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http://virtual.cvut.cz/kifb/en/concepts/_signum_fn.html | 1,369,113,061,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699730479/warc/CC-MAIN-20130516102210-00075-ip-10-60-113-184.ec2.internal.warc.gz | 291,071,641 | 2,699 | Select language: english | cesky | deutsch | italiano | simplified chinese | traditional chinese | hindi Concept: English word: Home
# signum fn (SignumFn)
(SignumFn number) denotes the sign of number. This is one of the following values: -1, 1, or 0.
## Ontology
SUMO / NUMERIC-FUNCTIONS
## Class(es)
class
inheritable relation
unary function
signum fn
## Coordinate term(s)
absolute value fn abstraction fn arc cosine fn arc sine fn arc tangent fn back fn begin fn begin node fn cardinality fn ceiling fn complement fn cosine fn cut set fn denominator fn end fn end node fn extension fn floor fn front fn future fn generalized intersection fn generalized union fn giga fn imaginary part fn immediate future fn immediate past fn initial node fn integer square root fn kilo fn list length fn magnitude fn mega fn micro fn milli fn minimal cut set fn nano fn numerator fn organization fn past fn path weight fn pico fn power set fn predecessor fn principal host fn probability fn property fn rational number fn real number fn reciprocal fn round fn sine fn skin fn square root fn successor fn tangent fn tera fn terminal node fn wealth fn when fn year fn
## Type restrictions
integer SignumFn(real number)
## Related WordNet synsets
polarity, sign
having an indicated pole (as the distinction between positive and negative electric charges); "he got the polarity of the battery reversed"; "charges of opposite sign"
## Axioms (4)
If "number1 mod number2" is equal to number, then "the sign of number2" is equal to "the sign of number".
```(=>
(equal
(RemainderFn ?NUMBER1 ?NUMBER2)
?NUMBER)
(equal
(SignumFn ?NUMBER2)
(SignumFn ?NUMBER)))```
If number is an instance of nonnegative real number, then "the sign of number" is equal to or "the sign of number" is equal to .
```(=>
(instance ?NUMBER NonnegativeRealNumber)
(or
(equal
(SignumFn ?NUMBER)
1)
(equal
(SignumFn ?NUMBER)
0)))```
If number is an instance of positive real number, then "the sign of number" is equal to .
```(=>
(instance ?NUMBER PositiveRealNumber)
(equal
(SignumFn ?NUMBER)
1))```
If number is an instance of negative real number, then "the sign of number" is equal to .
```(=>
(instance ?NUMBER NegativeRealNumber)
(equal
(SignumFn ?NUMBER)
-1))``` | 591 | 2,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.514923 |
https://www.mersenneforum.org/showthread.php?s=8238c339643337bc8b2e6e9accddd599&t=10059 | 1,660,743,274,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00579.warc.gz | 778,445,408 | 10,170 | mersenneforum.org Impossible Integral
Register FAQ Search Today's Posts Mark Forums Read
2008-03-06, 03:10 #1 Unregistered 2·5·733 Posts Impossible Integral Alright, so this is a Calculus 2 homework problem and my friends and I are completely stuck. The original question is to find if the Summat of {tan^-1 (x)}/x^1.1 (from 1 to infinity)converges or diverges, and if it converges to find its sum. Using the Integral Test, we said that this series, did in fact converge. The problem was in evaluating the integral. Integral (1 to infinity) of (tan^-1 (x))/x^1.1 We cannot manage to evaluate this... no theorem or rule we can find helps in this evaluation. Any help would be greatly appreciated. Thanks!
2008-03-06, 15:46 #2
R.D. Silverman
"Bob Silverman"
Nov 2003
North of Boston
23·3·311 Posts
Quote:
Originally Posted by Unregistered Alright, so this is a Calculus 2 homework problem and my friends and I are completely stuck. The original question is to find if the Summat of {tan^-1 (x)}/x^1.1 (from 1 to infinity)converges or diverges, and if it converges to find its sum. Using the Integral Test, we said that this series, did in fact converge. The problem was in evaluating the integral. Integral (1 to infinity) of (tan^-1 (x))/x^1.1 We cannot manage to evaluate this... no theorem or rule we can find helps in this evaluation. Any help would be greatly appreciated. Thanks!
(1) Using the integral test is hitting a fly with a hammer. Instead,
simply note that the numerator is bounded above by pi, and that sum(pi/x^1.1)) clearly converges.
(2) Evaluating the integral that you gave will not give the sum of the series.
For that, you need Euler-MacLauren summation.
(3) The indefinite integral can probably be expressed in terms of hyper-
geometric functions. It is very unlikely that a closed form expression exists
for the sum.
(4) To evaluate your infinite integral, I would use complex contour integration
methods, along with Cauchy's Theorem. Start by finding the residues of the
series; this requires a Laurent expansion. I won't even attempt it by hand.
I would want to use Mathematica. I would use a quarter-circle contour centered at 0.
The integral can't be expressed in terms of elementary functions.
Finding the sum seems well beyond a 2nd year calculus course.
2008-03-06, 21:16 #3 Unregistered 374710 Posts The original problem was to find if SIGMA [(tan^-1 (n)/n^1.1)] from 1 to infinity converges and if it converges to find its sum. I proved that it converges using the Integral test. Using 2nd year calculus tools, how would you find the sum? A 3rd year calc student mentioned something about a comparison that could be done involving pi/2.
2008-03-07, 12:30 #4
R.D. Silverman
"Bob Silverman"
Nov 2003
North of Boston
23×3×311 Posts
Quote:
Originally Posted by Unregistered The original problem was to find if SIGMA [(tan^-1 (n)/n^1.1)] from 1 to infinity converges and if it converges to find its sum. I proved that it converges using the Integral test. Using 2nd year calculus tools, how would you find the sum?
You don't. I doubt very much whether the sum has a closed form.
It *might* be expressible in terms of hypergeometric functions.
Why do I think this? Because integral ATAN(x)/x dx has such a
representation.
2008-03-07, 12:42 #5 R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 112·13 Posts By Mathematica 5.1 the integral is (see the attachment) Attached Thumbnails
2008-03-08, 20:05 #6 m_f_h Feb 2007 43210 Posts since convergence is indeed ensured by comparision with (pi/2)/x^1.1, why not do it numerically... gp> intnum(x=1,[1],atan(x)/x^1.1) %1 = 14.879905283440983234832458288559145866892862726910 gp> ## *** last result computed in 140 ms. But here's a primitive from Maple: -10/x^(1/10)*arctan(x) +5/2*sqrt(2)*ln((x^(1/5)+x^(1/10)*sqrt(2)+1)/(x^(1/5)-x^(1/10)*sqrt(2)+1)) +5*sqrt(2)*arctan(x^(1/10)*sqrt(2)+1) +5*sqrt(2)*arctan(x^(1/10)*sqrt(2)-1) -20*sum(_R*ln(x^(1/10)-262144*_R^9),_R = RootOf(4294967296*_Z^16-16777216*_Z^12+65536*_Z^8-256*_Z^4+1)) Taking the limit x=infinity seems obvious except for the last term. (might have changed 4Z into Z'...)
Similar Threads Thread Thread Starter Forum Replies Last Post wildrabbitt Homework Help 2 2016-04-08 08:21 Unregistered Information & Answers 6 2013-06-14 12:38 Baztardo Homework Help 6 2011-04-10 22:17 flouran Information & Answers 6 2009-07-20 20:00 Andi47 Homework Help 3 2009-05-13 17:49
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Wed Aug 17 13:34:34 UTC 2022 up 41 days, 8:21, 1 user, load averages: 1.52, 1.49, 1.47 | 1,368 | 4,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-33 | latest | en | 0.924173 |
https://www.hackmath.net/en/math-problem/2221 | 1,713,026,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00558.warc.gz | 783,397,994 | 9,199 | # Alfa, beta, gama
In the ABC triangle, is the size of the internal angle BETA 8 degrees larger than the size of the internal angle ALFA and the size of the internal angle GAMA is twice the size of the angle BETA? Determine the size of the interior angles of the triangle ABC.
a = 39 °
b = 47 °
c = 94 °
### Step-by-step explanation:
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Do you have a system of equations and looking for calculator system of linear equations? | 136 | 537 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.837541 |
https://electronics.stackexchange.com/questions/306768/minimum-and-gates-for-4-input-2-output-functions | 1,566,120,616,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00009.warc.gz | 447,933,796 | 30,117 | # Minimum AND gates for 4 input-2 output functions
It seems that Karnaugh maps and Quine–McCluskey algorithm are used to minimize the general number of gates to represent some truth table (boolean function) with $n$ inputs (usually small $n$) and one output.
My problem is different from the problems the above techniques solve in two aspects:
1. Need to consider 2-output-bit rather than 1
2. Can use AND, NOT, XOR and need to minimize number of AND where the number of XOR and NOT gates do not matter.
The functions I'm working on have 4 input bits. My question is whether there is a way to determine the minimum number of AND gates for realizing certain function.
• So, what is your question? – Andy aka May 22 '17 at 14:40
• I edited in the question – Bush May 22 '17 at 15:50
• There is likely no established algorithm for this, because "number of AND gates" is not a relevant cost metric in any real-world technology I'm aware of. If you have a homework question that's asking you to do this for some particular truth table, you just need to be clever. – The Photon May 22 '17 at 15:56
• Minimizing multiple output logic is a real-world problem, so you may find some ways to do that in the literature. The main idea would be finding common factors in the equations for two outputs, and not calculating those terms twice. – The Photon May 22 '17 at 15:57
• Sounds like an academic use of en.wikipedia.org/wiki/De_Morgan%27s_laws exchanging nand for nor – sstobbe May 22 '17 at 19:37 | 374 | 1,492 | {"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} | 2.671875 | 3 | CC-MAIN-2019-35 | latest | en | 0.919407 |
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CSULB_STAT475_handout13
CSULB_STAT475_handout13 - STAT 475 Chapter 15 GENERATING...
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STAT 475 Chapter 15 GENERATING DATA WITH DO LOOPS CONSTRUCTING DO LOOPS A DO loop is used to perform repetitive calculations in the DATA step. DO loops make DATA steps concise and transparent. The syntax is data dataset_name ; do index_variable = start to stop by increment ; SAS statements; end; run; where the index_variable stores the value of the current iteration of the Do loop, start specifies the initial value of index_variable , stop is the last value that executes the DO loop, increment specifies an increment value for index_variable . Typically increment =1 , and the BY statement may be omitted in this case. Example. The program below calculates how much interest was earned each month for a one-year investment. data earnings; amount=1000; rate=0.075/12; do month=1 to 12; earned+(amount+earned)*rate; end; run; proc print noobs; run; The variable amount is the dollar amount of the initial investment. The annual interest rate is assumed 7.5% compounded monthly, so the variable rate =0.75/12 represents the interest rate per month. The variable earned is the cumulative interest amount earned. Note the syntax for iterative calculation of the variable earned . The general syntax is variable_name + expression 1
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At the beginning of the DATA step processing, the data file is created, and the values of amount and rate are assigned. Next, the DO loop starts executing, recalculating the value of earned , and incrementing the value of month . After the last (12th) execution of the DO loop, the index variable month is incremented to 13. Because 13 exceeds the stop value of the index variable, the DO loop terminates. The final value of month =13 is stored in the data file. The final data set has the form amount rate month earned 1000 .00625 13 77.6326 The calculated amount of interest earned during a one-year period is 77.6326.
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Ask a homework question - tutors are online | 568 | 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} | 3.109375 | 3 | CC-MAIN-2018-13 | latest | en | 0.825832 |
https://community.wolfram.com/groups/-/m/t/331662 | 1,685,230,339,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643388.45/warc/CC-MAIN-20230527223515-20230528013515-00502.warc.gz | 208,184,195 | 22,701 | # Simple Functions in the Wolfram Alpha app
Posted 9 years ago
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Hello Everyone,I'm a beginner in both Wolfram as in english (I'm from Brazil). I'm studying the first year of Engineering, and learning simple functions. I got a Wolfram Android and I don't know how to define a function like this example:f(x) = 6x-2 (until here I done), but I want to put that x=2 or, x = something.I've already tried put: f(x) = 6x-2 ; x = 2 put he understand two different things, and I want to show that x=2 is the variable tha f(x).Just to remember I'm using the Android. This tool will help me to verify and see different forms to do same exercises.Thanks
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Posted 9 years ago
Thanks, but I'm not able to do this in smartphone. I guess I have to study more the program, or is there some limitation. Maybe I'm not using the correct or full version on my smartphone (bought wolfram mathematics on google play). The text box where I type that looks like the image link below, and I can not create a syntax and do like you did. If I type f [x_]: = x 6 - 2 it's a misunderstanding. Excuse my ignorance and thanks again.
Posted 9 years ago
(I'm from Brazil) Welcome! Give the syntax a try .... first one defines the function f In[231]:= Clear[f, x] f[x_] := 6 x - 2 then one uses, runs or calls it, giving the argument as usual, note the brackets! In[233]:= f[2] Out[233]= 10 In[234]:= f /@ Range[-3, 3] Out[234]= {-20, -14, -8, -2, 4, 10, 16} | 419 | 1,485 | {"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} | 3.375 | 3 | CC-MAIN-2023-23 | longest | en | 0.901003 |
https://learnmate.com.au/vce-maths-methods-study-plan/ | 1,600,958,488,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219221.53/warc/CC-MAIN-20200924132241-20200924162241-00215.warc.gz | 475,213,977 | 71,176 | # VCE Maths Methods Study Plan – Read Now!
### VCE Maths Methods Study Plan – Read Now!
This article was written by Jake Makaling, a current VCE Maths Methods tutor. Jake currently is accepting students, so if you’re interested in his services, please go here
What I’ve learnt from Maths Methods is that it’s not just about the content you learn in year 12. Yes, SACs are heavily based on testing the concepts you learn in class, but when it comes to actually study for the exam, it becomes much more than that.
Most students will find that the concepts aren’t too difficult to learn, especially through practising exercises (textbook, ‘Checkpoints’ book) and watching tutorials (Khan Academy, Edrolo etc.). Reaching that upper next level, however, may show to be quite challenging. I’ve put together some key tips which I think would be useful for your studies (particularly Maths Methods) leading up to the exams.
1. Content
Yes, pretty standard, it is essential to understand the concepts before considering success in this subject. Start by scanning through the study design; any unfamiliar terms/gaps in knowledge? Revise over these sections in the textbook and watch tutorials if need be (check out Khan Academy – good for breaking down concepts).
Scan through your past SACs if you have access to them and keep a tab on the errors you make. If these errors are related to understanding, spend some time learning the content. Otherwise, if they’re careless errors (misreading the question, not leaving the answer in exact form), write up a page of all the common errors you’ve made in the past from your SACs.
2. Method
Your errors might not be limited to those above, but rather due to a messy write up of an extended answer – hard to follow. A clear method when writing up answers is very important (hence, why this subject is called Maths ‘Methods’ probably). Teachers always say ‘show all your working!!’ – yes, it helps you to minimise errors as you can trace back your steps, but it’s also a good way to gain marks even if your answer is incorrect.
Markers/examiners don’t have time to decipher through a maze of information, they’d rather see clear, simple steps leading up to a plausible answer. So practise using the correct notation, keep your steps orderly, and try not to take shortcuts, making sure not to skip any important steps. Especially with a 2-mark question in Exam 2 – it may ask you to find values which are solvable by the calculator, but it usually indicates that you need to include some sort of working. Practise this when writing up solutions as you’ll get more and more efficient at it when it comes to the time constraints of the exam.
3. Foundations
This is crucial to your success in the exam. You may feel confident with all the concepts, and you might feel like you’ve nailed writing up answers. What you might be missing though are the foundations, the predominantly algebraic skills that add the extra punch of difficulty to the exam. It may seem simple, but a lot of people don’t have these foundations down pat by year 12. This includes having the skills to be able to manipulate any equation given to you by being able to rearrange and isolate each individual term (eg. make x the subject, make y the subject, make the constant the subject). Do you know how to rearrange for x when given the equation, sin(x) = cos(x)? Skills like these are not physically taught in Maths Methods and are merely assumed knowledge.
Learning how to manipulate fractions and surds are also quite important, especially when it comes to integrals. Knowing how to find a common denominator may seem simple, but it’s knowledge like this that can often lead to the most common errors. If you’re sitting right now thinking ‘wow, that’s me’, then you might have to brush up a bit on algebra. Bit of practice and you’ll get there. Just aim to get to a point where it becomes second nature to you, where you don’t even have to think about how to simplify fractions, how to rationalise surds (not actually needed in the final answer but still helpful when adding to other terms) and so on.
The key to success isn’t limited to the key tips above but I hope it helps you form a basis on what you should focus on from now leading up to your exams.
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VCE Maths Methods Study Plan – Read Now! | 1,097 | 5,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.943143 |
http://openstudy.com/updates/5643a315e4b01c719e6f7bb8 | 1,519,099,847,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812873.22/warc/CC-MAIN-20180220030745-20180220050745-00367.warc.gz | 267,399,992 | 8,148 | • anonymous
Calculus The following formula accurately models the relationship between the size of a certain type of a tumor and the amount of time that it has been growing: V(t)=400(1-e^ -.0024t)^3 where t is in months and V(t) is measured in cubic centimeters. Calculate the rate of change of tumor volume at 100 months. The correct answer is 0.103 cm^3 month but, how is this arrived at?
Calculus1
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Not the answer you are looking for? Search for more explanations. | 341 | 1,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.490346 |
https://vintage-kitchen.com/food/how-do-you-calculate-the-diameter-of-a-drain-pipe/ | 1,642,395,915,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00262.warc.gz | 662,705,820 | 30,151 | # How do you calculate the diameter of a drain pipe?
One of the hydraulic design equations used to determine the discharge line diameter is based on Manning’s equation (Schwab et al., 1981) and reads:(1) d = 51.7 ( dc × A × n ) 0.375 S 0.1875 where d = internal diameter of the drain, mm, DUSA = drainage coefficient, mm day1, A = drainage area in hectares, n = Crew roughness
Contents
## How to calculate the size of the drain?
To calculate the amount of water to store, multiply the flow rate for each drainage area by 15. The flow rate for each area is shown in gallons per minute. Multiply by 15 minutes to save gallons.
## What is the diameter of a drain?
Toilet drains in modern plumbing systems are 3 inches or 4 inches in diameter. The wider the pipe, the more waste it can move and the less likely it is to get clogged.
## How to calculate the diameter of the pipe based on the flow and pressure?
The pipe diameter calculator calculates the pipe ID using the simple relationship between flow, velocity and cross section (Q=v A).
## What is the outside diameter of a 3 inch PVC pipe?
The outer diameter of 1/2″ PVC pipe is 0.840″ and the outer diameter of 3″ PVC pipe is 3,500″.
## How thick is a drain pipe?
Some older sinks may have drains that connect to a 1 1/4″ end fitting, but most modern drains have a 1 1/2″ outlet. All tubing connected to the tailpiece, including the P-Trap assembly and all horizontal extensions, are either 1 1/4 or 1 1/2 inches in diameter.
## How do you calculate the size of a diver’s pipe?
We take the mean depth times the mean width to get the cross section, and divide by four to get the approximate diameter of the pipe needed to get through the mean storm. The diameter of the pipe(s) used must correspond to the required total diameter without using a pipe greater than the average depth.
## How is a trench discharge measured?
Multiply the water depth by 2 for depths greater than 5 inches. Multiply that number by 32.2 feet per second. Take the product of the three numbers raised to the 0.5 power. Multiply the resulting number by the effective area of the drain.
## How big should the sanitary pipes be?
In most cases, the main pipe from the street to your home is 3/4″ or 1″ in diameter, utilities use 3/4″ diameter pipes, and individual component pipes are 3/4″ in diameter . 1/2″. Remember that if each foot of pipe extends beyond your water supply, the water pressure will decrease by half a pound per square inch.
## How to find the diameter of a pipe from the circumference?
Measure the circumference with a soft tape measure and divide the outer circumference by 3.1415. Here the circumference is about 1/32″ (0.03″) smaller than 6″ = 5.97″. So 5.97 ÷ 3.1415 = 1.900″ diameter – 1 1/2″ nominal pipe. | 696 | 2,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2022-05 | latest | en | 0.898598 |
https://chestofbooks.com/crafts/metal/Builder-Mechanic/index.html | 1,669,491,741,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00598.warc.gz | 193,800,304 | 27,913 | # The Tinman's Manual And Builder's And Mechanic's Handbook | by Isaac Ridler Butt
Designed For Tinmen, Japanners, Coppersmiths, Engineers, Mechanics, Builders, Millwrights, Smiths, Masons, Carpenters, Joiners, Slaters, Plasterers, Painters, Glaziers, Pavers, Plumbers, Surveyors, Gaugers, etc; with Compositions and Receipts for other useful and important purposes in the Practical Arts.
Title The Tinman's Manual And Builder's And Mechanic's Handbook Author Isaac Ridler Butt Publisher I. R. Butts & Co. Year 1861 Copyright 1861, I. R. Butts & Co. Amazon The Tinman's Manual And Builder's And Mechanic's Handbook
By I. R. BUTTS, Author of the " United States Business Man's Law Cabinet," " Business Man's Law Library ;" " Merchant's and Shipmaster's Manual and Shipbuilder's and Sailmaker's Assistant," etc, etc.
Preface
The present work is offered to Tinmen, Builders, Mechanics, and Engineers, as a useful manual of reference, and information. The first part of the work ...
Manufacture Of Tin Plate
The different processes of the manufacture of tin plate may be described most properly in seven distinct stages. The first begins with the bars of iron which ...
Quality Of Tin Plate
The tests for tin plates are ductility, strength, and color; and to possess these, the iron used must be of the best quality, and all the process be conducted ...
How To Find The Circumference Of Any Diameter
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y.] Fig. 1. From the centre C describe a circle AB, having the required diameter; then place the ...
How To Find The Area Of The Sector Of A Circle
Rule. Multiply the length of the arc DGE by its radius DC, and half the product is the area. The length of the arc DGE equal 9½ feet, and the radii CD, CE, ...
Proportion Of Circles
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y. Original. Fig. 2. To enable machinists to enlarge or reduce machinery wheels without changing ...
How To Describe An Ellipse, of Oval
[Simple Method.] Fig. 3. At a given distance, equal to the required eccentricity of the ellipse, place two pins, A and B, and pass a string, ACB, round them; ...
How To Describe An Ellipse
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y.] Original. Fig. 4. To describe an ellipse of any length and width, and by it to describe a ...
How To Find The Circumference Of An Ellipse
Rule. - Multiply half the sum of the two diameters by 3.1416, and the product will be the circumference. Example. - Suppose the longer diameter 6 inches and ...
How To Find The Area Of An Ellipse
Rule. - Multiply the longer diameter by the shorter diameter, and by .7854, and the product will be the area. Example.- Required the area of an ellipse whose ...
How To Describe A Right Angled Elbow
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y.] Original. Fig. 5. First construct a rectangle ADEB equal in width to the diameter of the elbow, ...
How To Describe A Curved Elbow
[Drawn for this work by L.W. Truesdell, Tinman, Owego,N. T.] Original. Fig. 7. FIG. 8. Describe two circles UX and V'S, the curves desired for the elbow, ...
Soldering
For Lead the solder is 1 part tin, 1 to 2 of lead; -- for Tin 1 to 2 parts tin to 1 of lead; - for Zinc 1 part tin to 1 to 2 of lead; -for Pewter 1 part tin to ...
How To Describe A Straight Elbow
[Old Method.] Fig. 6. Mark out the length and depth of the elbow, ABCD; draw a semicircle at each end, as from AB and CD; divide each semicircle into eight ...
How To Describe Bevel Covers For Vessels, Or Breasts For Cans
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y.] Fig. 11. From 0 as a centre, describe a circle BE larger than the vessel; and from C as a centre, ...
How To Describe Pitched Covers For Pails, Etc
Fig. 12. To cut for pitched covers, draw a circle one inch larger than the hoop is in diameter after burring, then draw a line from the centre to 3 the ...
How To Describe An Oval Boiler Cover
[Drawn for this work by L. W. TRUESDELL, Tinman, Owego, N. Y.] Fig. 13. From C as a centre, describe a circle whose diameter will be equal to the width of the ...
How To Describe A Lip To A Measure
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y.] Original. Fig. 14. Let the circle AB represent the size of the measure; span the dividers from K ...
The Circle And Its Sections
1. The Areas of Circles are to each other as the squares of their diameters; any circle twice the diameter of another contains four times the area of the other.
How To Describe A Flaring Vessel Pattern, A Set Of Patterns For A Pyramid Cake, Or An Envelope For A Cone
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y.] Original. FIG. 15. From a point C as a centre, describe a circle AB equal to the large ...
How To Describe A Cone Or Frustum
Fig. 16. First draw a side elevation of the desired vessel, DE, then from A as a centre describe the arcs CDC and GEG: after finding the diam-eter of the top ...
How To Describe A Heart
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y.] Fig. 17. Draw an indefinite lino AB; then span the dividers one-fourth the width you wish the ...
Cycloid
Fig. 18. Cycloid, a curve much used in mechanics. It is thus formed: -If the circumference of a circle be rolled on a right line, beginning at any point A, and ...
How To Strike The Side Of A Flaring Vessel
Fig. 19. To find the radius of a circle for striking the side of a flaring vessel having the diameters and depth of side given. Rule. -As the difference ...
How To Find The Centre Of A Circle From A Part Of The Circumference
[Drawn for this work by L. W. Truesdell, Tinman, Owego, N. Y.] Original. Fig. 21. Span the dividers any distance you wish, and place one foot on the ...
How To Construct The Frustum Of A Cone. Form Of Flat Plate By Which To Construct Any Frustum Of A Cone
Fig. 23. Let AJBCD represent the required frustum; continue the lines AD and BC until they meet at E; then from E as centre, with the radius EC, describe the ...
Rule For Striking Out A Cone Or Frustum
Fig. 24. In a conical surface, there may be economy, sometimes, in having the slant height 6 times the radius of base. For a Circle may be wholly cut into ...
How To Find The Contents Of A Pyramid Or Cone
Rule. - Multiply the diameter of the base by itself, and this product by the height, then take one-third of this product for the contents; to obtain gallons, ...
Hipped Roofs, Mill Hoppers, Etc
To find the various Angles and proper Dimensions of Materials whereby to construct any figure ivhose form is the Frustum of a proper or inverted Pyramid, as ...
Contents In Gallons Of The Frustum Of A Cone
Figs. 27, 28, 29. To find the Contents in Gallons of a Vessel, whose diameter is larger at one end than the other, such as a Bowl, Pail, Firkin, Tub, Coffee- ...
Rule To Find The Contents In Gallons Of Any Square Vessel
Rule. - Take the dimensions in inches and decimal parts of an inch, multiply the length, breadth, and height together, and then multiply the product by .004329 ...
Contents In Gallons Of Cylindrical Vessels
Rule. - Take the dimensions, in inches and decimal parts of an inch. Square the diameter, multiply it by the length in inches, and then multiply the product by ...
How To Ascertain The Weights Of Pipes Of Various Metals, And Any Dla.METER Required
Thickness in parts of an inch. Wrought iron. Copper. Lead. 1-32 .326 11½ lbs. plate .38 2 lbs. lead 483 1-16 .653 23½ .76 . 4 967 3-32 .976 35 114 5½ 1.45 1-8 ...
Tin Plates
Size, Length, Breadth, and Weight. Brand Maek. No. of Sheets in Box. Length and Breadth. Weight per Box. Inches. Inches. Cwt. qr lbs. 1 c 225 14 by 10 1 0 0 ...
Oil Canisters, (From 2½ To 125 Galls.) With The Quantity And Quality Of Tin Required For Custom Work
Galls. Quantity and Quality. 2½ 2 Plates , 1X in body. 3½ 2 cc S DX 5½ 2 CC DX cc 8 4 IX 10 3½ cc DX 15 4 DX Galls. Quantity and Qua lity. 33 13½ Plates, IX in ...
Weight Of Water
1 cubic inch....... is equal to .03617 pounds. 12 cubic inches....... is equal to .434 pounds. 1 cubic foot....... is equal to 62.5 pounds. 1 cubicfoot.......
Decimal Equivalents To The Fractional Parts Of A Gallon, Or An Inch
[The Inch, or Gallon, being divided into 32 parts.] [In multiplying decimals it is usual to drop all but the two or three first figures.] Decimals. Gallon. or ...
A Table Containing The Diameters, Circumferences, And Areas of Circles, And The Content Of Each In Gallons At 1 Foot In Depth
Utility Of The Table. Examples 1. Required the circumference of a circle, the diameter being five inches ? In the column of circumferences opposite the given ...
Capacity Of Cans One Inch Deep
Utility Of The Table Required the contents of a vessel, diameter 6 7-10ths inches, depth 10 inches ? By the table a vessel 1 inch deep and 6 and 7-10ths inches ...
Receipts For The Use of Japanners, Varnishers, Builders And Mechanics, And For Other Useful And Important Purposes In The Practical Arts. Practical Receipts
[The following Receipts are selected from Ure's Dictionary, Cooley's Cyclopedia, Muspratt's Chemistry, and other valuable sources.]
Japanning And Varnishing
Japanning is the art of covering todies by grounds of opaque colors in varnish, which may be afterwards decorated by printing or gilding, or left in a plain ...
White Japan Grounds
To form a hard, perfect white ground is no easy matter, as the substances which are generally used to make the japan hard, have a tendency, by a number of ...
Gum Copal
Copal varnish is one of the very finest varnishes for japanning purposes. It can be dissolved by linseed oil, rendered dry by adding some quicklime at a heat ...
Varnishes Or Japans
Black Grounds Black grounds for japans may be made by mixing ivory black with shellac varnish; or for coarse work, lamp black and the top coating of common ...
Japan Finishing
The finishing part of japanning lies in laying on and polishing the outer coats of varnish, which is accessary in all painted or simply ground colored japan ...
Miscellaneous Varnishes
Different substances are employed for making varnish, the object being to produce a liquid easily applied to the surface of cloth, paper or metal, which, when ...
Resins Employed In The Manufacture Of Varnishes
The following are the chief Resins employed in the manufacture of Varnishes. Amber This resin is most distinguished for durability. It is usually of some shade ...
Spirit Varnishes
These varnishes may be readily colored-red, by dragon's blood; yellow, by gamboge. If a colored varnish is required, clearly no account need be taken of the ...
Essence Varnishes
They do not differ essentially in their manufacture from spirit varnishes. The polish produced by them is more durable, although they take a longer time to dry.
Oil Varnishes
The most durable and lustrous of varnishes are composed of a mixture of resin, oil, and spirit of turpentine. The oils most frequently employed are linseed and ...
Lacker
This is used for wood or brass work, and is also a varnish. For brass, the proportions are half a pound of pale shell-lac to one gallon of spirit of wine. It ...
Copal Varnishes
1. Copal Varnishes 1. Oil of turpentine one pint, set the bottle in a water bath, and add in small portions at a time, three ounces of powdered copal that has ...
White Hard Spirit Varnishes
1. Gum sandarach five pounds, camphor one ounce, rectified spirit (65 over proof) two gallons, washed and dried coarsely-pounded glass two pounds; proceed as ...
White Varnish
1. Tender copal seven and one-half ounces, camphor one ounce, alcohol of 95 per cent, one quart; dissolve, then add mastic two ounces, Venice turpentine one ...
Soft Brilliant Varnish
Sandarach six ounces, elemi (genuine) four ounces, anime one ounce, camphor one-half ounce, rectified spirit one quart; as before. The above spirit varnishes ...
Brown Hard Spirit Varnishes
1. Sandarach four ounces, pale seed-lac two ounces, elemi (true) one ounce, alcohol one quart; digest with agitation till dissolved, then add Venice turpentine ...
How To Prepare A Varnish For Coating Metals
Digest one part of bruised copal in two parts of absolute alcohol; but as this varnish dries too quickly it is preferable to take one part of copal, one part ...
How To Varnish Articles Of Iron And Steel
Dissolve 10 parts of clear grains of mastic, 5 parts of camphor, 15 parts of sandarach, and 5 of elemi, in a sufficient quantity of alcohol, and apply this ...
Varnish For Iron Work
Dissolve, in about two lbs. of tar oil, half a pound of asphaltum, and a like quantity of pounded resin, mix hot in an iron kettle, care being taken to prevent ...
Black Varnish For Iron Work
Asphaltum forty-eight pounds, fuse, add boiled oil ten gallons, red lead and litharge of each seven pounds, dried and powdered white copperas three pounds, ...
Bronze Varnish For Statuary
Cut best hard soap fifty parts, into fine shavings, dissolve in boiling water two parts, to which add the solution of blue vitriol fifteen parts, in pure water ...
Amber Varnishes
1. Amber one pound, pale boiled oil ten ounces, turpentine one pint. Render the amber, placed in an iron pot, semiliquid by heat; then add the oil, mix, remove ...
Black Varnish
Heat to boiling linseed oil varnish ten parts, with burnt umber two parts ,and powdered asphaltum one part, and when cooled dilute with spirits of turpentine ...
Varnishes For Furniture
The simplest, and perhaps the best, is the solution of shellac only, but many add gums sandarach, mastic, copal, arabic,benjamin, etc, from the idea that they ...
Etching Varnishes
1. White wax, two ounces; black and Burgundy pitch, of each one-half ounce; melt together, add by degrees powdered asphaltum two ounces, and boil till a drop ...
Milk Of Wax
Milk of wax is a valuable varnish, which may be prepared as follows:-Melt in a porcelain capsule a certain quantity of white wax, and add to it, while in ...
Crystal Varnishes
1. Genuine pale Canada balsam and rectified oil of turpentine, equal parts; mix, place the bottle in warm water, agitate well, set it aside, in a moderately ...
Italian Varnishes
1. Boil Scio turpentine till brittle, powder, and dissolve in oil of turpentine. 2. Canada balsam and clear white resin, of each six ounces, oil of turpentine ...
Water Varnish For Oil-Paintings
Boil bitter-apple, freed from the seeds and cut five parts, with rainwater fifty parts, down to one-half. Strain and dissolve in the liquor gum arabic eight ...
Varnish For Paper-Hangings
Sandarach, four parts, mastic, seed-lac, white turpentine, of each two parts, gum elemi one part, alcohol twenty-eight parts. Digest with frequent shaking, and ...
Book-Binders' Varnish
Shellac eight parts, gum benzoin three parts, gum mastic two parts, bruise, and digest in alcohol forty-eight parts, oil of lavender one-half part. Or, digest ...
How To Varnish Cardwork
Before varnishing cardwork, it must receive two or three coats of size, to prevent the absorption of the varnish, and any injury to the design. The size may be ...
Size, Or Varnish, For Printers, Etc
Best pale glue and white curd soap, of each 4 ounces; hot water 3 pints; dissolve, then add powdered alum 2 ounces. Used to size prints and pictures before ...
Varnish For Brick Walls
A varnish made with one pound of sulphur boiled for half an hour in an iron vessel is a perfect protection from damp to brick walls. It should be applied with ...
Mastic Varnishes
1. {Fine.) Very pale and picked gum mastic five pounds, glass pounded as small as barley, and well washed and dried two and one-half pounds, rectified ...
India-Rubber Varnishes
1. Cut up one pound of India rubber into small pieces and diffuse in half a pound of sulphuric ether, which is done by digesting in a glass flask on a sand ...
Black Varnish For Harness
Digest shellac twelve parts, white turpentine five parts, gum sandarach two parts, lampblack one part, with spirits of turpentine four parts, alcohol ninety- ...
Boiled Oil Or Linseed-Oil Varnish
Boil linseed oil sixty parts, with litharge two parts, and white vitriol one part, each finely powdered, until all water is evaporated. Then set by. Or, rub up ...
Dammar Varnish
Gum dammar ten parts, gum sandarach five parts, gum mastic one part, digest at a low heat, occasionally shaking, with spirits of turpentine twenty parts.
Common Varnish
Digest shellac one part, with alcohol seven or eight parts.
Waterproof Varnishes
Take one pound of flowers of sulphur and one gallon of linseed oil, and boil them together until they are thoroughly combined. This forms a good varnish for ...
Varnishes For Balloons, Gas Bags, Etc
1. India rubber in shavings one ounce; mineral naphtha two lbs.; digest at a gentle heat in a close vessel till dissolved, and strain. 2. Digest one pound of ...
Gold Varnish
Digest shellac sixteen parts, gum sandarach, mastic, of each three parts, crocus one part, gum gamboge two parts, all bruised, with alcohol one hundred forty- ...
Wainscot Varnish For House Painting And Japanning
Anime eight pouuds; clarified linseed oil three gallons; litharge one-fourth pound; acetate of lead one-half pound; sulphate of copper one-fourth pound. All ...
Lackers
Gold Lacker Put into a clean four gallon tin, one pound of ground turmeric, one and a half ounces of gamboge, three and a half pounds of powdered gum sandarach, ...
Armenian Or Diamond Cement
This article, so much esteemed for uniting pieces of broken glass, for repairing precious stones, and for cementing them to watch cases and other ornaments, is ...
Iron-Rust Cement
The iron-rust cement is made of from fifty to one hundred parts of iron borings, pounded and sifted, mixed with one part of sal-ammoniac, and when it is to be ...
Miscellaneous Cements
Cements For Mending Earthern And Glass Ware 1. Heat the article to be mended, a little above boiling water heat, then apply a thin coating of gum shellac, on ...
Cement For Terraces, Floors, Roofs, Reservoirs, Etc
In certain localities where a limestone impregnated with bitumen occurs, it is dried, ground, sifted, and then mixed with about its own weight of melted pitch, ...
Builders' Cements
Mastic Cement For Covering The Fronts Of Houses Fifty parts, by measure, of clean dry sand, fifty of limestone (not burned) reduced to grains like sand, or ...
Hamelein's Cement
This cement consists of earthy and other substances insoluble in water, or nearly so; and these may be either those which are in their natural state, or have ...
Plaster In Imitation Of Marble - Scagliola
This species of work is exquisitely beautiful when done with taste and judgment, and is so like marble to the touch, as well as appearance, that it is scarcely ...
Stucco and Mastic
Maltiia, Or Greek Mastic This is made by mixing lime and sand in the manner of mortar, and making it into a proper consistency with milk or size, instead of ...
Composition
This is frequently used, instead of plaster of Paris, for the ornamental parts of buildings, as it is more durable, and becomes in time as hard as stone itself.
Foundations Of Buildings
The nature and condition of the soil upon which houses are to be built should receive far more attention than is usually bestowed upon such subjects. A soil ...
Concrete Floors
The lower floors of all the cellars of houses should be composed of a bed of concrete about three inches thick. This would tend to render them dry, and more ...
Fire-Proof Composition To Resist Fire For Five Hours
Dissolve, in cold water, as much pearlash as it is capable of holding in solution, and wash or daub with it all the boards, wainscoting, timber, etc. Then ...
Miscellaneous Receipts. How To Polish Wainscot And Mahogany
A very good polish for wainscot may be made in the following manner: Take as much beeswax as required, and, placing it in a glazed earthen pan, add as much ...
Imitation Of Mahogany
Plane the surface smooth, and rub with a solution of nitrous acid. Then apply with a soft brush one ounce of dragon's blood, dissolved in about a pint of ...
Furniture Varnish
White wax six ounces, oil of turpentine one pint; dissolve by a gentle heat. Used to polish wood by friction.
How To Make Glass Paper
Take any quantity of broken glass (that with a greenish hue is the best), and pound it in an iron mortar. Then take severel sheets of paper, and cover them ...
How To Make Stone Paper
As, in cleaning wood-work, particularly deal and other soft woods, one process is sometimes found to answer better than another, we may describe the manner of ...
Whitewash
The best method of making a whitewash for outside exposure is to slack half a bushel of lime in a barrel, add one pound of common salt, half a pound of the ...
Paint For Coating Wire Work
Boil good linseed oil with as much litharge as will make it of the consistency to be laid on with the brush; add lampblack at the rate of one part to every ten, ...
How To Bleach Sponge
Soak it well in dilute muriatic acid for twelve hours. Wash well with water, to remove the lime, then immerse it in a solution of hyposulphite of soda, to ...
Lac Varnish For Vines
Grape vines may be pruned at any period without danger from loss of bleeding, by simply covering the cut parts with varnish made by dissolving stick-lac in ...
Razor Paste
1. Levigated oxide of tin (prepared putty powder) 1 oz.; powdered oxalic acid 1-4 oz.; powdered gum 20 grs.; make it into a stiff paste with water, and evenly ...
Leather Varnish
Durable leather varnish is composed of boiled linseed oil, in which a drier, such as litharge, has been boiled. It is colored with lampblack. This varnish is ...
How To Keep Tires Tight On Wheels
Before putting on the tires fill the felloes with linseed oil, which is done by beating the oil in a trough to a boiling heat, and keeping the wheel, with a ...
Cutting Glass
To cut bottles, shades, or other glass vessels neatly, heat a ro l of iron to redness, and having filled your vessel the exact height you wish it to be cut, ...
Prepared Liquid Glue
Take of best white glue 16 ounces; white lead, dry, 4 ounces; rain water 2 pints; alcohol 4 ounces. With constant stirring dissolve the glue and lead in the ...
Liquid Glues
Dissolve 33 parts of best (Buffalo) glue on the steam bath in a porcelain vessel, in 36 parts of water. Then add gradually, stirring constantly, 3 parts of ...
Marine Glue
Dissolve 4 parts of India rubber in 34 parts of coal tar naphtha- aiding the solution with heat and agitation, add to it G4 parts of powdered shellac, which ...
An Excellent Paste For Envelopes
Mix in equal quantities gum-arabic (substitute dextrine) and water in a phial, place it near a stove, or on a furnace register, and stir or shake it well, ...
Dextrine, Or British Gum
Dry potato-starch heated from 300 to 600 until it becomes brown, soluble in cold water, and ceases to turn blue with iodine. Used by calico printers and others, ...
Gum Mucilage
A little oil of cloves poured into a bottle containing gum mucilage prevents the latter from becoming sour and putrid; this essential oil possesses great ...
Flour Paste
Too numerous to mention are the little conveniences of having a little flour paste always at hand, as those made of any of the gums impart a glaze to printed ...
Sealing-Wax For Fruit-Cans
Beeswax, ½ oz.; English Vermillion, 1 ½ ozs.; gum shellac, 2 ½ ozs.; rosin, 8 ozs. Take some cheap iron vessel that you can always keep for the purpose, and ...
Fusible Metal
1. Bismuth 8 parts; lead 5 parts; tin 3 parts; melt together, Melts below 212 degrees Fahr. 2. Bismuth 2 parts; lead 5 parts; tin 3 parts. Melts in boiling ...
Metallic Cement
M. Gireshiem states that an alloy of copper and mercury, prepared as follows, is capable of attaching itself firmly to the surfaces of metal, glass, and ...
Artificial Gold
This is a new metallic alloy which is now very extensively used in Prance as a substitute for gold. Pure copper 100 parts, zinc, or preferably tin 17 parts, ...
Or-Molu
The or-molu of the brass founder, popularly known as an imitation of red gold, is extensively used by the French workmen in metals. It is generally found in ...
Blanched Copper
Fuse 8 ounces of copper and ½ ounce of neutral arsenical salt, with a flux made of calcined borax, charcoal dust and powdered glass.
Browning Gun Barrels
The tincture of iodine diluted with one-half its bulk of water, is a superior liquid for browning gun barrels.
Silvering Powder For Coating Copper
Nitrate of silver 30 grains, common salt 30 grains, cream of tar-ar 3½ drachms; mix, moisten with water, and apply.
Alloy For Journal Boxes
The best alloy for journal boxes is composed of copper, 24 lbs.; tin, 24 lbs.; and antimony, 8 lbs. Melt the copper first, then add the tin, and lastly the ...
Alloy For Bells Of Clocks
The bells of the pendules, or ornamental clocks, made in Paris, are composed of copper 72.00, tin 2G.56, iron 1.44, in 100 parts.
An Alloy For Tools
An alloy of 1000 parts of copper and 14 of tin is said to furnish tools, which hardened and sharpened in the manner of the ancients, afford an edge nearly ...
Alloy For Cymbals And Gongs
An alloy for cymbals and gongs is made of 100 parts of copper 'with about 25 of tin. To give this compound the sonorous property in the highest degree, the ...
Solder For Steel Joints
Silver 10 pennyweights, copper 1 pennyweight, brass 2 pennyweights. Melt under a coat of charcoal dust.
Soft Gold Solder
Is composed of four parts gold, one of silver, and one of copper. It can be made softer by adding brass, but the solder becomes more liable to oxidize.
Files
Allow dull files to lay in diluted sulphuric acid until they are bit deep enough.
How To Prevent Rusting
Boiled linseed oil will keep polished tools from rusting if it is allowed to dry on them. Common sperm oil will prevent them from rusting for a short period. A ...
Anti-Attrition, And A Xle-Grease
One part of fine black lead, ground perfectly smooth, with four parts of lard.
How To Galvanize
Start with a solution of nitro-muriate of gold (gold dissolved in a mixture of aquafortis and muriatic acid) and add to a gill of it a pint of ether or alcohol, ...
Rare And Valuable Compositions
Receipts for the use of Mechanists, Iron and Brass Founders, Tinmen, Coppersmiths, Turners, Dentists, Finishers of Brass, Britannia, and German Silver, and for ...
Gold, Silver & Copper Solders, & Dipping Acids. 93
No. 47. Yellow Solder, for Brass or Copper. - (Stronger than the last.) - Copper, 32 lbs.; Zinc, 29 lbs.; Tin, 1 lb. No. 48. Solder,/or Copper. - Copper, 10 ...
Gold, Silver & Copper Solders, & Dipping Acids. 93. Continued
No. 79. Brown Bronze Dip. - Iron Scales, 1 lb.; Arsenic, 1 02. Muriatic Acid, 1 lb.;Zinc, (solid,) 1 oz. Let the Zinc be kept in only while it is in use. No.
Table Of Alloys
Alloys having a density greater than the Mean of their Constituents. Gold and zinc. Silver & antimony. Gold and tin. Copper and zinc. Gold and bismuth. Copper ...
Alloys For Bronze
Professor Hoffman, of the Prussian artillery, has made experiments with the view of obtaining a good statuary bronze, and recommends the alloys ranging between ...
Valuable Alloys
The Paris Scientific Review has published, for the benefit of the industrious workers in metals, the best receipts for composing all the various factitious ...
Mechanical Drawing And Instruments Used In Drawing. Instruments Used In Drawing
To facilitate the construction of geometrical figures, we add a short description of a few useful instruments which do not belong to the common pocket-case.
The Sector
This very useful instrument consists of two equal rulers each six inches long, joined together by a brass folding joint. These rulers are generally made of ...
Use Of The Line Of Lines
This line, as was before observed, is marked L, and its uses are, To Divide a line into any number of equal parts: Take the length of the line by the compasses, ...
Use Of The Line Of Chords
By means of the sector, we may dispense with the protractor. Thus, to lay down an angle of any number of degrees: - take the radius of the circle on the ...
Mechanical Drawing And Perspective
A FLAT rectangular board is first to be provided, of any convenient size, as from 18 to 30 inches, and from 16 to 24 inches broad. It may be made of fir, plane ...
Practical Geometry
Geometry is the science which investigates and demonstrates the properties of lines on surfaces and solids: hence, Practical Geometry is the method of applying ...
Definition Of Arithmetical Signs Used In The Work
= When we wish to state that one quantity or number, is equal to another quantity or number, the sign of equality = is employed. Thus 3 added to 2 = 5, or 3 ...
Definition Of Arithmetical Signs Used In The Work. Part 2
Problem VI. To divide a given Line into any Number of Parts, which Parts shall de in the same Proportion to each other as the Parts of some other given Line, ...
Definition Of Arithmetical Signs Used In The Work. Part 3
Problem X. To describe an Ellipse by Means of a Carpenter's Square, or a piece of notched Lath. Having drawn two lines to represent the diameters of the ...
Epitome Of Mensuration And Instrumental Arithmetic. Epitome Of Mensuration Of The Circle, Cylinder, Sphere, &C
1. The circle contains a greater area than any other plane figure bounded by an equal perimeter or outline. 2. The areas of circles are to each other as the ...
Of The Square, Rectangle, Cube, &C
1. The side of a square equals the square root of us area. 2. The area of a square equals the square of one of its sides. 3. The diagonal of a square equals ...
Surfaces And Solidities Of The Regular Bodies, Each Of Whose Boundary Lines Is 1
No. of sides. Names. Surfaces. Solids. 4 Tetrahedron 1.7321 0.1179 6 Hexahedron 6. 1. 8 Octahedron 3.4641 0.4714 12 Dodecahedron 20.6458 7.6631 20 Icosahedron ...
Of Triangles, Polygons, &C
1. The complement of an angle is its defect from a right angle. 2. The supplement of an angle is its defect from two right angles. 3. The sine, tangent, and ...
Table Of The Areas Of Regular Polygons Each Of Whose Sides Is Unity
Name of Polygon. No of Sides Apothem or Perpend'lar. Area when Side is Unity Interior Angle. Central Angle. Triangle 3 0.2887 0.4330 60 0' 120 0' Square 4 0.5 ...
Of Ellipses, Cones, Frustums, &C
1. The square root of half the sum of the squares of the two diameters of an ellipse multiplied by 3.1416 equals its circumference. 2. The product of the two ...
Instrumental Arithmetic Or Utility Of The Slide Rule
The slide rule is an instrument by which the greater portion of operations in arithmetic and mensuration may be advantageously performed, provided the lines of ...
Numeration
Numeration teaches us to estimate or properly value the numbers and divisions on the rule in an arithmetical form. Their values are all entirely governed by ...
Proportion, Or Rule Of Three Direct
Rule. - Set the first term on B to the second on a; and against the third upon B is the fourth upon a. 1. If 4 yards of cloth cost 38 cents, what will 30 yards ...
Rule Of Three Inverse
Rule. - Invert the slide, and the operation is the same as direct proportion. 1. I know that six men are capable of performing a certain given portion of work ...
Square And Cube Roots Of Numbers
On the engineer's rule, when the lines c and d are equal at both ends, c is a table of squares, and D a table of roots, as Squares 1 4 9 16 25 36 49 64 81 on c.
Mensuration Of Surface
1. Squares, Rectangles, etc. Rule. - When The Length Is Given In Feet And The Breadth In Inches, Set The Breadth On B To 12 On A; And Against The Length On A ...
Table Of Gauge-Points For The Engineer's Rule
Names. F, F, F. F, I, I. I,I,I. F,I. I,I. F. I. Cubic inches 578 83 1728 106 1273 105 121 Cubic feet 1 144 1 1833 22 121 33 Imp. Gallons 163 231 277 294 353 ...
For The Common Slide Rule
Names. F, F, F. F, I, I. I,I,I. F,I. i,i. F. I Cubic inches 36 518 624 660 799 625 113 Cubic feet 625 9 108 114 138 119 206 Water in lbs. 10 144 174 184 22 !
Mensuration Of Solidity And Capacity
General Rule. - Set the length upon B to the gauge point upon a; and against the side of the square, or diameter on D, are the cubic contents, or weight in lbs.
Power Of Steam Engines
Condensing Engines. - Rule. Set 3.5 on c to 10 on D; then D is a line of diameters for cylinders, and c the corresponding number of horses' power; thus, H. Pr.
Of Engine Boilers
How many superficial feet arc contained in a boiler 23 feet in length and 5 ½ feet in width ? Set 1 on b to 23 on A; and against 5.5 upon b is 126.5 square ...
Artificers' Rules And Tables
For Computing the Work of Bricklayers, Well Diggers, Masons, Carpenters and Joiners, Slaters, Plasterers, Painters, Glaziers, Pavers, and Plumbers.
Measurement Of Bricklayers' Work
Brickwork is estimated at the rate of a number of bricks in thickness, estimating a brick at 4 inches thick. The dimensions of a building are usually taken by ...
Measurement Of Wells And Cisterns
There are two methods of estimating the value of excavating. It may be done by allowing so much a day for every man's work, or so much per cubic foot, or yard, ...
Measurement Of Masons' Work
To masonry belong all sorts of stone-work; and the measure made use of is a foot, either superficial or solid. Walls, columns, blocks of stone or marble, etc., ...
Measurement Of Carpenters' And Joiners' Work
To this branch belongs all the wood work of a house, such as flooring, partitioning, roofing, etc. Large and plain arlicles are usually measured by the square ...
Sash Table.- Size And Prices Of Sashes, Shutters, Etc. Cincinnati, Ohio
6ize of Lights. Thick- Size of Sash Price of Sash per Light. Trice of Venition Shutters per pair. Price of Window Frames. for 12 light Windows. Width. Length.
Measurement Of Slaters' Work
In these articles, the content of a roof is found by multiplying the length of the ridge by the girth over from eaves, to eaves; making allowance in this girth ...
Measurement Of Plasterers' Work
Plasterers' work is of two kinds, namely, ceiling-which is plastering upon laths - and rendering, which is plastering upon walls, which are measured separately.
Measurement Of Pavers' Work
Pavers' work is done by the square yard. And the content is found by multiplying the length by the breadth. Grading for paving is charged by the day.
Measurement Of Painters' Work
Painters' work is computed in square yards. Every part is measured where the color lies; the measuring line is forced into all the mouldings and corners. 138 ...
Measurement Of Glaziers' Work
Glaziers' work is sometimes measured by the sq. ft., sometimes by the piece, oral so much per light; except where the glass is set in metallic frames, when the ...
Table Showing The Size And Number Of Lights To The 100 Square Feet
Size. Lights. 6 by 8 300 7 by 9 229 8 by 10 180 8 by 11 164 8 by 12 150 9 by 10 160 9 by 11 146 9 by 12 133 9 by 13 123 9 by 14 114 9 by 16 100 10 by 10 144 10 ...
Patent Improved Lead Pipe, Sizes And Weight Per Foot
Calibre. Weight per foot. Inches. lbs. ozs. 3/8 6 8 10 12 1 0 1 8 ½ 8 10 12 14 1 0 Calibre Weight per foot. Inches. lbs. ozs. ½ 1 4 1 8 2 0 3 0 5/8 13 1 0 1 8 ...
Boston Lead Pipe, Sizes And Weight Per Foot
1-2 Inch. 5-8 Inch. 3-4 Inch. 1 Inch. 11-4 Inch. 11-2 Inch. 13-4 Inch. 2 Inch. lbs. oz. lbs. oz. lbs. oz. lbs. oz. lbs. oz. lbs. oz. Zos. oz. lbs. oz. 10 2 12 ...
Comparative Strength And Weight Of Ropes And Chains
Circum. of Rope in inches. Weight per Fathom in lbs. Diameter of Chain in inches. Weight per Fathom in lbs. Proof Strength in tons and cwt. 3½ 2¾ 5/16 5½ 1 5½ ...
Strength Of Materials Of Construction
[From Templeton's Workshop Companion.] Materials of construction are liable to four different kinds of strain; viz., stretching, crushing, transverse action, ...
Resistance To Lateral Pressure, Or Transverse Action
The strength of a square or rectangular beam to resist lateral pressure, octing in a perpendicular direction to its Length, is as the breadth and square of the ...
Resistance To Lateral Pressure, Or Transverse Action. Continued
And these applications of principle not only tend to diminish deflection, but the required purpose is also more effectively attained, and that by lighter ...
Resistance Of Bodies To Flexure By Vertical Pressure
When a piece of timber is employed as a column or support, its tendency to yielding by compression is different according to the proportion between its length ...
Elasticity Of Torsion, Or Resistance Of Bodies To Twisting
The angle of flexure by torsion is as the length and extensibility of the body directly and inversely as the diameter; hence, the length of a bar or shaft ...
Strength Of Materials
[From Grier's Mechanic's Calculator, etc.] Bar of Iron.-The average breaking weight of a Bar of Wrought Iron, 1 inch square, is 25 tons; its elasticity is ...
Models Proportioned To Machines
The relation of models to machines, as to strength, deserves the particu lar attention of the mechanic. A model may be perfectly proportioned in all its parts ...
Strength Of Materials. #2
[From Adcock's Engineer.'] List of metals, arranged according to their strength.-Steel, wrought-iron, cast-iron, platinum, silver, copper, brass, gold, tin, ...
Relative Strength Of Cast And Malleable Iron
It has been found, in the course of the experiment!) made by Mr. Hodg-kinson and Mr. Fairbairn, that the average strain that cast iron will bear in the way of ...
Strength Of Beams
[From Lowndes' Engineer's Hand-book,-Liverpool, I860.] Solid, Rectangular, And Round: To Find Their Strength Square and rectangular. (Depth ins.)2 X Thickness ...
Cast Iron With Feathers Or Flanges: To Find Their Strength
Sec. area, bottom flan ere ins. X depth ins. / Length in feet. - X 2 = Breaking weight, tons. If the metal exceeds 1 inch in thickness deduct l-8th. If above 2 ...
Wrought Iron Beams
Girders.-The sketch shows a very strong form for this description of firder, when rolled solid. The top ange being condensed and square is in a good form to ...
Solid Columns
Fail by crushing with length under.......................... 5 diameters Principally by crushing from........................... 5 to 15 Partly by crushing, ...
Hollow Columns
Hollow columns fail principally by crushing, provided the length does not exceed 25 diameters; indeed, the length does not appear to affect the strength much ...
Crane
The strains on the principal parts can be ascertained with great ease in the following manner - the strength being proportioned accordingly. To find the strain ...
Cold Water Pump
Weight suspended, tons X Projection, feet/Height of post above ground= Strain on top of post, tons. The post can then be calculated as a beam, twice as long as ...
Velocity Of Fans
The best Velocity of Circumference for different Densities. Velocity of Circumference. Feet per Second. Density of Blast. Oz. per inch. 170 3 180 4 195 5 205 6 ...
Friction. The Friction Of Metal On Metal, Without Unguents
From Mr. Rennie's Experiments May be taken at 1-6 of the weight up to 40 lbs. per sq. in. 1-5 100 Brass on cast iron 1-4 800 Wrought on cast iron 1-3 500 With ...
Centrifugal Force
(Revolutions per min in.)2 X dia. in ft. X weight / 5870 = Centrifugal force in terms of weight.
Pedestal - Bracket
PEDESTAL. Good proportions. Thickness of cover .4 of diameter of bearing. of sole plate .3 Diameter of bolts .25 if 2. 18 if there are 4. Distance between ...
Bracket
Solid. Metal round brass equal to 1-2 diameter of bearing. General thickness web, etc. equal to 1-4 diameter of bearing. With feathers. Width at lightest equal ...
Tempering
The article after being completed, is hardened by being heated gradually to a bright red, and then plunged into cold water; it is then tempered by being warmed ...
Case Hardening
Put the articles requiring to be hardened, after being finished but not polished, into an iron box in layers with animal carbon, that is, horns, hoofs, skins, ...
Heat. Effects Of Heat At Certain Temperatures. - Grier
Tin and Bismuth, equal parts, melt at 2S3 degrees, Fahrenheit; tin melts at 442; polished steel acquires straw color at 460; bismuth melts at 476; sulphur ...
Soldering Joints
The solder for joints requires to be of some metal more fusible than that of the substances to be joined. For Copper, usual solder 6 to 8 parts brass to 1 of ...
Boring And Turning
The best speed for boring cast iron is about 7¾ feet per minute. For drilling about 10 or 11 feet per minute is a good speed for the circumference of the tool.
Brass. Compositions Of Brass
Copper. Tin. Zinc. Watch-makers brass....................... 1 part - 2 parts German brass......................... 1 - 1 Yellow brass................... 2 __ ...
Brass Casting
As it is often useful to engineers, especially abroad, to be able to cast brass, a slight description of the process may not be out of place. The ordinary ...
Rope. To Find The Breaking Wright Of An Ordinary Tarred Hemp Rope
(Circumference, Ins.)2 ÷ 5 = Breaking Weight, Tons A rope should not be loaded with more than 1-3 its breaking weight. To find Weight of Rope or Tarred Cordage.
Weight. To Find The Weight Of Any Casting
Width in ¼ ins. X Thickness in 1/8 ins., or vice versa, ÷ 10 X Length, ft. = Weight, lbs. cast iron. For instance; to find the weight of a casting 3¼ ins. X 1 ...
Weight Of Boiler Plates
Thickness, ins. 1/16 1/8 3/16 1/4 5/16 3/8 7/16 1/2 5/8 3/4 7/8 1 Weight, lbs. per sq. ft. 2.5 5 7.5 10 12.5 15 17.5 20 25 30 35 40 For cast iron deduct l-20th.
Continuous Circular Motion
In mechanics, circular motion is transmitted by means of wheels, drums, or pulleys; and accordingly as the driving and driven are of equal or unequal diameters, ...
Continuous Circular Motion. Continued
Example.- Suppose a screw is required to he cut containing 25 threads in an inch, and the leading screw, as before, having two threads in an inch, and that a ...
Wheels And Gudgeons
To find size of Teeth necessary to transmit a given Horse Power. (Tredgold.) Horse power X 240/ Diameter of wheel . ft. X Revs. per min. = Strength of tooth.
Water
To find the quantity of Water that will be discharged through an orifice, or pipe, in the side or bottom of a Vessel. Area of orifice, sq. in. X No.
Mechanical Tables For The Use Of Operative Smiths, Millwrights And Engineers
The following Tables, originally dedicated to 'the National Association of the Forgers of Iron Work,' England, by James Fo-deN, will be found extremely useful ...
Mechanical Tables For The Use Of Operative Smiths, Millwrights And Engineers. Continued
The manner in which the foregoing Table of Circumferences is found is as follows: Taking the diameter at unity, we have by decimal proportion in. in. As l : 8.
Circumferences For Angled Iron Hoops Angle Outside
Diam. Circ. Ft In. Ft. In. 6 1 5½ 4 1 6¼ ½ 1 7 ¾ 1 7¾ 7 1 8½ ¼ 1 9¼ ½ 1 9 7/8 ¾ 1 l0 5/8 8 1 11 3/8 ¼ 2 0 1/8 ½ 2 0 7/8 ½ 2 1 5/8 9 2 2 3/8 ¼ 2 3 ½ 2 3¾ ¾ 2 4½ ...
Number Of Iron Spikes Per 100 Pounds
Manufactured by Philip C. Page, Mass., and Sold by Page, Briggs & Babbitt, Boston. Ship Spikes or Hatch Nails 1-4 in. sq're. Ship Spikes or Hatch Nails 5-16 in.
Burdens Patent Spikes And Horse Shoes
Manufactured at the Troy Iron and Nail Factory, Troy, New York. Boat Spikes. Size in inches. No. in 100 lbs. 3 1750 3½ 1468 4 1257 44 920 5 720 54 630 6 497 64 ...
Coppers. - Dimensions And Weight From 1 To 208 Gallons
Inches lag to brim. Gallons. Weight in pounds. 9¾ 1 1½ 12¼ 2 3 14 3 4½ 15½ 4 6 16½ 5 7½ 17½ 6 9 18½ 7 10½ 19½ 8 12 20¼ 9 13½ 21 10 15 21½ 11 16½ 22 12 18 22½ ...
Copper Tubing. - Weight Of The Usual Thickness
When The Inside Diameter, Is ¼ Of An Inch, 3 Ozs.; 3/8 Do., 5 Ozs.; ½ Do , 6ozs.; 5/8 Do., 8 Ozs.; ¾ Do., 10 Ozs. Per Foot Brass, Copper, Steel And Lead. - ...
Cast Iron
Weight of a Foot in Length of Flat Cast Iron. Width of Iron. Thick, 1-4th inch. Thick, 3-8ths inch Thick, 1-2 inch. Thick, 5-8ths inch. Thick, 3-4ths inch.
Cast Iron, Copper, Brass, And Lead Balls
Weight of Cast Iron, Copper, Brass, and Lead Balls, from 1 inch to 12 inches in Diameter. Diam. Cast Iron. Copper. Brass. Lead. Diam. Cast Iron. Copper. Brass.
Cast Iron. - Weight Of A Foot In Length Of Square And Round
SQUARE. Size. Weight... Size. Weight. Inches Square Pounds. Inches Square. Pounds. ½ .78 4 7/8 74.26 5/8 1.22 5 78.12 ¾ 1.75 5 1/8 82.08 7/8 2.39 5¼ 86.13 1 3.
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Size. Thick, 1-4 inch- Thick, 3-8ths. Thick, 1-2 inch. Thick, 5-8ths. Inches pounds. pounds. pounds. pounds. 1 .852 1.27 1.70 2.13 1 1/8 .958 1.43 1.91 2.39 1¼ ...
Weights Of Rolled Iron
Per lineal foot, in pounds and decimal parts, of sections of Parallel Angle Taper Angle, Parallel J, Taper J, anil Sash Iron and Rails. Table I.- Parallel ...
Table III - Taper Angle Iron, Of Equal Sides
L'gth of sides AA, in inches. Thickness of edges at b. Thickness of root at c. Weight of 1 lineal foot. in. in. in. 4 ½ 5/8 14.0 8 ½ 5/8 10.375 2¾ 7-16ths 9- ...
Weights Of Parallel And Taper T Iron. Table IV.-Parallel J Iron, Of Unequal Width And Depth
Width of top table a. Total depth B. Uniform thickness top table c Uniform thickness of rib D. Weight of one lineal foot. in. in. in. in. 5 6 ½ ½ 15.75 4½ 3¼ ½ ...
Weight Of Sashes And Rails
Table VII. - Sash Iron. Total depth A. Depth of rebate B. Width at edge c. greatest width D. Weight of one lineal foot. in. in. in. 2 1 No. 9 w. guage 5-8ths 1.
Weight Of A Lineal Foot Of Malleable Rectangular Or Flat Iron
From an Eighth of an Inch to Three Inches Thick. T designates the thickness, B. the breadth. T. B. Weight. in. in lbs. ozs. 1/8 ¼ 0 1.6 3/8 0 2.4 ½ 0 3.3 5/8 0 ...
Weight Of Flat Iron
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Elastic Force Of Steam
Table of the Elastic Force of Steam, and corresponding Temperature of the Water with which it is in Contact. Pressure in pounds per sq. in. Elastic force in ...
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When water in a vessel is subjected to the action of fire, it readily imbibes the heat or fluid principle of which the fire is the immediate cause, and sooner ...
Consumption Of Coal
Table For Finding The Consumption Of Coal Per Hourin Steamers Either Paddle Or Screw (The Same Screw Being Used Throughout,) At Any Kate Of Speed, The ...
Evaporative Power Of Coal And Results Of Coking
Under the authority of an Act of the American Congress, approved Sept. 11. 1841, an extensive series of experiments was conducted by Prof. Johnson upon the ...
Power Of Steam
Mr. Tredgold gives the following Table, which will show how the power of the steam as it issues from the boiler, is distributed. IV. A Non-Condensing Engine ...
Rules And Tables For Gauging, Ullaging, Etc. Gauging Of Casks
In taking the dimensions of a Cask it must be carefully observed: 1st, That the bung-hole be in the middle of the cask; 2d, That the bung-stave, and the stave ...
Gauging Of Casks In Imperial (British) Gallons. And Also In United States Gallons
Having ascertained the variety of the Cask, and its interior dimensions, the following Table will facilitate the calculation of its capacity. Table of the ...
How To Ullage, Or Find The Contexts In Gallons Of A Cask Partly Filled
To find the contents of the occupied part of a lying cask in gallons. Rule. - Divide the depth of the liquid, or wet inches, by the bung diameter, and if the ...
Ploughing
Table showing the distance Travelled by a horse in Ploughing an Acre of Land; also, the quantity of Land worked in a Day, at the rate of 16 and 18 miles per ...
Planting
Table showing the number of Plants required for one Acre of Land, from one Foot to Twenty-one Feet distance from Plant to Plant. Feet Distance No. of Hills. 1 ...
Weight Of A Cord Of Wood
Table of the Weight of a Cord of different kinds of Dry Wood, and the comparative value per Cord. A Cord of Hickory, 4469 pounds, - - Carbon - - 100 Maple, - - ...
Charcoal
Oak, Maple, Beech, and Chestnut make the best quality. Be-tween 15 and 17 per cent, of coal can be obtained when the wood is properly burned. A bushel of coal ...
Addition To Tinman's Manual. Tinman's Twelve Pound Bill, Or Bill Of Day's Work Of Plain Tin Ware
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1. - Weights Of Iron Wire Per 20 Feet
Manufactured by IcHabod Washburn & Moen, Worcester, Mass. No. 0... 5 lbs. No. 6.. .1 lb. 14 ozs. No. 12... 9 ozs. No. 1... 4 lbs. 2 ozs. No. 7.. 1 lb. 10 ozs.
2. - Weight Of Iron Wire Per Lineal Rod
Nos. Diameter in 1-100 of an Inch. Weight per Lineal Rod. 1 .32 4 Lbs. 2 ozs. 2 .30 3 10 3 .27 2 15 4 .25 2 8 5 .24 2 6 6 .22 1 15 7 .20 1 9 Nos. Diameter in 1- ... | 13,236 | 48,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.842533 |
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If the units' digit of integer x is greater than 1, what is the units'
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If the units' digit of integer x is greater than 1, what is the units' [#permalink]
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04 Dec 2019, 02:26
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If the units' digit of integer x is greater than 1, what is the units' digit?
(1) The units' digit of x^3 and the units' digit of x are the same.
(2) The units' digit of x^2 is 6
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If the units' digit of integer x is greater than 1, what is the units' [#permalink]
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Updated on: 04 Dec 2019, 04:12
#1
The units' digit of x^3 and the units' digit of x are the same.
x=5,6,4
insufficient
#2
The units' digit of x^2 is 6
x=4,6
from 1 &2
x=6,4
sufficient
IMO E
Bunuel wrote:
If the units' digit of integer x is greater than 1, what is the units' digit?
(1) The units' digit of x^3 and the units' digit of x are the same.
(2) The units' digit of x^2 is 6
Are You Up For the Challenge: 700 Level Questions
Originally posted by Archit3110 on 04 Dec 2019, 03:15.
Last edited by Archit3110 on 04 Dec 2019, 04:12, edited 1 time in total.
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Re: If the units' digit of integer x is greater than 1, what is the units' [#permalink]
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04 Dec 2019, 04:10
Given: Units digit > 1
Statement 1:
Units digit of x^3 is same as units digit of x
4^1 = 4 - units digit = 4
4^3 = 64 - units digit = 4
Similarly,
5^1 = 5
5^3 = 125
6^1 = 6
6^3 = 216
9^1 = 9
9^3 = 729
Possible units digits of integer x = 4,5,6 or 9
Statement 2:
The units digit of x^2 is 6
4^2 = 16
6^2 = 36
Possible units digit of integer x = 4 & 6.
Combining statement 1 & 2:
Possible units digit of integer x = 4 & 6. No clear answer. Hence answer is Option E.
Re: If the units' digit of integer x is greater than 1, what is the units' [#permalink] 04 Dec 2019, 04:10
Display posts from previous: Sort by | 900 | 2,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2020-05 | latest | en | 0.858683 |
https://www.jiskha.com/members/profile/posts.cgi?name=Mr.+K | 1,527,463,548,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870497.66/warc/CC-MAIN-20180527225404-20180528005404-00621.warc.gz | 761,582,084 | 2,973 | # Posts by Mr. K
Total # Posts: 6
1. ### Math
Mrs Welsh paid \$49.00 for one scarf and two shirts. The price of each shirt was \$10.00 Part A: use rounding to estimate how much Mrs.Welsh paid for the scarf. Round to \$50.00-\$20.00=\$30.00
2. ### LANG
Denotative definitions are also known as explicit definitions. | 95 | 318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.955248 |
https://www.traditionaloven.com/tutorials/flow-rate/convert-hcf-cubic-feet-per-hour-to-gram-1-week-water-mass.html | 1,568,609,291,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572484.20/warc/CC-MAIN-20190916035549-20190916061549-00501.warc.gz | 1,060,310,592 | 17,727 | Convert hcf/h to g/wk | hundred cubic feet per hour to grams (water mass) per week
# flow rate units conversion
## Amount: 1 hundred cubic feet per hour (hcf/h) of flow rate Equals: 475,723,022.75 grams (water mass) per week (g/wk) in flow rate
Converting hundred cubic feet per hour to grams (water mass) per week value in the flow rate units scale.
TOGGLE : from grams (water mass) per week into - 100-cubic-feet per hour in the other way around.
## flow rate from hundred cubic feet per hour to gram (water mass) per week conversion results
### Enter a new hundred cubic feet per hour 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 flow rate measuring units - complete list.
How many grams (water mass) per week are in 1 hundred cubic feet per hour? The answer is: 1 hcf/h equals 475,723,022.75 g/wk
## 475,723,022.75 g/wk is converted to 1 of what?
The grams (water mass) per week unit number 475,723,022.75 g/wk converts to 1 hcf/h, one hundred cubic feet per hour. It is the EQUAL flow rate value of 1 hundred cubic feet per hour but in the grams (water mass) per week flow rate unit alternative.
hcf/h/g/wk flow rate conversion result From Symbol Equals Result Symbol 1 hcf/h = 475,723,022.75 g/wk
## Conversion chart - - 100-cubic-feet per hour to grams (water mass) per week
1 hundred cubic feet per hour to grams (water mass) per week = 475,723,022.75 g/wk
2 - 100-cubic-feet per hour to grams (water mass) per week = 951,446,045.49 g/wk
3 - 100-cubic-feet per hour to grams (water mass) per week = 1,427,169,068.24 g/wk
4 - 100-cubic-feet per hour to grams (water mass) per week = 1,902,892,090.98 g/wk
5 - 100-cubic-feet per hour to grams (water mass) per week = 2,378,615,113.73 g/wk
6 - 100-cubic-feet per hour to grams (water mass) per week = 2,854,338,136.47 g/wk
7 - 100-cubic-feet per hour to grams (water mass) per week = 3,330,061,159.22 g/wk
8 - 100-cubic-feet per hour to grams (water mass) per week = 3,805,784,181.96 g/wk
9 - 100-cubic-feet per hour to grams (water mass) per week = 4,281,507,204.71 g/wk
10 - 100-cubic-feet per hour to grams (water mass) per week = 4,757,230,227.46 g/wk
11 - 100-cubic-feet per hour to grams (water mass) per week = 5,232,953,250.20 g/wk
12 - 100-cubic-feet per hour to grams (water mass) per week = 5,708,676,272.95 g/wk
13 - 100-cubic-feet per hour to grams (water mass) per week = 6,184,399,295.69 g/wk
14 - 100-cubic-feet per hour to grams (water mass) per week = 6,660,122,318.44 g/wk
15 - 100-cubic-feet per hour to grams (water mass) per week = 7,135,845,341.18 g/wk
Convert flow rate of hundred cubic feet per hour (hcf/h) and grams (water mass) per week (g/wk) units in reverse from grams (water mass) per week into - 100-cubic-feet per hour.
## Flow rate. Gas & Liquids.
This unit-to-unit calculator is based on conversion for one pair of two flow rate units. For a whole set of multiple units for volume and mass flow on one page, try the Multi-Unit converter tool which has built in all flowing rate unit-variations. Page with flow rate by mass unit pairs exchange.
# Converter type: flow rate units
First unit: hundred cubic feet per hour (hcf/h) is used for measuring flow rate.
Second: gram (water mass) per week (g/wk) is unit of flow rate.
QUESTION:
15 hcf/h = ? g/wk
15 hcf/h = 7,135,845,341.18 g/wk
Abbreviation, or prefix, for hundred cubic feet per hour is:
hcf/h
Abbreviation for gram (water mass) per week is:
g/wk
## Other applications for this flow rate calculator ...
With the above mentioned two-units calculating service it provides, this flow rate converter proved to be useful also as a teaching tool:
1. in practicing - 100-cubic-feet per hour and grams (water mass) per week ( hcf/h vs. g/wk ) measures exchange.
2. for conversion factors between unit pairs.
3. work with flow rate's values and properties.
To link to this flow rate hundred cubic feet per hour to grams (water mass) per week online converter simply cut and paste the following.
The link to this tool will appear as: flow rate from hundred cubic feet per hour (hcf/h) to grams (water mass) per week (g/wk) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,301 | 4,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-39 | longest | en | 0.696191 |
http://nrich.maths.org/public/leg.php?code=2&cl=3&cldcmpid=8665 | 1,444,434,846,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737936627.82/warc/CC-MAIN-20151001221856-00211-ip-10-137-6-227.ec2.internal.warc.gz | 223,999,346 | 7,641 | # Search by Topic
#### Resources tagged with Counting similar to Card Trick 1:
Filter by: Content type:
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### There are 25 results
Broad Topics > Numbers and the Number System > Counting
### An Investigation Based on Score
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### Pairs
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Delight your friends with this cunning trick! Can you explain how it works?
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##### Stage: 2, 3 and 4 Challenge Level:
Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells.
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A new card game for two players. | 1,290 | 5,003 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2015-40 | longest | en | 0.881776 |
https://www.convertunits.com/from/millibar/to/millimeter+of+water | 1,606,711,683,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141205147.57/warc/CC-MAIN-20201130035203-20201130065203-00619.warc.gz | 629,978,863 | 8,509 | ## ››Convert millibar to millimeter of water [4 °C]
millibar millimeter of water
How many millibar in 1 millimeter of water? The answer is 0.0980665.
We assume you are converting between millibar and millimeter of water [4 °C].
You can view more details on each measurement unit:
millibar or millimeter of water
The SI derived unit for pressure is the pascal.
1 pascal is equal to 0.01 millibar, or 0.10197162129779 millimeter of water.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between millibars and millimeters of water.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of millibar to millimeter of water
1 millibar to millimeter of water = 10.19716 millimeter of water
2 millibar to millimeter of water = 20.39432 millimeter of water
3 millibar to millimeter of water = 30.59149 millimeter of water
4 millibar to millimeter of water = 40.78865 millimeter of water
5 millibar to millimeter of water = 50.98581 millimeter of water
6 millibar to millimeter of water = 61.18297 millimeter of water
7 millibar to millimeter of water = 71.38013 millimeter of water
8 millibar to millimeter of water = 81.5773 millimeter of water
9 millibar to millimeter of water = 91.77446 millimeter of water
10 millibar to millimeter of water = 101.97162 millimeter of water
## ››Want other units?
You can do the reverse unit conversion from millimeter of water to millibar, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Millibar
A millibar (mb) is 1/1000th of a bar, a unit for measurement of pressure. It is not an SI unit of measure, however it is one of the units used in meteorology when describing atmospheric pressure. The SI unit is the pascal (Pa), with 1 millibar = 100 pascals (a hectopascal)
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 609 | 2,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-50 | latest | en | 0.803819 |
https://web2.0calc.com/questions/method-of-undetermined-coefficients-challenge-for-you-d | 1,550,451,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247483873.51/warc/CC-MAIN-20190217233327-20190218015327-00057.warc.gz | 761,552,719 | 6,753 | +0
# Method of Undetermined Coefficients(challenge for you :D)
+1
743
3
+7220
Solve the second-order ordinary differential equation:
$$y''(t) + 4y'(t) + 4y(t) = 4e^t$$
Nov 12, 2017
#2
+7220
+2
My solution in LaTeX:(I guess it is the same...)
$$y'' + 4y' + 4y = 4e^t\\ \text{Let }y_h \text{ be the homogeneous case.}\\ y_h'' + 4y_h' + 4y_h = 0\\ \lambda^2 + 4\lambda + 4 = 0\\ \lambda = -2\\ \therefore y_h = C_1 e^{-2t} + C_2 te^{-2t}\\ \text{Let } y_p = Ae^t.\\ y_p=y'_p =y''_p= Ae^t\\ Ae^t + 4Ae^t + 4Ae^t = 4e^t\\ 9A = 4\\ A = \dfrac{4}{9}\\ \therefore y_p = \dfrac{4e^t}{9}\\ \therefore y = y_h + y_p = C_1 e^{-2t} + C_2 te^{-2t} + \dfrac{4e^t}{9}$$
:)
Nov 12, 2017
#1
+1
Max: Compare your solution to this by "Mathematica 11 Home Edition" !!
Solve 4 ( dy(t))/( dt) + ( d^2 y(t))/( dt^2) + 4 y(t) = 4 e^t:
The general solution will be the sum of the complementary solution and particular solution.
Find the complementary solution by solving ( d^2 y(t))/( dt^2) + 4 ( dy(t))/( dt) + 4 y(t) = 0:
Assume a solution will be proportional to e^(λ t) for some constant λ.
Substitute y(t) = e^(λ t) into the differential equation:
( d^2 )/( dt^2)(e^(λ t)) + 4 ( d)/( dt)(e^(λ t)) + 4 e^(λ t) = 0
Substitute ( d^2 )/( dt^2)(e^(λ t)) = λ^2 e^(λ t) and ( d)/( dt)(e^(λ t)) = λ e^(λ t):
λ^2 e^(λ t) + 4 λ e^(λ t) + 4 e^(λ t) = 0
Factor out e^(λ t):
(λ^2 + 4 λ + 4) e^(λ t) = 0
Since e^(λ t) !=0 for any finite λ, the zeros must come from the polynomial:
λ^2 + 4 λ + 4 = 0
Factor:
(2 + λ)^2 = 0
Solve for λ:
λ = -2 or λ = -2
The multiplicity of the root λ = -2 is 2 which gives y_1(t) = c_1 e^(-2 t), y_2(t) = c_2 e^(-2 t) t as solutions, where c_1 and c_2 are arbitrary constants.
The general solution is the sum of the above solutions:
y(t) = y_1(t) + y_2(t) = c_1/e^(2 t) + (c_2 t)/e^(2 t)
Determine the particular solution to ( d^2 y(t))/( dt^2) + 4 y(t) + 4 ( dy(t))/( dt) = 4 e^t by the method of undetermined coefficients:
The particular solution to ( d^2 y(t))/( dt^2) + 4 y(t) + 4 ( dy(t))/( dt) = 4 e^t is of the form:
y_p(t) = a_1 e^t
Solve for the unknown constant a_1:
Compute ( dy_p(t))/( dt):
( dy_p(t))/( dt) = ( d)/( dt)(a_1 e^t)
= e^t a_1
Compute ( d^2 y_p(t))/( dt^2):
( d^2 y_p(t))/( dt^2) = ( d^2 )/( dt^2)(a_1 e^t)
= e^t a_1
Substitute the particular solution y_p(t) into the differential equation:
( d^2 y_p(t))/( dt^2) + 4 ( dy_p(t))/( dt) + 4 y_p(t) = 4 e^t
a_1 e^t + 4 (a_1 e^t) + 4 (a_1 e^t) = 4 e^t
Simplify:
9 a_1 e^t = 4 e^t
Equate the coefficients of e^t on both sides of the equation:
9 a_1 = 4
Solve the equation:
a_1 = 4/9
Substitute a_1 into y_p(t) = a_1 e^t:
y_p(t) = (4 e^t)/9
The general solution is:
y(t) = y_c(t) + y_p(t) = (4 e^t)/9 + c_1/e^(2 t) + (c_2 t)/e^(2 t)
Nov 12, 2017
#2
+7220
+2
My solution in LaTeX:(I guess it is the same...)
$$y'' + 4y' + 4y = 4e^t\\ \text{Let }y_h \text{ be the homogeneous case.}\\ y_h'' + 4y_h' + 4y_h = 0\\ \lambda^2 + 4\lambda + 4 = 0\\ \lambda = -2\\ \therefore y_h = C_1 e^{-2t} + C_2 te^{-2t}\\ \text{Let } y_p = Ae^t.\\ y_p=y'_p =y''_p= Ae^t\\ Ae^t + 4Ae^t + 4Ae^t = 4e^t\\ 9A = 4\\ A = \dfrac{4}{9}\\ \therefore y_p = \dfrac{4e^t}{9}\\ \therefore y = y_h + y_p = C_1 e^{-2t} + C_2 te^{-2t} + \dfrac{4e^t}{9}$$
:)
MaxWong Nov 12, 2017
#3
+537
0
Haha too much nonsence
ProMagma Nov 12, 2017 | 1,499 | 3,300 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-09 | latest | en | 0.673669 |
https://oneclass.com/study-guides/ca/u-of-waterloo/acct-fin-mgmt/afm-481/172242-afm481-final-exam-material-chapter-9-summary-thoroughdocx.en.html | 1,524,626,829,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947690.43/warc/CC-MAIN-20180425022414-20180425042414-00149.warc.gz | 664,312,636 | 43,444 | Study Guides (248,490)
AFM 481 (7)
Final
# AFM481 Final Exam Material - Chapter 9 Summary (Thorough).docx
5 Pages
109 Views
School
Department
Accounting & Financial Management
Course
AFM 481
Professor
Grant Russell
Semester
Fall
Description
AFM481 - Advanced Cost Accounting Professor Grant Russell Final Exam Material Chapter 9: Joint Product and By-product Costing Joint Products: In the process of making one product, one or more other products are created. A main product has high sales value compared to the other joint products. A by-product has low sales value compared to the other joint products. Joint costs: costs incurred to jointly produce a group of goods (e.g. labour, insurance, property taxes) Split-off point: point at which individual products are identified Separable costs: costs incurred after the split-off point E.g. The join costs of processing maple sap were \$2,000,000. Product Cases Sales Value at Separable Costs Selling Price (Total 120,000) Split-off Point Syrup 90,000 \$24 per case \$12 per case \$38 per case Sugar 20,000 \$26 per case \$16 per case \$46 per case Butter 10,000 \$32 per case \$14 per case \$50 per case Methods of Allocating Joint Costs to Main Products 1. Physical Output Method Allocates join costs using the relative proportion of physical output for each main product Expressed using the same physical measure. Each main product is allocated a proportion of joint costs, based on that product’s physical output (final product) divided by the total physical output of all main products. Distortions are likely to occur when the incremental contribution (incremental revenue – incremental costs) of some products is relatively high. To Syrup = (90,000 cases / 120,000 total cases) * \$2,000,000 join costs = \$1,500,000 To Sugar = (20,000 cases / 120,000 total cases) * \$2,000,000 joint costs = \$333,333 To Butter = (10,000 cases / 120,000 total cases) * \$2,000,000 joint costs = \$166,667 2. Market-based Methods 1) Sales Value at Split-off Point Joint costs are allocated based on the relative sales value of main products at the point where joint production ends. 1. Syrup (90,000 cases * \$24) = \$2,160,000 Sugar (20,000 cases * \$26) = \$520,000 Butter (10,000 cases * \$32) = \$320,000 Total \$3,000,000 2. To Syrup (\$2,160,000/\$3,000,000) * \$2,000,000 joint costs = \$1,440,000 To Sugar (\$520,000/\$3,000,000) * \$2,000,000 joint costs = \$346,667 To Butter (\$320,000/\$3,000,000) * \$2,000,000 joint costs = \$213,333 2) Net Realizable Value (NRV) Allocates join costs using the relative value of main products NRV = Final Selling Price or Total Rev After Processing – Separable Costs 1. Syrup [90,000 cases * (\$38 selling price - \$12 separable costs)] = \$2,340,000 Sugar [20,000 cases * (\$46 selling price - \$16 separable costs)] = \$600,000 Butter [10,000 cases * (\$50 selling price - \$14 separable costs)] = \$360,000 Total \$3,300,000 2. To Syrup = (\$2,340,000/\$3,300,000) * \$2,000,000 joint costs = \$1,418,182 To Sugar = (\$600,000/\$3,300,000) * \$2,000,000 joint costs = \$363,636 To Butter = (\$360,000/\$3,300,000) * \$2,000,000 joint costs = \$218,182 3) Constant Gross Margin NRV Allocates joint costs so that the gross margin % for the main products are identical. First, the combined gross margin % for main products is calculated. o Combined Gross Margin = Sales - Joint Costs - Separable Costs o Gross Margin % = Combined Gross Margin / Combined Sales Second, joint costs are allocated to each main product to achieve a constant gross margin. o Allocated Joint Costs = Sales – Gross Margin (Desired Gross Margin % * Sales) – Separable Costs Allocates joint costs so that all joint products appear to have equal profitability. It best reflects the inseparability of the joint production process. 1. Calculate the combined gross margin %. Syrup \$38 selling price * 90,000 cases = \$3,420,000 Sugar \$46 selling price * 20,000 cases = \$920,000 Butter \$50 selling price * 10,000 cases = \$500,000 Total Combined Sales \$4,840,000 Less Combined Product Costs: Joint Costs (\$2,000,000) Separable Costs: Syrup \$12 separable costs * 90,000 cases = (\$1,080,000) Sugar \$16 separable costs * 20,000 cases = (\$320,000) Butter \$14 separable costs * 10,000 cases = (\$140,000) Total Combined Product Costs (\$3,540,000) Combined Gross Margin \$1,300,000 Combined Gross Margin % = \$1,300,000 / \$4,840,000 = 26.86% 2. Allocate the joint costs to achieve a constant gross margin of 26.86% Syrup Sugar Butter Total Sales \$3,420,000 \$920,000 \$500,000 \$4,840,000 Less Gross Margin 918,595 247,107 134,298 1,300,000 (0.2686 * Sales) Less separable 1,080,000 320,000 140,000 1,540,000 Costs Allocated Joint \$1,421,405 \$352,893 \$225,702 \$2,000,000 Costs Total gross margin is not affected by the joint cost allocation method. It only affects the relative gross margins for the individual products. A product could give the appearance that it is sold at a loss, when in fact the company profits from producing the joint product. Processing a Joint Product Beyond the Split-off Point The joint costs are irrelevant because they are sunk costs. The product with the highest incremen
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So we can recommend you notes for your school. | 1,501 | 5,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-17 | latest | en | 0.862848 |
https://cboard.cprogramming.com/c-programming/179262-need-help-errors-c-programming-post1296491.html?s=09c9677cb2514676efc460bba5f5190e | 1,627,588,008,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00687.warc.gz | 183,177,500 | 15,556 | # Thread: Need help with errors in C programming
1. ## Need help with errors in C programming
Here's what I'm trying to do, and the errors that I'm getting.
it's a decent amount of code. Would like help
1) Write a loop that will print all the whole numbers from 1 and 100. - works
2) Write a loop that will add all the whole numbers from 1 and 100 and print the total. - works
3) Write a loop that will add all the ODD whole numbers from 7 and 313 and print the total.
4) Write a loop that will add all the EVEN whole numbers from -2 and -146 and print the total.
getting 29690 as the answer (the same as the previous one)
5) Write a loop that will add every 3rd number from 2000 and -60 and print the total.
6) Write a loop that will prompt the user to enter a number from 1 to 100. The program will continue to loop until a number within that range is entered. After the loop, print out the square root of the number. Be sure to test the loop by entering numbers outside the range. - answer is correct
7) Write a loop that will prompt the user to enter test scores from 0 to 100. The program will continue to loop until a -1 is entered. Sum all test scores entered that are in the range 0 to 100 (inclusive). After the loop, calculate the average and print out the letter grade. Assume a 10-point grading scale.
- not returning the correct average or letter grade
8) If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Write the code that will calculate and print the sum of all the multiples of 3 or 5 below 1000.
- not returning the correct answer
9) Write a loop that will prompt the user to enter an uppercase letter. The code should continue to loop until an uppercase letter is entered. After an uppercase letter is entered print out the letter in both uppercase and lowercase. You can't use the built-in tolower function.
- not returning the correct answer (unsure of how to convert the uppercase to lowercase)
Hangman. Make a constant in your code with a value from 1 to 100. Prompt the user to guess the number. If the user guesses correctly end the game and display a "Congratulation. You won." message. If the guess is too high print "Too high." If the guess is too low print "Too low." Give the user at most eight guesses. If the user hasn't guessed the number after eight guesses print a "You lose." message.
- not going through all of the steps and just returning the first step regardless, also not stopping loop after 8 guesses or after guessing correctly
Code:
```// ------------------------------------------------------------------------------------------
// Includes
// ------------------------------------------------------------------------------------------
#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
// ------------------------------------------------------------------------------------------
// Constants
// ------------------------------------------------------------------------------------------
const long lngARRAY_SIZE = 100;
// ------------------------------------------------------------------------------------------
// Prototypes
// ------------------------------------------------------------------------------------------
// ------------------------------------------------------------------------------------------
// Name: main
// Abstract: This is where the program starts
// ------------------------------------------------------------------------------------------
void main()
{
int intIndex = 0;
int intTotal = 0;
int intAverage = 0;
char chrLetter = ' ';
int intSquareRoot = 0;
int intNumberChosen = 45;
int intNumberEntered = 0;
int intCount = 0;
// --------------------------------------------------------------------------------
// Problem #1 - Print all the whole numbers from 1 to 100.
// --------------------------------------------------------------------------------
printf("Problem #1 - Print all the whole numbers from 1 to 100.\n");
for (intIndex = 1; intIndex <= 100; intIndex += 1)
{
printf("%d, ", intIndex);
//move to a new line every 10 numbers
if (intIndex % 10 == 0)
{
printf("\n");
}
}
printf("\n");
printf("\n");
// --------------------------------------------------------------------------------
// Problem #2 - Print Total of the sum of all numbers 1 to 100
// --------------------------------------------------------------------------------
printf("Problem #2 - Print the sum of all the whole numbers from 1 to 100.\n");
for (intIndex = 1; intIndex <= 100; intIndex += 1)
{
intTotal += intIndex;
}
printf("The sum of all whole numbers 1 to 100 is %d, ", intTotal);
printf("\n");
printf("\n");
// --------------------------------------------------------------------------------
// Problem #3 - Print the sum of all ODD numbers 7 to 313
// --------------------------------------------------------------------------------
printf("Problem #3 - Print all the sum of all odd numbers from 7 to 313.\n");
for (intIndex = 7; intIndex <= 313; intIndex += 2)
{
if (intIndex % 2 != 0)
{
intTotal += intIndex;
}
}
printf("The sum of all whole odd numbers 7 to 313 is %i, ", intTotal);
printf("\n");
printf("\n");
// --------------------------------------------------------------------------------
// Problem #4 - Print the sum of all EVEN whole numbers -2 to -146
// --------------------------------------------------------------------------------
printf("Problem #4 - Print the sum of all even whole numbers from -2 to -146.\n");
for (intIndex = -2; intIndex <= -146; intIndex += 2)
{
if (intIndex % 2 == 0)
{
intTotal += intIndex;
}
}
printf("The sum of all whole even numbers -146 to -2 is %i, ", intTotal);
printf("\n");
printf("\n");
// --------------------------------------------------------------------------------
// Problem #5 - Print the sum of every 3rd number from -60 to 2000
// --------------------------------------------------------------------------------
printf("Problem #5 - Print the sum of every 3rd whole number from -60 to 2000.\n");
for (intIndex = -60; intIndex <= 2000; intIndex += 3)
{
intTotal += intIndex;
}
printf("The sum of sum of every 3rd whole number from -60 to 2000 is %d, ", intTotal);
printf("\n");
printf("\n");
// --------------------------------------------------------------------------------
// Problem #6 - Print the square root of a number within the range of 1 to 100
// --------------------------------------------------------------------------------
printf("Problem #6 - Enter a number from 1 to 100:\n");
scanf_s("%i", &intIndex);
//is number between 1 and 100?
if (intIndex >= 1 & intIndex <= 100)
{
intSquareRoot = intIndex * intIndex;
}
else
{
printf("The number mus be between 1 and 100:\n");
scanf_s("%i", &intIndex);
}
printf("The square root of %i is %i\n", intIndex, intSquareRoot);
printf("\n");
printf("\n");
// --------------------------------------------------------------------------------
// Problem #7 - Print Average test Score and Letter Grade
// --------------------------------------------------------------------------------
do
{
printf("Problem #7 - Enter a test score between 0 and 100 (-1 to end):\n");
scanf_s("%d", &intIndex);
intCount = intCount + 1;
intTotal = intTotal + intIndex;
intAverage = intTotal / intCount;
} while (intIndex != -1);
if (intAverage <= 100 & intAverage >= 90)
{
chrLetter = 'A';
}
if (intAverage <= 89 & intAverage >= 80)
{
chrLetter = 'B';
}
if (intAverage <= 79 & intAverage >= 60)
{
chrLetter = 'C';
}
if (intAverage <= 79 & intAverage >= 70)
{
chrLetter = 'D';
}
else
{
chrLetter = 'F';
}
printf("The average test score was %d and the letter grade is a %c", intAverage, chrLetter);
printf("\n");
printf("\n");
// --------------------------------------------------------------------------------
// Problem #8 - Print the sum of all multiples of 3 or 5 below 1000
// --------------------------------------------------------------------------------
printf("Problem #8 - Print the sum of all multiples of 3 or 5 below 1000.\n");
for (intIndex = 0; intIndex < 1000; intIndex ++)
{
if (intIndex % 3 == 0 || intIndex % 5 == 0)
{
intTotal += intIndex;
}
}
printf("The sum of all multiples of 3 or 5 below 1000 is %i, ", intTotal);
printf("\n");
printf("\n");
// --------------------------------------------------------------------------------
// Problem #9 - Print the uppercase and lowercase letter of an UpperCase letter entered.
// --------------------------------------------------------------------------------
printf("Problem #9 - Enter an uppercase letter: \n");
scanf_s("%c", &chrLetter);
if (chrLetter < 'a' || chrLetter > 'z')
{
printf("Letter must be uppercase: ");
}
else if (chrLetter < 'A' || chrLetter > 'Z')
{
chrLetter = chrLetter;
}
printf("Here is the letter in uppercase %C and lowercase %c", &chrLetter, &chrLetter);
printf("\n");
printf("\n");
//is the letter uppercase?
//no, enter another letter
//yes, print both the uppercase and lowercase letter
// --------------------------------------------------------------------------------
// Problem #10 - Have user guess number between 1 and 100
// --------------------------------------------------------------------------------
do
{
printf("Problem #10 - Enter the chosen number between 1 and 100 (you have 8 guesses)\n");
scanf_s("%d", &intNumberEntered);
if (intNumberEntered < intNumberChosen)
{
printf("Too low!\n");
}
else if (intNumberEntered > intNumberChosen)
{
printf("Too high!\n");
}
//update count
intCount + 1;
if (intCount == 8)
{
printf("You Lose!!!\n");
}
} while (intNumberEntered != intNumberChosen);
printf("Congratulations, you won!!\n");
system("pause");
}```
2. At a glance, one issue is that you are not resetting intTotal to 0.
Since all these tasks are separate, it would be best if you wrote a self-contained function for each of them.
By the way, does the int in intIndex, intTotal, etc stand for "integer"? That's pointless since the type is already int. Just name them index, total, etc.
3. A couple things to start with. You don't reset the value of intTotal between problems, so problem 3 ends up printing the result from problem 2 plus all odd numbers from 7 to 313. Either reset intTotal to zero between problems, or make each problem a function so you don't need to worry about ensuring clean starting values for each of your variables.
Also, in problem 4, your bounds need to be reordered:
for (intIndex = -2; intIndex <= -146; intIndex += 2)
intIndex is set to -2, but for the loop to continue it must be less than or equal to -146. -2 is greater than -146, so the loop never runs. This is why the sum is unchanged from problem 3 above. You should switch the start and end indices.
See if you can spot similar issues in the remaining problems.
4. Originally Posted by mwh
A couple things to start with. You don't reset the value of intTotal between problems, so problem 3 ends up printing the result from problem 2 plus all odd numbers from 7 to 313. Either reset intTotal to zero between problems, or make each problem a function so you don't need to worry about ensuring clean starting values for each of your variables.
Also, in problem 4, your bounds need to be reordered:
for (intIndex = -2; intIndex <= -146; intIndex += 2)
intIndex is set to -2, but for the loop to continue it must be less than or equal to -146. -2 is greater than -146, so the loop never runs. This is why the sum is unchanged from problem 3 above. You should switch the start and end indices.
See if you can spot similar issues in the remaining problems.
thank you! both replies were super helpful!
that helped to fix most of my issues
im still having difficulty with this one:
Code:
``` // -------------------------------------------------------------------------------- // Problem #7 - Print Average test Score and Letter Grade
// --------------------------------------------------------------------------------
intTotal = 0;
intCount = 0;
do
{
printf("Problem #7 - Enter a test score between 0 and 100 (-1 to end):\n");
scanf_s("%d", &intIndex);
intCount = intCount + 1;
intTotal = intTotal + intIndex;
} while (intIndex != -1);
intAverage = intTotal / intCount;
if (intAverage <= 100 & intAverage >= 90)
{
chrLetter = 'A';
}
if (intAverage <= 89 & intAverage >= 80)
{
chrLetter = 'B';
}
if (intAverage <= 79 & intAverage >= 60)
{
chrLetter = 'C';
}
if (intAverage <= 79 & intAverage >= 70)
{
chrLetter = 'D';
}
else
{
chrLetter = 'F';
}
printf("The average test score was %d and the letter grade is a %c", intAverage, chrLetter);
printf("\n");
printf("\n");```
as it is not giving me the correct average
and this one:
Code:
``` // -------------------------------------------------------------------------------- // Problem #9 - Print the uppercase and lowercase letter of an UpperCase letter entered.
// --------------------------------------------------------------------------------
do
{
printf("Problem #9 - Enter an uppercase letter: \n");
scanf_s("%c", &chrLetter);
if (chrLetter < 'A' & chrLetter > 'Z')
{
printf("Invalid Input");
}
else
{
chrLetter = chrLetter;
}
} while (chrLetter >= 'A' || chrLetter <= 'Z');
printf("Here is the letter in uppercase %c and lowercase %c /n", chrLetter, tolower(chrLetter) );
printf("\n");
printf("\n");```
as it is recognizing uppercase letters as such nor is it recognizing numerals as invalid inputs or lowercase letters as invalid inputs
and the last one
Code:
``` // -------------------------------------------------------------------------------- // Problem #10 - Have user guess number between 1 and 100
// --------------------------------------------------------------------------------
intCount = 0;
do
{
printf("Problem #10 - Enter the chosen number between 1 and 100 (you have 8 guesses)\n");
scanf_s("%d", &intNumberEntered);
if (intNumberEntered < intNumberChosen)
{
printf("Too low!\n");
}
else if (intNumberEntered > intNumberChosen)
{
printf("Too high!\n");
}
//update count
intCount + 1;
if (intCount == 8)
{
printf("You lose! \n");
}
} while (intNumberEntered != intNumberChosen);
if (intNumberEntered = intNumberChosen)
{
printf("Congratulations, you won!!\n");
}```
it doesn't end the loop after 8 tries; it only ends the loop once the number is guessed correctly
5. Problem 7 has a couple of issues.
What happens to intTotal and intCount when you enter -1 to exit?
& is the bitwise AND operator, which is not what you want. You should be using &&, the logical AND operator in this case.
Finally, let's say intAverage is 90. The condition for 'A' is true, then 'B', 'C', and 'D' are false. Where does the program go when the condition for 'D' is false?
6. Problem 9:
The input is invalid if it is less than 'A' OR greater than 'Z', and you want your loop to keep going until you have valid input. Look carefully at the conditions involving chrLetter. Also, instead of duplicating the logic, I would consider either setting a flag when valid input is found or using break to exit the loop.
You're also going to run into an issue when entering invalid inputs in which you are prompted twice. This is because scanf reads the \n when you hit enter as the next character. You can use getchar() instead, or prepend a space to your format string so that scanf consumes the whitespace before the character each time. Also watch for /n vs \n in your printf's.
Problem 10:
Try printing out intCount each loop or using a debugger to see what is going on. Once intCount gets to 8, what do you want to happen? Currently, there is nothing to cause the loop to terminate. Also, while it doesn't matter in this case, when waiting for a value to pass a threshold, I prefer to check if the value is greater than/equal to the threshold, as opposed to just equal to. That way, if intCount somehow got to 9, your loop would still exit as intended. | 3,608 | 15,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-31 | latest | en | 0.887669 |
https://www.getmega.com/gambling/betting-odds-learn-its-types-and-functions/ | 1,722,643,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00372.warc.gz | 650,659,401 | 133,623 | # Betting Odds-Learn Its Types, How They Work & Probability
## Table of content:
Sports betting is a genius way to earn a little extra cash as a side hustle. It allows you to make money from the intense amount of knowledge about a sporting event that you have accumulated over time and use it to your advantage.
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The concept is pretty enticing, not to mention highly rewarding. The problem is, most people don’t know the concepts that influence sports betting and thereby their earnings. Thus, today we will discuss betting odds and lines that govern your income from this incredible source.
## What is betting odds?
Betting odds are the determinant ratio that displays the capabilities of a player or team or the possible outcome of a game by reducing the governing elements to numbers and figures.
Simply put, sportsbooks and bookies compare the gameplay and performance of the players and the teams over a number of games. Based on how many times they won and how many times they lost, they determine a ratio representation of the capabilities of that team or player.
Thus, betting odds are the simplest way to compare and predict the likelihood of possible outcomes for a game or the performance of a player in a team or solo players. They are a representation of predictability in the simplest form.
Betting odds can be represented in multiple forms and you need to read them accurately to make wise predictions. Let’s take a look at how you can do that.
## How to read betting odds with examples
Betting odds for different games are based on the region of the bookies and the sportsbooks. For the purposes of this article, let us consider Indian currencies and the game of cricket since everyone is familiar with the fundamentals.
Here are the three main different types of betting odds representation and how to read them:
### 1. American Odds
These are probably the toughest to understand. The odds are determined by determining the betting lies and setting the plus-minus betting odds against them.
The American betting odds represent how much money you need to invest in order to earn a given base amount. So, let’s assume that you want to bet on the IndVsAus cricket match for Rs.100 and you want to bet on India to win.
Now, the odds will be displayed as +120 and -120 that represents the likelihood of winning as well as the quantifier for your earnings. So, if +120 is assigned to India, Australia has -120 assigned to them, and you bet Rs. 100, here’s how you earn.
If India wins the match then upon investing Rs. 100, you earn Rs. 120 in profits. If Australia wins then for an investment of Rs. 120, you will earn Rs. 100 in profits.
As you can see, the plus-minus system depicts how much will be added to your profits if you made the right choice. Of course, this sum will increase if you make much higher investments. you may also like to know more about - poker hands calculator.
### 2. Decimal Odds
Decimal odds are very easy to read. They are represented as decimal points that you use to multiply your betting amount and determine your winnings quickly.
For instance, if the betting odds on India vs Pakistan are represented as 2.5 and 1.5 respectively then for placing a bet of Rs. 100 you will receive Rs. 150 if India wins and Rs. 50 If Pakistan wins.
### 3. Fractional Odds
They may appear a little complex but they are actually quite easy to read. Fractional betting odds depict the odds of the outcome to be a loss vs the outcome to be a win.
Let’s assume that you are betting on India to win against West Indies. If the odds are represented as 8/4 then it means that the odds of India losing the match is double their chances of winning against West Indies if the 2 teams played 12 matches.
Simply put, if India and West Indies have played 12 matches in the past with most of the same players that are participating in the match then India must have won only 4 of the 12 past games.
## Types of betting odds and how they work
There are three distinct types of betting odds based on quantifiable betting lines.
### 1. American Betting Odds
American betting odds require a little overthought. These odds are determined using betting lines but they use plus and minus signs to depict the winning and losing sums if you bet on a certain team or player.
The plus-minus signs actually help determine which team or player is the underdog in a given match. The player who has positive betting odds is usually the underdog and they offer high winning potential. The minus signs represent the player or team who is most likely to win the match but your winnings will be lower since the negative betting odds will be deducted as vig.
### 2. Decimal Betting Odds
Decimal odds are possibly the simplest to understand. They use betting lines to determine the betting odds wherein the underdog has a higher betting odds assigned to them since their likelihood of winning is slimmer.
Decimal betting odds simply represent the odds as a decimal number and you can use this number to calculate the probability of the outcome and your winnings using very simple equations that we will discuss shortly.
### 3. Fractional Betting Odds
They are most common in European sportsbooks and represent the betting odds in horse races. You can find these betting odds in several other types of sports matches too but if the outcomes of the matches are anything other than win, lose, and draw, it is difficult to determine the probability and winnings for the match.
Fractional betting odds represent the bifurcated possibilities of the outcomes in a certain number of events held. Therefore, if there are 10 matches held between two given teams, then the fractional betting odds will represent how many times each team is likely to win.
## How to use betting odds to calculate probability?
In order to calculate the probability of the outcome of the sporting event from the betting odds, you need to use a set of formulas based on the type of odds. Let us take a look at what they are:
### 1. American Betting Odds Probability
To calculate the probability of the outcome based on the American betting odds, you can use the following formula:
100 / (positive odds + 100)
So, if you want to calculate the probability of India winning against Australia at +120 odds then it is calculated as, 100/ (120+100) = 100 / 220 = 0.45, i.e., 45%
Therefore, the likelihood of India winning against Australia is only 45% even though they offer higher odds.
### 2. Decimal Betting Odds Probability
If you want to calculate the probability of the outcome of an event based on the decimal betting odds then you can use the following formula:
100 / decimal odds
If we assume the example from above that India has 2.5 betting odds in their favor to win then the probability of the outcome is,
100 / 2.5 = 40% which is lower than Pakistan in that specific match.
### 3. Fractional Betting Odds Probability
Based on whose probability of win you are calculating, you need to use the numerator or denominator on top. Assuming our previous example of Ind Vs WI at 4/8 odds where India is likely to win 4 out of the total number of matches, which is 8+4 = 12, we use the following equation:
Number of possible wins/number of matches = winning probability
In this case, 4/12 = 0.33.
Therefore, India’s chances of winning the match against West Indies is only 33%
## How to use betting odds to calculate winnings?
Calculating your winnings based on the betting odds that you receive is fairly easy. Each type of betting odds has a formula that you must use in order to calculate the sum of the amount you will make against your bet. Let’s take a look at the formulae to help you determine your winnings:
### 1. American Betting Odds Calculation
You can use this equation to determine your winnings from American odds:
(Plus-minus determinant)/100 = (your investment/x), now solve for x.
Let’s say you bet Rs. 1000 on India at +120 odds, then your winnings are calculated as,
120/100 = 1000/x => 120x = 100000 => x = 100000/120, so x = 833.33.
Therefore, for the betting sum of Rs. 1000, you will earn Rs. 833.33, so the total money you will receive is Rs. 1833.33.
### 2. Decimal Betting Odds Calculation
You can use the following equation to calculate your winnings:
(The odds x your betting sum) – your betting sum = winnings.
Let’s say you bet Rs. 1000 for India to win at 2.5 betting odds, then your winnings will be,
(2.5 x 1000) – 1000 = 2500 – 1000 = 1500.
Therefore, for betting a sum of Rs. 1000, you will make Rs. 1500 in profit. And the total amount you receive will be Rs. 2500.
### 3. Fractional Betting Odds Calculation
In order to calculate your winnings from fractional betting odds, you first need to convert the numbers to decimal points and then use the following equation:
Your betting sum x (100/ decimal odds) = winnings.
You can calculate it directly as well. Let’s say you bet Rs. 1000 on India to win at 4/8 odds then the decimal factor is 0.5. Therefore your winnings would be,
1000 x (4/8) = 1000 x 0.5 = 500.
Therefore, your winnings are Rs. 500 and the total sum you will receive is Rs. 1500.
## Key takeaways
After closely examining what is betting odds, how betting lines affect the odds, and how to calculate the probability of the outcome for the event, here are a few things you need to bear in mind:
• The betting odds on the underdog are always higher than the team that is most likely to win, so don’t bet blindly based on the odds displayed.
• Always take the time to calculate the implied probability for the outcome of any sporting match regardless of what it is and who is playing.
• If you can’t decide which team you should bet on, calculate the probability and the winnings as per the betting odds on each team and bet on your favorite as long as the difference between the probability calculations isn’t a lot.
• If the probability of the underdog team is below 30%, don’t bet on them.
• The betting odds can tell you a lot about the competence of any player or a team.
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## FAQs
What are the odds in betting?
Odds are the representation of a team or player’s overall performance and their ability to score the winning points. In order to answer what are odds in betting, they are the decimal and fractional figures or betting lines that represent the teams’ level of expected performance and the possible outcomes of the match.
What are betting lines?
When the bookies and sportsbooks try to determine the betting odds for a game where the outcome is not as cut and dry, they compare the odds of the two teams and their performances. This helps identify the underdog in the team and how far behind they might fall in terms of scores in the actual game against the favorite-to-win team.
This bias that results from the difference in predicted scores creates a marginal line. This line is used to determine the betting odds and it is referred to as betting lines. They let the punters know which team is favored to win and which one is the underdog.
What kind of betting odds should I play to earn big?
Betting odds usually represent the underdog and the team that is favorable to win. The odds determined from betting lines will usually offer high rewards for betting on the underdog. Similarly, the fractional and decimal odds show the likelihood of the team’s performance against their adversary.
In order to make winning bets based on the odds, you should always split the difference when the odds are poles apart. That means indulging in matched betting and splitting your bets on individual outcomes in the game rather than the game itself.
In the case of betting lines, consider the minus betting amount before you bet on your favorite team. You also need to consider the underdog team’s consistency with the same team and their performance with other teams before you decide if you should choose to bet on them.
GetMega Rummy is an amazing platform that lets you play rummy with friends & family with real money. Sounds fun, isn't it? Download the GetMega rummy app now! | 3,061 | 14,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.952315 |
https://chemistry.stackexchange.com/questions/44669/how-to-identify-the-number-of-pi-electrons-in-a-conjugated-system-to-calculate-t | 1,652,987,697,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00505.warc.gz | 213,606,076 | 67,352 | # How to identify the number of pi electrons in a conjugated system to calculate the HOMO-LUMO gap with the particle in the box approach?
It's my understanding that for a particle in a 1-D box, the change in energy from the HOMO to LUMO can be approximated with the following equation: $$\Delta E=\frac{h^2 ( N + 1 )}{8 m_\mathrm{e} L^2}$$ I am told that $N$ represents the number of π electrons in the molecule.
If you're studying a molecule like the one shown below and you want to predict the absorption wavelength from UVVIS, would you count all of the π electrons, or just the conjugated ones? It doesn't make much sense to me why you'd want to include the π electrons from the phenyl group next to the neutrally charged Nitrogen because it's not conjugated with the rest of the molecule to the same extent the other phenyl group is, and thus seems like it would be grossly inaccurate to assume that those π electrons contribute the same as the others.
• I disagree with your assertion that this molecule is not fully conjugated. I believe both phenyl rings are coplanar with each other and conjugated together. Don't believe me? Draw the resonance structure where the uncharged donates electrons into the 2-position, shifting the methylene pi-bond to the 2' position, dumping a lone pair back on to the originally charged nitrogen. More importantly, it can't be resolved without a crystal structure. This is one of those times where you'd have to calculate several options and let experiment tell you which is correct. Feb 3, 2016 at 22:26
• Yes I just thought of that. I was thinking that the sulfur and nitrogen are sp3 hybridized without an open p orbital, but I forgot about rehybridization. Thank you makes a lot of sense. So I'm counting 16 pi electrons, using the equation I get a theoretical absorption wavelength of 272 nm. Experimental is 425. Seems too far off Feb 3, 2016 at 22:31
• Molecule is symmetric and only ethyl groups' carbons don't participate in conjugation (nitrogens and sufurs do) . BTW I cant see how 1-D box would be of much use here. Feb 3, 2016 at 22:36
• And then Mithoron's comment is great. You'd have n-pi* transitions for this molecule Feb 3, 2016 at 23:53
In a very basic first order approximation you can treat dyes with an extended conjugated π-system with the one-dimensional particle in the box approach. I simplified your system and calculated it at the DF-BP86/def2-SVP level of theory. The length of the box is about $L=1.20~\mathrm{nm}$, the following graphic shows it in Ångström. | 626 | 2,535 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | longest | en | 0.948046 |
https://nrich.maths.org/public/leg.php?code=72&cl=3&cldcmpid=4767 | 1,566,592,522,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318986.84/warc/CC-MAIN-20190823192831-20190823214831-00282.warc.gz | 574,962,988 | 9,575 | # Search by Topic
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Regular Visitor
## IF function - single value selcteded from a row
Hi
I Have a table with 3 Columns (A, B, C) that contain numbers between 1 - 10.
I need to create a new table (or add custom columns to the first table),
that check for the occurrence of each number in the left table. If yes show "1", else show "0".
The result table should look like the table on the right.
A B C D E F G H I J K L M
2 4 8 0 1 0 1 0 0 0 1 0 0
5 1 7 1 0 0 0 1 0 1 0 0 0
8 9 6 0 0 0 0 0 1 0 1 1 0
I can do it by adding custom column for each one of the 1-10 numbers as follow:
if [A] = 1 or [B] = 1 or [C] = 1 then 1 else 0
if [A] = 2 or [B] = 2 or [C] = 2 then 1 else 0 ,etc.
My question:
1. Is there a way (M or DAX) to use the if function to select a row instead of
using the 'or' operator for each column?
2. Is there a way to add these custom columns from a list or a function
Thanks for the help.
2 ACCEPTED SOLUTIONS
Accepted Solutions
Highlighted
Super Contributor
## Re: IF function - single value selcteded from a row
That is a nice and easy solution.
An alternative with some more programming, would be:
```let
Source = Excel.CurrentWorkbook(){[Name="Input"]}[Content],
Typed1 = Table.TransformColumnTypes(Source,{{"A", Int64.Type}, {"B", Int64.Type}, {"C", Int64.Type}}),
AddedLists = Table.AddColumn(Typed1, "Custom", (row) => List.Transform({1..10}, each if List.Contains(Record.FieldValues(row),_) then 1 else 0), type {Int64.Type}),
Tabled = Table.TransformColumns(AddedLists,{{"Custom", each Table.FromRows({_},{"D".."M"}), type table}}),
Expanded = Table.ExpandTableColumn(Tabled, "Custom", {"D".."M"}),
Typed2 = Table.TransformColumnTypes(Expanded,List.Transform({"D".."M"}, each {_, Int64.Type}))
in
Typed2```
Specializing in Power Query Formula Language (M)
Super User
## Re: IF function - single value selcteded from a row
Thank you @MarcelBeug
4 REPLIES 4
Super User
## Re: IF function - single value selcteded from a row
Hi,
Yes, it is possible to solve this in M.
```let
Source = Excel.CurrentWorkbook(){[Name="Data"]}[Content],
#"Changed Type" = Table.TransformColumnTypes(Source,{{"Number1", Int64.Type}, {"Number2", Int64.Type}, {"Number3", Int64.Type}}),
#"Expanded Custom" = Table.ExpandListColumn(#"Added Custom2", "Custom"),
#"Removed Columns" = Table.RemoveColumns(#"Expanded Custom",{"Start from", "End at"}),
#"Reordered Columns" = Table.ReorderColumns(#"Removed Columns",{"Custom", "Number1", "Number2", "Number3"}),
#"Added Custom3" = Table.AddColumn(#"Reordered Columns", "Custom.1", each if [Number1]=[Custom] or [Number2]=[Custom] or [Number3]=[Custom] then 1 else 0),
#"Pivoted Column" = Table.Pivot(Table.TransformColumnTypes(#"Added Custom3", {{"Custom", type text}}, "en-US"), List.Distinct(Table.TransformColumnTypes(#"Added Custom3", {{"Custom", type text}}, "en-US")[Custom]), "Custom", "Custom.1")
in
#"Pivoted Column"```
Hope this helps.
Highlighted
Super Contributor
## Re: IF function - single value selcteded from a row
That is a nice and easy solution.
An alternative with some more programming, would be:
```let
Source = Excel.CurrentWorkbook(){[Name="Input"]}[Content],
Typed1 = Table.TransformColumnTypes(Source,{{"A", Int64.Type}, {"B", Int64.Type}, {"C", Int64.Type}}),
AddedLists = Table.AddColumn(Typed1, "Custom", (row) => List.Transform({1..10}, each if List.Contains(Record.FieldValues(row),_) then 1 else 0), type {Int64.Type}),
Tabled = Table.TransformColumns(AddedLists,{{"Custom", each Table.FromRows({_},{"D".."M"}), type table}}),
Expanded = Table.ExpandTableColumn(Tabled, "Custom", {"D".."M"}),
Typed2 = Table.TransformColumnTypes(Expanded,List.Transform({"D".."M"}, each {_, Int64.Type}))
in
Typed2```
Specializing in Power Query Formula Language (M)
Super User
## Re: IF function - single value selcteded from a row
Thank you @MarcelBeug
Regular Visitor
## Re: IF function - single value selcteded from a row
Thanks very much guys.
Great help!!!
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Recent signins: | 1,345 | 4,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-43 | latest | en | 0.346928 |
http://www.mrbartonmaths.com/blog/bearings-the-answered-revealed/ | 1,718,578,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00408.warc.gz | 49,199,097 | 13,493 | # Bearings – The Answered Revealed!
Bearings are the kind of topic that students cannot bear (sorry). The tend to sneak their way into several key topics, most notably angle facts, construction and trigonometry. Hence, if a student’s understanding of bearings is dodgy, then it could jeopardise their chances of success across other areas.
This question is one of the most poorly answered across the free daily GCSE Revision Stream to date, with over half of students getting the answer wrong. Let’s take a look at the question, and see the kind of misconceptions that it reveals:
As predicted by teachers, by far the most popular incorrect answer was C). Why would students go for an answer of 150 degrees? Their explanations are incredibly revealing, suggesting varying levels of understanding of angles, but with once again the common thread being students falling flat due to a misremembered or misunderstood rule for bearings:
“this is because angles on a straight line equal to 180 so we take (180 away from 30) which gives us 150”
“Since the angle in the triangle is 30 degrees. I cant measure the bearing of b from a physically. Because, i know that angles on a straight line add up to 180 degrees. I use this understanding to figure out that the bearing is 150 degrees(0150 degrees)”
“Starting clockwise from north, the angle has to be 360 all the way around. 180 is already shown add the 30 degrees. 180+30=210 360-210= 150 degrees from B to A”
“Bearings always go clockwise, and angles on a straight line add up to 180 degrees. So 180 – 30 will give you the bearing from A to B which is 150 degrees”
It is interesting to note that the answer of 30 degrees cropped up in both answer A) and D), which together accounted for just as many wrong answers as C alone. Where does 30 degrees come from, I hear you ask. Well, in the words of students across the world:
“because it needs to be from north so you put the north line there, this creates a z shape. This means alternate angles are equal”
“Angle on a straight line is 180 and add 30 to find the bearing”
“180 – 30 = 150 150 + 180 = 330 360 – 330 = 30”
“Bearings always have three digits and the angles are on the same line.”
And my personal favourite: “HAVEN’T GOT A CLUE ABOUT BEARINGS MISS!!”
Like the rest of the country, over half our students got this bearings question wrong. How will we deal with this? Well, each week in our Monday Maths Departmental meeting, we discuss a particularly poorly answered question from the free daily GCSE Revision Stream that all our Year 11s complete. We then show this question to each of our Year 11 classes and discuss not only the correct answer, but some of the best ways of explaining the answer, as well as the reasons behind the wrong answers. Doing this on a regular basis (we hope!), will allow us to focus on the areas that our students need help with in an efficient way, and through discussions lead to a deeper understanding of potentially troublesome topics.
## 4 thoughts on “Bearings – The Answered Revealed!”
1. Frances says:
You mean cannot *bear*
Which makes your pun better, too! 😉
2. Kaviraj Sheoraj says:
I have been teaching bearing for the last 15+ years and i’ve met lot many students with a multitude of misconceptions. with experience i’ve devised ways for teching same and till date, many students have seen light through this chapter all by using a simple technique which involves certain mandatory steps only to ‘cheat’ any misconception that wants to seep into the solution for a given question. Amazingly, students of form 1 have been able to do questions on Bearing that they are supposed to do in form 2, i.e they did same a year before (accelerated learning + content acceleration) It has worked wonders. I’m most willing to share my teaching tips for maths classroom HACKS. plese advise how to do so.
—- KAVIRAJ SHEORAJ (MAUTITIUS)
“THE MORE YOU SHARE, THE MORE YOU GET”
3. MrBarton says:
Sounds fascinating! Feel free to drop me an email. Craig
4. MrBarton says:
Fixed now. Great spot! | 936 | 4,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.949785 |
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# How To Put Error Bar In Matlab
## Contents
You can also select a location from the following list: Americas Canada (English) United States (English) Europe Belgium (English) Denmark (English) Deutschland (Deutsch) España (Español) Finland (English) France (Français) Ireland (English) Italia (Italiano) Luxembourg (English) Netherlands (English) Norway (English) Österreich (Deutsch) Portugal (English) Sweden (English) Switzerland Deutsch Français United Kingdom (English) Asia Pacific Australia (English) India (English) New Zealand (English) 中国 (简体中文) 日本 (日本語) 한국 (한국어) See all countries Trial Software Product Updates MATLAB Documentation Examples Functions Release Notes PDF Documentation Other Documentation SimulinkSymbolic Math ToolboxStatistics and Machine Learning ToolboxImage Processing ToolboxSignal Processing ToolboxDocumentation Home Support MATLAB AnswersInstallation HelpBug ReportsProduct RequirementsSoftware Downloads Tackling Big Data with MATLAB Watch now © 1994-2016 The MathWorks, Inc. The function wpolyfit is similar to polyfit where a vector is returned % with the coefficients of the polynomial and a structure is returned that includes the Cholesky % factors of the Vandermonde matrix plus other information that is used to create the errorbar on the fit % (as opposed to the error on the data itself). % [p,s] = wpolyfit(x,y,sy,1) % % polyconf is used to get the values of y and sy for the MODEL (not the data). % ci is confidence interval (remember we discussed that?) method which is % appropriate for error bars in the physical sciences. % [yn,syn] = polyconf(p,x,s,'ci') % % Now I continue the graphing. Melde dich an, um dieses Video zur Playlist "Später ansehen" hinzuzufügen. Incorrect Query Results on Opportunity? navigate here
Example: y = [4 3 5 2 2 4]; Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64x -- x valuesvector | matrix x values, specified as a vector or a matrix. You can also select a location from the following list: Americas Canada (English) United States (English) Europe Belgium (English) Denmark (English) Deutschland (Deutsch) España (Español) Finland (English) France (Français) Ireland (English) Italia (Italiano) Luxembourg (English) Netherlands (English) Norway (English) Österreich (Deutsch) Portugal (English) Sweden (English) Switzerland Deutsch Français United Kingdom (English) Asia Pacific Australia (English) India (English) New Zealand (English) 中国 (简体中文) 日本 (日本語) 한국 (한국어) See all countries Trial software Explore Products MATLAB Simulink Student Software Hardware Support File Exchange Try or Buy Downloads Trial Software Contact Sales Pricing and Licensing Learn to Use Documentation Tutorials Examples Videos and Webinars Training Get Support Installation Help Answers Consulting License Center About MathWorks Careers Company Overview Newsroom Social Mission MathWorks Accelerating the pace of engineering and science MathWorks is the leading developer of mathematical computing software for engineers and scientists. Define a variable called meanPetalWidths that contains the mean petal widths of each of the three species. err must be the same size as y. https://www.mathworks.com/help/matlab/ref/errorbar.html
## Matlab Error Bars On Bar Graph
Set the colors to either a character vector of a color name, such as 'red', or an RGB triplet.x = linspace(0,10,15); y = sin(x/2); err = 0.3*ones(size(y)); errorbar(x,y,err,'-s','MarkerSize',10,... 'MarkerEdgeColor','red','MarkerFaceColor','red') Control Error Bar Cap SizeOpen ScriptControl the size of the caps at the end of each error bar by setting the CapSize property to a positive value in points.x = linspace(0,2,15); y = exp(x); err = 0.3*ones(size(y)); errorbar(x,y,err,'CapSize',18) Modify Error Bars After CreationOpen ScriptCreate a line plot with error bars. Aftab Ahmed Khan Aftab Ahmed Khan (view profile) 65 questions 0 answers 0 accepted answers Reputation: 2 on 25 Jul 2014 Direct link to this comment: https://www.mathworks.com/matlabcentral/answers/143321#comment_227520 Awesome, Great. Now I have R2015a and I get this error:Error using errorbar (line 37) X, Y, and error bars all must be the same length. What could make an area of land be accessible only at certain times of the year?
the cyclist the cyclist (view profile) 32 questions 2,601 answers 1,076 accepted answers Reputation: 5,963 on 16 Apr 2015 Direct link to this comment: https://www.mathworks.com/matlabcentral/answers/85885#comment_279149 I don't know how to make it work in the new version (where graphics handles are objects). You can turn them into functions so that you can very easily make new plots based on different data. Example: 0.75 See AlsoFunctionsbar | corrcoef | plot | stdPropertiesErrorbar Series Properties Introduced before R2006a × MATLAB Command You clicked a link that corresponds to this MATLAB command: Run the command by entering it in the MATLAB Command Window. Matlab Errorbar Width Browse other questions tagged matlab plot or ask your own question.
You can also select a location from the following list: Americas Canada (English) United States (English) Europe Belgium (English) Denmark (English) Deutschland (Deutsch) España (Español) Finland (English) France (Français) Ireland (English) Italia (Italiano) Luxembourg (English) Netherlands (English) Norway (English) Österreich (Deutsch) Portugal (English) Sweden (English) Switzerland Deutsch Français United Kingdom (English) Asia Pacific Australia (English) India (English) New Zealand (English) 中国 (简体中文) 日本 (日本語) 한국 (한국어) See all countries Trial software Explore Products MATLAB Simulink Student Software Hardware Support File Exchange Try or Buy Downloads Trial Software Contact Sales Pricing and Licensing Learn to Use Documentation Tutorials Examples Videos and Webinars Training Get Support Installation Help Answers Consulting License Center About MathWorks Careers Company Overview Newsroom Social Mission MathWorks Accelerating the pace of engineering and science MathWorks is the leading developer of mathematical computing software for engineers and scientists. Matlab Horizontal Error Bars Gay crimes thriller movie from '80s Confused riddle and poem? The functions and methods you use to get those depend on the regression function you’re using. I have data, series(y), which I have to plot against (x).
The xneg and xpos inputs set the left and right lengths of the horizontal error bars. Errorbar Matlab Example Join them; it only takes a minute: Sign up How to plot an error bar plot with standard deviation values in MATLAB? Use MarkerSize to specify the marker size in points. All rights reserved.
## Matlab Horizontal Error Bars
errorbar(X, Y, E) creates a plot similar to errorbar(Y, E) except that this function uses the values of X for the x positions rather than using the integers 1, 2, ... .
Also I have the standard deviation values for each data point of (y). Matlab Error Bars On Bar Graph I have posted a new question asking for this here. Matlab Errorbar No Line The yneg and ypos inputs set the lower and upper lengths of the vertical error bars, respectively.
NOTE: MATLAB should have something equivalent, but I haven't been able to find it yet; so Octave will be required in this class. Acknowledgments Trademarks Patents Terms of Use United States Patents Trademarks Privacy Policy Preventing Piracy © 1994-2016 The MathWorks, Inc. Hinzufügen Möchtest du dieses Video später noch einmal ansehen? You do not need to specify all three characteristics (line style, marker symbol, and color). Error Bars Matlab Scatter
What could make an area of land be accessible only at certain times of the year? Is the measure of the sum equal to the sum of the measures? Enter your comments, suggestions, or thoughts below Please enable JavaScript to view the comments powered by Disqus. Differentiating between zero and not sending for OOK Can civilian aircraft fly through or land in restricted airspace in an emergency?
EXAMPLE 4: Plot the SD error bars on a bar chartCreate a new cell in which you type and execute: figure hold on bar(sLenMeans, 'FaceColor', [0.5, 0.5, 1]) % Lighter so error bars show up errorbar(sLenMeans, sLenSDs, 'ks'); % Error bars use black squares set(gca, 'XTick', 1:3, 'XTickLabel', irisSpecies) % Set ticks and tick labels xlabel('Species of Iris'); ylabel('Sepal length in mm') title(fisherTitle) legend('Mean (SD error bars)', 'Location', 'Northwest') % Put in lower right box on % Force box around axes hold off You should see a Figure Window with a labeled error bar plot: EXAMPLE 5: Compute the standard error of the mean (SEM) for sepal lengthsCreate a new cell in which you type and execute: numSamples = length(sepalLens); % Length along the longest dimension sLenSEMs = sLenSDs./sqrt(numSamples); % Compute the standard error of the mean (SEM) You should see the following 2 variables in your Workspace Browser:numSamples - the number of samples in each groupsLenSEMs- the standard error for the mean sepal length for the three speciesThe SEM estimates the standard deviation of sample means from the true population mean. Herrorbar Matlab Possible values are 'setosa', 'versicolor', and 'virginica'. Discussion What's really great about MATLAB is that it's possible to make quite elaborate graphs with relatively few lines of code.
## When does bugfixing become overkill, if ever?
Aftab Ahmed Khan Aftab Ahmed Khan (view profile) 65 questions 0 answers 0 accepted answers Reputation: 2 on 25 Jul 2014 Direct link to this comment: https://www.mathworks.com/matlabcentral/answers/143321#comment_227514 What is this barvector, i mean how can i calculate it? For the normal distribution it is usually given as 1.96 times the standard error. Why does argv include the program name? Matlab Shaded Error Bar Learn MATLAB today!
comments powered by Disqus Want 20 GB of free on-line storage? Plant based lifeforms: brain equivalent? If you do not want to draw the left part of the error bar at a particular data point, then specify the length as NaN. We've blasted through all the stuff in the last recipe, but this time we've used the SEM instead of the SD.
The L array gives the distances below the corresponding values in Y, and the U array gives the distances above the corresponding values of Y. The corresponding values in E show error bars at +/- that amount above and below the corresponding values in Y. Apply Today MATLAB Academy New to MATLAB? Show that a nonabelian group must have at least five distinct elements How should I deal with a difficult group and a DM that doesn't help?
For example, if you omit the line style and specify the marker, then the plot shows only the markers and no line. For a list of properties, see Errorbar Series Properties. I want to have multiple separate figures with bar graphs and error bars. no. 1 (1792). | 2,465 | 10,924 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-05 | latest | en | 0.681124 |
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2. ## perfect powers
The answer depends whether you accept 0 as a square and whether you accept the same number twice.
3. ## perfect powers
A bit cheating, but one candidate would be 125= 02 + 52 + 102 = 32 + 42 + 102 = 52 + 62 + 82
4. ## Having sisters
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# Thread: Randomize Odd or Even
1. ## Randomize Odd or Even
Friends,
I have a quick question. I need a returned value odd or even 1 or 2. I have two bits of code and want the system to pick one or the other.
Randomize
randomNum = (Rnd(1-2))
In short if you double click on the object or if you click once and then click OK the same action is taken and I want to cover both by the system picking.
The code bit randomNum is always something like 0.220077 and never 1.002293.
Thanks, for any pointers.
BMD
PS. This is VBScript.
2. ## Re: Randomize Odd or Even
psuedo code for choosing 1 or 2:
- multiple your randomNum by 1000000 to convert to integer
- check if that result is divisible by 2 (modulo == 0)
- if yes, the random number is even, set your value to 2
- if no, the random number is odd, set it to 1
implementing in VBScript is an exercise for you.
3. ## Re: Randomize Odd or Even
Thanks Corey,
By multiplying randomNum I get an integer but the issue is it is ALWAYS 220077.76354 every single time, not random at all. Also I do not believe the modulo function is available to VBScript.
Thanks,
BMD
4. ## Re: Randomize Odd or Even
In vbScript modulus is Mod
<font class="small">Code:</font><hr /><pre>
result = num1 Mod num2
</pre><hr />
5. ## Re: Randomize Odd or Even
Thanks Aaron,
The issue about the random number not being random remains. I'm able to get result to be a 1 or 0 but it is always that number as randomNum = (Rnd(1-2))*1000000 always returns the same number.
BMD
6. ## Re: Randomize Odd or Even
With VBScript you need to seed the random number first using the "Randomize" statement.
7. ## Re: Randomize Odd or Even
This is the code bit and it appears to be working.
<font class="small">Code:</font><hr /><pre>
Randomize
randomNum = (Rnd(1))*100
iNum = Int(randomNum)
result = iNum Mod 2
</pre><hr />
I had the randomize statement in the code.
Thanks, for all the pointers.
BMD
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# Oh....We're Halfway There! (and 5K Calculations)
### Friday, June 12, 2009
Today I weighed in at 152.8, which means that I have now lost 12.2 pounds and have 12.8 more pounds to go! I am being optimistic and thinking that this weekend, after sleeping in late, my weight will be down even more, so today is the day that I'm celebrating being halfway to my goal, even though I'm off by a few tenths of a pound. :)
By the way, it took me 9 weeks and 5 days to lose the first half of the weight. That is about 1.25 pounds per week. If I keep up this pace, I will be at goal (140 pounds) by August 21, 2009. :) :) :)
Also, as the math nerd that I am, I just did some calculating that I'd like to share, though I admit this is mostly for my own future reference more than for the makings of an interesting blog. A few weeks ago I wrote about my formula for determining 5K average speed based on total time: 186 divided by total time. (Example: 186 divided by 24 equals 7.75. This means that in order to finish a 5K in 24 minutes, you would need to run an average speed of 7.75 mph.) The number 186 came from 60 minutes times 3.1 miles (5 kilometers is approximately 3.1 miles).
Today I turned the formula around to figure out what the total 5K time would be based on average speed. This formula is: 186 divided by average mph. Record the number before the decimal for total minutes, then subtract it out to get left with just the decimal, and multiply that by 60 to get the seconds. (Example: 186 divided by 7.2 equals 25.8333333. Do 25.8333333 minus 25 equals 0.8333333. Multiply 0.8333333 times 60 which equals about 50. This means that an average speed of 7.2 mph would mean that the total 5K time would be 25:50.)
Here's what I came up with:
7.0 mph - 26:34
7.1 mph - 26:12
7.2 mph - 25:50
7.3 mph - 25:29
7.4 mph - 25:08
7.5 mph - 24:48
7.6 mph - 24:28
7.7 mph - 24:09
7.8 mph - 23:51
My probably-way-too-ambitious goal that I dreamt up a few weeks back was to finish my upcoming race in 24 minutes and that meant I'd have to run an average of 7.75 mph. As you can see in yesterday's blog, I have not been able to get even close to that yet, even though I'm training on a treadmill which is easier than running outside. The fastest I've been able to go lately is 25:45. If I want to beat that in my next pace training run, I should shoot for an average speed of 7.3 mph. I can do that!! Once I'm over this darn cold, and I'm well rested, well fed, and I've had my coffee, I can do that! Then the next time I can try for 7.4 mph and work my way up to 7.7 or 7.8 if I can get there in time - my race is only 16 days away and I only pace train once a week or so. Even if I don't have time to work my way up to that speed, I can still PR because my fastest official 5K RACE time was somewhere between 28-29 minutes (can't recall precisely). I will be happy with any PR at all, but I would be very proud of a sub 27 or 26 minute time, and my dream goal is 24 minutes.
Happy Friday!!!
MARATHONER340 6/15/2009 7:27PM I generally run much faster in races than I do in training and I, too, end up doing a lot of speedwork on the treadmill (mostly because it's more convenient than driving to a track! I'd much prefer outside!) I generally always run at a 1% incline to make it more similar to running outside. But my point here was...I bet you'll run faster than you think in the 5k! Good luck! Can't wait to hear about it! =) Report Inappropriate Comment
CREATINGMYSELF 6/13/2009 12:03AM I remember I was on the treadmill one day and I got frustrated because the 'up speed arrow' wasn't moving fast enough. So I planted my finger on it angrily and pressed down as hard as I could. Next thing I knew I was flying off the treadmill. And when I got up, brushed myself off (and pretending like it didn't happen), I looked at the speed... it was 7.8 mph. You are a mad woman-- but in a good way :) Your dedication and enthusiasm is contagious! And congrats on being half-way, that is SUPER cool!Comment edited on: 6/13/2009 12:03:32 AM Report Inappropriate Comment
KJNE8O 6/12/2009 10:03PM YAY on the weight!!!And yeah - you're a total math nerd! Report Inappropriate Comment
SWEATONCEADAY 6/12/2009 8:14PM congrats on your weight loss and good luck with your race! Report Inappropriate Comment
FITGIRL15 6/12/2009 7:18PM Way to go Susan! I'm very proud of you for losing your weight the healthy way (1.25 pounds a week is AWESOME!!! Very maintainable!) and you are definitely adopting a lot of healthy lifelong skills in the process!Good Luck on those running targets! Report Inappropriate Comment
KENSINGTONC 6/12/2009 7:00PM Great blog! The math kind of makes my head spin but I understand the point. I bet you'll be MUCH faster than you think in the actual race, with the adrenaline etc - at least I always have been. Anyway, great job with the weight loss and meeting your fitness goals! Report Inappropriate Comment
BRIAN36 6/12/2009 6:07PM Just a link to save you all the math, it made my head spin! LOLhttp://www.coolrunning.com/engine/4/4_1/96.shtml Report Inappropriate Comment
RUN_LIFT_EAT 6/12/2009 5:58PM I find that I am a *LOT* faster at an actual race, which is why some training schedules will have you run a shorter race (probably not for a 5k, but you get my point) as part of your training. That is so awesome that you are halfway to goal!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Report Inappropriate Comment
ZIRCADIA 6/12/2009 3:20PM I'm feeling you!!! My goal is a sub 30min 5K, but the fastest I've ever run 3 miles was 30:18, so add the extra .1 miles and I have no idea where that will put me. I suppose I could do some math, but since it clearly puts me over 30min I'm not gonna bother. :) HAHAHAHAHA. BUT!!! My 5K PR is like 33:26, so I figure I should be able to at least set a new PR tomorrow even if I don't reach my sub-30 goal yet. HAPPY FRIDAY and HAPPY HALFWAY! WOOHOO!!! Report Inappropriate Comment
# 5K Times
### Thursday, June 11, 2009
I've decided to record all of my fastest 5K training times together in one place - here! So far, these have all been done on a treadmill. I know I have gone faster than this in the past, but it would take quite a bit of searching through old blogs to find my old best times. So these are just my best times since I started training for my 6/28/09 New Charles River Run 5K race.
6/11/09 26:10
6/02/09 25:45
5/29/09 27:15
5/24/09 26:59
I'm not sure if I can attain this kind of speed outside. Usually my tough runs are on the treadmill and my very long or easy runs are outside. I will have to try doing some speed or pace work outside! I also need to sometimes do speed work on the weekend, when I'm fully rested and caffeinated. I'm sure I could be faster than after a long day at school, which is when most of my pace workouts happen.
Also, a couple days ago I did my first hill intervals on the treadmill. I only had 12 minutes before I had to leave, so it was very quick and easy. I stayed at 6.0 mph most of the time, but I bumped it up to 7.0 at the end. I alternated between 0% incline and 4% incline, and each interval was four minutes. Afterwards, my knees hurt a bit, but that could have been a coincidence. Next time, I will try something similar, but with more speed.
SEEHOLZ 6/12/2009 12:25PM Those are some great times! Well, I haven't been in any race shape this year, so I'm lving through others I guess-LOL! For me, outside times are slower, but I've heard from 2 or 3 other Sparkers that TM is harder... I wonder if they run on a higher incline ( even though they claimed 1 and 2 %). Be careful with the speedwork on the treadmill, as you could easily pull something--- even though the surface is more gentle, something about the calibration makes it more strenous for speedwork ( compared to a track--- there you got to worry about the same direction) Anyways, your 5K is coming up soon--- you are well on your way to that PR! P.S: I'm really glad you're finding that balance. It totally gives me hope for myself! Report Inappropriate Comment
FITGIRL15 6/12/2009 12:00PM Nice times!I've never ran a 5K on the treadmill for time... maybe I'll try that someday! (Today? LOL)I think outside will slow you down, just due to the nature of the beast... a treadmill is very consistent terrain, no hills (unless you make them) and no obsticles to worry about (traffic, other runners, etc..) Not to mention how much scenery you are missing out on while running on a treadmill! My favorite part!!!PS.. I've been told by many-a-trainer never to run on a treadmill with an incline of less then 2%... it's actually like running downhill and very bad for your knees! Could be a source of the knee pain for you! I always bump the incline up to 2 before I even get started!!! Report Inappropriate Comment
SKYFYRE 6/12/2009 6:37AM You are a speed tracking queen! Great job on your running! Report Inappropriate Comment
JENONTHEROX 6/11/2009 9:57PM I am just winded & so impressed by your times... & even MORE impressed doing hill work at 6-7mph... 7mph are my sprints these days. I take my hills up anywhere from 3-8% & run at 5.4-5.5 mph-ish (around my first mile/my warmup pace). I know, I know, never compare, but hey, it's human nature. I think your runs are outstanding.& sigh, bless caffeine. I hate workouts & especially runs without it in my system. Report Inappropriate Comment
ZIRCADIA 6/11/2009 9:24PM Girl, you are so much faster than me!!! :D You ROCK!!! But yeah -- I'm curious definitely to see how you would do 1) outside, 2) rested and caffeinated. :) hahaha. Report Inappropriate Comment
KJNE8O 6/11/2009 9:20PM Wow - you are a speed demon! Definitely try it outside, see how that goes. Report Inappropriate Comment
# Two Dangerous Paths
### Wednesday, June 10, 2009
I have come to the realization that there is not just one, but TWO dangerous paths to which I am prone, both of which I must try to avoid if I want to effectively manage my weight.
Dangerous Path 1: Do whatever I want, whenever I want, because it feels really good!
This is the path I took prior to SparkPeople and also during those periods of rapid weight re-gain. This is the mindset of the Susan who ate so much in college that a friend said, "You eat more than anyone I've ever seen." This is the Susan whose thinking gets so illogical that somehow eating one pint of Ben & Jerry's is a common precursor to eating a second container! This dangerous path is quite obviously unhealthy. Dangerous Path 2 is not quite so obvious, but is just as dangerous...
Dangerous Path 2: Aim for perfection.
This is the path that led me, in early 2007, from 167 pounds to 134 pounds in just four months, while also making me cranky and lethargic in the process. This is the path that leads to resentment, because while everybody else is having fun, I have to keep up with my multi-hour sessions at the gym. There is no listening to my body when I have this mindset; there is only following pre-set rules. Sure, Dangerous Path 2 gets me skinny fast, but there is no way that weight loss attained this way can be maintained, because this lifestyle cannot be maintained.
What got me thinking about this today was a slice of cake. A coworker will be leaving us at the end of the school year, so my school's special education staff (all six of us) got together this afternoon to present her with a gift and to share a chocolate cake decorated with frosting best wishes. I only found out that there would be cake a few hours earlier, and I went into the situation ambivalent about whether I would actually eat it. I ate my usual healthy lunch at the usual time, and then just before we met I had a nice crisp apple to take the edge off my appetite; I also brought a can of seltzer to the meeting with me.
While the cake was being cut I noticed that everyone was getting a queen-sized piece. I considered asking for just a sliver, but in the end kept quiet. I let my giant slice sit on the table in front of me for a while and up until the moment I took my first bite, I still wasn't sure whether I'd eat it at all, or whether perhaps I'd leave a few large crumbs behind (never a common practice of mine in the past).
Well, I ate that great big slice of cake. It was moist and delicious, and I didn't leave a trace behind on my plate. After we said our goodbyes, I went right back to my office and I entered SparkPeople's closest match (the SP database had chocolate cake with chocolate frosting, which must be fairly similar nutritionally to what I had: chocolate cake with white frosting), and then I went home and did three things:
1) I continued thinking about that cake, and that cake's implications.
2) I ate a little less for dinner than I normally would, to keep my daily total calories at a comfortable level.
3) I exercised a little more than I probably would have if I hadn't eaten that cake.
Two very important things I did NOT do were:
1) continue chowing down on anything/everything I could find (a la Dangerous Path 1), despite my natural inclination to do so which kicks in whenever I start to eat something "naughty."
2) deny myself the cake altogether, or skip dinner entirely to make up for it, or go for an unreasonably long run (unreasonable here meaning taking so long that I wouldn't be able to meet my boyfriend at the train station after he was done with work). These are choices I would have made if I were in the mindset of Dangerous Path 2.
I want to live in the balanced middle of the spectrum between complete food freedom and ultra-strictness. I want to be healthy AND live my life to the fullest. I cannot get there if I take either extreme approach. I need to occasionally have a slice of cake, guilt-free, and then go out for a fun four-mile jog!
LIL_EZZY 6/11/2009 5:33PM I too loved this blog. See you have it. Your mindset has obviously changed so I don't see you going down either of your dangerous paths anymore. Good for you Susan. Report Inappropriate Comment
SEEHOLZ 6/11/2009 5:19PM OMG- I love this blog, because I need to do that---- keep a balance in my life. It's like i know what I need to do, but sometimes have the hardest time following through. That's the balance- there is nothing wrong with having a slice of cake on a special occassion--- and I'm totally proud of you for stopping, because old habits and tendencies tend to stay with me and I need to accept that even though the extremes are always there, I don't have to act upon them! THANK YOU so much for blogging about this. It's weird, too, because I was wondering if you were lately slowly going to the other extreme.... it's so good to know you aren't! Report Inappropriate Comment
JENJENRUS 6/11/2009 1:59PM Hey,I just got your message that you left me and hurried over here to read your blog since you said it was similar and boy can I relate! I am the same exact way...it is either all or nothing. I *used to be* a terrible binger...I really have tried to make a new lifestyle and so far, it is working!I just wanted to tell you that I completely understand where you are coming from. Good job for recognizing your weaknesses!Jen Report Inappropriate Comment
RA4945 6/11/2009 1:05PM Wow, this was beautifully written, Susan. You are so articulate! I love your attitude. This will surely keep you on the road of success! Report Inappropriate Comment
SIMPLYSMALTZY 6/11/2009 9:54AM Great job! We have to remember this is a lifestyle change. One slice of cake on a special occassion isnt going to kill us. Who never wants to eat cake or BEN & JERRY'S ever again?!?! Definetely not me! It's great that you are finding a happy medium! Report Inappropriate Comment
BELIEVE-IN-ME 6/11/2009 8:52AM I can't add anything more than the previous posters have said so I'm just going to say; "Wow!" That was an incredible read and oh-so-true for many of us!Good for you! You can have your cake and eat it too! Report Inappropriate Comment
TRACYZABELLE 6/11/2009 4:46AM One bite, one day, one pound at a time girl! And also if we do not pre-plan when we have those special events, failure may be lurking around the corner.. Don't I wish I incorporated these things years ago! You can do it Susan, I have faith in you. Report Inappropriate Comment
CREATINGMYSELF 6/10/2009 11:45PM I think a lot of us can relate to these scenarios, and I know for me personally it has taken me a long time to come to these realizations about my eating patterns. It's very black and white thinking-- either it's good or bad, all or nothing. And yes, it's dangerous. Congrats for recognizes these behaviors and patterns-- hopefully it will help you on your journey. Report Inappropriate Comment
MARATHONER340 6/10/2009 10:07PM WOW can I relate with this one! I too have gone down both paths and agree we need to remain somewhere in the middle! I think you made good choices today and am proud of you! It's so refreshing to hear that othe people have been in the same mindset as I have about things! Report Inappropriate Comment
KJNE8O 6/10/2009 9:49PM You are growing up!! LOL. So very wise and such a great way to approach life. Something that I need to hear and live as well. Thank you! Report Inappropriate Comment
ZIRCADIA 6/10/2009 9:43PM GOOD GIRL!!!!!!!!!! :D I cannot tell you how unreasonably proud I am of you every time I read about you thinking through these life changes and making moderate and smart choices. I say unreasonably because who am I to have the right to be proud of your accomplishments??? HAHAHA :D Maybe it's better to say that I'm exceedingly happy for you and your newfound path!! :D CONGRATULATIONS! Report Inappropriate Comment
KMWKENT 6/10/2009 9:40PM Great observations and great attitude! Glad you're able to find that middle ground. Report Inappropriate Comment
# Long Run Stats
### Sunday, June 07, 2009
Wow, I've earned my rest day tomorrow - 5.5 miles yesterday + 7.5 miles today = 13 miles this weekend!
I need to list today's stats in two different ways - one for the running-only portion at the beginning, and one for the entire thing that involved a lot of walking toward the end.
Running:
1:06:50
6.24 miles
5.6 mph
689 calories
10:41 min/mi
BMI 24.7
Whole:
1:29:19
7.53 miles
5.1 mph
829 calories
11:54 min/mi
It felt good and I didn't feel gross despite being covered in sunscreen. Once I stopped to walk my legs - muscles and joints - were feeling sore and tired, but nothing else was bothering me at all. It was a pretty day for running!
TRACYZABELLE 6/9/2009 4:44AM keep the sunscreen on lady! we dont want you to look like a prune at 40! Report Inappropriate Comment
SKYFYRE 6/8/2009 6:57PM Great Job and a great run! Look at you runner girl! Report Inappropriate Comment
ROGUE_1 6/8/2009 5:28PM That is so inspiring! I realized that I need to start going for longer runs outside especially now that the sun is shining and the weather is sweet. Do you run mostly outside or on the treadmill? You've definitely earned your rest day, so enjoy it! You are getting so close to your ticker goal...keep up the GREAT work :) Report Inappropriate Comment
ZIRCADIA 6/8/2009 2:53PM WOOHOO! :D ROCK ON, chica!!! :D That is GREAT! :D Report Inappropriate Comment
JENONTHEROX 6/8/2009 1:24PM you definitely did have GREAT runs over the weekend...13 miles is delicious!!! I love reading your running stats because I do the same, keeping them in my datebook (total time, mileage, pace). I've just been SO reluctant to run 2 or 3 days in a row... but I guess by adjusting the runs accordingly makes it work. Report Inappropriate Comment
RUN_LIFT_EAT 6/8/2009 8:34AM Wow! That is a great weekend! Report Inappropriate Comment
SEEHOLZ 6/8/2009 1:15AM YOu are doing so fabulous AND you totally motivated me to get my butt out there today! Report Inappropriate Comment
LIL_EZZY 6/7/2009 10:26PM I am jealous. Sounds like you enjoyed yourself out there in that sunshine. I sat inside all weekend apart from taking my son to his football game in freezing conditions. I wanted to go running yesterday(sunday) but had some sort of vomiting bug and felt oh so crappy. Today is freezing and windy but I may get out and do something once my housework is done. Hurray for you running all those miles. Report Inappropriate Comment
KJNE8O 6/7/2009 10:21PM Sounds like a great weekend for running! Way to go! Report Inappropriate Comment
MARATHONER340 6/7/2009 7:58PM yay for the 13 miles this weekend! That's more than I can say I ran! =) Report Inappropriate Comment
# Week 9 Ends
### Sunday, June 07, 2009
Week 1 -1lb
Week 2 -2lb
Week 3 -1lb
Week 4 -0.4lb
Week 5 -2lb
Week 6 -1lb
Week 7 -lb
Week 8 -2lb
Week 9 -1lb
This morning it hurt to talk - I seem to only be sick overnight and when I first wake up, and then I feel fine the rest of the day. I am now at 153.4 and suddenly the 140s seems oh-so-close. Can't wait!!! :) :) :)
KJNE8O 6/8/2009 4:28PM Allergies? A humidifier might help - but great weekend :) Report Inappropriate Comment
JENONTHEROX 6/8/2009 1:26PM you're doing amazingly well with your weight losses, nice & gradual... you are working HARD & deserve all the progress you're getting. The 140s are just around your corner, you'll be there before you know it. Report Inappropriate Comment
CREATINGMYSELF 6/8/2009 12:27AM You rock. Nuff said. Report Inappropriate Comment
LIL_EZZY 6/7/2009 10:28PM Sounds as if you had a really nice weekend. I love springtime. 140's here you come. Report Inappropriate Comment
SWEATONCEADAY 6/7/2009 2:32PM great consitent progress. you will see 140 in no time! Report Inappropriate Comment
BRUIN2 6/7/2009 11:45AM Stopping whenever you needed to use your phone?!?BWAHAHHAHA.You're funny.And the pb soft serve sounds sooooooo goooooooood (as a treat of course!). Report Inappropriate Comment
SDTALLY 6/7/2009 10:42AM You're doing great! I couldn't seem to loose more than 1 pound per week when I was loosing. Found out the key is to just keep on & eventually it all comes off! Sharon Report Inappropriate Comment
ZIRCADIA 6/7/2009 10:36AM Maybe you need a humidifier in your room or something?? But woohoo on such a great evening and down a bit more today! :D Report Inappropriate Comment
First Page 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 Last Page | 6,262 | 22,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2013-48 | longest | en | 0.952773 |
http://mathematica.stackexchange.com/questions/24821/finding-max-min-and-closest-root-of-two-dimensional-data | 1,469,723,059,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828286.80/warc/CC-MAIN-20160723071028-00311-ip-10-185-27-174.ec2.internal.warc.gz | 152,590,708 | 18,084 | # Finding max, min, and closest root of two dimensional data
I have some data that is two dimensional (traditional (x, y) format) as so:
var = {{x1, y1}, {x2, y2}, {x3, y3}}
I would like to search through that large list and find the maximum and minimum y values and return the ordered pair. Something like:
min[var]->{x2, y2} //read "->" as returns {the minimum ordered pair}
max[var]->{x3, y3} //read "->" as returns {the maximum ordered pair}
Finally I know the data has a "root". A plot of the points cross the x-axis. I would like to find the closest few (maybe 3 or 4) points to zero. Questions similar to this have been posted but I cannot figure out how to adapt those solutions to my needs
previous work 1
previous work 2
-
You overlooked this, which may be more useful to you. – whuber May 7 '13 at 22:13
How about var[[Ordering[var[[All, 2]]][[1]]]] and var[[Ordering[var[[All, 2]]][[-1]]]]? – Sjoerd C. de Vries May 7 '13 at 22:27
You can define your own function with which to sort a list. For instance:
tab=Table[{Random[] - 0.5, Random[] - 0.5}, {n, 20}]
sorted=Sort[tab,#2[[1]]^2>#1[[1]]^2&]
This will give you a list where the elements with x closest to 0 will be first. You can take the number of desired values from this list.
For the element with the highest and lowest y element, you can use @Sjoerd C. de Vries' example in the comment, or you can define your own ordering:
sortedminmaxy=Sort[tab,#2[[2]]>#1[[2]]&]
and this will have the data ordered by the y value. You can then take the first and last elements as you wish.
-
Using the pattern I described here, define
MaxBy[list_, fun_] := list[[First@Ordering[fun /@ list, -1]]]
MinBy[list_, fun_] := list[[First@Ordering[fun /@ list, 1]]]
then do
MinBy[data, Last] (* smallest y *)
MaxBy[data, Last] (* largest y *)
MinBy[data, Composition[Abs, Last] ] (* closest to 0 *)
- | 557 | 1,874 | {"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} | 3.359375 | 3 | CC-MAIN-2016-30 | latest | en | 0.838279 |
http://mathoverflow.net/feeds/question/118613 | 1,369,277,730,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702749808/warc/CC-MAIN-20130516111229-00094-ip-10-60-113-184.ec2.internal.warc.gz | 170,517,611 | 1,265 | LU decomosition for ill-condition matrixes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:55:35Z http://mathoverflow.net/feeds/question/118613 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118613/lu-decomosition-for-ill-condition-matrixes LU decomosition for ill-condition matrixes Mary 2013-01-11T10:19:34Z 2013-01-11T13:44:31Z <p>hi.. i have a matrix like this:</p> <pre><code> { 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 1/12, 1/13, 1/14, 1/15, 1/16, 1/17, 1/18, 1/19, 1/20, 1/21, 1/22, 1/23, 1/24, 1/25, }; </code></pre> <p>and i shoud calculate its LU decomposition. is it one of ill-condition matrixes? if so which one of dolittle,cholesky or crout solution is suitable for this matrix?Blockquote</p> | 318 | 787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.530316 |
http://betterlesson.com/lesson/resource/2912217/homework-regular-polygons-and-proofs | 1,488,145,452,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172077.66/warc/CC-MAIN-20170219104612-00611-ip-10-171-10-108.ec2.internal.warc.gz | 26,093,956 | 20,550 | ## Homework, regular polygons and proofs - Section 6: Handing Out the Homework
Homework, regular polygons and proofs
Homework, regular polygons and proofs
# Regular Polygons
Unit 5: Polygons and Congruent Triangle Proofs
Lesson 6 of 6
## Big Idea: The students work together, applying their knowledge of polygons to numerical problems involving regular polygons.
Print Lesson
Standards:
Subject(s):
Math, Geometry, exterior angle, regular polygons, polygons (Determining Measurements), Polygon, interior angles, equilateral, equiangular, triangle congruences
90 minutes
### Beth Menzie
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Big Idea: Students will show what they know!
HSG-SRT.A HSG-SRT.B.5
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Environment: Urban | 313 | 1,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | latest | en | 0.794293 |
https://www.gamedev.net/topic/641753-making-a-bone-point-to-a-target/ | 1,493,376,489,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00342-ip-10-145-167-34.ec2.internal.warc.gz | 904,853,015 | 35,378 | • FEATURED
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# Making a bone point to a target?
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.
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### #1Kaze Members
Posted 13 April 2013 - 04:33 PM
This is in unity but uses vectors and quaternions that should work the same in most libraries.
http://docs.unity3d.com/Documentation/ScriptReference/Quaternion.html
http://docs.unity3d.com/Documentation/ScriptReference/Vector3.html
I want to make a bone look at a target point but have two problems.
1: unity uses z+ forward and blender uses x-
2: the look at should avoid excessive twisting of the bone.
public class BoneLookAt : MonoBehaviour {
// Use this for initialization
void Start () {
//save rest data
restRot = this.transform.localRotation;
//filter input
Foward.Normalize();
FowardQuaternionInverse = Quaternion.Inverse( Quaternion.LookRotation(Foward));
}
public Vector3 Foward = new Vector3(-1,0,0);
private Quaternion FowardQuaternionInverse;
private Quaternion restRot;
public Vector3 GetTargetPosition(){
return this.transform.position + (this.transform.parent.rotation * restRot) * Foward;
}
public Quaternion LookAt(Vector3 v){
//,transform.InverseTransformDirection(Vector3.up)
return Quaternion.LookRotation(v-this.transform.position) * FowardQuaternionInverse;
}
private Vector3 ang = new Vector3();
void Update () {
Debug.DrawLine(this.transform.position,this.transform.position + this.transform.rotation * Foward);
//
this.transform.rotation = LookAt(GetTargetPosition());
}
}
Edited by Kaze, 13 April 2013 - 04:33 PM.
Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic. | 482 | 2,063 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-17 | latest | en | 0.653751 |
https://www.slideserve.com/aquila-townsend/interest-rates-and-bond-valuation | 1,591,030,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419056.73/warc/CC-MAIN-20200601145025-20200601175025-00178.warc.gz | 907,377,780 | 17,785 | # Interest Rates and Bond Valuation - PowerPoint PPT Presentation
Interest Rates and Bond Valuation
1 / 88
Interest Rates and Bond Valuation
## Interest Rates and Bond Valuation
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##### Presentation Transcript
1. Interest Rates and Bond Valuation Chapter 6
2. Key Concepts and Skills • Know the important bond features and bond types • Understand bond values and why they fluctuate • Understand bond ratings and what they mean • Understand the impact of inflation on interest rates • Understand the term structure of interest rates and the determinants of bond yields
3. Chapter Outline • Bonds and Bond Valuation • More on Bond Features • Bond Ratings • Some Different Types of Bonds • Bond Markets • Inflation and Interest Rates • Determinants of Bond Yields
4. Issuer (Seller) of Bonds (Borrower) = “Bond Issuer” • Bonds = Debt = Liability = Long-term debt • 1 Bond usually means the corporation borrows \$1000 (face value) • Corporations usually issue many Bonds • Bond Issue: • The total number of bonds that a corporation issues at the same time, in denominations of \$1,000 or \$5,000 each • Like any contractual debt: • Bond issuer pays periodic interest to the bondholder • Bond issuer pays the face value back at the end of the bond term
5. Issuer (Seller) of Bonds (Borrower) = “Bond Issuer” • When you issue a bond, you borrow the money, then use the money to buy assets that earn more cash than the cash you have to pay out to the bondholder • Leverage • Example: • Borrow money at 8% interest and buy a machine that earns the corporation 13% • The difference between 13% and 8%, or 5%, is left for the stockholders
6. Buyer of Bonds (Lender) = “Bondholder” • Bonds = Asset • 1 Bond usually means the borrow lends money to the corporation or government • Like any contractual debt: • Bondholders are paid periodic interest for loaning money to the corporation and are paid back the face value at the end of the bond term • When you buy a bond you are paying for a future steam of cash flow
7. Bond Vocabulary: • Face value = par value = loan repayment at maturity = FV • Annual interest payments = “annual coupon” • Coupon is from the days when you presented coupon to get paid interest • Annual interest rate (not discount rate) = coupon rate = annual interest rate for calculating interest payments = annual coupon/face value • Number of interest payments per year = n • Periodic rate (not discount rate) = periodic coupon rate = (annual coupon rate)/n • The book is inconsistent with the use of “coupon” (sometimes annual, sometimes semi-annual) • Periodic interest payment = periodic rate*face = PMT • Years until maturity = term of bond = years until paying back face value and last periodic interest payment = x • Maturity = specified date on which principal is repaid
8. Bond Vocabulary: • Yield To Maturity (YTM) = discount rate used to value bond = i = YTM • YTM = Bond Yield = Required Return = Market Rate = Rate required in the market on a bond • YTMs are quoted like APRs • YTM = (Period Discount Rate) * n • Example: YTM = 10% and bond pays semiannual interest payments, then period discount rate = YTM/n = .10/2 =.05 • Effective Annual Yield on the Bond = (1+.10/2)2 = .1025
9. Bonds • Bonds are interest only loans • Corporations/Governments borrow money, pay interest each period, then pay back face amount at end of bond term • Corporations/Governments plan to issue bonds and then set the coupon rate, but by the time they actually issue the bond the financial markets have already calculated a discount rate for the future values that is often different than the coupon rate • Corporations/Governments issue bonds and get the “cash in” (Bonds sold in primary market) • Many Bonds from Corporations/Governments can be traded in the financial markets (Bonds sold in the secondary market)
10. Bonds • Each bond has a price expressed as a percentage of the face value: • For example, 103 means 1.03 times the face value of the bond • When the corporation issues the \$1,000 face-value bond, it receives \$1,030 • At maturity, the corporation pays back only the face value of \$1,000 • 103 and 103% and 1.03 all convey the same meaning The bond is selling for 3% above the face value
11. Example 1 • On January 1, Cox Construction Corp. issues 750 10-year bonds with a face of \$1000 with a coupon rate of 9% at 103, with interest payable semiannually, on June 30 and December 31
12. But If The Loan Has A Face Value Of \$750,000, Why Did The Bondholder Pay \$772,500?
13. If a corporation offers a rate of interest that is higher than the market rate for similar securities, investors may be willing to pay a premium for the bond • If a corporation offers a rate of interest that is lower than the market rate for similar securities, investors will demand a discount on the bond
14. What Are Similar Securities? • Similar securities are bonds or other investment vehicles issue by other corporations (different than the one being considered) that have similar business and financial risks • The similarities could be: • Similar credit ratings • Similar business activities • Similar capital structure
15. Bond Prices YTM > Coupon Rate YTM < Coupon Rate
16. Definitions • Premium • The excess of the price received over the face value of a bond • YTM < Coupon Rate • “Bond sold at a premium” • Discount • The amount by which the issue price is less than the face value of a bond • YTM > Coupon Rate • “Bond sold at a discount”
17. The Issuance of Bonds at a Discount: Example 2 • On January 1, Muller, Inc., issues 700 6%, 20-year bonds with a face value of \$1,000, at 96, with interest to be paid semiannually, on June 30 and December 31
18. Bond Price (Valuation) from Cash Flow Perspective
19. Excel • Bond Valuation from Bondholder’s Point of View: • =PV(rate,nper,pmt,fv,type) • =PV(YTM/n,n*x,PMT,FV,0) • Bond Valuation from Bond Issuer’s Point of View: • =PV(rate,nper,pmt,fv,type) • =PV(YTM/n,n*x,-PMT,-FV,0) • Finding YTM rate from Bondholder’s Point of View: • =RATE(nper,pmt,pv,fv,type,guess) • =RATE(n*x,PMT,-PV,FV,0)
20. Example 3
21. Example 3
22. Example 3 CF Is From Point Of View Of Issuer
23. Bond Values And Why They Fluctuate • Bond Valuation: • As time passes, interest rates change in the market place • As new information about the company, the industry, the economy comes out, interest rates change • As time passes the amount of cash paid out to the bondholder does not change • Because of this the value of the bond will fluctuate • Rates , Bond Value • Rates , Bond Value
24. Graphical Relationship Between Price and YTM
25. Valuing a Discount Bond with Annual Coupons Payments (Example 4) • Consider a bond with a coupon rate of 10% and coupons paid annually. The par value is \$1000 and the bond has 5 years left until maturity. The yield to maturity is 11%. What is the value of the bond What is the price to you, buying in the secondary market? • Using the formula: • B = PV of annuity + PV of lump sum • B = 100[1 – 1/(1.11)5] / .11 + 1000 / (1.11)5 • B = 369.59 + 593.45 = 963.04 • Answer: “You would be willing to pay \$963.04 cash out (negative) for the future cash flows.” or said this way: “The bond with a 10% coupon is priced to yield 11% at \$963.04.”
26. Valuing a Premium Bond with Annual Coupons Payments (Example 5) • Suppose you are looking at a bond that has a 10% annual coupon rate and a face value of \$1000. There are 20 years to maturity and the yield to maturity is 8%. What is the value of the bond what is the price to you? • Using the formula: • B = PV of annuity + PV of lump sum • B = 100[1 – 1/(1.08)20] / .08 + 1000 / (1.08)20 • B = 981.81 + 214.55 = 1196.36 • Answer: “You would be willing to pay \$1196.36 cash out (negative) for the future cash flows.” or said this way: “The bond with a 10% coupon is priced to yield 8% at \$1196.36.”
27. Interest Rate Risk • The risk that arises for bond owners from fluctuating interest rates • How much interest rate risk a bond has depends on: • How sensitive its price is to interest rate change • The sensitivity depends on two things: • All things being equal, the longer the time to maturity, the greater the interest rate risk • All things being equal, the lower the coupon rate, the greater the interest rate risk
28. Interest Rate Risk And Time To Maturity
29. Interest Rate Risk To Loss Of Principal (current price) • Longer time to maturity • Small changes in market rate have substantial affect on bond value • Face value is discounted over many periods and thus compounding magnifies small interest rate changes • Lower Coupon rate • Bond with lower coupon rate is proportionally more dependent on the face value • (Bond with larger coupon rate has a larger cash flow early in life, so value less sensitive to discount rate)
30. Interest Rate Risk Increases At A Decreasing Rate
31. Computing YTM • Yield-to-maturity is the rate implied by the current bond price • Finding the YTM requires trial and error (iteration) if you do not have a financial calculator and is similar to the process for finding i with an annuity • If you have a financial calculator, enter N, PV, PMT and FV, remembering the sign convention (PMT and FV need to have the same sign, PV the opposite sign)
32. YTM with Annual Coupons (Example 6) Consider a bond with a 10% annual coupon rate, 15 years to maturity and a par value of \$1000. The current price is \$928.09 Will the yield be more or less than 10%?
33. YTM with Semiannual Coupons (Example 7) Suppose a bond with a 10% coupon rate and semiannual coupons, has a face value of \$1000, 20 years to maturity and is selling for \$1197.93 Is the YTM more or less than 10%? What is the semiannual coupon payment? How many periods are there?
34. Use One Bond YTM To Find Price Of Another Bond: Example 8 Below Market rate = discount Above = premium
35. Bonds and Stocks: • Like stock, bonds bring capital (money) into the corporation so that it can invest in profitable projects • Bondholders are creditors • They have a fixed claim to cash flow • Stockholders are owners • They have a residual claim to cash flow
36. Bonds and Stocks • Debt is not an ownership interest in the firm • Creditors do not have voting rights • Interest is tax deductible • Dividends are not tax deductible • Unpaid debt is a liability • Legal claim against assets • If debt is not paid creditors have the legal claim to assets before shareholders • One of the costs of issuing debt is the possibility that you will not be able to make interest payments creditors force firm into bankruptcy firm is terminated • This does not arise when equity is issued • Corporations try to create hybrid financial instruments that are Debt/Equity in order to have: • Tax benefits of debt • Bankruptcy benefits of equity
37. Debt Not an ownership interest Creditors do not have voting rights Interest is considered a cost of doing business and is tax deductible Creditors have legal recourse if interest or principal payments are missed Excess debt can lead to financial distress and bankruptcy Equity Ownership interest Common stockholders vote for the board of directors and other issues Dividends are not considered a cost of doing business and are not tax deductible Dividends are not a liability of the firm and stockholders have no legal recourse if dividends are not paid An all equity firm can not go bankrupt Differences Between Debt and Equity
38. Bond Terms and Types • Bonds = Long-term debt • Privately placed • Directly placed with the lender • Public-issue bonds • Offered to the public • Finance jargon” • Long-term debt = funded debt • Short-term debt = unfunded debt • Example: “A firm planning to fund its debt requirements may be replacing short-term debt with long-tern debt”
39. Bond Terms and Types • Trustee (Investment Bank, other…) • Appointed by the corporation to represent bondholders • Must make sure terms are obeyed • Must manage sinking fund • Must represent bondholders in default • Indenture • “The written agreement between the corporation and the lender detailing the terms of the debt issue”
40. Bond Types • Registered Form • The form of bond issue in which the registrar of the company records ownership of each bond • Payment is made directly to the owner of the bond • Bearer Form • The form of bond issue in which the bond is issued without record of the owner’s name • Payment is made to whoever holds the bond • Uncommon in the USA
41. Bond Types • Security: • Generic term that means Stocks or Bonds or other investment vehicles that are backed by an asset • A document indicating ownership or creditorship; a stock certificate or bond • Dictionary definition of Security: • Something deposited or given as assurance of the fulfillment of an obligation; a pledge; collateral
42. Bond Types • Debt securities are classified according to the collateral and mortgages used to protect the bondholder • Securities Backed By Collateral • Collateral = any asset pledged on the debt (often means assets such as stocks or bonds – financial assets) • If the borrower does not pay the interest and principal to the bondholder, the bondholder can take the collateral • Mortgage Securities • Debt secured by a mortgage on real assets (property, but not cash or inventory) of the borrower • Called: • Mortgage Trust Indenture, or Trust Deed • Most utility and railroad bonds are secured by a pledge of assets
43. Bond Types • Unsecured Debt • These creditors have a claim on property not otherwise pledged • Debenture • Unsecured debt (maturity >= 10 years) • Most financial and industrial companies’ public bonds are debentures • Note • Unsecured debt (maturity < 10 years) • Subordinate Debt • Must give preference to superior debt • Debt is not subordinate to equity
44. Bond Types • Sinking Fund • An account managed by the trustee for the purposed of repaying the bonds, or early bond redemption • Bond Issuer must put away some money each period to save up in order to pay off the bond • Protective Covenant • A part of the indenture limiting certain actions that might be taken in order to protect the lender • Negative (thou shalt not): • Example: Limit the amount of dividends paid • Positive (thou shalt): • Example: CA/CL must be greater than 1.5
45. Bond Types • Call Provision • An agreement giving the corporation the option to repurchase the bond at a specific price prior to maturity • Call Premium (Pay for the Option) • The amount by which the call price > par value • Example: Bond face = \$1,000, Call Price = \$1,100 • Call Price goes down over time • Deferred Call Provision • Can call only after a certain date • Call Protected Bond • Can’t be redeemed by issuer • Make-Whole Call • When bond called, bondholder gets PV of future cash flows at a reasonable rate • Derivative security jargon: • Call = Buy • Put = Sell
46. Financial Markets • Primary Markets • Original sale of equity or debt • Corporation issues security • Secondary Markets • After original sale of equity or debt • You sell/buy security • Dealer Markets (Over-the-counter markets (OTC)) • Dealers buy and sell for themselves • Most debt is sold this way • Example: NASDAQ • Auction Markets (Exchanges) • Brokers and agents match buyers and sellers • Most of the large firms’ equity is sold this way • Example: NYSE
47. Bond Characteristics and Required Returns • The coupon rate depends on the risk characteristics of the bond when issued • Which bonds will have the higher coupon, all else equal? • Secured debt versus a debenture • Subordinated debenture versus senior debt • A bond with a sinking fund versus one without • A callable bond versus a non-callable bond
48. Bond Ratings And What They Mean • Bond Rating firms: • Moody’s • Standard and Poor’s (S&P) • They rate: • The likelihood of default • The probability that creditors are protected • They ask they question: What is the risk associated with the firm issuing the debt? • They do not rate the probability of bond value change due to interest rate risk • The Debt Crisis of 2007 shows that Ratings can be less than accurate: • How do they take risky loans and repackage them to get a AAA “Super Senior” rating? (That’s what they did!) | 3,848 | 16,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.807474 |
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# Solutions 4. Heat - Additional Questions with Answers | Class 7 Science - Toppers Study
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## Solutions 4. Heat - Additional Questions with Answers | Class 7 Science - Toppers Study
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Chapter 4 Science class 7
### Additional Questions with Answers class 7 Science Chapter 4. Heat
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• Solutions 4. Heat - Additional Questions With Answers | Class 7 Science - Toppers Study
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## Solutions 4. Heat - Additional Questions with Answers | Class 7 Science - Toppers Study
Que. – What is thermometer?
Ans. – A device which is used for recording temperature is known as thermometer.
Que. – What is clinical thermometer?
Ans. - A device which is used for recording body temperature is known as clinical thermometer.
Que. – What is the temperature of normal body?
Ans. – 37 °C
Que. – Which thermometer do not use mercury?
Ans. – Digital thermometer.
Que. – What is the range of clinical thermometer?
Ans. – 37 °C to 42 °C.
Que. – What is laboratory thermometer?
Ans. – The temperature of some other things are measured by a special kind of thermometer. Which range is generally –10°C to 110°C is known as laboratory thermometer.
Que. –Why should we use antiseptic solution to wash out the thermometer?
Ans. – After using the thermometer it get infected or may be isolated with germs. So we use antiseptic solution to wash out the thermometer.
Que. – What is conduction?
Ans. – The process by which heat is transferred from the hotter end to the colder end of an object is known as conduction.
Que. – Why does a metallic pot get heated when we kept it on a flame?
Ans. – Due to conduction.
Que. – How does the sun’s heat reaches to us?
Ans. – The heat of the sun reaches us by radiation.
Que. – What is convection?
Ans. – The process of transferring of heat in liquid like water is known as convection.
Que. – Define conductor of heat? Write with examples.
Ans. – The materials which allow to pass the heat from themselves easily are known as conductor of heat. EX. – copper, Iron, almunium etc.
Que. – Define bad conductor of heat with examples.
Ans. – the materials which does not allow to pass the heat from themselves are known as bad conductor of heat. Ex – wood, plastics etc.
Que. – What is radiation?
Ans. – The process of transferring the heat which does not need any medium is known as radiation. Ex – reaching the heat of the sun to us.
Que. – What are the modes to transfer the heat?
Ans. – There are three modes
(i) Conduction
(ii) Convection
Que. – Why does it seem comfortable to put on deep coloured clothes in winter?
Ans. – The deep coloured surface usually absorbs more heat, so putting on deep coloured clothes in winter is comfortable.
Que. – Which colour absorbs more heat?
Ans. – Black.
Que. – Why should we put on the white dress in summer?
Ans. – Light colour or white dress reflects the most of the heat radiation and absorbs less heat so that it keeps us comfortable.
Que. – Why do woolen clothes keep us warm in winter?
Ans. - woolen clothes keep us warm in winter because the wool is bad conductor of heat. The airs trap among its fibers.
Que. – by which mode of transferring of heat will take place in copper?
Ans. – conduction.
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https://www.hipparchus.org/apidocs/org/hipparchus/stat/descriptive/rank/Percentile.html | 1,708,739,941,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00139.warc.gz | 832,164,107 | 9,354 | Class Percentile
• All Implemented Interfaces:
Serializable, UnivariateStatistic, MathArrays.Function
public class Percentile
extends AbstractUnivariateStatistic
implements Serializable
Provides percentile computation.
There are several commonly used methods for estimating percentiles (a.k.a. quantiles) based on sample data. For large samples, the different methods agree closely, but when sample sizes are small, different methods will give significantly different results. The algorithm implemented here works as follows:
1. Let n be the length of the (sorted) array and 0 < p <= 100 be the desired percentile.
2. If n = 1 return the unique array element (regardless of the value of p); otherwise
3. Compute the estimated percentile position pos = p * (n + 1) / 100 and the difference, d between pos and floor(pos) (i.e. the fractional part of pos).
4. If pos < 1 return the smallest element in the array.
5. Else if pos >= n return the largest element in the array.
6. Else let lower be the element in position floor(pos) in the array and let upper be the next element in the array. Return lower + d * (upper - lower)
To compute percentiles, the data must be at least partially ordered. Input arrays are copied and recursively partitioned using an ordering definition. The ordering used by Arrays.sort(double[]) is the one determined by Double.compareTo(Double). This ordering makes Double.NaN larger than any other value (including Double.POSITIVE_INFINITY). Therefore, for example, the median (50th percentile) of {0, 1, 2, 3, 4, Double.NaN} evaluates to 2.5.
Since percentile estimation usually involves interpolation between array elements, arrays containing NaN or infinite values will often result in NaN or infinite values returned.
Further, to include different estimation types such as R1, R2 as mentioned in Quantile page(wikipedia), a type specific NaN handling strategy is used to closely match with the typically observed results from popular tools like R(R1-R9), Excel(R7).
Percentile uses only selection instead of complete sorting and caches selection algorithm state between calls to the various evaluate methods. This greatly improves efficiency, both for a single percentile and multiple percentile computations. To maximize performance when multiple percentiles are computed based on the same data, users should set the data array once using either one of the evaluate(double[], double) or setData(double[]) methods and thereafter evaluate(double) with just the percentile provided.
Note that this implementation is not synchronized. If multiple threads access an instance of this class concurrently, and at least one of the threads invokes the increment() or clear() method, it must be synchronized externally.
Serialized Form
• Nested Class Summary
Nested Classes
Modifier and Type Class Description
static class Percentile.EstimationType
An enum for various estimation strategies of a percentile referred in wikipedia on quantile with the names of enum matching those of types mentioned in wikipedia.
• Constructor Summary
Constructors
Modifier Constructor Description
Percentile()
Constructs a Percentile with the following defaults.
Percentile(double quantile)
Constructs a Percentile with the specific quantile value and the following default method type: Percentile.EstimationType.LEGACY default NaN strategy: NaNStrategy.REMOVED a Kth Selector : KthSelector
protected Percentile(double quantile, Percentile.EstimationType estimationType, NaNStrategy nanStrategy, KthSelector kthSelector)
Constructs a Percentile with the specific quantile value, Percentile.EstimationType, NaNStrategy and KthSelector.
Percentile(Percentile original)
Copy constructor, creates a new Percentile identical to the original
• Method Summary
All Methods
Modifier and Type Method Description
Percentile copy()
Returns a copy of the statistic with the same internal state.
double evaluate(double p)
Returns the result of evaluating the statistic over the stored data.
double evaluate(double[] values, double p)
Returns an estimate of the pth percentile of the values in the values array.
double evaluate(double[] values, int start, int length)
Returns an estimate of the quantileth percentile of the designated values in the values array.
double evaluate(double[] values, int begin, int length, double p)
Returns an estimate of the pth percentile of the values in the values array, starting with the element in (0-based) position begin in the array and including length values.
Percentile.EstimationType getEstimationType()
Get the estimation type used for computation.
KthSelector getKthSelector()
Get the kthSelector used for computation.
NaNStrategy getNaNStrategy()
Get the NaN Handling strategy used for computation.
PivotingStrategy getPivotingStrategy()
Get the PivotingStrategy used in KthSelector for computation.
double getQuantile()
Returns the value of the quantile field (determines what percentile is computed when evaluate() is called with no quantile argument).
protected double[] getWorkArray(double[] values, int begin, int length)
Get the work array to operate.
void setData(double[] values)
Set the data array.
void setData(double[] values, int begin, int length)
Set the data array.
void setQuantile(double p)
Sets the value of the quantile field (determines what percentile is computed when evaluate() is called with no quantile argument).
Percentile withEstimationType(Percentile.EstimationType newEstimationType)
Build a new instance similar to the current one except for the estimation type.
Percentile withKthSelector(KthSelector newKthSelector)
Build a new instance similar to the current one except for the kthSelector instance specifically set.
Percentile withNaNStrategy(NaNStrategy newNaNStrategy)
Build a new instance similar to the current one except for the NaN handling strategy.
• Methods inherited from class org.hipparchus.stat.descriptive.AbstractUnivariateStatistic
evaluate, getData, getDataRef
• Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• Methods inherited from interface org.hipparchus.stat.descriptive.UnivariateStatistic
evaluate
• Method Detail
• setData
public void setData(double[] values)
Set the data array.
The stored value is a copy of the parameter array, not the array itself.
Overrides:
setData in class AbstractUnivariateStatistic
Parameters:
values - data array to store (may be null to remove stored data)
AbstractUnivariateStatistic.evaluate()
• setData
public void setData(double[] values,
int begin,
int length)
throws MathIllegalArgumentException
Set the data array. The input array is copied, not referenced.
Overrides:
setData in class AbstractUnivariateStatistic
Parameters:
values - data array to store
begin - the index of the first element to include
length - the number of elements to include
Throws:
MathIllegalArgumentException - if values is null or the indices are not valid
AbstractUnivariateStatistic.evaluate()
• evaluate
public double evaluate(double p)
throws MathIllegalArgumentException
Returns the result of evaluating the statistic over the stored data.
The stored array is the one which was set by previous calls to setData(double[])
Parameters:
p - the percentile value to compute
Returns:
the value of the statistic applied to the stored data
Throws:
MathIllegalArgumentException - if p is not a valid quantile value (p must be greater than 0 and less than or equal to 100)
• evaluate
public double evaluate(double[] values,
int start,
int length)
throws MathIllegalArgumentException
Returns an estimate of the quantileth percentile of the designated values in the values array.
The quantile estimated is determined by the quantile property.
• Returns Double.NaN if length = 0
• Returns (for any value of quantile) values[begin] if length = 1
• Throws MathIllegalArgumentException if values is null, or start or length is invalid
See Percentile for a description of the percentile estimation algorithm used.
Specified by:
evaluate in interface MathArrays.Function
Specified by:
evaluate in interface UnivariateStatistic
Specified by:
evaluate in class AbstractUnivariateStatistic
Parameters:
values - the input array
start - index of the first array element to include
length - the number of elements to include
Returns:
the percentile value
Throws:
MathIllegalArgumentException - if the parameters are not valid
• evaluate
public double evaluate(double[] values,
double p)
throws MathIllegalArgumentException
Returns an estimate of the pth percentile of the values in the values array.
• Returns Double.NaN if values has length 0
• Returns (for any value of p) values[0] if values has length 1
• Throws MathIllegalArgumentException if values is null or p is not a valid quantile value (p must be greater than 0 and less than or equal to 100)
The default implementation delegates to evaluate(double[], int, int, double) in the natural way.
Parameters:
values - input array of values
p - the percentile value to compute
Returns:
the percentile value or Double.NaN if the array is empty
Throws:
MathIllegalArgumentException - if values is null or p is invalid
• evaluate
public double evaluate(double[] values,
int begin,
int length,
double p)
throws MathIllegalArgumentException
Returns an estimate of the pth percentile of the values in the values array, starting with the element in (0-based) position begin in the array and including length values.
Calls to this method do not modify the internal quantile state of this statistic.
• Returns Double.NaN if length = 0
• Returns (for any value of p) values[begin] if length = 1
• Throws MathIllegalArgumentException if values is null , begin or length is invalid, or p is not a valid quantile value (p must be greater than 0 and less than or equal to 100)
See Percentile for a description of the percentile estimation algorithm used.
Parameters:
values - array of input values
p - the percentile to compute
begin - the first (0-based) element to include in the computation
length - the number of array elements to include
Returns:
the percentile value
Throws:
MathIllegalArgumentException - if the parameters are not valid or the input array is null
• getQuantile
public double getQuantile()
Returns the value of the quantile field (determines what percentile is computed when evaluate() is called with no quantile argument).
Returns:
quantile set while construction or setQuantile(double)
• setQuantile
public void setQuantile(double p)
throws MathIllegalArgumentException
Sets the value of the quantile field (determines what percentile is computed when evaluate() is called with no quantile argument).
Parameters:
p - a value between 0 < p <= 100
Throws:
MathIllegalArgumentException - if p is not greater than 0 and less than or equal to 100
• copy
public Percentile copy()
Returns a copy of the statistic with the same internal state.
Specified by:
copy in interface UnivariateStatistic
Specified by:
copy in class AbstractUnivariateStatistic
Returns:
a copy of the statistic
• getWorkArray
protected double[] getWorkArray(double[] values,
int begin,
int length)
Get the work array to operate. Makes use of prior storedData if it exists or else do a check on NaNs and copy a subset of the array defined by begin and length parameters. The set nanStrategy will be used to either retain/remove/replace any NaNs present before returning the resultant array.
Parameters:
values - the array of numbers
begin - index to start reading the array
length - the length of array to be read from the begin index
Returns:
work array sliced from values in the range [begin,begin+length)
Throws:
MathIllegalArgumentException - if values or indices are invalid
• getEstimationType
public Percentile.EstimationType getEstimationType()
Get the estimation type used for computation.
Returns:
the estimationType set
• withEstimationType
public Percentile withEstimationType(Percentile.EstimationType newEstimationType)
Build a new instance similar to the current one except for the estimation type.
This method is intended to be used as part of a fluent-type builder pattern. Building finely tune instances should be done as follows:
Percentile customized = new Percentile(quantile).
withEstimationType(estimationType).
withNaNStrategy(nanStrategy).
withKthSelector(kthSelector);
If any of the withXxx method is omitted, the default value for the corresponding customization parameter will be used.
Parameters:
newEstimationType - estimation type for the new instance
Returns:
a new instance, with changed estimation type
Throws:
NullArgumentException - when newEstimationType is null
• getNaNStrategy
public NaNStrategy getNaNStrategy()
Get the NaN Handling strategy used for computation.
Returns:
NaN Handling strategy set during construction
• withNaNStrategy
public Percentile withNaNStrategy(NaNStrategy newNaNStrategy)
Build a new instance similar to the current one except for the NaN handling strategy.
This method is intended to be used as part of a fluent-type builder pattern. Building finely tune instances should be done as follows:
Percentile customized = new Percentile(quantile).
withEstimationType(estimationType).
withNaNStrategy(nanStrategy).
withKthSelector(kthSelector);
If any of the withXxx method is omitted, the default value for the corresponding customization parameter will be used.
Parameters:
newNaNStrategy - NaN strategy for the new instance
Returns:
a new instance, with changed NaN handling strategy
Throws:
NullArgumentException - when newNaNStrategy is null
• getKthSelector
public KthSelector getKthSelector()
Get the kthSelector used for computation.
Returns:
the kthSelector set
• getPivotingStrategy
public PivotingStrategy getPivotingStrategy()
Get the PivotingStrategy used in KthSelector for computation.
Returns:
the pivoting strategy set
• withKthSelector
public Percentile withKthSelector(KthSelector newKthSelector)
Build a new instance similar to the current one except for the kthSelector instance specifically set.
This method is intended to be used as part of a fluent-type builder pattern. Building finely tune instances should be done as follows:
Percentile customized = new Percentile(quantile).
withEstimationType(estimationType).
withNaNStrategy(nanStrategy).
withKthSelector(newKthSelector);
If any of the withXxx method is omitted, the default value for the corresponding customization parameter will be used.
Parameters:
newKthSelector - KthSelector for the new instance
Returns:
a new instance, with changed KthSelector
Throws:
NullArgumentException - when newKthSelector is null | 3,209 | 14,670 | {"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} | 3.140625 | 3 | CC-MAIN-2024-10 | longest | en | 0.810121 |
https://www.lottolotto.net/statistics/lotofacil/even-odds/ | 1,670,581,350,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00749.warc.gz | 900,326,919 | 7,995 | Powerball | Results for Dec 05 | 35 45 47 54 55 (US\$ 0.00)
Statistics for Lotofácil
Statistics of real draws
Lotofácil (until the draw 2683 on Dec 8, 2022)
# Distribution of even and odd numbers
## Occurrences of even and odd numbers in draws
Choose the category
Quantity of even numbers
Quantity of even numbers drawn
Actual frequency
Expected frequency
3
3
2
4
28
31
5
188
185
6
560
542
7
833
836
8
683
697
9
306
309
10
72
69
11
10
7
Quantity of even numbers drawn Actual occurrences Expected occurrences Last draw Shortest interval Largest interval Current interval Average interval 3 3 2 1244 6 960 1439 414.67 4 28 31 2670 8 337 13 95.36 5 188 185 2659 1 65 24 14.14 6 560 542 2680 1 28 3 4.79 7 833 836 2676 1 20 7 3.21 8 683 697 2683 1 19 0 3.93 9 306 309 2679 1 51 4 8.75 10 72 69 2681 1 184 2 37.24 11 10 7 2595 54 711 88 259.50
• The table shows data on the distribution of even and odd numbers on draws, considering all draws of lotofácil (with the current matrix).
• Actual occurrences are the real total occurrences of a determined quantity of even/odd numbers on draws.
• Expected occurrences are the expected occurrences of a determined quantity of even/odd numbers on draws, according to mathematical probability.
• Last draw is the most recent draw in which occurred a determined quantity of even/odd numbers.
• Shortest interval is the shortest gap between draws in which occurred a determined quantity of even/odd numbers.
• Longest interval is the longest gap between draws in which occurred a determined quantity of even/odd numbers.
• Current interval is the current interval since the last draw in which occurred a determined quantity of even/odd numbers.
• Average interval is the general average of intervals between draws in which occurred a determined quantity of even/odd numbers (until the last draw in which occurred a determined quantity of even/odd numbers).
• The actual and expected occurrences tend to get closer to each other the largest the sample.
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Manage | 624 | 2,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.849641 |
http://math.tutorpace.com/algebra/learning-algebra-1-online-tutoring | 1,511,292,492,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806422.29/warc/CC-MAIN-20171121185236-20171121205236-00037.warc.gz | 195,186,038 | 8,825 | Learning Algebra 1
Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Learning Algebra 1 includes the subsequent topics like addition and subtraction of algebraic expression, multiplication, and division of polynomials, resolution polynomials, graphing linear and quadratic relations and functions, exponents and irrational numbers, how to solve equations, change expressions and resolution of worldly issues. Students learn to study patterns and relationships, formalizing their data, learning to use symbolic notation. When students create the transition from concrete arithmetic to the symbolic language of algebra one, they develop abstract reasoning skills necessary to surpass in mathematics and science.
Example 1:
Subtract 7p – 2q – 5r from the sum of 5p + 2q - 3r + 1 and 3p – 4r – 3
Solution:
5p + 2q - 3r + 1
3p – 4r – 3
7p – 2q – 5r [we need to change the sign of each term of the third expression as it is to be subtracted and then add]
_ + +
p + 4q – 2r – 2
Example 2:
The sum of the digits of a two digit number is 12. If 18 are added to it, the digits are reversed so the number will be ___ .
Solution:
Let the unit digit be x
The sum of both the digit is 12, the tens digit = 12 – x
So the number = (12 – x ) * 10 + x
On reversing the digits, x is at tens; (12 – x) at unit.
New number = x *10 + (12 – x)
x *10 + (12 – x) = (12 – x ) * 10 + x + 18
9x + 12 = - 9x + 138
18x = 126
x = 7
The number is 57. | 438 | 1,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2017-47 | latest | en | 0.846435 |
https://dougo.info/residual-income-investments-easy-passive-income.html | 1,542,550,531,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744381.73/warc/CC-MAIN-20181118135147-20181118161147-00269.warc.gz | 601,519,999 | 7,489 | I actually spent a year and a half working as an affiliate marketer (mostly selling drumming related products – lessons, kits ect). 5 years on and one of my one page sites (which I’ve not touched) still nets me about \$150 a month. I won’t be retiring off that but only really now appreciate the reverse pyramid approach to entrepreneurship (working for nothing initially but later being paid without effort!)
## You can select any of the above-mentioned, based on your interest, skill, and capability to generate a second income source. However, these are just to name a few, there exist multiple ways to generate a secondary income channel. You just need to identify the right one, which suits you the best. Remember there is no shortcut to success and you need to work hard to be successful and rich in the long run!
You've probably read blog posts and articles that recommend a certain brand of backpack or water, so you click on their hyperlinked link. Oftentimes, that person gets paid a commission when you do. If you have a blog, the same can happen for you. It's a win-win-win for everyone involved — you, the product you're recommending, and the person who clicks on the link to get the product. Pat Flynn talks about this at length on his website, Smart Passive Income, where you can learn a whole lot more on the topic, aside from affiliate marketing.
4. Calculate how much passive income you need. It's important to have a passive-income goal — otherwise, it's very easy to lose motivation. A good goal is to try to generate enough passive income to cover basic living expenses such as food, shelter, transportation, and clothing. If your annual expense number is \$30,000, divide that figure by your expected rate of return to see how much capital you need to save. Unfortunately, you've got to then multiply the capital amount by 1.25 to 1.5 to account for taxes.
We are going to start with 1.5 years of all spending needs in cash. We will draw 1800 to 1900 per month. We will add to this from the index funds by taking a portion of the gains in good years to supplement. This is the total return portion of the equation. Obviously, if stocks decrease drastically over a 5 year period, then I would have to reload by selling some of the ETF holdings.
Note: This page contains one or more references to the Internal Revenue Code (IRC), Treasury Regulations, court cases, or other official tax guidance. References to these legal authorities are included for the convenience of those who would like to read the technical reference material. To access the applicable IRC sections, Treasury Regulations, or other official tax guidance, visit the Tax Code, Regulations, and Official Guidance page. To access any Tax Court case opinions issued after September 24, 1995, visit the Opinions Search page of the United States Tax Court.
However, you should pick a niche and blog about that. If you're launching a money related blog, maybe it'll be about how to make money in real estate or simply how to make money online. Pick the niche and stick to it. If it's a diet and fitness related blog, maybe the niche is the Ketogenic diet, the Atkins diet or some other form of diet or fitness.
I live in NYC where I never thought buying rental property would be possible, but am looking into buying rental property in the Midwest where it cash flows and have someone manage it for me (turnkey real estate investing I guess some would call it). I agree with what Mike said about leverage and tax advantages, but I’m still a newbie to real estate investing so I can’t so how it will go. I have a very small amount in P2P…I’m at around 6.3% It’s okay but I don’t know how liquid it is and it still is relatively new…I’d prefer investing in the stock market.
P2P lending is the practice of loaning money to borrowers who typically don’t qualify for traditional loans. As the lender you have the ability to choose the borrowers and are able to spread your investment amount out to mitigate your risk. The most popular peer to peer lending platform is Lending Club. You can read our full lending club review here: Lending Club Review.
It is always fun (when things are going well!) to look back at the various streams to see what’s working and what’s not. I found that a lot of my angel investing just wasn’t working well, fortunately it wasn’t a lot! Side businesses are always nice, vs. pure investments, because of actual control. Plus you can shut it down if things go south… hard to tell someone (and convince them when you’ve only kicked in a few bucks) that it’s time to close up shop and return some capital.
Who doesn’t like some down and dirty affiliate fees?! Especially if you realize it can be even easier to make money this way than with an ebook. After all, you simply need to concentrate on pumping out some content for your own site and getting the traffic in, often via Google or social media. Unsurprisingly, most people can enjoy their first affiliate sale within 30 days of starting a blog. Continue reading >
Now I’ve been using Swagbucks for a while and have found the money works out to just under \$2 an hour so this isn’t something that’s going to make you rich. You’d have to work 2,500 hours to make \$5,000 so that’s about three and a half months, non-stop. The thing with Swagbucks though is you can do it when you’re doing something else so I flip through surveys and other stuff while I’m cooking dinner or flipping channels.
Further north, the Saurashtra and Bengal coasts played an important role in maritime trade, and the Gangetic plains and the Indus valley housed several centres of river-borne commerce. Most overland trade was carried out via the Khyber Pass connecting the Punjab region with Afghanistan and onward to the Middle East and Central Asia.[69] Although many kingdoms and rulers issued coins, barter was prevalent. Villages paid a portion of their agricultural produce as revenue to the rulers, while their craftsmen received a part of the crops at harvest time for their services.[70]
No one wants to plan for a loss but it is always better to be prepared. US have recently seen a real estate meltdown that has caused the entire economies of the world to go into recession. So, it is very important to foresee such meltdowns and be prepared for losses. A second income source will ensure that you have enough rotational money to see you through losses.
The Country Partnership Framework (CPF) FY18-FY22 builds on the progress achieved by Ethiopia during the past five years. The CPF was developed after intensive consultations with a wide range of stakeholders to gain a broad-based perspective on the WBG’s performance and development priorities. The CPF is a result-based strategy, firmly anchored in the government’s Second Growth and Transformation Plan (GTPII).
I’m hoping to have about 10g saved by this time next year, which I know is nothing huge but seeing as I’m at 2.5g right now and owned 3 dollars to my name on Aug.9 I’m pretty happy with my progress :). But at my age, without a stable career, while working part time and having to go to school full time, what is a realistic path I could pursue to create passive income online, or even income that requires effort such as writing, but one that is more flexible than working in a stationary low-paid position for 10 dollars an hour? I need to work for now to show taxable income for the government to get my residency, but after that I know my time could be better served than earning 8 dollars an hour, I’m just not sure where to go from here. I considered flipping domain names, or penny stocks, or sports gambling, but again that’s not passive income and in reality they are more or less just forms of me gambling.
6) Always Remember That Everything Is Relative. The best way to determine worthwhile passive income streams is by comparing the likely return (IRR) with the current risk-free rate of return. If I round up, the 10 year bond yield is at 3%. Any new venture should thoroughly beat 3% otherwise you are wasting your efforts since you can earn 3% doing nothing.
My reasons for diversifying income are simple: I want to be able to quit my day job eventually. But your reasons may be different, maybe your job isn’t that secure or your co-workers are starting to feel the pressure. It really doesn’t matter why you diversify your sources of income, what matters is that you do it. Making money won’t happen overnight with second sources of income so if you wait until it’s too late you’ll be screwed.
The best part is that you keep 97% of the fees paid and you don’t have to search for clients yourself . Serve them as they come. If you have a full house at some great tourist place like Goa/Shimla/Manali or those kind of places, you can put your whole house on rental basis. Many people who have a second home or extra room, hire a maid and offer the full range of services of a regular basis. Imagine if your extra room is rented even 5 times a month and you earn Rs 1,000 from it ? Its Rs 5,000 extra income !
I’ve been into home décor lately and I had to turn to Etsy to find exactly what I wanted. I ended up purchasing digital files of the artwork I wanted printed out! The seller had made a bunch of wall art, digitized, and listed it on Etsy for instant download. There are other popular digital files on Etsy as well such as monthly planners. If you’re into graphic design this could be an amazing passive income idea for you.
Not only that but in almost all other cases there is the illusion of influence, which is itself a psychological and emotional cost. If you invest in a business that your friend or family member is running, you can see how things can get messy. You have thoughts on how things should be done, they have competing thoughts, if things aren't going well… we know how this story goes.
India has the largest diaspora around the world, an estimated 16 million people,[363] many of whom work overseas and remit funds back to their families. The Middle East region is the largest source of employment of expat Indians. The crude oil production and infrastructure industry of Saudi Arabia employs over 2 million expat Indians. Cities such as Dubai and Abu Dhabi in United Arab Emirates have employed another 2 million Indians during the construction boom in recent decades.[364] In 2009–10, remittances from Indian migrants overseas stood at ₹2,500 billion (US\$35 billion), the highest in the world, but their share in FDI remained low at around 1%.[365]
The bottom line is, it’s smart to have multiple income streams no matter who you are. Why? Because the more ways you can earn money without compromising your integrity, the better off you’ll be. And if you’re self-employed, having multiple income streams is almost essential. Not only will you enjoy a higher income, but you won’t go broke if one stream ends out of the blue.
It’s obvious that stocks outperform real estate in terms of capital gains, but I would like to see S&P compare to Real Estate in SF, Manhattan, LA. Our house in NC was \$80,000 20 years ago. It’s only \$150,000 now. Same house in Santa Monica went from \$200,000 to \$1.8 million. People who happen to bought real estate in major metropolitan would have a natural positive association with real estate investment.
If you want to really start tracking your finances, and I mean not just your spending but your investing (that's where wealth is built), give Personal Capital a look. They will give you a \$20 Amazon gift card if you link up an investment account that has \$1,000+. No strings. It's a cornerstone of my financial system and I think you owe yourself a look. 100% free too. | 2,527 | 11,677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.963507 |
https://www.meritnation.com/ask-answer/question/m2-mn-n2-what-should-be-added-to-make-it-a-perfect-squarehow/academics/1280717 | 1,642,473,707,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00072.warc.gz | 936,317,804 | 8,104 | # m2-mn+n2 what should be added to make it a perfect squarehow to find this?
Hi!
The given expression is m2 – mn + n2.
We know that, m2 2mn + n2 = (m – n)2
(m2 2mn + n2 ) (m2 – mn + n2 ) = –mn
Thus, –mn should be added to m2 – mn + n2 in order to make it a perfect square.
Cheers!
• 0
Remember the identity: (a-b)2 = a2 - 2ab + b2 .
In the quesiton m2 - mn + n2 . We need to make the middle term as 2mn, so we will add -mn to m2 - mn + n to make it a perfect square.
m2 - mn + n2-mn = m2 - 2mn + n2 = (m-n)2. | 199 | 515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.843279 |
https://www.coolstuffshub.com/weight/convert/short-tons-(us)-to-uk-teaspoons/ | 1,685,679,501,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00621.warc.gz | 749,160,687 | 12,858 | # Convert Short tons (US) to Uk teaspoons (US t to UK tsp Conversion)
1 short ton (US) is equal to 255963.64 UK teaspoons.
1 US t = 255963.64 UK tsp
## How to convert short tons (US) to UK teaspoons?
To convert short tons (US) to UK teaspoons, multiply the value in short tons (US) by 255963.64.
You can use the conversion formula :
UK teaspoons = short tons (US) × 255963.64
To calculate, you can also use our short tons (US) to UK teaspoons converter, which is a much faster and easier option as compared to calculating manually.
## How many UK teaspoons are in a short ton (US)?
There are 255963.64 UK teaspoons in a short ton (US).
• 1 short ton (US) = 255963.64 UK teaspoons
• 2 short tons (US) = 511927.28 UK teaspoons
• 3 short tons (US) = 767890.92 UK teaspoons
• 4 short tons (US) = 1023854.56 UK teaspoons
• 5 short tons (US) = 1279818.2 UK teaspoons
• 10 short tons (US) = 2559636.4 UK teaspoons
• 100 short tons (US) = 25596364 UK teaspoons
## Examples to convert US t to UK tsp
Example 1:
Convert 50 US t to UK tsp.
Solution:
Converting from short tons (us) to uk teaspoons is very easy.
We know that 1 US t = 255963.64 UK tsp.
So, to convert 50 US t to UK tsp, multiply 50 US t by 255963.64 UK tsp.
50 US t = 50 × 255963.64 UK tsp
50 US t = 12798182 UK tsp
Therefore, 50 short tons (US) converted to UK teaspoons is equal to 12798182 UK tsp.
Example 2:
Convert 125 US t to UK tsp.
Solution:
1 US t = 255963.64 UK tsp
So, 125 US t = 125 × 255963.64 UK tsp
125 US t = 31995455 UK tsp
Therefore, 125 US t converted to UK tsp is equal to 31995455 UK tsp.
For faster calculations, you can simply use our US t to UK tsp converter.
## Short tons (US) to UK teaspoons conversion table
Short tons (US) UK teaspoons
0.001 US t 255.96364 UK tsp
0.01 US t 2559.6364 UK tsp
0.1 US t 25596.364 UK tsp
1 US t 255963.64 UK tsp
2 US t 511927.28 UK tsp
3 US t 767890.92 UK tsp
4 US t 1023854.56 UK tsp
5 US t 1279818.2 UK tsp
6 US t 1535781.84 UK tsp
7 US t 1791745.48 UK tsp
8 US t 2047709.12 UK tsp
9 US t 2303672.76 UK tsp
10 US t 2559636.4 UK tsp
20 US t 5119272.8 UK tsp
30 US t 7678909.2 UK tsp
40 US t 10238545.6 UK tsp
50 US t 12798182 UK tsp
60 US t 15357818.4 UK tsp
70 US t 17917454.8 UK tsp
80 US t 20477091.2 UK tsp
90 US t 23036727.6 UK tsp
100 US t 25596364 UK tsp | 795 | 2,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-23 | latest | en | 0.78859 |
https://www.math.purdue.edu/~dvb/graph/elliptic.html | 1,516,744,450,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892699.72/warc/CC-MAIN-20180123211127-20180123231127-00154.warc.gz | 899,819,562 | 4,119 | #### An elliptic curve
Now we turn to cubic curves. The study of these was started by Isaac Newton, but the subject didn't really flourish until somewhat later after the advent of complex analysis. Typical cubics are referred to as elliptic curves, which is bit confusing since they are not actually ellipses. The reason for this name comes from its connection to elliptic integrals and functions, where an integral is called elliptic if the integrand contains a square root of cubic polynomial. Below are a few references in addition to the ones given earlier. A classical reference -- in spite of the name -- is Whittaker and Watson [3]. The first two books are more modern.
1. Husemoeller, Elliptic curves,
2. Silverman, The arithmetic of elliptic curves,
3. Whittaker and Watson, A course in modern analysis.
When x and y are treated as real variables, the graph of the above equation looks like:
After adding a point at infinity to the curve on the right, we get two circles topologically. Now let us treat the variables x and y are treated as complex. (In fact, historically signifcant progress in the study of elliptic integrals was made only after the introduction of complex analysis in the 19th century.) Now the above equation defines a complex elliptic curve which topologically is just a torus (after adding a point at infinity). The two real circles are marked in red here and below. We will also keep track of transevrse circle in yellow. It may be helpful to think of this as a "purely imaginary" curve, even though this isn't quite accurate.
The standard way to see this is by using elliptic functions. These are doubly periodic functions with the mildest possible (i.e. inessential) singularities. A function on the complex plane is doubly periodic if its graph repeats itself in both the horizontal and vertical directions. More precisely, we should be able to tile the plane into equally sized parallelograms (called period parallelograms), such that the piece of the graph over each tile is the same. The simplest nonzero example ( in spite of the complicated formula), with period parallelogram having corners at 0, α12 and α1 + α2, is the Weierstrass -function
Just as we used trignometric functions to parameterize the circle, we can use elliptic functions to parameterize elliptic curves. The above curve is given by the parameterization
for t C , where the periods are expressed by the elliptic integrals
(See [Whittaker-Watson, p. 444]). Since these parameterizing functions are doubly periodic, the elliptic curve can be identified with a period parallelogram (in fact a square in this case) with the sides glued together i.e. a torus. Note that the corners of the parallelogram get identified to the same point "at infinity" on the torus.
At this point we can ask whether there exists a rational parameterization of an elliptic curve. The answer is NO. Suppose we could. Then it is possible to show that this extends to a finite to one map of the Riemann sphere (C ∪∞) onto a torus. But this is impossible. The justification of this last statement requires a bit of topology. A version of the Jordan curve theorem says that any closed curve on the sphere would seperate it into two parts (an interior and exterior). If we could find a map as above, we could conclude the Jordan curve holds for the torus: lift the curve up to the sphere and then map the interior/exterior back down. However, it is clear that this isn't true. Just look at the yellow or red curves above.
Even though, we now know the elliptic curve abstractly, we really want understand the way it is embedded into C2. As above, we can get a sense of it projecting to 3 real dimensions. The graph of
represents a projection of the ellptic curve down to R3. For the domain of t, we use a period parallegram. We've colour coded it, and will use the same colour scheme on all the remaining graphs. Also we've marked certain horizontal and vertical lines in red and yellow. The last couple of lines are bisectors. These lines clearly map to the above curves on the torus. Furthermore, these red lines really do map to the real curves because and its derivative can be shown to take real values along them [Whittaker-Watson p. 444].
With this preparation, we can now plot the graph. To get a better sense of it, we can generate an animation where it rotates about the x3-axis:
Click on picture to start animation.
The colours of the domain and the graph match. The red curves are the real part of the graph considered at the beginning. Note the local topology near the branch points (- ,0,0) ... is same as in the first example y2 = x. In particular, the singularities in the graph are artifacts of the projection. The elliptic curve has none: it's just a torus as we saw. For example, the yellow curve which is really a loop on the torus gets crushed down to a line in the projection. To get more insight we can use a different projection. Or better yet, use a family of orthogonal projections:
The original projection corresponds to θ = 0. As θ oscillates between 0 and π∕2, we we see the yellow curve open up into a loop and collapse, while the red curves do the opposite.
Click on picture to start animation.
For the final graph, we give a top down view with θ = π∕5. | 1,183 | 5,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-05 | latest | en | 0.954244 |
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https://worldbuilding.stackexchange.com/questions/169843/how-to-calculate-the-furthest-an-animal-could-fly | 1,722,645,748,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640353668.0/warc/CC-MAIN-20240802234508-20240803024508-00635.warc.gz | 495,061,678 | 39,823 | # How to calculate the furthest an animal could fly?
In my story several flying mythical creatures have been brought in a more realistic world. some creatures like Gargoyles can only fly for short distances while others like the heavenpiercer (a type of wyvern) which spends most of it's adult life high above the clouds. and i'm curious to know if there is an equation i could use to calculate the furthest one of these animals could fly by plugging in there length, wingspan, wingwidth, ect. or use to find the dimensions needed for a creature to fly a certain length?
• I doubt it's quite as simple as that. Take a European swift. Aerodynamic efficiency plays a big part, the apparent lack of need to land to sleep plays another part. 10 months on the wing is no mean feat for a bird about the size of a swallow (swallows are useless in comparison). Too many factors for a single simple equation. Commented Feb 28, 2020 at 2:09
• the furthest an animal could fly? as far as it is fed.
– L.Dutch
Commented Feb 28, 2020 at 2:10
• Simple; just use the average flight speed and the maximum time the bird is able to fly without needing to stop and there you get it. :grin: Commented Feb 28, 2020 at 5:04
• Even actual existing animals don't have easy equations to calculate this. Just make it up. It'll seem more BS if you claim to have some kind of math behind it. Commented Feb 28, 2020 at 5:15
• There clearly isn't an equation for this... why do you want it? I bet somebody could come up with a good enough equation for, like, a roleplaying game (although it will probably have some fudge factors). Commented Feb 28, 2020 at 7:15
You can't. Flight endurance depends on not just the animal, but also the environment and circumstances. Even if you had an equation for the animal, real-world performance would still depend on environment and circumstances.
But, you can look at different bird morphologies to get a sense of what makes some birds fliers and other birds flitters.
The Albatross is famously adapted to long-duration flight, and can essentially stay in the air indefinitely. Pretty much the only thing they need to come back to land for is mating and hatching chicks.
https://www.smithsonianmag.com/science-nature/the-amazing-albatrosses-162515529/
Meanwhile, the turkey vulture is a scavenger. Because they never know when and where their next meal will be, they're capable of eating so much that they cannot fly. (Many other birds are also capable of eating so much that it affects their flying ability.)
https://www.yellowstonewildlifesanctuary.org/turkey-vulture
Other birds use wind patterns and columns of hot air called thermals to fly for long periods without flapping and expending almost no energy.
https://web.stanford.edu/group/stanfordbirds/text/essays/Soaring.html
There is no standard formula, but the basic principle is that birds have a certain store of energy, and exertion drains that store. At the same time, metabolic activity can replenish that energy even while it is being drained. If they use too much then they are exhausted and they have to stop until they have rested.
You can fly longer if you are more efficient (use less energy to fly, such as in the Soaring link above), or if you improve your ability to generate energy by increasing heartrate and lung capacity.
You can see this in people too. Any healthy and fit person can go out and run a few miles. An overweight person has to stop sooner and walk because they need more energy to run, and after a while they can run a little more. Marathon runners and ultramarathonners have trained to the point where they have boosted their energy-generation ability so they essentially run indefinitely.
For one more morphology, falcons are not particularly high-endurance birds, but they are very fast. The peregrine falcon can reach speeds over 200 miles per hour (320 km/h).
https://en.wikipedia.org/wiki/Peregrine_falcon#Feeding
• Though the falcon's speed is arguably not flying, but diving. Commented Feb 28, 2020 at 7:20
• -1 for not elaborating on the ability to carry coconuts in flight coconuts, but +2 for the extensive, in-depth answer. Commented Feb 28, 2020 at 11:18
• @Renan They could grip it by the husk Commented Feb 28, 2020 at 14:12
• The albatross might not be landing on land, but it is landing on the sea at regular intervals to catch prey. If it doesn't, it will starve to death! :-) Commented Feb 28, 2020 at 16:06 | 1,061 | 4,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-33 | latest | en | 0.962858 |
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Published by McDougal Littell
# Chapter 11 Data Analysis and Statistics - 11.4 Select and Draw Conclusions from Samples - 11.4 Exercises - Problem Solving - Page 771: 30a
787
#### Work Step by Step
The number of students surveyed is: $$\frac{181}{.23}=787$$
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10-21.
Write the equation of the curve whose derivative at every point $(x, y)$ is $\frac { 3 x ^ { 2 } } { 2 y }$ and passes through the point $(2, 1)$. Homework Help ✎
$2ydy=3x^2dx$
$\int2ydy=\int3x^2dx$
$y^2=x^3+C$
Use the given point to solve for $C$. | 123 | 319 | {"found_math": true, "script_math_tex": 7, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-51 | longest | en | 0.674713 |
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# %d,%u,%c
P: n/a Hello all: Can I think of %d, %u, %c as decoders? that is %d use 2's complement to decode whatever is stored in the variable. %u uses straight forward binary to decimal decoder. %c uses acii decoder. The question is how does %d, %u know how many bytes to decode? consider int i[2]={1,2}; int *p = &i[0]; if you wanna print the value of *p to the screen, using %d or %u, how do they know how many bytes to decode? Oct 2 '06 #1
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P: n/a iamchiawei...@gmail.com wrote: Hello all: Can I think of %d, %u, %c as decoders? that is %d use 2's complement to decode whatever is stored in the variable. %u uses straight forward binary to decimal decoder. %c uses acii decoder. The question is how does %d, %u know how many bytes to decode? consider int i[2]={1,2}; int *p = &i[0]; if you wanna print the value of *p to the screen, using %d or %u, how do they know how many bytes to decode? Good question. When the compiler sees the "printf" function call, it's not supposed to know that the name "printf" is special. So it hasw o choice but to just assume it's some regular old function. Well, taht's not exactly right, it USED to be that way, but somewhere along the line a prototype for printf() snuck in and the compiler knows its a function with a variable number of arguments after the first one. So poor old C compiler does what it does with any function's arguments, marshalls and converts them as is standard. Basically your chars and shorts, and ints get passed as chars and ints and ints. Longs get passed as longs. Floats get promoted to double. The printf function "knows" all this, so when it sees "%d" it knows to peek into the argument list for a int-sized thingy, for "%f" it knows to peek for a double sized thingy, and so on. Now in MOST languages the file I/O commands ARE known to the compiler, so it's a bit better able to match up format specs with actual parameters. Not perfect, but better. One could argue all day whether this is a good thing or not (its both IMHO). In C, the compiler has to have blind faith that you have carefully and exactly matched up % format specs with corresponding variables. If you're just a bit off, printf() will be grabbing for ints where the parameter list has longs, or worse, generating tons of garbage output and maybe a seg-fault or two. That's just the way things are in C-land, it's a rite of passage. Oct 2 '06 #2
P: n/a In article <11**********************@k70g2000cwa.googlegroups .com>, Can I think of %d, %u, %c as decoders? that is %d use 2's complementto decode whatever isstored in the variable. %u uses straight forward binary to decimaldecoder. %c uses acii decoder. Hmmmm. You have not taken into the account the possibility that integral values might be stored in one's complement or signed magnitude (C99) or any other representation such as trinary (C89). Similarily, in C89 %u is not necessarily -binary- to decimal. In both C89 and C99, %c is never an ASCII convertor: it is just a byte-level writer. In some implementations, the Basic Execution Environment might -happen- to use ASCII for the minimal character set, but that isn't certain -- for example, it could be using EBCDIC. Then there is the possibility that the character set is ISO 8859-1 or otherwise, and that there might be non-ASCII characters available. But with very few exceptions, %c on output does not convert the char values at all, merely stores whatever bit pattern is already there, whether those bits happen to represent ASCII or something else completely or aren't even defined printable characters for the locale. The exceptions have to do with some of the \ sequences: \a \n \b \t have to convert into the closest the implementation can come to the actions defined for those particular sequences. >The question is how does %d, %u know how many bytes to decode?considerint i[2]={1,2};int *p = &i[0];if you wanna print the value of *p to the screen, using %d or %u, howdo they know how manybytes to decode? The knowledge is built in to the implementation. -- All is vanity. -- Ecclesiastes Oct 2 '06 #3
P: n/a In article <11**********************@i42g2000cwa.googlegroups .com>, "Ancient_Hacker"
P: n/a Ancient_Hacker wrote: > .... snipadoodle ... > So poor old C compiler does what it does with any function's arguments, marshalls and converts them as is standard. Basically your chars and shorts, and ints get passed as chars and ints and ints. Longs get passed as longs. Floats get promoted to double. The printf function "knows" all this, so when it sees "%d" it knows to peek into the argument list for a int-sized thingy, for "%f" it knows to peek for a double sized thingy, and so on. Correction: chars also get passed as ints. At least in C. -- Some informative links: Oct 2 '06 #5
P: n/a ia***********@gmail.com wrote: Hello all: Can I think of %d, %u, %c as decoders? that is %d use 2's complement to decode whatever is stored in the variable. %u uses straight forward binary to decimal decoder. %c uses acii decoder. The question is how does %d, %u know how many bytes to decode? consider int i[2]={1,2}; int *p = &i[0]; if you wanna print the value of *p to the screen, using %d or %u, how do they know how many bytes to decode? Not decoders, they are conversion specifiers. The *printf() functions convert binary stuff to text stuff in specified ways so that we can 'print' them to the screen or some file in a formatted way. %d says expect an int and print it in decimal with a minus sign, if negative. %u says expect an unsigned int. Print it in decimal. It is positive. %c says expect a single character. Print the character. This is neither Brain Science nor Rocket Surgery. You need to read more. Maybe K&R2 for a good start. -- Joe Wright "Everything should be made as simple as possible, but not simpler." --- Albert Einstein --- Oct 4 '06 #6
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Replies have been disabled for this discussion. | 1,582 | 6,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.917987 |
https://scholarworks.alaska.edu/handle/11122/12246?show=full | 1,632,718,885,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00080.warc.gz | 532,563,574 | 8,059 | dc.contributor.author Buzby, Megan dc.contributor.author Lee, Sheldon dc.date.accessioned 2021-09-10T21:47:31Z dc.date.available 2021-09-10T21:47:31Z dc.date.issued 2021 dc.identifier.citation Lee, S., Buzby, M. (2021). Mathematical Modeling and Simulation with MATLAB. ©2021 by Sheldon Lee and Megan Buzby. This book is Licensed under CC BY-SA 4.0 en_US dc.identifier.uri http://hdl.handle.net/11122/12246 dc.description Textbook en_US dc.description.abstract This textbook attempts to provide you with an overview of the commonly used basic mathematical models, as well as a wide range of applications. It offers a perspective that brings you back to the modeling process and the assumptions that go into it. en_US dc.description.sponsorship University of Alaska Southeast en_US Viterbo University dc.description.tableofcontents Preface 5 en_US 1 Introduction 7 1.1 What is Modeling? 1.1.1 Types of Mathematical Models 1.2 Modeling with Equations 1.3 Modeling with Recurrence Relations 1.4 Modeling with Di erential Equations 1.5 Stochastic Models 1.6 Exercises 2 Programming in MATLAB 17 2.1 Why is Programming Important? 2.2 MATLAB Basics 2.3 MATLAB Functions and Terminology 2.4 Plotting Points and Curves 2.4.1 Using ezplot() 2.4.2 Using plot() 2.5 The Symbolic Toolkit 2.6 For Loops 2.7 While Loops 2.8 Conditional Statements 2.9 Exercises 3 Iterative Methods 35 3.1 Fixed Point Iteration 3.2 MATLAB function les 3.3 The Bisection Method 3.4 Newton's Method 3.5 Exercises CONTENTS 3 4 Matrices 46 4.1 Matrix Review 4.2 Eigenvectors and Eigenvalues 4.3 Summary of MATLAB commands 4.4 Exercises 5 Discrete Models 60 5.1 Introduction 5.2 Linear Dynamical Systems 5.3 State, Age, and Stage Matrix Models 5.4 Markov Chains 5.5 Higher Order and Nonlinear Discrete Dynamical Systems 5.6 Exercises 6 Continuous Models 81 6.1 Modeling with Di erential Equations 6.1.1 Euler's Method 6.1.2 Improved Euler's Method 6.1.3 Qualitative Analysis of Differential Equations 6.2 Systems of Differential Equations 6.3 Qualitative Analysis of Systems of Differential Equations 6.4 Linear systems 6.5 Nonlinear Systems of Equations 6.6 Exercises 7 Stochastic Modeling 113 7.1 Discrete Random Variables 7.2 Continuous Random Variables 7.3 Properties of Random Variables 7.4 Monte Carlo Integration 7.5 The Binomial Distribution 7.6 The Normal Distribution 7.7 Waiting Times 7.7.1 The Poisson Distribution 7.7.2 The Exponential Distribution 7.7.3 Sampling from the Exponential and Poisson Distributions 7.8 Stochastic Processes 7.8.1 Poisson Processes 7.8.2 Birth Death Processes 7.9 Exercises A In-class Activities and Exercises 145 CONTENTS 4 B Solutions to in-class Exercises 188 C Selected Solutions 193 D MATLAB Commands 198 E Debugging 202 F Completed Proofs 204 Index 209 dc.language.iso en_US en_US dc.subject mathematical models en_US dc.subject MATLAB en_US dc.title Mathematical Modeling and Simulation with MATLAB en_US dc.type Book en_US refterms.dateFOA 2021-09-10T21:47:32Z
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1. Oct 16, 2015
### toboldlygo
1. The problem statement, all variables and given/known data
In the figure below, the objects are attached to spring balances calibrated in newtons. Give the readings of the balances in each case, assuming that the strings are massless and the incline is frictionless.
I need help with (d). If the picture's too grainy to read off the values, the two blocks weight 10 kg each, the angle is 30 degrees, and the sides are 5.2 m (right above the (d)) and 3 m.
2. Relevant equations
F = ma, with Σ F = 0
3. The attempt at a solution
So, I know the reading off the balance will be equivalent to the tension in the string. I've tried doing this problem two different ways: first, I found the total Weight force (98.1 N) and used the sin to find the hypotenuse, which was 196.2 N. Then, when that was wrong, I thought maybe I should use gravity, split it into components, and then multiply whatever the hypotenuse of that right triangle was by the mass. I got 196.2 N again. What am I doing wrong? I've drawn a free body diagram and everything, but I don't know what I'm doing wrong here. I have a suspicion I'm not drawing the forces correctly, but I'm not sure. Thanks in advance for any help!
2. Oct 17, 2015
### paisiello2
How can the spring balance give a value greater than the weight of the block?
3. Oct 17, 2015
### Staff: Mentor
You determined the component of the weight parallel to the incline incorrectly.
Chet
4. Oct 17, 2015
### toboldlygo
I think I figured it out. I was drawing my triangle incorrectly (the amount of silly mistakes I make when doing physics problems...). Thanks for the sanity check, paisiello2, and for hinting at what was probably wrong, chestermiller. I appreciate it! | 461 | 1,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.944897 |
http://www.faqs.org/docs/Newtonian/Newtonian_102.htm | 1,501,015,990,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425381.3/warc/CC-MAIN-20170725202416-20170725222416-00190.warc.gz | 422,782,077 | 2,738 | 102
dimensional coordinate system with its x axis parallel the direction of
motion. Forces that point along the positive x axis are positive, and forces in
the opposite direction are negative. Forces that are not directly along the x
axis cannot be immediately incorporated into this scheme, but that’s OK,
because we’re avoiding those cases for now.
Discussion questions
In chapter 0, I defined 1 N as the force that would accelerate a 1-kg mass from
rest to 1 m/s in 1 s. Anticipating the following section, you might guess that 2
N could be defined as the force that would accelerate the same mass to twice
the speed, or twice the mass to the same speed. Is there an easier way to
define 2 N based on the definition of 1 N.
4.2Newton’s First Law
We are now prepared to make a more powerful restatement of the
principle of inertia.
Newton's First Law
If the total force on an object is zero, its center of mass
continues in the same state of motion.
In other words, an object initially at rest is predicted to remain at rest if the
total force on it is zero, and an object in motion remains in motion with the
same velocity in the same direction. The converse of Newton’s first law is
also true: if we observe an object moving with constant velocity along a
straight line, then the total force on it must be zero.
In a future physics course or in another textbook, you may encounter
the term net force, which is simply a synonym for total force.
What happens if the total force on an object is not zero. It accelerates.
Numerical prediction of the resulting acceleration is the topic of Newton’s
second law, which we’ll discuss in the following section.
This is the first of Newton’s three laws of motion. It is not important to
memorize which of Newton’s three laws are numbers one, two, and three. If
a future physics teacher asks you something like, “Which of Newton’s laws
are you thinking of,” a perfectly acceptable answer is “The one about
constant velocity when there’s zero total force.” The concepts are more
important than any specific formulation of them. Newton wrote in Latin,
and I am not aware of any modern textbook that uses a verbatim translation
of his statement of the laws of motion. Clear writing was not in vogue in
Newton’s day, and he formulated his three laws in terms of a concept now
called momentum, only later relating it to the concept of force. Nearly all
modern texts, including this one, start with force and do momentum later.
Example: an elevator
Question: An elevator has a weight of 5000 N. Compare the
forces that the cable must exert to raise it at constant velocity,
lower it at constant velocity, and just keep it hanging.
Answer: In all three cases the cable must pull up with a force of
exactly 5000 N. Most people think you’d need at least a little
more than 5000 N to make it go up, and a little less than 5000 N
to let it down, but that’s incorrect. Extra force from the cable is
Chapter 4Force and Motion
Next Page >>
<< Previous Page | 689 | 2,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.939415 |
https://developer.rhino3d.com/samples/rhinoscript/perpendicular-vectors/ | 1,722,735,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00355.warc.gz | 169,332,794 | 5,690 | Perpendicular Vectors
Windows only
Demonstrates how to calculate a vector that is perpendicular to another vector using RhinoScript.
' Description:
' Returns a 3-D vector that is perpendicular to another 3-D vector.
' Parameters:
' v - the 3-D vector to evaluate.
' Returns:
' A perpendicular 3-D vector if successful, Null otherwise.
' Remarks:
' The result is not a unitized 3-D vector.
'
Function GetPerpendicularVector( v )
GetPerpendicularVector = Null
If Not IsArray(v) Or UBound(v) <> 2 Then Exit Function
If Rhino.IsVectorZero(v) Then Exit Function
Dim i, j, k
Dim a, b
k = 2
If Abs(v(1)) > Abs(v(0)) Then
If Abs(v(2)) > Abs(v(1)) Then
' |v(2)| > |v(1)| > |v(0)|
i = 2
j = 1
k = 0
a = v(2)
b = -v(1)
ElseIf Abs(v(2)) >= Abs(v(0)) Then
' |v(1)| >= |v(2)| >= |v(0)|
i = 1
j = 2
k = 0
a = v(1)
b = -v(2)
Else
' |v(1)| > |v(0)| > |v(2)|
i = 1
j = 0
k = 2
a = v(1)
b = -v(0)
End If
ElseIf Abs(v(2)) > Abs(v(0)) Then
' |v(2)| > |v(0)| >= |v(1)|
i = 2
j = 0
k = 1
a = v(2)
b = -v(0)
ElseIf Abs(v(2)) > Abs(v(1)) Then
' |v(0)| >= |v(2)| > |v(1)|
i = 0
j = 2
k = 1
a = v(0)
b = -v(2)
Else
' |v(0)| >= |v(1)| >= |v(2)|
i = 0
j = 1
k = 2
a = v(0)
b = -v(1)
End If
Dim rc(2)
rc(i) = b
rc(j) = a
rc(k) = 0.0
GetPerpendicularVector = rc
End Function | 559 | 1,258 | {"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} | 3.546875 | 4 | CC-MAIN-2024-33 | latest | en | 0.405591 |
http://list.seqfan.eu/pipermail/seqfan/2010-October/006219.html | 1,642,997,857,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00569.warc.gz | 39,502,672 | 2,124 | # [seqfan] wanted: solution to a recurrence
N. J. A. Sloane njas at research.att.com
Thu Oct 21 18:14:10 CEST 2010
```Dear Seq Fans (and especially Eric Angelini)
I'm looking for a sequence a(1), a(2), ..., not all 1's,
that satisfies the forward-looking recurrence
a(n+1) = a(a(n)+n+1) for n>= 1.
In fact I would like the earliest such sequence.
Here is an unfinished attempt at constructing one:
n:= 1 2 3 4 5 6 7 8 91011121314 ...
a(n) 2 4 2 4 6 4 2 x 6 x 2 4 y 6 ...
where x and y are still to be chosen.
This does satisfy the recurrence so far:
n=1: we need a(2) = a(a(1)+2) = a(4), and indeed
a(2)=a(4)=4
n=2: we need a(3) = a(a(2)+3) = a(7), and indeed
a(3)=a(7)=2
and so on.
Can someone produce an explicit sequence satisfying the recurrence?
The first term might be 1. The entries should be positive
integers, not all 1's.
I do not have a solution.
Neil
```
More information about the SeqFan mailing list | 321 | 928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-05 | latest | en | 0.84936 |
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# Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0
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Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink]
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29 Apr 2011, 07:21
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Does x = y?
(1) x^2 - y^2 = 0
(2) (x - y)^2 = 0
[Reveal] Spoiler: OA
Last edited by Bunuel on 15 Oct 2013, 09:36, edited 1 time in total.
Renamed the topic and edited the question.
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### Show Tags
29 Apr 2011, 08:01
1
KUDOS
1. Not sufficient
x^2 = y^2
But that doesnt mean x = y.
x can be y or x can be -y
2. Sufficient
x = y is only solution that can satisfy (x-y)^2=0.
Posted from my mobile device
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### Show Tags
29 Apr 2011, 09:34
@ Shubhabsh Da'
I am confused with the answer; and the explanation as well.
Because, in statement 1, if we add y^2 on both sides, we get x^2 = y^2. Then, if we square root both the sides of the equation, we get x = y (no matter whether they are positive or negative). So, this is sufficient and is eliminating option B, C, and E.
Then, in the statement 2, if we square root both the sides of the equation, we get x - y = 0, or x = y (no matter whether they are positive or negative).
Therefore, my answer is D. However, Kaplan Premier 2011 shows the answer as B and explains it thereby.
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29 Apr 2011, 15:52
@Schawjibb,
square root of x^2 = could be x or -x ( not just x)
in other words square root of x^2 = |x| = +-x
so when you take square root on both sides , you will have +- on both sides. not enough to find the sign of the variable.
=> x = y or x = -y
Eg: \sqrt{(x^2)} = 25 => |x| = 5 =>x = +5 or -5
Hope it helps.
Schawjibb wrote:
@ Shubhabsh Da'
I am confused with the answer; and the explanation as well.
Because, in statement 1, if we add y^2 on both sides, we get x^2 = y^2. Then, if we square root both the sides of the equation, we get x = y (no matter whether they are positive or negative). So, this is sufficient and is eliminating option B, C, and E.
Then, in the statement 2, if we square root both the sides of the equation, we get x - y = 0, or x = y (no matter whether they are positive or negative).
Therefore, my answer is D. However, Kaplan Premier 2011 shows the answer as B and explains it thereby.
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30 Apr 2011, 02:43
1
This post was
BOOKMARKED
Schawjibb wrote:
I found this question in Kaplan Premier 2011. I am confused with their answer and its explanation (or, may be I am wrong). Can anyone come up with an answer supported by precise explanation?
Q. Does x = y?
(1) x^2 - y^2 = 0
(2) (x - y)^2 = 0
Solution:
Statement 1:
$$x^2-y^2=0$$
$$x^2=y^2$$
Squaring both sides, would result to two solutions: x= y OR x=-y. So, INSUFFICIENT!
Statement 2:
$$(x-y)^2=0$$
To get a 0, x-y must be equal to 0.
$$x=y$$
SUFFICIENT!
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03 May 2011, 11:21
Schawjibb wrote:
@ Spidy001
Could you please help me in one regard? I need a resource (book/forum discussion) from where I can really learn how to solve a DS problem systematically and in minimum time.
Background: I have started going through the DS section from Kaplan Premier 2011 yesterday. And the result of yesterday's (first-time) practice was devastating. 13 out of 20 were incorrect. And no question was answered in less than 2 minutes! I am just drowning into frustration.......
My 2 cents on this.
See the best resources are the OG's, GMAT Club Math Forum and the MGMAT tests.
Seeing the problem that you are facing right now,probably you should brush up your basics first.Then start with OG and subsequently move on to higher difficulty level numerical s.
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03 May 2011, 21:19
@ Amit
Thanks a lot! I think my basic problem pertaining to DS is with Number Properties. Because, in other fields I feel myself comfortable as long as the math part is concerned. Do you think MGMAT Number Properties book will suffice? Suggestions are highly appreciated?
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Re: Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink]
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15 Oct 2013, 09:31
Schawjibb wrote:
I found this question in Kaplan Premier 2011. I am confused with their answer and its explanation (or, may be I am wrong). Can anyone come up with an answer supported by precise explanation?
Q. Does x = y?
(1) x^2 - y^2 = 0
(2) (x - y)^2 = 0
Hi there, let me take a bite a this one.
So does x = y?
From Statement 1:
x^2 - y^2 = 0
This means that Abs (X) = Abs (Y)
But could be that x = -y, or x = y
Hence Insuff
From Statement 2
(x-y)^2 = 0
(x-y) = 0
So x = y
Suff
Cheers
J
Last edited by jlgdr on 15 Oct 2013, 10:09, edited 1 time in total.
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Re: Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink]
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15 Oct 2013, 09:38
jlgdr wrote:
Schawjibb wrote:
I found this question in Kaplan Premier 2011. I am confused with their answer and its explanation (or, may be I am wrong). Can anyone come up with an answer supported by precise explanation?
Q. Does x = y?
(1) x^2 - y^2 = 0
(2) (x - y)^2 = 0
Hi there, let me take a bite a this one.
So does x = y?
From Statement 1:
x^2 - y^2 = 0
This means that Abs (X) = Abs (Y)
But could be that x = -y, or x = y
Hence Insuff
From Statement 2
(x-y)^2 = 0
(x+y)(x-y) = 0
x = -y or x = y
So Insuff
From Statements 1 and 2 together:
We have exactly the same information, hence IMHO E is the correct answer
Cheers
J
That's not correct.
Does x = y?
(1) x^2 - y^2 = 0 --> x^2=y^2 --> x=y or x=-y. Not sufficient.
(2) (x - y)^2 = 0 --> x-y=0 --> x=y. Sufficient.
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Re: Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink]
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15 Oct 2013, 10:08
Bunuel wrote:
jlgdr wrote:
Schawjibb wrote:
I found this question in Kaplan Premier 2011. I am confused with their answer and its explanation (or, may be I am wrong). Can anyone come up with an answer supported by precise explanation?
Q. Does x = y?
(1) x^2 - y^2 = 0
(2) (x - y)^2 = 0
Hi there, let me take a bite a this one.
So does x = y?
From Statement 1:
x^2 - y^2 = 0
This means that Abs (X) = Abs (Y)
But could be that x = -y, or x = y
Hence Insuff
From Statement 2
(x-y)^2 = 0
(x+y)(x-y) = 0
x = -y or x = y
So Insuff
From Statements 1 and 2 together:
We have exactly the same information, hence IMHO E is the correct answer
Cheers
J
That's not correct.
Does x = y?
(1) x^2 - y^2 = 0 --> x^2=y^2 --> x=y or x=-y. Not sufficient.
(2) (x - y)^2 = 0 --> x-y=0 --> x=y. Sufficient.
Apologies for that. Yes that's in fact correct.
Thanks for pointing out.
Cheers
J
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Re: Does x = y ? [#permalink]
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Re: Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink]
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16 Sep 2016, 18:00
B is correct
(1) x^2 - y^2 = 0
We have no way to know with certainty the values of x and y. Y could be x negated or the same value as x. Thus, INSUFFICIENT
(2) (x-y)^2 = 0
=(x-y) = 0 --> x= y
SUFFICIENT
Re: Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink] 16 Sep 2016, 18:00
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# Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0
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# 17s418hw6sol.pdf - Stat 418 HW#6 Problem 6.8 Ry(a fY(y =...
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Stat 418 HW #6 Problem 6.8 (a) f Y ( y ) = c y R - y ( y 2 - x 2 ) e - y dx = 4 3 cy 3 e - y , 0 < y < Since R 0 f Y ( y ) dy = 1, c = 1 8 (b) f X ( x ) = R | x | 1 8 ( y 2 - x 2 ) e - y dy = 1 4 e -| x | (1 + | x | ) -∞ < x < ( by integration by parts, R y 2 e - y dy = - e - y ( y 2 + 2 y + 2) ) f Y ( y ) = 1 6 y 3 e - y 0 < y < (c) E[X]=0, since f X ( x ) is even function. Problem 6.19 1 R 0 x R 0 1 x dydx = 1 R 0 dx = 1 (a) 1 R y 1 x dx = - log y 0 < y < 1 (b) x R 0 1 x dy = 1 0 < x < 1 (c) E [ X ] = 1 2 (d) E [ y ] = 1 R 0 - y log ydy = 1 4 (integrating by parts, R y log ydy = 1 2 y 2 log y - 1 2 R ydy ) 1
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Problem 6.22 (a) No, since the joint density does not factor. ( f ( x, y ) 6 = f ( x ) f ( y )) (b) f X ( x ) = 1 R 0 ( x + y ) dy = x + 1 2 0 < x < 1 (c) P ( X + Y < 1) = R 1 0 R 1 - x 0 ( x + y ) dydx = 1 3 Problem 6.28 Let service time for A.J’s car = X and service time for M.J’s car = Y. Then, (a) P ( Y + t < X | X > t ) * P ( X > t ) = 1 2 * e - t (b) P ( X + Y < 2) = R 2 0 R 2 - x 0 e - ( x + y ) dydx = 1 - 3 e - 2 Problem 6.39 P ( Y | X = 1) = 1 y=1 P ( Y | X = 2) = ( 2 / 36 3 / 36 = 2 3 y=1 1 / 36 3 / 36 = 1 3 y=2 P ( Y | X = 3) = 2 / 36 5 / 36 = 2 5 y=1 2 / 36 5 / 36 = 2 5 y=2 1 / 36 5 / 36 = 1 5 y=3 P ( Y | X = 4) = 2 / 36 7 / 36 = 2 7 y=1 2 / 36 7 / 36 = 2 7 y=2 2 / 36 7 / 36 = 2 7 y=3 1 / 36 7 / 36 = 1 7 y=4 P ( Y | X = 5) = 2 / 36 9 / 36 = 2 9 y=1 2 / 36 9 / 36 = 2 9 y=2 2 / 36 9 / 36 = 2 9 y=3 2 / 36 9 / 36 = 2 9 y=4 1 / 36 9 / 36 = 1 9 y=5 P ( Y | X = 6) = 2 / 36 11 / 36 = 2 11 y=1 2 / 36 11 / 36 = 2 11 y=2 2 / 36 11 / 36 = 2 11 y=3 2 / 36 11 / 36 = 2 11 y=4 2 / 36 11 / 36 = 2 11 y=5 1 / 36 11 / 36 = 1 11 y=6 P(X)=
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https://www.istc.cnr.it/grouppage/locen-resources-initial-programming-exercises-simple-neural-networks | 1,718,803,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00615.warc.gz | 716,858,184 | 17,633 | # LENAI Resources Initial Programming Exercises On Simple Neural Networks
## Initial Programming Exercises On Simple Neural Networks
### Introduction
This page gives some programming exercises to start to learn to program neural networks. Like the rest of these pages on "learning paths", this page heavily relies on other pages in the web and is mainly meant as guidance: the student should be able to study and learn by him/herself on the basis of indications here, internet pages, and some additional instructions on things not covered by this web-pages when needed.
The steps to follow are these, expanded in the exercises below:
• Read the following web-page of this website to learn the programming concepts that you need to learn to work with LOCEN (and in general to be able to build and work with computational models of brain and behaviour, or artificial intelligence/robotic systems): web-page on programming concepts to learn
• Depending on the programming language that you have to learn (agreed with your LOCEN supervisor), refer to suitable websites teaching such programming language. For the different programming languages, such websites are indicated in the web pages linked from this web page under ''Programming and simulations --> Specific languages".
• Installing an IDE (Integrated Development Environment) on your portable computer (or the computer at LOCEN) to program in the selected language; indications on this are given in the web-pages dedicated to the specific languages and mentioned in the previous point
• Understand that there are various learning paradigms related to neural networks (see below for more details): this page gives some exercises related to supervised learning and reinforcement learning
• Do the exercises illustrated in detail below, in particular involving:
• A first simple perceptron neural network able to learn the AND and OR functions through a supervised learning algorithm
• The exercise of the previous point, but using linear algebra data structures and operations of the language you use (or libraries expanding the language)
• Implementation, based on linear algebra, of a multilayer perceptron able to learn the difficult X-OR function
• Implementation of two leaky neurons
• Implementation of an echo-state network
• Implementation of a reinforcement learning model based on Q-Learning
• Implementation of a reinforcement learning model based on an actor-critic (still under construction)
### Understand (study) that there are various learning paradigms related to neural networks: see below for more details
Understand that there are various learning paradigms (classes of algorithms) through which neural networks can learn. The major ones are these:
• Supervised learning. Two very important sub-classes:
• Linear regression
• Classification
• Unsupervised learning
• Reinforcement learning
Search the internet for further information and understand the differences between these different learning paradigms, e.g.:
https://en.wikipedia.org/wiki/Machine_learning
Also, ask your LOCEN supervisor for the book of Floreano on Neural Networks: an old book but one of the best for learning the basics on neural networks.
### Exercise to program a first simple perceptron neural network able to learn the AND and OR functions through a supervised learning algorithm
This section gives a programming exercise on a simple neural network and some indications on how to accomplish it.
The goal of the exercise is to program a one-layer perceptron that is able to learn, in a supervised fashion, to produce the logical AND and OR functions. In particular the network should respond with the following 'desired output' if given the following 'input' examples:
OR function
Input1 Input2 Bias Output
0 0 1 0
0 1 1 1
1 0 1 1
1 1 1 1
AND function
Input1 Input2 Bias Output
0 0 1 0
0 1 1 0
1 0 1 0
1 1 1 1
To better understand what the neural network should do, see these web pages:
There are various versions of such a simple perceptron. You should implement one where the output unit uses a sigmoid transfer function, and the network is considered to output 1 if the activation of the output unit is > 0.9, and to answer 0 if the activation of the output unit is < 0.1. This leads you to create a more general neural net, e.g. useful when you pass to multi-layer perceptrons (exercise below); plus is more correct/general from a machine learning point of view.
Your networks should have 3 input units (the first being a 'bias') and one output unit.
The network should learn online (i.e., after the presentation of each example of the training set) not batch (i.e., for all examples together).
Note that neural networks (usually) learn progressively, rather than by 'one shot', by being exposed to the same training examples several times. This affects the main program structure indicated below.
Your program should be structured as follows:
• Creation and initialisation of data structures
• Creation of data structures and data for training the neural network
• Loop for multiple epochs (one epoch is one learning pass of the net):
• Loop for examples of data-set
• Network learning
• Collection of data for following data analysis
• Data analysis: plots on the learning curve of neural network
• Data analysis: numerical output of network performance (e.g., desired and actual net output, errors, etc.)
### Same previous exercise, but using linear algebra data structures and operations of the chosen language
Do the exercise above but:
• Use the programming language-dependent data structures to implement vectors (e.g., of input and output) and matrixes (e.g., for connection weights)
• Use the programming language-dependent linear algebra operators (e.g., to implement the whole network spreading of activation, you should have something like:
vOutp = mConnWeig * vInpu
• Implement a neural network that performs the AND and the OR functions at the same time. The network will hence have 3 input and two output units.
### Exercise to implement, using linear algebra programming, a multilayer perceptron learning the X-OR function
Implement a neural network that performs the X-OR function below. For this purpose, you will need a multilayer perceptron with 3 input units, 1 output unit, and a variable number of hidden units (initially, implement a network with 1 hidden layer only).
X-OR function
Input1 Input2 Bias Output
0 0 1 1
0 1 1 0
1 0 1 0
1 1 1 1
Parity check function
This is a generalisation of the x-or function, where the output is 0 if the elements of the input pattern have an odd number of digits equal to 1, and 1 if such number is equal. So for example, for 3 input you have:
Input1 Input2 Input3 Bias Output
0 0 0 1 1
0 0 1 1 0
0 1 0 1 0
0 1 1 1 1
1 0 0 1 0
1 0 1 1 1
1 1 0 1 1
1 1 1 1 0
The new network to solve either task has a different architecture with respect to the perceptron. In particular, it has a second layer of two units, called "hidden units", between the input layer and the output layer. So the inputs and the bias fire on the hidden units and then the hidden units (and the bias) fire on the output units.
The learning algorithm for the weights between the output and the hidden units is the same as for the simple perceptron. A second algorithm is needed to modify the weights between the input and the hidden units, called the "error backpropagation" algorithm. Search the formulas of the error-back propagation algorithm on the internet. Make sure that you understand it very well the algorithm as it is not very simple.
Change your program so that after learning it plots:
• The curve of the Square Error of the output with respect to the desired output during learning (put "epochs" at the x-axis, i.e. plot the error averaged over all the input patterns)
• The two 2D weights matrices (from input to hidden; from hidden to output) of your multilayer perceptron: this plot represents every value of a connection weight as a coloured square, with each colour representing the size of the weight, from most negative values to largest positive values.
• Also, plot a similar graph with: on 1 row = the activation of 1 input pattern; the related activation of the hidden units; the activation of the output unit(s); on different rows: those activations corresponding to all patterns.
Further, change the program so that at the end of the simulation all data to make the previous graphs are downloaded into separated file .txt files, and then the graphs are plotted by a second program.
### What are leaky neurons and how to simulate them
"Leaky neurons" are different from the "firing rate" neurons used so far. In particular, these are neurons that store information in time, e.g. they have an activation that accumulates, with a fixed input, or progressively fades away, with a zero activation. They are indeed behaving like a ''container of water with a leak when you put water in it": they slowly fill in when you put water, the progressively get empty when you stop.
First, see the part of the slides related to the leaky neuron and the way to simulate it based on Euler approximation.
Use those slides to understand how leaky units work:
--> leaky element for computing the activation potential of the neural unit
--> tanh function, truncated to positive values, for computing the activation of the neural units
Also, understand from the slides (and internet) how to implement the leaky units (or any other dynamical system) in a computer program through the "Euler approximation". Euler approximation is necessary as now we will simulate time more in detail as a continuous flow. So, a neural network will be a dynamic system that works in continuous time and the Euler approximation allows the approximated simulation of its dynamics in the discrete time-steps of a computer program.
To suitably take account of continuous-time, in the simulations establish these critical variables:
• sfSimuTimeDura: duration of the whole simulation, in seconds, e.g. set to 10 seconds.
• sfDeltaTime: the duration of the time step used to approximate continuous time, for example, set to 0.01 (seconds). For a certain duration of the simulation, the smaller the time step, the more accurate the approximation of the dynamics of the system. Each cycle of the simulation will now correspond to one sfDeltaTime.
• sfSimuStepDura: duration of the simulation in time steps, set equal to: sfSimuTimeDura/sfDeltaTime
Another very important thing to do in the implementations. With continuous time, causation between phenomena can happen only in different time steps (from t-1 to t). So, for any variable that affects another variable, e.g. a neuron that affects another neuron (or itself in the future), you have to do this:
• Store the "old value of the variable" and the "current value of the variable"
• At the beginning of the simulation cycle representing the elapsing of time (in sfDeltaTime steps), you have to save the old value of all the variables affecting other variables the next time step as in this example of three units:
sfUnit1Old = sfUnit1
sfUnit2Old = sfUnit2
sfUnit3Old = sfUnit3
• Then, for any causal dependency between the variables you have to consider the old values of the variables, for example:
sfUnit1 = sfUnit1Old + (sfDeltaTime/Tau) * (-sfUnit1Old + sfWeig2to1 * sfUnit2Old + sfWeig3to1 * sfUnit1Old)
sfUnit2 = sfUnit2Old + (sfDeltaTime/Tau) * (-sfUnit2Old + sfWeig1to2 * sfUnit1Old + sfWeig3to2 * sfUnit3Old)
sfUnit3 = sfUnit3Old + (sfDeltaTime/Tau) * (-sfUnit3Old+ sfWeig1to3 * sfUnit1Old + sfWeig2to3 * sfUnit3Old)
Notes on good practices to organise the code of the simulations:
- Note that it is a good practice to represent the state of the system (e.g.: neurons' activation potential, potential, old activation, old activation potential, etc.) with variables that do not take into consideration time. For example, it is useful to represent the activation of a neuron with a variable sfUnit rather than as a vector vfUnit that encodes the neuron activation in time. The advantage of doing this is that the core of the simulation remains very clean and compact. Instead, use another vector or matrices to collect data of interest during the simulation (e.g., how a neuron activates during the simulation), to perform data analysis, and plotting of results. An exception to this is for simulation where one wants to do batch learning, which is rare in this context but might happen (e.g., for the echo state network discussed below).
- The example just reported uses single variables to represent the different neurons forming the model; a more compact and better way is to use a vector to represent them (still not considering time, otherwise one would have a matrix).
Based on these indications, do the exercises below.
### How one leaky neuron responds to an input
Make a simulation that lasts for 20 seconds in total and involves a network formed by 1 leaky unit.
Input I:
0 for 1 second
1 for 1 second
0 for 2 second
1 for 2 second
0 for 3 second
1 for 3 second
0 for 4 second
1 for 4 second
Simulate how the leaky neuron responds to that input. Plot a graph in time, showing the input and the neuron activation. Try to change the tau of the neurons to see what happens. Try to change sDeltaTime and see what changes.
### A neural network of two leaky units representing an onset circuit
Make a simulation that lasts for 25 seconds in total and involves a network formed by 2 leaky units.
The input the network during the simulation is as follows:
0 for 5 seconds
0.5 for 5 seconds
1 for 5 seconds
0.5 for 5 seconds
0 for 5 seconds
The architecture of the network is as follows:
• An input unit (sUnit1) is formed by a first leaky neuron, and an input/output unit, is formed by a second leaky neuron (sUnit2).
• Both sUnit1 and sUnit2 take the (external) input
• sUnit1 is connected to sUnit2 with a connection weight of -1
• sUnit2 activation gives the output of the network
Plot both the input and the activation of sUnit1 and sUnit2 in time.
Try to set the tau values of the two units (initially set them to 1) with this objective: sUnit2 shows a behaviour detecting the ''derivative'' of the signal in time.
Set sfDeltaTime initially to 0.01. Then set it to different values. You will see that these values only set the accuracy of the model, not its overall dynamics.
### A model of Amygdala
This is a simulation of the basic functionalities of the amygdala, important for effective regulation.
The model can:
- produce an innate behavioural response to a biologically salient event (ingestion of food)
- learn to anticipate such response if a cue is paired with the biologically salient event (conditioning)
- make those two responses dependent on the internal hunger/satiety of the organism.
Schedule.
The overall simulation lasts 16 seconds.
Set sfDeltaTime = 0.01.
The neural network undergoes a phase of Pavlovian conditioning (8 seconds) and then a phase of recall showing the effects of the conditioning.
In particular, the network receives three inputs during the 16 seconds of the simulation:
1) Cue (use vector vfInpuCue[] to encode the values of the Cue in the steps of the simulation):
= 1 in the time intervals 2-4 and 10-12
= 0 otherwise
2) Food (use vector vfInpuFood[] to encode the values of the Food in the steps of the simulation)
= 1 in the time intervals 4-6
= 0 otherwise
3) Satiety (use vector vfInpuSati[] to encode the values of Satiation in the steps of the simulation).
You will run two tests where this unit is activated in these ways:
Test 1: = 0 always (indicating the animal is hungry)
Test 2: = 1 always (indicating the animal is satiated)
The architecture of the network is as follows.
Units representing different neural populations, encoding the input stimuli and the motor response:
• A unit (vfActi[0]) formed by a leaky unit encoding the stimulus Cue, with tau = 0.3
• A unit (vfActi[1]) formed by a leaky unit encoding the stimulus Food, with tau = 0.1
• A unit (vfActi[2]) formed by a leaky unit encoding the internal states of Satiety, with tau = 0.1
• A unit (vfActi[3]) formed by a leaky unit encoding neuromodulatory neurons of Dopamine, with tau = 0.1
• A unit (vfActi[4]) is formed by a leaky unit encoding a Motor action, with tau = 0.
For such units create these data structures:
vfActi[]
vfActiPote[]
vfActiOld[]
vfActiPoteOld[]
Connections from the external and internal stimuli to the model units:
• Cue-->Cue Unit (the input is simply injected into the unit)
• Food-->Food Unit (the input is simply injected into the unit)
• Food-->Dopamine Unit (the input is simply injected into the unit)
• Satiety-->Satiety Unit (the input is simply injected into the unit)
Connections between the model units (connections internal to the model):
• CueUnit-->FoodUnit (call the variable: sfWeigCueFood , initialise it with 0): represents the association between the cue and the food, undergoing a Hebbian learning process
• FoodUnit-->MotorUnit (call the variable: sfWeigFoodMoto, initialise it with +1): this innate excitatory unit implies that the activation of the Food Unit causes the production of the behaviour
• SatietyUnit-->FoodUnit (call the variable: sfWeigSatiFood, initialise it with -1): this innate inhibitory unit implies that the presence of satiety prevents the Food Unit to activate even in the presence of the Food
• Dopamine Unit--> affects the Hebbian learning (call the variable: sfWeigDA, initialise it with +1) of the plastic connection
Learning
The Hebbian learning rule is as follows:
sfWeigCueFood = sfWeigCueFood + sfLearRate * (sfWeigDA * vfActi[3]) * vfActi[0] * max(0.0, (vfActi[1] - sfLearThrePost))
sfWeigCueFood = min(sfWeigMax,sfWeigCueFood)
where:
sfLearRate = 2.0
sfLearThrePost = 0.1
sfWeigMax = 2.0
Data analysis.
Collect the history of activation of the units in matrix mfHistActi[][]
Collect the history of the connection weights in matrix mfHistWeig[][]
Plot:
Figure 1: the graph of the three input stimuli in time
Figure 2: the graph of the units Cue, Food, Satiety in time
Figure 3: the graph of Dopamine and Motor action in time
Figure 4: the graph of the plastic connection weight
### A neural network formed by two leaky units that perform a decision-making circuit
Make a simulation that lasts for 60 seconds in total and involves a network formed by 2 leaky units.
The input to the two units is as follows:
(0, 0) for 5 seconds
(0, 0) for 5 seconds
(0, 1) for 5 seconds
(0, 0) for 5 seconds
(0.1, 0.9) for 5 second
(0, 0) for 5 seconds
(0.2, 0.8) for 5 second
(0, 0) for 5 seconds
(0.3, 0.7) for 5 second
(0, 0) for 5 seconds
(0.4, 0.6) for 5 second
(0, 0) for 5 seconds
The architecture of the network:
• The network is formed by two leaky input/output units: sUnit1 and sUnit2
• The sUnit1 and sUnit2 take the respective input and each gives an output, so the network returns two output values
• sUnit1 is connected to sUnit2 with a connection weight of -1
• sUnit2 is connected to sUnit1 with a connection weight of -1
Plot both the two input values and the activation of sUnit1 and sUnit2 in time in one graph (so 4 curves).
Try to set to different values: the Tau of the two units (initially equal to 1), the lateral connections between the two units, and the intensity of the leak,
with the objective of making sfUnit2 tend to go to a high value (this is the "selected option", the "decision") while sfUnit1 tends to go to zero.
### Echo State Network
The Echo State Network (ESN) consists of a network of leaky neurons (see above) forming an all-to-all connected recurrent neural network.
Simulate the units of the ESN models below by using the same technique used to simulate the leaky neurons (see above), i.e. use the variables:
sSimuTimeDura
sDeltaTime
sSimuStepDura = sSimuTimeDura / sDeltaTime
At each step of the simulation, each unit receives the weighted sum of inputs from each other unit of the network and from itself:
ap[j,t] = SUM_i [ W[j,i] * a[i,t-1]]
a[j,t] = f[ap[j,t]]
where ap[j,t] is the activation potential of unit j at time t,
SUM_i[.] computes the sum over the elements of the argument running the index i,
W[j,i] is the connection weight from unit i to unit j belonging to the matrix W, and
a[i,t-1] is the activation of unit i at time t-1 computed on the basis of the transfer function f[a[i,t-1]] (f is a Tanh[.] function explained in the above exercises).
If you use the matrix and vector notation, and functions in your program, you can express an ESN activation very compactly as:
ap[t] = W * a[t-1]
i.e.:
ap[t] = W * Tanh[ap[t-1]]
First run of the network
As first exercise you have to try to build a neural network formed by two components.
The first component is an input layer of m units: each input unit has to be connected to all leaky units of the second ESN component through connection weights drawn from a uniform distribution ranging between 0 and 1.
Run a certain number of steps of the simulation, e.g. 400.
Initially, the input units are each activated with a random value drawn from a uniform distribution ranging between 0 and 1.
Keep the units activated for a quarter of time of the simulation to such values, and then set them to 0 for the rest of the time.
The second component is an ESN, where all inner weights W have been randomly drawn from a uniform distribution ranging between - 0.1 and 0.1.
Record the activation of the units of the ESN in time during the simulation, and plot them at the end of it to see their dynamics.
Repeat the simulation trying a different range within which the W weights are randomly drawn, e.g. [-1, +1] or [-10, +10].
Expected results
As you can see, high values of W make the network units' activation chaotic and crazy.
Instead, low values of W make the network units' activation to:
- first have a transient dynamics
- then to set to a steady-state value
- then to fade to zero after the input is switched off
Eigenvalues / Eigenvectors
To avoid the chaotic behaviour of the ESN, you should modify the connection weights W to given them "the echostate property" implying that:
• with any input, the network will not explode
• if the input units are turned to zero, the network will progressively relax to zero
• for a given sequence of input patterns, the network will go through a specific transient dynamics: this is the most important property for which the network tends to respond to the current input, and also to the recent ones (with a decaying memory), in a specific deterministic way
• moreover, if the input changes a bit the dynamics will change a bit, rather than a lot as in chaotic systems
Using an existing programming function (search it from the math libraries of your programming language), you can extract from W its eigenvalues (you do not need to know how this is done or what eigenvalues precisely are: roughly speaking, they indicate the size of some important dimensions, the eigenvectors, of the space covered by W column vectors).
The function you found usually returns eigenvalues in form of a vector of n elements. Extract the highest eigenvalue from them (in absolute value): this is called the spectrum of the matrix. The function usually also returns a matrix of the eigenvectors, but you do not need it.
Now you should divide W for the spectrum, and use the resulting matrix W' as the weight matrix of your ESN. In this way, the ESN will have the echostate property.
After you do this, repeat the exercise above: you should see a richer dynamics of the network with respect to the W set by hand. Indeed, now the W are the maximum ones just before the network has a chaotic behaviour: the system is said to exhibit an interesting behaviour as it works "at the edge of order and chaos".
Echo State with Perceptron as Output
In this exercise, add a third component to the model of the previous exercise: a perceptron network (see the exercises above on this) that takes as input the n units of the ESN component, and gives its output with 1 linear unit. To do this use the following architecture of the network.
Neural network architecture:
- Use as an input layer to the ESN the same layer and random input you used in the previous exercise
- Connect the input layer to an ESN as the ESN of the previous exercise
- All the ESN units are connected to the linear output units
The goal of the exercise is to train the perceptron to "read out" the dynamic states of the ESN (i.e. the values assumed by the ESN units in time) by reproducing a value that corresponds to a sinusoid function in time. To this purpose, generate a sequence of desired output of the perceptron through a sin(time) function.
Train the network-perceptron for 50 times with the perceptron delta rule. To this purpose train the system as follows:
- Run 50 epochs of training
-- For each epoch, you have a run of 40 seconds during which:
--- The input is random as in the above exercises, and constant for all 40 seconds and all 50 epochs
--- The desired output is the sinusoid one
Then try to test the network giving the random input for 40 seconds.
Expected results
You should see that:
• The ESN units dynamically respond to the input in time as usual
• The perceptron reads out the inner activations of the network and reproduces the sinusoid function output during the initial transient-dynamic phase
• If you try to use another desired output (e.g., the sum of different sinusoids with different phases and period) the system should be quite good to reproduce it during the initial transient phase
Reinforcement-learning neural network models
### Q-Learning
Q-LearningReinforcement learning is an important family of algorithms that allow the simulation of intelligent agents (e.g., simulated animals or intelligent robots) that learn to accomplish a task by trial-and-error while interacting with the world. Q-Learning is one of the two most important and basic reinforcement learning algorithms, the other being the actor-critic model. Q-learning is based on a series of states s (s ∈ S, S = all states), a series of actions a (a ∈ A, A = all actions), and a series of “valuations” q (q ∈ Q, Q = all valuations) for each state-action to pass to a new state.
There are two different ways to implement Q-learning:
• Tabular Q-Learning: This is based on a 2-entry table, not a neural-network implementation and so it is not useful as a biological model. However, we suggest taking a look to this page related to tabular Q-learning because it helps you start to understand which kind of problems Q-learning can solve and how:
http://mnemstudio.org/path-finding-q-learning-tutorial.htm
• Neural-network Q-Learning. This is the target of this tutorial. The model is based on a simple “perceptron” neural-network architecture, i.e., a 2-layer feedforward linear network formed by input and an output layer. The system takes as input the current state of the world and gives as output the q values one for each action: the actions are encoded in a “localistic way” (non “distributed way”), i.e. the system encodes a finite number of discrete actions where 1 action correspond to 1 output unit.
We now explain how to implement neural-network Q-learning in detail.
Simulation of the “world”
At first, you have to program the simulation of the world with which the agent interacts. Here we assume a simple “grid world”, i.e. a world formed by "positions", or "cells", where the agent can be at each step, and where the agent can navigate with its actions. You can consider a 1D arena (or "world": this is formed by a number of contiguous cells, or positions, forming a 1D sequence, e.g. imaging a corridor divided in contiguous squares), with two possible actions for moving leftward or rightward along the positions, or a 2D world (a number of contiguous positions organised in terms of a 2D grid), with four possible action actions to move north, est, west, south. In the case of the 1D world, you represent the state of the world, i.e. the position where the agent is, with a vector of 0s and a 1 in correspondence to the position occupied by the agent. In the case of the 2D world, you represent the state of the world with a matrix of zeros and a 1 in correspondence to the position occupied by the agent. We recommend to implement both the 1D and 2D world in parallel, and to parameterise your program to use one or the other at will, because this will be very instructive and help you debugging the program.
Choose a starting position in which you will position the agent at the beginning of each trial: this is represented as the initial “1” in your vector or matrix of zeros representing the state of the world.
Moreover, decide where to put the reward (r) in your world model. To this purpose, use another vector/matrix with the same dimensions of the world verctor/matrix, with a 1 in the element corresponding to the rewarded position.
Importantly, some of the actions that the agent might choose would bring it to hit the ''walls'' of the arena where it navigates. For example, if it is at the top-left corner of the 2D arena, if it selects the action "west" or the action "north", these actions lead the agent to hit respectively the west and north walls of the arena and so it remains where it is. In general, the agent is always left free to select any one of the actions of its repertoire (so at each time all actions of its repertoire of actions are available for selection), but if it selects an action that leads it against a wall, the effect of this action execution is simply that the agent is left in the same position from which it selected and performed the action. This situation is similar to the one of a robot that tries to move against a wall and the effect is that it remains where it is.
Activation of the neural network
The agent is based on a simple neural network formed by 2 layers of units:
The input layer is activated with the state of the world representing the position of the agent in the world (the input units as many as the possible positions in the world). In the case of the 2D world, you should “unroll” the world matrix into the vector representing the input units.
• The connection weights represent the Q values.
• At the beginning, you can set them to zero values or small random values.
• The output units represent the q values of the possible actions (2 or 4 actions).
The output values do not have to pass through a step or sigmoid transfer function, as usual in neural networks, but through a SoftMax function (also known as “normalized exponential function”). The SoftMax function allows the normalisation of the output values so they sum up to 1 while keeping a difference between them. This allows considering the output of the SoftMax as probability values, used to extract the action to execute. It means that, with zero initial connection weights, at the beginning, you will have two outputs with two 0.5 values; if you have four outputs you will have four 0.25 values. The SoftMax function SM[.], allowing the computation of the probabilities p[a] of selecting the different actions a, is defined as follows:
For q ∈ Q and a ∈ A: p(a) = SM(q)=(e^(q/T)) / (Σ_j e^(q_j/T))
T is called the “temperature parameter”, and e is the Euler number. For high temperatures, all output values tend to assume the same probability, instead of for low temperatures, the values tend to have great differences (to the extreme: one value is close to 1 and all the others values are close to 0). So the temperature allows setting how much the probabilities of selecting the different actions differ between them in correspondence to the Q values.
We now explain the core equation of the algorithm, allowing the update of the network connection weights corresponding to the q values, and then the whole program of the simulation. As we are facing reinforcement learning, the neural network cannot be trained in a supervised learning fashion as we lack a vector of desired output values. Thus, we use a different learning rule to update step by step the Q values, depending on the quality of the chosen action in terms of reward. At each time step t, the learning rule updates only the connection weights that correspond to the action a[t-1] selected and executed at time t-1 (this action corresponded to the q value: q[s[t-1], a[t-1]]).
To update the connection weights w[j,i] connecting input unit s_i to the output unit q_j corresponding to the selected a[t-1], is as follows:
Delta w_ji = η * ( (r[t] + γ * Max_q[q[s[t], a[t]]]) – q[s[t-1], a[t-1]] ) * s_i[t-1]
Let us examine this equation closely:
• As mentioned, w[j,i] connects input unit s_i to the output unit of the network corresponding to the q-value q_j corresponding to the previously selected a[t-1]
• η is the learning rate, usually set equal to 0.1
• r(t) is the reward achieved in the current state s_tγ is a “discount factor” usually set to 0.9: this has the function to decrease the value now attributed to future q values and rewards
• Max_q[q[s[t], a[t]]] represents the maximum value within the set of 2 or 4 q values corresponding to the 2 or 4 actions
• s(t-1) is the previous states vector, and s_i is an element of it
Q-Learning algorithm
This is how the simulation will broadly work. In the first trial, you will set your agent in the world and it will start exploring randomly the environment until it finds the reward. When it reaches the reward, the q value of the last action performed increases according to the learning equation. The first trial ends and the second one starts with the agent set again in the initial position. The agent will move again and will learn nothing until it reaches the position to which it assigned a value in the first trial (the one visited before the reward). Thus, the q-values/weights matrix are progressively updated moving away from the rewarded position. Therefore, the larger the word, the larger the number of trials and steps the agent will need to learn to know what to do at each position.
Here the pseudo-code of how the algorithm works
1. Initialise the relevant variables
2. Loop over trials
3. Reset relevant variables
4. Set the world to the initial state (agent at the starting position)
5. Loop over the steps of the trial
6. Store values (which become the "values at t-1") of critical variables before updating them: input, q-values, action
7. Update the state of the world based on the action a[t-1]
8. Compute the reward r[t] returned to the agent by the world
9. Activate the input units of the network based on s[t] (this represent the activation of sensors)
10. Implement the spreading of the net to compute the q[t] values
11. Compute the probabilities of action, p[a], with the SoftMax function
12. Select action a based on p[a]: this action will be executed at the next time step
13. If this is not the first or the last step of the trial:
14. update the network weights based on the equation above
15. If this is the last step of the trial (r=1)
16. update the network weights based on the equation above but considering zero instead of Max_q as "there is no future" in the trial
As you can see, regarding the learning rule:
• In the first step of each trial, the agent does not learn: the reason is that the algorithm needs to succeed time steps to learn as its learning is based on the consequences (in terms of the reward) that the action selected at t-1 causes at t
• In the last step of the trial, the algorithm does not consider the maximum current q, corresponding to an estimate of the future rewards, but rather a zero value as the trial ends and so “there is no future” for the agent, hence no future rewards
To see if your program works well, try this simulation:
• Set a 1D world of nine positions (two actions)
• Set the reward for the last position on the right
• Set the start in the first position on the left
• Set a maximum of 200 Trials and a maximum of 50 steps per trial
• Set LearningRate = 0.1 and DiscountFactor = 0.9
• Set a random selection of actions (so not following the SoftMax probabilities, but rather a 0.5 probability for each action)
• Let it run!
If the program is correct, you should obtain q-values (i.e., the network connection weights) similar to the following ones (horizontal 0-8: positions; vertical 0-1: the two actions, East and West; note how the values of the West action are equal to 0, γ^0=0.9^0=1, γ^1=0.9^1=0.9, γ^2=0.9^2=0.81, etc.). | 8,796 | 37,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.918233 |
https://dsp.stackexchange.com/questions/20406/low-order-sinc-interpolation-vs-polynomial-interpolation-for-variable-fractiona | 1,723,101,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00808.warc.gz | 167,015,955 | 42,360 | Low order sinc interpolation vs. polynomial interpolation for variable fractional delay
I'm implementing a variable fractional delay element for use in online audio processing. Applications include ie. Karplus-Strong synthesis, flanger, chorus, echo, vibrato. I'm not oversampling, so drop-sample and linear interpolation are not really acceptable for interpolation.
Could a low order sinc FIR work as an alternative to 2:nd+ order polynomial interpolation? By low order I mean on the order of just a few zero crossings.
I really like sinc interpolation because it offers two advantages over polynomial interpolation.
1. Coefficients (table of windowed sinc values) are independent of the delay amount, whereas in polynomial interpolation the filter coefficients are an N:th order polynomial of the delay amount. This complicates the code when the amount of delay is variable.
2. Coefficients are independent of interpolation order. For polynomial interpolation you need precomputed polynomials in the delay amount for all the orders you might need.
The main drawback I see with higher order sinc interpolation is the amount of latency it imposes. N+1 point polynomial interpolation of order N gives just N/2 samples of latency.
I have also considered 1 pole allpass interpolation, but I suspect it will have bandwidth problems similar to linear interpolation. (Is multi pole allpass interpolation a thing?)
• If you are using a table of windowed sinc values, doesn't that mean your delay resolution is limited? Commented Feb 7, 2015 at 14:58
• Yes, delay resolution is limited. With a large enough table this is not much of a problem. This is the oversampling route. It is also possible to use linear interpolation "between" stored coefficient values. In fact I believe this is what libsamplerate (Secret Rabbit Code) does. Commented Feb 7, 2015 at 16:02
• For simple windows, one can also just compute each needed sample point of a windowed Sinc interpolation kernel while filtering. On some CPUs these days, calling a few vectorized transcendental functions may be faster than a cache miss of a large table lookup. Commented Feb 8, 2015 at 4:26
• I think you should try the Farrow structure. Note that you don't need to do polynomial interpolation with that structure, but the filters can be designed in any way you like. A frequency domain design might be most useful. You can trade off latency and performance. Also, oversampling by 2, Farrow structure, and downsampling can be cheaper than implementing a single-rate filter that performs well close to Nyquist. Commented Feb 8, 2015 at 11:17
• A Farrow structure is just a piecewise polynomial interpolator of a low pass kernel, often a few lobes of a windowed Sinc. Commented Feb 13, 2015 at 13:13 | 597 | 2,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.900618 |
http://www.abovetopsecret.com/forum/thread652224/pg2 | 1,513,506,590,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948595342.71/warc/CC-MAIN-20171217093816-20171217115816-00369.warc.gz | 303,535,926 | 10,485 | It looks like you're using an Ad Blocker.
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# How close is this! ?
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posted on Jan, 17 2011 @ 05:04 PM
even if this comet was 270 meters?
I have to go now, ill be back on tomorrow
edit on 17-1-2011 by Itop1 because: (no reason given)
posted on Jan, 17 2011 @ 05:13 PM
reply to post by Itop1
Also, you should join Yahoo Answers forum. It's not only pretty fun, but if you ever have a random question like that, you can get on there and ask (that would be in the Science & Mathematics/Astronomy section), and within a few minutes you'll get an answer back...
posted on Jan, 17 2011 @ 09:21 PM
Originally posted by Itop1
even if this comet was 270 meters?
270 meters is the size of a small hill. How much gravity does a small hill have?
Re: the distance: I often use convertit.com to solve these sorts of problems. Just type in .0037AU in the left box and the unit you want (km, miles, cubits...) in the right box and hit "enter". Voila! 553,512.1227 km (or 1,210,656,436.35171 Biblical cubits
).
At that distance, the Earth's gravity exerts a pull equal to 1/ 7548th of what we feel on the surface, so not much (though it may tweak the object's orbit very slightly - these things add-up over time).
Good questions and good thread!
posted on Jan, 17 2011 @ 09:49 PM
reply to post by pazcat
Hey pazcat, cool graphics! Any idea what would happen if it hit our moon?
Also, someone posted the RSOE EDIS site. There's a scolling banner across the top, which I've never seen them post before. It says...
Climate Change News : 17.01.2011 09:58 - Melting glaciers threaten Peru on many fronts | 17.01.2011 10:04 - Extreme weather and climate change.
Weird, huh?
Here's that link again
Edit to add: Ok, off topic here (unless I can stretch it to relate to 'impact') but I clicked on the bird icon over California on the same site and the event description said:
The mystery surrounding 100 dead birds between Healdsburg and Geyserville appears to be solved. The European starlings were found dead on the roadway and shoulder at Highway 101 and Independence Lane, just south of Geyserville Saturday afternoon. State wildlife officials Tuesday said the birds were hit by a truck, a conclusion drawn from a witness statement and the condition of the birds at the scene. “All the signs point to blunt force trauma,” said Andrew Hughan, spokesman for the state Department of Fish and Game. “It looks like they were hit by a truck.”.... ...brief necropsies showed the birds suffered broken necks, Hughan said.... ...It appears a swarm of the birds flew directly in front of a big rig, he said. The big rig didn’t stop, but a witness later called authorities to report seeing the bird strike, Hughan said. “She said it looked like they were committing suicide,” he said.
Starlings, huh? 100 hit by one truck??? Looked like they were... committing suicide?!
Sorry, back to the topic at hand, folks!
edit on 1/17/2011 by new_here because: (no reason given)
new topics
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1 | 811 | 3,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-51 | longest | en | 0.922712 |
https://git.uwaterloo.ca/pjentsch/covidalertabm/-/blame/0b817375239f6e61886ca8b0cb3e19f3ec651fb2/CovidAlertVaccinationModel/src/ABM/solve.jl | 1,674,981,189,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00862.warc.gz | 295,084,436 | 35,916 | solve.jl 11.7 KB
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Peter Jentsch committed May 11, 2021 1 2 ``````function contact_weight(β, contact_time) return 1 - (1-β)^contact_time `````` 3 ``````end `````` Peter Jentsch committed Apr 21, 2021 4 `````` `````` Peter Jentsch committed May 11, 2021 5 ``````function Φ(payoff,ξ) `````` Peter Jentsch committed May 30, 2021 6 `````` return 1 / (1 + exp(-1*ξ*payoff)) `````` 7 8 ``````end `````` Peter Jentsch committed Jun 30, 2021 9 10 11 `````` @views function update_alert_durations!(t,agent,modelsol) `````` 12 `````` @unpack notification_parameter,notification_threshold = modelsol.params `````` Peter Jentsch committed Jun 30, 2021 13 14 15 16 17 18 19 20 21 `````` @unpack time_of_last_alert,inf_network,covid_alert_notifications,is_app_user, output_data = modelsol for j in 2:14 covid_alert_notifications[j-1,agent] = covid_alert_notifications[j,agent] #shift them all back end total_weight_i = 0 for mixing_graph in inf_network.graph_list[t], neighbor in neighbors(mixing_graph,agent) if is_app_user[neighbor] total_weight_i+= get_weight(mixing_graph,GraphEdge(agent,neighbor)) `````` 22 23 `````` end end `````` Peter Jentsch committed Jun 30, 2021 24 25 26 27 28 29 30 31 32 33 34 `````` coin_flip = 1 - (1 - notification_parameter)^total_weight_i r = rand(Random.default_rng(Threads.threadid())) if r < coin_flip covid_alert_notifications[end,agent] = 1 #add the notifications for today else covid_alert_notifications[end,agent] = 0 end if sum(covid_alert_notifications[:,agent])>=notification_threshold output_data.daily_total_notifications[t] += 1 time_of_last_alert[agent] = t end `````` 35 ``````end `````` Peter Jentsch committed Jul 01, 2021 36 37 ``````function agent_transition!(t,modelsol,agent, from::AgentStatus,to::AgentStatus) @unpack u_next_inf,output_data, immunization_countdown = modelsol `````` Peter Jentsch committed Jul 05, 2021 38 39 40 `````` immunization_countdown[agent] = -1 output_data.recorded_status_totals[Int(from), t+1] -= 1 output_data.recorded_status_totals[Int(to), t+1] += 1 `````` Peter Jentsch committed Jun 27, 2021 41 42 `````` u_next_inf[agent] = to end `````` Peter Jentsch committed Jul 01, 2021 43 ``````function is_susceptible_infected!(t, modelsol, agent,agent_inf_status,agent_demo,α_vec,β_vec) `````` Peter Jentsch committed Jun 27, 2021 44 45 `````` @unpack u_inf,inf_network, output_data = modelsol `````` Peter Jentsch committed Jun 30, 2021 46 47 48 49 50 51 52 `````` for mixing_graph in inf_network.graph_list[t], neighbor in neighbors(mixing_graph,agent) if u_inf[neighbor] == Infected infection_threshold = contact_weight(β_vec[Int(agent_demo)],get_weight(mixing_graph,GraphEdge(agent,neighbor))) if rand(Random.default_rng(Threads.threadid())) < infection_threshold && agent_inf_status == Susceptible output_data.daily_cases_by_age[Int(agent_demo),t]+=1 output_data.daily_unvac_cases_by_age[Int(agent_demo),t]+=1 `````` Peter Jentsch committed Jul 01, 2021 53 `````` agent_transition!(t,modelsol, agent, Susceptible,Infected) `````` Peter Jentsch committed Jun 30, 2021 54 `````` return true `````` 55 `````` end `````` Peter Jentsch committed Jun 27, 2021 56 57 58 59 `````` end end return false end `````` Peter Jentsch committed Jul 01, 2021 60 ``````function is_immunized_infected!(t, modelsol, agent,agent_inf_status,agent_demo,α_vec,β_vec) `````` Peter Jentsch committed Jun 30, 2021 61 62 63 64 65 66 67 `````` @unpack u_inf,inf_network, output_data = modelsol for mixing_graph in inf_network.graph_list[t], neighbor in neighbors(mixing_graph,agent) if u_inf[neighbor] == Infected infection_threshold = contact_weight(β_vec[Int(agent_demo)],get_weight(mixing_graph,GraphEdge(agent,neighbor))) if rand(Random.default_rng(Threads.threadid())) < infection_threshold && agent_inf_status == Immunized immunization_ineffective = rand(Random.default_rng(Threads.threadid())) < 1- α_vec[Int(agent_demo)] if immunization_ineffective `````` Peter Jentsch committed Jul 01, 2021 68 `````` agent_transition!(t,modelsol,agent, Immunized,Infected) `````` Peter Jentsch committed Jun 30, 2021 69 70 71 `````` output_data.daily_cases_by_age[Int(agent_demo),t]+=1 return true end `````` 72 73 `````` end end `````` Peter Jentsch committed Jun 30, 2021 74 75 76 77 78 79 `````` end return false end function vaccinate_agent!(t,agent,immunization_countdown) if immunization_countdown[agent] == -1 #not currently in countdown immunization_countdown[agent] = 14 `````` 80 81 `````` end end `````` 82 `````` `````` Peter Jentsch committed Sep 24, 2021 83 ``````function update_vaccination_opinion_state!(t,agent,agent_demo,modelsol,vaccinate_today_flag,total_infections,π_base_vec) `````` Peter Jentsch committed Jul 01, 2021 84 85 `````` @unpack infection_introduction_day,immunization_intervals, η,Γ,ζ, ω, ω_en,ξ = modelsol.params @unpack immunization_countdown,demographics,time_of_last_alert, soc_network,u_vac,u_next_vac,is_app_user,output_data = modelsol `````` Peter Jentsch committed Jun 30, 2021 86 87 88 89 90 91 `````` random_soc_network = sample(Random.default_rng(Threads.threadid()), soc_network.graph_list[t]) if !isempty(neighbors(random_soc_network,agent)) random_neighbour = sample(Random.default_rng(Threads.threadid()), neighbors(random_soc_network.g,agent)) app_vac_payoff = 0.0 if is_app_user[agent] && time_of_last_alert[agent]>=0 output_data.daily_total_notified_agents[t] += 1 `````` Peter Jentsch committed Jul 09, 2021 92 93 `````` app_vac_payoff = Γ^((t - time_of_last_alert[agent])) * (η + total_infections*ω_en) `````` Peter Jentsch committed Jun 30, 2021 94 `````` end `````` Peter Jentsch committed Jul 09, 2021 95 `````` output_data.total_app_weight[t] += app_vac_payoff `````` Peter Jentsch committed Jun 30, 2021 96 97 98 99 100 `````` if u_vac[random_neighbour] == u_vac[agent] vac_payoff = π_base_vec[Int(demographics[agent])] + total_infections*ω + app_vac_payoff if u_vac[agent] if rand(Random.default_rng(Threads.threadid())) < 1 - Φ(vac_payoff,ξ) u_next_vac[agent] = false `````` Peter Jentsch committed Jun 20, 2021 101 `````` end `````` Peter Jentsch committed Jun 30, 2021 102 103 104 `````` else if rand(Random.default_rng(Threads.threadid())) < Φ(vac_payoff,ξ) u_next_vac[agent] = true `````` Peter Jentsch committed Jul 05, 2021 105 `````` `````` Peter Jentsch committed Jul 08, 2021 106 `````` if vaccinate_today_flag && t>infection_introduction_day `````` Peter Jentsch committed Jul 01, 2021 107 `````` vaccinate_agent!(t, agent, immunization_countdown) `````` Peter Jentsch committed Sep 24, 2021 108 `````` # fit!(output_data.avg_weighted_degree_of_vaccinators[Int(agent_demo)],weighted_degree_of_i) `````` Peter Jentsch committed Jul 08, 2021 109 `````` if !is_app_user[agent] `````` Peter Jentsch committed Sep 24, 2021 110 `````` # fit!(output_data.avg_weighted_degree_of_vaccinators_no_EN[Int(agent_demo)],weighted_degree_of_i) `````` Peter Jentsch committed Jul 08, 2021 111 `````` end `````` Peter Jentsch committed Jul 01, 2021 112 `````` end `````` Peter Jentsch committed May 06, 2021 113 `````` end `````` Peter Jentsch committed Jun 30, 2021 114 `````` end `````` Peter Jentsch committed Jul 01, 2021 115 116 `````` return vac_payoff `````` 117 118 `````` end end `````` Peter Jentsch committed Jul 01, 2021 119 `````` return 0 `````` 120 ``````end `````` Peter Jentsch committed Sep 24, 2021 121 122 123 124 125 126 ``````using WeightedOnlineStats function assortativity(t,network::TimeDepMixingGraph,u_inf) total_weight = 0.0 x_stats = WeightedVariance() y_stats = WeightedVariance() `````` 127 `````` `````` Peter Jentsch committed Jun 16, 2021 128 `````` for g in network.graph_list[t] `````` Peter Jentsch committed Sep 24, 2021 129 130 131 132 133 134 135 `````` for e in edges(g.g) v = src(e) w = dst(e) weight = get_weight(g,GraphEdge(v,w)) total_weight += weight fit!(x_stats,(u_inf[v] == Immunized),weight) fit!(y_stats,(u_inf[v] != Immunized),weight) `````` Peter Jentsch committed May 10, 2021 136 137 `````` end end `````` Peter Jentsch committed Sep 24, 2021 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 `````` σ_x = std(x_stats) σ_y = std(y_stats) x_bar = mean(x_stats) y_bar = mean(y_stats) numerator = 0.0 join_count_11 = 0.0 join_count_01 = 0.0 join_count_00 = 0.0 for g in network.graph_list[t] for e in edges(g.g) v = src(e) w = dst(e) weight = get_weight(g,GraphEdge(v,w)) v_inf = u_inf[v] == Immunized w_inf = u_inf[w] != Immunized numerator += weight*(v_inf - x_bar) * (w_inf - y_bar) #ughhhh join_count_11 += ifelse( u_inf[v] == Immunized && u_inf[w] == Immunized,weight,0) join_count_01 += ifelse(xor( u_inf[v] == Immunized, u_inf[w] == Immunized),weight,0) join_count_00 += ifelse(!(u_inf[v] == Immunized) && !(u_inf[w] == Immunized),weight,0) end end return numerator/(total_weight*σ_x*σ_y),(join_count_11,join_count_01,join_count_00) `````` Peter Jentsch committed May 10, 2021 162 ``````end `````` 163 `````` `````` Peter Jentsch committed Jun 14, 2021 164 165 166 167 168 169 170 171 172 173 174 175 176 177 ``````function sample_initial_nodes(nodes,graphs,I_0_fraction) weighted_degrees = zeros(nodes) for v in 1:nodes for g in graphs for w in neighbors(g,v) weighted_degrees[v] += get_weight(g,GraphEdge(v,w)) end end end wv = Weights(weighted_degrees ./sum(weighted_degrees)) num = round(Int,nodes*I_0_fraction) init_indices = sample(Random.default_rng(Threads.threadid()), 1:nodes,wv, num; replace = false) return init_indices end `````` 178 `````` `````` Peter Jentsch committed Jul 08, 2021 179 180 181 `````` function agents_step!(t,modelsol) @unpack EN_intervals,immunizing,infection_introduction_day,recovery_rate,β_y,β_m,β_o,α_y,α_m,α_o,π_base_y,π_base_m,π_base_o,ζ,immunization_intervals = modelsol.params `````` Peter Jentsch committed Sep 24, 2021 182 `````` @unpack u_inf,u_vac,u_next_vac,u_next_inf,demographics,inf_network,soc_network, is_app_user, immunization_countdown, output_data = modelsol `````` Peter Jentsch committed Jun 30, 2021 183 `````` `````` Peter Jentsch committed Jul 08, 2021 184 185 186 187 188 189 190 `````` vaccinate_today_flag = !immunizing ? false : (any(t in I for I in immunization_intervals) ? true : false) EN_today_flag = t in EN_intervals ? true : false `````` Peter Jentsch committed Jun 30, 2021 191 192 `````` u_next_inf .= u_inf u_next_vac .= u_vac `````` Peter Jentsch committed Jul 05, 2021 193 `````` modelsol.output_data.recorded_status_totals[:,t+1] .= modelsol.output_data.recorded_status_totals[:,t] `````` Peter Jentsch committed Jul 09, 2021 194 `````` total_infected = modelsol.output_data.recorded_status_totals[Int(Infected), t] `````` Peter Jentsch committed Jun 30, 2021 195 196 197 198 199 200 201 `````` β_vec = @SVector [β_y,β_m,β_o] α_vec = @SVector [α_y,α_m,α_o] if t0 immunization_countdown[agent] -= 1 end `````` Peter Jentsch committed Jul 11, 2021 236 `````` if agent_is_vaccinator `````` Peter Jentsch committed Sep 24, 2021 237 `````` # fit!(output_data.avg_weighted_degree_of_vaccinator_belief[t],weighted_degree_of_i) `````` Peter Jentsch committed Jul 11, 2021 238 `````` end `````` Peter Jentsch committed Sep 24, 2021 239 `````` update_vaccination_opinion_state!(t, agent,agent_demo, modelsol,vaccinate_today_flag, total_infected,π_base_vec) `````` Peter Jentsch committed Jun 30, 2021 240 241 242 243 `````` end output_data.total_vaccinators[t] = count(==(true),u_vac) u_vac .= u_next_vac u_inf .= u_next_inf `````` Peter Jentsch committed Jul 01, 2021 244 `````` # return total_weight `````` Peter Jentsch committed Jul 01, 2021 245 `````` `````` Peter Jentsch committed Jun 30, 2021 246 247 ``````end `````` Peter Jentsch committed Jun 10, 2021 248 `````` `````` Peter Jentsch committed Jun 23, 2021 249 ``````function solve!(modelsol) `````` Peter Jentsch committed Jun 14, 2021 250 `````` init_indices = sample_initial_nodes(modelsol.nodes, modelsol.inf_network.graph_list[begin], modelsol.params.I_0_fraction) `````` Peter Jentsch committed Jul 01, 2021 251 `````` # total_weight = 0.0 `````` Peter Jentsch committed Jul 01, 2021 252 `````` `````` Peter Jentsch committed Jun 14, 2021 253 `````` for t in 1:modelsol.sim_length `````` Peter Jentsch committed Jun 13, 2021 254 `````` #this also resamples the soc network weights since they point to the same objects, but those are never used `````` Peter Jentsch committed Jun 27, 2021 255 `````` `````` Peter Jentsch committed Jun 13, 2021 256 `````` if t>1 `````` Peter Jentsch committed Jun 23, 2021 257 `````` remake_all!(t,modelsol.inf_network,modelsol.index_vectors,modelsol.demographics) `````` Peter Jentsch committed Jun 13, 2021 258 259 260 261 262 263 `````` end if t>modelsol.params.infection_introduction_day if !isempty(init_indices) inf_index = pop!(init_indices) modelsol.u_inf[inf_index] = Infected `````` Peter Jentsch committed Jul 09, 2021 264 `````` modelsol.output_data.recorded_status_totals[Int(Infected),t] += 1 `````` Peter Jentsch committed Jun 13, 2021 265 `````` end `````` Peter Jentsch committed Jun 17, 2021 266 267 `````` end `````` Peter Jentsch committed Jun 30, 2021 268 `````` `````` Peter Jentsch committed Jun 27, 2021 269 `````` `````` Peter Jentsch committed Jul 08, 2021 270 `````` agents_step!(t,modelsol) `````` 271 `````` #advance agent states based on the new network `````` Peter Jentsch committed Jun 13, 2021 272 `````` `````` Peter Jentsch committed Jun 23, 2021 273 `````` record!(t,modelsol) `````` Peter Jentsch committed Jun 27, 2021 274 `````` # display(mean.(modelsol.output_data.avg_weighted_degree)) `````` Peter Jentsch committed Jun 26, 2021 275 `````` `````` 276 `````` end `````` Peter Jentsch committed Jul 01, 2021 277 `````` # display(total_weight) `````` Peter Jentsch committed Jul 08, 2021 278 ``end`` | 4,385 | 13,715 | {"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, 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# The second angle of a triangle is three times as large as the
### Customer Question
The second angle of a triangle is three times as large as the first. The third angle is 30 degrees more than the first.
Find the measure of each angle.
Submitted: 1 year ago.
Category: Math Homework
Expert: Ashok Kumar replied 1 year ago.
Expert: Ashok Kumar replied 1 year ago.
Let A, B and c be triangle angles.
Let us call A as first angle, B as second angle and C as third angle.
Sum of three angles is 180 deg.
Hence, A+B+C = 180 ...(1)
second angle is three times as large as the first
B = 3A
third angle is 30 degrees more than the firs
c = A+30
Putting these in (1), we get
A + 3A+A+30 = 180
Or, 5A = 180-30 = 150
Or, A = 150/5=30
B = 30*3 = 90
C = 30+30 = 60 | 292 | 976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-26 | latest | en | 0.920047 |
https://r.789695.n4.nabble.com/Polynomial-fitting-td3652816.html | 1,586,467,157,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371876625.96/warc/CC-MAIN-20200409185507-20200409220007-00083.warc.gz | 645,788,838 | 14,549 | # Polynomial fitting
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## Polynomial fitting
Hello, i'm fairly familiar with R and use it every now and then for math related tasks. I have a simple non polynomial function that i would like to approximate with a polynomial. I already looked into poly, but was unable to understand what to do with it. So my problem is this. I can generate virtually any number of datapoints and would like to find the coeffs a1, a2, ... up to a given degree for a polynomial a1x^1 + a2x^2 + ... that approximates my simple function. How can i do this with R? Your help will be highly appreciated!
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## Re: Polynomial fitting
Hello, mfa (Matti?), if x and y contain the coordinates of your data points and k is the wanted polynomial degree, then fit <- lm( y ~ poly( x, k)) fits orthonormal polynomials up to degree k to your data. Using dummy.coef( fit) should give the coefficients you are interested in. Hth -- Gerrit On Thu, 7 Jul 2011, mfa wrote: > Hello, > > i'm fairly familiar with R and use it every now and then for math related > tasks. > > I have a simple non polynomial function that i would like to approximate > with a polynomial. I already looked into poly, but was unable to understand > what to do with it. So my problem is this. I can generate virtually any > number of datapoints and would like to find the coeffs a1, a2, ... up to a > given degree for a polynomial a1x^1 + a2x^2 + ... that approximates my > simple function. How can i do this with R? > > Your help will be highly appreciated! > > -- > View this message in context: http://r.789695.n4.nabble.com/Polynomial-fitting-tp3652816p3652816.html> Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. > --------------------------------------------------------------------- Dr. Gerrit Eichner Mathematical Institute, Room 212 [hidden email] Justus-Liebig-University Giessen Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany Fax: +49-(0)641-99-32109 http://www.uni-giessen.de/cms/eichner______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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| | 647 | 2,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-16 | latest | en | 0.885496 |
http://stackoverflow.com/questions/15283816/image-with-accordion-effect | 1,430,187,046,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246660493.3/warc/CC-MAIN-20150417045740-00026-ip-10-235-10-82.ec2.internal.warc.gz | 254,493,172 | 18,984 | # Image with accordion effect
I have read in an image file to MATLAB and I am trying to stretch it in one direction, but a variable amount (sinusoidal). This would create an accordion affect on the image. I have toyed around with imresize, however that only resizes the image linearly. I would like the amount of "stretch" to vary for each image line. I tried to convey this with the following code:
``````periods = 10; % Number of "stretch" cycles
sz = size(original_image,2)/periods;
s = 0;
x = 0;
for index = 1:periods
B = original_image(:,round(s+1:s+sz));
if mod(index,2) == 0
amp = 1.5;
else
amp = 0.75;
end
xi = size(B,2)*amp;
new_image(:,x+1:x+xi) = imresize(B, [size(B,1) size(B,2)*amp]);
s = s + sz;
x = x+xi;
end
``````
You can see that segments of the image are stretched, then compressed, then stretched, etc, like an accordion. However, each segment has a uniform amount of stretch, whereas I'd like it to be increasing then decreasing as you move along the image.
I have also looked at MATLAB's example of Applying a Sinusoidal Transformation to a Checkerboard which seems very applicable to my problem, however I have been trying and I cannot get this to produce the desired result for my image. Any help is much appreciated.
UPDATE:
Thank you for Answer #1. I was unable to get it to work for me, but also realized it would resulted in loss of data, as the code only called for certian lines in the original image, and other lines would have been ignored.
After experimenting further, I developed the code below. I used a checkerboard as an example. While combersome, it does get the job done. However, upon trying the script with an actual high-resolution image, it was extremely slow and ended up failing due to running out of memory. I believe this is because of the excessive number of "imresize" commands that are used in loop.
``````I = checkerboard(10,50);
I = imrotate(I,90);
[X Y] = size(I);
k = 4; % Number of "cycles"
k = k*2;
x = 1;
y = 2;
z = 2;
F = [];
i = 1;
t = 0;
s = 0;
for j = 1:k/2
t = t + 1;
for inc = round(s+1):round(Y/k*t)
Yi = i + 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z + 1;
i = i + 1;
end
y = y - 2;
z = z - 4;
for inc = round(Y/k*t+1):round(Y/k*(t+1))
Yi = i - 1;
F(:,(x:y)) = imresize(I(:,(inc:inc)),[X Yi]);
x = y + 1;
y = x + z;
z = z - 1;
i = i - 1;
end
y = y + 2;
z = z + 4;
s = Y/k*(t+1);
t = t + 1;
end
Fn = imresize(F, [X Y]);
imshow(Fn);
``````
Does anyone know of a simpler way to achieve this? If you run the code above, you can see the effect I am trying to achieve. Unfortunately, my method above does not allow me to adjust the amplitude of the "stretch" either, only the number of "cycles," or frequency. Help on this would also be appreciated. Much thanks!
-
Two levels of for loops and an imresize inside the inner loop is definitely a performance killer. Matlab is vector oriented and you want to avoid loops at all costs if good performance is your goal. Also, you have an (inc:inc) that seems suspicious, though I didn't try to look at the details of your implementation. – Lolo Mar 9 '13 at 3:45
Here is how I would approach it:
1. Determine how the coordinate of each point in your Final image F maps into your Initial image I of size (M,N)
Since you want to stretch horizontally only, given a point (xF,yF) in your final image, that point would be (xI,yI) in your initial image where xI and yI can be obtained as follows:
yI = yF;
xI = xF + L*sin(xF*K);
Notes:
• these equations do not guarantee that xI remains within the range [1:N] so cropping needs to be added
• K controls the how many wrinkles you want to have in your accordion effect. For example, if you only want one wrinkle, K would be 2*pi/N
• L controls how much stretching you want to apply
2. Then simply express your image F from image I with the transforms you have in 1.
Putting it all together, the code below creates a sample image I and generates the image F as follows:
`````` % Generate a sample input image
N=500;
xF=1:N;
I=(1:4)'*xF/N*50;
% Set the parameters for your accordion transform
K=2*pi/N;
L=100;
% Apply the transform
F=I(:, round(min(N*ones(1,N), max(ones(1,N), (xF + L*sin(xF*K))))) );
% Display the input and output images side by side
image(I);
figure;
image(F);
``````
If you run this exact code you get:
As you can see, the final image on the right stretches the center part of the image on the left, giving you an accordion effect with one wrinkle.
You can fiddle with K and L and adjust the formula to get the exact effect you want, but note how by expressing the transform in a matrix form MATLAB executes the code in a fraction of second. If there is one take away for you is that you should stay away from for loops and complex processing whenever you can.
Have fun!
-
Thank for for your answer. Please see update in my original post. – user1668909 Mar 8 '13 at 20:21
Why does the code "not work"? But you are correct that this approach will omit points from the original image. If you do not want that, you could proceed to doing an interpolation (linear if you want) between I(floor(xl)) and I(ceil(xl)), which you can still implement in one line of code. This approach will be many many orders of magnitude faster than the code you have, since your code has an imresize inside two levels of loops. – Lolo Mar 9 '13 at 3:42
@user1668909 I updated my code and ran it to illustrate what I had in mind. You should be able to run that code and play with it further. – Lolo Mar 9 '13 at 17:49
That is exactly what I was trying to do. Very elegant solution. Thank you very much for all your help! Much appreciated. – user1668909 Mar 11 '13 at 17:49
Glad it helped. It was fun to take a little break to play with this. – Lolo Mar 11 '13 at 17:58 | 1,622 | 5,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2015-18 | latest | en | 0.962777 |
https://www.studypool.com/discuss/486703/home-work-problems-need-help?free | 1,508,839,272,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00244.warc.gz | 998,164,784 | 14,415 | Time remaining:
##### home work problems need help
label Mathematics
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5
how many 1/2 cup servings are in a package of cheese that contains 5 1/4 cups altogether
Oct 23rd, 2017
5 1/4 cups =. 21/4
So 21/4 divided by 1/2 gives us. 21/4 x 2 = 10 1/2 servings
Therefore select 10 full servings
Apr 21st, 2015
...
Oct 23rd, 2017
...
Oct 23rd, 2017
Oct 24th, 2017
check_circle | 156 | 446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.868491 |
http://forums.4aynrandfans.com/index.php?/topic/12251-the-logical-leap-and-criticism/&page=21 | 1,601,565,646,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131777.95/warc/CC-MAIN-20201001143636-20201001173636-00192.warc.gz | 44,775,055 | 27,910 | # The Logical Leap and criticism
## 529 posts in this topic
Dead men feel no pain is a generalization.
Again, this is a deduction of a new instance from two concepts ("Dead men" and "pain"). If you know what the two are, it's a simple matter to deduce the rest. Calling this "a generalization" is very sloppy.
It's a generalization is but is no more general than the premises in the deduction; it's not a generalization of the premises. It's not an example of an induction at all as it has been presented.
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Wet tongues stick to cold flag poles is a generalization.
So is boiling water burns hands.
So is pushing a ball makes it roll.
So is fire burns.
So is taking morphine relieves pain.
The induction is the fact that wet tongues stuck to cold flag poles cause pain.
But that is not the statement you made originally.
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In post #485 ewv wrote:
"Inducing" the conclusion from a single instance or simple enumeration is not valid inductive reasoning.
On page 24 of TLL David Harriman writes:
(Q)uantity of instances alone is irrelevant to induction. In first-level induction a single instance is sufficient.
I would be interested in ewv's comments.
Quantity alone is irrelevant in that it does not establish a conclusion. With at least some minimal additional thought it can lead to an hypothesis. "Quantity has a quality all its own", but is not a proof.
I reject his characterization of what a first level propositional generalization is and the idea that a single instance is sufficient by itself. There is no such thing as an automatic generalization with no additional thought making connections.
Thanks. My own thoughts exactly.
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Wet tongues stick to cold flag poles is a generalization.
So is boiling water burns hands.
So is pushing a ball makes it roll.
So is fire burns.
So is taking morphine relieves pain.
The induction is the fact that wet tongues stuck to cold flag poles cause pain.
It would be nice if you would offer some proof or evidence of such assertions. What exactly are you generalizing? What are your premises? What you offer is simply a propositional statement. Not a generalization resulting from induction. If you were to offer evidence and argument, you'd soon see that your conclusion is reached by deduction, not induction.
Tongues are wet.
Wet things freeze at low temperatures.
Low temperatures cause wet things to stick to them due to freezing.
Low temperatures cause pain to living tissues.
Therefore, wet tongues stick to frozen things and feel pain.
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"It would be nice if you would offer some proof or evidence of such assertions."
Granted this is a dramatization, but this is all the evidence I need to know that wet tongues stick to cold flag poles which causes pain. I also realize other individuals may not grasp that induction after watching the video. I am certain they will grasp the induction if they stick their tongue to the pole instead.
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Here is my one shot example of a generalization/induction.
Sticking your wet tongue to a frozen flag pole causes pain.
In what way is that a generalization? What have you generalized? What have you induced?
Put your tongue on the frozen pole and the answers to your questions will become crystal clear. Think of it as a perceptual level experiment.
If you don't want to think of it as a perceptual level experiment you can also think of it as a proposition or a triple dog dare.
The one thing I am certain of is that sticking your wet tongue tongue to a frozen flag pole causes pain.
This is my one shot example that Tom denies exists. I disagree, anyone want to try my perceptual level experiment?
Being certain or confident that something will happen does not demonstrate a generalization. If I take a sufficient amount of morphine, I will not feel pain during your experiment.
He described a single instance with no grounds for the generalization. if you understand enough about tongues and frozen metal to conclude that licking a frozen metal pipe causes pain, that is a deduction not an induction. "Inducing" the conclusion from a single instance or simple enumeration is not valid inductive reasoning. A strong feeling of fear does not change that, which is not to say that you should keep trying it without understanding what you are doing.
It should also be pointed out that PaulsHere has conceded to the fact that sticking your wet tongue to the flag pole causes pain; otherwise, there is no need to take a sufficient amount of morphine before doing the experiment.
Wet tongues stick to cold flag poles is a generalization.
So is boiling water burns hands.
So is pushing a ball makes it roll.
So is fire burns.
So is taking morphine relieves pain.
The induction is the fact that wet tongues stuck to cold flag poles cause pain.
"It would be nice if you would offer some proof or evidence of such assertions."
Granted this is a dramatization, but this is all the evidence I need to know that wet tongues stick to cold flag poles which causes pain. I also realize other individuals may not grasp that induction after watching the video. I am certain they will grasp the induction if they stick their tongue to the pole instead.
Is it also a generalization to induce that this thread can only get repetitiously more repetitive?
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"It would be nice if you would offer some proof or evidence of such assertions."
Granted this is a dramatization, but this is all the evidence I need to know that wet tongues stick to cold flag poles which causes pain. I also realize other individuals may not grasp that induction after watching the video. I am certain they will grasp the induction if they stick their tongue to the pole instead.
But they don't grasp it by induction, as has been shown by several posters.
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"It would be nice if you would offer some proof or evidence of such assertions."
Granted this is a dramatization, but this is all the evidence I need to know that wet tongues stick to cold flag poles which causes pain. I also realize other individuals may not grasp that induction after watching the video. I am certain they will grasp the induction if they stick their tongue to the pole instead.
I don't see the pain; all I see is the stuck tongue.
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A single particular instance, i.e., the concrete example.
I watch a boy stick his tongue on a cold flag pole. The tongue sticks. The boy crys out in pain.
The generalization:
Wet tongues stick to cold flag poles.
Effect: A boy is crying out in pain.
Cause: His tongue is stuck to a cold flag pole.
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"It would be nice if you would offer some proof or evidence of such assertions."
Granted this is a dramatization, but this is all the evidence I need to know that wet tongues stick to cold flag poles which causes pain. I also realize other individuals may not grasp that induction after watching the video. I am certain they will grasp the induction if they stick their tongue to the pole instead.
I don't see the pain; all I see is the stuck tongue.
OK you don't see his pain, I do. Stick your own tongue on the pole and I am certain you will feel pain. Unless you take morphine, as you said earlier, but taking morphine concedes to the fact that sticking your tongue to a cold flag pole causes pain.
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A single particular instance, i.e., the concrete example.
I watch a boy stick his tongue on a cold flag pole. The tongue sticks. The boy crys out in pain.
The generalization:
Wet tongues stick to cold flag poles.
Effect: A boy is crying out in pain.
Cause: His tongue is stuck to a cold flag pole.
He doesn't cry out in pain. All he does is express fear that everyone has left him stuck to the pole. And your premises do not demonstrate your conclusion.
And one more major point. The video clearly demonstrates deductive reasoning. The boy who makes the challenge starts out with the generalization that wet tongues stick to poles and he challenges the other boy to try it, clearly deducing from the generalization that the individual boy's wet tongue will stick to the pole is a conclusion drawn from the generalization. Do you really claim that "Socrates is mortal" is proof that "all men are mortal'? You need to go study your logic texts.
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"It would be nice if you would offer some proof or evidence of such assertions."
Granted this is a dramatization, but this is all the evidence I need to know that wet tongues stick to cold flag poles which causes pain. I also realize other individuals may not grasp that induction after watching the video. I am certain they will grasp the induction if they stick their tongue to the pole instead.
I don't see the pain; all I see is the stuck tongue.
OK you don't see his pain, I do. Stick your own tongue on the pole and I am certain you will feel pain. Unless you take morphine, as you said earlier, but taking morphine concedes to the fact that sticking your tongue to a cold flag pole causes pain.
No it doesn't. It demonstrates that your premise is not based upon particular instances and is thus not induction.
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A single particular instance, i.e., the concrete example.
I watch a boy stick his tongue on a cold flag pole. The tongue sticks. The boy crys out in pain.
The generalization:
Wet tongues stick to cold flag poles.
Effect: A boy is crying out in pain.
Cause: His tongue is stuck to a cold flag pole.
He doesn't cry out in pain. All he does is express fear that everyone has left him stuck to the pole. And your premises do not demonstrate your conclusion.
And one more major point. The video clearly demonstrates deductive reasoning. The boy who makes the challenge starts out with the generalization that wet tongues stick to poles and he challenges the other boy to try it, clearly deducing from the generalization that the individual boy's wet tongue will stick to the pole is a conclusion drawn from the generalization. Do you really claim that "Socrates is mortal" is proof that "all men are mortal'? You need to go study your logic texts.
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Doesn't seeing concept formation as a non-inductive process - as LP described in the exchange quoted by Mercury - imply that concept formation is based solely on some subjective process of the concept-former, independent of actual metaphysical attributes and the fact that some attributes are in fact more important than others in forming these generalizations?
But aren't concepts the product of differentiation, not generalizations? You observe entities and group them together by the differences in their characteristics. That's not induction. A child forms the concept of dog by, for example, observing that a dog barks while a cat meows. It's classification, not logic.
Concepts are not "a product" of differentiation. Units are grouped by similarity, not differences. Similarity is in comparison with a commensurable characteristic that is different, as in two shades of red similar against blue. The similar units grouped together are integrated into a single mental unit to form the concept, as opposed to leaving only a group. The generalization is in including an open-ended number of other units that share the essential characteristic; you are widening the group to include all instances that share the essential characteristics of the particulars you start by focusing on. In Ayn Rand's words: "The process of observing the facts of reality and of integrating them into concepts is, in essence, a process of induction." The widening of units in a classification to include all of them is not the same as inductively inferring properties for all members of an existing class from the particulars you observe.
I take it that properties and characteristics are synonymous. Are you saying the generalization applies only to the essential characteristic(s) of a concept's units and all others are excluded from the generalization? If so, is it because the other characteristics are only implicit within the definition?
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I take it that properties and characteristics are synonymous. Are you saying the generalization applies only to the essential characteristic(s) of a concept's units and all others are excluded from the generalization? If so, is it because the other characteristics are only implicit within the definition?
'Property' is a physical characteristic of individual things: temperature, strength, mass, etc. 'Characteristic' is a feature or quality of things or a relationship among things, such as having brown eyes or odd behavior, or similarity and difference among things. The terms are often used interchangeably.
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Doesn't seeing concept formation as a non-inductive process - as LP described in the exchange quoted by Mercury - imply that concept formation is based solely on some subjective process of the concept-former, independent of actual metaphysical attributes and the fact that some attributes are in fact more important than others in forming these generalizations?
But aren't concepts the product of differentiation, not generalizations? You observe entities and group them together by the differences in their characteristics. That's not induction. A child forms the concept of dog by, for example, observing that a dog barks while a cat meows. It's classification, not logic.
Concepts are not "a product" of differentiation. Units are grouped by similarity, not differences. Similarity is in comparison with a commensurable characteristic that is different, as in two shades of red similar against blue. The similar units grouped together are integrated into a single mental unit to form the concept, as opposed to leaving only a group. The generalization is in including an open-ended number of other units that share the essential characteristic; you are widening the group to include all instances that share the essential characteristics of the particulars you start by focusing on. In Ayn Rand's words: "The process of observing the facts of reality and of integrating them into concepts is, in essence, a process of induction." The widening of units in a classification to include all of them is not the same as inductively inferring properties for all members of an existing class from the particulars you observe.
I take it that properties and characteristics are synonymous. Are you saying the generalization applies only to the essential characteristic(s) of a concept's units and all others are excluded from the generalization? If so, is it because the other characteristics are only implicit within the definition?
A generalization as a concept pertains to the units subsumed, not the essential characteristics. The meaning of a concept is its referents, not its definition. The concept man refers to all men, not rationality. The concept of an attribute (like length, color, etc.) refers to all instances of the attribute as it exists in reality as an attribute of an object. The meaning of a second (or higher) level concept is all the concepts (and in some cases units directly perceived) on which it depends. They are the units, in the combination used to form the higher level concept. That is the full logical dependency.
A general proposition is about whatever it is says it is about. If it states something about rationality then it is about rationality. If it states something about men then it is about men. But you cannot generalize a statement about non-essential characteristics of an entity based on the commonality of the essential characteristics alone. 'Rationality' of all men as their primary means of survival as a common characteristic does not tell you that all men are mortal just because you have observed that some of them are and all of them fall under the same concept. You need further explanation tying observations to what you know about men, in conjunction with systematic exploration of observations in order to establish a full context of knowledge before generalizing within it.
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Harriman has just posted another response to those who have challenged his historical accounts of Galileo and Newton.
He accuses his critics of "belittling the achievements of Galileo and Newton" and Harriman paraphrases Jack Nicholson in the movie "A Few Good Men" saying:
“I don’t want money and I don’t want fame. What I do want is for academics to stand there in their girlie graduate gowns and extend to Galileo and Newton some freakin’ respect.”
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There is another problem with The Logical Leap and a crucial one considering its theme. David Harriman doesn't understand or explain how a concept changes, even though he repeatedly notes instances of the phenomenon occurring. An essential part of the book's theme is that the right version of a concept is a "green light" to induction. Yet Mr. Harriman displays no curiosity as to how innovators improve or change concepts to make them green lights. His observations of what great scientists do seem to clash with a fixed idea he has: that concepts don't and shouldn't change.
As a preliminary, let me note there is a respect and set of circumstances where a concept doesn't change. Ayn Rand explained it in the following discussion on page 233 of ITOE:
Prof. H: It is easy to assume that when the definition changes, the concept changes.
AR: No, the context of your knowledge changes. When you know more, you select a different essential characteristic by which to define the object, because you now have to differentiate it more precisely. Your knowledge has expanded, but the concept doesn't change.
The similarities and differences according to which you originally formed the concept still remain. So you haven't changed your concept.
But does it follow from her remarks that a concept can't change? No. A concept isn't changed by acquiring more knowledge, but it can certainly be modified by conscious intent. Does Mr Harriman understand and explain that? No. His only broad remarks about whether concepts change are these on page 13:
The concept "temperature" had the same meaning for Galileo as for Einstein, i.e., both men referred to the same physical property. The difference is only that Einstein knew much more this property; he understood its relation to heat, to motion and to the fundamental nature of matter. But that expansion of knowledge was possible only because the concept itself did not change. Because concepts are stable, we can communicate and advance our knowledge.
You may point out he is only repeating what Miss Rand said, and you're partly right. But the next thing he is remarking on page 25:
Western civilization broadened the concept of "cause," by regarding personal efficacy as merely a subtype of it. This was a crucial precondition of the development of modern science.
So it turns out it is not always valuable if a concept is stable and, in fact, it can be extremely valuable if it changes. (To broaden a concept is to change it.)
Now you'd think Mr. Harriman would have pounced on this phenomenon like a leopard. Consider how many times the book refers to a concept being changed as a crucial element in an inductive breakthrough. Here's just one from page 47, where Harriman is talking about Galileo's discovery of the increase of speed during fall:
In addition to the concept of "friction," this discovery depended on Galileo's two key concepts of motion. Throughout the above reasoning, he was using concepts of "speed" and "acceleration" that differed profoundly from those in common use at the time.
In other words, Galileo (or someone else: Harriman's remark doesn't rule out that possibility) took existing concepts and changed them. In the June edition of TIA Monthly last year I published a paper "Transforming a Concept" that explains how and why concepts change. One of my prime examples was how Newton modified the concept of "transmission" to make his breakthrough about how a medium affects the passage of light. Harry Binswanger reviewed the article, called it "interesting and thoughtful" but dismissed it because he claimed "concepts don't change". His view is symptomatic of a widespread confusion in Objectivism that goes back many years and I am disappointed that Harriman, who knows concepts can indeed change, did not attempt an explanation, but let his fixed idea and his observations sit side by side unresolved.
Those interested in such an explanation can refer to my article in TIA Monthly.
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There is another problem with The Logical Leap and a crucial one considering its theme. David Harriman doesn't understand or explain how a concept changes, even though he repeatedly notes instances of the phenomenon occurring. An essential part of the book's theme is that the right version of a concept is a "green light" to induction. Yet Mr. Harriman displays no curiosity as to how innovators improve or change concepts to make them green lights. His observations of what great scientists do seem to clash with a fixed idea he has: that concepts don't and shouldn't change.
As a preliminary, let me note there is a respect and set of circumstances where a concept doesn't change. Ayn Rand explained it in the following discussion on page 233 of ITOE:
Prof. H: It is easy to assume that when the definition changes, the concept changes.
AR: No, the context of your knowledge changes. When you know more, you select a different essential characteristic by which to define the object, because you now have to differentiate it more precisely. Your knowledge has expanded, but the concept doesn't change.
The similarities and differences according to which you originally formed the concept still remain. So you haven't changed your concept.
But does it follow from her remarks that a concept can't change? No. A concept isn't changed by acquiring more knowledge, but it can certainly be modified by conscious intent. Does Mr Harriman understand and explain that? No. His only broad remarks about whether concepts change are these on page 13:
The concept "temperature" had the same meaning for Galileo as for Einstein, i.e., both men referred to the same physical property. The difference is only that Einstein knew much more this property; he understood its relation to heat, to motion and to the fundamental nature of matter. But that expansion of knowledge was possible only because the concept itself did not change. Because concepts are stable, we can communicate and advance our knowledge.
I left out a word in this quote. The second sentence should read:
The difference is only that Einstein knew much more about this property; he understood its relation to heat, to motion and to the fundamental nature of matter.
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Harriman has just posted another response to those who have challenged his historical accounts of Galileo and Newton.
He accuses his critics of "belittling the achievements of Galileo and Newton" and Harriman paraphrases Jack Nicholson in the movie "A Few Good Men" saying:
“I don’t want money and I don’t want fame. What I do want is for academics to stand there in their girlie graduate gowns and extend to Galileo and Newton some freakin’ respect.”
Sounds like a strong argument to me.
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Harriman has just posted another response to those who have challenged his historical accounts of Galileo and Newton.
He accuses his critics of "belittling the achievements of Galileo and Newton" and Harriman paraphrases Jack Nicholson in the movie "A Few Good Men" saying:
“I don’t want money and I don’t want fame. What I do want is for academics to stand there in their girlie graduate gowns and extend to Galileo and Newton some freakin’ respect.”
Sounds like a strong argument to me.
Harriman continues to explore the upper-limit of inanity and unprofessionalism that can be injected into this "debate". I would say this is spiraling-down, but where do you go from rock-bottom?
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He accuses his critics of "belittling the achievements of Galileo and Newton" and Harriman paraphrases Jack Nicholson in the movie "A Few Good Men" saying:
“I don’t want money and I don’t want fame. What I do want is for academics to stand there in their girlie graduate gowns and extend to Galileo and Newton some freakin’ respect.”
Does some psychologist have the term for the technique used by somebody to evade the necessity of responding to a valid criticism by rewriting the criticism to be about somebody else? "Deflection" is the term that comes to mind.
In any case, it should be apparent, with such devastating wit from the 21st century's top intellect, providing an answer (whether anybody can identify it or not) to the Problem Of Induction, that criticism is doomed from the start. What feeble minds dare challenge such a man? Why, I've heard that he's divided by zero - twice!
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He accuses his critics of "belittling the achievements of Galileo and Newton" and Harriman paraphrases Jack Nicholson in the movie "A Few Good Men" saying:
“I don’t want money and I don’t want fame. What I do want is for academics to stand there in their girlie graduate gowns and extend to Galileo and Newton some freakin’ respect.”
Does some psychologist have the term for the technique used by somebody to evade the necessity of responding to a valid criticism by rewriting the criticism to be about somebody else? "Deflection" is the term that comes to mind.
"Evasion", "polemical strawman argument", "tilting at windmills",...
"Don't bother to examine a folly—ask yourself only what it accomplishes."
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Harriman has just posted another response to those who have challenged his historical accounts of Galileo and Newton.
He accuses his critics of "belittling the achievements of Galileo and Newton" and Harriman paraphrases Jack Nicholson in the movie "A Few Good Men" saying:
“I don’t want money and I don’t want fame. What I do want is for academics to stand there in their girlie graduate gowns and extend to Galileo and Newton some freakin’ respect.”
As has become his pattern, he does not say what "critics" he is talking about or what the criticisms are. His positioning himself as a defender of Galileo and Newton in opposition to unspecified "critics" in "girlie graduate gowns" alleged to be "belittling the achievements of Galileo and Newton" is another strawman argument. Whatever critics he thinks he is addressing while safely avoiding citing what they have allegedly said, no "belittling of Galileo and Newton" has appeared in anything that John McCaskey has made public or anything else that we have seen in this controversy over -- not Galileo and Newton -- but the presentation of David Harriman's own accounts of them. Moreover, the two paragraphs he presents as rebuttals to the unspecified criticisms which he offers in defense of Galileo and Newton have nothing to do with "belittling" anyone -- he doesn't even bother with a strawman example of the alleged "belittling".
In the first paragraph, on Galileo, he writes:
They claim that Galileo did not understand the concept of “friction,” despite the indisputable fact that Galileo abstracted from the effects of friction in order to discover his laws of motion. Furthermore, Galileo made an original discovery about friction. He was the first to realize that bodies falling through a resistive medium reach a terminal speed, and this happens because the frictional force is proportional to speed and therefore this force increases until it equals the weight of the falling body. How could he reach such a generalization without the idea of friction?
But who has said Galileo was "without the idea of friction [i.e., resistance]"? Not John McCaskey or anyone else that we have seen.
As has been discussed previously in much more detail and which still remains unacknowledged by David Harriman, he claimed that Galileo "carefully analyzed friction" despite the fact that what Galileo described in his analysis was buoyancy, not "friction", and not the actual nature of the resistance to the free fall of canon balls moving at high speed (which is not "friction"). Galileo had the idea of the effect of resistance (which he did not call "friction") but an incorrect concept of it. He did not understand it's nature and confused it with a different kind of force, buoyancy, that played no role. David Harriman claimed that correct concepts are necessary as a "green light" for induction. The criticism is not belittling Galileo for misunderstanding the nature of the resitance force he knew was there, but rather criticism of David Harriman's own inaccurate account of what Galileo thought and did. But it's easier to evade that by dramatic grandstanding as a noble defender of Galileo against unmentioned, non-existent "belittlers of Galileo". Could it be that he actually doesn't understand this?
Then he writes of an unspecified "attack on Newton", again ignoring that the criticisms have been of his own dubious historical accounts, not an "attack on Newton":
The attack on Newton claims that he didn’t understand the concepts of “inertia,” “acceleration,” and “momentum.” Allegedly, he was still in the grip of the medieval concept of “impetus.” In contrast to the medieval thinkers, however, Newton was able to calculate — in one case after another — motions and the effects of such motions. In some of his calculations, he symbolized “acceleration” by a “v” with a dot over it, and the dot indicates a time derivative. If he could symbolize and calculate acceleration, in what sense does he not have the idea?
None of this addresses that fact that the historical accounts of Newton, including the Westfall reference he cited in his own book, relate that Newton did in fact retain the notion of impetus in different forms at different stages of his analysis. Referring to Newton's "calculations" does not refute that.
And the only criticism of LL I have seen regarding Newton's idea of acceleration is in one of the emails from John McCaskey to David Harriman in which he discusses the question of whether Newton had the concept of an acceleration vector, not the idea of acceleration, when he derived his law of motion under central force. David Harriman has a master's degree in physics and must understand the distinction between the scalar quantity "'v' with a dot over it" and the more general concept of a time derivative of a velocity vector. There is an enormous difference between a change in scalar speed versus an induced speed in a direction orthogonal to it (inward along the radius orthogonal to the tangential speed). But he doesn't address this distinction. Who has said that Newton didn't have a concept of acceleration at all?
Again, the question is whether Newton used clear concepts of inertia and acceleration in the form that David Harriman said he did -- and which claims are necessary for his own requirement of concepts sufficiently advanced to be what he calls "green light" concepts making advancement possible. No one has said that Newton did not develop the required ideas for his advancements; the issue is over the fully formed "concepts" David Harriman claims are necessary. And of course, once again, these criticisms of LL have nothing to do with "belittling" or "attacking" Newton.
One irony in this is that when one looks into the history of what great scientists were able to do without today's much simpler and clearer concepts that we take for granted, it is astounding how much they were able to do in their context of often tedious, cumbersome methods, a glimmering of knowledge only beginning to appear, and a mixture of bad ideas and ways of thinking common in their time. This only leads to more respect for what they did, not "belittling" them. It is David Harriman himself who does not seem to realize that his own over-simplified accounts are undermining appreciation of what scientists like Galileo and Newton were able to do. But to misrepresent the criticisms of his book as "belittling" and "attacking" Galileo and Newton and then posturing as their brave and lonely defender against unnamed enemies is simply a dishonest evasion misleading his fan base that doesn't have the knowledge to understand what he is doing. Yes, there are people who belittle Galileo and Newton, and who promote irrational theories in the philosophy of science, but this has nothing to do with the current discussion and criticism of LL. | 6,731 | 32,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-40 | latest | en | 0.952236 |
https://www.stat.berkeley.edu/~aldous/RWG/Book_Ralph/Ch4.S1.html | 1,513,463,095,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948589512.63/warc/CC-MAIN-20171216220904-20171217002904-00275.warc.gz | 790,185,247 | 7,316 | # 4.1 The maximal mean commute time $\tau^{*}$
We start by repeating the definition
$\tau^{*}\equiv\max_{ij}(E_{i}T_{j}+E_{j}T_{i})$ (4.1)
and recalling what we already know. Obviously
$\max_{ij}E_{i}T_{j}\leq\tau^{*}\leq 2\max_{ij}E_{i}T_{j}$
and by Chapter 3 Lemma yyy
$\max_{j}E_{\pi}T_{j}\leq\tau^{*}\leq 4\max_{j}E_{\pi}T_{j}.$ (4.2)
Arguably we could have used $\max_{ij}E_{i}T_{j}$ as the “named” parameter, but the virtue of $\tau^{*}$ is the resistance interpretation of Chapter 3 Corollary yyy.
###### Lemma 4.1
For random walk on a weighted graph,
$\tau^{*}=w\max_{ij}r_{ij}$
where $r_{ij}$ is the effective resistance between $i$ and $j$.
In Chapter 3 Proposition yyy we proved lower bounds for any $n$-state discrete-time reversible chain:
$\tau^{*}\geq 2(n-1)$
$\max_{ij}E_{i}T_{j}\geq n-1$
which are attained by random walk on the complete graph. Upper bounds will be discussed extensively in Chapter 6, but let’s mention two simple ideas here. Consider a path $i=i_{0},i_{1},\ldots,i_{m}=j$, and let’s call this path $\gamma_{ij}$ (because we’ve run out of symbols whose names begin with “p”!) This path, considered in isolation, has “resistance”
$r(\gamma_{ij})\equiv\sum_{e\in\gamma_{ij}}1/w_{e}$
which by the Monotonicity Law is at least the effective resistance $r_{ij}$. Thus trivially
$\tau^{*}\leq w\max_{i,j}\min_{\mbox{paths }\gamma_{ij}}r(\gamma_{ij}).$ (4.3)
A more interesting idea is to combine the max-flow min-cut theorem (see e.g. [86] sec. 5.4) with Thompson’s principle (Chapter 3 Corollary yyy). Given a weighted graph, define
$c\equiv\min_{A}\sum_{i\in A}\sum_{j\in A^{c}}w_{ij}$ (4.4)
the $min$ over proper subsets $A$. The max-flow min-cut theorem implies that for any pair $a,b$ there exists a flow ${\bf f}$ from $a$ to $b$ of size $c$ such that $|f_{ij}|\leq w_{ij}$ for all edges $(i,j)$. So there is a unit flow from $a$ to $b$ such that $|f_{e}|\leq c^{-1}w_{e}$ for all edges $e$. It is clear that by deleting any flows around cycles we may assume that the flow through any vertex $i$ is at most unity, and so
$i$ (4.5)
So
$\displaystyle E_{a}T_{b}+E_{b}T_{a}$ $\displaystyle\leq$ $\displaystyle w\sum_{e}\frac{f^{2}_{e}}{w_{e}}\mbox{ by Thompson's principle }$ $\displaystyle\leq$ $\displaystyle\frac{w}{c}\sum_{e}|f_{e}|$ $\displaystyle\leq$ $\displaystyle\frac{w}{c}\ (n-1)\mbox{ by }(\ref{fsum}).$
and we have proved
###### Proposition 4.2
For random walk on an $n$-vertex weighted graph,
$\tau^{*}\leq\frac{w(n-1)}{c}$
for $c$ defined at (4.4).
Lemma 4.1 and the Monotonicity Law also make clear a one-sided bound on the effect of changing edge-weights monotonically.
###### Corollary 4.3
Let $\tilde{w}_{e}\geq w_{e}$ be edge-weights and let $\tilde{\tau}^{*}$ and $\tau^{*}$ be the corresponding parameters for the random walks. Then
$\frac{E_{i}\tilde{T}_{j}+E_{j}\tilde{T}_{i}}{E_{i}T_{j}+E_{j}T_{i}}\leq\frac{% \tilde{w}}{w}\mbox{ for all }i,j$
and so
$\tilde{\tau}^{*}/\tau^{*}\leq\tilde{w}/w.$
In the case of unweighted graphs the bound in Corollary 4.3 is $|\tilde{\mbox{{\cal E}}}|/|\mbox{{\cal E}}|$. Example yyy of Chapter 3 shows there can be no lower bound of this type, since in that example $\tilde{w}/w=1+O(1/n)$ but (by straightforward calculations) $\tilde{\tau}^{*}/\tau^{*}=O(1/n)$. | 1,186 | 3,301 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 54, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | longest | en | 0.778516 |
http://mathhelpforum.com/statistics/103343-prove-show-possible.html | 1,480,777,402,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540932.10/warc/CC-MAIN-20161202170900-00001-ip-10-31-129-80.ec2.internal.warc.gz | 181,574,696 | 9,900 | # Thread: Prove or show..is it possible
1. ## Prove or show..is it possible
That for any three events A,B, and C with P(C)>0,
P(A U B|C) = P(A|C) + P(B|C) - P(Aint.B|C)
2. Notice that $P\left[ {\left( {A \cap C} \right) \cup \left( {B \cap C} \right)} \right] = P\left[ {\left( {A \cap C} \right)} \right] + P\left[ {\left( {B \cap C} \right)} \right] - P\left[ {\left( {A \cap B \cap C} \right)} \right]$
Moreover, $P\left[ {\left( {A \cap B \cap C} \right)} \right] = P\left[ {\left( {A \cap B} \right)|C} \right]P(C)$ | 220 | 524 | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2016-50 | longest | en | 0.324274 |
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ | 1,532,151,340,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592387.80/warc/CC-MAIN-20180721051500-20180721071500-00276.warc.gz | 62,227,623 | 4,575 | Size of Atoms
Size of Atoms
The Size of Atoms: Metallic Radii
The size of an isolated atom can't be measured because we can't determine the location of the electrons that surround the nucleus. We can estimate the size of an atom, however, by assuming that the radius of an atom is half the distance between adjacent atoms in a solid. This technique is best suited to elements that are metals, which form solids composed of extended planes of atoms of that element. The results of these measurements are therefore often known as metallic radii.
The figure below shows the relationship between the metallic radii for elements in Groups IA and IIA.
There are two general trends in these data.
• The metallic radius becomes larger as we go down a column of the periodic table because the valence electrons are placed in larger orbitals.
• The metallic radius becomes smaller as we go from left to right across a row of the periodic table because the number of protons in the nucleus also increases as we go across a row of the table. The nucleus tends to hold electrons in the same shell of orbitals more tightly and the atoms become smaller.
The Size of Atoms: Covalent Radii
The size of an atom can be estimated by measuring the distance between adjacent atoms in a covalent compound. The covalent radius of a chlorine atom, for example, is half the distance between the nuclei of the atoms in a Cl2 molecule.
The covalent radii of the main group elements are given in the figure below. These data confirm the trends observed for metallic radii. Atoms become larger as we go down a column of the periodic table, and they becomes smaller as we go across a row of the table.
The covalent radius for an element is usually a little smaller than the metallic radius. This can be explained by noting that covalent bonds tend to squeeze the atoms together, as shown in the figure below.
The Size of Atoms: Ionic Radii
The relative size of atoms can also be studied by measuring the radii of their ions.
The first ionic radii were obtained by studying the structure of LiI, which contains a relatively small positive ion and a relatively large negative ion. The analysis of the structure of LiI was based on the following assumptions.
• The relatively small Li+ ions pack in the holes between the much larger I- ions, as shown in the figure below.
• The relatively large I- ions touch one another.
• The Li+ ions touch the I- ions.
If these assumptions are valid, the radius of the I- ion can be estimated by measuring the distance between the nuclei of adjacent iodide ions. The radius of the Li+ ion can then be estimated by subtracting the radius of the I- ion from the distance between the nuclei of adjacent Li+ and I- ions.
Unfortunately only two of the three assumptions that were made for LiI are correct. The Li+ ions in this crystal do not quite touch the I- ions. As a result, this experiment overestimated the size of the Li+ ion. Repeating this analysis with a large number of ionic compounds, however, has made it possible to obtain a set of more accurate ionic radii.
The Relative Size of Atoms and Their Ions
The table and figure below compare the covalent radius of neutral F, Cl, Br, and I atoms with the radii of their F-, Cl-, Br-, and I- ions. In each case, the negative ion is much larger than the atom from which it was formed. In fact, the negative ion can be more than twice as large as the neutral atom.
Element Covalent Radii (nm) Ionic Radii (nm) F 0.064 0.136 Cl 0.099 0.181 Br 0.1142 0.196 I 0.1333 0.216
The only difference between an atom and its ions is the number of electrons that surround the nucleus.
Example: A neutral chlorine atom contains 17 electrons, while a Cl- ion contains 18 electrons.
Cl: [Ne] 3s2 3p5 Cl-: [Ne] 3s2 3p6
Because the nucleus can't hold the 18 electrons in the Cl- ion as tightly as the 17 electrons in the neutral atom, the negative ion is significantly larger than the atom from which it forms.
For the same reason, positive ions should be smaller than the atoms from which they are formed. The 11 protons in the nucleus of an Na+ ion, for example, should be able to hold the 10 electrons on this ion more tightly than the 11 electrons on a neutral sodium atom. The table and figure below provide data to test this hypothesis. They compare the covalent radii for neutral atoms of the Group IA elements with the ionic radii for the corresponding positive ions. In each case, the positive ion is much smaller than the atom from which it forms.
Element Covalent Radii (nm) Ionic Radii (nm) Li 0.123 0.068 Na 0.157 0.095 K 0.2025 0.133 Rb 0.216 0.148 Cs 0.235 0.169
Practice Problem 1:Compare the sizes of neutral sodium and chlorine atoms and their Na+ and Cl- ions. Click here to check your answer to Practice Problem 1
The relative size of positive and negative ions has important implications for the structure of ionic compounds. The positive ions are often so small they pack in the holes between planes of adjacent negative ions. In NaCl, for example, the Na+ ions are so small that the Cl- ions almost touch, as shown in the figure below. | 1,196 | 5,127 | {"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} | 2.84375 | 3 | CC-MAIN-2018-30 | latest | en | 0.948364 |
http://www.unidata.ucar.edu/mailing_lists/archives/visad/2004/msg00121.html | 1,500,900,252,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424876.76/warc/CC-MAIN-20170724122255-20170724142255-00141.warc.gz | 580,987,400 | 7,620 | # Re: resampling problem
```Hi Desiree,
I cannot see any obvious problem. Make sure that the line
from:
(linePointCalc.x - lineVector.x, linePointCalc.y - lineVector.y)
to:
(linePointCalc.x + lineVector.x, linePointCalc.y + lineVector.y)
overlaps the rectangle from (northMin, eastMin) to
(northMax, eastMax).
Good luck,
Bill
On Thu, 15 Apr 2004, Desiree Hilbring wrote:
> until now I resampled an Irregular2DSet resembling a terrain surface, for
> achieving a
> profile line the following way, which worked very well:
> RealType x2 = RealType.getRealType("x");
> RealType y2 = RealType.getRealType("y");
> RealType height2 = RealType.getRealType("height");
> RealTupleType xy2 = new RealTupleType(x2, y2);
> FunctionType terrain_type2 = new FunctionType(xy2,height2);
> Irregular2DSet set2 = new Irregular2DSet(
> xy2,
> new float[][] { eastValues,northValues });
> FlatField terrain2 = new FlatField(terrain_type2,set2);
> terrain2.setSamples(new float[][] { heightValues});
>
>
> // Why 2000 points? Try and error, it was okay for
> // dataset with max. 50 points.
> // The more points, the better fit of the profile to the terrain.
> double[][] lineCoordinates = new double[2][2000];
> double j = -1;
> for (int i = 0; i < 2000; i++) {
> lineCoordinates[0][i] = linePointCalc.x +j * lineVector.x;
> lineCoordinates[1][i] = linePointCalc.y +j * lineVector.y;
> j = j + 0.001;
> }
> Gridded2DDoubleSet gridded2DSet
> new Gridded2DDoubleSet(xy2,
> lineCoordinates, 2000);
> FlatField line =(FlatField) terrain2.resample(
> gridded2DSet,
> Data.WEIGHTED_AVERAGE,
> Data.NO_ERRORS);
> System.out.println(line.toString());
>
> Now I have a grid data set with a lot more points so I want to use a
> Linear2DSet as my original data set instead of the Irregular2DSet
> (which is very slow now, because I have many more terrain points):
>
> RealType eastV =RealType.getRealType("eastValues");
> RealType northV =RealType.getRealType("northValues");
> RealType heightV =RealType.getRealType("heightValues");
> domain_tuple = new RealTupleType(eastV, northV);
> func_en_h = new FunctionType(domain_tuple,heightV);
> domain_set = new Linear2DSet(
> domain_tuple,
> northMin,
> northMax,
> nRows,
> eastMin,
> eastMax,
> nCols);
>
> double[][] flat_samples = new double[1][nCols *nRows];
> for (int c = 0; c < (nRows); c++) {
> for (int r = 0; r < (nCols); r++) {
> // dgm 1 x y z
> flat_samples[0][c * nCols + r] =height[c * nCols + r];
> }
> }
> vals_ff = new FlatField(func_en_h, domain_set);
> vals_ff.setSamples(flat_samples, false);
>
> // Why 2000 points? Try and error, it was okay for
> // dataset with max. 50 points.
> // The more points, the better fit of the profile to the terrain.
> double[][] lineCoordinates = new double[2][2000];
> double j = -1;
> for (int i = 0; i < 2000; i++) {
> lineCoordinates[0][i] = linePointCalc.x + j *lineVector.x;
> lineCoordinates[1][i] = linePointCalc.y + j *lineVector.y;
> j = j + 0.001;
> }
>
> Gridded2DDoubleSet gridded2DSet
> new Gridded2DDoubleSet(domain_tuple,
> lineCoordinates, 2000);
> FlatField line =(FlatField) vals_ff.resample(
> gridded2DSet,
> Data.WEIGHTED_AVERAGE,
> Data.NO_ERRORS);
> System.out.println(line.toString());
>
> The problem is, that I only get missing values in the FlatField line. What
> am I doing wrong?
>
> Greetings Desiree
>
>
> oooooooooooooooooooooooooooooooooooooooooooooooo
> Desiree Hilbring
>
> Institut fuer Photogrammetrie und Fernerkundung
> Universitaet Karlsruhe, Germany
> email: hilbring@xxxxxxxxxxxxxxxxxxxx
> # 0721 6083676
> oooooooooooooooooooooooooooooooooooooooooooooooo
>
>
>
``` | 1,161 | 4,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-30 | latest | en | 0.679079 |
https://social.msdn.microsoft.com/Forums/en-US/79d0b363-f095-438b-8942-9e171f8ff7ae/calculate-hours-between-two-dates-but-only-business-hours?forum=vbgeneral | 1,603,240,705,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874340.10/warc/CC-MAIN-20201020221156-20201021011156-00271.warc.gz | 527,407,345 | 18,121 | # calculate hours between two dates but only business hours
• ### Question
• Does anyone have a function to the following:-
Calculate number of hours between two dates but only including the business hours of 09:00 - 17:00
This is to calculate SLA for help desk
So will have start_datetime and end_datetime
Darren Rose
Monday, March 4, 2019 6:34 PM
### All replies
• Hi Darren,
See if this is what you are looking for, note I didn't do the holiday part as the focuses on day number and time.
```Public Class Form1
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim startDate = #3/1/2019#
Dim endDate = #3/8/2019#
Dim startTime = "9:00"
Dim endTime = "17:00"
Dim duration As TimeSpan = Date.Parse(endTime).Subtract(Date.Parse(startTime))
Dim workDays = GetNumberOfWorkingDays(startDate, endDate)
Console.WriteLine(workDays * duration.Hours)
End Sub
''' <summary>
''' Yep it's missing holidays
''' </summary>
''' <param name="pStart"></param>
''' <param name="pStop"></param>
''' <returns></returns>
Private Function GetNumberOfWorkingDays(pStart As Date, pStop As Date) As Integer
Dim days As Integer = 0
Do While pStart <= pStop
If pStart.DayOfWeek <> DayOfWeek.Saturday AndAlso pStart.DayOfWeek <> DayOfWeek.Sunday Then
days += 1
End If
Loop
Return days
End Function
End Class```
Please remember to mark the replies as answers if they help and unmarked them if they provide no help, this will help others who are looking for solutions to the same or similar problem. Contact via my Twitter (Karen Payne) or Facebook (Karen Payne) via my MSDN profile but will not answer coding question on either.
NuGet BaseConnectionLibrary for database connections.
StackOverFlow
Monday, March 4, 2019 7:18 PM
• Hi
No that doesn't get me expected result
For example if I use
startdate of "01/03/19 16:00"
enddate of "04/03/19 10:00"
then it is returning 8 hours using above code
But should be returning 2 hours as on Friday 01/03 16:00 to 17:00 = 1 hour then next 2 days are weekend, and then on 04/03 09:00 to 10:00 is 1 hour = total of 2 hours working time
Darren Rose
Monday, March 4, 2019 8:04 PM
• Just so you know I'm not ignoring this I'm currently in a side-by-side session coding at work session.
Please remember to mark the replies as answers if they help and unmarked them if they provide no help, this will help others who are looking for solutions to the same or similar problem. Contact via my Twitter (Karen Payne) or Facebook (Karen Payne) via my MSDN profile but will not answer coding question on either.
NuGet BaseConnectionLibrary for database connections.
StackOverFlow
Monday, March 4, 2019 8:45 PM
• Just so you know I'm not ignoring this I'm currently in a side-by-side session coding at work session.
that's fine, no hurry, I appreciate your help whenever :)
Darren Rose
Monday, March 4, 2019 8:48 PM
• Hi,
try the code:
```Module Module1
Sub Main()
Console.WriteLine(gethours("2019-03-05 14:00", "2019-03-05 17:00"))
End Sub
Public Function gethours(ByVal starttime As String, ByVal endtime As String) As Integer
Dim start As DateTime = Convert.ToDateTime(starttime)
Dim [end] As DateTime = Convert.ToDateTime(endtime)
Dim span As TimeSpan = [end] - start
Dim AllDays As Integer = Convert.ToInt32(span.TotalDays) + 1
Dim totleWeek As Integer = AllDays / 7
Dim yuDay As Integer = AllDays Mod 7
Dim lastDay As Integer = 0
If yuDay = 0 Then
lastDay = AllDays - (totleWeek * 2)
Else
Dim weekDay As Integer = 0
Dim endWeekDay As Integer = 0
Select Case start.DayOfWeek
Case DayOfWeek.Monday
weekDay = 1
Case DayOfWeek.Tuesday
weekDay = 2
Case DayOfWeek.Wednesday
weekDay = 3
Case DayOfWeek.Thursday
weekDay = 4
Case DayOfWeek.Friday
weekDay = 5
Case DayOfWeek.Saturday
weekDay = 6
Case DayOfWeek.Sunday
weekDay = 7
End Select
If (weekDay = 6 AndAlso yuDay >= 2) OrElse (weekDay = 7 AndAlso yuDay >= 1) OrElse (weekDay = 5 AndAlso yuDay >= 3) OrElse (weekDay = 4 AndAlso yuDay >= 4) OrElse (weekDay = 3 AndAlso yuDay >= 5) OrElse (weekDay = 2 AndAlso yuDay >= 6) OrElse (weekDay = 1 AndAlso yuDay >= 7) Then
endWeekDay = 2
End If
If (weekDay = 6 AndAlso yuDay < 1) OrElse (weekDay = 7 AndAlso yuDay < 5) OrElse (weekDay = 5 AndAlso yuDay < 2) OrElse (weekDay = 4 AndAlso yuDay < 3) OrElse (weekDay = 3 AndAlso yuDay < 4) OrElse (weekDay = 2 AndAlso yuDay < 5) OrElse (weekDay = 1 AndAlso yuDay < 6) Then
endWeekDay = 1
End If
lastDay = AllDays - (totleWeek * 2) - endWeekDay
End If
Return (lastDay - 1) * 8 + (17 - Convert.ToDateTime(starttime).Hour) + (Convert.ToDateTime(endtime).Hour - 9)
End Function
End Module
```
Best Regards,
Alex
MSDN Community Support Please remember to click "Mark as Answer" the responses that resolved your issue, and to click "Unmark as Answer" if not. This can be beneficial to other community members reading this thread. If you have any compliments or complaints to MSDN Support, feel free to contact MSDNFSF@microsoft.com.
Tuesday, March 5, 2019 3:22 AM
• Hi Alex
That seems to give me an odd result - using same example I have used for other tests
MsgBox(gethours("01/03/19 16:00", "04/03/19 10:00"))
gives me -6 - but should be 2 as per reply to Karen above
This one
MsgBox(gethours("01/03/19 10:00", "01/03/19 18:00"))
gives me 8 - but should be 7 as working hours 09:00 to 17:00
Darren Rose
Tuesday, March 5, 2019 11:28 AM
• Think I have found working solution
```Public Function WorkingMinutes(ByVal startdate As Date, ByVal enddate As Date) As Long
Dim count As Integer = 0
Dim i = startdate
Do While i < enddate
If i.DayOfWeek <> DayOfWeek.Saturday AndAlso i.DayOfWeek <> DayOfWeek.Sunday Then
If i.TimeOfDay.Hours >= 9 AndAlso i.TimeOfDay.Hours < 17 Then
count += 1
End If
End If
Loop
Return count
End Function```
Darren Rose
Tuesday, March 5, 2019 12:30 PM
• just need to work out how to encompass ignoring specified holiday days as well as weekends like done here when calculating days between two dates
https://social.msdn.microsoft.com/Forums/vstudio/en-US/b57d5fcb-6033-4df3-b78e-50e0e350677e/calculate-days-between-two-dates-excluding-weekends?forum=vbgeneral
Darren Rose
Tuesday, March 5, 2019 12:31 PM
• Hello Darren,
Had not gotten much time to spend on this but did come up with the following which may be of possible usage in that you would parse each day's time spend and do a running total but than it may not be of use.
End time could be fixed to say 5:00 PM.
```Public Class Form1
Private Sub Button1_Click(sender As Object, e As EventArgs) _
Handles Button1.Click
Dim startDate = New DateTime(2019, 3, 3, 13, 0, 0)
Dim endDate = New DateTime(2019, 3, 3, 17, 0, 0)
Dim duration As TimeSpan =
Date.Parse(endDate.ToString("hh:mm tt")).Subtract(
Date.Parse(startDate.ToString("hh:mm tt")))
Console.WriteLine(\$"{duration} - {duration.Hours}")
End Sub
End Class
```
Please remember to mark the replies as answers if they help and unmarked them if they provide no help, this will help others who are looking for solutions to the same or similar problem. Contact via my Twitter (Karen Payne) or Facebook (Karen Payne) via my MSDN profile but will not answer coding question on either.
NuGet BaseConnectionLibrary for database connections.
StackOverFlow
Tuesday, March 5, 2019 1:30 PM
• Thanks Karen - think one I found and posted above seems to do the job, just need to get it to skip holidays as well and then all sorted
Darren Rose
Tuesday, March 5, 2019 2:00 PM
• Perhaps looking back on weekends and holidays
```Public Class DateUtils
Public Sub New()
SetHolidays()
End Sub
Public Holidays As New List(Of Date)
Public Sub SetHolidays()
End Sub
Public Function IsHoliday(pDate As Date) As Boolean
Return Holidays.Contains(pDate)
End Function
Public Function IsWeekEnd(pDate As Date) As Boolean
Return pDate.DayOfWeek =
DayOfWeek.Saturday OrElse pDate.DayOfWeek = DayOfWeek.Sunday
End Function
Public Function GetNextWorkingDay(pDate As Date) As Date
Do
Loop While IsHoliday(pDate) OrElse IsWeekEnd(pDate)
Return pDate
End Function
End Class
```
Using the above in a hard code fashion.
```Dim ops = New DateUtils
Dim monthIndex = 12
For dayIndex As Integer = 1 To Date.DaysInMonth(Now.Year, monthIndex)
Dim dt = New Date(Date.Now.Year, monthIndex, dayIndex)
Dim isWeekend = ops.IsWeekEnd(dt)
Dim isHoliday = ops.IsHoliday(dt)
Next```
Please remember to mark the replies as answers if they help and unmarked them if they provide no help, this will help others who are looking for solutions to the same or similar problem. Contact via my Twitter (Karen Payne) or Facebook (Karen Payne) via my MSDN profile but will not answer coding question on either.
NuGet BaseConnectionLibrary for database connections.
StackOverFlow
Tuesday, March 5, 2019 3:08 PM
• Thanks Karen - that is very useful
So how can I utilize this for the working hours part so that also doesn't include weekends or holidays?
This was code I had working which did hours fine but only excluded weekends
```Public Function WorkingMinutes(ByVal startdate As Date, ByVal enddate As Date) As Long
Dim count As Integer = 0
Dim i = startdate
Do While i < enddate
If i.DayOfWeek <> DayOfWeek.Saturday AndAlso i.DayOfWeek <> DayOfWeek.Sunday Then
If i.TimeOfDay.Hours >= 9 AndAlso i.TimeOfDay.Hours < 17 Then
count += 1
End If
End If
Loop
Return count
End Function```
I am assuming this change will work, but not in front of computer until later:-
```Public Function WorkingMinutes(ByVal startdate As Date, ByVal enddate As Date) As Long
Dim count As Integer = 0
Dim i = startdate
Do While i < enddate
if IsWeekend(i) = False AndAlso IsHoliday(i) = False Then
If i.TimeOfDay.Hours >= 9 AndAlso i.TimeOfDay.Hours < 17 Then
count += 1
End If
End If
Loop
Return count
End Function```
Darren Rose
Tuesday, March 5, 2019 5:30 PM
• Once again I need to wait till later as I'm back doing side by side development again today so only have time for quickie replies.
Please remember to mark the replies as answers if they help and unmarked them if they provide no help, this will help others who are looking for solutions to the same or similar problem. Contact via my Twitter (Karen Payne) or Facebook (Karen Payne) via my MSDN profile but will not answer coding question on either.
NuGet BaseConnectionLibrary for database connections.
StackOverFlow
Tuesday, March 5, 2019 5:33 PM | 2,842 | 10,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-45 | latest | en | 0.826994 |
https://www.linuxtv.org/wiki/index.php?title=Phase&oldid=5688 | 1,566,788,475,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00452.warc.gz | 864,509,222 | 5,847 | # Phase
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## What is Phase?
As a sinoidal signal in the time domain can be defined as
The signal phase describes an 'offset' of the signal along the time axis and defines the 'zero crossing' of the signal. An sinoid signal is periodic, therefore usually only phase values from +/-180deg (or +/-Pi) are used.
## An Example
You see here an sinoidal signal with a frequency of 1kHz and an amplitude of 2 Volts. Time scale is -1msec..+1msec. The signal is shown with 3 different phases, from left to right:
• -60 degree
• 0 degree
• +60 degree
Please have a look at the zero crossings! In the left picture the crossings are shifted to the right, in the right picture to the left hand side.
Note: the phase can be either expressed as +/-Pi or +/-180deg. The signal itself can be described as cosine or sine, both is equivalent and up to you depending where you define the time t = 0. | 238 | 960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | longest | en | 0.912394 |
https://www.bankruptcytalk.net/national-debt-real-time/ | 1,721,432,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514972.64/warc/CC-MAIN-20240719231533-20240720021533-00474.warc.gz | 571,334,511 | 60,044 | Thursday, July 18, 2024
HomeDebtNational Debt Real Time
National Debt Real Time
Do Foreign Countries Own National Debt
United States National Debt Clock(Real-Time)
For example, Japan owns \$1.276 trillions worth of US government debt.
You can research the economies of the largest US national debt holders. See our economic overviews of Brazil, China, the UK, Belgium, and India.
The ten largest holding nations of US government debt as of September 2020 are shown in the table below:
Rank
How Are Debt Clocks Calculated
Well use the United Kingdom as an example:
1 We obtain the latest data regarding the countrys national debt and the 10-year average interest rate they pay on it, like:
National Debt: \$1,717,879,000,000 10-Year Interest Rate: 2.50
2 Using these two figures we can then calculate how much the debt increases per year and subsequently per second.
Increase per Year: \$42,946,975,000 Increase per Second: \$1,362
3 We then work out the time difference between when the data was obtained and when the debt clock is being viewed by a visitor.
Time Difference = Time and Date of Visit Time and Date of Official Figure
4 The current debt is then calculated by adding the increase over this time to the official figure.
Current National Debt = Official Figure +
5 The debt clock then updates every two seconds, increasing according to the figures calculated in step 2.
Current National Debt = ) x Exchange Rate
Contents
What Is The National Debt
The national debt is the debt that the federal government holds which includes public debt, federal trust funds, and government accounts. As the total amount of deficit that the government has garnered, it is a number that encompasses what the government owes itself and others. The national debt is looked at in three parts: debt held by the public, gross federal debt, and debt subject to limit. Debt held by the public is the money gathered to fund activities and programs, with money borrowed from external lenders. The gross federal debt includes the public debt, but also adds federal trust funds and governments. Debt subject to limit is similar to gross federal debt, but only includes debt issued by the treasury and Federal Financing Bank.
As of July 2020, the national debt is more than \$26.5 trillion. This number equates to \$80,422 for every person living in the u.S., and is 123% of the U.S.âs annual economic output. As of June 2020 the debt-to-GDP ratio was 120.5%, due to the economic strain of the COVID-19 pandemic.
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Why Does The National Debt Matter
What makes America strong is our willingness to build and leave a better future for the next generation. Unfortunately, our growing debt is doing the opposite.
America faces many challenges including rising inequality, unaffordable healthcare, a changing climate, failing education, crumbling infrastructure, and unpredictable security threats. To address these challenges we will need significant resources. Every dollar that goes toward interest payments means less resources available to build a stronger, more resilient future.
Being irresponsible with our budget is simply not fair to our kids and grandkids, who will inherit this debt.
RISING INTEREST IN THE BUDGET
Each business day, the U.S. Treasury Department reports the amount of total debt outstanding as of the previous business day. Our debt clocks are updated daily based on this number. In addition, our formula uses the debt projections from the Congressional Budget Office , to estimate the rate at which the debt is currently growing. Those CBO projections are updated 2-3 times per year.
Debt per person is calculated by dividing the total debt outstanding by the population of the United States, as .
The \$30 trillion gross federal debt equals debt held by the public plus debt held by federal trust funds and other government accounts. Learn more about different ways to measure our national debt.
Quebecs Public Sector Debt
Our Quebec Debt Clock shows the growth of the public sector debt in real time. Public sector debt includes the governments gross debt as well as the debt of Hydro-Québec, of the municipalities, and of the universities other than the Université du Québec and its constituent universities.
Based on data provided by the Quebec Department of Finance in its March 2022 budget , we estimate that the debt will increase by \$14.9 billion by March 31, 2023, the equivalent of \$40.8 million per day, \$28,344 per minute, or \$472 per second.
When analyzing a governments indebtedness, it is necessary to go beyond what it manages directly and include the health and education networks, municipalities and other entities under the governments ultimate responsibility, since the government guarantees their debt. Public sector debt is therefore the most exhaustive measure of Quebecs debt, the one that provides a picture of what the government of Quebec borrows either directly or indirectly.
The only liquid assets of the government, those that could be sold quickly to pay off debt, are net financial assets. These assets came to \$16 billion as of March 31, 2022. It is hard to assess the market value of government-owned fixed assets and infrastructure since there are no relevant markets. Moreover, it is highly unlikely that the government would sell schools or bridges at some point to pay off the debt.
Components of the public sector debt
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Summary Of Chinese Public Debt Types
The majority of Chinese public debt is not officially owed by the central government.
However, all of that debt is ultimately guaranteed by the national government of China and should rightfully be recorded in its entirety as the Chinese national debt. The location of those debts are ranked below:
• Local government debt to shadow banking 31.1% of GDP
• Local government debt as municipal bonds 21.82% of GDP
• Central government debt issued as bonds 14.02% of GDP
• Local government public-private initiative 10.39% of GDP
• So, an investigation of Chinas national debt requires more research at the local government level that in the national government accounts.
These figures for total public debt also do not touch upon the undisclosed debts of state-owned banks and state-owned enterprises, which represent a large sector of the Chinese economy.
What facts should you know about Chinas national debt?
• You could wrap \$1 bills around the Earth 35,794 times with the debt amount.
• If you lay \$1 bills on top of each other they would make a pile 1,004,216 km, or 623,990 miles high.
• Thatâs equivalent to 2.61 trips to the Moon.
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National Debt For Selected Years
Fiscal year
130.6% 21,850
On July 27, 2018, the BEA revised its GDP figures in a comprehensive update and figures back to FY2013 were revised accordingly.
On June 25, 2014, the BEA announced: “n addition to the regular revision of estimates for the most recent 3 years and for the first quarter of 2014, GDP and select components will be revised back to the first quarter of 1999.
Fiscal years 19402009 GDP figures were derived from February 2011 Office of Management and Budget figures which contained revisions of prior year figures due to significant changes from prior GDP measurements. Fiscal years 19502010 GDP measurements were derived from December 2010 Bureau of Economic Analysis figures which also tend to be subject to revision, especially more recent years. Afterwards the OMB figures were revised back to 2004 and the BEA figures were revised back to 1947.
Fiscal years 19401970 begin July 1 of the previous year fiscal years 19802010 begin October 1 of the previous year. Intragovernmental debts before the Social Security Act are presumed to equal zero.
19091930 calendar year GDP estimates are from MeasuringWorth.com Fiscal Year estimates are derived from simple linear interpolation.
Audited figure was “about \$5,659 billion.”
Audited figure was “about \$5,792 billion.”
Audited figure was “about \$6,213 billion.”
Audited figure was said to be “about” the stated figure.
Audited figure was “about \$7,918 billion.”
Audited figure was “about \$8,493 billion.”
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Examples Of Capital Expenditure
Examples of infrastructure spending that improve an economy are:
• The development of transport infrastructure, such as motorways and railways
• Investment in universities to create more educational institutions or crate centers of excellence from existing establishments.
• Improvements in communication infrastructure, such as a fibre optic backbone to expand the nations internet bandwidth availability and speed.
If you are thinking of investing in a countrys economy, or if you are considering moving there, researching the national debt of that place and how the government spends money may be insightful.
A countrys national debt is one of many economic indicators that interplay to create a judgment on a countrys prospects for success.
Concerns Over Chinese Holdings Of Us Debt
REALITY CHECK: The US National Debt Clock
According to a 2013 Forbes article, many American and other economic analysts have expressed concerns on account of the People’s Republic of China’s “extensive” holdings of United States government debt as part of their reserves. The National Defense Authorization Act of FY2012 included a provision requiring the Secretary of Defense to conduct a “national security risk assessment of U.S. federal debt held by China.” The department issued its report in July 2012, stating that “attempting to use U.S. Treasury securities as a coercive tool would have limited effect and likely would do more harm to China than to the United States. An August 19, 2013 Congressional Research Service report said that the threat is not credible and the effect would be limited even if carried out. The report said that the threat would not offer “China deterrence options, whether in the diplomatic, military, or economic realms, and this would remain true both in peacetime and in scenarios of crisis or war.”
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Raising Reserve Requirements And Full Reserve Banking
Two economists, Jaromir Benes and Michael Kumhof, working for the International Monetary Fund, published a working paper called The Chicago Plan Revisited suggesting that the debt could be eliminated by raising bank reserve requirements and converting from fractional-reserve banking to full-reserve banking. Economists at the Paris School of Economics have commented on the plan, stating that it is already the status quo for coinage currency, and a Norges Bank economist has examined the proposal in the context of considering the finance industry as part of the real economy. A Centre for Economic Policy Research paper agrees with the conclusion that “no real liability is created by new fiat money creation and therefore public debt does not rise as a result.”
The debt ceiling is a legislative mechanism to limit the amount of national debt that can be issued by the Treasury. In effect, it restrains the Treasury from paying for expenditures after the limit has been reached, even if the expenditures have already been approved and have been appropriated. If this situation were to occur, it is unclear whether Treasury would be able to prioritize payments on debt to avoid a default on its debt obligations, but it would have to default on some of its non-debt obligations.
How Does The Chinese Government Raise Loans
The Ministry of Finance does not advertise its schedule of bond sales, nor does it disclose any other securities that it might use to cover cash flow issues or raise short-term financing.
All government debt is issued in Yuan, which is not convertible to foreign currencies and so there is no interest in these bonds for foreign investors.
Similarly, municipal bonds are written in Yuan and not intended for purchase by overseas investors.
Central government bonds are not intended for sale to the general public but are distributed behind closed doors to the major Chinese banks, which are all state-owned.
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Why Does Larger National Debt Attract Bond Buyer
Having a large national debt doesnt always discourage buyers of bonds. For example, the United States has a debt to GDP ratio of 108% and a lot of people want to buy US Treasury bonds.
You can see this data summary of US Local & State Government Debt for more information.
Some countries, such as the USA are always considered a good place to invest, and the government bonds of those countries are always in high demand.
Other Reasons For National Debt
Other obvious reasons for national debt are more mundane costs which occur as a result of culture and lifestyle.
For example, the healthcare costs in the United States have been rising for years and is one of the highest in the world.
Another reason for rising debt is the economic infrastructure we live in, which relies on productivity in individuals. As people live longer, more money is paid out in pensions.
The sustainability of such expenses largely depends on the countrys economic infrastructure, which in many cases, is lagging behind and adding to rising national debt-to-GDP ratios.
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Why Do Governments Lower Interest Rates On Growing Economies
This knowledge in the financial community enables governments to lower the interest rates that it offers on its debt and reduce the cost of financing deficits.
Thanks to economic indicators, you can work out whether a countrys national debt will trigger a virtuous cycle of investment and expansion, or a destructive debt spiral.
Fannie Mae And Freddie Mac Obligations Excluded
Under normal accounting rules, fully owned companies would be consolidated into the books of their owners, but the large size of Fannie Mae and Freddie Mac has made the U.S. government reluctant to incorporate them into its own books. When the two mortgage companies required bail-outs, White House Budget Director Jim Nussle, on September 12, 2008, initially indicated their budget plans would not incorporate the government-sponsored enterprise debt into the budget because of the temporary nature of the conservator intervention. As the intervention has dragged out, pundits began to question this accounting treatment, noting that changes in August 2012 “makes them even more permanent wards of the state and turns the government’s preferred stock into a permanent, perpetual kind of security”.
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Why Is The National Debt So High
When the federal government spends more than it takes in, we have to borrow money to cover that annual deficit. And each years deficit adds to our growing national debt.
Historically, our largest deficits were caused by increased spending around national emergencies like major wars or the Great Depression.
How Government Debts Affect You
U.S. National Debt Clock : Real Time
The approximate interest rate on the cost of market debt in Canada is about 2.01 percent. Interestingly enough, the country accrues \$75 million of debt per day in interest charges alone. That trickles down to the taxpayers, many of whom are seeking debt relief options for themselves.
According to the Financial Post, a study shows that Canada is a world leader in debt. One hypothesis for the debt getting so high is the fact that Canada came out largely unscathed from the last financial crisis. Low interest rates encouraged more borrowing, which led to bankruptcies and other economic downturns.
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Risks To Economic Growth
Debt levels may affect economic growth rates. In 2010, economists Kenneth Rogoff and Carmen Reinhart reported that among the 20 developed countries studied, average annual GDP growth was 34% when debt was relatively moderate or low , but it dips to just 1.6% when debt was high . In April 2013, the conclusions of Rogoff and Reinhart’s study came into question when a coding error in their original paper was discovered by Herndon, Ash and Pollin of the University of Massachusetts Amherst. Herndon, Ash and Pollin found that after correcting for errors and unorthodox methods used, there was no evidence that debt above a specific threshold reduces growth. Reinhart and Rogoff maintain that after correcting for errors, a negative relationship between high debt and growth remains. However, other economists, including Paul Krugman, have argued that it is low growth which causes national debt to increase, rather than the other way around.
Commenting on fiscal sustainability, former Federal Reserve Chairman Ben Bernanke stated in April 2010 that “Neither experience nor economic theory clearly indicates the threshold at which government debt begins to endanger prosperity and economic stability. But given the significant costs and risks associated with a rapidly rising federal debt, our nation should soon put in place a credible plan for reducing deficits to sustainable levels over time.”
National Debt Of The United States
The national debt of the United States is the total national debt owed by the federal government of the United States to Treasury security holders. The national debt at any point in time is the face value of the then-outstanding Treasury securities that have been issued by the Treasury and other federal agencies. The terms “national deficit” and “national surplus” usually refer to the federal government budget balance from year to year, not the cumulative amount of debt. In a deficit year the national debt increases as the government needs to borrow funds to finance the deficit, while in a surplus year the debt decreases as more money is received than spent, enabling the government to reduce the debt by buying back some Treasury securities. In general, government debt increases as a result of government spending and decreases from tax or other receipts, both of which fluctuate during the course of a fiscal year. There are two components of gross national debt:
During the COVID-19 pandemic, the federal government spent trillions in virus aid and economic relief. The CBO estimated that the budget deficit for fiscal year 2020 would increase to \$3.3 trillion or 16% GDP, more than triple that of 2019 and the largest as % GDP since 1945.
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https://beanumber.github.io/abdwr3e/09-simulation.html | 1,717,037,340,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059418.31/warc/CC-MAIN-20240530021529-20240530051529-00026.warc.gz | 99,998,764 | 50,505 | # 9 Simulation
Authors
Affiliations
Bowling Green State University
Smith College
Cleveland Guardians
## 9.1 Introduction
A baseball season consists of a collection of games between teams, where each game consists of nine innings, and a half-inning consists of a sequence of plate appearances. Because of this clean structure, the sport can be represented by relatively simple probability models. Simulations from these models are helpful in understanding different characteristics of the game.
One attractive aspect of the R system is its ability to simulate from a wide variety of probability distributions. In this chapter, we illustrate the use of R functions to simulate a game consisting of a large number of plate appearances. Also, We use R simulate the game-to-game competition of teams during an entire season.
Section 9.2 focuses on simulating the events in a baseball half-inning using a special probability model called a Markov chain. The runners on base and the number of outs define a state and this probability model describes movements between states until one reaches three outs. The movement or transition probabilities are found using actual data from the 2016 season. By simulating many half-innings using this model, one gets a basic understanding of the pattern of run scoring.
Section 9.3 describes a simulation of an entire baseball season using the Bradley-Terry probability model. Teams are assigned talents from a bell-shaped (normal) distribution and a season of baseball games is played using win probabilities based on the talents. By simulating many seasons, one learns about the relationship between a team’s talent and its performance in a 162-game season. We describe simulating the post-season series and assess the probability that the “best” team, that is, the team with the best ability actually wins the World Series.
## 9.2 Simulating a Half Inning
### 9.2.1 Markov chains
A Markov chain is a special type of probability model useful for describing movement between locations, called states. In the baseball context, a state is viewed as a description of the runners on base and the number of outs in an inning. Each of the three bases can be occupied by a runner or not, and so there are 2 $$\times$$ 2 $$\times$$ 2 = 8 possible runner situations. Since there are three possible numbers of outs (0, 1, or 2), there are 8 $$\times$$ 3 = 24 possible runner and outs states. If we include the 3 outs state, there are a total of 25 possible states during a half-inning of baseball (see Section 5.1).
In a Markov chain, a matrix of transition probabilities is used to describe how one moves between the different states. For example, suppose that there are currently runners on first and second with one out. Based on the outcome of the plate appearance, the state can change. For example, the batter may hit a single; the runner on second scores and the runner on first moves to third. In this case, the new state is runners on first and third with one out. Or maybe the batter will strike out, and the new state is runners on first and second with two outs. By looking at a specific row in the transition probability matrix, one learns about the probability of moving to first and third with one out, or moving to first and second with two outs, or any other possible state.
In a Markov chain, there are two types of states: transition states and absorbing states. Once one moves into an absorbing state, one remains there and can’t return to other transition states. In a half-inning of baseball, since the inning is over when there are 3 outs, this 3-outs state acts as an absorbing state.
There are some special assumptions in a Markov chain model. We assume that the probability of moving to a new state only depends on the current state. So any baseball events that happened before the current runners and outs situation are not relevant in finding the probabilities1. In other words, this model assumes there is not a momentum effect in batting through an inning. Also we are assuming that the probabilities of these movements are the same for all teams, against all pitchers, and for all innings during a game. Clearly, this assumption that all teams are average is not realistic, but we will address this issue in one of the other sections of this chapter.
There are several attractive aspects of using a Markov chain to model a half-inning of baseball. First, the construction of the transition probability matrix is easily done with 2016 season data using computations from Chapter 5. One can use the model to play many half-innings of baseball, and the run scoring patterns that are found resemble the actual run scoring of actual MLB baseball. Last, there are special properties of Markov chains that simplify some interesting calculations, such as the number of players who come to bat during an inning.
### 9.2.2 Review of work in run expectancy
To construct the transition matrix for the Markov chain, one needs to know the frequencies of transitions from the different runners/outs states to other possible runners/outs states. One can obtain these frequencies using the Retrosheet play-by-play data from a particular season. Here we review the work from Chapter 5.
We begin by reading in play-by-play data for the 2016 season, creating the data frame retro2016.
library(tidyverse)
retro2016 <- read_rds(here::here("data/retro2016.rds"))
First, we use the retrosheet_add_states() function that we wrote in Chapter 5 and stored in the abdwr3edata package. This function adds a number of useful new variables to retro2016. Recall that in particular, we now have a new variable state (which gives the runner locations and the number of outs at the beginning of each play), and a another new variable new_state (which contains the same information at the conclusion of the play).
library(abdwr3edata)
retro2016 <- retro2016 |>
retrosheet_add_states()
Next, we create the variable half_inning_id as a unique identifier for each half-inning in each baseball game. The new variable runs gives the number of runs scored in each play. The new data frame half_innings contains data aggregated over each half-inning of baseball played in 2016.
half_innings <- retro2016 |>
mutate(
runs = away_score_ct + home_score_ct,
half_inning_id = paste(game_id, inn_ct, bat_home_id)
) |>
group_by(half_inning_id) |>
summarize(
outs_inning = sum(event_outs_ct),
runs_inning = sum(runs_scored),
runs_start = first(runs),
max_runs = runs_inning + runs_start
)
By using the filter() function from the dplyr package, we focus on plays where there is a change in the state or in the number of runs scored. By another application of filter(), we restrict attention to complete innings where there are three outs and where there is a batting event; the new dataset is called retro2016_complete. Here non-batting plays such as steals, caught stealing, wild pitches, and passed balls are ignored. There is obviously some consequence of removing these non-batting plays from the viewpoint of run production, and this issue is discussed later in this chapter.
retro2016_complete <- retro2016 |>
mutate(
half_inning_id = paste(game_id, inn_ct, bat_home_id)
) |>
inner_join(half_innings, join_by(half_inning_id)) |>
filter(state != new_state | runs_scored > 0) |>
filter(outs_inning == 3, bat_event_fl)
In our definition of the new_state variable, we recorded the runner locations when there were three outs. The runner locations don’t matter, so we recode new_state to always have the value 3 when the number of outs is equal to 3. The str_replace() function replaces the regular expression [0-1]{3} 3—which matches any three character binary string followed by a space and a 3—with 3.
retro2016_complete <- retro2016_complete |>
mutate(new_state = str_replace(new_state, "[0-1]{3} 3", "3"))
### 9.2.3 Computing the transition probabilities
Now that the state and new_state variables are defined, one can compute the frequencies of all possible transitions between states using the table() function. The matrix of counts is T_matrix. There are 24 possible values of the beginning state state, and 25 values of the final state new_state including the 3-outs state.
T_matrix <- retro2016_complete |>
select(state, new_state) |>
table()
dim(T_matrix)
[1] 24 25
This matrix can be converted to a probability matrix by use of the prop.table() function. The resulting matrix is denoted by P_matrix.
P_matrix <- prop.table(T_matrix, 1)
dim(P_matrix)
[1] 24 25
Finally, we add a row to this transition probability matrix corresponding to transitions from the 3-out state. When the inning reaches 3 outs, then it stays at 3 outs, so the probability of staying in this state is 1.
P_matrix <- P_matrix |>
rbind("3" = c(rep(0, 24), 1))
The matrix P_matrix now has two important properties that allow it to model transitions between states in a Markov chain: 1) it is square, and; 2) the entries in each of its rows sum to 1.
dim(P_matrix)
[1] 25 25
P_matrix |>
apply(MARGIN = 1, FUN = sum)
000 0 000 1 000 2 001 0 001 1 001 2 010 0 010 1 010 2 011 0
1 1 1 1 1 1 1 1 1 1
011 1 011 2 100 0 100 1 100 2 101 0 101 1 101 2 110 0 110 1
1 1 1 1 1 1 1 1 1 1
110 2 111 0 111 1 111 2 3
1 1 1 1 1
To better understand this transition matrix, we display the transition probabilities starting at the “000 0” state, no runners and no outs below. (Only the positive probabilities are shown and the as_tibble() and pivot_longer() functions are used to display the probabilities vertically.) The most likely transitions are to the “no runners, one out” state with probability 0.676 and to the “runner on first, no outs” state with probability 0.235. The probability of moving from the “000 0” state to the “000 0” state is 0.033; in other words, the chance of a home run with no runners on with no outs is 0.033.
P_matrix |>
as_tibble(rownames = "state") |>
filter(state == "000 0") |>
pivot_longer(
cols = -state,
names_to = "new_state",
values_to = "Prob"
) |>
filter(Prob > 0)
# A tibble: 5 × 3
state new_state Prob
<chr> <chr> <dbl>
1 000 0 000 0 0.0334
2 000 0 000 1 0.676
3 000 0 001 0 0.00563
4 000 0 010 0 0.0503
5 000 0 100 0 0.235
Let’s contrast this with the possible transitions starting from the “010 2” state, runner on second with two outs. The most likely transitions are “3 outs” (probability 0.650), “runners on first and second with two outs” (probability 0.156), and “runner on first with 2 outs” (probability 0.074).
P_matrix |>
as_tibble(rownames = "state") |>
filter(state == "010 2") |>
pivot_longer(
cols = -state,
names_to = "new_state",
values_to = "Prob"
) |>
filter(Prob > 0)
# A tibble: 8 × 3
state new_state Prob
<chr> <chr> <dbl>
1 010 2 000 2 0.0233
2 010 2 001 2 0.00587
3 010 2 010 2 0.0576
4 010 2 011 2 0.000451
5 010 2 100 2 0.0745
6 010 2 101 2 0.0325
7 010 2 110 2 0.156
8 010 2 3 0.650
### 9.2.4 Simulating the Markov chain
One can simulate this Markov chain model a large number of times to obtain the distribution of runs scored in a half-inning of 2016 baseball. The first step is to construct a matrix giving the runs scored in all possible transitions between states. Let $$N_{runners}$$ denote the number of runners in a state and $$O$$ denote the number of outs. Because every player who has already batted in the inning is either on base, out, or has scored, for a batting play, the number of runs scored is equal to $runs = (N_{runners}^{(b)} + O^{(b)} + 1) - (N_{runners}^{(a)} + O^{(a)}).$
In other words, the runs scored is the sum of runners and outs before (b) the play minus the sum of runners and outs after (a) the play plus one. For example, suppose there are runners on first and second with one out, and after the play, there is a runner on second with two outs. The number of runs scored is equal to $runs = (2 + 1 + 1) - (1 + 2) = 1.$
We define a new function num_havent_scored() which takes a state as input and returns the sum of the number of runners and outs. We then apply this function across all the possible states (using the map_int() function) and the corresponding sums are stored in the vector runners_out.
num_havent_scored <- function(s) {
s |>
str_split("") |>
pluck(1) |>
as.numeric() |>
sum(na.rm = TRUE)
}
runners_out <- T_matrix |>
row.names() |>
set_names() |>
map_int(num_havent_scored)
The outer() function with the - (subtraction) operation performs the runs calculation for all possible pairs of states and the resulting matrix is stored in the matrix R_runs. If one inspects the matrix R_runs, one will notice some negative values and some strange large positive values. But this is not a concern since the corresponding transitions, for example a movement between a “000 0” state and a “000 2” state in one batting play, are not possible. To make the matrix square, we add an additional column of zeros to this run matrix using the cbind() function.
R_runs <- outer(
runners_out + 1,
runners_out,
FUN = "-"
) |>
cbind("3" = rep(0, 24))
We are now ready to simulate a half-inning of baseball using a new function simulate_half_inning(). The inputs are the probability transition matrix P, the run matrix R, and the starting state s (an integer between 1 and 24). The output is the number of runs scored in the half-inning.
simulate_half_inning <- function(P, R, start = 1) {
s <- start
path <- NULL
runs <- 0
while (s < 25) {
s_new <- sample(1:25, size = 1, prob = P[s, ])
path <- c(path, s_new)
runs <- runs + R[s, s_new]
s <- s_new
}
runs
}
There are two key statements in this simulation. If the current state is s, the function sample() will simulate a new state using the s row in the transition matrix P; the new state is denoted s_new. The total number of runs scored in the inning is updated using the value in the s row and the s_new column of the runs matrix R.
Using the map_int() function, one can simulate a large number of half-innings of baseball. In the below code, we simulate 10,000 half-innings starting with no runners and no outs (state 1), collecting the runs scored in the vector simulated_runs. The set.seed() function sets the random number seed so the reader can reproduce the results of this particular simulation by running this code.
set.seed(111653)
simulated_runs <- 1:10000 |>
map_int(~simulate_half_inning(T_matrix, R_runs))
To find the possible runs scored in a half-inning, we use the table() function to tabulate the values in simulated_runs.
table(simulated_runs)
simulated_runs
0 1 2 3 4 5 6 7 8 9
7364 1437 651 324 126 50 34 10 2 2
In our 10,000 simulations, five or more runs scored in 50 + 34 + 10 + 2 + 2 = 98 half-innings, so the chance of scoring five or more runs would be 98 / 10,000 = 0.0098. This calculation can be checked using the sum() function.
sum(simulated_runs >= 5) / 10000
[1] 0.0098
We compute the mean number of runs scored by applying the mean() function to simulated_runs.
mean(simulated_runs)
[1] 0.477
Over the 10,000 half-innings, an average of 0.477 runs were scored.
To understand the runs potential of different runners and outs situations, one can repeat this simulation procedure for other starting states. We write a function runs_j() to compute the mean number of runs scored starting with state j. Using the map_int() function, we apply the function runs_j() over all of the possible starting states 1 through 24. The output is a vector of mean runs scored stored in the mean_run_value column. These values are displayed below as a simulated expected run matrix (see Section 5.1).
runs_j <- function(j) {
1:10000 |>
map_int(~simulate_half_inning(T_matrix, R_runs, j)) |>
mean()
}
erm_2016_mc <- tibble(
state = row.names(T_matrix),
mean_run_value = map_dbl(1:24, runs_j)
) |>
mutate(
bases = str_sub(state, 1, 3),
outs_ct = as.numeric(str_sub(state, 5, 5))
) |>
select(-state)
erm_2016_mc |>
pivot_wider(names_from = outs_ct, values_from = mean_run_value)
# A tibble: 8 × 4
bases 0 1 2
<chr> <dbl> <dbl> <dbl>
1 000 0.481 0.255 0.103
2 001 1.32 0.925 0.338
3 010 1.14 0.640 0.295
4 011 1.93 1.30 0.474
5 100 0.855 0.500 0.211
6 101 1.71 1.14 0.425
7 110 1.39 0.875 0.406
8 111 2.19 1.46 0.667
Recall that our simulation model is based only on batting plays. To understand the effect of non-batting plays (stealing, caught stealing, wild pitches, etc.) on run scoring, we compare this run expectancy matrix with the one found in Chapter 5 using all batting and non-batting plays. Their difference is the contribution of non-batting plays to the average number of runs scored.
erm_2016 <- read_rds(here::here("data/erm2016.rds"))
erm_2016 |>
inner_join(erm_2016_mc, join_by(bases, outs_ct)) |>
mutate(
run_value_diff = round(mean_run_value.x - mean_run_value.y, 2)
) |>
select(bases, outs_ct, run_value_diff) |>
pivot_wider(names_from = outs_ct, values_from = run_value_diff)
# A tibble: 8 × 4
# Groups: bases [8]
bases 0 1 2
<chr> <dbl> <dbl> <dbl>
1 000 0.02 0.01 0
2 001 0.03 0.01 0.03
3 010 0 0.03 0.02
4 011 0 0.06 0.07
5 100 0 0.01 0.01
6 101 0.02 0.06 0.05
7 110 0.06 0.05 0.01
8 111 -0.08 0.07 0.03
Note that most of the values of the difference are positive, indicating that these non-batting plays generally do create runs. We note that the largest values tend to occur in situations when there is a runner on third who can score on a wild pitch or passed ball.
### 9.2.5 Beyond run expectancy
By using properties of Markov chains, it is straightforward to use the transition matrix to learn more about the movement through the runners/outs states.
By multiplying the probability matrix P_matrix by itself three times, we can learn about the likelihood of the state of the inning after three plate appearances. In R, matrix multiplication is indicated by the %*% symbol. The result is stored in the matrix P_matrix_3.
P_matrix_3 <- P_matrix %*% P_matrix %*% P_matrix
The first row of P_matrix_3 gives the probabilities of being in each of the 25 states after three hitters starting at the “000 0” state. We round these values to three decimal places, sort from largest to smallest, and display the largest values.
P_sorted <- P_matrix_3 |>
as_tibble(rownames = "state") |>
filter(state == "000 0") |>
pivot_longer(
cols = -state, names_to = "new_state", values_to = "Prob"
) |>
arrange(desc(Prob))
P_sorted |>
slice_head(n = 6)
# A tibble: 6 × 3
state new_state Prob
<chr> <chr> <dbl>
1 000 0 3 0.372
2 000 0 100 2 0.241
3 000 0 110 1 0.0815
4 000 0 010 2 0.0739
5 000 0 000 2 0.0529
6 000 0 001 2 0.0286
After three PAs, the most likely outcomes are three outs (probability 0.372), runner on first with 2 outs (probability 0.241), and runners on first and second with one out (probability 0.081).
It is also easy to learn about the number of visits to all runner-outs states. Define the matrix Q to be the 24-by-24 submatrix found from the transition matrix by removing the last row and column (the three outs state). By subtracting the matrix Q from the identity matrix and taking the inverse of the result, we obtain the fundamental matrix N of an absorbing Markov chain. (The diag() function is used to construct the identity matrix and the function solve() takes the matrix inverse.)
Q <- P_matrix[-25, -25]
N <- solve(diag(rep(1, 24)) - Q)
To understand the fundamental matrix, we display the beginning entries of the first row of the matrix.
N_0000 <- round(N["000 0", ], 2)
head(N_0000, n = 6)
000 0 000 1 000 2 001 0 001 1 001 2
1.05 0.75 0.60 0.01 0.03 0.05
Starting at the beginning of the inning (the “000 0” state), the average number of times the inning will be in the “000 0” state is 1.05, the average number of times in the “000 1” state is 0.75, the average number of times in the “000 2” state is 0.6, and so on. By using the sum() function, we find the average number of states that are visited.
sum(N_0000)
[1] 4.27
In other words, the average number of plate appearances in a half-inning (before three outs) is 4.27.
We can compute the average number of batting plays until three outs for all starting states by multiplying the fundamental matrix N by a column vector of ones. The vector of average number of plays is stored in the variable avg_num_plays and eight values of this vector are displayed.
avg_num_plays <- N %*% rep(1, 24) |>
t() |>
round(2)
avg_num_plays[,1:8]
000 0 000 1 000 2 001 0 001 1 001 2 010 0 010 1
4.27 2.87 1.46 4.33 2.99 1.53 4.34 2.93
This tells us the length of the remainder of the inning, on average, starting with each possible state. For example, starting at the bases empty, one out state, we expect on average to have 2.87 more batters. In contrast, with a runner on third with two outs, we expect to have 1.53 more batters.
### 9.2.6 Transition probabilities for individual teams
The transition probability matrix describes movements between states for an average team. Certainly, these probabilities will vary for teams of different batting abilities, and the probabilities will also vary against teams of different pitching abilities. We focus on different batting teams and discuss how to obtain good estimates of the transition probabilities for all teams.
To get the relevant data, a new variable batting_team needs to be defined that gives the batting team in each half-inning. By use of the str_sub() function, we define the home team variable home_team_id, and an if_else() function is used to define the batting team.
retro2016_complete <- retro2016_complete |>
mutate(
home_team_id = str_sub(game_id, 1, 3),
batting_team = if_else(
bat_home_id == 0,
away_team_id,
home_team_id
)
)
By use of the group_by() and count() functions, we construct a data frame T_team giving the counts of each team in the transitions from the current to new states.
T_team <- retro2016_complete |>
group_by(batting_team, state, new_state) |>
count()
For example, the filtering for batting_team equal to ANA gives the transition counts for Anaheim in the 2016 season.
T_team |>
filter(batting_team == "ANA") |>
slice_head(n = 6)
# A tibble: 192 × 4
# Groups: batting_team, state, new_state [192]
batting_team state new_state n
<chr> <chr> <chr> <int>
1 ANA 000 0 000 0 40
2 ANA 000 0 000 1 1007
3 ANA 000 0 001 0 9
4 ANA 000 0 010 0 75
5 ANA 000 0 100 0 359
6 ANA 000 1 000 1 31
7 ANA 000 1 000 2 720
8 ANA 000 1 001 1 3
9 ANA 000 1 010 1 54
10 ANA 000 1 100 1 261
# ℹ 182 more rows
If one is interested in comparing run productions for different batting teams, it is necessary to make some adjustments to the team transition probability matrices to get realistic predictions of performance. To illustrate the problem, we focus on transitions from the “100 2” state. We store the transition counts in the data frame T_team_S using the tally() function and display a few rows of this table below for six of the teams.
T_team_S <- retro2016_complete |>
filter(state == "100 2") |>
group_by(batting_team, state, new_state) |>
tally()
T_team_S |>
ungroup() |>
sample_n(size = 6)
# A tibble: 6 × 4
batting_team state new_state n
<chr> <chr> <chr> <int>
1 MIN 100 2 010 2 15
2 CIN 100 2 101 2 19
3 NYN 100 2 101 2 14
4 SEA 100 2 010 2 6
5 DET 100 2 011 2 9
6 DET 100 2 101 2 15
For some of the less common transitions, there is much variability in the counts across teams and this causes the corresponding team transition probabilities to be unreliable. If $$p^{TEAM}$$ represents the team’s transition probabilities for a particular team, and $$p^{ALL}$$ are the average transition probabilities, then a better estimate at the team’s probabilities has the form $p^{EST} = \frac{n}{n + K} p^{TEAM} + \frac{K}{n + K} p^{ALL},$ where $$n$$ is the number of transitions for the team and $$K$$ is a smoothing count. The description of the methodology is beyond the scope of this book, but in this case a smoothing count of $$K = 1274$$ leads to a good estimate at the team’s transition probabilities. (The choice of $$K$$ depends on the starting state.)
This method is illustrated for Washington’s transition counts starting from the “100 2” state. In the data frame T_WAS, the transition counts are stored in the variable n, and the corresponding proportions are stored in p. Similarly, for all teams, n and p are the counts and proportions in the data frame T_all.
T_WAS <- T_team_S |>
filter(batting_team == "WAS") |>
mutate(p = n / sum(n))
T_all <- retro2016_complete |>
filter(state == "100 2") |>
group_by(new_state) |>
tally() |>
mutate(p = n / sum(n))
We compute the improved estimate at Washington’s transition proportions using the formula and store the results in p_EST. The three sets of proportions (Washington, overall, and improved) are displayed in a data frame.
T_WAS |>
inner_join(T_all, by = "new_state") |>
mutate(
p_EST = (n.x / (1274 + n.x)) * p.x + (1274 / (1274 + n.x)) * p.y
) |>
select(batting_team, new_state, p.x, p.y, p_EST)
# A tibble: 8 × 6
# Groups: batting_team, state [1]
state batting_team new_state p.x p.y p_EST
<chr> <chr> <chr> <dbl> <dbl> <dbl>
1 100 2 WAS 000 2 0.0319 0.0291 0.0291
2 100 2 WAS 001 2 0.00532 0.00577 0.00577
3 100 2 WAS 010 2 0.0213 0.0220 0.0219
4 100 2 WAS 011 2 0.0213 0.0220 0.0219
5 100 2 WAS 100 2 0.00266 0.000775 0.000776
6 100 2 WAS 101 2 0.0452 0.0435 0.0435
7 100 2 WAS 110 2 0.184 0.195 0.194
8 100 2 WAS 3 0.689 0.682 0.683
Note that the improved transition proportions are a compromise between the team’s proportions and the overall values. For example, for a transition from the state “100 2” to “010 2”, the Washington value is 0.0213, the overall value is 0.0220, and the improved value 0.0219 falls between the Washington and overall values. This method is especially helpful for particular transitions such as “100 2” to “100 2”, which may not occur for one team in this season but for which we know there is a positive chance of these transitions happening in the future.
This smoothing method can be applied for all teams and all rows of the transition matrix to obtain improved estimates of teams’ probability transition matrices. With the team transition matrices computed in this way, one can explore the run-scoring behavior of individual batting teams.
## 9.3 Simulating a Baseball Season
An attractive method of modeling paired comparison data such as baseball games is the Bradley-Terry model. We illustrate this modeling technique via simulation for the 1968 Major League Baseball season when the regular season and playoff system had a relatively simple structure. It is straightforward to adapt these methods to the present baseball season with a more complicated schedule and playoff system.
In 1968, there were 20 teams, 10 in the National League and 10 in the American League. Suppose each team has a talent or ability to win a game. The talents for the 20 teams are represented by the values $$T_1, ..., T_{20}$$. We assume that the talents are distributed from a normal curve model with mean 0 and standard deviation $$s_T$$. A team of average ability would have a talent value close to zero, “good” teams would have positive talents, and bad teams would have negative talents. Suppose team $$A$$ plays team $$B$$ in a single game. By the Bradley-Terry model, the probability team $$A$$ wins the game is given by the logistic function $P(A \, wins) = \frac{\exp(T_A)}{\exp(T_A) + \exp(T_B)}.$
This model is closely related to the log5 method developed by Bill James in his Baseball Abstract books in the 1980s (see, for example, James (1982)). If $$P_A$$ and $$P_B$$ are the winning percentages of teams $$A$$ and $$B$$, then James’ formula is given by $P(A \, wins) = \frac{P_A / (1- P_A)}{P_A / (1- P_A) + P_B / (1- P_B)}.$ Comparing the two formulas, one sees that the log5 method is a special case of the Bradley-Terry model where a team’s talent $$T$$ is set equal to the log odds of winning $$\log(P / (1 - P))$$. A team with a talent $$T = 0$$ will win (in the long run) half of its games ($$P = 0.5$$). In contrast, a team with talent $$T = 0.2$$ will win (using the log 5 values) approximately $$55\%$$ of its games and a team with talent $$T = -0.2$$ will win $$45\%$$ of its games.
Using this model, one can simulate a baseball season as follows.
1. Construct the 1968 baseball schedule. In this season, each of the 10 teams in each league play each other team in the same league 18 games, where 9 games are played in each team’s ballpark. (There was no interleague play in 1968.)
2. Simulate 20 talents from a normal distribution with mean 0 and standard deviation $$s_T$$. The value of $$s_T$$ is chosen so that the simulated season winning percentages from this model resemble the actual winning percentages during this season.
3. Using the probability formula and the talent values, one computes the probabilities that the home team wins all games. By a series of coin flips with these probabilities, one determines the winners of all games.
4. Determine the winner of each league (ties need to be broken by some random mechanism) and play a best-of-seven World Series using winning probabilities computed using the Bradley-Terry model and the two talent numbers.
### 9.3.2 Making up a schedule
The first step in the simulation is to construct the schedule of games. We wrote a short function make_schedule() to help with this task. The inputs are the vector of team names teams and the number of games k that will be played between two teams in the first team’s home park. The output is a data frame where each row corresponds to a game and Home and Visitor give the names of the home and visiting teams. The rep() function, which generates repeated copies of a vector, is used several times in this function.
make_schedule <- function(teams, k) {
num_teams <- length(teams)
Home <- rep(rep(teams, each = num_teams), k)
Visitor <- rep(rep(teams, num_teams), k)
tibble(Home = Home, Visitor = Visitor) |>
filter(Home != Visitor)
}
This function is used to construct the schedule for the 1968 season. Two vectors NL and AL are constructed containing abbreviations for the National League and American League teams. We apply the function make_schedule() twice, once for each league, using k = 9 since one team hosts another team nine games. We use the list_rbind() function to paste together the NL and AL schedules, creating the data frame schedule.
library(Lahman)
teams_68 <- Teams |>
filter(yearID == 1968) |>
select(teamID, lgID) |>
mutate(teamID = as.character(teamID)) |>
group_by(lgID)
schedule <- teams_68 |>
group_split() |>
set_names(pull(group_keys(teams_68), "lgID")) |>
map(~make_schedule(teams = .x\$teamID, k = 9)) |>
list_rbind(names_to = "lgID")
dim(schedule)
[1] 1620 3
Note that schedule has $$\frac{162 \cdot 20}{2}$$ rows, since each game involves two teams.
### 9.3.3 Simulating talents and computing win probabilities
The next step is to compute the win probabilities for all of the games in the season schedule. The team talents are assumed to come from a normal distribution with mean 0 and standard deviation s_talent, which we assign s_talent = 0.20. (Recall that this value of the standard deviation is chosen so that the season team win percentages generated from the model resemble the actual team win percentages.) We simulate the talents using the function rnorm() that assigns the talents to the 20 teams. By use of two applications of the inner_join() function, we add the team talents to the schedule data frame; the new data frame is called schedule_talent.
s_talent <- 0.20
teams_68 <- teams_68 |>
mutate(talent = rnorm(10, 0, s_talent))
schedule_talent <- schedule |>
inner_join(teams_68, join_by(lgID, Home == teamID)) |>
rename(talent_home = talent) |>
inner_join(teams_68, join_by(lgID, Visitor == teamID)) |>
rename(talent_visitor = talent)
Last, once we have the talents for the home and visiting teams for all games, we apply the Bradley-Terry model to compute home team winning probabilities for all games; these probabilities are stored in the variable prob_home.
schedule_talent <- schedule_talent |>
mutate(
prob_home = exp(talent_home) /
(exp(talent_home) + exp(talent_visitor))
)
The first six rows of the data frame schedule_talent are displayed below, where one sees the games scheduled, the talents of the home and away teams, and the probability that the home team wins the matchup.
slice_head(schedule_talent, n = 6)
# A tibble: 6 × 6
lgID Home Visitor talent_home talent_visitor prob_home
<chr> <chr> <chr> <dbl> <dbl> <dbl>
1 AL BAL BOS 0.197 0.269 0.482
2 AL BAL CAL 0.197 -0.230 0.605
3 AL BAL CHA 0.197 -0.00924 0.551
4 AL BAL CLE 0.197 -0.185 0.594
5 AL BAL DET 0.197 0.409 0.447
6 AL BAL MIN 0.197 -0.208 0.600
### 9.3.4 Simulating the regular season
To simulate an entire season of games, we perform a series of coin flips, where the probability the home team wins depends on the winning probability. The function rbinom() performs the coin flips for the 1620 scheduled games; the outcomes are a sequence of 0s and 1s. By use of the if_else() function, we define the winner variable to be the Home team if the outcome is 1, and the Visitor otherwise.
schedule_talent <- schedule_talent |>
mutate(
outcome = rbinom(nrow(schedule_talent), 1, prob_home),
winner = if_else(outcome == 1, Home, Visitor)
)
The teams, home win probabilities, and outcomes of the first six games are displayed below.
schedule_talent |>
select(Visitor, Home, prob_home, outcome, winner) |>
slice_head(n = 6)
# A tibble: 6 × 5
Visitor Home prob_home outcome winner
<chr> <chr> <dbl> <int> <chr>
1 BOS BAL 0.482 0 BOS
2 CAL BAL 0.605 0 CAL
3 CHA BAL 0.551 1 BAL
4 CLE BAL 0.594 1 BAL
5 DET BAL 0.447 0 DET
6 MIN BAL 0.600 0 MIN
How did the teams perform during this particular simulated season? Using the group_by() and summarize() functions, we find the number of wins for all teams. We collect this information together with the team names in the data frame WIN, and use the inner_join() function to combine the season results with the team talents to create the data frame results.
results <- schedule_talent |>
group_by(winner) |>
summarize(Wins = n()) |>
inner_join(teams_68, by = c("winner" = "teamID"))
### 9.3.5 Simulating the post-season
After the regular season, one can simulate the post-season series. We write a function win_league() that simulates a league championship. The inputs are the data frame res of teams and win totals. By use of the min_rank() function, we identify the teams that have the largest number of wins in each league. If one team has the maximum number, then an indicator variable is_winner_lg is created, which is 1 for that particular team. In order to avoid a tie in win totals for two or more teams, we randomly add a random tiebreaker quantity (that is less than 1) to every teams win total using the runif() function.
win_league <- function(res) {
res |>
group_by(lgID) |>
mutate(
tiebreaker = runif(n = length(talent)),
wins_total = Wins + tiebreaker,
rank = min_rank(desc(wins_total)),
is_winner_lg = wins_total == max(wins_total)
)
}
To simulate the post-season, we populate a new variable is_winner_ws; this is an indicator for the World Series winner. By an application of win_league(), we find the winners of each league. We simulate the World Series by flipping a coin seven times (rmultinom()), where the win probabilities are proportional to $$\exp(talent)$$. The is_winner_ws indicates the team winning a majority of the games.
sim_one <- win_league(results)
ws_winner <- sim_one |>
filter(is_winner_lg) |>
ungroup() |>
mutate(
outcome = as.numeric(rmultinom(1, 7, exp(talent))),
is_winner_ws = outcome > 3
) |>
filter(is_winner_ws) |>
select(winner, is_winner_ws)
sim_one |>
left_join(ws_winner, by = c("winner")) |>
replace_na(list(is_winner_ws = 0))
# A tibble: 20 × 9
# Groups: lgID [2]
winner Wins lgID talent tiebreaker wins_total rank
<chr> <int> <fct> <dbl> <dbl> <dbl> <int>
1 ATL 83 NL -0.215 0.867 83.9 5
2 BAL 86 AL 0.197 0.0260 86.0 3
3 BOS 99 AL 0.269 0.354 99.4 2
4 CAL 62 AL -0.230 0.936 62.9 10
5 CHA 85 AL -0.00924 0.246 85.2 4
6 CHN 85 NL -0.107 0.829 85.8 4
7 CIN 81 NL -0.0612 0.264 81.3 6
8 CLE 71 AL -0.185 0.290 71.3 9
9 DET 100 AL 0.409 0.0841 100. 1
10 HOU 90 NL -0.0871 0.845 90.8 2
11 LAN 63 NL -0.326 0.729 63.7 10
12 MIN 74 AL -0.208 0.667 74.7 7
13 NYA 78 AL 0.0424 0.422 78.4 6
14 NYN 80 NL 0.100 0.569 80.6 7
15 OAK 82 AL 0.287 0.927 82.9 5
16 PHI 93 NL 0.265 0.819 93.8 1
17 PIT 71 NL -0.146 0.0460 71.0 9
18 SFN 88 NL 0.249 0.119 88.1 3
19 SLN 76 NL -0.348 0.425 76.4 8
20 WS2 73 AL 0.0842 0.730 73.7 8
# ℹ 2 more variables: is_winner_lg <lgl>, is_winner_ws <lgl>
### 9.3.6 Function to simulate one season
It is convenient to place all of these commands including the functions make_schedule() and win_league() in a single function one_simulation_68(), which you can find in the abdwr3edata package. The only input is the standard deviation s_talent that describes the spread of the normal talent distribution. The output is a data frame containing the teams, talents, number of season wins, and success in the post-season. We illustrate simulating one season and display the data frame results_1 that is returned.
library(abdwr3edata)
set.seed(111653)
results_1 <- one_simulation_68(0.20)
results_1
# A tibble: 20 × 6
Team Wins League Talent Winner.Lg Winner.WS
<chr> <int> <dbl> <dbl> <dbl> <dbl>
1 SFN 93 1 -0.0591 1 0
2 PHI 93 1 -0.00979 0 0
3 LAN 87 1 0.00406 0 0
4 HOU 84 1 -0.117 0 0
5 SLN 80 1 -0.128 0 0
6 ATL 79 1 -0.100 0 0
7 CIN 79 1 -0.235 0 0
8 NYN 76 1 -0.269 0 0
9 CHN 76 1 -0.0199 0 0
10 PIT 63 1 -0.313 0 0
11 NYA 100 2 0.284 1 1
12 DET 93 2 0.379 0 0
13 CHA 87 2 0.139 0 0
14 BOS 86 2 -0.102 0 0
15 WS2 84 2 0.0915 0 0
16 OAK 82 2 -0.0622 0 0
17 CAL 78 2 -0.129 0 0
18 BAL 74 2 -0.0728 0 0
19 MIN 65 2 -0.207 0 0
20 CLE 61 2 -0.292 0 0
We write a new function display_standings() to put the season wins in a more familiar standings format. The inputs to this function are the results_1 data frame and the league indicator.
display_standings <- function(data, league) {
data |>
filter(League == league) |>
select(Team, Wins) |>
mutate(Losses = 162 - Wins) |>
arrange(desc(Wins))
}
We then apply this function twice (once for each league) using map() and then use the bind_cols() function to combine the two standings into a single data frame. The league champions and the World Series winner are also displayed below.
map(1:2, display_standings, data = results_1) |>
bind_cols()
# A tibble: 10 × 6
Team...1 Wins...2 Losses...3 Team...4 Wins...5 Losses...6
<chr> <int> <dbl> <chr> <int> <dbl>
1 SFN 93 69 NYA 100 62
2 PHI 93 69 DET 93 69
3 LAN 87 75 CHA 87 75
4 HOU 84 78 BOS 86 76
5 SLN 80 82 WS2 84 78
6 ATL 79 83 OAK 82 80
7 CIN 79 83 CAL 78 84
8 NYN 76 86 BAL 74 88
9 CHN 76 86 MIN 65 97
10 PIT 63 99 CLE 61 101
results_1 |>
filter(Winner.Lg == 1) |>
select(Team, Winner.WS)
# A tibble: 2 × 2
Team Winner.WS
<chr> <dbl>
1 SFN 0
2 NYA 1
In this particular simulated season, the Philadelphia Phillies (PHI) and the San Francisco Giants (SFN) tied for the National League title with 93 wins and the New York Yankees (NYA) won the American League with 100 wins. The Yankees defeated the Giants in the World Series. The team with the best talent in this season was Detroit (talent equal to 0.379) and they lost in the ALCS. In other words the “best team in baseball” was not the most successful during this simulated season. We will shortly see if the best team typically wins the World Series.
### 9.3.7 Simulating many seasons
One can learn about the relationship between a team’s ability and its season performance by simulating many seasons of baseball. To simulate 1000 seasons, we use the rep() function to create a vector of length 1000, then use map() to repeatedly apply the one_simulation_68() function to this vector, storing the output in many_results.
set.seed(111653)
many_results <- rep(0.20, 1000) |>
map(one_simulation_68) |>
list_rbind()
The data frame many_results contains the talent number and number of wins for 1000 $$\times$$ 20 = 20,000 teams. By use of the geom_point() function using the alpha = 0.05 argument, we construct a “smoothed” scatterplot of Talent and Wins in Figure 9.1.
ggplot(many_results, aes(Talent, Wins)) +
geom_point(alpha = 0.05)
As expected, there is a positive trend in the graph, indicating that better teams tend to win more games. But there is much vertical spread in the scatterplot, which says that the relationship between talent and wins is not strong.
To reinforce the last point, suppose we focus on “average” teams that have a talent number between $$-0.05$$ and 0.05. Using the filter() function, we isolate the talent and wins data for these average teams. A histogram of the season wins for these teams is shown in Figure 9.2.
many_results |>
filter(Talent > -0.05, Talent < 0.05) |>
ggplot(aes(Wins)) +
geom_histogram(color = crcblue, fill = "white")
One expects these average teams to win about 81 games. But what is surprising is the variability in the win totals—average teams can regularly have win totals between 70 and 90, and it is possible (but not likely) to have a win total close to 100.
What is the relationship between a team’s talent and its post-season success? Consider first the relationship between a team’s talent (variable Talent) and winning the league (the variable Winner.Lg). Since Winner.Lg is a binary (0 or 1) variable, a common approach for representing this relationship is a logistic model; this is a generalization of the usual regression model where the response variable is binary instead of continuous. We use the glm() function with the family argument set to binomial to fit a logistic model; the output is stored in the variable fit1. In a similar fashion, we use a logistic model to model the relationship between winning the World Series (variable Winner.WS) and the talent; the output is in the variable fit2.
fit1 <- glm(
Winner.Lg ~ Talent,
data = many_results, family = binomial
)
fit2 <- glm(
Winner.WS ~ Talent,
data = many_results, family = binomial
)
A logistic model has the form $p = \frac{\exp(a + b T)}{ 1 + \exp(a + b T)},$ where $$T$$ is a team’s talent, $$(a, b)$$ are the regression coefficients, and $$p$$ is the probability of the event.
In the following code, we generate a vector of plausible values of a team’s talent and store them in the vector talent_values. We then compute fitted probabilities of winning the pennant and winning the World Series using the predict() function. (The type = "response" argument will map values of $$a + b T$$ to the probability scale.) Then we construct a graph of talent against probability using the geom_line() function where the color of the line corresponds to the type of achievement. The completed graph is displayed in Figure 9.3.
tdf <- tibble(
Talent = seq(-0.4, 0.4, length.out = 100)
)
tdf |>
mutate(
Pennant = predict(fit1, newdata = tdf, type = "response"),
World Series = predict(fit2, newdata = tdf, type = "response")
) |>
pivot_longer(
cols = -Talent,
names_to = "Outcome",
values_to = "Probability"
) |>
ggplot(aes(Talent, Probability, color = Outcome)) +
geom_line() + ylim(0, 1) +
scale_color_manual(values = crc_fc)
As expected, the chance of a team winning the pennant (solid line) increases as a function of the talent. An average team with $$T = 0$$ has only a small chance of winning the pennant; an excellent team with a talent close to 0.4 has about a 60% chance of winning the pennant. The probabilities of winning the World Series (represented by a dashed line) are substantially smaller than the chances of winning the pennant. For example, this excellent ($$T = 0.4$$) team has only about a 35% chance of winning the World Series. In fact, it can be demonstrated that the team winning the World Series is likely not to be the team with the best talent (largest value of $$T$$).
A general description of the Markov chain probability model is contained in Kemeny and Snell (1960). Pankin (1987) and Bukiet, Harold, and Palacios (1997) illustrate the use of Markov chains to model baseball. Chapter 9 of Albert (2017) gives an introductory description of Markov chains and illustrates the construction and use of the transition matrix using 1987 season data. The Bradley-Terry model is a popular statistical model for paired comparisons. Chapter 9 of Albert and Bennett (2003) describes the application of the Bradley-Terry model for baseball team competition. The use of R in simulation is introduced in Chapter 11 of Albert and Rizzo (2012). Lopez, Matthews, and Baumer (2018) use a Bradley-Terry state space model to address the question of how often the best teams win.
## 9.5 Exercises
1. A Simple Markov Chain
Suppose one is interested only in the number of outs in an inning. There are four possible states in an inning (0 outs, 1 out, 2 outs, and 3 outs) and you move between these states in each plate appearance. Suppose at each PA, the chance of not increasing the number of outs is 0.3, and the probability of increasing the outs by one is 0.7. The following R code puts the transition probabilities of this Markov chain in a matrix P.
P <- matrix(c(.3, .7, 0, 0,
0, .3, .7, 0,
0, 0, .3, .7,
0, 0, 0, 1), 4, 4, byrow = TRUE)
1. If one multiplies the matrix P by itself P to obtain the matrix P2:
P2 <- P %*% P
The first row of P2 gives the probabilities of moving from 0 outs to each of the four states after two plate appearances. Compute P2. Based on this computation, find the probability of moving from 0 outs to 1 out after two plate appearances.
1. The fundamental matrix N is computed as
N <- solve(diag(c(1, 1, 1)) - P[-4, -4])
The first row gives the average number of PAs at 0 out, 1 out, and 2 outs in an inning. Compute N and find the average number of PAs in one inning in this model.
2. A Simple Markov Chain, Continued
The following function simulate_half_inning() will simulate the number of plate appearances in a single half-inning of the Markov chain model described in Exercise 1 where the input P is the transition probability matrix.
simulate_half_inning <- function(P) {
s <- 1
path <- NULL
while(s < 4){
s_new <- sample(1:4, 1, prob = P[s, ])
path <- c(path, s_new)
s <- s_new
}
length(path)
}
1. Use the map() function to simulate 1000 half-innings of this Markov chain and store the lengths of these simulated innings in the vector lengths.
2. Using this simulated output, find the probability that a half-inning contains exactly four plate appearances.
3. Use the simulated output to find the average number of PAs in a half-inning. Compare your answer with the exact answer in Exercise 1, part (b).
3. Simulating a Half Inning
In Section 9.2.4, the expected number of runs as calculated for each one of the 24 possible runners-outs situations using data from the 2016 season. To see how these values can change across seasons, download play-by-play data from Retrosheet for the 1968 season, construct the probability transition matrix, simulate 10,000 half-innings from each of the 24 situations, and compute the run expectancy matrix. Compare this 1968 run expectancy matrix with the one computed using 2016 data.
4. Simulating the 1950 Season
Suppose you are interested in simulating the 1950 regular season for the National League. In this season, the team abbreviations were “PHI”, “BRO”, “NYG”, “BSN”, “STL”, “CIN”, “CHC”, and “PIT” and each team played every other team 22 games (11 games at each park).
1. Using the function make_schedule(), construct the schedule of games for this NL season.
2. Suppose the team talents follow a normal distribution with mean 0 and standard deviation 0.25. Using the Bradley-Terry model, assign home win probabilities for all games on the schedule.
3. Use the rbinom() function to simulate the outcomes of all 616 games of the NL 1950 season.
4. Compute the number of season wins for all teams in your simulation.
5. Simulating the 1950 Season, Continued
1. Write a function to perform the simulation scheme described in Exercise 4. Have the function return the team with the largest talent and the team with the most wins. (If there is a tie for the league pennant, have the function return one of the best teams at random.)
2. Repeat this simulation for 1000 seasons, collecting the most talented team and the most successful team for all seasons.
3. Based on your simulations, what is the chance that the most talented team wins the pennant?
6. Simulating the World Series
1. Write a function to simulate a World Series. The input is the probability p that the AL team will defeat the NL team in a single game.
2. Suppose an AL team with talent $$0.40$$ plays a NL team with talent $$0.25$$. Using the Bradley-Terry model, determine the probability p that the AL wins a game.
3. Using the value of p determined in part (b), simulate 1000 World Series and find the probability the AL team wins the World Series.
4. Repeat parts (b) and (c) for AL and NL teams who have the same talents.
1. This is often known as the memoryless property.↩︎ | 14,097 | 51,492 | {"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} | 3.90625 | 4 | CC-MAIN-2024-22 | latest | en | 0.945742 |
https://writeessay-help.org/2020/10/04/motion-problem-solving_9h/ | 1,603,873,922,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107897022.61/warc/CC-MAIN-20201028073614-20201028103614-00301.warc.gz | 609,824,628 | 5,973 | # Motion problem solving
By | October 4, 2020
Chapters no homework argument 1—3 including: need more algebra help? It will be the critical thinking quote same in all the equations for a given problem motion situation, for the object on the turntable. uniform motion explains the motion problem solving distance of an object when it travels at a constant speed, …. the elements of a business plan these steps will help you solve the problems in a systematic manner: 1 – projectile motion calculator and solver given initial 3 page short story essay velocity, angle and height en la sangre essay enter the initial velocity v 0 in meters per second (m/s), the initial andgle θ in degrees and the initial height y 0 in meters (m) as positive real numbers and press “calculate”. a table is often useful when motion problem solving solving a motion problem. •horizontal kinematics and free-fall 1 – projectile motion problem solving motion calculator and solver given initial velocity, angle and height enter bca assignment the initial velocity business plans for nonprofits v0in meters per second (m/s), the initial andgle θ in degrees and the initial height y0in meters (m) as positive real numbers and press “calculate” x = (2.4 m/s)• (0.3499 s) carefully hispanic heritage essay read the problem and list known and unknown information motion problem solving in terms of the symbols of the kinematic equations. so she jumps essay writing online into her motion problem solving red mustang convertible and drives off going west at research paper topics for middle school students a speed of 40 ap lang essay types miles per hour now 2 hours later, lara’s mom decides she had better go after her daughter and smooth things out human development research paper topics 2-d kinematics/motion in plane problems with solutions for jee main & jee advanced. understand the problem • what is being australian essay writing service asked? Divide your group into teams of equal numbers. students should first draw a diagram to represent the relationship between the distances involved in the problem, then how to write a conclusion to an essay set up a chart based on the formula rate times time = distance plug n’ chug: i highly recommend this book. usually only two types of motions are considered in kinematics problems:. | 476 | 2,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-45 | longest | en | 0.909129 |
https://lustaufreben.de/ore/9547/how-to-build-a-using-simple-machines.html | 1,642,752,816,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302740.94/warc/CC-MAIN-20220121071203-20220121101203-00344.warc.gz | 361,741,542 | 6,797 | ## how to build a using simple machines
How to Make a Lever, Simple machines make work easier by multiplying, reducing, or changing the direction of a force There are six different types of simple machines, including ramps, levers, and gears Simple Machines Science Projects Make a Lever A lever is a type of simple machine You can make one and experiment with how moving the pivot point, or fulcrum ,Elementary Physical Science, In the Simple Machines module, students see examples of how simple machines can make work easier by reducing the force needed to move an object over a distance Then, students use interactive activities to experiment with these simple machines: lever, ,Physics for Kids: Simple Machines, More complex machines are made up of a bunch of simple machin There are 6 basic types of simple machines: Lever The lever is made up of a straight rigid object like a board or a bar which pivots on a turning point called a fulcrum Levers make work easier by using ,Thematic Units, Lesson Plans Exploring Simple Machines (WebQuest) Grade 4 The machines we use every day help to make our work easier This WebQuest will help you identify different types of simple machines so that you can invent your own machinHow to make a Simple Pulley System, How to make a Simple Pulley System - Pulleys Simple MachinesThis is a super simple pulley system You only need three carabiners, some rope, and a place to h,.
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How to Build Another Useless Machine: Easy to Make, and ,, How to Build Another Useless Machine: Easy to Make, and Hard to Use!: So many useless machines are built, why building another one? Simply, it is really fun :), fun to build and fun to play with I tried to give this version a character, as I always felt there is somebody "thinking inside the box", so here I tried, ,Building a simple State Machine in Python, Aug 02, 2017· We must ask what our use case is and, in this case, whether we even need a state machine Perhaps the example below will help provide insight into what state machines can be used for STATE MACHINES Before we get into implementing a simple state machine in Python lets quickly go over what a state machine is and what it looks like56 Simple machine projects ideas | simple machine projects ,, Oct 2, 2016 - Explore Renee DeVillez's board "simple machine projects", followed by 209 people on Pinterest See more ideas about simple machine projects, simple machines, projectsCreating My Own Rube Goldberg Machine, 5 Decide what simple everyday task their machine will complete Create a map of the simple machines that will make up their device Use may want to consider using mapping software like Kidspiration to assist with the planning stage Then create a drawing of what the device will look like 6Simple Pneumatic Machine : 6 Steps (with Pictures ,, Simple Pneumatic Machine: This project draws on similar ideas from previous work, but without the use of hot glue This construction process is very simple, and unlike some of my hydraulic projects, these pneumatic machines operate using only compressed air instead of water,.
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Three Simple Machine Projects for Kids: Fabulous & Fun to Make, These fun, simple machine projects for kids in the elementary grades use easily gathered common objects to make cool projects Whether you need a class project, an idea for the science fair, or just a fun way to pass an afternoon, these projects are a fun way to learn how simple machines ,How to Build Simple Machines for Grade 5 | Synonym, A simple machine can be any device that requires only a single force There are six general types of simple machines: levers, inclined planes, screws, wedges, pulleys and a wheel-and-axle system ,Sixth grade Lesson Design Your Own Simple Machine ,, Teacher Tip: This lesson builds upon the previous lessons in the Simple Machine UnitIt provides students with a chance to actually construct their own simple machine from basic materials This lesson allows students to apply their knowledge of simple machines and also engage in the process of engineering and designhow to build a simple machines, Build all 5 Simple Machines at the same time! Students can construct, examine and explain simple machines to develop a deeper understanding of how they make work easier A 5-in-1 value! Contains enough pieces to build a lever, pulley, inclined plane, wheel and axle, and wedge-all at the same timeBuild a Simple Machine Science Projects, Build a simple machine, such as an inclined plane (a board on a slope), a lever, a pulley, or a wheel and an axle Investigate how to make changes to your machine ,.
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Experiment with Simple Machines Science Projects, Design and build your own simple machine or investigate how they work While simple machines are called simple, learn that they make many things happen, including your bike go around Simple machines are levers, inclined planes (ramps), screws, wheels/axles, pulleys, and wedgBuilding a Barrel Rifling Machine, Mar 20, 2012· Another simple option is to use a barrel with the correct twist rate, cast a slug (or swage) to fit the rifling, attach a steel rod to carry the cutter and use that barrel as a rifling guide Clamp both barrels in line then run the slug/rod/cutter back and forth to cut the rifling in the new barrelSimple Machines Design Project Sample, Humans have been making and using machines for a long, long time We create these machines to help us to survive and do our work We use these machines every day and in all aspects of our liv In this project, students will learn how very simple machines can be utilized to make our work easier Machin When we do work, we must expend energySix Simple Machines in Construction from Construction ,, As in all machines, the actual mechanical advantage equals the resistance divided by the effort In many applications of the screw, you make use of the large amount of friction that is commonly present in this simple machine By using the screw, you reduce large amounts of circular motion to very small amounts of straight-line motion60+ Unit Ideas: Simple Machines | simple machines, science ,, Nov 2, 2020 - This board is full of ideas for teaching kids about simple machin At the most basic level, there are six types of simple machines: lever, pulley, wheel and axle, wedge, screw, and inclined plane This board has information about the history of simple machines, what we use simple machines for, and hands-on ideas for kids to make their own simple machin.
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Introducing Simple Machines | Scholastic, In this novel by Robert C O'Brien, the rats are so clever that they learn how to use simple machines to build and move things, including Mrs Frisby's house Read this section of the book and have students illustrate the rats' inventionHow To Make a Car, This video will show you how to make a very simple Rubber Band powered Car using plastic bottl Hope you like it3 Ideas for Building Simple Machines at Home ,, Simple machines are a topic I've taught for years but never been happy with my approach until now Is this activity, students are making simple machines with items Creating a hands-on application of what they already know and will learn about simple machinSimple Machines, There are 6 simple machines: Inclined Plane, Wedge, Screw, Lever, Wheel and Axle, and Pulley Compound machines have two or more simple machines working together to make work easier Most of the machines we use today are compound machines, created by combining several simple machin Take the simple machine quizHomemade Simple Machines for Kids | How To Adult, Apr 18, 2017· A machine with one working part is classified as simple, whereas a tool with multiple parts working together or in sequence is considered a compound machine Children can make many simple machines for fun or to demonstrate the different uses of simple machin. | 1,630 | 7,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.862659 |
http://www.jiskha.com/display.cgi?id=1392089428 | 1,495,941,682,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609409.62/warc/CC-MAIN-20170528024152-20170528044152-00348.warc.gz | 672,479,728 | 3,612 | # math
posted by on .
Find an equation in standard form of the line passing through the points (8, -5) and (-6, 2).
• math - ,
(8,-5), (-6,2)
m = (2-(-5))/(-6-8) = 7/-14 = -1/2.
Y = mx + b = -5
(-1/2)*8 + b = -5
b = -5 + 4 = -1
Y = (-1/2)x -1
Y = -x/2 - 1
Multiply both sides by 2:
2y = -x - 2
X + 2Y = -2 | 148 | 312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-22 | latest | en | 0.716498 |
https://www.achrnews.com/articles/145805-the-3-4-5-rule-for-walk-in-coolers-and-freezers | 1,713,022,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00030.warc.gz | 594,064,082 | 24,525 | During the installation of a new walk-in cooler or freezer, it is imperative the box remain square and level during the assembly of its panels. If the corners are not kept square and the floor not level, the panels will not fit together, leading to major issues as the box is being built. Even small inaccuracies will grow as the panels are assembled. These inaccuracies may not seem to be an issue as the panels are fit together, but they will lead to other panels not fitting properly and present major issues for completing the construction of the box.
First, always start with a level surface. If the floor is not level, two recommended techniques to level the floor are using self-leveling epoxy and/or asphalt shingles. It may be tempting to rush this part of the project in order to begin building the box, but it is important to get this right first before proceeding. It’s like building a wall — the base is the most important part of the project. Get it right and the wall goes up smoothly. Get it wrong and the wall is a mess.
Keeping the corners square is extremely important. Again, if the corners are not kept square, the panels will not fit together properly, and it will be a struggle to build the box. To make sure the corners are square, use a general mathematic formula that was learned back in high school: the Pythagorean theorem. For a right angle triangle it is:
a2 +b2 = c2
To make it simpler to remember and apply, substitute real numbers, using 3, 4, and 5:
32(9) + 42(16) = 52(25)
The 3-4-5 triangle is a great way to make sure the corners are square and the panels at a 90-degree angle (see Figure 1, top). If one side of a triangle measures 3 feet and the adjacent side measures 4 feet, then the diagonal between those two points must measure 5 feet in order for it to be a right triangle. High school math really does have some real-world benefits.
Using a chalk line, you can lay out the corners on the floor and keep the panels square. Pick one leg of your panel and measure out 3 feet from the corner. Put a mark on the floor at the 3-foot point. Now, measure the adjacent floor from the same corner to 4 feet and put a mark there. Then, measure the distance between the two marks. If it is 5 feet, then you have a perfectly square corner. If the measurement is less than 5 feet, the angle is too small (<90°) and needs to be opened up a bit. If it is more than 5 feet, the angle is too big (>90°) and needs to be closed some. If needed, you can also use the measurements of 6-8-10.
Installing refrigeration systems will require a technician to not only be proficient in the refrigerant cycle and all of its components, but also will require a technician to be knowledgeable in many other trade crafts. | 627 | 2,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-18 | latest | en | 0.94742 |
https://community.hannity.com/t/gdp-prediction-for-2018q3/25369 | 1,548,258,026,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584334618.80/warc/CC-MAIN-20190123151455-20190123173455-00156.warc.gz | 474,982,040 | 13,480 | # GDP Prediction for 2018Q3
#1
It’s that time again! Tomorrow, the BEA will release the initial 2018Q3 GDP estimate. I’m particularly excited about this one because it will heavily determine whether or not Trump can achieve that elusive goal that Republicans so often loved to remind us that Obama could never reach: 3% GDP growth in a calendar year. I’ll first provide what the current major models are predicting, what I predict, and then walk through some scenarios to see how likely it is that Trump will get to 3% for 2018.
2018Q3 Major Model Estimates as of 10/25/2018:
Atlanta Fed [1]: 3.6%
NY Fed [2]: 2.13%
Moody’s [3]: 3.3%
Looking at other forecasts, consensus seems to be around 3.3% for Q3. For Q4, it looks to be about 3.1.
Oh man, pretty solid numbers! So that’s 2.2, 4.2 and possibly 3.3 for the first three quarters. So what does the economy have to achieve for that coveted 3%? Let’s see: there are four quarters in a year and we kind of have an idea of what three of those quarters are. To get 3% for the year (well, we’ll go with 2.95 because they round up to one digit after the decimal for these types of announcements), we do:
2.95 * 4 = 11.8
11.8 - (2.2 + 4.2 + 3.3) = 2.1
Trump would just need 2.1% for Q4 to achieve what Obama never could!!!
But not so fast. That’s not how annual GDP growth is calculated.
That’s right, annualized GDP is calculated by doing the following:
A = average GDP over the previous year’s four quarters, chained 2012 dollars
B = average GDP over this year’s four quarters, chained 2012 dollars
C = the difference between the two averages (B - A)
Annual GDP growth = C / A
Well, how do those quarterly GDP numbers come in to play then? Let’s take a look:
Quarter GDP Quarterly Change Quarterly Growth
2017Q1 17863.023 78.838 1.79%
2017Q2 17995.150 132.127 2.99%
2017Q3 18120.843 125.693 2.85%
2017Q4 18223.758 102.915 2.33%
2018Q1 18323.963 100.205 2.22%
2018Q2 18511.576 187.613 4.16%
The quarterly growth percentages basically represent the following:
If the economy grew by the Quarterly Change amount for all 4 quarters this year, we’d have this much annual GDP growth.
For example, the annual GDP growth for 2017 was 2.2%, but if you take the average of those 2017 quarterly figures, you get 2.49%. Because remember, to get the annual growth, you have to take C from above and divide by A.
So what does C have to be for 2018 to get 2.95% (i.e. 3%) growth? Well, let’s calculate A:
A = (17863.023 + 17995.150 + 18120.843 + 18223.758) / 4
A = 18050.6935
C = 0.0295 * A
C = 532.49545825
Ok, so that’s our magic number for 2018. Let’s say we go with the Atlanta Fed’s prediction of 3.6% for Q3. That would mean we’d have a Q3 Quarterly Change of 162.4562415 (A * 0.036 / 4) or a Q3 Quarterly GDP of 18674.0322415. What do we have to achieve in Q4 to get to 2.95?
A = 18050.6935
C = B - 18050.6935
532.49545825 = ((18323.963 + 18511.576 + (18674.0322415) + (18674.0322415 + x) / 4) - 18050.6935
18583.18895825 = (18323.963 + 18511.576 + (18674.0322415) + (18674.0322415 + x) / 4
74332.755833 = (18323.963 + 18511.576 + (18674.0322415) + (18674.0322415 + x)
149.15235 = x
If we have 3.6% in Q3, we need a Quarterly Change of 149.15235 in Q4 to get 2.95% annual growth. That comes out to be 3.31% growth required in Q4.
Let’s go with the consensus of 3.3% in Q3. We’d need 3.91% in Q4 to reach our 2.95%.
Let’s go with Adroit’s prediction of 3.1% in Q3. It jumps up to 4.31%.
What say you? What do you think Q3 will bring in and what are the chances that we’ll achieve 3% annual growth this year?
Trump Getting Rid of Regulations Helps the Economy More Than Most Anything He Has Done
#2
I’m going to say 2.8%. We’ve had multiple serious weather events and there seems to be some general fear about the economy starting to slow down.
#3
It’s going to be supercalifragilisticexpialidocious.
MAGA
#4
I hate to mess up your math but I believe GDP for the first quarter this year was revised down to 2%.
#5
Negative Ghost Rider.
1st quarter 2018: 2.2 percent [1][2][3].
#6
A gentlemen’s 3.0!
My eyes are more on Google and Amazon earnings, though. If they both miss, the market’s going to slide big tomorrow.
I love how Amazon tanked earlier in the week but jumped 7% today because people decided they’re optimistic for no reason (this is why day trading=gambling, btw).
#7
Trump supporters staying away. Must be all the math. Let me try again:
Third quarter GDP is going to be the bigliest GDP. Some are saying the most growth, perhaps ever. This year will reach historic numbers. MAGA.
#8
Excellent breakdown.
I’m going to follow Atlanta’s numbers and go 3.3%. I think the inpact of the 2 hurricanes are lagging enough to affect Q4 2018.
#9
Excellent thread. Great way to kick off the predictions.
3.14
I say that because it’s just a little higher than Adroit’s prediction and because my number is not repeatable.
#10
Trump likes pie. When you’re famous, they let you…never mind.
Sorry, just trying to draw in the red hatters.
#11
both down big after earnings.
AMZN down 8% GOOG 4%
earnings were better than expected but revenue missed and AMZN’s forecast for q4 was low
#12
Yup. The low AMZN revenue guidance is a broader issue for the market, I think.
Think today’s surge in the market will be a blip in a downward trend?
#13
2%. I think people be saving their money and cashing out of equities. Plus home sales and mortgages are down with mortgage rates increasing
#14
Yup. The low AMZN revenue guidance is a broader issue for the market, I think.
Think today’s surge in the market will be a blip in a downward trend?
#15
Not a Trump supporter but this was a very nice analysis. Well done! I’ll just guess 3.2%.
#16
Just curious, but did Amazon say why they had a lower forecast for 4Q?
#17
• Net sales are expected to be between \$66.5 billion and \$72.5 billion, or to grow between 10% and 20% compared with fourth quarter 2017. This guidance anticipates an unfavorable impact of approximately 80 basis points from foreign exchange rates.
#18
There is no way that a thread this good would last in the Politics forum. It’s more convenient to shelve it over here.
How would a Trump supporter even begin to disagree–I mean, without making fools of themselves . . . again?
#19
Depends on gdp. If gdp comes in high I think the market gets smoked. A weak number and the market will be ok imho
But I could be wrong it’s happened once or twice
#20
I’m gonna go ahead and guess that because Trump didn’t spill the beans like he did last time, we might even be like 2.9.
About 20 more minutes until release. | 1,940 | 6,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-04 | latest | en | 0.914751 |
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# The number of people flying first class on domestic flights
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The number of people flying first class on domestic flights [#permalink]
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25 Feb 2012, 02:45
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The number of people flying first class on domestic flights rose sharply in 1990, doubling the increase of the previous year.
A. doubling the increase of
B. doubling that of the increase in
C. double as much as the increase of
D. twice as many as the increase in
E. twice as many as the increase of
I found this question very tricky because I could make no sense of it
What do statements A and B mean? doubling the increase?
[Reveal] Spoiler: OA
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25 Feb 2012, 22:54
(A) doubling the increase of
phrase refers back to the subject of the main verb "number of people..."
(B) doubling that of the increase in
"that of" does not have a clear antecedent.
(C) double as much as the increase of
"twice as much as" is the correct form
(D) twice as many as the increase in
"many" cannot be used as comparative for the subject "number", needs to be "much".
(E) twice as many as the increase of
same as D
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26 Feb 2012, 22:08
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+1 for A.
I am not very good with the double vs twice questions . But the general principle is that twice is an adverb , while double is essentially an adjective.
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27 Feb 2012, 21:42
Slightly digressing... the number of is countable.. right? if yes then why much and not many? much is used for non countable..
Thanks
onedayill wrote:
(A) doubling the increase of
phrase refers back to the subject of the main verb "number of people..."
(B) doubling that of the increase in
"that of" does not have a clear antecedent.
(C) double as much as the increase of
"twice as much as" is the correct form
(D) twice as many as the increase in
"many" cannot be used as comparative for the subject "number", needs to be "much".
(E) twice as many as the increase of
same as D
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28 Feb 2012, 16:32
I agree with devinawilliam83. Double is the countable and can be used with idiom as many as. infact the use of "as much as" is incorrect here.
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02 Mar 2012, 19:35
Sarang wrote:
I agree with devinawilliam83. Double is the countable and can be used with idiom as many as. infact the use of "as much as" is incorrect here.
You are absolutely correct.
But what is the confusion ? C is not the correct choice because of the countable vs uncountable noun.
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03 Mar 2012, 02:41
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doubling the increase of -- correctly describe the result of rose
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04 Apr 2012, 22:49
No matter how many times i do it......... i always fall in same trap...........
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Re: The number of people flying first class on domestic flights [#permalink]
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1/// \file
2/// \ingroup tutorial_roofit
3/// \notebook -js
4/// Validation and MC studies:
5/// RooMCStudy - using separate fit and generator models, using the chi^2 calculator model
6/// Running a biased fit model against an optimal fit.
7///
8/// \macro_image
9/// \macro_code
10/// \macro_output
11///
12/// \date July 2008
13/// \author Wouter Verkerke
14
15#include "RooRealVar.h"
16#include "RooDataSet.h"
17#include "RooGaussian.h"
18#include "RooChebychev.h"
20#include "RooMCStudy.h"
21#include "RooChi2MCSModule.h"
22#include "RooPlot.h"
23#include "TCanvas.h"
24#include "TAxis.h"
25#include "TH1.h"
26#include "TDirectory.h"
27#include "TLegend.h"
28
29using namespace RooFit;
30
32{
33
34 // C r e a t e m o d e l
35 // -----------------------
36
37 // Observables, parameters
38 RooRealVar x("x", "x", -10, 10);
39 x.setBins(10);
40 RooRealVar mean("mean", "mean of gaussian", 0, -2., 1.8);
41 RooRealVar sigma("sigma", "width of gaussian", 5, 1, 10);
42
43 // Create Gaussian pdf
44 RooGaussian gauss("gauss", "gaussian PDF", x, mean, sigma);
45
46 // C r e a t e m a n a g e r w i t h c h i ^ 2 a d d - o n m o d u l e
47 // ----------------------------------------------------------------------------
48
49 // Create study manager for binned likelihood fits of a Gaussian pdf in 10 bins
50 RooMCStudy *mcs = new RooMCStudy(gauss, x, Silence(), Binned());
51
52 // Add chi^2 calculator module to mcs
53 RooChi2MCSModule chi2mod;
55
56 // Generate 1000 samples of 1000 events
57 mcs->generateAndFit(2000, 1000);
58
59 // Number of bins for chi2 plots
60 int nBins = 100;
61
62 // Fill histograms with distributions chi2 and prob(chi2,ndf) that
63 // are calculated by RooChiMCSModule
64 TH1 *hist_chi2 = mcs->fitParDataSet().createHistogram("chi2", AutoBinning(nBins));
65 hist_chi2->SetTitle("#chi^{2} values of all toy runs;#chi^{2}");
66 TH1 *hist_prob = mcs->fitParDataSet().createHistogram("prob", AutoBinning(nBins));
67 hist_prob->SetTitle("Corresponding #chi^{2} probability;Prob(#chi^{2},ndof)");
68
69
70 // C r e a t e m a n a g e r w i t h s e p a r a t e f i t m o d e l
71 // ----------------------------------------------------------------------------
72
73 // Create alternate pdf with shifted mean
74 RooRealVar mean2("mean2", "mean of gaussian 2", 2.);
75 RooGaussian gauss2("gauss2", "gaussian PDF2", x, mean2, sigma);
76
77 // Create study manager with separate generation and fit model. This configuration
78 // is set up to generate biased fits as the fit and generator model have different means,
79 // and the mean parameter is limited to [-2., 1.8], so it just misses the optimal
80 // mean value of 2 in the data.
81 RooMCStudy *mcs2 = new RooMCStudy(gauss2, x, FitModel(gauss), Silence(), Binned());
82
83 // Add chi^2 calculator module to mcs
84 RooChi2MCSModule chi2mod2;
86
87 // Generate 1000 samples of 1000 events
88 mcs2->generateAndFit(2000, 1000);
89
90 // Request a the pull plot of mean. The pulls will be one-sided because
91 // mean is limited to 1.8.
92 // Note that RooFit will have trouble to compute the pulls because the parameters
93 // are called mean in the fit, but mean2 in the generator model. It is not obvious
94 // that these are related. RooFit will nevertheless compute pulls, but complain that
95 // this is risky.
96 auto pullMeanFrame = mcs2->plotPull(mean);
97
98 // Fill histograms with distributions chi2 and prob(chi2,ndf) that
99 // are calculated by RooChiMCSModule
100 TH1 *hist2_chi2 = mcs2->fitParDataSet().createHistogram("chi2", AutoBinning(nBins));
101 TH1 *hist2_prob = mcs2->fitParDataSet().createHistogram("prob", AutoBinning(nBins));
102 hist2_chi2->SetLineColor(kRed);
103 hist2_prob->SetLineColor(kRed);
104
105 TLegend leg;
108 leg.SetBorderSize(0);
109 leg.SetFillStyle(0);
110
112 c->Divide(3);
113 c->cd(1);
115 hist_chi2->GetYaxis()->SetTitleOffset(1.4);
116 hist_chi2->Draw();
117 hist2_chi2->Draw("esame");
118 leg.DrawClone();
119 c->cd(2);
121 hist_prob->GetYaxis()->SetTitleOffset(1.4);
122 hist_prob->Draw();
123 hist2_prob->Draw("esame");
124 c->cd(3);
125 pullMeanFrame->Draw();
126
127
128 // Make RooMCStudy object available on command line after
129 // macro finishes
131}
#define c(i)
Definition RSha256.hxx:101
@ kRed
Definition Rtypes.h:66
#define gDirectory
Definition TDirectory.h:384
TH1 * createHistogram(const char *name, const RooAbsRealLValue &xvar, const RooCmdArg &arg1={}, const RooCmdArg &arg2={}, const RooCmdArg &arg3={}, const RooCmdArg &arg4={}, const RooCmdArg &arg5={}, const RooCmdArg &arg6={}, const RooCmdArg &arg7={}, const RooCmdArg &arg8={}) const
Calls createHistogram(const char *name, const RooAbsRealLValue& xvar, const RooLinkedList& argList) c...
RooChi2MCSModule is an add-on module to RooMCStudy that calculates the chi-squared of fitted p....
Plain Gaussian p.d.f.
Definition RooGaussian.h:24
Helper class to facilitate Monte Carlo studies such as 'goodness-of-fit' studies, that involve fittin...
Definition RooMCStudy.h:32
RooPlot * plotPull(const RooRealVar ¶m, const RooCmdArg &arg1, const RooCmdArg &arg2={}, const RooCmdArg &arg3={}, const RooCmdArg &arg4={}, const RooCmdArg &arg5={}, const RooCmdArg &arg6={}, const RooCmdArg &arg7={}, const RooCmdArg &arg8={})
Plot the distribution of pull values for the specified parameter on a newly created frame.
const RooDataSet & fitParDataSet()
Return a RooDataSet containing the post-fit parameters of each toy cycle.
bool generateAndFit(Int_t nSamples, Int_t nEvtPerSample=0, bool keepGenData=false, const char *asciiFilePat=nullptr)
Generate and fit 'nSamples' samples of 'nEvtPerSample' events.
Insert given RooMCStudy add-on module to the processing chain of this MCStudy object.
Variable that can be changed from the outside.
Definition RooRealVar.h:37
virtual void SetTitleOffset(Float_t offset=1)
Set distance between the axis and the axis title.
Definition TAttAxis.cxx:298
virtual void SetFillStyle(Style_t fstyle)
Set the fill area style.
Definition TAttFill.h:39
virtual void SetLineColor(Color_t lcolor)
Set the line color.
Definition TAttLine.h:40
The Canvas class.
Definition TCanvas.h:23
TH1 is the base class of all histogram classes in ROOT.
Definition TH1.h:59
void SetTitle(const char *title) override
Change/set the title.
Definition TH1.cxx:6709
TAxis * GetYaxis()
Definition TH1.h:325
void Draw(Option_t *option="") override
Draw this histogram with options.
Definition TH1.cxx:3066
This class displays a legend box (TPaveText) containing several legend entries.
Definition TLegend.h:23
TLegendEntry * AddEntry(const TObject *obj, const char *label="", Option_t *option="lpf")
Add a new entry to this legend.
Definition TLegend.cxx:320
RooCmdArg AutoBinning(Int_t nbins=100, double marginFactor=0.1)
RooCmdArg Silence(bool flag=true)
RooCmdArg Binned(bool flag=true)
const Double_t sigma
Double_t x[n]
Definition legend1.C:17
leg
Definition legend1.C:34
The namespace RooFit contains mostly switches that change the behaviour of functions of PDFs (or othe...
Definition JSONIO.h:26
void FitModel(RooWorkspace *, std::string data_name="obsData") | 2,047 | 7,151 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.354412 |
https://puredhamma.net/dhamma-and-philosophy/the-infinity-problem-in-buddhism/?highlight=infinity | 1,686,448,112,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00531.warc.gz | 546,582,931 | 76,344 | # The Infinity Problem in Buddhism
The infinity problem in Buddhism is the following. “Each of us” has been in the rebirth process for an infinite time. Thus, “each of us” has been exposed to Buddha Dhamma countless times, i.e., made infinite attempts to attain Nibbāna. How is it possible that all of “us” have not attained Nibbāna?
July 15, 2017; Revised February 5, 2018; Re-written March 23, 2021; re-written September 24, 2022
##### Introduction – “Infinite Monkey Problem”
1. The question is based on the following statement in several suttas in Anamatagga Saṃyutta starting with the “Tiṇakaṭṭha Sutta (SN 15.1)“: “There is no discernible beginning to the rebirth process.” In other words, we have had an infinite time to attain Nibbāna (because an infinite number of Buddhas must have been born too.) So, why have we all not attained Nibbāna yet?”
• This issue has been discussed in discussion forums without a conclusion. See “The problem of infinity in Buddhism” at Dhamma Wheel and “The infinity problem in Buddhism” at the Sutta Central forum in 2017.
• This question seems to have its origin in the “infinite monkey theorem,” which states that a monkey hitting keys at random on a typewriter keyboard for an infinite amount of time will almost surely type any given text, such as the complete works of William Shakespeare.
• By the way, this infinite monkey theorem is another evidence of how unimaginably large infinity is: “Infinity – How Big Is It?“.
##### Detailed Analysis of the Infinite Monkey Theorem
2. First, the monkey theorem is based on a monkey typing on a keyboard and generating random letters. It is assumed that the same monkey will keep typing on the keyboard non-stop for an infinite time.
• For example, the probability that the monkey will come up with the word “banana” would be less than 1 in 15 billion, but not zero. Thus it is a theoretically possible outcome, even though improbable.
• But the enormity comes to view when you realize that you have to get a WHOLE TEXT correctly without making too many errors at a stretch (in one continuous period.) For example, to get TWO words correctly, like “yellow banana,” has a probability that is the multiplication of the probabilities of getting each word right. The probability that a monkey gets those two words correctly is less than 1 in a billion-billion (1 in 10^18). That is extremely small.
3. The above Wikipedia article says: “..the probability that monkeys filling the entire observable universe would type a single complete work, such as Shakespeare’s Hamlet, is so tiny that the chance of it occurring during a period hundreds of thousands of orders of magnitude longer than the age of the universe is extremely low (but technically not zero).”
• That is because scientists estimate the age of our current universe to be only about 14 billion years. Infinity cannot be assigned a number. Any large number you can think about CAN NOT BE the largest number because you can just add 1 to that number to make it bigger. There is no ending! Thus, given an infinite time, it is theoretically possible that a monkey could type out the complete works of William Shakespeare.
• However, as we see below, the rebirth process involving a “lifestream” can not be compared to the same monkey typing on a keyboard for an eternity.
##### Two Relevant Issues
4. We will discuss TWO aspects of this issue.
• First, we will show that the infinite monkey theorem’s MECHANISM does not apply to the rebirth process. The rebirth process DOES NOT involve a “person/soul/ātman traveling the Saṁsāra (like a single monkey typing for an infinite time.) Nothing like a soul “moves” from this life to the next. Even the next moment in life arises based on causes and conditions based on the present moment, i.e., via the Paṭicca Samuppāda process.
• Second, we will show that even if an infinite number of living beings has attained Nibbāna, there will still be an infinite number left!
## First Issue
##### There Is No “Person” Traveling the Rebirth Process
5. During the rebirth process, various forms of life manifest. Even though we live human lives now, we have lived in most of the 31 realms described by the Buddha. We have been born a Deva, a Brahma, an animal, a hell-being, etc. countless times!
• The concept of a “lifestream” moving from life in one realm to another (rather than a soul incarnating or appearing in different forms) is what the Buddha described. See “What Reincarnates? – Concept of a Lifestream.”
• No “unchanging/permanent essence” like a soul moves from one life to the next. If that were the case, stopping the rebirth process and attaining Nibbāna would be impossible! How can a permanent entity cease to exist?
• What is taken to the next life is anusaya/gati/saṁyojana. None of those is permanent. Each one can change even momentarily!
• Furthermore, all those entities are associated with suffering. Elimination of anusaya/gati/saṁyojana is the end of suffering, not the end of an entity like a soul or an ātman. See “Yamaka Sutta (SN 22.85) – Arahanthood Is Not Annihilation but End of Suffering.”
6. When one understands that by comprehending “Paṭicca Samuppāda, Tilakkhana, Four Noble Truths,” one becomes a Sotapanna by removing significant parts of anusaya/gati/saṁyojana. That is getting rid of the wrong view of a “soul/ātman” or “sakkāya diṭṭhi.” See “Sakkāya Diṭṭhi and Tilakkhana.”
##### Path to Nibbāna Is Not a Mechanical Process
7. Therefore, reaching Nibbāna CAN NOT be considered a mechanical process and, thus, CAN NOT be compared to a monkey hitting arbitrary keys on a keyboard to generate Shakespeare’s Hamlet.
• We can consider a monkey typing a single letter to a living-being hearing the correct Buddha Dhamma. Therefore, we can call that a “single-shot” at Nibbāna, corresponding to a single keystroke by a monkey.
• However, a mathematician could still say that there will be an infinite number of such “single-shots” by a living being over an INFINITE time.
8. But the key issue is that it is NOT the same person who heard the correct Buddha Dhamma infinite times!
• Therefore, the problem is with the question itself. The infinity problem in Buddhism is phrased as follows: Each of us has been in the rebirth process for an infinite time. How is it possible that all of us have not attained Nibbāna?
• That question assumes that a fixed person/soul/ātman is repeatedly reborn! Paṭicca Samuppāda explains that there is no such permanent entity being reborn. See “What Reincarnates? – Concept of a Lifestream.”
• We went through the above discussion to show that the two processes cannot be equated. Now, we address the SECOND issue mentioned in #4 above.
## Second Issue
##### Infinite Number May Have Attained Nibbāna
9. It is indeed possible that an infinite number of living beings HAVE ATTAINED Nibbāna in the past.
• Not only that, an infinite number of living beings may have attained Buddhahood in the past. Of course, attaining the Buddhahood is infinitely more difficult than attaining Arahanthood.
• Therefore, the infinite set of living beings who have attained Nibbāna is “much larger” than the infinite set of living beings who have attained Buddhahood.
• The key to this puzzle is to realize that “many levels of infinity” exist. It has been revealed by mathematicians within the past hundred years, thanks to the pioneering work of the mathematician George Cantor. See “George Cantor – The Man Who Founded Set Theory.” The following video provides good insights too.
##### A Nice Visualization of Infinity Within Infinity – The Infinite Hotel Paradox
10. The following video explains why there can be “smaller infinities” types within infinity. In particular, the set of positive integers is a “smaller infinity.” Those who have attained Nibbāna fall under that category. Regardless of how many have attained Nibbāna, more could attain Nibbāna.
• The following video discusses an infinite number of buses filled with an infinite number of guests arriving at an infinite hotel. It is shown that the infinite hotel can accommodate all of them and more!
• I have set the video to stop around 2:10 minutes. It is enough to see that the hotel can accommodate an infinite number of guests at any time. During the presence of a Buddha Sāsana, only a finite number of living beings (humans, Devas, and Brahmas) attain Nibbāna.
• The rest of the video is more mathematical and shows that even an infinite number of buses with an infinite number of passengers in each bus can be accommodated! You can watch the whole video by clicking “watch on Youtube.”
##### Other Related Issues
11. Of course, several other questions now arise: Where do all these infinite numbers of living beings live? Do they all live in our Solar system? It will take many more future posts to explain these fully, but we can summarize them as follows.
• Brief answers to those questions are as follows: According to the Buddha, an uncountable number of planetary systems are populated with living beings. While an uncountable number of living beings live in our Solar system, there are an uncountable number of such planetary systems (cakkavāla) in the world. Each cluster of 10,000 such cakkavāla can have a Buddha appearing periodically. Thus, there could have been an infinite number of Buddhās.
• Such details are in suttas in the Tipiṭaka, mostly in the Digha Nikāya. I briefly discussed one sutta: “Buddhism and Evolution – Aggañña Sutta (DN 27).” | 2,262 | 9,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | longest | en | 0.942946 |
https://raw.githubusercontent.com/holoviz/holoviews/main/examples/reference/streams/bokeh/Bounds.ipynb | 1,726,011,753,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00547.warc.gz | 451,748,884 | 1,984 | { "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "#### **Title**: Bounds & selection stream example\n", "\n", "**Description**: A linked streams example demonstrating how to use Bounds and Selection streams together.\n", "\n", "**Dependencies** Bokeh\n", "\n", "**Backends** [Bokeh](./Bounds.ipynb), [Plotly](../plotly/Bounds.ipynb)" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "import numpy as np\n", "import holoviews as hv\n", "from holoviews import opts\n", "from holoviews import streams\n", "hv.extension('bokeh')" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "opts.defaults(opts.Histogram(framewise=True))\n", "\n", "# Declare distribution of Points\n", "points = hv.Points(np.random.multivariate_normal((0, 0), [[1, 0.1], [0.1, 1]], (1000,)))\n", "\n", "# Declare points selection selection\n", "sel = streams.Selection1D(source=points)\n", "\n", "# Declare DynamicMap computing mean y-value of selection\n", "mean_sel = hv.DynamicMap(lambda index: hv.HLine(points['y'][index].mean() if index else -10),\n", " kdims=[], streams=[sel])\n", "\n", "# Declare a Bounds stream and DynamicMap to get box_select geometry and draw it\n", "box = streams.BoundsXY(source=points, bounds=(0,0,0,0))\n", "bounds = hv.DynamicMap(lambda bounds: hv.Bounds(bounds), streams=[box])\n", "\n", "# Declare DynamicMap to apply bounds selection\n", "dmap = hv.DynamicMap(lambda bounds: points.select(x=(bounds[0], bounds[2]),\n", " y=(bounds[1], bounds[3])),\n", " streams=[box])\n", "\n", "# Compute histograms of selection along x-axis and y-axis\n", "yhist = hv.operation.histogram(dmap, bin_range=points.range('y'), dimension='y', dynamic=True, normed=False)\n", "xhist = hv.operation.histogram(dmap, bin_range=points.range('x'), dimension='x', dynamic=True, normed=False)\n", "\n", "# Combine components and display\n", "points * mean_sel * bounds << yhist << xhist" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "
" ] } ], "metadata": { "language_info": { "name": "python", "pygments_lexer": "ipython3" } }, "nbformat": 4, "nbformat_minor": 4 } | 631 | 2,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.536261 |
https://www.aqua-calc.com/convert/density/milligram-per-liter-to-slug-per-us-teaspoon | 1,656,860,748,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104244535.68/warc/CC-MAIN-20220703134535-20220703164535-00686.warc.gz | 681,816,848 | 8,766 | # Convert milligrams per liter to slugs per teaspoon
## [mg/l to sl/tsp] (mg:milligram, l:liter, sl:slug, tsp:teaspoon)
### Convert mg/l to sl/tsp
#### a density or mass density conversion table
to 100 with
milli-
gram per liter
×10−8,
slug per US teaspoon
milli-
gram per liter
×10−8,
slug per US teaspoon
milli-
gram per liter
×10−8,
slug per US teaspoon
milli-
gram per liter
×10−8,
slug per US teaspoon
milli-
gram per liter
×10−8,
slug per US teaspoon
10.0210.7411.4612.1812.7
20.1220.7421.4622.1822.8
30.1230.8431.5632.1832.8
40.1240.8441.5642.2842.8
50.2250.8451.5652.2852.9
60.2260.9461.6662.2862.9
70.2270.9471.6672.3872.9
80.3280.9481.6682.3883.0
90.3291.0491.7692.3893.0
100.3301.0501.7702.4903.0
110.4311.0511.7712.4913.1
120.4321.1521.8722.4923.1
130.4331.1531.8732.5933.1
140.5341.1541.8742.5943.2
150.5351.2551.9752.5953.2
160.5361.2561.9762.6963.2
170.6371.2571.9772.6973.3
180.6381.3582.0782.6983.3
190.6391.3592.0792.7993.3
200.7401.4602.0802.71003.4
### milligrams per liter to slugs per teaspoon conversion cards
• 1
through
20
milligrams per liter
• 1 mg/l to sl/tsp = 0 sl/tsp
• 2 mg/l to sl/tsp = 1.0 × 10-9 sl/tsp
• 3 mg/l to sl/tsp = 1.0 × 10-9 sl/tsp
• 4 mg/l to sl/tsp = 1.0 × 10-9 sl/tsp
• 5 mg/l to sl/tsp = 2.0 × 10-9 sl/tsp
• 6 mg/l to sl/tsp = 2.0 × 10-9 sl/tsp
• 7 mg/l to sl/tsp = 2.0 × 10-9 sl/tsp
• 8 mg/l to sl/tsp = 3.0 × 10-9 sl/tsp
• 9 mg/l to sl/tsp = 3.0 × 10-9 sl/tsp
• 10 mg/l to sl/tsp = 3.0 × 10-9 sl/tsp
• 11 mg/l to sl/tsp = 4.0 × 10-9 sl/tsp
• 12 mg/l to sl/tsp = 4.0 × 10-9 sl/tsp
• 13 mg/l to sl/tsp = 4.0 × 10-9 sl/tsp
• 14 mg/l to sl/tsp = 5.0 × 10-9 sl/tsp
• 15 mg/l to sl/tsp = 5.0 × 10-9 sl/tsp
• 16 mg/l to sl/tsp = 5.0 × 10-9 sl/tsp
• 17 mg/l to sl/tsp = 6.0 × 10-9 sl/tsp
• 18 mg/l to sl/tsp = 6.0 × 10-9 sl/tsp
• 19 mg/l to sl/tsp = 6.0 × 10-9 sl/tsp
• 20 mg/l to sl/tsp = 7.0 × 10-9 sl/tsp
• 21
through
40
milligrams per liter
• 21 mg/l to sl/tsp = 7.0 × 10-9 sl/tsp
• 22 mg/l to sl/tsp = 7.0 × 10-9 sl/tsp
• 23 mg/l to sl/tsp = 8.0 × 10-9 sl/tsp
• 24 mg/l to sl/tsp = 8.0 × 10-9 sl/tsp
• 25 mg/l to sl/tsp = 8.0 × 10-9 sl/tsp
• 26 mg/l to sl/tsp = 9.0 × 10-9 sl/tsp
• 27 mg/l to sl/tsp = 9.0 × 10-9 sl/tsp
• 28 mg/l to sl/tsp = 9.0 × 10-9 sl/tsp
• 29 mg/l to sl/tsp = 1.0 × 10-8 sl/tsp
• 30 mg/l to sl/tsp = 1.0 × 10-8 sl/tsp
• 31 mg/l to sl/tsp = 1.0 × 10-8 sl/tsp
• 32 mg/l to sl/tsp = 1.1 × 10-8 sl/tsp
• 33 mg/l to sl/tsp = 1.1 × 10-8 sl/tsp
• 34 mg/l to sl/tsp = 1.1 × 10-8 sl/tsp
• 35 mg/l to sl/tsp = 1.2 × 10-8 sl/tsp
• 36 mg/l to sl/tsp = 1.2 × 10-8 sl/tsp
• 37 mg/l to sl/tsp = 1.2 × 10-8 sl/tsp
• 38 mg/l to sl/tsp = 1.3 × 10-8 sl/tsp
• 39 mg/l to sl/tsp = 1.3 × 10-8 sl/tsp
• 40 mg/l to sl/tsp = 1.4 × 10-8 sl/tsp
• 41
through
60
milligrams per liter
• 41 mg/l to sl/tsp = 1.4 × 10-8 sl/tsp
• 42 mg/l to sl/tsp = 1.4 × 10-8 sl/tsp
• 43 mg/l to sl/tsp = 1.5 × 10-8 sl/tsp
• 44 mg/l to sl/tsp = 1.5 × 10-8 sl/tsp
• 45 mg/l to sl/tsp = 1.5 × 10-8 sl/tsp
• 46 mg/l to sl/tsp = 1.6 × 10-8 sl/tsp
• 47 mg/l to sl/tsp = 1.6 × 10-8 sl/tsp
• 48 mg/l to sl/tsp = 1.6 × 10-8 sl/tsp
• 49 mg/l to sl/tsp = 1.7 × 10-8 sl/tsp
• 50 mg/l to sl/tsp = 1.7 × 10-8 sl/tsp
• 51 mg/l to sl/tsp = 1.7 × 10-8 sl/tsp
• 52 mg/l to sl/tsp = 1.8 × 10-8 sl/tsp
• 53 mg/l to sl/tsp = 1.8 × 10-8 sl/tsp
• 54 mg/l to sl/tsp = 1.8 × 10-8 sl/tsp
• 55 mg/l to sl/tsp = 1.9 × 10-8 sl/tsp
• 56 mg/l to sl/tsp = 1.9 × 10-8 sl/tsp
• 57 mg/l to sl/tsp = 1.9 × 10-8 sl/tsp
• 58 mg/l to sl/tsp = 2.0 × 10-8 sl/tsp
• 59 mg/l to sl/tsp = 2.0 × 10-8 sl/tsp
• 60 mg/l to sl/tsp = 2.0 × 10-8 sl/tsp
• 61
through
80
milligrams per liter
• 61 mg/l to sl/tsp = 2.1 × 10-8 sl/tsp
• 62 mg/l to sl/tsp = 2.1 × 10-8 sl/tsp
• 63 mg/l to sl/tsp = 2.1 × 10-8 sl/tsp
• 64 mg/l to sl/tsp = 2.2 × 10-8 sl/tsp
• 65 mg/l to sl/tsp = 2.2 × 10-8 sl/tsp
• 66 mg/l to sl/tsp = 2.2 × 10-8 sl/tsp
• 67 mg/l to sl/tsp = 2.3 × 10-8 sl/tsp
• 68 mg/l to sl/tsp = 2.3 × 10-8 sl/tsp
• 69 mg/l to sl/tsp = 2.3 × 10-8 sl/tsp
• 70 mg/l to sl/tsp = 2.4 × 10-8 sl/tsp
• 71 mg/l to sl/tsp = 2.4 × 10-8 sl/tsp
• 72 mg/l to sl/tsp = 2.4 × 10-8 sl/tsp
• 73 mg/l to sl/tsp = 2.5 × 10-8 sl/tsp
• 74 mg/l to sl/tsp = 2.5 × 10-8 sl/tsp
• 75 mg/l to sl/tsp = 2.5 × 10-8 sl/tsp
• 76 mg/l to sl/tsp = 2.6 × 10-8 sl/tsp
• 77 mg/l to sl/tsp = 2.6 × 10-8 sl/tsp
• 78 mg/l to sl/tsp = 2.6 × 10-8 sl/tsp
• 79 mg/l to sl/tsp = 2.7 × 10-8 sl/tsp
• 80 mg/l to sl/tsp = 2.7 × 10-8 sl/tsp
• 81
through
100
milligrams per liter
• 81 mg/l to sl/tsp = 2.7 × 10-8 sl/tsp
• 82 mg/l to sl/tsp = 2.8 × 10-8 sl/tsp
• 83 mg/l to sl/tsp = 2.8 × 10-8 sl/tsp
• 84 mg/l to sl/tsp = 2.8 × 10-8 sl/tsp
• 85 mg/l to sl/tsp = 2.9 × 10-8 sl/tsp
• 86 mg/l to sl/tsp = 2.9 × 10-8 sl/tsp
• 87 mg/l to sl/tsp = 2.9 × 10-8 sl/tsp
• 88 mg/l to sl/tsp = 3.0 × 10-8 sl/tsp
• 89 mg/l to sl/tsp = 3.0 × 10-8 sl/tsp
• 90 mg/l to sl/tsp = 3.0 × 10-8 sl/tsp
• 91 mg/l to sl/tsp = 3.1 × 10-8 sl/tsp
• 92 mg/l to sl/tsp = 3.1 × 10-8 sl/tsp
• 93 mg/l to sl/tsp = 3.1 × 10-8 sl/tsp
• 94 mg/l to sl/tsp = 3.2 × 10-8 sl/tsp
• 95 mg/l to sl/tsp = 3.2 × 10-8 sl/tsp
• 96 mg/l to sl/tsp = 3.2 × 10-8 sl/tsp
• 97 mg/l to sl/tsp = 3.3 × 10-8 sl/tsp
• 98 mg/l to sl/tsp = 3.3 × 10-8 sl/tsp
• 99 mg/l to sl/tsp = 3.3 × 10-8 sl/tsp
• 100 mg/l to sl/tsp = 3.4 × 10-8 sl/tsp
• slug per teaspoon stands for slug per US teaspoon
• slugs per teaspoon stands for slugs per US teaspoon
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2025 foods that contain Betaine. List of these foods starting with the highest contents of Betaine and the lowest contents of Betaine
#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Special Grade Reef weighs 1 361.6 kg/m³ (85.00191 lb/ft³) with specific gravity of 1.3616 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
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Volume to weightweight to volume and cost conversions for Refrigerant R-500, liquid (R500) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
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t/US c to oz t/ft³ conversion table, t/US c to oz t/ft³ unit converter or convert between all units of density measurement.
#### Calculators
Humidity calculations using temperature, relative humidity, pressure | 3,454 | 7,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.267403 |
https://oeis.org/A324201 | 1,624,308,251,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488289268.76/warc/CC-MAIN-20210621181810-20210621211810-00534.warc.gz | 379,812,208 | 3,917 | 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!)
A324201 a(n) = A062457(A000043(n)) = prime(A000043(n))^A000043(n), where A000043 gives the exponent of the n-th Mersenne prime. 24
9, 125, 161051, 410338673, 925103102315013629321, 1271991467017507741703714391419, 49593099428404263766544428188098203, 165163983801975082169196428118414326197216835208154294976154161023 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS If there are no odd perfect numbers, then the terms give all solutions n > 1 to A323244(n) = 0. LINKS FORMULA a(n) = A062457(A000043(n)). A323244(a(n)) = 0. a(n) = A005940(1+A000396(n)). [Provided no odd perfect numbers exist] CROSSREFS Cf. A000043, A000396, A005940, A062457, A323244, A324200. Sequence in context: A192724 A078422 A291897 * A224495 A064199 A092343 Adjacent sequences: A324198 A324199 A324200 * A324202 A324203 A324204 KEYWORD nonn AUTHOR Antti Karttunen, Feb 18 2019 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified June 21 16:42 EDT 2021. Contains 345365 sequences. (Running on oeis4.) | 444 | 1,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.539869 |
https://www.iovi.com/books/algorithms-manual/leetcode/0044.html | 1,713,357,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817153.39/warc/CC-MAIN-20240417110701-20240417140701-00016.warc.gz | 749,486,122 | 14,895 | # 44. Wildcard Matching (Hard)
https://leetcode.com/problems/wildcard-matching/
Given an input string (`s`) and a pattern (`p`), implement wildcard pattern matching with support for `'?'` and `'*'`.
```'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
```
The matching should cover the entire input string (not partial).
Note:
• `s` could be empty and contains only lowercase letters `a-z`.
• `p` could be empty and contains only lowercase letters `a-z`, and characters like `?` or `*`.
Example 1:
```Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
```
Example 2:
```Input:
s = "aa"
p = "*"
Output: true
Explanation: '*' matches any sequence.
```
Example 3:
```Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.
```
Example 4:
```Input:
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".
```
Example 5:
```Input:
s = "acdcb"
p = "a*c?b"
Output: false
```
## Solutions
``````class Solution {
// Used to keep the matching results of already compared strings, otherwise,
// it will exceeds time limit.
Map<String, Boolean> track = new HashMap<>();
public boolean isMatch(String s, String p) {
if (s == null) {
s = "";
}
if (p == null) {
p = "";
}
String np = "";
char pre = 0;
// process p to merge consecutive *
for (int i = 0; i < p.length(); i++) {
// initialize the start condition
if (i == 0) {
np += p.charAt(0);
pre = p.charAt(0);
continue;
}
if (p.charAt(i) == pre && pre == '*') {
continue;
}
// if not consecutive *, move on
np += p.charAt(i);
pre = p.charAt(i);
}
return check(s, np);
}
private boolean check(String s, String p) {
/*
// if two string match, return immediately
if (isFound) {
return isFound;
}
*/
boolean ans = false;
if (s.length() == 0 && p.length() == 0) {
return true;
}
if (s.length() != 0 && p.length() == 0) {
return false;
}
if (track.containsKey(s + "#" + p)) {
return track.get(s + "#" + p);
}
if (s.length() == 0 && p.length() != 0) {
// After processing, p will not contain consecutive *, in other words, if p.length() > 1
// there must be some other chars that need to be matching 1 by 1. Since s.length() is 0,
// nothing to compare. Only if p.length() == 1 and p[0] = '*', p and s can match perfectly.
if (p.length() == 1 && p.charAt(0) == '*') {
ans = true;
} else {
ans = false;
}
track.put(s + "#" + p, ans);
return ans;
}
// To reach here, s.length and p.length both >= 1
// count chars except *
int charCount = 0;
for (int i = 0; i < p.length(); i++) {
if (p.charAt(i) != '*') {
charCount++;
}
}
// if p contains more regular chars, impossible to match them all
if (charCount > s.length()) {
ans = false;
track.put(s + "#" + p, ans);
return ans;
}
// if p[0] != '.' and '*', test if p[0] == s[0]
if (p.charAt(0) != '?' && p.charAt(0) != '*') {
if (s.charAt(0) != p.charAt(0)) {
// return false if not match
ans = false;
} else {
// check recursively if match
ans = check(s.substring(1), p.substring(1));
}
track.put(s + "#" + p, ans);
return ans;
}
// if p[0] == '.', no need to test if p[0] == s[0]
// move on to compare s[1:] p[1:]
if (p.charAt(0) == '?') {
ans = isMatch(s.substring(1), p.substring(1));
track.put(s + "#" + p, ans);
return ans;
}
// if p[0] == '*', three situations,
// 1. compare s[], p[1:] which means p[0] covers zero char
// 2. s[1:] p[1:] which p[0] covers 1 char and its coverage ends here.
// 3. compare s[1:], p[:] which means p[0] matches at least 2 chars and
// the followup letters will still be covered.
if (p.charAt(0) == '*') {
// p[0] counts for 0 char
ans = ans || check(s, p.substring(1));
// p[0] counts for 1 char
ans = ans || check(s.substring(1), p.substring(1));
// p[0] counts for at least 2 char
ans = ans || check(s.substring(1), p);
}
// any true results of above three matching test will result in a final true
track.put(s + "#" + p, ans);
return ans;
}
}
`````` | 1,223 | 4,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.528216 |
http://indoweb.ga/vyvym/how-to-calculate-average-inflation-rate-using-cpi-1616.php | 1,555,828,592,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530253.25/warc/CC-MAIN-20190421060341-20190421082341-00350.warc.gz | 93,995,602 | 7,019 | Select Page
# How to calculate average inflation rate using cpi
With a calculator, enter "1. Thanks for letting us know. Because of the price differences would be If prices go in product costs over a specific time period, and it larger than "B" and we to minimize these variables. The Consumer Price Index CPI is a measure of changes down and we experienced Price of compounding interest, so you may not simply divide the indicator of the cost of negative number. If you don't know it, you can find it here: used when you added the until prices increased Inflated by. Pros and Cons of Bankruptcy. Households measured in the CPI-W must have earned at least half their income from clerical or hourly wage jobs and and not merely the change of a single item, you living and economic growth. What is Velocity of Money.
## Getting Started With the Calculations
In September of the CPI cost less than a few specializes in personal finance and. Or if you believe a is an indicator that measures words you may prefer just paid by consumers for goods and services over a set format or Average Annual Inflation Rates by Decade. In this example, raise 1. Things You'll Need Note pad. For example, if one loaf of bread was in your to calculate the difference in bread must be part of the list of current prices. If you already know how index was We will look cents a few years ago, home value. For those of you who are interested in understanding how a CPI value and take calculator provided below, to compute. If you don't care how it's done and just want first list, one loaf of prices between two different dates use the CPI Inflation Calculator. The Consumer Price Index CPI the calculation process is executed, an economy functions, it would be interesting to know how to calculate inflation rate. References University of Oregon: What phenomenal rise in the prices you can directly use the costs several dollars now. .
The base year serves as only relevant if calculated for other years are compared. The change in CPI is ago the Consumer Price Index a specifically quantifiable amount of time. Not Helpful 2 Helpful 8. Ayub Jul 17, Also avoid Bad question Other. To calculate it, we can. Add together the current prices.
1. Video of the Day
In September of the CPI index was Divide current prices used when you added the. I agree that my data list of items as you a reliable source like the prices of past items together. Hence, with this formula we Rate it uses the most for any given year as compares it to data from exactly 12 months prior using is available. Pull data on each of prices would be Add together by the old prices. If you are using a record of your own purchases. The base year serves as single item that you purchased the current prices. Answer this question Flag as the items purchased previously. Find the current price of What is Quantitative Easing. Using the previous example, current the benchmark against which all in the past. Then I found wikiHow answers, food, housing, clothing, transportation and.
1. Calculating Consumer Price Index (CPI)
Elizabeth B. Appelbaum, "The Consumer Price Index and Inflation - Calculate and Graph Inflation Rates," Convergence (December ). Hence, with this formula we can calculate the inflation rate for any given year as long as the CPI of that and the preceding year is available. In a Nutshell. The Consumer Price Index (CPI) is an indicator that measures the average change in prices paid by consumers for goods and .
1. How to Calculate Annual Inflation Over Multiple Years
We will look at all a measure of inflation. It may be useful, for the sake of comparison, to make sure the prices used are based upon the same brands and from the same. This site uses cookies e. A Anonymous Feb We appreciate determine the change in prices. How to calculate the real four steps in more detail below. If prices go down and we experienced Price Deflation then "A" would be larger than "B" and we would end up with a negative number. Are there any examples of calculating CPI. It is widely used as.
1. Calculating the Inflation Rate
Economists use the CPI to the prices of the same a specifically quantifiable amount of an indicator of economic expansion. By looking at the above only relevant if calculated for used to calculate the rate it went from to Cost. Households measured in the CPI-W track changes in the cost of living, as well as or hourly wage jobs and. The Weighted CPI assigns weights year would work well for. Grocery receipts from the past to each category depending on its importance. It will help you get at each store and from how the inflationary forces, which lower dollar value are measured of prices over time is an economy is heading. The change in CPI is to download to your smartphone dieting and excessive exercise are. | 1,004 | 4,807 | {"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} | 2.921875 | 3 | CC-MAIN-2019-18 | latest | en | 0.948507 |
https://www.talkstats.com/threads/on-multiple-linear-regression-and-partial-correlation.71171/#post-205922 | 1,638,275,976,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00033.warc.gz | 1,099,020,610 | 10,414 | # On Multiple Linear Regression and Partial Correlation
#### Etubom
##### New Member
I am conducting a study that involves A as the dependent variable (the DV) and B,C and D as the independent variables (the IVs).
In the first place, for hypothesis one, I have conducted a standard multiple regression of B,C and D (the IVs) on A (the DV).
After this, for hypothesis two, I have conducted partial correlation of A (the DV) and B partialling C and D.
For hypothesis three, I have conducted partial correlation of A (the DV) and C partialling B and D.
for hypothesis four, I have conducted partial correlation of A (the DV) and D partialling B and C.
Please confirm for me whether I am right to conduct the partial correlations after the multiple regression or otherwise. Please, I expect your reply.Thanks
#### hlsmith
##### Less is more. Stay pure. Stay poor.
You can typically run a single regression and just look at variance inflation factor or tolerance statistic to understand multicollinearity.
Last edited:
#### Etubom
##### New Member
Thanks, hismith , I mean how to order the different hypotheses. Should the Multiple regression hypothesis come before the partial correlation hypotheses?
#### ondansetron
##### TS Contributor
OP: please provide more details for what you wish to achieve.
If I recall, partialling out would be that you regress B on C and D and save the residuals from this regression as Resid(B).
Repeat this so you regress C on B and D to get Resid(C). Then do this once more so you have D regressed on B and C to get Resid(D).
If you now regress A on Resid(B), A on Resid(C), and A on Resid(D) each in a simple regression, the coefficients should be equal to the coefficients obtained by regressing A on B,C, and D in a multiple linear regression. This is because each of the coefficients in MLR are interpreted as the effect of an X variable on Y after accounting for the other variables. By regressing B on C and D in an MLR and saving residuals, we found the part of B that is unrelated to the group of C and D (we have "accounted" for C and D).
Not sure if this helps at all, but do let us know more details for what you want to do.
#### hlsmith
##### Less is more. Stay pure. Stay poor.
Yes, there are many directions this could go without further description of the purpose. Are you trying to do structural equation modelling, partial R**2, mediation analysis (direct and non direct effects), etc.? | 554 | 2,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-49 | latest | en | 0.925989 |
https://proficientwritershub.com/a-clinical-trial-is-conducted-to-evaluate-the-efficacy-of-a-new-drug-for-prevention-of-hypertension-in-patients-with/ | 1,723,331,272,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00503.warc.gz | 371,638,863 | 17,353 | A clinical trial is conducted to evaluate the efficacy of a new drug for prevention of hypertension in patients with…
A clinical trial is conducted to evaluate the efficacy of a new drug for prevention of hypertension in patients with pre-hypertension (defined as systolic blood pressure between 120–139 mmHg or diastolic blood pressure between 80–89 mmHg). A total of 20 patients are randomized to receive the new drug or a currently available drug for treatment of high blood pressure. Participants are followed for up to 12 months, and time to progression to hypertension is measured. The experiences of participants in each arm of the trial are shown below. New Drug Currently Available Drug Hypertension Free of Hypertension Hypertension Free of Hypertension 7 8 6 8 8 8 7 9 10 8 9 11 9 10 11 11 11 12 12 12 Estimate the survival (time to progression to hypertension) functions for each treatment group using the Kaplan-Meier approach.   New Drug Complete the table below. Time, Months Number at Risk Nt Number of Events (Hypertension) Dt Number Censored Ct Survival Probability St+1 = St*((Nt-Dt)/Nt) 0 10 7 8 9 10 11 12 Currently Available Drug Complete the table below. Time, Weeks Number at Risk Nt Number of Events (Hypertension) Dt Number Censored Ct Survival Probability St+1 = St*((Nt-Dt)/Nt) 0 10 6 7 8 9 10 11 12 To answer the question as to whether or not there is a difference in time to progression, a Chi square statistic is computed. The critical value for rejection of the null hypothesis is 3.84. The computed Chi square is 0.335. Based on comparing the computed Chi square and the critical Chi square, which of the following is (are) true? A. There is not statistically significant evidence to show that the time to progression is different between groups. B. There is statistically significant evidence to show that the time to progression is different between groups. C. The time to progression is essentially the same for each group. D. a and c. The hazard ratio risk of progression to hypertension is 0.658. Based on this computation, which of the following is (are) true? A. The risk of progression to hypertension is reduced by 34.2% in patients assigned to the new drug as compared to the currently available drug. B. The risk of progression to hypertension is 1.52 times higher in patient’s current drug as compared to the new drug. C. The risk of progression to hypertension is 5.12 times higher in patient’s current drug as compared to the new drug D. a and b | 603 | 2,503 | {"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} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.904141 |
https://q-and-answers.com/mathematics/question515382322 | 1,656,363,201,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00151.warc.gz | 512,666,777 | 16,532 | # HELP ASAP!!! Solve the following system of equations by substitution. Show all steps. f(x) = -x^2 + 2x + 3 g(x) = -2x + 3
jeffcarpenter · 05.06.2022 09:50
25.06.2019 23:00
chicken chicken chicken
step-by-step explanation:
25.06.2019 11:30
a
b
d
step-by-step explanation:
25.06.2019 08:30
∠1≅∠2 by the alternate exterior angles theorem.
step-by-step explanation:
given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f
we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.
we have to fill the missing statement.
transitivity property states that if a = b and b = c, then a = c.
now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .
hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.
the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4
∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.
hence, the missing statement is ∠1≅∠2.
so, ∠1≅∠2 by the alternate exterior angles theorem.
05.06.2022 09:50
(0,3) and (4,-5)
Step-by-step explanation:
f(x) just means y and so does g(x).
Both equations are equal to y. So they are equal to each other. Set equal to each other to solve. After finding x, don't forget to find the y that goes with each x. See image.
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22.06.2019 05:00 | 885 | 2,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-27 | latest | en | 0.868072 |
https://www.lmfdb.org/Character/Dirichlet/2015/393 | 1,575,907,612,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00472.warc.gz | 767,398,926 | 6,213 | Properties
Conductor 2015 Order 60 Real No Primitive Yes Parity Even Orbit Label 2015.gu
Learn more about
Show commands for: SageMath / Pari/GP
sage: from dirichlet_conrey import DirichletGroup_conrey # requires nonstandard Sage package to be installed
sage: H = DirichletGroup_conrey(2015)
sage: chi = H[393]
pari: [g,chi] = znchar(Mod(393,2015))
Basic properties
sage: chi.conductor() pari: znconreyconductor(g,chi) Conductor = 2015 sage: chi.multiplicative_order() pari: charorder(g,chi) Order = 60 Real = No sage: chi.is_primitive() pari: #znconreyconductor(g,chi)==1 \\ if not primitive returns [cond,factorization] Primitive = Yes sage: chi.is_odd() pari: zncharisodd(g,chi) Parity = Even Orbit label = 2015.gu Orbit index = 177
Galois orbit
sage: chi.sage_character().galois_orbit()
pari: order = charorder(g,chi)
pari: [ charpow(g,chi, k % order) | k <-[1..order-1], gcd(k,order)==1 ]
Values on generators
$$(807,1861,716)$$ → $$(-i,e\left(\frac{1}{3}\right),e\left(\frac{29}{30}\right))$$
Values
-1 1 2 3 4 6 7 8 9 11 12 14 $$1$$ $$1$$ $$e\left(\frac{17}{60}\right)$$ $$e\left(\frac{11}{20}\right)$$ $$e\left(\frac{17}{30}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{29}{60}\right)$$ $$e\left(\frac{17}{20}\right)$$ $$e\left(\frac{1}{10}\right)$$ $$e\left(\frac{17}{30}\right)$$ $$e\left(\frac{7}{60}\right)$$ $$e\left(\frac{23}{30}\right)$$
value at e.g. 2
Related number fields
Field of values $$\Q(\zeta_{60})$$ | 543 | 1,460 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-51 | latest | en | 0.314388 |
https://us.metamath.org/mpeuni/relcoss.html | 1,713,252,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00797.warc.gz | 569,366,967 | 3,886 | Mathbox for Peter Mazsa < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > relcoss Structured version Visualization version GIF version
Theorem relcoss 35660
Description: Cosets by 𝑅 is a relation. (Contributed by Peter Mazsa, 27-Dec-2018.)
Assertion
Ref Expression
relcoss Rel ≀ 𝑅
Proof of Theorem relcoss
Dummy variables 𝑢 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 df-coss 35651 . 2 𝑅 = {⟨𝑥, 𝑦⟩ ∣ ∃𝑢(𝑢𝑅𝑥𝑢𝑅𝑦)}
21relopabi 5687 1 Rel ≀ 𝑅
Colors of variables: wff setvar class Syntax hints: ∧ wa 398 ∃wex 1774 class class class wbr 5057 Rel wrel 5553 ≀ ccoss 35445 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1790 ax-4 1804 ax-5 1905 ax-6 1964 ax-7 2009 ax-8 2110 ax-9 2118 ax-10 2139 ax-11 2154 ax-12 2170 ax-ext 2791 This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-3an 1084 df-tru 1534 df-ex 1775 df-nf 1779 df-sb 2064 df-clab 2798 df-cleq 2812 df-clel 2891 df-nfc 2961 df-rab 3145 df-v 3495 df-dif 3937 df-un 3939 df-in 3941 df-ss 3950 df-nul 4290 df-if 4466 df-sn 4560 df-pr 4562 df-op 4566 df-opab 5120 df-xp 5554 df-rel 5555 df-coss 35651 This theorem is referenced by: relcoels 35661 cocossss 35673 cnvcosseq 35674 refrelcoss3 35695 symrelcoss3 35697 1cosscnvxrn 35707 eleccossin 35715 cosselrels 35728 cnvrefrelcoss2 35765 eqvrelcoss3 35845
Copyright terms: Public domain W3C validator | 718 | 1,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.160442 |
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Question Video: Finding the Equation That Represents Given Mapping Diagrams Mathematics • Third Year of Preparatory School
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Given that π
is a relation from π to π, where π β π and π β π, which of the following equations correctly expresses relation π
? [A] π = π + 1 [B] π = 2π + 2 [C] π = 2π β 2 [D] π = 2π + 2 [E] π = 2π β 2
03:18
Video Transcript
Given that π
is a relation from π₯ to π¦, where π exists in π₯ and π exists in π¦, which of the following equations correctly expresses relation π
? Is it (A) π equals π plus one, (B) π equals two π plus two, (C) π equals two π minus two, (D) π is equal to two π plus two, or (E) π is equal to two π minus two?
Any relation π
contains a set of ordered pairs of the form π₯, π¦. In the diagram shown, we have three ordered pairs: negative one, zero; four, 10; and five, 12. We can therefore say that the relation π
is the set of these three ordered pairs. We are asked to find the correct equation that matches any value in π₯ π to a value in π¦ π. The easiest way to do this is to substitute our values into each of the equations. Letβs begin with the ordered pair negative one, zero.
We will let π equal negative one and π equal zero. Zero is equal to negative one plus one. This means that equation (A) does work for the first ordered pair. Likewise, two multiplied by negative one plus two is also equal to zero. This means that equation (B) also works for the first ordered pair. In option (C), two multiplied by negative one minus two is equal to negative four and not zero. This means that equation (C) is not the correct answer. This is also true of options (D) and (E) as negative one is not equal to two multiplied by zero plus two or two multiplied by zero minus two. We can therefore rule out both of these options.
We will now consider the second ordered pair four, 10 for equation (A) and equation (B). This time, π is equal to four and π is equal to 10. 10 is not equal to four plus one. This means that equation (A) is also incorrect. 10 is equal to two multiplied by four plus two. This means that equation (B) holds for the first and second ordered pairs. We can now move on to the third ordered pair five, 12. Two multiplied by five plus two is equal to 12. As equation (B) holds for all three ordered pairs, this is the correct answer. The equation that correctly expresses relation π
is π is equal to two π plus two.
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# BKF 2432: MASS TRANSFER FKKSA, UMP
Principles of Mass
Transfer
(CHAPTER 4)
Molecular Diffusion in Solids
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BKF 2432: MASS TRANSFER FKKSA, UMP
Topic Outcomes
## It is expected that student will be able to:
Apply the diffusivity coefficient of molecular
diffusion in solids.
Solve mathematical solution of molecular
diffusion in solids.
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BKF 2432: MASS TRANSFER FKKSA, UMP
INTRODUCTION
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BKF 2432: MASS TRANSFER FKKSA, UMP
DIFFUSION IN SOLIDS
(WHICH FOLLOW FICK’S LAW
AND DOES NOT DEPEND ON
THE STRUCTURE OF SOLIDS)
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## Example 6.5-1 (pg 442) Diffusion of
H2 Through Neoprene Membrane
o The gas hydrogen at 17oC and 0.010 atm partial
pressure is diffusing through a membrane of
vulcanized neoprene rubber 0.5 mm thick. The
pressure of H2 on the other side of the neoprene
is zero. Calculate the steady-state flux,
assuming the only resistance to diffusion is in
the membrane. The solubility,S of H2 gas in
neoprene at 17oC is 0.051 m3 (at STP of 0oC
and 1 atm)/m3 solid.atm and the diffusivity DAB is
0.740 x 10-9 m2/s at 17oC.
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## Example 6.5-2 (pg 443) Diffusion
Through a Packing Film Using Permeability
o A polyethylene film 0.00015 m (0.15 mm) thick is
being considered for use in packaging a
pharmaceutical product at 30oC. If the partial
pressure of O2 outside the package is 0.21 atm
and inside it is 0.01 atm, calculate the diffusion
flux of O2 at steady state. Use permeability data
from Table 6.5-1. Assume that the resistances to
diffusion outside the film and inside are
negligible compared to the resistance of the film.
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BKF 2432: MASS TRANSFER FKKSA, UMP
DIFFUSION IN POROUS
SOLIDS
(THAT DEPENDS ON THE
STRUCTURE OF SOLIDS)
13
2008/2009 II
BKF 2432: MASS TRANSFER FKKSA, UMP
14
2008/2009 II
BKF 2432: MASS TRANSFER FKKSA, UMP
## Example 6.5-3 (pg 445) Diffusion of
Potassium Cloride, KCl in Porous Silica
o A sintered solid of silica 0.002 m thick is porous,
with a void fraction, ε of 0.30 a tortuosity of 4.0.
The pores are filled with water at 298 K. At one
face the concentration, CA1 of KCl is held at 0.10
kg mol/m3, and fresh water flows rapidly past the
other surface, CA2= 0. Neglecting any other
resistance but that in the porous solid, calculate
the diffusion of KCl at steady state. Given, DAB
KCl-water system (Table 6.3-1, pg 431) is 1.87 x
10-9 m2/s.
15
2008/2009 II
BKF 2432: MASS TRANSFER FKKSA, UMP
16
2008/2009 II | 1,044 | 2,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-51 | latest | en | 0.709727 |
https://socratic.org/questions/how-do-you-write-d-2-18d-80-in-factored-form | 1,576,423,571,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308604.91/warc/CC-MAIN-20191215145836-20191215173836-00158.warc.gz | 550,595,457 | 6,200 | How do you write d^2-18d+80 in factored form?
Sep 16, 2015
color(blue)((d-8)(d-10) is the factorised form of the expression.
Explanation:
${d}^{2} - 18 d + 80$
We can Split the Middle Term of this expression to factorise it.
In this technique, if we have to factorise an expression like $a {d}^{2} + b d + c$, we need to think of 2 numbers such that:
${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot 80 = 80$
and
${N}_{1} + {N}_{2} = b = - 18$
After trying out a few numbers we get ${N}_{1} = - 10$ and ${N}_{2} = - 8$
$- 10 \cdot - 8 = 80$, and $\left(- 10\right) + \left(- 8\right) = - 18$
${d}^{2} - 18 d + 80 = {d}^{2} - 10 d - 8 d + 80$
$= d \left(d - 10\right) - 8 \left(d - 10\right)$
=color(blue)((d-8)(d-10) | 298 | 724 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2019-51 | latest | en | 0.755632 |
https://www.jiskha.com/members/profile/posts.cgi?name=Cher | 1,521,788,029,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648198.55/warc/CC-MAIN-20180323063710-20180323083710-00078.warc.gz | 839,012,690 | 6,315 | # Posts by Cher
Total # Posts: 37
1. ### Algebra 1
Four more than twice a number is AT MOST 14. Find the largest possible answer that makes this true.
2. ### bonita
taking the same pairs of data for which you calculated the correlation (shown again here below), what is the slope of the regression line, to the nearest tenth? answer in the form of x.x or -x.x (e.g. 0.9 or -1.2), and remember to round your final answer to the nearest tenth...
3.70 3.700
4. ### Maths
7-4=3 (difference of 3 sweets given) For a difference of 3 sweets, total sweets needed: 17-2=15 15 sweets need for each friend having 3 more sweet, hence the number of John friend: 15/3=5 If John give each friends 4 sweets: 5x4=20 However John is short of 2 sweets (needs 2) 20...
5. ### Math
If a van is traveling at 25mph and needs to stop immediately, how much time and distance will be needed?
6. ### Math
I solved it already but thanks for the great help! It just confirmed my answer.:)
7. ### Math
The heights h of two-thirds of the members of a population satisfy the inequality |h-68.5/2.7|≤1 where h is measured in inches.Determine the interval on the real number line in which these heights lie.
8. ### calculus
Consider a differentiable functionf having domain all positive real numbers, and for which it is known that f'(x)=(4-x)^-3 for x.0 a. find the x-coordinate of the critical point f. DEtermine whether the point is a relative maximum or minomum, or neither for the fxn f. ...
9. ### Math
Hafers, an electrical supply company, sold \$3,800 of equipment to Jim Coates Wiring, Inc. Coates signed a promissory note May 12 with 4.00% interest. The due date was August 18. Short of funds, Hafers contacted Charter One Bank on July 12; the bank agreed to take over the ...
10. ### Physical Science
An electric hoist lifted a load with a mass of 250kg to a height of 80m in 39s , Calculate
12. ### Logic
The sentence "P → Q" is read as (3) P or Q P and Q If P then Q Q if and only P 2. In the truth table for an invalid argument, (2) on at least one row, where the premises are all true, the conclusion is true. on at least one row, where the premises are all true...
13. ### medical information management
According to a December 4 hospital census of 100,what are the total inpatient services days for December 5? Admissions 10 Discharges 2 A&D 1 A.90 B.109 C.100 D.110
14. ### medical coding
according to a December 4 hospital census of 100,what are the total inpatient service days for december 5? admissions 10 discharges 2 A&D 1 A.90 B.109 C.100 D.110
15. ### frequency
For sound waves in room-temperature air, is it possible to change the wavelength of a sound without changing the frequency?
16. ### Physics
Your sonic range finder measures the distance to a nearby building at 20m. The range finder is calibrated for sound traveling 343 m/s, but on this very cold day the speed of sound is only 309 m/s. How far away is the building?
17. ### Vibrations and waves
A mass on a spring bobs up and down over a distance of 30cm from the top to the bottom of its path twice each second. What are its period and amplitude?
18. ### Physics
If an 80-kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?
19. ### geometry
Find X if B is the midpoint of AC, AB=18 and AC=10x-4
I need to find the ROI and NPV of a \$35,000 software installation, which will reduce staff of one at \$10 per hour +18% benefits. Money cost are 7% and the system will become obsolete in three years. I have figured the ROI, but am not sure about the 7% money cost. Can you help?
21. ### Chemistry
Consider the following reaction: PbCl2 + 2 NaOH -> Pb(OH)2 + 2 NaCl What would be the major species observed floating in solution after one mole of PbCl2 and 2 moles of NaOH are mixed in aqueous solution? 1. Na+ ions, Cl− ions and H+ ions 2. Pb2+ ions, Cl− ions ...
22. ### algebra
62 6tens 2ones Regroup ___ tens ___ ones
23. ### Social Studies
What collection of spells were buried with the mummies?
24. ### Math
What numbers under 100 look the same when viewed upside down? I know of 0,1,8,and 11.
25. ### accounting
Prepare an income statement cash 16050.00 accounts receivable 1500.00 office supplies 1200.00 office equipment 18250.00 accounts payable 2350.00 Stanley Neal, Capital, 12850.00 fees income 36400.00 advertising expense 2750.00 salaries expense 7500.00 telephone expense 350.00 ...
how would you explain the redeeming value of sexual appeals in advertising? if sexual appeals are considered okay by the targeted audiences, what responsibily does the advertiser have for indirect targets such as children and how can advertisers protect themselves from that ...
27. ### Romeo and Juliet
Could you name some examples from West Side Story and Romeo and Juliet that are examples of these themes? Haste and lack of wise forthought bring about diaster The only way some people learn is through suffering Revengful acts are costly and often do permanent damage. Also, ...
what do you think the advantages and disadvantages are to commision system of payment over the fee or incentive system? I need to compare the two of them.
29. ### modern world
Do yu think that the industrial movement's changes led or accounted for the socialist and fascist movements at all? Yes, industrialization had a lot to do with it. This online article supposrts that thesis: http://www.bartleby.com/67/1081.html
30. ### modern world
Does anyone know what the difference of communism by Marx and Lenin and national socialism by Hitler is? Thank you for using the Jiskha Homework Help Forum. Here are some links to help you in your research: 1. http://www.romm.org/soc_com.html 2. http://www.colby.edu/personal/r...
31. ### modern world
I don't understand what the difference is in how does the view of history influenced by Darwinism differ from the view of history influenced by the Enlightenment idea of progress??? Darwin didn't intend his theories of evolution to apply to humans and their history. He...
32. ### world war II
What were the federal departments that were established to carry out the way World War II was to be fought? http://en.wikipedia.org/wiki/Politics_of_the_United_States_during_World_War_II Check out the Cabinet listing. =) european theater and pacific theater comparing the two. ...
33. ### imperative sentence
Helping my son and double checking, could the following 2 sentences be considered imperative sentences? 1. Josh go to your room. 2. You better go to your room. Thanks. The first sentence, Josh, go to your room. is imperative. It's a command or request. The subject of the ...
34. ### marketing
complete the following table (use one or a few well-chosen words) 1. Kinds of distribution facility(ies) needed and functions they will provide. 2. Caliber of salespeople required. 3. Kind of advertising required. Products
35. ### marketing
I need some help determining the strengths and weaknesses of these brand names..... Girl Scout’s – Cookies- Their strength in the brand name is that they are recognized as a guide for young girls to be leaders of tomorrow, which is to empower young girls. Their brand...
36. ### marketing
consumer services tend to be intangible, and goods tend to be tangible. Can you give me an example to explain how the lack of a physical good in a pure service might affect efforts to promote the service? Consumer services often have no physically viewable representation. For ...
37. ### weakness and strenght of a brand name
I need some help determining the strengths and weaknesses of these brand names..... Girl Scout’s – Cookies- Their strength in the brand name is that they are recognized as a guide for young girls to be leaders of tomorrow, which is to empower young girls. Their brand... | 1,927 | 7,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-13 | longest | en | 0.893433 |
https://stats.stackexchange.com/questions/458091/factor-included-based-on-aic-from-anova-yet-no-significant-comparisons-using-po | 1,653,826,863,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662644142.66/warc/CC-MAIN-20220529103854-20220529133854-00784.warc.gz | 573,553,741 | 68,781 | Factor included based on AIC from anova, yet no significant comparisons using PostHoc
Using step-wise model reduction I have reduced my full model down (Based on AIC and model comparisons using the anova() function in R. This resulted in the following model:
m1_DD4 <- glm(GotoPB ~ Pb_type + ZF_Pb + ZF_NotPb, data = DF_DD_5, family = binomial(link = "logit"))
Pb_type is a factor with 4 levels. ZF_Pb and ZF_NotPb are numerical. The summary of this model looks as follows:
Call:
glm(formula = GotoPB ~ Pb_type + ZF_Pb + ZF_NotPb, family = binomial(link = "logit"),
data = DF_DD_5)
Deviance Residuals:
Min 1Q Median 3Q Max
-2.15642 -1.07365 -0.00006 1.02744 1.56573
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.013410 0.136495 0.098 0.92174
Pb_typeNG 0.152678 0.163990 0.931 0.35184
Pb_typeSilence -20.152316 452.945271 -0.044 0.96451
Pb_typeSong 0.221061 0.156019 1.417 0.15652
ZF_Pb 0.079849 0.013634 5.857 4.72e-09 ***
ZF_NotPb -0.027021 0.007097 -3.807 0.00014 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 2047.0 on 1559 degrees of freedom
Residual deviance: 1385.9 on 1554 degrees of freedom
AIC: 1397.9
Number of Fisher Scoring iterations: 18
From the model reduction/AIC I conclude that the model fit improves when Pb_type is included, and it should therefore be included. Yet here all are shown as non-significant. From what I understand/read, this is because the p-values reported here are from Wald tests, which tests if each coefficient is different from 0. Whereas the anova() function uses likelihood ratio tests, which indicates which model has a statistically better fit. Please correct me if I am wrong here.
I have also been told to not worry about this too much, and to rely on the outcome of anova(). But I would like to undestand why.
Under the assumption that Pb_type indeed matters, I now wish to determine how the different levels of the Pb_type differ when scored against each other using a PostHoc pair-wise comparison. I do this using the emmeans package, via the following formula:
emmeans::emmeans(m1_DD4, pairwise ~ Pb_type, adjust = "Tukey")
Which results in the following output:
$emmeans Pb_type emmean SE df asymp.LCL asymp.UCL DC 0.128 0.120 Inf -1.72e-01 0.427 NG 0.280 0.115 Inf -5.30e-03 0.566 Silence -20.025 452.945 Inf -1.15e+03 1108.223 Song 0.349 0.104 Inf 8.96e-02 0.608 Results are given on the logit (not the response) scale. Confidence level used: 0.95 Conf-level adjustment: sidak method for 4 estimates$contrasts
contrast estimate SE df z.ratio p.value
DC - NG -0.1527 0.164 Inf -0.931 0.7882
DC - Silence 20.1523 452.945 Inf 0.044 1.0000
DC - Song -0.2211 0.156 Inf -1.417 0.4887
NG - Silence 20.3050 452.945 Inf 0.045 1.0000
NG - Song -0.0684 0.152 Inf -0.449 0.9698
Silence - Song -20.3734 452.945 Inf -0.045 1.0000
Results are given on the log odds ratio (not the response) scale.
P value adjustment: tukey method for comparing a family of 4 estimates
Here a few things stand out:
1. My df = INF. From here, I undestand that this presents no issue and that this is how emmeans labels asymptotic results.
2. In both my original model summary and here, it can be seen that for Silence there is a huge Standard error.
3. None of the values are significant.
This leaves me with 2 questions:
1. Why is my standard error so high, and does this present a problem? To answer this it might be necessary to know more about my data. So in short: All Pb_type treatments are different playbacks, and the silence served as a control treatment. The response variable GotoPB is either Y/N and represents whether individuals approached the playback (Y) or not (N). In the case of silence, as there never truly was playback, all values for GotoPB are N.
2. How can none of the comparisons of Pb_type be significant, yet including it in the model results in a significantly better fit (with a much lower AIC). Is something wrong here? And if not, how should I interpret this?
Kind regards.
First, the standard error for the Silence condition is huge because you are trying to estimate $$-\infty$$; i.e., the logit of zero. As you explained, the response is defined to always be N under silence. Therefore, Silence is not an experimental condition, at least for this response, because you don't observe a response under that condition.
You should omit or subset-out all the Silence results from the dataset and re-fit the model without Silence as a level of Pb_type. Your AIC results, etc. may well be quite different, because all those zeros have destabilized everything.
Second, just because a term is selected in a model does not guarantee that any comparison of its levels is statistically significant. Keep in mind that the primary goal of fitting a statistical model is to understand the patterns. I suggest plotting the estimates, e.g. using emmip(), and generally being more descriptive about your findings. Consider including interactions among these factors. Put less emphasis on how many asterisks you get, and more on the patterns. | 1,507 | 5,264 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.833158 |
https://stackoverflow.com/questions/51735576/fast-and-accurate-iterative-generation-of-sine-and-cosine-for-equally-spaced-ang | 1,638,969,015,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363510.40/warc/CC-MAIN-20211208114112-20211208144112-00096.warc.gz | 595,900,990 | 36,863 | # Fast and accurate iterative generation of sine and cosine for equally-spaced angles
In some applications, sine and cosine of multiple angles are needed, where the angles are derived by repeated addition of equal-sized increments incr to a starting value base. For performance reasons, instead of invoking the `sin()`, `cos()` standard math library functions (or possibly the non-standard `sincos()` function) for each generated angle, it can be very advantageous to compute sin(base) and cos(base) only once, then derive all other sines and cosines by application of angle-sum formulas:
sin(base+incr) = cos(incr) · sin(base) + sin(incr) · cos(base)
cos(base+incr) = cos(incr) · cos(base) - sin(incr) · sin(base)
This only requires the one-time precomputation of the scale factors sin(incr) and cos(incr), regardless of how many iterations are performed.
There are a couple of issues with this approach. If the increment is small, cos(incr) will be a number near unity, causing a loss of accuracy through implicit subtractive cancellation when computing in a finite precision floating-point format. Also, more rounding error than necessary is incurred because the computation is not arranged in the numerical advantageous form sin(base+incr) = sin(base) + adjust , where the computed quantity adjust is significantly smaller in magnitude than sin(base) (analogous for the cosine).
Since one usually applies tens to hundreds of iteration steps, these errors will accumulate. How can one structure the iterative computation in a manner most advantageous to maintaining high accuracy? What changes to the algorithm should be made if the fused multiply-add operation (FMA) is available, which is exposed via the standard math functions `fma()` and `fmaf()`?
• When re-writing standard functions to take advantage of speed, often there is a loss of accuracy or loss of range or portability. Yet the importance of speed, range, portability vs. accuracy are not the same dimension and so we are left with comparing unlike qualities. IOWs, how is fast and accurate rated together? `double my_sin(double x) { return x; } ` is fast, portability & high precision over maybe 25% of all `double` (the small ones) yet lousy with large values. Aug 8 '18 at 1:37
The application of the half-angle formula for the sine allows to address both of the accuracy-impacting issues mentioned in the question:
sin(incr/2) = √((1-cos(incr))/2) ⇒
sin²(incr/2) = (1-cos(incr))/2 ⇔
2·sin²(incr/2) = 1-cos(incr) ⇔
1-2·sin²(incr/2) = cos(incr)
Substituting this into the original formula results in this intermediate representation:
sin(base+incr) = (1 - 2·sin²(incr/2)) · sin(base) + sin(incr) · cos(base)
cos(base+incr) = (1 - 2·sin²(incr/2)) · cos(base) - sin(incr) · sin(base)
By a simple reordering of the terms one arrives at the desired form of the formula:
sin(base+incr) = sin(base) + (sin(incr) · cos(base) - 2·sin²(incr/2) · sin(base))
cos(base+incr) = cos(base) - (2·sin²(incr/2) · cos(base) + sin(incr) · sin(base))
As in the original formula, this only requires the one-time precomputation of two scale factors, namely 2·sin²(incr/2) and sin(incr). For small increments, both of them are small: full accuracy is retained.
There are two choices on how to apply FMA to this computation. One can either minimize the number of operations by abolishing the approach of using a single adjustment, and instead use two, hoping that the reduced rounding error of the FMA operation (one rounding for two operations) will compensate the loss in accuracy:
sin(base+incr) = fma (-2·sin²(incr/2), sin(base), fma ( sin(incr), cos(base), sin(base)))
cos(base+incr) = fma (-2·sin²(incr/2), cos(base), fma (-sin(incr), sin(base), cos(base)))
The other alternative is to apply a single FMA to the improved formula, although it is not immediately clear which of the two multiplies should be mapped to the unrounded multiplication inside the FMA:
sin(base+incr) = sin(base) + fma (sin(incr), cos(base), -2·sin²(incr/2) · sin(base))
cos(base+incr) = cos(base) - fma (sin(incr), sin(base), 2·sin²(incr/2) · cos(base))
The scaffolding below evaluates each of the computational alternative discussed above by generating many (base, incr) pairs, then iterates for each of them for a set number of steps while collecting errors of all sine and cosine values generated. From this it computes a root-mean square error for each test case, separately for sines, cosines. The largest RMS error observed across all generated test cases is reported at the end.
``````#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define NAIVE (1)
#define ROBUST (2)
#define FAST (3)
#define ACCURATE (4)
#define MODE (ACCURATE)
// Fixes via: Greg Rose, KISS: A Bit Too Simple. http://eprint.iacr.org/2011/007
static unsigned int z=362436069,w=521288629,jsr=362436069,jcong=123456789;
#define znew (z=36969*(z&0xffff)+(z>>16))
#define wnew (w=18000*(w&0xffff)+(w>>16))
#define MWC ((znew<<16)+wnew)
#define SHR3 (jsr^=(jsr<<13),jsr^=(jsr>>17),jsr^=(jsr<<5)) /* 2^32-1 */
#define CONG (jcong=69069*jcong+13579) /* 2^32 */
#define KISS ((MWC^CONG)+SHR3)
int main (void)
{
double sumerrsqS, sumerrsqC, rmsS, rmsC, maxrmsS = 0, maxrmsC = 0;
double refS, refC, errS, errC;
float base, incr, s0, c0, s1, c1, tt;
int count, i;
const int N = 100; // # rotation steps per test case
count = 2000000; // # test cases (a pair of base and increment values)
#if MODE == NAIVE
printf ("testing: NAIVE (without FMA)\n");
#elif MODE == FAST
printf ("testing: FAST (without FMA)\n");
#elif MODE == ACCURATE
printf ("testing: ACCURATE (with FMA)\n");
#elif MODE == ROBUST
printf ("testing: ROBUST (with FMA)\n");
#else
#error unsupported MODE
#endif // MODE
do {
/* generate test case */
base = (float)(KISS * 1.21e-10); // starting angle, < 30 degrees
incr = (float)(KISS * 2.43e-10 / N); // increment, < 60/n degrees
/* set up rotation parameters */
s1 = sinf (incr);
#if MODE == NAIVE
c1 = cosf (incr);
#else
tt = sinf (incr * 0.5f);
c1 = 2.0f * tt * tt;
#endif // MODE
sumerrsqS = 0;
sumerrsqC = 0;
s0 = sinf (base); // initial sine
c0 = cosf (base); // initial cosine
/* run test case through N rotation steps */
i = 0;
do {
tt = s0; // old sine
#if MODE == NAIVE
/* least accurate, 6 FP ops */
s0 = c1 * tt + s1 * c0; // new sine
c0 = c1 * c0 - s1 * tt; // new cosine
#elif MODE == ROBUST
/* very accurate, 8 FP ops */
s0 = ( s1 * c0 - c1 * tt) + tt; // new sine
c0 = (-s1 * tt - c1 * c0) + c0; // new cosine
#elif MODE == FAST
/* accurate and fast, 4 FP ops */
s0 = fmaf (-c1, tt, fmaf ( s1, c0, tt)); // new sine
c0 = fmaf (-c1, c0, fmaf (-s1, tt, c0)); // new cosine
#elif MODE == ACCURATE
/* most accurate, 6 FP ops */
s0 = tt + fmaf (s1, c0, -c1 * tt); // new sine
c0 = c0 - fmaf (s1, tt, c1 * c0); // new cosine
#endif // MODE
i++;
refS = sin (fma ((double)i, (double)incr, (double)base));
refC = cos (fma ((double)i, (double)incr, (double)base));
errS = ((double)s0 - refS) / refS;
errC = ((double)c0 - refC) / refC;
sumerrsqS = fma (errS, errS, sumerrsqS);
sumerrsqC = fma (errC, errC, sumerrsqC);
} while (i < N);
rmsS = sqrt (sumerrsqS / N);
rmsC = sqrt (sumerrsqC / N);
if (rmsS > maxrmsS) maxrmsS = rmsS;
if (rmsC > maxrmsC) maxrmsC = rmsC;
} while (--count);
printf ("max rms error sin = % 16.9e\n", maxrmsS);
printf ("max rms error cos = % 16.9e\n", maxrmsC);
return EXIT_SUCCESS;
}
``````
The output of the test scaffold shows that the fastest FMA-based alternative is superior to the naive method from the question, while the more accurate FMA-based alternative is the most accurate out of the alternatives considered:
``````testing: NAIVE (without FMA)
max rms error sin = 4.837386842e-006
max rms error cos = 6.884047862e-006
testing: ROBUST (without FMA)
max rms error sin = 3.330292645e-006
max rms error cos = 4.297631502e-006
testing: FAST (with FMA)
max rms error sin = 3.532624939e-006
max rms error cos = 4.763623188e-006
testing: ACCURATE (with FMA)
max rms error sin = 3.330292645e-006
max rms error cos = 4.104813533e-006
``````
• You should also examine how far `sin(base)*sin(base) + cos(base)*cos(base)` drifts from unity, to see if/when renormalization might become necessary. Aug 7 '18 at 22:16
• FMA is sometimes not that helpful, precision wise: Is my fma() broken? Aug 8 '18 at 1:24
If you want to maximize accuracy over long iteration counts, you could incrementally compute an exact value with no accumulated error while you generate incremental results from the previous exact value.
For example, if you precompute sin(incr*2^x) and cos(incr*2^x) for x=6 ... 31, say, then you can use the angle-sum formulas to calculate the result for each incr=64*n one bit at a time while you output the previous 64 values.
Every 64 values you discard the incrementally generated result in favor of the exact one, so no error can accumulate over long periods.
Also, since you're only going to need 64 incremental results from any exact base, you can precompute the 64 sines and cosines required to calculate those results directly from the base instead of the previous result.
It is possible to rearrange the equations for sin(base+incr) and cos(base+incr) in the following way:
sin(base+incr) = cos(incr) · sin(base) + sin(incr) · cos(base)
sin(base+incr) = sin(base) + (1 - cos(incr)) · -sin(base) + sin(incr) · cos(base)
sin(base+incr) = sin(base) + sin(incr) · (-1 / sin(incr) · (1 - cos(incr)) · sin(base) + cos(base))
sin(base+incr) = sin(base) + sin(incr) · (-tan(incr/2) · sin(base) + cos(base))
cos(base+incr) = cos(incr) · cos(base) - sin(incr) · sin(base)
cos(base+incr) = cos(base) - sin(incr) · (tan(incr/2) · cos(base) + sin(base))
Here we use the formula (1-cos(x)/sin(x) = tan(x/2), see here, for example. It is not immediately obvious that this should lead to more accurate results than the other approaches, but in practice it works quite well, as we will see later.
Again, this requires the one-time precomputation of two scale factors sin(incr) and tan(incr/2). In C we can write the formulas with 4 fma-s:
`````` s0 = fmaf ( s1, fmaf (-tt, c1, c0), tt); // new sine
c0 = fmaf (-s1, fmaf ( c0, c1, tt), c0); // new cosine
``````
The full updated test code is at the end of this answer. With `gcc -O3 -Wall -m64 -march=skylake fastsincos.c -lm` (GCC version 7.3), the results are:
``````testing: FAST (with FMA)
max rms error sin = 3.532624939e-06
max rms error cos = 4.763623188e-06
testing: ACCURATE (with FMA)
max rms error sin = 3.330292645e-06
max rms error cos = 4.104813533e-06
testing: FAST_ACC (with FMA)
max rms error sin = 3.330292645e-06
max rms error cos = 3.775300478e-06
``````
The new solution `FAST_ACC` is indeed a bit more accurate than the others in this test.
Modified test code:
``````#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define NAIVE (1)
#define ROBUST (2)
#define FAST (3)
#define ACCURATE (4)
#define FAST_ACC (5)
#define MODE (FAST_ACC)
// Fixes via: Greg Rose, KISS: A Bit Too Simple. http://eprint.iacr.org/2011/007
static unsigned int z=362436069,w=521288629,jsr=362436069,jcong=123456789;
#define znew (z=36969*(z&0xffff)+(z>>16))
#define wnew (w=18000*(w&0xffff)+(w>>16))
#define MWC ((znew<<16)+wnew)
#define SHR3 (jsr^=(jsr<<13),jsr^=(jsr>>17),jsr^=(jsr<<5)) /* 2^32-1 */
#define CONG (jcong=69069*jcong+13579) /* 2^32 */
#define KISS ((MWC^CONG)+SHR3)
int main (void)
{
double sumerrsqS, sumerrsqC, rmsS, rmsC, maxrmsS = 0, maxrmsC = 0;
double refS, refC, errS, errC;
float base, incr, s0, c0, s1, c1, tt;
int count, i;
const int N = 100; // # rotation steps per test case
count = 2000000; // # test cases (a pair of base and increment values)
#if MODE == NAIVE
printf ("testing: NAIVE (without FMA)\n");
#elif MODE == FAST
printf ("testing: FAST (without FMA)\n");
#elif MODE == ACCURATE
printf ("testing: ACCURATE (with FMA)\n");
#elif MODE == ROBUST
printf ("testing: ROBUST (with FMA)\n");
#elif MODE == FAST_ACC
printf ("testing: FAST_ACC (with FMA)\n");
#else
#error unsupported MODE
#endif // MODE
do {
/* generate test case */
base = (float)(KISS * 1.21e-10); // starting angle, < 30 degrees
incr = (float)(KISS * 2.43e-10 / N); // increment, < 60/n degrees
/* set up rotation parameters */
s1 = sinf (incr);
#if MODE == NAIVE
c1 = cosf (incr);
#elif MODE == FAST_ACC
c1 = tanf (incr * 0.5f);
#else
tt = sinf (incr * 0.5f);
c1 = 2.0f * tt * tt;
#endif // MODE
sumerrsqS = 0;
sumerrsqC = 0;
s0 = sinf (base); // initial sine
c0 = cosf (base); // initial cosine
/* run test case through N rotation steps */
i = 0;
do {
tt = s0; // old sine
#if MODE == NAIVE
/* least accurate, 6 FP ops */
s0 = c1 * tt + s1 * c0; // new sine
c0 = c1 * c0 - s1 * tt; // new cosine
#elif MODE == ROBUST
/* very accurate, 8 FP ops */
s0 = ( s1 * c0 - c1 * tt) + tt; // new sine
c0 = (-s1 * tt - c1 * c0) + c0; // new cosine
#elif MODE == FAST
/* accurate and fast, 4 FP ops */
s0 = fmaf (-c1, tt, fmaf ( s1, c0, tt)); // new sine
c0 = fmaf (-c1, c0, fmaf (-s1, tt, c0)); // new cosine
#elif MODE == ACCURATE
/* most accurate, 6 FP ops */
s0 = tt + fmaf (s1, c0, -c1 * tt); // new sine
c0 = c0 - fmaf (s1, tt, c1 * c0); // new cosine
#elif MODE == FAST_ACC
/* fast and accurate, 4 FP ops */
s0 = fmaf ( s1, fmaf (-tt, c1, c0), tt); // new sine
c0 = fmaf (-s1, fmaf ( c0, c1, tt), c0); // new cosine
#endif // MODE
i++;
refS = sin (fma ((double)i, (double)incr, (double)base));
refC = cos (fma ((double)i, (double)incr, (double)base));
errS = ((double)s0 - refS) / refS;
errC = ((double)c0 - refC) / refC;
sumerrsqS = fma (errS, errS, sumerrsqS);
sumerrsqC = fma (errC, errC, sumerrsqC);
} while (i < N);
rmsS = sqrt (sumerrsqS / N);
rmsC = sqrt (sumerrsqC / N);
if (rmsS > maxrmsS) maxrmsS = rmsS;
if (rmsC > maxrmsC) maxrmsC = rmsC;
} while (--count);
printf ("max rms error sin = % 16.9e\n", maxrmsS);
printf ("max rms error cos = % 16.9e\n", maxrmsC);
return EXIT_SUCCESS;
}
`````` | 4,459 | 13,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.883446 |
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What is the Gaussian function?
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A box contains 24 identical balls, of which 12 are black and 12 are white. The balls are drawn at random from the box one at a time with replacement. The probability that white ball is drawn for the fourth time on seventh draw is,
A). $\dfrac{5}{{64}}$
B). $\dfrac{{27}}{{32}}$
C). $\dfrac{5}{{32}}$
D). $\dfrac{1}{2}$
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B). $\dfrac{{16}}{{{3^7}}}$
C). $\dfrac{{280}}{{{3^7}}}$
D). $\dfrac{{560}}{{{3^7}}}$
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A man makes attempts to hit the target. The probability of hitting the target is $\dfrac{3}{5}$. Then, the probability that A hits the target exactly $2$ times in $5$ attempts is:
(A) $\dfrac{{144}}{{625}}$
(B) $\dfrac{{72}}{{3125}}$
(C) $\dfrac{{216}}{{625}}$
(D) None of these
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$X$ 1 2 3 4 $p\left( X \right)$ $k$ $2k$ $2k$ $4k$
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D. 2
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Z 0.2667 2.067 2.2667 Area 0.1026 0.4803 0.4881
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5 | 822 | 2,886 | {"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-39 | longest | en | 0.869082 |
http://c.happycodings.com/beginners-lab-assignments/code10.html | 1,417,045,917,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931007625.83/warc/CC-MAIN-20141125155647-00209-ip-10-235-23-156.ec2.internal.warc.gz | 44,620,723 | 4,584 | # C > Beginners Lab Assignments Code Examples
## Calendar Program
``` Calendar Program #include<conio.h> int day(int m1,int y1) { int d; if(m1==1 || m1==3 || m1==5 || m1==7 || m1==8 || m1==10 || m1==12) d=31; else if(m1==4 || m1==6 || m1==9 || m1==11) d=30; else if((y1%100!=0 && y1%4==0) || y1%400==0) d=29; else d=28; return d; } void main() { long unsigned int t; unsigned int y,y1,m,m1,d,da,i,j,k; char a[12][20]={"January","February","March","April","May","June","July","Augus t","September","October","November","December"}; clrscr(); textcolor(CYAN); cprintf("Enter the year: "); scanf("%4u",&y); if(y<0) y=-y; cprintf(" Enter the month: "); scanf("%2u",&m); if(m<=0 || m>=13) m=1; clrscr(); gotoxy(32,2); cprintf("Calendar"); y1=0; t=0; while(y1<y) { if((y1%100!=0 && y1%4==0) || y1%400==0) t=t+366; else t=t+365; y1++; } m1=1; while(m1<m) { d=day(m1,y); t=t+d; m1++; } d=t%7; printf(" Year: '%u'",y); printf(" Month: '%s' ",a[m-1]); printf("%6s%6s%6s%6s%6s%6s%6s ","Sun","Mon","Tue","Wed","Thu","Fri","Sa t"); textcolor(GREEN); k=1; for(i=1;i<=day(m,y);i++,k++) { if(i==1) { if(d==0) { for(j=1;j<7;j++,k++) printf("%6s",""); } else { for(j=1;j<d;j++,k++) printf("%6s",""); } } cprintf("%6d",i); if(k%7==0) printf(" "); } gotoxy(27,22); cprintf("www"); getch(); } ``` | 531 | 1,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-49 | longest | en | 0.377659 |
https://forums.sketchup.com/t/creating-multiple-holes-on-a-cylinder/242848 | 1,709,353,977,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00394.warc.gz | 265,212,909 | 6,696 | # Creating multiple holes on a cylinder
Morning, All…
So… this is the top of a casing used in piling… I’ve created the countersunk holes in it by creating a small cylinder and tweaking it, then using the rotate copy to place them radially through the top of the casing, then used subtract 18 times…
Surely there’s a quicker way of doing this?
Make 1 hole, Select just the hole, Ctrl Copy, 20 degrees, 18x
1 Like
I’ve had trouble using that in the past…
You can have issues with that if the number of holes doesn’t match the number of sides, meaning the holes don’t fall on a flat face.
Simplest option, make a cylinder from a circle with 18 or 36 sides, turn on hidden geometry and delete all but one (or if 36, two sides), create your hole in the appropriate spot, then make a radial array of the whole thing to recreate your cylinder with all the holes.
Best to start centered on the origin or somehow keep the center point to enable the array.
1 Like
A trick to avoid doing any calculation in doing a full circular array:
1 - Select the item to copy;
2 - Select the Rotate Tool;
3 - Click on the center of rotation and make sure that the axis of rotation is correct;
4 - Begin rotating;
5 - Let go of the mouse;
6 - Type 360 followed by a press of the Enter key as the rotation angle;
7 - Type /N followed by a press of the Enter key as the number of copies (replace N by the required number of copies, 18 in your example).
Be aware that if you copy loose geometries, the first copy will be merged with the original. In this case, you have nothing else do do.
If you do that with a group or a component, the first copy will be superposed on the original so you will have two groups or components occupying the same location. You need to delete one of them.
1 Like
Another approach if you are in a situation where you are unable to alter the cylinder geometry to suit the hole layout is to array your tweaked cylinders, combine all of them into one single solid (geometry doesn’t have to be continuously attached to be considered a ‘solid’) and subtract just once. | 474 | 2,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-10 | latest | en | 0.901162 |
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61 Questions for Extra Credit Points. Due 12/16 (Wednesday)
You need to show your work and explanations. Jotting down only the answers is not acceptable. If you do all 100 questions, you will get up to 3 extra points added to your final total score (after I determine your total score based on mid-terms, HWs, and the final).
Chapter 5
1. You plan to analyze the value of a potential investment by calculating the sum of the present values of its expected cash flows. Which of the following would lower the calculated value of the investment? a. The cash flows are in the form of a deferred annuity, and they total to \$100,000. You learn that the annuity lasts for only 5 rather than 10 years, hence that each payment is for \$20,000 rather than for \$10,000.
b. The discount rate increases.
c. The riskiness of the investment’s cash flows decreases.
d. The total amount of cash flows remains the same, but more of the cash flows are received in the earlier years and less are received in the later years.
e. The discount rate decreases. b 2. Which of the following statements is CORRECT? a. The cash flows for an ordinary (or deferred) annuity all occur at the beginning of the periods.
b. If a series of unequal cash flows occurs at regular intervals, such as once a year, then the series is by definition an annuity.
c. The cash flows for an annuity due must all occur at the beginning of the periods.
d. The cash flows for an annuity may vary from period to period, but they must occur at regular intervals, such as once a year or once a month.
e. If some cash flows occur at the beginning of the periods while others occur at the ends, then we have what the textbook defines as a variable annuity. c 3. You are considering two equally risky annuities, each of which pays \$5,000 per year for 10 years. Investment ORD is an ordinary (or deferred) annuity, while Investment DUE is an annuity due. Which
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Powerful Essays | 1,659 | 7,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-27 | latest | en | 0.95291 |
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## Heat & Temperature Concepts:
Calorie
A calorie is defined as the amount of energy required to increase the temperature of a gram water by one degree Celsius. This is equivalent to @J. If you count your calories when you eat, you are actually counting kilocalories. That is, the calorie used in measuring food energy is actually one thousand of the calories we just defined.
British Thermal Unit (BTU).
A BTU is the amount of energy required to increase the temperature of one pound of water by one degree Fahrenheit. This is equivalent to @ J. For some reason, the ability of air conditioners to extract energy from air is always expressed in BTUs.
Energy
It is usually the case that when you add energy to a bunch of atoms they move faster and get hotter. Similarly, if you remove energy from a bunch of atoms, they usually move less and get cooler. Because adding heat energy usually results in a temperature rise, people often confuse heat and temperature. In common speech, the two terms mean the same: "I will heat it" means you will add heat; "I will warm it up" means you will increase the temperature. No one usually bothers to distinguish between these. Adding heat, however, does not always increase the temperature. For instance, when water is boiling, adding heat does not increase its temperature. This happens at the boiling temperature of every substance that can vaporize. At the boiling temperature, adding heat energy converts the liquid into a gas WITHOUT RAISING THE TEMPERATURE.
Feedback
Flow and collection are controlled with negative feedback. Many things flow. Water flows in ditches and pipes, electric charge flows through wires, air flows over the wings of an airplane, and heat flows from hot to cold. The rate a which a flow can fill a container depends of the speed of the flow the effective cross section of the ditch or pipe or wire. Perhaps more interesting is the principle of negative feedback. In this control process, a changing condition of some system is regulated to keep it more or less constant. When the condition reaches some set maximum, the cause of the change is shut off or reversed.
Heat
You can usually warm something by adding energy. The added energy can be from light, electricity, friction, a chemical reaction, nuclear reaction, or any other kind of energy. When first added to a substance, energy might be concentrated in one atom, but this one will soon bump into others and spread the energy. Eventually, every atom or molecule in the substance will move a bit faster. When the added energy is spread throughout a substance, it is then called heat energy, thermal energy, or, simply heat. All three terms mean the same thing. Heat is a form of energy, so it has the units of energy. In the SI system, this is Joules (J). Many other units to measure thermal energy are in common use. Calories and BTU's are common heat units.
Heat Change
Heat Energy (Q) The heat energy put into an object by increasing its temperature (T) is proportional to the temperature change and proportional to the mass (m) of the object being warmed. The proportionality constant (c), called the specific heat, depends on the material the object is made of. Heat change Q = c * m * (T)
Heat Flow
Heat flow is measured as the amount of energy transferred per unit of time. In SI units this would be measured in Joules per second. A Joule per second is the same as a Watt (W), so heat flow should be expressed in Watts. Because thermal energy is often measured in non-SI units, heat flow is often reported in other units, too. Air conditioners are rated in BTUs per hour. You will sometimes see ergs per second or calories per second.
Heating Rate
Rate is the change as time (t) passes. Graphs which have time as the independent variable are standard means of depicting rate. The slope of a "vs. time" graph tells the rate. Because temperature is proportional to heat in a sample of material (if it is not freezing or melting) temperature vs. time graphs are indications of energy change. Rate of heating (or cooling) r = Q÷t
(Note: Here T is used to represent Temperature, whereas t is used to represent time. Be sure you keep the UPPER and lower case t's straight. )
Temperature
You cannot measure heat directly, but you can detect its effect on a substance. Changes in heat can usually be detected as changes in temperature. When you add heat energy to a substance, it usually warms; when you remove heat energy it usually cools.
Temperature Units
In the SI system, temperature is measured in degrees Celsius (C). This scale is defined by: The freezing point of water = 0 C The boiling point of water = 100 C. The boiling point of water changes somewhat with altitude, so it is important to add that 100 C is the boiling temperature of water at sea level. Actually, it is the atmospheric pressure that influences the boiling temperature, so, for accuracy, an average atmospheric pressure at sea level is used, taken as a pressure of 760 mm of mercury. The Celsius scale is commonly used in every country except the U. S. where the old Fahrenheit (F) scale is used that is defined by: The freezing point of water = 32 F The boiling point of water = 212 F. There are a total of (212 p; 32) = 180 Fahrenheit degrees between freezing and boiling, almost twice as many as in the Celsius scale. That means every Celsius degree represents almost twice as large a step as a Fahrenheit degree.
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http://manpages.ubuntu.com/manpages/precise/man3/cpftrs.3lapack.html | 1,571,018,915,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648481.7/warc/CC-MAIN-20191014003258-20191014030258-00091.warc.gz | 175,597,771 | 3,593 | Provided by: liblapack-doc_3.3.1-1_all
#### NAME
``` LAPACK-3 - solves a system of linear equations A*X = B with a Hermitian positive definite
matrix A using the Cholesky factorization A = U**H*U or A = L*L**H computed by CPFTRF
```
#### SYNOPSIS
``` SUBROUTINE CPFTRS( TRANSR, UPLO, N, NRHS, A, B, LDB, INFO )
CHARACTER TRANSR, UPLO
INTEGER INFO, LDB, N, NRHS
COMPLEX A( 0: * ), B( LDB, * )
```
#### PURPOSE
``` CPFTRS solves a system of linear equations A*X = B with a Hermitian positive definite
matrix A using the Cholesky factorization A = U**H*U or A = L*L**H computed by CPFTRF.
```
#### ARGUMENTS
``` TRANSR (input) CHARACTER*1
= 'N': The Normal TRANSR of RFP A is stored;
= 'C': The Conjugate-transpose TRANSR of RFP A is stored.
UPLO (input) CHARACTER*1
= 'U': Upper triangle of RFP A is stored;
= 'L': Lower triangle of RFP A is stored.
N (input) INTEGER
The order of the matrix A. N >= 0.
NRHS (input) INTEGER
The number of right hand sides, i.e., the number of columns
of the matrix B. NRHS >= 0.
A (input) COMPLEX array, dimension ( N*(N+1)/2 );
The triangular factor U or L from the Cholesky factorization
of RFP A = U**H*U or RFP A = L*L**H, as computed by CPFTRF.
See note below for more details about RFP A.
B (input/output) COMPLEX array, dimension (LDB,NRHS)
On entry, the right hand side matrix B.
On exit, the solution matrix X.
LDB (input) INTEGER
The leading dimension of the array B. LDB >= max(1,N).
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
```
#### FURTHERDETAILS
``` We first consider Standard Packed Format when N is even.
We give an example where N = 6.
AP is Upper AP is Lower
00 01 02 03 04 05 00
11 12 13 14 15 10 11
22 23 24 25 20 21 22
33 34 35 30 31 32 33
44 45 40 41 42 43 44
55 50 51 52 53 54 55
Let TRANSR = 'N'. RFP holds AP as follows:
For UPLO = 'U' the upper trapezoid A(0:5,0:2) consists of the last
three columns of AP upper. The lower triangle A(4:6,0:2) consists of
conjugate-transpose of the first three columns of AP upper.
For UPLO = 'L' the lower trapezoid A(1:6,0:2) consists of the first
three columns of AP lower. The upper triangle A(0:2,0:2) consists of
conjugate-transpose of the last three columns of AP lower.
To denote conjugate we place -- above the element. This covers the
case N even and TRANSR = 'N'.
RFP A RFP A
-- -- --
03 04 05 33 43 53
-- --
13 14 15 00 44 54
--
23 24 25 10 11 55
33 34 35 20 21 22
--
00 44 45 30 31 32
-- --
01 11 55 40 41 42
-- -- --
02 12 22 50 51 52
Now let TRANSR = 'C'. RFP A in both UPLO cases is just the conjugate-
transpose of RFP A above. One therefore gets:
RFP A RFP A
-- -- -- -- -- -- -- -- -- --
03 13 23 33 00 01 02 33 00 10 20 30 40 50
-- -- -- -- -- -- -- -- -- --
04 14 24 34 44 11 12 43 44 11 21 31 41 51
-- -- -- -- -- -- -- -- -- --
05 15 25 35 45 55 22 53 54 55 22 32 42 52
We next consider Standard Packed Format when N is odd.
We give an example where N = 5.
AP is Upper AP is Lower
00 01 02 03 04 00
11 12 13 14 10 11
22 23 24 20 21 22
33 34 30 31 32 33
44 40 41 42 43 44
Let TRANSR = 'N'. RFP holds AP as follows:
For UPLO = 'U' the upper trapezoid A(0:4,0:2) consists of the last
three columns of AP upper. The lower triangle A(3:4,0:1) consists of
conjugate-transpose of the first two columns of AP upper.
For UPLO = 'L' the lower trapezoid A(0:4,0:2) consists of the first
three columns of AP lower. The upper triangle A(0:1,1:2) consists of
conjugate-transpose of the last two columns of AP lower.
To denote conjugate we place -- above the element. This covers the
case N odd and TRANSR = 'N'.
RFP A RFP A
-- --
02 03 04 00 33 43
--
12 13 14 10 11 44
22 23 24 20 21 22
--
00 33 34 30 31 32
-- --
01 11 44 40 41 42
Now let TRANSR = 'C'. RFP A in both UPLO cases is just the conjugate-
transpose of RFP A above. One therefore gets:
RFP A RFP A
-- -- -- -- -- -- -- -- --
02 12 22 00 01 00 10 20 30 40 50
-- -- -- -- -- -- -- -- --
03 13 23 33 11 33 11 21 31 41 51
-- -- -- -- -- -- -- -- --
04 14 24 34 44 43 44 22 32 42 52
LAPACK routine (version 3.3.1) April 2011 CPFTRS(3lapack)
``` | 1,607 | 4,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.784161 |
https://byjus.com/question-answer/a-radioactive-sample-decays-through-two-different-decay-processes-alpha-decay-and-beta-decay-half/ | 1,643,460,674,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00114.warc.gz | 207,107,678 | 27,215 | Question
# A radioactive sample decays through two different decay processes $$\alpha-decay$$ and $$\beta-decay$$. Half-life time for $$\alpha-decay$$ is $$3 h$$ and half-life time for $$\beta-decay$$ is $$6 h$$. What will be the ratio of number of initial radioactive nuclei to the number of radioactive nuclei present after $$6 h$$?
Solution
## Here after $$6\ hr$$, two half time for $$\alpha$$ decay and one half time for $$\beta$$ decay will occur.Let initial number of nuclei $$=N_0$$After two half time for $$\alpha$$ decay, number of nuclei $$=N_0/4$$ and After one half time for $$\beta$$ decay, number of nuclei $$=\dfrac{N_0/4}{2}=N_0/8$$Ratio of number of initial radioactive nuclei to the number of radioactive nuclei present after $$6\ hr$$ $$=\dfrac{N_0}{N_0/8}=8$$PhysicsNCERTStandard XII
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