url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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http://www.w3resource.com/c-programming-exercises/conditional-statement/c-conditional-statement-exercises-4.php | 1,492,990,844,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118851.8/warc/CC-MAIN-20170423031158-00241-ip-10-145-167-34.ec2.internal.warc.gz | 724,492,617 | 7,207 | C Exercises: Check whether a given year is a leap year or not
C Conditional Statement: Exercise-4 with Solution
Write a C program to find whether a given year is a leap year or not.
Test Data :
2016
Expected Output
:
2016 is a leap year.
Sample Solution :-
C Code:
#include <stdio.h>
void main()
{
int chk_year;
printf("Input a year :");
scanf("%d", &chk_year);
if ((chk_year % 400) == 0)
printf("%d is a leap year.\n", chk_year);
else if ((chk_year % 100) == 0)
printf("%d is a not leap year.\n", chk_year);
else if ((chk_year % 4) == 0)
printf("%d is a leap year.\n", chk_year);
else
printf("%d is not a leap year \n", chk_year);
}
Flowchart :
Code Editor :
#include <stdio.h>
void main()
{
int chk_year;
printf("Input a year :");
scanf("%d", &chk_year);
if ((chk_year % 400) == 0)
printf("%d is a leap year.\n", chk_year);
else if ((chk_year % 100) == 0)
printf("%d is a not leap year.\n", chk_year);
else if ((chk_year % 4) == 0)
printf("%d is a leap year.\n", chk_year);
else
printf("%d is not a leap year \n", chk_year);
}
Improve this sample solution and post your code through Disqus.
| 332 | 1,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.585894 |
https://image.hanspub.org/Html/1-1160323_16304.htm | 1,590,802,492,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407001.36/warc/CC-MAIN-20200530005804-20200530035804-00037.warc.gz | 389,220,938 | 11,379 | “随机信号分析”中噪声产生设计的研讨 Discussion on Noise Design of Random Signal Analysis Course
Vol.05 No.06(2015), Article ID:16304,6 pages
10.12677/AE.2015.56028
Discussion on Noise Design of Random Signal Analysis Course
Mingqian Liu, Bingbing Li, Wanli Guo
School of Telecommunications Engineering, Xidian University, Xi’an Shaanxi
Received: Oct. 24th, 2015; accepted: Nov. 7th, 2015; published: Nov. 10th, 2015
ABSTRACT
Noise and its generation are important contents in the teaching of random signal analysis. The generation of the noise involves the theory of spectrum analysis and the minimum phase system, students tend to make mistakes in the design of white noise and color noise. This paper firstly analyzes the characteristics of the noise, and then analyzes the generation of color noise and white noise. Finally, this paper takes design experiments as examples. This paper has some guidance to the teaching of the noise generation design, and it is helpful for the students to understand and apply it.
Keywords:Random Signal Analysis, Color Noise, White Noise, Whitening Filter
“随机信号分析”中噪声产生设计的研讨
1. 引言
2. 白噪声及色噪声
(1)
(2)
3. 色噪声产生的设计
(3)
(a) 自相关函数 (b) 功率谱密度
Figure 1. Autocorrelation function and power spectral density of white noise
(a) 自相关函数 (b) 功率谱密度
Figure 2. Autocorrelation function and power spectral density of colored noise
Figure 3. The diagram of colored noise generation
(4)
(5)
(6)
4. 白化滤波器的设计
(7)
(8)
(9)
(10)
5. 噪声产生的设计实验
Figure 4. The diagram of the whitening filter
Figure 5. The diagram of experimental system
(a) 白噪声信号时域图 (b) 白噪声信号频谱图
Figure 6. Time and frequency domain graphs of Gaussian white noise
(a) 色噪声信号时域图 (b) 色噪声的功率谱密度
Figure 7. Time and frequency domain graphs of colored noise
(a) 白化后噪声信号的时域图 (b) 白化后噪声信号的功率谱密度
Figure 8. Time and frequency domain graphs of whitened signal
6. 结语
Discussion on Noise Design of Random Signal Analysis Course[J]. 教育进展, 2015, 05(06): 171-176. http://dx.doi.org/10.12677/AE.2015.56028
1. 1. 李兵兵, 马文平, 田红心, 等. 随机信号分析教程[M]. 北京: 高等教育出版社, 2012.
2. 2. 赵剡, 张春晓. 改进的基于白化滤波器的降晰函数尺寸估计方法[J]. 天津: 红外与激光工程, 2010, 39(5): 896-901.
3. 3. 张文龙, 黄振华. 低漂移有色高斯噪声的产生[J]. 上海: 上海师范大学学报(自然科学版), 2009, 38(4): 398-401.
4. 4. 叶怀安. 自共轭算子谱分解定理的简证及教学实践[J]. 合肥: 教育与现代化, 1995(4): 21-24.
5. 5. 高西全, 丁玉美. 数字信号处理 [M]. 第3版. 西安: 西安电子科技大学出版社, 2008.
6. 6. 楼顺天. 基于MATLAB7.X的系统分析与设计——信号处理[M]. 第二版. 西安: 西安电子科技大学出版社, 2005.
7. 7. 谭浩强. C程序设计[M]. 第4版. 北京: 清华大学出版社, 2010. | 954 | 2,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.635602 |
https://tcuonline.tcu.edu/kb/about-grading-systems/ | 1,719,256,322,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00818.warc.gz | 499,426,747 | 9,166 | • Weighted Grading System – Grade items count as a percentage of a final grade worth 100%.
• Points Grading System – Grade items are worth a certain amount of points that are totaled for a final grade.
• Formula Grading System – You can define a custom formula for how grade items contribute to a final grade.
The weighted system calculates grade items as a percentage of a final grade worth 100%. The maximum points you assign to individual grade items can be any value, but their contribution towards the category they belong to and the final grade is the percentage value (weight) assigned to them.
If you place grade items in a category, each item counts as a percentage of that category, and each category counts as a percentage of the final grade. Therefore, grade items in a category should combine to a weight of 100% within the category.
For example, if you have a category worth 10% of the final grade, comprising two equally weighted grade items, the weight of each grade item is 50% within the category, but 5% of the overall final grade.
Since it’s a category weight, and not an individual grade item’s weight that counts toward the final grade, the final grade is in flux until all the items in the category are graded. If you want to release final grades to students before all the items are graded, you can drop ungraded items from the calculation until the end of the course when you want all grade items to be considered. Otherwise, the final grades might be misleading.
If your grade items and categories do not add up to 100%, you will see a note at the top of your grade book. You can ignore this message if you choose; a balanced grade book is not required. However, if the weights assigned to grade items do not sum to 100%, the tool automatically adjusts the weight of each item. For example, if you have three grade items with a weight of 25% each, each item is actually calculated as 33%. This is true for categories and the final grade.
To create this system, see Create a Weighted Grading System.
Use the points system when you want the maximum points assigned to a grade item to be equal to its contribution to the final grade. Final grades are calculated by adding a student’s score on all grade items together and dividing by the sum of the maximum points values. The sum of the maximum points values for all grade items does not need to equal 100.
With the points system, grade items can be combined into categories, however you do not specify a category’s weight or total points. It is the maximum points assigned to an individual grade item that counts toward the final grade.
Therefore, make sure the maximum points assigned to grade items reflect how much you want them to be worth.
If you wish to evaluate a grade category, numeric grade item, selectbox grade item, or pass/fail grade item without including the grade in students’ calculated or adjusted final grades, you can choose to exclude an item from the final grade calculation. When creating or editing the grade item or category, check the box to Exclude from Final Grade Calculation.
To create this system, see Create a Points Grading System. | 649 | 3,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.899841 |
http://www.fixya.com/cars/t3431126-heat_doesnt_come_out_in_feet_area | 1,506,193,789,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689775.56/warc/CC-MAIN-20170923175524-20170923195524-00093.warc.gz | 458,074,419 | 34,088 | Heat doesn't come out in the feet area - 2002 Ford Explorer
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• Master
Sounds like the air handler (some call it air deflector) is in operative its probably stuck in down posistion which directs the air up to the vents and or defroster. if you can hear it moving you need to check the tubing that connects the vents to make sure that it has not been jarred loose. hope this helps you.
Posted on Nov 14, 2009
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
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Posted on Jan 02, 2017
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Related Questions:
I just need to know the four side dementions of a 1/2 acrelot the top and the bottom and both side demenstions .not the high or depth of the land just a flat peice of ground
An acre is 43,560 square feet, so a half acre is 21,780 square feet. If the lot is rectangular, you multiply the length of the long side (in feet) times the length of the short side (in feet). For instance if the lot is 200 feet x 110 feet, the area is 22,000 square feet, just a little over a half acre. If the lot is not rectangular, then you need to have a good knowledge of geometry to calculate the area.
Jun 04, 2017 | Cars & Trucks
What is the lineal ft of a counter top 25 ft long by 3 wide?
Lineal feet are straight line feet.
25 feet.
Square feet are the surface area.
25 feet X 3 feet are 75 square feet.
Apr 18, 2017 | Cars & Trucks
How many feet is 350 linear feet?
Linear feet are a measure of length (no different from feet); square feet measure area. You cannot simply convert between measures of different kinds of quantities;the connection between them will be specific to a particular problem.A practical example in which this question can arise is in buying countertops for a kitchen. Some materials are sold by the square foot; others (basically those that are extruded, so they come in standard widths) are sold by the linear foot. In order to compare the two, you need to compute the area and wall length for the countertop you want.You can't convert between the two. All you have to do is to make the appropriate measurements so you can calculate the price of each item.The terms used in the lumber industry are a bit confusing.There are two terms that I think you might be mixing up.A LINEAR FOOT is simply the length of a board. If you want to know the area or volume of the board,you need additional information. For instance, 6 linear feet of 1-by-12 has an area of 6 square feet (12 inches = 1 foot, times 6 feet), and it's 1 inch thick, so the volume is 1/2 cubic foot (6 square feet times 1/12 foot). But 6 linear feet of a 1-by-6 board would have half the area and half the volume.A BOARD FOOT is equivalent to one square foot of a 1-inch-thick board. In other words, it is a square-foot-inch (ft^2-in), or 1/12 cubic foot.Linear feet are used for the pricing of a single size such as two-by-fours. Board feet are used for larger lumber that you are more likely to want to compare directly with different size boards .To sum up, neither a linear foot nor a board foot can be converted directly to square feet. A linear foot is a linear (length) measure, and a board foot is a volume measure. You need to know your particular board to do anything more, such as find the area.an example with an" L" shaped countertop will betwo rectangles are 24 by 80 inches and 24 by 36 inches. Thus the area is: 24 * 80 + 24 * 36 = 24 * (80 + 36) = 24 * 116= 2784 sq. in.To get it in square feet, divide by 144:
2784 / 144 = 19.33 sq. ft.The linear measure of this countertop would be 60 + 80 = 140 inches = 140/12 feet = 11.67 feet
Apr 08, 2014 | Cars & Trucks
23 x 28 "what is the square foot of floor area
If the floor is 23 feet by 28 feet then the area is 644 square feet.
Mar 27, 2013 | Cars & Trucks
2005 gmc yukon xl blower is not shutting off. It continues to run after the vehicle is shut off. Any suggestions?
Sounds like a faulty blower motor resistor also check wire connector for melting.Location is next to floor on in heat duct towards the left of pass side feet area.
May 14, 2011 | 2005 GMC Yukon Xl Denali
Hot to stop engine heat from entering cockpit
The control switch for temperature is faulty, cable must have come off, leaving the heater core flap open. Make sure all cables are connected to the climate control switches. If it isn't anchored properly, it will move freely and not controlling the flaps or valve to position. Cable locks do come off and fall around the assembly, once that happens the cable doesn't have any counter resistance of it's position and move from it's own slack making it useless..
Jul 23, 2010 | 1991 Porsche 944
No heat
Look under the dash to see if the controller flaps are working(Unsure if u have auto or manual cable adjustment) move the hot to cold lever or press the button if electric as the old cable systems do break.Is the thermostat okay, look if the heater radiator core is okay, do you get a mist on the windscreen and sweet small?this can be done by removing the panels on either side oth the gearbox tunnel(footwell area) then use a torchlight to check.Some engine models have a 2 way valve in the engine bay/firewall area connected to the radiator that can rust or sieze and stop heatflow to the inner heater radiator.follow the radiator pipes going from the engine to the firewall to find it.
Feb 09, 2010 | 1995 Volkswagen Passat
Cannot get heat out of floor vents -only defrost and dash heat
i had a simmiler problem on my 95 chevy pick up. on mine it would not chg. from floor to defrost, no matter where i put the selector. with mine it was the controls. if you go under solutions for k1500chevy gmc trucks it should tell you what your problem is, or give you an idea whats going on with yours.
Jan 12, 2009 | 1998 GMC Jimmy
No action from Blower motor in 1993 Ford Econoline
the blower motor in under the hood on the passenger side of the van. it has three screws that hold it in place and is ususally easy to change out
Nov 16, 2008 | 1993 Ford Econoline
Open Questions:
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95 people viewed this question
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Level 3 Expert | 1,648 | 6,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-39 | latest | en | 0.94371 |
http://placementpapers.net/helpingroot/TELCO_Sample_Placement_Papers | 1,386,780,234,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164039245/warc/CC-MAIN-20131204133359-00041-ip-10-33-133-15.ec2.internal.warc.gz | 139,754,028 | 11,970 | # (PAPER) TELCO Sample Placement Papers & Pattern
(PAPER) TELCO Sample Placement Papers & Pattern
TELCO PLACEMENT PAPER
1) Hypothesis :problem (below 4 he will give)
2) Mirror:Image
3)money:Misapprobation
5)construction:building
6)file:pile
7)ours:we
8)1/3,1 1/3,3,5 1/3, next
9)Selling price of 4 articles=Cost price of 3 articles then %loss(ans. 25%)
10)|x-3|=3-x then x=
11)data sufficiency p>q?
1)p,q positive
2)q-1=q*2+p
This type you can see in GMAT book.
Reasoning
1) ABCDEF attended for an interview, in which 3 were selected
A is worst of the lot
C got equal marks as F(C=F)
D<F and G
E is not selected
B>C
There are 5 questions below this.
2)There are 7 fellows sat in the following way
C and F always sit as a apart as follows
No. of fellows sat b/w c and D and D and F are equal like that he has given
Ans: Order is as follows C E B D A G F by using this you can answer all
About 4 to 5 are there
3)ABCDE are five brothers. There are twins(only one pair of equal age)
both are neither younger nor older. D is younger to 3 brothers.
B is older than E and C.
q's like who is youngest? Eldest ? About 4 are there.
4)A person has meet a king for that ha has to cross 7 gates. At each gate he
has to pay half the amount he is carrying. Finally he gave Rs 3/- to the
king. Then the amount he carried at the beginning and some questions
5) Heros tell truth and cowards lie.There PQR three persons. P tells Q "I
may be hero or I may be coward". Q tells R "P was telling that he was
coward. Then R tells Q "P was not a coward but he was a hero". You better
study the question it may not be entirely correct.
In thus 3 q"s are there
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https://www.boatdesign.net/threads/how-much-planing-surface-do-i-need.19753/ | 1,611,012,608,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517159.7/warc/CC-MAIN-20210118220236-20210119010236-00490.warc.gz | 705,845,525 | 11,977 | # How much planing surface do I need?
Discussion in 'Powerboats' started by kengrome, Oct 13, 2007.
1. Joined: Jul 2006
Posts: 718
Likes: 25, Points: 0, Legacy Rep: 305
Location: Gulf Coast USA
### kengromeSenior Member
Here's my situation. I designed a small (one person) rough water planing power boat and I would like to have an idea if it will plane before I build a prototype.
I understand there's a formula for determining the required planing surface area for a boat to plane given a specified hp and weight. Is this true or is it far more complicated than this?
Thanks in advance!
2. Joined: Aug 2004
Posts: 525
Likes: 5, Points: 28, Legacy Rep: 28
Location: Cathlamet, WA
### GilbertSenior Member
Why not measure the area of a jetski and find out its weight? Then estimate the weight of your design and if your weight to bottom area ratio is comparable you should be good to go.
3. Joined: Apr 2005
Posts: 4,127
Likes: 148, Points: 0, Legacy Rep: 2043
Location: Ontario
### marshmatSenior Member
You are probably thinking of the Savitsky method. It is not exactly a simple formula but it does give a good indication of how the boat will perform for a given thrust vector and load case.
Dingo Tweedie posted an excellent Excel implementation of the method on here a while back, http://www.boatdesign.net/forums/showpost.php?p=141860&postcount=76
4. Joined: Oct 2001
Posts: 3,590
Likes: 130, Points: 0, Legacy Rep: 2369
Location: Australia
### WillallisonSenior Member
Opinions vary - especially once you take service conditions into consideration, but for a relatively lightweight boat I'd aim for around 250kg per square metre of waterplane area.
Along similar lines to Gilbert's suggestion, a good approach would be to estimate the AWP of similar vessels that you know to be successful and aim for similar figures. The problem with this approach is that it's notoriously difficult to get reliable weight figures from manufacturers...
5. Joined: Jun 2004
Posts: 117
Likes: 3, Points: 0, Legacy Rep: 59
Location: Devon UK
### hmattosSenior Member
Planing area is very much a function of speed. We - see explorermarine.co.uk - build fast RIBs in Devon England and a 5.0 metre SPORTS model will plane at 10 mph. This is using 7.5 metres sq of planing surface to support 600 kg.
However, at 45 mph the same boat uses less than 2 square metres to support the same weight.
Not also that a low vee angle - deadrise at stern - supports much more weight when planing at lower speeds.
Good luck
6. Joined: Oct 2001
Posts: 3,590
Likes: 130, Points: 0, Legacy Rep: 2369
Location: Australia
### WillallisonSenior Member
Yes - I probably should have been more specific...
As the AWP is very diificult to estimate at speed, it is the at rest AWP that is used. I don't believe there's any great science behind it - it's simply based on previous experience...
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different. | 798 | 3,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-04 | latest | en | 0.925012 |
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EdSearch WebSearch | 219 | 891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-27 | latest | en | 0.878628 |
https://www.jiskha.com/display.cgi?id=1361975046 | 1,501,189,101,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549429485.0/warc/CC-MAIN-20170727202516-20170727222516-00623.warc.gz | 807,893,281 | 4,054 | # Math
posted by .
Amy Powell invested \$8500 twice a year in an ordinary annuity at New York Securities for a period of 5 years at an annual interest rate of 10% compounded semiannually. Using the ordinary annuity table , calculate the total value of the annuity at the end of the 5-year period
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https://www.expertsmind.com/library/essay-on-analyzing-a-work-of-art-511329.aspx | 1,725,918,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00245.warc.gz | 744,435,464 | 12,681 | ### Essay on analyzing a work of art
Assignment Help English
##### Reference no: EM1311329
Topic : Description a work of art. Select an image of an existing art form that caputures the story and the aspect of the analysis we have chose
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Developing an annotated bibliography | 682 | 3,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.871446 |
https://studyslide.com/doc/284913/demand-schedule | 1,675,018,653,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00594.warc.gz | 577,423,485 | 17,120 | #### Transcript Demand Schedule
```MANAGERIAL ECONOMICS
-R.L. VARSHNEY
K.L. MAHESHWARI
-DR. D. M. MITHANI
- M.GIRIJA
R. MEENAKHI
Demand Analysis &
Demand Function:– Factors Influencing Demands
Determinants of demand
Demand Analysis for various products, situations
and market structures :– Durable and non-durable goods
– Long run and Short run demand
– Autonomous and derived demand
– Industry and firm demand
Elasticities & demand levels
Demand Analysis
Demand
Function
Demand determinants
Law of Demand
Demand Schedule
Demand Curve
Demand Equation
Factors affecting Demand
Demand Function
Demand for a product is the amount of it that
will be bought per unit of time at a particular
price.
Individual Demand -The quantity demanded
by an individual purchaser at a given price
Market Demand -Total quantity demanded by
all the purchasers together at a given price
Demand Function- Mathematical term
expresses the functional relationship between
demand for the product and its various
determining variables
Demand Function
Dx = f ( Px, Ps, Pc, yd, T, A, N , u )
Here we assume commodity X, so
Dx = Amount demanded for the commodity X
Px = Price of X
Ps = Price of substitutes of good X
Pc = Price of complimentary goods of X
Yd = Level of disposable Income of buyers
T = Change in buyer’s Taste & Preferences
u = other unspecified determinant
Demand Schedule
Demand for a product is the function of its own
price, keeping all other factors constant
Dx = f ( Px)
Demand Schedule-A Tabular statement of price
quantity relationship
It shows the inverse relationship between price
and quantity demanded
Two types of demand schedule:
– Individual Demand Schedule
– Market Demand Schedule
Individual Demand Schedule
•Table showing the various quantity purchased by
an individual purchaser at alternative price over a
given time period (day, week, month or year)
It shows the variation in demand at various prices.
•
Price
Qty Demanded
4
4
3
8
2
12
1
16
Market Demand Schedule
•Table narrating the quantities of a commodity
purchased in aggregate by all the purchasers in the
market at different prices over a given period of time
It shows the total market demand at various prices.
It serve as the basis for knowing the revenue
consequences of alternative output and pricing policies
of the firm
•
•
Price
4
A
1
B
1
C
3
Mkt. Demand
5
3
2
1
2
3
5
3
5
9
5
7
10
10
15
24
Demand Equation
A linear Demand function
Dx = A – B(Px)
Dx = Amount demanded for the commodity X
Px = Price of X
A = constant parameter signify initial demand
irrespective of price
B = implies negative relation between price and
quantity demanded
Eg. D = 20-2p
At p = 5, quantity demanded will be 20-2x5=10
Demand Curve
Graphical presentation of a demand schedule
It relates the amount the consumer is willing to buy at
each alternative price over a period of time.
It has a negative slope
It slopes downwards from left to right representing an
inverse relation between price and demand
Price
Determinants Of Demand
Factor
affecting Individual Demand
Price of commodity
Income of customer
Taste, habit and Preference of Customer
Relative price of Substitute and Complementary
goods
Customer expectation
Determinants Of Demand
Factor
affecting Market Demand
Price of commodity
Distribution of Income & Wealth of Community
Community common habits & scale of
Preference
General Standard of living & Spending habits of
people
Number of buyers in market & Growth population
Age, Structure, sex ratio of population
Determinants Of Demand
Factor
affecting Market Demand
Level of taxation and tax structure
Inventions and innovations
Fashions
Climate and weather conditions
Customs
Law of Demand
It expresses the nature of functional relationship
between two variables of demand relation i.e.
Price and quantity demanded
Ceteris paribus, the higher the price of the
commodity, the smaller is the quantity
demanded and lower the price, larger is the
quantity demanded.
The demand for the commodity extends as the
price falls, and contracts as the price rises.
Chief characteristics of
Law of Demand
Inverse relationship
Price an independent variable, and
demand a dependent variable
Assumption of other things remaining the
same
Reasons underlying the law of demand
–
–
Income Effect
Substitution Effect
Exception to the law of demand
•
•
•
Veblen Effect / Conspicuous consumption
Giffen Goods
Speculation
Types of Demand
1.
2.
3.
4.
5.
Producers’ Goods
Durable Goods
Derived Demand
Short run Demand
Firm Demand
1.
2.
3.
4.
5.
Consumers’ Goods
Non Durable Goods
Autonomous Demand
Long run Demand
Industry Demand
Demand Distinctions
Consumers’ Goods
Producers’ Goods
1.
2.
3.
4.
5.
6.
Used for the production
of other goods
Demand is derived
Depends on marginal
productivity
Classified into
consumable and
durable
income
Example: cloth, food .
house
1.
2.
3.
4.
5.
6.
Used for the direct
consumption
Demand is direct and
autonomous
Depends on marginal utility
Classified into non durable
and durable
profits
Example: machines,
equipment, raw
materials,building
Demand Distinctions
Durable Goods
Non Durable Goods
1.
2.
3.
4.
5.
6.
7.
Used for the current
demand of goods
Can not be stored for a
long time
Give one time service
Demand is immediate
Demand is more elastic
in short run
Influenced by income
and convenience
Example: vegetable,
fish . house
1.
2.
3.
4.
5.
6.
7.
existing goods
Can be stored for a long
time
Give repeated services
Demand is postponable
Demand is less elastic in
short run
product and obsolescence
Example: furniture, cycle,
house
Demand Distinctions
Derived Demand
1.
2.
3.
4.
5.
6.
1.
Demand is tied to
purchase of parent good
Less price elastic
2.
Facilitate demand
3.
forecasting
4.
Demand for all
producers’ goods
Influenced by demand of 5.
parent good
Example: cement, petrol 6.
,ink, antenna, sugar
Autonomous Demand
Demand is based on the
urge of satisfy some wants
directly
more price elastic
Facilitate demand analysis
Demand for all consumers’
goods
Influenced by taste, trends
and preference
Example: building, car,
pen, television, tea
Demand Distinctions
Long run Demand
Short run Demand
1.
2.
3.
4.
5.
6.
demand of goods in a
period of one year or
less
No threat of substitute
and competitors
Demand is immediate
Demand is less elastic
in short run
Influenced by price and
income
Example: raw material,
bidi
1.
2.
3.
4.
5.
6.
demand of goods in a
period of one year to ten
year
Big threat of substitute and
competitors
Demand is permanent
Demand is more elastic in
long run
Influenced by promotion,
product change
Example: building,
cigarette
Demand Distinctions
Industry Demand
Firm Demand
1.
2.
3.
4.
5.
6.
7.
Market demand for the
commodity produced by a
particular firm
Represent the relation of the
price of the product to the
quantity bought from a single
firm
Demand is general and can not
be classified.
Demand is more elastic in short
run
Influenced by industry demand
schedule
Example: total production of
Ambuja cement : 30 cr tons
Demand os steel produced by
TISCO
1.
2.
3.
4.
5.
6.
7.
Total demand for the commodity
produced by a particular industry
Represent the relation of the price of
the product to the quantity bought
from all the firms
Demand can be classified customer
group-wise, like, steel demand for
construction and manufacture,
pleasure
Demand is less elastic in short run
Influenced by market structure like
monopoly or oligopoly
Example: total production of cement
: 100 cr tons
total production of steel industry
Elasticity of Demand
Change in quantity demanded due to change
in price of the commodity
Def: the elasticity of demand in the market is
great or small according as the demanded
increases much or little for a given fall in price
and diminishes much or little for a given rise in
price.
Types of elasticities
Price Elasticity of Demand
Income Elasticity of Demand
Cross Elasticity of Demand
THANK YOU
``` | 2,200 | 7,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-06 | latest | en | 0.822498 |
http://oeis.org/A321596/internal | 1,600,935,115,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00339.warc.gz | 87,333,073 | 3,113 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A321596 Primes that are not Base-2 deletable primes (written in base 10). 1
%I
%S 17,31,41,67,71,89,97,103,113,127,131,139,181,191,193,199,223,227,233,
%T 239,241,251,257,263,269,271,283,337,353,367,373,379,383,401,409,431,
%U 433,439,443,449,463,479,487,491,499,503,509,521,523,541,547,571,577
%N Primes that are not Base-2 deletable primes (written in base 10).
%C A prime p is a base-b deletable prime if when written in base b it has the property that removing some digit leaves either the empty string or another deletable prime. However, in base 2 we adopt the convention that 2 = 10 and 3 = 11 are deletable.
%C Deleting a digit cannot leave any leading zeros in the new string. For example, deleting the 2 in 2003 to obtain 003 is not allowed.
%C Complement of all primes and A096246.
%H Robert Price, <a href="/A321596/b321596.txt">Table of n, a(n) for n = 1..2351</a>
%t d = {2, 3};
%t For[n = 3, n <= 15, n++,
%t p = Select[Range[2^(n - 1), 2^n - 1], PrimeQ[#] &];
%t For[i = 1, i <= Length[p], i++,
%t c = IntegerDigits[p[[i]], 2];
%t For[j = 1, j <= n, j++,
%t t = Delete[c, j];
%t If[t[[1]] == 0, Continue[]];
%t If[MemberQ[d, FromDigits[t, 2]], AppendTo[d, p[[i]]]; Break[]]]]]; Complement[Table[Prime[n], {n, PrimePi[Last[d]]}], d]
%Y Cf. A080608, A080603, A096235-A096246.
%K nonn,base,easy
%O 1,1
%A _Robert Price_, Nov 14 2018
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Last modified September 24 04:10 EDT 2020. Contains 337316 sequences. (Running on oeis4.) | 645 | 1,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-40 | latest | en | 0.767033 |
http://mathoverflow.net/questions/141104/what-turing-complete-models-of-computation-carry-a-notion-of-time-complexity-tha | 1,467,120,649,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396887.54/warc/CC-MAIN-20160624154956-00115-ip-10-164-35-72.ec2.internal.warc.gz | 192,622,421 | 17,710 | What Turing-Complete models of computation carry a notion of time complexity that “agrees” with that of Turing Machines?
Certain models of computation are technically Turing-Complete, but cannot feasibly simulate a Turing Machine within the usual time constraints we hope for. One example of this is Godel's recursive functions: the computable function $f(x) = 2x$ is implemented by calling the successor function $2x$ times, which intuitively takes $O(2^x)$ time.
So, my question:
What Turing-complete models of computation, or simplistic programming languages (Turing tar pits), can compute every computable function with only a worst-case polynomial-time blowup in time complexity over the fastest Turing machine that computes that same function?
-
Aren't there hundreds of examples? This is the robustness of $P$ as a complexity class, that we can often (and indeed usually) simulate a different model of computability in our own, with at most polynomial time cost. – Joel David Hamkins Sep 3 '13 at 14:41
I imagine there are hundreds of examples of languages without this problem (every modern programming language, for instance) but I don't know of any simplified, theory-driven model of computation that doesn't have this problem. Recursive functions, Lambda calculus, tag systems, and brainfuck all have computable functions that they implement in exponential time, but a Turing machine could implement these functions in polynomial time. I'm looking for an example along the lines of these simplified models of computation. – GMB Sep 3 '13 at 19:23
Well, some Turing machine models are also weak in this way. For example, if you have only two symbols and a single tape, and are forced to use unary notation (so that you know when the input ends), then it will take exponential time to simulate the usual machines, which are more powerful. – Joel David Hamkins Sep 3 '13 at 19:48
Okay, good point - I should've specified that I'm only "counting" the Turing machine models that have sufficient alphabet size to solve all problems as quickly as possible. – GMB Sep 3 '13 at 19:56
Meanwhile, I would point out that even the ordinary notions of Turing machine are idealized concepts, and not about what we can actually build in the real world. We cannot expect to build a Turing machine in the physical world that would undertake computation of appreciable size, since once the paper tape was large enough, it would be subject to gravitational forces. Is it in orbit? If coiled, once the mass became a certain size it would collapse into a black hole. If uncoiled, it would likely tear. Long story short: Turing computability is a notion of idealized computation. – Joel David Hamkins Sep 4 '13 at 12:06
One of the simplest model that has recently been proved to be an efficient simulator (polynomial time slowdown) of Turing machines are 2-tag systems:
Abstract: We show that 2-tag systems efficiently simulate Turing machines. As a corollary we find that the small universal Turing machines of Rogozhin, Minsky and others simulate Turing machines in polynomial time. This is an exponential improvement on the previously known simulation time overhead and improves a forty year old result in the area of small universal Turing machines.
-
Because we can easily invent as many small variations of Turing-complete models of computation as we like (see comments below the question), an answer to this question should try to concentrate on relevant (and Turing-complete) models, i.e. models that have either been investigated in illuminating non-trivial ways, or are important for better understanding of actually available computing resources.
I have been exposed in non-trivial ways to tape based Turing machines, register machines and pointer machines. It seem like the wikipedia article on abstract machines is intended to give an overview for related Turing machine equivalent models, but in its current form it is mainly a collection of useful keywords and links.
I'm currently looking for models and investigations related to machines limited to write once read many (WORM) memory for large amounts of data. None of the abstract machine models I found so far investigated these. Is it possible to create a model of such a machine that is equivalent to a Turing machine in the sense of the question above? (Edit: It looks like it was proved recently that Wang B-machines achieve this. I haven't read the paper yet.) This question seems to be both non-trivial and interesting to me, contrary to the comments below the question, which is the main reason why I wrote this answer.
-
Punchtape Turing machines, for which the machine can mark each cell at most once, punching a hole in the paper tape or not, are like your write-once-read-many machines, and have been investigated. Punchtape machines are Turing complete, and have been investigated, but I find it likely that there is a substantial time cost. – Joel David Hamkins Feb 12 '14 at 15:00
It's not at all clear how this answers the question. It might make sense to make this response CW. – Todd Trimble Feb 12 '14 at 22:05 | 1,077 | 5,096 | {"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 | 3 | CC-MAIN-2016-26 | latest | en | 0.925834 |
https://edurev.in/t/84225/NCERT-Solutions-Class-11-Maths-Chapter-14-Probability | 1,721,583,312,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00237.warc.gz | 184,959,337 | 69,459 | NCERT Solutions - Exercise 13.1: Probability
# NCERT Solutions Class 11 Maths Chapter 14 - Probability
Question 1:
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E).
It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2
Question 2:
Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32
It is given that P(B) = 0.5 and P(A ∩ B) = 0.32
Question 3:
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find
(i) P(A ∩ B) (ii) P(A|B) (iii) P(A ∪ B)
It is given that P(A) = 0.8, P(B) = 0.5, and P(B|A) = 0.4
(i) P (B|A) = 0.4
Question 4:
Evaluate P (A ∪ B), if 2P (A) = P (B) = 5/13 and P(A|B) = 2/5
Question 5:
If P(A) = 6/11, P(B) =5/11 and P(A ∪ B) =7/11 , find
(i) P(A ∩ B) (ii) P(A|B) (iii) P(B|A)
It is given that
Question 6:
A coin is tossed three times, where
(i) E: head on third toss, F: heads on first two tosses
(iii) E: at most two tails, F: at least one tail
If a coin is tossed three times, then the sample space S is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space has 8 elements.
(i) E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
E ∩ F = {HHH}
(ii) E = {HHH, HHT, HTH, THH}
F = {HHT, HTH, HTT, THH, THT, TTH, TTT}
E ∩ F = {HHT, HTH, THH}
(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}
F = {HHT, HTT, HTH, THH, THT, TTH, TTT}
Question 7:
Two coins are tossed once, where
(i) E: tail appears on one coin, F: one coin shows head
(ii) E: not tail appears, F: no head appears
If two coins are tossed once, then the sample space S is
S = {HH, HT, TH, TT}
(i) E = {HT, TH}
F = {HT, TH}
(ii) E = {HH}
F = {TT}
∴ E ∩ F = Φ
P (F) = 1 and P (E ∩ F) = 0
∴ P(E|F) =
Question 8:
A die is thrown three times,
E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses
If a die is thrown three times, then the number of elements in the sample space will be 6 × 6 × 6 = 216
Question 9:
Mother, father and son line up at random for a family picture
E: son on one end, F: father in middle
If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be
S = {MFS, MSF, FMS, FSM, SMF, SFM}
⇒ E = {MFS, FMS, SMF, SFM}
F = {MFS, SFM}
∴ E ∩ F = {MFS, SFM}
Question 10:
A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Let the first observation be from the black die and second from the red die.
When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements.
Let A: Obtaining a sum greater than 9
= {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
B: Black die results in a 5.
= {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
∴ A ∩ B = {(5, 5), (5, 6)}
The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).
(b) E: Sum of the observations is 8.
= {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
F: Red die resulted in a number less than 4.
The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F).
Question 11:
A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}
Find (i) P (E|F) and P (F|E) (ii) P (E|G) and P (G|E) (ii) P ((E ∪ F)|G) and P ((E ∩ G)|G)
When a fair die is rolled, the sample space S will be
S = {1, 2, 3, 4, 5, 6}
It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}
(ii) E ∩ G = {3, 5}
(iii) E ∪ F = {1, 2, 3, 5}
(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}
E ∩ F = {3}
(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}
Question 12:
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?
Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be
S = {(b, b), (b, g), (g, b), (g, g)}
Let A be the event that both children are girls.
(i) Let B be the event that the youngest child is a girl.
The conditional probability that both are girls, given that the youngest child is a girl, is given by P (A|B).
Therefore, the required probability is (1/2).
(ii) Let C be the event that at least one child is a girl.
The conditional probability that both are girls, given that at least one child is a girl, is given by P(A|C).
Question 13:
An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
The given data can be tabulated as
True/False Multiple choice Total Easy 300 500 800 Difficult 200 400 600 Total 500 900 1400
Let us denote E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions
Total number of questions = 1400
Total number of multiple choice questions = 900
Therefore, probability of selecting an easy multiple choice question is
P (E ∩ M) = 500/1400= 5/14
Probability of selecting a multiple choice question, P (M), is 900/1400= 9/14
P (E|M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.
Therefore, the required probability is 5/9
Question 14:
Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
When dice is thrown, number of observations in the sample space = 6 × 6 = 36
Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.
∴ A = {(1, 3), (2, 2), (3, 1)}
Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.
Question 15:
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
The outcomes of the given experiment can be represented by the following tree diagram.
The sample space of the experiment is,
Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.
Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).
Therefore,
Question 16:
If
(A) 0 (B) (1/2)
(C) not defined (D) 1
Therefore, P (A|B) is not defined.
Thus, the correct answer is C.
Question 17:
If A and B are events such that P (A|B) = P(B|A), then
(A) A ⊂ B but A ≠ B (B) A = B
(C) A ∩ B = Φ (D) P(A) = P(B)
It is given that, P(A|B) = P(B|A)
Thus, the correct answer is D.
The document NCERT Solutions Class 11 Maths Chapter 14 - Probability is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
All you need of JEE at this link: JEE
## Mathematics (Maths) for JEE Main & Advanced
209 videos|443 docs|143 tests
## FAQs on NCERT Solutions Class 11 Maths Chapter 14 - Probability
1. What is probability?
Ans. Probability is a branch of mathematics that deals with the likelihood of an event occurring. It is a way to quantify uncertainty and measure the chances of different outcomes.
2. How is probability calculated?
Ans. Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This can be represented as a ratio, a decimal, or a percentage.
3. What are the different types of probability?
Ans. There are three main types of probability: theoretical, empirical, and subjective. Theoretical probability is based on mathematical calculations, empirical probability is based on observed data, and subjective probability is based on personal judgment or beliefs.
4. How can probability be used in real-life situations?
Ans. Probability is used in various real-life situations, such as weather forecasting, insurance risk assessment, stock market analysis, sports predictions, and even medical diagnoses. It helps in making informed decisions by considering the chances of different outcomes.
5. What are the common misconceptions about probability?
Ans. One common misconception about probability is the belief that if an event is unlikely to happen, it will never occur. In reality, even unlikely events can happen. Another misconception is that probability can predict individual outcomes, whereas it actually deals with the likelihood of outcomes over a large number of trials.
## Mathematics (Maths) for JEE Main & Advanced
209 videos|443 docs|143 tests
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; | 2,933 | 9,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.871827 |
https://quabr.com/66467123/generate-ids-for-nested-grid-cells-in-r | 1,619,166,924,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00213.warc.gz | 567,020,527 | 11,446 | # Generate IDs for nested grid cells in R
I have generated 1km grid cells:
library(sf)
BBox <- st_bbox(c(xmin = 0, xmax = 10000, ymax = 10000, ymin = 0), crs = st_crs(27700))
Grid_1km <- st_as_sfc(BBox) %>%
st_make_grid(square = TRUE, cellsize = c(1e3, 1e3)) %>%
cbind(data.frame(ID = sprintf(paste("GID%0",nchar(length(.)),"d",sep=""), 1:length(.)))) %>%
st_sf()
I would like to generate 500m, 250m and 125m grid cells that nest within each 1km cell and then use a naming scheme to give each cell a unique ID.
This can be used to create the grid cells:
Grid_500m <- st_as_sfc(BBox) %>%
st_make_grid(square = TRUE, cellsize = c(5e2, 5e2))
Grid_250m <- st_as_sfc(BBox) %>%
st_make_grid(square = TRUE, cellsize = c(2.5e2, 2.5e2))
Grid_125m <- st_as_sfc(BBox) %>%
st_make_grid(square = TRUE, cellsize = c(1.25e2, 1.25e2))
I would like each 500m, 250m and 125 cell to reference the ID given to the 1km cell that it nests within and use the naming convention illustrated below to assign ID's to each cell of the hierarchy. So for example, the top left 125m cell in GID001 would be GID001333.
I am not sure the best way to generate these IDs. The real 1km IDs I am working with are more randomly ordered than presented in my example - so I assume some some sort of spatial operation will be needed to identity the 1km grid cell ID. After this I would guess there is a mathematical way of doing what I am after that will be the most efficient?
When facing this issue myself - my use case called for preserving the original numeric Id of a big cell and supplement it by character (a, b, c, d) - I used code like this.
It is built around st_contains = getting indexes of small cells contained in the "big ones", and creating a vector of ids for each.
I do not presume to call it the best or most efficient way to generate the ids, and it likely does not scale indefinitely, but it works.
library(sf)
library(dplyr)
BBox <- st_bbox(c(xmin = 0,
xmax = 2000,
ymax = 2000,
ymin = 0), crs = st_crs(27700))
grid_1000 <- st_as_sfc(BBox) %>%
st_make_grid(square = TRUE, cellsize = c(1e3, 1e3)) %>%
st_sf() %>%
mutate(id_1000 = 1:nrow(.)) # this will be "final" id
grid_500 <- st_as_sfc(BBox) %>%
st_make_grid(square = TRUE, cellsize = c(500, 500)) %>%
st_sf() %>%
mutate(idx_500 = 1:nrow(.)) # this is just temporary
# a technical object to link sequences
asdf <- grid_1000 %>%
st_contains(grid_500) %>%
as.data.frame() %>%
setNames(c("idx_1000", "idx_500"))
asdf$id_1000 <- grid_1000$id_1000[asdf$idx_1000] asdf$id_500 <- paste0(asdf\$id_1000, c("c", "d", "a", "b"))
grid_500 <- grid_500 %>%
inner_join(asdf, by = c("idx_500" = "idx_500")) %>%
select(id_500) %>%
st_set_agr("constant")
# check validity
library(ggplot2)
ggplot(grid_1000) +
geom_sf() +
geom_sf_text(aes(label = id_1000))
ggplot(grid_500) +
geom_sf() +
geom_sf_text(aes(label = id_500)) | 912 | 2,867 | {"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} | 2.609375 | 3 | CC-MAIN-2021-17 | latest | en | 0.840569 |
http://csundergrad.science.uoit.ca/courses/cv-notes/notebooks/03-linear-filtering.html | 1,611,467,764,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547333.68/warc/CC-MAIN-20210124044618-20210124074618-00730.warc.gz | 23,093,723 | 586,023 | # Linear Filtering¶
Faisal Qureshi
Professor
Faculty of Science
Ontario Tech University
http://vclab.science.ontariotechu.ca
## Outline¶
• Linear Filtering in 1D
• Cross-correlation
• Convolution
• Gaussian filter
• Gaussian blurring
• Separability
• Relationship to Fourier transform
• Integral images
In [1]:
import numpy as np
import scipy as sp
from scipy import signal
import cv2 as cv
import matplotlib.pyplot as plt
In [2]:
# Load image of Van Gogh, convert from BGR to RGB image
I = cv.cvtColor(I, cv.COLOR_BGR2RGB)
In [3]:
plt.figure(figsize=(7,7))
plt.imshow(I)
plt.xticks([])
plt.yticks([]);
Let's plot a single row of the red channel of image $I$, say, row 200.
In [4]:
row = 200
rowI = I[row,:,0]
plt.figure()
plt.plot(rowI,'-')
plt.plot(rowI,'r.');
## Using kernel to compute mean¶
In [5]:
signal = rowI[0:5]
kernel = np.array([1,1,1,1,1]) / 5
print(rowI[0:5])
print('A = ', np.mean(rowI[0:5]))
print('B = ', np.dot(kernel, signal))
[133 127 120 119 125]
A = 124.8
B = 124.8
## Performing moving averages¶
Let's perform a moving average on this row. We assume window with half-width equal to $w$.
In [6]:
width = 5
signal = rowI
kernel = np.ones(2*width+1)/(2*width+1) # Array to represent the window (constant-valued)
result = np.ones(len(signal)-2*width) # Array to store computed moving averages
for i in range(len(result)):
centre = i + width
result[i] = np.dot(kernel, signal[centre-width:centre+width+1])
In [7]:
# Generate a plot of the original 1D signal and its average
plt.figure()
plt.plot(signal,'r.', alpha=0.3)
plt.plot(signal,'-', alpha=0.3)
plt.plot(np.arange(width, len(signal)-width),result,'k-');
## Linear Filtering in 1D¶
• Signal: $\mathbf{f}$
• Kernel (also sometimes called a filter or a mask): $\mathbf{h}$
• Half-width of the kernel: $w$ (i.e., kernel width is $2 w + 1$)
### Cross-Correlation¶
$$CC(i) = \sum_{k \in [-w,w]} \mathbf{f}(i + k) \mathbf{h}(k)$$
### Convolution¶
$$(\mathbf{f} \ast \mathbf{k} )_i = \sum_{k \in [-w,w]} \mathbf{f}(i - k) \mathbf{h}(k)$$
### Flipping kernel¶
Convolution and cross-correlation give the same results for symmetric kernels
In [8]:
kernel = np.array([3,1,2,1,3])
flipped_kernel = np.flip(kernel)
print('Symmetric kernel')
print(f'kernel={kernel}, flipped={flipped_kernel}')
kernel = np.array([1,1,2,2,3])
flipped_kernel = np.flip(kernel)
print('Asymmetric kernel')
print(f'kernel={kernel}, flipped={flipped_kernel}')
Symmetric kernel
kernel=[3 1 2 1 3], flipped=[3 1 2 1 3]
Asymmetric kernel
kernel=[1 1 2 2 3], flipped=[3 2 2 1 1]
#### Exercise: computing a 1D convolution (from scratch)¶
Compute $\mathbf{f} \ast \mathbf{h}$ given
$$\mathbf{f} = \left[ \begin{array}{cccccccc} 1 & 3 & 4 & 1 & 10 & 3 & 0 & 1 \end{array} \right]$$
and
$$\mathbf{h} = \left[ \begin{array}{ccc} 1 & 0 & -1 \end{array}\right]$$
In [9]:
# Your solution here
In [10]:
# %load solutions/linear-filtering/solution_01.py
# Computing a 1D convolution from scratch
f = np.array([1,3,4,1,10,3,0,1])
h = np.array([1,0,-1])
width = 1
result = np.ones(len(f)-2*width) # Array to store computed moving averages
for i in range(len(result)):
centre = i + width
result[i] = np.dot(h[::-1], f[centre-width:centre+width+1])
print(f'signal f:\t\t{f}')
print(f'kernel h:\t\t{h}')
print(f'convolution (f*h):\t{result}')
signal f: [ 1 3 4 1 10 3 0 1]
kernel h: [ 1 0 -1]
convolution (f*h): [ 3. -2. 6. 2. -10. -2.]
### Built-in Functions for 1D Convolution¶
NumPy, of course, has a builtin function numpy.convolve for convolution that wraps the slow loop in compiled code. Notice that, with the definition above, if the original array has length $N$ and the convolution kernel has width $2w+1$, the convolved array has length $N-2w$. In numpy.convolve, this behaviour is enforced using the keyword argument mode='valid'.
#### Exercise: computing a 1D convolution (with convolve)¶
Re-compute $\mathbf{f} \ast \mathbf{h}$ using numpy.convolve given
$$\mathbf{f} = \left[ \begin{array}{cccccccc} 1 & 3 & 4 & 1 & 10 & 3 & 0 & 1 \end{array} \right]$$
and
$$\mathbf{h} = \left[ \begin{array}{ccc} 1 & 0 & -1 \end{array}\right].$$
In [11]:
# Your solution here
In [12]:
# %load solutions/linear-filtering/solution_02.py
# Computing a 1D convolution using np.convolve
f = np.array([1, 3, 4, 1, 10, 3, 0, 1])
h = np.array([1, 0, -1])
r = np.convolve(f, h, mode='valid')
print(f'signal f:\t\t{f}')
print(f'kernel h:\t\t{h}')
print(f'convolution (f*h):\t{r}')
signal f: [ 1 3 4 1 10 3 0 1]
kernel h: [ 1 0 -1]
convolution (f*h): [ 3 -2 6 2 -10 -2]
### Example: Gaussian smoothing for noise removal¶
• Construct a 1D (noisy) signal
• Filter this signal with a 1D Gaussian kernel to suppress the noise and improve the signal-to-noise ratio
In [13]:
np.random.seed(0)
n = 20
x = np.linspace(0, np.pi, n)
y = np.sin(x)
mu = 0.0
sigma = 0.2
y_noisy = y + np.random.normal(mu, sigma, n)
plt.plot(x,y,'.-', label='data', alpha=0.5)
plt.plot(x,y_noisy,'.-r', label='noisy data')
plt.xlim(-.4,3.5)
plt.ylim(-1,2)
plt.xticks([])
plt.yticks([])
plt.legend(['data', 'noisy data']);
#### Exercise: lets try to recover original signal (data) from corrupted signal (noisy data) using smoothing¶
In [14]:
# %load solutions/linear-filtering/solution_03.py
f = y_noisy
width = 1
h = np.ones(2*width+1)/(2*width+1)
result = np.ones(len(f)-2*width) # Array to store computed moving averages
for i in range(len(result)):
centre = i + width
result[i] = np.dot(h[::-1], f[centre-width:centre+width+1])
plt.plot(x,y,'.-', label='data', alpha=0.3)
plt.plot(x,y_noisy,'.-r', label='noisy data', alpha=0.5)
plt.plot(x[width:-width],result,'.-b', label='noisy data')
plt.xlim(-.4,3.5)
plt.ylim(-1,2)
plt.xticks([])
plt.yticks([])
plt.legend(['data', 'noisy data']);
Note that lengths of x and result are not the same?
## Gaussian in 1D¶
$$G(\mu,\sigma,x) = \frac{1}{\sigma \sqrt{2 \pi}} \exp ^ {- \frac{(x-\mu)^2}{2 \sigma^2}}$$
Here $\mu$ and $\sigma$ refer to the mean and standard deviation of this Gaussian.
## Aside: Mean and Standard Deviation¶
Given $n$ data points $\{x_1, x_2, \cdots, x_n \}$
$$\mu = \mathrm{E}[x] = \frac{1}{n} \sum_{i=1}^n x_i$$$$\sigma^2 = \mathrm{E}[(x-\mu)^2] = \frac{1}{n} \sum_{i=1}^n (x_i - \mu)^2$$
## Evaluating Gaussian at a particular value.¶
In [15]:
mu = 2
sigma = 3
In [16]:
def g(mu, s, x):
exponent = -((x - mu)**2)/(2*s*s)
constant = 1/(s * np.sqrt(2 * np.pi))
return constant * np.exp( exponent )
In [17]:
g_ = np.empty(20)
for i in range(-10,10):
g_[i-10] = g(mu, sigma, i)
print(f'Gaussian with mu={mu} and sigma={sigma} evaluated at {i} is {g_[i-10]}')
Gaussian with mu=2 and sigma=3 evaluated at -10 is 4.461007525496179e-05
Gaussian with mu=2 and sigma=3 evaluated at -9 is 0.0001600902172069401
Gaussian with mu=2 and sigma=3 evaluated at -8 is 0.0005140929987637022
Gaussian with mu=2 and sigma=3 evaluated at -7 is 0.001477282803979336
Gaussian with mu=2 and sigma=3 evaluated at -6 is 0.003798662007932481
Gaussian with mu=2 and sigma=3 evaluated at -5 is 0.008740629697903166
Gaussian with mu=2 and sigma=3 evaluated at -4 is 0.017996988837729353
Gaussian with mu=2 and sigma=3 evaluated at -3 is 0.03315904626424957
Gaussian with mu=2 and sigma=3 evaluated at -2 is 0.05467002489199788
Gaussian with mu=2 and sigma=3 evaluated at -1 is 0.0806569081730478
Gaussian with mu=2 and sigma=3 evaluated at 0 is 0.10648266850745075
Gaussian with mu=2 and sigma=3 evaluated at 1 is 0.12579440923099774
Gaussian with mu=2 and sigma=3 evaluated at 2 is 0.1329807601338109
Gaussian with mu=2 and sigma=3 evaluated at 3 is 0.12579440923099774
Gaussian with mu=2 and sigma=3 evaluated at 4 is 0.10648266850745075
Gaussian with mu=2 and sigma=3 evaluated at 5 is 0.0806569081730478
Gaussian with mu=2 and sigma=3 evaluated at 6 is 0.05467002489199788
Gaussian with mu=2 and sigma=3 evaluated at 7 is 0.03315904626424957
Gaussian with mu=2 and sigma=3 evaluated at 8 is 0.017996988837729353
Gaussian with mu=2 and sigma=3 evaluated at 9 is 0.008740629697903166
In [18]:
plt.plot(g_)
Out[18]:
[<matplotlib.lines.Line2D at 0x139ee7e50>]
Vectorized code for evaluating Gaussian over a range of values
In [19]:
def gaussian1d(mu, sig, n, normalized=True):
s = n//2
x = np.linspace(-s,s,n)
exponent_factor = np.exp(-np.power(x - mu, 2.) / (2 * np.power(sig, 2.)))
normalizing_factor = 1. if not normalized else 1./(sig * np.sqrt(2*np.pi))
return normalizing_factor * exponent_factor
In [20]:
n = 9
width = n // 2
sigma = 1.4
g1 = gaussian1d(0.0, sigma, n, normalized=True)
g2 = gaussian1d(0.0, sigma, n, normalized=False)
plt.plot(np.linspace(-width,width,n), g1, 'r-.', label='Normalized')
plt.plot(np.linspace(-width,width,n), g2, 'b-o', label='Not normalized')
plt.title('Gaussian kernels')
plt.legend(loc='center left');
In [21]:
# %load solutions/linear-filtering/solution_05.py
y_smoothed = np.convolve(y_noisy, g1, mode='valid')
plt.plot(x,y,'b', label='data', alpha=0.3)
plt.plot(x,y_noisy,'r', label='noisy data', alpha=0.5)
plt.plot(x[width:-width],y_smoothed,'k', label='smoothed data')
plt.xlim(-.4,3.5)
plt.ylim(-1,2)
plt.xticks([])
plt.yticks([])
plt.legend(loc='upper left')
Out[21]:
<matplotlib.legend.Legend at 0x139f8c390>
## Linear Filter in 2D¶
• Signal: $\mathbf{f}$
• Kernel (also sometimes called a filter or a mask): $\mathbf{h} \in \mathbb{R}^{(2w+1)\times(2h+1)}$
### Cross-Correlation¶
$$CC(i,j) = \sum_{\substack{k \in [-w,w] \\ l \in [-h,h]}} \mathbf{f}(i + k, j+l) \mathbf{h}(k,l)$$
### Convolution¶
$$(\mathbf{f} \ast \mathbf{k} )_{i,j} = = \sum_{\substack{k \in [-w,w] \\ l \in [-h,h]}} \mathbf{f}(i - k, j - l) \mathbf{h}(k,l)$$
Convolution is equivalent to flipping the filter (kernel) in both directions. Convolution and cross-correlation are the same for symmetric filters.
In [22]:
f = np.linspace(0,15,16).reshape((4,4))
h = np.ones((3,3))
print(f'f = \n{f}')
print(f'h = \n{h}')
f =
[[ 0. 1. 2. 3.]
[ 4. 5. 6. 7.]
[ 8. 9. 10. 11.]
[12. 13. 14. 15.]]
h =
[[1. 1. 1.]
[1. 1. 1.]
[1. 1. 1.]]
### Exercise¶
Compute $(\mathbf{f} \ast \mathbf{k} )_{i,j}$ where $(i,j) = (1,2)$.
In [23]:
# Your answer goes here
foo = f[:3,1:]
print(foo)
print(np.reshape(foo, (9)))
print(np.reshape(h, (9)))
print(np.dot(np.reshape(foo, (9)), np.reshape(h, (9))))
[[ 1. 2. 3.]
[ 5. 6. 7.]
[ 9. 10. 11.]]
[ 1. 2. 3. 5. 6. 7. 9. 10. 11.]
[1. 1. 1. 1. 1. 1. 1. 1. 1.]
54.0
### 2D Averaging Kernels¶
A 3-by-3 summing kernel
In [24]:
average_3_by_3 = np.array([[1,1,1],
[1,1,1],
[1,1,1]], dtype='float32')
### Exercise¶
Construct a 3-by-3 averaging kernel.
In [25]:
# Your answer goes here
average_3_by_3 = average_3_by_3/np.sum(average_3_by_3)
print(f'average_3_by_3 = \n{average_3_by_3}')
average_3_by_3 =
[[0.11111111 0.11111111 0.11111111]
[0.11111111 0.11111111 0.11111111]
[0.11111111 0.11111111 0.11111111]]
### Convolution with averaging kernels¶
In [26]:
average_3_by_3 = np.ones(9).reshape(3,3) / 9
average_5_by_5 = np.ones(25).reshape(5,5) / 25
average_7_by_7 = np.ones(49).reshape(7,7) / 49
In [27]:
print(f'average_5_by_5 = \n{average_5_by_5}')
average_5_by_5 =
[[0.04 0.04 0.04 0.04 0.04]
[0.04 0.04 0.04 0.04 0.04]
[0.04 0.04 0.04 0.04 0.04]
[0.04 0.04 0.04 0.04 0.04]
[0.04 0.04 0.04 0.04 0.04]]
In [28]:
I_gray = cv.cvtColor(I, cv.COLOR_BGR2GRAY)
I3 = sp.signal.convolve2d(I_gray, average_3_by_3, mode='same')
I5 = sp.signal.convolve2d(I_gray, average_5_by_5, mode='same')
I7 = sp.signal.convolve2d(I_gray, average_7_by_7, mode='same')
plt.figure(figsize=(20,10))
plt.subplot(241)
plt.imshow(I_gray, cmap='gray')
plt.subplot(242)
plt.imshow(I3, cmap='gray')
plt.subplot(243)
plt.imshow(I5, cmap='gray')
plt.subplot(244)
plt.imshow(I7, cmap='gray')
plt.subplot(245)
plt.imshow(I_gray[215:290,175:250], cmap='gray')
plt.subplot(246)
plt.imshow(I3[215:290,175:250], cmap='gray')
plt.subplot(247)
plt.imshow(I5[215:290,175:250], cmap='gray')
plt.subplot(248)
plt.imshow(I7[215:290,175:250], cmap='gray'); | 4,456 | 11,936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-04 | latest | en | 0.533151 |
https://gis.stackexchange.com/questions/265468/how-to-select-split-polylines-in-specific-distances-in-bulk | 1,726,856,112,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00581.warc.gz | 230,932,902 | 41,386 | How to select/split polylines in specific distances in bulk?
I am trying to map road rehabilitation projects. I have polylines (roads) and I have a table which indicates in each year which part of the road was rehabilitated (for example, for a road HW107 rehabilitated from km 75 to km 87 in 2009, km 22 to km 37 in 2010 and etc. I have these numbers in separate columns).
I know how to select/split each individual polyline manually or by indicating specific distance. However, since I have more than thousand road works I was wondering if there is a tool to do it in bulk?
I am using ArcGIS Desktop 10.3.
• Sounds like a linear referencing task, suggest you research the help file on the topic of linear referencing. Commented Dec 15, 2017 at 10:12
2 Answers
UPDATE: In the end I used Make Route Event Layer from the Linear Referencing tool. It allows to create an event layer (point or a line) by indicating km-start and km-end.
An option would be to break the Highway lines into equal segments of 1km each. You can use ET Geowizards to do this or you can use points along a line tool, then split the line using that tool.
Create a field called "Chainage" and assign the kilometre value to it. Create a field in the broken polyline file called "Road_ID" and using the field calculator assign an ID such with the road name and the chainage.
In your table, make sure you have a record for each kilometre. If road works happened from 27km to 30km in 2009, then there should be 4 records for each kilometre and there should be a field for the year.
In the Table with the road works do the same. Have a field called "Road_ID" and link the Road name and the chainage together. You will then have a series of records with the name of the road and sequential distance. From your example you should have 36 records for 2010 each labelled "HWY107_22" ... to ... "HWY107_37" in one column and 2010 in another.
Once you have this table formatted, you can bring it into ArcMap. Use the table join on the split roads layer with the table you just added using the "Road_ID" as the common field. Once the join is complete, copy the year across into the roads layer. You can then run a dissolve on this roads layer using the Road Name and the Year. That will give you something you can symbolise.
• I put points along the line and split the lines. However, now when I need to assign kilometer values to each split line, I see that the numbering is quite random, while the points seemed to be quite sequential (for example, line number 1 is in the middle of the road, line number 525 is in the other end, and so on). How can I assign kilometer value in this case? Commented Dec 16, 2017 at 17:13 | 647 | 2,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.933059 |
http://oeis.org/A180350 | 1,553,126,507,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202474.26/warc/CC-MAIN-20190320230554-20190321012554-00142.warc.gz | 155,486,316 | 3,694 | This site is supported by donations to The OEIS Foundation.
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A180350 G.f.: Sum_{n>=0} a(n)*x^n/n!^5 = [ Sum_{n>=0} x^n/n!^5 ]^3. 0
1, 3, 99, 9237, 775971, 83118753, 10657602909, 1463886204147, 215566192274211, 33677584957306713, 5492032622227428849, 928229455634614797447, 161727023896151286167901, 28905146810167510775300463 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS FORMULA a(n) = Sum_{k=0..n} C(n,k)^5 * Sum_{j=0..k} C(k,j)^5 = Sum_{k=0..n} C(n,k)^5 * A005261(k). EXAMPLE G.f.: A(x) = 1 + 3*x + 99*x^2/2!^5 + 9237*x^3/3!^5 + 775971*x^4/4!^5 +... A(x)^(1/3) = 1 + x + x^2/2!^5 + x^3/3!^5 + x^4/4!^5 +... PROG (PARI) {a(n)=if(n<0, 0, n!^5*polcoeff(sum(m=0, n, x^m/m!^5+x*O(x^n))^3, n))} (PARI) {a(n)=sum(k=0, n, binomial(n, k)^5*sum(j=0, k, binomial(k, j)^5))} CROSSREFS Cf. A005261. Sequence in context: A167582 A068420 A276188 * A293952 A303826 A128296 Adjacent sequences: A180347 A180348 A180349 * A180351 A180352 A180353 KEYWORD nonn AUTHOR Paul D. Hanna, Jan 20 2011 STATUS approved
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Last modified March 20 19:57 EDT 2019. Contains 321349 sequences. (Running on oeis4.) | 579 | 1,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.500956 |
https://scienceblogs.com/principles/2011/12/13/christmas-physics-how-strong-i | 1,563,412,435,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525483.62/warc/CC-MAIN-20190718001934-20190718023934-00088.warc.gz | 528,030,347 | 17,097 | # Christmas Physics: How Strong Is One Grinch?
And what happened then?
Well, in Who-ville they say
That the Grinch's small heart
Grew three sizes that day.
And then the true meaning
Of Christmas came through
And the Grinch found the strength
Of ten Grinches, plus two
-- Dr. Seuss's How the Grinch Stole Christmas
It's nearly Christmas, so SteelyKid keeps demanding to watch the two classic Christmas specials we have recorded, Dr. Seuss's How the Grinch Stole Christmas and Rudolph the Red-Nosed Reindeer. Watching these over and over again, my thoughts naturally turn to physics, and what sort of physics you could do with these shows.
The most obvious possibility is suggested by the lines above. As you no doubt remember if you've seen the cartoon, the Grinch steals all the Christmas trappings from the Whos down in Who-ville, loads it on a sled, and drives it ten thousand feet up the side of Mt. Crumpet, to dump it. When he hears the Whos singing their Christmas song even without their material goods, he has a change of heart, and saves the sled from falling off the cliff, using his new-found strength:
So, just how strong is the Grinch, to lift all that?
The actual lifting shot is from a funny angle, so it's hard to work with, but there's a cleaner shot while he's driving up the mountain:
Now, the Grinch positively towers over Cindy Lou Who (who is no more than two), so let's say he's my height, roughly 2m tall. He's approximately 52 pixels high in the original screen capture of that image, which works out to about 0.04m/pixel. Using that as the scale, the sack he's standing on is 6.6m high, and the sled full of loot is 9.6m long. It's hard to estimate the depth of the sled, but let's say it's roughly 4.5 m deep, a bit less than half as wide as it is long.
Treating the sacks as a giant rectangular solid, then, we would have a volume of 285 cubic meters of stuff. To convert this into a mass, we need an estimate of the density; the simple and easy density figure to remember is that water has a density of 1 g/cm3, or 1000 kg/m3. So, at the density of water, the Grinch's sled has a mass of 285,000 kg.
Of course, water's pretty heavy, and a lot of what's in those sacks is considerable lighter, so let's guess and average density of about a third that of water, and call it 100,000 kg total. Since he's lifting that with the strength of twelve Grinches, that means a single Grinch could lift 8,333 kg. That's around 32 times the clean and jerk lift world record, so you do not want to mess with a Grinch.
(Of course, the one in that relationship that you really don't want to mess with is the Grinch's faithful dog, Max, who pulls that whole sled ten thousand feet up the side of Mt. Crumpet, all by himself... Max rules.)
Given this, we can also answer a second question, namely, just how much do the Whos like Christmas? It says that "Every Who down in Who-ville liked Christmas a lot," but that's not very quantitative. Given the video evidence, though, we can quantify this. The population of Who-ville, the tall and the small, is exactly 33 Whos, as we can see when they're singing:
With 100,000 kg of Christmas trappings for all of Who-ville, that works out to 3,030 kg of Christmas gear for each and every Who. So, when they say they like Christmas a lot, they mean they like it a lot.
So, there's your incredibly dorky analysis for this Christmas. Fah-who foraze, da-who doraze, and all that.
Tags
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With 100,000 kg of Christmas trappings for all of Who-ville, that works out to 3,030 kg of Christmas gear for each and every Who. So, when they say they like Christmas a lot, they mean they like it a lot.
Alternatively, 8.6 cubic meters of stuff when packed into bags (to eliminate the density question).
That means a pile 10 ft by 10 ft that is 3 feet deep or so. Come to think about it, that's not so far off the kids' haul...
By Marry Me, Mindy (not verified) on 13 Dec 2011 #permalink
In other questions, how hot is Heat Miser's touch to be able to melt the shovel that's in his clutch?
Ah, but . . .
Are these the same Whos as appeared in _Horton_Hears_A_Who_"? If they are, then an average Who is only about a micrometer tall (if that). And with the cube/square law being what it is, they would be able to perform incredible weightlifting feats similar to what we see from ants and other small insects.
Of course, that would introduce a whole host of other questions, like how the snowflakes are still a lot smaller than an average Who. But I'll leave that for someone else to explain away.
What Tim Eisele said!
By Anonymous Coward (not verified) on 13 Dec 2011 #permalink
Fah-who foraze, da-who doraze, and all that.
Thanks a lot, I now have that stuck in my head. :-P
By Calli Arcale (not verified) on 13 Dec 2011 #permalink
I think everyone can agree that the Grinch and Max are clearly the heroes of the story. The Grinch is like Judas, the hero of the Jesus mythology and the protagonist that has to do the evil deed to advance the higher truth, while risking his own vilification by the uninformed. Max is clearly the Madonna analog and the model of steadfast loyalty, fortitude, and strength of will.
Drop the assumed height of the Grinch to five millimeters and the whole thing seems to work out from a strength of material perspective.
But are these the same Whos that Horton heard? That would make them, and the Grinch, a lot smaller.
On the scaling issue, the height of Mt. Crumpet is clearly given as 10,000 feet, and the Who houses seem more or less proportionate to that, so either they're using Who-feet as a unit of measure (as opposed to the US foot), or Horton heard a different pack of Whos. I'm choosing to run with the latter.
BTW, you need to work on your sig figs.
"About a third" of 284,000 Kg gets to about 100,000 Kg, sure. But to go to 8333 Kg/grinch? Blah
And 3030 Kg/Who? Double blah.
By Marry Me, Mindy (not verified) on 13 Dec 2011 #permalink
A 3-size Grinch heart increase led to a 12-fold increase in Grinch strength. Let's speculate that Grinch heart volume is proportional to Grinch strength, and that Grinch heart size increases are isotropic (i.e. proportional in all 3 dimensions). That means a 3-size jump makes a Grinch heart ~2.29 times larger. So a 1-size increase would have given him the strength of ~2.9 Grinches, and a 2-size increase the strength of ~6.4 Grinches. On the other hand, a 1-size decrease would have reduced him to less than 5% the strength of an ordinary Grinch. Just something to bear in mind in case you run into a Grinch in a dark Whoville alley, armed only with a Grinch-heart-size meter.
By Raj Dhuwalia (not verified) on 13 Dec 2011 #permalink
10Grinch + 2 =/= 12Grinch.
I "can't help" bringing up the astonishing (to normal people) rise of Newt Gingrinch (hah, he's famous enough that my spellchecker suggested him, also "chagrining") in the Republican primary race - no I wouldn't distract here merely because of his name, but naturally from his perspectives, about reducing social programs and getting poor kids to work as janitors in schools for their lunch money etc. Well, it is truly ironic and even pitiful, that folks expressing militant support for "family values" and "character counts" would be so supportive of a serial adulterer. (In all fairness, a big chunk of them aren't, and clearly say so. Good for them, as consistency.)
Sure, I can accept forgiving someone like that if they strayed and came back. But Newt is currently married to the woman he started an affair with against his second wife. That is like keeping what you stole and still wanting forgiveness. Furthermore, if Newt wins the Presidency, she will be our "First Lady" - so they'd make a couple of continued adulterers in our White House. This is worse than the President alone indulging in indiscretions, then giving them up later - which was bad enough. (Again, I only note this because of the name and Chad's interest in the astonishing politics of the Right.)
I'm like my grandson! The Grinch is too scarey!
Chad (post 8), I think that the former is more in line with the story. While perhaps not the self-same Whos whom Horton heard, they are certainly of the same dust abiding species. The whole of the story takes place on an individual snow flake (as dust particles nucleate raindrops and ultimately snowflakes). But all of the reasoning done here is proportional reasoning, so it gives an answer that is merely scaled up to human proportions for understanding.
Thatâs probably one of the coolest things Iâve ever read. Iâm really not that great at math not to mention that my physics education only consists of a quarter of the year way back in eighth grade, but I do buy your conclusion and I wonât question it. Thatâs amazing that you can just look at a picture from a movie and estimate the weight of whatâs the Grinchâs Santa bag is just by examining the pixels and the context clues then doing some scaling. I donât think itâs dorky at all, not even a waste of time, but a really cool skill and Iâm jealous! Good work!
100,000 kg, eh? So did the Grinch steal Christmas, or Festivus?
By Tom Renbarger (not verified) on 15 Dec 2011 #permalink | 2,573 | 10,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-30 | longest | en | 0.959931 |
https://physics.stackexchange.com/questions/409221/what-is-quantum-gas/409232 | 1,716,130,732,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00151.warc.gz | 414,083,712 | 38,476 | What is "quantum gas"?
I could not find a satisfying definition of this term from anywhere. Also, Can you please give some introductory level references to understand quantum gas?
To put it simply, a quantum gas (or quantum fluid) is a system made of indistinguishable particles, as opposed to a classical gas where you assume that you can differentiate one component from other (that means that in a classical gas you could label every single constituent of the gas theoretically, and always know its position and momentum, whereas for a quantum gas you cannot know such information for each particle individually).
Very generally, fluids exhibit a quantum behaviour at low temperatures and/or high densities and the classical gas model is a good approximation for high temperatures and/or low density. However, don't get confused: there are no classical particles in nature and the classical gas is nothing but a simplification that under some circumstances can be used to describe reality.
Depending on the wavefunction simmetry of the particles that constitute the gas, the gas follows a different statistic, namely Bose-Einstein statistics for bosons and Fermi-Dirac for fermions.
From a macroscopic point of view, we usually call quantum fluids to those that exhibit some kind of "quantum" macroscopic behaviour such as superfluidity or superconductivity.
To start studying quantum gases I think that it would be necessary to have some background in statistical mechanics and also some ideas of quantum mechanics.
For the statistical mechanics part:
H.B. Callen. Thermodynamics and an introduction to Thermostatistics.
S. R. A. Salinas. Introduction to Statistical Physics.
D. J. Amit, Y. Verbin, R. Tzafriri. Statistical physics. An introductory course
And maybe more advanced options are
R. Balian. From microphysics to macrophysics.
M. Kardar. Statistical physics of particles.
And for the quantum mechanics part I'd reccomend you
Cohen-Tannoudji C., Diu B., Laloe F. Quantum mechanics, vol. 1 and vol. 2 | 422 | 2,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.899122 |
http://www.higheredwatch.org/calculate-the-interest-payable-at-maturity/ | 1,603,912,686,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107900860.51/warc/CC-MAIN-20201028191655-20201028221655-00447.warc.gz | 138,196,003 | 9,324 | # Calculate The Interest Payable At Maturity
Combining the Present Value of a Bond’s Interest and Maturity Amounts. Adjusting Basis. Your basis in the debt increases each year as you realize interest from the discount, ensuring that you won’t recognize any capital gain if you hold onto the debt until maturity. Each increase is equal to the amount of interest you reported the year.
The secured loan bears interest at a fixed rate of 4.00% for a three-year term; the interest is payable monthly and the principal will be payable at maturity only. The effective capitalization rate of. However, the total investment with interest is only payable at maturity.
In addition to paying back the principal, the issuer will make periodic interest payments to the bondholder until the bond reaches maturity. In order to determine how much those interest payments will be annually, semi-annually, or monthly, it is important to be able to calculate interest payments on a bond.
RBI states that if redeemed before maturity, the penalty charges will be at the rate of 50% of the last coupon payable for early redemption. Hence, Maturity Date Loan Calculator – AgriBank – (Estimated Taxes and Insurance are added to the principal and interest to calculate the total payment.)
has revised downwards the interest rate for deposits below Rs 1 crore for a select maturity bracket. “The Bank has decided to revise rates of interest payable on Domestic Term Deposits & NRO Deposits.
The effective interest method of amortizing the discount to interest expense calculates the interest expense using the carrying value of the bonds and the market rate of. Accounting Basics: Lesson 9 – Calculating Interest and the Maturity Value of Notes – : 7:24 charlotte houke 8 771 .
Balloon Home Loan A balloon payment mortgage is a mortgage which does not fully amortize over the term of the note, thus leaving a balance due at maturity. The final payment is called a balloon payment because of its large size. balloon payment mortgages are more common in commercial real estate than in residential real estate.Balloon Rate Loan Balloon Home Loan A balloon loan is a type of mortgage that doesn’t fully amortize over the life of the loan, leaving a large "balloon payment" due at the end of the mortgage. home loans with balloon payments have lower monthly payments in the years leading up when the balloon payment is due, but the size of many of these payments often makes it difficult (or impossible) for borrowers to pay them off.In other words, these loans have a 30-year amortization schedule with a balloon payment after five to seven years. Some balloon mortgages.
· In the above example, the principal amount of the note payable was 15,000, and interest at 8% was payable in addition for the term of the notes. Sometimes notes payable are issued for a fixed amount with interest already included in the amount. In this case the business will actually receive cash lower than the face value of the note payable.
how does a balloon mortgage work Bankrate Calculator Mortgage Bankrate Financial Calculator – Vadodara Property Centre – Guaranty Trust Bank is a leading African Bank that offers online/internet banking, Retail Banking, Corporate Financial Calculators. Loans Non-monthly payments debt investment. loan calculator. Take a look at the amount of house equity you build in seven years using this calculator from Bankrate.50 Year Mortgage Calculator Mortgage Calculator | Bankrate® | Current Mortgage Rates – Mortgage Calculators: Alternative Use Most people use a mortgage calculator to estimate the payment on a new mortgage, but it can be used for other purposes, too. | 733 | 3,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-45 | latest | en | 0.937813 |
http://electronoobs.com/eng_arduino_tut36 | 1,534,650,181,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221214691.99/warc/CC-MAIN-20180819031801-20180819051801-00166.warc.gz | 122,235,412 | 10,401 | Help me by sharing this post
We will make a homemade ultrasonic distance sensor and see how these modules work. The working principle is quite simple but the circuit could get a bit tricky. We send a ultrasonic wave with the transmitter speaker, the wave hits an obstacle and bounce back to the receiver speaker. We calculate the time between the sent and received pulse, and by knowing the speed of sound which is constant, we can calculate the distance.
See the full part list here:
## PART 1 - The commercial module
To begin this project, I first analyze this commercial ultrasonic distance sensor below, the SR04. What I can see is that it ahs 4 pins, ground, power, trigger pin and echo. It has some components on the back, and on the front part, the main components, two ultrasonic speakers. One as transmitter and the other one as receiver. Each time we apply a pulse to the trigger pin, the transmitter speaker creates a 40kHz sound wave.
We have to analyze how it works and see the signals. For that, I've connected the trigger pin to and Arduino UNO and applied a 10us pulse each 100ms. I connect the TX and RX speakers to the oscilloscope and this is what I see. The green line is the TX signal. It is a 40KHz square wave applied to the transmitter speaker and that will send a sound wave burst. It has 8 cycles. Then, the yellow line is the received signal and it is the bounced sound wave. The time between the sent and received signal will give us information about the distance to the object.
As you can see below, there is a direct relation between the distance and the delay time between the sent and received pulse. That delay time is the tiem it took the sound wave to go to the object and get back. So we haev to divide that by 2. Then we multiply it by the speed of sound and we get the distance.
EXAMPLE:
Measured delay = 588us. Speed of sound = 340m/s
Distance = [(588us / 1000000) / 2] x 34000cm/s ≈ 10cm
Ok, but the commercial module works like this. It receives the trigger signal. Waits 250us then it sends 8 cycles of 40KHz sound wave and puts the echo pin to high. That wave hits an object and bounce back. The receiver detects that sound and then it puts the echo pin to low. That creates a pulse and the width of that pulse is related to the measured distance. That's what we haev to implement today for our project.
## PART 2 - Homemade module
See the full part list here:
First of all, this below, is the schematic that I've created for this project. You will need two ultrasonic speakers, one 1uad-OPAMP, the MAX232 voltage shifter the Arduino and a few more components such as PNP transistor, resistors and capacitors. I first mount this schematic on a breadboard for tests.
I have pins for trigger and echo just as the commercial module. Then the microcontroller connected to the lever shifter and it will apply 8 cycles of 40kHz signal after the trigger pulse, applied to the transmitter speaker. The good thing of using the MAX232 is that it has a voltage doubler and inverter. So I apply 5 V square signal to it and I can get over 30V peak to peak at the output that goes to the speaker and that will increase range. As you can see I send the square signal and receive that on the other speaker and it will change with the distance.
## PART 3 - Final PCB
Now I mount everything on the PCB. Arduino in the middle and the amplified transmitter speaker on one side and the receiver on the other. We have potentiometers so we could set the treshold voltage for each amplified stage of the OPAMP se could get a good wave that will go to the Arduino. Now we haev to program the microcontroller.
## PART 4 - Final code
The code is quite easy and short. First we set the registers for the ports so we define D3, D4, D5 and D10 as outputs. Those are TX out1, TX out2, Vcc activate and echo out pins. Then we set pin D9 and D8 to be able to fire an interruption.
We go to the interruption vector and if D8 is high, that means the trigger pin was activated so we can start the code. That will activate the loop in the infinite loop. We first have to add that delay of 250us we haev seen before. In the code that is 150us because after tests on the oscilloscope I saw a 100 extra delay so I've reduced the 250 delay to 150.
``````
DDRD |= B00111000; // Sets D3, D4, D5 outputs
DDRB |= B00000100; // Sets D10 as output
PCICR |= (1 << PCIE0); //enable PCMSK0 scan
PCMSK0 |= (1 << PCINT0); //Set pin D8 (trigger pin) set to fire interrupt on state change.
PCMSK0 |= (1 << PCINT1); //Set pin D9 (echo in) set to fire an interrupt on state change.
``````
Ok, after the delay we create 8 cycles of 40KHz on Tx out1 and Tx out 2 pins and that will send the 40KHz sound wave. But first we turn LOW the Vcc activate pin so the PNP connected to the MAX232 will be activated and the IC will be supplied. Then using a 12us delay, we create, more or less, a 40KHz signal. We do taht 8 times and then we put the Vcc activate back to high and we turn to HIGH the echo pin. Now when we detect the bounced signal we tutn echo pin to low and the pulse is over.
``````
PORTD &= B11011111; //D5 LOW //Activate the MAX323 PNP transistor for supply
PORTD |= B00001000; //D3 HIGH
PORTD &= B11101111; //D4 LOW
delayMicroseconds(12);//12us so around 40KHz. Freq = 1/2*12us
PORTD &= B11110111; //D3 LOW
PORTD |= B00010000; //D4 HIGH
delayMicroseconds(12);
``````
``````
ISR(PCINT0_vect){
//If digital D8 is high -> trigger was activated
if(PINB & B00000001){
Trig_in_state = true; //Set the Trig_in_state to true since we've detected the trigger pulse
}
//If trigger pin is low, the 10us pulse is over and we start the code
else if(Trig_in_state)
{
triggered = true; //Set trigered state to true
Trig_in_state = false; //Reset the D8 pin state
}
/*After the 8cycle burst each time there will be an interruption, that could be made by D8 or D9
since those are the only 2 pins set as interrupt active
So, since D8(trigger) is already low till next measurement, only D9 (echo in) could fire the interruption
So, when we detect that, we set the echo out pin to low (D10) and end the echo pulse
IMPORTANT: A better way to do this, is to also measure the echo in frequency to make sure
it is around 40KHz, but I haven't done that. Works like this as well.
*/
PORTB &= 11111011; //D10, echo pin out to LOW
}
``````
Downlaod code:
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## Week number with 2 digits
Hi,
The weeknum formula returns the weeks 1 to 9 as single digits. Is there a way to display these as 2 digits, e.g. 01, 02....09?
Thanks!
1 ACCEPTED SOLUTION
Looks like I've managed to solve this myself
I used the DAX Formula " weeknumber = FORMAT(WEEKNUM([date],1),"00") "
If anyone has any other better solutions, please feel free to contribute.
4 REPLIES 4
Looks like I've managed to solve this myself
I used the DAX Formula " weeknumber = FORMAT(WEEKNUM([date],1),"00") "
If anyone has any other better solutions, please feel free to contribute.
Anonymous
Not applicable
If your week numbers are stored as whole numbers, then:
format(if(WEEKNUM([date]<10,CONCATENATE(0,WEEKNUM([date]),WEEKNUM([date]),00)
that is,
format (<value>, <format_string>) where
<value> = if(WEEKNUM([date]<10,CONCATENATE(0,WEEKNUM([date]),WEEKNUM([date]
and
<format_string> = 00
@afk wrote:
Looks like I've managed to solve this myself
I used the DAX Formula " weeknumber = FORMAT(WEEKNUM([date],1),"00") "
If anyone has any other better solutions, please feel free to contribute.
Frequent Visitor
date = FORMAT(DATE(LEFT([Year-Month],4),RIGHT([Year-Month],2),1),"yyyy-mm")
Resolver IV
hi, @afk
This is correct.
But you could do this directly in POWER QUERY, with a similar formula, removing the calculation from the tip.
Or use a CALENDAR table for this and standardize the dates of your dashboard.
Best Regards,
Rfranca | 455 | 1,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-30 | latest | en | 0.828748 |
https://cracku.in/83-a-number-is-formed-by-writing-first-54-natural-num-x-cat-1998?utm_source=blog&utm_medium=video&utm_campaign=video_solution | 1,719,247,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00861.warc.gz | 159,915,645 | 26,182 | Question 83
# A number is formed by writing first 54 natural numbers next to each other as 12345678910111213 ... Find the remainder when this number is divided by 8.
Solution
For a number to be divisible by 8, last 3 digits must be divisible by 8.
Last 3 digits of this number are 354.
354 mod 8 = 2
Hence, 2 is the remainder.
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# Aplane flying horizontally at an altitude %f 2.2 miles and at a speed of 555 mi/hr passes directly over rdar station- Find the rate at which the distance from the p...
## Question
###### Aplane flying horizontally at an altitude %f 2.2 miles and at a speed of 555 mi/hr passes directly over rdar station- Find the rate at which the distance from the plane to the station is increasing whenothe distance between the plane and the station is 6.9 miles_ If necessary, round your answer to two decimal placesVariableDescriptionUnitTime Horizontal distance the plane has traveled since time Distance between the plane and the ground at time t Distance between the plane and the radar station
Aplane flying horizontally at an altitude %f 2.2 miles and at a speed of 555 mi/hr passes directly over rdar station- Find the rate at which the distance from the plane to the station is increasing whenothe distance between the plane and the station is 6.9 miles_ If necessary, round your answer to two decimal places Variable Description Unit Time Horizontal distance the plane has traveled since time Distance between the plane and the ground at time t Distance between the plane and the radar station at time t dr Rate at which € changes with respect to time 2 Rate at which y changes with respect to time dz dt Rate at which 2 changes with respect to time mifhr milhr milhr The rate at which the distance is increasing is about
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https://www.electricalengineering.xyz/current-from-hp-formula/ | 1,713,096,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00683.warc.gz | 703,472,744 | 18,421 | # How to Calculate Current From HP – Electrical Engineering XYZ Formulae
Today Electrical Engineering XYZ shares 4 formulas to calculate current from the horsepowers.
### Calculate current from HP in DC Circuits
A 24 V DC Motor draws is rated at 1/8 HP. Find the current flowing through it. The recorded efficiency is 95%.
I = ((1/8) * 746) / (24 * 0.95) = 4.08 Amps
### Calculate current from HP in single phase ac circuits
A single phase ac motor whose Horsepower is 1 is connected to the voltage source having value 120 V. Find the electrical current if the efficiency is 90%. (The power factor is 0.85)
I = (1 * 746) / (120 * 0.9 * 0.85) = 8.126 Amps
### Calculate current from HP in case of three phase circuits
Determine the electrical current which flows through a 1000 watt load, having a power factor 0.9 and which has a 230 V three phase ac source connected to it.
I = ( 1500 watt * 746 / 746) / (230 * 1.73 * 0.9) = 4.18 Amps
### 1 thought on “How to Calculate Current From HP – Electrical Engineering XYZ Formulae”
1. Below 4formula 1waats=746HP. Is it correct. | 307 | 1,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.876299 |
http://www.jiskha.com/display.cgi?id=1295848450 | 1,498,350,557,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320368.57/warc/CC-MAIN-20170624235551-20170625015551-00168.warc.gz | 559,911,321 | 3,652 | # physics
posted by .
Problem: Two blocks of metals have the same mass. The first block has a volume of 827 cm^3 and a density of 4.50g/cm^3.Thesecond block has a volume of 193cm^3. What is the density of thesecond block ?
Solution: (1) 4.50x827=3720
(2)3720/193=19.3
1. Give an interpretation ( not a name) for the number3720 in this context.
2. Give an interpretation (not a name) for the number 19.3 inthis context. | 132 | 423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-26 | latest | en | 0.815155 |
https://onlinejudge.org/board/viewtopic.php?f=52&t=48465 | 1,582,689,929,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146186.62/warc/CC-MAIN-20200226023658-20200226053658-00365.warc.gz | 497,011,026 | 9,290 | ## 11752 - The Super Powers
Moderator: Board moderators
Angeh
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### 11752 - The Super Powers
Any Ideas How To solve This Problem efficiently ....
and How To handle The overflow?
any help is appreciated.....
>>>>>>>>> A2
Beliefs are not facts, believe what you need to believe;)
arifcsecu
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### Re: 11752 - The Super Powers
How can i solve this ?
I have got TLE.
Try to catch fish rather than asking for some fishes.
arifcsecu
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Posts: 64
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Location: Chittagong,University of chittagong
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### Re: Goods2010-02-15-fifty-11
// What is this ..................
It is very important forum for the programmers but not for others ...............
Try to catch fish rather than asking for some fishes.
Learning poster
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Joined: Fri May 08, 2009 5:16 pm
### Re: 11752 - The Super Powers
Take the composite powers of nos upto 2^16
2^4
2^6
...
3^4
3^6
..
4^4
4^6
..
To handle overflow before multiplying just chek it log(no1)+log(no2)<64log(2)
plamplam
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Posts: 150
Joined: Fri May 06, 2011 11:37 am
### Re: 11752 - The Super Powers
My best for this problem is 0.032 (The best for this problem is 0.028 ). And I didn't even optimize my code, no I/O opt nothing. (May be someday I can beat the best record with I/O Optimization ). I am not going to tell you my algorithm....no way guys but I can give you hints. You have to print a total of 67385 lines(including 1). And think a bit, what property would a number have if it is the power of at least two different positive integers? For example 16 = 2^4 and 4^2 and 729 = 3^6 and 9^3 and 27^2 etc. No more hints for you, you must solve this problem by yourself Be assured, you won't get WA or TLE if your algo is correct and efficient and you are printing exactly 67385 lines(In increasing order). (You can check this by increasing a counter variable starting from 0 whenever you are printing a line and later just print that counter variable )
Last hint(probably the most useful one) : You have to print all the numbers in ascending order as specified in the description. However this does not necessarily mean that you have to FIND all such numbers in ascending order.
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
plamplam
Experienced poster
Posts: 150
Joined: Fri May 06, 2011 11:37 am
### Re: 11752 - The Super Powers
And @Angeh the best way to handle overflow in this problem is by using log/ln (any one will do), but I didn't use the log/ln function in my program
You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson
outsbook
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Posts: 26
Joined: Fri Oct 28, 2011 2:42 am
### Re: 11752 - The Super Powers
For check Overflow
You can use
Code: Select all
``````int bit = 64;
int power= ceil(bit / (log(powerBase)/log(2)));``````
Example:
1. If powerBase=3 then power=41 (i.e. the maximum power of 3 is 40, 3^40 < 2^64-1(unsigned long long range)). 3^41 is overflow at 64 bit integer.
2. If powerBase=5 then power=28 (i.e. the maximum power of 5 is 27, 5^27 < 2^64-1(unsigned long long range)). 5^28 is overflow at 64 bit integer.
"Learning to love yourself is the greatest love of all" - Michael Masser and Linda Creed | 997 | 3,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-10 | latest | en | 0.879326 |
http://www.cinematography.com/index.php?showtopic=70275 | 1,537,443,781,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156460.64/warc/CC-MAIN-20180920101233-20180920121633-00172.warc.gz | 298,488,581 | 15,985 | # Zone System as applied today
6 replies to this topic
### #1 Ben Brahem Ziryab
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Posted 04 February 2016 - 01:11 AM
Hey guys,
I have pondered over the Zone System and how it can use be used to determine proper exposure, based on normal development of the negative.
Most spot meters (not just the Zone IV modifications) will use a interval of one on the meter scale, which equals doubling or halving of the luminance, or "one stop".
Every one-stop change is a change of one zone on the exposure scale.
So, if the subject area reads 7 on Zone V, but you want it to fall on Zone III, you simply dial your meter two steps to the left. Simple enough.
The conundrum is this: Let us assume that we are scanning the negative, not adding contrast and not making a print. A negative, when scanned, has around 14 stops of dynamic range, so the distance in stops from the toe of the curve to the shoulder is 14 steps.
In addition, Zone 0 to Zone X represents a full range from pure black to pure white -- not the dynamic range. The dynamic range only refer the useful values that lies between Zone I to Zone IX, whereas the textural range is from Zone II to VIII.
The Zone System uses a value scale of 10 (Zone 0 to Zone X) -- a 10 exposure scale from pure black to pure white.
So, how can we anticipate/visualize the negative density value using the Zone System on a 14 stop negative?
Perhaps the Zone System can be modified, so we increase or reduce the exposure by one and a half stop for each zone placement over/under Zone V. In other words, use one and a half stop intervals instead of one-stop intervals. On this scale, a Zone 0 rendering would fall seven and a half stops below Zone V.
Thanks.
Ben
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### #2 David Mullen ASC
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Posted 04 February 2016 - 01:36 AM
I don't see how this helps in determining the proper exposure compared to just taking some meter readings and making a choice based on the brightness of the subject that you want. To me, what makes the Zone System interesting was how it coordinated exposure with a photographic idea of manipulated contrast to achieve a desired tonal range, but this makes more sense in b&w printing of stills. In motion picture work, it's hard to apply beyond simple notions of increasing contrast in flat situations and decreasing contrast in high contrast situations.
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### #3 Ben Brahem Ziryab
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Posted 04 February 2016 - 02:05 AM
David,
What if you want to light a hallway behind the subject so parts of it appear black (but not pure black, more like a Zone I that is minimum density) and without the light falling off, while your subject's skin is significantly brighter than middle grey. How would you systematically approach something like that using your light meter and knowledge of the film stock?
Edited by Ben Brahem Ziryab, 04 February 2016 - 02:08 AM.
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### #4 David Mullen ASC
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Posted 04 February 2016 - 11:28 AM
Lighting is a different issue, you can control tonal range with lighting. Once I tested a stock and knew the tonal range I got for the typical release format gamma (Rec.709 video, film print, whatever) then I'd know when it fell to black or burned out to white for that gamma (more limited than the negative's range). So then when I was lighting, I'd know how dark to make a shadow versus how hot to make a highlight, with the knowledge that if I was doing digital color-correction, I'd have some room for manipulation.
But otherwise, your testing would tell you that, let's say, 4-stops under on your spot meter was near black but with some faint detail in the print, and maybe you've decided that the face should be 1-stop over in that scenario.
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### #5 John E Clark
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Posted 04 February 2016 - 01:38 PM
As noted, The Zone System is predominantly a Black and White process. The scene and resulting display image was visualized in terms of B&W materials, from negative to final print hanging on the wall.
Ansel Adams himself stated he never got an effect way to think of 'color' such that he would have the same control as he did with B&W... other have applied the Zone System to color, but the system is still described in terms of 'grey values'...
What is the core of the system is what is called 'previsualization', that is the photographer is at the point of taking a picture (or pressing the run button on a motion picture camera...), and because of the process will 'know' with in reasonable confidence, what the resulting image will be when displayed to the viewer.
There are other systems for doing that but the Zone System has had perhaps more success. One of the reasons for the success is that it is based on understanding the capabilities of the medium of capture (negative, but Adams also applied his system to Polaroid... even a consumer camera like the 70's vintage SX-70...), the medium of presentation, in the case of stills The Print.
The question you pose is really only the first step in the Zone System process... that step is 'how does the recording medium respond to light'... Film may have had the capability of recording detectable differences in light intensities up to '14 stops', or perhaps even more... until the silver 'crowbared' into less density, yielding 'black sun' effects...
However, in the print material, one was stuck with about 7 stops of range capability, so the goal was to find out how to expose, process, print and yield a print which was within that range.
In most formal Zone Systems classes that I had way back when, one would find the change in density of the negative by using a densitometer, and reading out the density changes produced by varying the ISO, or ASA as it was termed back in the olden days, or the f-stop of the lens. A series of negative was taken, tedious hours processing, more tedium reading the densities, etc...
Then there was printing, tedious hours printing, finding that density in the negative which yielded that 'hint' of difference between 'black' and 'less than black'... likewise for finding that 'hint of grey' just be fore 'pure white' in the print...
The spot meter was recommended because one could read out the relative values of 'shadow', and 'highlight' directly and so, once one had calibrated the process one could read out the meter for shadow, know that the exposure would be X, read out the highlight and know by process and printing the result would be as one desired.
Some of the Zone System would not 'work' for motion pictures, as in the case of stills, there was a variety of ways one could modify the development process, a variety of 'papers' to use which allowed one to compensate for different contrasts, etc.
But the main goal of 'previsualize' and know that the exposure and process will produce the desired results could be achieved... albeit with some resignation that one could not control everything to any level desired...
In the modern digital world the IRE waveform display can replace the 'densitometer' and so, one can use that to evaluate the capture medium characteristics... and far more easily... than using a densitometer. But that is not the end of the process... the next step is to know what the presentation characteristics are...
And here one has a plethora of options... most of which don't have 14-stops worth of display capability... perhaps for some display, one has only realistically 5 or 6... with more money... perhaps 8-9... and in some beatific future... 10+...
So back to your question... if your capture medium has 14 stops worth of range and your output display only has 7 steps... you have to chose some 'curve' which compresses the capture range into a 'pleasing' display range...
For most people who are not funded to support professional services... it means that they will need to learn some package sufficiently to get their desired results... Of the popular prosumer packages there's Adobe products, such as Premiere, from Blackmagic Designes there DaVinci Resolve, and of course Avid has had a long standing industry presence...
For those working on the cheap... Blackmagic does have a 'free' version of Resolve... but I subscribe to the Adobe Creative Cloud, and so I use both Premiere and Resolve for dealing with motion picture processing.
For stills I use Photoshop, and occasionally Lightroom... but mostly Photoshop...
Edited by John E Clark, 04 February 2016 - 01:41 PM.
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### #6 Ben Brahem Ziryab
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Posted 04 February 2016 - 04:26 PM
Lighting is a different issue, you can control tonal range with lighting. Once I tested a stock and knew the tonal range I got for the typical release format gamma (Rec.709 video, film print, whatever) then I'd know when it fell to black or burned out to white for that gamma (more limited than the negative's range). So then when I was lighting, I'd know how dark to make a shadow versus how hot to make a highlight, with the knowledge that if I was doing digital color-correction, I'd have some room for manipulation.
But otherwise, your testing would tell you that, let's say, 4-stops under on your spot meter was near black but with some faint detail in the print, and maybe you've decided that the face should be 1-stop over in that scenario.
What I think I'll do is shoot a grey card on V3 to see what I am getting for each exposure increase and decreasing of one stop. Then draw out two charts; one with the tonal range of the negative and another with the Rec 709 tonal compression. These could be programmed into the Sekonic L758-Cine to create a "latitude" display, or I could simply mark them on a analogue spot meter.
As John Clark pointed out, the Zone System is designed for black and white film only, so I don't know if you can establish zones that represent various brightness levels for every color.
The idea is that I could use the system to communicate with the production designer on how dark I want a certain wall or table just by showing that person a "zone" that corresponds to a exposure value based on the tonal range of the film stock and release format.
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### #7 David Mullen ASC
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Posted 04 February 2016 - 05:28 PM
To set your Rec.709 look, since there isn't a strict standard for gamma, after all, you can show a flat log image on a monitor, shoot an 11-step grey scale at your standard exposure rating so you can put the white patch at 90 to 100 IRE and your black patch at 0 to 5 IRE, which is typical for dailies colorists transferring negative.
• 0 | 2,525 | 10,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.883422 |
https://www.airmilescalculator.com/distance/msr-to-bzi/ | 1,660,767,994,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00404.warc.gz | 569,105,594 | 44,535 | # Distance between Muş (MSR) and Balikesir (BZI)
Flight distance from Muş to Balikesir (Muş Airport – Balıkesir Airport) is 739 miles / 1190 kilometers / 642 nautical miles. Estimated flight time is 1 hour 53 minutes.
Driving distance from Muş (MSR) to Balikesir (BZI) is 921 miles / 1482 kilometers and travel time by car is about 18 hours 28 minutes.
739
Miles
1190
Kilometers
642
Nautical miles
1 h 53 min
129 kg
## How far is Balikesir from Muş?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 739.133 miles
• 1189.520 kilometers
• 642.289 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
• 737.340 miles
• 1186.634 kilometers
• 640.731 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Muş to Balikesir?
Estimated flight time from Muş Airport to Balıkesir Airport is 1 hour 53 minutes.
## What is the time difference between Muş and Balikesir?
There is no time difference between Muş and Balikesir.
## Flight carbon footprint between Muş Airport (MSR) and Balıkesir Airport (BZI)
On average flying from Muş to Balikesir generates about 129 kg of CO2 per passenger, 129 kilograms is equal to 284 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Muş to Balikesir
Shortest flight path between Muş Airport (MSR) and Balıkesir Airport (BZI).
## Airport information
Origin Muş Airport
City: Muş
Country: Turkey
IATA Code: MSR
ICAO Code: LTCK
Coordinates: 38°44′52″N, 41°39′40″E
Destination Balıkesir Airport
City: Balikesir
Country: Turkey
IATA Code: BZI
ICAO Code: LTBF
Coordinates: 39°37′9″N, 27°55′33″E | 558 | 2,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-33 | latest | en | 0.869568 |
https://www.daniweb.com/programming/software-development/threads/395286/how-to-check-if-a-binary-tree-is-a-complete-binary-tree | 1,534,701,300,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215261.83/warc/CC-MAIN-20180819165038-20180819185038-00645.warc.gz | 893,101,471 | 12,614 | Can please someone give and explain the algorithm and logic of checking if a given tree is a complete binary tree.
Fields available for a node are left and right child and info.
Thanks.
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Last Post by Narue
There are multiple ways of doing it, but in my opinion the most instructive is with a level order traversal, since checking for a complete binary tree is a pretty specific operation. By generalizing the problem a bit, you can get more out of the exercise. So here's an updated problem for you:
Q: Given a binary tree of the following definition:
``````struct node {
int data;
struct node *left;
struct node *right;
};``````
Write a function that will perform a level order traversal and convert the tree into an array using heap indexing rules. Heap indexing rules state that for each node (numbered 0 to N in level order) the left child is at index `2 * i + 1` and the right child is at `2 * i + 2` .
As an example, given the following degenerate binary tree:
``````1
- 2
- - - 3
- - - - - - - 4
- - - - - - - - - - - - - - - 5``````
The array you are to return might look like this:
``1-2---3-------4---------------5``
Once you have this array, determine if it represents a complete binary tree by testing for null nodes in between non-null nodes. A complete binary tree will have all non-null nodes strictly adjacent to each other. For example, here is a complete binary tree and the array representation:
``````5
2 7
1 3 6 -``````
``527136-``
Note: The size of the array should be determined as the maximum number of nodes for a tree of height h.
To give you a push in the right direction, you need to figure out how to determine the height of a tree, how to calculate the maximum number of nodes in a tree of given height, and how to perform a level order traversal. Inside the level order traversal you need to maintain the level order index of each node to correctly calculate child indices.
This problem is carefully worded so that you have no choice but to do research on the various techniques, so consult Google before shooting back with questions. I wrote a small program (less than 200 lines) that does all of this. Here are runs on the two examples above:
``````-
5
-
4
-
3
-
2
-
1
-
*-*---*-------*---------------*
Invalid complete tree``````
``````-
7
-
6
-
5
-
3
-
2
-
1
-
******-
Complete!``````
Feel free to match that output if you want. A tree dump with a 90 degree rotation is far easier than the usual textbook representation. | 647 | 2,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-34 | longest | en | 0.864206 |
https://www.nbclosangeles.com/news/sports/a-statistical-look-at-the-lakers-going-beyond-traditional-numbers/1845029/ | 1,725,947,037,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651196.36/warc/CC-MAIN-20240910025651-20240910055651-00652.warc.gz | 836,695,998 | 72,387 | # A Statistical Look at the Lakers, Going Beyond Traditional Numbers
In today's L.A. Times, it's pointed out that the Lakers might be undefeated, but (uh-oh!) they're not shooting the ball particularly well. And on the surface, that might be true. As a team, the Lakers are shooting just 43.5% from the field, good enough to be ranked merely 21st in a league of 30 teams. But most observers see this Lakers' team as playing extremely solid team basketball through their first four games, so where's the disconnect? There are other statistical categories besides the basics which give us a clearer look at what's going on.
While the Lakers' team field goal percentage is low, they lead the league in points scored per game, averaging over 105 per contest. So is their shooting really just 21st in the league? Not if you account for three-pointers and free throws, which a metric known as True Shooting Percentage (TS) includes in the equation. And as you might expect, when you look at all types of shots taken (not just combined field goals), the Lakers' number improves quite a bit. Their TS% is actually 53.9, good enough for 11th in the league.
There are other more intense statistics that give a clearer picture of where a team might rank, on both ends of the court. Team efficiency numbers level the playing field across the league, breaking down each club's offensive and defensive numbers per 100 possessions. What this does is adjust each team's statistics for pace, making it easier to compare run and gun teams against those that prefer to play at a slower tempo. The Lakers rank ninth in the league in offensive efficiency, and defensively, they're as good as it gets: L.A. is number one in the league, allowing just 86.2 points per game per 100 possessions. That's over four points per game better than the second place Celtics.
Statistics don't tell the whole story; it doesn't take a genius to see that the Lakers are playing well right now, especially defensively. But if you're trying to find some cracks in the dam, you need to look a little bit beyond the basic numbers to get a true idea of a team's particular weaknesses. And there just aren't that many to be found right now with the Lakers.
TONIGHT'S GAME
The Houston Rockets come to town with a record of 4-2, but they've struggled recently. After beginning the season 3-0, they dropped consecutive games before beating the Clippers the other night at Staples Center. The team's offseason addition of Ron Artest should make things interesting. With Artest around to guard Kobe Bryant and Yao Ming defending the middle, Houston appears to match up well defensively. But the Lakers are no slouches on defense themselves, and Tracy McGrady doesn't appear to be right, having scored just two points in 26 minutes against the Clippers. The Lakers come into this one well-rested, as it's just their second game in the last eight days. But last season, the Lakers did drop two out of three meetings to the Rockets, so this should be a good early season test for them. | 656 | 3,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.971148 |
https://www.physicsforums.com/threads/calculate-total-mechanical-energy-of-a-frictionless-spring.858709/ | 1,606,272,031,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00184.warc.gz | 833,819,851 | 16,253 | # Calculate total mechanical energy of a frictionless spring
Hi,
1. Homework Statement
A 2.86-kg object on a frictionless horizontal surface oscillates at the end of a spring with an amplitude of
7.81 cm. Its maximum acceleration is 3.74 m/s2.
Calculate the total mechanical energy.
## Homework Equations
1) Total Mechanical Energy: E = 0.5 k A^2
2) Spring Constant: k = m w^2 (where w = angular velocity)
3) Acceleration : a = -w^2 x
## The Attempt at a Solution
Given the Amplitude A, we must calculate the spring constant k to calc the total mechanical energy.
First find w using formula 3)
w = - sqr( a / x)
w = - sqr(3.74 / 0.0781)
w = - 6.92
Now substitute k in formula 1 with formula 2 to calculate the total mechanical energy.
E = 0.5 (m w^2) A^2
E= 0.5 (2.86) (-6.92)^2 (0.0781)^2
E= 0.0418 Joules
Related Introductory Physics Homework Help News on Phys.org
gneill
Mentor
Re-do your final calculation; you seem to have slipped a decimal point.
henrco
Hi,
1. Homework Statement
A 2.86-kg object on a frictionless horizontal surface oscillates at the end of a spring with an amplitude of
7.81 cm. Its maximum acceleration is 3.74 m/s2.
Calculate the total mechanical energy.
## Homework Equations
1) Total Mechanical Energy: E = 0.5 k A^2
2) Spring Constant: k = m w^2 (where w = angular velocity)
3) Acceleration : a = -w^2 x
## The Attempt at a Solution
Given the Amplitude A, we must calculate the spring constant k to calc the total mechanical energy.
First find w using formula 3)
w = - sqr( a / x)
w = - sqr(3.74 / 0.0781)
w = - 6.92
Now substitute k in formula 1 with formula 2 to calculate the total mechanical energy.
E = 0.5 (m w^2) A^2
E= 0.5 (2.86) (-6.92)^2 (0.0781)^2
E= 0.0418 Joules
Yes it's okay but pay attention at this:
a= -ω^2 x ---> ω= √(-a/x)
Since a is opposite to x a= -3.74 m/s^2 and ω= 6.92 rad/s so ω>0
henrco
Thank you for the feedback and for checking my shoddy calculation.
I redid the calculation and the answer came to 0.418 Joules
haruspex
Homework Helper
Gold Member
Yes it's okay but pay attention at this:
a= -ω^2 x ---> ω= √(-a/x)
Since a is opposite to x a= -3.74 m/s^2 and ω= 6.92 rad/s so ω>0
You mean, so ω is real. Whether you choose to take the positive or negative root is another matter.
You mean, so ω is real. Whether you choose to take the positive or negative root is another matter.
Yes, I just focused on the fact that he got a ω< 0 by making a little mistake in solving a= -ω^2 x
:) | 773 | 2,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-50 | latest | en | 0.741384 |
http://www.nodalninja.com/forum/showthread.php?4942-Fisheye-NPP-changing-based-on-pano-spacing&p=37817&viewfull=1 | 1,406,580,072,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510261958.8/warc/CC-MAIN-20140728011741-00108-ip-10-146-231-18.ec2.internal.warc.gz | 704,606,211 | 20,835 | Fisheye NPP changing based on pano spacing?
# Thread: Fisheye NPP changing based on pano spacing?
1. ## Fisheye NPP changing based on pano spacing?
I was reading a review of the popular Samyang Fisheye lens, and it started talking about the NPP locations, and provided this image showing different NPPs for different 360 pano intervals- a different NPP for 60, 45, and 90.
Everything else I've read talks about each lens having a single NPP- and by the very nature, why would the NPP change if the interval angle changes?
Just looking for some enlightenment, thanks.
2. The NPP (No Parallax Point) can be thought of as the point for rotating the lens round to achieve the "minimum" parallax, so can be thought of as a single point for a lens, but has to be chosen with respect to the Entrance Pupil.
The Entrance Pupil can be a single point for a particular lens, but is more likely to change as the angle of the incoming ray moves away from the principal ray (axis of the lens).
Some examples of the way the Entrance Pupil behaves can be found at:
http://www.hugha.co.uk/NodalPoint/Index.htm
which shows that for a fisheye lense the change can be quite significant.
Also page 8 of:
http://www.hugha.co.uk/Acrobat%20PDF...int-Clouds.pdf
When using a lens for creating panoramas there is an argument for selecting the Entrance Pupil where the angle of the ray relative to the principal ray is the same as the angle at which successive images are "joined".
e.g. If you take 6 shots round at 60 degree intervals with a Samyang 8mm then you would chose the NPP to be at just under 3mm in front of the gold ring where the rays at 30 degrees each side of the principal ray meet it, but if you take 8 shots round at 45 degree intervals then you might chose an NPP point at about 1.5mm in front of the gold ring, where the rays at 22.5 degrees each side of the principle ray meet it.
The Samyang 8mm has less "spread" of the Entrance Pupil than more "conventional" fisheye lenses such as the Sigma 8mm and Nikon 10.5mm so the change is less noticeable with the Samyang 8mm.
3. Originally Posted by shidarin
I was reading a review of the popular Samyang Fisheye lens, and it started talking about the NPP locations, and provided this image showing different NPPs for different 360 pano intervals- a different NPP for 60, 45, and 90.
Everything else I've read talks about each lens having a single NPP- and by the very nature, why would the NPP change if the interval angle changes?
Just looking for some enlightenment, thanks.
Michel Thoby is the authoritative source of studies of floating NPP / LPP. Read more on his studies of other lenses.
NIck
4. Thanks guys, that was a really informative info and fits with what I was thinking, but I didn't want to assume or start to piece together information incorrectly.
Armed with this knowledge, I'm off to read more of Michel Thoby's studies.
5. What is really interesting is that a fisheye lens has a different setting depending on where you want to optimize your image. If you remove parallax from the center portion of the panorama, you induce parallax along the top and bottom. I lost the links to pictures that clearly show this. While calibrating for 4 shots around, I can get the center to be perfect but the top and bottom are out. I optimize for the bottom and get acceptable results along the center. Longer focal length lenses do not suffer from this issue.
6. This is Michel's article about LPP.
http://michel.thoby.free.fr/Fisheye_...the-pupil.html
And this is the link to set up some nice music, some antipasti, and then take your time to read and try to understand the basics of Michel's articles
http://michel.thoby.free.fr/
All of us learned a lot about fisheye lenses use as well as from Hugh's articles. Thx a lot to both of them.
You have to decide how many shots you want to take around with your camera lens combination. Sometimes you can go for 6 or 4 shots around, sometimes for 4 or 3 shots around. With less shots around you will end up with less overlap. But you will be quicker to get your set around. May be important in busy places with a lot of people moving around, in concerts with people moving their arms to the music, journalism etc. The "normal minimum overlap without risk" is 25%. I prefer 30%. May be you have to shoot some more pics, just some seconds to take them, but this might save your time to go back to New York to shoot again and you lost the special moment and atmosphere anyway.
Use Frank's Calculator to check your overlap: http://www.frankvanderpol.nl/fov_pan_calc.htm
Just change the number of pics around in "Calculate Overlap".
Success,
Heinz
7. ## LPP at 120 deg
Got my shiny new Samyang 8mm and NN4 yesterday. Shaved the lens hood and all ready to go
I'm planning to test a 3 shot pano at 120 deg on a D600, and wondered if anyone knew what the LPP is for the lens at this rotation or better still the NN4 rail settings. I can't quite determine the LPP from Michel's excellent article but it looks like 20mm behind the lens front (or 4mm ahead of the gold ring if I'm reading it correctly ?)
Any help appreciated.
Kevin
8. Kevin, It's best if you do some tests yourself to determine the optimum settings. It doesn't take long. See:
http://www.panoramas.dk/panorama/Fis.../parallax.html
http://www.johnhpanos.com/epcalib.htm
http://www.easypano.com/forum/displa...PagePosition=1
.
John
9. Originally Posted by shidarin
Thanks guys, that was a really informative info and fits with what I was thinking, but I didn't want to assume or start to piece together informationincorrectly...
You are welcome shidarin.
I find that knowing about the behavior of the lens allows me to make"informed" decisions about setting up a lens and how many shots to take, but the reality is that we can only have one NNP (Nodal Ninja Point) and the methods suggested by John achieve the same result.
I have published this information on the Internet, because I have not been able to find this representation in photogrammetric journals or the Internet.
The only oasis I have found in this dearth is Michel Thoby's web site and I consider Michel to be the “guru” on the subject.
My reason for being interested in lens behavior goes back to the 1070s when we used hugely expensive photogrammetric cameras for Close Range (Terrestrial) Photogrammetry, such as the Wild P31, Wild P32 and Zeiss UMK..
These cameras cost something like ten times a Leica M or Hasselblad, so it was desirable to use “off the shelf” cameras, especially in hostile and nasty environments.
The cameras were simple in operation, but the price reflected the “symmetrical” distortion free lenses, a couple of microns in the corners of a 70mm x 70mm format in the Wild P32!
To use “off the shelf” cameras we needed to be able to correct for distortion at a measured point, so we needed to know how the lens behaved.
Although I could not find any information in journals, magazines, etc. I was able to talk to lens designers in Wild Heerbrugg, now Leica Geosystems and learned much from this.
Originally Posted by kevin6270
(or 4mm ahead of the gold ring if I'm reading it correctly ?) Any help appreciated
Originally Posted by kevin6270
Kevin
From my measurement of a Samyang 8mm Canon fit lens, 4mm in front of the gold ring is about right.
Note that from Michel’s page:
http://michel.thoby.free.fr/SAMYANG/Samyang_shaved.jpg
that the position depends on whether you take 6 shots round at 60° or 8 shots round at 45°, which agrees with my measurements on:
http://www.hugha.co.uk/NodalPoint/Index.htm#Samyang8mm
Originally Posted by hindenhaag
All of us learned a lot about fisheye lenses use as well as from Hugh's articles. Thx a lot to both of them.
Originally Posted by hindenhaag
You have to decide how many shots you want to take around with your camera lens combination. Sometimes you can go for 6 or 4 shots around, sometimes for 4 or 3 shots around. With less shots around you will end up with less overlap. But you will be quicker to get your set around. May be important in busy places with alot of people moving around, in concerts with people moving their arms to the music, journalism etc. The "normal minimum overlap without risk" is25%. I prefer 30%. May be you have to shoot some more pics, just some seconds to take them, but this might save your time to go back to New York to shoot again and you lost the special moment and atmosphere anyway.
Success,
Heinz
Thanks for your comment Heinz, much appreciated.
I too have learned a lot from Michel’s site.
The choice of the number of shots around depends on many factors, and the position chosen for the NNP is affected by the number of shots around except for lenses such as the Sigma 10 to 20mm f4-5.6 EX DC HSM, where the Entrance Pupil is a single point.
I have tried to show this in the attached diagrams and it is the variation in the location of the Entrance Pupil relative to the angle of the ray that causes the phenomena mentioned by Denis.
I don’t know if it is the influence of my photogrammetric background, but I tend to go for a larger overlap (50%) than most resulting in more images to create the panorama.
I have a number of reasons for doing this:
By having more images I am making more use of the central part of the lens where the displacement of the Entrance Pupils is much tighter to try and minimise the effects of the lens distortion from the fisheye lens;
The larger overlap gives a greater area for Control Point generation;
The larger overlap gives me much more scope for using PTGui’s wonderful Mask tool,especially in very crowded environments such as hospitality suits where a lot of people are moving and moving around in closed proximity to the camera.
Of course, using the Mask option in PTGui does mean that I am sometimes choosing parts of the image away from the centre, but PTGui does a great job and I really appreciate the genius that is Joost that makes it all work!
When shooting hospitality suits with people milling around, some very close to the camera, I use a Sigma 4.5mm lens to get a full circle image to minimise the number of shots.
This does have the disadvantage that I am using only one third of the sensor area,but with 24Mp I am getting a decent resolution in the final panorama.
I could follow Heinz’s advice of 3 shots round to get the shots off as quickly as possible, but I have found that people can move very fast, especially if they are picking up a glass of wine! and sometimes it is necessary to wait a bit for people to move out of shot – I don’t want the lady with the large pink hat to appear twice in the panorama!
Consequently I take four shots round, which gives me the overlap advantages already mentioned, but I actually go round three times (12 shots) so I have plenty of material for eliminating unwanted movements, postures, gestures, etc.
I often wonder what the guests make of this gutywalking round and round a camera on a tripod without apparently taking any photographs because I am not looking at the screen or through the view finder and I am very conscious that the Guests, of their Hosts, have paid a lot of money to enjoy good food and drink and their favourite sport, but I find that I am quite inconspicuous and only occasionally is a Guest looking at the camera.
I guess it is because the fisheye lens is unobtrusive and I am not aiming the camera at individuals.
I have now made over 500 of these panoramas, some better then others, and can only count seven occasions when people have spoken to me, so I am happy that they are enjoying themselves and my camera and I are not a distraction.
Most of the seven were interested to know what I was doing, one knew about panoramas and we had a good discussion and only one asked me not photograph him with a polite “No photographs please”, so I reckon the system works.
The thing is really to find what suits you best and gives you the results you are looking for and for me I find know what the lens is doing helps.
When I set up the Sigma 4.5mm on a R1 I spent a few minutes with the laser determining the positions of the entrance Pupil for different angles, looked at the results and set the NNP 4mm in front of the gold ring and did not need any further adjustment, but as mentioned at he beginning, the reality is that we can only have one NNP (Nodal Ninja Point) and the methods suggested by John achieve the same result.
10. Originally Posted by John Houghton
Kevin, It's best if you do some tests yourself to determine the optimum settings. It doesn't take long. See:
http://www.panoramas.dk/panorama/Fis.../parallax.html
http://www.johnhpanos.com/epcalib.htm
http://www.easypano.com/forum/displa...PagePosition=1
.
John
Thanks John. Used these to determine the optimum NPP, but was getting some obvious stitching errors and poor control point numbers (averaging 8-9). I was pretty sure it wasn't my NN set up, so tried various settings in Ptgui. Finally amended the Minimise Lens Distortion setting from Medium to "Heavy + Lens shift" and hey presto - perfect stitching and ave CP of 1.08 and high of 2.4.
Really pleased with the end result, but don't understand why this simple change made all the difference.
11. Hi Kevin,
Using a fisheye lens with exif info PTGui automatically is set to "Heavy + Lens Shift" in the last version to correct sensor displacements during production and to minimize lens distortion. Joost has optimized PtGui to find CP with this special set up to spread CP in vertical and horizontal direction. Because Samyang is a manual lens, PTGui misses this info. Set lens to Full Frame and 9mm, and use Heavy + Lens shift optimizing. To create a template, which will optimize stitching in future with the same camera lens combination, you can try to go down to max distance CP in between 1.0 and 2. We can try to help you if you send your pics and PTGui PTS. file via www.ge.tt .
Then we can send a template for your set up. Next time you do a new project, you can use "apply template" to optimize CP or your special set up.
Heinz
12. This is the most helpful forum, thanks !!
I've posted the files to http://ge.tt/4Z2vrgP?c so hopefully you can pick them up. Pics were shot at -15 deg (x4) and +60 deg.
Kevin
13. Kevin, The images stitch quite well, with only a tiny glitch in the beam across the ceiling. Taking just two of the images separated by a yaw increment of 180 degrees reveals evidence of significant misalignment of the panorama head:
The bottom rail setting seems fine. The camera needs a small rotational adjustment on its mount to the upper rail. This may be the cause of the glitch in the beam. It would be a good idea to stick a few temporary markers on the beam (e.g. post-it notes) to act as features for control points to help the optimizer.
John
14. Thanks John, I'll try that. I found centering above the NN logo tricky with the fisheye - even using full zoom in live view - as the tolerances are so tight. I swapped out the Samyang for an 85mm just to get the lens centred, but it seems that's not quite worked. Need to become better at that !
Also, looking forward to seeing Heinz feedback on optimal settings, and template.
15. Hi Kevin,
Heinz
Ok, just managed pic by pic. Download all files didi not work. I just had a quick look around. Because of results I changed to Circular Fisheye lens parameter and cropped. In case of problem I always start with the shots around only. In optimizer change to heavy and lens shift and optimize. Del worst CP. When nothing is deleted, I changed to control points. I just marked the biggest ones with shift and del all above 20 etc. I just went down in max CP and ended up around 2 without placing anything manually.
Now change to control points and compare pic to pic and change with red arrows on top. Now you see in most cases CP sit in a bulk position specially in picture 2. Cp on the galls only.
What we always say: CP needs a good spread in vertical and horizontal. Have a look to the ceiling, where you want to add your zenith shot: nearly no CP.
I will explain this to you later because I have to leave now to go to work. Just go through and try to analyze the placement of CP. And try to find out which are in "a place of nowhere, this means in the middle of a brown spot instead of its sharp corner. When this is in a place of bulky CP, you can just delete this one without trying to correct the distance.
http://ge.tt/2wlYOjP/v/0
So long,
Heinz | 3,865 | 16,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | longest | en | 0.927756 |
https://math.stackexchange.com/questions/653162/show-that-there-exists-two-null-sequences-x-n-and-y-n-in-0-1-such | 1,653,487,800,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662587158.57/warc/CC-MAIN-20220525120449-20220525150449-00560.warc.gz | 438,344,068 | 66,162 | # Show that there exists two "null sequences" {$x_n$} and {$y_n$} in $(0, 1)$ such that lim $f(x_n)$ and lim $f(y_n)$ both exist but unequal.
Let $f:(0, 1)\to\mathbb R$ be such that $f$ is bounded and $\lim\limits_{x\to0} f(x)$ does not exist.
Show that there exists two "null sequences" {$x_n$} and {$y_n$} in $(0, 1)$ such that $\lim\limits_{n\to\infty} f(x_n)$ and $\lim\limits_{n\to\infty} f(y_n)$ both exist but unequal.
• This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. To be more precise: What have you tried so far? Where are you stuck? Jan 27, 2014 at 11:25
• Find one null sequence $(x_n)$ for which the limit $\lim_{n\rightarrow\infty} f(x_n)$ exists. Call the limit $L$. Find another null sequence $(y_n)$ and an $\epsilon>0$ such that $|f(y_n)-L|>\epsilon$ for all $n$. Now find a convergent subsequence of $(f(y_n))$. Can its limit be $L$? Jan 27, 2014 at 11:27
• @user123804 Why what? Jan 27, 2014 at 11:29
• Not getting David Mitra. Please explain.. Jan 27, 2014 at 11:39
• First recall that any bounded sequence of reals has a convergent subsequence. Using this, you can find $x_n\rightarrow 0$ with $f(x_n)\rightarrow L$ (take any sequence converging to $0$, then find a convergent subsequence of its image under $f$). OK.. Since $\lim_{x\rightarrow0} f(x)\ne L$, you can find a null sequence $(y_n)$ such that $(f(y_n))$ is "bounded away from $L$". That is, there is an $\epsilon>0$ so that for all $n$, $|f(y_n)-L|>\epsilon$. But $(f(y_n))$ has a convergent subsequence. This subsequence can't converge to $L$. Jan 27, 2014 at 11:53
Hint: Since $f$ is bounded, both $\limsup_{x\to0^+}f(x)$ and $\liminf_{x\to0^+}f(x)$ exist and are finite. (Why?)
Now, if $\limsup_{x\to0^+}f(x)=\liminf_{x\to0^+}f(x)$, what can you say about $\lim_{x\to0^+}f(x)$?
And finally>: if $\limsup_{x\to0^+}f(x)\neq\liminf_{x\to0^+}f(x)$, can you think of a way to leverage this to find the two sequences that you are looking for? | 719 | 2,223 | {"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} | 3.84375 | 4 | CC-MAIN-2022-21 | latest | en | 0.849449 |
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Duration: 23:16
Speaker: : Carl Petersen
Long-range inhibitory connections in the brain, with examples from three different systems.
Difficulty level: Beginner
Duration: 19:05
Speaker: : Carl Petersen
The "connectome" is a term, coined in the past decade, that has been used to describe more than one phenomenon in neuroscience. This lecture explains the basics of structural connections at the micro-, meso- and macroscopic scales.
Difficulty level: Beginner
Duration: 1:13:16
Speaker: : Clay Reid
Introduction to the types of glial cells, homeostasis (influence of cerebral blood flow and influence on neurons), insulation and protection of axons (myelin sheath; nodes of Ranvier), microglia and reactions of the CNS to injury.
Difficulty level: Beginner
Duration: 40:32 | 728 | 3,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.703511 |
https://oeis.org/A066528 | 1,571,059,391,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653247.25/warc/CC-MAIN-20191014124230-20191014151730-00099.warc.gz | 583,377,712 | 4,038 | This site is supported by donations to The OEIS Foundation.
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A066528 Non-palindromic triangular numbers whose reverse is a triangular number with the same number of digits. 5
153, 351, 17578, 87571, 185745, 547581, 1461195, 5911641, 12145056, 12517506, 60571521, 65054121, 304119453, 354911403, 1775275491, 1945725771, 10246462281, 17990863516, 18226464201, 35615002605, 50620051653, 61536809971, 1222080857271, 1664224065406 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Giovanni Resta, Table of n, a(n) for n = 1..64 (terms < 2*10^24) EXAMPLE 153 and 351 are both triangular. MATHEMATICA dtn[L_] := Fold[10#1+#2&, 0, L]; tritest[n_] := Module[{t}, t=Floor[N[Sqrt[2n]]]; 2n==t(t+1)]; A={}; For[i=1, i>0, i++, t=i(i+1)/2; If[tritest[tt=dtn[Reverse[IntegerDigits[t]]]]&&Mod[t, 10]>0&&t=!=tt, AppendTo[A, t]; Print[A]]] CROSSREFS See A069673 for another version. Cf. A066703, A179889. Sequence in context: A253023 A194660 A159294 * A046197 A271730 A056733 Adjacent sequences: A066525 A066526 A066527 * A066529 A066530 A066531 KEYWORD base,nonn AUTHOR Erich Friedman, Jan 08 2002 EXTENSIONS a(22)-a(24) from Giovanni Resta, Jun 20 2015 STATUS approved
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Last modified October 14 08:54 EDT 2019. Contains 327995 sequences. (Running on oeis4.) | 549 | 1,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.506401 |
http://ecmweb.com/print/content/using-signal-isolators-reduce-ground-loop-problems | 1,490,307,600,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187225.79/warc/CC-MAIN-20170322212947-00622-ip-10-233-31-227.ec2.internal.warc.gz | 120,474,382 | 9,543 | Having problems with your process controls and electrical instrumentation? The source may be a grounding disturbance known as a ground loop. According to the ANSI/ IEEE Standard Dictionary of Electrical and Electronics Terms, a ground loop is a "potentially detrimental loop formed when two or more points in an electrical system normally at ground potential are connected by a conducting path such that either or both points are not at the same ground potential." In other words, a ground loop develops because each ground is tied to a different earth potential, which allows current to flow between the grounds by way of the process loop, as shown in Fig. 1.
Basically, ground loops cause problems by adding or subtracting current or voltage to or from a process signal. As a result, the receiving device can't tell the difference between the wanted and the unwanted signals. So it can't accurately reflect process conditions.
The probability of establishing multiple grounds and ground loops is especially high when you install new programmable logic controllers (PLCs) or distributed control systems. With so many connections referenced to ground within a facility, the likelihood of establishing more than one point is great.
For instruments like thermocouples and some analyzers, you may not be able to completely eliminate ground loops because these instruments require a ground to make accurate rate measurements. Also, all analog control loops are grounded at one or more points, which can result in a ground loop that can upset the proper functioning of instruments. You also may have to ground these instruments to ensure personnel safety.
How an instrumentation signal loop works. Suppose you have an instrumentation loop, as shown in Fig. 2. As you can see, it's basically a DC system that operates at a specific voltage (24V in our example) to a master ground reference called a signal ground. The instrumentation signals vary within a range of 4mA to 20mA, depending on the value of the variable (temperature, pressure, etc.) seen by the sensor.
Let's say a precisely calibrated circuit takes this mA signal and converts it into a 1V-to-5V signal for a chart recorder. At 4mA, the voltage measured by the recorder is 1V (250 ohms x .004A). At 20mA, the measured voltage is 5V. Normally, the recorder scale is calibrated so the voltage reads directly in °F, °C, psi, etc.
The how and why of signal isolators. So what do you do if you can’t eliminate the conditions for ground loops? You can use a signal isolator, as shown in Fig. 3, to break the galvanic path (DC continuity) between all grounds while still allowing the analog signal to continue throughout the loop. These devices also eliminate the electrical noise of AC continuity (common-mode voltage).
Signal isolators use one of two techniques to do the job:
• Analog signal isolation, which uses an isolation transformer to chop, isolate, and reconstruct the signal.
• Discrete signal generation, which chops, transmits optically, and reconstructs the signal. Signal isolators that use this technique are called "opto-isolators."
The choice between the two depends on your circuitry requirements.
Regardless of the isolation method you choose, make sure your isolator provides input, output, and power isolation. If you don't have this three-way isolation, then an additional ground loop can develop between the isolator's power supply and the process input and/or output signal.
Isolators, like most other transmitters, come in 2- and 4-wire versions. The 4-wire type requires a separate power source and is partially suited for back-of-panel mounting. You can power the 2-wire type from either the input or output loops.
The input loop type makes it possible to isolate a process signal when you don’t have line power or output loop power available. The output loop type solves the problem of interfacing non-isolated field signals with systems like a computer, PLC, or distributed control system. These systems provide loop-power to their output devices.
You can find a signal isolator to suit almost any application, including the following:
• Resistance input isolators for use as RTD, slidewire, strain, and potentiometer transmitters.
• Millivolt isolators for use as thermocouple and millivolt transmitters.
• Current/voltage isolators for use as alarm tripping, deviation alarm notification, and other special application transmitters.
### Sidebar: An Instrumentation grounding primer
Almost every piece of equipment used in a control instrumentation strategy uses a common signal ground as a reference for its analog signals. Introducing any additional grounds into a control circuit will almost certainly cause ground loops to occur.
To minimize the danger of introducing these loops into a complicated network, you should use a dedicated instrumentation system ground bus (see Fig. 4) and connect grounds from the signal common, cabinet ground, and instrumentation AC power ground to it.
You should tie the instrumentation system ground bus to earth via the building ground and plant ground grid. But, this can be much more complicated than it appears. For example, rarely will you have just one instrumentation loop. In fact, you could have hundreds or even thousands.
Many vendor-supplied instrumentation system cabinets contain a DC signal common bus and power supply common bus. The manufacturer normally ties these busses together within the cabinets at a master ground bus.
The cabinet ground is a safety ground that protects equipment and personnel from accidental shock hazards. It also provides a direct drain line for any static charges or electromagnetic interference (EMI) that may affect the cabinets. This cabinet ground remains separate from the DC signal ground until it terminates at the master ground bus.
The AC service ground is a single-point ground termination of the system AC power. This ground connects to the neutral-to-ground bond at the main AC power isolation transformer. It also terminates at a single point on the plant ground grid (usually the grounding electrode). | 1,205 | 6,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-13 | latest | en | 0.947636 |
http://www.blurtit.com/u/3142022/ | 1,498,676,564,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323730.30/warc/CC-MAIN-20170628185804-20170628205804-00113.warc.gz | 475,034,515 | 6,277 | Chrystian Tys voted up Virginia Lou's answer
Hi Taylor Lynn,
I found an equation which should be what you need:
The velocity (v) at splat time is the square root of twice the acceleration of gravity (g) multiplied by the distance fallen (h). Or,
v = sqrt (2 x g x h). So,
v^2 = 2 x g x h, and then solving for h,
h = v^2/2g | 98 | 329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.863797 |
http://www.ask.com/web?qsrc=6&o=15734&oo=15734&l=dir&gc=1&q=Matrix+Math | 1,485,282,615,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285001.96/warc/CC-MAIN-20170116095125-00037-ip-10-171-10-70.ec2.internal.warc.gz | 330,032,118 | 13,917 | Web Results
## Matrix (mathematics) - Wikipedia
en.wikipedia.org/wiki/Matrix_(mathematics)
In mathematics, a matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. For example, the ...
## Matrices - Math is Fun
www.mathsisfun.com/algebra/matrix-introduction.html
Math explained in easy language, plus puzzles, games, quizzes, worksheets ... Matrices. A Matrix is an array of numbers: A Matrix (This one has 2 Rows and 3 ...
## Intro to matrices | Introduction to matrices | Matrices | Precalculus ...
Take the red pill and enter the Matrix! ... Next tutorial. Representing linear systems of equations with augmented matrices. Current time:0:00Total duration:4 :29 ...
## Matrices | Precalculus | Khan Academy
Learn what matrices are and about their various uses: solving systems of equations, transforming shapes and vectors, and representing real-world situations.
## Matrices - Cool math Algebra Help Lessons - What's a Matrix?
www.coolmath.com/algebra/24-matrices/01-whats-a-matrix-01
You've already seen glimpses of matrices -- determinants (for Cramer's Rule) and Gaussian elimination... Now, we'll see what else we can do with them.
## S.O.S. Math - Matrix Algebra
www.sosmath.com/matrix/matrix.html
Application of Invertible Matrices: Coding · Complex numbers as Matrices · Markov Chains. Systems of Linear Equations. System of Equations: An Introduction ...
## Introduction and Basic Operations - SOS Math
www.sosmath.com/matrix/matrix0/matrix0.html
Matrices, though they may appear weird objects at first, are a very important tool in expressing and discussing problems which arise from real life cases. Our first ...
## Matrices and Determinants
www.maths.surrey.ac.uk/explore/emmaspages/option1.html
Introduction and Examples; Matrix Addition and Subtraction; Matrix Multiplication; The Transpose of a Matrix; The Determinant of a Matrix; The Inverse of Matrix ...
## Introduction to Matrices / Matrix Size - Purplemath
www.purplemath.com/modules/matrices.htm
Defines matrices and basic matrix terms, illustrating these terms with worked solutions to typical homework exercises.
## Matrix -- from Wolfram MathWorld
mathworld.wolfram.com/Matrix.html
In particular, every linear transformation can be represented by a matrix, and every matrix corresponds to a unique linear transformation. The matrix, and its ... | 541 | 2,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-04 | latest | en | 0.736128 |
https://docs.chemgymrl.com/en/web-update/lesson_1_extract.html | 1,702,098,643,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00779.warc.gz | 256,765,517 | 6,326 | chemgymrl.com
# Extraction Bench: Lesson 1
## Using a non-polar solute to extract a solute from water
In this tutorial, I am going to walk you through how our extraction environment works and hopefully give some insight into how an RL agent might interact with the environment. In this extraction we are going to be using water to extract sodium and chlorine from oil. We are going to be using this jupyter notebook in order to interact with the environment.
```import gymnasium as gym
import chemistrylab
import matplotlib,time
import numpy as np
from matplotlib import pyplot as plt
from chemistrylab.util import Visualization
from IPython.display import display,clear_output
Visualization.use_mpl_light(size=2)
```
So start by running the code cell below you should see a series of graphs like these appear:
```env = gym.make('WaterOilExtract-v0')
env.reset()
rgb = env.render()
plt.imshow(rgb)
plt.axis("off")
plt.show()
```
These graphs show the contents of each of our containers and the level of separation between the materials. The graphs to the right then show the layers of materials forming in the container.
When we start the environment we will see that we have a container filled with oil, Na and Cl. Using a polar solvent we can get the sodium and chlorine to diffuse from the oil into that solvent, in this case we can use water as our polar solvent!
### Step 1: Pour Water into the extraction vessel
`Action 33 H2O Vessel: pour by volume ([0.4]) -> extraction_vessel `
We can then see that storage vessel is now filled with the oil poured from the condensation vessel.
```env.reset()
total_reward = 0
#Pour in water
obs,rew,d,*_ = env.step(33)
total_reward += rew
obs,rew,d,*_ = env.step(39)
total_reward += rew
obs,rew,d,*_ = env.step(39)
total_reward += rew
rgb = env.render()
plt.imshow(rgb)
plt.axis("off")
plt.show()
```
### Step 2: Mix the extraction vessel
`Action 9 extraction_vessel: mix ([-1.]) -> None`
Now that we’ve added the water we need to mix the vessel to get the solutes to transfer into the oil, so let’s mix the vessel! As seen in the graph below we can see that based on the layer representation that we have mixed the oil and the water.
```obs,rew,d,*_ = env.step(9)
total_reward += rew
rgb = env.render()
plt.imshow(rgb)
plt.axis("off")
plt.show()
```
### Step 3: Wait for the layers to separate
`Action 39 extraction_vessel: mix ([0.16]) -> None`
Now that we have done some mixing we need to wait for the oil to settle to the top of the water so we can drain the water. Keep repeating the following command until the graph settles.
```obs,rew,d,*_ = env.step(39)
total_reward += rew
obs,rew,d,*_ = env.step(39)
total_reward += rew
rgb = env.render()
plt.imshow(rgb)
plt.axis("off")
plt.show()
```
### Step 4: Pouring out the Saltwater
`Action 4 extraction_vessel: drain by pixel ([10]) -> Beaker 1`:
Now that the water and oil have settled we want to drain out our water into beaker 1 so that we can pour out our oil out as waste.
```#Pouring
for i in range(4):
obs,rew,d,*_=env.step(4)
total_reward += rew
rgb = env.render()
plt.imshow(rgb)
plt.axis("off")
plt.show()
```
### Step 5: Pouring the oil into the waste vessel
`Action 24 extraction_vessel: pour by volume ([1.]) -> Waste Vessel`
Now we just have to empty out the extraction vessel into the waste vessel to get rid of the oil and we are done
```obs,rew,d,*_= env.step(24)
total_reward += rew
rgb = env.render()
plt.imshow(rgb)
plt.axis("off")
plt.show()
```
### Ending the Experiment
```obs,rew,d,*_ = env.step(40)
total_reward += rew
print(total_reward,d)
```
```0.8213018800367559 True
```
I hope this tutorial helped with your understanding of how an agent might interact with the extraction environment! | 964 | 3,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.830541 |
http://gmatclub.com/forum/a-recent-new-york-times-editorial-criticized-the-citys-19436.html?fl=similar | 1,484,575,997,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00007-ip-10-171-10-70.ec2.internal.warc.gz | 121,419,257 | 58,525 | A recent New York Times editorial criticized the citys : GMAT Sentence Correction (SC)
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# A recent New York Times editorial criticized the citys
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GMAT Club Legend
Joined: 07 Jul 2004
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Location: Singapore
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A recent New York Times editorial criticized the citys [#permalink]
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31 Aug 2005, 05:54
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A recent New York Times editorial criticized the city’s election board for, first of all, failing to replace outmoded voting machines prone to breakdowns, and secondarily, for their failure to investigate allegations of corruption involving board members.
(A) secondarily, for their failure to
(B) secondly, for their failure to
(C) secondly, that they failed and did not
(D) second, that they failed to
(E) second, for failing to
If you have any questions
you can ask an expert
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Joined: 01 Jun 2005
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31 Aug 2005, 05:58
I will go with E.
pronoun problem. the reference of "their" is ambiguous. Does it refer to "New York Times editorial" or "city's election board"
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### Show Tags
01 Sep 2005, 08:26
OA is E.
01 Sep 2005, 08:26
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# A recent New York Times editorial criticized the citys
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 812 | 3,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-04 | latest | en | 0.875709 |
http://nacretv.easylia.com/uamkx/1ea173-standard-error-of-ols-estimator | 1,618,768,523,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038507477.62/warc/CC-MAIN-20210418163541-20210418193541-00144.warc.gz | 76,711,388 | 3,878 | 0 βˆ The OLS coefficient estimator βˆ 1 is unbiased, meaning that . OLS assumption is violated), then it will be difficult to trust the standard errors of the OLS estimates. In econometrics, Ordinary Least Squares (OLS) method is widely used to estimate the parameters of a linear regression model. (26) The standard deviations sd(βˆ 0) and sd(βˆ1) of the OLS estimators … Ordinary Least Squares is the most common estimation method for linear models—and that’s true for a good reason.As long as your model satisfies the OLS assumptions for linear regression, you can rest easy knowing that you’re getting the best possible estimates.. Regression is a powerful analysis that can analyze multiple variables simultaneously to answer complex research questions. But why residuals autocorrelation would affect the coefficient standard errors? Variance of the OLS estimator The variance is in general different for the two parameters of the simple regression model. Ordinary least squares estimation and time series data One of the assumptions underlying ordinary least squares (OLS) estimation is that the errors be uncorrelated. 1) 1 E(βˆ =βThe OLS coefficient estimator βˆ 0 is unbiased, meaning that . The Assumption of Homoscedasticity (OLS Assumption 5) – If errors are heteroscedastic (i.e. Again, this variation leads to uncertainty of those estimators which we … The standard errors describe the accuracy of an estimator (the smaller the better). For the validity of OLS estimates, there are assumptions made while running linear regression models. In this case n = 122; you would need to be given n 1 and n 2.Otherwise assume n1 and n 2 are equal then n 1 = 61 and n 2 = 61. From the Wikipedia article on autocorrelation: While it does not bias the OLS coefficient estimates, the standard errors tend to be underestimated (and the t-scores overestimated) when the autocorrelations of the errors … 6.5 The Distribution of the OLS Estimators in Multiple Regression. Introduction to Properties of OLS Estimators. Linear regression models have several applications in real life. Var (βˆ 0) is given by (without proof): Var (βˆ 0) = σ2 Ns2 x ∑N i=1 x2 i. As in simple linear regression, different samples will produce different values of the OLS estimators in the multiple regression model. ECONOMICS 351* -- NOTE 4 M.G. Of course, this assumption can easily be violated for time series data, since it is quite reasonable to think that a … Kesalahan Standar Estimasi adalah standar deviasi di sekitar garis estimasi regresi yang mengukur variabilitas nilai Y aktual dari Y prediksi, disimbolkan dengan S YX.Meskipun metode kuadrat-terkecil (OLS) menghasilkan garis estimasi dengan jumlah variasi minimum (kecuali jika koefisien determinasi r 2 = 1) persamaan regresi bukanlah prediktor yang sempurna. 0) 0 E(βˆ =β• Definition of unbiasedness: The coefficient estimator is unbiased if and only if ; i.e., its mean or expectation is equal to the true coefficient β Abbott ¾ PROPERTY 2: Unbiasedness of βˆ 1 and . A1. The standard errors are measures of the sampling variability of the least squares estimates $$\widehat{\beta}_1$$ and $$\widehat{\beta}_2$$ in repeated samples - if we Calculating the unknown betas by Ordinary Least Squares is a mathematical approximation method that needs no statistical assumptions. We obtain $$\hat \beta = \left(\mathbf X' \mathbf X\right) ^{-1} \mathbf X'\mathbf y$$ Hence, the confidence intervals will be either too narrow or too wide.
## standard error of ols estimator
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https://studentsfocus.com/cs8451-daa-question-papers-design-and-analysis-of-algorithms-previous-year-question-papers-cse-4th-sem/ | 1,680,197,726,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00650.warc.gz | 614,153,206 | 12,106 | CS8451 DAA Question Papers
Anna University Regulation 2017 CSE CS8451 DAA Question Papers for previous years are provided below. Previous Year Question Papers for CSE 4th SEM CS8451 Design and Analysis of Algorithms, Engineering are listed down for students to make perfect utilization and score maximum marks with our study materials.
### Anna University Regulation 2017 (CSE) 4th SEM CS8451 DAA – Design and Analysis of Algorithms question paper
1. Define time complexity and space complexity. Write an algorithm for adding n natural numbers and find the space required by that algorithm
2. List the steps to write an Algorithm
3. Illustrate an algorithm for (i) Finding factorial of n number. (ii).Sorting the Elements.
4. Evaluate an algorithm for computing gcd(m,n) using Euclid’s algorithm
5. Design the equality gcd(m,n)=gcd(n,m mod n) for every pair of positive integers m and n.
6. List out the steps that need to design an algorithm.
7. Examine an algorithm to convert a binary number to a decimal number.
8. Identify how you will measure input size of algorithms.
9. Explain how many algorithms can you write for solving find the prime numbers. Compare which is the simplest and the most efficient.
10. Explain the various types of problems that can be solved using algorithm.
11. Apply the common technique for proving the correctness of an algorithm.
12. Define the term Algorithm
13. Define Big ‘Oh’ notation.
14. Formulate the order of growth. Compare the order of growth n! and 2n.
15. Differentiate between Best, average and worst case efficiency.
16. Discuss the concepts of asymptotic notations and its properties.
17. Analyze the order of growth.
(i).F(n) = 2n2 + 5 and g(n) = 7n. Use the Ω (g(n)) notation.
18. Evaluate the recurrence relations.
(i). x (n) = x (n-1) + 5 for n>1.
(ii). X (n) = x(n/3) +1 for n >1,x(1) =1. (Solve for n = 3k)
19. Discuss the General plan for analyzing efficiency of Non recursive & Recursive algorithms Understand
20. Discuss the following questions by consider the definition based algorithm for adding two n by n matrices.
1. What is basic operation?
2. How many times it is performed as a function of the matrix order n?
3. How many times it is performed as a function of the total number of elements in the input matrices?
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CS8451 Design and Analysis of Algorithms Engineering previous year question papers free download
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https://ibacnet.org/quadratic-formula-practice-worksheet-algebra-2/ | 1,569,253,193,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514577363.98/warc/CC-MAIN-20190923150847-20190923172847-00288.warc.gz | 503,670,385 | 31,053 | # Quadratic Formula Practice Worksheet Algebra 2
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https://unlearningmath.com/1.8-cm-to-mm/ | 1,721,234,490,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00065.warc.gz | 511,798,393 | 10,820 | # 1.8 cm to mm (Centimeter to Millimeter)
By / Under Centimeter To Millimeter / Published on
Discover the easy conversion process from 1.8 cm to mm and other key informations related to practical use and relevance in daily life.
## An optimal guide to the conversion of 1.8 cm to mm
1.8 cm equals 18 mm. The conversion from centimeters to millimeters is a straightforward process that is commonplace in both academic and everyday life. Whether it's mathematics or everyday measurements, understanding these conversions can make a significant difference.
Centimeters and millimeters, part of the metric system, are widely used around the globe, making them universal units of measurement. Their daily application ranges from the simplest tasks such as measure a piece of paper, to the most complex such as architectural blueprints, or medical and scientific studies.
Statistically speaking, 95% of the world's countries use the metric system. This system simplifies calculations tenfold, as it is a base-10 mathematical system - the foundation for the conversion from 1.8 cm to mm. To demonstrate, the relationship between centimeters and millimeters is akin to dollars and cents. Just as a dollar has 100 cents, a centimeter has 10 millimeters.
Conversion from centimeters to millimeters becomes simple arithmetic once this concept is understood. For every 1 centimeter, there are 10 millimeters. So, for the conversion of 1.8 cm to mm, we multiply the value in centimeters (1.8) by 10. The result is 18 millimeters.
This conversion is relevant to numerous industries. For instance, industries like manufacturing, construction, and scientific research depend on accurate measurements. In the healthcare domain, a small misstep can have significant consequences. For instance, the diameter of medical equipment like catheters, which have to be precise to ensure patient safety.
External link: Metric Conversion Guide This comprehensive resource provides additional insights into the metric system, including the conversion of 1.8 cm to mm.
How many millimeters are in 1.8 centimeters?
There are 18 millimeters in 1.8 centimeters. This is calculated by multiplying 1.8 by 10.
Why is it important to understand conversions like 1.8 cm to mm?
Conversions like 1.8 cm to mm are essential as they heighten simplicity and precision. This enables smooth communication and operations in various sectors including medicine, engineering and mathematics.
What tools can help with conversions like 1.8 cm to mm?
A ruler or tape measure can physically aid in visualizing the conversion. However, in the digital realm, online converting calculators and apps are available to make conversions immediate and accurate.
Converting 1.8 cm to mm is a straightforward task when the principles of the metric system are understood. A basic understanding of these units of measurement and their conversion can simplify and optimize many tasks in daily life.
Millimeter: 0 | 620 | 2,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.906295 |
https://nursingassignmentcrackers.com/nrs-493-topic-1-capstone-change-project-topics/ | 1,659,922,306,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570741.21/warc/CC-MAIN-20220808001418-20220808031418-00772.warc.gz | 397,559,522 | 10,223 | # NRS 493 Topic 1 Capstone Change Project Topics
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## NRS 493 Topic 1 Capstone Change Project Topics
NRS 493 Topic 1 Capstone Change Project Topics
Work with your preceptor to perform a needs assessment of the organization and community for your practicum. Review the needs assessment to identify possible project topics. In preparation for the capstone change project proposal, compile a list of three to five possible topics for your project and submit to the assignment instructor in LoudCloud.
You are not required to submit this assignment to LopesWrite.
Name: Assignment Rubric
NRS 493 Topic 1 Capstone Change Project Topics
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Excellent Good Fair Poor Summarize your interpretation of the frequency data provided in the output for respondent’s age, highest school grade completed, and family income from prior month. 32 (32%) – 35 (35%) The response accurately and clearly explains, in detail, a summary of the frequency distributions for the variables presented. The response accurately and clearly explains, in detail, the number of times the value occurs in the data. The response accurately and clearly explains, in detail, the appearance of the data, the range of data values, and an explanation of extreme values in describing intervals that sufficiently provides an analysis that fully supports the categorization of each variable value. The response includes relevant, specific, and appropriate examples that fully support the explanations provided for each of the areas described. 28 (28%) – 31 (31%) The response accurately summarizes the frequency distributions for the variables presented. The response accurately explains the number of times the value occurs in the data. The response accurately explains the appearance of the data, the range of data values, and explains extreme values in describing intervals that provides an analysis which supports the categorization of each variable value. The response includes relevant, specific, and accurate examples that support the explanations provided for each of the areas described. 25 (25%) – 27 (27%) The response inaccurately or vaguely summarizes the frequency distributions for the variables presented. The response inaccurately or vaguely explains the number of times the value occurs in the data. The response inaccurately or vaguely explains the appearance of the data, the range of data values, and inaccurately or vaguely explains extreme values. An analysis that may support the categorization of each variable value is inaccurate or vague. The response includes inaccurate and irrelevant examples that may support the explanations provided for each of the areas described. 0 (0%) – 24 (24%) The response inaccurately and vaguely summarizes the frequency distributions for the variables presented, or it is missing. The response inaccurately and vaguely explains the number of times the value occurs in the data, or it is missing. The response inaccurately and vaguely explains the appearance of the data, the range of data values, and an explanation of extreme values, or it is missing. An analysis that does not support the categorization of each variable values is provided, or it is missing. The response includes inaccurate and vague examples that do not support the explanations provided for each of the areas described, or it is missing. Summarize your interpretation of the descriptive statistics provided in the output for respondent’s age, highest school grade completed, race and ethnicity, currently employed, and family income from prior month. 45 (45%) – 50 (50%) The response accurately and clearly summarizes in detail the interpretation of the descriptive statistics provided. The response accurately and clearly evaluates in detail each of the variables presented, including an accurate and complete description of the sample size, the mean, the median, standard deviation, and the size and spread of the data. 40 (40%) – 44 (44%) The response accurately summarizes the interpretation of the descriptive statistics provided. The response accurately explains evaluates each of the variables presented, including an accurate description of the sample size, the mean, the median, standard deviation, and the size and spread of the data. 35 (35%) – 39 (39%) The response inaccurately or vaguely summarizes the interpretation of the descriptive statistics provided. The response inaccurately or vaguely evaluates each of the variables presented, including an inaccurate or vague description of the sample size, the mean, the median, the standard deviation, and the size and spread of the data. 0 (0%) – 34 (34%) The response inaccurately and vaguely summarizes the interpretation of the descriptive statistics provided, or it is missing. The response inaccurately and vaguely evaluates each of the variables presented, including an inaccurate and vague description of the sample size, the mean, the median, the standard deviation, and the size and spread of the data, or it is missing. Written Expression and Formatting – Paragraph Development and Organization: Paragraphs make clear points that support well-developed ideas, flow logically, and demonstrate continuity of ideas. Sentences are carefully focused—neither long and rambling nor short and lacking substance. A clear and comprehensive purpose statement and introduction is provided which delineates all required criteria. 5 (5%) – 5 (5%) Paragraphs and sentences follow writing standards for flow, continuity, and clarity. A clear and comprehensive purpose statement, introduction, and conclusion is provided which delineates all required criteria. 4 (4%) – 4 (4%) Paragraphs and sentences follow writing standards for flow, continuity, and clarity 80% of the time. Purpose, introduction, and conclusion of the assignment is stated, yet is brief and not descriptive. 3 (3%) – 3 (3%) Paragraphs and sentences follow writing standards for flow, continuity, and clarity 60%–79% of the time. Purpose, introduction, and conclusion of the assignment is vague or off topic. 0 (0%) – 2 (2%) Paragraphs and sentences follow writing standards for flow, continuity, and clarity < 60% of the time. No purpose statement, introduction, or conclusion was provided. Written Expression and Formatting – English writing standards: Correct grammar, mechanics, and proper punctuation 5 (5%) – 5 (5%) Uses correct grammar, spelling, and punctuation with no errors. 4 (4%) – 4 (4%) Contains a few (1 or 2) grammar, spelling, and punctuation errors. 3 (3%) – 3 (3%) Contains several (3 or 4) grammar, spelling, and punctuation errors. 0 (0%) – 2 (2%) Contains many (≥ 5) grammar, spelling, and punctuation errors that interfere with the reader’s understanding. Written Expression and Formatting – The paper follows correct APA format for title page, headings, font, spacing, margins, indentations, page numbers, parenthetical/in-text citations, and reference list. 5 (5%) – 5 (5%) Uses correct APA format with no errors. 4 (4%) – 4 (4%) Contains a few (1 or 2) APA format errors. 3 (3%) – 3 (3%) Contains several (3 or 4) APA format errors. 0 (0%) – 2 (2%) Contains many (≥ 5) APA format errors. Total Points: 100
Name: Assignment Rubric
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https://www.slideserve.com/varuna/discounted-cash-flow-analysis-aka-engineering-economy | 1,580,327,997,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251802249.87/warc/CC-MAIN-20200129194333-20200129223333-00120.warc.gz | 1,085,122,383 | 13,716 | # Discounted Cash Flow Analysis (aka Engineering Economy) - PowerPoint PPT Presentation
Discounted Cash Flow Analysis (aka Engineering Economy)
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Discounted Cash Flow Analysis (aka Engineering Economy)
## Discounted Cash Flow Analysis (aka Engineering Economy)
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##### Presentation Transcript
1. Cash Flow + R P F time - Discounted Cash Flow Analysis(aka Engineering Economy) • Objective: • To provide economic comparison of benefits and costs that occur over time • Assumptions: • Future benefits and costs can be predicted • All Benefits, Costs measured in money • Single point of view (generally, the project owner)
2. Issue - Value over time • Money now has a different value than the same amount at a different date • Comparable to – but not equal to – interest rate • Proper name: Discount Rate, r because future benefits/costs are reduced… (that is, “discounted”) to compare with present
3. Formulas for N Periods • Single amounts a) Future Amount = P (1 + r)N = P (caf) caf = Compound Amount Factor b) Present Amount = F/caf 1/caf = Present Worth Factor • Finite Series c) F = i R (1 + r)i = R [(1 + r)N - 1] / r d) R = P (crf) = [P*r (1+r) N ] / [(1 + r) N - 1] crf = Capital Recovery Factor
4. Formulas for N Periods (cont’) • Infinite Series 1 << (1 + r)N => (1+ r)N / [(1 + r)N - 1] => 1 => crf = r • Small Periods (1 + r)N ==> erN
5. Present Value Analysis • Present Value Analysis puts all future amounts on a common basis, typically “the present” • It may be some other reference year that is convenient, such as year of proposed investment • “Net” Present Value is the total of the present values of all future amounts, typically: Revenues – Costs = Net Amount • Referred to as “NPV”
6. Example Application of Present Value Analysis All by spreadsheets! Example:
7. Higher Discount rates => smaller value of future benefits discourages projects with long pay back periods project advocates try to minimize discount rate Examples: Massive Dams ; Airports; Aircraft projects Argument over Discount rates Often very difficult politically Not hard from technical perspective Generally higher than politicians want, for example, US Government base case before 2000 was 7%, but under current administration it is ~5% (see later slides) MORE DISCUSSION NEXT WEEK Effect of Different Discount Rates
8. US Govt base position on Discount rate (OMB Circular A-94,1992,revised yearly) 1. Base-Case Analysis. Constant-dollar benefit-cost analyses of proposed investments and regulations should report net present value and other outcomes determined using a real discount rate of 7 percent. This rate approximates the marginal pretax rate of return on an average investment in the private sector in recent years. Significant changes in this rate will be reflected in future updates of this Circular. 2. Other Discount Rates. Analyses should show the sensitivity of the discounted net present value and other outcomes to variations in the discount rate. The importance of these alternative calculations will depend on the specific economic characteristics of the program under analysis. For example, in analyzing a regulatory proposal whose main cost is to reduce business investment, net present value should also be calculated using a higher discount rate than 7 percent. http://www.whitehouse.gov/omb/circulars/a094/a094.html (accessed pre 2000)
9. Graphical view of Effect of Different Discount Rates and Lengths of Time
10. Discount Rate Approximation • To appreciate effect of discounting: “Rule of 72” or “Rule of 70” erN = 2.0 when rN = 0.72 (actually = 0.693) • Therefore, present amount doubles when future amount halves rN = 72 with r expressed in percent • Examples • When would \$1000 invested at 10% double? • What is, at 9%, the value of \$1000 in 8 years?
11. Longer Periods of Benefits Increase Present Values Increment depends on discount rate What length of time matters? For US Government Rate, not much over 30 years For Rates commonly used in business (15 to 20%), anything over 20 years has little value Exception: If future benefits grow exponentially, they may compensate for discounting of future revenues Effect of Different Time Horizons
12. Current US Government position on Discount rate (OMB Circular A-94) Provides 2 rates: Nominal represents future purchasing power; reflects inflation Real represents constant-dollar; assumes no inflation => Difference implies assumed rate of inflation These rates change yearly (as of last several years) These rates differ according to period – lower rates for shorter periods Note: assumed rate of inflation differs by period (See next slides)
13. Discount rates by OMB Circ. A-94, Appendix C[Ref: www.whitehouse.gov/omb/circulars/a094/a094_appx-c.html]
14. Guidelines appear precise (2 significant figures) But they are highly variable from year to year (see next slide), so analysis done in year N can be significantly off in year N+1 Would seem to be political, as evidenced by comparing rates from last 6 years with those given before 2000 (see slide 8) However…
15. Previous Discount rates by OMB. A-94
16. Formulas Simple Especially by Spreadsheets Discount rate is key issue High rates recommended (but see later presentations) Longer term benefits not large Summary | 1,249 | 5,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-05 | latest | en | 0.835862 |
http://mathhelpforum.com/pre-calculus/16828-composition-functions.html | 1,505,888,547,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686465.34/warc/CC-MAIN-20170920052220-20170920072220-00523.warc.gz | 205,409,346 | 10,671 | 1. ## Composition of Functions
I need help understanding whats going on...
Let f(x)=(x^2)-1 and g(x)=3x.
Find f * g.
1.(f * g)(x)=f(g(x))
2.---------=f(3x)
3.---------=((3x)^2)-1
4.---------=(9x^2)-1
I primarily dont understand line 3. How can 3x((x^2)-1)=((3x)^2)-1 ?
Perhaps you have a better way of explaining.
Trying to teach myself Algebra 2.
2. $f(g(x))=f(3x)=(3x)^2-1=9x^2-1$
3. I can see that. I just dont understand the third line of my original post.
How do they get to that conclusion?
4. $f(3x)$ means that in the function expression $x$ is replaced by $3x$.
So $f(3x)=(3x)^2-1$, not $3x^2-1$
5. Originally Posted by ceasar_19134
I need help understanding whats going on...
Let f(x)=(x^2)-1 and g(x)=3x.
Find f * g.
$(f\circ{g})(x)=[g(x)]^2-1$
But $g(x)=3x$, so $(f\circ{g})(x)=(3x)^2-1={\color{blue}9x^2-1}~\blacksquare$
Did you get it now?
6. Okay I get line 3 now, but how did you get (9x)^2 out of (3x)^2?
7. Originally Posted by ceasar_19134
Okay I get line 3 now, but how did you get (9x)^2 out of (3x)^2?
Nobody has got $(9x^2)$ from $(3x^2)$ what they
have done is: $(3x)^2=(3)^2(x)^2=9x^2$
RonL | 442 | 1,134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-39 | longest | en | 0.815125 |
http://www.overclock.net/t/951398/promising-algorithm-needs-evaluation | 1,503,064,962,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104636.62/warc/CC-MAIN-20170818121545-20170818141545-00120.warc.gz | 650,437,023 | 63,609 | Overclock.net › Forums › Software, Programming and Coding › Networking & Security › Promising algorithm needs evaluation
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# Promising algorithm needs evaluation
Hi, forum
I posted some time ago, while developing an encryption algorithm, and now that it is somewhat finished, I'd love to get some feedback, or evaluation, if possible
Now, here are some general statements:
The "one-time pad" algorithm is unbreakable if the key length >= message length (proven).
In the other case, where the keys is shorter, it is breakable,
because patterns may emmerge in cyphertext, where identical group of
characters has been encrypted by identical fraction of the key,
leading to the supposed lenght of the key and so on.
That makes is practically a Vigenere (right?).
My algorithm:
It uses key of different size, from 128bit to 200,000-bit and above.
What I do is, I take the positiong of the current bit in the message,
starting at 0, I take the binary of that number, and count the 1s- if
they are even, return 0, if odd- return 1. I'll call that "hash", though its not correct.
Now, this single bit I "XNOR" with the corresponding bit of the key at
the same location, and then XOR the result with the bit in the
message.
EXAMPLE:
Key: 10111000 00001111.10111000 00001111
"hashes": 01101001 10010110 10010110 01101001
Message: 01110000 01011100.01110000 10011011(as you will note, the
two halves of the message (where the key repeats after the 8-th bit)
order to expose a pattern)
Key after XNOR operation: 00101110 01100110 11011001 00011001
Key: 00101110 01100110 11011001 00011001
Message:01110000 01011100.01110000 10011011
Final: 01011110 00111010 00101001 10000010
No pattern exposed. I can prepare a more complex sample, if you'd
like?
In the final message, 1s and 0s seem as equally well spread as in the
original.
So, what is available in the end is:
Encrypted message: 01011110 00111010 00101001 10000010
"hashes": 01101001 10010110 10010110 01101001
Can someone get the original message from this data?
(original message was: 01110000 01011100 01110000 10011011 )
Any opinions?
Thank you all!
Edited by ronnin426850 - 2/26/11 at 2:52pm
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Bump. Any1 to share a critic?
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Maybe he can help?
Quote:
Originally Posted by 78@pwnt4lif3 Uploaded with ImageShack.us Maybe he can help?
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Quote:
EXAMPLE: Key: 10111000 00001111.10111000 00001111 "hashes": 01101001 10010110 10010110 01101001 Message: 01110000 01011100.01110000 10011011(as you will note, the two halves of the message (where the key repeats after the 8-th bit) start with identical bytes in order to expose a pattern)
this makes no sense to me.
How are you getting your hash here?
The message doesn't repeat (nor does the key) until the 16th bit.
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Quote:
Originally Posted by serge2k this makes no sense to me. How are you getting your hash here? The message doesn't repeat (nor does the key) until the 16th bit.
I get the hash in a rather lame way, actually, but it still helps to remove the pattern.
For the specified number, if digits>2, I sum all digits in a loop, until I get 2 digit nr.
Then I remove all 0s and remove the 1s by pairs. Return the remainder. If 1s are ever, the remainder is 0. If not, its 1.
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So this is just a stream cipher then?
So never reuse the keys. It's vulnerable to a known plaintext attack if you do (compromising all data encrypted with the same key).
The other issue is, the hash is static and adds nothing. If you XNOR the final message with the hash you get the message XOR the key.
Code:
``````Key: 10111000 00001111.10111000 00001111
Hash: 01101001 10010110 10010110 01101001
Key XNOR Hash: 00101110 01100110 11010001 10011001
Message: 01110000 01011100.01110000 10011011
Final: 01011110 00111010 10100001 00000010
Key ^ Message: 11001000 01010011 11001000 10010100
Final XNOR Hash: 11001000 01010011 11001000 10010100``````
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Quote:
Originally Posted by serge2k So this is just a stream cipher then? So never reuse the keys. It's vulnerable to a known plaintext attack if you do (compromising all data encrypted with the same key). The other issue is, the hash is static and adds nothing. If you XNOR the final message with the hash you get the message XOR the key. Code: ``````Key: 10111000 00001111.10111000 00001111 Hash: 01101001 10010110 10010110 01101001 Key XNOR Hash: 00101110 01100110 11010001 10011001 Message: 01110000 01011100.01110000 10011011 Final: 01011110 00111010 10100001 00000010 Key ^ Message: 11001000 01010011 11001000 10010100 Final XNOR Hash: 11001000 01010011 11001000 10010100``````
Yes, found that out today. Also my numbers are a bit wrong, which led me to the confusin. Working on it
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• Promising algorithm needs evaluation
Overclock.net › Forums › Software, Programming and Coding › Networking & Security › Promising algorithm needs evaluation | 4,390 | 12,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.832833 |
https://uclouvain.be/en-cours-2021-ltarc1261 | 1,656,239,370,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00731.warc.gz | 624,355,911 | 8,707 | # Structural analysis 2
ltarc1261 2021-2022 Tournai
Structural analysis 2
5.00 credits
20.0 h + 45.0 h
Q1
Teacher(s)
Language
French
Prerequisites
The prerequisite(s) for this Teaching Unit (Unité d’enseignement – UE) for the programmes/courses that offer this Teaching Unit are specified at the end of this sheet.
Main themes
This teaching unit provides an introduction to the analysis of load bearing structures. It forms part of the continuous process of learning about structures and their behaviour.
The course introduces fundamental concepts which enable students to
• formulate all the stages in the analysis of a structure : production of a static diagram, assessment of stress and internal loads.
• maintain a dialogue with an engineer specialised in this field.
The following topics are covered:
• Extended, compressed and bent structures
• Tensile (cables) and compressed (arches) structures
• Isostatic and hyperstatic structures
• Stability of form (slender elements) and stability of the whole (bracing).
Learning outcomes
At the end of this learning unit, the student is able to : 1 Specific learning outcomes: By the end of the course, students are able to 1. undertake an overall analysis of a structure, i.e.: formulate the vertical and horizontal stresses acting on a structure produce the static diagram which shows this formulate the conditions of overall stability formulate the conditions of stability / instability of an isolated structural element analyse the structural behaviour of supports and assemblies. 2. use graphic and analytical methods applied to principles of balance, the determination of internal loads and associated constraints, the determination of deformations in the context of compressed, extended and bent structures (isostatic and hyperstatic structures). 3. identify the influence of hyperstaticity on the mechanical behaviour of a structure. 4. develop a logical procedure which on one hand, summarises acquired knowledge and demonstrates mastery of basic concepts and on the other hand, makes a link with other disciplines, particularly the architectural project. Contribution to the learning outcomes reference framework: With regard to the learning outcomes reference framework of the Bachelor's degree in Architecture, this teaching unit contributes to the development, the acquisition and the assessment of the following learning outcomes: Make use of other subjects Interpret the knowledge of other subjects Make use of other subjects to ask questions about the design and implementation of an architectural project Use the technical dimension Be familiar with and describe the main technical principles of building Observe and assess the main construction principles of a building Acquire an instinctive understanding of structures to use in producing a creative work of architecture
Bibliography
Allen E., Zalewski W., Form and Forces, Designing efficient, expressive structures, Boston, Wiley, 2010
Muttoni A., L'art des structures, Lausanne, PPUR, 2004
Salvadori M., Comment ça tient ?, Editions Parenthèses, 2005
Studer M-A. & Frey Fr., Introduction à l'analyse des structures, Lausanne, PPUR, 1997
Schodek D., Bechthold M., Structures, sixth edition, Pearson Prentice Hall, 2008
Gordon J., Structures et matériaux, Pour la science, Belin, 1994
Teaching materials
• Présentations PowerPoint du cours théorique
• Enoncés d'exercices sur les différents thèmes abordés en cours
• Notes personnelles de l'étudiant.e
Faculty or entity
#### Programmes / formations proposant cette unité d'enseignement (UE)
Title of the programme
Sigle
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Prerequisites
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Bachelor in Architecture (Tournai) | 793 | 3,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-27 | latest | en | 0.855126 |
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First, the calculator gets the numerator and denominator of a decimal fraction, that is, 0.375 and 1. Both are multiplied by the necessary power of 10 to get rid of the decimal point, i.e., to get integers, that is, 375 and 1000. Then the calculator looks for the greatest common divisor (GCD) of these two integers. Both integers are then divided by it to get smaller values. Let's do it with our example
The GCD of 375 and 1000 is 125. 375 divided by 125 equals 3, 1000 divided by 125 equals 8. The resulting fraction is 3/8.
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PLANETCALC, Fraction to Decimal and Decimal to Fraction converter | 431 | 1,872 | {"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} | 3.28125 | 3 | CC-MAIN-2022-05 | longest | en | 0.819541 |
http://doctintuc.info/lesson-104-problem-solving-hyperbolas-46/ | 1,571,371,686,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00478.warc.gz | 56,060,660 | 8,944 | # LESSON 10.4 PROBLEM SOLVING HYPERBOLAS
Share buttons are a little bit lower. Identify the Vertices and Foci of the hyperbola. Be particularly careful with hyperbolas that have a horizontal transverse axis such as the one shown below:. This half, when reflected over a horizontal line, will result in a complete graph of the hyperbola. A similar result occurs with a hyperbola. For example, let’s look at how the equation of the ellipse would be graphed on a graphing calculator. You saw what happens when an ellipse is not graphed correctly.
Identify the Vertices and Foci of the hyperbola. Note, however, that a, b and c are related differently for hyperbolas than for ellipses. Now we continue onto the hyperbola, which in. This half, when reflected over a horizontal line, will result in a complete graph of the hyperbola. Definition A hyperbola is the set of all points in a plane, the difference of whose distances from two distinct fixed point.
Overview In Section 9.
Identify the Vertices and Foci of the hyperbola Hyperbolas. Quadratic Relations and Conic Sections For example, let’s look at how the equation of the ellipse would be graphed on a graphing calculator.
Help for Exercise 49 on page Because hyperbolas are not functions, their equations cannot be directly graphed on a graphing calculator. So, the vertices occur at —2, 0 and 2, 0 the endpoints of the conjugate axis occur at 0, —4 and 0, 4and you can sketch the rectangle shown in Figure 9.
Now we continue onto the hyperbola, which in.
My presentations Profile Feedback Log out. Hyperbola A hyperbola is the set of all points in a plane whose distances from two fixed points in the plane 110.4 a. Classify conics from their general equations.
You must draw a half that does represent a function. Definition A hyperbola is the set of all points in a plane, the difference of whose distances from two distinct fixed point.
To make this website work, we log user data and share it with processors. The bottom half represented byneeds to be included to complete the graph as shown at right.
# | CK Foundation
If we only took the positive square root,and graphed the function on a lwsson calculator, we would get the graph on the left:. Divide each side by Subtract 16 from each side and factor. By the Midpoint Formula, the center of the hyperbola occurs at the point 2, 2. So, the graph is a parabola.
Download ppt “Hyperbolas and Rotation of Conics”. A vertical line test will confirm this result.
DSGE MODELS DISSERTATION
Write in completed square form. This half, when reflected over a horizontal line, will result in a complete graph of the hyperbola. Be particularly careful with hyperbolas that have a horizontal transverse axis such as the one shown below:. Hyperbolws must graph the equation of a hyperbola in two separate pieces. Auth with social network: Transverse axis is horizontal.
## Hyperbolas and Rotation of Conics
A similar result occurs with a hyperbola. You saw proble happens when an ellipse is not graphed correctly. A similar situation occurs when graphing an ellipse. You should solvinb the bottom half is a reflection of about the x -axis. The difference is that for an ellipse, the sum of the distances between the foci and a point on the ellipse is constant; whereas for a hyperbola, the difference of the distances between the foci and a point on the hyperbola is constant. | 785 | 3,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-43 | latest | en | 0.897142 |
https://jameshoward.us/2015/07/30/comments-on-logarithmic-measurements/ | 1,726,659,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00044.warc.gz | 291,101,100 | 22,424 | ### Thursday July 30, 2015
###### • college algebra • data science • logarithms • math education • mathematics • measurement • scientific computing • trigonometry •
Logarithmic scales are excellent for relating different measurements that exist on different scales. But not all measures are suitable for logarithmic measures. The amount of time your coworkers spend on their commutes is probably not appropriate. Your time may be 20 minutes and the next person is 30 minutes. Even if the third person’s unfortunate commute is 80 minutes, these all exist on the same scale. Nobody’s commute ranges into days or weeks.
It does not make sense to measure the height of people, which is more or less, normally distributed, because the difference between someone who is 1.6 meters (m) and someone who is 1.9m is better represented by the raw difference. If individuals’s heights varied from, for instance, 1.5m through 200m, then a logarithmic scale might make more sense.
A person’s height is a fairly constrained measurement. Length, more broadly, can be vary on many different scales. And space is big.1 My commute is 50km, but the distance from the Earth to the Sun is approximately 150 million kilometers and the distance from the Earth to the Andromeda is $2.4 \times 10^{19}$ kilometers. That’s a lot. And that is something we should measure on a logarithmic scale.
Image by Brett L. / Flickr.
1. Adams, Douglas. The Hitchhiker’s Guide to the Galaxy: The Original Radio Scripts. London: Pan, 1985, p. 39. | 360 | 1,518 | {"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} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.886037 |
https://www.thestudentroom.co.uk/showthread.php?t=3172 | 1,586,323,609,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371810617.95/warc/CC-MAIN-20200408041431-20200408071931-00311.warc.gz | 1,192,565,357 | 35,488 | # Chemistry Paper 6b Edexcel
Watch
This discussion is closed.
16 years ago
#1
2AI) A Fluoride of Phosphorous Y, contains 24.6% by mass of phosphrous and has a molar mass of 126gmol-1 Deduce the Molecular formula.
2c) when hydrogen fluoride is dissolved in water a solution of a weak acid is formed. write the expression for the acid disocciation constant, Ka, for hydrogen fluroide. Calculate the value of Ka, with units given that 0.150 mol dm-3 solution of hydrogen flouride has a pH of 2.04.
1b) what indicator did you use and why?
1c)Suggest a Chemical test to confirm the presence of ammonia in solution X.
0
16 years ago
#2
Hi there, I just did Edexcel Chem 6B as well. Here are the answers I got for the questions you posted:
2a) i) Ar(P) = 31 Ar(F) = 19
24.6% P thus 100 - 24.6 = 75.4% F
24.6/31 = 0.794 75.4/19 = 3.968
0.794/0.794 = 1 3.968/0.794 = 5
Thus molecular formula = PF5
This is confirmed by the molar mass of 126: 31 + 5(19).
2c) Ka = [H+][F-]/[HF]
Ka = [H+]^2/[HF]
[H+] = 10^-2.04 = 0.00912 mol dm^-3
Ka = 0.00912^2/0.150 = 5.55 x 10^-4 mol dm^3
1b) Methyl orange. Turns from colourless to orange (orange below pH 5), fits vertical pH range between 5 and 3.5.
1c) Use red litmus. Turns from red to blue with the presence of ammonia.
0
16 years ago
#3
What did you think of the exam?
0
16 years ago
#4
1b) Methyl orange. Turns from colourless to orange (orange below pH 5), fits vertical pH range between 5 and 3.5.
1c) Use red litmus. Turns from red to blue with the presence of ammonia.
That aint accurate
Methyl Orange goes from RED to YELLOW
Use DAMP RED LITMUS On top of where NH3 Gas is released and gas turns BLUE indicating ALKALINE gas for NH3.
0
16 years ago
#5
Ok, fair enough about the colour changes. The dampness for the litmus paper doesn't have to be explicit though.
0
16 years ago
#6
this was my test: soak X in cotton wool, and another with HJCL, bring them together, white fumes should form.
Is that OK?
0
16 years ago
#7
easiest synoptic i have seen for a while, even tho im **** at chem Question 2 was a gimme, and so was the majority of 4 although i think i screwed up the Kc part of it by doing pressing the wrong buttons on my calulator even tho im pretty sure i had the right working out - d'oh! Whats the grade boundaries for unit 6 - ive heard about 77/100 for an A?
0
16 years ago
#8
Originally posted by Unregistered
easiest synoptic i have seen for a while, even tho im **** at chem Question 2 was a gimme, and so was the majority of 4 although i think i screwed up the Kc part of it by doing pressing the wrong buttons on my calulator even tho im pretty sure i had the right working out - d'oh! Whats the grade boundaries for unit 6 - ive heard about 77/100 for an A?
Yeah, considering they have such high grade boundaries for this paper, it's really hard to get A.
0
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Transformations are changes done in the shapes on a coordinate plane by rotation,reflection or translation. In the 19th century, Felix Klein proposed a new perspective on geometry known as transformational geometry. Most of the proofs in geometry are based on the transformations of objects. We can alter any image in a coordinate plane using transformations. The graphics used in video games are better understood with the rules of transformations applied. Let us learn to identify the transformations, understand the rules of transformations of functions, and explore the types of transformations.
1 What are Transformations? 2 Types of Transformations 3 Rules For Transformations 4 Formula of Transformations 5 FAQs on Transformations
## What are Transformations in Math?
A function, f, that maps to itself is called the transformation, i.e., f: X → X. The pre-image X becomes the image X after the transformation. This transformation can be any or the combination of operations like translation, rotation, reflection, and dilation. The translation is moving a function in a specific direction, rotation is spinning the function about a point, reflection is the mirror image of the function, and dilation is the scaling of a function. Transformations in Math describe how two-dimensional figures move around a coordinate plane.
## Types of Transformations
There are four common types of transformations - translation, rotation, reflection, and dilation. From the definition of the transformation, we have a rotation about any point, reflection over any line, and translation along any vector. These are rigid transformations wherein the image is congruent to its pre-image. They are also known as isometric transformations. Dilation is performed at about any point and it is non-isometric. Here the image is similar to its pre-image.
TransformationFunctionResult
RotationRotates or turns the pre-image around an axisNo change in size or shape
ReflectionFlips the pre-image and produces the mirror-imageNo change in size or shape or orientation
TranslationSlides or moves the pre-imageNo change in size or shape; Changes only the direction of the shape
DilationStretches or shrinks the pre-imageExpands or contracts the shape
## Rules for Transformations
Consider a function f(x). On a coordinate grid, we use the x-axis and y-axis to measure the movement. Here are the rules for transformations of function that could be applied to the graphs of functions. Transformations can be represented algebraically and graphically. Transformations are commonly found in algebraic functions. We can use the formula of transformations in graphical functions to obtain the graph just by transforming the basic or the parent function, and thereby move the graph around, rather than tabulating the coordinate values. Transformations help us visualize and learn the equations in algebra.
### Transformation of Translation
Translation of a 2-d shape causes sliding of that shape. To describe the position of the blue figure relative to the red figure, let’s observe the relative positions of their vertices. We need to find the positions of A′, B′, and C′ comparing its position with respect to the points A, B, and C. We find that A′, B′, and C′ are:
• 8 units to the left of A, B, and C respectively.
• 3 units below A, B, and C respectively.
This translation can algebraically be translated as 8 units left and 3 units down. i.e. (x,y) → (x-8, y-3)
(Video) Transformations of Graphs - Corbettmaths
We can apply the transformation rules to graphs of quadratic functions. This pre-image in the first function shows the function f(x) = x2. The transformation f(x) = (x+2)2 shifts the parabola 2 steps right.
### Transformation of Reflection
The type of transformation that occurs when each point in the shape is reflected over a line is called the reflection. When the points are reflected over a line, the image is at the same distance from the line as the pre-image but on the other side of the line. Every point (p,q) is reflected onto an image point (q,p). If point A is 3 units away from the line of reflection to the right of the line, then point A' will be 3 units away from the line of reflection to the left of the line. Thus the line of reflection acts as a perpendicular bisector between the corresponding points of the image and the pre-image.
Here is the graph of a quadratic function that shows the transformation of reflection. The function f(x) = x3. The transformation of f(x) is g(x) = - x3 that is the reflection of the f(x) about the x-axis.
### Transformation of Rotation
The transformation that rotates each point in the shape at a certain number of degrees around that point is called rotation. The shape rotates counter-clockwise when the number of degrees is positive and rotates clockwise when the number of degrees is negative. The general rule of transformation of rotation about the origin is as follows.
To rotate 90º: (x,y) → (-y, x)
To rotate 180º (x,y) → (-x,-y)
To rotate 270º (x,y) → (y, -x)
(Video) Translations Reflections and Rotations - Geometric Transformations!
In the function graph below, we observe the transformation of rotation wherein the pre-image is rotated to 180º at the center of rotation at (0,1). Let us observe the rule of rotation being applied here from (x,y) to each vertex. The transformation that is taken place here is from (x,y) → (-x, 2-y)
(-2,4) →(2,-2), (-3,1) → (3,1) and (0,1 ) → (0,1)
### Transformation of Dilation
The transformation that causes the 2-d shape to stretch or shrink vertically or horizontally by a constant factor is called the dilation. The vertical stretch is given by the equation y = a.f(x). If a > 1, the function stretches with respect to the y-axis. If a < 1 the function shrinks with respect to the y-axis. The horizontal stretch is given by y = f.(ax). If a > 1, the function shrinks with respect to the x-axis. If a < 1, the function stretches with respect to the x-axis.
Consider a parent function y = x2 +x. After a horizontal shrink by a factor of 5, the altered function becomes 5x2+5x at point (1/5x, y). After a horizontal stretch by factor 1/5, the transformed function becomes 1/5 x2 +1/5 x at a point (5x,y). After a vertical shrink by a factor of 5, the altered function becomes 5x2+5x at point (x, 5y). After a vertical stretch by factor 1/5, the transformed function becomes 1/5 x2 +1/5 x at a point (x, 1/5y).
## Formula of Transformations
Let us consider the graph f(x) = x2
• Suppose we need to graph f(x) = x2-3, we shift the vertex 3 units down.
• Suppose we need to graph f(x) = 3x2+ 2, we shift the vertex two units up and stretch vertically by a factor of three.
• Suppose we need to graph f(x) = 2(x-1)2, we shift the vertex one unit to the right and stretch vertically by a factor of 2
• Thus, we get the general formula of transformations as
f(x) =a(bx-h)n+k
where k is the vertical shift,
h is the horizontal shift,
(Video) Graph Transformations | Grade 7-9 Maths Series | GCSE Maths Tutor
a is the vertical stretch and
b is the horizontal stretch.
Likewise, f(x) can be transformed in many ways. This table shows the resultant graph after the transformation is applied on f(x).
Transformation of f(x)Transformation of graphChange (x,y) to
-f(x)reflect f(x) over the x-axis(x,-y)
f(-x)reflect f(x) over the y-axis(-x,y)
f(x)+ashift f(x) up by a units(x,y+a)
f(x) - ashift f(x) down by a units(x,y-a)
f(x+a)shift f(x) left by a units(x-a,y)
f(x-a)shift f(x) right by a units(x+a,y)
a. f(x)stretch/shrink f(x) vertically(x,ay)
f(ax)stretch/shrink f(x) horizontally(x/a,y)
Important Notes on Transformations
• Transformations in geometry can be combined. A shape can be reflected or rotated or translated or dilated or can have a combination of these transformations.
• Transformations are expressed algebraically in the graph functions.
☛ Related Topics:
• Graphing functions
## FAQs on Transformations
### What Are Transformations?
The transformations are the alterations done to a function by translation, reflection, rotation, and dilation. The original image known as the pre-image is altered to get the image.
### What are the 4 Types of Transformations?
Translation, reflection, rotation, and dilation are the 4 types of transformations. The translation is moving the shape in a particular direction, reflection is producing the mirror image of the shape, rotation flips the shape about a point in degrees, and dilation is stretching or shrinking the shape by a constant factor.
### Why are Transformations Important in Math?
Math is about identifying patterns and understanding the relationships between concepts to work out a solution to a problem. Transformations are important in Math to mainly know the congruence and similarity of figures in a plane. We could alter the position of a point, or a line, or a 2-d shape using the 4 transformations.
(Video) Transformations of Functions | Precalculus
### How do you Plot a Reflection and What are the Transformations That Take Place?
When we reflect a point across the x-axis, the y-coordinate is transformed and the x-coordinate remains the same. x-coordinate will have the same sign, but the sign of the y-coordinate changes.
### What is the Formula For Transformations?
The general formula of transformations is f(x) =a(bx-h)n+k
where k is the vertical shift,
h is the horizontal shift,
a is the vertical stretch and
b is the horizontal stretch.
### How do Transformations Work?
The transformations allow us to change the graph of the function to slide, stretch or shrink, rotate.
### How do you do Transformations on a Graph?
The following steps are to be followed while we do transformations on a graph.
• Find the parent function f(x) and identify the sequence of the transformations to be made.
• Determine if it is to be reflected over the x-axis or y-axis, to be shifted vertically or horizontally, to be rotated about degrees at a point, or to be stretched or shrunk about the axes using the scaling factor.
• Key-in the coordinates to the parent function following the rules of transformations.
• Plot the transformed function accordingly.
### What is the Rule for the Transformations?
Suppose a parent function is f(x), f(x)+a shifts f(x) up by a units, f(x+a) shifts f(x) left by a units, -f(x) reflects f(x) over the x-axis, f(-x) reflects f(x) over the y-axis, a. f(x) stretches or shrinks f(x) vertically and f(a.x) stretches or shrinks f(x) horizontally. These are a few rules for the transformations of graphs.
## Videos
1. Introduction to transformations | Transformations | Geometry | Khan Academy
2. A-Level Maths: B9-13 [Graph Transformations: Examples of Transforming y = x^2]
(TLMaths)
3. Transformations - Rotate 90 Degrees Around The Origin
(mrmaisonet)
4. Graphing Techniques Q&A (1 of 3: Horizontal transformation example)
(Eddie Woo)
5. Graphing and describing transformations of a quadratic equation
(Brian McLogan)
6. How To Change The Subject of a Formula - GCSE Maths
(The Organic Chemistry Tutor)
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posted by .
A cone has a radius of 6 inches and a height of 9 inches
a) calculate the volume of the cone. use pi. as my answer for part a i got 339.92 cubic inches,,,,,,, Is my answer correct ??
I need help doing part b and c.
b) calculate a possible radius and height of a cylinder that has the same volume as the cone.
c) If the radius of the cylinder is the same as the radius of the cone, what is the height of the cylinder?
• math -
v = pi/3 r^2 h = pi/3 * 36 * 9 = 108pi
for cylinder,
v = pi r^2 h, so a cylinder 1/3 as high would have the same volume as the cone.
see above. same radius, 1/3 height.
• math -
thank you very much - Steve
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More Similar Questions | 755 | 2,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-34 | latest | en | 0.907802 |
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-------------------------------------
(6a^3 - 5a^2 + 10a - 3) divided by (3a - 1)
-------------------------------------
(10x^3 - 3x^2 - 6x -4) divided by (5x - 1)
-------------------------------------
Use the formula A= lw and long division to find l when A = 10a^3 + 22a^2 - 2a - 12 and w = 2a + 4
-------------------------------------
thanks so much!
2. Hello, omgitzbella!
You need to learn/practice long division . . .
$\displaystyle (6a^3 - 5a^2 + 10a - 3) \:\div \3a - 1)$
Code:
2a² - a + 3
---------------------
3a - 1 ) 6a³ - 5a² + 10a - 3
6a³ - 2a²
---------
-3a² + 10a
-3a² + a
-----------
9a - 3
9a - 3
------- | 254 | 688 | {"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} | 3.8125 | 4 | CC-MAIN-2018-22 | latest | en | 0.351121 |
http://www.pansoft.ch/2019/08/09/get-the-scoop-on-what-is-i-in-math-before-youre-too-late-8/ | 1,566,628,640,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00408.warc.gz | 288,749,479 | 9,616 | # Get the Scoop on What Is I in Math Before You’re Too Late
## How to Find What Is I in Math Online
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## Most Noticeable What Is I in Math
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## What Is I in Math Secrets
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## The Bad Side of What Is I in Math
Once the options are selected, click the Start” button and begin answering questions. A useful operation we can perform on two sets is called the Cartesian product. Select the amount of questions then click the Start button.
The key feature here is the important index. The access code will be provided in class. Some personal data, including your name, may appear in the log. | 1,065 | 5,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-35 | latest | en | 0.940736 |
https://minuteshours.com/21-51-minutes-in-hours-and-minutes | 1,620,896,552,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990584.33/warc/CC-MAIN-20210513080742-20210513110742-00080.warc.gz | 424,344,073 | 5,000 | # 21.51 minutes in hours and minutes
## Result
21.51 minutes equals 0 hours and 21.51 minutes
You can also convert 21.51 minutes to hours.
## Converter
Twenty-one point five one minutes is equal to zero hours and twenty-one point five one minutes. | 64 | 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} | 2.515625 | 3 | CC-MAIN-2021-21 | latest | en | 0.854213 |
http://www.jiskha.com/display.cgi?id=1362675410 | 1,498,504,925,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320863.60/warc/CC-MAIN-20170626184725-20170626204725-00181.warc.gz | 576,838,549 | 3,844 | # geometry.
posted by .
In the diagram, a trapezoid is shown with x=36 and y=56. Find the area of the trapezoid. It's an isosceles triangle with base angles of 60°.
• geometry. -
I need the diagram or a description of what is labelled x and y.
• geometry. -
assuming x and y are the bases, then since y-x = 20, there is a right triangle on each end with base 10, height 10√3.
So, now you know the bases and the height. Plug and chug. | 127 | 440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-26 | latest | en | 0.923336 |
https://www.slideshare.net/MsKendall/algebra-i-ccp-quarter-3-benchmark-review-2013-2 | 1,516,348,522,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887832.51/warc/CC-MAIN-20180119065719-20180119085719-00634.warc.gz | 994,075,611 | 34,005 | Successfully reported this slideshow.
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# Algebra i ccp quarter 3 benchmark review 2013 (2)
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### Algebra i ccp quarter 3 benchmark review 2013 (2)
1. 1. Algebra I CCP Quarter 3 Benchmark Review – Packet #1Find the equation of each line in slope-intercept form (y = mx + b).1. 2.3. 4.5. ` 6.
2. 2. Graph the equations of the lines below in the form given. 21. y= x 2. y = -x – 2 33. 4x – 3y = 12 4. x=4
3. 3. 15. y= (x – 2) 6. y – 2 = -(x + 4) 4Find the x and y intercepts of the following linear equations and then graph.1. 6x – 12y = -36
4. 4. Find the x and y intercepts of the following linear equations and then graph.2. x + 2y = 53. 0.8x + 0.3y = 2.4
5. 5. Find the equation of a line given the slope and a point. Put in both point-slope and slope-intercept forms.1. m=3 (1, 4)2. m = 2 (2, −3)3. m = −1 (−4, 2)
6. 6. Find the equation of a line given 2 points. Put in slope-intercept, point-slopeand Standard forms.1. (1, −5) (4, 7)2. (−3, 4) (0, 7)3. (5, −2) (−1, 0)
7. 7. Rewrite the equation in slope-intercept form (y = mx + b). Solve for y.1. 4x – y + 3 = 0 2. 2x + 4y = -123. 2x – 4y = 20 4. 4x – 8y = -16Rewrite each of the following linear equations in Standard Form. 31. y= x–1 2. y = -x – 7 4 13. y – 5 = -2(x + 3) 4. y+2= (x + 5) 2
8. 8. Parallel and Perpendicular Lines1. Find the equation of the line perpendicular to 4y - 2x = 6 and passes through the point (3, 1).2. Find the equation of the line parallel to 2y + 3x = 5 and passes through the point (−2, 5).3. Write the equation of the line that is perpendicular to the given line and passes through the given point. Then find the parallel line going through the point too.
9. 9. Solve each of the following linear systems by graphing.1. 2x + y = 4 x+y=22. x − 2y = 2 2x + 5y = −5
10. 10. Solve each of the following linear systems using the substitution method.1. 2x + 4y = 8 5x + y = -72. x + 2y = 3 2x + 4y = 6 13. y=- x-1 3 4x - 3y = 18 | 805 | 2,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-05 | latest | en | 0.801127 |
https://physics.stackexchange.com/questions/88191/azimuthal-quantum-number-ell-and-magnetic-quantum-number-m-are-from-angular | 1,585,389,856,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370490497.6/warc/CC-MAIN-20200328074047-20200328104047-00555.warc.gz | 663,206,825 | 32,989 | # Azimuthal quantum number $\ell$ and magnetic quantum number $m$ are from angular momentum?
Azimuthal quantum number $\ell$, and magnetic quantum number $m$ are defined when we do derivation for $L^2f=\ell(\ell+1){\hbar}^2f$ and $L_zf=\ell{\hbar}f$.
This is my own conclusion after studying introduction to quantum mechanics by Griffiths. Correct?
But I don't know why does $L_zf=\ell{\hbar}f$ become $L_zf=m{\hbar}f$? I wonder why when we operator $L_+$ on an eigenstate, the only change is that $m$ increases by one not for $\ell$, e.g.$$L_+ Y_{2}^{1}= AY_{2}^{2},$$ why not $\ell$ change?
This is not generally(as in, it can be but is not always) true: $$L_zf=l{\hbar}f.$$
What is true is: $$L_z f = m\hbar f$$
But $m$ can take values between $l$ and $-l$ in integer steps. Look through the derivation again now that you know this and see if that helps.
Your second equation is not always true, as Bobby said. What is true is this:
$L^2$ is an operator which when acting on an eigenstate $|\ell,m\rangle$ gives $\ell(\ell+1)\hbar^2|\ell,m\rangle$, while $L_z$ is an operator which when acting on an eigenstate $|\ell,m\rangle$ gives $m\hbar|\ell,m\rangle$. Here the familiar quantum numbers are really just eigenvalues of the operators (important!), and what you're saying is the eigenvalue corresponds to the physics you're interested in.
The answer to your second question is provided by the definition of the angular momentum ladder operators: $L_{\pm} := L_x\pm iL_y$. The only way to evaluate these is to make it into a combination of $L_z$ operators which imply the final state is raised/lowered in the spin number.
• I'm a little rusty on the exact calculation, but it should involve using commutators and seeing if you can subtract out the $L_x$ and $L_y$ parts. A bit of a long but important exercise. If you want you can also think that the angular momentum state fixes the angular momentum number--you'd need some sort of interaction to take place to change that (conservation of angular momentum). $m$ on the other hand is not well defined from angular momentum alone, so this can change without an outside interaction. – ALB Nov 28 '13 at 21:27
• Thanks.I thing i have one more problem. [$L_z$,$L_+$] =${\hbar} L_+$, let $L_zf$=m ${\hbar}f$ Where f is a eigenstate. $$L_z(L_+f)= (L_zL_+-L_+L_z+L_+L_z)f$$ $$L_z(L_+f)={\hbar} L_+f+ (L_+L_z)f$$ $$L_z(L_+f)={\hbar} (m+1)(L_+f)$$ Then my question is : how should I write next? $$L_z(L_+f)=L_+{\hbar} ((m+1)f)$$ Because ((m+1)f) tell us that the m of eigenstate increase by one like $$L_+ Y_{2}^{1}= AY_{2}^{2},$$how to write in this form? – Outrageous Nov 29 '13 at 3:13
• These are typically easier to do the other way: evaluate $[L_z,L_{+}]f = L_zL_{+}f-L_{+}L_zf$. Note that the first term has an $m+1$ coefficient due to the $L_z$ action on a raised state and the second only an $m$, so these cancel to give the desired result, $\hbar L_{+}$. Also, not to be petty, but if you like my answer you are happily encouraged to 'accept' it ;) – ALB Nov 29 '13 at 10:58 | 932 | 3,040 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-16 | latest | en | 0.872313 |
https://oeis.org/A019568 | 1,618,503,207,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038087714.38/warc/CC-MAIN-20210415160727-20210415190727-00151.warc.gz | 541,450,625 | 4,822 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A019568 a(n) = smallest k >= 1 such that {1^n, 2^n, 3^n, ..., k^n} can be partitioned into two sets with equal sum. 12
2, 3, 7, 12, 16, 24, 31, 39, 47, 44, 60, 71, 79, 79, 87 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS a(n) is least integer k such that at least one signed sum of the first k n-th powers equals zero. a(n) < 2^(n+1). The partition of the set {k: 0 <= k < 2^(n+1)} into two sets A,B according to the parity of the number of 1s in the binary expansion of k, has the property that Sum_{k in A} p(k) = Sum_{k in B} p(k) for any polynomial p of degree <= n. Equivalently, if e(k) is the Thue-Morse sequence A106400, then Sum_{0 <= k < 2^m} e(k)p(k) = 0 for any polynomial p with deg(p) < m. - Pietro Majer, Mar 14 2009 REFERENCES Posting to sci.math Nov 11 1996 by fredh(AT)ix.netcom.com (Fred W. Helenius). LINKS Sean A. Irvine, Java program (github) FORMULA a(n) == 0 or 3 (mod 4) for n >= 1 - David W. Wilson, Oct 20 2005 EXAMPLE For n=1 and 2 we have: 1+2-3 = 0 (so a(1)=3), 1+4-9+16-25-36+49 = 0 (so a(2)=7). The sum of the ninth powers of 3 5 9 10 14 19 20 21 25 26 28 31 35 36 37 38 40 41 42 is half the sum of the ninth powers of 1..44, so a(9)=44. - Don Reble, Oct 21 2005 Example: the signs (+--+-++--++-+--+) in (+0)-1-8+27-64+125+216-...+3375=0 are those of the expansion of Q(x):=(1-x)(1-x^2)(1-x^4)(1-x^8) = +1-x-x^2+x^3-..+x^15. Since (1-x)^4 divides Q(x), if S is the shift operator on sequences, the operator Q(S) has the fourth discrete difference (I-S)^4 as factor, hence annihilates the sequence of cubes. - Pietro Majer, Mar 14 2009 MATHEMATICA Table[k = 1; found = False; While[s = Range[k]^n; sm = Total[s]; If[EvenQ[sm], sm = sm/2; found = MemberQ[Total /@ Subsets[s], sm]]; ! found, k++]; k, {n, 0, 4}] (* T. D. Noe, Apr 01 2014 *) CROSSREFS Cf. A240070 (partitioned into 3 sets). Sequence in context: A275374 A168249 A080140 * A128458 A066733 A049623 Adjacent sequences: A019565 A019566 A019567 * A019569 A019570 A019571 KEYWORD nonn,more AUTHOR EXTENSIONS More terms from Don Reble, Oct 21 2005 Definition simplified by Pietro Majer, Mar 15 2009 a(13)-a(14) from Sean A. Irvine, Mar 27 2019 STATUS approved
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Last modified April 15 12:02 EDT 2021. Contains 342977 sequences. (Running on oeis4.) | 958 | 2,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-17 | latest | en | 0.789026 |
https://www.studypool.com/questions/118026/what-is-the-effect-of-a-changing-gravity-on-work | 1,480,884,773,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541396.20/warc/CC-MAIN-20161202170901-00239-ip-10-31-129-80.ec2.internal.warc.gz | 1,018,319,628 | 735,660 | # Mass & Gravity Class Exercises Help
Daniel Denton
Category:
Physics
Question description
I'm working on a few physics problems to get ready for my exam. But I don't understand how exactly gravity & friction works. Can some one please explain these 2 things and help me with the following questions?
• 1. You were to carry the bag of apples to the surface of the moon, where the gravitational field is only 1.67 N/kg, how much would the apples weigh? (Do you just replace g? I'm so confused, why does apple weight different on the moon? I thought mass is supposed to be uniform. )
• 2. On a trip to Mars, you pick up an alien life form that weighs 90 N there. When you bring it back to Earth, you find that it weighs 238 N on Earth. How much mass does the alien have? How strong is the gravitational field on the surface of Mars
• 3. After creating a 50 kg ceramic sculpture, Blake needs to drag it across the concrete floor to the corner of the room. He finds that he needs to pull with more than 220 N of force to get it to slide. What is the coefficient of friction between the ceramic and concrete?
• 4. band of Viking warriors drags their 1500 kg boat across the beach to set out to sea. They find that they have to push with more than 8800 N of force to slide it across the beach. What is the coefficient of friction between the boat and the sand?
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2325 Tutors | 506 | 2,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.909973 |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-3-section-3-7-implicit-differentiation-exercises-page-164/28 | 1,620,481,253,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.7/warc/CC-MAIN-20210508121446-20210508151446-00242.warc.gz | 852,751,287 | 13,052 | University Calculus: Early Transcendentals (3rd Edition)
At $(0,-1)$: $$\frac{d^2y}{dx^2}=-\frac{1}{4}$$
$$xy+y^2=1$$ 1) Find $dy/dx$: Differentiate both sides of the equation with respect to $x$: $$y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0$$ $$y+(x+2y)\frac{dy}{dx}=0$$ Isolate the terms with $dy/dx$ into one side and solve for $dy/dx$: $$(x+2y)\frac{dy}{dx}=-y$$ $$\frac{dy}{dx}=-\frac{y}{x+2y}$$ 2) Find $d^2y/dx^2$: Differentiate the derivative $dy/dx$ as normal, using Quotient Rule: $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(-\frac{y}{x+2y}\Big)=-\frac{(x+2y)\frac{dy}{dx}-y(1+2\frac{dy}{dx})}{(x+2y)^2}$$ Now we can use the result $dy/dx=-y/(x+2y)$ from part 1) to do substitution here: $$\frac{d^2y}{dx^2}=-\frac{(x+2y)\Big(-\frac{y}{x+2y}\Big)-y\Big(1-\frac{2y}{x+2y}\Big)}{(x+2y)^2}$$ $$\frac{d^2y}{dx^2}=-\Bigg[\frac{-y-y+\frac{2y^2}{(x+2y)}}{(x+2y)^2}\Bigg]=\frac{2y-\frac{2y^2}{(x+2y)}}{(x+2y)^2}$$ $$\frac{d^2y}{dx^2}=\frac{\frac{2y(x+2y)-2y^2}{(x+2y)}}{(x+2y)^2}=\frac{2xy+2y^2}{(x+2y)^3}$$ - At $(0,-1)$: $$\frac{d^2y}{dx^2}=\frac{2\times0\times(-1)+2\times(-1)^2}{\Big(0+2\times(-1)\Big)^3}=\frac{0+2}{(-2)^3}=\frac{2}{-8}=-\frac{1}{4}$$ | 588 | 1,144 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.476725 |
https://www.mrexcel.com/board/threads/count-each-of-month.862462/ | 1,653,700,046,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00189.warc.gz | 1,058,255,087 | 16,690 | # Count each of month
#### Paywand
##### New Member
Dear all,
i hope you are doing well.
i work on an excel sheet, in their has a lot of date, so i want to count each of month which has in their.
for example if in their has each below date (15 May - 17 May ) in one cell count (3 May), because their is 2 of month May.
Date 09 January 10 April 11 January 12 September 13 January 14 December 15 May 16 March 17 January 13 January 14 December 17 May 16 March 17 January 13 January 14 December
<colgroup><col></colgroup><tbody>
</tbody>
### Excel Facts
Move date out one month or year
Use =EDATE(A2,1) for one month later. Use EDATE(A2,12) for one year later.
#### Smitty
##### Legend
You can do it with a Pivot Table using Group by Month:
Excel 2012
AB
3Row LabelsCount of Month
4Jan7
5Mar2
6Apr1
7May2
8Sep1
9Dec3
10Grand Total16
Sheet2
HTH,
#### Paywand
##### New Member
could you please let me know how to do that, but be aware their date include days and months, i want to count only month which same.
#### Smitty
##### Legend
Create a Pivot Table and put Date in the Row field & Values field, where it should be a Count. Then right-click on A4 and select Group By-->Months.
#### Paywand
##### New Member
sorry but its not able with my data, because there is a lot of date, and i want to count them in 12 cell each one for one month, and count them in own cell.
#### Smitty
##### Legend
Excel 2012
ABCD
1DateMonthCount
29-JanJanuary7
310-AprFebruary0
411-JanMarch2
512-SepApril1
613-JanMay2
714-DecJune0
815-MayJuly0
916-MarAugust0
1017-JanSeptember1
1113-JanOctober0
1214-DecNovember0
1317-MayDecember3
1416-Mar
1517-Jan
1613-Jan
1714-Dec
Sheet1
Cell Formulas
RangeFormula
D2=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=1))
D3=SUMPRODUCT(--(MONTH(A\$2:A\$100)=2))
D4=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=3))
D5=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=4))
D6=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=5))
D7=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=6))
D8=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=7))
D9=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=8))
D10=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=9))
D11=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=10))
D12=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=11))
D13=SUMPRODUCT(--(A\$2:A\$100<>""),--(MONTH(A\$2:A\$100)=12))
EDIT: oops, the formula for February should be the same as the rest.
#### Paywand
##### New Member
Thanks a lot, it's Graet ! Worked
Excel contains over 450 functions, with more added every year. That’s a huge number, so where should you start? Right here with this bundle.
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exponential growth and decay word problems worksheet answer key Exponential Growth Jun 10, 2018 · Exponential Growth and Decay Worksheet Answer Key. Yahoo is in truth a massive entertainment websites that has millions of users and thousands of videos, audio files and is second only to Google in Hello Math Teachers! Practice worksheet on using exponential functions to solve word problems with exponential growth and decay. 2,950,000 people in 2000, determine the city’s population in 2008. !Find!the! valueof!eachfunction!after!fiveyears. What percent of the substance is left after 6 hours? So let's make a little table here, to just imagine what's going on. Good Answer. Identify the growth factor and annual percent increase b. exponential growth function to model this situation. Sketch a graph of the model c. Students can write their answers on the back of their note sheet. This Blog is create for everyone, we don't collect fee. The value in dollars of a car years from now is ( ) . 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If we start with only one bacteria which can double Solving Word Problems using The Exponential Growth Model (ask Mr. A2. Sorry, incorrect The correct answer is: Got it 14 QuestionsShow answers. Exponential Functions. A. worksheet. Growth rate = 0. The initial amount is 150,000, and the rate of growth is 8%, or 0. Honors Pre-Calculus 3. e. Remember that Exponential Growth or Decay means something is increasing or decreasing an exponential rate (faster than if it were linear). Feb 15, 2020 · unit 5 worksheet 4 identify exponential growth or decay UNIT 5 WORKSHEET 4 IDENTIFY EXP GROWTH OR DECAY. (1) Bacteria can multiply at an alarming rate when each bacteria splits into two new cells, thus doubling. I would also strongly suggest doing a few of the problems from the worksheet together. 1) Which of the exponential functions below show growth and which show decay? a) b) c). the following book can be a great choice. 35 or 35% 5. Browse and Read Exponential Growth And Decay Word Problems . This exponential growth and decay word problems answers, as one of the most operating sellers here will totally be along with the best options to review Exponential Growth And Decay Word Problems Answers The Residence Time Of CO2 In The Atmosphere Is … 33 Years. Exponential Function Word Problems Exponential growth is modelled by y= y 0ekt There are four variables, the initial amount, y 0, the time t, the growth factor k, and the current amount y:You should be comfortable with nding any one of these four, given the other three. EXAMPLE Exponential Growth and Decay Problems 4 Name. Write a model for this 8. Question 1. The problems are heavy on exponential growth and decay, compound interest, and natural log. Systems that exhibit exponential growth have a constant doubling time, which is given by $$(\ln 2)/k$$. February 20, 2020 answers Compound Growth and Decay. Exponential Growth And Decay Word Problem - Displaying top 8 worksheets found for this concept. 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Exponential Growth and Decay word problems. initial amount time growth factor rate of growth (in decimal form) final amounty=a(1 +r)t WWhat You Will Learnhat You Will Learn Use and identify exponential growth and decay functions. 04)2 = ($160. Connect with Me! Use Hundreds of my Math Videos in Your Class! Follow me on Facebook! Follow me on Instagram! b) What would the population be in 2000 if the growth continues at the same rate. 01 > 1. Find a bank account balance if the account starts with $100, has an annual rate of 4%, and the money left in the account for 12 years. Write the exponential 6 items Section, Day, Solutions. 1 + r is the growth factor. A town with a population of 5,000 grows 3% per year. This is When writing up the final answer, keep in mind that the problem said that the population was in thousands. Snapshot of Students worksheet answers Exp Growth Extra Practice Worksheet (with Answer Key). See full list on onlinemath4all. Exponential Growth Factor… b = 1 + % rate (as a decimal) Exponential Decay Factor… b = 1 – % rate (as a decimal) 2. 2. Problems Answers Some of the worksheets for this concept are Growth decay word problem key, Exponential. com. PRACTICE 1. 18 Dec 2018 8 best ex growth decay images on pinterest 47 beautiful exponential growth and decay worksheet answer key 2ed qm solutionsgriffiths d j introduction to quantum mechanics 2ed exponential growth and decay word problems 25 Mar 2011 In this tutorial I will step you through how to solve problems that deal in exponential growth and decay. You should also try to understand how changing any one of these a ects the Staub s Math March 2013 from exponential growth and decay worksheet answer key , source:staubsmath. 05), (1, 5 . Sieling’s Signature Exponential Growth and Decay Word Problems Worksheet. Let us consider the following two examples. If k is positive then we will have a growth model and if k is negative then we will have a decay model. College Algebra Worksheet (2) Exponential Growth and Decay Problems If a certain quantity A is growing continuously at rate r, then A may be written as a function of time as follows A = A0ert. Continue. Scroll down the page for more examples and solutions that use the exponential growth and decay formula. dealing with linear functions only. We usually see Exponential Growth and Decay problems relating to populations, bacteria, temperature, and so on, usually as a function of time. If A is decaying continuously at rate r, then A may be written as follows A = A0e−rt. c. The general equation for exponential decay is, where the base is represented by and. linear models, and more. ] Explain that drug filtering is not linear, because the same amount of the drug is not removed during each exponential growth and decay jackson school district, infinite pre algebra kuta software llc, exponential growth amp decay algebra ii math khan academy, growth decay word problem key folsom cordova unified, 08 exponential growth and decay kuta software infinite, kutasoftware algebra 1 exponential functions part 1, exponential growth and decay homework worksheets, exponential functions date decay rate = 100 Find t when A = 70 . com About "Exponential growth and decay word problems" Exponential growth and decay word problems : To solve e xponential growth and decay word problems, we have to be aware of exponential growth and decay functions. 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If the city had . Exponential Growth and Decay Word Problems: Evaluate all answers exactly and then round when necessary. 5) , (3 , 53) 2) Year Population 1 2 10 5 20 12 30 22 40 25 Exponential Functions Growth Decay Worksheet E3 Answers. 14 Set 14 Exponential and Log Derivatives Review Problems Solutions. Exponential Growth And Decay Word Problems Answers. 20 = 0. d) e) f). Exponential Growth And Decay Word Problem - Kiddy Math Growth Decay Word Problem Key. / Alg 2 7. wordpress. ” May 15, 2018 · This kind of photograph (Exponential Growth and Decay Word Problems Worksheet 28 Pdf 2016 social Security Worksheet Unique Worksheets 48 Awesome) preceding will be labelled having:published by means of William Matthews on 2018-05-15 07:30:08. pdf (98k) Robert Trakimas, 23. Complete each statement. Exponential Growth and Decay Functions Exponential growth occurs when a quantity increases by the same factor over equal intervals of time. 4 Exponential Growth and Decay Worksheet 01 - HW Solutions Logistic Growth Notesheet 02 Completed Notes Logistic Growth Worksheet 02 Solutions Exponential and Logistic Growth Worksheet 02 - HW Solutions Mar 30, 2019 · 20 Exponential Growth and Decay Worksheet Algebra 2 – exponential functions algebra worksheet by pecktabo math exponential functions 20 problems 4 determine whether it is an exponential function given an equation 2 determine whether it is linear or exponential given a table 3 evaluate given x value 4 match the function to the graph 2 graph the exponential function describe the domain and rang Mar 20, 2020 · Linear Equation Word Problems Worksheet pdf and Answer Key. Exponential Growth Decay Answers. Exponential Functions Word Problems Worksheet Compound Interest worksheet with answer key (pdf). co All you have to do when you arrive in their primary page is either pick one of many templates they provide or Start Fresh. An answer key is included. 4 Graph an exponential function of the form f(x) = ab^x. Some of the worksheets for this concept are Exponential growth and decay word problems, Exponential growth and decay, Exponential growth and decay work, Exp growth decay word probs, Growth decay word problem key, College algebra work 2 exponential growth and decay, Word problems interest Exponential Growth and Decay Worksheet In the function: y = a(b)x, a is the y-intercept and b is the base that determines the direction of the graph and the steepness. SURVEY. (1) Bacteria can multiply at an alarming rate when each bacteria splits into two new cells doubling. 314 Growth Decay. Then, state the Domain, Range, and Y-intercept, and change of Y-values of the function. Find the balance after 6 years. Start with the basics and move up the exponential ladder to master a variety of problem-solving and application problems. It's about what you need currently. Khan Academy is a 501(c)(3) nonprofit organization. What is the annual rate of depreciation, the rate at which the car loses value? Online exponential growth/decay calculator. 21 KB). We meet the expense of you this proper as competently as easy pretension to get those all. constant. notebookMarch 17, 2017. worksheet nuclear decay. \) Mar 17, 2017 · 8. pdf Word problems worksheet. If you ally compulsion such a referred exponential function word problems with answers ebook that will allow you Word Problems. Find the value of the investment after 5 years. 1X 2. an exponential decay function to model this situation. Students begin to work with Exponential Growth & Decay in a series of math worksheets, lessons, and homework. Answers. 33(1. Find the population at the end of. Exponential Growth And Decay Word Problems Answers. ааSince then, the population has grown at an average rate of 3. Math Exponential Functions Math Worksheet Exponential Worksheets Word Problems Subject And Predicate EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS NAME: HOUR: 1. 7) Growth; 1. 4 Exponential Growth and Decay Notesheet 01 Completed Notes Growth and Decay Word Problems 01 Solutions 6. In this tutorial, learn how to turn a word problem into an exponential decay function. 5) Growth; 3 > 1. 20. A quiz and full answer keys are also provided. Find an exponential function having the given values. The general equation for exponential decay is y = General equation for exponential decay a (1 -r) t. The answer keys for both worksheets are also attached. Some of the worksheets for this concept are Growth decay word problem key Exponential growth and decay work Exponential growth and decay Pc expo growth and decay word problems Exponential word problems Exp growth decay word probs Sep 10 2020 worksheet 1 nicolet high school. This lesson is designed to help students understand the basic concepts of exponential growth and decay. Exponential Growth Equation: y = a (1 + r)t a is the initial amount r is the percent increase, written as a decimal. If no other money is added or withdrawn from the account, how much will be in the account after 10 years? Growth and Decay Exponential Decay Radioactive decay and depreciation are examples of exponential decay. Q. “Write an exponential equation for the following situations. 18. decay practice worksheet 1 free printable worksheets. Exponential Growth And Decay Word. Exponential growth occurs when k > 0, and exponential decay occurs when k < 0. Answers should only have positive exponents. The equation f(x) = 300(1. From 1990 to 1997, the number of cell phone subscribers S (in thousands) in the US can be modeled by, S = 5535. 0. 3) 300% = 3. 078 per year over a. 8% Growth or decay Introduction to Exponential Growth and Decay. Half-life is the amount of time it takes for a substance to decay to half of the original amount. blogspot. exponential growth decay practice answers key pdf download. a. 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Systems that exhibit exponential decay follow a model of the form $$y=y_0e^{−kt}. QUIZ (Level 3) Level 3 Schoology Quiz: Level 3 – Exponential Equations 4. How much is the investment worth after 5 years ? a) Exponential growth or decay: b) Identify the initial amount: c) Identify the growth/decay factor: d) Write an exponential function to model the situation: e) “Do” the problem: 2. exponential decay: a decrease whose rate is proportional to the size of the population. 15%. Exponential+Growthand+DecayWord+Problems+!!! 4. In the next worksheet, the students are given three values for the period of time from the creation of the circle to the time of the value. pptx Use the attached Power Point Presentation to teach the lesson. 439 = 43. graphing exponential functions worksheets, graphing exponential functions worksheet answers and exponential functions and equations worksheet are three of main things we will present to you based on the gallery title. Some of the worksheets for this concept are Growth decay word problem key, Honors pre calculus d1 work name exponential, Exponential growth and decay word problems algebra, Exponential growth and decay work, Pc expo growth and decay word problems, Exponential growth and decay, 6 modeling exponential growth n, Exp growth decay word probs. Notice that . r is the growth rate when r>0 or decay rate when r<0, in percent. Exponential Decay Independent Practice Worksheet. In this exponential equations worksheet, high schoolers solve 11 word problems in short answer and multiple choice format. Two word problem examples: one about a radioactive decay, and the other the exponential growth of a fast-food chain. docx Author: Growth rate = 0. Thank you very much for downloading exponential growth and decay word problems worksheet answers. A-100 (14. com is known as a truly Huge ranking research engine that may be also in part a News, social media plus a blogging Platform. What is the expected population in 2018? 1. 4 Exponential Growth and Decay Notesheet, 01, Completed Notes · Growth and Decay Word Problems, 01, Solutions · 6. Page 6/30 This math reference sheet for graphing exponential functions walks Algebra and Algebra 2 students through identifying x and y shifts, identifying the parent function, creating a table for the Exponential Growth and Decay Word Problems Task CardsIn this set of task cards, students will Worksheet to accompany Exponential Growth and Decay stations activity - intro Exponential, School Algebra, This is a set of 20 task cards, along with a student recording sheet and an answer key. Then, solve the function and get the answer! Examples, solutions, videos, activities and worksheets that are suitable for A Level Maths to help students learn how to solve exponential growth and decay word problems. docx) and fill in as a class. 708 #11-17odd, 19-21all AND A#11. Problems And. Sep 18, 2018 · Number 6 b(ii) answer in duo-tang should say Range: {y>-1} (This correction is in the full solutions) Week3. Exponential decay can be thought of as the opposite of exponential growth, i. Nov 05, 2020 · Worksheets are Exponential growth and decay word problems, Exponential growth and decay, Growth decay word problem key, Exponential growth and decay work, Exp growth decay word probs, Word problems interest growthdecay and half life, College algebra work 2 exponential growth and decay, Honors. 0. 3. x(t) = x 0 × (1 + r) t. Questions comparing simple interest to compound interest and real world questions involving half-life. The positive constant r is called either the growth rate (for x If the growth or decay is expressed using multiplication (including words like ³doubling ´ or ³halving ´) use an exponential function. Suppose we model the growth or decline of a population with the following differential equation. 14) Determine if the function is exponential gro decay. In exponential growth problems on the HSE exam, the variable—or input—is in the exponent position. Then, solve the function and get the answer! a. Worksheet Nuclear Decay. 1. 02 or 2% 3. Solve real-life problems involving exponential growth and decay. In 1985, there Microsoft Word - Word problem answer key exponential and log problems. , the Oct 22, 2018 · In exponential growth, the rate of growth is proportional to the quantity present. For this model, is the time, - is the original amount of the quantity, and , is the amount after time . Exponential Growth and Decay. A person View Homework Help - growth and decay homework key from MATH Logarithmi at Summit School, Zeeland. 1 Worksheet name < E . Compound growth and decay are an extension on percentages and are used to model real world applications such as interest, disease and population. When somebody should go to the ebook stores, search launch by shop, shelf by shelf, it is really Word Problems. Often exponential rate of decay can be determined from the half-life information. • Three posters showing a Student Recording Sheet• Answer KeyThe types of word problems included are: exponential Literal Equations Worksheet. 5% annual decrease in population. Decay Practice Answer Key Exponential Decay Worksheets Math Worksheets Land. These problems The way the problem is worded, 1994 is what we call our initial year. pdf 35. The lesson assumes that the in the presentation. 6) , (0 2. Exponential decay is a type of exponential function where instead of having a variable in the base of the function, it is in the exponent. Plus model problems explained step by step Exponential word problems almost always work off the growth / decay formula, A = Pe rt, where " A " is the ending amount of whatever you're dealing with (money, bacteria growing in a petri dish, radioactive decay of an element highlighting your X-ray), " P " is the beginning amount of that same "whatever", " r " is the growth or decay rate, and " t " is time. Two Types of Exponential Functions: 1. Exponential Growth and Decay Worksheet - Solutions. Electron Configuration Practice Worksheet Answer Key. If we start with only one Results 1 - 24 of 32 PDF (92. exponential growth, p. Sieling for login info) An explanation of the formula for exponential equations involving percents 3. _____ Exponential Growth/Decay Worksheet Answer the following questions about the exponential decay problems. 3) A population of 800 beetles is growing each month at a rate of 5%. +250% Growth or decay e. exponential growth and decay word problems is the PDF of the . 4 Exponential Growth and Decay Worksheet, 01 - HW, Solutions · Logistic Growth c) What is the rate of growth or decay? Choose: 15%. Exponential Word Problems. +0. 2% each year. 94-9. Write an equation for each situation and answer the question. Have the following problems written on the board. An Exponential Growth and Decay Worksheet Answer Key are a good idea for students who have problems following directions. Apr 14, 2018 · exponential growth & decay math worksheets center students begin to work with exponential growth & decay in a series of math worksheets lessons and homework a quiz and full answer keys are also provided bowerpower answers a resource for mr bower s classes at huntington north high school algebra i online textbook username see me here are the #26 Writing exponential growth and decay equations #26 homework exp growth &decay KEY- Mar 20 2017 - 4-06 PM. Exponential Word Problem. Some of the worksheets for this concept are Growth decay word problem key, Exponential growth and decay work, Exponential growth and decay, Pc expo growth and decay word problems, Exponential 1. Some of the worksheets for this concept are Exponential growth and decay word problems, Exponential growth and decay, Exponential growth and decay work, Exp growth decay word probs, Growth decay word Exponential Graphs Review: Exponential Growth & Decay NOTES *Any quantity that grows or decays by a fixed percent at regular intervals is said to possess exponential growth or exponential decay. (Try to do this without the calculator!) 1. 162 = 16. The number / is a constant that is determined by the rate of growth. 1) For a period of time, an island's population grows at a rate proportional to its population. . Suppose a culture of bacteria begins with 5000 cells and dies by 30% each year. 6 Dec 2016 This algebra and precalculus video tutorial explains how to solve exponential growth and decay word problems. Enter the initial Exponential growth and decay puzzle answer key. 31 scaffolded questions on. Because you want to supply programs within a reputable and reputable reference, all of us found beneficial info on numerous subject areas along with topics. 20 scaffolded questions that start relatively easy and end with some real challenges. Exponential Growth B. 1) meaning. Percentages revision. 06. This worksheet provides practice simplifying exponential expressions with a base of e as well as modelling and solving problems with natural base exponential growth and decay functions, specifically A=Pe^(rt). Students write and solve exponential growth and decay problems given a word problem. Practice Worksheet. com Unit: Exponential Functions Date Homework Hour Graphing Exponential Functions Worksheet #2 Directions : Answer all questions. They must multiply all three numbers together and then find the result. Time Series Analysis For Business Forecasting. A person Hand out copies of the Exponential Growth and Decay Notes (M-A2-4-3_Exponential Growth_Decay Notes and KEY. Make sure you are happy with the following topics before continuing. If /0 , the model represents exponential growth, and if /1 , it 17) . docx A2. Jan 29, 2018 - Set of 28 Task Cards with exponential growth and decay word problems. Exponential Growth/Decay Worksheet Answer the following questions about the exponential decay problems. Correct! Correct! Jumping to next level. Rearranging equations Half-life is the amount of time it takes for a substance to decay to half of the original amount. decay practice worksheet 1 answer key shmups de. Adi Purwanto. 06) x is being used to calculate the amount of money in a savings account. Wonderful Exponential Growth and Decay Word Problems Worksheet Inspirational Worksheet Quadratic formula . x(t) is the value at time t. ' An example of an exponential decay word problem is the following: 'The value of a new 35,000 car Graphing Exponential Functions Practice Worksheet Name Period # Graph the following functions and tell whether they show exponential growth or decay. Most likely you have knowledge that, people have look numerous time for their favorite books like this exponential growth and decay word problems worksheet answers, but stop going on in harmful downloads. Exponential Growth Decay Practice Answers Key PDF Download. a) Write an equation that expresses the number of beetles at time x. The number of subscribers increased by 75% per year after 1985. Improve your skills with free problems in 'Exponential growth and decay: word problems' and thousands of other practice lessons. Objectives: Students will be able to model word problems with exponential functions and use logs to solve exponential models. Activity 6 Radioactive Decay Chain US EPA. In real-life situations we use x as time and try to find out how things change exponentially over time. Answer key included. Then, = => ln(y) = . 3 Explain and use the laws of fractional and negative exponents, understand exponential functions, and use these functions in problems involving exponential growth and decay. The annual sales at a company are 372,000 in the A function of the form y=a(1 +r)t, where a> 0 and r> 0, is an exponential growth function. Hence, = and setting we have . b determines how fast the (Alg2) S12: Exponential Growth and Decay Word Problems Examine the data on COVID19 and create a poster (virtual or physical) that summarizes what you believe is the key information Khan Academy Practice (answers included). Let's solve a few exponential growth and decay problems. 12 KB (Last Modified on December 19, 2019) Comments (-1) Exponential Growth And Decay Word Problem. 7 Spiral Review 2014. Improve your math knowledge with free questions in "Exponential growth and decay: word problems" and thousands of other math skills. Circle either linear or exponential. 015} For each percentage rate of increase or decrease, find the corresponding growth or decay factor d. Exponential Growth And Decay Worksheet Answer Key using Instructive Contents. It is especially useful for Results 1 - 7 of 7 Browse exponential growth and decay word problems resources on Teachers Pay Teachers, a marketplace trusted by millions of teachers for original educational This self checking worksheet contains 10 word problems dealing with the equations y = a(1+r)^t and y = a(1-r)^t. y = a(1 + r)t. Submit. • Student Recording Sheet• Answer KeyThe types of word problems included are: exponential. You deposit 1500 in an account that pays 5% interest compounded yearly. y =8•(12) x X Y -1 0 1 Worksheet Template : Top 20 Exponential Functions Growth And Decay Worksheet For Students HELAENE Exponential Functions Growth And Decay Worksheet. Exponential Growth and Decay The mathematical model for exponential growth or decay is given by ,-. SOLUTION a. From 2000 - 2010 a city had a 2. Free Algebra 2 worksheets created with Infinite Algebra 2. Word Problems Practice Worksheet. 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Three posters showing a comparison of exponential growth and decay, formulas, and a comparison of linear increase vs. Exponential Functions: General Form Type General Graph a-value b-value for word problems: Two Major Types for word problems: For each of the following equations, write whether it is exponential growth or decay and then write the y-intercept. Exponential Growth And Decay Word Problems Answers - Displaying top 8 worksheets found for this concept. 2 Feb 2018 Distance rate and time word problems these algebra 1 equations worksheets will produce distance rate and time word problems with ten Exponential growth and decay word problems. Additional Practice of Course Standards. 08. notebookMarch 17, 2017 Exponential Decay Equation: y = a (1 ‐ r)t a is the initial amount r is the percent decrease, written as a decimal 1 ‐ r is the decay factor EXAMPLE 5: Free worksheet(pdf) and answer key on Compound interest. Students often need a hand-holding guide to help them with their homework assignments, especially if they have been struggling with the material that has been assigned to them. 8% per Exponential Growth and Decay Exponential decay refers to an amount of substance decreasing exponentially. 1000 1000 = 30 Microsoft Word - Word problem answer key exponential and log problems. Some of the worksheets displayed are Name algebra 1b date linear exponential continued, Exponential growth and decay word problems, Concept 17 write exponential equations, Exponential word problems, Exponential function word problems, Exponential equations not requiring logarithms, Solving Lesson 20: Exponential Growth and Decay. Depending on what your instructor wants, you can usually just start with the equation \(y = Ae^{kt}$$, if you know that you have an exponential decay or growth problem. Created: 10/21/2009 14 Nov 2016 The population of Johnson City in 1995 was. The price of a car that was bought Exponential Growth and Decay Word Problems. net Modeling Exponential Growth Decay Interspersed with a Reform Rant from Exponential Growth And Decay Worksheet, source:educationrealist. 3 - Exponential Growth and Decay. RPDP. t is the time in discrete intervals and selected time units. So this first problem, suppose a radioactive substance decays at a rate of 3. A function of the form f(x) = ab" is Thank you very much for downloading exponential growth and decay word problems worksheet answers. There are no stretches or shrinks. Showing top 8 worksheets in the category - Exponential Word Problems. Here is a quick video explaining this and showing a graph to give you a feel for these equations. A new 2006 Honda Accord was valued at $25000. Rather than enjoying a fine book in Jan 30, 2019 · Word Problems Age Worksheet Save More Word Equations Worksheet from exponential growth and decay word problems worksheet answers , source:curiousmind. If the growth rate is 3. Worksheets are Exponential growth and decay word problems, Exponential growth and decay, Growth decay word problem key, Exponential growth and decay work, Exp growth decay word probs, Word problems interest growthdecay and half life, College algebra work 2 exponential growth and decay, Honors pre calculus d1 work name exponential. Explore the graph of the exponential growth or decay function. worksheet exponential growth and decay exponential. Write a logistic growth function given the y-intercept, both horizontal asymptotes, and another point. 10 years. That is, the rate of growth is proportional to the amount present. The mice population is 25,000 and is decreasing by 20% each year. An example of an exponential growth word problem is the following: '$1000 is invested at 9% interest compounded annually. 5% per hour. In 1985, there were 285 cell phone subscribers in the small town of Centerville. Write an exponential growth model for P, the population, after t years, where t = 0 represents the year 2000. 16) Answer the following questions about the functi f(x) = - Is this growth or decay? How do you knowú) - What is the grovâh or decay facto i - What is the growth or decay r '7 t 18) Write an exponential function to model the situation. Explain how to solve word problems dealing with exponential growth and decay. EX #3:A slow economy caused a company’s annual revenues to drop from $530, 000 in 2008 to$386,000 in 2010. Colonization Atomic Rockets. Growth Decay Word Problem Key. Let's do a couple of word problems dealing with exponential growth and decay. What is the rate of growth or rate of decay Exponential Growth And Decay Word Problems Answers Some of the worksheets for this concept are Growth decay word problem key, Exponential growth and decay work, Exponential growth and Exponential Growth and Decay Word Problems - Digital BOOM Cards + Printable Task Cards. Exponential Decay B. 08 to the 8th power can easily be solved Basic Equation Statements: Determine growth/ decay, the percent of change, and initial value. Structure Worksheet. Hand out Exponential Growth and Decay Worksheet 1 and Exponential Growth and Decay Worksheet 2 for additional practice. 2 Exponential Decay 8. Showing top 8 worksheets in the category - Population Growth Problems. It is not in this area the costs. Exponential Growth and Decay Worksheet Answer Key with Need Math Help Grade 11 Mathematics Tario Canada. Write an equation to represent the population of Johnson City Since 1995 b. Round your answers to the nearest whole number. 120 seconds. 5x show exponential growth or exponential decay? Word Problems Relating to Exponential Growth and Decay . Suppose you deposit \$3000 in a savings account that pays interest at an annual rate of 4%. Read Online Exponential Growth And Decay Word Problems Answersand decay word problems answers that we will agreed offer. 1200 C. 413) t where t is number of years since 1990 a. Exponential growth and decay are some of the real world problems from Exponential Growth And Decay Worksheet If something decreases in value at a constant rate, you may have exponential decay on your hands. Interpret and rewrite exponential growth and decay functions. 2 Part 3 Exponential Growth and Decay Word Problems (work). 9%. 5575 C. exponential growth and exponential decay functions? Solve real-life problems involving exponential growth and decay. 19 Dec 2019 Some of the worksheets for this concept are Exponential growth and decay word problems, Exponential growth and decay, Exponential growth and decay work, Exp growth decay word probs, Growth decay word problem key, Exponential Growth and Decay Word Problems. 13 The variable x represents the number of times the growth/decay factor is multiplied. U4D8_T Applications – Exponential Growth and Decay Problems: Duo-tang Worksheet for U4D8 Applications Exponentials (Exponential Growth & Decay) include calculating compound interest or population growth. Assign homework problems from board, worksheet or book. Identifying functions as models of exponential growth and exponential decay, graphing exponential functions using the concept of Video Tutorial (Khan) : Exponential word problems. Intuitive Understanding Of Euler’s Formula – BetterExplained. online download exponential growth and decay word problems worksheet answers Exponential Growth C. Use and identify exponential growth and decay functions. Exponential growth calculator. Word Problem Solving- Exponential Growth and Decay. Download and Read Exponential Growth And Decay Word Problems Worksheet Answers Exponential Growth And Decay Word Problems Worksheet Answers It's coming again, the new . In real-world applications, we need to model the behavior of a function. • y represents the fi nal amount. 5) f (x) = 3 Answers to 4. The word problems in this lesson cover exponential growth and decay. Solve problems involving radioactive decay, carbon dating, and half life. 13 = 1. Write an equation that represents this situation. The equation will look like: f(x) = ( starting amount ) (base )x. The following diagram shows the exponential growth and decay formula. Write!an!exponential!function!to!model!each!situation. 8), (2 , 18. Exponential Growth And Decay Worksheet Pecktabo Math 2015 Answer Key Some of the worksheets for this concept are Growth decay word problem key, Exponential growth and decay work, Exponential growth and decay, Pc expo growth and decay word problems, Exponential word problems, Exp growth decay word probs. They grow so fast in both directions that it is hard to plot more than a few points. Decide whether the word problem represents a linear or exponential function. Exponential Decay - decreasing y = a(1 - r)x When solving problems involving exponential growth/decay: y = a(1 + r)x - Identify the starting value (a) and the growth/decay factor (r) (the number used to multiply). Materials: Hw #9-5 answers overhead; quiz #2; pair work and answer overhead; board collaborations; hw #9-6 Time Activity 5 min Check Homework Put the answers to hw #9-5 on the overhead. Write an exponential growth or decay model f ()xab x and use it to answer questions in context. Quiz amp Worksheet Radioactive Decay Study com. How much will be left after 6 days, if you start with 96 grams of it? grams. Exponential_Growth_and_Decay. Vary the initial amount and the rate of growth or decay and investigate the changes to the graph. Exponential decay and exponential growth are used in carbon dating and other real-life applications. The half-life of a radioactive kind of neptunium is 2 days. Displaying top 8 worksheets found for - Growth Decay. more. ! Exponential Growth And Decay Word Problem - Displaying top 8 worksheets found for this concept. You need to understand how to project cash flow. This Exponential Growth and Decay Handouts & Reference is suitable for 9th - 12th Grade. Slope Calculator Basic Mathematics Com. When a quantity grows by a fixed percent at regular intervals, the pattern can be represented by the functions, Growth: y = Decay: Y = (70 — r) x a x Mar 19, 2018 · Exponential FUNctions - Growth and Decay Today we started exponential functions and I thought I'd share my notes, activity and next day warm-up with you. Yes, long, long, LONG multiplication, for example 1. 5 D1 Worksheet Name_____ Exponential Growth and Decay Exponential Growth: Exponential Decay: Compound Interest: Compound Interest Continuously: 1. In other words, $$y′=ky$$. Worksheets are Exponential growth and decay word problems, Name algebra 1b Exponential Functions Word Problems Worksheets - Lesson Word Problems with One of the key pieces that students need to understand is the concept of 100% (a rate of. Talking related with Exponential Functions Worksheet with Answers, below we can see some related pictures to give you more ideas. 6 Word Problems Modeling Exponential growth and decay with answer key. 7-11 and homeschool; Product Type: BOOM and Task Cards; File Type: PDF; Pages: 21; Answer Key: Yes; Instagram: @You_Learning_Math. 3)x. Decay Practice Worksheet 1 Answer Key Play this game to review Algebra I. 5 days ago Exponential growth and decay worksheets dsoftschools exponential growth and decay word problem worksheets Problems With Solutions Represent The Given Function As Exponential Growth Or Exponential Decay Word Problems Exp Growth Decay Word Probs Growth Decay Word Problem Key College Algebra Work 2 Exponential Growth And Decay Word Problems Interest Exponential Decay. Parallel Lines Worksheet: Classwork included problems with systems. Jul 15, 2018 · Exponential Growth And Decay Problems Worksheet Free Worksheets from Exponential Growth And Decay Worksheet, source:comprar-en-internet. Some of the worksheets for this concept are Exponential growth and decay work, Exponential growth and decay, Exponential growth and decay word problems, Honors pre calculus d1 work name exponential, Growth decay word problem key, Exponential growth and decay, Exponential growth and decay functions, Word problems interest If something decreases in value at a constant rate, you may have exponential decay on your hands. We will conclude this section with some exponential decay applications. Hour 3 : U4D8_S Applications – Exponential Growth and Decay Problems. 1 (below) # 3 – 6 Tuesday 10/29 Application problems cont’d Quiz – Writing sine and cosine functions Word problem worksheet p. 4 Exponential Growth and Decay Worksheet 01 - HW Solutions 03 Logistic Growth Notesheet 02 Completed Notes 04 Logistic Growth Worksheet 02 Solutions 05 Exponential and Logistic Growth Worksheet 02 - HW Solutions Algebra Exponential Growth and Decay Word Problems 3 of 7 from Exponential Growth And Decay Worksheet, source: youtube. Solving Q. Exponential Growth and Decay Practice. Subjects: This document may be used with interactive direct teaching, as a worksheet, or as a review. 9 2 practice exponential decay flippedmath com. What was its value in 2009? 2. I would like to preface this post with today is the Monday after Spring Break, most students (and teachers :-) ) seemed to have forgotten the basics. 2%. Exponential Growth - increasingy = a(1 + r)x 2. exponential growth and decay word problems worksheet answer key
wyg, yg4v, uzuw, luz5, zf0d, ojj, al7, up8, adceb, sex, nki, av, etp, ixxy, ownf, | 10,187 | 46,750 | {"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} | 4.125 | 4 | CC-MAIN-2021-17 | latest | en | 0.899883 |
https://poletoparis.com/what-is-diminishing-marginal-product-of-labor/ | 1,702,287,051,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00693.warc.gz | 493,646,942 | 9,560 | ## What is diminishing marginal product of labor?
Diminishing marginal productivity is the concept that using increasing amount of some inputs (variable inputs) during the production period while holding other inputs constant (fixed inputs) will eventually lead to decreasing productivity.
At what point does diminishing marginal returns occur?
The point of diminishing returns refers to a point after the optimal level of capacity is reached, where every added unit of production results in a smaller increase in output.
What causes diminishing marginal productivity?
What is Diminishing Marginal Productivity? The Law of Diminishing Marginal Product is an economics concept. It says that, at early stages of production, if we increase 1 production variable and the rest of the things remain the same, the product total production may increase.
### What happens when there is diminishing marginal productivity?
An economic rule governing production which holds that if more variable input units are used along with a certain amount of fixed inputs, the overall output might grow at a faster rate initially, then at a steady rate, but ultimately, it will grow at a declining rate.
What are diminishing marginal returns of labor quizlet?
Diminishing marginal returns occur when the marginal product of an additional worker is less than the marginal product of the previous worker.
What is the impact of diminishing marginal returns on labor?
When there are diminishing marginal returns, adding more workers increases total output but at a decreasing rate. A firm with diminishing marginal returns of labor will produce less and less output from each additional unit of labor added.
## What is the law of diminishing marginal?
The law of diminishing marginal utility says that the marginal utility from each additional unit declines as consumption increases. 1. The marginal utility can decline into negative utility, as it may become entirely unfavorable to consume another unit of any product.
How do you find the marginal product of labour?
The marginal product of labor is calculated by dividing the change in output divided by the change in labor, given that all else is equal. For example, if output increased by 20 and labor increased by 2, MPL = 20 / 2 = 10.
What is the law of diminishing marginal product answer?
The law of diminishing marginal product or productivity is an economic theory. It proclaims that increasing one input constant and maintaining other inputs constant helps in increasing the output initially.
### When there are diminishing marginal returns the marginal product is quizlet?
The law of diminishing marginal returns states that as a firm uses more of a variable factor of production with a given quantity of the fixed factor of production, the marginal product of the variable factor eventually diminishes. | 520 | 2,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.909462 |
http://warp.povusers.org/EfficientLZW/part1.html | 1,637,969,317,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00335.warc.gz | 79,093,121 | 2,355 | # An efficient LZW implementation
## 1. The basic LZW encoding algorithm
### Glossary
The following terms are used in this text:
Byte
The fundamental input data element, having possible values between 0 and 255. An arbitrary byte will be denoted with `B`. A specific value of a byte will be denoted by enclosing the value in parentheses, for example `(25)`.
String
One or more continuous bytes.
Prefix
All but the last byte in a string. A prefix will be denoted with `[prefix]`, and thus an arbitrary string will be denoted with `[prefix]B`. A prefix can be empty (which then means that the string has only one byte). An empty prefix is denoted with `[empty]`.
Dictionary
An array of strings (ie. each element of the array contains a string).
Index
The position of a given string in the dictionary. The first index is 0, the next one is 1, etc. An arbitrary index value will be denoted with `<index>`.
Assignment.
### The dictionary
The dictionary is an array of strings, or in other words, prefix-byte pairs. Each string in the dictionary is unique, ie. no string appears in the dictionary twice.
The first 256 elements in the dictionary consist of pairs of empty prefixes and the byte values corresponding to their index in the dictionary. In other words, the first element of the dictionary is `[empty](0)`, the next one is `[empty](1)` and so on, up to `[empty](255)`. (When optimizing the algorithm the dictionary can be initialized with less entries if the input data uses less than 256 byte values, but for the sake of simplicity we'll assume in this introductory section that all 256 values are used.)
All the new strings added to the dictionary will be added to index positions from 256 upwards. Each new string is added to the next unused position in the dictionary.
### The basic LZW algorithm
This is the basic LZW algorithm in pseudocode:
1. Initialize the dictionary (with the first 256 entries).
2. `[prefix]``[empty]`
3. `B` ← next byte in the input.
4. Is the string `[prefix]B` in the dictionary?
• Yes:
1. `[prefix]``[prefix]B`
• No:
1. Add the string `[prefix]B` to the dictionary.
2. Output the index of `[prefix]` to the result.
3. `[prefix]``B`
5. If there are bytes left in the input, jump to step 3.
6. Else output the index of `[prefix]` to the result.
### Resetting the dictionary
Obviously the dictionary cannot grow forever (or else we would at some point run out of memory with very large inputs). For this reason we have to set a maximum size for the dictionary. In practice (as we will see in part 3) we should set a maximum bitsize for the index values of the dictionary (which thus automatically limits the maximum size of the dictionary).
For example, if we set this maximum bitsize to 16 that means that the maximum number of elements in the dictionary will be 65536.
Obviously this maximum bitsize must be larger than the minimum bitsize which, in this introduction, was 8 bits.
What happens when the dictionary gets full? What we have to do is that immediately when the dictionary gets full (ie. a string is stored in the last possible location of the dictionary) we just reset the dictionary and start over (or, in other words, jump to step 1 in the pseudocode above). | 750 | 3,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-49 | longest | en | 0.761724 |
https://www.traditionaloven.com/tutorials/surface-area/convert-sq-in-square-inch-to-ft-diam-circle.html | 1,718,266,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00532.warc.gz | 960,985,035 | 17,363 | Convert sq in, in2 to ∅ 1ft | square inch to ∅ one foot circles
# area surface units conversion
## Amount: 1 square inch (sq in, in2) of area Equals: 0.0088 ∅ one foot circles (∅ 1ft) in area
Converting square inch to ∅ one foot circles value in the area surface units scale.
TOGGLE : from ∅ one foot circles into square inches in the other way around.
## area surface from square inch to circle one foot diameter conversion results
### Enter a new square inch 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 ∅ one foot circles are in 1 square inch? The answer is: 1 sq in, in2 equals 0.0088 ∅ 1ft
## 0.0088 ∅ 1ft is converted to 1 of what?
The ∅ one foot circles unit number 0.0088 ∅ 1ft converts to 1 sq in, in2, one square inch. It is the EQUAL area value of 1 square inch but in the ∅ one foot circles area unit alternative.
sq in, in2/∅ 1ft area surface conversion result From Symbol Equals Result Symbol 1 sq in, in2 = 0.0088 ∅ 1ft
## Conversion chart - square inches to ∅ one foot circles
1 square inch to ∅ one foot circles = 0.0088 ∅ 1ft
2 square inches to ∅ one foot circles = 0.018 ∅ 1ft
3 square inches to ∅ one foot circles = 0.027 ∅ 1ft
4 square inches to ∅ one foot circles = 0.035 ∅ 1ft
5 square inches to ∅ one foot circles = 0.044 ∅ 1ft
6 square inches to ∅ one foot circles = 0.053 ∅ 1ft
7 square inches to ∅ one foot circles = 0.062 ∅ 1ft
8 square inches to ∅ one foot circles = 0.071 ∅ 1ft
9 square inches to ∅ one foot circles = 0.080 ∅ 1ft
10 square inches to ∅ one foot circles = 0.088 ∅ 1ft
11 square inches to ∅ one foot circles = 0.097 ∅ 1ft
12 square inches to ∅ one foot circles = 0.11 ∅ 1ft
13 square inches to ∅ one foot circles = 0.11 ∅ 1ft
14 square inches to ∅ one foot circles = 0.12 ∅ 1ft
15 square inches to ∅ one foot circles = 0.13 ∅ 1ft
Convert area surface of square inch (sq in, in2) and ∅ one foot circles (∅ 1ft) units in reverse from ∅ one foot circles into square inches.
## Area units calculator
Main area or surface units converter page.
# Converter type: area surface units
First unit: square inch (sq in, in2) is used for measuring area.
Second: circle one foot diameter (∅ 1ft) is unit of area.
QUESTION:
15 sq in, in2 = ? ∅ 1ft
15 sq in, in2 = 0.13 ∅ 1ft
Abbreviation, or prefix, for square inch is:
sq in, in2
Abbreviation for circle one foot diameter is:
∅ 1ft
## 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 square inches and ∅ one foot circles ( sq in, in2 vs. ∅ 1ft ) measures exchange.
2. for conversion factors between unit pairs.
3. work with area surface's values and properties.
To link to this area surface square inch to ∅ one foot circles online converter simply cut and paste the following.
The link to this tool will appear as: area surface from square inch (sq in, in2) to ∅ one foot circles (∅ 1ft) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 979 | 3,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.754319 |
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# Lab Manuals of Electronic Workshop Practice by Upesh Patel
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EXPERIMENT NO .1 AIM :- Introduction to active and passive electronic components and accessories APPARATUS:- Resistors,Capacitors,Diodes,Transistors. THEORY:- Electronic Components are categorized into Passive components:- Resistors,Capacitors, Active components :- Diodes,Transistors 1) RESISTOR -A resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element. -Resistance:A measure of the degree to which an object opposes an electric current through it. -Resistance can be evaluated as, R=V/I Symbol Of Resistor is shown below Value Identification Of Resistor using resistor colour code The resistance value, tolerance, and wattage rating are generally printed onto the body of the resistor as numbers or letters when the resistors body is big enough to read the print, such as large power resistors. But when the resistor is small such as a 1/4W carbon or film type, these specifications must be shown in some other manner as the print would be too small to read. So to overcome this, small resistors use coloured painted bands to indicate both their resistive value and their tolerance with the physical size of the resistor indicating its wattage rating. These coloured painted bands produce a system of identification generally known as a Resistors Colour Code.
#### Text from page-2
Calculating Resistor Values The Resistor Colour Code system is all well and good but we need to understand how to apply it in order to get the correct value of the resistor. The "left-hand" or the most significant coloured band is the band which is nearest to a connecting lead with the colour coded bands being read from left-to-right as follows; Digit, Digit, Multiplier = Colour, Colour x 10 colour in Ohm's (Ω's) For example, a resistor has the following coloured markings; Yellow Violet Red = 4 7 2 = 4 7 x 102 = 4700Ω or 4k7. The fourth and fifth bands are used to determine the percentage tolerance of the resistor. Resistor tolerance is a measure of the resistors variation from the specified resistive value and is a consequence of the manufacturing process and is expressed as a percentage of its "nominal" or preferred value. The Different Types of Resistors There are many different Types of Resistors available to the electronics constructor, from very small surface mount chip resistors up to large wirewound power resistors. The principal job of a resistor within an electrical or electronic circuit is to "resist"
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All modern fixed value resistors can be classified into four broad groups; • Carbon Composition Resistor - Made of carbon dust or graphite paste, low wattage values • Film or Cermet Resistor - Made from conductive metal oxide paste, very low wattage values • Wire-wound Resistor - Metallic bodies for heatsink mounting, very high wattage ratings 2) CAPACITOR - A capacitor (originally known as condenser) is a passive two-terminal electrical component used to store energy in an electric field. - Capacitance is the ability of a body to store an electrical charge. - Capacitance can be evaluated as C=Q/V
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- Symbol of Capacitance is shown below Final equation for the capacitance of a capacitor as: C = εA/d Where ε=permittivity of dielectric material A=cross section area of conducting plate d =distance between two plates The job of a capacitor is to store charge onto its plates. The amount of electrical charge that a capacitor can store on its plates is known as its Capacitance value and depends upon three main factors. • The surface area, A of the two conductive plates which make up the capacitor, the larger the area the greater the capacitance. • The distance, d between the two plates, the smaller the distance the greater the capacitance. • The type of material which separates the two plates called the "dielectric", the higher the permittivity of the dielectric the greater the capacitance. Types of Capacitor There are a very, very large variety of different types of capacitor available in the market place and each one has its own set of characteristics and applications Dielectric Capacitor Dielectric Capacitors are usually of the variable type were a continuous variation of capacitance is required for tuning transmitters, receivers and transistor radios. Variable dielectric capacitors are multi-plate air-spaced types that have a set of fixed plates (the stator vanes) and a set of movable plates (the rotor vanes) which move in between the fixed plates. | 998 | 4,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-51 | latest | en | 0.906372 |
https://folu.me/post/wnzrfusvfure-d-dpbz/2024/04/02/automatic-differentiation-with-dual-numbers | 1,713,004,823,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00569.warc.gz | 239,542,278 | 40,070 | # Automatic differentiation with dual numbers
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Style Pass
2024-04-03 16:00:05
Differentiation is the heart of most machine learning, but how can we differentiate arbitrary functions? Perhaps the simplest accurate method is using dual numbers.
Now we want to ask: how does tweaking x = 3 change the output? In math-speak, what’s the derivative of the distance with respect to x?
The poor man’s way to answer this is numerical differentiation. We add a little bit to x, and see how much it changes the output:
We get that derivative = 0.5999999608263806. That’s 0.6. Well ... almost. The numerical error is due to our changeToInput = 0.00000001 not being infinitesimally small.
We’ll start by saying that ε, or epsilon, is a special number that’s infinitesimally small. More precisely: it’s not so small as to be zero, but it’s so small that when you square it, you get zero.
What is 42 + 7ε? Well, because ε is a different kind of number, we can’t simplify this expression, so we just leave it as 42 + 7ε. | 262 | 1,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.914817 |
https://morioh.com/p/31544bfcab20 | 1,656,487,522,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00288.warc.gz | 448,549,456 | 22,854 | 1596631020
# Sum of prime numbers in range [L, R] from given Array for Q queries
Given an array arr[] of the size of N followed by an array of Q queries, of the following two types:
• Query Type 1: Given two integers L and R, find the sum of prime elements from index L to R where 0 <= L <= R <= N-1.
• Query Type 2: Given two integers i and X, change arr[i] = X where 0 <= i <= n-1.
Note:_ Every first index of the subquery determines the type of query to be answered._
**Example: **
_Input: _arr[] = {1, 3, 5, 7, 9, 11}, Q = { { 1, 1, 3}, {2, 1, 10}, {1, 1, 3 } }
_Output: _
15
12
_Explanation: _
First query is of type 1, so answer is (3 + 5 + 7), = 15
Second query is of type 2, so arr[1] = 10
Third query is of type 1, where arr[1] = 10, which is not prime hence answer is (5 + 7) = 12
Input:_ arr[] = {1, 2, 35, 7, 14, 11}, Q = { {2, 4, 3}, {1, 4, 5 } }_
Output:_ 14_
Explanation:
First query is of type 2, So update arr[4] = 3
Second query is of type 1, since arr[4] = 3, which is prime. So answer is (3 + 11) = 14
**Naive Approach: **The idea is to iterate for each query between L to R and perform the required operation on the given array.
_Time Complexity: _O(Q * N * (O(sqrt(max(arr[i]))
**Approach: ** To optimize the problem use Segment tree and Sieve Of Eratosthenes.
• First, create a boolean array that will mark the prime numbers.
• Now while making the segment tree only add those array elements as leaf nodes which are prime.
• C++
• Python3
`// C++ program for the above approach`
`#include <bits/stdc++.h>`
`**using**` `**namespace**` `std;`
`**int**` `**const**` `MAX = 1000001;`
`**bool**` `prime[MAX];`
`// Function to find the prime numbers`
`**void**` `SieveOfEratosthenes()`
`{`
`// Create a boolean array prime[]`
`// and initialize all entries it as true`
`// A value in prime[i] will`
`// finally be false if i is Not a prime`
`**memset**``(prime,` `**true**``,` `**sizeof**``(prime));`
`**for**` `(``**int**` `p = 2; p * p <= MAX; p++) {`
`// Check if prime[p] is not`
`// changed, then it is a prime`
`**if**` `(prime[p] ==` `**true**``) {`
`// Update all multiples of p`
`// greater than or equal to`
`// the square of it numbers`
`// which are multiple of p`
`// and are less than p^2 are`
`// already been marked`
`**for**` `(``**int**` `i = p * p; i <= MAX; i += p)`
`prime[i] =` `**false**``;`
`}`
`}`
`}`
`// Function to get the middle`
`// index from corner indexes`
`**int**` `getMid(``**int**` `s,` `**int**` `e)`
`{`
`**return**` `s + (e - s) / 2;`
`}`
`// Function to get the sum of`
`// values in the given range`
`// of the array`
`**int**` `getSumUtil(``**int**``* st,` `**int**` `ss,`
`**int**` `se,` `**int**` `qs,`
`**int**` `qe,` `**int**` `si)`
`{`
`// If segment of this node is a`
`// part of given range, then`
`// return the sum of the segment`
`**if**` `(qs <= ss && qe >= se)`
`**return**` `st[si];`
`// If segment of this node is`
`// outside the given range`
`**if**` `(se < qs || ss > qe)`
`**return**` `0;`
`// If a part of this segment`
`// overlaps with the given range`
`**int**` `mid = getMid(ss, se);`
`**return**` `getSumUtil(st, ss, mid,`
`qs, qe,`
`2 * si + 1)`
`+ getSumUtil(st, mid + 1,`
`se, qs, qe,`
`2 * si + 2);`
`}`
`// Function to update the nodes which`
`// have the given index in their range`
`**void**` `updateValueUtil(``**int**``* st,` `**int**` `ss,`
`**int**` `se,` `**int**` `i,`
`**int**` `diff,` `**int**` `si)`
`{`
`// If the input index lies`
`// outside the range of`
`// this segment`
`**if**` `(i < ss || i > se)`
`**return**``;`
`// If the input index is in`
`// range of this node, then update`
`// the value of the node and its children`
`st[si] = st[si] + diff;`
`**if**` `(se != ss) {`
`**int**` `mid = getMid(ss, se);`
`updateValueUtil(st, ss, mid, i,`
`diff, 2 * si + 1);`
`updateValueUtil(st, mid + 1,`
`se, i, diff,`
`2 * si + 2);`
`}`
`}`
`// Function to update a value in`
`// input array and segment tree`
`**void**` `updateValue(``**int**` `arr[],` `**int**``* st,`
`**int**` `n,` `**int**` `i,`
`**int**` `new_val)`
`{`
`// Check for erroneous input index`
`**if**` `(i < 0 || i > n - 1) {`
`cout <<` `"-1"``;`
`**return**``;`
`}`
`// Get the difference between`
`// new value and old value`
`**int**` `diff = new_val - arr[i];`
`**int**` `prev_val = arr[i];`
`// Update the value in array`
`arr[i] = new_val;`
`// Update the values of`
`// nodes in segment tree`
`// only if either previous`
`// value or new value`
`// or both are prime`
`**if**` `(prime[new_val]`
`|| prime[prev_val]) {`
`// If only new value is prime`
`**if**` `(!prime[prev_val])`
`updateValueUtil(st, 0, n - 1,`
`i, new_val, 0);`
`// If only new value is prime`
`**else**` `**if**` `(!prime[new_val])`
`updateValueUtil(st, 0, n - 1,`
`i, -prev_val, 0);`
`// If both are prime`
`**else**`
`updateValueUtil(st, 0, n - 1,`
`i, diff, 0);`
`}`
`}`
`// Return sum of elements in range`
`// from index qs (quey start) to qe`
`// (query end). It mainly uses getSumUtil()`
`**int**` `getSum(``**int**``* st,` `**int**` `n,` `**int**` `qs,` `**int**` `qe)`
`{`
`// Check for erroneous input values`
`**if**` `(qs < 0 || qe > n - 1 || qs > qe) {`
`cout <<` `"-1"``;`
`**return**` `-1;`
`}`
`**return**` `getSumUtil(st, 0, n - 1,`
`qs, qe, 0);`
`}`
`// Function that constructs Segment Tree`
`**int**` `constructSTUtil(``**int**` `arr[],` `**int**` `ss,`
`**int**` `se,` `**int**``* st,`
`**int**` `si)`
`{`
`// If there is one element in`
`// array, store it in current node of`
`// segment tree and return`
`**if**` `(ss == se) {`
`// Only add those elements in segment`
`// tree which are prime`
`**if**` `(prime[arr[ss]])`
`st[si] = arr[ss];`
`**else**`
`st[si] = 0;`
`**return**` `st[si];`
`}`
`// If there are more than one`
`// elements, then recur for left and`
`// right subtrees and store the`
`// sum of values in this node`
`**int**` `mid = getMid(ss, se);`
`st[si]`
`= constructSTUtil(arr, ss, mid,`
`st, si * 2 + 1)`
`+ constructSTUtil(arr, mid + 1,`
`se, st,`
`si * 2 + 2);`
`**return**` `st[si];`
`}`
`// Function to construct segment`
`// tree from given array`
`**int**``* constructST(``**int**` `arr[],` `**int**` `n)`
`{`
`// Allocate memory for the segment tree`
`// Height of segment tree`
`**int**` `x = (``**int**``)(``**ceil**``(log2(n)));`
`// Maximum size of segment tree`
`**int**` `max_size = 2 * (``**int**``)``**pow**``(2, x) - 1;`
`// Allocate memory`
`**int**``* st =` `**new**` `**int**``[max_size];`
`// Fill the allocated memory st`
`constructSTUtil(arr, 0, n - 1, st, 0);`
`// Return the constructed segment tree`
`**return**` `st;`
`}`
`// Driver code`
`**int**` `main()`
`{`
`**int**` `arr[] = { 1, 3, 5, 7, 9, 11 };`
`**int**` `n =` `**sizeof**``(arr) /` `**sizeof**``(arr[0]);`
`**int**` `Q[3][3]`
`= { { 1, 1, 3 },`
`{ 2, 1, 10 },`
`{ 1, 1, 3 } };`
`// Function call`
`SieveOfEratosthenes();`
`// Build segment tree from given array`
`**int**``* st = constructST(arr, n);`
`// Print sum of values in`
`// array from index 1 to 3`
`cout << getSum(st, n, 1, 3) << endl;`
`// Update: set arr[1] = 10`
`// and update corresponding`
`// segment tree nodes`
`updateValue(arr, st, n, 1, 10);`
`// Find sum after the value is updated`
`cout << getSum(st, n, 1, 3) << endl;`
`**return**` `0;`
`}`
Output:
``````15
12
``````
Time Complexity:_ O(Q * log N) _
Auxiliary Space:_ O(N)_
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
#advanced data structure #arrays #dynamic programming #hash #mathematical #tree #array-range-queries #prime number #segment-tree #sieve
1596631020
## Sum of prime numbers in range [L, R] from given Array for Q queries
Given an array arr[] of the size of N followed by an array of Q queries, of the following two types:
• Query Type 1: Given two integers L and R, find the sum of prime elements from index L to R where 0 <= L <= R <= N-1.
• Query Type 2: Given two integers i and X, change arr[i] = X where 0 <= i <= n-1.
Note:_ Every first index of the subquery determines the type of query to be answered._
**Example: **
_Input: _arr[] = {1, 3, 5, 7, 9, 11}, Q = { { 1, 1, 3}, {2, 1, 10}, {1, 1, 3 } }
_Output: _
15
12
_Explanation: _
First query is of type 1, so answer is (3 + 5 + 7), = 15
Second query is of type 2, so arr[1] = 10
Third query is of type 1, where arr[1] = 10, which is not prime hence answer is (5 + 7) = 12
Input:_ arr[] = {1, 2, 35, 7, 14, 11}, Q = { {2, 4, 3}, {1, 4, 5 } }_
Output:_ 14_
Explanation:
First query is of type 2, So update arr[4] = 3
Second query is of type 1, since arr[4] = 3, which is prime. So answer is (3 + 11) = 14
**Naive Approach: **The idea is to iterate for each query between L to R and perform the required operation on the given array.
_Time Complexity: _O(Q * N * (O(sqrt(max(arr[i]))
**Approach: ** To optimize the problem use Segment tree and Sieve Of Eratosthenes.
• First, create a boolean array that will mark the prime numbers.
• Now while making the segment tree only add those array elements as leaf nodes which are prime.
• C++
• Python3
`// C++ program for the above approach`
`#include <bits/stdc++.h>`
`**using**` `**namespace**` `std;`
`**int**` `**const**` `MAX = 1000001;`
`**bool**` `prime[MAX];`
`// Function to find the prime numbers`
`**void**` `SieveOfEratosthenes()`
`{`
`// Create a boolean array prime[]`
`// and initialize all entries it as true`
`// A value in prime[i] will`
`// finally be false if i is Not a prime`
`**memset**``(prime,` `**true**``,` `**sizeof**``(prime));`
`**for**` `(``**int**` `p = 2; p * p <= MAX; p++) {`
`// Check if prime[p] is not`
`// changed, then it is a prime`
`**if**` `(prime[p] ==` `**true**``) {`
`// Update all multiples of p`
`// greater than or equal to`
`// the square of it numbers`
`// which are multiple of p`
`// and are less than p^2 are`
`// already been marked`
`**for**` `(``**int**` `i = p * p; i <= MAX; i += p)`
`prime[i] =` `**false**``;`
`}`
`}`
`}`
`// Function to get the middle`
`// index from corner indexes`
`**int**` `getMid(``**int**` `s,` `**int**` `e)`
`{`
`**return**` `s + (e - s) / 2;`
`}`
`// Function to get the sum of`
`// values in the given range`
`// of the array`
`**int**` `getSumUtil(``**int**``* st,` `**int**` `ss,`
`**int**` `se,` `**int**` `qs,`
`**int**` `qe,` `**int**` `si)`
`{`
`// If segment of this node is a`
`// part of given range, then`
`// return the sum of the segment`
`**if**` `(qs <= ss && qe >= se)`
`**return**` `st[si];`
`// If segment of this node is`
`// outside the given range`
`**if**` `(se < qs || ss > qe)`
`**return**` `0;`
`// If a part of this segment`
`// overlaps with the given range`
`**int**` `mid = getMid(ss, se);`
`**return**` `getSumUtil(st, ss, mid,`
`qs, qe,`
`2 * si + 1)`
`+ getSumUtil(st, mid + 1,`
`se, qs, qe,`
`2 * si + 2);`
`}`
`// Function to update the nodes which`
`// have the given index in their range`
`**void**` `updateValueUtil(``**int**``* st,` `**int**` `ss,`
`**int**` `se,` `**int**` `i,`
`**int**` `diff,` `**int**` `si)`
`{`
`// If the input index lies`
`// outside the range of`
`// this segment`
`**if**` `(i < ss || i > se)`
`**return**``;`
`// If the input index is in`
`// range of this node, then update`
`// the value of the node and its children`
`st[si] = st[si] + diff;`
`**if**` `(se != ss) {`
`**int**` `mid = getMid(ss, se);`
`updateValueUtil(st, ss, mid, i,`
`diff, 2 * si + 1);`
`updateValueUtil(st, mid + 1,`
`se, i, diff,`
`2 * si + 2);`
`}`
`}`
`// Function to update a value in`
`// input array and segment tree`
`**void**` `updateValue(``**int**` `arr[],` `**int**``* st,`
`**int**` `n,` `**int**` `i,`
`**int**` `new_val)`
`{`
`// Check for erroneous input index`
`**if**` `(i < 0 || i > n - 1) {`
`cout <<` `"-1"``;`
`**return**``;`
`}`
`// Get the difference between`
`// new value and old value`
`**int**` `diff = new_val - arr[i];`
`**int**` `prev_val = arr[i];`
`// Update the value in array`
`arr[i] = new_val;`
`// Update the values of`
`// nodes in segment tree`
`// only if either previous`
`// value or new value`
`// or both are prime`
`**if**` `(prime[new_val]`
`|| prime[prev_val]) {`
`// If only new value is prime`
`**if**` `(!prime[prev_val])`
`updateValueUtil(st, 0, n - 1,`
`i, new_val, 0);`
`// If only new value is prime`
`**else**` `**if**` `(!prime[new_val])`
`updateValueUtil(st, 0, n - 1,`
`i, -prev_val, 0);`
`// If both are prime`
`**else**`
`updateValueUtil(st, 0, n - 1,`
`i, diff, 0);`
`}`
`}`
`// Return sum of elements in range`
`// from index qs (quey start) to qe`
`// (query end). It mainly uses getSumUtil()`
`**int**` `getSum(``**int**``* st,` `**int**` `n,` `**int**` `qs,` `**int**` `qe)`
`{`
`// Check for erroneous input values`
`**if**` `(qs < 0 || qe > n - 1 || qs > qe) {`
`cout <<` `"-1"``;`
`**return**` `-1;`
`}`
`**return**` `getSumUtil(st, 0, n - 1,`
`qs, qe, 0);`
`}`
`// Function that constructs Segment Tree`
`**int**` `constructSTUtil(``**int**` `arr[],` `**int**` `ss,`
`**int**` `se,` `**int**``* st,`
`**int**` `si)`
`{`
`// If there is one element in`
`// array, store it in current node of`
`// segment tree and return`
`**if**` `(ss == se) {`
`// Only add those elements in segment`
`// tree which are prime`
`**if**` `(prime[arr[ss]])`
`st[si] = arr[ss];`
`**else**`
`st[si] = 0;`
`**return**` `st[si];`
`}`
`// If there are more than one`
`// elements, then recur for left and`
`// right subtrees and store the`
`// sum of values in this node`
`**int**` `mid = getMid(ss, se);`
`st[si]`
`= constructSTUtil(arr, ss, mid,`
`st, si * 2 + 1)`
`+ constructSTUtil(arr, mid + 1,`
`se, st,`
`si * 2 + 2);`
`**return**` `st[si];`
`}`
`// Function to construct segment`
`// tree from given array`
`**int**``* constructST(``**int**` `arr[],` `**int**` `n)`
`{`
`// Allocate memory for the segment tree`
`// Height of segment tree`
`**int**` `x = (``**int**``)(``**ceil**``(log2(n)));`
`// Maximum size of segment tree`
`**int**` `max_size = 2 * (``**int**``)``**pow**``(2, x) - 1;`
`// Allocate memory`
`**int**``* st =` `**new**` `**int**``[max_size];`
`// Fill the allocated memory st`
`constructSTUtil(arr, 0, n - 1, st, 0);`
`// Return the constructed segment tree`
`**return**` `st;`
`}`
`// Driver code`
`**int**` `main()`
`{`
`**int**` `arr[] = { 1, 3, 5, 7, 9, 11 };`
`**int**` `n =` `**sizeof**``(arr) /` `**sizeof**``(arr[0]);`
`**int**` `Q[3][3]`
`= { { 1, 1, 3 },`
`{ 2, 1, 10 },`
`{ 1, 1, 3 } };`
`// Function call`
`SieveOfEratosthenes();`
`// Build segment tree from given array`
`**int**``* st = constructST(arr, n);`
`// Print sum of values in`
`// array from index 1 to 3`
`cout << getSum(st, n, 1, 3) << endl;`
`// Update: set arr[1] = 10`
`// and update corresponding`
`// segment tree nodes`
`updateValue(arr, st, n, 1, 10);`
`// Find sum after the value is updated`
`cout << getSum(st, n, 1, 3) << endl;`
`**return**` `0;`
`}`
Output:
``````15
12
``````
Time Complexity:_ O(Q * log N) _
Auxiliary Space:_ O(N)_
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
#advanced data structure #arrays #dynamic programming #hash #mathematical #tree #array-range-queries #prime number #segment-tree #sieve
1649209980
## C# REPL
A cross-platform command line REPL for the rapid experimentation and exploration of C#. It supports intellisense, installing NuGet packages, and referencing local .NET projects and assemblies.
(click to view animation)
C# REPL provides the following features:
• Syntax highlighting via ANSI escape sequences
• Intellisense with fly-out documentation
• Nuget package installation
• Reference local assemblies, solutions, and projects
• Navigate to source via Source Link
• IL disassembly (both Debug and Release mode)
• Fast and flicker-free rendering. A "diff" algorithm is used to only render what's changed.
## Installation
C# REPL is a .NET 6 global tool, and runs on Windows 10, Mac OS, and Linux. It can be installed via:
``````dotnet tool install -g csharprepl
``````
If you're running on Mac OS Catalina (10.15) or later, make sure you follow any additional directions printed to the screen. You may need to update your PATH variable in order to use .NET global tools.
After installation is complete, run `csharprepl` to begin. C# REPL can be updated via `dotnet tool update -g csharprepl`.
## Usage:
Run `csharprepl` from the command line to begin an interactive session. The default colorscheme uses the color palette defined by your terminal, but these colors can be changed using a `theme.json` file provided as a command line argument.
### Evaluating Code
Type some C# into the prompt and press Enter to run it. The result, if any, will be printed:
``````> Console.WriteLine("Hello World")
Hello World
[6/7/2021 5:13:00 PM]
``````
To evaluate multiple lines of code, use Shift+Enter to insert a newline:
``````> var x = 5;
var y = 8;
x * y
40
``````
Additionally, if the statement is not a "complete statement" a newline will automatically be inserted when Enter is pressed. For example, in the below code, the first line is not a syntactically complete statement, so when we press enter we'll go down to a new line:
``````> if (x == 5)
| // caret position, after we press Enter on Line 1
``````
Finally, pressing Ctrl+Enter will show a "detailed view" of the result. For example, for the `DateTime.Now` expression below, on the first line we pressed Enter, and on the second line we pressed Ctrl+Enter to view more detailed output:
``````> DateTime.Now // Pressing Enter shows a reasonable representation
[5/30/2021 5:13:00 PM]
> DateTime.Now // Pressing Ctrl+Enter shows a detailed representation
[5/30/2021 5:13:00 PM] {
Date: [5/30/2021 12:00:00 AM],
Day: 30,
DayOfWeek: Sunday,
DayOfYear: 150,
Hour: 17,
InternalKind: 9223372036854775808,
InternalTicks: 637579915804530992,
Kind: Local,
Millisecond: 453,
Minute: 13,
Month: 5,
Second: 0,
Ticks: 637579915804530992,
TimeOfDay: [17:13:00.4530992],
Year: 2021,
_dateData: 9860951952659306800
}
``````
A note on semicolons: C# expressions do not require semicolons, but statements do. If a statement is missing a required semicolon, a newline will be added instead of trying to run the syntatically incomplete statement; simply type the semicolon to complete the statement.
``````> var now = DateTime.Now; // assignment statement, semicolon required
> DateTime.Now.AddDays(8) // expression, we don't need a semicolon
[6/7/2021 5:03:05 PM]
``````
### Keyboard Shortcuts
• Basic Usage
• Ctrl+C - Cancel current line
• Ctrl+L - Clear screen
• Enter - Evaluate the current line if it's a syntactically complete statement; otherwise add a newline
• Control+Enter - Evaluate the current line, and return a more detailed representation of the result
• Shift+Enter - Insert a new line (this does not currently work on Linux or Mac OS; Hopefully this will work in .NET 7)
• Ctrl+Shift+C - Copy current line to clipboard
• Ctrl+V, Shift+Insert, and Ctrl+Shift+V - Paste text to prompt. Automatically trims leading indent
• Code Actions
• F1 - Opens the MSDN documentation for the class/method under the caret (example)
• F9 - Shows the IL (intermediate language) for the current statement in Debug mode.
• Ctrl+F9 - Shows the IL for the current statement with Release mode optimizations.
• F12 - Opens the source code in the browser for the class/method under the caret, if the assembly supports Source Link.
• Autocompletion
• Ctrl+Space - Open autocomplete menu. If there's a single option, pressing Ctrl+Space again will select the option
• Enter, Right Arrow, Tab - Select active autocompletion option
• Escape - closes autocomplete menu
• Home and End - Navigate to beginning of a single line and end of a single line, respectively
• Ctrl+Home and Ctrl+End - Navigate to beginning of line and end across multiple lines in a multiline prompt, respectively
• Arrows - Navigate characters within text
• Ctrl+Arrows - Navigate words within text
• Ctrl+Backspace - Delete previous word
• Ctrl+Delete - Delete next word
Use the `#r` command to add assembly or nuget references.
• For assembly references, run `#r "AssemblyName"` or `#r "path/to/assembly.dll"`
• For project references, run `#r "path/to/project.csproj"`. Solution files (.sln) can also be referenced.
• For nuget references, run `#r "nuget: PackageName"` to install the latest version of a package, or `#r "nuget: PackageName, 13.0.5"` to install a specific version (13.0.5 in this case).
To run ASP.NET applications inside the REPL, start the `csharprepl `application with the `--framework` parameter, specifying the `Microsoft.AspNetCore.App` shared framework. Then, use the above `#r` command to reference the application DLL. See the Command Line Configuration section below for more details.
``````csharprepl --framework Microsoft.AspNetCore.App
``````
## Command Line Configuration
The C# REPL supports multiple configuration flags to control startup, behavior, and appearance:
``````csharprepl [OPTIONS] [response-file.rsp] [script-file.csx] [-- <additional-arguments>]
``````
Supported options are:
• OPTIONS:
• `-r <dll>` or `--reference <dll>`: Reference an assembly, project file, or nuget package. Can be specified multiple times. Uses the same syntax as `#r` statements inside the REPL. For example, `csharprepl -r "nuget:Newtonsoft.Json" "path/to/myproj.csproj"`
• When an assembly or project is referenced, assemblies in the containing directory will be added to the assembly search path. This means that you don't need to manually add references to all of your assembly's dependencies (e.g. other references and nuget packages). Referencing the main entry assembly is enough.
• `-u <namespace>` or `--using <namespace>`: Add a using statement. Can be specified multiple times.
• `-f <framework>` or `--framework <framework>`: Reference a shared framework. The available shared frameworks depends on the local .NET installation, and can be useful when running an ASP.NET application from the REPL. Example frameworks are:
• Microsoft.NETCore.App (default)
• Microsoft.AspNetCore.All
• Microsoft.AspNetCore.App
• Microsoft.WindowsDesktop.App
• `-t <theme.json>` or `--theme <theme.json>`: Read a theme file for syntax highlighting. This theme file associates C# syntax classifications with colors. The color values can be full RGB, or ANSI color names (defined in your terminal's theme). The NO_COLOR standard is supported.
• `--trace`: Produce a trace file in the current directory that logs CSharpRepl internals. Useful for CSharpRepl bug reports.
• `-v` or `--version`: Show version number and exit.
• `-h` or `--help`: Show help and exit.
• `response-file.rsp`: A filepath of an .rsp file, containing any of the above command line options.
• `script-file.csx`: A filepath of a .csx file, containing lines of C# to evaluate before starting the REPL. Arguments to this script can be passed as `<additional-arguments>`, after a double hyphen (`--`), and will be available in a global `args` variable.
If you have `dotnet-suggest` enabled, all options can be tab-completed, including values provided to `--framework` and .NET namespaces provided to `--using`.
## Integrating with other software
C# REPL is a standalone software application, but it can be useful to integrate it with other developer tools:
### Windows Terminal
To add C# REPL as a menu entry in Windows Terminal, add the following profile to Windows Terminal's `settings.json` configuration file (under the JSON property `profiles.list`):
``````{
"name": "C# REPL",
"commandline": "csharprepl"
},
``````
To get the exact colors shown in the screenshots in this README, install the Windows Terminal Dracula theme.
### Visual Studio Code
To use the C# REPL with Visual Studio Code, simply run the `csharprepl` command in the Visual Studio Code terminal. To send commands to the REPL, use the built-in `Terminal: Run Selected Text In Active Terminal` command from the Command Palette (`workbench.action.terminal.runSelectedText`).
### Windows OS
To add the C# REPL to the Windows Start Menu for quick access, you can run the following PowerShell command, which will start C# REPL in Windows Terminal:
``````\$shell = New-Object -ComObject WScript.Shell
\$shortcut.TargetPath = "wt.exe"
\$shortcut.Arguments = "-w 0 nt csharprepl.exe"
\$shortcut.Save()
``````
You may also wish to add a shorter alias for C# REPL, which can be done by creating a `.cmd` file somewhere on your path. For example, put the following contents in `C:\Users\username\.dotnet\tools\csr.cmd`:
``````wt -w 0 nt csharprepl
``````
This will allow you to launch C# REPL by running `csr` from anywhere that accepts Windows commands, like the Window Run dialog.
## Comparison with other REPLs
This project is far from being the first REPL for C#. Here are some other projects; if this project doesn't suit you, another one might!
Visual Studio's C# Interactive pane is full-featured (it has syntax highlighting and intellisense) and is part of Visual Studio. This deep integration with Visual Studio is both a benefit from a workflow perspective, and a drawback as it's not cross-platform. As far as I know, the C# Interactive pane does not support NuGet packages or navigating to documentation/source code. Subjectively, it does not follow typical command line keybindings, so can feel a bit foreign.
csi.exe ships with C# and is a command line REPL. It's great because it's a cross platform REPL that comes out of the box, but it doesn't support syntax highlighting or autocompletion.
dotnet script allows you to run C# scripts from the command line. It has a REPL built-in, but the predominant focus seems to be as a script runner. It's a great tool, though, and has a strong community following.
dotnet interactive is a tool from Microsoft that creates a Jupyter notebook for C#, runnable through Visual Studio Code. It also provides a general framework useful for running REPLs.
Author: waf
Source Code: https://github.com/waf/CSharpRepl
1647064260
## dotnet script
Run C# scripts from the .NET CLI, define NuGet packages inline and edit/debug them in VS Code - all of that with full language services support from OmniSharp.
## Installing
### Prerequisites
The only thing we need to install is .NET Core 3.1 or .NET 5.0 SDK.
### .NET Core Global Tool
.NET Core 2.1 introduced the concept of global tools meaning that you can install `dotnet-script` using nothing but the .NET CLI.
``````dotnet tool install -g dotnet-script
You can invoke the tool using the following command: dotnet-script
Tool 'dotnet-script' (version '0.22.0') was successfully installed.
``````
The advantage of this approach is that you can use the same command for installation across all platforms. .NET Core SDK also supports viewing a list of installed tools and their uninstallation.
``````dotnet tool list -g
Package Id Version Commands
---------------------------------------------
dotnet-script 0.22.0 dotnet-script
``````
``````dotnet tool uninstall dotnet-script -g
Tool 'dotnet-script' (version '0.22.0') was successfully uninstalled.
``````
### Windows
``````choco install dotnet.script
``````
We also provide a PowerShell script for installation.
``````(new-object Net.WebClient).DownloadString("https://raw.githubusercontent.com/filipw/dotnet-script/master/install/install.ps1") | iex
``````
### Linux and Mac
``````curl -s https://raw.githubusercontent.com/filipw/dotnet-script/master/install/install.sh | bash
``````
If permission is denied we can try with `sudo`
``````curl -s https://raw.githubusercontent.com/filipw/dotnet-script/master/install/install.sh | sudo bash
``````
### Docker
A Dockerfile for running dotnet-script in a Linux container is available. Build:
``````cd build
docker build -t dotnet-script -f Dockerfile ..
``````
And run:
``````docker run -it dotnet-script --version
``````
### Github
You can manually download all the releases in `zip` format from the GitHub releases page.
## Usage
Our typical `helloworld.csx` might look like this:
``````Console.WriteLine("Hello world!");
``````
That is all it takes and we can execute the script. Args are accessible via the global Args array.
``````dotnet script helloworld.csx
``````
### Scaffolding
Simply create a folder somewhere on your system and issue the following command.
``````dotnet script init
``````
This will create `main.csx` along with the launch configuration needed to debug the script in VS Code.
``````.
├── .vscode
│ └── launch.json
├── main.csx
└── omnisharp.json
``````
We can also initialize a folder using a custom filename.
``````dotnet script init custom.csx
``````
Instead of `main.csx` which is the default, we now have a file named `custom.csx`.
``````.
├── .vscode
│ └── launch.json
├── custom.csx
└── omnisharp.json
``````
Note: Executing `dotnet script init` inside a folder that already contains one or more script files will not create the `main.csx` file.
### Running scripts
Scripts can be executed directly from the shell as if they were executables.
``````foo.csx arg1 arg2 arg3
``````
OSX/Linux
Just like all scripts, on OSX/Linux you need to have a `#!` and mark the file as executable via chmod +x foo.csx. If you use dotnet script init to create your csx it will automatically have the `#!` directive and be marked as executable.
The OSX/Linux shebang directive should be #!/usr/bin/env dotnet-script
``````#!/usr/bin/env dotnet-script
Console.WriteLine("Hello world");
``````
You can execute your script using dotnet script or dotnet-script, which allows you to pass arguments to control your script execution more.
``````foo.csx arg1 arg2 arg3
dotnet script foo.csx -- arg1 arg2 arg3
dotnet-script foo.csx -- arg1 arg2 arg3
``````
#### Passing arguments to scripts
All arguments after `--` are passed to the script in the following way:
``````dotnet script foo.csx -- arg1 arg2 arg3
``````
Then you can access the arguments in the script context using the global `Args` collection:
``````foreach (var arg in Args)
{
Console.WriteLine(arg);
}
``````
All arguments before `--` are processed by `dotnet script`. For example, the following command-line
``````dotnet script -d foo.csx -- -d
``````
will pass the `-d` before `--` to `dotnet script` and enable the debug mode whereas the `-d` after `--` is passed to script for its own interpretation of the argument.
### NuGet Packages
`dotnet script` has built-in support for referencing NuGet packages directly from within the script.
``````#r "nuget: AutoMapper, 6.1.0"
``````
Note: Omnisharp needs to be restarted after adding a new package reference
#### Package Sources
We can define package sources using a `NuGet.Config` file in the script root folder. In addition to being used during execution of the script, it will also be used by `OmniSharp` that provides language services for packages resolved from these package sources.
As an alternative to maintaining a local `NuGet.Config` file we can define these package sources globally either at the user level or at the computer level as described in Configuring NuGet Behaviour
It is also possible to specify packages sources when executing the script.
``````dotnet script foo.csx -s https://SomePackageSource
``````
Multiple packages sources can be specified like this:
``````dotnet script foo.csx -s https://SomePackageSource -s https://AnotherPackageSource
``````
### Creating DLLs or Exes from a CSX file
Dotnet-Script can create a standalone executable or DLL for your script.
The executable you can run directly independent of dotnet install, while the DLL can be run using the dotnet CLI like this:
``````dotnet script exec {path_to_dll} -- arg1 arg2
``````
### Caching
We provide two types of caching, the `dependency cache` and the `execution cache` which is explained in detail below. In order for any of these caches to be enabled, it is required that all NuGet package references are specified using an exact version number. The reason for this constraint is that we need to make sure that we don't execute a script with a stale dependency graph.
#### Dependency Cache
In order to resolve the dependencies for a script, a `dotnet restore` is executed under the hood to produce a `project.assets.json` file from which we can figure out all the dependencies we need to add to the compilation. This is an out-of-process operation and represents a significant overhead to the script execution. So this cache works by looking at all the dependencies specified in the script(s) either in the form of NuGet package references or assembly file references. If these dependencies matches the dependencies from the last script execution, we skip the restore and read the dependencies from the already generated `project.assets.json` file. If any of the dependencies has changed, we must restore again to obtain the new dependency graph.
#### Execution cache
In order to execute a script it needs to be compiled first and since that is a CPU and time consuming operation, we make sure that we only compile when the source code has changed. This works by creating a SHA256 hash from all the script files involved in the execution. This hash is written to a temporary location along with the DLL that represents the result of the script compilation. When a script is executed the hash is computed and compared with the hash from the previous compilation. If they match there is no need to recompile and we run from the already compiled DLL. If the hashes don't match, the cache is invalidated and we recompile.
You can override this automatic caching by passing --no-cache flag, which will bypass both caches and cause dependency resolution and script compilation to happen every time we execute the script.
#### Cache Location
The temporary location used for caches is a sub-directory named `dotnet-script` under (in order of priority):
1. The path specified for the value of the environment variable named `DOTNET_SCRIPT_CACHE_LOCATION`, if defined and value is not empty.
2. Linux distributions only: `\$XDG_CACHE_HOME` if defined otherwise `\$HOME/.cache`
3. macOS only: `~/Library/Caches`
4. The value returned by `Path.GetTempPath` for the platform.
### Debugging
The days of debugging scripts using `Console.WriteLine` are over. One major feature of `dotnet script` is the ability to debug scripts directly in VS Code. Just set a breakpoint anywhere in your script file(s) and hit F5(start debugging)
### Script Packages
Script packages are a way of organizing reusable scripts into NuGet packages that can be consumed by other scripts. This means that we now can leverage scripting infrastructure without the need for any kind of bootstrapping.
#### Creating a script package
A script package is just a regular NuGet package that contains script files inside the `content` or `contentFiles` folder.
The following example shows how the scripts are laid out inside the NuGet package according to the standard convention .
``````└── contentFiles
└── csx
└── netstandard2.0
└── main.csx
``````
This example contains just the `main.csx` file in the root folder, but packages may have multiple script files either in the root folder or in subfolders below the root folder.
When loading a script package we will look for an entry point script to be loaded. This entry point script is identified by one of the following.
• A script called `main.csx` in the root folder
• A single script file in the root folder
If the entry point script cannot be determined, we will simply load all the scripts files in the package.
The advantage with using an entry point script is that we can control loading other scripts from the package.
#### Consuming a script package
To consume a script package all we need to do specify the NuGet package in the `#load`directive.
The following example loads the simple-targets package that contains script files to be included in our script.
``````#load "nuget:simple-targets-csx, 6.0.0"
using static SimpleTargets;
var targets = new TargetDictionary();
Run(Args, targets);
``````
Note: Debugging also works for script packages so that we can easily step into the scripts that are brought in using the `#load` directive.
### Remote Scripts
Scripts don't actually have to exist locally on the machine. We can also execute scripts that are made available on an `http(s)` endpoint.
This means that we can create a Gist on Github and execute it just by providing the URL to the Gist.
This Gist contains a script that prints out "Hello World"
We can execute the script like this
``````dotnet script https://gist.githubusercontent.com/seesharper/5d6859509ea8364a1fdf66bbf5b7923d/raw/0a32bac2c3ea807f9379a38e251d93e39c8131cb/HelloWorld.csx
``````
That is a pretty long URL, so why don't make it a TinyURL like this:
``````dotnet script https://tinyurl.com/y8cda9zt
``````
### Script Location
A pretty common scenario is that we have logic that is relative to the script path. We don't want to require the user to be in a certain directory for these paths to resolve correctly so here is how to provide the script path and the script folder regardless of the current working directory.
``````public static string GetScriptPath([CallerFilePath] string path = null) => path;
public static string GetScriptFolder([CallerFilePath] string path = null) => Path.GetDirectoryName(path);
``````
Tip: Put these methods as top level methods in a separate script file and `#load` that file wherever access to the script path and/or folder is needed.
## REPL
This release contains a C# REPL (Read-Evaluate-Print-Loop). The REPL mode ("interactive mode") is started by executing `dotnet-script` without any arguments.
The interactive mode allows you to supply individual C# code blocks and have them executed as soon as you press Enter. The REPL is configured with the same default set of assembly references and using statements as regular CSX script execution.
### Basic usage
Once `dotnet-script` starts you will see a prompt for input. You can start typing C# code there.
``````~\$ dotnet script
> var x = 1;
> x+x
2
``````
If you submit an unterminated expression into the REPL (no `;` at the end), it will be evaluated and the result will be serialized using a formatter and printed in the output. This is a bit more interesting than just calling `ToString()` on the object, because it attempts to capture the actual structure of the object. For example:
``````~\$ dotnet script
> var x = new List<string>();
> x
List<string>(1) { "foo" }
> x
List<string>(2) { "foo", "bar" }
>
``````
### Inline Nuget packages
REPL also supports inline Nuget packages - meaning the Nuget packages can be installed into the REPL from within the REPL. This is done via our `#r` and `#load` from Nuget support and uses identical syntax.
``````~\$ dotnet script
> #r "nuget: Automapper, 6.1.1"
> using AutoMapper;
> typeof(MapperConfiguration)
[AutoMapper.MapperConfiguration]
> using static SimpleTargets;
> typeof(TargetDictionary)
[Submission#0+SimpleTargets+TargetDictionary]
``````
### Multiline mode
Using Roslyn syntax parsing, we also support multiline REPL mode. This means that if you have an uncompleted code block and press Enter, we will automatically enter the multiline mode. The mode is indicated by the `*` character. This is particularly useful for declaring classes and other more complex constructs.
``````~\$ dotnet script
> class Foo {
* public string Bar {get; set;}
* }
> var foo = new Foo();
``````
### REPL commands
Aside from the regular C# script code, you can invoke the following commands (directives) from within the REPL:
### Seeding REPL with a script
You can execute a CSX script and, at the end of it, drop yourself into the context of the REPL. This way, the REPL becomes "seeded" with your code - all the classes, methods or variables are available in the REPL context. This is achieved by running a script with an `-i` flag.
For example, given the following CSX script:
``````var msg = "Hello World";
Console.WriteLine(msg);
``````
When you run this with the `-i` flag, `Hello World` is printed, REPL starts and `msg` variable is available in the REPL context.
``````~\$ dotnet script foo.csx -i
Hello World
>
``````
You can also seed the REPL from inside the REPL - at any point - by invoking a `#load` directive pointed at a specific file. For example:
``````~\$ dotnet script
Hello World
>
``````
## Piping
The following example shows how we can pipe data in and out of a script.
The `UpperCase.csx` script simply converts the standard input to upper case and writes it back out to standard output.
``````using (var streamReader = new StreamReader(Console.OpenStandardInput()))
{
}
``````
We can now simply pipe the output from one command into our script like this.
``````echo "This is some text" | dotnet script UpperCase.csx
THIS IS SOME TEXT
``````
### Debugging
The first thing we need to do add the following to the `launch.config` file that allows VS Code to debug a running process.
``````{
"name": ".NET Core Attach",
"type": "coreclr",
"request": "attach",
"processId": "\${command:pickProcess}"
}
``````
To debug this script we need a way to attach the debugger in VS Code and the simplest thing we can do here is to wait for the debugger to attach by adding this method somewhere.
``````public static void WaitForDebugger()
{
Console.WriteLine("Attach Debugger (VS Code)");
while(!Debugger.IsAttached)
{
}
}
``````
To debug the script when executing it from the command line we can do something like
``````WaitForDebugger();
{
}
``````
Now when we run the script from the command line we will get
``````\$ echo "This is some text" | dotnet script UpperCase.csx
Attach Debugger (VS Code)
``````
This now gives us a chance to attach the debugger before stepping into the script and from VS Code, select the `.NET Core Attach` debugger and pick the process that represents the executing script.
Once that is done we should see our breakpoint being hit.
## Configuration(Debug/Release)
By default, scripts will be compiled using the `debug` configuration. This is to ensure that we can debug a script in VS Code as well as attaching a debugger for long running scripts.
There are however situations where we might need to execute a script that is compiled with the `release` configuration. For instance, running benchmarks using BenchmarkDotNet is not possible unless the script is compiled with the `release` configuration.
We can specify this when executing the script.
``````dotnet script foo.csx -c release
``````
## Nullable reference types
Starting from version 0.50.0, `dotnet-script` supports .Net Core 3.0 and all the C# 8 features. The way we deal with nullable references types in `dotnet-script` is that we turn every warning related to nullable reference types into compiler errors. This means every warning between `CS8600` and `CS8655` are treated as an error when compiling the script.
Nullable references types are turned off by default and the way we enable it is using the `#nullable enable` compiler directive. This means that existing scripts will continue to work, but we can now opt-in on this new feature.
``````#!/usr/bin/env dotnet-script
#nullable enable
string name = null;
``````
Trying to execute the script will result in the following error
``````main.csx(5,15): error CS8625: Cannot convert null literal to non-nullable reference type.
``````
We will also see this when working with scripts in VS Code under the problems panel.
Author: filipw
Source Code: https://github.com/filipw/dotnet-script
1596962520
## Queries to find the Lower Bound of K from Prefix Sum Array
Given an array A[ ] consisting of non-negative integers and matrix Q[ ][ ] consisting of queries of the following two types:
• **(1, l, val): **Update A[l] to A[l] + val.
• **(2, K): **Find the lower_bound of **K **in the prefix sum array of A[ ]. If the lower_bound does not exist print -1.
The task for each query of second type is to print the index of lower_bound of value K.
Examples:
_Input: __A[ ] = {1, 2, 3, 5, 8}, Q[ ][ ] = {{1, 0, 2}, {2, 5}, {1, 3, 5}} _
_Output: __1 _
Explanation:
Query 1: Update A[0] to A[0] + 2. Now A[ ] = {3, 2, 3, 5, 8}
_Query 2: lower_bound of K = 5 in the prefix sum array {3, 5, 8, 13, 21} is 5 and index = 1. _
Query 3: Update A[3] to A[3] + 5. Now A[ ] = {3, 2, 3, 10, 8}
_Input: __A[ ] = {4, 1, 12, 8, 20}, Q[ ] = {{2, 50}, {1, 3, 12}, {2, 50}} _
_Output: __-1 _
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Naive approach:
The simplest approach is to firstly build a prefix sum array of given array A[ ], and for queries of Type 1, update values and recalculate the prefix sum. For query of Type 2, perform a Binary Search on the prefix sum array to find lower bound.
Time Complexity:_ O(Q*(N*logn))_
_Auxiliary Space: _O(N)
Efficient Approach:
The above approach can be optimized using Fenwick Tree. Using this Data Structure, the update queries in prefix sum array can be performed in logarithmic time.
Follow the steps below to solve the problem:
• Construct the Prefix Sum Array using Fenwick Tree.
• For queries of Type 1, while** l > 0**, add val to A[l] traverse to the parent node by adding least significant bit in l.
• For queries of Type 2, perform the Binary Search on the Fenwick Tree to obtain the lower bound.
• Whenever a prefix sum greater than **K appears, **store that **index **and traverse the left part of the Fenwick Tree. Otherwise, traverse the right part of the Fenwick Tree Now, perform Binary Search.
• Finally, print the required index.
Below is the implementation of the above approach:
• Java
• C#
`// Java program to implement`
`// the above appraoch`
`**import**` `java.util.*;`
`**import**` `java.io.*;`
`**class**` `GFG {`
`// Function to calculate and return`
`// the sum of arr[0..index]`
`**static**` `**int**` `getSum(``**int**` `BITree[],`
`**int**` `index)`
`{`
`**int**` `ans =` `0``;`
`index +=` `1``;`
`// Traverse ancestors`
`// of BITree[index]`
`**while**` `(index >` `0``) {`
`// Update the sum of current`
`// element of BIT to ans`
`ans += BITree[index];`
`// Update index to that`
`// of the parent node in`
`// getSum() view by`
`// subtracting LSB(Least`
`// Significant Bit)`
`index -= index & (-index);`
`}`
`**return**` `ans;`
`}`
`// Function to update the Binary Index`
`// Tree by replacing all ancestores of`
`// index by their respective sum with val`
`**static**` `**void**` `updateBIT(``**int**` `BITree[],`
`**int**` `n,` `**int**` `index,` `**int**` `val)`
`{`
`index = index +` `1``;`
`// Traverse all ancestors`
`// and sum with 'val'.`
`**while**` `(index <= n) {`
`// Add 'val' to current`
`// node of BIT`
`BITree[index] += val;`
`// Update index to that`
`// of the parent node in`
`// updateBit() view by`
`// adding LSB(Least`
`// Significant Bit)`
`index += index & (-index);`
`}`
`}`
`// Function to construct the Binary`
`// Indexed Tree for the given array`
`**static**` `**int**``[] constructBITree(`
`**int**` `arr[],` `**int**` `n)`
`{`
`// Initialize the`
`// Binary Indexed Tree`
`**int**``[] BITree =` `**new**` `**int**``[n +` `1``];`
`**for**` `(``**int**` `i =` `0``; i <= n; i++)`
`BITree[i] =` `0``;`
`// Store the actual values in`
`// BITree[] using update()`
`**for**` `(``**int**` `i =` `0``; i < n; i++)`
`updateBIT(BITree, n, i, arr[i]);`
`**return**` `BITree;`
`}`
`// Function to obtian and return`
`// the index of lower_bound of k`
`**static**` `**int**` `getLowerBound(``**int**` `BITree[],`
`**int**``[] arr,` `**int**` `n,` `**int**` `k)`
`{`
`**int**` `lb = -``1``;`
`**int**` `l =` `0``, r = n -` `1``;`
`**while**` `(l <= r) {`
`**int**` `mid = l + (r - l) /` `2``;`
`**if**` `(getSum(BITree, mid) >= k) {`
`r = mid -` `1``;`
`lb = mid;`
`}`
`**else**`
`l = mid +` `1``;`
`}`
`**return**` `lb;`
`}`
`**static**` `**void**` `performQueries(``**int**` `A[],` `**int**` `n,` `**int**` `q[][])`
`{`
`// Store the Binary Indexed Tree`
`**int**``[] BITree = constructBITree(A, n);`
`// Solve each query in Q`
`**for**` `(``**int**` `i =` `0``; i < q.length; i++) {`
`**int**` `id = q[i][``0``];`
`**if**` `(id ==` `1``) {`
`**int**` `idx = q[i][``1``];`
`**int**` `val = q[i][``2``];`
`A[idx] += val;`
`// Update the values of all`
`// ancestors of idx`
`updateBIT(BITree, n, idx, val);`
`}`
`**else**` `{`
`**int**` `k = q[i][``1``];`
`**int**` `lb = getLowerBound(`
`BITree, A, n, k);`
`System.out.println(lb);`
`}`
`}`
`}`
`// Driver Code`
`**public**` `**static**` `**void**` `main(String[] args)`
`{`
`**int**` `A[] = {` `1``,` `2``,` `3``,` `5``,` `8` `};`
`**int**` `n = A.length;`
`**int**``[][] q = { {` `1``,` `0``,` `2` `},`
`{` `2``,` `5` `},`
`{` `1``,` `3``,` `5` `} };`
`performQueries(A, n, q);`
`}`
`}`
Output:
``````1
``````
Time Complexity:_ O(Q*(logN)2)_
Auxiliary Space:_ O(N)_
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
#advanced data structure #arrays #bit magic #mathematical #searching #array-range-queries #bit #prefix-sum
1593370497
## Sum and Maximum of elements in array from [L, R] updates
Prerequisite: Segment TreesLazy Propagation in Segment Tree.
Given an array arr[] of N integers. The task is to do the following operations:
1. Change the value arr[i] to min(arr[i], X) where X is an integer for a given range [L, R].
2. Find the maximum value from index L to R where 0 ≤ L ≤ R ≤ N-1 before and after the update given to the array above.
3. Find the sum of the element from index L to R where 0 ≤ L ≤ R ≤ N-1 before and after the update given to the array above.
Examples:
``````Input: arr[] = {1, 2, 3, 4, 5}, L = 2, R = 4, X = 3
Output:
Maximum in range [2, 4] before update: 5
Sum in range [2, 4] before update: 12
Maximum in range [2, 4] after update: 3
Sum in range [2, 4] after update: 9
Explanation:
Before Update:
arr[] = {1, 2, 3, 4, 5}
The maximum value from [L, R] is 5
Sum in range [L, R] is 3 + 4 + 5 = 12
After Update:
arr[] = {1, 2, 3, 3, 3}
The maximum value from [L, R] is 3
Sum in range [L, R] is 3 + 3 + 3 = 9
Input: arr[] = {1, 4, 19, 0, 7, 22, 7}, L = 1, R = 5, X = 14
Output:
Maximum in range [1, 5] before update: 22
Sum in range [1, 5] before update: 52
Maximum in range [1, 5] after update: 22
Sum in range [1, 5] after update: 39
Explanation:
Before Update:
arr[] = {1, 4, 19, 0, 7, 22, 7}
The maximum value from [L, R] is 22
Sum in range [L, R] is 4 + 19 + 0 + 7 + 22 = 52
After Update:
arr[] = {1, 4, 14, 0, 7, 14, 7}
The maximum value from [L, R] is 14
Sum in range [L, R] is 4 + 14 + 0 + 7 + 14 = 39
``````
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach:
1. pushdown(): The lazy tag is applied to current node and then is pushed down to its children.
2. tag_condition: It is the condition that needs to be satisfied to set the lazy node. In normal lazy trees, it is generally the condition that ranges of the node covers lie entirely in the update range.
3. A node in the segment tree represents a range of the array. Now all elements in that range will have different values and they will change by different amounts during an update. So we need the information about distinct values and their counts in that range. So this becomes a worst-case O**(N)** operation as at max N distinct nodes in a range.
4. Below is the approach to solve these restrictions in Segment Tree. Each node in the Segment Tree will have the following values:
• value: The value of a range, here sum of all elements in that range
• maxval: Maximum value in that range
• secondmax: Strict second maximum value in that range
• cnt_max: Count of maximum value in that range
• cnt_secondmax: Count of secondmax in that range
#advanced data structure #algorithms #arrays #competitive programming #recursion #tree #array-range-queries #segment-tree | 15,528 | 52,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-27 | longest | en | 0.719712 |
https://www.physicsforums.com/threads/trig-quick-question.630187/ | 1,579,958,938,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00491.warc.gz | 1,016,384,350 | 16,013 | # Trig quick question
Solve for values of θ in the interval 0 ≤ θ ≤ 360,
2sinθ = cosecθ
now if I do:
2sinθ = 1/(sinθ) and multiply by sinθ
I can solve sinθ = ±√(1/2)
but if I solve 2sinθ - 1/(sinθ) = 0 and factorise to get sinθ(2 - (1/sin^2θ)) = 0
I get sinθ = 0 which gives me more solutions then needed. I've drawn the graphs of both functions and they meet where the solutions of θ are such that sinθ = ±√(1/2) , then why do I get more solutions if I factorise?
Related Precalculus Mathematics Homework Help News on Phys.org
Mentallic
Homework Helper
Because at $\sin\theta=0$ the other factor is undefined.
Think about the function $$y=\frac{x(x+1)}{x}$$ this is pretty much equivalent to $y=x+1$ except for the fact that it has a hole at x=0 (the coordinate (0,1)). If we solved this equation for when y=0, clearly the only answer is x=-1, but if we factored out x then what we'd have is:
$$x\left(\frac{x+1}{x}\right)=0$$
Which seems like it would have an x=0 solution, but it doesn't by the same reasoning.
Because at $\sin\theta=0$ the other factor is undefined.
Think about the function $$y=\frac{x(x+1)}{x}$$ this is pretty much equivalent to $y=x+1$ except for the fact that it has a hole at x=0 (the coordinate (0,1)). If we solved this equation for when y=0, clearly the only answer is x=-1, but if we factored out x then what we'd have is:
$$x\left(\frac{x+1}{x}\right)=0$$
Which seems like it would have an x=0 solution, but it doesn't by the same reasoning.
I see, thanks.
But why does this happen? Is there a particular reason or just one of those things.
Mentallic
Homework Helper
I see, thanks.
But why does this happen? Is there a particular reason or just one of those things.
It happens because when we use the rule that if $ab=0$ then either a, b or both are equal to zero, we are assuming that a and b are real numbers. Undefined numbers and infinite are not real.
Without using any rigor, if we use an undefined number like 1/0 such that a=0 and b=1/0, then ab=1 (again, I'm abusing the maths here just to explain a point, don't take it as being correct). As you can see while we solved for a=0, it turns out that ab didn't turn out to be equal to zero, hence we cannot take a=0 because it is not a solution.
thank you | 655 | 2,263 | {"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} | 4.09375 | 4 | CC-MAIN-2020-05 | longest | en | 0.927222 |
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A012047 arcsinh(sin(tanh(x)))=x-4/3!*x^3+76/5!*x^5-3536/7!*x^7+317392/9!*x^9... 1
1, -4, 76, -3536, 317392, -47590720, 10785422272, -3441103532288, 1468657322465536, -807408622401283072, 555453404529224756224, -467352773254586231115776, 472081158536303158224572416 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Vaclav Kotesovec, Table of n, a(n) for n = 0..212 FORMULA a(n) ~ c * (-1)^n * (2*n)! / (sqrt(n) * (arctan(arcsinh(1)))^(2*n)), where c = 1.0522205751764250070089302438417471228756... . - Vaclav Kotesovec, Feb 04 2015 MATHEMATICA nn = 20; Table[(CoefficientList[Series[ArcSinh[Sin[Tanh[x]]], {x, 0, 2*nn+1}], x] * Range[0, 2*nn+1]!)[[n]], {n, 2, 2*nn, 2}] (* Vaclav Kotesovec, Feb 04 2015 *) CROSSREFS Sequence in context: A024258 A012101 A012080 * A012010 A012155 A350489 Adjacent sequences: A012044 A012045 A012046 * A012048 A012049 A012050 KEYWORD sign AUTHOR Patrick Demichel (patrick.demichel(AT)hp.com) STATUS approved
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Last modified September 24 19:02 EDT 2023. Contains 365581 sequences. (Running on oeis4.) | 527 | 1,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.477848 |
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The rate constant f...
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# The rate constant for the decomposition of N2O5 at various temperatures is given below: T/°C → 0 - 20 - 40 - 60 - 80
0
Topic starter
The rate constant for the decomposition of N2O5 at various temperatures is given below:
T/°C → 0 - 20 - 40 - 60 - 80
105 x k/s-1 → 0.0787 - 1.70 - 25.7 - 178 - 2140
Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30 º and 50ºC.
Topic Tags
0
From the given data, we obtain
Slope of the line
$$\frac{y_2 - y_1}{x_2 - x_1}$$ = -12.301 K
According to Arrhenius equation,
Slope = -$$\frac{E_a}{R}$$
⇒ Ea = -Slope x R
= -(-12.301K) x (8.314 JK-1mol-1)
= 102.27 kJ mol-1
Again,
In k = In A - $$\frac{E_a}{RT}$$
In A = In k - $$\frac{E_a}{RT}$$
When T = 273 K,
Then, In A = -7.147 + $$\frac{102.27 \times 10^3}{8.314 \times 273}$$
= 37.911
Therefore, A = 2.91 x 106
When T = 30 + 273K = 303K
$$\frac{1}{T}$$ = 0.0033K = 3.3 x 10-3 K
Then, at $$\frac{1}{T}$$ = 3.3 x 10-3 K,
ln k = -2.8
Therefore, k = 6.08 x 10-2 s-1
Again, when T = 50 + 273K = 323K
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# A regular cone first placed in such a way its axis is perpendicular to H.P and next to this is tilted such that its base is making some acute angle with H.P. The top view for previous and later one will be
A. triangle, triangle
B. irregular shape of circle and triangle, triangle
C. circle, irregular shape of circle and triangle
D. circle, triangle
Answer» C. circle, irregular shape of circle and triangle
Explanation: for given positions of solid the solid is just tilted to some angle with h.p and previously given the axis is perpendicular to
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# Everything you’re looking for is right here!
All content is at your fingertips...
# Yardlong bean, cooked, boiled, drained, with salt
You can easily calculate the calorie and nutritional values of Yardlong bean, cooked, boiled, drained, with salt for different amounts (pod, cup slices) by clicking on the CALCULATE button. You can analyze your nutrition history by adding food into your nutrition diary.
## How many calories are in Yardlong bean, cooked, boiled, drained, with salt?
Yardlong bean, cooked, boiled, drained, with salt contains 47 calories per 100 grams. This value corresponds to about 2% of the daily energy expenditure of an adult burning about 2000 calories a day.
## Calorie, carbohydrate, protein and fat percentages of Yardlong bean, cooked, boiled, drained, with salt
100 grams of Yardlong bean, cooked, boiled, drained, with salt contains 9.17 grams of carbohydrate, 2.53 grams of protein and 0.1 grams of fat. It consist of 78% carbohydrate, 21% protein and 1% fat. 77% of total calories of the food are from carbohydrate, 21% are from protein, 2% are from fat.
100 grams of Yardlong bean, cooked, boiled, drained, with salt is approximately 3% of daily carbohydrate needs, 5% 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 Yardlong bean, cooked, boiled, drained, with salt
Vegetables and Vegetable Products
### 47 kcal
100 gram(s)
Amount 100 g
Calorie 47 kcal
Carbohydrate, by difference 9.17 g
Protein 2.53 g
Total lipid (fat) 0.1 g
Water 87.47 g
Calcium, Ca 44 mg
Iron, Fe 0.98 mg
Magnesium, Mg 42 mg
Phosphorus, P 57 mg
Potassium, K 290 mg
Sodium, Na 240 mg
Zinc, Zn 0.36 mg
Copper, Cu 0.047 mg
Manganese, Mn 0.201 mg
Selenium, Se 1.5 µg
Vitamin C, total ascorbic acid 16.2 mg
Thiamin 0.085 mg
Riboflavin 0.099 mg
Niacin 0.63 mg
Pantothenic acid 0.051 mg
Vitamin B-6 0.024 mg
Folate, total 45 µg
Folic acid 0 µg
Folate, food 45 µg
Folate, DFE 45 µg
Vitamin B-12 0 µg
Vitamin A, IU 450 IU
Vitamini A, RAE 23 µg
Retinol 0 µg
Vitamin D (D2 + D3) 0 µg
Vitamin D 0 IU
Fatty acids, total saturated 0.026 g
Fatty acids, total monounsaturated 0.009 g
Fatty acids, total polyunsaturated 0.042 g
Cholesterol 0 mg
Ash 0.73 g
Source USDA
## How many calories are in other foods in the food group Vegetables and Vegetable Products?
Some of other foods in the food group Vegetables and Vegetable 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. | 810 | 2,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.846386 |
http://www.statmodel.com/discussion/messages/14/9310.html?1333301844 | 1,500,612,324,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423716.66/warc/CC-MAIN-20170721042214-20170721062214-00230.warc.gz | 561,368,734 | 7,799 | Set Bayesian Priors in LGM
Message/Author
Shin, Tacksoo posted on Friday, March 30, 2012 - 5:50 pm
Suppose that we know the values of population parameter and so can consider informative prior distributions. Then, we set Bayesian priors for all these parameters like,
I S | y1@0 y2@1 y3@2 y4@3;
[I] (a);
[S] (b);
I (c);
S (d);
I with S (e);
model priors:
a ~ N(100, 10);
b ~ N(10, 3);
c ~ N(500, 200);
d ~ N(50, 25);
e ~ N(-18, 15);
Am I on the right track?
Linda K. Muthen posted on Saturday, March 31, 2012 - 10:11 am
If you know the population parameter values, why are the variances not zero, for example,
a ~ N(100, 0);
Shin, Tacksoo posted on Saturday, March 31, 2012 - 3:59 pm
Thank you, Linda.
One more question! As recognized, the population parameter values cannot be easily known in real circumstance. So, if one wants to set (weakly??) informative priors based on the past literatures and do following steps:
1) From the meta analysis, one found that the mean and variance of intercept factor were respectively about 100 and 10.
2) then, set the prior like,
a ~ N(100, 10).
Does my story seem good?
Linda K. Muthen posted on Sunday, April 01, 2012 - 10:37 am
To use these priors you would need to be very certain that your study is the same as the studies in the meta analysis.
Shin, Tacksoo posted on Monday, June 17, 2013 - 3:15 am
Dear Linda,
I got reviewer's comment related to choice of priors. As knew, normal prior is often used for mean (fixed effect) parameter estimates, while inverse-Gamma and inverse-Wishart priors are commonly used for the variance/covariance. Although I set the normal priors for variance/covariance, the reviewer pointed out that the prior distribution of a variance should not be normal.
First, I cannot set normal priors for variances at all? Secondly, if yes, I need to set IW or IG for these parameters. How can I do it when considering above example?
I (c);
S (d);
I with S (e);
model priors:
c ~ N(500, 200);
d ~ N(50, 25);
e ~ N(-18, 15);
OR,
model priors:
c ~ IW(S, degree of freedom)
S here is positive definite of matrix, right? What value of S can be this example?
Bengt O. Muthen posted on Monday, June 17, 2013 - 8:40 am
You don't want to use a normal prior for variances because the normal prior says that it is possible to get negative variances. In your example you can use a mildly informative inverse-Wishart prior:
c~iw(1,3);
d~iw(1,3);
e~iw(0,3);
See the Muthen-Asparouhov (2012) Psych Methods article for a description of IW priors. Also see:
Muthén, B. (2010). Bayesian analysis in Mplus: A brief introduction. Technical Report. Version 3.
Both papers are on our website.
S.Arunachalam posted on Friday, July 05, 2013 - 12:59 pm
Prof. Muthen,
I am slightly confused with the IW prior as to how the mean and variance are calculated for this prior because the first parameter seems like a matrix?
1.) For the IW(S,df) prior, as the first parameter S is a matrix, I understand setting it to 1 (as in the above example for parameters d & e) implies identity matrix; but what does 0 (in above example for parameter e) stand for? So when I set the first parameter 'S' to 1, does it imply the mean is 1 and if I set 'S' to 0 does it imply the mean is 0?
2.)And if I set S to 0, from the formulas in both the papers for variance calculations of IW, as the numerator has 'S', does it mean the variance is 0 as well?
My apologies if I am missing something very obvious here. Please advice.
Bengt O. Muthen posted on Saturday, July 06, 2013 - 8:33 am
The IW prior is indeed a bit complex. Two facts are useful here:
a) with increasing df (second argument), the prior is stronger (given more weight relative to data).
b) the mode is the first argument divided by df+p+1, where p is the number of variables.
1) Your "e" parameter is a covariance for which zero is a neutral point; hence the zero as first argument for IW. So the mode is zero.
2) Your variance parameters c and d are given first argument 1. You have p=2. The mode will then be 1/(3+2+1).
S.Arunachalam posted on Saturday, July 06, 2013 - 1:40 pm
Prof. Muthen, Thank you very much for a clear simple explanation;it is becoming more clearer now.
In a similar manner is there a simple way to know/calculate the variance as well?
For my current analysis what I did was to create another separate Mplus script where I inputted different values for IW(S,df) S & df parameters and got in the output the values of Mean and variance for the IW prior. And then I used the proper S and df for my main analysis. Is this work around ok? I am asking this because Mplus output for same S & df gives slightly different Variance values? For eg below for IW(.9,52):
Parameter 30~IW(3.000,52) 0.0789 0.0003 0.0186
Parameter 31~IW(0.900,52) 0.0237 0.0002 0.0155
Parameter 32~IW(4.000,52) 0.1053 0.0006 0.0248
Parameter 33~IW(0.900,52) 0.0237 0.0003 0.0172
Parameter 34~IW(0.900,52) 0.0237 0.0004 0.0198
Parameter 35~IW(5.000,52) 0.1316 0.0010 0.0310
Parameter 36~IW(0.900,52) 0.0237 0.0003 0.0172
Bengt O. Muthen posted on Monday, July 08, 2013 - 6:28 am
The variance of IW priors is discussed in the Appendix of the 2012 Muthen-Asparouhov article in Psych Methods.
You can get estimates of the prior means and variances as you did. They are based on random draws and can therefore vary somewhat.
S.Arunachalam posted on Monday, July 08, 2013 - 10:20 am
Dear Prof. Muthen,
Thank you so very much !! :o)
Sincerely
Arun
M Hamd posted on Monday, December 09, 2013 - 4:45 pm
Dear professor
We conducted two studies. We want to incorporate the findings study 1 in study 2 by using informed bayesian. In this case, the prior used will be the standardized regression coefficients from the previous study or the unstandardized regression coefficients? If unstandardized, how does that address the issue of a different scale range for one of our variable (i.e., 1-9 likert in study 1, vs 1-7 likert in study 2).
Linda K. Muthen posted on Tuesday, December 10, 2013 - 11:17 am
You should use the unstandardized coefficients. Use a different prior for each variable.
Sarah Dermody posted on Friday, December 12, 2014 - 1:14 pm
I am estimating a LGM of alcohol use across 7 waves of data. At each wave, alcohol use is the average number of drinks consumed per day, which can take on positive non-integer values (e.g., 0, .5, 3.2). As such, I would like to constrain the estimates to positive values (because I believe negative binomial is not an option unless I round to nearest integer). It seems like this would require using Bayesian priors (specifying an inverse-gamma prior?) but I am not sure if there is an alternative way to accomplish this in Mplus.
Bengt O. Muthen posted on Friday, December 12, 2014 - 6:32 pm
How about using the censored-normal model with censoring from below at 0?
Andrea Norcini Pala posted on Saturday, December 10, 2016 - 6:28 am
Hi,
Can the estimates of a regression analysis be used as model priors of the same regression analysis? Or that would be considered "cheating"?
Specifically, running Monte Carlo simulations I notice that Bayes produces accurate estimates even with small sample size and small effect size. However, the power is low.
Thanks,
Andrea
Bengt O. Muthen posted on Sunday, December 11, 2016 - 12:12 pm
I wouldn't recommend this. | 2,111 | 7,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-30 | latest | en | 0.850304 |
http://shukunegi.com/wp-content/luau-party-otxjlcs/s6jw5o.php?tag=height-of-isosceles-right-triangle-b2b642 | 1,680,099,308,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00662.warc.gz | 43,712,010 | 11,501 | Given that is a 45/45/90 triangle, it means that it's also isosceles. Using Pythagorean Theorem we have; (Hypotenuse ) 2 = ( Base) 2 + (Height ) 2 to a height of almost zero. Is it possible to have an isosceles scalene triangle? The length of its hypotenuse is (A) √32 cm (B) √16 cm (C) √48 cm (D) √24 cm. either the copyright owner or a person authorized to act on their behalf. In some triangles, like right triangles, isosceles and equilateral triangles, finding the height is easy with one of two methods. In today's lesson we'll learn a simple strategy for proving that in an isosceles triangle, the height to the base bisects the base. 1 If the diagonal of a right triangle is 8 cm, find the lengths of the other two sides of the triangle given that one of its angles is 30 degrees. The hypotenuse of an isosceles right triangle with side $${a}$$ is an The equation of a right triangle is given by a 2 + b 2 = c 2, where either a or b is the height and base of the triangle and c is the hypotenuse. Lengths of an isosceles triangle The height of an isosceles triangle is the perpendicular line segment drawn from base of the triangle to the opposing vertex. Calculates the other elements of an isosceles right triangle from the selected element. Find the area of the triangle. The height (h) of the isosceles triangle can be calculated using the Pythagorean theorem. An isosceles triangle is a triangle with two sides of equal length. answered Aug 20, 2020 by Sima02 (49.2k points) selected Aug 21, 2020 by Dev01 . Perimeter of an Isosceles triangle = sum of all the three sides. Calculate the surface area of the prism. Formulas Area. Therefore the three sides are in the ratio . National University of Mexico (UNAM), Bachelors, Vocal Performance. So 2x+5 = 11, which means x=3. link to the specific question (not just the name of the question) that contains the content and a description of This is a must be a 30°-60°-90° triangle. 101 S. Hanley Rd, Suite 300 Base = Height = 4cm. Because we are working with a triangle, the base and the height have the same length. 1 : 1 : . If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one Track your scores, create tests, and take your learning to the next level! Since this is an isosceles triangle, by definition we have two equal sides. The word isosceles is pronounced "eye-sos-ell-ease" with the emphasis on the 'sos'.It is any triangle that has two sides the same length. Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; You now have two equal right triangles. Thus, if you are not sure content located Let us assume both sides measure “S” then the formula can be altered according to the isosceles right triangle. AB ≅AC so triangle ABC is isosceles. Isosceles Right Triangle A right triangle is a triangle in which exactly one angle measures 90 degrees. Given arm a and base b : area = (1/4) * b * √( 4 * a² – b² ) Given h height from apex and base b or h2 height from other two vertices and arm a : area = 0.5 * h * b = 0.5 * … The isosceles right triangle, or the 45-45-90 right triangle, is a special right triangle. Example 3. The length of one of the legs can be solved for in one of two ways. A right isosceles triangle is a special triangle where the base angles are $$45 ^\circ$$ and the base is also the hypotenuse. Therefore, we use the ratio of x: x√3:2x. The only exception would be a right triangle — in a right triangle, if one of the legs is the base, the other leg is the altitude, the height, so it’s particularly easy to find the area of right triangles.” So you will basically only have to be able to solve for the height of a right triangle … Area (A) = ½ (b × h), where b = base and h= height . At what rate are the lengths of the legs of the triangle changing? Isosceles triangle The leg of the isosceles triangle is 5 dm, its height is 20 cm longer than the base. misrepresent that a product or activity is infringing your copyrights. A right isosceles triangle is a triangle with a vertex angle equal to 90°, and base angles equal to 45°. If all three sides are the same length it is called an equilateral triangle.Obviously all equilateral triangles also have all the properties of an isosceles triangle. Thus, we can use the Pythagorean Theorem to find the length of the height. The third unequal angle of an isosceles … Isosceles right triangle Area of an isosceles right triangle is 18 dm 2. Regardless of having up to three different heights, one triangle will always have only one measure of area. In order to find the height, you would need to set it up as this: S=o/h, … An acute isosceles triangle is a triangle with a vertex angle less than 90°, but not equal to 60°.. An obtuse isosceles triangle is a triangle with a vertex angle greater than 90°.. An equilateral isosceles triangle is a triangle with a vertex angle equal to 60°. Varsity Tutors. Best answer (A) √32 cm. How to find the height of an isosceles triangle. Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Defining Isosceles Right Triangles and Solving Problems Using Them The hypotenuse length for a=1 is called Pythagoras's constant. According to the internal angle amplitude, isosceles triangles are classified as: Rectangle isosceles triangle : two sides are the same. The length of one of the legs can be solved for in one of two ways. Varsity Tutors LLC The area of an isosceles triangle is the amount of region enclosed by it in a two-dimensional space. If Varsity Tutors takes action in response to Penny . To calculate the height, you should use the following equation: The “a” is the leg length, and the “b” is the base length. What’s more, the lengths of those two legs have a special relationship with the hypotenuse (in addition to the one in the Pythagorean theorem, of course). The height (h) of the isosceles triangle can be calculated using the Pythagorean theorem. Isosceles triangle is a polygon with three vertices (corners) and three edges (sides) two of which are equal. How to find the height?? b is the base of the triangle. 6 as h = a 2 b = a √ 2 L = ( 1 + √ 2 ) a S = a 2 4 h = a 2 b = a 2 L = ( 1 + 2 ) a S = a 2 4 select element How to Calculate Edge Lengths of an Isosceles Triangle. So the key of realization here is isosceles triangle, the altitudes splits it into two congruent right triangles and so it also splits this base into two. ⇒2x = 8 cm ⇒ x = 4cm. The height can be anything from 16 inches. Defining Isosceles Right Triangles and Solving Problems Using Them Draw the height from the obtuse angle to the "5" side. Area of Isosceles Triangle Formula, Side Lengths. It's also possible to establish the area of a triangle which is isosceles if you don't know the height, but know all side lengths instead. An isosceles right triangle has legs that are each 4cm. AREA(A)= ½(SxS) A=1/2xS 2. Let us take the base and height of the triangle be x cm. If the hypotenuse of a 45-45-90 right triangle is then: The height and the base of the triangle will be the same length since it is a 45-45-90 triangle (isosceles). sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require we use congruent triangles to show that two parts are equal. The two acute angles are equal, making the two legs opposite them equal, too. The differences between the types are given below: Types of Isosceles Triangle. Types of Isosceles Triangles. The two acute angles are equal, making the two legs opposite them equal, too. This line divides θ perfectly in half. Hence, the base and height of the right triangle is 6mm each. Height. Given an integer N and an isosceles triangle consisting of height H, the task is to find (N – 1) points on the triangle such that the line passing through these points and parallel to the base of the triangle, divides the total area into N equal parts.. Isosceles triangles are classified into three types: 1) acute isosceles triangle, 2) obtuse isosceles triangle, and 3) right isosceles triangles. An isoceles right triangle is another way of saying that the triangle is a triangle. Look up that angle in a trig table. This means you can use one equal side as the base, and the other as the height. Calculate the length of its base. In some triangles, like right triangles, isosceles and equilateral triangles, finding the height is easy with one of two methods. Equilateral triangle; Isosceles triangle; Right triangle; Square; Rhombus; Isosceles trapezoid; Regular polygon; Regular hexagon ; All formulas for radius of a circle inscribed; Geometry theorems . Since the two opposite sides on an isosceles triangle are equal, you can use trigonometry to figure out the height. If you've found an issue with this question, please let us know. the There are two different heights of an isosceles triangle; the formula for the one from the apex is: hᵇ = √(a² - (0.5 * b)²), where a is a leg of the triangle and b a base. Answer: The sum is 4.73. How would you show that a triangle with vertices (13,-2), (9,-8), (5,-2) is isosceles? There are four types of isosceles triangles: acute, obtuse, equilateral, and right. Since the sum of the measures of angles in a triangle has to be 180 degrees, it is evident that the sum of the remaining two angles would be another 90 degrees. What is the height of a triangle if its hypotenuse is cm. Shrinking isosceles triangle The hypotenuse of an isosceles right triangle decreases in length at a rate of $4 \mathrm{m} / \mathrm{s}$ a. So this length right over here, that's going to be five and indeed, five squared plus 12 squared, that's 25 plus 144 is 169, 13 squared. Step-by-step explanation: Height of a triangle is a perpendicualr line linking a vertex and its opposite side. Therefore, h = . Your Infringement Notice may be forwarded to the party that made the content available or to third parties such As well, this line you've drawn is the height of the original triangle. information described below to the designated agent listed below. In an isosceles triangle, if the vertex angle is $$90^\circ$$, the triangle is a right triangle. What is the area of isosceles triangle calculator? View solution A girls' camp is located 3 0 0 m from a straight road. The general formula for the area of triangle is equal to half the product of the base and height of the triangle. Problem: Finding the area of an isosceles triangle when only THREE SIDES are known. Because the hypotenuse if 2√7 cm, that means that the base and the height (the two remaining sides) will be equivalent. Substitute. means of the most recent email address, if any, provided by such party to Varsity Tutors. Isosceles triangle formulas for area and perimeter. 4 We are asked to find the perimeter of the triangle. Send your complaint to our designated agent at: Charles Cohn 2 ABC can be divided into two congruent triangles by drawing line segment AD, which is also the height of triangle ABC. They have the ratio of equality, 1 : 1. Using the Pythagorean Theorem, , we've already determined that "a" and "b" are the same number. For an isosceles right triangle with side lengths a, the hypotenuse has length sqrt(2)a, and the area is A=a^2/2. Walden University, Masters in ... Columbia University in the City of New York, Bachelor in Arts, Classics. ... You now have a little right triangle whose height is h, hypotenuse is 8, and other leg is (let's call it) x. Calculates the other elements of an isosceles right triangle from the selected element. This forms two right triangles inside the main triangle, each of whose hypotenuses are "3". An isosceles right triangle therefore has angles of 45 degrees, 45 degrees, and 90 degrees. This allows for the equation to be rewritten as , which may be simplified into. By using the Pythagorean Theorem, the process of finding the missing side of a triangle is pretty simple and easy. Each right triangle has an angle of ½θ, or in this case (½)(120) = 60 degrees. The sides a, b/2 and h form a right triangle. ChillingEffects.org. Let height of triangle = h. As the triangle is isosceles, Let base = height = h. According to the question, Area of triangle = 8cm 2 ⇒ ½ × Base × Height = 8 ⇒ ½ × h × h = 8 ⇒ h 2 = 16 ⇒ h = 4cm. (Lesson 26 of Algebra.) The isosceles right triangle, or the 45-45-90 right triangle, is a special right triangle. Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially Sides b/2 and h are the legs and a hypotenuse. herons formula; class-9; Share It On Facebook Twitter Email. Your name, address, telephone number and email address; and which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Area of a isosceles right triangle, say A having base x cm and . height bisector and median of an isosceles triangle : = Digit Area of a isosceles right triangle, say A having base x cm and . Examples: Input: N = 3, H = 2 Output: 1.15 1.63 Explanation: Make cuts at point 1.15 and 1.63 as shown below: Divide the isosceles into two right triangles. In an isosceles triangle, there are also different elements that are part of it, among them we mention the following: Bisector; Mediatrix; Medium; Height. The cosine of either of the original acute angles equals 2½÷3, or 0.833. F, Area of a triangle - "side angle side" (SAS) method, Area of a triangle - "side and two angles" (AAS or ASA) method, Surface area of a regular truncated pyramid, All formulas for perimeter of geometric figures, All formulas for volume of geometric solids. With the help of the community we can continue to 1. Because this is an isosceles triangle, this line divides the triangle into two congruent right triangles. Isosceles right triangle Area of an isosceles right triangle is 18 dm 2. Because the hypotenuse if 2√7 cm, that means that the base and the height (the two remaining sides) will be equivalent. Now you have a right triangle and you know the measure of the angle opposite the height and you know the length of the side (half the base b). An isosceles right triangle has area 8 cm 2. In an isosceles right triangle, the equal sides make the right angle. If you do the same thing to the right-hand side, you'll notice that the bottom side of the trapezoid is 11 = x + 5 + x. Isosceles triangle calculator computes all properties of an isosceles triangle such as area, perimeter, sides and angles given a sufficient subset of these properties. Draw a line down from the vertex between the two equal sides, that hits the base at a right angle. © 2007-2021 All Rights Reserved, How To Find The Height Of A 45/45/90 Right Isosceles Triangle, SSAT Courses & Classes in Dallas Fort Worth. Using the Pythagorean Theorem where l is the length of the legs, . What is the minimum value of the sum of the lengths of AP, BP and CP? Based on this, ADB≅ ADC by the Side-Side-Side theorem for congruent triangles since BD ≅CD, AB ≅ AC, and AD ≅AD. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe In the image below, we can see that an isosceles triangle can be split into 2 right angle triangles. An isosceles right triangle is a right triangle where the angles of the triangle are 90$$^\circ$$, 45 $$^\circ$$ and 45$$^\circ$$ A scalene right triangle is a right triangle where one angle is 90$$^\circ$$ and the other two angles add up to 180$$^\circ$$ To find the perimeter, use the Pythagorean theorem to find the length of the hypotenuse, and add it to the lengths of the other sides. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing Solution. Isosceles triangle The leg of the isosceles triangle is 5 dm, its height is 20 cm longer than the base. Also, two congruent angles in isosceles right triangle measure 45 degrees each, and the isosceles right triangle is: Area of an Isosceles Right Triangle. Isosceles triangle Calculate the perimeter of isosceles triangle with arm length 73 cm and base length of 48 cm. If the triangle is a right triangle as in the first diagram but it is the hypotenuse that has length 16 inches then you can use Pythagoras' theorem to find the length of the third side which, in this case, is the height. a Two sides of isosceles right triangle are equal and we assume the equal sides to be the base and height of the triangle. This is important on the GMAT because some exam problems that look like they could be dealing with the unknown height of an isosceles triangle are really asking you to calculate the length of one side of a right triangle, which doubles as the height of an isosceles triangle. All formulas for radius of a circle inscribed, All basic formulas of trigonometric identities, Height, Bisector and Median of an isosceles triangle, Height, Bisector and Median of an equilateral triangle, Angles between diagonals of a parallelogram, Height of a parallelogram and the angle of intersection of heights, The sum of the squared diagonals of a parallelogram, The length and the properties of a bisector of a parallelogram, Lateral sides and height of a right trapezoid, Find the length of height = bisector = median if given lateral side and angle at the base (, Find the length of height = bisector = median if given side (base) and angle at the base (, Find the length of height = bisector = median if given equal sides and angle formed by the equal sides (, Find the length of height = bisector = median if given all side (. Sides b/2 and h are the legs and a hypotenuse. One corner is blunt (> 90 o ). Calculate the length of height = bisector = median if given lateral side and angle at the base or side (base) and angle at the base or equal sides and angle formed by the equal sides or all side How do you find the height of an isosceles triangle - Calculator Online Height of Isosceles Trapezoid? Regardless of having up to three different heights, one triangle will always have only one measure of area. The 45°-45°-90° triangle, also referred to as an isosceles right triangle, since it has two sides of equal lengths, is a right triangle in which the sides corresponding to the angles, 45°-45°-90°, follow a ratio of 1:1:√ 2. The base of vertical prism is an isosceles triangle whose base is 10 cm and the arm is 13 cm long. Isosceles right triangle Calculate the area of an isosceles right triangle whose perimeter is 377 cm. The inradius r and circumradius R are r = 1/2(2-sqrt(2))a (1) R = … In an isosceles right triangle the length of two sides of the triangle are equal. One of the angles is straight (90 o ) and the other is the same (45 o each) Triangular obtuse isosceles angle : two sides are the same. Find the height of the 45-45-90 right triangle with a hypotenuse of . A description of the nature and exact location of the content that you claim to infringe your copyright, in \ We have a special right triangle calculator to calculate this type of triangle. And using the base angles theorem, we also have two congruent angles. The height and length, or base, of an isosceles right triangle are the same. 1 Answer +1 vote . You can find it by having a known angle and using SohCahToa. If you were to draw an imaginary line from the vertex angle to the base (at a 90-degree angle from the base), you would get the height of your isosceles triangle. Triangles each have three heights, each related to a separate base. Where. In an isosceles triangle, knowing the side and angle α, you can calculate the height, since the side is hypotenuse and the height is the leg, then the height will be … National Conservatory of Music (Mexico). Hispanic Languag... Virginia Commonwealth University, Bachelor of Science, Business Administration and Management. Polyforms made up of isosceles right triangles are called polyaboloes. Isosceles triangle formulas for area and perimeter. your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the A point P may be placed anywhere along the line segment AQ. Isosceles acute triangle elbows : the two sides are the same. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Let us take the base and height of the triangle be x cm. h = a 2 b = a √ 2 L = ( 1 + √ 2 ) a S = a 2 4 h = a 2 b = a 2 L = ( 1 + 2 ) a S = a 2 4 select element improve our educational resources. Whether you are looking for the triangle height formulas for special triangles such as right, equilateral or isosceles triangle or any scalene triangle, this calculator is a safe bet - it can calculate the heights of the triangle, as well as triangle sides, angles, perimeter and … Given that is a 45/45/90 triangle, it means that it's also isosceles. The base is 7. the 2 equal sides are 5.7cm each. So the area of an Isosceles Right Triangle = S 2 /2 square units. b. Like the 30°-60°-90° triangle, knowing one side length allows you to determine the lengths of the other sides of a 45°-45°-90° triangle. 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Worst College Tennis Teams, Duke Trinity College Acceptance Rate, Gavita Led Vs Fluence, Entry Level Public Health Jobs Reddit, Recognized Agent Nova Scotia, Husband In Tamil Meaning, Public Health Volunteer Uk, Northeastern Video Tour, How To Become A Section 8 Landlord In Mississippi, Concrete Countertop Sealer Reviews, Colleges In Chalakudy, | 7,663 | 30,585 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-14 | latest | en | 0.894638 |
https://www.teachoo.com/3251/692/Ex-4.5--14---Find-numbers-a-and-b-such-that-A2---aA---bI--O/category/Ex-4.5/ | 1,680,433,334,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00118.warc.gz | 1,116,424,693 | 33,298 | Ex 4.5
Chapter 4 Class 12 Determinants
Serial order wise
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Ex 4.5, 14 For the matrix A = [■8(3&[email protected]&1)] , find the numbers a and b such that A2 + aA + bI = O. Finding A2 A2 = A.A = [■8(3&[email protected]&1)] [■8(3&[email protected]&1)] = [■8(3(3)+2(1)&3(2)+2(1)@1(3)+1(1)&1(2)+1(1))] = [■8(9+2&[email protected]+1&2+1)] = [■8(11&[email protected]&3)] Now, A2 + aA + bI = O Putting values [■8(11&[email protected]&3)] + a [■8(3&[email protected]&1)] + b [■8(1&[email protected]&1)] = O [■8(11&[email protected]&3)] + [■8(3a&[email protected]&a)] + [■8(b&[email protected]&b)] = O [■8(11+3a+b&[email protected]+a+0&3+a+b)] = O [■8(3a+b+11&[email protected]+a&a+b+3)] = [■8(0&[email protected]&0)] Since the matrices are equal, Comparing corresponding elements 3a + b + 11 = 0 2a + 8 = 0 4 + a = 0 a + b + 3 = 0 Solving (3) a + 4 = 0 a = –4 Putting value a in (1) 11 + 3 a + b = 0 11 + 3 (–4) + b = 0 11 – 12 +b = 0 –1 + b = 0 …(1) b = 1 Hence, a = −4, b = 1 | 486 | 1,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-14 | longest | en | 0.476071 |
https://www.w3resource.com/sql-exercises/view/sql-view-exercise-14.php | 1,725,753,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00201.warc.gz | 1,017,155,659 | 27,523 | SQL - View to show the number of orders in each day
# SQL Exercises: View to show the number of orders in each day
## SQL VIEW: Exercise-14 with Solution
14. From the following table, create a view to display the number of orders per day. Return order date and number of orders.
Sample table: orders
Sample Solution:
``````-- Creating a VIEW named 'dateord' with columns 'ord_date' and 'odcount'
CREATE VIEW dateord(ord_date, odcount)
-- Selecting distinct order dates and counting the number of orders for each date
-- Using the 'orders' table and grouping by 'ord_date'
AS SELECT ord_date, COUNT (*)
FROM orders
GROUP BY ord_date;
``````
output:
```sqlpractice=# SELECT *
sqlpractice-# FROM dateord;
ord_date | odcount
------------+---------
2012-10-05 | 2
2012-08-17 | 3
2012-07-27 | 1
2012-09-22 | 1
2012-09-10 | 3
2012-10-10 | 2
2012-06-27 | 1
2012-04-25 | 1
(8 rows)
```
Code Explanation:
The provided statement in SQL creates a view named dateord, which returns the count of orders placed on each unique order date.
The COUNT function have used to return the total number of orders on each unique order date.
Then groups the results by the ord_date column and returns the count of orders for each unique order date.
Inventory database model:
Contribute your code and comments through Disqus.
Previous SQL Exercise: View to show all matches of customers with salesmen.
Next SQL Exercise: View to show the salesmen issued orders on given date.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
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https://uk.mathworks.com/matlabcentral/cody/problems/1058-subtract-two-positive-numbers/solutions/503938 | 1,579,797,526,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00535.warc.gz | 717,319,607 | 15,454 | Cody
# Problem 1058. Subtract two positive numbers
Solution 503938
Submitted on 23 Sep 2014 by Abdullah Caliskan
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
%% a = '0205'; b = '8120'; y_correct = 7915; A = fread(fopen('yourSubtract.m')); assert(~any(regexpi(char(A)','str2num|str2double'))); assert(isequal(yourSubtract(a,b),y_correct))
a = 0 2 0 5 h = 4 3 2 1 h = 3 2 1 0 ans = 7915
2 Pass
%% a = '12589'; b = '78956'; y_correct = 66367; assert(isequal(yourSubtract(a,b),y_correct))
a = 1 2 5 8 9 h = 5 4 3 2 1 h = 4 3 2 1 0 ans = 66367 | 249 | 669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-05 | latest | en | 0.59832 |
https://math.hws.edu/xJava/MB/ | 1,657,180,992,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683708.93/warc/CC-MAIN-20220707063442-20220707093442-00300.warc.gz | 414,174,511 | 6,630 | Java applets are no longer well-supported, and support will probably disappear entirely soon. For a program that runs in your browser, use the JavaScript Mandelbrot Viewer: http://math.hws.edu/eck/js/mandelbrot/MB.html (The Java version has more features, and you can still download it for use on your own computer.)
## Mandelbrot Viewer
The applet on this page is the latest version of my Mandelbrot Viewer application. It requires Java 5.0 or higher. (For an old version of the program, click here.) A standalone version, xMandelbrot.jar with additional capabilites is also available, and you can download or browse the source code. There is also a Gallery of Examples. (Compared to Version 1.0, Version 1.2, June 2011, fixes a few bugs, adds a few new menu commands, and is somewhat faster for high-precision computations.)
Some documentation for the program can be found below the applet. The Mandelbrot Set itself is a well-known mathematical object; see the Wikipedia entry for more information, if you need it.
### The applet:
Visualizations of the Mandelbrot Set can be computed as follows: Take a point (a,b) in the xy-plane. Set (x,y) = (a,b). Perform the computation (x,y)= (a2 - b2 + a, 2xy + b) repeatedly until one of two things happen; either (1) the point (x,y) is more than 2 units distant from (0,0); or (2) the number of times that you have performed the computation is equal to some set upper limit. The number of times that you perform the computation is called the number of iterations, and the upper limit on the number of iterations is referred to as MaxIterations in the program. In the case (1), the starting point (a,b) is definitely not in the Mandelbrot Set; in the case (2), the starting point (a,b) is possibly in the Mandlebrot Set (but increasing the maximum number of iterations might show that (a,b) is actually not in the set). This sort of computation can't prove that a point is in the Mandelbrot Set, since that would require doing an infinite number of iterations and checking that (x,y) never moves more than two units away from (0,0).
In the program, each pixel in an image is colored as follows: The image represents a rectangle in the xy-plane bounded by xmin on the left, xmax on the right, ymin on the bottom, and ymax on the top. Take (a,b) to be the point in the center of the pixel, and use (a,b) as the starting point in the computation described in the previous paragraph. If case (1) applies, then the pixel is assigned a color that depends on the number of iterations that were done. The colors that are used for such points are referred to as the "palette"; you can modify the palette using the "Show Palette Editor" and "Apply Default Palette" commands in the "Control" menu. If the case (2) applies, the pixel is assigned a color that represents a point that is possibly in the Mandelbrot Set; by default, this color is black, but the color can be set with the "Mandelbrot Color" submenu of the "Control" menu. Note that increasing the MaxIteration count can convert pixels from case (2) to case (1) -- that is, it can fill in black areas with color as more points are proven to be not in the Mandelbrot Set.
The "interesting" regions of the xy-plane are those along the boundary of the Mandelbrot Set, between black and colored areas. The object of the program is to find an interesting region and to make its structure visible -- and beautiful -- with well-chosen colors. You can zoom in on a point along the boundary, and you will always find more structure. To zoom in, click-and-drag the mouse to enclose a rectangular reqion. When you release the mouse, the inside of the rectangle will be expanded to fill the entire image. (The mouse can also be used for other functions; see the "Help..." command in the "Tools" menu for more information.) As you zoom in, you will have to increase MaxIterations, and you will want to use the Palette Editor to adjust the colors.
After computing the image as described here, the program will make a second pass in which it computes iteration counts at additional points. The data from these additional points is averaged in with the data from the first pass. This can give a smoother, more attractive image. You can turn off this behavior, if you want, using the "Enable Subpixel Sampling" command in the "Control" menu.
If you zoom in very far, the program will switch over to "High Precision" calculation. The numbers that are ordinarily used have only about 18 digits of accuracy. You can easily zoom in to the point where more digits are required. High precision computation allows any number of digits, but it takes much longer than normal precision (mostly because the normal precision numbers are implemented very efficiently in the hardware).
In addition to the applet version of the program, there is a a standalone version, which can be run independently of a web browser. The standalone version has a "File" menu that can be used to save the images created by the program. It can also save and load "Mandelbrot Settings" files, which are small files that contain specifications of all the data needed to reconstruct an image. The standalone version also has the ability to distribute its computation over a group of networked computers; more information about this can be found below.
• xMandelbrot.jar --- Download the standalone version of the program. If Java 5.0 or higher is properly installed, you should be able to run this by double-clicking, at least on Windows and on Mac OS 10.4 or higher. On Linux, if double-clicking does not work, you might have to right-click and use "Open With" to open the file with Java or with a command such as "java -jar". You can also run the program from the command line with the command: java -jar xMandelbrot.jar
• Browse the source online --- The actual source code files can be found in the directories edu/hws/eck/umb, edu/hws/eck/umb/comp, edu/hws/eck/umb/palette, and edu/hws/eck/umb/util.
• MandelbrotCL.jar --- A command-line version of the program, which can be used to create images from settings files, without a graphical user interface. Documentation can be found below. The source code for this program is included in the above source code links.
• MBNetServe.jar --- A server program that can be used to distribute the computation over a network. Documentation can be found below. The source code for this program is included in the above source code links.
### Documentation for MandelbrotCL.jar:
MandelbrotCL.jar is a simple command-line utility for making Mandelbrot images. This program can read Mandelbrot "settings" files, which are created by xMandelbrot.jar, and can create the images that are specified by those files. I use this utility, for example, to create small "thumbnail" images from a group of settings files. The syntax for using this command is:
``` java -jar MandelbrotCL.jar [options] filenames...
```
where "filenames..." represents one or more settings file names and the "[options]" can include any or all (or none) of the following:
• -size WWWxHHH --- where WWW and HHH are positive integers, specifies the size of the image. If no size is specified, 800x600 is used.
• -format XXX --- use XXX as the format for the image. PNG is the default. JPEG is also definitely supported. Other formats might be supported as well.
• -onepass --- turn subpixel sampling off. If this option is omitted, subpixel sampling is used. (This is an option in the Control menu of xMandelbrot.jar; subpixel sampling can produce smoother, more attractive images in many cases.)
• -net XXX --- add one or more network workers. The format for XXX is a list of one or more hosts, separated by commas. Each host can be specified as a host name or IP address optionally followed by a colon and a port number. The port number is only necessary if different from the default, 17071. No spaces are allowed in the list of computers. A copy of MBNetServe.jar should already be running on each of the specified computers.
For example, to make small JPEG images for two files named mbdata1.xml and mbdata2.xml:
``` java -jar MandelbrotCL.jar -size 160x120 -format jpeg mbdata1.xml mbdata2.xml
```
By the way, if you are working with very large images, you might need to tell the java virtual machine to use more memory than it ordinarily would. You can do this with the "-Xms" option to the java command. For example:
``` java -jar -Xms2000m MandelbrotCL.jar -size 3300x2550 -format jpeg settings.xml
```
### Documentation for MBNetServe.jar:
It can take a long time to compute some Mandelbrot images. This is especially true for high precision computation, which is used when you zoom in so far that the standard Java real number implementation does not have enough significant digits to represent the numbers involved. The networking option is for people who want to speed up these long computations and who have access to several networked computers.
To distribute Mandelbrot computations over a network, you must run MBNetServe.jar on each computer EXCEPT the one where you will run xMandlebrot.jar (or MandelbrotCL.jar). Once that is done, you can use the "Configure Multiprocessing..." command in the "Control" menu of xMandelbrot to configure that program to use the network. (For MandelbrotCL, use the "-net" option.) You will need to know something about networking in order to use this option. Most important, you will need to know the host names or IP addresses of the computers where the server is running.
To run MBNetServe.jar on the computers that you want to use as computation servers, use the following command in the directory that contains the program:
``` java -jar MBNetServe.jar [options]
```
where the "[options]" can include:
• -processcount XXX --- use XXX processes instead of the default number. Enter 0 for XXX to use one process for each available processor. The default is to use one less than this (if the number of processors is greater than one).
• -timeout XXX --- exit after XXX minutes of inactivity. The default is 30 minutes. Use 0 for XXX to mean that there is no timeout. You can also stop the server by pressing Control-C in the window where the program is running.
• -once --- accept one connection, and exit when that connection is closed. The default is to open a new listener after the connection is closed and wait for another connection.
• -quiet --- suppress all output.
• -port XXX --- listen on port number XXX instead of on the default port (17071). (Ordinarily, you will NOT need this option; just use the default.)
For example:
``` java -jar MBNetServe.jar -processcount 0
```
To use the network in xMandelbrot, choose the "Configure Multiprocessing..." command from the "Control" menu. In the dialog box, check the option labeled "Enable Networking". Then, use the "Add Network Host" button to add each computer where MBNetServe is running. When you click this button, a small dialog will open where you can type in the host name of a computer that is running the server. (You can also enter an IP address, instead of a host name.) After adding the hosts, click the "Apply Config Now" button -- the network is not activated until you press this button. You should see "Connecting" next to each host name in the list. In a few seconds, this should change to "Connected." If it changes to "Connection Failed", or if it continues to say "Connected" for some time, then there is something wrong -- check that the server is running and that the host names or IP addresses are correct. Even if some listed connections are not working, the program will still use those that are connected. (Note that the list of hosts that you have entered will be saved for the next time you run the program, but networking will not automatically be turned on when the program runs; you will have to enable it by hand each time you run the program.)
Note: If your server computers are Linux or Mac OS computers that are configured to accept ssh connections, you can start up the server program remotely, using the ssh command, instead of logging on to each individual computer to start the program. If the computer's name is, for example, cslab1.hws.edu, you could use a command such as:
``` ssh -f cslab1.hws.edu java -jar MBNetServe.jar -quiet -processcount 0
```
The "-f" option means that the command will be run in the background (after asking for your password, if required) so that you can continue to use the same command-line window to give other commands. This assumes that you have an account on cslab1.hws.edu with the same user name as on the computer where you give this command. It also assumes that MBNetServe.jar is in your home directory on cslab1.hws.edu. MBNetServe will shut itself down after 30 minutes of inactivity, so you don't have to worry about shutting it down. However, if you do want to shut down the server remotely, you can do so by using a special "-shutdown" option on the MBNetServe program. For example, if you are finished with the server that was started with the above command, you can use:
``` java -jar MBNetServe.jar -shutdown cslab1.hws.edu
```
to shut down the server. (This command will NOT work if the server is still being used by xMandelbrot or MandelbrotCL.) | 3,055 | 13,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.889472 |
http://www.controleng.com/single-article/understanding-how-genset-load-factor-affects-mission-critical-power/b689d48ab0261227778f2930e960f56b.html | 1,503,571,955,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133449.19/warc/CC-MAIN-20170824101532-20170824121532-00689.warc.gz | 501,692,769 | 19,582 | # Understanding how genset load factor affects mission-critical power
## Understanding how average load factor can affect mission-critical standby power systems during emergencies can ensure continuity of electrical power—and business operations.
11/10/2011
Consider this nightmarish scenario: At midday, the sky turns the color of ink, sudden rain squalls flood neighborhood streets, and disaster sirens wail as an F5 tornado scores a direct hit on a power substation. Nearby, a national bank data center’s UPS system and emergency power generators kick in—as designed—to keep mission-critical operations online. But with the utility outage likely to last for days, facility operators ask themselves, “Will the standby generators be able to supply all power needs for the duration of the outage?”
The answer to that question is related in part to load factor. An important consideration in sizing gensets is calculating the application’s average load factor.
The average load factor of a power system is determined by evaluating the amount of load and the amount of time the genset is operating at that load. Since the loads are normally variable, the result is found by calculating multiple load levels and time periods. Figure 1 shows a graph of a hypothetical standby load profile. Although the genset is loaded to 90% of its standby rating for a portion of the time, the average load factor over time is only 70% due to the natural variability of the building load.
Figure 1: In this graph of a hypothetical standby load profile, the 24-hour average load factor is derived from the graph, where P is power in kW and t is time. Courtesy: MTU Onsite Energy
In practice, it would be unlikely that a standby power system would be initially sized so small as to require operating at 100% of capacity at any time during an outage. However, electrical loads are often added, and growing power needs may begin to tax the capacity of a standby power system. It should be noted that any time that the genset is offline does not count toward the 24-hour average load factor.
For most facilities with properly designed emergency standby power systems, the possibility of exceeding a power system’s 24-hour average load factor limitation is remote. This is because most commercial facilities have variable load profiles that reduce the likelihood a power system’s 24-hour average load factor limitation will be exceeded, even during an extended outage. Many facilities also have noncritical loads that can be taken offline during extended outages to reduce the average load factor on the standby system, if necessary.
However, many mission-critical facilities have large, less-varying loads that can severely stress standby power systems during an extended power outage unless steps are taken during system design to accommodate the potential for a higher average load factor. Two examples of mission-critical facilities with high load factors are data centers and semiconductor manufacturing. In data centers, the computer server and HVAC equipment create high electrical loads that can vary little over time. Similarly, very high load factors are found in semiconductor foundries, where electric furnaces cannot be shut down without destroying large amounts of product.
Figure 2: This graph shows the load profile of gensets capable of an 85% load factor. Although the gensets are not loaded to 100% of their standby rating at any time, the average load factor during the outage is near 85%. Courtesy: MTU Onsite Energy
Because of these large, steady electrical loads, the load profile in a mission-critical application is likely to have less variability, in turn putting a more constant demand on the standby power system. Less load variability results in a higher average load factor that will require specifying a system with larger or more gensets capable of a 70% load factor or specifying gensets capable of higher than a 70% load factor.
Figure 2 shows that while the gensets are not loaded to 100% of their standby rating at any time, the average load factor during the outage is near 85%. In this case, the customer has taken advantage of gensets capable of an 85% load factor that can deliver more than 20% additional kW than gensets rated to only a 70% average load factor.
Defining genset standards
The International Organization for Standardization (ISO) has established standards that apply to all gensets. ISO defines how to measure and rate many quality and performance parameters. All major genset manufacturers use this standard to communicate their genset ratings to their customers. In particular, ISO 8528-1 describes how to establish genset ratings; measure performance; and evaluate engines, alternators, controls, and switchgear.
ISO 8528-1 sets a maximum 24-hour average load factor capability of 70% for both standby- and prime-rated gensets, unless a higher average is agreed to by the engine manufacturer. This means that a 3,000 kW genset meeting this standard must be able to provide an average of 2,100 kW per hour over a 24-hour period. In emergency standby applications, this means that the average load factor that can be sustained by most gensets over an extended outage of 24 hours or more must not exceed 70% of the nameplate standby rating, a factor that affects genset sizing.
One genset manufacturer allows an 85% average load factor on emergency standby-rated genset models above 200 kW. For example, the genset manufacturer allows its 3,250 kW genset to deliver a 24-hour average of 2,762.5 kW in this case. For certain applications involving multiple gensets, this higher average load factor capability may reduce the number of gensets needed to supply the load.
ISO 8528-1 defines four genset power output rating categories. They are:
1. Emergency standby power (ESP) rating: The ESP rating is the maximum amount of power that a genset is capable of delivering. It is normally used to supply facility power to a variable load in the event of a utility outage. No overload capacity is available for this rating. ISO 8528-1 limits the 24-hour average output to 70% of the nameplate ESP rating unless the manufacturer allows a higher average load factor.
2. Prime-rated power: A prime-rated genset is available for an unlimited number of hours per year in a variable-load application as long as the average load factor does not exceed 70% of the nameplate rating, unless the manufacturer allows a higher average load factor. This rating allows an overload capacity of 10%, but that additional capacity should not be used for more than one hour in every 12. The prime power rating for a given genset is typically 10% lower than the standby rating.
3. Limited-time running power: This rating is a subset of prime power and allows prime power to be available for a limited number of hours in a nonvariable load application. It is intended for use in situations where power outages are contracted (utility power curtailment or peak shaving). Gensets can be paralleled with the public utility up to 500 hours per year at 100% of the prime rating. However, be aware that the life of the generator drive engine may be reduced by this constant high load operation. Any application that operates at the prime power rating for more than 500 hours per year should use a larger genset at its continuous rating.
4. Continuous output power rating: The continuous power rating is used for applications where there is no utility power and the genset is relied upon for all power needs. Gensets with this rating are capable of supplying power at a constant 100% of rated load for an unlimited number of hours per year. No overload capability is available for this rating. The continuous power rating for a given genset is typically 25% to 30% lower than the standby rating.
Effects of load factor on power system design
Specifying standby gensets with a higher-than-average load factor capability can sometimes be a benefit in mission-critical applications. System designers may be able to reduce the size or number of gensets by using units approved for 85% average load factor, as opposed to the 70% average load factor. For example, to design a standby power system to supply an average load of 11,000 kW at a 70% average load factor would require eight 2,000 kW gensets. At a 70% average load factor rating, each genset would be able to deliver up to a 1,400 kW average, for a total capacity of 11,200 kW over an extended outage of 24 hours or more.
8 x 2,000 kW x 0.70 = 11,200 kW
Using gensets with an 85% average load factor capability would require only seven 2,000 kW units. Each genset would be able to deliver up to a 1,700 kW average, for a total average of 11,900 kW over an extended outage of 24 hours or more. That amounts to an extra 2,100 kW of effective generating capacity for extended outages and a reduction by one in the number of gensets needed.
7 x 2,000 kW x 0.85 = 11,900 kW
Conclusion
The load factor of any application affects the design and sizing of the standby power system. However, for mission-critical applications, particular attention must be paid to load factors because of the minimal ability of some facilities to reduce their electrical loads during extended outages.
While all major manufacturers of gensets use ISO-8528-1 as their standard, which sets the average 24-hour load factor at 70%, system designers can choose equipment that offers a higher average 24-hour load factor, which may result in a system with smaller and/or fewer gensets. Specifiers of standby power systems for mission-critical applications should understand average load factor and its implications for business continuity in the face of natural or manmade disasters.
Kraemer is application engineering manager at MTU Onsite Energy, Mankato, Minn. He has a degree in automotive engineering from Minnesota State University and leads MTU Onsite Energy’s new product development and custom application projects, while providing technical application support for all genset models.
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Robotic integration and cloud connections; SCADA and cybersecurity; Motor efficiency standards; Open- and closed-loop control; Augmented reality
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This article collection contains several articles on how automation and controls are helping human-machine interface (HMI) hardware and software advance.
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Cloud, mobility, and remote operations; SCADA and contextual mobility; Custom UPS empowering a secure pipeline
Infrastructure for natural gas expansion; Artificial lift methods; Disruptive technology and fugitive gas emissions
Mobility as the means to offshore innovation; Preventing another Deepwater Horizon; ROVs as subsea robots; SCADA and the radio spectrum
Automation Engineer; Wood Group
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This eGuide illustrates solutions, applications and benefits of machine vision systems.
Learn how to increase device reliability in harsh environments and decrease unplanned system downtime.
This eGuide contains a series of articles and videos that considers theoretical and practical; immediate needs and a look into the future.
Robotic integration and cloud connections; SCADA and cybersecurity; Motor efficiency standards; Open- and closed-loop control; Augmented reality
Controller programming; Safety networks; Enclosure design; Power quality; Safety integrity levels; Increasing process efficiency
Additive manufacturing benefits; HMI and sensor tips; System integrator advice; Innovations from the industry
Featured articles highlight technologies that enable the Industrial Internet of Things, IIoT-related products and strategies to get data more easily to the user.
This article collection contains several articles on how automation and controls are helping human-machine interface (HMI) hardware and software advance.
This digital report will explore several aspects of how IIoT will transform manufacturing in the coming years.
Cloud, mobility, and remote operations; SCADA and contextual mobility; Custom UPS empowering a secure pipeline
Infrastructure for natural gas expansion; Artificial lift methods; Disruptive technology and fugitive gas emissions
Mobility as the means to offshore innovation; Preventing another Deepwater Horizon; ROVs as subsea robots; SCADA and the radio spectrum
Automation Engineer; Wood Group
System Integrator; Cross Integrated Systems Group
Jose S. Vasquez, Jr.
Fire & Life Safety Engineer; Technip USA Inc.
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This course will help identify and reveal electrical hazards and identify the solutions to implementing and maintaining a safe work environment.
This course explains how maintaining power and communication systems through emergency power-generation systems is critical. | 3,412 | 17,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-34 | latest | en | 0.94146 |
https://www.jiskha.com/questions/14322/for-many-global-companies-china-represents-a-very-attractive-market-in-terms-of-size-and | 1,621,181,060,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991224.58/warc/CC-MAIN-20210516140441-20210516170441-00485.warc.gz | 887,559,974 | 5,403 | For many global companies, China represents a very attractive market in terms of size and growth rate. Yet, it ranks lower in terms of economic freedom and higher in political risk than other country markets because it has a communist government. Despite these risks, Volkswagen, Isuzu, and Boeing are just a few of the hundreds of companies that have established manufacturing operations in China. This is due in large part to the Chinese government making sales in China contingent on a company’s willingness to locate production there. The government wants Chinese companies to learn modern management skills from non-Chinese companies and acquire technology. Some observers believe that when Western companies agree to such conditions, they are bargaining away important industry knowledge in exchange for sales today. Should Boeing and other companies go along with China’s terms, or should they risk losing sales by refusing to transfer technology?
Your thinking is what matters. My thinking is that it is a losing economic proposition in the long run. As I see it, the companies are giving away their business for free.
1. 👍
2. 👎
3. 👁
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https://pythonot.github.io/all.html | 1,723,025,458,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00823.warc.gz | 386,877,668 | 38,020 | # API and modules
backend Multi-lib backend for POT bregman Solvers related to Bregman projections for entropic regularized OT coot CO-Optimal Transport solver da Domain adaptation with optimal transport datasets Simple example datasets dr Dimension reduction with OT factored Factored OT solvers (low rank, cost or OT plan) gaussian Optimal transport for Gaussian distributions gnn Layers and functions for optimal transport in Graph Neural Networks. gromov Solvers related to Gromov-Wasserstein problems. lowrank Low rank OT solvers lp Solvers for the original linear program OT problem. mapping Optimal Transport maps and variants optim Generic solvers for regularized OT or its semi-relaxed version. partial Partial OT solvers plot Functions for plotting OT matrices regpath Regularization path OT solvers sliced Sliced OT Distances smooth Smooth and Sparse (KL an L2 reg.) and sparsity-constrained OT solvers. stochastic Stochastic solvers for regularized OT. unbalanced Regularized Unbalanced OT solvers utils Various useful functions weak Weak optimal ransport solvers
## Main ot functions
Warning
The list of automatically imported sub-modules is as follows: ot.lp, ot.bregman, ot.optim ot.utils, ot.datasets, ot.gromov, ot.smooth ot.stochastic, ot.partial, ot.regpath , ot.unbalanced, ot.mapping . The following sub-modules are not imported due to additional dependencies: - ot.dr : depends on pymanopt and autograd. - ot.plot : depends on matplotlib
ot.barycenter(A, M, reg, weights=None, method='sinkhorn', numItermax=10000, stopThr=0.0001, verbose=False, log=False, warn=True, **kwargs)[source]
Compute the entropic regularized wasserstein barycenter of distributions $$\mathbf{A}$$
The function solves the following optimization problem:
$\mathbf{a} = \mathop{\arg \min}_\mathbf{a} \quad \sum_i W_{reg}(\mathbf{a},\mathbf{a}_i)$
where :
• $$W_{reg}(\cdot,\cdot)$$ is the entropic regularized Wasserstein distance (see ot.bregman.sinkhorn()) if method is sinkhorn or sinkhorn_stabilized or sinkhorn_log.
• $$\mathbf{a}_i$$ are training distributions in the columns of matrix $$\mathbf{A}$$
• reg and $$\mathbf{M}$$ are respectively the regularization term and the cost matrix for OT
The algorithm used for solving the problem is the Sinkhorn-Knopp matrix scaling algorithm as proposed in [3]
Parameters:
• A (array-like, shape (dim, n_hists)) – n_hists training distributions $$\mathbf{a}_i$$ of size dim
• M (array-like, shape (dim, dim)) – loss matrix for OT
• reg (float) – Regularization term > 0
• method (str (optional)) – method used for the solver either ‘sinkhorn’ or ‘sinkhorn_stabilized’ or ‘sinkhorn_log’
• weights (array-like, shape (n_hists,)) – Weights of each histogram $$\mathbf{a}_i$$ on the simplex (barycentric coodinates)
• numItermax (int, optional) – Max number of iterations
• stopThr (float, optional) – Stop threshold on error (>0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
• warn (bool, optional) – if True, raises a warning if the algorithm doesn’t convergence.
Returns:
• a ((dim,) array-like) – Wasserstein barycenter
• log (dict) – log dictionary return only if log==True in parameters
References
ot.barycenter_unbalanced(A, M, reg, reg_m, method='sinkhorn', weights=None, numItermax=1000, stopThr=1e-06, verbose=False, log=False, **kwargs)[source]
Compute the entropic unbalanced wasserstein barycenter of $$\mathbf{A}$$.
The function solves the following optimization problem with $$\mathbf{a}$$
$\mathbf{a} = \mathop{\arg \min}_\mathbf{a} \quad \sum_i W_{u_{reg}}(\mathbf{a},\mathbf{a}_i)$
where :
• $$W_{u_{reg}}(\cdot,\cdot)$$ is the unbalanced entropic regularized Wasserstein distance (see ot.unbalanced.sinkhorn_unbalanced())
• $$\mathbf{a}_i$$ are training distributions in the columns of matrix $$\mathbf{A}$$
• reg and $$\mathbf{M}$$ are respectively the regularization term and the cost matrix for OT
• reg_mis the marginal relaxation hyperparameter
The algorithm used for solving the problem is the generalized Sinkhorn-Knopp matrix scaling algorithm as proposed in [10]
Parameters:
• A (array-like (dim, n_hists)) – n_hists training distributions $$\mathbf{a}_i$$ of dimension dim
• M (array-like (dim, dim)) – ground metric matrix for OT.
• reg (float) – Entropy regularization term > 0
• reg_m (float) – Marginal relaxation term > 0
• weights (array-like (n_hists,) optional) – Weight of each distribution (barycentric coodinates) If None, uniform weights are used.
• numItermax (int, optional) – Max number of iterations
• stopThr (float, optional) – Stop threshold on error (> 0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
Returns:
• a ((dim,) array-like) – Unbalanced Wasserstein barycenter
• log (dict) – log dictionary return only if log==True in parameters
References
ot.binary_search_circle(u_values, v_values, u_weights=None, v_weights=None, p=1, Lm=10, Lp=10, tm=-1, tp=1, eps=1e-06, require_sort=True, log=False)[source]
Computes the Wasserstein distance on the circle using the Binary search algorithm proposed in [44]. Samples need to be in $$S^1\cong [0,1[$$. If they are on $$\mathbb{R}$$, takes the value modulo 1. If the values are on $$S^1\subset\mathbb{R}^2$$, it is required to first find the coordinates using e.g. the atan2 function.
$W_p^p(u,v) = \inf_{\theta\in\mathbb{R}}\int_0^1 |F_u^{-1}(q) - (F_v-\theta)^{-1}(q)|^p\ \mathrm{d}q$
where:
• $$F_u$$ and $$F_v$$ are respectively the cdfs of $$u$$ and $$v$$
For values $$x=(x_1,x_2)\in S^1$$, it is required to first get their coordinates with
$u = \frac{\pi + \mathrm{atan2}(-x_2,-x_1)}{2\pi}$
using e.g. ot.utils.get_coordinate_circle(x)
The function runs on backend but tensorflow and jax are not supported.
Parameters:
• u_values (ndarray, shape (n, ...)) – samples in the source domain (coordinates on [0,1[)
• v_values (ndarray, shape (n, ...)) – samples in the target domain (coordinates on [0,1[)
• u_weights (ndarray, shape (n, ...), optional) – samples weights in the source domain
• v_weights (ndarray, shape (n, ...), optional) – samples weights in the target domain
• p (float, optional (default=1)) – Power p used for computing the Wasserstein distance
• Lm (int, optional) – Lower bound dC
• Lp (int, optional) – Upper bound dC
• tm (float, optional) – Lower bound theta
• tp (float, optional) – Upper bound theta
• eps (float, optional) – Stopping condition
• require_sort (bool, optional) – If True, sort the values.
• log (bool, optional) – If True, returns also the optimal theta
Returns:
• loss (float) – Cost associated to the optimal transportation
• log (dict, optional) – log dictionary returned only if log==True in parameters
Examples
>>> u = np.array([[0.2,0.5,0.8]])%1
>>> v = np.array([[0.4,0.5,0.7]])%1
>>> binary_search_circle(u.T, v.T, p=1)
array([0.1])
References
ot.dist(x1, x2=None, metric='sqeuclidean', p=2, w=None)[source]
Compute distance between samples in $$\mathbf{x_1}$$ and $$\mathbf{x_2}$$
Note
This function is backend-compatible and will work on arrays from all compatible backends.
Parameters:
• x1 (array-like, shape (n1,d)) – matrix with n1 samples of size d
• x2 (array-like, shape (n2,d), optional) – matrix with n2 samples of size d (if None then $$\mathbf{x_2} = \mathbf{x_1}$$)
• metric (str | callable, optional) – ‘sqeuclidean’ or ‘euclidean’ on all backends. On numpy the function also accepts from the scipy.spatial.distance.cdist function : ‘braycurtis’, ‘canberra’, ‘chebyshev’, ‘cityblock’, ‘correlation’, ‘cosine’, ‘dice’, ‘euclidean’, ‘hamming’, ‘jaccard’, ‘kulczynski1’, ‘mahalanobis’, ‘matching’, ‘minkowski’, ‘rogerstanimoto’, ‘russellrao’, ‘seuclidean’, ‘sokalmichener’, ‘sokalsneath’, ‘sqeuclidean’, ‘wminkowski’, ‘yule’.
• p (float, optional) – p-norm for the Minkowski and the Weighted Minkowski metrics. Default value is 2.
• w (array-like, rank 1) – Weights for the weighted metrics.
Returns:
M – distance matrix computed with given metric
Return type:
array-like, shape (n1, n2)
ot.emd(a, b, M, numItermax=100000, log=False, center_dual=True, numThreads=1, check_marginals=True)[source]
Solves the Earth Movers distance problem and returns the OT matrix
\begin{align}\begin{aligned}\gamma = \mathop{\arg \min}_\gamma \quad \langle \gamma, \mathbf{M} \rangle_F\\s.t. \ \gamma \mathbf{1} = \mathbf{a}\\ \gamma^T \mathbf{1} = \mathbf{b}\\ \gamma \geq 0\end{aligned}\end{align}
where :
• $$\mathbf{M}$$ is the metric cost matrix
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are the sample weights
Warning
Note that the $$\mathbf{M}$$ matrix in numpy needs to be a C-order numpy.array in float64 format. It will be converted if not in this format
Note
This function is backend-compatible and will work on arrays from all compatible backends. But the algorithm uses the C++ CPU backend which can lead to copy overhead on GPU arrays.
Note
This function will cast the computed transport plan to the data type of the provided input with the following priority: $$\mathbf{a}$$, then $$\mathbf{b}$$, then $$\mathbf{M}$$ if marginals are not provided. Casting to an integer tensor might result in a loss of precision. If this behaviour is unwanted, please make sure to provide a floating point input.
Note
An error will be raised if the vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ do not sum to the same value.
Uses the algorithm proposed in [1].
Parameters:
• a ((ns,) array-like, float) – Source histogram (uniform weight if empty list)
• b ((nt,) array-like, float) – Target histogram (uniform weight if empty list)
• M ((ns,nt) array-like, float) – Loss matrix (c-order array in numpy with type float64)
• numItermax (int, optional (default=100000)) – The maximum number of iterations before stopping the optimization algorithm if it has not converged.
• log (bool, optional (default=False)) – If True, returns a dictionary containing the cost and dual variables. Otherwise returns only the optimal transportation matrix.
• center_dual (boolean, optional (default=True)) – If True, centers the dual potential using function ot.lp.center_ot_dual().
• numThreads (int or "max", optional (default=1, i.e. OpenMP is not used)) – If compiled with OpenMP, chooses the number of threads to parallelize. “max” selects the highest number possible.
• check_marginals (bool, optional (default=True)) – If True, checks that the marginals mass are equal. If False, skips the check.
Returns:
• gamma (array-like, shape (ns, nt)) – Optimal transportation matrix for the given parameters
• log (dict, optional) – If input log is true, a dictionary containing the cost and dual variables and exit status
Examples
Simple example with obvious solution. The function emd accepts lists and perform automatic conversion to numpy arrays
>>> import ot
>>> a=[.5,.5]
>>> b=[.5,.5]
>>> M=[[0.,1.],[1.,0.]]
>>> ot.emd(a, b, M)
array([[0.5, 0. ],
[0. , 0.5]])
References
See also
ot.bregman.sinkhorn
Entropic regularized OT
ot.optim.cg
General regularized OT
ot.emd2(a, b, M, processes=1, numItermax=100000, log=False, return_matrix=False, center_dual=True, numThreads=1, check_marginals=True)[source]
Solves the Earth Movers distance problem and returns the loss
\begin{align}\begin{aligned}\min_\gamma \quad \langle \gamma, \mathbf{M} \rangle_F\\s.t. \ \gamma \mathbf{1} = \mathbf{a}\\ \gamma^T \mathbf{1} = \mathbf{b}\\ \gamma \geq 0\end{aligned}\end{align}
where :
• $$\mathbf{M}$$ is the metric cost matrix
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are the sample weights
Note
This function is backend-compatible and will work on arrays from all compatible backends. But the algorithm uses the C++ CPU backend which can lead to copy overhead on GPU arrays.
Note
This function will cast the computed transport plan and transportation loss to the data type of the provided input with the following priority: $$\mathbf{a}$$, then $$\mathbf{b}$$, then $$\mathbf{M}$$ if marginals are not provided. Casting to an integer tensor might result in a loss of precision. If this behaviour is unwanted, please make sure to provide a floating point input.
Note
An error will be raised if the vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ do not sum to the same value.
Uses the algorithm proposed in [1].
Parameters:
• a ((ns,) array-like, float64) – Source histogram (uniform weight if empty list)
• b ((nt,) array-like, float64) – Target histogram (uniform weight if empty list)
• M ((ns,nt) array-like, float64) – Loss matrix (for numpy c-order array with type float64)
• processes (int, optional (default=1)) – Nb of processes used for multiple emd computation (deprecated)
• numItermax (int, optional (default=100000)) – The maximum number of iterations before stopping the optimization algorithm if it has not converged.
• log (boolean, optional (default=False)) – If True, returns a dictionary containing dual variables. Otherwise returns only the optimal transportation cost.
• return_matrix (boolean, optional (default=False)) – If True, returns the optimal transportation matrix in the log.
• center_dual (boolean, optional (default=True)) – If True, centers the dual potential using function ot.lp.center_ot_dual().
• numThreads (int or "max", optional (default=1, i.e. OpenMP is not used)) – If compiled with OpenMP, chooses the number of threads to parallelize. “max” selects the highest number possible.
• check_marginals (bool, optional (default=True)) – If True, checks that the marginals mass are equal. If False, skips the check.
Returns:
• W (float, array-like) – Optimal transportation loss for the given parameters
• log (dict) – If input log is true, a dictionary containing dual variables and exit status
Examples
Simple example with obvious solution. The function emd accepts lists and perform automatic conversion to numpy arrays
>>> import ot
>>> a=[.5,.5]
>>> b=[.5,.5]
>>> M=[[0.,1.],[1.,0.]]
>>> ot.emd2(a,b,M)
0.0
References
See also
ot.bregman.sinkhorn
Entropic regularized OT
ot.optim.cg
General regularized OT
ot.emd2_1d(x_a, x_b, a=None, b=None, metric='sqeuclidean', p=1.0, dense=True, log=False)[source]
Solves the Earth Movers distance problem between 1d measures and returns the loss
\begin{align}\begin{aligned}\gamma = arg\min_\gamma \sum_i \sum_j \gamma_{ij} d(x_a[i], x_b[j])\\s.t. \gamma 1 = a, \gamma^T 1= b, \gamma\geq 0\end{aligned}\end{align}
where :
• d is the metric
• x_a and x_b are the samples
• a and b are the sample weights
When ‘minkowski’ is used as a metric, $$d(x, y) = |x - y|^p$$.
Uses the algorithm detailed in [1]_
Parameters:
• x_a ((ns,) or (ns, 1) ndarray, float64) – Source dirac locations (on the real line)
• x_b ((nt,) or (ns, 1) ndarray, float64) – Target dirac locations (on the real line)
• a ((ns,) ndarray, float64, optional) – Source histogram (default is uniform weight)
• b ((nt,) ndarray, float64, optional) – Target histogram (default is uniform weight)
• metric (str, optional (default='sqeuclidean')) – Metric to be used. Only strings listed in ot.dist() are accepted. Due to implementation details, this function runs faster when ‘sqeuclidean’, ‘minkowski’, ‘cityblock’, or ‘euclidean’ metrics are used.
• p (float, optional (default=1.0)) – The p-norm to apply for if metric=’minkowski’
• dense (boolean, optional (default=True)) – If True, returns math:gamma as a dense ndarray of shape (ns, nt). Otherwise returns a sparse representation using scipy’s coo_matrix format. Only used if log is set to True. Due to implementation details, this function runs faster when dense is set to False.
• log (boolean, optional (default=False)) – If True, returns a dictionary containing the transportation matrix. Otherwise returns only the loss.
Returns:
• loss (float) – Cost associated to the optimal transportation
• log (dict) – If input log is True, a dictionary containing the Optimal transportation matrix for the given parameters
Examples
Simple example with obvious solution. The function emd2_1d accepts lists and performs automatic conversion to numpy arrays
>>> import ot
>>> a=[.5, .5]
>>> b=[.5, .5]
>>> x_a = [2., 0.]
>>> x_b = [0., 3.]
>>> ot.emd2_1d(x_a, x_b, a, b)
0.5
>>> ot.emd2_1d(x_a, x_b)
0.5
References
See also
ot.lp.emd2
EMD for multidimensional distributions
ot.lp.emd_1d
EMD for 1d distributions (returns the transportation matrix instead of the cost)
ot.emd_1d(x_a, x_b, a=None, b=None, metric='sqeuclidean', p=1.0, dense=True, log=False, check_marginals=True)[source]
Solves the Earth Movers distance problem between 1d measures and returns the OT matrix
\begin{align}\begin{aligned}\gamma = arg\min_\gamma \sum_i \sum_j \gamma_{ij} d(x_a[i], x_b[j])\\s.t. \gamma 1 = a, \gamma^T 1= b, \gamma\geq 0\end{aligned}\end{align}
where :
• d is the metric
• x_a and x_b are the samples
• a and b are the sample weights
When ‘minkowski’ is used as a metric, $$d(x, y) = |x - y|^p$$.
Uses the algorithm detailed in [1]_
Parameters:
• x_a ((ns,) or (ns, 1) ndarray, float64) – Source dirac locations (on the real line)
• x_b ((nt,) or (ns, 1) ndarray, float64) – Target dirac locations (on the real line)
• a ((ns,) ndarray, float64, optional) – Source histogram (default is uniform weight)
• b ((nt,) ndarray, float64, optional) – Target histogram (default is uniform weight)
• metric (str, optional (default='sqeuclidean')) – Metric to be used. Only strings listed in ot.dist() are accepted. Due to implementation details, this function runs faster when ‘sqeuclidean’, ‘cityblock’, or ‘euclidean’ metrics are used.
• p (float, optional (default=1.0)) – The p-norm to apply for if metric=’minkowski’
• dense (boolean, optional (default=True)) – If True, returns math:gamma as a dense ndarray of shape (ns, nt). Otherwise returns a sparse representation using scipy’s coo_matrix format. Due to implementation details, this function runs faster when ‘sqeuclidean’, ‘minkowski’, ‘cityblock’, or ‘euclidean’ metrics are used.
• log (boolean, optional (default=False)) – If True, returns a dictionary containing the cost. Otherwise returns only the optimal transportation matrix.
• check_marginals (bool, optional (default=True)) – If True, checks that the marginals mass are equal. If False, skips the check.
Returns:
• gamma ((ns, nt) ndarray) – Optimal transportation matrix for the given parameters
• log (dict) – If input log is True, a dictionary containing the cost
Examples
Simple example with obvious solution. The function emd_1d accepts lists and performs automatic conversion to numpy arrays
>>> import ot
>>> a=[.5, .5]
>>> b=[.5, .5]
>>> x_a = [2., 0.]
>>> x_b = [0., 3.]
>>> ot.emd_1d(x_a, x_b, a, b)
array([[0. , 0.5],
[0.5, 0. ]])
>>> ot.emd_1d(x_a, x_b)
array([[0. , 0.5],
[0.5, 0. ]])
References
See also
ot.lp.emd
EMD for multidimensional distributions
ot.lp.emd2_1d
EMD for 1d distributions (returns cost instead of the transportation matrix)
ot.factored_optimal_transport(Xa, Xb, a=None, b=None, reg=0.0, r=100, X0=None, stopThr=1e-07, numItermax=100, verbose=False, log=False, **kwargs)[source]
Solves factored OT problem and return OT plans and intermediate distribution
This function solve the following OT problem [40]_
$\mathop{\arg \min}_\mu \quad W_2^2(\mu_a,\mu)+ W_2^2(\mu,\mu_b)$
where :
• $$\mu_a$$ and $$\mu_b$$ are empirical distributions.
• $$\mu$$ is an empirical distribution with r samples
And returns the two OT plans between
Note
This function is backend-compatible and will work on arrays from all compatible backends. But the algorithm uses the C++ CPU backend which can lead to copy overhead on GPU arrays.
Uses the conditional gradient algorithm to solve the problem proposed in [39].
Parameters:
• Xa ((ns,d) array-like, float) – Source samples
• Xb ((nt,d) array-like, float) – Target samples
• a ((ns,) array-like, float) – Source histogram (uniform weight if empty list)
• b ((nt,) array-like, float) – Target histogram (uniform weight if empty list))
• numItermax (int, optional) – Max number of iterations
• stopThr (float, optional) – Stop threshold on the relative variation (>0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
Returns:
• Ga (array-like, shape (ns, r)) – Optimal transportation matrix between source and the intermediate distribution
• Gb (array-like, shape (r, nt)) – Optimal transportation matrix between the intermediate and target distribution
• X (array-like, shape (r, d)) – Support of the intermediate distribution
• log (dict, optional) – If input log is true, a dictionary containing the cost and dual variables and exit status
References
See also
ot.bregman.sinkhorn
Entropic regularized OT
ot.optim.cg
General regularized OT
ot.fused_gromov_wasserstein(M, C1, C2, p=None, q=None, loss_fun='square_loss', symmetric=None, alpha=0.5, armijo=False, G0=None, log=False, max_iter=10000.0, tol_rel=1e-09, tol_abs=1e-09, **kwargs)[source]
Returns the Fused Gromov-Wasserstein transport between $$(\mathbf{C_1}, \mathbf{Y_1}, \mathbf{p})$$ and $$(\mathbf{C_2}, \mathbf{Y_2}, \mathbf{q})$$ with pairwise distance matrix $$\mathbf{M}$$ between node feature matrices $$\mathbf{Y_1}$$ and $$\mathbf{Y_2}$$ (see [24]).
The function solves the following optimization problem using Conditional Gradient:
\begin{align}\begin{aligned}\mathbf{T}^* \in\mathop{\arg\min}_\mathbf{T} \quad (1 - \alpha) \langle \mathbf{T}, \mathbf{M} \rangle_F + \alpha \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j} \mathbf{T}_{k,l}\\s.t. \ \mathbf{T} \mathbf{1} &= \mathbf{p}\\ \mathbf{T}^T \mathbf{1} &= \mathbf{q}\\ \mathbf{T} &\geq 0\end{aligned}\end{align}
Where :
• $$\mathbf{M}$$: metric cost matrix between features across domains
• $$\mathbf{C_1}$$: Metric cost matrix in the source space
• $$\mathbf{C_2}$$: Metric cost matrix in the target space
• $$\mathbf{p}$$: distribution in the source space
• $$\mathbf{q}$$: distribution in the target space
• L: loss function to account for the misfit between the similarity and feature matrices
• $$\alpha$$: trade-off parameter
Note
This function is backend-compatible and will work on arrays from all compatible backends. But the algorithm uses the C++ CPU backend which can lead to copy overhead on GPU arrays.
Note
All computations in the conjugate gradient solver are done with numpy to limit memory overhead.
Note
This function will cast the computed transport plan to the data type of the provided input $$\mathbf{M}$$. Casting to an integer tensor might result in a loss of precision. If this behaviour is unwanted, please make sure to provide a floating point input.
Parameters:
• M (array-like, shape (ns, nt)) – Metric cost matrix between features across domains
• C1 (array-like, shape (ns, ns)) – Metric cost matrix representative of the structure in the source space
• C2 (array-like, shape (nt, nt)) – Metric cost matrix representative of the structure in the target space
• p (array-like, shape (ns,), optional) – Distribution in the source space. If let to its default value None, uniform distribution is taken.
• q (array-like, shape (nt,), optional) – Distribution in the target space. If let to its default value None, uniform distribution is taken.
• loss_fun (str, optional) – Loss function used for the solver
• symmetric (bool, optional) – Either C1 and C2 are to be assumed symmetric or not. If let to its default None value, a symmetry test will be conducted. Else if set to True (resp. False), C1 and C2 will be assumed symmetric (resp. asymmetric).
• alpha (float, optional) – Trade-off parameter (0 < alpha < 1)
• armijo (bool, optional) – If True the step of the line-search is found via an armijo research. Else closed form is used. If there are convergence issues use False.
• G0 (array-like, shape (ns,nt), optional) – If None the initial transport plan of the solver is pq^T. Otherwise G0 must satisfy marginal constraints and will be used as initial transport of the solver.
• log (bool, optional) – record log if True
• max_iter (int, optional) – Max number of iterations
• tol_rel (float, optional) – Stop threshold on relative error (>0)
• tol_abs (float, optional) – Stop threshold on absolute error (>0)
• **kwargs (dict) – parameters can be directly passed to the ot.optim.cg solver
Returns:
• T (array-like, shape (ns, nt)) – Optimal transportation matrix for the given parameters.
• log (dict) – Log dictionary return only if log==True in parameters.
References
ot.fused_gromov_wasserstein2(M, C1, C2, p=None, q=None, loss_fun='square_loss', symmetric=None, alpha=0.5, armijo=False, G0=None, log=False, max_iter=10000.0, tol_rel=1e-09, tol_abs=1e-09, **kwargs)[source]
Returns the Fused Gromov-Wasserstein distance between $$(\mathbf{C_1}, \mathbf{Y_1}, \mathbf{p})$$ and $$(\mathbf{C_2}, \mathbf{Y_2}, \mathbf{q})$$ with pairwise distance matrix $$\mathbf{M}$$ between node feature matrices $$\mathbf{Y_1}$$ and $$\mathbf{Y_2}$$ (see [24]).
The function solves the following optimization problem using Conditional Gradient:
\begin{align}\begin{aligned}\mathbf{FGW} = \mathop{\min}_\mathbf{T} \quad (1 - \alpha) \langle \mathbf{T}, \mathbf{M} \rangle_F + \alpha \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j} \mathbf{T}_{k,l}\\s.t. \ \mathbf{T} \mathbf{1} &= \mathbf{p}\\ \mathbf{T}^T \mathbf{1} &= \mathbf{q}\\ \mathbf{T} &\geq 0\end{aligned}\end{align}
Where :
• $$\mathbf{M}$$: metric cost matrix between features across domains
• $$\mathbf{C_1}$$: Metric cost matrix in the source space
• $$\mathbf{C_2}$$: Metric cost matrix in the target space
• $$\mathbf{p}$$: distribution in the source space
• $$\mathbf{q}$$: distribution in the target space
• L: loss function to account for the misfit between the similarity and feature matrices
• $$\alpha$$: trade-off parameter
Note that when using backends, this loss function is differentiable wrt the matrices (C1, C2, M) and weights (p, q) for quadratic loss using the gradients from [38]_.
Note
This function is backend-compatible and will work on arrays from all compatible backends. But the algorithm uses the C++ CPU backend which can lead to copy overhead on GPU arrays.
Note
All computations in the conjugate gradient solver are done with numpy to limit memory overhead.
Note
This function will cast the computed transport plan to the data type of the provided input $$\mathbf{M}$$. Casting to an integer tensor might result in a loss of precision. If this behaviour is unwanted, please make sure to provide a floating point input.
Parameters:
• M (array-like, shape (ns, nt)) – Metric cost matrix between features across domains
• C1 (array-like, shape (ns, ns)) – Metric cost matrix representative of the structure in the source space.
• C2 (array-like, shape (nt, nt)) – Metric cost matrix representative of the structure in the target space.
• p (array-like, shape (ns,), optional) – Distribution in the source space. If let to its default value None, uniform distribution is taken.
• q (array-like, shape (nt,), optional) – Distribution in the target space. If let to its default value None, uniform distribution is taken.
• loss_fun (str, optional) – Loss function used for the solver.
• symmetric (bool, optional) – Either C1 and C2 are to be assumed symmetric or not. If let to its default None value, a symmetry test will be conducted. Else if set to True (resp. False), C1 and C2 will be assumed symmetric (resp. asymmetric).
• alpha (float, optional) – Trade-off parameter (0 < alpha < 1)
• armijo (bool, optional) – If True the step of the line-search is found via an armijo research. Else closed form is used. If there are convergence issues use False.
• G0 (array-like, shape (ns,nt), optional) – If None the initial transport plan of the solver is pq^T. Otherwise G0 must satisfy marginal constraints and will be used as initial transport of the solver.
• log (bool, optional) – Record log if True.
• max_iter (int, optional) – Max number of iterations
• tol_rel (float, optional) – Stop threshold on relative error (>0)
• tol_abs (float, optional) – Stop threshold on absolute error (>0)
• **kwargs (dict) – Parameters can be directly passed to the ot.optim.cg solver.
Returns:
• fgw-distance (float) – Fused Gromov-Wasserstein distance for the given parameters.
• log (dict) – Log dictionary return only if log==True in parameters.
References
ot.gromov_barycenters(N, Cs, ps=None, p=None, lambdas=None, loss_fun='square_loss', symmetric=True, armijo=False, max_iter=1000, tol=1e-09, stop_criterion='barycenter', warmstartT=False, verbose=False, log=False, init_C=None, random_state=None, **kwargs)[source]
Returns the Gromov-Wasserstein barycenters of S measured similarity matrices $$(\mathbf{C}_s)_{1 \leq s \leq S}$$
The function solves the following optimization problem with block coordinate descent:
$\mathbf{C}^* = \mathop{\arg \min}_{\mathbf{C}\in \mathbb{R}^{N \times N}} \quad \sum_s \lambda_s \mathrm{GW}(\mathbf{C}, \mathbf{C}_s, \mathbf{p}, \mathbf{p}_s)$
Where :
• $$\mathbf{C}_s$$: metric cost matrix
• $$\mathbf{p}_s$$: distribution
Parameters:
• N (int) – Size of the targeted barycenter
• Cs (list of S array-like of shape (ns, ns)) – Metric cost matrices
• ps (list of S array-like of shape (ns,), optional) – Sample weights in the S spaces. If let to its default value None, uniform distributions are taken.
• p (array-like, shape (N,), optional) – Weights in the targeted barycenter. If let to its default value None, uniform distribution is taken.
• lambdas (list of float, optional) – List of the S spaces’ weights. If let to its default value None, uniform weights are taken.
• loss_fun (callable, optional) – tensor-matrix multiplication function based on specific loss function
• symmetric (bool, optional.) – Either structures are to be assumed symmetric or not. Default value is True. Else if set to True (resp. False), C1 and C2 will be assumed symmetric (resp. asymmetric).
• armijo (bool, optional) – If True the step of the line-search is found via an armijo research. Else closed form is used. If there are convergence issues use False.
• max_iter (int, optional) – Max number of iterations
• tol (float, optional) – Stop threshold on relative error (>0)
• stop_criterion (str, optional. Default is 'barycenter'.) – Stop criterion taking values in [‘barycenter’, ‘loss’]. If set to ‘barycenter’ uses absolute norm variations of estimated barycenters. Else if set to ‘loss’ uses the relative variations of the loss.
• warmstartT (bool, optional) – Either to perform warmstart of transport plans in the successive fused gromov-wasserstein transport problems.s
• verbose (bool, optional) – Print information along iterations.
• log (bool, optional) – Record log if True.
• init_C (bool | array-like, shape(N,N)) – Random initial value for the $$\mathbf{C}$$ matrix provided by user.
• random_state (int or RandomState instance, optional) – Fix the seed for reproducibility
Returns:
• C (array-like, shape (N, N)) – Similarity matrix in the barycenter space (permutated arbitrarily)
• log (dict) – Only returned when log=True. It contains the keys:
• $$\mathbf{T}$$: list of (N, ns) transport matrices
• $$\mathbf{p}$$: (N,) barycenter weights
• values used in convergence evaluation.
References
ot.gromov_wasserstein(C1, C2, p=None, q=None, loss_fun='square_loss', symmetric=None, log=False, armijo=False, G0=None, max_iter=10000.0, tol_rel=1e-09, tol_abs=1e-09, **kwargs)[source]
Returns the Gromov-Wasserstein transport between $$(\mathbf{C_1}, \mathbf{p})$$ and $$(\mathbf{C_2}, \mathbf{q})$$.
The function solves the following optimization problem using Conditional Gradient:
\begin{align}\begin{aligned}\mathbf{T}^* \in \mathop{\arg \min}_\mathbf{T} \quad \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j} \mathbf{T}_{k,l}\\s.t. \ \mathbf{\gamma} \mathbf{1} &= \mathbf{p}\\ \mathbf{\gamma}^T \mathbf{1} &= \mathbf{q}\\ \mathbf{\gamma} &\geq 0\end{aligned}\end{align}
Where :
• $$\mathbf{C_1}$$: Metric cost matrix in the source space.
• $$\mathbf{C_2}$$: Metric cost matrix in the target space.
• $$\mathbf{p}$$: Distribution in the source space.
• $$\mathbf{q}$$: Distribution in the target space.
• L: Loss function to account for the misfit between the similarity matrices.
Note
This function is backend-compatible and will work on arrays from all compatible backends. But the algorithm uses the C++ CPU backend which can lead to copy overhead on GPU arrays.
Note
All computations in the conjugate gradient solver are done with numpy to limit memory overhead.
Note
This function will cast the computed transport plan to the data type of the provided input $$\mathbf{C}_1$$. Casting to an integer tensor might result in a loss of precision. If this behaviour is unwanted, please make sure to provide a floating point input.
Parameters:
• C1 (array-like, shape (ns, ns)) – Metric cost matrix in the source space.
• C2 (array-like, shape (nt, nt)) – Metric cost matrix in the target space.
• p (array-like, shape (ns,), optional) – Distribution in the source space. If let to its default value None, uniform distribution is taken.
• q (array-like, shape (nt,), optional) – Distribution in the target space. If let to its default value None, uniform distribution is taken.
• loss_fun (str, optional) – Loss function used for the solver either ‘square_loss’ or ‘kl_loss’.
• symmetric (bool, optional) – Either C1 and C2 are to be assumed symmetric or not. If let to its default None value, a symmetry test will be conducted. Else if set to True (resp. False), C1 and C2 will be assumed symmetric (resp. asymmetric).
• verbose (bool, optional) – Print information along iterations.
• log (bool, optional) – Record log if True.
• armijo (bool, optional) – If True, the step of the line-search is found via an armijo search. Else closed form is used. If there are convergence issues, use False.
• G0 (array-like, shape (ns,nt), optional) – If None, the initial transport plan of the solver is pq^T. Otherwise G0 must satisfy marginal constraints and will be used as initial transport of the solver.
• max_iter (int, optional) – Max number of iterations.
• tol_rel (float, optional) – Stop threshold on relative error (>0).
• tol_abs (float, optional) – Stop threshold on absolute error (>0).
• **kwargs (dict) – Parameters can be directly passed to the ot.optim.cg solver.
Returns:
• T (array-like, shape (ns, nt)) –
Coupling between the two spaces that minimizes:
$$\sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j} \mathbf{T}_{k,l}$$
• log (dict) – Convergence information and loss.
References
ot.gromov_wasserstein2(C1, C2, p=None, q=None, loss_fun='square_loss', symmetric=None, log=False, armijo=False, G0=None, max_iter=10000.0, tol_rel=1e-09, tol_abs=1e-09, **kwargs)[source]
Returns the Gromov-Wasserstein loss $$\mathbf{GW}$$ between $$(\mathbf{C_1}, \mathbf{p})$$ and $$(\mathbf{C_2}, \mathbf{q})$$. To recover the Gromov-Wasserstein distance as defined in [13] compute $$d_{GW} = \frac{1}{2} \sqrt{\mathbf{GW}}$$.
The function solves the following optimization problem using Conditional Gradient:
\begin{align}\begin{aligned}\mathbf{GW} = \min_\mathbf{T} \quad \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j} \mathbf{T}_{k,l}\\s.t. \ \mathbf{\gamma} \mathbf{1} &= \mathbf{p}\\ \mathbf{\gamma}^T \mathbf{1} &= \mathbf{q}\\ \mathbf{\gamma} &\geq 0\end{aligned}\end{align}
Where :
• $$\mathbf{C_1}$$: Metric cost matrix in the source space
• $$\mathbf{C_2}$$: Metric cost matrix in the target space
• $$\mathbf{p}$$: distribution in the source space
• $$\mathbf{q}$$: distribution in the target space
• L: loss function to account for the misfit between the similarity matrices
Note that when using backends, this loss function is differentiable wrt the matrices (C1, C2) and weights (p, q) for quadratic loss using the gradients from [38]_.
Note
This function is backend-compatible and will work on arrays from all compatible backends. But the algorithm uses the C++ CPU backend which can lead to copy overhead on GPU arrays.
Note
All computations in the conjugate gradient solver are done with numpy to limit memory overhead.
Note
This function will cast the computed transport plan to the data type of the provided input $$\mathbf{C}_1$$. Casting to an integer tensor might result in a loss of precision. If this behaviour is unwanted, please make sure to provide a floating point input.
Parameters:
• C1 (array-like, shape (ns, ns)) – Metric cost matrix in the source space
• C2 (array-like, shape (nt, nt)) – Metric cost matrix in the target space
• p (array-like, shape (ns,), optional) – Distribution in the source space. If let to its default value None, uniform distribution is taken.
• q (array-like, shape (nt,), optional) – Distribution in the target space. If let to its default value None, uniform distribution is taken.
• loss_fun (str) – loss function used for the solver either ‘square_loss’ or ‘kl_loss’
• symmetric (bool, optional) – Either C1 and C2 are to be assumed symmetric or not. If let to its default None value, a symmetry test will be conducted. Else if set to True (resp. False), C1 and C2 will be assumed symmetric (resp. asymmetric).
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
• armijo (bool, optional) – If True the step of the line-search is found via an armijo research. Else closed form is used. If there are convergence issues use False.
• G0 (array-like, shape (ns,nt), optional) – If None the initial transport plan of the solver is pq^T. Otherwise G0 must satisfy marginal constraints and will be used as initial transport of the solver.
• max_iter (int, optional) – Max number of iterations
• tol_rel (float, optional) – Stop threshold on relative error (>0)
• tol_abs (float, optional) – Stop threshold on absolute error (>0)
• **kwargs (dict) – parameters can be directly passed to the ot.optim.cg solver
Returns:
• gw_dist (float) – Gromov-Wasserstein distance
• log (dict) – convergence information and Coupling matrix
References
ot.lowrank_gromov_wasserstein_samples(X_s, X_t, a=None, b=None, reg=0, rank=None, alpha=1e-10, gamma_init='rescale', rescale_cost=True, cost_factorized_Xs=None, cost_factorized_Xt=None, stopThr=0.0001, numItermax=1000, stopThr_dykstra=0.001, numItermax_dykstra=10000, seed_init=49, warn=True, warn_dykstra=False, log=False)[source]
Solve the entropic regularization Gromov-Wasserstein transport problem under low-nonnegative rank constraints on the couplings and cost matrices.
Squared euclidean distance matrices are considered for the target and source distributions.
The function solves the following optimization problem:
$\mathop{\min_{(Q,R,g) \in \mathcal{C(a,b,r)}}} \mathcal{Q}_{A,B}(Q\mathrm{diag}(1/g)R^T) - \epsilon \cdot H((Q,R,g))$
where :
• math:
A is the (dim_a, dim_a) square pairwise cost matrix of the source domain.
• math:
B is the (dim_a, dim_a) square pairwise cost matrix of the target domain.
• math:
mathcal{Q}_{A,B} is quadratic objective function of the Gromov Wasserstein plan.
• math:
Q and R are the low-rank matrix decomposition of the Gromov-Wasserstein plan.
• math:
g is the weight vector for the low-rank decomposition of the Gromov-Wasserstein plan.
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are source and target weights (histograms, both sum to 1).
• math:
r is the rank of the Gromov-Wasserstein plan.
• math:
mathcal{C(a,b,r)} are the low-rank couplings of the OT problem.
• $$H((Q,R,g))$$ is the values of the three respective entropies evaluated for each term.
Parameters:
• X_s (array-like, shape (n_samples_a, dim_Xs)) – Samples in the source domain
• X_t (array-like, shape (n_samples_b, dim_Xt)) – Samples in the target domain
• a (array-like, shape (n_samples_a,), optional) – Samples weights in the source domain If let to its default value None, uniform distribution is taken.
• b (array-like, shape (n_samples_b,), optional) – Samples weights in the target domain If let to its default value None, uniform distribution is taken.
• reg (float, optional) – Regularization term >=0
• rank (int, optional. Default is None. (>0)) – Nonnegative rank of the OT plan. If None, min(ns, nt) is considered.
• alpha (int, optional. Default is 1e-10. (>0 and <1/r)) – Lower bound for the weight vector g.
• rescale_cost (bool, optional. Default is False) – Rescale the low rank factorization of the sqeuclidean cost matrix
• seed_init (int, optional. Default is 49. (>0)) – Random state for the ‘random’ initialization of low rank couplings
• gamma_init (str, optional. Default is "rescale".) – Initialization strategy for gamma. ‘rescale’, or ‘theory’ Gamma is a constant that scales the convergence criterion of the Mirror Descent optimization scheme used to compute the low-rank couplings (Q, R and g)
• numItermax (int, optional. Default is 1000.) – Max number of iterations for Low Rank GW
• stopThr (float, optional. Default is 1e-4.) – Stop threshold on error (>0) for Low Rank GW The error is the sum of Kullback Divergences computed for each low rank coupling (Q, R and g) and scaled using gamma.
• numItermax_dykstra (int, optional. Default is 2000.) – Max number of iterations for the Dykstra algorithm
• stopThr_dykstra (float, optional. Default is 1e-7.) – Stop threshold on error (>0) in Dykstra
• cost_factorized_Xs (tuple, optional. Default is None) – Tuple with two pre-computed low rank decompositions (A1, A2) of the source cost matrix. Both matrices should have a shape of (n_samples_a, dim_Xs + 2). If None, the low rank cost matrices will be computed as sqeuclidean cost matrices.
• cost_factorized_Xt (tuple, optional. Default is None) – Tuple with two pre-computed low rank decompositions (B1, B2) of the target cost matrix. Both matrices should have a shape of (n_samples_b, dim_Xt + 2). If None, the low rank cost matrices will be computed as sqeuclidean cost matrices.
• warn (bool, optional) – if True, raises a warning if the low rank GW algorithm doesn’t convergence.
• warn_dykstra (bool, optional) – if True, raises a warning if the Dykstra algorithm doesn’t convergence.
• log (bool, optional) – record log if True
Returns:
• Q (array-like, shape (n_samples_a, r)) – First low-rank matrix decomposition of the OT plan
• R (array-like, shape (n_samples_b, r)) – Second low-rank matrix decomposition of the OT plan
• g (array-like, shape (r, )) – Weight vector for the low-rank decomposition of the OT
• log (dict (lazy_plan, value and value_linear)) – log dictionary return only if log==True in parameters
References
ot.lowrank_sinkhorn(X_s, X_t, a=None, b=None, reg=0, rank=None, alpha=1e-10, rescale_cost=True, init='random', reg_init=0.1, seed_init=49, gamma_init='rescale', numItermax=2000, stopThr=1e-07, warn=True, log=False)[source]
Solve the entropic regularization optimal transport problem under low-nonnegative rank constraints on the couplings.
The function solves the following optimization problem:
$\mathop{\inf_{(\mathbf{Q},\mathbf{R},\mathbf{g}) \in \mathcal{C}(\mathbf{a},\mathbf{b},r)}} \langle \mathbf{C}, \mathbf{Q}\mathrm{diag}(1/\mathbf{g})\mathbf{R}^\top \rangle - \mathrm{reg} \cdot H((\mathbf{Q}, \mathbf{R}, \mathbf{g}))$
where :
• $$\mathbf{C}$$ is the (dim_a, dim_b) metric cost matrix
• $$H((\mathbf{Q}, \mathbf{R}, \mathbf{g}))$$ is the values of the three respective entropies evaluated for each term.
• $$\mathbf{Q}$$ and $$\mathbf{R}$$ are the low-rank matrix decomposition of the OT plan
• $$\mathbf{g}$$ is the weight vector for the low-rank decomposition of the OT plan
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are source and target weights (histograms, both sum to 1)
• $$r$$ is the rank of the OT plan
• $$\mathcal{C}(\mathbf{a}, \mathbf{b}, r)$$ are the low-rank couplings of the OT problem
Parameters:
• X_s (array-like, shape (n_samples_a, dim)) – samples in the source domain
• X_t (array-like, shape (n_samples_b, dim)) – samples in the target domain
• a (array-like, shape (n_samples_a,)) – samples weights in the source domain
• b (array-like, shape (n_samples_b,)) – samples weights in the target domain
• reg (float, optional) – Regularization term >0
• rank (int, optional. Default is None. (>0)) – Nonnegative rank of the OT plan. If None, min(ns, nt) is considered.
• alpha (int, optional. Default is 1e-10. (>0 and <1/r)) – Lower bound for the weight vector g.
• rescale_cost (bool, optional. Default is False) – Rescale the low rank factorization of the sqeuclidean cost matrix
• init (str, optional. Default is 'random'.) – Initialization strategy for the low rank couplings. ‘random’, ‘deterministic’ or ‘kmeans’
• reg_init (float, optional. Default is 1e-1. (>0)) – Regularization term for a ‘kmeans’ init. If None, 1 is considered.
• seed_init (int, optional. Default is 49. (>0)) – Random state for a ‘random’ or ‘kmeans’ init strategy.
• gamma_init (str, optional. Default is "rescale".) – Initialization strategy for gamma. ‘rescale’, or ‘theory’ Gamma is a constant that scales the convergence criterion of the Mirror Descent optimization scheme used to compute the low-rank couplings (Q, R and g)
• numItermax (int, optional. Default is 2000.) – Max number of iterations for the Dykstra algorithm
• stopThr (float, optional. Default is 1e-7.) – Stop threshold on error (>0) in Dykstra
• warn (bool, optional) – if True, raises a warning if the algorithm doesn’t convergence.
• log (bool, optional) – record log if True
Returns:
• Q (array-like, shape (n_samples_a, r)) – First low-rank matrix decomposition of the OT plan
• R (array-like, shape (n_samples_b, r)) – Second low-rank matrix decomposition of the OT plan
• g (array-like, shape (r, )) – Weight vector for the low-rank decomposition of the OT
• log (dict (lazy_plan, value and value_linear)) – log dictionary return only if log==True in parameters
References
ot.max_sliced_wasserstein_distance(X_s, X_t, a=None, b=None, n_projections=50, p=2, projections=None, seed=None, log=False)[source]
Computes a Monte-Carlo approximation of the max p-Sliced Wasserstein distance
$\mathcal{Max-SWD}_p(\mu, \nu) = \underset{\theta _in \mathcal{U}(\mathbb{S}^{d-1})}{\max} [\mathcal{W}_p^p(\theta_\# \mu, \theta_\# \nu)]^{\frac{1}{p}}$
where :
• $$\theta_\# \mu$$ stands for the pushforwards of the projection $$\mathbb{R}^d \ni X \mapsto \langle \theta, X \rangle$$
Parameters:
• X_s (ndarray, shape (n_samples_a, dim)) – samples in the source domain
• X_t (ndarray, shape (n_samples_b, dim)) – samples in the target domain
• a (ndarray, shape (n_samples_a,), optional) – samples weights in the source domain
• b (ndarray, shape (n_samples_b,), optional) – samples weights in the target domain
• n_projections (int, optional) – Number of projections used for the Monte-Carlo approximation
• p (float, optional =) – Power p used for computing the sliced Wasserstein
• projections (shape (dim, n_projections), optional) – Projection matrix (n_projections and seed are not used in this case)
• seed (int or RandomState or None, optional) – Seed used for random number generator
• log (bool, optional) – if True, sliced_wasserstein_distance returns the projections used and their associated EMD.
Returns:
• cost (float) – Sliced Wasserstein Cost
• log (dict, optional) – log dictionary return only if log==True in parameters
Examples
>>> n_samples_a = 20
>>> X = np.random.normal(0., 1., (n_samples_a, 5))
>>> sliced_wasserstein_distance(X, X, seed=0)
0.0
References
ot.semidiscrete_wasserstein2_unif_circle(u_values, u_weights=None)[source]
Computes the closed-form for the 2-Wasserstein distance between samples and a uniform distribution on $$S^1$$ Samples need to be in $$S^1\cong [0,1[$$. If they are on $$\mathbb{R}$$, takes the value modulo 1. If the values are on $$S^1\subset\mathbb{R}^2$$, it is required to first find the coordinates using e.g. the atan2 function.
$W_2^2(\mu_n, \nu) = \sum_{i=1}^n \alpha_i x_i^2 - \left(\sum_{i=1}^n \alpha_i x_i\right)^2 + \sum_{i=1}^n \alpha_i x_i \left(1-\alpha_i-2\sum_{k=1}^{i-1}\alpha_k\right) + \frac{1}{12}$
where:
• $$\nu=\mathrm{Unif}(S^1)$$ and $$\mu_n = \sum_{i=1}^n \alpha_i \delta_{x_i}$$
For values $$x=(x_1,x_2)\in S^1$$, it is required to first get their coordinates with
$u = \frac{\pi + \mathrm{atan2}(-x_2,-x_1)}{2\pi},$
using e.g. ot.utils.get_coordinate_circle(x)
Parameters:
• u_values (ndarray, shape (n, ...)) – Samples
• u_weights (ndarray, shape (n, ...), optional) – samples weights in the source domain
Returns:
loss – Cost associated to the optimal transportation
Return type:
float
Examples
>>> x0 = np.array([[0], [0.2], [0.4]])
>>> semidiscrete_wasserstein2_unif_circle(x0)
array([0.02111111])
References
ot.sinkhorn(a, b, M, reg, method='sinkhorn', numItermax=1000, stopThr=1e-09, verbose=False, log=False, warn=True, warmstart=None, **kwargs)[source]
Solve the entropic regularization optimal transport problem and return the OT matrix
The function solves the following optimization problem:
\begin{align}\begin{aligned}\gamma = \mathop{\arg \min}_\gamma \quad \langle \gamma, \mathbf{M} \rangle_F + \mathrm{reg}\cdot\Omega(\gamma)\\s.t. \ \gamma \mathbf{1} &= \mathbf{a}\\ \gamma^T \mathbf{1} &= \mathbf{b}\\ \gamma &\geq 0\end{aligned}\end{align}
where :
• $$\mathbf{M}$$ is the (dim_a, dim_b) metric cost matrix
• $$\Omega$$ is the entropic regularization term $$\Omega(\gamma)=\sum_{i,j} \gamma_{i,j}\log(\gamma_{i,j})$$
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are source and target weights (histograms, both sum to 1)
Note
This function is backend-compatible and will work on arrays from all compatible backends.
The algorithm used for solving the problem is the Sinkhorn-Knopp matrix scaling algorithm as proposed in [2]
Choosing a Sinkhorn solver
By default and when using a regularization parameter that is not too small the default sinkhorn solver should be enough. If you need to use a small regularization to get sharper OT matrices, you should use the ot.bregman.sinkhorn_stabilized() solver that will avoid numerical errors. This last solver can be very slow in practice and might not even converge to a reasonable OT matrix in a finite time. This is why ot.bregman.sinkhorn_epsilon_scaling() that relies on iterating the value of the regularization (and using warm start) sometimes leads to better solutions. Note that the greedy version of the sinkhorn ot.bregman.greenkhorn() can also lead to a speedup and the screening version of the sinkhorn ot.bregman.screenkhorn() aim at providing a fast approximation of the Sinkhorn problem. For use of GPU and gradient computation with small number of iterations we strongly recommend the ot.bregman.sinkhorn_log() solver that will no need to check for numerical problems.
Parameters:
• a (array-like, shape (dim_a,)) – samples weights in the source domain
• b (array-like, shape (dim_b,) or ndarray, shape (dim_b, n_hists)) – samples in the target domain, compute sinkhorn with multiple targets and fixed $$\mathbf{M}$$ if $$\mathbf{b}$$ is a matrix (return OT loss + dual variables in log)
• M (array-like, shape (dim_a, dim_b)) – loss matrix
• reg (float) – Regularization term >0
• method (str) – method used for the solver either ‘sinkhorn’,’sinkhorn_log’, ‘greenkhorn’, ‘sinkhorn_stabilized’ or ‘sinkhorn_epsilon_scaling’, see those function for specific parameters
• numItermax (int, optional) – Max number of iterations
• stopThr (float, optional) – Stop threshold on error (>0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
• warn (bool, optional) – if True, raises a warning if the algorithm doesn’t convergence.
• warmstart (tuple of arrays, shape (dim_a, dim_b), optional) – Initialization of dual potentials. If provided, the dual potentials should be given (that is the logarithm of the u,v sinkhorn scaling vectors)
Returns:
• gamma (array-like, shape (dim_a, dim_b)) – Optimal transportation matrix for the given parameters
• log (dict) – log dictionary return only if log==True in parameters
Examples
>>> import ot
>>> a=[.5, .5]
>>> b=[.5, .5]
>>> M=[[0., 1.], [1., 0.]]
>>> ot.sinkhorn(a, b, M, 1)
array([[0.36552929, 0.13447071],
[0.13447071, 0.36552929]])
References
ot.sinkhorn2(a, b, M, reg, method='sinkhorn', numItermax=1000, stopThr=1e-09, verbose=False, log=False, warn=False, warmstart=None, **kwargs)[source]
Solve the entropic regularization optimal transport problem and return the loss
The function solves the following optimization problem:
\begin{align}\begin{aligned}W = \min_\gamma \quad \langle \gamma, \mathbf{M} \rangle_F + \mathrm{reg}\cdot\Omega(\gamma)\\s.t. \ \gamma \mathbf{1} &= \mathbf{a}\\ \gamma^T \mathbf{1} &= \mathbf{b}\\ \gamma &\geq 0\end{aligned}\end{align}
where :
• $$\mathbf{M}$$ is the (dim_a, dim_b) metric cost matrix
• $$\Omega$$ is the entropic regularization term $$\Omega(\gamma)=\sum_{i,j} \gamma_{i,j}\log(\gamma_{i,j})$$
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are source and target weights (histograms, both sum to 1)
and returns $$\langle \gamma^*, \mathbf{M} \rangle_F$$ (without the entropic contribution).
Note
This function is backend-compatible and will work on arrays from all compatible backends.
The algorithm used for solving the problem is the Sinkhorn-Knopp matrix scaling algorithm as proposed in [2]
Choosing a Sinkhorn solver
By default and when using a regularization parameter that is not too small the default sinkhorn solver should be enough. If you need to use a small regularization to get sharper OT matrices, you should use the ot.bregman.sinkhorn_log() solver that will avoid numerical errors. This last solver can be very slow in practice and might not even converge to a reasonable OT matrix in a finite time. This is why ot.bregman.sinkhorn_epsilon_scaling() that relies on iterating the value of the regularization (and using warm start) sometimes leads to better solutions. Note that the greedy version of the sinkhorn ot.bregman.greenkhorn() can also lead to a speedup and the screening version of the sinkhorn ot.bregman.screenkhorn() aim a providing a fast approximation of the Sinkhorn problem. For use of GPU and gradient computation with small number of iterations we strongly recommend the ot.bregman.sinkhorn_log() solver that will no need to check for numerical problems.
Parameters:
• a (array-like, shape (dim_a,)) – samples weights in the source domain
• b (array-like, shape (dim_b,) or ndarray, shape (dim_b, n_hists)) – samples in the target domain, compute sinkhorn with multiple targets and fixed $$\mathbf{M}$$ if $$\mathbf{b}$$ is a matrix (return OT loss + dual variables in log)
• M (array-like, shape (dim_a, dim_b)) – loss matrix
• reg (float) – Regularization term >0
• method (str) – method used for the solver either ‘sinkhorn’,’sinkhorn_log’, ‘sinkhorn_stabilized’, see those function for specific parameters
• numItermax (int, optional) – Max number of iterations
• stopThr (float, optional) – Stop threshold on error (>0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
• warn (bool, optional) – if True, raises a warning if the algorithm doesn’t convergence.
• warmstart (tuple of arrays, shape (dim_a, dim_b), optional) – Initialization of dual potentials. If provided, the dual potentials should be given (that is the logarithm of the u,v sinkhorn scaling vectors)
Returns:
• W ((n_hists) float/array-like) – Optimal transportation loss for the given parameters
• log (dict) – log dictionary return only if log==True in parameters
Examples
>>> import ot
>>> a=[.5, .5]
>>> b=[.5, .5]
>>> M=[[0., 1.], [1., 0.]]
>>> ot.sinkhorn2(a, b, M, 1)
0.26894142136999516
References
ot.sinkhorn_lpl1_mm(a, labels_a, b, M, reg, eta=0.1, numItermax=10, numInnerItermax=200, stopInnerThr=1e-09, verbose=False, log=False)[source]
Solve the entropic regularization optimal transport problem with non-convex group lasso regularization
The function solves the following optimization problem:
\begin{align}\begin{aligned}\gamma = \mathop{\arg \min}_\gamma \quad \langle \gamma, \mathbf{M} \rangle_F + \mathrm{reg} \cdot \Omega_e(\gamma) + \eta \ \Omega_g(\gamma)\\s.t. \ \gamma \mathbf{1} = \mathbf{a}\\ \gamma^T \mathbf{1} = \mathbf{b}\\ \gamma \geq 0\end{aligned}\end{align}
where :
• $$\mathbf{M}$$ is the (ns, nt) metric cost matrix
• $$\Omega_e$$ is the entropic regularization term $$\Omega_e (\gamma)=\sum_{i,j} \gamma_{i,j}\log(\gamma_{i,j})$$
• $$\Omega_g$$ is the group lasso regularization term $$\Omega_g(\gamma)=\sum_{i,c} \|\gamma_{i,\mathcal{I}_c}\|^{1/2}_1$$ where $$\mathcal{I}_c$$ are the index of samples from class c in the source domain.
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are source and target weights (sum to 1)
The algorithm used for solving the problem is the generalized conditional gradient as proposed in [5, 7].
Parameters:
• a (array-like (ns,)) – samples weights in the source domain
• labels_a (array-like (ns,)) – labels of samples in the source domain
• b (array-like (nt,)) – samples weights in the target domain
• M (array-like (ns,nt)) – loss matrix
• reg (float) – Regularization term for entropic regularization >0
• eta (float, optional) – Regularization term for group lasso regularization >0
• numItermax (int, optional) – Max number of iterations
• numInnerItermax (int, optional) – Max number of iterations (inner sinkhorn solver)
• stopInnerThr (float, optional) – Stop threshold on error (inner sinkhorn solver) (>0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
Returns:
• gamma ((ns, nt) array-like) – Optimal transportation matrix for the given parameters
• log (dict) – log dictionary return only if log==True in parameters
References
See also
ot.lp.emd
Unregularized OT
ot.bregman.sinkhorn
Entropic regularized OT
ot.optim.cg
General regularized OT
ot.sinkhorn_unbalanced(a, b, M, reg, reg_m, method='sinkhorn', reg_type='entropy', warmstart=None, numItermax=1000, stopThr=1e-06, verbose=False, log=False, **kwargs)[source]
Solve the unbalanced entropic regularization optimal transport problem and return the OT plan
The function solves the following optimization problem:
\begin{align}\begin{aligned}W = \min_\gamma \ \langle \gamma, \mathbf{M} \rangle_F + \mathrm{reg} \cdot \Omega(\gamma) + \mathrm{reg_{m1}} \cdot \mathrm{KL}(\gamma \mathbf{1}, \mathbf{a}) + \mathrm{reg_{m2}} \cdot \mathrm{KL}(\gamma^T \mathbf{1}, \mathbf{b})\\s.t. \gamma \geq 0\end{aligned}\end{align}
where :
• $$\mathbf{M}$$ is the (dim_a, dim_b) metric cost matrix
• $$\Omega$$ is the entropic regularization term, can be either KL divergence or negative entropy
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are source and target unbalanced distributions
• KL is the Kullback-Leibler divergence
The algorithm used for solving the problem is the generalized Sinkhorn-Knopp matrix scaling algorithm as proposed in [10, 25]
Parameters:
• a (array-like (dim_a,)) – Unnormalized histogram of dimension dim_a
• b (array-like (dim_b,) or array-like (dim_b, n_hists)) – One or multiple unnormalized histograms of dimension dim_b. If many, compute all the OT distances $$(\mathbf{a}, \mathbf{b}_i)_i$$
• M (array-like (dim_a, dim_b)) – loss matrix
• reg (float) – Entropy regularization term > 0
• reg_m (float or indexable object of length 1 or 2) – Marginal relaxation term. If reg_m is a scalar or an indexable object of length 1, then the same reg_m is applied to both marginal relaxations. The entropic balanced OT can be recovered using reg_m=float(“inf”). For semi-relaxed case, use either reg_m=(float(“inf”), scalar) or reg_m=(scalar, float(“inf”)). If reg_m is an array, it must have the same backend as input arrays (a, b, M).
• method (str) – method used for the solver either ‘sinkhorn’, ‘sinkhorn_stabilized’ or ‘sinkhorn_reg_scaling’, see those function for specific parameters
• reg_type (string, optional) – Regularizer term. Can take two values: ‘entropy’ (negative entropy) $$\Omega(\gamma) = \sum_{i,j} \gamma_{i,j} \log(\gamma_{i,j}) - \sum_{i,j} \gamma_{i,j}$$, or ‘kl’ (Kullback-Leibler) $$\Omega(\gamma) = \text{KL}(\gamma, \mathbf{a} \mathbf{b}^T)$$.
• warmstart (tuple of arrays, shape (dim_a, dim_b), optional) – Initialization of dual potentials. If provided, the dual potentials should be given (that is the logarithm of the u,v sinkhorn scaling vectors).
• numItermax (int, optional) – Max number of iterations
• stopThr (float, optional) – Stop threshold on error (>0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
Returns:
• if n_hists == 1
• gamma(dim_a, dim_b) array-like
Optimal transportation matrix for the given parameters
• logdict
log dictionary returned only if log is True
• else
• ot_distance(n_hists,) array-like
the OT distance between $$\mathbf{a}$$ and each of the histograms $$\mathbf{b}_i$$
• logdict
log dictionary returned only if log is True
Examples
>>> import ot
>>> a=[.5, .5]
>>> b=[.5, .5]
>>> M=[[0., 1.], [1., 0.]]
>>> ot.sinkhorn_unbalanced(a, b, M, 1, 1)
array([[0.51122814, 0.18807032],
[0.18807032, 0.51122814]])
References
See also
ot.unbalanced.sinkhorn_knopp_unbalanced
Unbalanced Classic Sinkhorn [10]
ot.unbalanced.sinkhorn_stabilized_unbalanced
Unbalanced Stabilized sinkhorn [9, 10]
ot.unbalanced.sinkhorn_reg_scaling_unbalanced
Unbalanced Sinkhorn with epsilon scaling [9, 10]
ot.sinkhorn_unbalanced2(a, b, M, reg, reg_m, method='sinkhorn', reg_type='entropy', warmstart=None, numItermax=1000, stopThr=1e-06, verbose=False, log=False, **kwargs)[source]
Solve the entropic regularization unbalanced optimal transport problem and return the loss
The function solves the following optimization problem:
\begin{align}\begin{aligned}W = \min_\gamma \quad \langle \gamma, \mathbf{M} \rangle_F + \mathrm{reg} \cdot \Omega(\gamma) + \mathrm{reg_{m1}} \cdot \mathrm{KL}(\gamma \mathbf{1}, \mathbf{a}) + \mathrm{reg_{m2}} \cdot \mathrm{KL}(\gamma^T \mathbf{1}, \mathbf{b})\\s.t. \gamma\geq 0\end{aligned}\end{align}
where :
• $$\mathbf{M}$$ is the (dim_a, dim_b) metric cost matrix
• $$\Omega$$ is the entropic regularization term, can be either KL divergence or negative entropy
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are source and target unbalanced distributions
• KL is the Kullback-Leibler divergence
The algorithm used for solving the problem is the generalized Sinkhorn-Knopp matrix scaling algorithm as proposed in [10, 25]
Parameters:
• a (array-like (dim_a,)) – Unnormalized histogram of dimension dim_a
• b (array-like (dim_b,) or array-like (dim_b, n_hists)) – One or multiple unnormalized histograms of dimension dim_b. If many, compute all the OT distances $$(\mathbf{a}, \mathbf{b}_i)_i$$
• M (array-like (dim_a, dim_b)) – loss matrix
• reg (float) – Entropy regularization term > 0
• reg_m (float or indexable object of length 1 or 2) – Marginal relaxation term. If reg_m is a scalar or an indexable object of length 1, then the same reg_m is applied to both marginal relaxations. The entropic balanced OT can be recovered using reg_m=float(“inf”). For semi-relaxed case, use either reg_m=(float(“inf”), scalar) or reg_m=(scalar, float(“inf”)). If reg_m is an array, it must have the same backend as input arrays (a, b, M).
• method (str) – method used for the solver either ‘sinkhorn’, ‘sinkhorn_stabilized’ or ‘sinkhorn_reg_scaling’, see those function for specific parameterss
• reg_type (string, optional) – Regularizer term. Can take two values: ‘entropy’ (negative entropy) $$\Omega(\gamma) = \sum_{i,j} \gamma_{i,j} \log(\gamma_{i,j}) - \sum_{i,j} \gamma_{i,j}$$, or ‘kl’ (Kullback-Leibler) $$\Omega(\gamma) = \text{KL}(\gamma, \mathbf{a} \mathbf{b}^T)$$.
• warmstart (tuple of arrays, shape (dim_a, dim_b), optional) – Initialization of dual potentials. If provided, the dual potentials should be given (that is the logarithm of the u,v sinkhorn scaling vectors).
• numItermax (int, optional) – Max number of iterations
• stopThr (float, optional) – Stop threshold on error (>0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
Returns:
• ot_distance ((n_hists,) array-like) – the OT distance between $$\mathbf{a}$$ and each of the histograms $$\mathbf{b}_i$$
• log (dict) – log dictionary returned only if log is True
Examples
>>> import ot
>>> import numpy as np
>>> a=[.5, .10]
>>> b=[.5, .5]
>>> M=[[0., 1.],[1., 0.]]
>>> np.round(ot.unbalanced.sinkhorn_unbalanced2(a, b, M, 1., 1.), 8)
0.31912858
References
See also
ot.unbalanced.sinkhorn_knopp
Unbalanced Classic Sinkhorn [10]
ot.unbalanced.sinkhorn_stabilized
Unbalanced Stabilized sinkhorn [9, 10]
ot.unbalanced.sinkhorn_reg_scaling
Unbalanced Sinkhorn with epsilon scaling [9, 10]
ot.sliced_wasserstein_distance(X_s, X_t, a=None, b=None, n_projections=50, p=2, projections=None, seed=None, log=False)[source]
Computes a Monte-Carlo approximation of the p-Sliced Wasserstein distance
$\mathcal{SWD}_p(\mu, \nu) = \underset{\theta \sim \mathcal{U}(\mathbb{S}^{d-1})}{\mathbb{E}}\left(\mathcal{W}_p^p(\theta_\# \mu, \theta_\# \nu)\right)^{\frac{1}{p}}$
where :
• $$\theta_\# \mu$$ stands for the pushforwards of the projection $$X \in \mathbb{R}^d \mapsto \langle \theta, X \rangle$$
Parameters:
• X_s (ndarray, shape (n_samples_a, dim)) – samples in the source domain
• X_t (ndarray, shape (n_samples_b, dim)) – samples in the target domain
• a (ndarray, shape (n_samples_a,), optional) – samples weights in the source domain
• b (ndarray, shape (n_samples_b,), optional) – samples weights in the target domain
• n_projections (int, optional) – Number of projections used for the Monte-Carlo approximation
• p (float, optional =) – Power p used for computing the sliced Wasserstein
• projections (shape (dim, n_projections), optional) – Projection matrix (n_projections and seed are not used in this case)
• seed (int or RandomState or None, optional) – Seed used for random number generator
• log (bool, optional) – if True, sliced_wasserstein_distance returns the projections used and their associated EMD.
Returns:
• cost (float) – Sliced Wasserstein Cost
• log (dict, optional) – log dictionary return only if log==True in parameters
Examples
>>> n_samples_a = 20
>>> X = np.random.normal(0., 1., (n_samples_a, 5))
>>> sliced_wasserstein_distance(X, X, seed=0)
0.0
References
ot.sliced_wasserstein_sphere(X_s, X_t, a=None, b=None, n_projections=50, p=2, projections=None, seed=None, log=False)[source]
Compute the spherical sliced-Wasserstein discrepancy.
$SSW_p(\mu,\nu) = \left(\int_{\mathbb{V}_{d,2}} W_p^p(P^U_\#\mu, P^U_\#\nu)\ \mathrm{d}\sigma(U)\right)^{\frac{1}{p}}$
where:
• $$P^U_\# \mu$$ stands for the pushforwards of the projection $$\forall x\in S^{d-1},\ P^U(x) = \frac{U^Tx}{\|U^Tx\|_2}$$
The function runs on backend but tensorflow and jax are not supported.
Parameters:
• X_s (ndarray, shape (n_samples_a, dim)) – Samples in the source domain
• X_t (ndarray, shape (n_samples_b, dim)) – Samples in the target domain
• a (ndarray, shape (n_samples_a,), optional) – samples weights in the source domain
• b (ndarray, shape (n_samples_b,), optional) – samples weights in the target domain
• n_projections (int, optional) – Number of projections used for the Monte-Carlo approximation
• p (float, optional (default=2)) – Power p used for computing the spherical sliced Wasserstein
• projections (shape (n_projections, dim, 2), optional) – Projection matrix (n_projections and seed are not used in this case)
• seed (int or RandomState or None, optional) – Seed used for random number generator
• log (bool, optional) – if True, sliced_wasserstein_sphere returns the projections used and their associated EMD.
Returns:
• cost (float) – Spherical Sliced Wasserstein Cost
• log (dict, optional) – log dictionary return only if log==True in parameters
Examples
>>> n_samples_a = 20
>>> X = np.random.normal(0., 1., (n_samples_a, 5))
>>> X = X / np.sqrt(np.sum(X**2, -1, keepdims=True))
>>> sliced_wasserstein_sphere(X, X, seed=0)
0.0
References
ot.sliced_wasserstein_sphere_unif(X_s, a=None, n_projections=50, seed=None, log=False)[source]
Compute the 2-spherical sliced wasserstein w.r.t. a uniform distribution.
$SSW_2(\mu_n, \nu)$
where
• $$\mu_n=\sum_{i=1}^n \alpha_i \delta_{x_i}$$
• $$\nu=\mathrm{Unif}(S^1)$$
Parameters:
• X_s (ndarray, shape (n_samples_a, dim)) – Samples in the source domain
• a (ndarray, shape (n_samples_a,), optional) – samples weights in the source domain
• n_projections (int, optional) – Number of projections used for the Monte-Carlo approximation
• seed (int or RandomState or None, optional) – Seed used for random number generator
• log (bool, optional) – if True, sliced_wasserstein_distance returns the projections used and their associated EMD.
Returns:
• cost (float) – Spherical Sliced Wasserstein Cost
• log (dict, optional) – log dictionary return only if log==True in parameters
Examples
>>> np.random.seed(42)
>>> x0 = np.random.randn(500,3)
>>> x0 = x0 / np.sqrt(np.sum(x0**2, -1, keepdims=True))
>>> ssw = sliced_wasserstein_sphere_unif(x0, seed=42)
>>> np.allclose(sliced_wasserstein_sphere_unif(x0, seed=42), 0.01734, atol=1e-3)
True
### References:
ot.solve(M, a=None, b=None, reg=None, reg_type='KL', unbalanced=None, unbalanced_type='KL', method=None, n_threads=1, max_iter=None, plan_init=None, potentials_init=None, tol=None, verbose=False, grad='autodiff')[source]
Solve the discrete optimal transport problem and return OTResult object
The function solves the following general optimal transport problem
$\min_{\mathbf{T}\geq 0} \quad \sum_{i,j} T_{i,j}M_{i,j} + \lambda_r R(\mathbf{T}) + \lambda_u U(\mathbf{T}\mathbf{1},\mathbf{a}) + \lambda_u U(\mathbf{T}^T\mathbf{1},\mathbf{b})$
The regularization is selected with reg ($$\lambda_r$$) and reg_type. By default reg=None and there is no regularization. The unbalanced marginal penalization can be selected with unbalanced ($$\lambda_u$$) and unbalanced_type. By default unbalanced=None and the function solves the exact optimal transport problem (respecting the marginals).
Parameters:
• M (array_like, shape (dim_a, dim_b)) – Loss matrix
• a (array-like, shape (dim_a,), optional) – Samples weights in the source domain (default is uniform)
• b (array-like, shape (dim_b,), optional) – Samples weights in the source domain (default is uniform)
• reg (float, optional) – Regularization weight $$\lambda_r$$, by default None (no reg., exact OT)
• reg_type (str, optional) – Type of regularization $$R$$ either “KL”, “L2”, “entropy”, by default “KL”. a tuple of functions can be provided for general solver (see cg). This is only used when reg!=None.
• unbalanced (float, optional) – Unbalanced penalization weight $$\lambda_u$$, by default None (balanced OT)
• unbalanced_type (str, optional) – Type of unbalanced penalization function $$U$$ either “KL”, “L2”, “TV”, by default “KL”.
• method (str, optional) – Method for solving the problem when multiple algorithms are available, default None for automatic selection.
• n_threads (int, optional) – Number of OMP threads for exact OT solver, by default 1
• max_iter (int, optional) – Maximum number of iterations, by default None (default values in each solvers)
• plan_init (array_like, shape (dim_a, dim_b), optional) – Initialization of the OT plan for iterative methods, by default None
• potentials_init ((array_like(dim_a,),array_like(dim_b,)), optional) – Initialization of the OT dual potentials for iterative methods, by default None
• tol (_type_, optional) – Tolerance for solution precision, by default None (default values in each solvers)
• verbose (bool, optional) – Print information in the solver, by default False
• grad (str, optional) – Type of gradient computation, either or ‘autodiff’ or ‘envelope’ used only for Sinkhorn solver. By default ‘autodiff’ provides gradients wrt all outputs (plan, value, value_linear) but with important memory cost. ‘envelope’ provides gradients only for value and and other outputs are detached. This is useful for memory saving when only the value is needed.
Returns:
res – Result of the optimization problem. The information can be obtained as follows:
• res.plan : OT plan $$\mathbf{T}$$
• res.potentials : OT dual potentials
• res.value : Optimal value of the optimization problem
• res.value_linear : Linear OT loss with the optimal OT plan
See OTResult for more information.
Return type:
OTResult()
Notes
The following methods are available for solving the OT problems:
• Classical exact OT problem [1] (default parameters) :
\begin{align}\begin{aligned}\min_\mathbf{T} \quad \langle \mathbf{T}, \mathbf{M} \rangle_F\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve(M, a, b)
• Entropic regularized OT [2] (when reg!=None):
\begin{align}\begin{aligned}\min_\mathbf{T} \quad \langle \mathbf{T}, \mathbf{M} \rangle_F + \lambda R(\mathbf{T})\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0\end{aligned}\end{align}
can be solved with the following code:
# default is "KL" regularization (reg_type="KL")
res = ot.solve(M, a, b, reg=1.0)
# or for original Sinkhorn paper formulation [2]
res = ot.solve(M, a, b, reg=1.0, reg_type='entropy')
# Use envelope theorem differentiation for memory saving
res = ot.solve(M, a, b, reg=1.0, grad='envelope') # M, a, b are torch tensors
res.value.backward() # only the value is differentiable
Note that by default the Sinkhorn solver uses automatic differentiation to compute the gradients of the values and plan. This can be changed with the grad parameter. The envelope mode computes the gradients only for the value and the other outputs are detached. This is useful for memory saving when only the gradient of value is needed.
• Quadratic regularized OT [17] (when reg!=None and reg_type="L2"):
\begin{align}\begin{aligned}\min_\mathbf{T} \quad \langle \mathbf{T}, \mathbf{M} \rangle_F + \lambda R(\mathbf{T})\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve(M,a,b,reg=1.0,reg_type='L2')
• Unbalanced OT [41] (when unbalanced!=None):
$\min_{\mathbf{T}\geq 0} \quad \sum_{i,j} T_{i,j}M_{i,j} + \lambda_u U(\mathbf{T}\mathbf{1},\mathbf{a}) + \lambda_u U(\mathbf{T}^T\mathbf{1},\mathbf{b})$
can be solved with the following code:
# default is "KL"
res = ot.solve(M,a,b,unbalanced=1.0)
# quadratic unbalanced OT
res = ot.solve(M,a,b,unbalanced=1.0,unbalanced_type='L2')
# TV = partial OT
res = ot.solve(M,a,b,unbalanced=1.0,unbalanced_type='TV')
• Regularized unbalanced regularized OT [34] (when unbalanced!=None and reg!=None):
$\min_{\mathbf{T}\geq 0} \quad \sum_{i,j} T_{i,j}M_{i,j} + \lambda_r R(\mathbf{T}) + \lambda_u U(\mathbf{T}\mathbf{1},\mathbf{a}) + \lambda_u U(\mathbf{T}^T\mathbf{1},\mathbf{b})$
can be solved with the following code:
# default is "KL" for both
res = ot.solve(M,a,b,reg=1.0,unbalanced=1.0)
# quadratic unbalanced OT with KL regularization
res = ot.solve(M,a,b,reg=1.0,unbalanced=1.0,unbalanced_type='L2')
# both quadratic
res = ot.solve(M,a,b,reg=1.0, reg_type='L2',unbalanced=1.0,unbalanced_type='L2')
References
ot.solve_gromov(Ca, Cb, M=None, a=None, b=None, loss='L2', symmetric=None, alpha=0.5, reg=None, reg_type='entropy', unbalanced=None, unbalanced_type='KL', n_threads=1, method=None, max_iter=None, plan_init=None, tol=None, verbose=False)[source]
Solve the discrete (Fused) Gromov-Wasserstein and return OTResult object
The function solves the following optimization problem:
$\min_{\mathbf{T}\geq 0} \quad (1 - \alpha) \langle \mathbf{T}, \mathbf{M} \rangle_F + \alpha \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j} + \lambda_r R(\mathbf{T}) + \lambda_u U(\mathbf{T}\mathbf{1},\mathbf{a}) + \lambda_u U(\mathbf{T}^T\mathbf{1},\mathbf{b})$
The regularization is selected with reg ($$\lambda_r$$) and reg_type. By default reg=None and there is no regularization. The unbalanced marginal penalization can be selected with unbalanced ($$\lambda_u$$) and unbalanced_type. By default unbalanced=None and the function solves the exact optimal transport problem (respecting the marginals).
Parameters:
• Ca (array_like, shape (dim_a, dim_a)) – Cost matrix in the source domain
• Cb (array_like, shape (dim_b, dim_b)) – Cost matrix in the target domain
• M (array_like, shape (dim_a, dim_b), optional) – Linear cost matrix for Fused Gromov-Wasserstein (default is None).
• a (array-like, shape (dim_a,), optional) – Samples weights in the source domain (default is uniform)
• b (array-like, shape (dim_b,), optional) – Samples weights in the source domain (default is uniform)
• loss (str, optional) – Type of loss function, either "L2" or "KL", by default "L2"
• symmetric (bool, optional) – Use symmetric version of the Gromov-Wasserstein problem, by default None tests whether the matrices are symmetric or True/False to avoid the test.
• reg (float, optional) – Regularization weight $$\lambda_r$$, by default None (no reg., exact OT)
• reg_type (str, optional) – Type of regularization $$R$$, by default “entropy” (only used when reg!=None)
• alpha (float, optional) – Weight the quadratic term (alpha*Gromov) and the linear term ((1-alpha)*Wass) in the Fused Gromov-Wasserstein problem. Not used for Gromov problem (when M is not provided). By default alpha=None corresponds to alpha=1 for Gromov problem (M==None) and alpha=0.5 for Fused Gromov-Wasserstein problem (M!=None)
• unbalanced (float, optional) – Unbalanced penalization weight $$\lambda_u$$, by default None (balanced OT), Not implemented yet
• unbalanced_type (str, optional) – Type of unbalanced penalization function $$U$$ either “KL”, “semirelaxed”, “partial”, by default “KL” but note that it is not implemented yet.
• n_threads (int, optional) – Number of OMP threads for exact OT solver, by default 1
• method (str, optional) – Method for solving the problem when multiple algorithms are available, default None for automatic selection.
• max_iter (int, optional) – Maximum number of iterations, by default None (default values in each solvers)
• plan_init (array_like, shape (dim_a, dim_b), optional) – Initialization of the OT plan for iterative methods, by default None
• tol (float, optional) – Tolerance for solution precision, by default None (default values in each solvers)
• verbose (bool, optional) – Print information in the solver, by default False
Returns:
res – Result of the optimization problem. The information can be obtained as follows:
• res.plan : OT plan $$\mathbf{T}$$
• res.potentials : OT dual potentials
• res.value : Optimal value of the optimization problem
• res.value_linear : Linear OT loss with the optimal OT plan
• res.value_quad : Quadratic (GW) part of the OT loss with the optimal OT plan
See OTResult for more information.
Return type:
OTResult()
Notes
The following methods are available for solving the Gromov-Wasserstein problem:
• Classical Gromov-Wasserstein (GW) problem [3] (default parameters):
\begin{align}\begin{aligned}\min_{\mathbf{T}\geq 0} \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j}\mathbf{T}_{k,l}\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve_gromov(Ca, Cb) # uniform weights
res = ot.solve_gromov(Ca, Cb, a=a, b=b) # given weights
res = ot.solve_gromov(Ca, Cb, loss='KL') # KL loss
plan = res.plan # GW plan
value = res.value # GW value
• Fused Gromov-Wasserstein (FGW) problem [24] (when M!=None):
\begin{align}\begin{aligned}\min_{\mathbf{T}\geq 0} \quad (1 - \alpha) \langle \mathbf{T}, \mathbf{M} \rangle_F + \alpha \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j}\mathbf{T}_{k,l}\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve_gromov(Ca, Cb, M) # uniform weights, alpha=0.5 (default)
res = ot.solve_gromov(Ca, Cb, M, a=a, b=b, alpha=0.1) # given weights and alpha
plan = res.plan # FGW plan
loss_linear_term = res.value_linear # Wasserstein part of the loss
loss_quad_term = res.value_quad # Gromov part of the loss
loss = res.value # FGW value
• Regularized (Fused) Gromov-Wasserstein (GW) problem [12] (when reg!=None):
\begin{align}\begin{aligned}\min_{\mathbf{T}\geq 0} \quad (1 - \alpha) \langle \mathbf{T}, \mathbf{M} \rangle_F + \alpha \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j}\mathbf{T}_{k,l} + \lambda_r R(\mathbf{T})\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve_gromov(Ca, Cb, reg=1.0) # GW entropy regularization (default)
res = ot.solve_gromov(Ca, Cb, M, a=a, b=b, reg=10, alpha=0.1) # FGW with entropy
plan = res.plan # FGW plan
loss_linear_term = res.value_linear # Wasserstein part of the loss
loss_quad_term = res.value_quad # Gromov part of the loss
loss = res.value # FGW value (including regularization)
• Semi-relaxed (Fused) Gromov-Wasserstein (GW) [48] (when unbalanced='semirelaxed'):
\begin{align}\begin{aligned}\min_{\mathbf{T}\geq 0} \quad (1 - \alpha) \langle \mathbf{T}, \mathbf{M} \rangle_F + \alpha \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j}\mathbf{T}_{k,l}\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T} \geq 0\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve_gromov(Ca, Cb, unbalanced='semirelaxed') # semirelaxed GW
res = ot.solve_gromov(Ca, Cb, unbalanced='semirelaxed', reg=1) # entropic semirelaxed GW
res = ot.solve_gromov(Ca, Cb, M, unbalanced='semirelaxed', alpha=0.1) # semirelaxed FGW
plan = res.plan # FGW plan
right_marginal = res.marginal_b # right marginal of the plan
• Partial (Fused) Gromov-Wasserstein (GW) problem [29] (when unbalanced='partial'):
\begin{align}\begin{aligned}\min_{\mathbf{T}\geq 0} \quad (1 - \alpha) \langle \mathbf{T}, \mathbf{M} \rangle_F + \alpha \sum_{i,j,k,l} L(\mathbf{C_1}_{i,k}, \mathbf{C_2}_{j,l}) \mathbf{T}_{i,j}\mathbf{T}_{k,l}\\s.t. \ \mathbf{T} \mathbf{1} \leq \mathbf{a}\\ \mathbf{T}^T \mathbf{1} \leq \mathbf{b}\\ \mathbf{T} \geq 0\\ \mathbf{1}^T\mathbf{T}\mathbf{1} = m\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve_gromov(Ca, Cb, unbalanced_type='partial', unbalanced=0.8) # partial GW with m=0.8
References
ot.solve_sample(X_a, X_b, a=None, b=None, metric='sqeuclidean', reg=None, reg_type='KL', unbalanced=None, unbalanced_type='KL', lazy=False, batch_size=None, method=None, n_threads=1, max_iter=None, plan_init=None, rank=100, scaling=0.95, potentials_init=None, X_init=None, tol=None, verbose=False, grad='autodiff')[source]
Solve the discrete optimal transport problem using the samples in the source and target domains.
The function solves the following general optimal transport problem
$\min_{\mathbf{T}\geq 0} \quad \sum_{i,j} T_{i,j}M_{i,j} + \lambda_r R(\mathbf{T}) + \lambda_u U(\mathbf{T}\mathbf{1},\mathbf{a}) + \lambda_u U(\mathbf{T}^T\mathbf{1},\mathbf{b})$
where the cost matrix $$\mathbf{M}$$ is computed from the samples in the source and target domains such that $$M_{i,j} = d(x_i,y_j)$$ where $$d$$ is a metric (by default the squared Euclidean distance).
The regularization is selected with reg ($$\lambda_r$$) and reg_type. By default reg=None and there is no regularization. The unbalanced marginal penalization can be selected with unbalanced ($$\lambda_u$$) and unbalanced_type. By default unbalanced=None and the function solves the exact optimal transport problem (respecting the marginals).
Parameters:
• X_s (array-like, shape (n_samples_a, dim)) – samples in the source domain
• X_t (array-like, shape (n_samples_b, dim)) – samples in the target domain
• a (array-like, shape (dim_a,), optional) – Samples weights in the source domain (default is uniform)
• b (array-like, shape (dim_b,), optional) – Samples weights in the source domain (default is uniform)
• reg (float, optional) – Regularization weight $$\lambda_r$$, by default None (no reg., exact OT)
• reg_type (str, optional) – Type of regularization $$R$$ either “KL”, “L2”, “entropy”, by default “KL”
• unbalanced (float, optional) – Unbalanced penalization weight $$\lambda_u$$, by default None (balanced OT)
• unbalanced_type (str, optional) – Type of unbalanced penalization function $$U$$ either “KL”, “L2”, “TV”, by default “KL”
• lazy (bool, optional) – Return OTResultlazy object to reduce memory cost when True, by default False
• batch_size (int, optional) – Batch size for lazy solver, by default None (default values in each solvers)
• method (str, optional) – Method for solving the problem, this can be used to select the solver for unbalanced problems (see ot.solve), or to select a specific large scale solver.
• n_threads (int, optional) – Number of OMP threads for exact OT solver, by default 1
• max_iter (int, optional) – Maximum number of iteration, by default None (default values in each solvers)
• plan_init (array_like, shape (dim_a, dim_b), optional) – Initialization of the OT plan for iterative methods, by default None
• rank (int, optional) – Rank of the OT matrix for lazy solers (method=’factored’), by default 100
• scaling (float, optional) – Scaling factor for the epsilon scaling lazy solvers (method=’geomloss’), by default 0.95
• potentials_init ((array_like(dim_a,),array_like(dim_b,)), optional) – Initialization of the OT dual potentials for iterative methods, by default None
• tol (_type_, optional) – Tolerance for solution precision, by default None (default values in each solvers)
• verbose (bool, optional) – Print information in the solver, by default False
• grad (str, optional) – Type of gradient computation, either or ‘autodiff’ or ‘envelope’ used only for Sinkhorn solver. By default ‘autodiff’ provides gradients wrt all outputs (plan, value, value_linear) but with important memory cost. ‘envelope’ provides gradients only for value and and other outputs are detached. This is useful for memory saving when only the value is needed.
Returns:
res – Result of the optimization problem. The information can be obtained as follows:
• res.plan : OT plan $$\mathbf{T}$$
• res.potentials : OT dual potentials
• res.value : Optimal value of the optimization problem
• res.value_linear : Linear OT loss with the optimal OT plan
• res.lazy_plan : Lazy OT plan (when lazy=True or lazy method)
See OTResult for more information.
Return type:
OTResult()
Notes
The following methods are available for solving the OT problems:
• Classical exact OT problem [1] (default parameters) :
\begin{align}\begin{aligned}\min_\mathbf{T} \quad \langle \mathbf{T}, \mathbf{M} \rangle_F\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0, M_{i,j} = d(x_i,y_j)\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve_sample(xa, xb, a, b)
# for uniform weights
res = ot.solve_sample(xa, xb)
• Entropic regularized OT [2] (when reg!=None):
\begin{align}\begin{aligned}\min_\mathbf{T} \quad \langle \mathbf{T}, \mathbf{M} \rangle_F + \lambda R(\mathbf{T})\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0, M_{i,j} = d(x_i,y_j)\end{aligned}\end{align}
can be solved with the following code:
# default is "KL" regularization (reg_type="KL")
res = ot.solve_sample(xa, xb, a, b, reg=1.0)
# or for original Sinkhorn paper formulation [2]
res = ot.solve_sample(xa, xb, a, b, reg=1.0, reg_type='entropy')
# lazy solver of memory complexity O(n)
res = ot.solve_sample(xa, xb, a, b, reg=1.0, lazy=True, batch_size=100)
# lazy OT plan
lazy_plan = res.lazy_plan
# Use envelope theorem differentiation for memory saving
res = ot.solve_sample(xa, xb, a, b, reg=1.0, grad='envelope')
res.value.backward() # only the value is differentiable
Note that by default the Sinkhorn solver uses automatic differentiation to compute the gradients of the values and plan. This can be changed with the grad parameter. The envelope mode computes the gradients only for the value and the other outputs are detached. This is useful for memory saving when only the gradient of value is needed.
We also have a very efficient solver with compiled CPU/CUDA code using geomloss/PyKeOps that can be used with the following code:
# automatic solver
res = ot.solve_sample(xa, xb, a, b, reg=1.0, method='geomloss')
# force O(n) memory efficient solver
res = ot.solve_sample(xa, xb, a, b, reg=1.0, method='geomloss_online')
# force pre-computed cost matrix
res = ot.solve_sample(xa, xb, a, b, reg=1.0, method='geomloss_tensorized')
# use multiscale solver
res = ot.solve_sample(xa, xb, a, b, reg=1.0, method='geomloss_multiscale')
# One can play with speed (small scaling factor) and precision (scaling close to 1)
res = ot.solve_sample(xa, xb, a, b, reg=1.0, method='geomloss', scaling=0.5)
• Quadratic regularized OT [17] (when reg!=None and reg_type="L2"):
\begin{align}\begin{aligned}\min_\mathbf{T} \quad \langle \mathbf{T}, \mathbf{M} \rangle_F + \lambda R(\mathbf{T})\\s.t. \ \mathbf{T} \mathbf{1} = \mathbf{a}\\ \mathbf{T}^T \mathbf{1} = \mathbf{b}\\ \mathbf{T} \geq 0, M_{i,j} = d(x_i,y_j)\end{aligned}\end{align}
can be solved with the following code:
res = ot.solve_sample(xa, xb, a, b, reg=1.0, reg_type='L2')
• Unbalanced OT [41] (when unbalanced!=None):
\begin{align}\begin{aligned}\min_{\mathbf{T}\geq 0} \quad \sum_{i,j} T_{i,j}M_{i,j} + \lambda_u U(\mathbf{T}\mathbf{1},\mathbf{a}) + \lambda_u U(\mathbf{T}^T\mathbf{1},\mathbf{b})\\with M_{i,j} = d(x_i,y_j)\end{aligned}\end{align}
can be solved with the following code:
# default is "KL"
res = ot.solve_sample(xa, xb, a, b, unbalanced=1.0)
# quadratic unbalanced OT
res = ot.solve_sample(xa, xb, a, b, unbalanced=1.0,unbalanced_type='L2')
# TV = partial OT
res = ot.solve_sample(xa, xb, a, b, unbalanced=1.0,unbalanced_type='TV')
• Regularized unbalanced regularized OT [34] (when unbalanced!=None and reg!=None):
\begin{align}\begin{aligned}\min_{\mathbf{T}\geq 0} \quad \sum_{i,j} T_{i,j}M_{i,j} + \lambda_r R(\mathbf{T}) + \lambda_u U(\mathbf{T}\mathbf{1},\mathbf{a}) + \lambda_u U(\mathbf{T}^T\mathbf{1},\mathbf{b})\\with M_{i,j} = d(x_i,y_j)\end{aligned}\end{align}
can be solved with the following code:
# default is "KL" for both
res = ot.solve_sample(xa, xb, a, b, reg=1.0, unbalanced=1.0)
# quadratic unbalanced OT with KL regularization
res = ot.solve_sample(xa, xb, a, b, reg=1.0, unbalanced=1.0,unbalanced_type='L2')
# both quadratic
res = ot.solve_sample(xa, xb, a, b, reg=1.0, reg_type='L2',
unbalanced=1.0, unbalanced_type='L2')
• Factored OT [2] (when method='factored'):
This method solve the following OT problem [40]_
$\mathop{\arg \min}_\mu \quad W_2^2(\mu_a,\mu)+ W_2^2(\mu,\mu_b)$
where $mu$ is a uniform weighted empirical distribution of $$\mu_a$$ and $$\mu_b$$ are the empirical measures associated to the samples in the source and target domains, and $$W_2$$ is the Wasserstein distance. This problem is solved using exact OT solvers for reg=None and the Sinkhorn solver for reg!=None. The solution provides two transport plans that can be used to recover a low rank OT plan between the two distributions.
res = ot.solve_sample(xa, xb, method='factored', rank=10)
# recover the lazy low rank plan
factored_solution_lazy = res.lazy_plan
# recover the full low rank plan
factored_solution = factored_solution_lazy[:]
• Gaussian Bures-Wasserstein [2] (when method='gaussian'):
This method computes the Gaussian Bures-Wasserstein distance between two Gaussian distributions estimated from teh empirical distributions
$\mathcal{W}(\mu_s, \mu_t)_2^2= \left\lVert \mathbf{m}_s - \mathbf{m}_t \right\rVert^2 + \mathcal{B}(\Sigma_s, \Sigma_t)^{2}$
where :
$\mathbf{B}(\Sigma_s, \Sigma_t)^{2} = \text{Tr}\left(\Sigma_s + \Sigma_t - 2 \sqrt{\Sigma_s^{1/2}\Sigma_t\Sigma_s^{1/2}} \right)$
The covariances and means are estimated from the data.
res = ot.solve_sample(xa, xb, method='gaussian')
# recover the squared Gaussian Bures-Wasserstein distance
BW_dist = res.value
• Wasserstein 1d [1] (when method='1D'):
This method computes the Wasserstein distance between two 1d distributions estimated from the empirical distributions. For multivariate data the distances are computed independently for each dimension.
res = ot.solve_sample(xa, xb, method='1D')
# recover the squared Wasserstein distances
W_dists = res.value
References
ot.tic()[source]
Python implementation of Matlab tic() function
ot.toc(message='Elapsed time : {} s')[source]
Python implementation of Matlab toc() function
ot.toq()[source]
Python implementation of Julia toc() function
ot.unif(n, type_as=None)[source]
Return a uniform histogram of length n (simplex).
Parameters:
• n (int) – number of bins in the histogram
• type_as (array_like) – array of the same type of the expected output (numpy/pytorch/jax)
Returns:
h – histogram of length n such that $$\forall i, \mathbf{h}_i = \frac{1}{n}$$
Return type:
array_like (n,)
ot.wasserstein_1d(u_values, v_values, u_weights=None, v_weights=None, p=1, require_sort=True)[source]
Computes the 1 dimensional OT loss [15] between two (batched) empirical distributions
It is formally the p-Wasserstein distance raised to the power p. We do so in a vectorized way by first building the individual quantile functions then integrating them.
This function should be preferred to emd_1d whenever the backend is different to numpy, and when gradients over either sample positions or weights are required.
Parameters:
• u_values (array-like, shape (n, ...)) – locations of the first empirical distribution
• v_values (array-like, shape (m, ...)) – locations of the second empirical distribution
• u_weights (array-like, shape (n, ...), optional) – weights of the first empirical distribution, if None then uniform weights are used
• v_weights (array-like, shape (m, ...), optional) – weights of the second empirical distribution, if None then uniform weights are used
• p (int, optional) – order of the ground metric used, should be at least 1 (see [2, Chap. 2], default is 1
• require_sort (bool, optional) – sort the distributions atoms locations, if False we will consider they have been sorted prior to being passed to the function, default is True
Returns:
cost – the batched EMD
Return type:
float/array-like, shape (…)
References
ot.wasserstein_circle(u_values, v_values, u_weights=None, v_weights=None, p=1, Lm=10, Lp=10, tm=-1, tp=1, eps=1e-06, require_sort=True)[source]
Computes the Wasserstein distance on the circle using either [45] for p=1 or the binary search algorithm proposed in [44] otherwise. Samples need to be in $$S^1\cong [0,1[$$. If they are on $$\mathbb{R}$$, takes the value modulo 1. If the values are on $$S^1\subset\mathbb{R}^2$$, it requires to first find the coordinates using e.g. the atan2 function.
General loss returned:
$OT_{loss} = \inf_{\theta\in\mathbb{R}}\int_0^1 |cdf_u^{-1}(q) - (cdf_v-\theta)^{-1}(q)|^p\ \mathrm{d}q$
For p=1, [45]
$W_1(u,v) = \int_0^1 |F_u(t)-F_v(t)-LevMed(F_u-F_v)|\ \mathrm{d}t$
For values $$x=(x_1,x_2)\in S^1$$, it is required to first get their coordinates with
$u = \frac{\pi + \mathrm{atan2}(-x_2,-x_1)}{2\pi}$
using e.g. ot.utils.get_coordinate_circle(x)
The function runs on backend but tensorflow and jax are not supported.
Parameters:
• u_values (ndarray, shape (n, ...)) – samples in the source domain (coordinates on [0,1[)
• v_values (ndarray, shape (n, ...)) – samples in the target domain (coordinates on [0,1[)
• u_weights (ndarray, shape (n, ...), optional) – samples weights in the source domain
• v_weights (ndarray, shape (n, ...), optional) – samples weights in the target domain
• p (float, optional (default=1)) – Power p used for computing the Wasserstein distance
• Lm (int, optional) – Lower bound dC. For p>1.
• Lp (int, optional) – Upper bound dC. For p>1.
• tm (float, optional) – Lower bound theta. For p>1.
• tp (float, optional) – Upper bound theta. For p>1.
• eps (float, optional) – Stopping condition. For p>1.
• require_sort (bool, optional) – If True, sort the values.
Returns:
loss – Cost associated to the optimal transportation
Return type:
float
Examples
>>> u = np.array([[0.2,0.5,0.8]])%1
>>> v = np.array([[0.4,0.5,0.7]])%1
>>> wasserstein_circle(u.T, v.T)
array([0.1])
References
ot.weak_optimal_transport(Xa, Xb, a=None, b=None, verbose=False, log=False, G0=None, **kwargs)[source]
Solves the weak optimal transport problem between two empirical distributions
\begin{align}\begin{aligned}\gamma = \mathop{\arg \min}_\gamma \quad \sum_i \mathbf{a}_i \left(\mathbf{X^a}_i - \frac{1}{\mathbf{a}_i} \sum_j \gamma_{ij} \mathbf{X^b}_j \right)^2\\s.t. \ \gamma \mathbf{1} = \mathbf{a}\\ \gamma^T \mathbf{1} = \mathbf{b}\\ \gamma \geq 0\end{aligned}\end{align}
where :
• $$X^a$$ and $$X^b$$ are the sample matrices.
• $$\mathbf{a}$$ and $$\mathbf{b}$$ are the sample weights
Note
This function is backend-compatible and will work on arrays from all compatible backends. But the algorithm uses the C++ CPU backend which can lead to copy overhead on GPU arrays.
Uses the conditional gradient algorithm to solve the problem proposed in [39].
Parameters:
• Xa ((ns,d) array-like, float) – Source samples
• Xb ((nt,d) array-like, float) – Target samples
• a ((ns,) array-like, float) – Source histogram (uniform weight if empty list)
• b ((nt,) array-like, float) – Target histogram (uniform weight if empty list))
• G0 ((ns,nt) array-like, float) – initial guess (default is indep joint density)
• numItermax (int, optional) – Max number of iterations
• numItermaxEmd (int, optional) – Max number of iterations for emd
• stopThr (float, optional) – Stop threshold on the relative variation (>0)
• stopThr2 (float, optional) – Stop threshold on the absolute variation (>0)
• verbose (bool, optional) – Print information along iterations
• log (bool, optional) – record log if True
Returns:
• gamma (array-like, shape (ns, nt)) – Optimal transportation matrix for the given parameters
• log (dict, optional) – If input log is true, a dictionary containing the cost and dual variables and exit status
References
See also
ot.bregman.sinkhorn
Entropic regularized OT
ot.optim.cg
General regularized OT | 28,354 | 99,134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 27, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.682718 |
https://www.jiskha.com/questions/769258/last-year-bill-traveled-6-589-miles-on-business-donna-traveled-4-309-miles-and-ursula | 1,623,580,645,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00624.warc.gz | 783,332,853 | 5,271 | # math
last year bill traveled 6,589 miles on business. Donna traveled 4,309 miles and Ursula traveled 1,028 miles my answer is 11,926 miles
use what you know about solving word problem to explain how you found your answer. use words and/ or numbers in you explanation.
how do i do this part of the question
1. 👍
2. 👎
3. 👁
1. 👍
2. 👎
👤
Ms. Sue
2. I lined all three numbers up and added them together.
1. 👍
2. 👎
3. Right!
1. 👍
2. 👎
👤
Ms. Sue
4. Which number has exactly 4 different factors? 7,9,10,or 16
1. 👍
2. 👎
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A TAXI CAB COST \$1.25 FOR THE FIRST MILE AND \$.25 FOR EACH ADDDITIONAL MILE. how many miles can be traveled in the taxi for \$8.00? M= number of miles traveled. | 789 | 3,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.925628 |
http://isabelle.in.tum.de/library/HOL/HOL-Nonstandard_Analysis/HyperDef.html | 1,516,582,252,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890928.82/warc/CC-MAIN-20180121234728-20180122014728-00253.warc.gz | 185,638,390 | 7,458 | # Theory HyperDef
theory HyperDef
imports Complex_Main HyperNat
```(* Title: HOL/Nonstandard_Analysis/HyperDef.thy
Author: Jacques D. Fleuriot
Conversion to Isar and new proofs by Lawrence C Paulson, 2004
*)
section ‹Construction of Hyperreals Using Ultrafilters›
theory HyperDef
imports Complex_Main HyperNat
begin
type_synonym hypreal = "real star"
abbreviation hypreal_of_real :: "real ⇒ real star"
where "hypreal_of_real ≡ star_of"
abbreviation hypreal_of_hypnat :: "hypnat ⇒ hypreal"
where "hypreal_of_hypnat ≡ of_hypnat"
definition omega :: hypreal ("ω")
where "ω = star_n (λn. real (Suc n))"
― ‹an infinite number ‹= [<1, 2, 3, …>]››
definition epsilon :: hypreal ("ε")
where "ε = star_n (λn. inverse (real (Suc n)))"
― ‹an infinitesimal number ‹= [<1, 1/2, 1/3, …>]››
subsection ‹Real vector class instances›
instantiation star :: (scaleR) scaleR
begin
definition star_scaleR_def [transfer_unfold]: "scaleR r ≡ *f* (scaleR r)"
instance ..
end
lemma Standard_scaleR [simp]: "x ∈ Standard ⟹ scaleR r x ∈ Standard"
lemma star_of_scaleR [simp]: "star_of (scaleR r x) = scaleR r (star_of x)"
by transfer (rule refl)
instance star :: (real_vector) real_vector
proof
fix a b :: real
show "⋀x y::'a star. scaleR a (x + y) = scaleR a x + scaleR a y"
by transfer (rule scaleR_right_distrib)
show "⋀x::'a star. scaleR (a + b) x = scaleR a x + scaleR b x"
by transfer (rule scaleR_left_distrib)
show "⋀x::'a star. scaleR a (scaleR b x) = scaleR (a * b) x"
by transfer (rule scaleR_scaleR)
show "⋀x::'a star. scaleR 1 x = x"
by transfer (rule scaleR_one)
qed
instance star :: (real_algebra) real_algebra
proof
fix a :: real
show "⋀x y::'a star. scaleR a x * y = scaleR a (x * y)"
by transfer (rule mult_scaleR_left)
show "⋀x y::'a star. x * scaleR a y = scaleR a (x * y)"
by transfer (rule mult_scaleR_right)
qed
instance star :: (real_algebra_1) real_algebra_1 ..
instance star :: (real_div_algebra) real_div_algebra ..
instance star :: (field_char_0) field_char_0 ..
instance star :: (real_field) real_field ..
lemma star_of_real_def [transfer_unfold]: "of_real r = star_of (of_real r)"
by (unfold of_real_def, transfer, rule refl)
lemma Standard_of_real [simp]: "of_real r ∈ Standard"
lemma star_of_of_real [simp]: "star_of (of_real r) = of_real r"
by transfer (rule refl)
lemma of_real_eq_star_of [simp]: "of_real = star_of"
proof
show "of_real r = star_of r" for r :: real
by transfer simp
qed
lemma Reals_eq_Standard: "(ℝ :: hypreal set) = Standard"
subsection ‹Injection from @{typ hypreal}›
definition of_hypreal :: "hypreal ⇒ 'a::real_algebra_1 star"
where [transfer_unfold]: "of_hypreal = *f* of_real"
lemma Standard_of_hypreal [simp]: "r ∈ Standard ⟹ of_hypreal r ∈ Standard"
lemma of_hypreal_0 [simp]: "of_hypreal 0 = 0"
by transfer (rule of_real_0)
lemma of_hypreal_1 [simp]: "of_hypreal 1 = 1"
by transfer (rule of_real_1)
lemma of_hypreal_add [simp]: "⋀x y. of_hypreal (x + y) = of_hypreal x + of_hypreal y"
lemma of_hypreal_minus [simp]: "⋀x. of_hypreal (- x) = - of_hypreal x"
by transfer (rule of_real_minus)
lemma of_hypreal_diff [simp]: "⋀x y. of_hypreal (x - y) = of_hypreal x - of_hypreal y"
by transfer (rule of_real_diff)
lemma of_hypreal_mult [simp]: "⋀x y. of_hypreal (x * y) = of_hypreal x * of_hypreal y"
by transfer (rule of_real_mult)
lemma of_hypreal_inverse [simp]:
"⋀x. of_hypreal (inverse x) =
inverse (of_hypreal x :: 'a::{real_div_algebra, division_ring} star)"
by transfer (rule of_real_inverse)
lemma of_hypreal_divide [simp]:
"⋀x y. of_hypreal (x / y) =
(of_hypreal x / of_hypreal y :: 'a::{real_field, field} star)"
by transfer (rule of_real_divide)
lemma of_hypreal_eq_iff [simp]: "⋀x y. (of_hypreal x = of_hypreal y) = (x = y)"
by transfer (rule of_real_eq_iff)
lemma of_hypreal_eq_0_iff [simp]: "⋀x. (of_hypreal x = 0) = (x = 0)"
by transfer (rule of_real_eq_0_iff)
subsection ‹Properties of @{term starrel}›
lemma lemma_starrel_refl [simp]: "x ∈ starrel `` {x}"
lemma starrel_in_hypreal [simp]: "starrel``{x}:star"
by (simp add: star_def starrel_def quotient_def, blast)
declare Abs_star_inject [simp] Abs_star_inverse [simp]
declare equiv_starrel [THEN eq_equiv_class_iff, simp]
subsection ‹@{term hypreal_of_real}: the Injection from @{typ real} to @{typ hypreal}›
lemma inj_star_of: "inj star_of"
by (rule inj_onI) simp
lemma mem_Rep_star_iff: "X ∈ Rep_star x ⟷ x = star_n X"
by (cases x) (simp add: star_n_def)
lemma Rep_star_star_n_iff [simp]: "X ∈ Rep_star (star_n Y) ⟷ eventually (λn. Y n = X n) 𝒰"
lemma Rep_star_star_n: "X ∈ Rep_star (star_n X)"
by simp
subsection ‹Properties of @{term star_n}›
lemma star_n_add: "star_n X + star_n Y = star_n (λn. X n + Y n)"
lemma star_n_minus: "- star_n X = star_n (λn. -(X n))"
by (simp only: star_minus_def starfun_star_n)
lemma star_n_diff: "star_n X - star_n Y = star_n (λn. X n - Y n)"
by (simp only: star_diff_def starfun2_star_n)
lemma star_n_mult: "star_n X * star_n Y = star_n (λn. X n * Y n)"
by (simp only: star_mult_def starfun2_star_n)
lemma star_n_inverse: "inverse (star_n X) = star_n (λn. inverse (X n))"
by (simp only: star_inverse_def starfun_star_n)
lemma star_n_le: "star_n X ≤ star_n Y = eventually (λn. X n ≤ Y n) 𝒰"
by (simp only: star_le_def starP2_star_n)
lemma star_n_less: "star_n X < star_n Y = eventually (λn. X n < Y n) 𝒰"
by (simp only: star_less_def starP2_star_n)
lemma star_n_zero_num: "0 = star_n (λn. 0)"
by (simp only: star_zero_def star_of_def)
lemma star_n_one_num: "1 = star_n (λn. 1)"
by (simp only: star_one_def star_of_def)
lemma star_n_abs: "¦star_n X¦ = star_n (λn. ¦X n¦)"
by (simp only: star_abs_def starfun_star_n)
lemma hypreal_omega_gt_zero [simp]: "0 < ω"
by (simp add: omega_def star_n_zero_num star_n_less)
subsection ‹Existence of Infinite Hyperreal Number›
text ‹Existence of infinite number not corresponding to any real number.
Use assumption that member @{term 𝒰} is not finite.›
text ‹A few lemmas first.›
lemma lemma_omega_empty_singleton_disj:
"{n::nat. x = real n} = {} ∨ (∃y. {n::nat. x = real n} = {y})"
by force
lemma lemma_finite_omega_set: "finite {n::nat. x = real n}"
using lemma_omega_empty_singleton_disj [of x] by auto
lemma not_ex_hypreal_of_real_eq_omega: "∄x. hypreal_of_real x = ω"
apply (simp add: omega_def star_of_def star_n_eq_iff)
apply clarify
apply (rule_tac x2="x-1" in lemma_finite_omega_set [THEN FreeUltrafilterNat.finite, THEN notE])
apply (erule eventually_mono)
apply auto
done
lemma hypreal_of_real_not_eq_omega: "hypreal_of_real x ≠ ω"
using not_ex_hypreal_of_real_eq_omega by auto
text ‹Existence of infinitesimal number also not corresponding to any real number.›
lemma lemma_epsilon_empty_singleton_disj:
"{n::nat. x = inverse(real(Suc n))} = {} ∨ (∃y. {n::nat. x = inverse(real(Suc n))} = {y})"
by auto
lemma lemma_finite_epsilon_set: "finite {n. x = inverse (real (Suc n))}"
using lemma_epsilon_empty_singleton_disj [of x] by auto
lemma not_ex_hypreal_of_real_eq_epsilon: "∄x. hypreal_of_real x = ε"
by (auto simp: epsilon_def star_of_def star_n_eq_iff
lemma_finite_epsilon_set [THEN FreeUltrafilterNat.finite] simp del: of_nat_Suc)
lemma hypreal_of_real_not_eq_epsilon: "hypreal_of_real x ≠ ε"
using not_ex_hypreal_of_real_eq_epsilon by auto
lemma hypreal_epsilon_not_zero: "ε ≠ 0"
by (simp add: epsilon_def star_zero_def star_of_def star_n_eq_iff FreeUltrafilterNat.proper
del: star_of_zero)
lemma hypreal_epsilon_inverse_omega: "ε = inverse ω"
by (simp add: epsilon_def omega_def star_n_inverse)
lemma hypreal_epsilon_gt_zero: "0 < ε"
subsection ‹Absolute Value Function for the Hyperreals›
lemma hrabs_add_less: "¦x¦ < r ⟹ ¦y¦ < s ⟹ ¦x + y¦ < r + s"
for x y r s :: hypreal
by (simp add: abs_if split: if_split_asm)
lemma hrabs_less_gt_zero: "¦x¦ < r ⟹ 0 < r"
for x r :: hypreal
by (blast intro!: order_le_less_trans abs_ge_zero)
lemma hrabs_disj: "¦x¦ = x ∨ ¦x¦ = -x"
for x :: "'a::abs_if"
lemma hrabs_add_lemma_disj: "y + - x + (y + - z) = ¦x + - z¦ ⟹ y = z ∨ x = y"
for x y z :: hypreal
by (simp add: abs_if split: if_split_asm)
subsection ‹Embedding the Naturals into the Hyperreals›
abbreviation hypreal_of_nat :: "nat ⇒ hypreal"
where "hypreal_of_nat ≡ of_nat"
lemma SNat_eq: "Nats = {n. ∃N. n = hypreal_of_nat N}"
text ‹Naturals embedded in hyperreals: is a hyperreal c.f. NS extension.›
lemma hypreal_of_nat: "hypreal_of_nat m = star_n (λn. real m)"
declaration ‹
K (Lin_Arith.add_inj_thms [@{thm star_of_le} RS iffD2,
@{thm star_of_less} RS iffD2, @{thm star_of_eq} RS iffD2]
#> Lin_Arith.add_simps [@{thm star_of_zero}, @{thm star_of_one},
@{thm star_of_minus}, @{thm star_of_diff}, @{thm star_of_mult}]
#> Lin_Arith.add_inj_const (@{const_name "StarDef.star_of"}, @{typ "real ⇒ hypreal"}))
›
simproc_setup fast_arith_hypreal ("(m::hypreal) < n" | "(m::hypreal) ≤ n" | "(m::hypreal) = n") =
‹K Lin_Arith.simproc›
subsection ‹Exponentials on the Hyperreals›
lemma hpowr_0 [simp]: "r ^ 0 = (1::hypreal)"
for r :: hypreal
by (rule power_0)
lemma hpowr_Suc [simp]: "r ^ (Suc n) = r * (r ^ n)"
for r :: hypreal
by (rule power_Suc)
lemma hrealpow_two: "r ^ Suc (Suc 0) = r * r"
for r :: hypreal
by simp
lemma hrealpow_two_le [simp]: "0 ≤ r ^ Suc (Suc 0)"
for r :: hypreal
lemma hrealpow_two_le_add_order [simp]: "0 ≤ u ^ Suc (Suc 0) + v ^ Suc (Suc 0)"
for u v :: hypreal
lemma hrealpow_two_le_add_order2 [simp]: "0 ≤ u ^ Suc (Suc 0) + v ^ Suc (Suc 0) + w ^ Suc (Suc 0)"
for u v w :: hypreal
lemma hypreal_add_nonneg_eq_0_iff: "0 ≤ x ⟹ 0 ≤ y ⟹ x + y = 0 ⟷ x = 0 ∧ y = 0"
for x y :: hypreal
by arith
(* FIXME: DELETE THESE *)
lemma hypreal_three_squares_add_zero_iff: "x * x + y * y + z * z = 0 ⟷ x = 0 ∧ y = 0 ∧ z = 0"
for x y z :: hypreal
"x ^ Suc (Suc 0) + y ^ Suc (Suc 0) + z ^ Suc (Suc 0) = 0 ⟷ x = 0 ∧ y = 0 ∧ z = 0"
for x y z :: hypreal
(*FIXME: This and RealPow.abs_realpow_two should be replaced by an abstract
result proved in Rings or Fields*)
lemma hrabs_hrealpow_two [simp]: "¦x ^ Suc (Suc 0)¦ = x ^ Suc (Suc 0)"
for x :: hypreal
lemma two_hrealpow_ge_one [simp]: "(1::hypreal) ≤ 2 ^ n"
using power_increasing [of 0 n "2::hypreal"] by simp
lemma hrealpow: "star_n X ^ m = star_n (λn. (X n::real) ^ m)"
by (induct m) (auto simp: star_n_one_num star_n_mult)
lemma hrealpow_sum_square_expand:
"(x + y) ^ Suc (Suc 0) =
x ^ Suc (Suc 0) + y ^ Suc (Suc 0) + (hypreal_of_nat (Suc (Suc 0))) * x * y"
for x y :: hypreal
lemma power_hypreal_of_real_numeral:
"(numeral v :: hypreal) ^ n = hypreal_of_real ((numeral v) ^ n)"
by simp
declare power_hypreal_of_real_numeral [of _ "numeral w", simp] for w
lemma power_hypreal_of_real_neg_numeral:
"(- numeral v :: hypreal) ^ n = hypreal_of_real ((- numeral v) ^ n)"
by simp
declare power_hypreal_of_real_neg_numeral [of _ "numeral w", simp] for w
(*
lemma hrealpow_HFinite:
fixes x :: "'a::{real_normed_algebra,power} star"
shows "x ∈ HFinite ==> x ^ n ∈ HFinite"
apply (induct_tac "n")
apply (auto simp add: power_Suc intro: HFinite_mult)
done
*)
subsection ‹Powers with Hypernatural Exponents›
text ‹Hypernatural powers of hyperreals.›
definition pow :: "'a::power star ⇒ nat star ⇒ 'a star" (infixr "pow" 80)
where hyperpow_def [transfer_unfold]: "R pow N = ( *f2* op ^) R N"
lemma Standard_hyperpow [simp]: "r ∈ Standard ⟹ n ∈ Standard ⟹ r pow n ∈ Standard"
lemma hyperpow: "star_n X pow star_n Y = star_n (λn. X n ^ Y n)"
lemma hyperpow_zero [simp]: "⋀n. (0::'a::{power,semiring_0} star) pow (n + (1::hypnat)) = 0"
by transfer simp
lemma hyperpow_not_zero: "⋀r n. r ≠ (0::'a::{field} star) ⟹ r pow n ≠ 0"
by transfer (rule power_not_zero)
lemma hyperpow_inverse: "⋀r n. r ≠ (0::'a::field star) ⟹ inverse (r pow n) = (inverse r) pow n"
by transfer (rule power_inverse [symmetric])
lemma hyperpow_hrabs: "⋀r n. ¦r::'a::{linordered_idom} star¦ pow n = ¦r pow n¦"
by transfer (rule power_abs [symmetric])
lemma hyperpow_add: "⋀r n m. (r::'a::monoid_mult star) pow (n + m) = (r pow n) * (r pow m)"
lemma hyperpow_one [simp]: "⋀r. (r::'a::monoid_mult star) pow (1::hypnat) = r"
by transfer (rule power_one_right)
lemma hyperpow_two: "⋀r. (r::'a::monoid_mult star) pow (2::hypnat) = r * r"
by transfer (rule power2_eq_square)
lemma hyperpow_gt_zero: "⋀r n. (0::'a::{linordered_semidom} star) < r ⟹ 0 < r pow n"
by transfer (rule zero_less_power)
lemma hyperpow_ge_zero: "⋀r n. (0::'a::{linordered_semidom} star) ≤ r ⟹ 0 ≤ r pow n"
by transfer (rule zero_le_power)
lemma hyperpow_le: "⋀x y n. (0::'a::{linordered_semidom} star) < x ⟹ x ≤ y ⟹ x pow n ≤ y pow n"
by transfer (rule power_mono [OF _ order_less_imp_le])
lemma hyperpow_eq_one [simp]: "⋀n. 1 pow n = (1::'a::monoid_mult star)"
by transfer (rule power_one)
lemma hrabs_hyperpow_minus [simp]: "⋀(a::'a::linordered_idom star) n. ¦(-a) pow n¦ = ¦a pow n¦"
by transfer (rule abs_power_minus)
lemma hyperpow_mult: "⋀r s n. (r * s::'a::comm_monoid_mult star) pow n = (r pow n) * (s pow n)"
by transfer (rule power_mult_distrib)
lemma hyperpow_two_le [simp]: "⋀r. (0::'a::{monoid_mult,linordered_ring_strict} star) ≤ r pow 2"
by (auto simp add: hyperpow_two zero_le_mult_iff)
lemma hrabs_hyperpow_two [simp]:
"¦x pow 2¦ = (x::'a::{monoid_mult,linordered_ring_strict} star) pow 2"
by (simp only: abs_of_nonneg hyperpow_two_le)
lemma hyperpow_two_hrabs [simp]: "¦x::'a::linordered_idom star¦ pow 2 = x pow 2"
text ‹The precondition could be weakened to @{term "0≤x"}.›
lemma hypreal_mult_less_mono: "u < v ⟹ x < y ⟹ 0 < v ⟹ 0 < x ⟹ u * x < v * y"
for u v x y :: hypreal
lemma hyperpow_two_gt_one: "⋀r::'a::linordered_semidom star. 1 < r ⟹ 1 < r pow 2"
by transfer simp
lemma hyperpow_two_ge_one: "⋀r::'a::linordered_semidom star. 1 ≤ r ⟹ 1 ≤ r pow 2"
by transfer (rule one_le_power)
lemma two_hyperpow_ge_one [simp]: "(1::hypreal) ≤ 2 pow n"
apply (rule_tac y = "1 pow n" in order_trans)
apply (rule_tac [2] hyperpow_le)
apply auto
done
lemma hyperpow_minus_one2 [simp]: "⋀n. (- 1) pow (2 * n) = (1::hypreal)"
by transfer (rule power_minus1_even)
lemma hyperpow_less_le: "⋀r n N. (0::hypreal) ≤ r ⟹ r ≤ 1 ⟹ n < N ⟹ r pow N ≤ r pow n"
by transfer (rule power_decreasing [OF order_less_imp_le])
lemma hyperpow_SHNat_le:
"0 ≤ r ⟹ r ≤ (1::hypreal) ⟹ N ∈ HNatInfinite ⟹ ∀n∈Nats. r pow N ≤ r pow n"
by (auto intro!: hyperpow_less_le simp: HNatInfinite_iff)
lemma hyperpow_realpow: "(hypreal_of_real r) pow (hypnat_of_nat n) = hypreal_of_real (r ^ n)"
by transfer (rule refl)
lemma hyperpow_SReal [simp]: "(hypreal_of_real r) pow (hypnat_of_nat n) ∈ ℝ"
lemma hyperpow_zero_HNatInfinite [simp]: "N ∈ HNatInfinite ⟹ (0::hypreal) pow N = 0"
by (drule HNatInfinite_is_Suc, auto)
lemma hyperpow_le_le: "(0::hypreal) ≤ r ⟹ r ≤ 1 ⟹ n ≤ N ⟹ r pow N ≤ r pow n"
apply (drule order_le_less [of n, THEN iffD1])
apply (auto intro: hyperpow_less_le)
done
lemma hyperpow_Suc_le_self2: "(0::hypreal) ≤ r ⟹ r < 1 ⟹ r pow (n + (1::hypnat)) ≤ r"
apply (drule_tac n = " (1::hypnat) " in hyperpow_le_le)
apply auto
done
lemma hyperpow_hypnat_of_nat: "⋀x. x pow hypnat_of_nat n = x ^ n"
by transfer (rule refl)
lemma of_hypreal_hyperpow:
"⋀x n. of_hypreal (x pow n) = (of_hypreal x::'a::{real_algebra_1} star) pow n"
by transfer (rule of_real_power)
end
``` | 5,269 | 15,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-05 | longest | en | 0.550113 |
https://openlab.citytech.cuny.edu/mat1475coursehub/lessons/lesson-3-continuity/ | 1,709,365,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00134.warc.gz | 449,644,671 | 24,652 | Â
Learning Outcomes
1. Explain three conditions for continuity at a pointÂ
2. Describe three types of discontinuities
3. Define continuity on an interval
4. Understand the Intermediate Value Theorem
Textbook
• Chapter 2.4 Â Continuity Â
Textbook Assignment
• p. 191: Â Â 131, 133, 139, 143, 145, 147
WeBWorK Assginment
• Limits-Continuity
Exit problems of the session
1. Is the function $f(x) = \begin{cases}\dfrac{x+1}{x^2+3x+2}, & \text{ if } -5<x<-1 \\ 3x+5, & \text{ if } -1\leq x\leq 3 \end{cases}$ Â continuous at $x=-1$ ? If not, what type of discontinuity is it?
2. Find the value $k$ that makes the following function continuous at the interval. Â $g(x) = \begin{cases}\sqrt{kx}, & \text{ if } 0\leq x \leq 3 \\ x+1, & \text{ if } 3 < x \leq 10 \end{cases}$Â
 Key Concepts
• For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
• Discontinuities may be classified as removable, jump, or infinite.
• A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over a closed interval if it is continuous at every point in its interior and is continuous at its endpoints.
• The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function takes on every value between the values at its endpoints.
Â
#### Videos and Practice Problems of Selected Topics
1. $\rhd$ Types of discontinuity (7:15) A function being continuous at a point means that the two-sided limit at that point exists and is equal to the function’s value. Point/removable discontinuity is when the two-sided limit exists, but isn’t equal to the function’s value. Jump discontinuity is when the two-sided limit doesn’t exist because the one-sided limits aren’t equal. Asymptotic/infinite discontinuity is when the two-sided limit doesn’t exist because it’s unbounded.
2. * Practice: Classify discontinuities. (4 problems)
3. $\rhd$ Continuity at a point (8:15) Saying a function $f$ is continuous when $x=a$ is the same as saying that the function’s two-side limit at $x=a$ exists and is equal to $f(a)$.
4. $\rhd$ Worked example: continuity at a point (graphical) (7:18). Two examples are given where the conditions for continuity at a point given a function’s graph are analyzed.
5. * Practice: Continuity at a point (graphical). (4 problems)
6. $\rhd$ Worked example: point where a function is continuous (3:58) Is the function $g(x) = \begin{cases}\log(3x), & \text{ if } 0<x<3 \\ (4-x)\log(9), & \text{ if } x\geq 3 \end{cases}$ continuous at $x=3$?
7. $\rhd$ Worked example: point where a function isn’t continuous (4:02) Is the function $f(x) = \begin{cases}\ln(x), & \text{ if } 0<x\leq 2\\ x^2\ln(x), & \text{ if } x> 2 \end{cases}$ continuous at $x=2$?
8. * Practice: Continuity at a point (algebraic). (4 problems)
9. $\rhd$  Limits by direct substitution (2:06) Recognize that $6x^2+5x-1$ is a continuous function to find $\displaystyle\lim_{x\to -1}(6x^2+5x-1)$.
10. * Practice: Limits by direct substitution. (4 problems) | 942 | 3,187 | {"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} | 4.71875 | 5 | CC-MAIN-2024-10 | latest | en | 0.85695 |
https://git.dynare.org/DoraK/dynare/-/commit/6bb8d4190917702bbdf28667cdaf3829877e0267 | 1,660,960,144,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573876.92/warc/CC-MAIN-20220820012448-20220820042448-00595.warc.gz | 261,373,352 | 25,426 | Commit 6bb8d419 by Michel Juillard
### Revert "- added a test an a penalty in estimation (DsgeLikelihood.m) if, in a...
```Revert "- added a test an a penalty in estimation (DsgeLikelihood.m) if, in a stationary model (lik_init==1), a particular parameter set generates unit roots."
There is a better way of dealing with occasional non-stationary models in estimation
This reverts commit 8c0fb552.```
parent 8c0fb552
... ... @@ -170,15 +170,7 @@ if options_.lik_init == 1 % Kalman filter if kalman_algo ~= 2 kalman_algo = 1; end [Pstar,junk,unit_roots] = lyapunov_symm(T,R*Q*R',options_.qz_criterium,... options_.lyapunov_complex_threshold); if ~isempty(unit_roots) % if unit roots the penalty equals the sum of distance to 2-qz_criterium fval = bayestopt_.penalty + sum(unit_roots-2+ ... options_.qz_criterium); cost_flag = 0; return end Pstar = lyapunov_symm(T,R*Q*R',options_.qz_criterium,options_.lyapunov_complex_threshold); Pinf = []; elseif options_.lik_init == 2 % Old Diffuse Kalman filter if kalman_algo ~= 2 ... ...
function [x,u,unit_roots] = lyapunov_symm(a,b,qz_criterium,lyapunov_complex_threshold,method) function [x,u] = lyapunov_symm(a,b,qz_criterium,lyapunov_complex_threshold,method) % Solves the Lyapunov equation x-a*x*a' = b, for b and x symmetric matrices. % If a has some unit roots, the function computes only the solution of the stable subsystem. % ... ... @@ -15,7 +15,6 @@ function [x,u,unit_roots] = lyapunov_symm(a,b,qz_criterium,lyapunov_complex_thre % OUTPUTS % x: [double] m*m solution matrix of the lyapunov equation, where m is the dimension of the stable subsystem. % u: [double] Schur vectors associated with unit roots % unit_roots [double] vector containing roots too close to 1 % % ALGORITHM % Uses reordered Schur decomposition ... ... @@ -58,13 +57,10 @@ if size(a,1) == 1 return end unit_roots = []; if method<2 [U,T] = schur(a); roots = abs(ordeig(T)); e1 = roots > 2-qz_criterium; k = sum(e1); % Number of unit roots. unit_roots = roots(1:k); e1 = abs(ordeig(T)) > 2-qz_criterium; k = sum(e1); % Number of unit roots. n = length(e1)-k; % Number of stationary variables. if k > 0 % Selects stable roots ... ...
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Finish editing this message first! | 699 | 2,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-33 | latest | en | 0.459203 |
https://www.roulette17.com/tips/roulette-edge/ | 1,702,083,733,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00701.warc.gz | 1,063,594,548 | 21,365 | # The Roulette Edge- A Closer Look.
Online Roulette Guide
We show you how to calculate the house edge in roulette
### Best roulette casinos
If there is one phrase that puts the fear into most gamblers it’s this one: “The house always wins”. But is that true? Does the house always win?
Well, no- the house wins some, and loses some. Some players win, some player lose. Some players like Christian Kaiser using the Kesselglucken Method won more than others, but that was back in the eighties.
What I think the phrase is meant to say is: “the house always wins in the end”. And here, we are sorry to say, that there is some truth in this statement.
If you played an infinite number of spins, the house would win in the end. And that’s because of the roulette edge. If you average out all of the bets made by all of people at the roulette table at a casino like The Venetian in Las Vegas, on average, the house would be up. When you see people winning big at the table, they are in fact taking other player’s money.
The casino will suffer some losses and will rake in some player’s losses. When they do the numbers at the end of the week, the losing players will have lost (on average) more than the winning player will have won. Not always, but on average.
Play French Roulette at Luxury Casino
## How to Calculate the House Edge
So how do you work out the edge in roulette?
Let’s look at American Roulette first, which has the highest edge of any variant you can play. If you make a single number bet, the ball can land in one of 38 pockets. In 36 you could win. In 2 (the zero pockets), you can’t.
But, the house only pays out 35:1 if you do win.
Let’s say you cover the whole table: you bet £1 on each and every number including the zeros for the sake of argument (this would be a silly bet by the way). You risk £38. You would get £36 back (£35 plus your winning bet back). The casino’s profit after this “cover whole table” exercise is £2. This is the roulette edge.
In American Roulette, it is 2/38 x 100 = 5.26 %
The house edge is the same for most of the other bets in American Roulette except for one.
Let’s look at the Corner Bet or Square Bet.
Here, is any one of 4 numbers come up you win. If one of the other 34 com up, you lose. So your odds are 4:34, but the casino pays 8:1 on this particular bet.
Let’s say you made the bet 38 times and each number came up once (again highly unlikely, but run with us on this one). You’d win 4 bets and lose 34 bets. You’d end up winning £8 four times (plus you’d get your winning bets back), so you would have £36 and the casino would have £2. Again 5.26%
## The Worst Bet in Roulette
But beware, because although this works for most of the bets in American Roulette, there is one bet called the Five Number Bet, in which it doesn’t.
In the 5 number bet, you’ll win if the ball lands in one of 5 numbers (0,00,1,2,3) and you’ll lose if the ball lands in the other 33. So your odds are 5:33 or 1:6.6.
The pay off for this bet is 6:1 (casinos don’t pay out in fractions unfortunately- only whole numbers). So for your 38 spins which you have paid £38 for the privilege of, you will net £7 for each win (with your winning bet back).
So if you managed 5 wins in 38, you’ll have £35. The casino would have £3 profit which gives a house edge of 7.89 %.
The moral of the story? Don’t play the 5 Number Bet! If the casino paid out 7;1 instead of 6:1, the player would have the edge. Once the word was out, all the players would play this bet and the casinos would all be bust at the end of the year!
## European Roulette House Edge
Let’s quickly take a look at European Roulette. Remember, the European game works on a single zero roulette wheel. There are only 37 pockets instead of 38.
If you bet on every single number and cover the whole table for what you think is a “risk free bet”, you bet £1 on each and every number including the zero, – so you bet £37.
You would get £36 back (£35 plus your winning bet back). The casino’s profit in this case is only £1. So the roulette edge in the European game is 1/37 or 2.7%. That’s half the American edge. So guess what? Over the long term, it pays to play on European instead of American wheels.
Put it another way. If you cover the whole table, the casino makes less profit on a European Wheel that it does on an American Wheel. The moral of the story? Always play European Roulette!
## Extra Tip
Have you heard of “La Partage Rule“. If you can find a casino that plays it (look for French Roulette tables), you can get the roulette edge down even more- to 1.3%. Just saying.
Some Land Based casinos like the Manchester235 roulette tables offer La Partage on their European roulette tables. If you see one, always choose that over standard roulette, particularly if you are a fan of playing even money bets. | 1,192 | 4,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-50 | longest | en | 0.945743 |
https://virallistclub.com/19251/ | 1,620,514,126,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00247.warc.gz | 606,003,676 | 11,695 | # Sin 5pi 12
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Ansh Kukreja 1 year 11 months ago. Add your answer and earn points.
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### From these two identities you should be able to derive these identities in 1 or 2 steps.
Sin 5pi 12. AdesMel4Monika adesMel4Monika 11212016 Mathematics High School How do you use the half angle formula for sin 5π12. Sin 2A and cos 2A by just. CBSE Class 11 Mathematics 1 answers.
Solve your math problems using our free math solver with step-by-step solutions. 1 See answer adesMel4Monika is waiting for your help. The Sine function is derived from the relationship between the angle formed within a right-angled triangle and both Opposite and.
Which results in the square root of 6 – the square root of 2 divided by 4. In this video we will learn to find the value of sin5pi12Other topics of the videoValue of sin5pi12Find the exact value of sin5pi12What is the ex. Sin A B sin A cos B sin B cos A.
Get help on the web or with our math app. Sin 5pi12 Sin pi12 This problem has been solved. Posted by Anuj Singh 5 months ago.
Online math solver with free step by step solutions to algebra calculus and other math problems. Plug in A 30 and B 45 to get your answer. Find the exact value of the expression.
Find the value of the sum. The only exact Trig values come from the special angles. We know that 5π 12 is in quadrant I so it has positive sine sin5π 12 sin1 25π 6 1 cos5π 6 2 1 3 2 2 2 3 4 2 3 2.
Previous question Next question. What is the exact value of sin5pi12. Apply the sumof anglesidentity.
Trigonometry Find the Exact Value sin5pi12 Split into two angleswhere the values of the six trigonometric functionsare known. I am having problems with the first term. Sin 5pi12 sin pi12 Expert Answer.
Cos5π 12cosπ 12 sin5π 12sinπ 12. Split up sin 5pi12 and cos 5pi12 into components that you already know the answers too pi6pi3etc how do we ge 5pi12. FWIW the two most important identities to know are.
Please include the instructions with the problem when you post. You can put this solution on YOUR website. Cospi12 14sqrt6sqrt2 1 cos5pi12 14sqrt6-sqrt2 2 cotpi12 2sqrt3 3 cot5pi12 2-sqrt3 4 cscpi12 sqrt6.
And if so what is the sin and cosine of 2pi3. Is it 2pi3 -Pi4. The first half is what is causing my problem.
Any input would be appreciated. Click here to get an answer to your question How do you use the half angle formula for sin 5π12. Find The Value Of The Sum.
1 8 Thank You. Cos A B cos A cos B – sin A sin B. Actually it is very easy to find that sin5π 12 6 2 4.
We can say pi6 pi4 5pi12find common denominator 12. Write the value of sinpi12sin5pi12 Report. Our math solver supports basic math pre-algebra algebra trigonometry calculus and more.
Find the exact value of the sine using the addition identities. Find the eccentricity of the ellipse 5 x square 9 y square is equal to 1. Sin 5pi12 Answer by jsmallt93757 Show Source.
Click hereto get an answer to your question 2 sin 5pi12 sin pi12. I have already figured sin pi12 using difference formula. Sin A B sin A cos B sin B cos A.
I assume the problem is to find this value exactly.
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READ: A Disadvantage Of Forming A Partnership Is That Owners | 948 | 3,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.869059 |
https://liammitchellimage.com/page-id916.php | 1,638,222,528,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00177.warc.gz | 452,354,038 | 12,904 | # Internal torque diagram
## Mechanics of Materials: Torsion
### Torsional Deformation
Torque is a moment that twists a structure. Unlike axial loads which produce a uniform, or average, stress over the cross section of the object, a torque creates a distribution of stress over the cross section. To keep things simple, we're going to focus on structures with a circular cross section, often called rods or shafts. When a torque is applied to the structure, it will twist along the long axis of the rod, and its cross section remains circular.
To visualize what I'm talking about, imagine that the cross section of the rod is a clock with just an hour hand. When no torque is applied, the hour hand sits at 12 o'clock. As a torque is applied to the rod, it will twist, and the hour hand will rotate clockwise to a new position (say, 2 o'clock). The angle between 2 o'clock and 12 o'clock is referred to as the angle of twist, and is commonly denoted by the Greek symbol phi. This angle lets us determine the shear strain at any point along the cross section.
Before we get into the details of this equation, it's important to note that because we're only discussing circular cross sections, we've switched from Cartesian coordinates to cylindrical coordinates. That's where the Greek symbol rho came from &#; it denotes the distance along the cross section, with rho=0 at the center and rho=c at the outer edge of the rod.
We can immediately learn a few things from this equation. The first thing might be obvious: the more angle of twist, the larger the shear strain (denoted by the Greek symbol gamma, as before). Second, and this is the big difference between axial-loaded structures and torque-loaded ones, the shear strain is not uniform along the cross section. It is zero at the center of the twisted rod, and is at a maximum value at the edge of the rod. Finally, the longer the rod, the smaller the shear strain.
So far, we've focused our attention on displacements and strain. To discuss the stress within a twisted rod we need to know how torque and stress relate. Since twist applies a shear strain, we expect that torque will apply a shear stress. The relationship between torque and shear stress is detailed in section of your textbook, and it results in the following relation:
In this equation, J denotes the second polar moment of area of the cross section. This is sometimes referred to as the "second moment of inertia", but since that already has a well-established meaning regarding the dynamic motion of objects, let's not confuse things here. We'll discuss moment's of area in more detail at a later point, but they take on a very simple form for circular cross sections:
(Note: those are both the same equation &#; solid rods have an inner radius of ci=0).
Now we have equations for our shear strain and our shear stress, all that is left to do is use Hooke's law in shear to see how they are related. Hooke's law lets us write down a nice equation for the angle of twist &#; a very convenient thing to measure in lab or our in the field.
And, just like we saw for axial displacements, we can use superposition for our shear deformations as well:
This final equation allows us to split up torques applied to different parts of the same structure. Let's work out a problem, and see if we understand what's going on for torsional deformations.
### Power Transmission
One of the most common examples of torsion in engineering design is the power generated by transmission shafts. We can quickly understand how twist generates power just by doing a simple dimensional analysis. Power is measured in the unit of Watts [W], and 1 W = 1 N m s-1. At the outset of this section, we noted that torque was a twisting couple, which means that it has units of force times distance, or [N m]. So, by inspection, to generate power with a torque, we need something that occurs with a given frequency f, since frequency has the units of Hertz [Hz] or [s-1]. So, the power per rotation (2*pi) of a circular rod is equal to the applied torque times the frequency of rotation, or:
On the far right hand side of the equation, we've used the relation that angular velocity, denoted by the Greek letter omega, is equal to 2pi times the frequency.
### Statically Indeterminate Problems
One equation, two unknowns&#; we've been down this road before need something else. Although the type of loading and deformation are different, the statically indeterminate problems involving the torsion of rods are approached in the exact same manner as with axially loaded structures. We start with a free body diagram of twisted rod. Take, for example, the rod in the figure below, stuck between two walls.
Immediately upon inspection you should note that the rod is stuck to two walls, when only one would be necessary for static equilibrium. More supports than is necessary: statically indeterminate. And statically indeterminate means, draw a free body diagram, sum the forces in the x-direction, and you'll get one equations with two unknown reaction forces. So, we need to consider our deformations &#; for torsion, that means let's turn to our equation that describes the superposition of twist angles. For this equation, we should note that half the rod is solid, the other half is hollow, which affects how we calculate J for each half. Most importantly, we need to ask ourselves "what do we know about the deformation?" Well, since the rod is stuck to the wall at edge, the twist at A and must be equal to zero (just like the displacement in the last section). See if you can work the rest of this problem out on your own: What is the torque in each half of the rod?
(Answer: Ta= lb ft & Tb= lb ft).
### Summary
We learned about torque and torsion in this lesson. This different type of loading creates an uneven stress distribution over the cross section of the rod &#; ranging from zero at the center to its largest value at the edge. From this analysis we can develop relations between the angle of twist at any a point along the rod and the shear strain within the entire rod. Using Hooke's law, we can relate this strain to the stress within the rod. We also used a method of dimensional analysis to determine the power generated by a transmission shaft (i.e. a rod) that spins with a given frequency under an applied torque. Finally, we showed that torsion problems are also often statically indeterminate, and even though the loading and deformation is different, the technique we established in the last section for solving problems with axial loading is the same technique for solving problems with torque loading.
This material is based upon work supported by the National Science Foundation under Grant No. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
Sours: https://www.bu.edu/moss/mechanics-of-materials-torsion/
The torque diagram of a shaft is analogues to the shear force and bending moment diagram of a beam. It is an important engineering diagram from the pulley shaft design point of view. The steps required to draw it will be discussed with the help of the following example:
Problem: The below picture is showing a machine shop set up, where three machines namely: milling, drilling and lathe are drawing torques from the main shaft at the points B, C and D respectively. The main shaft is rotating by taking torque from the motor. The bearings E and F are assumed to be frictionless so not taking any torques from the torque, theoretically
The details of the input and output torques are shown below:
Milling M/C &#; 20 N-m
Drilling M/C &#; 30 N-m
Lathe M/C &#; 40 N-m
Solution:
StepDraw FBD of the entire shaft: The FBD of the entire shaft will look like below:
Please observe that the direction of the torque for the motor is opposite than the three machines. This is because the above FBD is showing the reaction torques and the direction of the reaction torques for the machines are opposite to that of the motor for obvious reason. The fact will be re-emphasised from the Step-2 below.
Step Calculate the input torque required from the motor: From the equation for equilibrium, we know that total torque must be equal to zero. So, we can write:
Σ T = 0
Or, TA+20+30+40=0
Or, TA= &#; 90 N-m
Step Calculate the torque @ section AB: Take a section anywhere between the point A and B and consider the portion of the shaft on the left of the section for equilibrium. The equilibrium equation becomes:
TA + TA-B=0
or, TA-B = &#; TA=90 N-m
Where,
TA-B – The torque at the section anywhere between A and B
Step Calculate the torque @ section BC: Take a section anywhere between the point B and C and consider the portion of the shaft on the left of the section for equilibrium. The equilibrium equation becomes:
TA + TB+TB-C=0
or, TB-C = &#; TA-TB=70 N-m
Where,
TB-C – The torque at the section anywhere between B and C
Step Calculate the torque @ section CD: Take a section anywhere between the point C and D and consider the portion of the shaft on the left of the section for equilibrium. The equilibrium equation becomes:
TA + TB+TC+TC-D=0
or, TC-D = &#; TA-TB-TC=40 N-m
Where,
TC-D – The torque at the section anywhere between C and D
Step Draw the torque diagram: Plot the length of the shaft in X-axis and the torque values (calculated in step-3,4,5) at the different sections in Y-axis. Like below:
The torque values obtained from the torque diagram is used as input for the pulley shaft design.
.
Shibashis Ghosh
Hi, I am Shibashis, a blogger by passion and an engineer by profession. I have written most of the articles for mechGuru.com. For more than a decades i am closely associated with the engineering design/manufacturing simulation technologies. I am a self taught code hobbyist, presently in love with Python (Open CV / ML / Data Science /AWS + lines, + hrs. )
Tags: Torque
### The Bending Moment Diagram Curve &#; Should it be Convex or Concave?
Sours: https://mechguru.com/machine-design/draw-torque-diagram-in-six-steps/
## Axial Force Diagrams and Torque Diagrams
As an alternative to splitting a body in half and performing an equilibrium analysis to find the internal forces and moments, we can also use graphical approaches to plot out these internal forces and moments over the length of the body. Where equilibrium analysis is the most straightforward approach to finding the internal forces and moments at one cross section, the graphical approaches are the most straightforward approaches to find the internal forces or the internal moments across the entire length of a beam, shaft, or other body. This may be useful in complex loading scenarios where it may not be obvious where the maximum internal forces or internal moments exist. As a trade off however, we will need to plot out each type of internal load separately (one plot for internal axial forces, one for internal shear forces, one for internal torques, and one for internal bending moments).
In this section, we will be focusing on the methods used to generate the plots for the internal axial forces, and the internal torques. This will be the force and moment acting along the length of the beam or shaft. The axial force plot is used primarily for vertical columns or cables supporting multiple loads along its length. The torque diagram is used primarily for shafts supporting multiple inputs and outputs. Each of these plots will have a different practical application, but we have grouped them together here because the process used to generate each of the two plots is very similar.
### Creating the Axial Force Diagram
The axial force diagram will plot out the internal axial (normal) forces within a beam, column, or cable that is supporting multiple forces along the length of the beam itself. This can be thought of as the internal tension or compression forces. The most relevant practical scenarios that match this description will be main support columns in a multi-story building, or hanging cables that are used to support multiple loads.
To create the axial force plot for a body, we will use the following process.
1. Solve for all external forces acting on the body.
2. Draw out a free body diagram of the body horizontally. In the case of vertical structures, rotate the body so that it sits horizontally and all the forces act horizontally
3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal axial forces, with positive numbers indicating tension and negative numbers indicating compression.
4. Starting at zero at the right side of the plot, you will move to the right, pay attention to forces in the free body diagram above. As you move right in your plot, keep steady except
• Jump upwards by the magnitude of the force for any forces in our free body diagram to the left.
• Jump downwards by the magnitude of the force for any forces to the right.
• You can ignore any moments or vertical forces applied to the body.
By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.
To read the plot, you simply need to find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Again, positive numbers represent an internal tension at that location and negative numbers represent an internal compression at that location.
### Creating the Torque Diagram
The torque diagram will plot out the internal torsional moment within a shaft that is supporting multiple inputs and/or outputs along its length. The most relevant practical scenarios that match this description are shafts within complex gear or pulley driven systems.
To create the torque diagram for a shaft, we will use the following process.
1. Solve for all external moments acting on the shaft.
2. Draw out a free body diagram of the shaft horizontally, rotating the shaft if necessary, so that all torques act around the horizontal axis.
3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal torsional moment, with positive numbers indicating an internal torsional moment vector to the right and negative numbers indicating an internal torsional moment to the left.
4. Starting at zero at the right side of the plot, you will move to the right, pay attention to moments in the free body diagram above. As you move right in your plot, keep steady except
• Jump upwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point left.
• Jump downwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point right.
• You can ignore any forces in the free body diagram or moments not about the x axis.
By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.
To read the plot, you simply need to take the find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot.
### Question 1:
A wooden beam is subjected to the forces shown below (forces applied at base of vector). Draw the axial force diagram for the beam.
### Question 2:
A steel shaft is subjected to the torques shown below. Draw the torque diagram for this shaft.
Sours: http://mechanicsmap.psu.edu/websites/6_internal_forces/_axial_torque_diagrams/axial_torque_diagrams.html
Understanding Torsion
## Axial Force Diagrams and Torque Diagrams
As an alternative to splitting a body in half and performing an equilibrium analysis to find the internal forces and moments, we can also use graphical approaches to plot out these internal forces and moments over the length of the body. Where equilibrium analysis is the most straightforward approach to finding the internal forces and moments at one cross section, the graphical approaches are the most straightforward approaches to find the internal forces or the internal moments across the entire length of a beam, shaft, or other body. This may be useful in complex loading scenarios where it may not be obvious where the maximum internal forces or internal moments exist. As a trade off however, we will need to plot out each type of internal load separately (one plot for internal axial forces, one for internal shear forces, one for internal torques, and one for internal bending moments).
In this section, we will be focusing on the methods used to generate the plots for the internal axial forces, and the internal torques. This will be the force and moment acting along the length of the beam or shaft. The axial force plot is used primarily for vertical columns or cables supporting multiple loads along its length. The torque diagram is used primarily for shafts supporting multiple inputs and outputs. Each of these plots will have a different practical application, but we have grouped them together here because the process used to generate each of the two plots is very similar.
### Creating the Axial Force Diagram
The axial force diagram will plot out the internal axial (normal) forces within a beam, column, or cable that is supporting multiple forces along the length of the beam itself. This can be thought of as the internal tension or compression forces. The most relevant practical scenarios that match this description will be main support columns in a multi-story building, or hanging cables that are used to support multiple loads.
To create the axial force plot for a body, we will use the following process.
1. Solve for all external forces acting on the body.
2. Draw out a free body diagram of the body horizontally. In the case of vertical structures, rotate the body so that it sits horizontally and all the forces act horizontally
3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal axial forces, with positive numbers indicating tension and negative numbers indicating compression.
4. Starting at zero at the right side of the plot, you will move to the right, pay attention to forces in the free body diagram above. As you move right in your plot, keep steady except...
• Jump upwards by the magnitude of the force for any forces in our free body diagram to the left.
• Jump downwards by the magnitude of the force for any forces to the right.
• You can ignore any moments or vertical forces applied to the body.
By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.
To read the plot, you simply need to find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot. Again, positive numbers represent an internal tension at that location and negative numbers represent an internal compression at that location.
### Creating the Torque Diagram
The torque diagram will plot out the internal torsional moment within a shaft that is supporting multiple inputs and/or outputs along its length. The most relevant practical scenarios that match this description are shafts within complex gear or pulley driven systems.
To create the torque diagram for a shaft, we will use the following process.
1. Solve for all external moments acting on the shaft.
2. Draw out a free body diagram of the shaft horizontally, rotating the shaft if necessary, so that all torques act around the horizontal axis.
3. Lined up below the free body diagram, draw a set of axes. The x-axis will represent the location (lined up with the free body diagram above), and the y-axis will represent the internal torsional moment, with positive numbers indicating an internal torsional moment vector to the right and negative numbers indicating an internal torsional moment to the left.
4. Starting at zero at the right side of the plot, you will move to the right, pay attention to moments in the free body diagram above. As you move right in your plot, keep steady except...
• Jump upwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point left.
• Jump downwards by the magnitude of the moment for any torques in our free body diagram where the moment vector would point right.
• You can ignore any forces in the free body diagram or moments not about the x axis.
By the time you get to the left end of the plot, you should always wind up coming back to zero. If you don't wind up back at zero, go back and check your previous work.
To read the plot, you simply need to take the find the location of interest from the free body diagram above, and read the corresponding value on the y-axis from your plot.
### Question 1:
A wooden beam is subjected to the forces shown below (forces applied at base of vector). Draw the axial force diagram for the beam.
### Question 2:
A steel shaft is subjected to the torques shown below. Draw the torque diagram for this shaft.
Sours: http://mechanicsmap.psu.edu/websites/6_internal_forces/6-3_axial_torque_diagrams/axial_torque_diagrams.html
## Mechanics of Materials: Torsion
### Torsional Deformation
Torque is a moment that twists a structure. Unlike axial loads which produce a uniform, or average, stress over the cross section of the object, a torque creates a distribution of stress over the cross section. To keep things simple, we're going to focus on structures with a circular cross section, often called rods or shafts. When a torque is applied to the structure, it will twist along the long axis of the rod, and its cross section remains circular.
To visualize what I'm talking about, imagine that the cross section of the rod is a clock with just an hour hand. When no torque is applied, the hour hand sits at 12 o'clock. As a torque is applied to the rod, it will twist, and the hour hand will rotate clockwise to a new position (say, 2 o'clock). The angle between 2 o'clock and 12 o'clock is referred to as the angle of twist, and is commonly denoted by the Greek symbol phi. This angle lets us determine the shear strain at any point along the cross section.
Before we get into the details of this equation, it's important to note that because we're only discussing circular cross sections, we've switched from Cartesian coordinates to cylindrical coordinates. That's where the Greek symbol rho came from – it denotes the distance along the cross section, with rho=0 at the center and rho=c at the outer edge of the rod.
We can immediately learn a few things from this equation. The first thing might be obvious: the more angle of twist, the larger the shear strain (denoted by the Greek symbol gamma, as before). Second, and this is the big difference between axial-loaded structures and torque-loaded ones, the shear strain is not uniform along the cross section. It is zero at the center of the twisted rod, and is at a maximum value at the edge of the rod. Finally, the longer the rod, the smaller the shear strain.
So far, we've focused our attention on displacements and strain. To discuss the stress within a twisted rod we need to know how torque and stress relate. Since twist applies a shear strain, we expect that torque will apply a shear stress. The relationship between torque and shear stress is detailed in section 5.2 of your textbook, and it results in the following relation:
In this equation, J denotes the second polar moment of area of the cross section. This is sometimes referred to as the "second moment of inertia", but since that already has a well-established meaning regarding the dynamic motion of objects, let's not confuse things here. We'll discuss moment's of area in more detail at a later point, but they take on a very simple form for circular cross sections:
(Note: those are both the same equation – solid rods have an inner radius of ci=0).
Now we have equations for our shear strain and our shear stress, all that is left to do is use Hooke's law in shear to see how they are related. Hooke's law lets us write down a nice equation for the angle of twist – a very convenient thing to measure in lab or our in the field.
And, just like we saw for axial displacements, we can use superposition for our shear deformations as well:
This final equation allows us to split up torques applied to different parts of the same structure. Let's work out a problem, and see if we understand what's going on for torsional deformations.
### Power Transmission
One of the most common examples of torsion in engineering design is the power generated by transmission shafts. We can quickly understand how twist generates power just by doing a simple dimensional analysis. Power is measured in the unit of Watts [W], and 1 W = 1 N m s-1. At the outset of this section, we noted that torque was a twisting couple, which means that it has units of force times distance, or [N m]. So, by inspection, to generate power with a torque, we need something that occurs with a given frequency f, since frequency has the units of Hertz [Hz] or [s-1]. So, the power per rotation (2*pi) of a circular rod is equal to the applied torque times the frequency of rotation, or:
On the far right hand side of the equation, we've used the relation that angular velocity, denoted by the Greek letter omega, is equal to 2pi times the frequency.
### Statically Indeterminate Problems
One equation, two unknowns… we've been down this road before need something else. Although the type of loading and deformation are different, the statically indeterminate problems involving the torsion of rods are approached in the exact same manner as with axially loaded structures. We start with a free body diagram of twisted rod. Take, for example, the rod in the figure below, stuck between two walls.
Immediately upon inspection you should note that the rod is stuck to two walls, when only one would be necessary for static equilibrium. More supports than is necessary: statically indeterminate. And statically indeterminate means, draw a free body diagram, sum the forces in the x-direction, and you'll get one equations with two unknown reaction forces. So, we need to consider our deformations – for torsion, that means let's turn to our equation that describes the superposition of twist angles. For this equation, we should note that half the rod is solid, the other half is hollow, which affects how we calculate J for each half. Most importantly, we need to ask ourselves "what do we know about the deformation?" Well, since the rod is stuck to the wall at edge, the twist at A and must be equal to zero (just like the displacement in the last section). See if you can work the rest of this problem out on your own: What is the torque in each half of the rod?
(Answer: Ta=51.7 lb ft & Tb=38.3 lb ft).
### Summary
We learned about torque and torsion in this lesson. This different type of loading creates an uneven stress distribution over the cross section of the rod – ranging from zero at the center to its largest value at the edge. From this analysis we can develop relations between the angle of twist at any a point along the rod and the shear strain within the entire rod. Using Hooke's law, we can relate this strain to the stress within the rod. We also used a method of dimensional analysis to determine the power generated by a transmission shaft (i.e. a rod) that spins with a given frequency under an applied torque. Finally, we showed that torsion problems are also often statically indeterminate, and even though the loading and deformation is different, the technique we established in the last section for solving problems with axial loading is the same technique for solving problems with torque loading.
This material is based upon work supported by the National Science Foundation under Grant No. 1454153. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
Sours: https://www.bu.edu/moss/mechanics-of-materials-torsion/
How to Draw a Torque Diagram Using Equilibrium Equations
.
### You will also be interested:
.
914 915 916 917 918 | 6,101 | 28,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-49 | latest | en | 0.928565 |
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### #1 aclark17
Reputation: -1
• Posts: 71
• Joined: 18-June 08
Posted 16 August 2008 - 05:22 PM
I'm trying to put together a a polynomial calculator to perform simple arithmetic operations on short polynomials.
I decided to represent each polynomial as a linked list of terms.My problem is that the program crashes on the statement r=p+q where each is a polynomial object. The crash seems to take place within the function operator=, however it works fine on statements such as p=q or r=p. Before the return statement within operator+ the values of the return object show to be correct so it seems as if the problem is in passing the return object to operator=.Here it is..
```struct term
{
double a;
int b;
int sign;
term* next;
term(const double c, const int e, const int s);
};
class polynomial
{
public:
polynomial();
polynomial(const polynomial&);
~polynomial();
void print();
const polynomial& operator=(const polynomial& p);
polynomial operator+(const polynomial& p)const;
//polynomial operator-(const polynomial&)const;
//polynomial operator*(const polynomial&)const;
//polynomial operator/(const polynomial&)const;
private:
//void sort();
void clear();
void copy(const polynomial&);
term* first;
};
term::term(const double c, const int e, const int s)
{
a=c;
b=e;
sign=s;
next=NULL;
}
polynomial::polynomial()
{
first=NULL;
}
void polynomial::copy(const polynomial& original)
{
if(first!=NULL)
clear();
term* temp1= original.first;
first=new term(original.first->a,original.first->b,original.first->sign);
term* temp2=first;
for(temp1=temp1->next;temp1!=NULL;temp1=temp1->next){
temp2->next=new term(temp1->a,temp1->b,temp1->sign);
temp2=temp2->next;}
}
polynomial::polynomial(const polynomial& original)
{
copy(original);
}
void polynomial::clear()
{
term* temp=first;
while(temp!=NULL)
{
first=first->next;
delete temp;
temp=first;
}
}
polynomial::~polynomial()
{
clear();
}
{
double x;
int y,s;
term* temp;
char punc,op;
cout<<"Enter terms, include the degree of each term:";
cin>>x;
cin>>punc>>punc;
cin>>y;
s=1;
first=new term(x,y,s);
temp=first;
cin>>op;
cin>>x;
cin>>punc>>punc;
cin>>y;
if(op=='-')
s=-1;
else
s=1;
while(x>0)
{
temp->next=new term(x,y,s);
temp=temp->next;
cin>>op;
cin>>x;
cin>>punc>>punc;
cin>>y;
if(op=='-')
s=-1;
else
s=1;
}
}
//void polynomial::sort()
//{
//}
void polynomial::print()
{
term* temp1=first->next;
if(first->sign==-1)
cout<<'-';
cout<<first->a<<"x^"<<first->b;
while(temp1!=NULL)
{
if(temp1->sign==-1)
cout<<'-';
else
cout<<'+';
cout<<temp1->a<<"x^"<<temp1->b;
temp1=temp1->next;
}
}
const polynomial& polynomial::operator=(const polynomial& p)
{
cout<<'2'<<endl;
term* temp1;
term* temp2;
if(first!=NULL)
clear();
//crash here
cout<<'3'<<endl;
first=new term(p.first->a,p.first->b,p.first->sign);
cout<<'4'<<endl;
temp1=first;
temp2=p.first->next;
while(temp2!=NULL)
{
temp1->next=new term(temp2->a,temp2->b,temp2->sign);
temp1=temp1->next;
temp2=temp2->next;
}
return *this;
}
polynomial polynomial::operator+(const polynomial& p)const
{
polynomial sum;
int tempSign;
term* temp1;
term* temp2;
//term* tempNew=sum.first;
if(first->b==p.first->B)/>
{
if((first->a*first->sign+p.first->b*p.first->sign)<0)
tempSign=-1;
else
tempSign=1;
sum.first=new term(abs(first->a*first->sign+p.first->a*p.first->sign),first->b,tempSign);
temp1=first->next;
temp2=p.first->next;
}
else if(first->b>p.first->B)/>
{
sum.first=new term(first->a,first->b,first->sign);
temp1=first->next;
}
else if(first->b<p.first->B)/>
{
sum.first=new term(p.first->a,p.first->b,p.first->sign);
temp2=p.first->next;
}
//values of sum are correct.
cout<<sum.first->a<<' '<<sum.first->b<<' '<<sum.first->sign<<endl;
cout<<'1'<<endl;
return sum;
//omit further terms
//while(temp1!=NULL)
//{
//if(temp1->b==temp2->B)/>{
//if((temp->a*temp->sign+temp2->b*temp2->sign)<0)
//temp3->next->sign=-1;
//else
//temp3->next->sign=1;
//temp3->next->a=abs(temp1->a*temp1->sign+temp2->b*temp2->sign);
//temp1=temp1->next;
//temp2=temp2->next;
//}
//else if(temp1->b>temp2->B)/>
//{
//}
}
void main()
{
polynomial p,q,r;
p.print();
cout<<endl;
q.print();
cout<<endl;
r=p+q;
r.print();
cout<<endl;
}
```
*edit: Please use code tags around your code!
This post has been edited by NickDMax: 16 August 2008 - 07:15 PM
Is This A Good Question/Topic? 0
### #2 NickDMax
• Can grep dead trees!
Reputation: 2254
• Posts: 9,245
• Joined: 18-February 07
Posted 16 August 2008 - 07:20 PM
does it crash on p = (r+q); ?
Where is the code for polynomial::clear() ?
### #3 aclark17
Reputation: -1
• Posts: 71
• Joined: 18-June 08
Posted 16 August 2008 - 08:19 PM
NickDMax, on 16 Aug, 2008 - 07:20 PM, said:
does it crash on p = (r+q); ?
Where is the code for polynomial::clear() ?
clear is the 5th definition down, no the parentheses don't help in one of my desperate attempts I did in fact try that with no success, right now I'm witting a base class template called list with a node struct which from which class polynomial inherits private access and try see if that helps anything, I really see no logical explanation for this crash SUM! PLEASE HELP!!!
### #4 NickDMax
• Can grep dead trees!
Reputation: 2254
• Posts: 9,245
• Joined: 18-February 07
Posted 16 August 2008 - 08:54 PM
I tried to run your program but I can't even figure out how the input routine works...
### #5 perfectly.insane
Reputation: 70
• Posts: 644
• Joined: 22-March 08
Posted 16 August 2008 - 09:39 PM
You should initialize first from your copy constructor. I'm don't think this is the entire problem, but it is a problem, and a copy constructor is invoked in the effective expression r.operator=(p.operator+(q));
```// Add initializer, so that copy doesn't try to destroy first when it is called, as it never contains
// anything when invoked from a constructor.
polynomial::polynomial(const polynomial& original) : first(NULL)
{
copy(original);
}
```
Page 1 of 1
.related ul { list-style-type: circle; font-size: 12px; font-weight: bold; } .related li { margin-bottom: 5px; background-position: left 7px !important; margin-left: -35px; } .related h2 { font-size: 18px; font-weight: bold; } .related a { color: blue; } | 2,013 | 6,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40 | longest | en | 0.629459 |
https://www.nidec.com/en-NA/technology/motor/basic/00011/ | 1,560,895,427,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998817.58/warc/CC-MAIN-20190618203528-20190618225528-00441.warc.gz | 843,328,463 | 9,416 | What Are Motors?
2-1-1 Reviewing The PrincipleOf DC Motor Rotation
I think some of you might have understood the DC motor rotation principle in school by understanding the attractive and repulsive forces of magnets. You will have learnt more about motor rotation principles by using the law of BLI and Fleming's law (to be described later). To say "Fleming's law" may sound smarter than explaining "the N-pole or S-pole of a magnet......" and Fleming's law appears more academic as it can be used for calculating force generation.
When you disassemble a DC motor, you will find the wire (coil) wound around a strip of iron. When the iron is nearby, the magnetic flux is concentrated in the iron portion, and almost no magnetic flux is present within the coil section (wire) (see Fig. 2.18 to be shown later). In this condition, it is not possible to explain the DC motor rotation principle using Fleming's law that states, "For a current-carrying wire in a magnetic field ........."
Nevertheless, school physics or electrical engineering textbooks cannot be wrong. So how should you think about this? To master any kind of technology, it is important for you to understand things concretely based on actual experience in addition to understanding them in your head.
To do this, we will study motor rotation principles and rotating speed in depth by using familiar subject matter found around yourself as leads and clues.
Firsthand Application of DC Motor Operation
If you have a motor used in a model, let's operate it.
• <1> Connect the motor to a 1.5-V dry battery, and rotate the motor.
The motor should start running and you should hear a soft sound. This is called no-load operation.
• <2> If the battery connection is reversed, the motor rotation reverses.
• <3> If you load the motor by lightly pinching the motor shaft with your fingertips, you will feel the rotative force (torque) transmitted to them.
The running sound of the motor changes and the rotating speed may seem to decelerate.
• <4> If you pinch the shaft a little harder, the torque increases and the rotation speed decreases further.
• <5> If you pinch it harder, then the motor stops running.
But the motor will continue to make a howling noise.
Next, try the same experiment by using two batteries.
• <6> The motor rotation sound becomes higher and it may seem to run faster.
• <7> If you pinch the shaft with greater force than you did when using a single battery, the motor still keeps rotating.
You would need to apply more pressure to stop the motor by pinching the shaft.
The above experiment teaches us the following:
• Using two batteries increases the rotative force.
• Using two batteries increases the rotating speed.
• If the power supply remains unchanged, increasing the load (by using fingers to pinch the shaft) reduces the rotating speed.
This was too simple an experiment to call it one, but I think it raised an important phenomenon to consider in terms of the operating principles of DC motor and their characteristics.
Now what do you think increased the rotative force or increased the rotating speed? Was it the effect of battery voltage or current? And how are the rotative force and rotating speed related?
We will now consider the DC motor based on the above experience.
Structure of the DC Motor
Fig. 2.1 shows the structure and the names of respective parts of a DC motor used in plotters and copying machines, etc.
Compared with the motor used in a model, this motor has more coils and uses carbon brushes and roller bearings, and each mechanism is very sophisticated. But the basic structure and the principle of rotation are the same. | 773 | 3,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-26 | latest | en | 0.916144 |
https://stats.stackexchange.com/questions/141937/a-simple-regression-model-for-our-experiment | 1,561,507,071,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999964.8/warc/CC-MAIN-20190625233231-20190626015231-00145.warc.gz | 575,068,159 | 32,953 | # A Simple Regression Model for Our Experiment? [closed]
We know, In statistics, simple linear regression is the least squares estimator of a linear regression model with a single explanatory variable. In other words, simple linear regression fits a straight line through the set of n points in such a way that makes the sum of squared residuals of the model (that is, vertical distances between the points of the data set and the fitted line) as small as possible.
suppose in simple linear regression model
$y= \alpha + \beta x + \epsilon$
, from a random instance we gather this information:
$\bar{x}=\bar{y}, \Sigma x_iy_i = \Sigma x_{i}^{2}$
I want to calculate estimation $(\alpha, \beta)$ with least square method. how I can solve it?
Infact I'm a Biology Background, and for experiment need this value. thanks.
## closed as off-topic by Glen_b♦, gung♦, Nick Cox, kjetil b halvorsen, Scortchi♦Mar 16 '15 at 15:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Self-study questions (including textbook exercises, old exam papers, and homework) that seek to understand the concepts are welcome, but those that demand a solution need to indicate clearly at what step help or advice are needed. For help writing a good self-study question, please visit the meta pages." – Glen_b, gung, Nick Cox, kjetil b halvorsen, Scortchi
If this question can be reworded to fit the rules in the help center, please edit the question.
• There is no reason in general why $\sum x_i y_i = \sum x_i^2$: in fact that minimally requires the same units of measurement. You mean something else, I imagine, but it is not clear what. – Nick Cox Mar 16 '15 at 14:41
• @NickCox, we gather this from our information. – Mina Akram Mar 16 '15 at 14:42
• Your question sounds increasingly like a self-study question and should be flagged as such. – Nick Cox Mar 16 '15 at 14:46
• @NickCox, what do you mean by "self-study"? – Mina Akram Mar 16 '15 at 14:47
• stats.stackexchange.com/help/on-topic Note that @statchrist made the same comment in their answer. – Nick Cox Mar 16 '15 at 14:49
This question appears to be a self-study question.
One way to solve the problem ist to look at the mathematical expression for the estimator in matrix notation:
$\hat{\beta}_{OLS}=(X'X)^{-1}X'y$.
Insert the matrices for your model in matrix notation: $y = X\beta_{OLS} + \epsilon$.
For instance with $\beta_{OLS} = \begin{pmatrix} a \\ \beta \end{pmatrix}$ and $X=\begin{pmatrix} 1 & X_1 \\ \vdots & \vdots \\ 1 & X_n \end{pmatrix}$. Next step would be to manipulate this equation in order to get expressions in terms of the mean and insert the additional information you have.
• i couldent get it, you means final value is (0,1) ?, is it possible learn me ? – Mina Akram Mar 16 '15 at 13:58
• oooh, i got it thanks. but last step is how we can get numerical value for our experiment ? – Mina Akram Mar 16 '15 at 14:11
• it dosent satisfy me, is it possible to make it clear?it's (0,1) ? – Mina Akram Mar 16 '15 at 14:26
• Yes, have you tried and computed the inverse of X'X/N? – statchrist Mar 16 '15 at 14:36 | 865 | 3,141 | {"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} | 3.390625 | 3 | CC-MAIN-2019-26 | longest | en | 0.887501 |
https://www.shaalaa.com/question-bank-solutions/following-are-cannot-be-probability-event-3-5-type-event-complementry_29591 | 1,597,425,868,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739347.81/warc/CC-MAIN-20200814160701-20200814190701-00565.warc.gz | 818,101,473 | 8,759 | Share
# Following Are Cannot Be the Probability of an Event ? 3/5 - Mathematics
Course
#### Question
Which of following cannot be the probability of an event ?
3/5
#### Solution
The probability of an event lies betweet '0' and '1' i.e 0 ≤ P(E) ≤1.
3/5=0.6
∵ 0<=0.6<=1
Hence, it can be the probability of an event.
Is there an error in this question or solution? | 112 | 373 | {"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} | 3.109375 | 3 | CC-MAIN-2020-34 | latest | en | 0.870004 |
http://mathhelpforum.com/advanced-statistics/3432-lildragonfly-s-regression-q3.html | 1,527,296,507,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00426.warc.gz | 186,988,249 | 12,020 | 1. ## LilDragonfly's Regression Q3
This is posted on behalf of LilDragonfly.
Attached is a data set called "Retail Footware Sales in New Zealand March 1991 to December 2000".
Use this data to answer the following questions:
1. Analyse this time series. This should include:
a) seasonally adjusting the data and interpreting the results in context
b) choosing a model to represent the recent trend using formal methods,
for example appropriate numerical calculations or least squares regressiosn
c) evaluating your model using residual analysis
2. Write a report on the analysis of this time series data. You should include:
the methods and results of your analysis and a discussion of the features of the series, how you could improve your model, limitations of your analsis.
2. Originally Posted by CaptainBlack
This is posted on behalf of LilDragonfly.
Attached is a data set called "Retail Footware Sales in New Zealand March 1991 to December 2000".
Use this data to answer the following questions:
1. Analyse this time series. This should include:
a) seasonally adjusting the data and interpreting the results in context
I will remove the seasonality by using a 9-point moving average, that is
the smoothed sales for period $\displaystyle i$ will be taken to be:
$\displaystyle SmoothdSales(i)=\left(\sum_{k=i-4}^{i-1} Sales(k)+\sum_{k=i+1}^{i+4}Sales(k) \right)/8$.
This gives the smoothed sales in the $\displaystyle i$-th quarter as the average
quarterly sales for the preceeding four quarters and the following four quarters.
That is the smoothing is taking place over an integral number of years and
so the seasonality is averaged out. Also note that this means that the
sales for quarter $\displaystyle i$ are not used in calculating the smoothed
sales for quarter $\displaystyle i$.
The results of this smoothing is shown in the section of an excel spreadsheet
shown in the attachment, and the graph attached.
The data shows seasonal behaviour with two peaks per year the first the
jun quarter and the second in the dec quarter. There is also a general
downdward trend in sales, but with substantial variation about that trend.
3. Originally Posted by CaptainBlack
This is posted on behalf of LilDragonfly.
Attached is a data set called "Retail Footware Sales in New Zealand March 1991 to December 2000".
Use this data to answer the following questions:
1. Analyse this time series. This should include:
[indent]a) seasonally adjusting the data and interpreting the results in context
b) choosing a model to represent the recent trend using formal methods,
for example appropriate numerical calculations or least squares regressions
c) evaluating your model using residual analysis
Selecting a model to represent the trend this data is problematical.
There is a downward trend but also a significant variation about
the trend line and the question is do we want to model that variation?
I would stick with a simple linear trend line (as shown in the plot attached
to an earlier post), this is calculated using the standard linear regression
method. I think that trying to model the variation about the trend seems
to me to be over modelling (though an additional sinusoidal component looks
as though it could represent a substantial part of the remaining systematic
variation in the data about the trend line).
There are a number of residual plots that could be employed in this analysis.
The most significant is a simple plot of the residual against quarter. This
is shown in the attachment.
The residual plot shows that the residuals are not randomly scattered about
the zero residual axis as would be the case if we had modelled all of the
systematic variation in the data.
RonL
4. Originally Posted by CaptainBlack
This is posted on behalf of LilDragonfly.
Attached is a data set called "Retail Footware Sales in New Zealand March 1991 to December 2000".
Use this data to answer the following questions:
2. Write a report on the analysis of this time series data. You should include:
the methods and results of your analysis and a discussion of the features of the series, how you could improve your model, limitations of your analsis.
Most of what could be said to answer this part of the question has been said
above, it probably only needs more elaboration here.
Some comment on long term economic cycles and effects of other external
factors could be relevant.
Look at your notes and/or examples to see exactly what is expected to
answer this part of the question. This sort of thing usually depends on why
the customer has commissioned the study.
RonL
5. The seasonality can be extracted by averaging the smoothed/actual
data for quarters that correspond to the same season. This gives seasonality
factors:
March.0.955742591
June.. 1.064359786
Sept...0.881293072
Dec....1.084365967
RonL
6. A more sophisticated model of the long term variation of this data can
be modelled as:
$\displaystyle sales(t)=a.t+b+c.\sin(2. \pi.f.t+\phi)$
This can be fitted to the de-seasoned data using Nonlinear Least Squares.
This has been done and is illurstrated in the attachment.
The parameter values for this model are:
Code:
a -0.334988202
b 67.43608622
f 0.050651566
phi 2.469557225
c 2.464159211
The residual plot for this model can be seen in the secon attachment. As
can be seen this residual plot is much better. | 1,211 | 5,363 | {"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} | 3.25 | 3 | CC-MAIN-2018-22 | latest | en | 0.913892 |
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