the formula.
Simplify.
Ellipses have two vertices.
:
:
:
:
Find the foci.
The first focus of an ellipse can be found by adding to .
Substitute the known values of , , and into the formula.
Simplify.
The second vertex of an ellipse can be found by subtracting from .
Substitute the known values of , , and into the formula.
Simplify.
Ellipses have two foci.
:
:
:
:
Find the eccentricity.
Find the eccentricity by using the following formula.
Substitute the values of and into the formula.
Simplify the numerator.
Raise to the power of .
Raise to the power of .
Multiply by .
Subtract from .
Rewrite as .
Pull terms out from under the radical, assuming positive real numbers.
These values represent the important values for graphing and analyzing an ellipse.
Center:
:
:
:
:
Eccentricity:
Do you know how to Graph ((x+1)^2)/25+((y-2)^2)/9=1? If not, you can write to our math experts in our application. The best solution for your task you can find above on this page.
### Name
Name one billion six hundred thirty million three hundred seventy-seven thousand eight hundred ninety-nine
### Interesting facts
• 1630377899 has 8 divisors, whose sum is 1667980512
• The reverse of 1630377899 is 9987730361
• Previous prime number is 71
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 10
• Sum of Digits 53
• Digital Root 8
### Name
Name two billion one hundred twenty million six hundred seventy-seven thousand thirty-one
### Interesting facts
• 2120677031 has 8 divisors, whose sum is 2314946304
• The reverse of 2120677031 is 1307760212
• Previous prime number is 1583
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 10
• Sum of Digits 29
• Digital Root 2
### Name
Name one hundred sixty-one million seven hundred forty-one thousand three hundred fifty-seven
### Interesting facts
• 161741357 has 8 divisors, whose sum is 163160640
• The reverse of 161741357 is 753147161
• Previous prime number is 179
### Basic properties
• Is Prime? no
• Number parity odd
• Number length 9
• Sum of Digits 35
• Digital Root 8 | 860 | 3,426 | {"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} | 3.6875 | 4 | CC-MAIN-2024-10 | latest | en | 0.868528 |
http://www.instructables.com/id/voltmeter-with-arduino/ | 1,498,233,322,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320070.48/warc/CC-MAIN-20170623151757-20170623171757-00298.warc.gz | 548,409,382 | 14,460 | The idea comes, because I'm traveling and i forget my voltmeter in another home, and i'm really needing one. But my Arduino kit aways are with me and with a few minutes I made my voltmeter.
Necessary items:
-An arduino board
-Resistores -LCD 16x2(optionally)
- Wires
-Batteries to measure
- A little math
## Step 1: Hardware: Voltage Divider
First thing to do is choose what the max voltage that the voltmeter will work.
Reading the AnalogReference on http://arduino.cc/en/Reference/AnalogReference .
We see a warning: "Don't use anything less than 0V or more than 5V for external reference voltage on the AREF pin!"
So we can't put directly an battery with more than 5v on analog pin of arduino, but with a little math and physics we "can" do this.
Using laws of resistors we will do a voltage divider
I decide to use 15v with max voltage to measure, and we need to put 5v max in the analog pin, to do this divisor, i put 2 resistors of 10k ohms in parallel to have 5k ohms and put another resistor with 10k ohms in series, and my divisor is done ! now, the voltage will be divided by 3, so in the math calculus on arduino, we will have to multiply to 3...
Be careful choosing the max, because if you pass of this, you can burn your arduino board.
## Step 2: Programming and Set It Up
The only important part of the code is the :
float volt = ((sensor1 *5)/ 1023)*3;
look that the only thing i'm doing is changing the analog read to volts and multiplying by 3.
And is done! But warning is on your own risk!
The hardware have no protection to overcharge, reverse voltage and some little things, but is fully useful.
<p>epic fail this project, you fail</p>
<p>Wow this is an awesome gadget, it's great that you can DIY so many things with arduino! Nice job showing each step! Welcome to instructables!</p>
<p>15K = 15 V ??</p><p>i don't understand </p><p>thanks!<br>marC:)</p>
15k ohms *
<p>Nice project! Congradulation! But how do you select the resistor value? :)</p><p>thanks!<br>MarC:)</p>
On step1 the picture shows that vout = (r2/(r1+r2))*vin<br>And i choose set my Vin as 15v and the vout must be 5v.<br>I havent a resistor of 5k so i put 2 of 10k in parallel.<br>I try calculate with the resistors that I have, <br>R1 = 10k<br>R2 = 5k<br>This satisfies the calculus .<br>Or doing direct, 10k +5k =15k, 15k/ 5k =3 , so the voltage divider will divide for 3 <br>Lots of resistors can be used to divide to 3, I only choose the easy for me. <br> | 700 | 2,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-26 | longest | en | 0.886842 |
https://lists.boost.org/ublas/2015/03/5720.php | 1,669,575,169,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00342.warc.gz | 407,132,025 | 4,028 | Ublas :
Subject: [ublas] Fwd: [GSOC2015] uBLAS Matrix Solver Project
Date: 2015-03-06 10:30:32
Hi,
My name is Raj and I am Phd student in Computer Graphics. I am interested
in tackling the problem of uBLAS Matrix Solver and in order to write my
proposal, I am looking for inputs for which of the following algorithms
will be most useful for prospective users in boost-numeric library. Here is
a categorical list of all the prospective ones which will bring uBLAS
*David Bellot* : As a potential mentor, do you have any specific additions
or deletions for this list? This could also be useful for other candidates
pursuing this project.
*DENSE SOLVERS AND DECOMPOSITION* :
1) *QR Decomposition* - *(Must have)* For orthogonalization of column
spaces and solutions to linear systems. (Bonus : Also rank revealing..)
2) *Cholesky Decomposition* - *(Must have)* For symmetric Positive Definite
systems often encountered in PDE for FEM Systems...
3) *Householder Method* - Conversion to tridiagonal form for eigen solvers.
*SPARSE SOLVERS AND PRECONDITIONERS* :
1) *Conjugate Gradient* - *(Must have)* For symmetric Positive Definite
systems, this is the kryvlov space method of choice. Both general and
preconditioned variants need to be implemented for convergence issues. 2)
*BiCGSTAB* *(Needs introspection)* - For non symmetric systems..
3) *Incomplete Cholesky Decomposition* *(Good to have)* - For symmetric
Positive definite sparse matrices, to be used as preconditioner as
extension to (1) for preconditioned CG Methods ...
4) *Jacobi Preconditioner* *(Must have)* - As prerequisite for step(1).
*EIGEN DECOMPOSITION MODULES (ONLY FOR DENSE MODULES)**:*
1) *Symmetric Eigen Values* - *(Must have)* Like SSYEV Module in Lapack -
That is first reduction to a tridiagonal form using Householder then using
QR Algorithm for Eigen Value computation.
2) *NonSymmetric Eigen Values* - *(Good to have)* Like SGEEV module in
Lapack - using Schur decompositions as an intermediate step in the above
algorithm.
3) *Generalized Eigen Values* - *(needs introspection)* I use this in my
research a lot and its a good thing to have..
** Computing Eigen Decomposition of sparse modules needs special robust
numerical treatment using implicitly restarted arnoldi iterations and may
be treated as optional extensions. | 568 | 2,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | longest | en | 0.724604 |
https://uk.mathworks.com/matlabcentral/cody/problems/43134-vector-multiplication/solutions/1973963 | 1,603,660,109,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889651.52/warc/CC-MAIN-20201025183844-20201025213844-00038.warc.gz | 585,957,660 | 17,103 | Cody
# Problem 43134. Vector Multiplication
Solution 1973963
Submitted on 13 Oct 2019 by Kruti Rajhans
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]; y=[5 6]; z=[5 7]; y_correct = [ 50 210]; assert(isequal(vec_mul(x,y,z),y_correct))
2 Pass
x = [1 1 1 ]; y=[1 1 1 ]; z=[1 1 1]; y_correct = [1 1 1]; assert(isequal(vec_mul(x,y,z),y_correct))
3 Pass
x = [1 2]; y=[1 2]; z=[1 2]; y_correct = [1 8]; assert(isequal(vec_mul(x,y,z),y_correct))
4 Pass
x = 1; y=1; z=1; y_correct = 1; assert(isequal(vec_mul(x,y,z),y_correct))
### Community Treasure Hunt
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Start Hunting! | 268 | 787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-45 | latest | en | 0.549867 |
http://www.ipaulownia.es/sa283/per-running-meter-weight-find_6254.html | 1,656,475,690,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00627.warc.gz | 84,013,621 | 9,393 | # per running meter weight find
### Weight Calculator - Civil Engineering
Mild Steel Sheet Weight Calculator – Online MS Sheet Weight Calculator; Length Conversion Tool – Online Length Converter ... all types of ms angles, c channels weight for per 1 meter. Reply Link. Arun August 11, ... how to find running meter weight in kg for …sp.info Weight Calculator - htextrusion.comWeight Calculator Enter the Specific Values of A&B to get the weight per meter of the rods in Brass and Copper Round rodssp.info What are the weights of 16mm, 12mm, 20mm, 25mm and …Weight of any dia meter of bar calculated by this formula =(Area of the bar*Density of the steel) For example : weight of 12 mm dia bar per meter length =(3.142*(.12)^2/4)*7850 =0.887 kgsSee more resultssp.info Online calculator: Price based on running meterRunning, or linear meter, is term often used in fabric manufacturing. For roll of fabric price is given for running meter and the width of roll is known. Now, the task is to calculate price of fabric slice of certain length and width, given the price of running meter and width of roll.
### How to convert Square meter to Running meter?, Yahoo …
Dec 13, 2007 · Now there are certain firms which want me to quote them in Running Meter (1RM=1M) the paintwork i do is in microns, ranging from 65-125 microns. Running Meter and A meter are the same, just used differently in engineering projects to mean the same thing. ie., 1000cm or 10000mmsp.info How to Calculate Steel Pipe Weight per Foot/Meter by Size ...How to calculate steel pipe per foot or meter. Basically, there are two ways to calculate steel pipe weight per foot or per meter. One is to calculate according a steel pipe weight formula. Another way is to find from steel pipe weight chart. How to calculate the weight of steel pipe by formula. The steel pipe unit weight (kg/m or lb/ft) shall ...sp.info How do you find reinforcement bar running meter weight ...Both are measurements of volume, and one cubic meter is always equal to one thousand liters: 1 cubic meter = 1,000 liters Specific Gravity is the weight of a certain amount of a substance compared ...sp.info Hollow section weight calculator - Online Structural DesignCircular and rectangular hollow section weight per meter calculator Hollow section weight calculator The webpage is not working since JavaScript is not enabled.
### How to convert rainforcement calculation: Running meter to kg
all steel running meter to per kg 6mm steel----0.22 running meter per kg 8mm steel----o.39 running meter per kg 10mm steel---0.62 running meter per kg 12mm steel---0.89 running meter per kg 16mmsteel----1.58 running meter per kg.sp.info what is the weight of 8mm dia steel with 1 meter length ...what is the weight of 8mm dia steel with 1 meter length. Is there any formula to calculate the weight for any given diametersp.infoPipe running meter weight calculation. Reference ...sp.info Weight Metres per Bar Dia Metre per Tonne (Kg)Weight per Metre (Kg) Metres per Tonne 6mm 0.222 4504 8mm 0.395 2531 10mm 0.616 1623 12mm 0.888 1126 16mm 1.579 633 20mm 2.466 406 25mm 3.854 …
### FLAT BAR - Weight in kg per meter
sp.info M.S. TMT BARS Size in mm Weight in Kgs. Per Feet …M.S. TMT BARS Size in mm Weight in Kgs. Per Feet Weight in Kgs. Per Mtr. 6 (R) 0.067 0.220 8 0.120 0.395 10 0.188 0.617 12 0.270 0.885 16 0.480 1.578 20 0.751 2.466 16 0.480 1.578 16 0.480 1.578 20 0.751 2.466 25 1.174 3.852 32 1.925 6.313sp.info Steel Flats - WeightTypically weight of mild steel flats. The weight of a mild steel flat can be calculated as W = 0.0075 w t (1)sp.info Mechanical Charts - Weight of Equal Steel Anglescalculators, engineering calculators.... Enter your search terms Submit search form
### ISMC weight chart, I beam weight chart in kg
Indian standard structural steel weight chart, ismb weight calculation formula, ismb weight chart, ismb 150 weight, isa weight chart, ismb beam weight chartsp.info Running Calorie Calculator - High accuracy calculationCalculate my cardio calories for 2 X 15 min runs to and from the gym (full body resitance training at gym 40mins). [5] 2019/10/31 21:32 Male / 30 years old level / Self-employed people / Useful / Purpose of useSome results are removed in response to a notice of local law requirement. For more information, please see here.sp.infoRunning Calorie Calculator - High accuracy calculationThis calculator said 843kcal burned while the treadmill said 675kcal for 60min of running at 10.8km/hour.sp.info Calculation ideal weight
### STANDARD SECTIONAL WEIGHT OF COMMONLY …
STANDARD SECTIONAL WEIGHT OF COMMONLY USED CHANNELS Size Weight per Designation meter (W) a x b x t mm kg ISJC 100 100 x 45 5.8 ISJC 125 125 x 50 7.9sp.info Pick a table to convert kilograms per meter [kg/m]Pick a table with which to convert kilograms per meter to any other unit of measurement of linear density. In addition, the page shows conversion formulas and correlations between kilograms per meter and all other units of measurement of linear density. Alternatively, convert between all units of measurement of linear density with a single click.sp.info What is "Running Meter" and how it is calculated for ...Apr 08, 2013 · A running meter means a length of pipe one meter long. Source(s): Running meters just means the length of the installed pipe from point A to point B. It includes bends and joints. If you know the cost per meter of the insulation for one size pipe it is hard to see how you can calculate it for another size.sp.info How to convert Square meter to Running meter?, Yahoo AnswersDec 13, 2007 · How to convert Square meter to Running meter? ... For a given diameter, how do we calculate the weight of steel per meter length of the steel bar and vice versa? For any steel reinforcement bar, weight per running meter is equal to d2/162 Kg, where d is diameter of the bar in mm. For example, 10 mm diameter bar will weigh 10x10/162 = 0.617 Kg/m.
### Calculate Pipe Weight - Engineering ToolBox
Weight of Empty Pipe. Weight of empty pipe per unit length can be calculated as. ... This calculator can be used to calculate the weight of a pipe with - or without - liquid. The calculator is generic and can be used for both SI and Imperial units as long as the use of units are consistent.sp.info Online calculator: Relations between the mass, the length ...The SI derived unit for area density is kilograms per square meter (or grams per square meter). For example, typical values of cotton density used for shirts are 60-100 g/m2, for jeans - 400 g/m2. Running, or linear meter, is term often used in fabric manufacturing. For roll of …sp.info Calculating Weld Volume and Weight - TWICalculating the volume of a weld is one of the first steps to be taken when estimating the cost of making a weld. With this information, and knowing the deposition rate of the process, it is possible to determine the arc time (the length of time that an arc is burning and depositing weld metal) and the amount of welding consumables required to fill the joint.sp.info running meter - definition - EnglishThe deviation from the plane must not exceed 5 mm per running meter. EurLex-2 the turnover in local currency and the volume in running meters of the product concerned sold for export to the Community during the period 1 October 2005 — 30 September 2006,Some results are removed in response to a notice of local law requirement. For more information, please see here.
## Message
You may also leave contact information, we will contact you as soon as possible! | 1,852 | 7,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-27 | latest | en | 0.840322 |
https://www.bpadjogja.info/iwcf-formula-sheet-50/ | 1,628,182,809,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046156141.29/warc/CC-MAIN-20210805161906-20210805191906-00279.warc.gz | 660,389,648 | 12,053 | # IWCF FORMULA SHEET PDF
Iwcf Drilling Well Control surface/subsea Kill Sheet, API Formula Sheet, Free Download, IWCF/IADC Exercise Test & Exam Paper, API Field Units Formula Sheet. Intertek Consulting & Training Unpublished work. All rights reserved. Revised Feb. 3, Intertek Formula Sheet. Pressure Calculations (psi). 1. IWCF also shares excellent kill sheets used for well control situation to everybody . There are both surface and sub sea BOP version. Additionally, IWCF provides.
Author: Vozilkree Akidal Country: Samoa Language: English (Spanish) Genre: Marketing Published (Last): 2 January 2011 Pages: 242 PDF File Size: 18.50 Mb ePub File Size: 18.25 Mb ISBN: 974-8-13686-122-7 Downloads: 25819 Price: Free* [*Free Regsitration Required] Uploader: Gokus
This section is where you calculate the pressure drop per every strokes of the mud pump as you start pumping heavier kill weight mud down the hole. Additionally, IWCF provides the files in several units. You need to understand how the u-tube is always balanced even though you may see different pressure readings on the drill pipe and pressure gauges.
Mud Increment is determined lwcf the following equation: Additionally, these ebooks are available to download from IWCF website.
Since the length of Heavy Weight Drill Pipe and Drill Collars are given to you, all you need to do is subtract these two values from the total MD of the well to find the drill pipe length. Brine Density with Temperature Correction Calculat I had probably 10 of these problems on my IWCF Level 4 exam and they made up a very large percentage of my final score. Ben Dinsmore 1 year ago. Ben Xheet 3 years ago. As I mentioned with the formulas above, the more you can think your way through a problem and understand what is being asked, the easier it will be for you to answer the question.
ENERGY ESSENTIALS FOR WITCHES AND SPELLCASTERS PDF
Try to think your way through what the problem is asking and what you actually need to solve for to get the answer. Fabrizio 1 year ago. On the right hand side start with your ICP and decrease the pressure by If you use m then you will not have the correct annulus volume.
In the left hand column start at 0 and increase in stroke increments until you get to the total number of strokes needed to displace the drill string in our example. Surface leak-off pressure with 11 ppg mud psi. Detecting a Plugged Bit: Try solving as many problems as you can without referring to the formula sheet. The solution of this part is also provided. Above all thank you Could you please send me some Drillingformula excels ,well wheet and IWCF documents to my email?
The second part is about areas and volume. Detecting a Plugged Choke: Good morning, on which iwcf version are your questions based?
Understanding this concept is very beneficial when you start studying the practice formul. Not only did I not know much about lining up a choke manifold or standpipe manifold, I knew virtually nothing about what RPM to drill at, how much pressure to maintain on the drillpipe or Fofmula on the mud pumpshow much weight to maintain on the drill bit while drilling, etc. Slow pump rate data: View all posts by DrillingFormulas.
The most effective way I found to do this was to hit the material in small minutes blocks a few times each day. In almost every IWCF well control situation, the casing gauge reads higher than the drill pipe pressure.
ESTENOSIS PEPTICA BENIGNA PDF
## IWCF Well Control Calculator Android App
As you learn more and more about the various equipment and their components, the next group of questions in the IWCF Study Pack become easier and easier. It will be tremendously helpful! Today, I find ofrmula very useful well control documents which are used to teach students at the university.
Geoff 2 years ago. Tagged drilling calculationIWCFiwcf well control. I was against this. SF is important because it will allow the bottom hole pressure to be over formation pressure so the well is not in underbalance condition while conducting later steps. Prev Article Next Article.
### IWCF Drilling Calculation Part 1 – 3 Review – Drilling Formulas and Drilling Calculations
Dealing with a Tripped Mud Pump: The last two parts are about introduction to well control relating to kick prevention and detection, primary well control and secondary well control. This article will demonstrate how to perform volume metric well control.
Trey Pace 2 years ago. The next part is about the circulating system on the rig.
Your kind cooperation and assistance is highly appreciated as usual. April 1, at 3: Basic knowledge in well control 2.
Every once in a while I would run through these cards until it became second nature to identify the problem and what would need to be done to solve it. | 1,054 | 4,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-31 | latest | en | 0.911981 |
http://www.kirupa.com/forum/showthread.php?301085-Uneven-terrain-collision-detection&s=320fb5b2a68bbc3ee51ec36a218ee7e2 | 1,386,222,276,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163040059/warc/CC-MAIN-20131204131720-00022-ip-10-33-133-15.ec2.internal.warc.gz | 424,566,711 | 10,635 | # Thread: Uneven terrain collision detection
1. 65
posts
flash n00b
## Uneven terrain collision detection
I making a lander style game where you fly around and save penguins (dont ask!) and then you fly your helicopter to a landing pad. There is along way to go, but the collision detection wont work and im out of ideas! Here is the problem:
Here is the swf
Ive used this as the basis of the hit collection, a sentry/perimeter setup. There is a perimeter set around the helicopter, and one set around the ground. If they two hit each other, then some code moves them apart. It works really well. But sometimes, the helicopter gets stuck inside the the ground (see swf). Here is the code that does the testing:
PHP Code:
``` ground_collision = function(l, o) { var c=0; var xs=0; var ys=0; var diff=0; for (i in l.perim) { l.perim.localToGlobal(p={x:l.perim[i]._x,y:l.perim[i]._y}) if(o.hitTest(p.x, p.y, true)){ _root.fuel -= 5; _root.points -= 1; xMovement *= -1; gravity *= -1; } ys += p.y; xs += p.x; c++; }} ```
As you can see, it simply reverses the gravity and xmovement so that the helicopter goes away from the ground. If it moves slowly, this works brilliantly. However, if you hit it with any speed, the helicopter goes right through. I know why this is, I just cant fix it
Can anyone suggest a better method?
2. 45
posts
Registered User
I made a platform game as a final year project at uni.. The way I did it was, calculate the ySpeed, and then:
Code:
```groundHit=false;
for i=Math.round(yPosition);i<=yPosition+Math.round(ySpeed);i++
if(ground_collision(xPosition,i))
{
yPosition=i;
groundHit=true;
break;
}
if(!groundHit)
yPosition=yPosition+ySpeed;```
No idea if an approach like this is applicable to flash at all? (my game was written with DirectX and C++). Another approach could be to simply stop ground collisions from happening if the sprite is heading upwards?
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• | 609 | 2,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.788201 |
https://www.webassign.net/features/textbooks/harmathap12/details.html?toc=1&l=subject | 1,601,267,671,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401585213.82/warc/CC-MAIN-20200928041630-20200928071630-00647.warc.gz | 1,088,234,699 | 21,034 | # Mathematical Applications for the Management, Life, and Social Sciences 12th edition
Ronald J. Harshbarger and James J. Reynolds
Publisher: Cengage Learning
## Personal Study Plan Module
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• Chapter 1: Linear Equations and Functions
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• 2: Review Exercises (15)
• 2: Chapter Test
• 2: Extended Applications and Group Projects (1)
• Chapter 3: Matrices
• 3.1: Matrices (29)
• 3.2: Multiplication of Matrices (41)
• 3.3: Gauss-Jordan Elimination: Solving Systems of Equations (39)
• 3.4: Inverse of a Square Matrix; Matrix Equations (41)
• 3.5: Applications of Matrices: Leontief Input-Output Models (24)
• 3: Warm-Up Exercises (5)
• 3: Review Exercises (16)
• 3: Chapter Test
• 3: Extended Applications and Group Projects
• Chapter 4: Inequalities and Linear Programming
• 4.1: Linear Inequalities in Two Variables (32)
• 4.2: Linear Programming: Graphical Methods (28)
• 4.3: The Simplex Method: Maximization (34)
• 4.4: The Simplex Method: Duality and Minimization (18)
• 4.5: The Simplex Method with Mixed Constraints (27)
• 4: Warm-Up Exercises (5)
• 4: Review Exercises (16)
• 4: Chapter Test
• 4: Extended Applications and Group Projects (1)
• Chapter 5: Exponential and Logarithmic Functions
• 5.1: Exponential Functions (40)
• 5.2: Logarithmic Functions and Their Properties (48)
• 5.3: Equations and Applications with Exponential and Logarithmic Functions (40)
• 5: Warm-Up Exercises (4)
• 5: Review Exercises (15)
• 5: Chapter Test
• 5: Extended Applications and Group Projects (1)
• Chapter 6: Mathematics of Finance
• 6.1: Simple Interest; Sequences (35)
• 6.2: Compound Interest; Geometric Sequences (52)
• 6.3: Future Values of Annuities (38)
• 6.4: Present Values of Annuities (36)
• 6.5: Loans and Amortization (31)
• 6: Warm-Up Exercises (7)
• 6: Review Exercises (15)
• 6: Chapter Test
• 6: Extended Applications and Group Projects (1)
• Chapter 7: Introduction to Probability
• 7.1: Probability; Odds (41)
• 7.2: Unions and Intersections of Events: One-Trial Experiments (33)
• 7.3: Conditional Probability: The Product Rule (42)
• 7.4: Probability Trees and Bayes' Formula (24)
• 7.5: Counting: Permutations and Combinations (35)
• 7.6: Permutations, Combinations, and Probability (31)
• 7.7: Markov Chains (25)
• 7: Warm-Up Exercises (6)
• 7: Review Exercises (16)
• 7: Chapter Test
• 7: Extended Applications and Group Projects
• Chapter 8: Further Topics in Probability; Data Description
• 8.1: Binomial Probability Experiments (20)
• 8.2: Data Description (30)
• 8.3: Discrete Probability Distributions; The Binomial Distribution (31)
• 8.4: Normal Probability Distribution (29)
• 8.5: The Normal Curve Approximation to the Binomial Distribution (22)
• 8: Warm-Up Exercises (6)
• 8: Review Exercises (14)
• 8: Chapter Test
• 8: Extended Applications and Group Projects
• Chapter 9: Derivatives
• 9.1: Limits (43)
• 9.2: Continuous Functions; Limits at Infinity (39)
• 9.3: Rates of Change and Derivatives (33)
• 9.4: Derivative Formulas (41)
• 9.5: The Product Rule and the Quotient Rule (40)
• 9.6: The Chain Rule and the Power Rule (44)
• 9.7: Using Derivative Formulas (35)
• 9.8: Higher-Order Derivatives (33)
• 9.9: Applications: Marginals and Derivatives (27)
• 9: Warm-Up Exercises (9)
• 9: Review Exercises (15)
• 9: Chapter Test
• 9: Extended Applications and Group Projects (1)
• Chapter 10: Applications of Derivatives
• 10.1: Relative Maxima and Minima: Curve Sketching (44)
• 10.2: Concavity; Points of Inflection (34)
• 10.3: Optimization in Business and Economics (38)
• 10.4: Applications of Maxima and Minima (26)
• 10.5: Rational Functions: More Curve Sketching (28)
• 10: Warm-Up Exercises (8)
• 10: Review Exercises (15)
• 10: Chapter Test
• 10: Extended Applications and Group Projects (1)
• Chapter 11: Derivatives Continued
• 11.1: Derivatives of Logarithmic Functions (34)
• 11.2: Derivatives of Exponential Functions (43)
• 11.3: Implicit Differentiation (43)
• 11.4: Related Rates (26)
• 11.5: Applications in Business and Economics (18)
• 11: Warm-Up Exercises (8)
• 11: Review Exercises (15)
• 11: Chapter Test
• 11: Extended Applications and Group Projects (1)
• Chapter 12: Indefinite Integrals
• 12.1: Indefinite Integrals (34)
• 12.2: The Power Rule (39)
• 12.3: Integrals Involving Exponential and Logarithmic Functions (37)
• 12.4: Applications of the Indefinite Integral in Business and Economics (26)
• 12.5: Differential Equations (36)
• 12: Warm-Up Exercises (8)
• 12: Review Exercises (15)
• 12: Chapter Test
• 12: Extended Applications and Group Projects (1)
• Chapter 13: Definite Integrals: Techniques of Integration
• 13.1: Area Under a Curve (29)
• 13.2: The Definite Integral: The Fundamental Theorem of Calculus (44)
• 13.3: Area Between Two Curves (41)
• 13.4: Applications of Definite Integrals in Business and Economics (30)
• 13.5: Using Tables of Integrals (30)
• 13.6: Integration by Parts (25)
• 13.7: Improper Integrals and Their Applications (34)
• 13.8: Numerical Integration Methods: The Trapezoidal Rule and Simpson's Rule (22)
• 13: Warm-Up Exercises (5)
• 13: Review Exercises (15)
• 13: Chapter Test
• 13: Extended Applications and Group Projects (1)
• Chapter 14: Functions of Two or More Variables
• 14.1: Functions of Two or More Variables (25)
• 14.2: Partial Differentiation (33)
• 14.3: Applications of Functions of Two Variables in Business and Economics (19)
• 14.4: Maxima and Minima (20)
• 14.5: Maxima and Minima of Functions Subject to Constraints: Lagrange Multipliers (18)
• 14: Warm-Up Exercises (9)
• 14: Review Exercises (15)
• 14: Chapter Test
• 14: Extended Applications and Group Projects
• Chapter 0: Algebraic Concepts
• 0.1: Sets (40)
• 0.2: The Real Numbers (42)
• 0.3: Integral Exponents (37)
• 0.4: Radicals and Rational Exponents (44)
• 0.5: Operations with Algebraic Expressions (43)
• 0.6: Factoring (38)
• 0.7: Algebraic Fractions (34)
• 0: Review Exercises (15)
• 0: Chapter Test
• 0: Extended Applications and Group Project
Mathematical Applications for the Management, Life, and Social Sciences, 12th edition, engages students with its concept-based approach, multiple presentation methods, and relevant applications throughout. Intended for two-semester applied calculus or combined finite mathematics and applied calculus courses, the book places concepts in real-life contexts to help students strengthen their understanding. A focus on modeling (with modeling problems clearly labeled in the examples and problems, so they can be treated as optional) and flexible content organization accommodate different teaching approaches, enabling instructors to decide the order in which topics are presented and the degree to which they are emphasized.
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• Master It tutorials (labeled MIs) guide students through the steps they must take to work through given problems.
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## Questions Available within WebAssign
Most questions from this textbook are available in WebAssign. The online questions are identical to the textbook questions except for minor wording changes necessary for Web use. Whenever possible, variables, numbers, or words have been randomized so that each student receives a unique version of the question. This list is updated nightly.
##### Question Group Key
MI - Master It Tutorial
MI.SA - Stand Alone Master It
##### Question Availability Color Key
BLACK questions are available now
GRAY questions are under development
Group Quantity Questions
Chapter 0: Algebraic Concepts
0.R 15 006 012 017 023 029 036 043 047 052 060 072 078 085 090 095
0.1 40 001 004 006 007 008 009 010 012 012.EP 014 015 018 018.EP 019 019.EP 020 020.EP 024 026 028 030 032 034 035.MI 035.MI.SA 036 036.EP 038 038.EP 040 040.EP 045.MI 045.MI.SA 047 048 049 051 052 054 054.EP
0.2 42 001 003 004 005 006 008 008.EP 010 010.EP 012 012.EP 013 013.EP 015 016.MI 016.MI.SA 017 018 021 022.MI 022.MI.SA 023 024 029 031 032 034 034.EP 036 036.EP 037.MI 037.MI.SA 039 040 042 044 051 051.EP 052 053 055 501.XP
0.3 37 001 002 003 004 008 010 012 014 015 016.MI 016.MI.SA 018 021 024 026 027 029 032.MI 032.MI.SA 033 035 038 039 040 042.MI 042.MI.SA 044 045 048 051 054 056 057 061 063 065 067
0.4 44 001 002 003 004.MI 004.MI.SA 006 008 010 012 013 016 017.MI 017.MI.SA 019 021 022 023 026 028 031 032 034.MI 034.MI.SA 037.MI 037.MI.SA 040 041 044.MI 044.MI.SA 045 049 052 054.MI 054.MI.SA 056 058 060 063 064 066 068 069 072 073
0.5 43 002 004 007 010 011 014 015.MI 015.MI.SA 016 018 019 021 022 025 028 029 032 033 036.MI 036.MI.SA 037 040 042 043 047 050 052 054 056.MI 056.MI.SA 059.MI 059.MI.SA 061.MI 061.MI.SA 064 065 067 069 070 071 072 073 074
0.6 38 001 002 004 005.MI 005.MI.SA 007 010 013 014 018 019.MI 019.MI.SA 020 022 024 025 028 030 032 034 036 037 039 040 041 046.MI 046.MI.SA 047 050 052.MI 052.MI.SA 055 058 060 061.MI 061.MI.SA 062 063
0.7 34 001 002 004 005 007 009 012 013 014.MI 014.MI.SA 017.MI 017.MI.SA 018 019 022 023 026 028 031 033 035.MI 035.MI.SA 038 041 043 044 047.MI 047.MI.SA 049 050 051 054 055 058
Chapter Demo
Demo.4 26 001 003 004 005 007 009 011 014.MI 015 017 019 021.EP 023 025 026.MI 027.EP 030.MI 031 033 043 047 050 051.MI 053 055 056
Chapter 1: Linear Equations and Functions
1.PJT 1 001
1.R 14 002 010 016 023 027 032 037 043 047 053 058 061 066 072
1.WU 8 001abc 001defg 002 003 004 005abc 005def 006
1.1 44 002 003 005.MI 005.MI.SA 007 009 010 013 016 018.MI 018.MI.SA 019 021 023 024 027 029 031 033.MI 033.MI.SA 035 037 039 039.EP 040 040.EP 041 041.EP 045 047.MI 047.MI.SA 049 051.MI 051.MI.SA 053 054 056 057 059 061 062 066 068 501.XP
1.2 38 001 003 005 006 007 008 009 011 015 017 019.MI 019.MI.SA 020 021 025.MI 025.MI.SA 027 029 032 033.MI 033.MI.SA 035 037.MI 037.MI.SA 039 041 044 047 049 050 050.EP 051 052 053.MI 053.MI.SA 054 058 059
1.3 46 001 003 005.MI 005.MI.SA 007 009 011 013 015.MI 015.MI.SA 017 019 021 023.MI 023.MI.SA 025 027.MI 027.MI.SA 029.MI 029.MI.SA 031 033 035.MI 035.MI.SA 036 037 040 041.MI 041.MI.SA 043 043.EP 045.MI 045.MI.SA 047 047.EP 049.MI 049.MI.SA 051 052 053 055 057 057.EP 063 064 065
1.4 19 007 017.MI 017.MI.SA 024 027 029 030 031 035.MI 035.MI.SA 037 039 041 042 043 044 045 047 048
1.5 43 001 002 003.MI 003.MI.SA 005 007.MI 007.MI.SA 009 011 012 013.MI 013.MI.SA 014 015 016 017.MI 017.MI.SA 020.MI 020.MI.SA 022 023 025 027.MI 027.MI.SA 031 032 035.MI 035.MI.SA 037.MI 037.MI.SA 039 040 043.MI 043.MI.SA 044.MI 044.MI.SA 045 048.MI 048.MI.SA 051 501.XP 502.XP.MI 502.XP.MI.SA
1.6 46 001.MI 001.MI.SA 003 004.MI 004.MI.SA 005 007 008 009 011.MI 011.MI.SA 013 014 015 017.MI 017.MI.SA 018 019 021 023 026 027 031 033 035 035.EP 036.MI 036.MI.SA 039 040 041 044.MI 044.MI.SA 045 047 048 049 049.EP 050.MI 050.MI.SA 051 055 055.EP 057 058.MI 058.MI.SA
Chapter 2: Quadratic and Other Special Functions
2.PJT 1 002
2.R 15 001 008 014 018 022 029 032 039 042 046 049 055 059 065 067
2.WU 6 001 002 003 004 005 006
2.1 55 001.MI 001.MI.SA 003 005 007.MI 007.MI.SA 009 009.EP 011 011.EP 013 014 015.MI 015.MI.SA 017 019 020 021.MI 021.MI.SA 023 025 027 029.MI 029.MI.SA 032 033.MI 033.MI.SA 036 037 039 043.MI 043.MI.SA 045 045.EP 047.MI 047.MI.SA 049.MI 049.MI.SA 051 051.EP 053.MI 053.MI.SA 054.MI 054.MI.SA 055 057.MI 057.MI.SA 059 059.EP 062 063 064.MI 064.MI.SA 066.MI 066.MI.SA
2.2 31 001 003 005 006 007.MI 007.MI.SA 009 011 011.EP 012 012.EP 013 015 017 018 021.MI 021.MI.SA 022 025 029 031 032 033 037.MI 037.MI.SA 039 041 043 044 045 048
2.3 28 001 003 004.MI 004.MI.SA 006 007 009 009.EP 010 011 012.MI 012.MI.SA 014 017 017.EP 018 018.EP 020 023 025 027 027.EP 028 029 031 031.EP 032.MI 032.MI.SA
2.4 33 002 005 008 010 011 013 016 019 021 023 025 027.MI 027.MI.SA 029 031 033 034.MI 034.MI.SA 035 036 037 040 041 041.EP 042 043 045 047 047.EP 050 051 052 054
2.5 21 001 003 005 007 009.MI 009.MI.SA 011.MI 011.MI.SA 014 015 016 018 019 021 022 024 026 027 030 035 036
Chapter 3: Matrices
3.R 16 001 006 011 016 019 021 026 031 035 038 043 046 049 050 054 058
3.WU 5 001abc 001d 002 003 004
3.1 29 001 002 004 007 009 010 011 013 014 015 017 019 020 023 025 027 029 031 032 033 035 039 040 042 044 045 047 051 501.XP
3.2 41 001.MI 001.MI.SA 003.MI 003.MI.SA 005 006 007 009 011 013.MI 013.MI.SA 015.MI 015.MI.SA 017 019.MI 019.MI.SA 022 022.EP 025.MI 025.MI.SA 026 027 031 033 038 039 041 043 045 046.MI 046.MI.SA 048 049.MI 049.MI.SA 050.MI 050.MI.SA 051.MI 051.MI.SA 053.MI 053.MI.SA 055
3.3 39 002 003 005 007.MI 007.MI.SA 009 011 013 015 017 018 020 023.MI 023.MI.SA 025 027 028 029 031 033 037 039 041 042 045 047 049 051.MI 051.MI.SA 054 055 057 057.EP 059 059.EP 061 061.EP 062 062.EP
3.4 41 002 002.EP 003 003.EP 004 004.EP 005 007 010 011 013 015.MI 015.MI.SA 017 020 023 024 025 029.MI 029.MI.SA 031 032 034 035 037 039 041 043 047 051 052 053.MI 053.MI.SA 055 056.MI 056.MI.SA 057 058 059.MI 059.MI.SA 061
3.5 24 001 003.MI 003.MI.SA 004 005 006 007 008 009 010 011.MI 011.MI.SA 015 018 021 023 026.MI 026.MI.SA 029 033 035 038 040 042
Chapter 4: Inequalities and Linear Programming
4.PJT 1 001
4.R 16 002 005 008 010 013 017 020 022 025 027 030 033 037 040 043 045
4.WU 5 001 002 003 004 005
4.1 32 001.MI 001.MI.SA 003 003.EP 005 006 006.EP 007 011 013.MI 013.MI.SA 015 015.EP 017 017.EP 018 018.EP 019 019.EP 021 021.EP 024 024.EP 027 028 029.MI 029.MI.SA 031 032 034 035 036
4.2 28 001.MI 001.MI.SA 002 005 007 009 012 013.MI 013.MI.SA 015 017 020 022 023 025 026 027.MI 027.MI.SA 029 031 033 038 039 040 041 043 044 045
4.3 34 003.MI 003.MI.SA 005 007 009 011 013 015 017 019.MI 019.MI.SA 021 024 025 026 027 028 030 032 033 035 039.MI 039.MI.SA 041 042 043 045 046 048 048.EP 050 052.MI 052.MI.SA 053
4.4 18 001.MI 001.MI.SA 004 005 007.MI 007.MI.SA 011 013 015 017 021 023.MI 023.MI.SA 025 028 029 032 034
4.5 27 001 001.EP 003 005.MI 005.MI.SA 007 009 010 011.MI 011.MI.SA 013 015 018 019 020 021 023 025 027.MI 027.MI.SA 029 030 031 032 035 037 038
Chapter 5: Exponential and Logarithmic Functions
5.PJT 1 002
5.R 15 002 006 010 014 017 021 029 032 035 037 042 045 050 054 058
5.WU 4 001 002 003 004
5.1 40 002 004 005 006 007 008 009.MI 009.MI.SA 011 013 015 017 018 020 021 024 025 026 028 029 033.MI 033.MI.SA 034 035 036 039 041 041and43 042 042and44 043 044 046 047 048.MI 048.MI.SA 049 052 501.XP 502.XP
5.2 48 001 003 004 005.MI 005.MI.SA 006 007 008 009 011 013 015 017 018 019 021 023 025 027 029 031 033 035.MI 035.MI.SA 036 037 039.MI 039.MI.SA 041 043 044 049 050 051.MI 051.MI.SA 052 053 055 060 062 063 064 067.MI 067.MI.SA 068 071 074 077
5.3 40 001.MI 001.MI.SA 003 005 007 009 011 013 015 016 017 018 019 021.MI 021.MI.SA 025.MI 025.MI.SA 026 027.MI 027.MI.SA 029 031 033 034 035 036 037 040 043 044 045 047 051 055 058.MI 058.MI.SA 059 062 067 068
Chapter 6: Mathematics of Finance
6.PJT 1 001
6.R 15 003 008 010 014 017 021 026 029 034 040 043 049 055 061 065
6.WU 7 001 002 003 004 005 006 007
6.1 35 001 002 003 005 007 009.MI 009.MI.SA 011 013 016.MI 016.MI.SA 017 019 021 021.EP 023 025 027 028 031 033 035 037 038.MI 038.MI.SA 039 041 043 044 046.MI 046.MI.SA 048 049 049.EP 052
6.2 52 001 002 003 005 007 008 009 010.MI 010.MI.SA 011 013 014.MI 014.MI.SA 017 017.EP 018 020 021 021.EP 023 025 028 028.EP 031 032 033 034 037.MI 037.MI.SA 038 041 043 043.EP 044 045 047 049 051.MI 051.MI.SA 053 055 057 059 061 061.EP 063 063.EP 065.MI 065.MI.SA 067 069 071
6.3 38 001 002 005 007.MI 007.MI.SA 008 009 010 011 011.EP 012 012.EP 013 014 015.MI 015.MI.SA 017 018 019 020 023 024 025 027.MI 027.MI.SA 030 033 035 037 039 041 041.EP 042 042.EP 043.MI 043.MI.SA 044 044.EP
6.4 36 001 002 003.MI 003.MI.SA 004 005 006 007 008 009 011 014 015 017 019 020 021.MI 021.MI.SA 023 025 028 029 030 033 035 037 040 041.MI 041.MI.SA 043 046 047 049.MI 049.MI.SA 051 054
6.5 31 001 003 005.MI 005.MI.SA 006 007 009 011 012 013 015.MI 015.MI.SA 017 018 020 021 023.MI 023.MI.SA 025 025.EP 029 029.EP 030 032 032.EP 036.MI 036.MI.SA 038 039 039.EP 040
Chapter 7: Introduction to Probability
7.R 16 001 006 008 011 014 016 019 023 025 027 029 033 036 041 049 052
7.WU 6 001 002 003 004 005 006
7.1 41 001 003 005 006 007 008 009 011.MI 011.MI.SA 015 016 017 018 019 021 023 025 027.MI 027.MI.SA 029 030 033 036 038 039 040 044.MI 044.MI.SA 045 046 048 050 050.EP 052 053 055 057.MI 057.MI.SA 060 062 064
7.2 33 001 003.MI 003.MI.SA 004 005 007 009 010 012 016 017 019 021 023 025.MI 025.MI.SA 028 029 031 031.EP 033 035 037 038.MI 038.MI.SA 039 040 042 044 047 048.MI 048.MI.SA 501.XP
7.3 42 001 001.EP 003 003.EP 004 005 007 010.MI 010.MI.SA 011 014 015 017 018 019 021 023 025 027 029 031.MI 031.MI.SA 033 035 037 039 041 043 044 045 047.MI 047.MI.SA 050 051 051.EP 053 053.EP 055 055.EP 058 061.MI 061.MI.SA
7.4 24 001 003.MI 003.MI.SA 005 007 008.MI 008.MI.SA 010 012 013.MI 013.MI.SA 015 017 018 020 021 022.MI 022.MI.SA 023 023.EP 024 029 501.XP 502.XP
7.5 35 001 003 005 007.MI 007.MI.SA 009 011 012.MI 012.MI.SA 015 017 019 021 023 026 028.MI 028.MI.SA 030 031 033 034 034.EP 036 036.EP 037 039 040 042 044.MI 044.MI.SA 045 047 047.EP 050 501.XP
7.6 31 001 003 004.MI 004.MI.SA 005 007 008 009 011 011.EP 012.MI 012.MI.SA 013 015 017 019 020 021 021.EP 022.MI 022.MI.SA 023 025 025.EP 027 030 032.MI 032.MI.SA 034 034.EP 035
7.7 25 001 005 009 011 013 015 017 019 020.MI 020.MI.SA 021 023 024.MI 024.MI.SA 025 026 027 029 033.MI 033.MI.SA 034 036 037 038 039
Chapter 8: Further Topics in Probability; Data Description
8.R 14 003 005 009 014 018 022 025 032 034 039 043 047 052 054
8.WU 6 001 002 003 004 005 006
8.1 20 001 003.MI 003.MI.SA 005 009 011.MI 011.MI.SA 013 015 017.MI 017.MI.SA 020 021 023.MI 023.MI.SA 025 027 029 031.MI 031.MI.SA
8.2 30 001 005 007 009 011 013 015 017.MI 017.MI.SA 019 021 022 023 025 027 029 031 033.MI 033.MI.SA 035 038 040 042 044.MI 044.MI.SA 048 050 051 501.XP 502.XP
8.3 31 001 003 005 005.EP 009 011 013 015 017 019.MI 019.MI.SA 021 023 026 028 029 033 035 037.MI 037.MI.SA 039 041 043 047 048.MI 048.MI.SA 049 050 051 053.MI 053.MI.SA
8.4 29 001 003 005 006 007 009 011 013 014 015.MI 015.MI.SA 017 019 021 023 025 028 030 031.MI 031.MI.SA 032 033 034 036 037 038.MI 038.MI.SA 039 041
8.5 22 001 001.EP 005 007.MI 007.MI.SA 009 011 013 015 017 019 020 023 025.MI 025.MI.SA 027 029 032 034.MI 034.MI.SA 036 038
Chapter 9: Derivatives
9.PJT 1 002
9.R 15 005 010 021 027 035 042 048 052 067 074 082 087 094 099 107
9.WU 9 001 002 003 004 005 006 007 008 009
9.1 43 002 004 005 006 008 010 011 012 013 016.MI 016.MI.SA 018 020 021.MI 021.MI.SA 022 024 025 026 027 029 031 033 035 037 040.MI 040.MI.SA 042 044 046 048 051 053 055.MI 055.MI.SA 057 058.MI 058.MI.SA 061 063 065 067 068
9.2 39 001 002 003 003.EP 005 005.EP 007 007.EP 008 008.EP 009 011.MI 011.MI.SA 013 014 015 016 018 020 021.MI 021.MI.SA 023 025 027 029 031 034 036 039 039.EP 040.MI 040.MI.SA 043 045 045.EP 047.MI 047.MI.SA 049 501.XP
9.3 33 001 003 006 007.MI 007.MI.SA 008 009 011 013 015 017 020 022 023 025 027.MI 027.MI.SA 029 030 031 033 035 037 039 040.MI 040.MI.SA 041 042.MI 042.MI.SA 045 047 048 052
9.4 41 001 003 004 005 007 009 011 014.MI 014.MI.SA 015 017 019 021 021.EP 023 025 026.MI 026.MI.SA 027 027.EP 030.MI 030.MI.SA 031 033 035 038 039 043 046 047 050 051.MI 051.MI.SA 052 053 055 056 058.MI 058.MI.SA 060 061
9.5 40 002.MI 002.MI.SA 003 004 005 008 009 010 011 011.EP 013 013.EP 015 015.EP 017.MI 017.MI.SA 020 021 022 023 023.EP 025 025.EP 026 030 031 034 036 038 039 041.MI 041.MI.SA 043 045.MI 045.MI.SA 046 049 049.EP 051 054
9.6 44 001.MI 001.MI.SA 002 003.MI 003.MI.SA 005 007 009 011.MI 011.MI.SA 013 015 016 018 020.MI 020.MI.SA 021.MI 021.MI.SA 023 025.MI 025.MI.SA 027 027.EP 028 031.MI 031.MI.SA 033.MI 033.MI.SA 035.MI 035.MI.SA 037 038 039 039.EP 042 044 045 047 048.MI 048.MI.SA 049 049.EP 052 501.XP
9.7 35 001 003.MI 003.MI.SA 005 006 007 010 011 013 015 017.MI 017.MI.SA 018 018.EP 020 021.MI 021.MI.SA 023 023.EP 025 025.EP 027 029 029.EP 031 033 035 036 037.MI 037.MI.SA 039 040.MI 040.MI.SA 041 043
9.8 33 001 003 004 005.MI 005.MI.SA 006 007 009 011 013 015.MI 015.MI.SA 017 019 019.EP 021 021.EP 024.MI 024.MI.SA 025 026 030 031 033 035.MI 035.MI.SA 036 036.EP 037 038 041 042 046
9.9 27 001 003 004 005 006 009.MI 009.MI.SA 011 012 015 017 020.MI 020.MI.SA 023 025 027 028 029 031 034.MI 034.MI.SA 035 036.MI 036.MI.SA 037 037.EP 038
Chapter 10: Applications of Derivatives
10.PJT 1 001
10.R 15 002 007 012 015 020 023 029 033 036 038 042 045 049 053 055
10.WU 8 001 002 003 004 005 006 007 008
10.1 44 001 001and03 002 002and04 003 004 005 007 009 011 014 015 017 019 021 025.MI 025.MI.SA 027 027.EP 029 031 033 035 036 038 040 042 043 045 048 049.MI 049.MI.SA 051.MI 051.MI.SA 052 053 053.EP 054 055 057 058 060.MI 060.MI.SA 062
10.2 34 001 001.EP 003 003.EP 004 004.EP 005 007 009 011 013 013.EP 015 016 017 020.MI 020.MI.SA 021 023 026 028 029 031 034 035 037 038.MI 038.MI.SA 039 040.MI 040.MI.SA 042 046 501.XP
10.3 38 001 001.EP 003 003.EP 004 005 006 007.MI 007.MI.SA 009 012 013 013.EP 014 014.EP 017 019.MI 019.MI.SA 021 021.EP 022 027 029 031 032 033.MI 033.MI.SA 035 037 039 039.EP 040 040.EP 042 043.MI 043.MI.SA 045 047
10.4 26 001 003.MI 003.MI.SA 005 007 008 009 011 013.MI 013.MI.SA 015.MI 015.MI.SA 017 018 019 019.EP 020 021 022 023 025 029 031.MI 031.MI.SA 032 032.EP
10.5 28 001 004 005.MI 005.MI.SA 007 009 011 013 015 017 019.MI 019.MI.SA 021 023.MI 023.MI.SA 024 026 027 030 032 035.MI 035.MI.SA 037 040.MI 040.MI.SA 041 042 043
Chapter 11: Derivatives Continued
11.PJT 1 002
11.R 15 002 006 009 010 014 017 019 021 024 027 030 031 033 037 040
11.WU 8 001 002 003 004 005abc 005def 006 007
11.1 34 001 003 005 007 009 011 013 015 017 019.MI 019.MI.SA 021 023 025 027 027.EP 029 031 031.EP 033 034 037 039 040 042 043 045 046 047 049.MI 049.MI.SA 050 051 052
11.2 43 001 003 005 007.MI 007.MI.SA 009 011 011.EP 012 013 014 015 016 018 019 021 023 025 027.MI 027.MI.SA 029 030 031 033 033.EP 038 039 039.EP 040 043 044.MI 044.MI.SA 045 047 049 051 053 056 058 059 060 062 065
11.3 43 001 003 006 007.MI 007.MI.SA 009 011 013 013.EP 015 017 019 020 022 023 024 025 027.MI 027.MI.SA 028 029 030 031 033 033.EP 035 037 037.EP 039 041 041.EP 043 045 048 052 054 055 057 058.MI 058.MI.SA 059 061 062
11.4 26 001 003 005 007 009.MI 009.MI.SA 011 013 015 016 017 019 020.MI 020.MI.SA 021 022 023.MI 023.MI.SA 025 027 028.MI 028.MI.SA 030 033 035 037
11.5 18 001 002 003 004 006.MI 006.MI.SA 007 008 009 011 015 017 019.MI 019.MI.SA 020 021 024.MI 024.MI.SA
Chapter 12: Indefinite Integrals
12.PJT 1 001
12.R 15 003 006 008 012 014 018 021 023 027 028 034 038 040 042 046
12.WU 8 001 002 003 004 005 006 007 008
12.1 34 001 003 005 007 008 009 011 013 015 017 019 021 024 025 027 029.MI 029.MI.SA 031 031.EP 033 035 037 039 041.MI 041.MI.SA 043 045 047 048.MI 048.MI.SA 049 054 055 056
12.2 39 001 003 004 005 007 008 009.MI 009.MI.SA 012 013 015 016 017 019 021 023 025 025.EP 026 027 027.EP 029 031.MI 031.MI.SA 032 033 035 037 040 042 043 047.MI 047.MI.SA 048 051 053 055 056 056.EP
12.3 37 001 004 005 005.EP 007 009.MI 009.MI.SA 010 011 012 013 015 015.EP 017 017.EP 018 019 019.EP 020 021 022 023 026.MI 026.MI.SA 027 029 032 035 037 041 043.MI 043.MI.SA 047 049 050 051 054
12.4 26 001 003.MI 003.MI.SA 004 005 006 007 009.MI 009.MI.SA 011 012 014 015 017 019.MI 019.MI.SA 021 022 023 024 025 026 027.MI 027.MI.SA 028 501.XP
12.5 36 001 003 003.EP 005 007 009 011.MI 011.MI.SA 013 013.EP 015 017 019 022 024.MI 024.MI.SA 025 028 030 030.EP 031 033 034 034.EP 035 035.EP 038 041 043 045 047 049 051 056.MI 056.MI.SA 061
Chapter 13: Definite Integrals: Techniques of Integration
13.PJT 1 002
13.R 15 003 006 010 015 019 024 030 032 037 040 043 047 051 053 055
13.WU 5 001 002 003 004 005
13.1 29 001.MI 001.MI.SA 001and05 003 003.EP 003and07 005 005.EP 007.MI 007.MI.SA 009 011 013 015 016.MI 016.MI.SA 018 020 022 024.MI 024.MI.SA 026 029 033 034 036 037 038 039
13.2 44 001 003 005 007 008 009 011 012 013.MI 013.MI.SA 015 019 022 025 027 029 031.MI 031.MI.SA 032 034 037 038 039 041 043 044.MI 044.MI.SA 045 050 051 053 055 057 058 059 060 062 063 065 067.MI 067.MI.SA 069 071 072
13.3 41 001 002 003 005 007 008 009.MI 009.MI.SA 011 013.MI 013.MI.SA 014 016 016.EP 017 019 021 021.EP 024 024.EP 025 027.MI 027.MI.SA 029 030 030.EP 032 032.EP 033 036 037 038 039 040 040.EP 041 043.EP 043.MI 043.MI.SA 046 047
13.4 30 001 003 003.EP 004 005 006 007 009 009.EP 011 013.MI 013.MI.SA 015 017 017.EP 019 019.EP 021 023 023.EP 025.MI 025.MI.SA 027 029 029.EP 031 033 033.EP 035.MI 035.MI.SA
13.5 30 001 003 005 007 008 009 011 013 014 015 017 018.MI 018.MI.SA 020 022 023 025 027 028.MI 028.MI.SA 029 031 033 036 037 038 039 040 041.MI 041.MI.SA
13.6 25 001 003 005.MI 005.MI.SA 007 007.EP 009 011 013 015 016 016.EP 017.MI 017.MI.SA 019 021 022 025 029 031 032.MI 032.MI.SA 033 035 036
13.7 34 001 003 003.EP 005 005.EP 006 008 010 011 013 015 017 018 019 020.MI 020.MI.SA 021 022 023 026 029 033.MI 033.MI.SA 036 038 039 040.MI 040.MI.SA 043 044 045 046 048 050
13.8 22 001 003 005 007 010 011 013.MI 013.MI.SA 015 017 019 021 023 024.MI 024.MI.SA 025 026.MI 026.MI.SA 029 030 031 032
Chapter 14: Functions of Two or More Variables
14.R 15 002 004 006 007 011 013 016 018 019 021 023 027 031 033 035
14.WU 9 001 002 003 004 005 006 007 008 009
14.1 25 002 003 005.MI 005.MI.SA 007 009 012 013 015.MI 015.MI.SA 017 019 021 023 025 026.MI 026.MI.SA 027 028 030 033 035 036 037 038
14.2 33 001.MI 001.MI.SA 003 004 005 008 009 011 012 013 017.MI 017.MI.SA 019 021 023 025 027 029 031 033 035 037 039.MI 039.MI.SA 041 043 045 048 049 050 051 053 054
14.3 19 001 002 003.MI 003.MI.SA 006 007 009 011 013.MI 013.MI.SA 014 017 019 020.MI 020.MI.SA 023 025 027 029
14.4 20 001 003 004 005 007 009 011.MI 011.MI.SA 013 015 016 019.MI 019.MI.SA 021 023 024 025 027.MI 027.MI.SA 029
14.5 18 001 003 004.MI 004.MI.SA 005 007 008 010 011 012 014.MI 014.MI.SA 015 017 021 023.MI 023.MI.SA 025
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https://www.studysmarter.co.uk/explanations/engineering/professional-engineering/calibration/ | 1,719,096,887,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00103.warc.gz | 857,466,343 | 68,289 | # Calibration
Dive into the crucial world of calibration in engineering, an indispensable process ensuring the accuracy and reliability of measurements. This comprehensive guide will illuminate the understanding of calibration, its undeniable importance, and its practical applications in the engineering field. Uncover contemporary techniques, the key standards, and the specialised equipment used in professional engineering. With practical examples and insightful explanations, you will effectively master the concept and practice of calibration in engineering.
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## Understanding Calibration in Engineering
Calibration resides at the heart of engineering. It's an integral process contributing to the accuracy and precision of measuring tools and equipment.
Calibration, in essence, is the process of comparing the readings from the instrument being evaluated against the known output from an accurate standard device.
This ensures instruments produce reliable and valid results, thereby enabling meaningful interpretations and informed decision-making.
### Defining Calibration: Calibration Meaning in Professional Engineering
Diving deeper into understanding calibration, we'll explore its essential role in professional engineering.
In engineering settings, calibration refers to the procedure of adjusting an instrument to maintain its accuracy. This accuracy is kept relative to an unchanging standard.
Typically, the process might involve these steps:
• Identify calibration standards
• Perform initial measurement
• Final accuracy check
Tools calibrated could be anything from pressure sensors to oscilloscopes, and the calibration frequency will depend on the usage and elemental exposure. In the realm of engineering, three principal types of calibration are common:
Zero Calibration: Zeroing out errors by comparing and adjusting to the measurement of 'nothing' or a zero state Span Calibration: Comparing and aligning the instrument readings to the full scale of the measuring range Field Calibration: Calibration performed within the operating environment of the instrument
For instance, when calibrating a temperature sensor, initially the readings from the device are compared against the output of an accurate thermometer. If discrepancies are found, the temperature sensor is adjusted. The process is repeated until the temperature sensor registers the correct value, hence, achieving calibration.
#### Importance of Calibration in Engineering
The importance of calibration cannot be understated in engineering. It ensures the consistency of results derived from instruments. In more practical terms, calibration offers these benefits:
• Maintains quality control
• Reduces operational errors
• Enhances safety protocols
• Supports customer's trust and compliance
For instance, calibration is pivotal in manufacturing - an ill-calibrated machine may produce defective products, leading to financial losses. In aviation, improper calibration of altimeters could pose severe risks. Thus, adhering to rigorous calibration standards and procedures is essential in engineering enterprises. It’s not just about taking accurate measurements but also about assuring safety, reliability, and credibility.
Measurements made by an instrument are subject to some degree of uncertainty. In essence, to make them meaningful, these uncertainties or errors need to be controlled and minimized using precise calibration. The process, by adjusting the instrument's readings to align with a reference standard, ensures reduced biases, increased reliability, and improved measurement quality.
## Practical Applications of Calibration: Calibration Examples
Delving into the practical realm of calibration, there is a multitude of applications that reach well beyond the confines of any lab or classroom. These span industries like automotive, aerospace, manufacturing, and health care. Understanding these examples will concrete your knowledge of calibration's uses.
### Real-life Calibration Examples in Engineering
In engineering, calibration plays a crucial role in daily processes. Let's unpack some tangible examples of its application and why it's crucial for engineers to understand and commit to routine calibration. Firstly, in the automotive industry, calibration is routine and critical. For instance, modern cars are comprised of numerous systems that require consistent calibration to ensure optimal performance. These include engine control units (ECUs), fuel injectors, and onboard sensors.
An engine control unit (ECU) acts as the car's computer, coordinating various parts like the engine, transmission, and sensors, ensuring smooth performance. Hence, incorrect calibration could lead to poor fuel efficiency or damage.
• In thermodynamics, engineers frequently use temperature sensors. Calibration of these sensors is vital to ensure accurate readings, which in turn affect processes like energy production or heat transfer in equipment.
• In electrical circuits, oscilloscopes are used as signal-measuring devices. Calibration ensures they consistently provide accurate readings of a signal's frequency, shape, and amplitude.
• In material testing, accurate force measurements are paramount, so force gauges need precise calibration to assure reliable data. Improper calibration could result in product failures, injury, or significant financial loss.
#### Industry-based Examples of Calibration Usage
Let's now turn our attention to specific industry-based examples to gain an even clearer perception of practical calibration usage.
• In the manufacturing industry, measuring equipment like vernier callipers or magnetic resonance imaging (MRI) machines in the healthcare sector require rigorous calibration. These tools, if inaccurately calibrated, can lead to costly errors or, in the case of healthcare, serious harm to patients.
• In the energy sector, calibration of meters is essential for exact measurements of energy use. Incorrectly calibrated meters could lead to inaccurately high bills for users or incorrect data for energy providers, disrupting supply and demand equilibrium.
• In food and beverage industries, equipment that measures factors like temperature or pH may need regular calibration to ensure product quality and safety standards.
• In the aviation industry, calibration of tools like altimeters, airspeed indicators, and temperature probes is essential. If these devices give incorrect readings, they can compromise flight safety.
The importance of calibration is universal across industries - to ensure streamlined operations, guarantee product quality, and maintain safety regulations. Every tool or piece of equipment in an engineering environment, whether directly or indirectly measuring, interacting, or influencing another, will have its underpinning in precise and accurate calibration. To underscore this, calibration is not just a chore on a check-list. Instead, it's the foundation railing the frontline in the perpetual pursuit of precision and excellence in engineering practice.
## Mastering Calibration: Calibration Techniques in Engineering
The discipline of calibration is an essential cog in the engineering machinery. It encompasses multiple techniques, specifically tailored to meet the demands of different measurement systems and industries. Decades of engineering practice and innovation have culminated in the following significant calibration techniques.
### Popular Calibration Techniques in Engineering
The landscape of calibration in engineering is as diverse as the engineering field itself. Different instruments require unique calibration methods to suit their specific measurement functions and complexities. Here, you'll explore some of these techniques widely used in engineering. First up is Calibration using Comparators. These devices provide an accurate standard for comparison against the instrument under calibration. Comparators come in various forms, such as electrical, mechanical, or pneumatic, based on the type of measurement.
A comparator essentially compares the output of an instrument against the known measure from an accurate standard device.
Next is End-to-End Calibration. This method tests and calibrates the entire measurement system simultaneously, including sensors, data acquisition devices, and software for data calculation and presentation. The Field Calibration technique involves calibrating devices in their working environments. This method considers the potentially varying environmental factors impacting instrument accuracy like temperature, pressure, and humidity. Another important technique is Auto-Calibration. Certain advanced instruments can conduct self-calibration by comparing their readings with internal standards. Meanwhile, Software Calibration involves adjusting the software parameters that interpret and present the sensor's readings, affecting the overall accuracy of the device. The use of software calibration methods is becoming increasingly common with advancements in digital technology. In addition to these, some devices require Multi-point Calibration. Here, multiple points across the instrument's measurement range are calibrated. This method is commonly used when devices measure over large ranges or have nonlinear responses. Let's dive into the methodical arrangement of this calibration tapestry, starting with the steps involved in conducting these techniques.
#### Steps Involved in Calibration Techniques
Regardless of the calibration technique employed, a general series of steps typically guides the calibration process. Firstly, Preparation entails gathering the instrument's specifications, operational guidelines, and output data. Depending on these, suitable calibration standards, devices, and techniques are chosen. During the Initial Measurement phase, the device's current output is assessed and recorded to understand its current state. Measurements are typically gathered multiple times for consistency and error minimisation. Next is the crucial Comparison phase. In this step, the output from the measuring device is compared against the known output from a reference standard. This process is often iterated to improve the accuracy of readings further. The following step is Adjustment, where the instrument output is modified to match the reference standards. This might involve tweaking physical parts or adjusting software parameters. In certain cases, another round of measurements (similar to the 'Initial Measurements' stage) is taken after adjustments. These Post-adjustment Measurements help verify that the modifications have brought the output in line with the standard. Finally, the Documentation phase involves recording all calibration results. Details such as measurement errors before and after adjustment, date of calibration, and next scheduled calibration are commonly documented. This systematic chain of steps allows precise calibration of devices across industries. They offer a structured guideline but may be modified based on specific equipment or calibration technique requirements. Optimising the usage and understanding of these calibration techniques requires continuous learning and updating. These ever-evolving methods serve as the backbone for engineering, ultimately strengthening the precision, accuracy, and reliability of measurements - the cornerstones of this technical discipline.
## Guided by the Rule: Calibration Standards in Engineering
Navigating the field of calibration in engineering, it becomes clear that it is a discipline structured and ruled by standards. These standards not only provide the reference points for all calibration activities but also ensure consistency, compatibility, and reliability across different instruments, industries, and regions. Delving into these calibration standards can open up impressive insights into the rigour and precision that defines the engineering field.
### Importance of Calibration Standards in Professional Engineering
Calibration is essentially a comparison process where a measurement device's output is compared against a known standard. The hidebound existence of a standard is fundamental to calibration. In professional engineering, the importance of these standards extends beyond just being a reference point. To start with, these standards provide a universal basis for measurements, enabling compatibility across different equipment, manufacturers, and countries. This compatibility is crucial in today's globalised world, where components are manufactured and sourced from various parts of the world. Without calibration to common standards, reliably assembling these diverse components into a functioning system would be extemporaneously chaotic. Moreover, regular calibration to these standards ensures the reliability and accuracy of measurements from different devices. Imagine a scenario where multiple devices give differing measurements of the same parameter. This could lead to confusion, errors, and inefficiency. Accurate measurements are the cornerstone of engineering, from designing components to controlling processes. Hence, calibration to standards ensures that all measurement devices 'speak the same language', reinforcing accuracy and consistency. Finally, these standards enable traceability of measurements, a concept considered a hallmark of good practice in professional engineering. Traceability refers to the ability to relate a measurement directly back to a known standard, usually through an unbroken chain of comparisons. This proves that a particular device's measurements are accurate as per the defined standard. It escalates credibility and often complies with regulatory requirements in many industries.
'Traceability' in the context of calibration, refers to the unbroken chain of comparisons relating a measurement back to a known standard.
#### Different Calibration Standards used in Engineering
Given the critical role of calibration standards, various sets have been developed globally to cater to different fields and measurements. These standards might encompass physical constants, like the speed of light, scalar quantities like mass or time, or complex parameters like electrical measurements. One fundamental standard is the International System of Units (SI). This is maintained and disseminated by the International Bureau of Weights and Measures (BIPM) and is accepted globally. It includes standards for basic physical quantities like length (metre), time (second), mass (kilogram), temperature (Kelvin), electric current (Ampere), luminous intensity (Candela), and amount of substance (mole).
The International System of Units (SI) is a globally accepted system of units for basic physical quantities. It provides the fundamental reference for measurements in scientific, technical, and everyday contexts.
Additionally, numerous national and international institutions develop, maintain, and disseminate various other calibration standards. Some of these include the National Institute of Standards and Technology (NIST) in the United States, the Physikalisch-Technische Bundesanstalt (PTB) in Germany, and the National Physical Laboratory in the UK. Manufacturing and industry-specific standards have also been developed. These often cater to more specific parameters and industries, such as automotive, aerospace, or medical equipment. Examples include standards for calibrating torque wrenches in the automotive industry or standards for medical devices like blood pressure monitors. In the digital age, software-based calibration standards are gaining prominence. These standards, like the IEEE 1588 Precision Time Protocol (PTP), cater to digital and networked devices, with a focus on synchronisation and timing issues.
IEEE 1588 Precision Time Protocol (PTP) is a standard for clock synchronisation in networked digital systems. By providing a single time reference, it ensures all devices in a network are synchronised, vital for operations in sectors like telecommunications, manufacturing, and finance.
These are just some examples of the vast array of calibration standards available worldwide, each serving its unique role in the vast arena of engineering. Despite their differences, what unites all these standards is their common purpose of ensuring precise, accurate, and standardised measurements - illustrating the precision that underpins engineering.
## Tools for Accuracy: Calibration Equipment in Engineering
Within the intricate realm of engineering, calibration plays a pivotal role, and the need for precise accuracy drives this process. The equipment employed to maintain such precision becomes essential to perform a calibration. To delineate this matter, let us delve into the notable equipment, tools and implements used in professional calibration in the world of engineering.
### Notable Calibration Equipment used in Professional Engineering
To meet the variegated requirements of calibration in diverse engineering sectors, an assortment of equipment is employed. These tools offer tailored solutions to calibrate varied instruments, ensuring their accuracy and reliability. The list of notable calibration equipment includes:
• Pressure Calibrators: These are used in calibrating pressure-sensitive instruments such as pressure sensors, transducers, and gauges. They are capable of generating precise pressure outputs for comparison against the instrument being calibrated.
• Temperature Calibrators: These are essential for calibrations involving temperature-sensitive devices such as thermometers and thermostats. They can simulate accurate temperature inputs and check the device's corresponding response.
• Electrical Calibrators: These are necessary for instruments dealing with electrical parameters like voltage, current, resistance, or power. Electrical calibrators generate specific signals to assess the accuracy of the instrument's readings.
• Flow Calibrators: Instruments such as flow meters require these calibrators. They can generate precise flow rates to facilitate calibration.
• RF (Radio Frequency) Calibrators: These calibrators work on devices like frequency meters or RF power meters. They produce known RF signals for calibration.
• Humidity Calibrators: Employed for calibrating humidity sensors or hygrometers, they can accurately simulate levels of humidity.
• Analyzer Calibrators: These are used for instruments like gas analyzers. These calibrators can generate precise solutions of different gases for calibration purposes.
• Sound Calibrators: Devices that measure sound parameters, such as dB meters, require sound calibrators. They generate a precise sound level required for calibration.
These are just some examples of the sundry calibrators in use, each dealing with a specific type of measurement. While these calibration instruments share a common purpose of ensuring the accuracy and reliability of engineering equipment, each uses a distinctive means to achieve this purpose.
#### How to Use Calibration Equipment: A Practical Guide
To deploy calibration equipment appropriately for a converged device utility, a practical understanding of the equipment's operation becomes indispensable. Although every calibrator's outward aspects may vary depending on its purpose, there are some general steps that apply across the board. To begin with, you need to ascertain that the calibration equipment itself is accurate. The calibrator is usually calibrated first using a higher accuracy reference standard. This is an essential step to ensure subsequent calibration with the equipment is on point. Next, you need to know the equipment specifications you aim to calibrate. Understanding the manufacturer's instructions and technical parameters for the device helps in setting up the calibration process correctly. Afterward, install or setup the calibration equipment and the device to be calibrated in accordance to the manufacturer's instructions. This might involve connecting any probes or cables, adjusting the environment, or fitting the calibrator to the device. Now, you can use the calibration equipment to generate precise measurements. Depending on the type of calibrator, this might involve setting a specific pressure, temperature, electrical signal, or flow rate. The calibration equipment will then compare the device's output against the generated measurement.
When comparing the measured value, the electrical calibration equipment often uses a balanced bridge circuit. For an ideal case, the balance equation is given by $$\frac{R_{1}}{R_{2}}= \frac{R_{3}}{R_{4}}$$ where $$R_{1}$$, $$R_{2}$$, $$R_{3}$$ , and $$R_{4}$$ are resistive elements in a Wheatstone Bridge circuit.
If any discrepancies occur in this stage, you'll need to adjust the device under test either manually or software-wise until it matches the calibrator's output. Be aware that safety guidelines should be adhered to when performing these adjustments, particularly for devices dealing with high voltages, temperatures, or pressures. Finally, all calibration results should be documented diligently. This includes noting down any discrepancies found, the adjustments made, and the final measurement. The calibration status of the equipment, calibration date, and next due date are all typically noted down. Learning to use calibration equipment skilfully and safely can make a considerable difference in any engineering field. By gaining mastery of these tools, you can ensure the accuracy and reliability of test and measurement equipment, consequently contributing to the broader engineering project.
## Calibration - Key takeaways
• Calibration: It refers to the process of adjusting instruments to ensure accuracy and consistency of measurements.
• Engine Control Unit (ECU): Acts as the car's computer, coordinating various parts like the engine, transmission, and sensors. Incorrect calibration could lead to poor fuel efficiency or damage.
• Calibration in various sectors: In automotive, thermodynamics, manufacturing, energy, food and beverage industries, and aviation, accurate calibration of respective tools and devices is essential for optimal performance, safety, and quality control.
• Calibration Techniques: Examples include Calibration using Comparators, End-to-End Calibration, Field Calibration, Auto-Calibration, Software Calibration, and Multi-point Calibration.
• Calibration Standards: These provide the reference points for all calibration activities and ensure consistency, compatibility, and reliability across different instruments, industries, and regions. Examples include the International System of Units (SI), national and international institutions like NIST, PTB, and standards for industry-specific measurements.
#### Flashcards in Calibration 15
###### Learn with 15 Calibration flashcards in the free StudySmarter app
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What is calibration? Please write in UK English.
Calibration in engineering is the process of configuring an instrument to provide accurate results by comparing its measurements to established standards. The aim is to minimise any measurement uncertainty by ensuring the precision of the instrument's readings.
What does calibration mean? Please write in UK English.
Calibration in engineering refers to the process of setting or correcting a measuring device or base level, usually by adjusting it to match or conform to a dependably known and unvarying measure. It ensures that systems provide accurate readings for effective operation.
"What is the calibration of equipment?"
Calibration of equipment in engineering is a process that verifies the accuracy and precision of the tools, machines, or devices against a recognised standard to ensure reliable results and optimal functionality. It usually involves adjusting the equipment's performance for accuracy and reliability.
Why do we calibrate equipment?
We calibrate equipment to ensure its accurate and consistent functionality. Calibration maintains precision, reduces measurement errors, improves quality control and helps meet industry standards and regulatory compliance. It also extends the equipment's lifespan and improves safety.
Why is calibration important?
Calibration is important because it ensures the accuracy of measurement instruments, helping maintain consistency in data and improving quality control. It reduces errors in measurement, aiding in decision-making and enhancing efficiency and productivity. Calibration also supports compliance with industry standards and regulations.
## Test your knowledge with multiple choice flashcards
What is calibration in engineering?
What are the three principal types of calibration in engineering?
Why is calibration important in engineering?
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• Checked by StudySmarter Editorial Team | 4,426 | 25,845 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.910596 |
https://speakerdeck.com/ericmann/wordpress-meet-ai | 1,716,773,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00054.warc.gz | 453,849,893 | 16,679 | Eric Mann
April 23, 2021
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# WordPress, Meet AI
With new advances in machine learning, advanced integrations with AI platforms are now available to everyone! You can easily build AI into your WordPress site without a Ph.D. or advanced knowledge of linear algebra or the algorithms that make machine learning work.
In this talk, we’ll cover some simple integrations with commonly available tools to make your WordPress installation truly “smart.” No prior experience in machine learning is required, just come prepared to learn, ask questions, and get your hands dirty with tools like AWS Recognition.
April 23, 2021
## Transcript
4. ### CREATE POPULATION EVALUATE FITNESS MUTATE REPRODUCE Randomly-initialized strings can eventually
converge to meet a target string over several iterations of a random optimization algorithm over a large set of candidates. Evaluate how close each candidate is to the optimum Evaluate Fitness Initialize a collection of pseudrandom candidates Create Population Create a new candidate for the next generation Reproduce Mutate the new candidate so we can further optimize Mutate Genetic Algorithms
6. ### Ant Colony Optimization Traveling Salesman 1 Drop an “ant” at
a random city. Random Starting Point 2 The ant visits each city in turn at random, weighted by the attractiveness of each destination. “Attractiveness”-based Route 3 An ant deposits pheromone on each edge it travels, with an amount inversely proportional to the total distance travelled in its route. Calculate Pheromone Trail 4 The total pheromone for the next iteration is part left over from the first and part added to by the ants - repeat all steps until we converge. Evaporate and Repeat
8. ### I'm an avid photographer, and I'm primarily found shooting with
my DSLR or my instant film camera that I carry around for casual use. While nothing beats my DSLR in power and convenience, there's something magical about my instant film camera. Perhaps it's that you're shooting on actual film, or maybe it's that every shot you take is a unique physical artifact (which is special in today's world of Instagram and Facebook, where photos are a dime a dozen). All I know for sure is that they are incredibly fun to use and peoples' eyes light up when you pull one of these out at a party.
9. ### I'm an avid photographer, and I'm primarily found shooting with
my DSLR or my instant film camera that I carry around for casual use. While nothing beats my DSLR in power and convenience, there's something magical about my instant film camera. Perhaps it's that you're shooting on actual film, or maybe it's that every shot you take is a unique physical artifact (which is special in today's world of Instagram and Facebook, where photos are a dime a dozen). All I know for sure is that they are incredibly fun to use and peoples' eyes light up when you pull one of these out at a party.
10. ### Leverage computer vision to provide automatic “alt” text for images.
Image Descriptions Identify and center cropping and resize operations on the focal point of an image. Smart Crop Automat tagging and grouping images to simplify media management. Image Tagging ClassifAI is Open Source It’s free to use and leverages the free tiers of both IBM Watson and Azure Machine Learning! AWS support is on the horizon, and the plugin is easily extensible if you want to roll your own. | 727 | 3,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.904324 |
https://math.answers.com/Q/What_is_14p_plus_14q_plus_p_minus_7q_minus_6p | 1,643,178,837,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00698.warc.gz | 419,859,466 | 59,953 | 0
# What is 14p plus 14q plus p minus 7q minus 6p?
Wiki User
2012-01-25 16:43:22
14p + 14q + p - 7q - 6p = 9p + 7q
Wiki User
2012-01-25 16:43:22
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373 Reviews | 118 | 220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.527926 |
http://www.electronicsandcommunications.com/2012/07/region-of-convergence-of-z-transform.html | 1,542,398,205,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743184.39/warc/CC-MAIN-20181116194306-20181116220306-00350.warc.gz | 403,301,453 | 14,649 | ## Pages
### Region of Convergence of Z Transform Properties
The z-transform exists when the infinite sum converges.
The sum may not converge for all values of ‘z’. The value of ‘z’ for which the sum converges is called Region of Convergence (ROC).
PROPERTIES OF REGION OF CONVERGENCE
1. The ROC is a concentric ring or a circle in the z-plane centered at the origin.
2. The ROC cannot contain any poles.
3. If x(n) is a finite duration causal sequence, the ROC is entire z-plane except at z=0.
If x(n) is a finite duration anti causal sequence, then the ROC is the entire z-plane except at z=∞.
If x(n) is a finite duration 2-sided sequence, then the ROC will the entire z-plane except at z=0 and z=∞.
4.If x(n) is a right sided sequence and if the circle |z|=r0 is in the ROC, then all finite values of ‘z’ for which |z|>ro will also be in ROC.
5. If x(n) is a left sided sequence and if the circle |z|=r0 is in the ROC, then all values of z for which 0<|z|<ro will be in ROC.
6. If x(n) is a 2-sided sequence and if the circle |z|=r0 is in the ROC, then the ROC will consists of a ring in the z-plane that includes the circle |z|=r0
7. If the z-Transform X(z) of x(n) is rational, then its ROC is bounded by poles or extends to ‘∞’.
8. If the z-Transform X(z) of x(n) is rational and if x(n) is right sided, then ROC is the region in the z-plane outside the outermost pole. In other words, Outside the radius of circle = the largest magnitude of pole of x(z). If x(n) is causal then ROC also includes Z= ∞.
9. If the z-Transform X(z) of x(n) is rational and if x(n) is left sided, then ROC is the region in the z-plane inside the outermost ‘non zero pole’.
In other words, inside the circle of radius = the smallest magnitude of pole of x(z) other than at z=0 and extending inwards to and possibly including z=0. If x(n) is anti-causal, ROC includes z=0.
10.If x(n) is a finite duration 2-sided sequence, then ROC will consists of a circular ring in the z-plane bounded on the interior and exterior by a pole and not containing any pole.
11. The ROC of an LTI (Linear Time Invariant) system contains the unit circle.
12. ROC must be a connected region. | 594 | 2,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-47 | longest | en | 0.873435 |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/9/lesson/9.4.3/problem/9-135 | 1,720,833,138,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00509.warc.gz | 258,398,666 | 15,085 | ### Home > APCALC > Chapter 9 > Lesson 9.4.3 > Problem9-135
9-135.
For each of the series below, write an equivalent expression using sigma notation.
1. $5 + 10 + 20 + 40 + ... + 5(2)^{n-1}$
$\displaystyle \sum_{i=1}^{n}5(2)^{i-1}$
1. $1+\frac { 2 } { 3 } + \frac { 4 } { 9 } + \frac { 8 } { 27 } + \ldots$
$a_n=\Big(\frac{2}{3}\Big)^n\text{ starting with }n=0.$ | 159 | 368 | {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-30 | latest | en | 0.265855 |
https://support.nag.com/numeric/nl/nagdoc_25/nagdoc_fl25/examples/source/c02agfe.f90.html | 1,695,540,884,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506623.27/warc/CC-MAIN-20230924055210-20230924085210-00486.warc.gz | 616,930,709 | 3,019 | ```! C02AGF Example Program Text
! Mark 25 Release. NAG Copyright 2014.
Module c02agfe_mod
! C02AGF Example Program Module:
! Parameters
! .. Implicit None Statement ..
Implicit None
! .. Accessibility Statements ..
Private
! .. Parameters ..
Integer, Parameter, Public :: nin = 5, nout = 6
Logical, Parameter, Public :: scal = .True.
End Module c02agfe_mod
Program c02agfe
! C02AGF Example Main Program
! .. Use Statements ..
Use c02agfe_mod, Only: nout
! .. Implicit None Statement ..
Implicit None
! .. Executable Statements ..
Write (nout,*) 'C02AGF Example Program Results'
Call ex1
Call ex2
Contains
Subroutine ex1
! .. Use Statements ..
Use nag_library, Only: c02agf, nag_wp
Use c02agfe_mod, Only: nin, scal
! .. Local Scalars ..
Real (Kind=nag_wp) :: zi, zr
Integer :: i, ifail, n, nroot
! .. Local Arrays ..
Real (Kind=nag_wp), Allocatable :: a(:), w(:), z(:,:)
! .. Intrinsic Procedures ..
Intrinsic :: abs
! .. Executable Statements ..
Write (nout,*)
Write (nout,*)
Write (nout,*) 'Example 1'
! Skip heading in data file
Allocate (a(0:n),w(2*(n+1)),z(2,n))
Write (nout,*)
Write (nout,99999) 'Degree of polynomial = ', n
ifail = 0
Call c02agf(a,n,scal,z,w,ifail)
Write (nout,99998) 'Computed roots of polynomial'
nroot = 1
Do While (nroot<=n)
zr = z(1,nroot)
zi = z(2,nroot)
If (zi==0.0E0_nag_wp) Then
Write (nout,99997) 'z = ', zr
nroot = nroot + 1
Else
Write (nout,99997) 'z = ', zr, ' +/- ', abs(zi), '*i'
nroot = nroot + 2
End If
End Do
99999 Format (/1X,A,I4)
99998 Format (/1X,A/)
99997 Format (1X,A,1P,E12.4,A,E12.4,A)
End Subroutine ex1
Subroutine ex2
! .. Use Statements ..
Use nag_library, Only: a02abf, c02agf, nag_wp, x02ajf, x02alf
Use c02agfe_mod, Only: nin, scal
! .. Local Scalars ..
Real (Kind=nag_wp) :: deltac, deltai, di, eps, epsbar, &
f, r1, r2, r3, rmax
Integer :: i, ifail, j, jmin, n
! .. Local Arrays ..
Real (Kind=nag_wp), Allocatable :: a(:), abar(:), r(:), w(:), z(:,:), &
zbar(:,:)
Integer, Allocatable :: m(:)
! .. Intrinsic Procedures ..
Intrinsic :: abs, max, min
! .. Executable Statements ..
Write (nout,*)
Write (nout,*)
Write (nout,*) 'Example 2'
! Skip heading in data file
Allocate (a(0:n),abar(0:n),r(n),w(2*(n+1)),z(2,n),zbar(2,n),m(n))
! Read in the coefficients of the original polynomial.
! Compute the roots of the original polynomial.
ifail = 0
Call c02agf(a,n,scal,z,w,ifail)
! Form the coefficients of the perturbed polynomial.
eps = x02ajf()
epsbar = 3.0_nag_wp*eps
Do i = 0, n
If (a(i)/=0.0_nag_wp) Then
f = 1.0_nag_wp + epsbar
epsbar = -epsbar
abar(i) = f*a(i)
Else
abar(i) = 0.0E0_nag_wp
End If
End Do
! Compute the roots of the perturbed polynomial.
ifail = 0
Call c02agf(abar,n,scal,zbar,w,ifail)
! Perform error analysis.
! Initialize markers to 0 (unmarked).
m(1:n) = 0
rmax = x02alf()
! Loop over all unperturbed roots (stored in Z).
Do i = 1, n
deltai = rmax
r1 = a02abf(z(1,i),z(2,i))
! Loop over all perturbed roots (stored in ZBAR).
Do j = 1, n
! Compare the current unperturbed root to all unmarked
! perturbed roots.
If (m(j)==0) Then
r2 = a02abf(zbar(1,j),zbar(2,j))
deltac = abs(r1-r2)
If (deltac<deltai) Then
deltai = deltac
jmin = j
End If
End If
End Do
! Mark the selected perturbed root.
m(jmin) = 1
! Compute the relative error.
If (r1/=0.0E0_nag_wp) Then
r3 = a02abf(zbar(1,jmin),zbar(2,jmin))
di = min(r1,r3)
r(i) = max(deltai/max(di,deltai/rmax),eps)
Else
r(i) = 0.0_nag_wp
End If
End Do
Write (nout,*)
Write (nout,99999) 'Degree of polynomial = ', n
Write (nout,*)
Write (nout,*) 'Computed roots of polynomial ', ' Error estimates'
Write (nout,*) ' ', &
' (machine-dependent)'
Write (nout,*)
Do i = 1, n
Write (nout,99998) 'z = ', z(1,i), z(2,i), '*i', r(i)
End Do
99999 Format (1X,A,I4)
99998 Format (1X,A,1P,E12.4,Sp,E12.4,A,5X,Ss,E9.1)
End Subroutine ex2
End Program c02agfe
``` | 1,444 | 4,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.467158 |
https://www.physicsforums.com/threads/markov-chain-sum-of-n-dice-rolls.565476/ | 1,527,303,798,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00336.warc.gz | 816,212,661 | 14,771 | # Markov chain, sum of N dice rolls
1. Jan 6, 2012
### simba_
Question : Let Xn be the maximum score obtained after n throws of a fair dice
a) Prove that Xn is a markov chain and write down the transition matrix
Im having a problem starting the transition matrix
im assuming the states are meant to be the sum. then do you write out the transition matrix for the first 2 throws and have this matrix to the power of n-1?
2. Jan 6, 2012
### Stephen Tashi
I suppose if you are studying markov chains with an infinite number of states, you could try interpreting "maximum score" to mean some sort of sum. However, it seems to me that the problem intends the state of the process on the nth roll to be $max \{ R_1,R_2,...R_n\}$ and not $R_1 + R_2 + ... + R_n$. So if make 3 rolls and they are {3,5,4} the state of the process is $X_3 = 5$
3. Jan 7, 2012
### simba_
So the transition matrix is an upper triangular matrix to the power of n-1 with the diagonal entries 1/6, 2/6, 3/6, 4/6, 5/6, 6/6 respectively?
4. Jan 7, 2012
### Stephen Tashi
That is incorrect terminology. To compute things about the state at step n in the process, one may raise the transition matrix to a power, but the transition matrix itself, in simple examples, is not a function of n.
Yes.
5. Jan 7, 2012 | 373 | 1,290 | {"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} | 3.40625 | 3 | CC-MAIN-2018-22 | latest | en | 0.875068 |
http://wowmoney.eu/poker/gambling-probability-roulette.php | 1,591,326,220,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348492427.71/warc/CC-MAIN-20200605014501-20200605044501-00533.warc.gz | 130,186,182 | 12,160 | 4692
# Gambling probability roulette
The mathematics of gambling are a collection of probability applications encountered in games of chance and can be included in game theory. From a mathematical point of view, the games of chance are experiments generating various types of aleatory events, the probability of which can be calculated by using the properties of probability on a finite . To understand how to play roulette, like all gambling games, you need to know how to play WOWMONEY.EU get the most out of your roulette play it is critical to know how often you can expect to win and how much of an advantage you need to overcome. Learn Roulette Odds - All you need to know about the odds of roulette games. Use our trusted, expert guide so you can play better, both live and online.
## Roulette Odds & Probabilities – All You Need to Know
Applied probability Gambling mathematics Gambling. The winning odds for those bets is You want to know the probability of the coin landing on heads. Using probability in real life. The Modern Pocket Hoyle:
It Is Interesting about casino
• Croupier's clothes are given out by the casino. There are no pockets there, so you can not hide or steal chips.
• The longest game of poker in the casino is 90 hours. The record belongs to the Irishman Phil Laak. Phil not only set a record, but also won 7 thousand dollars.
## 14 Gambling Probability Examples
Probability is a branch of mathematics, and a lot of people have trouble with math. The first concept to understand is that probability is something that applies to random events. And the first thing to know is that a probability is always a number between 0 and 1. An event that will always occur has a probability of 1. Anything that might or might not happen can be measured with a number between 0 and 1. And that number is easily calculated. You take the number of ways an event can happen and divide it by the total number of events possible.
You flip a coin. There are 2 possible outcomes, both of which are equally likely. You want to know the probability of the coin landing on heads. That can also be expressed as 0. All probabilities are, by their nature, fractions. But there are multiple ways of expressing a fraction. I gave examples in 1 related to the toss of a coin.
Or you could express it as 1. But most people are comfortable expressing it as a percentage. Suppose you want to know the probability of getting heads twice in a row. In this case, we have a 0. The die has 6 possible results, 1, 2, 3, 4, 5, or 6. A standard deck of cards consists of 52 cards. The cards are divided into 4 different suits—clubs, diamonds, hearts, and spades. The cards are also divided into 13 different ranks: But one of the interesting things about card games is that the dealing of cards changes the probability of getting subsequent cards.
You have 3 aces and a deuce. You only have one opponent, and he has 4 cards, none of which is an ace. An American roulette wheel has 38 pockets in which the ball can land. This can also be expressed as 37 to 1. This is a great example, because it demonstrates how the casino gets its edge over the player.
The first and most obvious victims of the government's lies are the 40,000 or so Americans who this year will become HIV-positive, overwhelmingly gay men or poor, inner-city drug users and their sexual partners.
Недолго думая, девушка избавляется от трусиков, демонстрируя писечку, и задумывает показать себя в роли темной красотки. А вы сами разве собираетесь ей что - то запрещать и в начинаниях ее мешать. Наташа явно засмущалась. In a relationship, will consider proposals for the road I like black coffee - hot and alluring, with a little bitterness, searing, with a unique flavor.
Девушки и женщины в голом виде приводят к особенному возбуждению, после чего хочется заниматься с ними сексом до бесконечности. Может она и вела себя с ним как послушная сучка, но она была живой.
### Details
If you are about to wager real money on casino games, it is of the utmost importance that you familiarise yourself with the chances of winning each game has. This is especially relevant for roulette — a game where each bet has a specific mathematical probability of winning. Of course, those odds are affected by the type of roulette you are playing and the extra rules that apply to it.
As you probably already know, European roulette has better odds for the player than American roulette. Understanding those odds and probabilities is paramount if you want to start winning big at roulette. This is why we have prepared this informative article for you.
This article covers only the odds and probabilities of the various roulette bets and not all the rules of the game in general. If you want to learn the absolute basics of the game, then we recommend that you first read our roulette rules article.
In the game of roulette, there are two main groups of bets — Outside Bets and Inside Bets. The names are derived from the bet positions on the layout of the table. Outside bets have higher chances of winning, but much smaller payouts.
Inside bets, on the other hand, have less probability of winning, but considerably bigger payouts. People who like to play it safe prefer the outside bets, because they have much better odds of winning.
## Gambling guidelines
Gaming grant information, documents, and forms including: Gaming grant applications must be submitted online. Go to the Online Service to start your online application look for the "Apply Online" link. Use the Gaming Account Summary Report form to report all deposits to, and all disbursements from, your organization's gaming account s during your fiscal year.
Organizations must submit this form within 90 days following their fiscal year-end. Use the Gaming Event Revenue Report form to report all revenues and expenses associated with a licensed gambling event held by your organization. Organizations must submit this form within 90 days after the expiry of each gambling event licence.
## Video
### Free Roulette
The thrill of watching the spinning red and black Roulette wheel has long served to grip many avid gamblers around the g... | 1,340 | 6,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.950374 |
https://mathcenter.oxford.emory.edu/site/math125/probSetChineseRemainderTheorem/ | 1,723,662,859,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00647.warc.gz | 296,740,947 | 2,750 | ## Exercises - Primality Testing and Carmichael Numbers
1. Solve the system of congruences: $$\begin{array}{rcl} x \equiv 6\pmod{7}\\ x \equiv 4\pmod{8} \end{array}$$
By the first congruence, $x = 7k+6$ for some integer $k$. Substituting this into the second congruence leads to $7k+6 \equiv 4\pmod{8}$, and then $7k \equiv -2\pmod{8}$.
Solving this in the usual way (Euclidean Algorithm and "back-solving"), we find $k \equiv 2\pmod{8}$. So $k = 8m+2$ for some integer $m$. Substituting this back into the equation for $x$ above, we have $x = 7(8m+2)+6$ which simplifies to $x = 56m + 20$. Thus,
$$x \equiv 20\pmod{56}$$
2. Solve the system of congruences: $$\begin{array}{rcl} 2x \equiv 5\pmod{7}\\ 3x \equiv 4\pmod{8} \end{array}$$
$$x \equiv 20\pmod{56}$$
3. Solve the system of congruences: $$\begin{array}{rcl} x \equiv 3\pmod{4}\\ x \equiv 0\pmod{6} \end{array}$$
There are no solutions, as the first implies $x$ is odd, the second that $x$ is even.
4. Solve the system of congruences: $$\begin{array}{rcl} x \equiv 2\pmod{5}\\ x \equiv 3\pmod{7}\\ x \equiv 10\pmod{11} \end{array}$$
$x \equiv 87\pmod{385}$
5. Solve the system of congruences: $$\begin{array}{rcl} 2x \equiv 6\pmod{14}\\ 3x \equiv 9\pmod{15}\\ 5x \equiv 20\pmod{60} \end{array}$$
$x \equiv 388\pmod{420}$
6. Solve the system of congruences: $$\begin{array}{rcl} x \equiv 3\pmod{8}\\ x \equiv 7\pmod{12} \end{array}$$
Note the moduli are not relatively prime. Can you figure out what $x$ would be$\pmod{8}$ and$\pmod{3}$? | 599 | 1,508 | {"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} | 4.625 | 5 | CC-MAIN-2024-33 | latest | en | 0.578981 |
https://www.clutchprep.com/chemistry/practice-problems/71272/using-hess-law-calculate-the-enthalpy-of-formation-for-c3h8-g-using-the-followin | 1,643,273,759,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00010.warc.gz | 757,330,920 | 22,201 | # Problem: Using Hess' Law, calculate the enthalpy of formation for C 3H8(g) using the following thermodynamic data.C(s) + O2(g) → CO2(g) ΔH = -393.5 kj/molH2(g) + 1/2O2(g) → H2O(I) ΔH = -285.8 kj/molC3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(I) ΔH = 2199.0 kJ/mol
###### FREE Expert Solution
87% (347 ratings)
###### Problem Details
Using Hess' Law, calculate the enthalpy of formation for C 3H8(g) using the following thermodynamic data.
C(s) + O2(g) → CO2(g) ΔH = -393.5 kj/mol
H2(g) + 1/2O2(g) → H2O(I) ΔH = -285.8 kj/mol
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(I) ΔH = 2199.0 kJ/mol
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Hess's Law concept. You can view video lessons to learn Hess's Law Or if you need more Hess's Law practice, you can also practice Hess's Law practice problems .
What is the difficulty of this problem?
Our tutors rated the difficulty of Using Hess' Law, calculate the enthalpy of formation for C 3... as medium difficulty.
How long does this problem take to solve?
Our expert Chemistry tutor, Sabrina took 7 minutes to solve this problem. You can follow their steps in the video explanation above.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Collins' class at AUSTINCC. | 420 | 1,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-05 | latest | en | 0.883167 |
https://www.cs.miami.edu/home/burt/learning/Csc609.011/Perfect/ | 1,638,088,675,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00057.warc.gz | 796,049,689 | 1,481 | ### Perfect Secrecy
A cryptosystem has perfect secrecy if for any message `x` and any encipherment `y`, `p(x|y)=p(x)`.
This implies that there must be for any message, cipher pair at least one key that connects them. Hence:
``` |K| >= |C| >= |P|
```
In the boundary case of equality we have the following theorem:
Theorem(Shannon perfect secrecy)
Suppose a cryptosystem with `|K|=|C|=|P|`. The cryptosystem has perfect secrecy if and only if
• each key is used with equal probability `1/|K|`,
• for every plaintext `x` and ciphertext `y` there is a unique key `k` such that `e_k(x)=y`.
Proof
Suppose perfect secrecy, i.e. `p(x|y)=p(x)` for all `x` and `y`. Unless `p(x)=0`, there must be enough keys so that any ciphertext can be decoded as a given plaintext, that is, `|K|>=|C|`, but by supposition, equality must hold. Hence there is a unique key for every `x` `y` pair.
Suppose keys `k_1, k_2, ... ` are the unique keys such that `d_k_i(y)=x_i`. Using Bayes law:
``` p(x_i|y) = ( p(y|x_i) p(x_i) ) / p(y)
```
Using the assumption of perfect secrecy, we have:
``` p(y|x_i) = p(y)
```
hence each `k_i` must occur with the same probability
We now assume that `p(k)=1/|K|` and that there is a unique key relating any plaintext-ciphertext pair.
``` p(x|y) = ( p(y|x) p(x) ) / p(y)
```
By the uniqueness of keys, `p(y|x)=1/|K|`. We also calculate
``` p(y) = sum_k p(k) p(d_k(y))
= 1/|K| sum_k p(d_k(y))
= 1/|K| sum_x p(x)
= 1/|K|
```
Cancelling the `1/|K|` gives the result `p(x|y) = p(x)`, that is, perfect secrecy. | 503 | 1,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-49 | longest | en | 0.830886 |
https://www.convert-measurement-units.com/convert+Kilodalton+to+Quintal.php | 1,719,153,182,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00115.warc.gz | 616,645,471 | 13,651 | Convert kDa to Quintal (Kilodalton to Quintal)
## Kilodalton into Quintal
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Kilodalton+to+Quintal.php
# Convert Kilodalton to Quintal (kDa to Quintal):
1. Choose the right category from the selection list, in this case 'Mass / Weight'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Kilodalton [kDa]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Quintal'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '997 Kilodalton'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Kilodalton' or 'kDa'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Mass / Weight'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '85 kDa to Quintal' or '52 kDa into Quintal' or '8 Kilodalton -> Quintal' or '5 kDa = Quintal' or '59 Kilodalton to Quintal' or '79 Kilodalton into Quintal'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(74 * 35) kDa'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '997 Kilodalton + 2991 Quintal' or '69mm x 16cm x 54dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 5.581 750 072 662 9×1026. For this form of presentation, the number will be segmented into an exponent, here 26, and the actual number, here 5.581 750 072 662 9. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 5.581 750 072 662 9E+26. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 558 175 007 266 290 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 892 | 3,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-26 | latest | en | 0.837197 |
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May 26 answered Rijndael S-Box Permutation Group May 26 comment “Casual” mathematical facts with practical consequences @Jim: Yup, that's why I used the word "seem". Anyway, I think it's pretty clever. May 25 comment “Casual” mathematical facts with practical consequences The puzzle is to explain why these two differently-sized rectangles seem to be both partitioned in to the same 4 polygons. Knuth claims this was a favorite puzzle of Lewis Carroll (See "Concrete Mathematics", Graham, Knuth, Patashnik, pg 293) May 25 answered “Casual” mathematical facts with practical consequences May 25 revised “Casual” mathematical facts with practical consequences added 3 characters in body May 24 comment Is the factorization problem harder than RSA factorization ($n = pq$)? The answer is trivially "yes" since you didn't limit computational power in any way, and factorization is certainly decidable. May 23 awarded Nice Question May 23 revised How valid is the concern over narrow pipe cryptographic hash function designs? added 328 characters in body May 23 comment How valid is the concern over narrow pipe cryptographic hash function designs? @Pete: Point well-taken. However, I prefer remaining anonymous on these forums, so I'll have to live with the loss-of-value engendered as a result. May 23 answered How valid is the concern over narrow pipe cryptographic hash function designs? May 23 comment Partial sum over $M$, of ${m+j \choose M} {1-M \choose m+i-M}$? Although I don't know if your summation is handled by the methods in the following book, you might want to check it out. The methods here are often called "revolutionary" as regards summations of binomial coefficients: math.upenn.edu/~wilf/AeqB.html May 21 comment Subset sum problem is NP-complete? Yes, thanks. Corrected. May 21 revised Subset sum problem is NP-complete? added 2 characters in body May 21 answered Subset sum problem is NP-complete? May 21 revised What are a , b and c? edited tags May 21 answered Can a number have a prime factor that isn't a part of the number's prime factorization? May 20 comment Another pigeonhole principle question @Arvin: Yes, c(7,3) is correct. I was leaving that part for you to work out, but I see other respondents did the whole problem shortly after. Ah well. May 20 awarded Good Question May 20 answered Another pigeonhole principle question May 20 revised Preparation for Putnam? edited tags | 570 | 2,491 | {"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} | 2.609375 | 3 | CC-MAIN-2016-18 | latest | en | 0.941303 |
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Trang chủ » Dynamic Positioning Test Questions? Top 61 Best Answers
# Dynamic Positioning Test Questions? Top 61 Best Answers
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Below is a multiple question test to check your knowledge on DP systems. 7.1. General Questions on DP What is the difference between ‘DP current’ and real current? ‘DP current’ is the current value that the DP system shows on the DP screen. It is equal to the total thruster force minus the Wind Force. DP current = Thruster force …
## Dynamic Positioning Systems, Principles, Design and Applications
Dynamic Positioning Systems, Principles, Design and Applications
Dynamic Positioning Systems, Principles, Design and Applications
## What is dynamic positioning (DP)?
Dynamic Positioning (DP) is a vessel capability provided via an integration of a variety of individual systems and functions*. A computer control system automatically maintains a vessel’s position and heading by using her own propellers and thrusters.
Dynamic positioning (DP) is a computer-controlled system to automatically maintain a vessel ‘s position and heading by using its own propellers and thrusters.
What is dynamic positioning?
Introduction to Dynamic Positioning 1.1. Dynamic Positioning Systems For many offshore operations it is necessary to keep a vessel at a fix position and heading. Traditionally this has been done using an anchor spread. Nowadays, Dynamic Positioning (DP) systems are replacing anchors.
What is a positioning system (DP)?
The positioning system, usually a GPS, monitors the position of the vessel. When the vessel moves off the intended position the DP computer will calculate the required thrust which will then be applying by the thrusters in order to maintain the position of the vessel. 1.2. The use of DP systems
What is a dynamic positioning operator (DPO)?
Dynamic positioning systems require specially trained and skilled personnel to handle the systems at sea. The person who mans the entire system and ensures the smooth functioning of a DP vessel is known as a DP operator. The dynamic positioning operator (DPO) needs to be trained appropriately before taking up the reins of the demanding profession.
What are the applications of Dynamic Positioning System in maritime industry?
As the applications of Dynamic Positioning System increase in the maritime industry, different types of ships are now being fitted with DP systems to improve control and handling over vessels at sea.
## How do schools use the dynamic placement test?
Schools, universities and companies use the test for a variety of purposes, including testing incoming students pre- or post-arrival, assigning learners to specific courses, and filtering job applicants based on their English. Find out more about how these institutions (and others) use the Dynamic Placement Test.
Schools, universities and companies use the test for a variety of purposes, including testing incoming students pre- or post-arrival, assigning learners to specific courses, and filtering job applicants based on their English. Find out more about how these institutions (and others) use the Dynamic Placement Test.
## What is the importance of capability curve in DP operation?
The capability curves allows to find the limiting wind speed for safe DP operation of the vessel.
To prepare a DP capability curve for a certain vessel wind, current and waves are normally taken to act in the same direction. What is consequence analysis DP? Consequence analysis is used to verify the safety of a DP operation. It is used to check whether there is sufficient running power and thruster capacity available to retain sufficient …
## How does the system track the position of the object?
Once the initial position is inputed to the system it able to track the any movements and determine the relative position from a start point. The systems is accurate for a limited time period.
Position tracking often involves triangulation, a process that pinpoints a location using three points of reference. Radio frequencies between these geographical landmarks (towers or satellites) are averaged to a specific point—the location of the position tracking device.
What is position tracking and how does it work?
Position tracking systems are also adept for mapping rooms, facilities, and transit routes for different electronics. Some companies, like Accuware, use a hybridized system for tracking and sensing. This mapping-and-positioning method employs a camera as a sensor. The camera also enables continuous tracking in a 3D space.
How does an optical tracking system work?
By using a model of this configuration of each object, the optical tracking system is able to distinguish between objects and determine the 3D position and orientation of each object. Optical tracking has proved to be a valuable alternative to tracking systems based on other technologies, such as magnetic, acoustic, gyroscopic and mechanical.
What is a tracking system?
A tracking system, also known as a locating system, is used for the observing of persons or objects on the move and supplying a timely ordered sequence of location data for further processing. There are myriads of tracking systems.
How do the tags track the position of the object?
The tags triangulate their 3D position using the anchors placed around the perimeter. A wireless technology called Ultra Wideband has enabled the position tracking to reach a precision of under 100 mm. By using sensor fusion and high speed algorithms, the tracking precision can reach 5 mm level with update speeds of 200 Hz or 5 ms latency .
References:
Online English Placement Test | Dynamic Placement Test
Dynamic Positioning (DP) investigations – FORCE …
Guidelines on testing dp systems ver1 2 – Dynamic …
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dynamic positioning test questions
You have just come across an article on the topic dynamic positioning test questions. If you found this article useful, please share it. Thank you very much. | 1,330 | 6,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-14 | latest | en | 0.861974 |
https://math.stackexchange.com/questions/3496344/inflation-property-of-artin-l-series | 1,723,287,315,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00201.warc.gz | 304,063,727 | 37,051 | # Inflation property of Artin $L$-series
I have been looking at several different proofs that Artin $$L$$-series of Abelian extensions coincide with Hecke $$L$$-series.
In Serge Lang's $$\textit{Algebraic Number Fields}$$ (XII §2) and Jürgen Neukirch's $$\textit{Algebraische Zahlentheorie}$$ (VII §10) they use a property which I paraphrase in the following manner:
Let $$E/K$$ be an Abelian extension, $$G:=\textrm{Gal}(E/K)$$, and let $$\chi$$ be a simple character of $$G$$. We then have $$G/\textrm{Ker}(\chi) \cong \textrm{Gal}(E_{\chi}/K)$$, where $$E_{\chi}$$ is the fixed subfield corresponding to $$\textrm{Ker}(\chi) \vartriangleleft G$$. By inflation, we may view $$\chi$$ is a character of $$\textrm{Gal}(E_{\chi}/K)$$, and we have: $$L(E_{\chi}/K,\chi,s) = L(E/K,\chi,s)$$
I have some scruples with this argument.
I understand the property of inflation to mean the following:
Let $$E/K$$ be a Galois extension, $$G:=\textrm{Gal}(E/K)$$, and let $$E'/K$$ be a bigger Galois extension ($$E \subset E'$$), $$G':=\textrm{Gal}(E'/K)$$, and let $$\chi$$ be a simple character of $$G$$. We then have: $$L(E'/K,\chi',s) = L(E/K,\chi,s)$$ $$\chi' = \chi \circ \pi$$, where $$\pi: G' \to G$$ is the canonical projection.
That is, inflation allows us to pass $$\textit{from a smaller Galois extension to a bigger one}$$.
But Lang and Neukirch (et al.) seem to be going the other way: They take a character of a bigger Galois extension and pass to a smaller one.
But this is manifestly impossible. Take $$\textit{e.g.}$$ $$\mathbb{Q} / \mathbb{Q}$$ for the smaller extension and $$\mathbb{Q}(i) / \mathbb{Q}$$ for the bigger one. The above would imply that: $$L(\mathbb{Q}(i) / \mathbb{Q}, \chi, s) = L( \mathbb{Q} / \mathbb{Q} , \chi , s) = \zeta(s)$$ On the left, we could let $$\chi$$ be the non-trivial sign character. On the right, we must necessarily have the trivial character.
How is this to be understood?
The converse to the inflation is that, if $$\rho$$ is a representation of $$G=Gal(E/K)$$ then $$H=\ker(\rho)$$ is normal and $$\tilde{\rho}(gH)=\rho(g)$$ is a (faithful) representation of $$G/H=Gal(E/K)/Gal(E/E^H)=Gal(E^H/K)$$ and $$L(E/K,\rho,s)=L(E^H/K,\tilde{\rho},s)$$ | 748 | 2,200 | {"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": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.776135 |
https://forums.tomsguide.com/threads/obama-says-steve-jobs-deserves-to-be-rich.5603/page-10 | 1,709,094,075,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474690.22/warc/CC-MAIN-20240228012542-20240228042542-00810.warc.gz | 264,944,512 | 30,453 | # Obama Says: Steve Jobs Deserves to be Rich
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[citation][nom]10tacle[/nom]I call 'em like I see 'em. Nothing personal.[/citation]
Do you know who else calls people names when they're not getting their way? Small children.
If you want to sit at the grown up table and talk about national policy, then you need to use your grown up words.
Calling people names doesn't support your argument.
[citation]Uhm, no, them's the facts. You need to learn basic math and percentages (and how not to add them up in an attempt to debate). http/www.ntu.org/tax-basics/who- [...] taxes.html[/citation]
Thank you for providing your source. You seem to be confused by what these numbers actually mean. Since the total percentages do not have a sum of 100, it is implied that these numbers refer to a percentile and not a percentage. The term percentile and the related term percentile rank are often used in descriptive statistics; so the 10th percentile is the value (or score) below which 10 percent of the category (in this case, annual income) is determined.
There are actually a number of ways to calculate this. Sadly, NTU didn't provide a description.
The most common percentile is the "Nearest Rank" method: p-th percentile of N ordered values is obtained by first calculating the rank:
n = (N/100)p + 1/2
In this case, rounding to the nearest integer, and taking the value that corresponds to that rank.
This is the simplest for of a percentile, but there is also weighted and Linear Interpolation Between Closest Ranks methods that are common. Since there are several values needed to calculate an income tax percentile, the Linear interpolation method was most likely required.
But why?
Well for one thing, not everyone within our population has an income, or an income which is taxable. Children, the unemployed, and even those that work in fields where their income is considered non-taxable. In 2007 I was in a combat zone and was exempt from paying income tax.
Yeah, that's right. This person that you so charmingly characterized as a "Libtard" is a combat veteran and lifetime member of the NRA. But you'd find it much more convenient to dismiss me as some dope smoking hippie or something instead of considering, that maybe, just maybe you don't have all of the data needed to determine something as complex as tax policy.
So what data is missing besides non-taxable income earners?
Well, the NTU also fails to specify what the actual p value is for a given income. How many people actually earn more than say, \$380,000 matters a great deal. If more people earn this amount than those earning \$100,000 for example, both could be taxed at an equal rate, yet the percentile would deceptively indicate that the person earning \$380,000 actually pays a higher income tax, when in reality it would just be in reference to the population that earns that amount.
Fortunately, we don't have to guess, because the data is available, thanks to the wonders of the government-funded world wide web.
http/www.irs.gov/pub/irs-drop/rp-09-50.pdf
Look at page seven:
There are six tax brackets ranging from %10 - %35, and in four categories: Single, Married Jointly, Married Separate, and Head of Household.
There is another side to this: Capital Gains Tax. But you may or may not know about that, so I'll leave it alone for right now.
Income is taxed at a given bracket only for each dollar earned within that bracket. So for example, if a single taxpayer earned \$10,000 in 2009 they would be taxed 10% of each dollar earned from the 1st dollar to the 8,350th dollar (10% × \$8,350 = \$835.00), then 15% of each dollar earned from the 8,351st dollar to the 10,000th dollar (15% of \$1,650 = \$247.50), totaling \$1,082.50. This amount is lower than if the individual had been taxed at 15% on the full \$10,000. This is because the individual's marginal rate has no effect on the income taxed at a lower bracket. This ensures that every rise in a person's pre-tax salary results in an increase of his after-tax salary.
Are you still with me? Do you understand why this matters?
It matters because the taxes paid by Steve Jobs, and Bill Gates, are at the same rate as someone that owns a bar or restaurant and takes home more than \$373,000 annually. Though it should be noted that again, the dollars earned up to \$373,000 are all in different tax brackets, and only the amount earned above \$373,000 is taxed at the top rate of %35.
Again, this tax rate is historically low:
http/www.irs.gov/pub/irs-soi/02inpetr.pdf
Take a look at that, in 1952-1953 people earning more than \$400,000 paid an income tax of 92%! A rate nearly 3x that of today. This was during the cold war, when suspicion of communism got people thrown before congress and charged with treason, yet our own federal government was taxing the wealthy at a rate of \$0.89 for every dollar earned over \$400,000 annually. Yet there was no Tea-Party hysterics, no misspelled signs claiming that Ike and Truman weren't "Real Americans". In fact, the 1950's and 1960's were pretty good decades for most Americans, and the economy was growing.
Let's move on, shall we?
[citation]WTF are you talking about? English please - and clear English. Mindless liberal psychobabble won't cut it with me.[/citation]
Is anything you find difficult to understand automatically categorized as "Mindless liberal psychobabble"? I've been very clear in both this and my other comments. If you are finding something difficult to understand, please just ask for specifics and I'll gladly provide you with the best answer that I can. But instead you are just calling me names, like a child does when they are upset. This is a conversation for grown-ups and I expect you to act like one.
[citation]What? RACISM??!! Where in my post? WHERE?[/citation]
Well if the pointed hat and white sheets fit...
I mentioned reading books on the subject of taxes as an alternative to watching Fox News, and challenged you to address a specific case of President Obama raising your taxes. You replied with several all-caps words; complaining about lottery winners inability to retain their wealth, and then accusing Mexican immigrants of taking away business opportunities from citizens born in the United States.
When no-one at the table is discussing an issue, and you choose to insert it; that reflects badly on you. The observation of racism is a direct result of your completely off-topic tangent regarding Mexican immigration, so please don't try to act as though it came out of nowhere when I pointed this out.
[citation]What's under that beanie cap of yours? Am I "racists" because the current president is BLACK and I disagree with his socialistic near Marxist policies??!![/citation]
Could you please tell me what legislation he has signed into law which is categorically "near Marxist"? Words have meaning. I rebutted your comment regarding Mexican immigrants, and you are now acting confused and assuming that I am referring to the race of the president. This feels like a dodge, and not a very good one.
[citation]Go ahead and spit it out. All you psychos on the left can do is claim racism at the drop of a hat. Sorry pal, that dog won't hunt with me. You can return to your Daily Kook, Huff & Puffington Post, and other lib wingnut sites, thanks. It's been fun, but you belong in the small ponds with the small minds.[/citation]
Calling someone names doesn't support your argument, or provide you with a credible argument against those you disagree with. What exactly have I said that you find particularly "liberal"? You can pretend that your rant about Mexicans wasn't racist, but you were the one that brought it up in the first place. If you want to convince anyone, you'll need more than generic talking points. Address the facts directly, if you have the courage to.
#### idiom
##### Distinguished
[citation][nom]10tacle[/nom]I call 'em like I see 'em. Nothing personal.[/citation]
Do you know who else calls people names when they're not getting their way? Small children.
If you want to sit at the grown up table and talk about national policy, then you need to use your grown up words.
Calling people names doesn't support your argument.
[citation][nom]10tacle[/nom]Uhm, no, them's the facts. You need to learn basic math and percentages (and how not to add them up in an attempt to debate). http/www.ntu.org/tax-basics/who- [...] taxes.html[/citation]
Thank you for providing your source. You seem to be confused by what these numbers actually mean. Since the total percentages do not have a sum of 100, it is implied that these numbers refer to a percentile and not a percentage. The term percentile and the related term percentile rank are often used in descriptive statistics; so the 10th percentile is the value (or score) below which 10 percent of the category (in this case, annual income) is determined.
There are actually a number of ways to calculate this. Sadly, NTU didn't provide a description.
The most common percentile is the "Nearest Rank" method: p-th percentile of N ordered values is obtained by first calculating the rank:
n = (N/100)p + 1/2
In this case, rounding to the nearest integer, and taking the value that corresponds to that rank.
This is the simplest for of a percentile, but there is also weighted and Linear Interpolation Between Closest Ranks methods that are common. Since there are several values needed to calculate an income tax percentile, the Linear interpolation method was most likely required.
But why?
Well for one thing, not everyone within our population has an income, or an income which is taxable. Children, the unemployed, and even those that work in fields where their income is considered non-taxable. In 2007 I was in a combat zone and was exempt from paying income tax.
Yeah, that's right. This person that you so charmingly characterized as a "Libtard" is a combat veteran and lifetime member of the NRA. But you'd find it much more convenient to dismiss me as some dope smoking hippie or something instead of considering, that maybe, just maybe you don't have all of the data needed to determine something as complex as tax policy.
So what data is missing besides non-taxable income earners?
Well, the NTU also fails to specify what the actual p value is for a given income. How many people actually earn more than say, \$380,000 matters a great deal. If more people earn this amount than those earning \$100,000 for example, both could be taxed at an equal rate, yet the percentile would deceptively indicate that the person earning \$380,000 actually pays a higher income tax, when in reality it would just be in reference to the population that earns that amount.
Fortunately, we don't have to guess, because the data is available, thanks to the wonders of the government-funded world wide web.
http/www.irs.gov/pub/irs-drop/rp-09-50.pdf
Look at page seven:
There are six tax brackets ranging from %10 - %35, and in four categories: Single, Married Jointly, Married Separate, and Head of Household.
There is another side to this: Capital Gains Tax. But you may or may not know about that, so I'll leave it alone for right now.
Income is taxed at a given bracket only for each dollar earned within that bracket. So for example, if a single taxpayer earned \$10,000 in 2009 they would be taxed 10% of each dollar earned from the 1st dollar to the 8,350th dollar (10% × \$8,350 = \$835.00), then 15% of each dollar earned from the 8,351st dollar to the 10,000th dollar (15% of \$1,650 = \$247.50), totaling \$1,082.50. This amount is lower than if the individual had been taxed at 15% on the full \$10,000. This is because the individual's marginal rate has no effect on the income taxed at a lower bracket. This ensures that every rise in a person's pre-tax salary results in an increase of his after-tax salary.
Are you still with me? Do you understand why this matters?
It matters because the taxes paid by Steve Jobs, and Bill Gates, are at the same rate as someone that owns a bar or restaurant and takes home more than \$373,000 annually. Though it should be noted that again, the dollars earned up to \$373,000 are all in different tax brackets, and only the amount earned above \$373,000 is taxed at the top rate of %35.
Again, this tax rate is historically low:
http/www.irs.gov/pub/irs-soi/02inpetr.pdf
Take a look at that, in 1952-1953 people earning more than \$400,000 paid an income tax of 92%! A rate nearly 3x that of today. This was during the cold war, when suspicion of communism got people thrown before congress and charged with treason, yet our own federal government was taxing the wealthy at a rate of \$0.89 for every dollar earned over \$400,000 annually. Yet there was no Tea-Party hysterics, no misspelled signs claiming that Ike and Truman weren't "Real Americans". In fact, the 1950's and 1960's were pretty good decades for most Americans, and the economy was growing.
Let's move on, shall we?
[citation][nom]10tacle[/nom]WTF are you talking about? English please - and clear English. Mindless liberal psychobabble won't cut it with me.[/citation]
Is anything you find difficult to understand automatically categorized as "Mindless liberal psychobabble"? I've been very clear in both this and my other comments. If you are finding something difficult to understand, please just ask for specifics and I'll gladly provide you with the best answer that I can. But instead you are just calling me names, like a child does when they are upset. This is a conversation for grown-ups and I expect you to act like one.
[citation][nom]10tacle[/nom]What? RACISM??!! Where in my post? WHERE?[/citation]
Well if the pointed hat and white sheets fit...
I mentioned reading books on the subject of taxes as an alternative to watching Fox News, and challenged you to address a specific case of President Obama raising your taxes. You replied with several all-caps words; complaining about lottery winners inability to retain their wealth, and then accusing Mexican immigrants of taking away business opportunities from citizens born in the United States.
When no-one at the table is discussing an issue, and you choose to insert it; that reflects badly on you. The observation of racism is a direct result of your completely off-topic tangent regarding Mexican immigration, so please don't try to act as though it came out of nowhere when I pointed this out.
[citation][nom]10tacle[/nom]What's under that beanie cap of yours? Am I "racists" because the current president is BLACK and I disagree with his socialistic near Marxist policies??!![/citation]
Could you please tell me what legislation he has signed into law which is categorically "near Marxist"? Words have meaning. I rebutted your comment regarding Mexican immigrants, and you are now acting confused and assuming that I am referring to the race of the president. This feels like a dodge, and not a very good one.
[citation][nom]10tacle[/nom]Go ahead and spit it out. All you psychos on the left can do is claim racism at the drop of a hat. Sorry pal, that dog won't hunt with me. You can return to your Daily Kook, Huff & Puffington Post, and other lib wingnut sites, thanks. It's been fun, but you belong in the small ponds with the small minds.[/citation]
Calling someone names doesn't support your argument, or provide you with a credible argument against those you disagree with. What exactly have I said that you find particularly "liberal"? You can pretend that your rant about Mexicans wasn't racist, but you were the one that brought it up in the first place. If you want to convince anyone, you'll need more than generic talking points. Address the facts directly, if you have the courage to.
G
#### Guest
##### Guest
Ah yes, the politics of pure envy. The hatred of those who are successful (rich). I have to laugh to keep from crying because it's so pathetic. People who think that because someone like Jobs is successful (read:"rich") must mean that 1k, 100k or 1 million people are poor because they don't have HIS money. Whether you like Apple or not, it cannot be argued that they create and market products that many people buy. But the genuine hatred, contempt and envy for people who are successful (again, read: "rich") is so pathetic it makes me cry from what our nation of once hard-working, smart-minded, resourceful people has become.
#### f-14
##### Distinguished
gonna do some pimple math for a couple ppl who either saw something in the wrong light or twisted it to the wrong way.
10tacle was citing IRS figures and those figures were of the 100% of taxes owed what each tax bracket actually paid of the amount the IRS said they owed before deductions and loopholes removed tax liability. so if they owed 1m in taxes they only paid 400,000 of it. taxable income is limited in 2008 to only 380,354 dollars even if it's donald trump and it's his best year ever he only gets taxed on that 380,354 which is why the IRS labeled it adjusted gross income =A.G.I. as specified by the * at the bottom of the chart. that limit was 410,096 in 2007 even if you made a gajizillion dollars you only pay on that amount.
this math is so simple you should have understood it in a glance.
income earned \$380,354 being the max liability you can be taxed on whether your steve jobs, donald trump or bill gates, you may have earned more then that, but taxation stops at 380,354 as the max limit taxable. say the tax rate is 35% at that bracket taxes would be \$133,123.90
of that \$133,123.90 owed to the IRS after deductions and tax loopholes that top 1% of income earners making \$380,354 only paid on average 38.02 percent of the total the IRS said they owed initialy meaning of 133,123.90 owed x 38.02% = \$50,613.71 was what they paid after deductions and loopholes instead of the 133,123.90 the IRS initial said those top 1% income earners owed each and every single one of them initially.
idk how you could screw up numbers and math that bad when it was written and explained by the IRS in terms most people with a high school degree could understand (i say most because those athletes who can't read end up in college or the NFL/NBA with a diploma or teachers who want to just collect a check will graduate anybody for the last 40 years, you can quote the US ARMY on that one).
#### eddieroolz
##### Distinguished
Moderator
By that logic the US should refund all money it took out of Microsoft and Bill Gates, because he was merely living the American Dream too.
Jobs is the worst type of CEO and millionaire out there. Cold heart and no sympathy to any cause. People give Microsoft too little cred for their actions.
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168 | 4,378 | 19,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-10 | latest | en | 0.92882 |
http://mathhelpforum.com/calculus/171940-find-co-ords-parallel-tangent-quad-function.html | 1,526,874,490,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863923.6/warc/CC-MAIN-20180521023747-20180521043747-00558.warc.gz | 199,074,716 | 11,749 | 1. Find Co-ords of parallel tangent to quad function
Looking for the co-ordinates of the tangent parallel to this when x=0.6
2x^3-6x^2+10x+9
so far got - x=0.6 y=13.272 gradient = 4.96 (of original function)
how do i find the parallel tangent and then its co-ordinates.
2. You mean you're looking for the equation of the tangent of the curve?
3. A 2nd tangent parallel to one at (0.6,13.272)
4. Originally Posted by stophe73
Looking for the co-ordinates of the tangent parallel to this when x=0.6
2x^3-6x^2+10x+9
so far got - x=0.6 y=13.272 gradient = 4.96 (of original function)
how do i find the parallel tangent and then its co-ordinates.
Originally Posted by stophe73
A 2nd tangent parallel to one at (0.6,13.272)
I don't get this either. You are looking for "the co-ordinates of the tangent parallel to this" y =
2x^3-6x^2+10x+9 at the point (0.6, 13.272)? We can certainly get the tangent line to this curve at x = 0.6. Is this what you are looking for? What do you mean by "2nd tangent?" This is badly worded.
-Dan
5. this function(graph is s shaped 2 knees? turns?) has a tangent to it at co ords 0.6,13.272 with gradient 4.96.
I'm looking for another tangent with a parallel gradient , but different co-ordinates.
6. I can only interpret this as meaning that you are looking for a value of x, other than .6 that has the same slope.
With $\displaystyle f(x)= 2x^3-6x^2+10x+9$, $\displaystyle f'(x)= 6x^2-12x+ 10$. When $\displaystyle x= .6$, that is
$\displaystyle f'(.6)= 6(.6)^2- 12(.6)+ 10= 4.96$
so now you are looking for another value of x that satisfies $\displaystyle f'(x)= 6x^2- 12x+10= 4.96$
That is the same as solving $\displaystyle 6x^2- 12x+ 5.04$ and that is especially easy to solve because you already know that x= .6 is a solution: divide $\displaystyle 6x^2- 12x+ 5.04$ by $\displaystyle x- .6$.
7. The question asks for the x,y on the same 'line' where the same gradient / slope occurs. i.e the co-ordinates where a parallel tangent exists.
So how would I carry out it out for the y, co-ordinate? ('original' y by calculation 13.272)
8. From the post of HallsofIvy, you should see that you will get another value of x where a tangent will have the same gradient as the tangent at x = 0.6.
Once you get the other x coordinate, you can find the other coordinate.
9. You mean with y=mx+c , how do i find c for the "2nd tangent" , sorry about the confusing grammar.
10. No, I mean you look for the other value of x which satisfies the equation given by HallsofIvy:
$\displaystyle 6x^2 -12x +5.04 = 0$
One solution is 0.6, as given by your question. There is another one. Find it, then use that x to get the y-coordinate.
11. I see...but how do i find the y co-ord and the intercept?
12. You do just like you did for the first one, you use the value of that new x and plug it in the original equation.
Then, you take the point and the gradient of the tangent to get directly the equation of the line. | 897 | 2,950 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.910004 |
https://www.physicsforums.com/threads/heine-borel-theorem-in-r-n-stromberg-theorem-3-40.1041583/ | 1,716,335,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00234.warc.gz | 851,454,730 | 17,908 | # Heine-Borel Theorem in R^n .... Stromberg, Theorem 3.40 .... ....
• MHB
• Math Amateur
In summary: Yes, I was talking about \mathscr{U} = \{ B_k (0) \ : \ k \in \mathbb{N} \} as given by Stromberg ... but I did not say so explicitly ... so strictly speaking ... yes ... what I said was not correct ... indeed reading literally ... it didn't make sense."
Math Amateur
Gold Member
MHB
I am reading Karl R. Stromberg's book: "An Introduction to Classical Real Analysis". ... ...
I am focused on Chapter 3: Limits and Continuity ... ...
I need help in order to fully understand the proof of Theorem 3.40 on page 104 ... ... Theorem 3.40 and its proof read as follows:
View attachment 9141
In the above proof by Stromberg we read the following:
" ... ... To see that $$\displaystyle S$$ is bounded, consider $$\displaystyle \mathscr{U} = \{ B_k (0) \ : \ k \in \mathbb{N} \}$$. Plainly $$\displaystyle \mathscr{U}$$ is an open cover of $$\displaystyle \mathbb{R}^n$$ and, in particular of $$\displaystyle S$$. Thus there is a $$\displaystyle k_0 \in \mathbb{N}$$ such that $$\displaystyle S \subset B_{ k_0 } (0)$$; whence $$\displaystyle \text{diam}S \leq 2 k_0 \lt \infty$$. ... ... "
I am troubled by the apparent (to me) lack of a rigorous argument to demonstrate that $$\displaystyle S$$ is bounded ...It appears to me that $$\displaystyle \mathbb{R}^n$$ is "infinitely large", that is unbounded ... and hence $$\displaystyle S$$ can be "infinitely large" and hence unbounded ... for example if $$\displaystyle S$$ is the $$\displaystyle x_1$$ axis of $$\displaystyle \mathbb{R}^n$$ then $$\displaystyle S \subset \mathbb{R}^n$$ and $$\displaystyle S$$ is unbounded ...
Can someone please demonstrate rigorously that $$\displaystyle S$$ is bounded ...
*** EDIT ***Just a thought after reflection ... even though open covers of \mathbb{R}^n and S may be "infinite" (in dimension) because S is compact it must have a finite subcover ... and hence S is bounded ... is that the correct argument?***************************************************************************************************Help will be appreciated ...
Peter
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• Stromberg - Theorem 3.40 ... ... .png
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"Because S is compact it must have a finite subcover."
No, that is not correct. In fact, it doesn't make sense. To talk about a "subcover" you must first have specified a "cover". Are you talking about the one given in the argument, $B_k(0)$, the collection of open balls with center at 0 and radius k= 1, 2, 3, ...?
Notice that these balls are each contained in the next so that any point in the union of a finite number of them is in the largest, the largest "k". Since every point is S is some finite distance from 0 every point of S is in one of those so it is an open cover for S.
Since S is compact there exist a finite
subcover of that cover. There are only a finite number of such "$B_k(0)$" so there is a largest "k". S is covered by that single set so the distance between any two points in S is less than 2k.
HallsofIvy said:
"Because S is compact it must have a finite subcover."
No, that is not correct. In fact, it doesn't make sense. To talk about a "subcover" you must first have specified a "cover". Are you talking about the one given in the argument, $B_k(0)$, the collection of open balls with center at 0 and radius k= 1, 2, 3, ...?
Notice that these balls are each contained in the next so that any point in the union of a finite number of them is in the largest, the largest "k". Since every point is S is some finite distance from 0 every point of S is in one of those so it is an open cover for S.
Since S is compact there exist a finite
subcover of that cover. There are only a finite number of such "$B_k(0)$" so there is a largest "k". S is covered by that single set so the distance between any two points in S is less than 2k.
Thanks for the help, HallsofIvy ... ...
You write:
" ... ... No, that is not correct. In fact, it doesn't make sense. To talk about a "subcover" you must first have specified a "cover". Are you talking about the one given in the argument ... "Yes, I was talking about $$\displaystyle \mathscr{U} = \{ B_k (0) \ : \ k \in \mathbb{N} \}$$ as given by Stromberg ... but I did not say so explicitly ... so strictly speaking ... yes ... what I said was not correct ... indeed reading literally ... it didn't make sense ... as you say ...
Thanks again ...
Peter
## 1. What is the Heine-Borel Theorem in R^n?
The Heine-Borel Theorem in R^n is a fundamental result in real analysis that states that a subset of R^n is compact if and only if it is closed and bounded. This theorem is named after mathematicians Eduard Heine and Émile Borel, who independently proved it in the late 19th century.
## 2. How is the Heine-Borel Theorem used in mathematics?
The Heine-Borel Theorem is used in various areas of mathematics, including real analysis, topology, and functional analysis. It is particularly useful in proving the existence of solutions to equations and in establishing the convergence of sequences and series.
## 3. What is the significance of the Heine-Borel Theorem in R^n?
The Heine-Borel Theorem is significant because it provides a necessary and sufficient condition for a subset of R^n to be compact. This allows mathematicians to easily identify and work with compact sets, which have many important properties and applications in mathematics.
## 4. How does Stromberg's Theorem 3.40 relate to the Heine-Borel Theorem?
Stromberg's Theorem 3.40 is a generalization of the Heine-Borel Theorem in R^n. It states that a subset of a metric space is compact if and only if it is sequentially compact, meaning that every sequence in the subset has a convergent subsequence. This theorem encompasses the Heine-Borel Theorem as a special case when the metric space is R^n.
## 5. Can the Heine-Borel Theorem be extended to other spaces?
Yes, the Heine-Borel Theorem has been extended to other spaces, such as metric spaces, topological spaces, and normed vector spaces. These extensions are known as the Heine-Borel Theorem for general spaces, and they provide necessary and sufficient conditions for a subset to be compact in these different types of spaces.
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1K | 1,786 | 6,710 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-22 | latest | en | 0.875942 |
https://math.stackexchange.com/questions/956599/joint-probability-of-iid-samples-where-pdf-is-not-continuous-at-some-points | 1,560,940,916,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998959.46/warc/CC-MAIN-20190619103826-20190619125826-00145.warc.gz | 516,028,121 | 36,056 | # Joint probability of iid samples where pdf is not continuous at some points
Suppose $y_i$ is a random variable generated by sampling $y_i' \sim \mathcal{N}(0,1)$ and setting $y_i = y_i'$ if $c_1 \leq y_i' \leq c_2$, $y_i = c_1$ if $y_i' < c_1$, and $y_i = c_2$ if $y_i' > c_2$. In other words, take a normally distributed sample $y_i'$ and threshold it at some lower and upper limit to give $y_i$. The "pdf" of $y_i$ then looks like a normal distribution for $c_1 < y_i < c_2$, and is $0$ everywhere else except at $c_1,c_2$ where it is some other value.
Let $Y_j \in \mathbb{R}^T$ be a vector of iid samples $(y_1, \ldots ,y_T)$ where each $y_i$ is distributed in this way, and suppose I have a set of vectors $\{Y_j\}, j = 1, \ldots ,N$.
I want to calculate $p(Y_j)$ = $p(y_{j,1},\ldots ,y_{j,T})$ for each $Y_j$ in a way that I can compare this value for different $Y_j$ in a meaningful way. If there was no discontinuity in the pdf of $y_i$ this would be easy - just multiply the pdf of the iid components: $$p(Y_j) = \prod_{i=1}^T p(y_{j,i})$$
I'm confused as to what to do when the pdf of each $y_i$ is not continuous. If I work it out as a product like this, what value do I use for $p(y_i)$ if $y_i = c_1$ or $c_2$? If I use for instance: $$p(c_2) = \int_{c_2}^{\infty} \mathcal{N}(y_i; 0,1) \mathbb{d} y_i$$
This integral could be much lower than $p(c_2 - \epsilon)$ even though the "probability of seeing $c_2$" is much higher than a value slightly less than $c_2$. When I take the product I think the value for $p(y_i = c_2)$ should be larger than for $p(y_i = c_2 - \epsilon)$, otherwise comparing different $p(Y_j$) is meaningless.
My actual situation is that I have a set of measurements $Y_j$ in a file, and need to calculate $p(Y_j)$ for each.
Not really an answer but worth mentioning:
The rv $y_i$ as described in top of your question has no PDF.
If it would have - let's call if $f$ - then $P(y_i\in A)=\int_Af(y)dy$ should be true for each measurable set $A$.
However, if $\emptyset\neq A\subseteq\{c_1,c_2\}$ then the LHS is positive and the RHS is zero.
• Seems like an answer to me, probably even the answer. – Did Oct 3 '14 at 8:41
• What does this mean for $P(Y_j)$? – akxlr Oct 3 '14 at 8:47
• To avoid misunderstandings: what exactly is meant by $P(Y_j)$ (or $p(Y_j)$ in the question)? You are saying that you want to calculate it without mentioning what it is. – drhab Oct 3 '14 at 8:53
• I mean it to be the probability of observing the set of $T$ measurements $y_{j,1}, \ldots ,y_{j,T}$ - i.e. each $Y_j$ is a vector of $T$ iid measurements and I want to work out the probability of observing all of them assuming they are independent. – akxlr Oct 3 '14 at 9:13
• Probability of "observing..." sounds vague. Its probability can only be calculated if "observing..." is some sort of event, i.e. a measurable subset of a probability space. Can you describe this event in clear terms of probability theory? – drhab Oct 3 '14 at 9:23 | 956 | 2,971 | {"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} | 3.5625 | 4 | CC-MAIN-2019-26 | latest | en | 0.889458 |
https://www.clutchprep.com/physics/practice-problems/147264/if-the-force-on-an-object-is-in-the-negative-direction-the-work-it-does-on-the-o | 1,611,606,309,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703644033.96/warc/CC-MAIN-20210125185643-20210125215643-00487.warc.gz | 709,693,070 | 30,698 | Intro to Calculating Work Video Lessons
Concept
# Problem: If the force on an object is in the negative direction, the work it does on the object must be a. positive.b. negative.c. The work could be either positive or negative, depending on the direction the object moves.
###### FREE Expert Solution
Work is given by:
$\overline{){\mathbf{W}}{\mathbf{=}}{\mathbf{F}}{\mathbf{•}}{\mathbf{d}}}$
96% (444 ratings)
###### Problem Details
If the force on an object is in the negative direction, the work it does on the object must be
a. positive.
b. negative.
c. The work could be either positive or negative, depending on the direction the object moves. | 167 | 660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-04 | latest | en | 0.74842 |
https://www.physicsforums.com/threads/limit-on-faradays-cage-rearranging-charges.812912/ | 1,718,786,525,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00723.warc.gz | 823,114,829 | 15,033 | # Limit on Faraday's cage (rearranging charges)
• olu
In summary, the conversation discusses the concept of Faraday's cage and how applying a magnetic field can cause the charges in the cage to rearrange, resulting in a net field of zero inside the cage. However, there is a question about whether it is possible to apply a strong enough field to cancel out all the charges and still have a non-zero net field inside the cage. The conversation also suggests providing specific numbers and orders of magnitude to better understand the situation.
olu
Having a question regarding Faraday's cage,Applying a magnetic field over the cage, will cause the charges in the conducting cage to rearrange, thus causing another field which opposes the first field. This will give a net field inside the cage of zero (this is why we are safe in a car during a lightning).
However, since we have a finite amount of charges in the conductive cage. Is it possible to apply a field strong enough so that all the charges have rearranged to oppose the field, but with the fact that all the charges is not enough charges to cancel the field, and thus giving a net field inside the cage which is not zero?
Oscar
olu said:
we have a finite amount of charges in the conductive cage.
Could you put a number on that? It could help you to get a handle on what you are saying. What field strengths are you considering? If this isn't a fruitless 'irresistible force and immovable object' type question then you need to supply some orders of magnitude for the causes and effects that you are discussing.
## What is a Faraday's cage?
A Faraday's cage is a metallic enclosure that is designed to block external electric fields. It works by redistributing the charges on its surface, cancelling out the electric field inside the cage.
## What is the limit on rearranging charges in a Faraday's cage?
The limit on rearranging charges in a Faraday's cage is that the electric field inside the cage cannot be reduced to zero. This is due to the fact that charges can only be redistributed, not created or destroyed.
## How is the limit on rearranging charges related to the size of the cage?
The limit on rearranging charges is directly related to the size of the Faraday's cage. As the size of the cage increases, the amount of charges that can be redistributed also increases, allowing for a stronger cancellation of the electric field.
## Can the limit on rearranging charges be exceeded?
No, the limit on rearranging charges in a Faraday's cage cannot be exceeded. It is a fundamental law of physics that charges cannot be created or destroyed, only redistributed. Therefore, the electric field inside the cage can never be completely eliminated.
## What factors can affect the effectiveness of a Faraday's cage in blocking electric fields?
The effectiveness of a Faraday's cage in blocking electric fields can be affected by factors such as the thickness and conductivity of the metal used, the size of the cage, and the strength of the external electric field. Additionally, any gaps or openings in the cage can also reduce its effectiveness.
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https://money.com/debt-income-ratio-calculator/?ref=/new-mortgage-refinance-programs/ | 1,638,483,801,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362297.22/warc/CC-MAIN-20211202205828-20211202235828-00049.warc.gz | 433,251,090 | 90,211 | Many companies featured on Money advertise with us. Opinions are our own, but compensation and
Debt-to-income (DTI) ratio compares how much you earn to your total monthly debt payments. Understanding your DTI is crucial if you are thinking about buying a home or refinancing a mortgage. Crunch the numbers with Money’s DTI ratio calculator and find out if you’re ready to apply for a home loan.
Please enter monthly payments for all fields
i
Gross income is the amount you receive before taxes and other deductions.
Money’s calculator results are for illustrative purposes only and not guaranteed. Money uses regional averages, so your mortgage payment may vary.
i
We try to keep our information current and accurate. However, interest rates are subject to market fluctuations and vary based on your qualifications. Calculator results assume a good credit score and factor-in regional averages; your actual interest rate may differ. Calculator results are for educational and informational purposes only and are not guaranteed. You should consult a licensed financial professional before making any personal financial decisions.
Debt to income ratio
0 %
What is Debt to income?
Simply put, it is the percentage of your income that you use to pay your debts. When you're looking to buy a home, most banks are looking for a debt to income ratio of 40% of less.
3What is Today's Rate?
Rate for yesterday Nov 30 was
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## What Is Debt-to-Income Ratio?
The debt-to-income (DTI) ratio is a key financial metric that lets lenders know how much of a borrower’s gross monthly income goes into paying off their current debt. Gross monthly income refers to the sum total of your monthly earnings before taxes and deductions. A low DTI indicates that the consumer is a low-risk borrower while a high one is taken to mean that the person is at a higher risk of defaulting on their debts.
## How To Calculate Debt-To-Income Ratio?
To calculate your debt-to-income ratio, first add up your monthly bills, such as rent or mortgage, student loan payments, car payments, minimum credit card payments, and other regular payments. Then, divide the total by your gross monthly income. This includes wages, salaries, freelance income, overtime pay, commissions, tips and other allowances, etc.
### What is included in debt-to-income ratio?
Your DTI ratio should include all revolving and installment debts — car loans, personal loans, student loans, mortgage loans, credit card debt, and any other debt that shows up on a credit report. Certain financial obligations like child support and alimony should also be included.
Monthly expenses like health insurance premiums, transportation costs, 401k or IRA contributions, and bills for utilities and services (electricity, water, gas, internet, and cable, etc.) are generally not included. However, if you have long-overdue bills for these types of accounts, they might eventually be passed on to a collection agency. The debt may be included in the calculation if that is the case.
There are two types of DTI ratios that lenders look at when considering a mortgage application: front-end and back-end.
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### Front-end ratio
The front-end-DTI ratio, also called the housing ratio, only looks at how much of an applicant’s gross income is spent on housing costs, including principal, interest, taxes, and insurance.
### Back-end ratio
The back-end-DTI ratio considers what portion of your income is needed to cover your monthly debt obligations, including future mortgage payments and housing expenses. This is the number most lenders focus on, as it gives a broad picture of an applicant’s monthly spending and the relationship between income and overall debt.
A general rule would be to work towards a back-end ratio of 36% or lower, with a front-end ratio that does not exceed 28%. Mortgage expenses should not take up more than 28% of your income.
#### When to include your spouse’s debt
Including your spouse’s debt depends on whether you’ll be applying for the mortgage jointly or as an individual. Certain states operate under community property rules. These rules establish that both spouses are under equal obligation to repay debts incurred during the marriage, so it is not an option to exclude a spouse’s debt from the DTI ratio.
States where community property rules apply are:
• Arizona
• California
• Idaho
• Louisiana
• New Mexico
• Texas
• Washington
• Wisconsin
In the rest of the country (including Alaska, which allows couples to opt out of community property rules) common-law rules apply. Couples are not legally obligated to equally share all debt acquired while married. This means they can apply for a loan as individuals and the spouse’s income and debt will bear no influence in the lender’s evaluation.
## How Does Your DTI Ratio Affect You?
Your debt-to-income ratio mainly affects the type of home-related loans you are able to take out. Lenders factor DTI, among other things, into mortgage loans, mortgage refinancing, and home equity products. You can calculate these using our free mortgage calculator.
### DTI ratio and mortgage loans
DTI ratio directly factors into whether a mortgage lender will approve your loan or not. When buying your first home, your DTI is calculated with the estimated payments, taxes, and fees from the purchase. Depending on your credit score, savings, and down payment, lenders may accept higher ratios.
Lender limits can vary considerably, depending on the type of loan and overall financial profile of a prospective applicant, but there are guidelines in place that can serve as a frame of reference. Since the Federal National Mortgage Association (commonly known as Fannie Mae) raised their DTI limit in 2007, the maximum limit for most lenders will not exceed 50%.
Prospective borrowers should strive for a DTI of at least 43%, or the maximum allowed to access Qualified Mortgage loans. These loans comply with federal guidelines that were created to prevent high-risk transactions between lenders and borrowers.
For some examples of what this looks like in real life, let’s look at some lenders:
• Quicken Loans sets their DTI limit at 50% for most of their loans, making an exception for VA loans, for which the DTI ratio limit can go up to 60%. Consider one of the best VA loans if you are considering this type of mortgage.
• Veterans United recommends a DTI of 41% or lower, with mortgage debt included in the DTI calculation. Higher ratios may still be allowed, but borrowers with a DTI of 41% or higher will need to compensate by having a residual income that exceeds Veterans United’s guidelines by at least 20%.
• Better Mortgage offers loans to candidates with a DTI ratio as high as 47%
• Rocket Mortgage sets the limit at 50%
Loans guaranteed by the Federal Government have their own set of limits, as well:
• USDA loans set their limit at 29% for front-end-ratio and 41% for back-end-ratio, but allow each lender to approve candidates with higher percentages if there are compensating factors (such as supplemental income, generous savings, or strong credit history) that vouch for the applicant’s ability to repay.
• FHA loan limits can go up to 50%, but it depends a lot on the strength of other compensating factors, too. A low credit score can mean that your DTI ratio cannot exceed 45% in order to qualify, while a higher credit score will typically allow greater flexibility.
To make sure you get a good deal on your mortgage, pick one of the best mortgage lenders of the year
### DTI ratio and mortgage refinance
Creditors will also consider your DTI ratio when applying for a mortgage refinance. As with mortgage loans, a higher DTI will make it much harder to get approved for refinancing your home loan. Check our refinance calculator to determine if refinancing your mortgage is the right choice for you.
• For cash-out refinance, Chase recommends that consumers have a DTI of 40% or lower.
• Rocket Mortgage states that most lenders prefer consumers which have a DTI of 50% or lower when applying for mortgage refinance.
### DTI ratio and home equity
DTI ratio affects how much of your home equity you can access. In addition to loan-to-value and combined loan-to-value ratios, lenders will consider your DTI when you apply for a home equity loan or line of credit.
Home equity loans have more stringent requirements than mortgages. Borrowers must have a 43% DTI or lower to qualify, in most cases, and some lenders may even require DTIs as low as 36%. Here are some examples:
• Because of the stricter requirements for home equity loans, Quicken Loans recommends that potential borrowers maintain a DTI of 43% or lower.
• Veteran’s United does not impose a maximum DTI ratio for Veterans and military members. However, those with a DTI above 41% may encounter additional financial scrutiny.
• Rocket Mortgage will not offer home equity loans to anyone with a DTI higher than 43%.
### DTI ratio and credit scores
Your DTI never directly affects your credit score or credit report. Credit-reporting agencies may know your income but they don’t include it in their calculations. Your creditworthiness is still factored into your home loan application. However, borrowers with a high DTI ratio may have a high credit utilization ratio — which accounts for 30 percent of your credit score. Lowering your credit utilization ratio will help boost your credit score and lower your DTI ratio because you are paying down more debt.
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## How Can You Lower Your DTI Ratio?
There are several strategies to lower your DTI. The goal is not only to reduce overall debt but also how much you’ll pay on a monthly basis.
1. Start a monthly budget to get a better overview of your spending habits and see where it’s possible to limit spending. If a pen and paper aren’t for you, there are secure budgeting apps that can streamline this process.
3. Target debt with a high ‘bill-to-balance’ ratio to reduce your DTI the most for the least amount of cash paid. Experts also recommend paying off your auto loan before applying for a mortgage.
4. Refinance or consolidate your debt and use balance transfers to lower the interest rate on your loans. If refinancing, We recommend that you choose one of the best mortgage refinance companies of the year.
5. Boost your income with a source of side income, such as renting out a spare room or delivering food.
6. Negotiate a higher salary, if possible, to improve DTI by raising your income.
7. Look into loan forgiveness for federal loans, especially for student loans.
8. Add a co-borrower with a lower DTI to your loan application — but make sure that their credit score doesn’t work against you.
## How COVID-19 Has Affected the Mortgage Market
The Federal Reserve regulates interest rates in response to economic activity. It had been slowly lowering interest rates since 2019 until the outbreak of COVID-19 forced the central bank to move at a more drastic pace, in an attempt to offset the economic impact of the pandemic.
This has represented a big boost to the housing market, with many homeowners taking advantage of the low interest rates to refinance their mortgages. Meanwhile, potential homebuyers are motivated to shop for offers they would not have been able to afford before. Nobody can time the market, but these rates have been projected to remain in place until at least 2023, as part of an effort to re-strengthen the economy.
Mortgage rates remain near historic lows in 2021, which continues to help new buyers break into the market. However, they have been on an upwards trend since the beginning of the year, as predicted by economists.
Monthly house prices have also increased to an average annual rate of 10 percent, further above their already high levels relative to housing rents before the COVID-19 pandemic.
## Debt to Income Ratio FAQ
### Why is the DTI ratio important?
Lenders use DTI ratios to determine whether someone is financially secure enough to manage monthly loan payments. A high DTI ratio is seen as riskier for default. A low DTI ratio indicates that the borrower will be able to make payments even with a hit to their income.
### What’s a good debt-to-income ratio?
While a good DTI ratio should fall between 36% to 43% — the lower, the better. A DTI higher than 43% is seen as a sign of financial stress. While it does not disqualify the borrower, it will make it more difficult to get a good loan offer. | 2,678 | 12,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | longest | en | 0.948933 |
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version 1.0.0.0 (3.62 KB) by
Updated 24 Oct 2013
[m] = conjGrad('calcAx', init, b, options, ...)
provide a minimizer 'm' of the criterion J(x) = 1/2 x^tAx + b^tx as
the solution of the linear system Ax = b, computed by a conjugate
absolutely not restricted, to inverse problems where the criterion
take the form
J(x) = ||y - Hx||^2 + l||Dx||^2, and a algorithm is available to
compute Hx or H^te (x is the unkown, H the direct model, y is data
and D the regularisation.
In these cases, A = 2(H^tH + lD^tD) and b = H^ty.
Anyway, the code is quite general since it accept any function that
compute the product Ax, whatever A.
PARAMETERS
'calcAx' - the name of the function to compute the matrix vector
product Ax. Consequently, it's not necessary to compute and store
matrix A, only the product is necessary (think about the
convolution). The name can be anything, but must correspond to a
callable function. The first argument of 'calcAx' MUST BE the
vector x.
init - the starting point of the optimisation x^(0)
b - the term b = H^t y. Must be provided.
options - the options of the algorithm. See section below.
... - all the remaning argument are passed to the function 'calcAx'
which is call like this calcAx(x,...)
OPTIONS
The variable options is matlab structure. The field are listed
below. Options with '*' are necessary. Others are optional.
thresold (*) - the stoping criterion
maxIter (*) - the maximum number of iteration (the algorithm is
automatically stopped when the iteration equal the dimension of x)
numfig - this option must be a integer. In these case, the current
minimizer and residual are displayed as image of real number in the
figure 'numfig'.
An example is
cgoptions.thresold = 1e-6; cgoptions.maxIter = 50;
Copyright (C) 2013 by François Orieux <francois.orieux@gmail.com>
### Cite As
François Orieux (2021). Preconditionnate conjugate gradient (https://www.mathworks.com/matlabcentral/fileexchange/44074-preconditionnate-conjugate-gradient), MATLAB Central File Exchange. Retrieved .
##### MATLAB Release Compatibility
Created with R2010a
Compatible with any release
##### Platform Compatibility
Windows macOS Linux | 578 | 2,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-31 | longest | en | 0.787945 |
https://excel.tips.net/T002609_Sorting_Letters_and_Numbers | 1,563,852,709,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528687.63/warc/CC-MAIN-20190723022935-20190723044935-00342.warc.gz | 387,444,233 | 10,802 | Please Note: This article is written for users of the following Microsoft Excel versions: 97, 2000, 2002, and 2003. If you are using a later version (Excel 2007 or later), this tip may not work for you. For a version of this tip written specifically for later versions of Excel, click here: Sorting Letters and Numbers.
# Sorting Letters and Numbers
by Allen Wyatt
(last updated January 16, 2016)
Let's say that you have a worksheet in which a particular column contains entries such as F1, F2, F3, etc., all the way up to F149. If you need to sort the data in the worksheet based on the contents of this column, the results may disappoint you. Because the first character in each cell is a letter, Excel sorts the column as text.
The upshot is that the cells are sorted in the order F1, F10, F100, F101, F102, etc. In this arrangement, F2 doesn't show up until the sixty-second entry in the sorted list. The reason this happens is because the cells are treated as text. As text, all the cells starting with F1 (there are 61 of them) come before the cells starting with F2.
The only way around this situation is to make sure that the numbers in the cells are front-padded with zeros. In other words, you shouldn't use F1, but F001. You can use the following formula to convert the old format numbers to the new format:
```=LEFT(C1,1) & RIGHT("000" & RIGHT(C1,LEN(C1)-1),3)
```
Now, when you sort by the newly formatted entries, you get the desired results: F001, F002, F003, etc.
ExcelTips is your source for cost-effective Microsoft Excel training. This tip (2609) applies to Microsoft Excel 97, 2000, 2002, and 2003. You can find a version of this tip for the ribbon interface of Excel (Excel 2007 and later) here: Sorting Letters and Numbers.
##### Author Bio
Allen Wyatt
With more than 50 non-fiction books and numerous magazine articles to his credit, Allen Wyatt is an internationally recognized author. He is president of Sharon Parq Associates, a computer and publishing services company. ...
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What is nine more than 0?
2016-01-17 13:09:19
Nancy
Harris Dimoliatis There are two ways to approach this. One would be to be add a column and use the formula
=IF(LEN(A3)=15,RIGHT(A3,3),RIGHT(A3,2))
(assuming your data is in Column A) Sort all columns by the new column choosing sort anything that looks like a number.
Another way would be to copy the column and parse using text to columns and the deliminter "=". Sort all data by the number column. This will take care of numbers of any length. The first one will need nested if you have numbers less than 2 digits long or more than 3
2016-01-16 06:34:26
Michael (Micky) Avidan
@yaser abu odeh,
Assuming your list resides in column "A" starting in cell A1:
1) In cell B1 type: =MID(A1,9,2) and copy down as long as your list.
2) Sort BOTH columns in ascending order according column "B".
*** Now you can delete the helper column "B".
--------------------------
Michael (Micky) Avidan
“Microsoft® Answers" - Wiki author & Forums Moderator
“Microsoft®” MVP – Excel (2009-2016)
ISRAEL
2016-01-16 00:16:58
yaser abu odeh
how can i sort the following according to the two letters inside:-
61384589TB0439
61388315BG0121
61238348FX0139
61238348RI0139
61388627BS0139
61388627FG0241
61388634BG0121
61874018ET0186
61654324BS0142
61884324ET0141
61888004FX0146
6138321CBG0121
61388003SS0326
2016-01-14 13:54:10
Harris Dimoliatis
I have exactly this problem, but i didn't get the tip.
I have to sort this and i need a "O" after the "="
category_id=168
category_id=169
category_id=17
category_id=171
category_id=173
category_id=174
category_id=175
category_id=176
category_id=177
category_id=178
category_id=18
category_id=182
category_id=183
category_id=184
##### This Site
Got a version of Excel that uses the menu interface (Excel 97, Excel 2000, Excel 2002, or Excel 2003)? This site is for you! If you use a later version of Excel, visit our ExcelTips site focusing on the ribbon interface. | 1,526 | 6,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-30 | longest | en | 0.910187 |
https://ask.sagemath.org/question/40271/problem-with-trig_simplify/ | 1,713,872,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00796.warc.gz | 98,017,533 | 15,806 | # Problem with trig_simplify()
I am testing Sage with basic simbolic expressions appliying derivatives to some functions f(t). When I try to simplify the trigonometric final expressión Sage Notebook does not recognize the partial derivative symbol. I do not understand the new variable psi(t) that appears in the solution.
Here is the Notebook Sage code:
var('t,alpha,beta,gamma');
alpha(t)=function('alpha',t); beta(t)=function('beta',t); gamma(t)=function('gamma',t);
mi(t)=sin(beta)^2*diff(alpha(t), t) + (cos(beta)*diff(alpha(t), t) +diff(gamma(t), t))*cos(beta);
print(mi(t).expand());
print(mi(t).expand().trig_simplify());
Here is the result:
cos(beta(t))^2*diff(alpha(t), t) + sin(beta(t))^2*diff(alpha(t), t) +
cos(beta(t))*diff(gamma(t), t)
cos(beta(t))*gamma(t)*psi(t) + diff(alpha(t), t)
edit retag close merge delete
have you tried psi? ? It will tell you what the psi function is.
( 2017-12-21 23:49:22 +0200 )edit
Notebook tells it is psi(t). But it is amazing!! It is not defined. And seems to susbtitute 'diff(gamma(t),t)' that it is supposed to be the right answer.
Right answer must be: cos(beta(t))*diff(gamma(t), t) + diff(alpha(t), t)
( 2017-12-22 09:22:14 +0200 )edit
Sage just prefers writing out the differential in terms of the digamma function, which is documented to be the logarithmic derivative of the gamma function:
sage: var('t')
t
sage: diff(gamma(t),t)
psi(t)*gamma(t)
sage: psi(1)
-euler_gamma
I'm not sure what you mean by "it is not defined". Are you using a very old version of Sage? In 8.1, the symbol psi in the global scope is bound to the digamma function.
( 2017-12-22 22:03:59 +0200 )edit
I did not want to use mathematical function gamma. I only wanted to define an angle variable. I have just used another name for the symbolic angle variable 'rho' and now it works.
Here it is the code:
var('t,alpha,beta,rho');
alpha(t)=function('alpha',t); beta(t)=function('beta',t); rho(t)=function('rho',t);
mi(t)=sin(beta)^2*diff(alpha(t), t) + (cos(beta)*diff(alpha(t), t) +diff(rho(t), t))*cos(beta);
show(mi(t).expand());
show(mi(t).trig_simplify());
And the expected result:
cos(beta(t))^2*diff(alpha(t), t) + sin(beta(t))^2*diff(alpha(t), t) +
cos(beta(t))*diff(rho(t), t)
cos(beta(t))*diff(rho(t), t) + diff(alpha(t), t)
Thanks Nbruin
( 2017-12-23 12:16:12 +0200 )edit
@ceiar: what version of Sage are you using? (What do you get if you type version() in Sage?)
( 2017-12-26 22:13:52 +0200 )edit
Sort by » oldest newest most voted
There is no need to declare alpha, beta, gamma as symbolic variables before defining them as symbolic functions. So you could simplify the declaration of variables and functions as follows.
sage: t = SR.var('t')
sage: alpha(t) = function('alpha')(t)
sage: beta(t) = function('beta')(t)
sage: gamma(t) = function('gamma')(t)
Then you define:
sage: mi(t) = sin(beta)^2*diff(alpha(t), t) + (cos(beta)*diff(alpha(t), t) +diff(gamma(t), t))*cos(beta)
sage: mi(t)
sin(beta(t))^2*diff(alpha(t), t) + (cos(beta(t))*diff(alpha(t), t) + diff(gamma(t), t))*cos(beta(t))
Then you expand and apply trig_simplify:
sage: mi_e = mi(t).expand()
sage: mi_e_ts = mi_e.trig_simplify()
This is what you get:
sage: mi_e
cos(beta(t))^2*diff(alpha(t), t) + sin(beta(t))^2*diff(alpha(t), t) + cos(beta(t))*diff(gamma(t), t)
sage: mi_e_ts
cos(beta(t))*gamma(t)*psi(t) + diff(alpha(t), t)
The result involves psi, so we investigate what it is. Sage allows you to use blah? to get the documentation of blah. Use this as follows:
sage: psi?
Signature: psi(x, *args, **kwds)
Docstring:
The digamma function, psi(x), is the logarithmic derivative of the
gamma function.
psi(x) = frac{d}{dx} log(Gamma(x)) =
frac{Gamma'(x)}{Gamma(x)}
We represent the n-th derivative of the digamma function with
psi(n, x) or psi(n, x).
EXAMPLES:
sage: psi(x)
psi(x)
sage: psi(.5)
-1.96351002602142
sage: psi(3)
-euler_gamma + 3/2
sage: psi(1, 5)
1/6*pi^2 - 205/144
sage: psi(1, x)
psi(1, x)
sage: psi(1, x).derivative(x)
psi(2, x)
sage: psi(3, hold=True)
psi(3)
sage: psi(1, 5, hold=True)
psi(1, 5)
Init docstring: x.__init__(...) initializes x; see help(type(x)) for signature
File: /path/to/sage-8.1/local/lib/python2.7/site-packages/sage/functions/other.py
Type: function
more
I did not know this way for define symbolic functions. Thanks
( 2018-01-10 10:12:28 +0200 )edit
If an answer solves your question, please accept it by clicking the "accept" button (the one with a check mark, below the upvote button, score, and downvote button, at the top left of the answer).
This will mark the question as solved in the list of questions on the main page of Ask Sage, as well as in lists of questions related to a particular query or keyword.
( 2018-01-11 07:23:20 +0200 )edit | 1,446 | 4,779 | {"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} | 3.3125 | 3 | CC-MAIN-2024-18 | latest | en | 0.789242 |
https://oeis.org/A074985 | 1,723,255,452,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782288.54/warc/CC-MAIN-20240809235615-20240810025615-00825.warc.gz | 338,174,700 | 4,683 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A074985 Squares of semiprimes (A001358). 10
16, 36, 81, 100, 196, 225, 441, 484, 625, 676, 1089, 1156, 1225, 1444, 1521, 2116, 2401, 2601, 3025, 3249, 3364, 3844, 4225, 4761, 5476, 5929, 6724, 7225, 7396, 7569, 8281, 8649, 8836, 9025, 11236, 12321, 13225, 13924, 14161, 14641, 14884, 15129 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Disjoint union of 4th powers of primes, A030514, and squares of squarefree semiprimes, A085986. - M. F. Hasler, Nov 12 2021 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 FORMULA a(n) ~ (n log n/log log n)^2. - Charles R Greathouse IV, Oct 16 2015 Sum_{n>=1} 1/a(n) = (P(2)^2 + P(4))/2 = (A085548^2 + A085964)/2 = 0.1407604343..., where P is the prime zeta function. - Amiram Eldar, Oct 30 2020 EXAMPLE 4 is divisible by 2 (twice) and 4*4 = 16. 6 is divisible by exactly 2 and 3 and 6*6 = 36. MAPLE readlib(issqr): ts_kv_sp := proc(n); if (numtheory[bigomega](n)=4 and issqr(n)='true') then RETURN(n); fi; end: seq(ts_kv_sp(i), i=1..50000); MATHEMATICA Select[Range[200], PrimeOmega[#]==2&]^2 (* Harvey P. Dale, Oct 03 2011 *) PROG (Haskell) a074985 = a000290 . a001358 -- Reinhard Zumkeller, Aug 02 2012 (PARI) is(n)=if(issquare(n, &n), isprimepower(n)==2 || factor(n)[, 2]==[1, 1]~, 0) \\ Charles R Greathouse IV, Oct 16 2015 (PARI) list(lim)=lim=sqrtint(lim\1); my(v=List()); forprime(p=2, sqrtint(lim), forprime(q=p, lim\p, listput(v, (p*q)^2))); Set(v) \\ Charles R Greathouse IV, Nov 13 2021 CROSSREFS Cf. A001358, A085548, A085964. Cf. A030514 (4th powers of primes), A085986 (squares of squarefree semiprimes). Sequence in context: A125240 A050775 A022040 * A229134 A069262 A076956 Adjacent sequences: A074982 A074983 A074984 * A074986 A074987 A074988 KEYWORD easy,nonn AUTHOR Jani Melik, Oct 07 2002 STATUS approved
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Last modified August 9 21:53 EDT 2024. Contains 375044 sequences. (Running on oeis4.) | 871 | 2,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.508369 |
https://math.stackexchange.com/questions/1598318/basis-for-traceless-symmetric-matrices | 1,566,119,028,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00131.warc.gz | 551,260,987 | 30,339 | # Basis for traceless, symmetric matrices?
Consider, for example, the set of of all symmetric, traceless $4 \times 4$ matrices. I'm trying to find a correctly normalized basis for this set. So far, I have
$$s(1)=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(2)=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(3)=\left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right);$$ $$s(4)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right);s(5)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right);s(6)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right);$$ $$s(7)=\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);s(8)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);s(9)=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right);$$ $$s(10)= \frac{1}{\sqrt{6}}\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -3 \\ \end{array} \right),$$
Is this correct? If yes, these are a basis for the $10$-dimensional representation of $SU(4)$, but how do I need to normalize these matrices propely in the convention, where the Dynkin index $Tr(T_a T_b)$ of the lowest-dimensional representation is $\frac{1}{2} \delta_{ab}$?
• The space of symmetric $4 \times$ matrices is $10$-dimensional; the space of traceless symmetric $4 \times 4$ matrices is $9$-dimensional. Note that $$s(10) = -\tfrac{1}{\sqrt{6}}\bigl(s(7) + s(8) + s(9)\bigr).$$That aside, each matrix satisfies $\operatorname{Tr}(T_{a}T_{a}) = 2$, not $\frac{1}{2}$, so it appears each needs to be multiplied by $\frac{1}{2}$. Beyond that, I'm afraid I can't help.... – Andrew D. Hwang Jan 3 '16 at 15:09
• @AndrewD.Hwang Ah of course! It turns out this was my main problem and you solved it. Thanks! – jak Jan 3 '16 at 15:13
• According to Wikipedia, you want traceless Hermitian $4 \times 4$ matrices (a.k.a., the Lie algebra of $SU(4)$). The space of these is $15$-dimensional: The six complex entries above the diagonal are "free", and three of the four real diagonal entries are free. The corresponding basis matrices are $s(1)$, ..., $s(6)$; $i$ times corresponding skew-symmetric matrices; and $s(7)$, $s(8)$, $s(9)$. Again, $s(10)$ is redundant. – Andrew D. Hwang Jan 4 '16 at 16:06 | 1,187 | 2,756 | {"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} | 3.875 | 4 | CC-MAIN-2019-35 | latest | en | 0.494012 |
https://studylib.net/doc/10928637/symplectic-geometry-of-homological-algebra-maxim-kontsevi... | 1,624,327,151,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00247.warc.gz | 496,142,140 | 16,840 | # Symplectic geometry of homological algebra Maxim Kontsevich June 10, 2009
```Symplectic geometry of homological algebra
Maxim Kontsevich
June 10, 2009
Derived non-commutative algebraic geometry
With any scheme X over ground field k we can associate a k-linear triangulated category Perf(X) of perfect complexes, i.e. the full subcategory of
the unbounded derived category of quasi-coherent sheaves on X, consisting
of objects which are locally (in Zariski topology) quasi-isomorphic to finite
complexes of free sheaves of finite rank.
The category Perf(X) is essentially small, admits a natural enhancement
to a differential graded (dg in short) category up to a homotopy equivalence,
and is Karoubi (e.g. idempotent) closed. The main idea of derived noncommutative algebraic geometry is to treat any Karoubi closed small dg
category as the category of perfect complexes on a “space”.
By a foundamental result of A. Bondal and M. Van den Bergh, any
separated scheme of finite type is affine in the derived sense, i.e. Perf(X) is
generated by just one object. Equivalently,
Perf(X) ∼ Perf(A)
for some dg algebra A, where prefect A-modules are direct summands in the
homotopy sense of modules M which are free finitely generated Z-graded Amodules, with generators m1 , . . . , mN of certain degree deg(mi ) ∈ Z, such
that dmi ∈ ⊕j<i A · mj for all i. Algebra A associated with X is not unique,
it is defined up to a derived Morita equivalence.
Some basic properties of schemes one can formulate purely in derived
terms.
Definition 1. Dg algebra A is called smooth if A ∈ Perf(A ⊗ Aop ). It
is compact if dim H • (A, d) < ∞. This properties are preserved under the
derived Morita equivalence.
1
For a separated scheme X of finite type the properties of smoothness and
properness are equivalent to the corresponding properties of a dg algebra A
with Perf(A) ∼ Perf(X). Smooth and compact dg algebras are expected to
be the “ideal” objects of derived geometry, similar to smooth projective varieties in the usual algebraic geometry. For a smooth algebra A the homotopy
category Fin(A) of dg-modules with finite-dimensional total cohomology is
contained in Perf(A), and for compact A the category Perf(A) is contained
in Fin(A). One can define two notions of a Calabi-Yau algebra of dimension
D ∈ Z. In the smooth case it says that A! := HomA⊗Aop −mod (A, A ⊗ Aop ) is
quasi-isomorphic to A[−D] as A ⊗ Aop -module (it corresponds to the triviality of the canonical bundle for smooth schemes). Similarly, in the compact
case we demand that A∗ = Homk−mod (A, k) is quasi-ismorphic to A[D], as a
bimodule (it corresponds for schemes to the condition that X has Gorenstein
singularities and the dualizing sheaf is trivial).
The notion of smoothness for dg algebras is itself not perfect, as e.g. it
includes somewhat pathological example k[x, (1/(x − xi )i∈S ] where S ⊂ k is
an infinite subset. It seems that the right analog of smooth shemes (of finite
type) is encoded in the following notion of dg algebra of finite type due to
B. Toën and M. Vaquie:
Definition 2. A dg algebra A is called of finite type if it is a homotopy
retract in the homotopy category of dg algebras of the free finitely generated
algbera khx1 , . . . , xN i, deg(xi ) ∈ Z with the differential of the form
dxi ∈ khx1 , . . . , xi−1 i, i = 1, . . . , N .
Any dg algebra of finite type is smooth, and any smooth compact dg
algebra is of finite type. It is also convenient to replace a free graded algebra
in the definition of finite type by the algebra of paths in a finite Z-graded
quiver.
A large class of small triangulated categories (including many examples from representation theory) can be interpreted as the categories of
perfect complexes on a space of finite type with a given “support”. In
terms of dg algebras, in order to specify the support one should pick a
perfect complex M ∈ Perf(A). The corresponding category is the full subcategory of Perf(A) generated by M , and is equivalent to Perf(B) where
B = EndA−mod (M, M )op . One can say in non-commutative terms what is
bSupp M of
the “complement” X − Supp(M ) and the “formal completion” X
X at Supp(M ). The complement is given by the localization of Perf(X) =
Perf(A) at M , and is again of finite type. By Drinfeld’s construction, in
terms of dg quivers it means that we add a new free generator hM ∈
2
Hom−1 (M, M ) with dhM = idM . The formal completion is given by algebra C = EndB−mod (M, M )op . E.g. when A = k[x] and M = k with x
acting trivially, we have B = H • (S 1 , k) (the exterior algebra in one variable
in degree +1), and C = k[[x]].
Examples of categories of finite type
Algebraic geometry: For any smooth scheme X the category Perf(X) '
Db (Coh(X)) is of finite type.
Topology: Let X be now a space homotopy equivalent to a finite connected CW complex. Define AX := Chains(Ω(X, x0 )), the dg algebra of
chains (graded in non-positive degrees) of the monoid of based loops in X,
with the product induced from the composition of loops. This algebra is
of finite type as can be seen directly from the following description of a
quasi-isomorphic algebra.
Let us assume for simplicity that X is simplicial subcomplex in a standard simplex ∆K for some K ∈ Z≥0 . We associate with such X a finite
dg quiver QX . Its vertices are vi , i = 0, . . . , K for i ∈ X. The arrows are
ai0 ,...,ik for k > 0, where (i0 , . . . , ik ) is a face of X, and i0 < i1 < · · · < ik .
The arrow ai0 ,...,ik has degree (1 − k) and goes from vi0 to vik . We define
the differential in QX by
dai0 ,...,ik =
k−1
X
(−1)j ai0 ,...,ij · aij ,...,ik
j=1
Then we have to “invert” all arrows of degree 0, i.e. add inverse arrows ai0 ,i1
for all egdes (i0 , i1 ) in X. It can be done either directly (but then we obtain
a non-free quiver), or in a more pedantical way which gives a free quiver.
In general, if want to invert a arrow aEF in a dg quiver connecting verices
E and F , with deg aEF = 0 and daEF = 0, one can proceed as follows.
To say that aEF is an isomorphism is the same as to say that the cone
C := Cone(aEF : E → F ) is zero. Hence we should add an endmorphism
hC of the cone of degree −1 whose differential is the identity morphism.
Describing hC as 2 × 2 matrix one obtains the following. One has to add 4
arrows
−1
−2
h0F E , h−1
EE , hF F , hEF
with degrees indicated by the upper index, with differentials
0
dh0F E = 0, dh−1
EE = idE −aEF · hF E ,
−2
−1
−1
0
dh−1
F F = idF −hF E · aEF , dhEF = aEF · hF F − hEE · aEF .
3
Theorem 1. The quiver QX localized in either way, is dg equivalent to AX .
In particular, if X is space of type K(Γ, 1) then AX is homotopy equivalent to an ordinary algebra in degree 0, the group ring k[Γ]. In particular,
such an algebra is of finite type. In the case char(k) = 0 one can also allow torsion, i.e. consider orbispaces, hence Γ can be an arithmetic group, a
mapping class group, etc.
The full subcategory of finite-dimensional dg modules Fin(AX ) ⊂ Perf(AX )
is the triangulated category of sheaves whose cohomology are finite rank local systems on X. If we invert not all arrows of degree 0 in QX for simplicial
X ⊂ ∆K , we can obtain categories of complexes of sheaves with cohomology constructible with respect to a given CW-stratification, and even more
general categories.
Algebraic geometry II: The last example of a category of finite type
Theorem 2. (V.Lunts) For any separated scheme X of finite type the category Db (Coh(X)) (with its natural dg enhancement) is of finite type.
Morally one should interpret Perf(X) as the category of perfect complexes on a smooth derived noncommutative space Y with support on a
closed subset Z. Then the category Db (Coh(X)) can be thought as the
bZ . It turns out
category of perfect complexes on the formal neighborhood X
that for the case of usual schemes this neighbourhood coincides with Y itself. The informal reason is that the “transversal coordinates” to Z in Y
are of strictly negative degrees, hence the formal power series coincide with
Fukaya categories
Let (X, ω) be a compact symplectic C ∞ manifold with c1 (TX ) = 0
The idea of K. Fukaya is that one should associate with (X, ω) a compact
A∞ Calabi-Yau category over a non-archimedean field (Novikov ring)
X
N ov :=
ai T Ei , ai ∈ Q, Ei ∈ R, Ei → +∞ ,
i
where numbers Ei have the meaning of areas of pseudo-holomorphic discs.
The objects of F(X) in the classical limit T → 0 should be oriented Lagrangain spin manifolds L ⊂ X (maybe endowed with a local system). There
are several modifications of the original definition:
4
• one can allow manifolds with c1 6= 0 (in this case one get only a Z/2Zgraded category),
• on can allow X to have a pseudo-convex boundary (see the discussion
of the Stein case below),
• (Landau-Ginzburg model), X is endowed with a potential W : X → C
satisfying some conditions at infinity (then the corresponding FukayaSeidel category is not a Calabi-Yau one),
• allow X to have holes inside, then one get so called “wrapped” Fukaya
category with infinite-dimensional Hom-spaces.
Fukaya categories of Stein manifolds
The simplest and the most important case is when X is compact complex
manifold with real boundary such that there exists a strictly plurisubharmonic function f : X → R≤0 with f|∂X = 0 and no critical points on ∂X.
Seidel in his book gave a complete definition of the Fukaya category of
Stein manifold in terms of Lefschetz fibrations. The additional data necessary for Z-grading is a trivialization of the square of the canonical bundle.
One can analyze his construction and associate certain algebra A of finite
type (over Z) such that the Fukaya category constructed by Seidel is a full
subcategory of Fin(A). We propose to consider A (or category Perf(A) and
not Fin(A)) to be a more foundamental object, and to formulate all the theory in such terms. For example, for X = T ∗ Y where Y is a compact oriented
manifold, the algebra A is Chains(Ω(Y, y0 )) contains information about the
foundamental group of Y , whereas the category of finite-dimensional representations could be very poor for non-residually finite group π1 (Y ).
Also we propose a slightly different viewpoint on AX . Namely, one can
make X smaller and smaller without changing A, and eventually contract
X to a singular Lagrangian submanifold L ⊂ X. Hence we can say that
A = AL depends only on L (up to derived Morita equivalence). One can
think for example about L being a 3-valent graph embedded in an open
complex curve X as a homotopy retract. If X is endowed with a potential,
we should contract X to a noncompact L such that Re(W )|L : L → R
is a proper map to [c,P
+∞), c ∈ R, e.g. L = Rn for X = Cn with the
holomorphic potential ni=1 zi2 .
We expect that Fin(AX ) is the global category associated with a constructible sheaf (in homotopy sense) EL of smooth compact dg categories on
5
L depending only on the local geometry. In terms of dg algebras, AX is a
homotopy colimit of a finite diagram of local algebras. For example, if L is
smooth and oriented and spin, the sheaf EL is the constant sheaf of Perf(Z),
and the global algebra is the algebra Chains(Ω(L, x0 )) considered before.
In terms of topological field theory, the stalks of EL are possible boundary
terms for the theory of pseudo-holomorphic discs in X with boundary on L.
In codimension 1 singular Lagrangian L looks generically as the product
of a smooth manifold with the union of three rays {z ∈ C | z 3 ∈ R≥0 },
endowed with a natural cyclic order. The stalk of the sheaf EL at such a
point is Perf(A2 ), the category of representations of quiver A2 (two vertices
and one directed egde). The symmetry group of Perf(A2 ) after factoring by
the central subgroup of shifts by 2Z is equal to Z/3Z. Explicitly it can be
done by the following modification of the quiver at triple points. Namely,
consider the quiver with three vertices (corresponding to 3 objects E, F, G),
a closed arrow F → G of degree 0, two arrows E → F, E → G of degreess
−1 and 0 respectively (with differential saying that we have a morphism
E → Cone(F → G). We say that E is quasi-isomorohic to Cone(F → G),
i.e.
Cone(E → Cone(F → G)) = 0 .
This can be done explicitly by constructing a homotopy to the identity of the
above object, which is a 3 × 3-matrix. Combining all equations together we
get a quiver with 3 vertices and 12 arrows which gives a heavy but explicit
finite type model for exact triangles.
A natural example of a Lagrangian submanifold with triple point singularities comes from any union of transversally intersecting Lagarngain
submanifolds Li ⊂ X, i = 1, . . . , k. For any point x of intersection (or selfintersection) we should remove small discs in two branches of Lagrangian
manifolds intersecting at x, and glue a small ball with two collars. The set
of triple points forms a sphere.
Global algebra AL of finite type is Calabi-Yau if L is compact, and not
Calabi-Yau in general for non-compact L. There are many examples of
(compact and non-compact) sungular Lagrangian manifolds such that
Perf(AL ) ' Db (Coh(X))
for some scheme X of finite type over Z (maybe singular and/or noncompact). In the pictures at the end we collected several examples of this
“limiting mirror symmetry”. Categories of type AL one can consider as
“non-commutative spaces of finite type” defined combinatorially, without
6
parameters. Among other examples one can list toric varieties, maximally
degenerate stable curves, etc.
Deforming degenerate Fukaya categories
Let us assume that X is compact, in fact a complex projective manifold,
and take a complement X o to an ample divisor. New manifold X o is Stein,
and can be contracted to a singular Lagrangian L ⊂ X o . The advantage of
X o is that is has no continuous parameters as a symplectic manifold. As
was advocated by P. Seidel several years ago, one can think of F(X) as a
deformation of F(X o ). For example, if X is a two-dimensional torus (elliptic
curve) and X o is the complement to a finite set, then F(X o ) is equivalent
to Perf(Y0 ) where Y0 is a degenerate elliptic curve, a chain of copies of P1 .
In algebraic terms, holomorphic discs in X give a solution of the MaurerCartan equation
b N ov
dγ + [γ, γ]/2 = 0, γ ∈ C • (AL , AL )⊗m
where C • (AL , AL ) = Cone(AL → Der(AL )) is the cohomological Hochschild
complex of smooth algebra AL , and mN ov is the maximal ideal in the ring
of integers in the Novikov field N ov.
Analogy with algebraic geometry suggests that different choices of open
o
X ⊂ X should lead to dg algebras of finite type endowed with deformations over mN ov such that algebras became (in certain sense) derived Morita
equivalent after the localization to N ov. We expect that such a formulation
will handle the cases when the deformed Fukaya category is too small, e.g.
when the mirror family consists of non-algebraic varieties (e.g. non-algebraic
K3 surfaces or complex tori).
The group of connected components of X (with appropriate modications
for the potential/Landau-Ginzburg/wrapped cases) acts by automorphisms
of dg category F(X) over the local field N ov. One can ask whetehr this
group coincides with the whole group of automorphisms. To our knowledge,
there is no counterexamples to it! In principle, one can extend the
P group
by taking the product of X with the Landau-Ginzburg model (Cn , ni=1 zi2 )
which is undistinguishable categorically from a point. So, a more realistic
conjecture is that the automorphism group of Fukaya category coincides with
the stabilized symplectomorphism group. Why anything like this should be
true?
7
There is an analogous statement in the (commutative) algebraic geometry. The group of automorphisms of a maximally degenerating Calabi-Yau
variety Y over a local non-archimedean field K maps naturally to the group
of integral piece-wise linear homeomorphisms of certain polytope (called the
skeleton, and usally homeomorphic to a sphere). The skeleton lies intrisically
in the Berkovich spectrum Y an where the latter is defined as the colimit of
sets of points X(K 0 ) over all non-archimedean field extensions K 0 ⊃ K. The
Berkovich spectrum is a very hairy but Hausdorff topological space, and the
skeleton is a naturally defined homotopy retract of Y an .
We expect that one can define some notion of analytic spectrum for a dg
algebra over a non-archimedean field, and its skeleton should be probably
a piecewise symplectic manifold (maybe infinite-dimensional). For Fukaya
type categories this skeleton should be the original symplectic manifold.
8
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Yokan, prepared from adzuki beans and sugar
You can easily calculate the calorie and nutritional values of Yokan, prepared from adzuki beans and sugar for different amounts (slice) by clicking on the CALCULATE button. You can analyze your nutrition history by adding food into your nutrition diary.
How many calories are in Yokan, prepared from adzuki beans and sugar?
Yokan, prepared from adzuki beans and sugar contains 260 calories per 100 grams. This value corresponds to about 13% of the daily energy expenditure of an adult burning about 2000 calories a day.
Calorie, carbohydrate, protein and fat percentages of Yokan, prepared from adzuki beans and sugar
100 grams of Yokan, prepared from adzuki beans and sugar contains 60.72 grams of carbohydrate, 3.29 grams of protein and 0.12 grams of fat. It consist of 95% carbohydrate, 5% protein and 0% fat. 94% of total calories of the food are from carbohydrate, 5% are from protein, 0% are from fat.
100 grams of Yokan, prepared from adzuki beans and sugar is approximately 20% of daily carbohydrate needs, 7% of protein needs and 0% of fat needs of an adult consuming about 2000 calories of energy a day.
Recommended percentages are 50-55% carbohydrate, 15-20% protein and 20-30% fat. Total nutrition content should be close to these percentages for healthy eating.
Calculate nutritional values of Yokan, prepared from adzuki beans and sugar
Legumes and Legume Products
260 kcal
100 gram(s)
Amount 100 g
Calorie 260 kcal
Carbohydrate, by difference 60.72 g
Protein 3.29 g
Total lipid (fat) 0.12 g
Water 35.45 g
Calcium, Ca 27 mg
Iron, Fe 1.16 mg
Magnesium, Mg 18 mg
Phosphorus, P 40 mg
Potassium, K 45 mg
Sodium, Na 83 mg
Zinc, Zn 0.07 mg
Copper, Cu 0.029 mg
Manganese, Mn 0.14 mg
Selenium, Se 2.3 µg
Vitamin C, total ascorbic acid 0 mg
Thiamin 0.005 mg
Riboflavin 0.004 mg
Niacin 0.057 mg
Pantothenic acid 0.099 mg
Vitamin B-6 0.008 mg
Folate, total 8 µg
Folic acid 0 µg
Folate, food 8 µg
Folate, DFE 8 µg
Vitamin B-12 0 µg
Vitamin A, IU 1 IU
Vitamini A, RAE 0 µg
Retinol 0 µg
Vitamin D (D2 + D3) 0 µg
Vitamin D 0 IU
Fatty acids, total saturated 0.043 g
Fatty acids, total monounsaturated 0.011 g
Fatty acids, total polyunsaturated 0.026 g
Cholesterol 0 mg
Ash 0.42 g
Source USDA
How many calories are in other foods in the food group Legumes and Legume Products?
Some of other foods in the food group Legumes and Legume Products are listed below. You can learn how many calories are contained in food by clicking on the relevant food and reach other nutritional values, especially carbohydrate, protein and fat. | 792 | 2,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-40 | latest | en | 0.836168 |
https://math.stackexchange.com/questions/2147690/rubiks-cube-group-elements-and-their-distance-from-the-solved-position?noredirect=1 | 1,580,263,470,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783621.89/warc/CC-MAIN-20200129010251-20200129040251-00509.warc.gz | 534,325,832 | 32,705 | # Rubik's cube group elements and their distance from the solved position
The Rubik's cube group has 6 2 generators (see EDIT). Let's define the "distance" between two positions dist(A,B) as the shortest sequence of generator moves that takes you from A to B. Now if dist(X,Y) = 1, X and Y can't have the same distance from the solved position.
1. How do I prove this?
2. Is this true for all groups?
3. Does this statement have a name, or does it immediately follow from some well-known theorem?
EDIT
As Robert Chamberlain points out in the comments: a minimal generating set of the Rubik's cube group is composed of 2 elements, not 6 as I previously thought.
• I would like to point out that your questions depend not only on the group, but also the choice of generators. If the set of generators is the whole group for example then $\mathrm{dist}(X,Y)=1$ for any $X,Y$ so the statement would not hold. – Robert Chamberlain Feb 16 '17 at 19:47
• There is a notion of parity in the rubik's cube. If you can show that each generator changes the edge parity or that each generator changes the corner parity, then two positions $X,Y$ with $\mathrm{dist}(X,Y)=1$ will have different edge / corner parity which is impossible if they are the same distance from the solved position. As this isn't a full answer I'll leave it to you or someone else to fill the gaps. – Robert Chamberlain Feb 16 '17 at 19:54
• @RobertChamberlain maybe I wasn't clear enough: by "generators" I mean the elements of a minimal set of generators. There are more than one such sets, but all of them have size 6. – UndefinedBehavior Feb 16 '17 at 20:46
• Actually, the Rubik's cube group has a generating set of size 2. For an example of a group and minimal generating set where this doesn't hold, consider $\mathbb{Z}_5$ as generated by $1$ and 'solved position' $0$. $\mathrm{dist}(2,3)=1$ but $2$ and $3$ are both distance $2$ from $0$. You can check this works for any cyclic group of odd order. – Robert Chamberlain Feb 16 '17 at 20:53
• @RobertChamberlain tbh I didn't know that the Rubik's cube is 2-generated. So my previous comment should read: by "generators" I mean the elements of a minimal set of generators. There could be more than one such sets, but all of them have size 2. Anyway your example is not really correct: as the generator is {1} dist(2,0) is 3, not 2, as you have to "apply" the generator "1" 3 times to go from 2 to 0. – UndefinedBehavior Feb 16 '17 at 21:42
This depends crucially on the choice of generating set; in fact, as I'll show below, there are size-2 generating sets both with and without (a version of) your property. The observation you need is this, which has nothing to do with Rubik's cubes:
Lemma: Suppose $G$ is a finite group, $H < G$ is an index-2 subgroup, and $S$ is a generating set of $G$ that is disjoint from $H$. Then no $g \in G$ can be written both as a product of an odd number of generators and as a product of an even number of generators.
Proof: The product of an even number of generators is in $H$, and the product of an odd number isn't.
In fact, conversely, you can see that if some generating set has the property claimed in the lemma, then the set of $g \in G$ that can be written as a product of an even number of generators must be an index-2 subgroup that contains none of the generators. So index-2 subgroups are the only things that can produce these parity restrictions.
In the case of the Rubik's cube group $G$, let $H$ consist of operations that (ignoring orientations) are even permutations of both edges and corners. (This is the only index-2 subgroup of $G$; you can prove this by showing it equals the derived subgroup.) None of the standard six generators are in this subgroup, so the standard generating set has the desired property.
On the other hand, suppose $S = \{g, h\}$ generates $G$. At least one of the two is outside of $H$; say $g$ is. Then if $h \in H$, we get another generating set $\{g, gh\}$ with both elements outside of $H$. Conversely, if $h \notin H$, then $\{g, gh\}$ is a generating set with $gh \in H$. So given that $G$ has some 2-element generating set (which it does), it must have some with the parity property and some without.
I don't think this is easily provable. Here are some thoughts.
Let $G = \{L,R,U,D,F,B\}$ denote the set of generators. Let $S$ denote the solved position. We will think of $X$ as also denoting the element of the Rubik's group that takes you from the solved position to $X$ (similarly for $Y$). Let $n = dist(S,X)$. Because $dist(X,Y) = 1$, we have $$X = g_n \cdots g_2g_1\\ Y = g_{n+1}g_n \cdots g_2g_1 = g_{n+1}X$$ with $g_i \in G$. Now, suppose that $dist(S,Y) = n$. We then have $$Y = h_n\cdots h_2h_1$$ for $h_i \in G$. Thus, we have $$X = g_{n+1}^{-1}h_n\cdots h_2h_1$$ Now, we have $$S = g_n\cdots g_2g_1h_1^{-1}h_2^{-1} \cdots h_n^{-1}g_{n+1}$$ Now, this is what I think is the missing ingredient:
Conjecture: If $k$ is odd, we cannot have $g_k \cdots g_1 = S$ with $g_1 \in S$. That is: all relations on the generators preserve parity of length.
If the above were true, we would have reached a contradiction.
Edit: Based on Chamberlain's link, it suffices to note that any generator switches the "parity" of the edges (and simultaneously, that of the corners). With that, we see that any word of odd length on the generators must switch the parity, and can therefore not be the identity element. | 1,493 | 5,428 | {"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} | 3.546875 | 4 | CC-MAIN-2020-05 | latest | en | 0.924146 |
https://metanumbers.com/19280 | 1,721,022,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00347.warc.gz | 350,496,200 | 7,525 | # 19280 (number)
19280 is an even five-digits composite number following 19279 and preceding 19281. In scientific notation, it is written as 1.928 × 104. The sum of its digits is 20. It has a total of 6 prime factors and 20 positive divisors. There are 7,680 positive integers (up to 19280) that are relatively prime to 19280.
## Basic properties
• Is Prime? no
• Number parity even
• Number length 5
• Sum of Digits 20
• Digital Root 2
## Name
Name nineteen thousand two hundred eighty
## Notation
Scientific notation 1.928 × 104 19.28 × 103
## Prime Factorization of 19280
Prime Factorization 24 × 5 × 241
Composite number
Distinct Factors Total Factors Radical ω 3 Total number of distinct prime factors Ω 6 Total number of prime factors rad 2410 Product of the distinct prime numbers λ 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 19280 is 24 × 5 × 241. Since it has a total of 6 prime factors, 19280 is a composite number.
## Divisors of 19280
1, 2, 4, 5, 8, 10, 16, 20, 40, 80, 241, 482, 964, 1205, 1928, 2410, 3856, 4820, 9640, 19280
20 divisors
Even divisors 16 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ 20 Total number of the positive divisors of n σ 45012 Sum of all the positive divisors of n s 25732 Sum of the proper positive divisors of n A 2250.6 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 138.852 Returns the nth root of the product of n divisors H 8.5666 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 19280 can be divided by 20 positive divisors (out of which 16 are even, and 4 are odd). The sum of these divisors (counting 19280) is 45012, the average is 2250.6.
## Other Arithmetic Functions (n = 19280)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 7680 Total number of positive integers not greater than n that are coprime to n λ 240 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 2194 Total number of primes less than or equal to n r2 16 The number of ways n can be represented as the sum of 2 squares
There are 7,680 positive integers (less than 19280) that are coprime with 19280. And there are approximately 2,194 prime numbers less than or equal to 19280.
## Divisibility of 19280
m n mod m
2 0
3 2
4 0
5 0
6 2
7 2
8 0
9 2
The number 19280 is divisible by 2, 4, 5 and 8.
• Refactorable
• Abundant
• Polite
## Base conversion 19280
Base System Value
2 Binary 100101101010000
3 Ternary 222110002
4 Quaternary 10231100
5 Quinary 1104110
6 Senary 225132
8 Octal 45520
10 Decimal 19280
12 Duodecimal b1a8
16 Hexadecimal 4b50
20 Vigesimal 2840
36 Base36 evk
## Basic calculations (n = 19280)
### Multiplication
n×y
n×2 38560 57840 77120 96400
### Division
n÷y
n÷2 9640 6426.67 4820 3856
### Exponentiation
ny
n2 371718400 7166730752000 138174568898560000 2664005688364236800000
### Nth Root
y√n
2√n 138.852 26.8145 11.7836 7.19484
## 19280 as geometric shapes
### Circle
Radius = n
Diameter 38560 121140 1.16779e+09
### Sphere
Radius = n
Volume 3.00199e+13 4.67115e+09 121140
### Square
Length = n
Perimeter 77120 3.71718e+08 27266
### Cube
Length = n
Surface area 2.23031e+09 7.16673e+12 33393.9
### Equilateral Triangle
Length = n
Perimeter 57840 1.60959e+08 16697
### Triangular Pyramid
Length = n
Surface area 6.43835e+08 8.44607e+11 15742.1
## Cryptographic Hash Functions
md5 7efcd8c929bcbfd5b016c00a711f5759 a7daf410e4edddd28e91fa193c27ed2a6f7a0c1c c90b545c08aeacc643504c7a0a165cc1517fc246be87fb5ea8572316a2a604c6 97d89f284af7de2e033d26c7a31f2ad49691a5f659095121404e9ef7b6dafa4ae74261994d139dd1f5a7a7e6140be30cb9d3ce6639f340423489481de76cf6a4 b6fc9875fdc549b07e5ebe71da19e247f491b22d | 1,441 | 4,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.824998 |
https://bettersheets.co/formulas/edate | 1,723,019,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00586.warc.gz | 104,386,571 | 8,929 | Formulas > =EDATE()
# How To Use EDATE() Function in Google Sheets
Description
Returns a date a specified number of months before or after another date.
What are the common questions about the EDATE formula?
• How does the formula work? Enter a date and the months before or after.
• What is the syntax of the formula? EDATE(start_date, months)
• Can the formula be used to calculate dates in the past or future? Yes
• Can the formula handle leap years? Yes
• What is the maximum number of months that can be added or subtracted using the formula? There is no known limit
• Can the formula handle non-standard date formats? Your dates must be structured appropriately.
How can the EDATE formula be used appropriately?
The EDATE formula can be used appropriately by following the correct syntax and specifying the input parameters in the correct order. The formula can be used to calculate due dates, maturity dates, expiration dates, and other dates based on a specific number of months in the future or past.
How can the EDATE formula be commonly mistyped?
The EDATE formula can be mistyped by using incorrect syntax or input parameters. For example, the formula may not work if the cell references are not correct or if the number of months is not entered as a valid number.
Also the EDATE formula can be mistyped as EDITE, or EDTE, or DATE, or EDATES, or EDITES, or EDUTE, or DATEE, or DATEAFTER or BEFOREDATE, or "E DATE".
What are some common ways the EDATE formula is used inappropriately?
The EDATE formula may be used inappropriately if it is used to calculate dates based on an incorrect number of months or if the formula is not applied consistently across a range of cells. Additionally, the formula may not be appropriate if the date range spans multiple eons or if the date format is not standard.
What are some common pitfalls when using the EDATE formula?
Some common pitfalls when using the EDATE formula include:
• Incorrect syntax or input parameters
• Not using the formula consistently across a range of cells
• Not accounting for leap years or non-standard date formats
• Not specifying the date range correctly
What are common mistakes when using the EDATE Formula?
Common mistakes when using the EDATE formula include:
Entering the number of months as a text value instead of a numeric value
Not using the correct cell references for the input parameters
Using the formula on a range of cells with different date formats
What are common misconceptions people might have with the EDATE Formula?
One common misconception about the EDATE formula is that it can be used to calculate dates based on a specific day of the month (e.g., the 15th of each month). The EDATE formula only calculates dates based on a specific number of months in the future or past, so it cannot be used to calculate dates based on a specific day of the month. Another misconception is that the formula can handle non-standard date formats, but this is not always the case. It is important to use the correct date format when using the EDATE formula.
How To Actually Use EDATE() in Sheets
`EDATE(start_date, months)`
1Better Sheets Tutorial
How to use the EDATE function to add or subtract months in Sheets
The EDATE function takes a starting date and a number of months as its input. It then increments (or decrements) the date by the number of months you specify. Learn how to use it correctly, as dates require special care when used in formulas. They don't behave as regular numbers do in spreadsheets.
Generate a EDATE() formula for your needs with AI
Whatever you need to do in sheets, you can generate a formula. Use the Better Sheets Formula generator to create a formula for any need. Completely free for members.
Asa
Looking for more help inside sheets get the free Add-on: Asa. Ask Sheets Anything. Go ahead, ask it any problem you migth have. Bring your own APIKEY and generate formulas inside of Google Sheets.
Better Sheets | 843 | 3,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.851634 |
https://gamedev.stackexchange.com/questions/143445/the-rotation-along-z-axis-is-not-changing-to-zero | 1,709,410,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00833.warc.gz | 262,542,332 | 33,998 | # The rotation along z axis is not changing to zero? [duplicate]
Code:
function Start(){
Cursor.lockState=CursorLockMode.Locked;
}
function OnGUI () {
transform.Rotate(0,Input.GetAxis("Mouse X"),0);
transform.Rotate(Input.GetAxis("Mouse Y"),0,0);
transform.rotation.z=0;
}
But the rotation along z axis is not changing to zero.
• OnGUI is called for rendering and handling GUI events, so you should better put this in the Update function anyways. And put the "transform.rotation.z=0;" line in the Start() function as it doesn't have to be called every frame but only once. Jul 6, 2017 at 7:18
• @Shashimee transform.rotation.z = 0 can't just move to Start() — the lines before it can modify z, for 2 reasons that may not be obvious. 1) Rotation axes aren't globally orthogonal: combinations of x&y rotations can yield z rotations (see link above for examples) 2) transform.rotation.z is not a z axis rotation angle. It's one of the imaginary components of a quaternion, representing the z component of the rotation axis scaled by the sine of half the rotation angle. It's not meaningful to change alone. transform.eulerAngles is closer, but still not a universal fix. Rotations are weird Jul 6, 2017 at 23:37
First of all, i believe you meant to use Update instead of OnGui;
Since altering a transform's position/rotation have implicity implications in phisics, every time you alter them, there's a bunch of callbacks inside Unity, mainly, OnTransformChanged()/OnRotationChanged(). So, every time you change the transform, do it only once, that's why Unity will not allow you to assign their values individually like this:
transform.position = new Vector3(0,0,0); // for example, instead of
transform.position.x = 0;
transform.position.x = 0;
transform.position.x = 0;
And you also need to use
transform.rotation = Quaternion.Euler (new Vector3 (bla,ble,0) );
to change a transforms angle. | 452 | 1,908 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-10 | latest | en | 0.841242 |
http://cvsweb.netbsd.org/bsdweb.cgi/src/lib/libm/src/s_atan.c?rev=1.5&content-type=text/x-cvsweb-markup | 1,508,365,184,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823153.58/warc/CC-MAIN-20171018214541-20171018234541-00795.warc.gz | 79,486,343 | 3,383 | Return to s_atan.c CVS log Up to [cvs.NetBSD.org] / src / lib / libm / src
File: [cvs.NetBSD.org] / src / lib / libm / src / s_atan.c (download) Revision 1.5, Wed Aug 10 20:31:55 1994 UTC (23 years, 2 months ago) by jtc Branch: MAIN Changes since 1.4: +7 -11 lines ```Float versions of math functions. From Ian Taylor (ian@cygnus.com), with minor changes by me. ```
```/* @(#)s_atan.c 5.1 93/09/24 */
/*
* ====================================================
*
* Developed at SunPro, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
#ifndef lint
static char rcsid[] = "\$Id: s_atan.c,v 1.5 1994/08/10 20:31:55 jtc Exp \$";
#endif
/* atan(x)
* Method
* 1. Reduce x to positive by atan(x) = -atan(-x).
* 2. According to the integer k=4t+0.25 chopped, t=x, the argument
* is further reduced to one of the following intervals and the
* arctangent of t is evaluated by the corresponding formula:
*
* [0,7/16] atan(x) = t-t^3*(a1+t^2*(a2+...(a10+t^2*a11)...)
* [7/16,11/16] atan(x) = atan(1/2) + atan( (t-0.5)/(1+t/2) )
* [11/16.19/16] atan(x) = atan( 1 ) + atan( (t-1)/(1+t) )
* [19/16,39/16] atan(x) = atan(3/2) + atan( (t-1.5)/(1+1.5t) )
* [39/16,INF] atan(x) = atan(INF) + atan( -1/t )
*
* Constants:
* The hexadecimal values are the intended ones for the following
* constants. The decimal values may be used, provided that the
* compiler will convert from decimal to binary accurately enough
* to produce the hexadecimal values shown.
*/
#include "math.h"
#include "math_private.h"
#ifdef __STDC__
static const double atanhi[] = {
#else
static double atanhi[] = {
#endif
4.63647609000806093515e-01, /* atan(0.5)hi 0x3FDDAC67, 0x0561BB4F */
7.85398163397448278999e-01, /* atan(1.0)hi 0x3FE921FB, 0x54442D18 */
9.82793723247329054082e-01, /* atan(1.5)hi 0x3FEF730B, 0xD281F69B */
1.57079632679489655800e+00, /* atan(inf)hi 0x3FF921FB, 0x54442D18 */
};
#ifdef __STDC__
static const double atanlo[] = {
#else
static double atanlo[] = {
#endif
2.26987774529616870924e-17, /* atan(0.5)lo 0x3C7A2B7F, 0x222F65E2 */
3.06161699786838301793e-17, /* atan(1.0)lo 0x3C81A626, 0x33145C07 */
1.39033110312309984516e-17, /* atan(1.5)lo 0x3C700788, 0x7AF0CBBD */
6.12323399573676603587e-17, /* atan(inf)lo 0x3C91A626, 0x33145C07 */
};
#ifdef __STDC__
static const double aT[] = {
#else
static double aT[] = {
#endif
3.33333333333329318027e-01, /* 0x3FD55555, 0x5555550D */
-1.99999999998764832476e-01, /* 0xBFC99999, 0x9998EBC4 */
1.42857142725034663711e-01, /* 0x3FC24924, 0x920083FF */
-1.11111104054623557880e-01, /* 0xBFBC71C6, 0xFE231671 */
9.09088713343650656196e-02, /* 0x3FB745CD, 0xC54C206E */
-7.69187620504482999495e-02, /* 0xBFB3B0F2, 0xAF749A6D */
6.66107313738753120669e-02, /* 0x3FB10D66, 0xA0D03D51 */
4.97687799461593236017e-02, /* 0x3FA97B4B, 0x24760DEB */
-3.65315727442169155270e-02, /* 0xBFA2B444, 0x2C6A6C2F */
};
#ifdef __STDC__
static const double
#else
static double
#endif
one = 1.0,
huge = 1.0e300;
#ifdef __STDC__
double atan(double x)
#else
double atan(x)
double x;
#endif
{
double w,s1,s2,z;
int ix,hx,id;
GET_HIGH_WORD(hx,x);
ix = hx&0x7fffffff;
if(ix>=0x44100000) { /* if |x| >= 2^66 */
unsigned int low;
GET_LOW_WORD(low,x);
if(ix>0x7ff00000||
(ix==0x7ff00000&&(low!=0)))
return x+x; /* NaN */
if(hx>0) return atanhi[3]+atanlo[3];
else return -atanhi[3]-atanlo[3];
} if (ix < 0x3fdc0000) { /* |x| < 0.4375 */
if (ix < 0x3e200000) { /* |x| < 2^-29 */
if(huge+x>one) return x; /* raise inexact */
}
id = -1;
} else {
x = fabs(x);
if (ix < 0x3ff30000) { /* |x| < 1.1875 */
if (ix < 0x3fe60000) { /* 7/16 <=|x|<11/16 */
id = 0; x = (2.0*x-one)/(2.0+x);
} else { /* 11/16<=|x|< 19/16 */
id = 1; x = (x-one)/(x+one);
}
} else {
if (ix < 0x40038000) { /* |x| < 2.4375 */
id = 2; x = (x-1.5)/(one+1.5*x);
} else { /* 2.4375 <= |x| < 2^66 */
id = 3; x = -1.0/x;
}
}}
/* end of argument reduction */
z = x*x;
w = z*z;
/* break sum from i=0 to 10 aT[i]z**(i+1) into odd and even poly */
s1 = z*(aT[0]+w*(aT[2]+w*(aT[4]+w*(aT[6]+w*(aT[8]+w*aT[10])))));
s2 = w*(aT[1]+w*(aT[3]+w*(aT[5]+w*(aT[7]+w*aT[9]))));
if (id<0) return x - x*(s1+s2);
else {
z = atanhi[id] - ((x*(s1+s2) - atanlo[id]) - x);
return (hx<0)? -z:z;
}
}
``` | 1,850 | 4,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-43 | latest | en | 0.600361 |
https://help.scilab.org/docs/6.0.0/ru_RU/leastsq.html | 1,620,567,011,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988986.98/warc/CC-MAIN-20210509122756-20210509152756-00131.warc.gz | 325,862,518 | 12,450 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
Change language to: English - Français - Português - 日本語 -
See the recommended documentation of this function
# leastsq
Solves non-linear least squares problems
### Syntax
```fopt=leastsq(fun, x0)
fopt=leastsq(fun, x0)
fopt=leastsq(fun, dfun, x0)
fopt=leastsq(fun, cstr, x0)
fopt=leastsq(fun, dfun, cstr, x0)
fopt=leastsq(fun, dfun, cstr, x0, algo)
fopt=leastsq([iprint], fun [,dfun] [,cstr],x0 [,algo],[df0,[mem]],[stop])
[fopt,xopt] = leastsq(...)
[fopt,xopt,gopt] = = leastsq(...)```
### Arguments
fopt
value of the function `f(x)=||fun(x)||^2` at `xopt`
xopt
best value of `x` found to minimize `||fun(x)||^2`
gopt
gradient of `f` at `xopt`
fun
a scilab function or a list defining a function from `R^n` to `R^m` (see more details in DESCRIPTION).
x0
real vector (initial guess of the variable to be minimized).
dfun
a scilab function or a string defining the Jacobian matrix of `fun` (see more details in DESCRIPTION).
cstr
bound constraints on `x`. They must be introduced by the string keyword `'b'` followed by the lower bound `binf` then by the upper bound `bsup` (so `cstr` appears as `'b',binf,bsup` in the syntax). Those bounds are real vectors with same dimension than `x0` (-%inf and +%inf may be used for dimension which are unrestricted).
algo
a string with possible values: `'qn'` or `'gc'` or `'nd'`. These strings stand for quasi-Newton (default), conjugate gradient or non-differentiable respectively. Note that `'nd'` does not accept bounds on `x`.
iprint
scalar argument used to set the trace mode. `iprint=0` nothing (except errors) is reported, `iprint=1` initial and final reports, `iprint=2` adds a report per iteration, `iprint>2` add reports on linear search. Warning, most of these reports are written on the Scilab standard output.
df0
real scalar. Guessed decreasing of `||fun||^2` at first iteration. (`df0=1` is the default value).
mem
integer, number of variables used to approximate the Hessian (second derivatives) of `f` when `algo``='qn'`. Default value is 10.
stop
sequence of optional parameters controlling the convergence of the algorithm. They are introduced by the keyword `'ar'`, the sequence being of the form `'ar',nap, [iter [,epsg [,epsf [,epsx]]]]`
nap
maximum number of calls to `fun` allowed.
iter
maximum number of iterations allowed.
epsg
epsf
threshold controlling decreasing of `f`
epsx
threshold controlling variation of `x`. This vector (possibly matrix) of same size as `x0` can be used to scale `x`.
### Description
The `leastsq` function solves the problem
where `f` is a function from `R^n` to `R^m`. Bound constraints cab be imposed on `x`.
### How to provide fun and dfun
`fun` can be a scilab function (case 1) or a fortran or a C routine linked to scilab (case 2).
case 1:
When `fun` is a Scilab function, its calling sequence must be:
```y=fun(x)
```
In the case where the cost function needs extra parameters, its header must be:
```y=f(x,a1,a2,...)
```
In this case, we provide `fun` as a list, which contains `list(f,a1,a2,...)`.
case 2:
When `fun` is a Fortran or C routine, it must be `list(fun_name,m[,a1,a2,...])` in the syntax of `leastsq`, where `fun_name` is a 1-by-1 matrix of strings, the name of the routine which must be linked to Scilab (see link). The header must be, in Fortran:
```subroutine fun(m, n, x, params, y)
integer m,n
double precision x(n), params(*), y(m)
```
and in C:
```void fun(int *m, int *n, double *x, double *params, double *y)
```
where `n` is the dimension of vector `x`, `m` the dimension of vector `y`, with `y=fun(x)`, and `params` is a vector which contains the optional parameters `a1, a2, ...`. Each parameter may be a vector, for instance if `a1` has 3 components, the description of `a2` begin from `params(4)` (in fortran), and from `params[3]` (in C). Note that even if `fun` does not need supplementary parameters you must anyway write the fortran code with a `params` argument (which is then unused in the subroutine core).
By default, the algorithm uses a finite difference approximation of the Jacobian matrix. The Jacobian matrix can be provided by defining the function `dfun`, where to the optimizer it may be given as a usual scilab function or as a fortran or a C routine linked to scilab.
case 1:
when `dfun` is a scilab function, its calling sequence must be:
``` y=dfun(x)
```
where `y(i,j)=dfi/dxj`. If extra parameters are required by `fun`, i.e. if arguments `a1,a2,...` are required, they are passed also to `dfun`, which must have header
``` y=dfun(x,a1,a2,...)
```
Note that, even if `dfun` needs extra parameters, it must appear simply as `dfun` in the syntax of `leastsq`.
case 2:
When `dfun` is defined by a Fortran or C routine it must be a string, the name of the function linked to Scilab. The calling sequences must be, in Fortran:
```subroutine dfun(m, n, x, params, y)
integer m,n
double precision x(n), params(*), y(m,n)
```
in C:
``` void fun(int *m, int *n, double *x, double *params, double *y)
```
In the C case `y(i,j)=dfi/dxj` must be stored in `y[m*(j-1)+i-1]`.
### Remarks
Like datafit, `leastsq` is a front end onto the optim function. If you want to try the Levenberg-Marquard method instead, use lsqrsolve.
A least squares problem may be solved directly with the optim function ; in this case the function NDcost may be useful to compute the derivatives (see the NDcost help page which provides a simple example for parameters identification of a differential equation).
### Examples
We will show different calling possibilities of leastsq on one (trivial) example which is non linear but does not really need to be solved with leastsq (applying log linearizes the model and the problem may be solved with linear algebra). In this example we look for the 2 parameters x(1) and x(2) of a simple exponential decay model (x(1) being the unknown initial value and x(2) the decay constant):
```function y=yth(t, x)
y = x(1)*exp(-x(2)*t)
endfunction
// we have the m measures (ti, yi):
m = 10;
tm = [0.25, 0.5, 0.75, 1.0, 1.25, 1.5, 1.75, 2.0, 2.25, 2.5]';
ym = [0.79, 0.59, 0.47, 0.36, 0.29, 0.23, 0.17, 0.15, 0.12, 0.08]';
// measure weights (here all equal to 1...)
wm = ones(m,1);
// and we want to find the parameters x such that the model fits the given
// data in the least square sense:
//
// minimize f(x) = sum_i wm(i)^2 ( yth(tm(i),x) - ym(i) )^2
// initial parameters guess
x0 = [1.5 ; 0.8];
// in the first examples, we define the function fun and dfun
// in scilab language
function e=myfun(x, tm, ym, wm)
e = wm.*( yth(tm, x) - ym )
endfunction
function g=mydfun(x, tm, ym, wm)
v = wm.*exp(-x(2)*tm)
g = [v , -x(1)*tm.*v]
endfunction
// now we could call leastsq:
// 1- the simplest call
[f,xopt, gopt] = leastsq(list(myfun,tm,ym,wm),x0)
// 2- we provide the Jacobian
[f,xopt, gopt] = leastsq(list(myfun,tm,ym,wm),mydfun,x0)
// a small graphic (before showing other calling features)
tt = linspace(0,1.1*max(tm),100)';
yy = yth(tt, xopt);
scf();
plot(tm, ym, "kx")
plot(tt, yy, "b-")
legend(["measure points", "fitted curve"]);
xtitle("a simple fit with leastsq")
// 3- how to get some information (we use iprint=1)
[f,xopt, gopt] = leastsq(1,list(myfun,tm,ym,wm),mydfun,x0)
[f,xopt, gopt] = leastsq(1,list(myfun,tm,ym,wm),mydfun,x0,"gc")
// 5- how to provide bound constraints (not useful here !)
xinf = [-%inf,-%inf];
xsup = [%inf, %inf];
// without Jacobian:
[f,xopt, gopt] = leastsq(list(myfun,tm,ym,wm),"b",xinf,xsup,x0)
// with Jacobian :
[f,xopt, gopt] = leastsq(list(myfun,tm,ym,wm),mydfun,"b",xinf,xsup,x0)
// 6- playing with some stopping parameters of the algorithm
// (allows only 40 function calls, 8 iterations and set epsg=0.01, epsf=0.1)
[f,xopt, gopt] = leastsq(1,list(myfun,tm,ym,wm),mydfun,x0,"ar",40,8,0.01,0.1)```
### Examples with compiled functions
Now we want to define fun and dfun in Fortran, then in C. Note that the "compile and link to scilab" method used here is believed to be OS independent (but there are some requirements, in particular you need a C and a fortran compiler, and they must be compatible with the ones used to build your scilab binary).
Let us begin by an example with fun and dfun in fortran
```// 7-1/ Let 's Scilab write the fortran code (in the TMPDIR directory):
f_code = [" subroutine myfun(m,n,x,param,f)"
"* param(i) = tm(i), param(m+i) = ym(i), param(2m+i) = wm(i)"
" implicit none"
" integer n,m"
" double precision x(n), param(*), f(m)"
" integer i"
" do i = 1,m"
" f(i) = param(2*m+i)*( x(1)*exp(-x(2)*param(i)) - param(m+i) )"
" enddo"
" end ! subroutine fun"
""
" subroutine mydfun(m,n,x,param,df)"
"* param(i) = tm(i), param(m+i) = ym(i), param(2m+i) = wm(i)"
" implicit none"
" integer n,m"
" double precision x(n), param(*), df(m,n)"
" integer i"
" do i = 1,m"
" df(i,1) = param(2*m+i)*exp(-x(2)*param(i))"
" df(i,2) = -x(1)*param(i)*df(i,1)"
" enddo"
" end ! subroutine dfun"];
cd TMPDIR;
mputl(f_code,TMPDIR+'/myfun.f')
// 7-2/ compiles it. You need a fortran compiler !
names = ["myfun" "mydfun"]
// 7-4/ ready for the leastsq call: be carreful do not forget to
// give the dimension m after the routine name !
[f,xopt, gopt] = leastsq(list("myfun",m,tm,ym,wm),x0) // without Jacobian
[f,xopt, gopt] = leastsq(list("myfun",m,tm,ym,wm),"mydfun",x0) // with Jacobian```
Last example: fun and dfun in C.
```// 8-1/ Let 's Scilab write the C code (in the TMPDIR directory):
c_code = ["#include <math.h>"
"void myfunc(int *m,int *n, double *x, double *param, double *f)"
"{"
" /* param[i] = tm[i], param[m+i] = ym[i], param[2m+i] = wm[i] */"
" int i;"
" for ( i = 0 ; i < *m ; i++ )"
" f[i] = param[2*(*m)+i]*( x[0]*exp(-x[1]*param[i]) - param[(*m)+i] );"
" return;"
"}"
""
"void mydfunc(int *m,int *n, double *x, double *param, double *df)"
"{"
" /* param[i] = tm[i], param[m+i] = ym[i], param[2m+i] = wm[i] */"
" int i;"
" for ( i = 0 ; i < *m ; i++ )"
" {"
" df[i] = param[2*(*m)+i]*exp(-x[1]*param[i]);"
" df[i+(*m)] = -x[0]*param[i]*df[i];"
" }"
" return;"
"}"];
mputl(c_code,TMPDIR+'/myfunc.c')
// 8-2/ compiles it. You need a C compiler !
names = ["myfunc" "mydfunc"]
// 8-4/ ready for the leastsq call
[f,xopt, gopt] = leastsq(list("myfunc",m,tm,ym,wm),"mydfunc",x0)```
• lsqrsolve — minimize the sum of the squares of nonlinear functions, levenberg-marquardt algorithm
• optim — non-linear optimization routine
• NDcost — generic external for optim computing gradient using finite differences
• datafit — Parameter identification based on measured data
• external — объект Scilab'а, внешняя функция или подпрограмма
• qpsolve — linear quadratic programming solver | 3,466 | 10,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-21 | longest | en | 0.661847 |
https://www.educart.co/cbse/class-12-physics-chapter-wise-lesson-plans | 1,726,296,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00847.warc.gz | 682,868,034 | 30,890 | # Lesson Plan for Physics Class 12 PDF
## Download the Class 12 Physics Lesson Plans PDFs
Are you afraid of physics? Or is it like a horror movie you are afraid to watch? No worries; Educart has come up with lesson plans to ensure a smooth study. Teachers can plan their lectures accordingly with the help of lesson plans, as they are equipped with a time duration for every chapter, avoiding any rush while teaching. Students can incorporate the lesson plan for class 12 physics into their study schedule to have hassle-free learning.
The lesson plan aims to align with the NCERT syllabus, providing a structured approach to cover all essential topics. Creating an effective lesson plan for physics in Class 12 CBSE is crucial to ensuring you grasp complex concepts and succeed in the exams. Whether you are a teacher or a student, this plan will help you organise your lectures.
Download links for chapter-wise lesson plans are provided below. The collection is from multiple sources, such as Diksha Platform, Jawahar Navodaya Vidyalaya, and Salwan Public School. We hope to make the work of school teachers easier.
## CBSE Class 12 Physics Lesson Plans: Syllabus
The Class 12 Physics syllabus contains nine units (14 chapters), and each unit carries a different weight of marks in the exam. Here is a detailed breakdown of the syllabus, along with some of its important topics:
#### Unit 1: Electrostatics
• Chapter 1: Electric Charges and Fields: It includes electric charges, conservation of charge, Coulomb's law force between two point charges, forces between multiple charges, the superposition principle, and continuous charge distribution.
• Chapter 2: Electrostatic Potential and Capacitance: Conductors and Insulators, Free Charges and Bound Charges Inside a Conductor Dielectrics and electric polarisation, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor.
#### Unit 2: Current Electricity
• Chapter 3: Current Electricity: Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility, and their relation with electric current; Ohm's law; V-I characteristics (linear and non-linear); electrical energy and power; electrical resistivity and conductivity; temperature dependence of resistance.
#### Unit 3: Magnetic Effects of Current and Magnetism
• Chapter 4: Moving Charges and Magnetism: Concept of Magnetic Field, Oersted's Experiment Biot-Savart law and its application to the current-carrying circular loop. Ampere's law and its applications to infinitely long straight wires Straight solenoid
• Chapter 5: Magnetism and Matter: Magnetic Properties of Materials—Para-, Dia-, and Ferromagnetic Substances with Examples Magnetization of materials is the effect of temperature on magnetic properties.
#### Unit 4: Electromagnetic Induction and Alternating Currents
• Chapter 6: Electromagnetic Induction: Faraday's Laws, Induced EMF and Current; Lenz's Law, Self and Mutual Induction
• Chapter 7: Alternating Current: LCR series circuits (phasors only), resonance, power in AC circuits, power factor, wattless current. AC generator, transformer.
#### Unit 5: Electromagnetic Waves
• Chapter 8: Electromagnetic Waves: The electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, and gamma rays) includes elementary facts about their uses.
#### Unit 6: Optics
• Chapter 9: Ray Optics and Optical Instruments: Reflection of Light, Spherical Mirrors, Mirror Formula, Refraction of Light, Total Internal Reflection and Optical Fibers, Refraction at Spherical Surfaces, Lenses, Thin Lens Formula, etc.
• Chapter 10: Wave Optics: Wavefront and Huygens Principle, Reflection and Refraction of a Plane Wave at a Plane Surface Using Wavefronts, Young's Double Slit Experiment.
#### Unit 7: The Dual Nature of Radiation and Matter
• Chapter 11: Dual Nature of Radiation and Matter: Dual Nature of Radiation, Photoelectric Effect, Hertz and Lenard's Observations, Einstein's Photoelectric Equation, and the Particle Nature of Light Experimental study of the photoelectric effect
#### Unit 8: Atoms and Nuclei
• Chapter 12: Atoms: Alpha-particle scattering experiment; Rutherford's model of the atom; Bohr model of the hydrogen atom.
• Chapter 13: Nuclei: Composition and Size of the Nucleus, Nuclear Force Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion.
#### Unit 9: Electronic Devices
Chapter 14: Semiconductors: Semiconductor Diode Energy bands in conductors, semiconductors, and insulators (qualitative ideas only) Intrinsic and extrinsic semiconductors—p and n-type, p-n junctions.
## Why use a lesson plan for physics?
Following a structured lesson plan offers numerous advantages for both teachers and students, mainly in subjects like physics. Lesson plans can enhance teaching styles and students’s understanding. Here are some benefits of using the lesson plan for class 12 physics:
• A lesson plan provides structured learning; this structure helps students learn more easily and understand each topic.
• It helps with time management and prevents the rush to finish the syllabus by providing allocated and appropriate time to cover each topic.
• Chapters are broken down into sections and provide detailed explanations and step-by-step solutions to facilitate understanding and help in understanding fundamental concepts.
• With the help of lesson plans, teachers can allocate appropriate time to each topic, avoiding the risk of skipping important topics.
• Instead of rote memorization, lesson plans include active learning strategies such as project-based learning, activities, and resources.
• It will help substitute teachers by following the guidelines in your absence.
## How do I use a lesson plan effectively?
Physics is a complex subject, so for the best preparation, you also require some of the best strategic and effective tips. Let's see some of them:
• Consistency is the key. Regularly solve the problems from both the textbooks and the solutions to improve your understanding of the concept.
• Read NCERT and reference books to clarify concepts.
• Physics requires deep understanding, so focus on understanding the concepts rather than memorising the steps. This will help in solving complex problems easily.
• Studying in groups can help you gain different insights and understand various problem-solving techniques.
• Seek help when needed, and don't hesitate to ask for help from teachers if you encounter difficulties.
• Use a sample paper or the previous year's question paper; this will help you analyse your preparation and also provide you with an idea of the actual question paper.
Lesson plan for physics Class 12 helps students and teachers stay organised; they are equipped with chapter-wise lesson maps and ensure that every student and teacher is equipped with the right study and teaching material.
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## Homework Assignments Online Free
I am trying another different application, and I do not know how to get my web site to work. 2. I have created a new web site for this web, but the page does not work properly. 3. I want to add the new site in the new web site, so I tried to add the forms to the website and then the page is saved, but the formComputer Homework Assignments are a great way to obtain a comprehensive understanding of the requirements for the assignment. The requirements can be well understood with the help of two homology classes: the homology class of a finite set of homology groups, and the homology Class for a set of homologies. The homology class (or homology Class) of a finite and a finite set can be constructed through the use of the homology formulae in the homology classes of a finite group. The homological Formulae are useful for understanding the structure of the sets and the homological Formulas are useful for developing the understanding of homology. A finite set of Homology Forms is more general than a finite group, and it is more general to a set of Homologies than a set of Groups. The homologies can also be constructed using the Homology Formulae, since the Homology Forms are related to the Homology Class of a finite G-group. The Homology Classes of a set of Classes of a finite Set are a generalization of the Homology Classes for a set. The Homological Forms of a set is a generalization to the Homological Forms for a set, but it can be generalized to the HomoHomology Formulism. The HomoHomological Formulism is a formulism of the Homological Formulants that consists of the HomoClass and HomoClass of a set. It is a form of the Homologisms of the Homologic Forms of a Set. The Homologisms are an important part of thehomology Classes for the sets. A set of Homologisms is a set of the Homologies of a set and can be constructed from the Homology Homology of a set through the Homology Fixing and Homological Formulas. An Homology Fixation is the homology Fixing of a set, and an Homological Formula is the homological Fixing of my response set. The homologisms can be made from the Homologism Fixation through the homology Forms of a G-group, but it is not possible to make homologisms from the Homological forms of the set through the homological Forms of the G-group through the Homological Fixing and the Homological Classes of a G Group. Further, the Homologies of the Homologically Forms of a group are not important anymore, but they are not important in the Homology of the G Group. It can be proved that the Homology classes of the members of a G group can be written as the homology Classes of the Homologists of the G G-group and the Homology Groups in the G-Group.
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The Homology Classes are the homology Groups of the Homomorphisms of the G Groups. The Homomorphisms can be seen as the homomorphisms that are homomorphisms of a G G- Group. The Homomorphic Forms can be determined from the homological Classes of the homomorphism from the Homomorphic Forms of the Homlicted Groups. In addition to the Homomorphism of a GG-Group, there are possible HomoHomomorphisms of GG-Groups. The Homologically Forms are the Homological Homological Forms, defined by the HomologicalForms of a G–G–Group. Homological Forms can be defined by the homological HomologicalFormulae of the Homomorphic Homological Forms. A Homology Class is an algebraic class of a set whose members are homomorphism of two sets. The Homologists of a Homology Class are the Homologists of the Homologist of the G–G-G–Group, the Homological Functions of the Homitored Homology Classes, and the Homologizing Functions of the G‐G-G-Group. The Homotopy Classes of the Group of Homological Functions are the Homotopy Class of the Homotopic Homological Functions. There are several ways to discover the Homology Categories of a set: The homology Classes can be used to determine the Homology Formulas of a G – G–G – G–Group. The Homologic Forms can be used for determining the Homological Formulas for a G–Group, and can be used in the Homological Framework of the G – G – G Group. Algebraic Homology Classes An algebraComputer Homework Assignments D. V. Mathers In my first year of college, I was hooked on a student application and a few things. I’ve moved on and I’m glad I did. I was the only undergrad student in the program at the time, but the other sophomore students were in the same class. I did not have a lot of experience in the field of computer science. My computer science course taught me a lot about the subject. I knew a lot of the math was confusing and I had a lot of trouble with it. I knew that I could use a little help with some of this stuff, but I didn’t know a lot.
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## Coding Lesson Plans For Elementary
In the fall of 2004, I was trying to get into a computer science course. I wanted something different. I wanted it to be more of a computer science study. I wanted more information, but not much of it. I wanted my computer science classes to have more knowledge. I wanted click resources class to have a lot more work and more information. It was a really good day. I was very happy with my class. I was able to get into the class. I think I was pretty good at it. On the way to the class, I ended up having to go to my first computer science class. I didn”t get to see any of the | 1,639 | 7,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-49 | latest | en | 0.939545 |
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## Mathcad 15 Getting Vectors From For Loops
So I have a hw problem that doesn't need to be solved this way but I have a feeling I will be using the Modified bishops method on a future project so I wanna take my time on this. Basically, I have two equations :
This is a function to find the Factor of Safety for a slope under certain conditions. In the actual psi equation, the F is The same FS in the upper eq. This requires an iterative process where you guess a F run it through psi and then the FS function. You then compare F to the resulting FS to see how close they are. You continue guessing F until F and FS converge to a similar number. The summations with the variable I represent the 7 different vertical slices that are being summed to form the whole slope. I want to write a program where I can input F as a sequence like 1,2 .. n. It then should pick the first F and run it through psi summing over the 7 slices. This should output one character into the psi resultant vector. Overall once it has run through all F, I want the resultant psi to be an nx1 vector with each psi value pertaining to each F. Then I want it to run the FS equation, similar to the psi I want this to sum over the seven slices for each psi input. At the end, I should have a resulting FS vector with one FS value for each F value input. Here is what I have so far:
I am an absolute beginner with coding so any help would be greatly appreciated!
6 REPLIES 6
21-Topaz I
(To:AS_10102578)
Hi,
Can you please include the sheet with the posting.
Cheers
Terry
24-Ruby II
(To:AS_10102578)
21-Topaz I
Hi,
We need to know the values of "l", "c", "w", "u", phi, and alpha to help you.
Cheers
Terry
21-Topaz I
(To:AS_10102578)
Hi,
Set up a solve block that will iterate to a solution automatically.
23-Emerald II
(To:AS_10102578)
@AS_10102578 wrote:
So I have a hw problem that doesn't need to be solved this way but I have a feeling I will be using the Modified bishops method on a future project so I wanna take my time on this. Basically, I have two equations :
T At the end, I should have a resulting FS vector with one FS value for each F value input. Here is what I have so far:
I don't have your version of Mathcad (I've only got Prime Express 7). However, I can make a few comments on the code you've shown.
You are expecting ψF to be a vector and use it as such in the 2nd expression for FS. However, ψF doesn't exist as far as FS is concerned. ψF is a local variable within the first FS expression and doesn't exist outside of it. So what you needed to have done was assign the result to variable ψF instead of FS - this ψF is a different object to the ψF defined within the first expression.
A second problem you've got is that the ψF in the 1st expression is *not* a vector. All the expression is doing is assigning the summation result to ψF., overwriting any previous values. Consequently, the first FS simply has the value of ψF for F = 6. You need a second iterator to assign the results to a new element in ψF (you could use F, but only if any future F you have in mind continues to pull its values from an integer range).
The third problem is that you need an iterator (range variable or for/while loop) in the second expression to select each value of ψF; otherwise, you're just dividing the (intended) vector ψF into the top summation expression.
Unfortunately, my version of Mathcad doesn't allow programming, but I can show a (crude) version of your code that uses functions.
Often, but not always, you can use vectorize operator to carry out parallel operations without using an iterator. Occasionally, you can write an expression so that Mathcad automatically iterates over an expression.
Stuart
23-Emerald III
(To:AS_10102578)
Apart from your iteration proposal and the solve block proposed elsewhere you can also do:
If you equate this to FS, you get:
So eventually:
If you define:
You can solve that simply with the Mathcad's built-in root() function:
But of course only if the vectors l, u, w alpha, and the values phi and c are known.
Success!
Luc
P.S. Of course you could also define the function f as:
but this function has a higher risk of running into errors because FS is the denominator of a ratio that fails if FS becomes 0.
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Other Versions
Spanish
Quantifying airflow through lumber stacks. April 2, 2002
Question
Can you predict the airflow through lumber stacks, if you know the kiln size, fan outputs, etc? There must be a formula for this. How does static pressure enter into the calculations?
Forum Responses
From contributor N:
The first step is to determine the fan performance. Since we are pushing air through a lumber stack, we encounter some resistance, which increases as the airflow increases. This is called static pressure. For my calculation, I normally use 1/2", but I know most people use a value between 1/4"-3/4". You can obtain these fan performance graphs from the manufacturer.
For the purpose of this example, we will use a kiln that is 27' wide and a stacking height of 12'. As a rule of thumb, half the "window" is lumber and half is free air on 4/4" lumber. We will be trying to get about 400 feet per minute airflow.
Window of open area: 27'x12'/2= 162 ft2
Air volume needed for 400 ft/min.: 400x162 = 64800 ft3
Our 42" fans will provide you with 20000 ft3/minute at 1/2" static pressure. Three fans should be sufficient for an airflow of nearly 400 feet per minute.
Total air volume from three 42" fans: 60000 ft3
Estimated airflow with three 42" fans: (60000/162)= 370 ft/min.
From contributor B:
This example only gets you to the theoretical air velocity through the sticker openings. It assumes your piles are perfectly piled, even ended, and the same width as the kiln, and that all the air goes through the sticker openings from one side of the load to the other.
In reality, you must allow for some losses, where air goes around or under the load. Even in a well-baffled kiln, there will be some losses around the ends of the boards. And if you have a pile of random length wood, or bundles or differing lengths, these can be even higher.
When we calculate estimated air velocity, we might assume as high as 30% losses, depending on the load configuration the customer is expecting. In a tiny kiln, your numbers will be better. In a large commercial kiln, these losses become a more important part of the equation.
From contributor D
Contributor B is right. I usually use 40-50% allowance depending on the product and stacking practices. One thing to remember - the velocity you need is related to the depth of the pile. For example, you don't need nearly as much velocity when you are blowing through 4' of 8' of lumber as when blowing through 20'. Also, species has a big effect.
From contributor N:
50% was a rule of thumb.
If one is to consider 4/4" lumber as 1 1/8" with 3/4" stickers, the actual opening is 40% of the face. (075 / (1.125+.75) = 40%.
This means that there is included 10% extra allowance for additional opening. Or in other words, 25% losses (25% of 40% is 10%). So I believe contributor B and I almost agree on the calculation.
I can only agree that the air velocity required will depend on the species, thickness, and depth of the kiln charge.
Although this may be controversial, I like to install a little extra fan capacity and then either install frequency drive (generally only affordable on larger kilns) or operate the fans in intervals. We have many customers who believe the interval-operated fans reduce energy consumption and improve drying quality.
Contrary to the first answer, you do not need to establish fan performance first. The first thing is to establish the airflow required or suggested for a particular species and thickness. This can range from 250 fpm for species like oak, 4/4 to 8/4, up to 600 fpm for hard maple 4/4-6/4, and even higher for many softwoods.
Next step is to figure out the area of the sticker openings that you have. Multiply the height of the sticker (inches) times the length of the lumber (feet). If you have several stacks, end to end, like both an 8' and 12' stack to give 20' overall, use 20'. Then divide the answer by 12 to get it into square feet. Then multiply this answer times the number of sticker openings. This gives the sticker area.
Next multiply the total area times the velocity to obtain cubic feet per minute GOING THROUGH THE LOAD.
As mentioned, there will also be some losses through the 4x4s, around the ends of the load, and so on. Therefore, you need to account for this loss as well. In a well-loaded kiln, you would have to add another 50% to the original cfm value.
Gene Wengert, forum technical advisor
From the original questioner:
My size is 18x12 area = 216-2 = 108. I not sure of my fan flow with my nile. Should there be an extra fan?
From contributor D:
18' is marginal with the standard two fans. It would be okay if you always dried air dried or thick stock. If you dry green lumber, especially stuff like maple or pine, a third fan would be a good idea, though when doing those species we recommend a smaller load anyway, so if you stack right, you can be fine with two fans. How's that for fuzzy?
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# Beginning in 1997, high school seniors in State Q have been
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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08 May 2014, 02:23
Beginning in 1997, high school seniors in State Q have been required to pass a comprehensive proficiency exam before they are allowed to graduate. The exam requirement was intended to ensure that a minimum level of academic quality will be achieved by the students in the state. In 1997, 20 percent of the seniors did not pass the exam and were, therefore, not allowed to graduate. In 1998, the number of seniors who passed the exam decreased by 10% from the previous year.
The argument above, if true, LEAST supports which of the following statement.
Given:
In 1997 : 100 STUDENTS = 80 PASSED | 20 FAILED
In 1998 : X students = 72 PASSED | X - 72 FAILED (X CANNOT BE LESS THAN 72)
A. If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period.
72 = 80/100 X = 90 thus X should be less than 90, A supports the argument.
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
X must be greater than 90 to have number of seniors passed below 80% in 1998. number of high school senior increased or decreased can't say.
72---90--(X)--100--(X)-- value of X can lie anywhere above 90.
C. Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent.
72---(X)--90--(X)---100 Condition unless the number of seniors lower than 100 in 1998 then 1997... the number of seniors passed in 1998 was lower than 80%... lets say if X is between 90 - 100 pass percentage will be below 80%. e.g. 72 passed out of 95 = pass percentage 75%; if X is below 90 say 80 then 72 passed out of 80 = 90% passed.
Thus conditions does not support both possibilities.
D. If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
Number of high school seniors failed in 1997 = 20 decreased by more than 10%: means number of high school seniors failed in 1998 less than 18.
in 1998 72 passed + (less than 18 failed = 17 failed) = 89 max limit. 72/89 * 100 = 80.89% approx. if we further reduce failed between 0 - 17 inclusive percentage will increase. thus this option supports the argument.
E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
72 = 70/100 X
X = 102.85 Thus X must be greater than 102.
e.g. if I take X 69%
72=69/100 X
X= 104
SIMILARLY for any percentage below 70 to keep passed student equal to 72. I will have to increase the value of X.
Therefore number of high school seniors in 1997 (100) < number of high school seniors in 1998 (102++)
Therefore E LEAST supported among all.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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13 Aug 2014, 04:25
I think it is better to reduce the percentages to absolute numbers in case of year 1997.
It makes things easier to contradict:
In 1997:
Total = 100
Pass = 80
Fail = 20
In 1998:
Total = X
Pass = 72 (= 90% of 80 = 72% of total students in 1997)
Fail = ??
In case of option B, just assume the total number of students in 1998 remained same, ie, 100. We see that the percentage of students who passed has dropped (to 72% in 1998) WITHOUT an increase in number of students as statement B supposes. However, B is still feasible. If we take total number of students in 1998 as 1000, B holds true.
You can work out the other options, and they come out as true.
I take that back.
Pass in 1998 = 72% of total students in 1997 ...............................(a)
According to E, Pass in 1998 < 70% of total students in 1998
Let Pass in 1998 = 70% of total students in 1998 = 0.7X
From (a),
0.7X = 0.72 * 100
X > 100, ie, X HAS to be greater than 100.
=> E is not supported at all.
Last edited by gaurav90 on 03 Nov 2014, 06:15, edited 1 time in total.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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24 Mar 2015, 09:03
omerrauf wrote:
Actually I see the light here. Sorry for my earlier posts. Come to think of it, this was pretty straightforward. It is highly confusing because you have to go through all the optins in any case in a question like this, since it asks for the least possible. So many numbers are being talked about and randomly, so I guess this was a reall cracker.
But here is the reason that I think. Here is how:
From Question Stem:
"In 1998, the number of seniors who passed the exam decreased by 10% from the previous year". This is the key statement. We already know that the number of students is less in 1998 than in 1997. Just re-read the statement and it is obvious.
Now Option B says:
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
We already know that the number of student decreased so B is totally out of the question, the least likely.
Can you explain why "the number of seniors who passed the exam decreased" means the TOTAL number of students is also less? It is not necessarily true.
Let's give a number and play it out based on premise
1997
Total no of students: 100
no of students who passed: 80
% of students who passed: 80%
no of students who failed: 20
% of students who failed: 20%
1998
Total no of students: ? (because we don't know this from stem)
no of students who passed: 72 (-10% than 1997)
if ? = 150 ( which is more than 100 in 1997)
% of students who passed = 72/150 * 100% = 48% (much lesser than 80%)
HENCE, the paragraph SUPPORTS B = If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period. Answer cannot be B IMO.
to test what you have said, let's decrease it
if ? = 90, % of students who passed = 80%
if ? = 80, % of students who passed = 90%
(as you can see as the number of students decreases, the % of students who passed increases, this actually SUPPORTS A and it is the opposite of your claim that if the number of student decreases, the % should increase NOT decrease. "In 1998, the number of seniors who passed the exam decreased by 10% from the previous year. This is the key statement. We already know that the number of students is less in 1998 than in 1997. Just re-read the statement and it is obvious."
I still go with E. Can someone give a better explanation why OA is B? I am not convinced!
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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13 Jun 2015, 00:28
Beginning in 1997, high school seniors in State Q have been required to pass a comprehensive proficiency exam before they are allowed to graduate. The exam requirement was intended to ensure that a minimum level of academic quality will be achieved by the students in the state. In 1997, 20 percent of the seniors did not pass the exam and were, therefore, not allowed to graduate. In 1998, the number of seniors who passed the exam decreased by 10% from the previous year.
The argument above, if true, LEAST supports which of the following statement.
A. If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period.
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
C. Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent.
D. If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
Pre think assumption More students graduated in 1998 from 1997 provided the total no.of students did not change from 1997 to 1998
[color=#6ecff6][color=#ffffff]A. If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period-[color=#ff0000]This supports the argument.For e.g lets say the no. of students in 1997 was 1000 and the percentage passed was 80% then 800 students passed in 1997.In 1998 lets say the no.of students was 900 and the percentage passed was 90% then 810 students passed hence the number of passed students increased in 1998 but the total no.of students has decreased.
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period-This clearly contradicts the argument by saying the percentage of students passed has decreased in 1998 from 1997.This provides least support hence could be the correct answer
[color=#ffffff]C. Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent
[/color][/color]-This scenario can play out if the number of students are equal or higher.
D. If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent-This could support the case if the total no.of students in 1997 and 1998 were equal say 1000 then the no.of students who passed in 1997 will be 800 and in 1998 will be 900 which is greater than 80%
E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.-Suppose the total no.of students in 1998 was 1000 and less than 70% passed e.g 690(69%) then it could be possible that the total no.of students in 1997 was 2000 and only 59% passed i.e 1180 then this scenario could be possible hence this statement somewhat supports the argument[/color][/color][/color]
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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22 Sep 2015, 22:57
B is the OE.
Premise: 20% students failed, 80% students passed.
Premise: the NUMBER of students decreased by 10% from previous year.
CASE 1: in 1997, total number of students: 100
80 passed, 20 failed.
No. of total students remain same in 1998 : So 0.1 of 80 = 8 decrease ---> 72 students passed. so % passed = 72
Hence no. of senior decreased.
CASE 2: in 1997, total number of students: 100
80 passed, 20 failed.
No. of students increased : So again, 0.1 of 80 = 8 decrease ---> 72 students passed. Assuming there are 110 students now, % of students passed < 72 and AGAIN, no. of seniors decreased.
CASE 3: in 1997, total number of students: 100
80 passed, 20 failed.
No. of students decreased : So again, 0.1 of 80 = 8 decrease ---> 72 students passed. Assuming there are 110 students now, % of students passed > 72 and AGAIN, no. of seniors decreased.
So under all circumstances, no. of seniors decrease or increase is NOT A SURE SHOT consideration.
B States otherwise. and hence B = OA
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Re: #Top150 CR: Beginning in 1997 high school seniors in State Q [#permalink]
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26 Feb 2016, 23:17
Hi,
davedekoos wrote:
pkm9995109794 did a great job of breaking down each answer choice, but unfortunately he made a couple errors which led to not being able to answer the question. I do agree however, that this is a messed up question.
Q argument may not be messed up but, the OA is actually messed up..
Quote:
But now we have a problem. Answer choice B is only partially supported by the argument, and answer choice E is contradicted by the argument. The way the question is phrased, "The argument above, if true, LEAST supports which of the following statements." could be interpreted as "Which of the following statement IS supported by the argument above, but supported the least". That way one could argue that B is the answer. But I don't think that is the correct interpretation of the question, and I'm not sure you can support less than by contradicting, which would suggest to me that the correct answer should be E, not B.
The Requirement -"The argument above, if true, LEAST supports which of the following statements."- does convey that the choice is likely to be supported in whatever little way.
So, B comes close to that requirement. But E is totally incorrect, overtaking B as the most appropriate choice but may not exactly fit into the requirement.
So my take would be that E was not intended by the source as it has turned out to be and, errorneously, 1997 has been mentioned as 1998 and vice-versa.
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Re: #Top150 CR: Beginning in 1997 high school seniors in State Q [#permalink]
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28 Feb 2016, 03:07
This question definitely needs fixing. As others have suggested, switching the years in E would easily make B the answer. If E is really the intended answer, then the question would need to clearly ask for an answer that "MUST BE FALSE." This is not something we'd usually see on the GMAT. In any case, when we're asked for an answer that is "least supported," there should really be four supported answers and one unsupported answer. The relative language "least" is just used to protect the test-writers in case we think of some slight exception or odd interpretation that might make two answers seem valid.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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20 Apr 2016, 06:44
you can think of a least possible scenario as an except scenario. courtesy of powerscore CR guide.
So Question rephrased : all choices support except one doesn't follow from the argument.
Year Pass Fail Total
97 80 20 100
98 72 ? ?
A. %HSS who passed increased from 97 to 98 --> total number of HSS decreased from 97 to 98
we know number of HSS who passed decreased from 80 to 72. Only way % can increase is the total decrease!
B. %HSS who passed decreased from 97 to 98 --> total number of HSS increased from 97 to 98
if you think of it this is the inverse logic of A. So this should not be supported.
the total number need not decrease for the %HSS to decrease. It can stay the same and still decrease.
from our example with total as 100 in 1998 as well. % of HSS who passed is 72%. Decreased but total stays the same.
C. if total number of seniors who passed in 1998 was greater than or equal to 80%--> number of seniors was lower in 1998 than in 1997.
Only way for % of seniors who passed be greater than 72% in 1998 is if the total number of people decrease. Similar logic to A.
D. if number of HSS who failed decreased by more than 10% from 1997 to 1998 --> % of HSS who passed in 1998 was greater than 80%.
So number of people who failed in 1998 decreased by more than 2 from our premise. So the number decreased to 17,16,15...etc.. So total is 72+17=89, or 88 , or 87....(72/90 =80%) so in all the scenario the percentage of people who passed will be greater than 80%.
E. % of people who passed was less than 70% in 1998--> number of seniors was higher in 97 than in 98.
How can % be less than 70% in 1998? only if total number increase.
if total decreased % will only increase! So option is definitely not supported.
E for me....
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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20 May 2016, 23:43
sanjuro9 wrote:
Beginning in 1997, high school seniors in State Q have been required to pass a comprehensive proficiency exam before they are allowed to graduate. The exam requirement was intended to ensure that a minimum level of academic quality will be achieved by the students in the state. In 1997, 20 percent of the seniors did not pass the exam and were, therefore, not allowed to graduate. In 1998, the number of seniors who passed the exam decreased by 10% from the previous year.
The argument above, if true, LEAST supports which of the following statement.
A. If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period.
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
C. Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent.
D. If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
Please provide an explanation with actual numbers.
OA:
[Reveal] Spoiler:
B
1997 80% passed, 20% failed
1998 number who passed decreased by 10% from number who passed in 1997
Both in 1997 and 1998 we don't know what was the number of students.
Lets start by taking a number for 1997, say 100 students appeared and 80 passed, 20 failed.
In 1998, we 'll have 10% fewer passing than 1997, so 72 passed the exam.
Now, lets look at the choices.
A - If the percentage of students who passed increased in 1998... so 72 is > 80% of x (x being number of students in 1998.
72 > 0.8x or x<90 so Choice A is valid. Note that the question stem asks for LEAST possible conclusion.
B - %passing decreased. so, 72/x*100 < 80 or 72/x < 0.8 or x>90. x could be 91 or 120. Don't know.
C - Not necessarily true. Number of students can remain the same, say 100 and the number of students passing is less than 80%, ie. 72%.
D - We know that the number of students passing the exam decreased by 10%, lets not evaluate this.
E - The percentage of passing students in 1998 would be less than 70% is when number of students in 1998 is greater than number of students in 1997. >102 to be precise.
As in case of B, the conclusion may or may not be true depending upon how much the percentage has decreased. In case of E the number of students in 1997 can not be greater than that in 1998.
This analysis would take more than 5-10 minutes. What's the source and are you sure the OA is B? What's the explanation at the source?
for answer option E), if 72 is less than 70%, then total number must be greater than 100. So, E is valid.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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19 Jun 2016, 09:44
Thank you for replying to my private message Chetan
This is exactly what my answer was appearing.. No matter what numbers I used I was getting option E as the least supported one.
I don't know why B is the OA.
Anyway I will go stick to my answer and your mathematical proof and mark Option E as correct.
Thanks again
chetan2u wrote:
Beginning in 1997, high school seniors in State Q have been required to pass a comprehensive proficiency exam before they are allowed to graduate. The exam requirement was intended to ensure that a minimum level of academic quality will be achieved by the students in the state. In 1997, 20 percent of the seniors did not pass the exam and were, therefore, not allowed to graduate. In 1998, the number of seniors who passed the exam decreased by 10% from the previous year.
The argument above, if true, LEAST supports which of the following statement.
A. If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period.
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
C. Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent.
D. If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
Please provide an explanation with actual numbers.
OA:
[Reveal] Spoiler:
B
Hi,
Responding to a PM...
Lets take a friendly number to check on choices..
Total seniors in 1997 = 100...
80 passed and 20 failed..
1998 - Total = T...
Passed = 90% of 80 = 72 and failed = F...
lets check the statements -
A. If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period.
% in 1997 = 80%, and % in $$1998 = \frac{72}{T}$$....
$$\frac{72}{T}$$ > 80%.... $$\frac{72}{T}> \frac{80}{100} ..... T < 72*\frac{100}{80}.... T<90$$...
so YES the number of high schools seniors decreased from 100 to LESS than 90 during that time period
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
% in 1997 = 80%, and % in 1998 = 72/T....
$$\frac{72}{T} < 80$$%....$$\frac{72}{T}< \frac{80}{100} ..... T > 72*\frac{100}{80}.... T>90...$$
so YES if the number of high schools seniors was 91 and
NO if it was 101 or 110 etc..
So this may not be TRUE everytime....
C. Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent.
let T< 100..... so 72/T <80/100..... T>90.....
so if T is between 90 and 100... ans is NO...% <80...
if T is <90... ans is YES > 80%
again Can be TRUE of FALSE..
D. If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
% in 1997 = 80%, and % in $$1998 = \frac{72}{T}$$....
so # failed in 1997 = 20, and in 1998 #<18, say 17, so T = 72+17 = 89<90..
$$\frac{72}{(<90)}=x$$ .... so x> 80%.
so YES the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
so $$\frac{72}{T}<\frac{70}{100}..............T> 72*\frac{100}{70}.......... T> 102...$$ so clearly ans is NO in every case..
Now we have B and C, which may be TRUE or FALSE, and E, which will always be FALSE..
so cleraly E is least supported...
OA given is B and many have found B to be correct..
But answer should be E, unless we mean LEAST supported means that the choice should be supported a bit but not completely..
And I do not think that should be the meaning
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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16 Oct 2016, 08:45
"E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
If the % of students who passed the exam in 1998 was less than 70% i.e. the total number of students would be more more than 100 because 72 is 72% of 100. For the % to be less than 70, the total number of students should be more than 100. True.
Can you explain this
If the % of students who passed the exam in 1998 was less than 70% i.e. the total number of students would be more more than 100 because 72 is 72% of 100. For the % to be less than 70, the total number of students should be more than 100.
The text marked in red refers to the number of students in 1998 or 1997 ?
It me it looks it should refer to the number in 1998.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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16 Oct 2016, 09:24
Capricorn369 wrote:
This is a tough one but i'll share my 2 cents.
1997 - 80% pass/ 20% fail.
1998 - 72% pass/ 28% fail. (becasue the number of seniors who passed the exam decreased by 10% from 1997)
After reading options we can observe that only option (B) talks inline with the facts - If the percentage of high school seniors who passed the exam decreased from 1997 to 1998. Rest all talks weird/inconsistent number or percentage.
Let me know your thoughts. Cheers!
Hey dude.
The question is which of the following is least supported by the argument, meaning option B is least consistent with the question, the rest can be correctly inferred from the argument. Go home and read the question first.
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Beginning in 1997, high school seniors in State Q have been [#permalink]
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04 Feb 2017, 08:55
To simplify the argument, let assume total number of seniors in 1997 is 100:
In 1997, 20% seniors did not pass the exam -> 80% pass -> it is 80 pass against 100 in total
In 1998, number of passed senior decreased 10% (of 80) -> it is 72 pass against "X" in total
=> If ratio of passed senior in 1998 equals to that in 1997, then X = 72/80% = 90
If X > 90, then 1998's ratio <80%
If X <90, then 1998's ratio > 80%
We have to eliminate all the answer choices which the above results support or partially support toward:
A/ If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period.
If 72/X > 80%, then X<100 <=> If X<90, then X<100 (yes, it is always true)
SUPPORT
B/ If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
If 72/X <80%, then X>100 <=> If X>90, then X >100 (Yes, It is true in many cases, but if 100>X>90 it is false)
PARTIALLY SUPPORT
C/ Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent.
It means if X>=100, then 72/X <80% <=> If X>=100, then X>90 (yes, it is always true)
SUPPORT
D/ If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
It means if the number of no-passed senior (say "n") <18, then (X-n)/X >80% (opp! with variety value of "X" and "n" we know that it can be true or false)
PARTIALLY SUPPORT
E/If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
If 72/X <70%, 100 > X <=> If X >102.8, then 100> X (wow! it is totally false)
It is such the time killer, all about math and easy to be confused, so I wont take time to resolve this kind of question in real test.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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04 Feb 2017, 11:28
+1 E.
Guessed right. :D
Nice explanations!
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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17 Mar 2017, 11:03
Beginning in 1997, high school seniors in State Q have been required to pass a comprehensive proficiency exam before they are allowed to graduate. The exam requirement was intended to ensure that a minimum level of academic quality will be achieved by the students in the state. In 1997, 20 percent of the seniors did not pass the exam and were, therefore, not allowed to graduate. In 1998, the number of seniors who passed the exam decreased by 10% from the previous year.
The argument above, if true, LEAST supports which of the following statement.
A. If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period.
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
C. Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent.
D. If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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17 Mar 2017, 11:06
Beginning in 1997, high school seniors in State Q have been required to pass a comprehensive proficiency exam before they are allowed to graduate. The exam requirement was intended to ensure that a minimum level of academic quality will be achieved by the students in the state. In 1997, 20 percent of the seniors did not pass the exam and were, therefore, not allowed to graduate. In 1998, the number of seniors who passed the exam decreased by 10% from the previous year.
The argument above, if true, LEAST supports which of the following statement.
A. If the percentage of high school seniors who passed the exam increased from 1997 to 1998 , the number of high schools seniors decreased during that time period.
B. If the percentage of high school seniors who passed the exam decreased from 1997 to 1998 , the number of high schools seniors increased during that time period.
C. Unless the number of high school seniors was lower in 1998 than in 1997, the number of seniors who passed the exam in 1998 was lower than 80 percent.
D. If the number of high school seniors who did not pass the exam decreased by more than 10 percent from 1997 to 1998, the percentage of high school seniors who passed the exam in 1998 was greater than 80 percent.
E. If the percentage of high school seniors who passed the exam in 1998 was less than 70 percent, the number of high school seniors in 1997 was higher than the number in 1998.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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21 Mar 2017, 21:54
Merging topics. Please, search questions before creating a discussion.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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28 Mar 2017, 12:05
Hi experts,
kindly help us with the above question
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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03 Apr 2017, 14:08
Expert's post
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Aketa, please see the above explanations if you haven't already -- and I particularly agree with DmitryFarber.
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Re: Beginning in 1997, high school seniors in State Q have been [#permalink]
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02 May 2017, 07:36
took 4 min 45 seconds but got correct answer
My go :
Passed / fail
1997 : 80/20 - suppose 100 people are there.
1998 : 72/ ? - Don't know how many people are there but we know only 72 people passed.
A B C D - can be true in some cases . but E can never be true .
Re: Beginning in 1997, high school seniors in State Q have been [#permalink] 02 May 2017, 07:36
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Decision making is the thought process of selecting a logical choice from the available options.
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This video involves the concept of factoring using stars. The stars total points are factored and you are able to create 1 dimensional stars with those factors.
Doodling in Math: Sick Number Games
This video is about turning numbers in specific triangles and creating patterns that would make it easy to guess the numbers that would follow.
Rhapsody on the Proof of Pi=4
This video involves breaking down a square with perimeter 4 into even smaller ones until you almost get the shape of a circle. Other shapes were used and were broken down until they became almost straight lines.
Part 2
The video I chose to discuss is Doodling in Math Class: Stars which explains how you can make factors of star points useful. It was the first video I watched so it completely intrigued me at first glance. The quick drawing with precision and accuracy made the presentation that was displayed appear visually entertaining. When I then looked at what was being said I became even more intrigued since it showed at first how stars can be easily made by using 2 of the same shapes. These shapes included a triangle, which came first, then a square and finally a hexagon. Then she went into factoring these stars and made 1 dimensional stars that correlated to the factors. It was like making factoring, which she said was barely discussed, into a more enlightening scope than a formal one.
Part 3
This video reflects to my math teaching since it can help to make people think outside of the box. Factoring might be simple but there is never a fun way to teach it. By creating stars and then factoring them then factoring can lead children to see the fun in math. People usually remember entertaining teachings but tend to forget the boring ones. It can even lead you to start your own ideas since one perspective can lead to many others. By looking at this video I was able to broaden my understanding of star points where they can be made using regular shapes instead of having to draw every single point one by one. This sort of correlates to Lockharts Lament since this sort of teaching of factoring would make children think and to not see things with a formal approach but a creative one. By doing this, it might spark creativity in them as well making them real mathematicians instead of just the copy and paste ones. | 480 | 2,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-26 | latest | en | 0.980178 |
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### Users who are viewing this thread
#### Michael Gauthier
##### New Member
I am working on a well that is about 20' deep and we are not getting enough pressure to make a lawn sprinkler work.
This system
20' well
above ground 3/4 HP pump
10 gal pressure tank
pressure switch is fixed type at 20 - 40 PSI
We drained the tank of all water, then pressurized it to 18 PSI.
The highest point in the yard is only 12' above the pump.
This setup shoots water out an impact sprinkler head only about 4'.
Do I need a different pressure switch, pump, tank, something else?
#### Reach4
##### Well-Known Member
Twelve feet above the pump?
Adjust the air precharge to 45 first. Maybe be rid of the pressure switch, bypass the pressure switch, or get a pressure switch you can adjust to 50/70.
I am not a pro.
#### Valveman
##### Cary Austin
Staff member
What kind of well? Are you driving a sand point?
#### Michael Gauthier
##### New Member
It is a water well with the intake a couple feet off the bottom.
Yes, the highest sprinkler head is 12' above the pump.
"Adjust the air precharge to 45 first.". Adjusting that air pressure higher will require the low point of the pressure switch to be adjusted to right?
Current:
Air Psi
18
Sw Psi
20-40
Proposed:
Air Psi
45
Sw Psi
50-70
#### Valveman
##### Cary Austin
Staff member
Ok then, if you can see down the well to the intake, is there any water in the well?
#### Reach4
##### Well-Known Member
When you adjust a common adjustable pressure switch, turn the 3/8 nut on the big spring CW about 3.5 turns for every 10 PSI of increase desired.
Make sure that the pump is able to shut off within a minute after you stop using water.
#### Michael Gauthier
##### New Member
Yes, there is lots of water in the well, it is about 3' above the intake.
On my previous question, I am thinking that if I increase the air pressure, I need to adjust the low switch pressure to be a a few psi higher than the air pressure. Is that a correct assumption?
#### Reach4
##### Well-Known Member
Yes, there is lots of water in the well, it is about 3' above the intake.
On my previous question, I am thinking that if I increase the air pressure, I need to adjust the low switch pressure to be a a few psi higher than the air pressure. Is that a correct assumption?
About 5 psi below cut-in for a jet pump.
Three feet is not very much.
#### Michael Gauthier
##### New Member
Thanks for the answer to the question on pressures.
It never get down to the intake, and this is the hottest and driest part of the years here in TX. It will be up another 2 feet or so when it rains again.
#### Valveman
##### Cary Austin
Staff member
If you know the foot valve is under water then you have a pump problem. Either the jet nozzle is clogged, the foot valve is stuck closed, there is a suction leak, or the pump is just not working properly. Test the pump circulating water in a bucket and see if it works at 20/40. The pump you have probably won't even build 40/60 or 50/70.
#### Michael Gauthier
##### New Member
The pump is new, so no clog there. The intake may be leaking so that is worth checking. Gonna spend \$12 for a new adjustable pressure switch too.
#### Valveman
##### Cary Austin
Staff member
Jump across the pressure switch and run it direct until you get it going. The pressure switch is not the problem.
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## 1. Question
Implement a basic calculator to evaluate a simple expression string.
The expression string may contain open`(`and closing parentheses`)`, the plus`+`or minus sign`-`,non-negativeintegers and empty spaces.
The expression string contains only non-negative integers,`+`,`-`,`*`,`/`operators , open`(`and closing parentheses`)`and empty spaces. The integer division should truncate toward zero.
You may assume that the given expression is always valid. All intermediate results will be in the range of`[-2147483648, 2147483647]`.
Some examples:
"1 + 1" = 2
" 6-4 / 2 " = 4
"2*(5+5*2)/3+(6/2+8)" = 21
"(2+6* 3+5- (3*14/7+2)*5)+3"=-12
Note:Do not use the`eval`built-in library function.
## 2. Implementation
(1) Stack
class Solution {
public int calculate(String s) {
if (s == null || s.length() == 0) {
return 0;
}
Stack<Integer> nums = new Stack();
Stack<Character> operators = new Stack();
int i = 0;
int num = 0;
int n = s.length();
while (i < n) {
char c = s.charAt(i);
if (Character.isDigit(c)) {
num = c - '0';
while ((i + 1) < n && Character.isDigit(s.charAt(i + 1))) {
num = 10 * num + s.charAt(i + 1) - '0';
++i;
}
nums.push(num);
}
else if (isOperator(c)) {
while (!operators.isEmpty() && hasPrecedence(c, operators.peek())) {
nums.push(calculate(operators.pop(), nums.pop(), nums.pop()));
}
operators.push(c);
}
else if (c == '(') {
operators.push(c);
}
else if (c == ')') {
while (!operators.isEmpty() && operators.peek() != '(') {
nums.push(calculate(operators.pop(), nums.pop(), nums.pop()));
}
operators.pop();
}
++i;
}
while (!operators.isEmpty()) {
nums.push(calculate(operators.pop(), nums.pop(), nums.pop()));
}
return nums.isEmpty() ? 0 : nums.pop();
}
public boolean isOperator(char c) {
return c == '+' || c == '-' || c == '*' || c == '/';
}
public int calculate(char operator, int num1, int num2) {
int res = 0;
switch(operator) {
case '+':
res = num1 + num2;
break;
case '-':
res = num2 - num1;
break;
case '*':
res = num1 * num2;
break;
case '/':
res = num2 / num1;
break;
}
return res;
}
// check if op2 has higher precendence than op1
public boolean hasPrecedence(char op1, char op2) {
if (op2 == ')' || op2 == '(') {
return false;
}
if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-')) {
return false;
}
return true;
}
}
## 3. Time & Space Complexity
Stack: 时间复杂度: O(n), 空间复杂度: O(n) | 717 | 2,362 | {"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} | 3.640625 | 4 | CC-MAIN-2023-06 | latest | en | 0.269465 |
http://math.stackexchange.com/questions/259980/find-all-ideals-with-given-norm | 1,466,801,587,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783391519.0/warc/CC-MAIN-20160624154951-00074-ip-10-164-35-72.ec2.internal.warc.gz | 196,008,631 | 16,424 | # Find all ideals with given norm
I'm working through a final exam from 2 years ago. First task was to find the ideal class group of $\Bbb{Q}(\sqrt{-73})$. That is not the difficult work. I can give the 4 representants of the group by $\omega$, $\frac{\omega+1}{2}$, $\frac{\omega+2}{7}$ and $\frac{\omega-2}{7}$ with $\omega=\sqrt{-73}$. This group is isomorphic to $C_4$. A generator for the group is given by $C=\frac{\omega-2}{7}$ (I hope this is true :) ) But now the following task:
Compute all ideals $\mathfrak{a}$ from $\Bbb{Z}[\sqrt{-73}]$ such that $N(\mathfrak{a})<15$ with $\mathfrak{a}\in C$ (here is N the Norm of the ideal).
Can someone help me with this question?! Thanks :)
-
I'm not sure what you mean with $\mathfrak a \in C$ since $C = \frac {\omega - 2 } 7$. I guess $C$ is the set of integral ideals. Using unique prime ideal factorization (the ring of integers of this field is $\mathbb Z[\sqrt{-73}]$), you can concentrate on prime ideals. Since they are the divisors of the principal ideals $(p)$ with $p \in \mathbb Z$ prime, I would deompose all the ideals $(p)$ up to a properly chosen bound. – Hans Giebenrath Dec 16 '12 at 15:06
with $\mathfrak{a}\in C$ i denote the ideals $\mathfrak{a}$ which lies in the equivalenceclass of $C$ – Trace Dec 16 '12 at 15:37
Now $C$ is an element of the field. Thus the class of $(C)$ (the ideal generated by $C$) is the trivial one. Therefore the class of $C$ is the set of all principal ideals. Do you want to compute all princial ideals with norm bounded by 15? Also your sentence about the ideal class group is confusing. All your representatives are principal ideals, thus they are equal in the ideal class group (trivial). – Hans Giebenrath Dec 16 '12 at 15:46
Okay i thought over $\Bbb{Z}+\Bbb{Z\omega}$,$2\Bbb{Z}+\Bbb{Z\omega+1}$,$7\Bbb{Z}+\Bbb{Z\omega+2}$ and $7\Bbb{Z}+\Bbb{Z\omega-2}$ resp. – Trace Dec 16 '12 at 16:06
And ...? Can you please tell us more. This is too little information. I don't know what you want to tell us. – Hans Giebenrath Dec 16 '12 at 16:18 | 649 | 2,045 | {"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} | 3.828125 | 4 | CC-MAIN-2016-26 | latest | en | 0.862779 |
https://www.mathcelebrity.com/community/threads/a-crate-contains-300-coins-and-stamps-the-coins-cost-3-each-and-the-stamps-cost-1-5-each-the-tot.2015/ | 1,721,382,573,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514900.59/warc/CC-MAIN-20240719074314-20240719104314-00648.warc.gz | 747,951,194 | 9,066 | # A crate contains 300 coins and stamps. The coins cost \$3 each and the stamps cost \$1.5 each. The tot
Discussion in 'Calculator Requests' started by math_celebrity, Sep 13, 2019.
1. ### math_celebrityAdministratorStaff Member
A crate contains 300 coins and stamps. The coins cost \$3 each and the stamps cost \$1.5 each. The total value of the items is \$825. How many coins are there?
Let c be the number of coins, and s be the number of stamps. We're given:
1. c + s = 300
2. 3c + 1.5s = 825
We have a set of simultaneous equations, or a system of equations. We can solve this 3 ways:
No matter which way we pick, we get:
s = 50
c = 250 | 190 | 645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-30 | latest | en | 0.889871 |
http://stackoverflow.com/questions/18074329/autobiographical-number-in-python | 1,435,969,800,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096290.39/warc/CC-MAIN-20150627031816-00209-ip-10-179-60-89.ec2.internal.warc.gz | 250,856,184 | 18,192 | Autobiographical number in python?
Hey all at stackoverflow! I am learning python and need to code a program which checks if a number is autobiographical. EG: 21200 is autobiographical as it has 2 0's 1 1's 2 2's 0 3's and 0 4's. This is what i have so far:
``````# Autobiographical numbers
a = input("Number: ")
abn =
if a == abn:
print(a, "is autobiographical")
else:
print(a, "is not autobiographical")
``````
as you can see i have left the abn variable open as i dont know exactly how to do it. I think i have to determine length of a with len(a) then use something like
``````[x[::1] for x in b]
``````
but i am not quite sure as i am pretty new to python. Thanks, A no0b at python.
-
– Dmitry Bychenko Aug 6 '13 at 7:46
I use the following code to test it. There should be a better solution.
``````from collections import Counter
def test(x):
if len(x) > 10:
return False
c = Counter(x)
for i, v in enumerate(x):
if int(v) != c.get(str(i), 0):
return False
return True
a = input('number: '): #2.x use raw_input
if test(a):
print(a, 'is')
else:
print(a, 'is not')
``````
Demo:
``````>>> auto
['1210', '2020', '21200', '3211000', '42101000']
>>> map(test, auto)
[True, True, True, True, True]
>>> auto = ['12321', '13213', '134', '1231', '123124543']
>>> map(test, auto)
[False, False, False, False, False]
``````
A much better solution from the wiki:
``````>>> def isSelfDescribing(n):
s = str(n)
return all(s.count(str(i)) == int(ch) for i, ch in enumerate(s))
``````
-
Thanks to all who replied! i loved the help from all of you and i appreciated the link to the site. I gave zhangyangyu first choice because he showed me how to get the desired output in python 3. Thank you! – NoviceProgrammer Aug 6 '13 at 8:17
As Dmitry said, Rosettacode has the answer: Self-describing numbers :
``````def isSelfDescribing(n):
s = str(n)
return all(s.count(str(i)) == int(ch) for i, ch in enumerate(s))
``````
My solution would be the following (despite being slower):
``````from collections import Counter
def isAutobiographical(n):
digits = map(int, str(x))
actualFrequency = Counter(digits)
claimedFrequency = dict((x,y) for x,y in enumerate(digits) if y > 0)
return actualFrequency == claimedFrequency
``````
-
``````import math
b = int(math.log(a,10))
*b is length of a. i.e., number of digits in a*
*in fact there is another tick:*
b = len(str(a))
*of course you need to check if a is a valid natural number*
``````
-
``````>>> def is_autobiographical(n):
s = str(n)
count_digits = ''.join([str(s.count(str(i))) for i in range(len(s))])
return s == count_digits
>>> is_autobiographical(21200)
True
>>> is_autobiographical(22)
False
``````
In your case you could use `abn = ''.join([str(str(a).count(str(i))) for i in range(len(str(a)))])` to fit your needs.
- | 835 | 2,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2015-27 | latest | en | 0.803389 |
https://www.slideshare.net/SpringerIndia/introduction-to-polarization-physics | 1,513,220,117,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948537139.36/warc/CC-MAIN-20171214020144-20171214040144-00220.warc.gz | 803,758,497 | 69,723 | Upcoming SlideShare
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# Introduction to polarization physics
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### Introduction to polarization physics
1. 1. Chapter 2 Spin in Strong Interactions In this chapter, the basic theoretical relations underlying the design and analysis of polarization experiments involving strong interactions are reported on classical ex- amples of pion–nucleon and nucleon–nucleon scattering. After the introduction of the density matrix and reaction matrix, we discuss such notions as the complete set of experiments and the equality of the polarization P in the direct reaction and the asymmetry A in the inverse reaction on the example of the simple pion–nucleon system. On the example of the nucleon–nucleon system, we present the method for explicitly constructing the reaction matrix, formulate the unitarity condition, and point to the possibilities of seeking the effects of parity and time reversal violation. The presentation is primarily based on the technique of nonrelativistic quantum me- chanics. Relativistic pion–nucleon and nucleon–nucleon elastic scattering matrices are discussed in the concluding sections. The inclusion of the relativistic effects insignificantly changes the nonrelativistic results. 2.1 Density Matrix As known, a particle with spin s is described by the wave function Ψ having 2s + 1 components Ψα, where α = s,s − 1,...,0,...,−(s − 1),−s. If only one of these components with a certain α value is nonzero, the particle is in a pure spin state. In this case, the mean value of any operator ˆO is given by the expression ˆO = Ψ ∗ α ˆO Ψα . However, components for several α values are most often nonzero in reality. In this case, the state is mixed. A simple example is a polarized proton beam. If its polarization is 100 %, this is a pure spin state: the spins of all protons are oriented identically. When the beam is partially polarized, the protons are in a mixed spin state. This means that the spins of some protons are directed upwards and the spins S.B. Nurushev et al., Introduction to Polarization Physics, Lecture Notes in Physics 859, DOI 10.1007/978-3-642-32163-4_2, © Moskovski Inzhenerno-Fisitscheski Institute, Moscow, Russia 2013 59
2. 2. 60 2 Spin in Strong Interactions of other protons are directed downwards. In this case, the mean value of an arbitrary spin operator ˆO in the mixed spin state Ψ is given by the expression ˆO = α Wα Ψ ∗ α ˆO Ψα , (2.1) where Wα is the weight of the pure state α. The Dirac brackets mean integration (summation) with respect to continuous (discrete) variables. Let us consider the set of other (2s + 1) components {χm}, which are the orthog- onal eigenfunctions of a certain spin operator as the basis functions. If this set is the complete orthogonal set, it can be used for the expansion Ψα = s m=−s Cα mχm. (2.2) The substitution of this expansion into Eq. (2.1) yields ˆO = αmn WαCα∗ m Cα n χ∗ m ˆOχn = mn α WαCα n Cα∗ m · χ∗ m ˆOχn = mn ρnmOmn, (2.3) where Omn is the matrix element of the operator ˆO and ρnm = α Wα Cα∗ m Cα n (2.4) is the matrix element of a certain operator ˆρ, which is called the density matrix. The operator ˆρ can be represented in the matrix form ˆρ = α WαΨαΨ + α . (2.5) Indeed, let us represent the matrix element of the operator ˆρ in the form ρmn = χ+ m ˆρχn = α Wα χ+ m ΨαΨ + α χn . (2.6) Since the eigenfunctions χ are orthogonal, expansion (2.2) provides χ+ m Ψα = n Cα n χ+ m χn = Cα m, Ψ + α χ+ m = n Cα∗ n χ+ n χm = Cα∗ m . (2.7) The substitution into Eq. (2.6) finally yields the expression ρmn = α WαCα mCα∗ n , (2.8) which coincides with Eq. (2.4). Using Eqs. (2.4) or (2.5), one can show that the matrix ρ is Hermitian: ˆρ+ = ρ; (2.9)
3. 3. 2.1 Density Matrix 61 i.e., the mean value of ρ is a real number. Let us define the sum of the diagonal elements of an arbitrary matrix C (its trace) as Tr C = i Cii. (2.10) If the operator ˆC is the product of two operators ˆA and ˆB, its matrix element is given by the expression Cij = m AimBmj . (2.11) In this case, the sum of the diagonal elements is Tr ˆC = mn AnmBmn = Tr ˆA ˆB. (2.12) Comparison with Eq. (2.3) provides ˆO = Tr ˆO ˆρ = Tr ˆρ ˆO. (2.13) The latter equality shows that two operators can be transposed in the trace even if they do not commute. However, in the general case, only a clockwise or counter- clockwise cyclic permutation without the transposition of the operators is possible (if the operators do not commute). Thus, to determine the mean value of the operator ˆO, it is necessary to multiply this operator by the density matrix and to calculate the sum of the diagonal elements of the resulting matrix. According to the theory of matrices, any matrix of rank m = n = (2s + 1) can be expanded in a complete set of (2s + 1)2 matrices {sν} of the same rank satisfying the orthogonality condition Tr sνsμ = δνμ(2s + 1). (2.14) Therefore, we can write the expansion ˆρ = (2s+1)2 μ=1 Cμsμ. (2.15) Multiplying this relation by sν from the right and calculating the trace, we obtain Tr ˆρsν = μ CμTr sμsν = (2s + 1) μ Cμδμν = (2s + 1)Cν. (2.16) The substitution of the coefficients Cμ determined from Eq. (2.16) into Eq. (2.15) finally gives ˆρ = 1 2s + 1 (2s+1)2 μ=1 Tr ( ˆρˆsμ)ˆsμ. (2.17)
4. 4. 62 2 Spin in Strong Interactions Substituting ˆO = ˆsμ (spin operator) into Eq. (2.13), we obtain the polarization vector P : 1 2 P = ˆsμ = Tr ( ˆρˆsμ). (2.18) Hence, the final expression for the density matrix in terms of observable P = ˆs has the form ˆρ = 1 2s + 1 (2s+1)2 μ=1 ˆsμ ˆsμ. (2.19) Thus, the density matrix is completely determined by the mean value of the op- erators ˆsμ. With the normalization of the mean value of the density matrix to unity, we obtain ˆO = Tr ˆρ ˆO/Tr ˆρ. (2.20) With this normalization condition, the final expression for the density matrix has the form ˆρ = 1 2s + 1 (2s+1)2 μ=1 Tr ˆρ ˆsμ ˆsμ. (2.21) The above presentation was based on works Martin and Spearman (1970) and Nurushev (1983). 2.2 Reaction Matrix When considering reactions involving particles with spin, we use the wave functions Ψ and Φ of the initial and final states, respectively. The transition matrix from the initial to final state, M, is defined as follows (Nurushev 1983): Φ = MΨ. (2.22) Then, we have two density matrices: ρi = α WαΨαΨ + α (2.23) for the initial state and ρf = α WαΦαΦ+ α (2.24) for the final state. Here, α stands for averaging over the initial spin states and summation over the final spin states. In view of relation (2.22), the relation between these two matrices is obtained from Eq. (2.24) in the form ρf = α WαMΨαΨ + α M+ = M α WαΨαΨ + α M+ (2.24a)
5. 5. 2.2 Reaction Matrix 63 or ρf = MρiM+ . (2.25) Hence, the solution of the problem of the interaction between particles is re- duced to the determination of ρi in terms of the mean values of the complete set of the spin matrices sν (the mean values sν are called the observables) and the oper- ator ˆM. Then, the density matrix of the final state is unambiguously determined by Eq. (2.25), and a necessary observable in the final state can be calculated. Let us find the operator ˆρf as a function of sν and M. To this end, we multiply Eq. (2.25) by sμ from the right and calculate the trace of the product of the matrices: Tr ˆρf sμ = Tr M ˆρiM+ sμ = Tr M 1 2s + 1 (2s+1)2 ν=1 Tr ˆρi sν isν M+ sμ = 1 2s + 1 Tr ˆρi (2s+1)2 ν=1 sν i · Tr MsνM+ sμ . (2.26) In view of the relation Tr ( ˆρf sμ) = sμ f Tr ˆρf , (2.27) we obtain sμ f · Tr ˆρf = 1 2s + 1 Tr ˆρi (2s+1)2 ν=1 sν i · Tr MsνM+ sμ . (2.28) Denoting the differential cross section as I = Tr ˆρf Tr ˆρi , (2.29) we arrive at the following expression for the mean value sμ f of the spin operator sμ final state: sμ f · I = 1 2s + 1 · (2s+1)2 ν=1 sν i · Tr MsνM+ sμ . (2.30) This expression allows one to calculate the mean value of any spin operator sμ in the final state using the known parameters of the initial state sν i and the scattering matrix M. Let us consider reactions of the type π + N = π + N, (2.31) or, in the spin notation, 0 + 1/2 → 0 + 1/2 (spins of the pion and nucleon are 0 and 1/2, respectively). In the corresponding two-dimensional spin space, the Pauli matrices σ (σx,σy,σz) together with the identity matrix 1 can be used as a complete
6. 6. 64 2 Spin in Strong Interactions set of spin operators. In this case, the density matrix of the initial state can be written in the form ˆρi = C0 · 1 + C1 · σ. (2.32) Let us determine the coefficients C0 and C1. From the normalization condition of the trace of the density matrix ˆρi to unity, we obtain Tr ˆρi = Co · Tr 1 + C1 · Tr σ = 2C0 = 1, (2.33) because Tr σ = 0. (2.34) Relation (2.34) is easily verified by writing the Pauli matrices in the explicit form σx = 0 1 1 0 , σy = 0 −i i 0 , σz = 1 0 0 −1 . (2.35) Let the nucleon in the initial state be polarized and have polarization vector Pt (the subscript t means the target). For the initial state, Pt = σt = Tr ˆρσt Tr ˆρ = Tr σt · 1 2 + C1 · σ = Tr σt (C1 · σ) = C1k · ek · Tr σkσi = C1k · ek · 2δik = 2C1t . (2.36) Here, ek are the unit coordinate vectors in the Cartesian coordinate system and we use the relation (σ · A)(σ · B) = A · B + iσ · (A × B), (2.37) which is easily proved using relations (2.35). Thus, we obtain C0 = 1 2 , C1 = 1 2 Pt . (2.38) Then, the density matrix of the initial state ˆρi, is represented in the form ˆρi = 1 2 (1 + Pt · σ). (2.39) Hence, the density matrix ˆρ is completely determined by the target polarization vector Pt (or the beam polarization vector PB in the case of the 1/2 + 0 → 1/2 + 0 reaction). Note that, in view of the properties of the Pauli matrices, expression (2.39) cannot include operators above the first order, because all such operators are reduced to an operator maximally of the first order. 2.3 R, P , and T Transformations In the next section, we consider nucleon–nucleon elastic scattering and, following Wolfenstein and Ashkin (1952), construct the elastic scattering matrix. Here, as a
7. 7. 2.3 R, P , and T Transformations 65 preparation to this consideration, we discuss the constraints on this matrix that fol- low from the physical requirements of the isotropy of space (R operation), space inversion (P operation), and time reversal (T operation). Below in this section, we follow Bilenky et al. (1964). In the interaction (or Heisenberg) representation, the S-matrix is defined in terms of the Dirac brackets as ψ(+∞) = S ψ(−∞) . (2.40) Here, |ψ(−∞) is the wave function of the system in the initial state at t → −∞ and |ψ(+∞) is the wave function of the system in the final state. According to this definition, the S-matrix transforms the initial state of two free nucleons to the final state with allowance for their interactions. Thus, the S-matrix contains all informa- tion on the interaction between the nucleons. If the nucleons do not interact, it is reasonable to set S = 1. Then, we formulate the necessary physical requirements. 1. R operation. Let |ψ(t) be the wave function of the system at time t in an arbitrary reference frame called base. We introduce the second reference frame R rotated by a certain angle and denote the wave function of the system in this ref- erence frame as |ψ(R,t) . The wave functions in two reference frames should be related by a unitary transformation (according to the requirement that the numbers of particles in both reference frames should be the same). The unitarity of the oper- ator implies the equality U+(R) = U−1(R). Hence, ψ(R,t) = U(R) ψ(t) . (2.41) The matrix U(R) is obviously a function of the rotation angles of the R frame with respect to the base frame. The multiplication of Eq. (2.40) by U(R) from the left gives ψ(R,+∞) = U(R)SU−1 (R) ψ(R,−∞) . (2.42) The wave functions |ψ(R,−∞) and |ψ(R,+∞) describe the initial and final states of the nucleons, respectively, in the rotated reference frame R. Hence, in this reference frame, by the definition of the S-matrix, which is determined only by the interaction dynamics but is independent of the choice of the reference frame, the wave functions should be related by the same S-matrix as in Eq. (2.40): ψ(R,+∞) = S ψ(R,−∞) . (2.43) Comparison of Eqs. (2.42) and (2.43) shows that U(R)SU−1 (R) = S. (2.44) Since the matrix U(R) is unitary, this equality can be represented in the other form U−1 (R)SU(R) = S. (2.45) Relation (2.44) (or (2.45)) expresses the invariance of strong interactions under rotations of the reference frame in physical space. An application of this relation will be considered elsewhere.
8. 8. 66 2 Spin in Strong Interactions 2. P operation. It is postulated that strong interactions are invariant under space inversion. This property is also called the parity conservation law. Let us consider the constraints imposed by this postulate on the S-matrix. As in the above consid- eration, we take the initial reference frame as base. As above, the wave function in this frame is denoted as |ψ(t) . We introduce the reference frame I, in which all coordinate axes are inverted, i.e., x → −x, y → −y, and z → −z. If the base frame is left-handed, the frame I is right-handed. Let U(I) be a unitary operator transforming the function |ψ(t) in the base frame to the wave function in the I frame: ψ(I,t) = U(I) ψ(t) . (2.46) In this case, both functions describe the same physical state, but in different coor- dinates. Hence, the S-matrices in two frames should be the same. Let us determine the matrix S(I). By the definition, ψ(I,+∞) = S(I) ψ(I,−∞) . (2.47) At the same time, from Eq. (2.46) we obtain ψ(I,+∞) = U(I) ψ(+∞) = U(I)S ψ(−∞) = U(I)SU−1 ψ(I,−∞) . (2.48) Comparison of Eqs. (2.47) and (2.48) shows that S(I) = U(I)SU−1 (I). (2.49) Thus, the postulate of the invariance of strong interactions under space inversion leads to the relation S = U−1 (I)SU(I). (2.50) These two cases imply that the invariance under the R and P transformations reduces to the commutativity of the S-matrix and the corresponding transformation U matrices. 3. T operation. The principle of the invariance of the strong interaction under time reversal is formulated as follows. Let us consider the Schrödinger equation in the interaction representation i ∂|ψ(t) ∂t = H(t) ψ(t) . (2.51) Changing t → −t and taking complex conjugation in this equation, we arrive at the equation i ∂|ψ(−t) ∗ ∂t = H∗ (−t) ψ(−t) ∗ . (2.52) Since H∗(−t) = H(t) in the general case, this new equation is not a Schrödinger equation. However, let us assume that there is a unitary operator U(T ) providing the transformation ψ(T,t) = U(T ) ψ(−t) ∗ . (2.53)
9. 9. 2.3 R, P , and T Transformations 67 The substitution of this relation into Eq. (2.52) yields i ∂|ψ(T,t) ∂t = U(T )H∗ (−t)U−1 (T ) ψ(T,t) . (2.54) Setting H(t) = U(T )H∗ (−t)U−1 (T ), (2.55) we arrive at the Schrödinger equation i ∂|ψ(T,t) ∂t = H(t) ψ(T,t) . (2.56) Thus, relation (2.55) presents a mathematical formulation of the physical postu- late of the invariance of the interaction under time reversal. According to Eq. (2.55), for each wave function |ψ(t) that is a solution of Schrödinger equation (2.51), there is another function |ψ(T,t) that satisfies Eq. (2.56) and describes the motion of the system in the reverse time direction. Let us obtain the requirement imposed on the S-matrix by the time reversibility condition. We have two Schrödinger equations with two wave functions, but with the same S-matrix, because the S-matrix is independent of the initial state of the system, but is completely determined by the dynamics of interaction. According to the above consideration and by analogy with relation (2.40), we write ψ(T,+∞) = S ψ(T,−∞) . Using relation (2.53), we represent this equality in the form U(T ) ψ(−∞) ∗ = SU(T ) ψ(+∞) ∗ . The multiplication of this relation by U−1(T ) from the left yields ψ(−∞) ∗ = U−1 (T )SU(T ) ψ(+∞) ∗ . (2.57) Let us take into account that the S-matrix is unitary, i.e., S+S = 1 and that S+ = ˜S∗ by the definition, where the asterisk and tilde mean the complex conju- gation and transposition of the matrix, respectively. Multiplying Eq. (2.40) by S+ from the left, taking complex conjugation, and taking into account the properties of the S-matrix, we obtain ψ(−∞) ∗ = ˜S ψ(+∞) ∗ . (2.58) The comparison of Eqs. (2.57) and (2.58) shows that U−1 (T )SU(T ) = ˜S. (2.59) This is the requirement imposed on the S-matrix by the time-reversal invariance of the interaction. The results are summarized as follows: • the invariance of the interaction under rotation of the reference frame, the corre- sponding R operation is given by relation (2.45): U−1(R)SU(R) = S;
10. 10. 68 2 Spin in Strong Interactions • the invariance of the interaction under space inversion, the corresponding P op- eration is given by relation (2.50): U−1(I)SU(I) = S; • the invariance of the interaction under time reversal, the corresponding T opera- tion is given by expression (2.59): U−1(T )SU(T ) = ˜S. In applications, another matrix M determined only by the interaction between the nucleons is used. The initial state of the particles at t → −∞ is denoted as |i , and the final state at t → +∞ is denoted as |f . The particles in the initial and final states do not interact and have the relative momenta p and p , total momenta Q and Q , and total energies E and E , respectively. The matrix M is expressed in terms of the S matrix as S − 1 = M, (2.60) or in terms of the matrix elements, taking into account the conservation of the energy and momentum: f |S|i = f |i − 2πiδ Q − Q δ E − E f |M|i . (2.61) Here, the δ functions ensure the conservation of the momentum and energy. The matrix M transforms the initial state |i to the final state |f and acts only in the spin space. By the definition of the matrix elements, f |M|i = χ + M p ,p χ . (2.62) Here, χ and χ are the spin wave functions of the initial and final states of the nucleons, respectively. The matrix M, as well as the S matrix, is determined by the dynamics of interaction and satisfies the requirements of the R, P , and T invari- ances. Let us consider their in more detail. From the postulate of the invariance of the interaction under rotation of the ref- erence frame, the following constraint on the S-matrix was obtained (see (2.45)): U−1 (R)SU(R) = S. Let us represent this relation in terms of the matrix elements in the base and rotated R frames f U−1 (R)SU(R) i = f,R|S|R,i = f |S|i . (2.63) Here, |R,i and |R,f are the wave functions in the rotated R frame. The total, Q, and relative, p, momenta in the base reference frame are related to the respective momenta QR and pR in the rotated reference frame as (QR)i = ailQi, (pR)i = ailpi. (2.64) Here, a is the rotation matrix from the base frame to the R frame and ali are its matrix elements (in this case, the cosines and sines of the rotation angle from the old to new reference frame). Relation (2.63) indicates that the respective elements of the S matrix in different frames are the same. According to M-matrix definition (2.61), this is valid for its matrix elements: χ + (R)M pR,pR χ(R) = χ + (p)M p ,p χ(p) , (2.65)
11. 11. 2.3 R, P , and T Transformations 69 where χ(R) and χ (R) are the spin wave functions in the rotated frame and χ and χ , in the base frame. Since these sets of functions describe the same spin state, but in different frames, they should be related by a unitary transformation: χ(R) = U(R)χ, χ (R) = U (R)χ . (2.66) The nucleon–nucleon scattering under consideration involves two spin particles in the initial and final states of the reaction. This means that the spin functions in the initial and final states of the reaction are the products of the functions of individual nucleons. As a result, the unitary operators U(R) and U (R) are the direct products of the matrices acting on the spin functions of the individual particles. The mean value of the spin operator in quantum mechanics is an observable, namely, the polarization vector (more precisely, the polarization vector is P = σ = 2s, where σ is the Pauli operator and s is the spin vector). The mean value of the spin operator should be transformed as a vector: χ+ (R)slχ(R) = aliχ+ siχ. Here, sl is the spin operator of one of the initial nucleons. Therefore, the matrix U(R) should satisfy the condition U−1 (R)slU(R) = alisi. (2.67) The same condition is obviously imposed on the matrix U (R). For a given rota- tion angle of the frame R, the matrix U can be reconstructed from these conditions. From Eqs. (2.65) and (2.66), it follows that U −1 (R)M pR,pR U (R) = M p ,p . (2.68) This is the mathematical expression of the postulate of the invariance of the in- teraction under space rotation. Then, we consider the P operation, i.e., space inversion. Under this operation, the momenta, being polar vectors, change signs, whereas the spin vector, being an axial vector, does not change sign. Therefore, sl = U−1 (I)slU(I), sl = U −1 (I)slU (I), (2.69) where the unitary matrix U(I)(U (I)) ensures the transformation of the wave func- tion of the initial (final) state from the base frame to the inverted frame I. This transformation is written as follows: χ(I) = U(I)χ, χ (I) = U (I)χ . The condition on the S-matrix provides χ + M p ,p χ = IiI∗ f χ + (I)M − ← p ,−p χ(I) . (2.70) Here, Ii and If are the internal parities of two initial and two final nucleons, respectively. Using the relation between the wave functions χ and χ , we obtain M p ,p = IiI∗ f U−1 (I)M −p ,−p U(I). (2.71)
12. 12. 70 2 Spin in Strong Interactions 2.4 Unitarity Condition The Schrödinger equation for the wave function Ψk has the form ∇2 Ψk + 2μ 2 (E − ˆu)Ψk = 0, (2.72) where μ is the reduced mass of colliding particles and ˆu is the potential energy of their interaction in the operator form (can include the spin and isospin operators). For the case of elastic scattering, Ψk can be represented in the form of the sum of two terms: Ψk = eik·r χ + 1 r eikr M k ,k χ. (2.73) The first term is an incident plane wave; the second term is the divergent scat- tering waves; k and k are the wave vectors of the initial (before the interaction) and final states, respectively; χ is the spinor function of the initial state; and M is the reaction matrix. Since the elastic scattering is considered in the center-of-mass frame, k = |k|, ˆu+ = ˆu. (2.74) The second condition is the Hermitian condition imposed on the interaction po- tential, because its eigenvalues should be real. Then, the interaction in the intermediate state of the colliding particles leads to the transition of the wave vector k to the wave vector k , determining a given direction (e.g., the direction to a recording detector). At large distances from the collision center, the following expansion can be used (to define the absolute phases, it is necessary to include a known, for example, elec- tromagnetic or weak interaction in addition to strong interactions): Ψ 0 k = eik·r = eikzcosθ = ∞ l=0 il (2l + 1)Pl(cosθ) sin(kr − 1 2 lπ) kr = 1 ikr ∞ l=0 (2l + 1) 2 Pl(cosθ)eikr − 1 ikr ∞ l=0 (−1)l (2l + 1) 2 Pl(cosθ)e−ikr . (2.75) It can be shown that ∞ l=0 (2l + 1) 2 Pl(cosθ) = δ(1 − cosθ), ∞ l=0 (−1)l (2l + 1) 2 Pl(cosθ) = δ(1 + cosθ). (2.76)
13. 13. 2.4 Unitarity Condition 71 These formulas are verified by multiplying the both sides by Pl(cosθ) and inte- grating with respect to cosθ. The validity of relations (2.76) is obvious in view of the orthogonality condition of the Legendre polynomials, 1 −1 Pl (cosθ)Pl(cosθ)d cosθ = 2 2l + 1 δll (2.77) and the definition of the Dirac δ functions f (x)δ(x − a)dx = f (a). (2.78) Taking into account Eq. (2.76), from Eq. (2.75) we obtain eik·r = 1 ikr eikr δ(1 − cosθ) − 1 ikr e−ikr δ(1 − cosθ). (2.79) Therefore, Ψk(r) = 1 r eikr · χ 1 ik δ(1 − cosθ) + M k ,k − 1 ikr e−ikr δ(1 + cosθ) · χ. (2.80) Similarly, Ψk (r) = 1 r e−ikr · χ+ M k ,k − 1 ik δ(1 − cosθ) + 1 ikr eikr δ(1 + cosθ) · χ+ . (2.81) From the Schrödinger equation ∇2 Ψk + 2μ 2 (E − ˆu)Ψk = 0, ∇2 Ψ + k + 2μ 2 · Ψ + k · (E − ˆu) = 0, (2.82) we can obtain the relation Ψ + k ∇2 Ψk − ∇2 Ψ + k Ψk = 0, (2.83) which can be modified to the form ∇ Ψ + k ∇Ψk − ∇Ψ + k Ψk = 0. (2.84) The integration over the volume V provides ∇ Ψ + k ∇Ψk − ∇Ψ + k Ψk dv = ∇ Ψ + k ∇Ψk − ∇Ψ + k Ψk ds = 0. (2.85) Using Eqs. (2.80) and (2.81), as well as the operator ∇ = ∂ ∂r , we obtain 1 kr2 δ(1 + cosθ)δ 1 + cosθ − 1 r2 M+ k ,k + i k δ 1 − cosθ · M k ,k − 1 k δ(1 − cosθ) ds = 0, (2.86) where ds = r2dωk . After the integration, we arrive at the formula 1 2i M k ,k − M+ k,k = k 4π M+ k ,k · M k ,k dωk . (2.87)
14. 14. 72 2 Spin in Strong Interactions This condition can be generalized to the case of inelastic reactions: 1 2i 1 ka Mab k ,k − 1 kb M+ ba k,k = C 1 4π M+ bc k ,k · Mac k ,k dωk , (2.88) where the summation over all possible reaction channels is implied. Relation (2.87) provides a number of important consequences. At k = k (elastic scattering at zero angle), this relation gives the so-called optical theorem Ima(0) = k 4π σTOT, (2.89) which is the relation between the imaginary part of the forward elastic scattering amplitude a(0) and the total cross section σTOT. The application of relation (2.88) to the pion–nucleon scattering matrix gives two relations Ima k ,k = k 4π a∗ k ,k · a k ,k + ib∗ k ,k × b k ,k · n × n dωk , (2.90) Reb k ,k = k 4π a∗ k ,k · b k ,k + b∗ k ,k · a k ,k + Re b∗ k ,k × b k ,k · n × n · n dωk . (2.91) Similar relations are also valid for nucleon–nucleon scattering. These relations are particularly suitable at low energies, when only the elastic channel is opened. As a result, the number of necessary experiments is halved (by two and five exper- iments for the πN and nucleon–nucleon scatterings, respectively). However, these statements should be considered carefully, because they are valid only under ideal conditions, which are almost inaccessible in actual experiments. The presentation in this section follows Nurushev (1983). 2.5 Pion–Nucleon Scattering Let us consider reactions of the type π + N = π + N, (2.92) or, in the spin notation, 0 + 1/2 → 0 + 1/2. In the corresponding two-dimensional spin space, the Pauli matrices σ(σx,σy,σz) together with the identity matrix 1 can be used as a complete set of spin operators. In this case, the density matrix of the initial state can be written in the form (see Eq. (2.32) in Sect. 2.2) ˆρi = C0 · 1 + C1 · σ. (2.93) Hence, the density matrix ˆρ is completely determined by the target, Pt , or beam, PB, polarization vector.
15. 15. 2.5 Pion–Nucleon Scattering 73 Then, it is necessary to determine the reaction matrix M. In the general case, it should be, first, a function of two variables, for example, momenta ki and kf before and after reactions, respectively, and, second, a two-row matrix in spin space. As the matrix, it can be expanded in the complete set consisting of the Pauli matrices and identity matrix: M ki,kf = a ki,kf · I + b ki,kf · σ. (2.94) According to the experimental data, we require that strong interactions satisfy the following conditions (Nurushev 1983): 1. Parity conservation law. Since the parities of the initial and final systems are identical, the M-matrix should be a scalar function of the initial energy and scat- tering angle of the particles. This means that the vector b, as well as σ, should also be axial. The only axial vector that can be composed of two vectors ki and kf has the form b (ki,kf ) = bn, (2.95) where n = ki ×kf /|ki ×kf | is the unit vector perpendicular to the reaction plane. The quantity a(ki,kf ) should obviously be a scalar function. 2. Time reversibility. Under the time reversal operation, ki → −kf and kf → −ki, so that n changes sign. Under this operation, σ also changes sign, so that the quantity σ · n is a scalar. For the (0 + 1/2) system, this requirement is satisfied simultaneously with requirement 1. However, for more complex systems (e.g., (1/2 + 1/2)), time reversibility gives rise to additional constraints. Thus, the πN elastic scattering matrix M(ki,kf ) has the form M(ki,kf ) = a(ki,kf ) + b(ki,kf ) · σ · n. (2.96) Here, b and a are called the amplitudes with and without spin flop, respectively, are complex quantities; therefore, four real functions (for a given initial energy and a given scattering angle) should be determined from experiments. A set of inde- pendent experiments necessary for the unambiguous determination of all reaction amplitudes is called the complete set of experiments. Hence, the complete set of ex- periments for the πN system should include at least four independent experiments. However, reality is much more complicated that the above description. First, since the experimentally measured quantities are quadratic combinations of the am- plitudes a and b, only the difference of the phases of the amplitudes a and b for strong interactions, rather than their absolute values, can be determined from an ex- periment. To determine the absolute values of the phases, it is necessary to include a known (for example, electromagnetic or weak) interaction in addition to strong interactions. Thus, the complete set of experiments should include more than four experiments. Second, if elastic scattering is the single allowed reaction channel, the unitarity condition gives rise to two additional relations between the amplitudes a and b; for this reason, to determine these amplitudes, it is sufficient to perform two independent experiments.
16. 16. 74 2 Spin in Strong Interactions However, at energies above the pion production threshold, the number of experi- ments in the complete set is larger than four. In reality, there are three reactions induced by charged pions: π+ + p → π+ + p(a), π− + p → π− + p(b), π− + p → π0 + n(c). (2.97) These three reactions are related by the requirement of the isotopic invariance of strong interactions. As a result, the matrices of these reactions are related as M(a) = M1, M(b) = 1 2 (M1 + M0), M(c) = 1 2 (M1 − M0). (2.98) Here, the matrices M1 and M0 correspond to the isotopic states of the pion– nucleon system with T = 3/2 and 1/2, respectively. These matrices are recon- structed similarly to the matrix M. Disregarding isotopic invariance, three reactions (2.97) are described by the set of 12 experiments. Allowance for isotopic invariance reduces this number to 8. For the system of two particles with spin 1/2 (example is pion + nucleon), we present the proof of the equality P = A, where P and A are the polarization of the particle and its asymmetry in the binary reactions, respectively; this relation is very important in applications (Bilenky et al. 1964). Let us consider the wave function of the system with the inverted direction of the wave vector that satisfies the Schrödinger equation with reversed time (t → −t). We represent it in the form Ψ−k = e−ik r + 1 r eikr M k ,−k = i kr e−ikr · δ 1 − cosθ + 1 r eikr M k ,−k − i k · δ 1 − cosθ . (2.99) The substitution of this wave function into the following Eq. (2.85) from Sect. 2.4: ∇ Ψ + k ∇Ψk − ∇Ψ + k Ψk dv = ∇ Ψ + k ∇Ψk − ∇Ψ + k Ψk ds = 0, gives M k ,k · δ 1 − cosθ + 1 ik · δ 1 − cosθ · δ(1 − cosθ) − M k ,−k · δ(1 + cosθ) dωk = 0. (2.100) After the integration, we obtain M k ,k = M −k,−k . (2.101) This relation is the condition of the time reversibility of the process.
17. 17. 2.5 Pion–Nucleon Scattering 75 Scattering matrix (2.96) constructed above satisfies this condition. Using the ex- plicit form of the scattering matrix, we prove the following statement widely used in applications. Let the polarization of the particle d be measured in the reaction a(0) + b(1/2) → c(0) + d(1/2) (2.102) (the spins of the particles are given in the parentheses). This is usually achieved by the rescattering of the particle d on a certain nucleus with a known analyzing power. To explain new terms, we make a brief digression. Let a beam with a given energy and unit polarization be scattered on a nuclear target at a given angle. The number of the scattered particles per unit flux of the incident beam whose polarization is directed upward from the scattering plane is denoted as N1. Under the same conditions, this number for the beam polarization downwards to the scattering plane is denoted as N2. In this notation, the analyzing power of the target is defined by the formula AN = N1 − N2 N1 + N2 . If the beam is partially polarized, i.e., P = 1, raw asymmetry (directly measured in experiment), can be defined as ε = P · A. The quantity ε is also called left–right asymmetry or, sometimes, “raw” asym- metry. According to the definition, the asymmetry ε coincides with the analyzing power AN at the 100 % polarization of the beam. We denote the polarization of the particle d as P . Let us consider the reverse reaction c(0) + d↑(1/2) → a(0) + b(1/2), (2.103) where the particle d is polarized. Let the left–right asymmetry AN is measured in the formation of the particle a (or b). Theorem: the polarization P of the particle d in reaction (2.102) is equal to the asymmetry A of the particle b (or a) in reaction (2.103). This statement is expressed by the equality P = AN . (2.104) Let us prove this statement. Indeed, by the definition of polarization (Bilenky et al. 1964). P = Tr (MρiM+σ) Tr (MρiM+) (2.105) since the particle b in initial system (2.102) is unpolarized, ρi = 1/2 and, calculat- ing P , we obtain P · I0 = 2Rea∗ b = Tr MM+ σ , (2.106) where I0 = |a|2 + |b|2 (2.107)
18. 18. 76 2 Spin in Strong Interactions is the differential cross section for reaction (2.102). Since the particle d in reaction (2.103) is polarized, ρi = 1 2 (1 + P0· · σ) and the scattering cross section is given by the expression If = Tr M ˆρiM+ = 1 2 Tr MM+ + 1 2 P0Tr MσM+ . (2.108) The direct calculation shows that MM+ = M+M and Tr MσM+ = Tr MM+ σ = 2I0P, (2.109) where P is determined by expression (2.106). Then, If = I0(1 + P0 · P). (2.110) By the definition of left–right asymmetry AN = 1 P0 If (+) − If (−) If (+) + If (−) = P, (2.111) quod erat demonstrandum. When deriving relation (2.111), we explicitly take into account the detailed bal- ance principle, i.e., the equality of the cross sections for the direct and inverse reac- tions. Note that the positive sign in relation (2.111) appears for a reaction in which the initial and final states have the same parity. For the case of different parities, the sign in relation (2.111) is negative. Theorem (2.104) is also proved for the case of the binary reaction when both initial (and final) particles have spin 1/2. This theorem is invalid for the inclusive reactions. A test of the relation P = A is simultaneously a test of the T invariance in strong interactions. At present, this problem is of very great interest in view of the creation of an absolute polarimeter for the RHIC collider (see Chap. 8 in the second part of the book, which is devoted to the polarimetry of beams). It is known that a reaction with the production of hyperons is a convenient reac- tion for verifying the relation P = A. If a target is unpolarized, the polarization of a hyperon is determined from its decay. If the target is polarized, left–right asymme- try can be measured with the same instruments. The same is true for the beam. An example of such interactions is the reaction π− + p → K0 + Λ(↑)(a), π− + p(↑) → K0 + Λ(b). (2.112) This reaction is very convenient because channels (a) and (b) can be measured simultaneously if a polarized target is used. In this case, it is very important to detect both K mesons and Λ hyperons. Averaging the experimental results over the target polarizations (as if the target polarization vanishes), one can determine the polarization of the Λ hyperons (channel (a)). Averaging over the polarization of the Λ hyperons for the polarized target (channel (b)), we determine asymmetry. Comparison of these two observables ensures the direct test of the equality P = AN . Such an experiment has not yet been carried out.
19. 19. 2.6 Nucleon–Nucleon Scattering 77 An important application of the relation P = AN is the measurement of asym- metry in the reaction π− + p(↑) → π0 + n. (2.113) For the direct measurement of the neutron polarization on an unpolarized target, it is necessary to scatter the neutron by another target and to detect the scattered neutron. This is a difficult experimental problem owing to loss in the yields for the second scattering process and low neutron detection efficiency. Therefore, the use of the polarized proton target made it possible to measure the neutron polarization in reaction (2.113). These measurements provided first doubts in the Regge pole model very popular in the 1960s. 2.6 Nucleon–Nucleon Scattering In this section, we present the method for determining the reaction matrix in the nonrelativistic case proposed in Wolfenstein and Ashkin (1952). We will see later that the relativistic approach does not change the results, but leads to the kinematic rotations of the observables lying in the reaction plane. The observables perpendic- ular to the reaction plane remain unchanged in this case. 2.6.1 Construction of the Reaction Matrix The system of two nucleons is described by two spin operators σ1 and σ2 and two identity operators I1 and I2 acting in the spin spaces of the first and second par- ticles, respectively. As a result, the scattering matrix is a four-dimensional matrix depending on the physical vectors σ1 and σ2, ki and kf (relative momenta of two nucleons in the initial and final states, respectively). When constructing the NN- scattering matrix, we follow Wolfenstein and Ashkin (1952). Since the initial (two- nucleon) and final (also two-nucleon) systems in our case have the same internal parity, the scattering matrix M(σ1,σ2;ki,kf ) should be a scalar function composed of the combination of spin operators and momenta. Using the spin operators, we can generally compose 16 combinations (complete set): 1 (scalar) (σ1 · σ2 − 1) (scalar) (σ1 + σ2) (axial vector) (σ1 − σ2) (axial vector) (σ1 × σ2) (axial vector) lαβ = (σ1ασ2β + σ1βσ2α) (symmetric tensor). (A) In view of the properties of the sigma operators, these combinations cannot in- clude terms of the orders higher than the first order.
20. 20. 78 2 Spin in Strong Interactions The combinations that can be composed of the momenta ki and kf are as follows: 1 (scalar) kf − ki = K (polar vector) kf × ki = n (axial vector) n × K = P (polar vector) KαKβ,nαnβ (symmetric tensors) PαPβ,KαPβ + KβPα (symmetric tensors). (B) Multiplying the quantities from sets (A) and (B), we take into account the re- quirement of the invariance of the matrix M(σ1,σ2;ki,kf ) under rotation and inver- sion of space. Thus, the following combinations can be included in the amplitudes of the scattering matrix: 1, (σ 1 · σ 2 −1), (σ1 + σ2) · n, (σ1 − σ2) · n, (2.114) (σ1 × σ2) · n, (2.115) αβ lαβKαKβ, αβ lαβnαnβ, αβ lαβPαPβ, αβ lαβ(KαPα + KβPα). (2.116) Combinations (2.116) can be represented in the form σ1 · Kσ2 · K, σ1 · nσ2 · n, σ1 · Pσ2 · P; (2.117) σ1 · Kσ2 · P + σ1 · Pσ2 · K. (2.118) Three vectors K, n, and P are mutually orthogonal. As a result, the sum of three terms in (2.117) is equal to the dot product σ1 · σ2, as can be verified by direct calculations. Hence, only two of three terms in (2.117) are independent. Now, we require the time-reversal invariance of these terms. Under time reversal t → −t, the spin operator and momentum are transformed as follows (prime marks the quantities with reversed time): σ = −σ, ki = −kf , kf = −ki. (2.119) Using Eq. (2.119) and the definition of vectors K, n, and P (see (B)), we can show that K = K, n = −n, and P = −P. (2.120) Under this transformation, terms (2.115) and (2.118) change sign and, corre- spondingly, are rejected. Finally, the nucleon–nucleon elastic scattering matrix is expressed in the form M(σ1,σ2;ki,kf ) = A + B(σ 1 · σ 2 −1) + C(σ1 + σ2) · n + D(σ1 − σ2) · n + E(σ1 · K)(σ2 · K) + F(σ1 · P)(σ2 · P). (2.121)
21. 21. 2.6 Nucleon–Nucleon Scattering 79 Let us introduce the triple of orthogonal unit vectors in the center-of-mass frame: n = k × k |k × k | , m = k − k |k − k| , l = k + k |k + k | , (2.122) where k = ki |ki| and k = kf |kf | are the unit vectors. It is convenient to introduce these unit vectors, because the vectors l and m in the nonrelativistic approximation coincide with the directions of the momenta of the scattered and recoil particles in the laboratory frame, respectively. The scattering matrix is rewritten in the new notation as M(σ1,σ2;ki,kf ) = a + b(σ 1 ·n)(σ 2 ·n) + c(σ1 + σ2) · n + d(σ1 − σ2) · n + e(σ1 · m)(σ2 · m) + f (σ1 · l)(σ2 · l). (2.123) The amplitudes a, b, c, d, e, and f are complex functions of the energy and scattering angle (kk ) = cosθ. The term with the amplitude d should be absent for nucleon–nucleon scattering. This is proved as follows (Bilenky et al. 1964). Two nucleons in the initial state have the internal parity (−1)l, total spin S, and total isospin T . According to the Pauli exclusion principle, the wave function of two nucleons should be antisymmetric under permutation, i.e., should change sign: Pi = (−1)l (−1)S+1 (−1)T +1 = −1. (2.124) A similar relation can also be obtained for the final nucleons with the orbital angular momentum l , spin S , and isospin T : Pf = (−1)l (−1)S +1 (−1)T +1 = −1. (2.125) To satisfy the condition Pi = Pf , we take into account that the parities of the nu- cleons in the interaction remain unchanged and the terms containing orbital angular momenta can be canceled. We also accept the hypothesis of the isotopic invariance of the strong interaction; for this reason, the terms with isotopic spin T can also be canceled. As a result, we arrive at the condition (−1)S = (−1)S . (2.126) Since the possible values of S and S are 0 (singlet state) and 1 (triplet state), then S = S . This means that transitions only within triplets and singlets separately are allowed in nucleon–nucleon scattering, whereas mixed singlet–triplet and in- verse transitions are forbidden. This leads to the exclusion of the term with d in the scattering matrix. The nucleon–nucleon scattering matrix has the final form M(σ1,σ2;ki,kf ) = a + b(σ 1 ·n)(σ 2 ·n) + c(σ1 + σ2) · n + e(σ1 · m)(σ2 · m) + f (σ1 · l)(σ2 · l). (2.127)
22. 22. 80 2 Spin in Strong Interactions The singlet and triplet projection operators are introduced as ˆS = 1 4 1 − (σ1 · σ2) , ˆT = 1 4 3 + (σ1 · σ2) . (2.128) Then, expression (2.127) can be represented in the form M(σ1,σ2;ki,kf ) = B ˆS + C(σ1 + σ2) · n + 1 2 G(σ1 · m)(σ2 · m) + (σ1 · l)(σ2 · l) + 1 2 H(σ1 · m)(σ2 · m) − (σ1 · l)(σ2 · l) + N(σ1 · n)(σ 2 ·n) ˆT. (2.129) The amplitude B corresponds to singlet scattering, whereas the remaining four amplitudes describe triplet scattering. The amplitudes in expressions (2.127) and (2.129) are related as: B = a − b − e − f, C = c, G = 2a + e + f, H = e − f, N = a + b. (2.130) For joint description of all possible types of nucleon–nucleon scattering (pp, nn, and np), the general matrix can be written taking into account isotopic invariance: M(σ1,σ2;ki,kf ) = M0 ˆT0 + M1 ˆT1. (2.131) Here, ˆT0 = 1 4 (1 − τ1 · τ2), ˆT1 = 1 4 (3 + τ1 · τ2) (2.132) are the isosinglet and isotriplet projection operators, respectively; and τ1 and τ2 are the isospin operators of the first and second nucleons, respectively. Each of the matrices M0 and M1 is a scattering matrix of form (2.129). The final wave function of the system of two nucleons can be written in the form χf = M σ1,σ2;k,k χiSχiT , (2.133) where χiS and χiT are the spin and isospin wave functions of the initial system of two nucleons. In view of Eqs. (2.129) and (2.132), the requirement of the antisym- metry of this function provides the following conditions on the amplitude under the change θ → π − θ: (a) the isotriplet amplitudes B, C, and H do not change their signs, whereas G and N change their signs; (b) on the contrary, the isosinglet amplitudes B, C, and H change their signs, whereas G and N do not change their signs. These relations allow one to investigate pp and nn scatterings only in the an- gular range 0 ≤ θ ≤ 90° . Moreover, the amplitude analysis for angles 0°, 90°, and 180° can be performed only with three rather than five amplitudes; this significantly reduces the number of necessary experimental observables. In the case of np scattering, where both isotopic matrices M0 and M1 are used, the measurements should be performed in a wider angular range, namely, 0 ≤ θ ≤ 180°.
23. 23. 2.6 Nucleon–Nucleon Scattering 81 2.6.2 Some Ways for Experimentally Seeking P - and T -noninvariant Terms in the Matrices of the Strong Interaction The nucleon–nucleon scattering matrix given by Eq. (2.129) can be written in the form M(0) = (u + v) + (u − v)(σ1 · n)(σ2 · n) + C (σ1 · n) + (σ2 · n) + (g − h)(σ1 · m)(σ2 · m) + (g + h)(σ1 · l)(σ2 · l), (2.134) where l, m, and n were defined in Eq. (2.122). The aim of the complete set of experiments on nucleon–nucleon scattering is to reconstruct the amplitudes u, v, c, g, and h from the experimental data. In the case of parity violation or time reversal violation, the scattering matrix contains additional terms, which are considered below. 2.6.2.1 Parity Violation The forward NN-scattering matrix in the case of parity violation can be written in the form M = M0 + (i/4)(Mos − Mso)(σ1 × σ2) · k + (i/4)(Mos + Mso)(σ1 − σ2) · k. (2.135) Here, M0 is given by expression (2.134) and the amplitudes Mos and Mso de- termine P -violating triplet–singlet and singlet–triplet transitions, respectively. The total cross section for the interaction between polarized particles corresponding to the matrix M is written in the form (Bilenky and Ryndin 1963; Philips 1963): σP1P2 = σ(0) P1P2 + (1/4)(σos − σso)(P1 × P2) · k + (1/4)(σos + σso)(P1 − P2) · k, (2.136) where P1 and P2 are the beam and target polarizations, respectively; and σos(σso) is the total cross section for the P -odd interaction with the triplet–singlet (singlet– triplet) transition. The cross section for the P -invariant interaction, σ (0) P1P2 , can be written in the form σ (0) P1P2 = σ0 + σ1(P1 · P2) + σ2(P1 · k)(P2 · k) = σ0 + σT P1T · P2T + σLP1L · P2L, (2.137) where the subscripts L and T mean the longitudinal and transverse polarization components of the beam. Formula (2.136) shows that, to detect a P -odd effect, it is necessary to measure the total cross section for the interaction of the longitudinally polarized beam with the unpolarized target or the unpolarized beam with the trans-
24. 24. 82 2 Spin in Strong Interactions versely polarized target (the third term). Such experiments have been performed, and they will be discussed in the next sections of this book. The second term in the formula (2.136) corresponds to simultaneous parity and time-reversal violation. To measure it, the polarized beam and polarized target, whose polarization vectors are perpendicular to each other and to the beam mo- mentum, should be used. Such experiments have not yet been carried out. 2.6.2.2 T -odd Terms If the interaction is invariant under space inversion, the term corresponding to the T -odd effect has the form M(1) = MT (σ1 · l)(σ2 · m) + (σ1 · m)(σ2 · l). (2.138) In the case of the matrix M(0), it can be shown that the polarization P of the final nucleon for the case of the unpolarized initial states is equal to left–right asymmetry AN in the scattering of the polarized nucleon on the unpolarized nucleon: P = AN . Note that this equality holds for the more general case of the direct a + b → c + d (2.139) and inverse c + d → a + b (2.140) reactions. In this case, the polarization P refers to the particle c in the direct reac- tion with the unpolarized particles a and b, whereas the asymmetry A refers to the particle a in the inverse reaction with polarized particles (Baz 1957). If the interaction contains the term M(1), the equality P = AN is violated and is replaced by the relation (for NN elastic scattering) σ0(P − A) = −8Im M∗ T · h . (2.141) This relation should be tested at the angles at which h is noticeably nonzero. Such a test is simplified if the unambiguous phase or amplitude analysis has been performed. 2.7 Complete Set of Experiments The idea of the complete set of experiments for the set of observables that com- pletely and unambiguously determine the reaction matrix elements was first pro- posed in Puzikov et al. (1957) and Smorodinskii (1960). In application to nucleon– nucleon elastic scattering, the possible ways for reconstructing the matrix elements were first proposed in Schumacher and Bethe (1961). We use these works to recon- struct the nucleon–nucleon scattering amplitudes.
25. 25. 2.7 Complete Set of Experiments 83 The nucleon–nucleon elastic scattering matrix was constructed in Sect. 2.6, where it was also shown that the total number of the independent complex am- plitudes necessary for describing the p + p → p + p reaction at a fixed angle and a fixed energy is five. This means that ten real quantities, namely, five absolute values of the amplitudes and five their phases, should be measured at a fixed angle and a fixed initial energy. Hence, these ten observables constitute a minimum set for the complete experiment. Pion–nucleon scattering is described by two amplitudes, and the complete set should consist of no less than four observables. The complete set for pion–pion scattering consists of only two observables. The number of the com- ponents of the complete experiments generally depends on the spin of the interacting particles. The number of observables in the complete set increases with the spin. It is seen that most observables in the complete experiment are associated with spin, i.e., spin carries rich information on interaction. Let us consider the examples of complete experiments at fixed angles and fixed initial energies. 2.7.1 Complete Experiment on pp Elastic Scattering at 0° in the Center-of-Mass Frame. The Total Cross Sections for Nucleon–Nucleon Interactions First, the total cross section σT for the interaction between two particles with spin 1/2 should be a scalar. Second, it should be a linear function of the polarizations of the initial particles, P1 and P2. Third, it should be composed of the kinematic quantities determining the reaction (Bilenky and Ryndin 1963; Philips 1963). Thus, σ = σ0 + σ1(P1 · P2) + σ2(P1 · k)(P2 · k). (2.142) Here, k is the unit vector in the incident beam direction and σ0, σ1, and σ2 are the experimentally measured parameters depending only on the initial beam energy. Their meaning is as follows. By the definition, the polarization is the mean value of the Pauli operator; hence, (P1 · P2) = (σ1 · σ2) = 2S2 − 3 , (P1 · k)(P2 · k) = (σ1 · k)(σ2 · k) = 2(S · k)2 − 1 . (2.143) Here, S = 1 2 (σ1 +σ2) is the total spin of two initial interacting nucleons. Accord- ing to Eqs. (2.142) and (2.143), (P1 · P2) = m wt m − 3ws , (P1 · k)(P2 · k) = m (−1)1+m wt m − ws , (2.144)
26. 26. 84 2 Spin in Strong Interactions where ws and wt are the probabilities of finding the system of two nucleons in the singlet and triplet states, respectively. Taking into account the normalization condition ws + m wt m = 1, it follows from Eq. (2.145) that ws = 1 4 1 − (P1 · P2) , wt 0 = 1 4 1 + (P1 · P2) − 2(P1 · k)(P2 · k) , wt + + wt − = 1 2 1 + (P1 · k)(P2 · k) . (2.145) Using expression (2.142) and the definition of the triplet projection operator, we can show that wt + = wt −. Representing the total cross section in the form of the sum of the weighted cross sections for the singlet and triplet states σ = ws σs + wt mσt m (2.146) and substituting the expressions for ws and wt m, we obtain σ = σ0 + 1 4 σt 0 − σs (P1 · P2) + 1 2 σt + − σt 0 (P1 · k)(P2 · k). (2.147) Comparison of Eqs. (2.142) and (2.147) provides the following relation between the coefficients: σ1 = 1 4 σt 0 − σs and σ2 = 1 2 σt + − σt 0 . (2.148) Thus, in an experiment with polarized nucleons, three total cross sections can be measured for the cases: (a) both nucleons are unpolarized, (b) both nucleons are polarized transversely to the beam, and (c) both nucleons are polarized along the beam. These three experiments constitute the complete set for determining the total cross sections for nucleon–nucleon interactions. As a result, σs, σt 0, and σt + can be reconstructed and their individual contributions to the usual (unpolarized) total cross section σ0 can be determined: σ0 = 1 4 σs + 1 4 σt 0 + 1 2 σt +. (2.149) The applications of the above relations are discussed in the section “Polarization experiments and results.” 2.7.2 Forward NN-Scattering Amplitudes In view of the symmetry condition, the forward scattering amplitudes satisfy the conditions c(0) = d(0) = 0, b(0) = e(0) and the forward scattering matrix has the form M(σ1,σ2;ki,kf ) = a(0) + e(0)(σ1 · σ2) + f (0) − e(0) (σ1 · k)(σ2 · k). (2.150)
27. 27. 2.7 Complete Set of Experiments 85 The above unitarity condition (see matrix relation (2.87) in Sect. 2.4) applied to matrix (2.150) leads to the following relations between the imaginary parts of the amplitudes and total cross sections (optical theorem): Ima(0) = k 4π σT , Ime(0) = k 4π σ1, Im f (0) − e(0) = k 4π σ2. (2.151) Here, k is the wave number in the center-of-mass frame. Thus, the measurements of three observables σ0, σ1, and σ2 allow one to reconstruct the imaginary parts of three amplitudes a, e, and f of the forward pp elastic scattering. The determination of three real parts of these amplitudes is necessary to com- plete the reconstruction of the scattering matrix. This requires the measurements of additional three parameters at very small angles (in the so-called Coulomb–nuclear interference region). One of them is the differential cross section. Such a measure- ment makes it possible to reconstruct the real part of the spin-independent ampli- tude a(0). The measurements of two other parameters, for example, σ1 and σ2, allow one to reconstruct the real parts of the amplitudes e and f through the dis- persion relations. Another way is to measure the spin–spin correlation parameters Aik (i,k = N,S,L) in the Coulomb–nuclear interference region. Here, N, S, and L denote the initial-proton polarizations that are (N) perpendicular to the reaction plane, (S) transverse, and longitudinal (L) with respect to the initial momentum in the reaction plane. Thus, the complete experiment on forward pp elastic scattering is finished. This approach is obviously universal and can be applied at any initial energy. Unfortunately, such a complete experiment has not yet been performed at any energy (except for the low-energy region, where the phase analysis has been per- formed). Relations (2.151) are useful in a number of cases. They impose additional con- ditions on the phases in the phase analysis, which can be substantial when choosing between several sets of phase solutions. The use of the dispersion relation allows the reconstruction of the real parts Re a(0), Ree(0), and Ref (0), if the imaginary parts of these amplitudes are known from the experimental data (these imaginary parts appear in the integrands in the dispersion relations). The differential cross section for forward proton–proton scattering is of great interest for theoretical analysis. At the same time, a method for its direct measurement is absent and it is necessary to use an extrapolation method; i.e., the differential cross sections for elastic scatter- ing are measured down to extremely small angles for which reliable data are yet obtained and, then, they are extrapolated to zero angle using a certain function, for example, an exponential. To test the resulting cross section, relations (2.151) are used as follows. The differential cross section for forward scattering can be written in the form dσ dΩ = k 4π 2 |a|2 + 2|e|2 + |f |2 . (2.152)
28. 28. 86 2 Spin in Strong Interactions This expression contains the squares of the real and imaginary parts of each of three amplitudes. The substitution of only imaginary parts from relations (2.151) gives the inequality dσ dΩ ≥ k 4π 2 (σT )2 + 2(σ1)2 + (σ1 + σ2)2 . (2.153) This is the test relation that provides a lower limit and is widely used to measure the differential cross sections for forward scattering. 2.7.3 Complete Set of Experiments on pp Elastic Scattering at 90° in the Center-of-Mass Frame One of the first attempts to solve this problem was made as early as in 1959 in Nurushev (1959). We briefly repeat the way proposed in that work. The following five parameters of pp elastic scattering were measured at an en- ergy of 660 MeV and an angle of 90° (Azhgirey et al. 1963): differential cross section I = (2.07 ± 0.03) mb/sr, spin correlation parameter Cnn = (0.93 ± 0.21), depolarization parameter D = (0.93 ± 0.17), transverse polarization rotation pa- rameter R = (0.26 ± 0.07), and longitudinal polarization rotation parameter A = (0.20 ± 0.06). Using these five observables, one can reconstruct three absolute val- ues and two relative phases of the nonzero amplitudes B, C and H (see Sect. 2.6 “Nucleon–nucleon scattering”). The phase of the amplitude B is taken to be zero. Thus, the amplitudes (dimensionless) normalized to the cross section are given by the formulas |b|2 = |B|2 4I = 1 2 (1 − Cnn), |c|2 = 2|C|2 I = 1 4 (1 + Cnn + 2D), |h|2 = |H|2 2I = 1 4 (1 + Cnn − 2D), sin(δC − δB) = − R + A 2dc , cos(δH − δB) = A − R 2bh . (2.154) The substitution of the numerical values of the observables gives (Kumekin et al. 1954) |b|2 = 0.35 ± 0.11, |c|2 = 1.00 ± 0.10, |h|2 = 0.02 ± 0.10, sinδc = 0.39 ± 0.21, cosδh = −0.36 ± 0.18. Comparison with similar data for lower energies shows (Nurushev 1959) that the contributions from the triplet amplitudes c and h prevail in the energy range under consideration, whereas the contribution from the singlet term b is smaller. The terms h (tensor interaction) and c (spin–orbit interaction) dominate in the upper and lower energy ranges, respectively.
29. 29. 2.7 Complete Set of Experiments 87 2.7.4 Complete Set of Experiments on pp Elastic Scattering at an Arbitrary Angle in the Center-of-Mass Frame In this section, we try to answer the question: How many and what particular ob- servables should be measured at a given initial energy in order to reconstruct the amplitude of nucleon–nucleon elastic scattering at an arbitrary angle θ in the center- of-mass frame? More briefly, how many measurements constitute the complete set of experiments? Unfortunately, the complete set of experiments has not yet been performed for any energy above 3 GeV. The direct reconstruction of the elements of the nucleon–nucleon scattering matrix from experimental data is the single method of analysis at energies above the meson production threshold. In such an approach, one common phase in the scattering matrix remains undetermined. It can be deter- mined at energies below the meson production threshold by means of the unitarity relation. The possibility of directly reconstructing the scattering amplitudes was dis- cussed in Puzikov et al. (1957), Smorodinskii (1960), and Schumacher and Bethe (1961). In the last work, 11 experimental observables (the differential cross section, polarization, and components of the depolarization, polarization transfer, and polar- ization correlation tensors) were used and the absolute values of five amplitudes, as well as four relative phases, were unambiguously reconstructed. In agreement with the expectations, one common phase was undetermined. According to Schumacher and Bethe (1961), the complete set includes tensors up to the second order. The pos- sibilities of simplifying the procedure for reconstructing the amplitudes with the use of the polarization tensors of the third and fourth orders were considered in Bilenky et al. (1965) and Vinternitts et al. (1965). It is not excluded that theoretical ideas can noticeably reduce the number of nec- essary experiments of the complete set at asymptotic energies. Below, we discuss the method for reconstructing the scalar amplitudes of nucleon–nucleon scattering in the relativistic case and present the particular sets of the complete set of experiments. In this presentation, we follow Bilenky et al. (1966). We write the nucleon–nucleon scattering matrix in the form (Wolfenstein and Ashkin 1952; Dalitz 1952) M p ,p = (u + v) + (u − v)(σ1 · n)(σ2 · n) + c(σ1 + σ2) · n + (g − h)(σ1 · m)(σ2 · m) + (g + h)(σ1 · l)(σ2 · l). (2.155) Here, the complex scalar scattering amplitudes u,v,c,g, and h are functions of the energy and scattering angle θ. Our main aim is to express these amplitudes in terms of experimental quantities. The unit vectors n, l, m are given by the expres- sions l = p + p |p + p| , m = p − p |p − p| , n = l × m = p × p |p × p | . (2.156) These unit vectors defined in the center-of-mass frame are mutually orthogo- nal. In the nonrelativistic approximation, the vectors l and m are directed along the
30. 30. 88 2 Spin in Strong Interactions scattered and recoil particle momenta in the laboratory frame, respectively. In the relativistic case, this relation is invalid and an additional rotation angle appears. The proton–proton scattering matrix should satisfy the Pauli exclusion principle: M p ,p = −Π(1,2)M −p ,p = −M p ,−p Π(1,2). (2.157) Here, Π(1,2) = 1 2 (1 + σ1 · σ2) (2.158) is the operator of the permutation of the spin variables. The substitution of pp- scattering matrix (2.155) into relation (2.157) indicates that the scalar scattering amplitudes satisfy the following symmetry conditions: u(π − θ) = −u(θ), h(π − θ) = h(θ), c(π − θ) = c(θ), v(π − θ) = −g(θ). (2.159) It follows from these relations that, for example, three amplitudes, c, h and, e.g., g, are nonzero for an angle of 90°. Their reconstruction was considered above. Neutron–proton elastic scattering was not discussed above. Such a discussion is most appropriate with the isotopic invariance hypothesis. In this case, two isotopic scattering matrices M1(p ,p) and M0(p ,p) with isospins 1 and 0, respectively, can be introduced. Both matrices are written in the form, where the scalar amplitudes have subscripts 1 and 0, and the matrix M1(p ,p) coincides with the pp-scattering matrix. In this case, the np-scattering matrix is determined by the expression Mnp p ,p = 1 2 M1 p ,p + M0 p ,p . (2.160) The generalized Pauli exclusion principle can be applied to the isotopic scattering matrices. Introducing the subscript i = 0,1, we write the Pauli condition as follows: Mi p ,p = (−1)i Π(1,2)Mi −p ,p = (−1)i Mi p ,−p Π(1,2). (2.161) This condition provides the following constraints on the scalar scattering ampli- tude in different isotopic states: ui(π − θ) = (−1)iui(θ), hi(π − θ) = (−1)i+1hi(θ), ci(π − θ) = (−1)i+1ci(θ), vi(π − θ) = (−1)igi(θ). (2.162) In the nonrelativistic approximation, the problem of the joint analysis of the pp- and np-scattering data for reconstructing the scattering matrix was considered in Kazarinov (1956) and Golovin et al. (1959). A theoretical analysis of collisions between particles is usually performed in the center-of-mass frame. However, observables such as cross sections and polarizations are measured in the laboratory frame. Therefore, it is necessary to determine the rules for the transition from one frame to the other taking into account the relativistic kinematics and the specificity of the spin transformations. Let us consider particular examples.
31. 31. 2.7 Complete Set of Experiments 89 According to the general rules, the mean value of any spin operator σ1L is expressed as σ1L aL = Tr σ1 aL R ρf /Tr ρf . (2.163) Here, ρf is the density matrix of the final state and aL is an arbitrary unit vector in the laboratory frame (L frame). It is intended to measure the projection of the po- larization vector of the first particle on this direction. The vector (aL)R = Rn(Ω )aL includes the relativistic transformation of the spin vector and is obtained from the vector aL by means of its rotation by the angle Ω = θ − 2θL about a vector perpen- dicular to the scattering plane. In the nonrelativistic limit, the center-of-mass scat- tering angle θ is equal to 2θL, where θL is the laboratory scattering angle. Hence, Ω = 0 in this case. For the measurement of the projection of the polarization vector of the recoil particle (particle 2) on the direction of the unit vector bL in the L frame, we have σ2L bL = Tr σ2 bL R ρf /Tr ρf . (2.164) It is intended to measure the projection of the polarization vector of the second particle on this direction. The vector (bL)R = Rn(Ω )bL includes the relativistic transformation of the spin vector and is obtained from the vector bL by means of its rotation by the angle Ω = 2ϕL − ϕ about a vector perpendicular to the scattering plane, where ϕ = π − θ and ϕL are the emission angles of the second particle in the center-of-mass frame and L frame, respectively. The correlation of polarization projections on the same unit vectors (aL,bL) can be measured in the experiment: σ1 · aL σ2 · bL = Tr σ1 · aL R σ2 · bL R ρf /Tr ρf . (2.165) Let us introduce the following three sets of the orthonormalized unit vectors in the laboratory frame: nL, kL, sL = nL × kL, (2.166) nL, kL, sL = nL × kL, (2.167) nL, kL, sL = nL × kL. (2.168) Here, kL, kL, kL are the unit vectors in the directions of the momenta of the inci- dent, scattered, and recoil nucleons, respectively, and nL = kL ×kL/| ← kL ×kL| is the unit vector perpendicular to the scattering plane; nL = n, where n is the unit vector perpendicular to the scattering plane in the center-of-mass frame. In the subsequent presentation, the initial, scattered, and recoil particles are described in coordinate systems (2.166), (2.167), and (2.168), respectively. When calculating observables, we use tensors up to the second rank. 1. The zero-rank tensor is the differential cross section. It is given by the expression σ0 = Tr ρf = 1 4 Tr MM+ = 2 |u|2 + |v|2 + |c|2 + |g|2 + |h|2 . (2.169) 2. The first-rank tensors:
32. 32. 90 2 Spin in Strong Interactions (a) The initial particles are unpolarized. The i-th component of the polarization of the scattered particle is measured. It is given by the expression P1iσ0 = 1 4 Tr σ1iMM+ . (2.170) (b) The initial particles are unpolarized. The i-th component of the polarization of the recoil particle is measured. It is given by the expression P2iσ0 = 1 4 Tr σ2iMM+ = P1iσ0. (2.171) (c) The incident (first) particle is polarized in the i-th direction. The target (sec- ond) particle is unpolarized. The i-th component of the asymmetry is mea- sured. It is given by the expression A1iσ0 = 1 4 Tr Mσ1iM+ . (2.172) (d) The incident (first) particle is unpolarized. The target (second) particle is polarized in the i-th direction. The i-th component of the asymmetry of the recoil particle is measured. It is given by the expression A2iσ0 = 1 4 Tr Mσ2iM+ = A1iσ0. (2.173) 3. The second-rank tensors: (a) The depolarization tensor Dik. The first particle is polarized (component i) in the initial state, and its polarization (component k) after scattering is mea- sured. The expression for the tensor Dik has the form Dikσ0 = 1 4 Tr σ1iMσ1kM+ . (2.174) (b) The depolarization tensor of the second particle D (2) ik . The second particle is polarized (component i, the first particle is unpolarized) in the initial state, and its polarization (component k) after scattering is measured. The expres- sion for the tensor D (2) ik has the form D (2) ik σ0 = 1 4 Tr σ2iMσ2kM+ = Dikσ0. (2.175) (c) The polarization transfer tensor from the first particle to the second Kik. The first particle is polarized (the second particle is unpolarized) in the initial state, and the polarization of the second particle after scattering is measured. The expression for the tensor Kik has the form Kikσ0 = 1 4 Tr σ1iMσ2kM+ . (2.176) (d) The polarization transfer tensor from the second particle to the first one K (2) ik . The second particle is polarized (the first particle is unpolarized) in the initial
33. 33. 2.7 Complete Set of Experiments 91 state, and the polarization of the first particle after scattering is measured. The expression for the tensor K (2) ik has the form K (2) ik σ0 = 1 4 Tr σ2iMσ1kM+ = Kikσ0. (2.177) (e) The polarization correlation tensor Cik. Both particles in the initial state are unpolarized, and the correlation of their polarizations after scattering is mea- sured. The expression for the tensor Cik has the form Cikσ0 = 1 4 Tr σ1iσ2kMM+ . (2.178) (f) The polarization correlation tensor C (2) ik . Both particles in the initial state are unpolarized, and the correlation of their polarizations after scattering is measured. The difference from (e) is that the Pauli operators are enclosed by M matrices. The expression for the tensor C (2) ik has the form C (2) ik σ0 = 1 4 Tr σ1iσ2kMM+ . (2.179) (g) The two-spin asymmetry tensor or asymmetry correlation tensor Aik. Both particles in the initial state are polarized: the first and second particles are polarized in the directions i and k, respectively. Asymmetry after scattering is measured. The expression for the tensor Aik has the form Aikσ0 = 1 4 Tr Mσ1iσ2kM+ . (2.180) (h) The two-spin asymmetry tensor A (2) ik . Both particles in the initial state are polarized: the first and second particles are polarized in the directions k and i, respectively. Asymmetry after scattering is measured. The expression for the tensor A (2) ik has the form A(2) ik σ0 = 1 4 Tr Mσ1kσ2iM+ = Aikσ0. (2.181) The requirements of the invariance of strong interactions under a number of con- tinuous (isotropy and uniformity of space) and discrete (space inversion and time reversal) transformations lead to relations between measured quantities. For exam- ple, Pi = Ai = Pni Cik p ,p = Aik −p,−p . (2.182) For the case of the beam with polarization P1 and the polarized target with polariza- tion P2 (or two colliding polarized proton beams with the indicated polarizations), the general requirements of the invariance of the scattering matrix under the trans- formations listed above in the parentheses provide the following expression for the differential cross section σ(P1,P2) = σ0 1 + P(P1 + P2) · nL + Ann(P1 · nL)(P2 · nL) + Ass(P1 · sL)(P2 · sL) + Akk(P1 · kL)(P2 · kL) + Ask (P1 · sL)(P2 · kL) + (P1 · kL)(P2 · sL) . (2.183)
34. 34. 92 2 Spin in Strong Interactions Here, P is the polarization appearing after the scattering of the unpolarized par- ticles and Aab = (aL)iAik(bL)k. (2.184) For the measurements of the depolarization parameters Dik or the polarization transfer tensor Kik, additional scattering is necessary in order to determine the po- larization of the scattered particles. Such experiments are very difficult because, first, luminosity in the second scattering process is low and, second, it is difficult to find analyzers with a high analyzing power at high energies. This is one of the main reasons why, for example, such experiments are not planned at the RHIC collider. In contrast to these parameters, the asymmetry correlation tensor Aik is directly measured at RHIC. For the measurements of, for example, the parameter All, the polarization of both initial particles should be oriented along their momentum; i.e., both beams should be longitudinally polarized. To measure, for example, Asl, one beam of particles should be polarized along the vector s, and the other, along the vector l. All these possibilities can be implemented at the RHIC polarized particle collider. As mentioned above, the measurements of observables are carried out in the laboratory frame, whereas all theoretical jobs with these quantities are performed in the center-of-mass frame. Let us determine the transformations between these two frames. First, we recall the relation between the triples of unit orthogonal vectors in the laboratory and center-of-mass frames: nL = n, kL = k, sL = s. (2.185) Here, k = p/|p| is the unit vector in the direction of the incident particle momen- tum in the center-of-mass frame, n is the unit vector perpendicular to the scattering plane in this frame, and s = n × k is the vector that is perpendicular to the mo- mentum of the initial particles and lies in the scattering plane. Thus, we obtain the relations Ass = C+ − Clm sinθ − C− cosθ, Akk = C+ + Clm sinθ + C− cosθ, Ask = −Clm cosθ + C− sinθ. (2.186) Here, C+ = 1 2 (Cll + Cmm), C− = 1 2 (Cll − Cmm). (2.187) The components Ann and Cnn perpendicular to the scattering plane are the same in both frames. The other components of the tensors satisfy the relations C+ = 1 2 (Ass + Akk), (2.188) Clm = −Ask cosθ + 1 2 (Akk − Ass)sinθ, (2.189) C− = Ask sinθ + 1 2 (Akk − Ass)cosθ. (2.190)
35. 35. 2.7 Complete Set of Experiments 93 Here, the polarization correlation tensors Cik are expressed in terms of the asym- metry correlation tensors Ann. This representation is reasonable. Indeed, the tensors Cik can also be measured with the use of unpolarized initial particles. However, this measurement requires the analysis of the final polarization (double scattering) and, therefore, analyzing scattering. As mentioned above, such experiments are very difficult. At the same time, it is easier to measure the parameters Ann (in single scat- tering) and to determine Cik in terms of these parameters by the above formulas. In the first experiments at low energies (beam kinetic energy 100–600 MeV), the spin correlation tensor Cnn was measured and other parameters were measured later. We consider these tensors: Cs s = sL Ri Cik sL Rk , Cs k = sL Ri Cik kL Rk , Ck s = kL Ri Cik sL Rk , Ck k = kL Ri Cik kL Rk . (2.191) Let us apply the rotation operator about the vector n to an arbitrary vector a and represent the result in terms of three vectors: Rn(Ω)a = (a · n)n(1 − cosθ) + a cosΩ + (n × a)sinΩ. (2.192) Using this relation and formulas (2.183), we can find that kL R = Rn Ω kL = l cosα + msinα, (2.193) sL R = Rn Ω sL = −l sinα + mcosα. (2.194) Here, the relativistic spin rotation angle is α = θ/2 − θL, where θ and θL are the particle scattering angles in the center-of-mass and laboratory frames, respectively. Similar transformation formulas can be obtained for the recoil particle: kL R = Rn Ω kL = −l sinα − mcosα , (2.195) sL R = Rn Ω sL = l cosα − msinα . (2.196) Here, the relativistic spin rotation angle is α = ϕ/2−ϕL, where ϕ and ϕL are the scattering angles of the recoil particle in the center-of-mass and laboratory frames, respectively. We emphasize two features. First, relativistic spin precession (Thomas preces- sion) concerns only the polarization components lying in the scattering plane and does not involve its normal component. Second, in the nonrelativistic limit, angles are α = α = 0 and Thomas precession at low energies does not affect observables. In the nonrelativistic case, we have the equalities kL R = l, sL R = m, kL R = − ← m, sL R = l. (2.197) Let us express experimentally measured quantities (2.189) in terms of the tensors in the center-of-mass frame using relations (2.191)–(2.194). The desired formulas have the form Cs s = −C+ sin α + α + Clm cos α − α − C− sin α − α . (2.198) Cs k = −C+ cos α + α + Clm sin α − α + C− cos α − α . (2.199)
36. 36. 94 2 Spin in Strong Interactions Ck s = C+ cos α + α + Clm sin α − α + C− cos α − α . (2.200) Ck k = −C+ sin α + α − Clm cos α − α + C− sin α − α . (2.201) It is easy to verify that the observables are related as (Cs s + Ck k )/(Cs k − Ck s ) = tan α + α . (2.202) Hence, the number of the independent observables is three rather than four. Let us solve the inverse problem, i.e., express three parameters C+,, C−, Clm in terms of the observables Cs s , Cs k , and Ck s . Using relations (2.196)–(2.199), we obtain C+ = (Ck s − Cs k )/2cos α + α , (2.203) C− = 1 2 (Cs k + Ck s )cos α − α − Cs s + 1 2 tan α + α (Ck s − Cs k ) sin α − α , (2.204) Clm = 1 2 (Cs k + Ck s )sin α − α + Cs s + 1 2 tan α + α (Ck s − Cs k ) cos α − α . (2.205) Now, we consider the case with the polarized beam and unpolarized target. The polarization components of the scattered particles are measured after collisions. In view of invariance, these components can be written in the form σ(P1) σ1 L · nL = σ0 P + Dnn(P1 · nL) , (2.206) σ(P1) σ1 L · kL = σ0 Dk k(P1 · kL) + Dk s(P1 · sL) , (2.207) σ(P1) σ1 L · sL = σ0 Ds k(P1 · kL) + Ds s(P1 · sL) . (2.208) Here, σ(P1) is the differential cross section for the scattering of particles with polarization P1 arbitrarily oriented in space on the unpolarized target; it is given by the expression σ(P1) = σ0 1 + P(P1 · n) . (2.209) Taking certain components of the initial and final polarizations, we arrive at the known Wolfenstein parameters: Dnn = D = (nL)iDik(nL)k, Ds s = R = sL Ri Dik(sL)k; (2.210) Ds k = A = sL Ri Dik(kL)k, Dk s = R = kL Ri Dik(sL)k; (2.211) Dk k = A = kL Ri Dik(kL)k. (2.212) The spin rotation parameters (A,R,A ,R ) contain the final unit vectors with the subscript R, which means the necessity of the inclusion of the relativistic spin rotation in the reaction plane.
37. 37. 2.7 Complete Set of Experiments 95 Referring readers interested in the details of the derivation of the following for- mulas to Bilenky et al. (1965), we present the expressions of physical observables in terms of the scattering amplitude in the relativistic case: σ0 = 2 |u|2 + |v|2 + |c|2 + |g|2 + |h|2 , (2.213) σ0Dnn = 2 |u|2 + |v|2 + |c|2 − |g|2 − |h|2 , (2.214) σ0Knn = 2 |u2 | − |v|2 + |c|2 + |g|2 − |h|2 , (2.215) σ0Cnn = 2 |u2 | − |v|2 + |c|2 − |g|2 + |h|2 , (2.216) σ0P = 4Recu∗ , (2.217) σ0D+ = 4Reuv∗ , (2.218) σ0D− = 4Regh∗ , (2.219) σ0Dlm = 4Imcv∗ , (2.220) σ0K+ = 4Reug∗ , (2.221) σ0K− = 4Revh∗ , (2.222) σ0Klm = 4Imcg∗ , (2.223) σ0C+ = 4Revg∗ , (2.224) σ0C− = 4Reuh∗ , (2.225) σ0Clm = −4Imch∗ . (2.226) These 14 experimental observables are used to reconstruct five amplitudes and their phases. In this case, one common phase remains undetermined. Fundamentally, it also can be reconstructed. However, this problem is not discussed here. One of the variants of reconstruction of the amplitudes is as follows (for details, see Bilenky et al. 1966): |g|2 = 1 8 σ0(1 + Knn − Dnn − Cnn); (2.227) |h|2 = 1 8 σ0(1 − Knn − Dnn + Cnn); (2.228) |v|2 = 1 8 σ0(1 − Knn + Dnn − Cnn); (2.229) |u|2 + |c|2 = 1 8 σ0(1 + Knn + Dnn + Cnn). (2.230) For further analysis, it is necessary to fix the phase of any of five amplitudes. For example, let the amplitude c be real positive. This means that the scattering matrix is determined up to the phase of the amplitude c. Under this condition, we obtain Reu = 1 4c σ0P, Imh = 1 4c σ0Clm, (2.231) Imv = − 1 4c σ0Dlm, Img = − 1 4c σ0Klm. (2.232)
38. 38. 96 2 Spin in Strong Interactions For further calculations, we write the following identity for any two complex quantities: |x|2 |y|2 − Rexy∗ 2 = |x|2 (Imy)2 + |y|2 (Imx)2 − 2Rexy∗ Imx Imy. (2.233) Taking x = g and y = h and using Eqs. (2.231) and (2.232), we obtain c2 = |g|2M2 − |h|2N2 − 2Regh∗MN |g|2|h|2 − (Regh∗)2 . (2.234) Here, the quantities |g|2, |h|2, and Regh∗ were determined above and M = 1 4 σ0Clm, N = − 1 4 σ0Klm. (2.235) It remains to determine the signs of the quantities Imu, Reh, Rev, Reg. (2.236) One relation can be written immediately in the form Regh∗ = Reg Reh + Img Imh = 1 4 σ0D−. (2.237) Using Eq. (2.218), we can determine the signs of Imu and Rev. Any unused equation from the set of Eqs. (2.213)–(2.226) can be used to eliminate the remaining ambiguity. Thus, the problem of the relativistic reconstruction of the nucleon–nucleon scat- tering matrix has been solved in the general form. Since the volume of the book is limited, we omit such interesting problems as the reconstruction of the amplitudes by means of the measurements of the polarization parameters of the recoil particles and the joint analysis of pp and np scatterings using isotopic invariance. 2.8 Partial Wave Analysis The results of polarization experiments on the elastic scattering of nucleons in the low-energy range (0.1–10 GeV) were considered using the partial wave analysis (Hoshizaki 1968; Matsuda 1993). In this method, the scattering amplitude is ex- panded in terms of the eigenfunctions of the complete set of conserving operators and the expansion coefficients are the elements of the scattering matrix S. These elements are expressed in terms of the phase shifts, which contain com- plete information on the interaction process. There are several reasons to apply this method. First, the number of the phases directly depends on the maximum orbital angular momentum Lmax in the interaction. According to nonrelativistic quantum mechanics, Lmax is related to the impact parameter b as (Lmax + 1 2 ) ≈ bpi, where pi is the incident particle momentum in the center-of-mass frame. According to this relation, Lmax increases with energy; thus, the phase analysis becomes impossible when the number of free parameters is equal to or larger than the number of ex- perimental points. The situation is further complicated at energies above the meson
39. 39. 2.8 Partial Wave Analysis 97 production threshold, when the phases become complex. This is the main reason why the phase analysis is not applied for high energies. Nevertheless, the phase analysis was widely used for low energies in 1950–1960 for several reasons. First, the number of the phases and, correspondingly, the num- ber of free parameters are small for low energies. For example, taking b = 1.5mπ c for the impact parameter, we can estimate Lmax = 1 and 3 at the kinetic energy T = 50 and 300 MeV, respectively. Hence, it is easy to perform the phase analysis. The second important reason to perform this analysis is that the phases, being de- pendent on the impact parameter, make it possible to scan the internal structure of the nucleon. If the particles pass at a large distance from each other, the interaction between them is weak and the phases are small. On the contrary, for the central col- lision (L = 0), the phases are expected to be large. Deviations from this picture are obviously possible, for example, in the presence of repulsion forces or when reso- nances are formed. The third reason to perform the phase analysis at low energies is that the angular dependence of any observable can be calculated (predicting its behavior) using a few phases and, then, the calculation can be compared with exper- imental data. Fourth, any theoretical model should be tested on the phase analysis data if the phase analysis is unambiguous. Let us express the scattering matrix elements in terms of the phase. In view of the unitarity of the S-matrix, it can be written in terms of the phase operator δ as follows: S = eiδ . (2.238) The nucleon–nucleon interaction matrix should conserve the total angular mo- mentum J , total spin S, and parity Π = (−1)l. The requirement of the antisymmetry under the permutation of two nucleons leads to the relation (−1)S+1+T +1 Π = (−1)S+T +L = −1. (2.239) Here, T is the isospin of the system of two nucleons. Relation (2.239) should be applied separately for the system of two nucleons in the initial and final states. Tak- ing into account this relation, the elements of the S-matrix can be characterized by three quantum numbers: the total angular momentum J , the total spin of the system of two nucleons S, and the orbital angular momentum L, because T is unambigu- ously determined from the above relation; i.e., T can be omitted when specifying the elements of the S-matrix. Let us consider the matrix M defined by the expression M(pf ,pi) = 2π ik θf ϕf |S − 1|θiϕi . (2.240) Here, |θiϕi and θf ϕf | are the wave functions of the initial and final system of two nucleons, respectively; and θn, ϕn, where n = i, f , are the angles of the mo- menta pi, pf . The elements of this matrix in spin space are given by the expression Sms M(pf ,pi) S ms = 2π ik θf ϕf ,Sms e2iδ − 1 S ms,θiϕi . (2.241)
40. 40. 98 2 Spin in Strong Interactions This expression can be rewritten in terms of the spherical functions of the angles using the properties of the completeness and orthogonality of the wave functions 2π ik θf ϕf |LmL LmLSmS|LSJmJ LSJmJ |e2iδ − 1 L S J mJ × L S J mJ L mLS mS L mL θiϕi . (2.242) Here, θϕ|LmL = YmL L (θ,ϕ). (2.243) Below, we use the following notation for the Clebsch–Gordan coefficients ap- pearing in expression (2.242): CLS(JmJ mLmS) = LmLSmS|LSJmJ . (2.244) The quantization axis is usually taken in the direction of the incident particle momentum; in this case, the angles θi and ϕi are zero and the angles θf and ϕf are the scattering angles of the final particles. After these simplifications, since S, J , and mJ are conserving quantum numbers, expression (2.242) can be represented in the form Sms M(pf ,pi) S ms = δSS 4π ik L L+S J=|L−S| L+S L =|J−S| 2L + 1 4π Y ms−ms L (θ,ϕ) × CLS J,ms,ms − ms,ms CLS J,ms,0,ms LSJms e2iδ − 1 L SJms . (2.245) The summation should be performed taking into account antisymmetry condition (2.239) and that the scattering matrix elements are independent of the projection of the total angular momentum, mJ , due to the isotropy of space. The nonzero elements of the matrix S − 1 are denoted as RL = L0LmJ e2iδ − 1 L0LmJ ; RLJ = L1JmJ e2iδ − 1 L1JmJ ; (2.246) −RJ ± = RJ = J ± 1,1,J,mJ e2iδ − 1 J ∓ 1,1,J,mJ . Time reversal invariance leads to the equality RJ + = RJ − = RJ . For further sim- plification, we introduce the following notation for the triplet and singlet elements: Mmsms = 1ms M 1ms , Mss = 00|M|00 . (2.247) As a result, the nonzero matrix elements for the singlet and triplet states can be written in the form Mss = 2π ik 2L + 1 4π RLY0 L(θ,ϕ) (2.248) for the singlet transitions and
41. 41. 2.8 Partial Wave Analysis 99 Mmsms = 2π ik L L+1 J=L−1 2L + 1 4π CL1 J,ms,ms − ms,ms CL1 J,ms,0,ms RLJ − 1 J=L±1 2L + 1 4π CL1 J,ms,ms − ms,ms CL1 J,ms,0,ms RJ × Y ms−ms L (θ,ϕ) (2.249) for the triplet transitions. Here, L = 2J − L in the second term. When applying these formulas to pp elastic scattering, it is necessary to take into account two cir- cumstances. First, two protons are identical; therefore, measuring instruments can- not determine which proton, from the beam or from the target, is detected. As a result, the number of counts is twice as large as that following from the above con- sideration. Thus, to compare with the theoretical cross section, the measured cross sections should be halved. The second circumstance is associated with the antisym- metry condition (Pauli exclusion principle). Owing to this circumstance, the partial amplitudes with spins S = 0 and 1 include only even and odd orbital angular mo- menta, respectively. Since the orbital angular momentum is not conserved, for a given total angular momentum J in the triplet state, there are two states differing in the orbital angular momentum. As an example, we point to the states 3P2–3F2, 3F4–3H4, etc. of the pp system. For each pair of such states, the mixed transitions 3P2 ↔3 F2, 3F4 ↔ 3H4 are possible in addition to the direct transitions 3P2 → 3P2, 3F2 → 3F2. Correspondingly, the phases describing the direct transitions should be supplemented by additional parameters for describing the mixed transitions. Such parameters are called mixing parameters and are usually denoted as εJ . To describe the mixed transitions in the absence of inelastic channels, the following two-dimensional symmetric unitary submatrix is introduced: SJ − 1 = RJ−1,J −RJ −RJ RJ+1,J . (2.250) An unambiguous method for parameterizing this matrix in terms of the phase is absent. One of the methods was proposed by Blatt and Weisskopf (1952) and Blatt and Biedenharn (1952). This method is the diagonalization of the matrix by means of a unitary transformation SJ = GSJ G−1 , (2.251) where SJ = e2iδJ−1,j 0 0 e2iδJ+1,j , G = cosεJ −sinεJ sinεJ cosεJ . (2.252) This set of phase shifts is called the proper phase shift and is convenient if the Coulomb interaction can be neglected. Another parameterization variant was proposed in Stapp et al. (1957) in the form SJ = ˜SJ ˜G ˜SJ , (2.253)
42. 42. 100 2 Spin in Strong Interactions where ˜SJ = ei ¯δJ−1,j 0 0 ei ¯δJ+1,j , ˜G = cos2¯εJ i sin2¯εJ i sin2¯εJ cos2¯εJ . (2.254) This parameterization is favorable over parameterization (2.252), because it pro- vides the estimation of the mixing parameters for low orbital angular momenta (where the nuclear interaction prevails over the Coulomb interaction) in the pure form (without the Coulomb contribution). An additional advantage of this param- eterization is that it allows one to more clearly separate the nuclear and Coulomb contributions. For this reason, this parameterization is more often used in the phase analysis. The relation between representations (2.251) and (2.253) can be found by equat- ing them to each other, because they are the elements of the same matrix. This relation is expressed as follows: δJ−1,J + δJ+1,J = ¯δJ−1,J + ¯δJ+1,J , sin(δJ−1,J − δJ+1,J ) = sin2¯εJ /sin2εJ , sin(¯δJ−1,J − ¯δJ+1,J ) = tan2¯εJ /tan2εJ . (2.255) Table 2.1 presents the elements of the matrix M in terms of the partial waves h. These matrix elements can also be used in the case of neutron–proton elastic scat- tering with three changes: (a) the Coulomb amplitudes are neglected, (b) all sums over the even or odd L values are extended over all L values (even and odd), and (c) the resulting sums are multiplied by a factor of 1/2. The partial nuclear amplitudes h are expressed in terms of the scattering phases by the formulas 2ikhl = e2i ¯δN l − 1 e2iΦl (2.256) for the singlet states and 2ikhlj = e2i ¯δN lj − 1 e2iΦl (2.257) for the triplet states. For the mixed singlet–triplet states, 2ikhj±1,j = cos2εN j e 2i ¯δN j±1,j − 1 e2iΦl (2.258) 2khj = sin2εN j e i(¯δN j−1,j +¯δN j+1,j ) , (2.259) where εN j is the mixing parameter for the total angular momentum j and the super- script N means that this parameter refers to pure nuclear scattering. The Coulomb amplitudes are defined as follows: fc(θ) = −n k(1 − cosθ) e−inlog[(1−cosθ)/2] , (2.260) where n = e2 v and v is the relative velocity in the center-of-mass frame. The sym- metrized and antisymmetrized Coulomb amplitudes used in the partial wave analy- sis are presented in Table 2.1.
43. 43. 2.8 Partial Wave Analysis 101 Table 2.1 Singlet–triplet matrix elements for pp elastic scattering in terms of the partial ampli- tudes h. The Coulomb interaction contributions are presented in the explicit form with allowance for the identity of protons 1 Mss = fc,s + 2 even l(2l + 1)hlPl 2 M11 = fc,a + odd l[(l + 2)hl,l+1 + (2l + 1)hl,l + (l − 1)hl,l−1 − √ (l + 1)(l + 2)hl+1 − √ (l − 1)lhl−1]Pl 3 M00 = fc,a + 2 odd l[(l + 1)hl,l+1 + lhl,l−1 + (l − 1)hl,l−1 + √ (l + 1)(l + 2)hl+1 + √ (l − 1)lhl−1]Pl 4 M01 = √ 2 odd l[−l+2 l+1 hl,l+1 + 2l+1 l(l+1) hl,l + l−1 l hl,l−1 + l+2 l+1 hl+1 − l−1 l hl−1]P 1 l 5 M10 = √ 2 odd l[hl,l+1 − hl,l−1 + l+2 l+1 hl+1 − l−1 l hl−1]P 1 l 6 M1−1 = odd l[ 1 l+1 hl,l+1 − 2l+1 l(l+1) hl,l + 1 l hl,l−1 − 1√ (l+1)(l+2) hl+1 − 1√ (l−1)l hl−1]P 2 l 7 M11 − M00 − M1−1 − √ 2ctgθ(M10 + M01) = 0 8 fc,s = fc(θ) + fc(π − θ),fc,a = fc(θ) − fc(π − θ), where fc is the Coulomb amplitude Pl, P 1 l , and P 2 l are the associated Legendre polynomials of the zeroth, first, and second orders, respectively Let us consider the separation of the Coulomb and nuclear contributions in the phase analysis. Since Coulomb forces are long-range and nuclear forces are short- range, i.e., they weakly overlap, it is usually accepted that the pure nuclear, ¯δN , and Coulomb, φ, phases are added. In this case, ¯δN L = ¯δL − φL, ¯δN JL = ¯δJL − φL, ¯εN J = ¯εJ . (2.261) These phases denoted by the overlined symbols with the superscript N are called pure nuclear phases. The Coulomb phases are calculated by the formula (Stapp et al. 1957): φL ≡ ηL − η0 = L x=1 atan n x . (2.262) Here, n = e2 v and v is the relative velocity. Let us introduce the Coulomb scatter- ing matrix by the formula Rc = Sc − 1. Then, the general reaction matrix is written in the form R = S − 1 = ε + Rc, α = S − Rc, (2.263) where ε is the mixing parameter for a given j value. The matrix α corresponding to pure nuclear scattering can be expanded in partial waves, whereas Rc is calculated exactly and is given by the expression
44. 44. 102 2 Spin in Strong Interactions Table 2.2 Expressions for experimentally measured quantities in terms of the amplitude of the pp elastic scattering matrix in the nonrelativistic case 1 σ0 = |a|2 + |b|2 + 2|c|2 + |e|2 + |f |2 2 σ0Dnn = |a|2 + |b|2 + 2|c|2 − |e|2 − |f |2 3 σ0Dll = |a|2 − |b|2 − |e|2 + |f |2 4 σ0Dmm = |a|2 − |b|2 + |e|2 − |f |2 5 σ0Dml = 2Imc∗(a − b) 6 σ0P0 = 2Rec∗(a + b) 7 σ0Cml = 2Imc∗(e − f ) 8 σ0Kml = 2Imc∗(e + f ) 9 σ0Cnn2 = Reab∗ + |c|2 − Reef ∗ 10 σ0Knn2 = Reab∗ + |c|2 + Reef ∗ 11 σ0Cll2 = Reaf ∗ − Rebe∗ 12 σ0Kll2 = Reaf ∗ + Rebe∗ 13 σ0Cmm2 = Reae∗ − Rebf ∗ 14 σ0Kmm2 = Reae∗ + Rebf ∗ f |Rc|i = ik 2π fc(θ), fc(θ) = − n k(1 − cosθ) exp −inlog 1 − cosθ 2 . (2.264) The partial amplitudes h are calculated using formulas (2.246), (2.253), and (2.261), as well as the relation α = 2ikh. These expressions have the form hL = 1 2ik exp 2iδN L − 1 exp(2iφL) (2.265) for the singlet state and hLJ = 1 2ik exp 2i ¯δN LJ − 1 exp(2iφL), hJ±1,J = 1 2ik cos2εN J exp 2i ¯δN J±1,J − 1 exp(2iφJ±1), hJ = 1 2ik sin2εN J exp i ¯δN J−1,J + i ¯δN J+1,J (2.266) for the triplet state. These relations make it possible to express the elements of the matrix M in terms of the phase shifts; hence, experimental observables are determined in terms of the phases. Thus, the phase analysis is possible. Table 2.2 presents the expressions for measured quantities in terms of the ampli- tudes a,b,c,e, and f expressed through linear relations with the matrix elements presented in Table 2.1. Hence, experimental data can be used either for the phase analysis or for direct reconstruction of the amplitudes.
45. 45. 2.9 Relativistic Pion–Nucleon Scattering Matrix 103 2.9 Relativistic Pion–Nucleon Scattering Matrix In the preceding sections, we consider pion–nucleon scattering in the nonrelativistic case, where the kinetic energy of a particle is much lower than its rest energy. This approach was appropriate in the early 1950s, when synchrocyclotrons accelerated protons up to energies 200–300 MeV. However, the kinetic energy of accelerated protons reached the rest energy in the mid-1950s and, then, became much higher than the rest energy. Theoreticians foresaw this situation and developed the covari- ant formulation for the density and scattering matrices, which made it possible to analyze processes at relativistic energies. The main results of such an analysis con- firmed the applicability of the nonrelativistic approach in the center-of-mass frame, and the relativistic corrections were reduced to an additional rotation angle. Such corrections refer to the observables that involve the polarization components in the scattering plane (parameters A and R), whereas the parameters that involved only polarization components perpendicular to the scattering plane (parameters P , DNN, and CNN) remain unchanged. Below, we apply the relativistic description to the reaction a(0) + b(1/2) = a(0) + b(1/2) (2.267) (the spins of the particles are given in the parentheses) proposed in Stapp (1956). Since the particle b is a Dirac particle with spin 1/2, it is described by a four- component wave function ψ. The wave function of a free incident (or initial) particle (with positive energy) can be written in the form ψ = exp(if · x) 2 i=1 AiUi, (2.268) whereas the wave function of an antiparticle (with negative energy) has the form ψ = exp(−if · x) 4 i=3 AiUi. (2.269) Here, f (f,f0) is the four-momentum of the particle in the base frame, where it is measured (for example, in the laboratory frame), such that f0 > 0; x are four- dimensional space coordinates. Each spinor Ui has four components Usi given by the expressions Usi(f ) = (∓if · γsi + m)/ 2m(f0 + m) . (2.270) Hereinafter, the upper sign (−) refers to the subscripts i = 1,2 (positive energy), whereas the lower sign (+), to the subscripts i = 3,4 (negative energy). The sub- scripts in the four-vector γ (−iβα,β) denote its matrix elements and m is the mass of the Dirac particle. The spinors are normalized in the covariant form as follows: U+ i (f )Uj (f ) = U∗ i (f )βUj (f ) = ±δij . (2.271) The sign (+) in the spinor means complex conjugation and transposition (in- terchange of the columns and rows of the matrix), i.e., Hermitian conjugation
46. 46. 104 2 Spin in Strong Interactions U+ = U∗β. Using formula (2.270), it is easy to verify that the spinors Ui satisfy the Dirac equation (±if · γ + m)Ui(f ) = 0. (2.272) To make the below expressions shorter, we introduce the notation γ (ν) = (γ · ν)/ (v · v). (2.273) The denominator on the right-hand side can be either a positive real number or a positive imaginary number. With this symbol, the Dirac equation is represented in the shorter form γ (f )Ui = ±Ui. (2.274) The wave function of the initial state of the pion–nucleon system was introduced above (see Eqs. (2.268) and (2.269)). Let Φ be the wave function of the final state of the same system. The relation between these two functions is determined by the reaction matrix S(f ,t,f ): Φ f = S f ,t,f Ψ (f ). (2.275) The theory of “holes” requires that the Dirac particle described by a plane wave with the momentum f at time T = −∞ and by a plane wave with the momentum f at time T = +∞ has the same sign of energy. This means that the transition of the particle to the antiparticle and vise versa is forbidden. This does not mean that the scattering matrix cannot describe the production of a particle. However, in this particular case, we analyze elastic processes and imply this exclusion, which is mathematically written in the form S f ,t,f = γ f S f ,t,f γ (f ). (2.276) Introducing the new symbol γ (u,w) = γ u/ |u · u| + w/ |w · w| = γ (u) + γ (w) (2.277) and taking into account the relations γ (u)γ (u) = 1 = γ (w)γ (w), (2.278) we obtain γ (u)γ (u,w) = γ (u,w)γ (w). (2.279) The new scattering matrix Sq(k ,t,k) is introduced through the relation S f ,t,f = γ f ,t Sq k ,t,k γ (t,f ). (2.280) This matrix Sq(k ,t,k) is the scattering matrix in the center-of-mass frame with the relative momenta k and k before and after scattering, respectively, and the total energy t in the center-of-mass frame. Then, the substitution of Eq. (2.280) into Eq. (2.276) provides the condition of the exclusion of the particle–antiparticle transition in the form Sq k ,t,k = γ (t)Sq k ,t,k γ (t). (2.281) | 29,726 | 98,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-51 | latest | en | 0.894049 |
https://stats.stackexchange.com/questions/422134/what-is-the-correct-number-of-observations-to-report-for-an-arima-arimax-model | 1,569,042,430,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574265.76/warc/CC-MAIN-20190921043014-20190921065014-00018.warc.gz | 660,064,209 | 32,123 | What is the correct number of observations to report for an ARIMA/ARIMAX model?
This might be due to my relative inexperience with time series modelling, but I am confused about the correct number of observations to report for an ARIMA/ARIMAX model. I couldn't find any post that directly gets at this (though Number of observations used for ARIMA modeling comes close).
Say I run the following model:
fit1 <- arima(lh, order = c(0,1,0))
And then check the number of “used” observations (wording from the documentation):
fit1$nobs length(lh) The number of observations is one less than the total length of the time series, because we difference it once (ARIMA(0,1,0)). Fair enough. But if I then add a lag: fit2 <- arima(lh, order = c(1,1,0)) fit2$nobs
The number of “used” observations is the same, which is confusing to me, since I would have expected to lose an additional observation in the beginning of the series. How can we have a value for the lag at the first observation? Same thing goes for MA terms:
fit3 <- arima(lh, order = c(0,1,1))
fit3\$nobs
How can we have a value for the lag of the error at the first observation? Clearly I’m missing something.
It gets even a little bit more confusing if I want to incorporate transfer functions with the arimax function from the TSA package, since arimax doesn’t return a nobs object nor does it have a nobs method.
I would greatly appreciate some help on this!
Best,
Bertel
The issue here is examining the # of estimable equations . When you introduce ar structure in the errors this CAN act to reduce the # of estimable equations. Lag structures in predictors have no effect if they are each less than or equal to the model-implied lag of Y . If they exceed the model-implied lag of Y based upon differencing in Y and the ar structure of the error process then the # of estimable equations is appropriately reduced by the differential.
Degrees of freedom = # of estimable equations less the # of parameters estimated
For example if we have NOB observations and have a first difference operator for the error structure we have NOB-1 estimable equations.
If we introduce one lag of X in the model this doesn't change the # of estimable equations. If we introduce a lag of 2 for the X variable this reduces the # of estimable equations to NOB-2
• This is incorrect: "observations" count data, not model degrees of freedom. – whuber Aug 14 at 12:37
• this is not a clear sentence , please clarify . I have added more content to my answer . – IrishStat Aug 14 at 12:48
• +1 Your clarification resolved my misunderstanding of your original answer--thank you. – whuber Aug 14 at 12:53
• @IrishStat : Thanks for the reply and the explanation! (and sorry for my slow response). So what would you report for N (the number of observations) in, let's say an ARIMA(3,1,1) model estimated on a time series of length 100? or an ARIMA(1,1,3)? Or would you not report N at all? – Bertel Aug 31 at 13:19
• The sample size is aLways N . The degrees of freedom associated with the error process is based upon the # of estimable relationships (SAY M) minus the # of estimated parameters (SAY J) . For a (3,1,1) model this would be J=4 AND M=N-4 thus DEGREES OF FREEDOM =M-J . ...For a (1,1,3) model this would be J=4 AND M=N-2 AND thus DEGREES OF FREEDOM =M-J . I would present ALL 4 i.e. N,M,J,DEGREES OF FREEDOM – IrishStat Aug 31 at 15:58 | 875 | 3,398 | {"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} | 3.15625 | 3 | CC-MAIN-2019-39 | latest | en | 0.897845 |
https://physics-network.org/how-do-you-calculate-specific-gravity-example/ | 1,670,310,821,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711074.68/warc/CC-MAIN-20221206060908-20221206090908-00678.warc.gz | 487,247,803 | 17,408 | # How do you calculate specific gravity example?
The specific gravity of a substance is characteristic; it is the same for different samples of a substance (if pure, the same in composition, and free from cavities or inclusions) and is used to help identify unknown substances.
## How do you solve specific gravity problems?
At 25°C (77°F or 298.15K), the weight of water is1 gram per cubic centimeter, or 1,000 kilograms per cubic meter.
## What is the formula for corrected specific gravity?
If the density of a substance is known, to calculate the specific gravity of a substance simply divide the density of the substance by the density of water or air. Because the density of water is 1000 kg/m3 or 1 g/cm3, it is simple to calculate. The density of air is 1.205 kg/m3.
## What is the specific gravity of water at 25 C?
Use the equation “m / v = D” where m is mass in grams or kilograms, v is volume in milliliters or liters, and D is density. For example, if you had a sample that was 8 grams and 9 milliliters, your equation would be: “8.00 g / 9.00 mL = 0.89 g/mL.”
## Why do we calculate specific gravity?
Specific gravity refers to the ratio of the density of an object and the reference material. Furthermore, the specific gravity can tell us if the object will sink or float in reference material. Besides, the reference material is water that always has a density of 1 gram per cubic centimeter or 1 gram per millimeter.
## How do you calculate specific gravity of water?
Density is defined as mass per unit volume. It has the SI unit kg m-3 or kg/m3 and is an absolute quantity. Specific gravity is the ratio of a material’s density with that of water at 4 °C (where it is most dense and is taken to have the value 999.974 kg m-3). It is therefore a relative quantity with no units.
## How do you calculate specific gravity of a liquid?
Specific gravity is defined as the ratio of the density of the solid part of a material to the density of water at 20°C. Typically, the specific gravity of soils is in the range 2.60 to about 2.80.
## What is the specific gravity of an object?
For example, the specific gravity of 0.84 corresponds to the density of 840 (0.84 x 1000) kg/cubic meters. Multiply the density by the acceleration of gravity (9.81) to calculate the specific weight. In our example, the specific weight is 840 x 9.81 = 8,240.4.
## Is specific gravity same as density?
The temperature of measurement should be specified and controlled in all specific gravity measurements because the specific gravity of a solution is affected by temperature. Standards of specific gravity are set by analysing standard mixes at 27 degrees Celsius (80.6 Fahrenheit).
## What is the value of specific gravity?
The density and specific gravity of milk is usually given at 15.6°C (60° F). (The specific gravity of water is usually expressed at 4°C). The specific gravity of some major milk constituents are: water: 1.00; fat: 0.93, protein: 1.346, lactose: 1.666, salts: 4.23, and SNF: 1.616.
## What is the volume of a solution that has a specific gravity of 1.2 and a mass of 185g?
Now we have the density and we have the mass of the solution, so we can find out the volume that will be mass divided by density here. The mass of the solution given 185 grams, divided by the density of 1.2 grams per milliliters and we get the volume of the solution- 154.17 milliliters.
## How do you convert specific gravity to weight?
Substances with a lesser density than water would have an SG of less than 1, and a denser substance more. Since water has a density of 1 gram per cubic centimeter, the equation for establishing a liquid’s specific gravity would be: Specific Gravity = (density of liquid) / (1 gm/cm³).
## Is specific gravity affected by temperature?
Specific gravity is defined as a ratio of a density of a fluid to the density of water at 4∘C. It has no unit.
## What is the specific gravity of milk?
Specific gravity values for a few common substances are: Au-19.3; mercury-13.6; alcohol-0.7893; benzene- 0.8786.
## What is specific gravity of sand?
Sand particles composed of quartz have a specific gravity ranging from 2.65 to 2.67. Inorganic clays generally range from 2.70 to 2.80. Soils with large amounts of organic matter or porous particles (such as diatomaceous earth) have specific gravities below 2.60.
## Why is the specific gravity of water 1?
What is the specific gravity of 10 Kg of water occupied in 10 m3 with respect to 200 g/m3? Explanation: Specific gravity = (10/10)/0.2 = 5.
## What is specific gravity and its unit?
The “Specific Gravity” of a substance is the ratio of its mass to that of an equal volume of water at the same temperature and pressure. This is also the ratio of the densities of the two substances: SG = (mass of a volume V of a material)/(mass of a volume V of water) = (rMaterial/rWater).
## What is the specific gravity of ice?
Specific gravity of ice = 0.9.
## What is the specific gravity of alcohol?
The specific gravity of a substance or liquid, including water, the reference liquid, is going to change depending on temperature and pressure. That is why a standard temperature and pressure are used in the calculation of specific gravity. If those outside influences are not regulated, specific gravity will change.
## What is the specific gravity of 10 kg of water?
Factors Affecting Specific Gravity The density of the substance that is being studied or considered. Mass (and weight) of the substance. Since specific gravity is measured under control, the temperature also affects it. We also consider the pressure conditions while talking about the specific gravity.
## How do you find specific gravity from mass and volume?
1 Answer. The concentration of a sample in a sample can be determined by means of density. Specific gravity is the density of solute divided by the density of solvent.
## What is the specific gravity of a substance that weights 75 g and occupies a volume of 150 ml?
Putting the values mass is 75. Gram, divided by a volume is 150 milliliters, so this will be equal to 0.5 gram per milliliter now specific gravity.
## Does specific gravity change with pressure?
Density = Mass / Volume Specific gravity is the density of a substance divided by the density of water. Since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density (but without any units). | 1,530 | 6,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-49 | latest | en | 0.902924 |
https://goprep.co/q15-an-analysis-of-monthly-wages-paid-to-workers-in-two-i-1nli7b | 1,611,718,742,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704820894.84/warc/CC-MAIN-20210127024104-20210127054104-00468.warc.gz | 368,610,723 | 30,427 | # An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:(i) Which firm A or B pays larger amount as monthly wages?(ii) Which firm, A or B, shows greater variability in individual wages?
Here
Mean monthly wages of firm A = 5253
No. of wage earners = 586
Total amount paid = 586 × 5253 = 3078258
Mean monthly wages of firm B = 5253
No. of wage earners = 648
Total amount paid = 648 × 5253 = 340 3944
(i) Hence the firm B pays larger amount as monthly wages.
(ii) Variance of firm A = 100
standard deviation (σ)= √100=10
Variance of firm B = 121
Standard deviation (σ)=√(121 )=11
Since the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.
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view all courses | 291 | 1,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-04 | latest | en | 0.894246 |
https://www.physicsforums.com/threads/size-of-a-magnetic-field.678022/ | 1,532,042,942,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591332.73/warc/CC-MAIN-20180719222958-20180720002958-00127.warc.gz | 965,313,890 | 12,898 | # Size of a Magnetic field?
1. Mar 12, 2013
### Yamspotato
Hello! I was just wondering, how can one determine the size of an electromagnetic or regular magnetic field?
For instance, if I had a regular bar magnet and I wanted to calculate to what distance the magnetic field could attract/repel another magnet, how would I go about doing that?
I know the measure of force would be Tesla, but the size of the field itself is what I'm not too sure about...
Thanks!
2. Mar 12, 2013
### Simon Bridge
Welcome to PF;
Put the magnet on the end of a Newtonmeter, read the scale, observe the orientation of the magnet. See the definition of the magnetic field.
The details depend on the magnetic field you want to measure.
Usually we would use an electronic device for small fields.
The SI unit for force is "Newton" - the unit for magnetic field strength is "Tesla".
We don't usually think of a magnetic field as having a "size".
Bear in mind that the bar magnet, in your example, will attract/repel another magnet to any distance at all. It's just that there are other forces around that can balance this out. | 255 | 1,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-30 | latest | en | 0.947744 |
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