url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
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ispcp-omega.net | 1,685,806,383,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649293.44/warc/CC-MAIN-20230603133129-20230603163129-00738.warc.gz | 339,579,509 | 17,007 | ### Are Merely Sucker A Good Online Chance?
Hi ZS, assuming that whether or one wins or loses on one scratch ticket (what is that, in either case?) is independent from winning or losing on every other scratch ticket, you treat each event as an independent event. Laws of probability tell us to multiply the various probabilities of independent affairs. It appears that the probability of [losing] on any particular scratch ticket must be 2/3. Useful and obviously the possibility of [losing] on 30 scratch tickets in a row (if that maybe what your problem is asking) must be (2/3)^30 = approximately 0.2 x 10^-6, which is about.0000052, or 52 via 10 million, which comes from 1 chance out of 192,307.
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Cape Town, South Africa | 711 | 3,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-23 | longest | en | 0.960232 |
https://www.jigsawacademy.com/blogs/business-analytics/types-of-outliers/ | 1,656,915,337,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104354651.73/warc/CC-MAIN-20220704050055-20220704080055-00210.warc.gz | 908,034,671 | 96,444 | # Types of Outliers – An Easy Guide in Just 2 Points
## Introduction
An outlier is an observation that falls at an abnormal distance from the other values that are there on the random sample from the population. The decision, however, rests on the analyst to understand which point will be considered to be abnormal or an outlier. The articles talk about the types of outliers.
## 1) What are outliers?
The outlier is a point of data that is different from the other observations. The outlier could be caused because of measurement variability. It could also be formed because of some experimental error. In the case of the latter, it gets altogether eliminated from the data set. An outlier is something that can affect the statistical analysis phenomenally.
The outlier values could be formed by chance in the distribution. In most cases, this could be because of a measurement error or if the population comes with a heavy-tailed distribution. In the former case, one will wish to discard them or use the test that is robust to the outlier. In the latter case, the analysis will show that the distribution is highly skewed and that one should be cautious when using the tools that assume a normal distribution.
Another cause of an outlier could be when two distributions are mixed. These could be two subpopulations that may be disconnected or it could indicate a trail that has an error of measurement. This gets modelled using the mixture model.
Outlier points could be faulty data or an error procedure. It could also be an area where some theory would be invalidated. If the sample size is large then some outlier is fine to have.
The outlier is an extreme observation and this could include the sample minimum or maximum or both of them. On the contrary, the sample minimum and maximum are not an outlier because these may not be far from the observation.
## 2) What is an outlier to explain the types of outliers?
Now that you know what an outlier is here are the types of outliers.
• Global Outliers or the point namely:
A data point gets considered to be a global outlier in case its value is very far away from the entire data set in what it is found. The global outlier is basically a sample point that is measured and which has a very high or a low value relative to the values that are present in the dataset.
• Contextual outliers or the conditional outlier
If a particular data point is different in the context that is specific to a condition but it is not different otherwise then this is called the contextual outlier. The data object attribute needs to get divided into two groups. The behavioural attributes are the object characteristics that are used in evaluating the outlier. It is difficult to spot the contextual outlier if you do not have any background information.
• Collective Outliers
If there is a data point collection that is totally different from the entire set of data then this is the collective outlier. A subset of the data point in the data set is different if the values as a group deviate from the data set totally. However, the values of these data points are not different in a global or a contextual sense.
## Conclusion
It is important to investigate outliers carefully. They may have some information about the process which is under investigation. Before you consider eliminating it you should first try to understand the reasons why they may have been here in the first place. In most cases, outliers could be a bad data point. Unfortunately, there is no strict statistical rule for outlier identification. This makes it highly subjective which is dependent on the analysts’ knowledge and the process of data collection.
If you are interested in making it big in the world of data and evolve as a Future Leader, you may consider our Integrated Program in Business Analytics, a 10-month online program, in collaboration with IIM Indore! | 762 | 3,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | longest | en | 0.945668 |
https://makeup-advice.com/etc/how-to-use-an-abacus.html | 1,624,446,447,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00264.warc.gz | 325,126,387 | 15,832 | # How To Use An Abacus?
## How do you work an abacus?
An abacus is like a manual calculator with sliding beads to represent numbers.
It has rows or columns of beads that represent the digits of your number.
You’ll have a ones place, a tens place, a hundreds place, a thousands place, and so on.
A Chinese abacus has columns of bead with an upper section and a lower section.
## How do you calculate with an abacus?
How to Count on an Abacus –
## How can I learn abacus at home?
Learning maths the easy way with abacus –
10 x 10 Abacus –
## What is the full name of Abacus?
The abacus (plural abaci or abacuses), also called a counting frame, is a calculating tool that was in use in the ancient Near East, Europe, China, and Russia, centuries before the adoption of the written Hindu–Arabic numeral system.
## Who created the Abacus?
May not answer this directly, but Tim Cranmer invented the Cranmer abacus. The oldest surviving counting board is the Salamis tablet (originally thought to be a gaming board), used by the Babylonians circa 300 B.C., discovered in 1846 on the island of Salamis. The abacus is still used in China and Japan.
## How high does an abacus count?
There is technically no limit as to how high you can count on an abacus. This one has over 20 digits.
## How many beads should an abacus have?
There are two beads on each rod in the upper deck and five beads each in the bottom. The beads are usually rounded and made of a hardwood. The beads are counted by moving them up or down towards the beam; beads moved toward the beam are counted, while those moved away from it are not.
## How many beads are in Abacus?
Before learning to use the abacus, realize there are different types of abacus’. For example, the classical abacus or Chinese abacus has five beads on the bottom and two beads at the top. The modern abacus, Japanese abacus, or soroban has four beads at the bottom and one bead at the top. | 456 | 1,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-25 | latest | en | 0.937457 |
https://howkgtolbs.com/convert/93.14-kg-to-lbs | 1,607,159,340,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747323.98/warc/CC-MAIN-20201205074417-20201205104417-00372.warc.gz | 330,594,658 | 11,521 | # 93.14 kg to lbs - 93.14 kilograms to pounds
## 93.14 kg to lbs
Do you want to learn how much is 93.14 kg equal to lbs and how to convert 93.14 kg to lbs? You couldn’t have chosen better. In this article you will find everything about kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to emphasize that whole this article is devoted to only one amount of kilograms - exactly one kilogram. So if you need to learn more about 93.14 kg to pound conversion - keep reading.
Before we go to the practice - this is 93.14 kg how much lbs calculation - we will tell you some theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 93.14 kg to lbs? 93.14 kilograms it is equal 205.3385508268 pounds, so 93.14 kg is equal 205.3385508268 lbs.
## 93.14 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in abbreviated form SI).
From time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.
The kilogram was defined first time in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but impractical to use.
Then, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was replaced by another definition.
Today the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It could be also divided to 100 decagrams and 1000 grams.
## 93.14 kilogram to pounds
You learned a little bit about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. We want to highlight that there are more than one kind of pound. What does it mean? For instance, there are also pound-force. In this article we are going to to concentrate only on pound-mass.
The pound is used in the British and United States customary systems of measurements. Naturally, this unit is in use also in another systems. The symbol of the pound is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound can be divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 5 kg?
93.14 kilogram is equal to 205.3385508268 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 93.14 kg in lbs
The most theoretical section is already behind us. In this part we will tell you how much is 93.14 kg to lbs. Now you know that 93.14 kg = x lbs. So it is time to know the answer. Let’s see:
93.14 kilogram = 205.3385508268 pounds.
It is a correct outcome of how much 93.14 kg to pound. It is possible to also round off the result. After rounding off your outcome is as following: 93.14 kg = 204.908 lbs.
You know 93.14 kg is how many lbs, so see how many kg 93.14 lbs: 93.14 pound = 0.45359237 kilograms.
Of course, this time you can also round it off. After it your outcome is exactly: 93.14 lb = 0.45 kgs.
We are also going to show you 93.14 kg to how many pounds and 93.14 pound how many kg results in tables. Have a look:
We are going to begin with a table for how much is 93.14 kg equal to pound.
### 93.14 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
93.14 205.3385508268 204.9080
Now look at a chart for how many kilograms 93.14 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
93.14 0.45359237 0.45
Now you know how many 93.14 kg to lbs and how many kilograms 93.14 pound, so it is time to go to the 93.14 kg to lbs formula.
### 93.14 kg to pounds
To convert 93.14 kg to us lbs you need a formula. We will show you a formula in two different versions. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 205.3385508268 outcome in pounds
The first formula give you the most exact result. In some situations even the smallest difference can be considerable. So if you want to get an exact outcome - first formula will be the best for you/option to convert how many pounds are equivalent to 93.14 kilogram.
So go to the second formula, which also enables conversions to learn how much 93.14 kilogram in pounds.
The second version of a formula is down below, look:
Number of kilograms * 2.2 = the outcome in pounds
As you can see, the second formula is simpler. It can be better solution if you need to make a conversion of 93.14 kilogram to pounds in easy way, for instance, during shopping. You only need to remember that final outcome will be not so accurate.
Now we are going to learn you how to use these two versions of a formula in practice. But before we will make a conversion of 93.14 kg to lbs we want to show you easier way to know 93.14 kg to how many lbs without any effort.
### 93.14 kg to lbs converter
Another way to know what is 93.14 kilogram equal to in pounds is to use 93.14 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. It is based on longer formula which we showed you above. Due to 93.14 kg pound calculator you can easily convert 93.14 kg to lbs. You only need to enter amount of kilograms which you need to calculate and click ‘calculate’ button. You will get the result in a flash.
So try to calculate 93.14 kg into lbs using 93.14 kg vs pound calculator. We entered 93.14 as an amount of kilograms. It is the result: 93.14 kilogram = 205.3385508268 pounds.
As you can see, our 93.14 kg vs lbs calculator is intuitive.
Now let’s move on to our primary issue - how to convert 93.14 kilograms to pounds on your own.
#### 93.14 kg to lbs conversion
We are going to begin 93.14 kilogram equals to how many pounds calculation with the first formula to get the most correct outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 205.3385508268 the result in pounds
So what have you do to check how many pounds equal to 93.14 kilogram? Just multiply amount of kilograms, this time 93.14, by 2.20462262. It gives 205.3385508268. So 93.14 kilogram is equal 205.3385508268.
It is also possible to round it off, for example, to two decimal places. It is equal 2.20. So 93.14 kilogram = 204.9080 pounds.
It is high time for an example from everyday life. Let’s calculate 93.14 kg gold in pounds. So 93.14 kg equal to how many lbs? As in the previous example - multiply 93.14 by 2.20462262. It is 205.3385508268. So equivalent of 93.14 kilograms to pounds, if it comes to gold, is 205.3385508268.
In this example it is also possible to round off the result. This is the result after rounding off, this time to one decimal place - 93.14 kilogram 204.908 pounds.
Now we can move on to examples calculated using short formula.
#### How many 93.14 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 204.908 the result in pounds
So 93.14 kg equal to how much lbs? As in the previous example you need to multiply number of kilogram, this time 93.14, by 2.2. Let’s see: 93.14 * 2.2 = 204.908. So 93.14 kilogram is equal 2.2 pounds.
Let’s make another calculation with use of shorer formula. Now convert something from everyday life, for example, 93.14 kg to lbs weight of strawberries.
So let’s calculate - 93.14 kilogram of strawberries * 2.2 = 204.908 pounds of strawberries. So 93.14 kg to pound mass is 204.908.
If you know how much is 93.14 kilogram weight in pounds and are able to convert it with use of two different versions of a formula, we can move on. Now we want to show you these outcomes in tables.
#### Convert 93.14 kilogram to pounds
We realize that outcomes presented in charts are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in tables for your convenience. Thanks to this you can easily compare 93.14 kg equivalent to lbs results.
Start with a 93.14 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
93.14 205.3385508268 204.9080
And now have a look at 93.14 kg equal pound chart for the second formula:
Kilograms Pounds
93.14 204.908
As you see, after rounding off, when it comes to how much 93.14 kilogram equals pounds, the outcomes are not different. The bigger amount the more considerable difference. Please note it when you need to make bigger amount than 93.14 kilograms pounds conversion.
#### How many kilograms 93.14 pound
Now you learned how to convert 93.14 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What about 93.14 kilogram to pounds and ounces conversion?
We want to show you how you can calculate it step by step. Let’s begin. How much is 93.14 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, in this case 93.14, by 2.20462262. So 93.14 * 2.20462262 = 205.3385508268. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To convert how much 93.14 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces.
So your result is equal 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then your outcome is equal 2 pounds and 33 ounces.
As you can see, conversion 93.14 kilogram in pounds and ounces simply.
The last calculation which we are going to show you is calculation of 93.14 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To calculate it you need another formula. Before we show you this formula, let’s see:
• 93.14 kilograms meters = 7.23301385 foot pounds,
• 93.14 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to calculate 93.14 foot pounds to kilograms meters you need to multiply 93.14 by 0.13825495. It is equal 0.13825495. So 93.14 foot pounds is equal 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 93.14 foot pounds is 0.14 kilogram meters.
We hope that this calculation was as easy as 93.14 kilogram into pounds conversions.
This article is a big compendium about kilogram, pound and 93.14 kg to lbs in calculation. Due to this conversion you know 93.14 kilogram is equivalent to how many pounds.
We showed you not only how to do a calculation 93.14 kilogram to metric pounds but also two another calculations - to know how many 93.14 kg in pounds and ounces and how many 93.14 foot pounds to kilograms meters.
We showed you also other way to make 93.14 kilogram how many pounds calculations, this is with use of 93.14 kg en pound calculator. It will be the best choice for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you are able to make 93.14 kilogram equal to how many pounds conversion - on your own or with use of our 93.14 kgs to pounds converter.
So what are you waiting for? Calculate 93.14 kilogram mass to pounds in the best way for you.
Do you need to do other than 93.14 kilogram as pounds calculation? For instance, for 15 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so simply as for 93.14 kilogram equal many pounds.
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
93.14 kg to lbs - 93.14 kilograms to pounds - Mass and Weight Conversion
## How to convert 93.14 kilograms (kg) to Pounds (lbs)
A POWERFUL TOOL for converting kilograms to pounds (kg to pounds). How to convert 93.14 kilograms (kg) to Pounds (lbs). How Heavy Is 93.14 Kilograms in Pounds? How do I convert kg to lbs? How Many Kilograms in a Pound?
### How do I convert kg to lbs?
##### Read more related articles:
93.01 kg to lbs = 205.052 93.02 kg to lbs = 205.074 93.03 kg to lbs = 205.096 93.04 kg to lbs = 205.118 93.05 kg to lbs = 205.14 93.06 kg to lbs = 205.162 93.07 kg to lbs = 205.184 93.08 kg to lbs = 205.206 93.09 kg to lbs = 205.228 93.1 kg to lbs = 205.25 93.11 kg to lbs = 205.272 93.12 kg to lbs = 205.294 93.13 kg to lbs = 205.316 93.14 kg to lbs = 205.339 93.15 kg to lbs = 205.361 93.16 kg to lbs = 205.383 93.17 kg to lbs = 205.405 93.18 kg to lbs = 205.427 93.19 kg to lbs = 205.449 93.2 kg to lbs = 205.471 93.21 kg to lbs = 205.493 93.22 kg to lbs = 205.515 93.23 kg to lbs = 205.537 93.24 kg to lbs = 205.559 93.25 kg to lbs = 205.581
93.26 kg to lbs = 205.603 93.27 kg to lbs = 205.625 93.28 kg to lbs = 205.647 93.29 kg to lbs = 205.669 93.3 kg to lbs = 205.691 93.31 kg to lbs = 205.713 93.32 kg to lbs = 205.735 93.33 kg to lbs = 205.757 93.34 kg to lbs = 205.779 93.35 kg to lbs = 205.802 93.36 kg to lbs = 205.824 93.37 kg to lbs = 205.846 93.38 kg to lbs = 205.868 93.39 kg to lbs = 205.89 93.4 kg to lbs = 205.912 93.41 kg to lbs = 205.934 93.42 kg to lbs = 205.956 93.43 kg to lbs = 205.978 93.44 kg to lbs = 206 93.45 kg to lbs = 206.022 93.46 kg to lbs = 206.044 93.47 kg to lbs = 206.066 93.48 kg to lbs = 206.088 93.49 kg to lbs = 206.11 93.5 kg to lbs = 206.132
93.51 kg to lbs = 206.154 93.52 kg to lbs = 206.176 93.53 kg to lbs = 206.198 93.54 kg to lbs = 206.22 93.55 kg to lbs = 206.242 93.56 kg to lbs = 206.264 93.57 kg to lbs = 206.287 93.58 kg to lbs = 206.309 93.59 kg to lbs = 206.331 93.6 kg to lbs = 206.353 93.61 kg to lbs = 206.375 93.62 kg to lbs = 206.397 93.63 kg to lbs = 206.419 93.64 kg to lbs = 206.441 93.65 kg to lbs = 206.463 93.66 kg to lbs = 206.485 93.67 kg to lbs = 206.507 93.68 kg to lbs = 206.529 93.69 kg to lbs = 206.551 93.7 kg to lbs = 206.573 93.71 kg to lbs = 206.595 93.72 kg to lbs = 206.617 93.73 kg to lbs = 206.639 93.74 kg to lbs = 206.661 93.75 kg to lbs = 206.683
93.76 kg to lbs = 206.705 93.77 kg to lbs = 206.727 93.78 kg to lbs = 206.75 93.79 kg to lbs = 206.772 93.8 kg to lbs = 206.794 93.81 kg to lbs = 206.816 93.82 kg to lbs = 206.838 93.83 kg to lbs = 206.86 93.84 kg to lbs = 206.882 93.85 kg to lbs = 206.904 93.86 kg to lbs = 206.926 93.87 kg to lbs = 206.948 93.88 kg to lbs = 206.97 93.89 kg to lbs = 206.992 93.9 kg to lbs = 207.014 93.91 kg to lbs = 207.036 93.92 kg to lbs = 207.058 93.93 kg to lbs = 207.08 93.94 kg to lbs = 207.102 93.95 kg to lbs = 207.124 93.96 kg to lbs = 207.146 93.97 kg to lbs = 207.168 93.98 kg to lbs = 207.19 93.99 kg to lbs = 207.212 94 kg to lbs = 207.235 | 4,662 | 15,813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-50 | latest | en | 0.939032 |
http://oeis.org/A141367 | 1,619,195,787,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039596883.98/warc/CC-MAIN-20210423161713-20210423191713-00607.warc.gz | 70,554,355 | 3,831 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A141367 G.f.: Sum_{n>=0} arcsinh(4^n*x)^n/n!, a power series in x having only integer coefficients. 2
1, 4, 128, 43680, 178946048, 9382409745280, 6558834518571089920, 62879485860387254833099776, 8439542720341303996200869198561280, 16110026846830031883594370688522189192189952 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS G. C. Greubel, Table of n, a(n) for n = 0..40 FORMULA a(n) = [x^n] [ sqrt(1+x^2) + x ]^(4^n). More generally, the following coefficient of x^n in the series: [x^n] Sum_{n>=0} arcsinh(q^n*x)^n/n! = [x^n] [ sqrt(1+x^2) + x ]^(q^n) is an integer for any even integer q. EXAMPLE G.f.: A(x) = 1 + 4*x + 128*x^2 + 43680*x^3 + 178946048*x^4 + ... MATHEMATICA Table[SeriesCoefficient[(Sqrt[1 + x^2] + x)^(4^n), {x, 0, n}], {n, 0, 25}] (* G. C. Greubel, Apr 15 2017 *) PROG (PARI) {a(n)=polcoeff(sum(k=0, n, asinh(4^k*x +x*O(x^n))^k/k!), n)} (PARI) {a(n)=polcoeff((x+sqrt(1+x^2 +x*O(x^n)))^(4^n), n)} CROSSREFS Cf. A136647, A141368. Sequence in context: A130318 A000318 A229385 * A141368 A146555 A305568 Adjacent sequences: A141364 A141365 A141366 * A141368 A141369 A141370 KEYWORD nonn AUTHOR Paul D. Hanna, Jul 02 2008 STATUS approved
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Last modified April 23 12:15 EDT 2021. Contains 343204 sequences. (Running on oeis4.) | 613 | 1,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-17 | latest | en | 0.548138 |
https://discourse.julialang.org/t/how-to-declare-parametric-with-return-type-of-a-function/40288 | 1,716,616,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00587.warc.gz | 174,032,756 | 5,265 | # How to declare parametric with return type of a function?
I have a method and I want to declare it with parametric and with return type.
What is the syntax?
``````f(a::T)::Bool where {T <: Number} = a == 1
``````
The error message is:
``````ERROR: UndefVarError: T not defined
Stacktrace:
[1] top-level scope at REPL[4]:1
``````
I know that there is another way to resolve this problem, but it’s a simplified version of my real problem. It is: I have a method declared with a return type and now I want to put a parametric argument.
The real question is: there is a syntax to declare both parametric argument and return type?
Thanks.
It’s just a slightly annoying parsing issue. Your definition is getting parsed as `f(a::T)::(Bool where {T <: Number})`, which isn’t what you want. You can do:
``````julia> (f(a::T)::Bool) where {T <: Number} = a == 1
f (generic function with 1 method)
julia> f(1.0)
true
``````
And by the way, you can see for yourself how the expression is parsed by quoting it, in case you’re ever curious:
``````julia> :(f(a::T)::Bool where {T <: Number} = a == 1)
:(f(a::T)::(Bool where T <: Number) = begin
#= REPL[3]:1 =#
a == 1
end)
``````
note how the `(Bool where T <: Number)` is grouped together.
5 Likes
Alternatively, you might find that the long-form function syntax is more readable here (and doesn’t have this parsing issue):
``````julia> function f(a::T)::Bool where {T <: Number}
a == 1
end
f (generic function with 1 method)
julia> f(1)
true
``````
3 Likes
It’s it. Thanks.
1 Like | 444 | 1,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.827698 |
http://math.stackexchange.com/questions/161770/differential-equations-and-family-of-function-solutions | 1,448,425,339,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398444228.5/warc/CC-MAIN-20151124205404-00193-ip-10-71-132-137.ec2.internal.warc.gz | 155,894,730 | 21,340 | Differential equations and family of function solutions
I can't follow what Stewart is doing in his book. I can easily follow his work but his conclusion doesn't make any sense to me.
"Show that ever member of the family of functions
$$y = \frac{1+ce^t}{1 - ce^t}$$ is a solution of the differential equation $$y' = \frac{1}{2} (y^2 - 1)$$
He starts by differentiating the right side of the first term and then just setting that equal to the $y'$ from the question.
I do not see how these are equal or what this means or what is going on at all really, and the book doens't feel the need to explain this anyways, so maybe it isn't important and maybe I just need to memorize that a solution is just the differential. But I don't see hwy this is important or how this helps anything.
For his final answer he gets $$y' = \frac{2ce^t}{(1-ce^t)^2}$$
-
Plug in the value of $y$ into $\frac{1}{2}(y^2-1)$. Verify that you get $\frac{2ce^t}{(1-ce^t)^2}$.
In other words, what you are doing is plugging in the proffered solutions into the equation and verifying that you get an equality.
It's exactly as if you had been asked to verify that $x=4$ is a solution to the equation $$x^2 - 5x + 6 = 2x^2 - 8x + 2.$$ All you need to do is plug in $x=4$ into the left hand side and compute, $4^2-5(4) + 6 = 16-20+6 = 2$; then plug it into the right hand side and compute, $2(4)^2 - 8(4) + 2 = 32 - 32 + 2 = 2$. And then say: "yes, it's a solution, because it satisfies the equality." You don't have to solve anything, just plug and verify.
-
Maybe I do not understand this correctly, but this is not the method that the book used is it? Also how is an equality verified? – user138246 Jun 22 '12 at 20:16
@Jordan: An equality is verified by evaluating each side of the equality and checking that the answer is the same. Exactly like I did in the example I gave you. As to how exactly the book does it, since as usual you did not bother to say which of the many books Stewart has written you are looking at, or where in that 700+ page book you are looking at, I can only try to guess (which is rather annoying). In my copy of the 4th edition of "Calculus, Early Transcendentals", this is Example 1 in Chapter 9, Section 1, page 584. (cont) – Arturo Magidin Jun 22 '12 at 20:27
@Jordan: What he does is exactly the method I say: first, he computes $y'$. Then he plugs in the value of $y$ into the right hand side, and simplifies. After he is done simplifying, he simply observes that what he obtained on the right hand side of $$y' = \frac{1}{2}(y^2-1)$$ by plugging in this value of $y$ is exactly the same thing he obtained by plugging $y$ into the left hand side of that equation (namely, computing $y'$). Since both sides give the same thing, then the given $y$ satisfies the equation. Exactly like I said. – Arturo Magidin Jun 22 '12 at 20:29
I have 7e, its on pg 583 Example 1. I did not specify that because I typed up what he did basically. I am just confused because you never used the quotient rule like he did. – user138246 Jun 22 '12 at 20:34
@Jordan. In a differential equation, the unknown is a function. So what you plug into the equation is a function. "Evaluating $y'$ at $y=\frac{1+ce^t}{1-ce^t}$", or "plugging in $y=\frac{1+ce^t}{1-ce^t}$ into $y'$" means evaluating $$\left(\frac{1+ce^t}{1-ce^t}\right)^'.$$ That is, finding the derivative of $y$ (the first thing Stewart did). "Plugging in $y=\frac{1+ce^t}{1-ce^t}$ into $\frac{1}{2}(y^2-1)$" means evaluating $$\frac{1}{2}\left(\left(\frac{1+ce^t}{1-ce^t}\right)^2-1\right)$$which is the second thing he did. – Arturo Magidin Jun 22 '12 at 20:41
From $y=\frac{1+ce^t}{1-ce^t}$, we get : $e^t=\frac{y-1}{c(y+1)}$ and derive this , you get : $e^t=\frac{2y'}{c(y+1)^2}$ then : $$e^t=\frac{y-1}{c(y+1)}=\frac{2y'}{c(y+1)^2}$$ who gives desired result
-
To show that a function is a solution to a differential equation, one must "plug it in" to the given equation. In this case, we have $$y=\frac{1+ce^t}{1-ce^t},$$ and must confirm that $$y'=\frac{1}{2}(y^2-1).$$
Indeed, on the one hand,
$\begin{eqnarray*} \frac{1}{2}(y^2-1) & = & \frac{1}{2}\left[\frac{1+2ce^t+c^2e^{2t}}{1-2ce^t+c^2e^{2t}}-1\right]\\ & = & \frac{1}{2}\left[\frac{1+2ce^t+c^2e^{2t}}{1-2ce^t+c^2e^{2t}}-\frac{1-2yce^t+c^2e^{2t}}{1-2ce^t+c^2e^{2t}}\right]\\ & = & \frac{2ce^t}{1-2ce^t+c^2e^{2t}}\\ & = & \frac{2ce^t}{(1-ce^t)^2}, \end{eqnarray*}$
and on the other, we have by quotient rule that
$\begin{eqnarray*} y' & = & \frac{dy}{dt}\\ & = & \frac{ce^t(1-ce^t)-(-ce^t)(1+ce^t)}{(1-ce^t)^2}\\ & = & \frac{ce^t-c^2e^{2t}+ce^t+c^2e^{2t}}{(1-ce^t)^2}\\ & = & \frac{2ce^t}{(1-ce^t)^2}. \end{eqnarray*}$
For contrast, suppose you'd been asked to verify whether $y=\sin t$ was a solution to $y'=\frac{1}{2}(y^2-1).$ Well, $y'=\cos t$, but $$\frac{1}{2}(y^2-1)=-\frac{1}{2}\cos^2t$$ by Pythagorean identity. But when $t=0$, we see that $y'=\cos 0=1$ and $\frac{1}{2}(y^2-1)= -\frac{1}{2}\cos^20=-\frac{1}{2}$, so if $y=\sin t$, then $y'\neq\frac{1}{2}(y^2-1)$. Thus, $y=\sin t$ is not a solution to the differential equation given.
-
How is that final quotient that you have equal to the differential equation? – user138246 Jun 22 '12 at 20:20
I'm not sure I understand what you mean. If you want to show that $a=b$, one way to do so is to show that $a=c$ and $b=c$. That's what I've done here. Or do you mean how did I get to the final quotient? I will add a step in case that's what you meant. Also, I'll add a further remark. – Cameron Buie Jun 22 '12 at 20:33
I understand how you got what you did, I just thought it was suppose to be an a = b sort of situation. – user138246 Jun 22 '12 at 20:37
Yes, it is. Let me know if my expanded answer clarifies things at all. – Cameron Buie Jun 22 '12 at 20:48
$\sin^2t+\cos^2t=1$. – Cameron Buie Jun 22 '12 at 20:52
To show that $y= \frac{1+ce^t}{1 - ce^t}$ is a solution of $y' = \frac{1}{2}(y^2-1)$, you just have to check that if you replace $y= \frac{1+ce^t}{1 - ce^t}$ into $y' = \frac{1}{2}(y^2-1)$ you get a valid equality.
- | 2,109 | 6,014 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2015-48 | longest | en | 0.978249 |
https://forums.developer.nvidia.com/t/higher-level-questions-of-optix-and-medical-imaging-applications/216537 | 1,656,632,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103915196.47/warc/CC-MAIN-20220630213820-20220701003820-00321.warc.gz | 310,980,180 | 6,460 | # Higher Level Questions of OptiX and Medical Imaging Applications
Hello there!
I’ve been orienting myself with OptiX, debating whether it’s appropriate to explore using in my work. I am asking if the following application would see a substantial increase using OptiX vs CUDA.
Presently I use CUDA to do the following program, which is very common in medical physics and radiation transport methods. It’s to calculate a line integral along a ray, where the summed values are sampled from a 3D Voxellized dataset
With parameter N_Steps, a Voxelized 3D data set, and N Rays
For each Ray (R_i)
integral[ray_i] = 0
For i to N_Steps
Calculate position = ray_i_start + (i / N_Steps) * (ray_i_end - ray_i_start)
value = sampling_fuction(voxellized_dataset, position)
integral[ray_i] += value * distance(ray_i_start, ray_i_end) / N_Steps
return integral[ray_i]
Where the sampling function can be trilinear or tricubic interpolation. Presently I do this using CUDA, and either perform the trilinear interpolation manually, or use a 3D texture for the voxellized data and perform a texture look up, tex3D(data, position). This is calculated using one thread for each ray.
Would OptiX lend itself to a substantial speed improvement over a pure CUDA implementation? I have seen work where the 3D data is a triangular mesh, and there is a definite speed up, as I understand OptiX is specifically designed for such applications. However for volumetric voxellized 3D data, I haven’t found much to indicate whether OptiX would be worth pursuing.
I would be very interested in learning whether a speed increase would be expected, and what hardware features OptiX provides vs CUDA that give that speed increase. For example, the texture lookup functions in CUDA typically give me 2-3x speed improvement over doing the trilinear interpolation manually (with some precision cost).
Hi anovack and thanks for your interest in the OptiX API. Though the algorithm you describe can quite easily be implemented in OptiX, the regular sampling of volume data along a ray does by itself not benefit from using OptiX. The OptiX volume sample in the SDK shows how integrating a volume density along a ray can be implemented using OptiX and NanoVDB (for the volume representation). The sampling is done using NanoVDB’s CUDA based 3D hierarchical DDA algorithm. The sample demonstrates how “empty space”, i.e. zero-density regions can be skipped using OptiX’ hardware accelerated ray-bounding-box intersection tests, but the performance gains of this technique are rather small for typical data.
1 Like
Thank you for your comment @fjargstorff ! I missed that example and will check it out.
Good to know it’s possible to do volumetric sampling even if there are no dramatic performance gains. | 624 | 2,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.884374 |
www.pebhmong.com | 1,679,901,371,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00529.warc.gz | 76,828,495 | 11,934 | ### Author Topic: I'm a Genius. Are you? (Read 8617 times)
0 Members and 1 Guest are viewing this topic.
#### lexicon
• Jr. Poster
• Posts: 3840
• Respect: +196
##### Re: I'm a Genius. Are you?
« Reply #15 on: September 08, 2016, 03:10:15 PM »
123 + 0?
Like this post: 0
#### bulbasaur
• Guest
##### Re: I'm a Genius. Are you?
« Reply #16 on: September 08, 2016, 03:22:41 PM »
More examples...
2+1=103
9+7=216
8+5=313
Just because you figured out Part 1, that doesn't mean that you actually answered Part 2. I don't want to tell too much because that would be giving away the answer.
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#### bulbasaur
• Guest
##### Re: I'm a Genius. Are you?
« Reply #17 on: September 08, 2016, 03:25:46 PM »
More examples...
22+10=1232
15+13=228
17+12=529
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#### bulbasaur
• Guest
##### Re: I'm a Genius. Are you?
« Reply #18 on: September 08, 2016, 03:33:07 PM »
The way this problem was originally proposed to me was starting with the question in Part 2 first. That makes the problem a little more challenging.
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#### bulbasaur
• Guest
##### Re: I'm a Genius. Are you?
« Reply #19 on: September 08, 2016, 04:43:40 PM »
It is not impressive to simply know the answer. It is more impressive to be able to solve it. Showing a completed 10,000 piece puzzle is not impressive if you didn't put it together yourself. The same thing goes for many things in life. Showing me a great art piece is fine, but that doesn't make you impressive or a genius. The person who made the art is impressive and a genius.
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#### bulbasaur
• Guest
##### Re: I'm a Genius. Are you?
« Reply #20 on: September 08, 2016, 04:47:49 PM »
Or, think of it this way....
You have a math book. The book comes with an answer guide. Just because you have all the answers, that does not make you good at the math that is in the book.
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#### bulbasaur
• Guest
##### Re: I'm a Genius. Are you?
« Reply #21 on: September 08, 2016, 05:40:51 PM »
I remember those several times where you couldn't think logically.
Dont you remember when we went to school together and you had to do a math problem on the board?
https://youtu.be/MTghfMCU1rY
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#### bulbasaur
• Guest
##### Re: I'm a Genius. Are you?
« Reply #22 on: September 08, 2016, 10:53:42 PM »
Hint: 12 and 11 is just one possible answer. There is more than one answer. You're supposed to find all possible answers.
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#### lexicon
• Jr. Poster
• Posts: 3840
• Respect: +196
##### Re: I'm a Genius. Are you?
« Reply #23 on: September 09, 2016, 08:03:11 AM »
Hint: 12 and 11 is just one possible answer. There is more than one answer. You're supposed to find all possible answers.
I figured as much. But all I can think of now is how many points will my RB's get me this Sunday?
I'll revisit this later today.
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#### lexicon
• Jr. Poster
• Posts: 3840
• Respect: +196
##### Re: I'm a Genius. Are you?
« Reply #24 on: September 09, 2016, 11:30:04 AM »
I think I have it.
In X + XX = 123, 123 dictates the value X and XX in only part A or SUBTRACTION. The value of X and XX remain as they are for part B ADDITION.
Since 123 = 3 digits = hundreds. Should 2 digits = tens?
(X*100)+(X*100)=123
1200+1100=123
1200-1100=100 SUBTRACTION
Overall 100+23=123
Examples;
2+1=103
103= 3 digits = hundreds
200-100=100
2+1=3
100+3=103
SO...
5+1=46
46= 2 digits= tens
50-10=40
5+1=6
40+6=46
« Last Edit: September 09, 2016, 12:17:50 PM by lexicon »
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#### Believe_N_Me
• Elite Poster
• Posts: 13518
• Respect: +444
##### Re: I'm a Genius. Are you?
« Reply #25 on: September 05, 2022, 04:30:58 AM »
So a friend at work posed this question to us. Apparently you need a high IQ to figure this out. I got it pretty quick, and some people were impressed. That being said, I don't think the problem requires genius intellect. They probably would have figured it out as well if given enough time. Anyways, here is the problem...
7+6=113
8+5=313
9+2=711
What's the pattern? Can you figure it out?
7+6 = 113 (1 = 7-6, 13 = 7+6)
8+5 = 313 (3 = 8-5, 13 = 8+5)
9+2 = 711 (7 = 9-2, 11 = 9+2)
continuing pattern:
5+2 = 37
6+4 = 210
7+6 = 113
8+8 = 016
Like this post: 0 | 1,408 | 4,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.887704 |
https://www.scribd.com/document/337817065/Written-Assignment | 1,563,485,987,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525829.33/warc/CC-MAIN-20190718211312-20190718233312-00466.warc.gz | 826,394,548 | 64,839 | You are on page 1of 18
# WRITTEN MODULE: 3
Question 1
I=
V
R
200
20
10 A
P=I 2 R
V2
R
( 10 ) ( 20 )=
(200)
20
2 KW =2 KW
Question 2
## Three phase voltage is achieved by connecting in 3 wires called
lines, therefore carry 3-phase (3-) alternating voltage.
In single phase each voltage line is taken with respect to
neutral because of this single phase (1-) voltage will always
be less than line voltage by 3 .
Single phase AC has advantage that it only requires 2 wires, it
is used when needed to take high voltage and plug something
## delicate like the computer, and for that to happen 3- power
has to be converted to 1- outlets.
3- power work because 3- conductors (lines) are carrying
currents of the same frequency, but reach their maximum peak
values at different times i.e. when the first conductor is at its
peak output, the other or second conductor is
2
1
3
cycle behind
## the cycle behind this means at
any given time one of the 3 conductors will be at its peak out
and current will flow consistently. 3- system is better than 1-
because it produce rotating magnetic field for 3- AC motor.
3- electric power is the most economic method of generation
and transmission of electricity, this is accomplished by placing
3 coils 120 apart and be rotated in the changing magnetic
field.
For transmission, current is kept low and voltage high in
distribution side voltage is kept low and current high.
Question 3
Apparent Power(s)
Is the product of voltage and current if the phase difference
between the voltage and current is ignored.
In AC circuits both dissipated and absorbed or returned power
is called apparent power, also the combination of reactive
power and true power is called apparent power, also the
combination of reactive power and true power is called
apparent power, in terms of RMS voltage and RMS current, also
their product is called apparent power. The units are VA
(Voltage*Current) where: 1VA=1V 1 A .
Formulas for apparent power.
S=VI
true power
Or
Apparent Power=
KVA= KW 2+ KVAR 2
## Active Power (P)
Alternative words are real power, actual power, true power or
watt-full power, in DC circuit, power supply is simply a product
of voltage across the load and the current flowing through it
i.e. P=VI, because in DC circuits there is no concepts of phase
angle between current and voltage in other words there is no
power factor in DC circuits, but when it comes to sinusoidal or
AC circuit situation becomes more complex because of the
phase difference between current and voltage, therefore the
average value of power, real Power=VIcos, is in fact one that
supplied to the load, in AC circuit when the circuit is pure
resistive, then the same formula used for power in DC as P=VI.
Formulas for real power.
P=VI
(in DC circuits)
P = VICos (in 1- AC circuits)
P= 3 VLIL Cos
(in 3- AC circuits)
P=3 V I cos
ph ph
P= S2Q2 P= VA 2VAR 2
Units for active power is (W/KW) also this can be written in the
form of words like apparent power. The power that stars
bounce back and forth between the source and the load is
known as reactive power, also the power that merely absorbed
and returned in load due to its reactive properties is referred to
as reactive power.
Reactive power is given by
P = VISin which can be (+ve) for inductive and negative for
Q = VISin
apparent power
Reactive Power=
VAR= VA 2P2
KVAR= K VA 2KW 2
## All these quantities are trigonometric related to each other,
graph below elaborates more, this triangle graph is called
power triangle.
## Pure inductive power
In pure inductive circuit, active power is zero because in
inductive circuit, current lagging from voltage by 90 from
current i.e. the phase difference between current and voltage
is 90., so if the current and voltage are 90 out of phase, then
power (P) will be zero because P = VICos , so if angle between
current and voltage is 90, Q=90 then P = VICos 90 =0
Question 4
(a)
R=50
XL =25
Z in complex form
Z= R+ jXL
= 50+j25
(b)
Vector Form
Question 5
Z1= 5+j20
Z2= 15+j25
15
R1
R2
V
AC
ZT=Z1+Z2
= 5+j20+15+j25
= 20+j45
If current in vector form is I=1 0
Hence:
V=IZ
= I (20+j45)
= (1 0 ) (49.2 66
=49.2 66
Question 6
100m
L1
L3
100m
L2
SIMetrix
AC
LT =
L1L2
+L
L1 + L2 3
LT =
100100
+250
100+100
AC
V
300 mH
100u
L
## By making current I=1 0 , V= IZ, first find inductive
reactance
XL = 2 F L
= 2 * 50* 0.3
= 94.2
Therefore: Inductive impedance
Z = 94.2 90
Therefore V = ZL *I or I*Z
= 1 0 * 94.2 90
= 94.2 90 V
So here voltage leads current by 90, it doesnt matter which
one you are defining as zero degrees between current and
current by 90.
Question 7
220
670
R1
R2
V
SIMetrix
## Current flowing through the circuit is the same.
So: IT = IR1 = VR1
RR1
18
= 230
= 0.078A
RT = R 1 + R 2
230+670
900
2
Hence: Power, PT = I RT
( 0.078 )2 ( 900 )
5.5W
Question 8
RMS is equal to peak voltage times 0.707 or peak voltage
divide by 2 .
So: RMS = Peak value * 0.707
or =
Peak value
2
Therefore:
Peak voltage of 240 VAC
339.5 V
= 240 2
Or
240
0.707
339.5 V
Peak current =
V Peak
R
339.5
100
3.4 A
Question 9
(a)
Inductive reactance XL =
2
6.28
X = 2 FL 2 500.02
(b)
In pure inductive circuit the power is zero because,
the voltage is leading the current by 90 i.e. the phase
difference between current and voltage is 90, so since
power, P = VICos
P = VICos VICos 90 0
(c)
The power factor of the circuit is going to be zero,
because my circuit is pure inductive- capacitive and since
what applies in pure capacitive circuit.
(d)
Capacitive reactance, XC =
=
159.2
1
L
1
2 FC
1
2 5020106
Question 10
XL = 2 FL
2 500.05
2 5050103
15.71
## (a) The angle between the currents is 120 since the
voltages are also 120 apart from each other, but equal
in magnitude.
(b) Peak Voltage = RMS Voltage * 2 1002 141..4 V
OR
(c)
100
=141. 4 V
0.707
X L=2 FL
Question 11
2 500.05
15.71
## When a resistor is placed in series with the power source and a
capacitor is placed in parallel to the same power source, as
shown on the circuit diagram below, this type of circuit forms a
low pass filter.
R
+
Vin
## This type of filter is called RC low pass filter again when an
inductor is placed in series with the power source and the
resistor is placed in parallel to that same power source, this
type of circuit forms a low pass filter, circuit below.
L
+
Vin
FCUT OFF=
R
2 L
FCUT OFF=
R
2 RC
## The disadvantage of RC low pass filter is that at high frequency
signal, the impedance of the circuit decreases because of the
capacitor that offer a very low impedance to a high frequency
signal.
In RL low pass filter inductor resist sudden changes in current,
because of this inductors will filter out certain high frequency
signals (depending on the inductance value) and let the lower
frequency pass through unchanged, so due to the inductors
## that offers a very high resistance at very high frequency
signals, this make the behaviour of this circuit to be more
reliable than RC low pass filter therefore is the one that I
suggest to be used because of its features.
On below circuit, all low frequencies will be blocked and only
high frequencies will be allowed to go to output.
Vin
Vout
Passive high pass filter
f=
1
2 T
1
2 RC
Question 12
To achieve low frequency RC low pass filter can be used.
Vin
Vout
C
A low pass filter passes all frequencies from zero to the cut off
frequency and blocks all frequencies above the cut off
frequency.
Passive low pass filter, resistor is in series with the output and
the capacitor is in parallel/ shunt configuration, at low
frequencies no current will flow through the capacitor because
of the high impedance of the capacitor, so current pass through
to the resistor to the load as frequency supply increases, the
## impedance capacitor decreases at beyond high frequencies the
resistance offer by capacitor will be zero, so the capacitor at
this stage will become a short circuit and all the frequency flow
through it back to the source, hence this type of filter is good
for low pass frequencies.
Question 13
Vin
I
Vr1
R1
200
Vout
V =I (R 1+ R 2)
R2
Vr2 800
0
V R 2=
V R2
R 1+R2
100800
200+ 800
80 V
Question 14
Rt
R1
R2
R3
R4
100
100
100
100
1
1 1 1 1
= + + +
R T R 1 R 2 R3 R4
1
1
1
1
+
+
+
100 100 100 100
0.04
RT =
1
=25
0.04
Question 15
KCL: The sum of the currents entering the node is equal to the
sum of currents leaving the node.
( I a+ I b ) ( I c + I d )=I e
## ( 24+35 )( 12+18 )=I e
I e= ( I a + I b ) ( I c + I d )
29 A
I a+ I b + I e =I c + I d
5930=I e
I e=29 A
Question 16
Ia
XMM1
Ib
R1
R3
10
30
R2
R4 =
Vs
50 V
20
## Using Wheatstone bridge:
Wheatstone bridge is the bridge for measuring ELECTRICAL
RESISTANCE that consists of a conductor joining two branches
of a circuit. By having a voltmeter placed as shown above this
means current flowing through that portion supposed to be
zero, if not R3 must be adjusted until no current flowing
through voltmeter, hence the circuit is balanced (Ia= Ib). This is
achieved when:
VR 2 VR 4
=
VR 1 VR 3
## Here R3 is a variable resistor in order to balance the circuit.
To find R4 using Wheatstone bridge:
VR 2 VR 4
=
VR 1 VR 3
Ia . R 2 Ib . R 4
=
Ia . R 1 Ib . R3
R2 R4
=
R1 R3
R 4=R 3.
R2
R1
(30)(20)
10
= 60
Question 17
CT
C1
C2
C3
47F
47F
47F
## Capacitors connected in parallel are the same as resistors
connected in series, therefore:
CT = C1 + C2 + C3
= 47uF +47uF + 47uF
= 141uF
Question 18
C1
C2
3mF
3mF
C5
C3
CT
4.5mF
6mF
C4
6mF
Question 19
R1
5k
Vs
R2
45 V
10k
R3
7.5k
R1
10V
2A
5K
E
I
R
VR1=IR1 .R1
RT = R1 +R2 + R3
= 5+10+7.5
= 22.5k
Therefore:
VR1=
R1
(VT )
RT
R2
20V
2A
10k
R3
15V
2A
7.5k
Total
45
2
22.5k
Volts
Amps
Omh
s
5
4 5
22.5
= 10V
Similar to VR2 and VR3
VR 2=
10
4 5
22.5
=20V
VR3=
7.5
45
22.5
=15V
Question 20
6V
12V
V1
V2
Va
30 V
I2
Vb
4V
V3
## Kirchhoffs voltage law state that:
The sum of all voltages drops around the loop is equal to zero.
Hence: Va - V1 + V2 Vb V3 = 0 or
Va + V2 = Vb + V1 + V3
Therefore Vb = Va + V2 (-V1 V3)
= 30 +12 (-6 - 4)
=52V | 3,240 | 10,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-30 | latest | en | 0.943165 |
https://math.stackexchange.com/questions/1191377/definition-of-a-minimal-set | 1,576,351,371,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541288287.53/warc/CC-MAIN-20191214174719-20191214202719-00305.warc.gz | 459,851,378 | 32,568 | # Definition of a Minimal Set
A few times while studying math I have encountered the notion of a "minimal set". For example, given some set of subsets, what is the "minimal" sigma algebra generated by that set of subsets? Or, in the example I am currently studying: one of the ZFC axioms ensures the existence of an inductive set. Therefore, we define the natural numbers to be the minimal inductive set. I am trying to think of some other examples, but I can't remember off the top of my head. Feel free to comment if you think of some other examples of minimal sets.
Here is my question. I have seen the "minimal" set defined different ways, and I am not sure which statement is definition and which is an implication of the definition.
The first definition I have seen is to call a set $M$ satisfying some property minimal if, for every other set $A$ satisfying the same property as the set $M$, we have $M \subseteq A$: \begin{equation*} M \textrm{ is the minimal set with property } P \iff (\forall A \textrm{ satisfying property P }) M \subseteq A \end{equation*}
The second definition I have seen is to define the minimal set $M$ satisfying some property to be the intersection of all other sets satisfying the same property: \begin{equation*} M \textrm{ is the minimal set in some class } C \iff M=\bigcap C \end{equation*}
The second definition seems more concise, however in the example I am studying right now (defining the set of natural numbers to be the minimal inductive set), I don't know if such a set $C$ (the set of all inductive sets) exists, so I am not even sure if the right hand side of definition 2 even makes any formal logical sense.
• The two are equivalent (assuming closure under intersections), so you can treat either as the definition as you prefer. – Qudit Mar 15 '15 at 20:45
• Thanks for your comment. I will try to prove that these two are equivalent. Typically, however, which statement is usually treated as the primary definition? – Mathemanic Mar 15 '15 at 20:47
• The first definition is generalizable to orders other than $\subseteq$, so it's better. – Git Gud Mar 15 '15 at 20:47
• @EthanAlvaree I'm not sure. I've definitely seen both as well. – Qudit Mar 15 '15 at 20:49
• @GitGud, I'm not sure what you mean by generalizable to orders other than $\subseteq$. Can you please explain more? – Mathemanic Mar 15 '15 at 20:50
The first definition is the more general one (and, as has been said in the comment, can be generalized to arbitrary partial orders besides $\subseteq$). The second definition is not always correct. However, if the class $C$ is nonempty has the property that an arbitrary intersection of members of $C$ is again in $C$, then the second definition is equivalent to the first definition.
Regarding your final problem about a minimal inductive set, note that $C$ need only be a nonempty class, not necessarily a set. In case you worry that $\bigcap$ is only definied for sets, not for (proper) classes of sets: No, the definition $$\tag1\bigcap C:=\left\{\,x\mid \forall c\in C\colon x\in c\,\right\}$$ is perfectly fine and defines a set for any nonempty(!) class $C$ though admittedly $(1)$ uses class builder, not set builder notation. But let $S\in C$ be an arbitrary set and define $$\tag2\bigcap C:=\left\{\,x\in S\mid \forall c\in C\colon x\in c\,\right\},$$ then the result does not depend on the choice of $S$ (why?) and as $(2)$ is an instance of the Axiom Schema of Comprehension, this shows that $\bigcap C$ is a set. (Then finally, as $C$ is closed under arbitrary intersection, we see that $\bigcap C$ is again an element of $C$ and surely the mnimal element in the sense of the first definition; If $C$ denotes the class of inductoive sets, then the usual formulation of the Axiom of Infnity can be rephrased as simply: $C$ is not empty - which is precisely what we need)
• Yes, I was worried about $\bigcap C$ being defined in set theory when $C$ is not a set. I think I understand, but please tell me if this is right. So $\bigcap C$ in both definitions $(1)$ and $(2)$ are equal to each other? But definition $(2)$ is preferred because it represents how we actually use the ZFC axioms to ensure existence of the minimal inductive set, correct? For instance, your set $S$ is the set whose existence is provided by the axiom of infinity, correct? I think I understand, but please let me know if I got this right. Thanks for your great answer! – Mathemanic Mar 15 '15 at 21:15
Given a set $S$ of sets you can consider the containment relation between sets. This relation is what is called a partial order, and makes $S$ into a poset. A minimal element $s\in S$ is an element that is not properly contained by another element in $S$. Of course when $S$ is closed under arbitrary intersections the minimal element of $S$ is the intersection of all the elements of $S$. | 1,238 | 4,852 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-51 | latest | en | 0.929152 |
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# Jacob Berry
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Darcy Cordell on 28 Sep 2022
Commented: James Tursa on 28 Sep 2022
I have a 2x2xN matrix, A, and a 2xN matrix, X. I want to multiply each Nth 2x2 matrix in A by the Nth 2x1 matrix in X. I can do this with:
for i = 1:N
B(:,i) = A(:,:,i)*X(:,i);
end
However, this is very slow, especially in my case where N is large.
Is there a way to do matrix multiplication "along a specified dimension" without a for loop?
Apologies if this is a duplicate question, I looked and could not find anything. Most of the other answers I found involve element-wise multiplication along a specified dimension, which is different than what I am doing.
Matt J on 28 Sep 2022
B=pagemtimes(A,X)
2 CommentsShow NoneHide None
Darcy Cordell on 28 Sep 2022
Exactly what I was looking for! The speed up is impressive. I would just add that in order for this to work, it is necessary to permute X to be 2x1xN using permute(X,[1,3,2]). Thanks!
James Tursa on 28 Sep 2022
Or reshape(X,2,1,[]); | 308 | 1,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.920903 |
https://www.mersenneforum.org/showthread.php?s=ab510eb1993af63325754f8993d3dc72&t=27144&goto=nextoldest | 1,669,831,139,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710765.76/warc/CC-MAIN-20221130160457-20221130190457-00425.warc.gz | 940,472,905 | 11,098 | mersenneforum.org (Not a) Formula for prime with 10^8 and 10^9 digits
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2021-09-19, 08:55 #1 jense 2185 Sep 2021 3 Posts (Not a) Formula for prime with 10^8 and 10^9 digits X=3(9^Y) +1 (10^X -1) ÷ 4.5 + ((10^X ) ×7)+5 While Y=108 or While Y=109 Result is prime with 100,000,000 digits or 1,000,000,000 digits ?
2021-09-19, 15:05 #2 Dobri "Καλός" May 2018 44010 Posts Is Y = 107 out of the spotlight?
2021-09-19, 15:55 #3 Dobri "Καλός" May 2018 23×5×11 Posts For Y = 1, n = 72,222,222,222,222,222,222,222,222,227 is a prime. For Y = 2, n = 72,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222, 222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222, 222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222, 222,222,222,222,222,222,222,222,227 = 251 × 5,426,180,741 × ... is composite. For Y = 3, n = 72,222,...,222,227 = 1,322,011 × ... is composite. ... Last fiddled with by Dobri on 2021-09-19 at 16:01
2021-09-19, 17:02 #4 jense 2185 Sep 2021 112 Posts X=7(9^Y)+1 ?
2021-09-19, 17:19 #5 Dobri "Καλός" May 2018 6708 Posts For Y = 1 and X=7(9^Y)+1, n = 72,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,222,227 is a prime. For Y = 2 and X=7(9^Y)+1, n = 72,222,...,222,227 = 12,391 × 173,429 × 9,639,563 × 53,804,376,883 × 72,037,926,256,979 × ... is composite. For Y = 3 and X=7(9^Y)+1, n = 72,222,...,222,227 = 1,352,773 × ... is composite. ... Last fiddled with by Dobri on 2021-09-19 at 17:24
2021-09-19, 18:05 #6
VBCurtis
"Curtis"
Feb 2005
Riverside, CA
555110 Posts
Quote:
Originally Posted by jense 2185 X=7(9^Y)+1 ?
How is it that you think you are smart enough to come up with a formula that turns out primes very often, but aren't smart enough to actually check if the (majority of) outputs are prime?
2021-09-19, 20:48 #7 jense 2185 Sep 2021 316 Posts Never claimed to be smart, but I hoped to be lucky. Also I don't have the hardware to factor the large numbers currently. Thanks
2021-09-20, 08:12 #8 kruoli "Oliver" Sep 2017 Porta Westfalica, DE 122110 Posts Everything what Dobri computed can be done on a Android phone from 2012. Yes, I tested it. So I assume you have the hardware, but you need to look up how to do these computations. Hint: Google Alpertron ECM.
2021-09-20, 09:05 #9
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
664010 Posts
Quote:
Originally Posted by jense 2185 ... I hoped to be lucky.
If a prime formula was really just that simple then there is a very strong possibility that it would have been discovered 400 years ago.
Today you could write a simple BASIC program to generate millions of formulae and hope to get "lucky" that one of them "works". I recommend you try it to see how easy it is to generate formulae. But I don't recommend you try posting all of them here and hoping others will verify/disprove all those formulae. That would be your job, to show it works, not our job to debunk endless lists of random formulae.
Last fiddled with by retina on 2021-09-20 at 09:05
2021-09-20, 13:27 #10
Dr Sardonicus
Feb 2017
Nowhere
2·11·277 Posts
Quote:
Originally Posted by jense 2185 Never claimed to be smart, but I hoped to be lucky. Also I don't have the hardware to factor the large numbers currently.
Instead of just hoping to "be lucky," it would be a good idea to at least try to check your formulas for divisibility by small primes. Your formula is
N = 2*(10^X - 1)/9 + 7*10^X + 5
This formula can be expressed more compactly as follows:
N = (65*10^X + 43)/9
1) Given a prime p, p = 2, 3, 5, 7, etc are there any values of X for which p divides N?
2) If so, characterize such X (this will be a congruence class)
3) Are there any such X of a particular form, say X = 3^(2*Y + 1) + 1?
It is easy to see "by formula" that p = 2, 5, 13, and 43 can never divide N. I supply the following table for when p divides N:
p = 2: impossible "by formula"
p = 3: X == 0 (mod 3)
p = 5: impossible "by formula"
p = 7: X == 1 (mod 6)
p = 11: X == 1 (mod 2)
p = 13: impossible "by formula"
p = 17: X == 11 (mod 16)
Up to p = 17, there are no Y for which 3^(2*Y + 1) + 1 can satisfy the congruence condition for X.
I am giving you the following homework assignment:
A) If you don't know how to do (2), learn how.
B) If you don't know how to do (3), learn how.
C) Find the smallest prime p which can divide N = (65*10^X + 43)/9 if X = 3^(2*Y + 1) + 1, and the congruence condition on Y for which this p divides N.
The kind of checking indicated above only involves modulo arithmetic to very small moduli.
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1 year, 9 months ago
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Just an idea.
Could be $$53.$$
$$53$$ is also a prime number.
As the product of the difference of $$3,11,29$$
$$11-3=8$$
$$29-11=18.$$
$$8\times18=144.$$
$$29,53,59$$
$$59-53=6$$
$$53-29=24.$$
$$24\times6=144!!$$
Though I am not very convinced with the logic.
- 1 year, 9 months ago
I don't think this is the answer since 29-11=18, 53-29=24, and 18x24=/=144.
- 1 year, 9 months ago
i know it.Thats why i said "iam not convinced with my solution".
- 1 year, 9 months ago | 526 | 1,406 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-26 | latest | en | 0.78205 |
https://mathspace.co/textbooks/syllabuses/Syllabus-453/topics/Topic-8414/subtopics/Subtopic-111453/ | 1,717,018,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059408.76/warc/CC-MAIN-20240529200239-20240529230239-00536.warc.gz | 321,731,133 | 65,095 | # Two step equations I
Lesson
We have already seen how to solve Single Step Equations. These ones seemed almost too simple to use a series of rules or steps to solve. Time to take it to the next level.
## Two Step Equations
A two step equation is one that will require two steps to solve. They generally have a multiplication/division and a subtraction/addition. The following are all two step equations.
$2x+5=11$2x+5=11, $\frac{1}{3}h-3=6$13h3=6, $19-2j=5$192j=5
Let's look at two methods for solving two step equations.
### Method 1 - Backtracking
Two step equations can be set up using a backtracking tool.
Start by writing the variable (variable) in a box. Then one step at a time mark in the operations that happen in order (according to Order of Operations). Remember we learnt how to set up equations in backtracking here.
#### Example
Solve: $-2x+4=8$2x+4=8
Think: First we need to set up the equation.
× $-2$−2 $+$+$4$4 $x$x $8$8
Do: Backtrack one step at a time, reversing each operation.
× $-2$−2 $+$+$4$4 $x$x $4$4 $8$8
× $-2$−2 $+$+$4$4 $-2$−2 $4$4 $8$8
So $x=-2$x=2.
With all equations we can check our solution. Does $x=-2$x=2 satisfy the equation $-2x+4=8$2x+4=8?
### Method 2 - Use formal algebraic techniques
This means using the "do the same to both sides" method, to isolate the $x$x. That is, get the $x$x on its own.
It is all about reversing the operations. So, we will need to remove the constant (number) term first. This is done by choosing the reverse operation.
$-2x+4=8$2x+4=8
$-2x+4$−2x+4 $=$= $8$8 The opposite of addition is subtraction. $-2x+4-4$−2x+4−4 $=$= $8-4$8−4 Start by subtracting $4$4 from both sides. $-2x$−2x $=$= $4$4 Simplify both sides of the equation $\frac{-2x}{-2}$−2x−2 $=$= $\frac{4}{-2}$4−2 Then identify the next operation that needs to be reversed. The opposite of multiplying by $-2$−2 is dividing by $-2$−2. $x$x $=$= $-2$−2 Simplify both sides to find $x$x | 672 | 1,966 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-22 | latest | en | 0.810977 |
https://www.essayhomeworkhelp.org/articles/sample-paper-on-mean-height-of-the-male-university-students-in-the-usa/ | 1,675,546,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00484.warc.gz | 757,110,450 | 18,388 | # Sample Paper on mean height of the male university students in the USA
Descriptive statistics
Executive summary
This project seeks to test the null hypothesis that the mean height of the male university students in the USA is always equal to 70 inches against an alternative hypothesis that this mean is not equal to 70 inches. In order to test this hypothesis, I will collect the height of 110 male university students in our university by simply asking them about their height. After collecting this data, then I will input it in an excel program to determine the mean, standard deviation and standard variance. I will also determine the mode, median and the sample’s range. The p-value will be 5% while the confidence level will be 95%. Once I calculate the mean, standard deviation and standard error, I will test the null hypothesis using the t distribution.
Introduction
It is hypothesized that the mean height of the male university students in the USA is 70 inches. In this test, I will be interested in testing this hypothesis. Therefore, in order to test it, I will randomly collect a sample height of 110 male students from our university by asking them about their heights. After collecting this sample, then I will input the data in excel program and calculate the mean, standard deviation and standard variance. I will also calculate the mode, median and range (Anderson, Sweeney and Williams 143). Then with the help of these data, I will test the null hypothesis using the t test.
Analysis
The p-value for this test is 0.05 (5%) while the confidence level is 95%.
For the confidence interval level, the upper and the lower limits will be µ ± 0.236318723. This means the upper limit will be 69.8 + 0.236318723 = 70.26838. On the other hand, the lower limit will be 69.8 – 0.236318723 = 69.33162. We can use this information to see what should happen to the null hypothesis by evaluating whether the hypothesized mean falls within this range or not, but let us use the t-distribution.
The null hypothesis states that the mean height of university students is equal to 70 inches. Conversely, the alternative hypothesis states that the mean height of university students is not equal to 70 inches.
H0: µ = 70;
H1: µ ≠ 70;
This test is a two-tailed test because the null hypothesis uses an equal sign while the alternative hypothesis negates it. At the same time, even though it has been hypothesized that the mean height is equal to 70 inches, it is also possible that this mean might in some instances be less than 70. This means that the test should be two-tailed.
The t-distribution utilizes the following formula to obtain the t value and I will use it to obtain my calculated t value. Then I will compare this value with the critical t value to determine whether I should retain the null hypothesis or reject it.
t = (µ -β)/S.E, where µ is the calculated sample mean, β s the hypothesized mean and S.E is the standard error (Anderson, Sweeney and Williams 143).
After collecting a sample data of 110 students and analyzing that data using excel program, I have obtained a sample mean of 69.8 inches, a standard deviation of 2.478531675 and a sample variance of 6.143119266. I will utilize these results to test my hypothesis.
Descriptive statistics Mean 69.8 Standard Error 0.236318723 Median 70 Mode 67 Standard Deviation 2.478531675 Sample Variance 6.143119266 Kurtosis -0.295639266 Skewness -0.064659862 Range 12 Minimum 64 Maximum 76 Sum 7678 Count 110 Confidence Level (95.0%) 0.468376012
(Davis and Pecar 100)
From the above table, it is clear that the sample’s mean is 69.8. To test the null hypothesis I should calculate the critical t value, which in this case is 1.981967. I should also calculate the t value using the formula above for t distribution and obtain -0.846314662. In this case, I will retain the null hypothesis if the calculated t value will be less than the critical t. On the contrary, I should reject it if it will be greater than the critical t value. Since the t value is less than the critical t then I should retain the null hypothesis (Davis and Pecar 100). This means that I should accept the hypothesis that the mean height of the male university students in the USA is 70 inches.
Summary/conclusion
In summary, the mean is 69.8, standard deviation 2.478531675 and variance 6.143119266. Furthermore, the critical t value is 1.981967 while the calculated t value is -0.846314662. This data demonstrates that I should accept the null hypothesis that claims that the mean height of the male university students in the USA is 70 inches because the hypothesized mean falls within the upper and the lower limits. At the same time, I should retain the null hypothesis because the value of the calculated t is less than the value of critical t. By accepting this hypothesis, I should reject the alternative hypothesis that states otherwise and conclude that the mean height of the male university students in the USA is 70 inches.
Works Cited
Anderson, David, Sweeney, Dennis, and Williams Thomas. Statistics for business and economics. Mason: Cengage learning. 2010.
Davis, Glyn, and Pecar Branko. Business Statistics Using Excel. Oxford: Oxford University Press, 2013. Print.
Appendix (data collected and used)
71 70 71 67 67 70 73 67 67 71 70 70 69 72 68 68 73 67 67 71 72 70 67 68 72 67 71 74 70 70 74 74 71 67 67 73 70 70 70 67 72 72 67 71 75 68 74 67 68 71 68 72 72 69 70 71 69 73 70 69 73 69 71 71 69 72 72 76 72 72 67 71 65 67 68 69 71 65 67 72 72 65 70 64 69 67 73 69 75 70 72 71 64 65 67 69 72 71 69 71 70 71 70 69 67 68 68 71 70 69 | 1,426 | 5,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.893833 |
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# C3 Differentiation Question Watch
1. (Original post by Zacken)
Yep, if the second derivative is less than 0 (i.e negative) then it's a maximum. Have you spotted why you got It should be - you probably forgot a minus whilst differentiating or something.
Yeah I did, thank you! I forgot to carry it onto the next line oops! And I didn't think differential sounded right, thanks for correcting me!
2. (Original post by Blake Jones)
Yeah I did, thank you! I forgot to carry it onto the next line oops! And I didn't think differential sounded right, thanks for correcting me!
Awesome, good job!
3. (Original post by Lollieboo)
Alright calm down cowboy. If you go to the 'report a post section', there is no option to report someone for giving a full solution or similar. So personally I only guidance of how to answer questions. No rules or strictly forbidding anything.
(Original post by RDKGames)
Yes there is. Not constructive.
Please do not post full solution unless as a last resort.
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 559 | 2,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-51 | latest | en | 0.919236 |
https://www.itetgroup.com/capricorn-aquarius-cusp-compatibility-with-cancer/ | 1,624,284,224,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00321.warc.gz | 756,658,729 | 9,936 | # Capricorn Aquarius Cusp Compatibility With Cancer
The Life Path number is the most vital since it is your key purpose in life and the main reason you were birthed; capricorn aquarius cusp compatibility with cancer the expression number is exactly how you reveal yourself to the globe and exactly how others see you.
The spirit’s urge number makes you satisfied in life and what brings you joy, and the birthday celebration number is all the abilities and talents and capacities that you come down into this lifetime with to support your life’s purpose.
So I’m going to reveal you just how to determine the Life Course number initially because that is the most crucial number in your graph, and it is your key purpose in life, so the Life Course number is identified by your day of birth.
So all we’re doing is we are simply adding every one of the numbers together in your day of birth until we damage them down and obtain a solitary number.
So in this example, we have December 14, 1995.
So we need to get a single number for each and every area.
First, we need to get a solitary number for the month.
We need to get a single figure for the day of the month, and we need to obtain a solitary number for the birth year, and then once we get a single number each of these, we can include these three together.
So December is the 12th month of the year, so we need to add the one in both to get a single number.
One plus two equates to three, and for the day of the month, we require to add the one in the 4 together, and that will certainly give us number five for the day of the month.
And after that we need to add 1995 together.
And that will provide us a solitary figure for the year, so one plus nine plus nine plus five equals twenty-four, and then we need to add the 2 in the four with each other because we have to damage it down to one digit for the year.
So two plus four amounts to 6.
So when we have a solitary digit for each one of these locations, a solitary number for the month, a single number for the day, and a single number for the year, then we’re mosting likely to add the 3 of these together, and in this instance.
It would certainly be 3 plus 5 plus 6 amounts to 14, and then we need to include the one and the 4 with each other since we’re trying to break every one of this down till we obtain a solitary number.
So if you include the one in the 4 together in 14, you get a number 5.
So in this circumstance, he or she’s life path number is a number five.
He or she is a 14 5, and it’s critical to bear in mind of the 2 last numbers that we included with each other to get the Life Path number five since these two last digits are crucial to the person’s life course number.
So, in the instance of a Life Course, the number 5 can be a fourteen 5 because the 4 amounts to number five.
They could also be a two and a 3 a twenty-three number 5. Still, those 2 numbers that you included with each other to get that Life Path number are significant because they inform you what energies you will need to utilize during your lifetime to achieve your life function. In this circumstance, capricorn aquarius cusp compatibility with cancer, he or she has a Life Path variety of number five.
Still, to meet their life path, the variety of a number five, they are mosting likely to need to utilize the energy of the number one and the number Four to attain their objective, so make note of those 2 last numbers that you combined to get your final Life Path number because those are extremely vital.
A fourteen-five is mosting likely to be very various than a twenty-three-five.
If you have any concerns concerning these two last numbers utilized to obtain your last Life Course, number remark listed below, and I will certainly try to address your questions currently.
I wished to show you this example because, in some scenarios, we do not damage down every one of the numbers to obtain a solitary digit.
So in this instance, we have December 14, 1992, and when we added every one of the numbers with each other in the month, the day, and the year, we finished up with a number 11 and 11 in numerology is a master number and the master numbers.
We do not include the two figures with each other, so there are 3 master numbers in numerology, and the three master numbers are 11, 22, and 33.
So after you have actually included every one of the digits together in your birth day and if you wind up with either an 11, a 22, or a 33, you will certainly not add these two digits together since you have a master number Life Path.
Number and the master numbers are various from the various other numbers in numerology due to the fact that they hold the dual numbers’ power, and we do not add the two digits together in these scenarios.
So if you have either an 11, a 22, or 33, you will certainly not add both figures with each other.
You will certainly maintain it as is, and you have a master number as a life course.
In this circumstance, the number that we included with each other to get the 11 were 8 and 3.
Those are significant numbers in this scenario because this master number 11 will certainly need to utilize the eight and the 3.
To obtain their number 11 life function, so in 83, 11 will be a lot various from on 92 11 since an 83 11 will certainly need to make use of the power of the 8 in the 3 to obtain their life objective. The 9211 will certainly Have to utilize the 9 and the two indicate get their 11 life objective.
So the birthday number is possibly one of the most available number to compute in your chart since for this number, all you have to do is include the numbers with each other of the day you were birthed on, so he or she was born on December 14, 1995.
So we will add the one and the 4 with each other due to the fact that those are the figures of the day. capricorn aquarius cusp compatibility with cancer
He or she was born, so 1 plus 4 amounts to 5.
So he or she’s birthday celebration number is a number 5.
Now, in this instance, December 11, 1995.
He or she was born upon the 11th day of the month, and 11 is a master number, so we do not add the two ones together since 11 is a master number, and there are 3 master numbers in numerology, 11, 22, and 33.
So if you were birthed on the 11th of a month or the 22nd of a month, you would not add both numbers with each other since your birthday celebration number is a master number.
So your birthday number is either a master number 11 or a master number 22. The master numbers are the only numbers in numerology that we do not include the 2 digits with each other to get a final number.
So you will maintain those two digits alone, and you will not include them together.
For the last two numbers, you have actually utilized your day of birth to compute those two numbers, however for the expression number and the soul’s desire number, you will certainly use the full name on your birth certificate.
You’re going to utilize your initial, middle, and surname on your birth certificate to determine your expression and your heart’s urge number, therefore we’re now going to make use of the Pythagorean number system to calculate these numbers.
And it’s called the Pythagorean system due to the fact that Pythagoras, a Greek mathematician, created it. He was the mathematician that developed the Pythagorean theory. He is the dad of numerology, and he found that all numbers hold energy. They all own a Specific resonance, and so after finding out that all numbers hold certain power, Pythagoras developed the Pythagorean number system. From that, we have modern numerology today.
It is a chart with every one of the letters of the alphabet and all letters representing a details number.
So basically, all letters in the alphabet have the power of a number.
And if you look at this graph, you can figure out what number each letter has the energy of.
So a has the energy of a number one B has the power of a number.
Two C has the power of a number three and so on, and so on.
So, for the expression number, all you have to do is include every one of the letters in your full birth name, so the first, middle, and surname on your birth certificate and decrease them down to one figure. capricorn aquarius cusp compatibility with cancer
So in this instance, we have Elvis Presley.
So what I did was I looked at the number graph, and I found the matching number per letter in Elvis’s name.
So E is a five, l is a 3 V is a four.
I am a nine, and I just found every one of the numbers matching to every one of the letters in his name; and I added every one of those numbers together, and the last number I got was an 81.
After that I added the eight and the one together due to the fact that we need to maintain breaking these down till we get a single-digit, and when I claimed the 8 and the one with each other, I obtained a nine, so Elvis’s expression number is a nine.
Currently the one situation where you would not remain to include these numbers with each other up until you got a solitary digit would certainly be if you got a master number, so the three master numbers are 11, 22, and 33.
If you got a master number, you would not proceed to include these numbers with each other.
You will certainly maintain them as either 11, 22, or 33.
So it’s the very same circumstance as it was with the life fifty percent number and the birthday celebration number.
The master numbers are one-of-a-kind, and we do not include the two figures together, so proceeding to the soles prompt number so in some cases the soles urge number can be called the single number, and it can additionally be called the heart’s desire number.
These words are used reciprocally however understand whenever you see the sole number or heart’s wish number that they are basically the very same thing as a soles prompt number.
For this number, we will include all of the vowels in your complete birth name.
Every one of the vowels in your initial, middle, and last name on your birth certification, and we’re going to minimize them to one digit.
So we’re mosting likely to utilize the very same Pythagorean graph that we did previously, and we’re going to look up all of the numbers that represent the vowels in your first, center, and surname.
So below we have Kate Middleton.
I just recently just did a video on her numerology, so I figured why not use her today.
So her very first middle and last name are Catherine, Elizabeth Middleton, so I searched for the numbers matching to only the vowels in her name.
So, as you can see, an equals 1 B amounts to 5, I amounts to nine, and E equals 5, and then I did that for every one of the vowels in her complete name.
And after that I simply included all of those numbers with each other, and that offered me 60, and then 60 minimizes to number 6 due to the fact that we remember we’re just attempting to break these numbers down until we obtain a solitary digit.
So in this situation, Kate’s Seoul’s urge number is a number 6.
Now you will not remain to damage down the numbers if you get an 11, a 22, or a 33, so, as I stated with all the various other numbers if you get among these numbers, this is a master number, and we do not include the 2 numbers with each other so 11, 22 and 33 are master numbers.
Capricorn Aquarius Cusp Compatibility With Cancer
If you obtain one of those as your hearts prompt numbers, you will not continue to add the numbers with each other; you will maintain them as double figures. | 2,620 | 11,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-25 | latest | en | 0.949382 |
https://www.studypool.com/discuss/5401199/C-program | 1,532,174,639,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592523.89/warc/CC-MAIN-20180721110117-20180721130117-00559.warc.gz | 979,206,816 | 23,299 | # C program
Anonymous
account_balance_wallet \$15
Question description
please see the attachment in order to know what are the requirements. you should work on the C program, all what you have to do is to type the requirements in the C program , they are really easy
ECE 102 Assignment #20 Monte Carlo method to estimate pi Review the notes for the description of this problem. Write a C program to implement the pseudocode as written. Add appropriate comments. Analysis: Input: the total number of darts (N, integer) Output: an estimate of pi (pi_estimate, double) Formula: pi_estimate = 4.0 *M / N, where M is the number of darts that fall within the unit circle constraint: N should be a large integer N > 500 intermediate variables: the number of darts inside the circle (M, integer) counter for the number of darts thrown (counter, integer) the position (coordinates) of a dart (x, y, double) Programming note: x =2* rand – 1; %randomly generate x coordinate between -1 and 1 y =2*rand – 1; %randomly generate y coordinate between -1 and 1 Pseudocode: 1. Get user input for the total number of darts, N 2. Error checking for N to be a large number N > 500 while (1) if N is less than or equal to 500 Display “error> N must be greater than 500” Get input for N else break; 3. Use a while loop to simulate the dart throwing and counting. a. Initialize the initial count to M = 0. M is the number of darts within the circle. b. Initialize the counter = 0 c. while counter is less than N Simulate throwing a dart (x=2*rand -1; y=2*rand -1;) //the location of the dart is (x, y) if (x, y) is within the circle (math relation: x2 + y2 <= 1) increment M by 1 4. Compute an estimate for pi: pi_estimate= 4.0*M/ N 5. Display the output: An estimate for pi is ___________ about _______ darts are thrown in this Monte pi_estimate N Carlo experiment
ECE 102 Assignment #21 Write a program that simulates the play of the game of pig. We will discuss this problem again on Wednesday, 11/15. Prelab: 1. Write down a detailed pseudocode. Hint: another running sum is needed to in order to get the total score for the entire game. Review the notes on how to calculate a running sum. 2. Trace the pseudocode to make sure the total score is calculated correctly. 3. Try your best to fix the pseudocode to make sure that it works correctly before writing the program. 4. Embed the while loop in your program for Assignment #19 in this assignment for the play of each turn.
Barbartos
School: UIUC
#20 and #21. That should be all the code; do you still want the writeup for all the questions?
/*
Ali Almusallam
This program simulates the game of pig for 5 turns*/
#include
#include
#include
#define ddie 6 /* number of faces on die */
#define turns 5 /* number of turns in a game of pig*/
int main(void)
{
int score = 0; /* player's total score*/
for...
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Wow this is really good.... didn't expect it. Sweet!!!!
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A363248 Nonprime base-10 palindromes whose arithmetic derivative is a base-10 palindrome. 0
0, 1, 4, 6, 9, 121, 222, 717, 989, 1331, 10201, 13231, 15251, 15751, 15851, 18281, 19291, 28882, 28982, 31613, 34043, 35653, 37073, 37673, 37873, 38383, 38683, 40304, 41814, 50405, 97079, 98789, 99899, 536635, 913319, 980089, 1030301, 1115111, 1226221, 1336331, 1794971, 2630362, 2882882, 3303033 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Nonprime members k of A002113 such that A003415(k) is also in A002113. A003415(p) = 1 is a palindrome for all primes p. It seems that most members of A363246 are primes. LINKS Table of n, a(n) for n=1..44. EXAMPLE a(7) = 222 is a term because it is a palindrome, is not prime, and its arithmetic derivative 191 is a palindrome. MAPLE ader:= proc(n) local t; n*add(t[2]/t[1], t=ifactors(n)[2]) end proc: rev:= proc(n) local L, i; L:= convert(n, base, 10); add(L[-i]*10^(i-1), i=1..nops(L)) end proc: palis:= proc(d) local x, y; if d::even then seq(10^(d/2)*x+rev(x), x=10^(d/2-1)..10^(d/2)-1) else seq(seq(10^((d+1)/2)*x+10^((d-1)/2)*y+rev(x), y=0..9), x=10^((d-3)/2) ..10^((d-1)/2)-1) fi end proc: palis(1):= \$0..9: filter:= proc(n) local d; if isprime(n) then return false fi; d:= ader(n); d = rev(d) end proc: select(filter, [seq(palis(i), i=1..7)]); CROSSREFS Cf. A002113, A003415. Complement of A002385 in A363246. Sequence in context: A245044 A104389 A115655 * A084350 A028279 A114743 Adjacent sequences: A363245 A363246 A363247 * A363249 A363250 A363251 KEYWORD nonn,base AUTHOR Robert Israel, May 23 2023 STATUS approved
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Last modified August 15 01:27 EDT 2024. Contains 375171 sequences. (Running on oeis4.) | 771 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.55631 |
https://web2.0calc.com/questions/what-is-the-value-of-z-in-the-triangle | 1,539,665,391,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509996.54/warc/CC-MAIN-20181016030831-20181016052331-00446.warc.gz | 824,442,218 | 5,520 | +0
# What is the value of z in the triangle?
0
1071
1
+164
What is the value of z in the triangle?
Gwendolynkristine Apr 28, 2017
#1
+7324
+1
cos θ = adjacent / hypotenuse
cos 37º = 10 / z
z cos 37º = 10
z = 10 / cos 37º
z ≈ 12.52 inches
hectictar Apr 28, 2017 | 126 | 272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.531876 |
https://silvercityindiaonline.com/qa/what-is-a-circle-graph-used-for.html | 1,624,009,252,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635920.39/warc/CC-MAIN-20210618073932-20210618103932-00335.warc.gz | 455,659,816 | 9,221 | # What Is A Circle Graph Used For?
## What is the difference between a pie chart and a circle graph?
Pie charts represent data in a circle, with “slices” corresponding to percentages of the whole, whereas bar graphs use bars of different lengths to represent data in a more flexible way..
## What graph category should you avoid?
There are some kinds of graphs which must be avoided. Those are – Pie charts, 3D and different tricks, radar graphs, donuts and surface graphs. The worthy data for the Pie charts is not easily available. The 3 D graphs are little bit confusing.
## What are the uses of graph?
Graphs are a common method to visually illustrate relationships in the data. The purpose of a graph is to present data that are too numerous or complicated to be described adequately in the text and in less space. Do not, however, use graphs for small amounts of data that could be conveyed succinctly in a sentence.
## How do you find the measure of an angle in a circle?
A circle has a total of 360 degrees all the way around the center, so if that central angle determining a sector has an angle measure of 60 degrees, then the sector takes up 60/360 or 1/6, of the degrees all the way around. In that case, the sector has 1/6 the area of the whole circle.
## What is a disadvantage of a bar graph?
Bar charts that attempt to represent wide ranges of numbers will struggle to efficiently communicate their message. … Bar graphs tend to be locked into a particular data set, making it hard to show multiple values or changes over time unless the chart is modified, such as by making the bars layered and three-dimensional.
## When would you use a pie chart?
Pie charts are generally used to show percentage or proportional data and usually the percentage represented by each category is provided next to the corresponding slice of pie. Pie charts are good for displaying data for around 6 categories or fewer.
## How do you describe a circle graph?
Initial Definition. A circle graph is a visual representation of data, made by dividing a circle into sectors that each represent parts of a whole. Usually the amounts in each sector are represented in percent, so that all of the amounts total 100%. Circle Graph Example.
## What is a round graph called?
A circle graph is also known as a pie chart. The graph is in the shape of a circle with different wedges that each represent a percentage of a total. These wedges often look like pieces of pie, which is why the circle graph is sometimes referred to as a pie chart.
## Is a circle a function?
A circle is a set of points in the plane. A function is a mapping from one set to another, so they’re completely different kinds of things, and a circle cannot be a function.
## What is a bar graph best used for?
Bar graphs are used to compare things between different groups or to track changes over time. However, when trying to measure change over time, bar graphs are best when the changes are larger.
## What percent of a circle is 1/6 of a circle?
15.91549430919 percent1 radian is equal to 0.95492965855137 1/6 circle, or 15.91549430919 percent.
## Why is a pie chart better than a bar graph?
The bar/column chart excels at showing discrete data while comparing one data-point vs. another, while the pie chart is the classic way to show how various parts makes up a whole. … The pie chart, on the other hand, is only useful to show relative values.
## What type of graph is used for percentages?
Pie chartsA pie chart typically represents numbers in percentages, used to visualize a part to whole relationship or a composition. Pie charts are not meant to compare individual sections to each other or to represent exact values (you should use a bar chart for that).
## How do you show data in a circle graph?
Finding a percentage of a total amount in a circle graphCircle graphs: A circle is divided into smaller portions. … To make a circle graph form the data in the table above.Step 1: Add up all the values in the table.Step 2: Next divide each value by the total and multiply by 100 to get a percent.Step 3: You can show this data by using a Cicle graph as follows:
## What are histograms best used for?
A histogram is used to summarize discrete or continuous data. In other words, it provides a visual interpretation. This requires focusing on the main points, factsof numerical data by showing the number of data points that fall within a specified range of values (called “bins”). It is similar to a vertical bar graph.
## When would you use a circle graph instead of a bar graph?
Circle graphs are most useful when comparing parts of a whole or total. Bar graphs also make comparisons easily. Unlike most circle graphs, bar graphs compare exact amounts. Circle graphs are used when dealing with percentages, and the percentages of the pieces add up to 100 percent.
## How do I know what graph to use?
You would use:Bar graphs to show numbers that are independent of each other. … Pie charts to show you how a whole is divided into different parts. … Line graphs show you how numbers have changed over time. … Cartesian graphs have numbers on both axes, which therefore allow you to show how changes in one thing affect another.
## What does a bar graph tell you?
A bar graph shows comparisons among discrete categories. One axis of the chart shows the specific categories being compared, and the other axis represents a measured value. Some bar graphs present bars clustered in groups of more than one, showing the values of more than one measured variable.
## What type of graph could be easily converted into a pie chart?
A circle graph/pie chart is constructed by converting the share of each component into a percentage of 360 degrees.
## What are the 6 types of graphs?
Different types of graphsLine graph. Line graphs illustrate how related data changes over a specific period of time. … Bar graph. Bar graphs offer a simple way to compare numeric values of any kind, including inventories, group sizes and financial predictions. … 3 . Pictograph. … Histogram. … Area graph. … Scatter plot. | 1,293 | 6,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-25 | latest | en | 0.924479 |
https://www.scribd.com/doc/70510168/Probability-and-Stochastic-Processes-Quiz-Sol | 1,510,980,703,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804610.37/warc/CC-MAIN-20171118040756-20171118060756-00202.warc.gz | 859,127,050 | 67,135 | Probability and Stochastic Processes
A Friendly Introduction for Electrical and Computer Engineers
Second Edition
Quiz Solutions
Roy D. Yates and David J. Goodman
May 22, 2004
• The MATLAB section quizzes at the end of each chapter use programs available for
download as the archive matcode.zip. This archive has programs of general pur-
pose programs for solving probability problems as well as specific .m files associated
with examples or quizzes in the text. Also available is a manual probmatlab.pdf
describing the general purpose .m files in matcode.zip.
• We have made a substantial effort to check the solution to every quiz. Nevertheless,
there is a nonzero probability (in fact, a probability close to unity) that errors will be
found. If you find errors or have suggestions or comments, please send email to
ryates@winlab.rutgers.edu.
When errors are found, corrected solutions will be posted at the website.
1
Quiz Solutions – Chapter 1
Quiz 1.1
In the Venn diagrams for parts (a)-(g) below, the shaded area represents the indicated
set.
M
O
T
M
O
T
M
O
T
(1) R = T
c
(2) M ∪ O (3) M ∩ O
M
O
T
M
O
T
M
O
T
(4) R ∪ M (4) R ∩ M (6) T
c
− M
Quiz 1.2
(1) A
1
= {vvv, vvd, vdv, vdd}
(2) B
1
= {dvv, dvd, ddv, ddd}
(3) A
2
= {vvv, vvd, dvv, dvd}
(4) B
2
= {vdv, vdd, ddv, ddd}
(5) A
3
= {vvv, ddd}
(6) B
3
= {vdv, dvd}
(7) A
4
= {vvv, vvd, vdv, dvv, vdd, dvd, ddv}
(8) B
4
= {ddd, ddv, dvd, vdd}
Recall that A
i
and B
i
are collectively exhaustive if A
i
∪ B
i
= S. Also, A
i
and B
i
are
mutually exclusive if A
i
∩ B
i
= φ. Since we have written down each pair A
i
and B
i
above,
we can simply check for these properties.
The pair A
1
and B
1
are mutually exclusive and collectively exhaustive. The pair A
2
and
B
2
are mutually exclusive and collectively exhaustive. The pair A
3
and B
3
are mutually
exclusive but not collectively exhaustive. The pair A
4
and B
4
are not mutually exclusive
since dvd belongs to A
4
and B
4
. However, A
4
and B
4
are collectively exhaustive.
2
Quiz 1.3
There are exactly 50 equally likely outcomes: s
51
through s
100
. Each of these outcomes
has probability 0.02.
(1) P[{s
79
}] = 0.02
(2) P[{s
100
}] = 0.02
(3) P[A] = P[{s
90
, . . . , s
100
}] = 11 ×0.02 = 0.22
(4) P[F] = P[{s
51
, . . . , s
59
}] = 9 ×0.02 = 0.18
(5) P[T ≥ 80] = P[{s
80
, . . . , s
100
}] = 21 ×0.02 = 0.42
(6) P[T < 90] = P[{s
51
, s
52
, . . . , s
89
}] = 39 ×0.02 = 0.78
(7) P[a C grade or better] = P[{s
70
, . . . , s
100
}] = 31 ×0.02 = 0.62
(8) P[student passes] = P[{s
60
, . . . , s
100
}] = 41 ×0.02 = 0.82
Quiz 1.4
We can describe this experiment by the event space consisting of the four possible
events V B, V L, DB, and DL. We represent these events in the table:
V D
L 0.35 ?
B ? ?
In a roundabout way, the problem statement tells us how to fill in the table. In particular,
P [V] = 0.7 = P [V L] + P [V B] (1)
P [L] = 0.6 = P [V L] + P [DL] (2)
Since P[V L] = 0.35, we can conclude that P[V B] = 0.35 and that P[DL] = 0.6 −
0.35 = 0.25. This allows us to fill in two more table entries:
V D
L 0.35 0.25
B 0.35 ?
The remaining table entry is filled in by observing that the probabilities must sum to 1.
This implies P[DB] = 0.05 and the complete table is
V D
L 0.35 0.25
B 0.35 0.05
Finding the various probabilities is now straightforward:
3
(1) P[DL] = 0.25
(2) P[D ∪ L] = P[V L] + P[DL] + P[DB] = 0.35 +0.25 +0.05 = 0.65.
(3) P[V B] = 0.35
(4) P[V ∪ L] = P[V] + P[L] − P[V L] = 0.7 +0.6 −0.35 = 0.95
(5) P[V ∪ D] = P[S] = 1
(6) P[LB] = P[LL
c
] = 0
Quiz 1.5
(1) The probability of exactly two voice calls is
P [N
V
= 2] = P [{vvd, vdv, dvv}] = 0.3 (1)
(2) The probability of at least one voice call is
P [N
V
≥ 1] = P [{vdd, dvd, ddv, vvd, vdv, dvv, vvv}] (2)
= 6(0.1) +0.2 = 0.8 (3)
An easier way to get the same answer is to observe that
P [N
V
≥ 1] = 1 − P [N
V
< 1] = 1 − P [N
V
= 0] = 1 − P [{ddd}] = 0.8 (4)
(3) The conditional probability of two voice calls followed by a data call given that there
were two voice calls is
P [{vvd} |N
V
= 2] =
P [{vvd} , N
V
= 2]
P [N
V
= 2]
=
P [{vvd}]
P [N
V
= 2]
=
0.1
0.3
=
1
3
(5)
(4) The conditional probability of two data calls followed by a voice call given there
were two voice calls is
P [{ddv} |N
V
= 2] =
P [{ddv} , N
V
= 2]
P [N
V
= 2]
= 0 (6)
The joint event of the outcome ddv and exactly two voice calls has probability zero
since there is only one voice call in the outcome ddv.
(5) The conditional probability of exactly two voice calls given at least one voice call is
P [N
V
= 2|N
v
≥ 1] =
P [N
V
= 2, N
V
≥ 1]
P [N
V
≥ 1]
=
P [N
V
= 2]
P [N
V
≥ 1]
=
0.3
0.8
=
3
8
(7)
(6) The conditional probability of at least one voice call given there were exactly two
voice calls is
P [N
V
≥ 1|N
V
= 2] =
P [N
V
≥ 1, N
V
= 2]
P [N
V
= 2]
=
P [N
V
= 2]
P [N
V
= 2]
= 1 (8)
Given that there were two voice calls, there must have been at least one voice call.
4
Quiz 1.6
In this experiment, there are four outcomes with probabilities
P[{vv}] = (0.8)
2
= 0.64 P[{vd}] = (0.8)(0.2) = 0.16
P[{dv}] = (0.2)(0.8) = 0.16 P[{dd}] = (0.2)
2
= 0.04
When checking the independence of any two events A and B, it’s wise to avoid intuition
and simply check whether P[AB] = P[A]P[B]. Using the probabilities of the outcomes,
we now can test for the independence of events.
(1) First, we calculate the probability of the joint event:
P [N
V
= 2, N
V
≥ 1] = P [N
V
= 2] = P [{vv}] = 0.64 (1)
Next, we observe that
P [N
V
≥ 1] = P [{vd, dv, vv}] = 0.96 (2)
Finally, we make the comparison
P [N
V
= 2] P [N
V
≥ 1] = (0.64)(0.96) = P [N
V
= 2, N
V
≥ 1] (3)
which shows the two events are dependent.
(2) The probability of the joint event is
P [N
V
≥ 1, C
1
= v] = P [{vd, vv}] = 0.80 (4)
From part (a), P[N
V
≥ 1] = 0.96. Further, P[C
1
= v] = 0.8 so that
P [N
V
≥ 1] P [C
1
= v] = (0.96)(0.8) = 0.768 = P [N
V
≥ 1, C
1
= v] (5)
Hence, the events are dependent.
(3) The problem statement that the calls were independent implies that the events the
second call is a voice call, {C
2
= v}, and the first call is a data call, {C
1
= d} are
independent events. Just to be sure, we can do the calculations to check:
P [C
1
= d, C
2
= v] = P [{dv}] = 0.16 (6)
Since P[C
1
= d]P[C
2
= v] = (0.2)(0.8) = 0.16, we confirm that the events are
independent. Note that this shouldn’t be surprising since we used the information that
the calls were independent in the problem statement to determine the probabilities of
the outcomes.
(4) The probability of the joint event is
P [C
2
= v, N
V
is even] = P [{vv}] = 0.64 (7)
Also, each event has probability
P [C
2
= v] = P [{dv, vv}] = 0.8, P [N
V
is even] = P [{dd, vv}] = 0.68 (8)
Thus, P[C
2
= v]P[N
V
is even] = (0.8)(0.68) = 0.544. Since P[C
2
= v, N
V
is even] =
0.544, the events are dependent.
5
Quiz 1.7
Let F
i
denote the event that that the user is found on page i . The tree for the experiment
is
¨
¨
¨
¨
¨
¨
F
1
0.8
F
c
1
0.2
¨
¨
¨
¨
¨
¨
F
2
0.8
F
c
2
0.2
¨
¨
¨
¨
¨
¨
F
3
0.8
F
c
3
0.2
The user is found unless all three paging attempts fail. Thus the probability the user is
found is
P [F] = 1 − P
_
F
c
1
F
c
2
F
c
3
_
= 1 −(0.2)
3
= 0.992 (1)
Quiz 1.8
(1) We can view choosing each bit in the code word as a subexperiment. Each subex-
periment has two possible outcomes: 0 and 1. Thus by the fundamental principle of
counting, there are 2 ×2 ×2 ×2 = 2
4
= 16 possible code words.
(2) An experiment that can yield all possible code words with two zeroes is to choose
which 2 bits (out of 4 bits) will be zero. The other two bits then must be ones. There
are
_
4
2
_
= 6 ways to do this. Hence, there are six code words with exactly two zeroes.
For this problem, it is also possible to simply enumerate the six code words:
1100, 1010, 1001, 0101, 0110, 0011.
(3) When the first bit must be a zero, then the first subexperiment of choosing the first
bit has only one outcome. For each of the next three bits, we have two choices. In
this case, there are 1 ×2 ×2 ×2 = 8 ways of choosing a code word.
(4) For the constant ratio code, we can specify a code word by choosing M of the bits to
be ones. The other N −M bits will be zeroes. The number of ways of choosing such
a code word is
_
N
M
_
. For N = 8 and M = 3, there are
_
8
3
_
= 56 code words.
Quiz 1.9
(1) In this problem, k bits received in error is the same as k failures in 100 trials. The
failure probability is = 1 − p and the success probability is 1 − = p. That is, the
probability of k bits in error and 100 −k correctly received bits is
P
_
S
k,100−k
_
=
_
100
k
_
k
(1 −)
100−k
(1)
6
For = 0.01,
P
_
S
0,100
_
= (1 −)
100
= (0.99)
100
= 0.3660 (2)
P
_
S
1,99
_
= 100(0.01)(0.99)
99
= 0.3700 (3)
P
_
S
2,98
_
= 4950(0.01)
2
(0.99)
9
8 = 0.1849 (4)
P
_
S
3,97
_
= 161, 700(0.01)
3
(0.99)
97
= 0.0610 (5)
(2) The probability a packet is decoded correctly is just
P [C] = P
_
S
0,100
_
+ P
_
S
1,99
_
+ P
_
S
2,98
_
+ P
_
S
3,97
_
= 0.9819 (6)
Quiz 1.10
Since the chip works only if all n transistors work, the transistors in the chip are like
devices in series. The probability that a chip works is P[C] = p
n
.
The module works if either 8 chips work or 9 chips work. Let C
k
denote the event that
exactly k chips work. Since transistor failures are independent of each other, chip failures
are also independent. Thus each P[C
k
] has the binomial probability
P [C
8
] =
_
9
8
_
(P [C])
8
(1 − P [C])
9−8
= 9p
8n
(1 − p
n
), (1)
P [C
9
] = (P [C])
9
= p
9n
. (2)
The probability a memory module works is
P [M] = P [C
8
] + P [C
9
] = p
8n
(9 −8p
n
) (3)
Quiz 1.11
R=rand(1,100);
X=(R<= 0.4) ...
+ (2*(R>0.4).*(R<=0.9)) ...
+ (3*(R>0.9));
Y=hist(X,1:3)
For a MATLAB simulation, we first gen-
erate a vector R of 100 random numbers.
Second, we generate vector X as a func-
tion of R to represent the 3 possible out-
comes of a flip. That is, X(i)=1 if flip i
was heads, X(i)=2 if flip i was tails, and
X(i)=3) is flip i landed on the edge.
To see how this works, we note there are three cases:
• If R(i) <= 0.4, then X(i)=1.
• If 0.4 < R(i) and R(i)<=0.9, then X(i)=2.
• If 0.9 < R(i), then X(i)=3.
These three cases will have probabilities 0.4, 0.5 and 0.1. Lastly, we use the hist function
to count how many occurences of each possible value of X(i).
7
Quiz Solutions – Chapter 2
Quiz 2.1
The sample space, probabilities and corresponding grades for the experiment are
Outcome P[·] G
BB 0.36 3.0
BC 0.24 2.5
CB 0.24 2.5
CC 0.16 2
Quiz 2.2
(1) To find c, we recall that the PMF must sum to 1. That is,
3
n=1
P
N
(n) = c
_
1 +
1
2
+
1
3
_
= 1 (1)
This implies c = 6/11. Now that we have found c, the remaining parts are straight-
forward.
(2) P[N = 1] = P
N
(1) = c = 6/11
(3) P[N ≥ 2] = P
N
(2) + P
N
(3) = c/2 +c/3 = 5/11
(4) P[N > 3] =
n=4
P
N
(n) = 0
Quiz 2.3
Decoding each transmitted bit is an independent trial where we call a bit error a “suc-
cess.” Each bit is in error, that is, the trial is a success, with probability p. Now we can
interpret each experiment in the generic context of independent trials.
(1) The random variable X is the number of trials up to and including the first success.
Similar to Example 2.11, X has the geometric PMF
P
X
(x) =
_
p(1 − p)
x−1
x = 1, 2, . . .
0 otherwise
(1)
(2) If p = 0.1, then the probability exactly 10 bits are sent is
P [X = 10] = P
X
(10) = (0.1)(0.9)
9
= 0.0387 (2)
8
The probability that at least 10 bits are sent is P[X ≥ 10] =
x=10
P
X
(x). This
sum is not too hard to calculate. However, its even easier to observe that X ≥ 10 if
the first 10 bits are transmitted correctly. That is,
P [X ≥ 10] = P [first 10 bits are correct] = (1 − p)
10
(3)
For p = 0.1, P[X ≥ 10] = 0.9
10
= 0.3487.
(3) The random variable Y is the number of successes in 100 independent trials. Just as
in Example 2.13, Y has the binomial PMF
P
Y
(y) =
_
100
y
_
p
y
(1 − p)
100−y
(4)
If p = 0.01, the probability of exactly 2 errors is
P [Y = 2] = P
Y
(2) =
_
100
2
_
(0.01)
2
(0.99)
98
= 0.1849 (5)
(4) The probability of no more than 2 errors is
P [Y ≤ 2] = P
Y
(0) + P
Y
(1) + P
Y
(2) (6)
= (0.99)
100
+100(0.01)(0.99)
99
+
_
100
2
_
(0.01)
2
(0.99)
98
(7)
= 0.9207 (8)
(5) Random variable Z is the number of trials up to and including the third success. Thus
Z has the Pascal PMF (see Example 2.15)
P
Z
(z) =
_
z −1
2
_
p
3
(1 − p)
z−3
(9)
Note that P
Z
(z) > 0 for z = 3, 4, 5, . . ..
(6) If p = 0.25, the probability that the third error occurs on bit 12 is
P
Z
(12) =
_
11
2
_
(0.25)
3
(0.75)
9
= 0.0645 (10)
Quiz 2.4
Each of these probabilities can be read off the CDF F
Y
(y). However, we must keep in
mind that when F
Y
(y) has a discontinuity at y
0
, F
Y
(y) takes the upper value F
Y
(y
+
0
).
(1) P[Y < 1] = F
Y
(1
) = 0
9
(2) P[Y ≤ 1] = F
Y
(1) = 0.6
(3) P[Y > 2] = 1 − P[Y ≤ 2] = 1 − F
Y
(2) = 1 −0.8 = 0.2
(4) P[Y ≥ 2] = 1 − P[Y < 2] = 1 − F
Y
(2
) = 1 −0.6 = 0.4
(5) P[Y = 1] = P[Y ≤ 1] − P[Y < 1] = F
Y
(1
+
) − F
Y
(1
) = 0.6
(6) P[Y = 3] = P[Y ≤ 3] − P[Y < 3] = F
Y
(3
+
) − F
Y
(3
) = 0.8 −0.8 = 0
Quiz 2.5
(1) With probability 0.7, a call is a voice call and C = 25. Otherwise, with probability
0.3, we have a data call and C = 40. This corresponds to the PMF
P
C
(c) =
0.7 c = 25
0.3 c = 40
0 otherwise
(1)
(2) The expected value of C is
E [C] = 25(0.7) +40(0.3) = 29.5 cents (2)
Quiz 2.6
(1) As a function of N, the cost T is
T = 25N +40(3 − N) = 120 −15N (1)
(2) To find the PMF of T, we can draw the following tree:
¨
¨
¨
¨
¨
¨
¨
N=0
0.1
r
r
r
r
r
r
r
N=3
0.3
\$N=1 0.3
N=2 0.3
•T=120
•T=105
•T=90
•T=75
From the tree, we can write down the PMF of T:
P
T
(t ) =
0.3 t = 75, 90, 105
0.1 t = 120
0 otherwise
(2)
From the PMF P
T
(t ), the expected value of T is
E [T] = 75P
T
(75) +90P
T
(90) +105P
T
(105) +120P
T
(120) (3)
= (75 +90 +105)(0.3) +120(0.1) = 62 (4)
10
Quiz 2.7
(1) Using Definition 2.14, the expected number of applications is
E [A] =
4
a=1
aP
A
(a) = 1(0.4) +2(0.3) +3(0.2) +4(0.1) = 2 (1)
(2) The number of memory chips is M = g(A) where
g(A) =
4 A = 1, 2
6 A = 3
8 A = 4
(2)
(3) By Theorem 2.10, the expected number of memory chips is
E [M] =
4
a=1
g(A)P
A
(a) = 4(0.4) +4(0.3) +6(0.2) +8(0.1) = 4.8 (3)
Since E[A] = 2, g(E[A]) = g(2) = 4. However, E[M] = 4.8 = g(E[A]). The two
quantities are different because g(A) is not of the form αA +β.
Quiz 2.8
The PMF P
N
(n) allows to calculate each of the desired quantities.
(1) The expected value of N is
E [N] =
2
n=0
nP
N
(n) = 0(0.1) +1(0.4) +2(0.5) = 1.4 (1)
(2) The second moment of N is
E
_
N
2
_
=
2
n=0
n
2
P
N
(n) = 0
2
(0.1) +1
2
(0.4) +2
2
(0.5) = 2.4 (2)
(3) The variance of N is
Var[N] = E
_
N
2
_
−(E [N])
2
= 2.4 −(1.4)
2
= 0.44 (3)
(4) The standard deviation is σ
N
=
Var[N] =
0.44 = 0.663.
11
Quiz 2.9
(1) From the problem statement, we learn that the conditional PMF of N given the event
I is
P
N|I
(n) =
_
0.02 n = 1, 2, . . . , 50
0 otherwise
(1)
(2) Also from the problem statement, the conditional PMF of N given the event T is
P
N|T
(n) =
_
0.2 n = 1, 2, 3, 4, 5
0 otherwise
(2)
(3) The problem statement tells us that P[T] = 1 − P[I ] = 3/4. From Theorem 1.10
(the law of total probability), we find the PMF of N is
P
N
(n) = P
N|T
(n) P [T] + P
N|I
(n) P [I ] (3)
=
0.2(0.75) +0.02(0.25) n = 1, 2, 3, 4, 5
0(0.75) +0.02(0.25) n = 6, 7, . . . , 50
0 otherwise
(4)
=
0.155 n = 1, 2, 3, 4, 5
0.005 n = 6, 7, . . . , 50
0 otherwise
(5)
(4) First we find
P [N ≤ 10] =
10
n=1
P
N
(n) = (0.155)(5) +(0.005)(5) = 0.80 (6)
By Theorem 2.17, the conditional PMF of N given N ≤ 10 is
P
N|N≤10
(n) =
_
P
N
(n)
P[N≤10]
n ≤ 10
0 otherwise
(7)
=
0.155/0.8 n = 1, 2, 3, 4, 5
0.005/0.8 n = 6, 7, 8, 9, 10
0 otherwise
(8)
=
0.19375 n = 1, 2, 3, 4, 5
0.00625 n = 6, 7, 8, 9, 10
0 otherwise
(9)
(5) Once we have the conditional PMF, calculating conditional expectations is easy.
E [N|N ≤ 10] =
n
nP
N|N≤10
(n) (10)
=
5
n=1
n(0.19375) +
10
n=6
n(0.00625) (11)
= 3.15625 (12)
12
0 50 100
0
2
4
6
8
10
0 500 1000
0
2
4
6
8
10
(a) samplemean(100) (b) samplemean(1000)
Figure 1: Two examples of the output of samplemean(k)
(6) To find the conditional variance, we first find the conditional second moment
E
_
N
2
|N ≤ 10
_
=
n
n
2
P
N|N≤10
(n) (13)
=
5
n=1
n
2
(0.19375) +
10
n=6
n
2
(0.00625) (14)
= 55(0.19375) +330(0.00625) = 12.71875 (15)
The conditional variance is
Var[N|N ≤ 10] = E
_
N
2
|N ≤ 10
_
−(E [N|N ≤ 10])
2
(16)
= 12.71875 −(3.15625)
2
= 2.75684 (17)
Quiz 2.10
The function samplemean(k) generates and plots five m
n
sequences for n = 1, 2, . . . , k.
The i th column M(:,i) of M holds a sequence m
1
, m
2
, . . . , m
k
.
function M=samplemean(k);
K=(1:k)’;
M=zeros(k,5);
for i=1:5,
X=duniformrv(0,10,k);
M(:,i)=cumsum(X)./K;
end;
plot(K,M);
Examples of the function calls (a) samplemean(100) and (b) samplemean(1000)
are shown in Figure 1. Each time samplemean(k) is called produces a random output.
What is observed in these figures is that for small n, m
n
is fairly random but as n gets
13
large, m
n
gets close to E[X] = 5. Although each sequence m
1
, m
2
, . . . that we generate is
random, the sequences always converges to E[X]. This random convergence is analyzed
in Chapter 7.
14
Quiz Solutions – Chapter 3
Quiz 3.1
The CDF of Y is
0 2 4
0
0.5
1
y
F
Y
(
y
)
F
Y
(y) =
0 y < 0
y/4 0 ≤ y ≤ 4
1 y > 4
(1)
From the CDF F
Y
(y), we can calculate the probabilities:
(1) P[Y ≤ −1] = F
Y
(−1) = 0
(2) P[Y ≤ 1] = F
Y
(1) = 1/4
(3) P[2 < Y ≤ 3] = F
Y
(3) − F
Y
(2) = 3/4 −2/4 = 1/4
(4) P[Y > 1.5] = 1 − P[Y ≤ 1.5] = 1 − F
Y
(1.5) = 1 −(1.5)/4 = 5/8
Quiz 3.2
(1) First we will find the constant c and then we will sketch the PDF. To find c, we use
the fact that
_
−∞
f
X
(x) dx = 1. We will evaluate this integral using integration by
parts:
_
−∞
f
X
(x) dx =
_
0
cxe
−x/2
dx (1)
= −2cxe
−x/2
¸
¸
¸
0
. ,, .
=0
+
_
0
2ce
−x/2
dx (2)
= −4ce
−x/2
¸
¸
¸
0
= 4c (3)
Thus c = 1/4 and X has the Erlang (n = 2, λ = 1/2) PDF
0 5 10 15
0
0.1
0.2
x
f
X
(
x
)
f
X
(x) =
_
(x/4)e
−x/2
x ≥ 0
0 otherwise
(4)
15
(2) To find the CDF F
X
(x), we first note X is a nonnegative random variable so that
F
X
(x) = 0 for all x < 0. For x ≥ 0,
F
X
(x) =
_
x
0
f
X
(y) dy =
_
x
0
y
4
e
−y/2
dy (5)
= −
y
2
e
−y/2
¸
¸
¸
x
0
_
x
0
1
2
e
−y/2
dy (6)
= 1 −
x
2
e
−x/2
−e
−x/2
(7)
The complete expression for the CDF is
0 5 10 15
0
0.5
1
x
F
X
(
x
)
F
X
(x) =
_
1 −
_
x
2
+1
_
e
−x/2
x ≥ 0
0 otherwise
(8)
(3) From the CDF F
X
(x),
P [0 ≤ X ≤ 4] = F
X
(4) − F
X
(0) = 1 −3e
−2
. (9)
(4) Similarly,
P [−2 ≤ X ≤ 2] = F
X
(2) − F
X
(−2) = 1 −3e
−1
. (10)
Quiz 3.3
The PDF of Y is
−2 0 2
0
1
2
3
y
f
Y
(
y
)
f
Y
(y) =
_
3y
2
/2 −1 ≤ y ≤ 1,
0 otherwise.
(1)
(1) The expected value of Y is
E [Y] =
_
−∞
y f
Y
(y) dy =
_
1
−1
(3/2)y
3
dy = (3/8)y
4
¸
¸
¸
1
−1
= 0. (2)
Note that the above calculation wasn’t really necessary because E[Y] = 0 whenever
the PDF f
Y
(y) is an even function (i.e., f
Y
(y) = f
Y
(−y)).
(2) The second moment of Y is
E
_
Y
2
_
=
_
−∞
y
2
f
Y
(y) dy =
_
1
−1
(3/2)y
4
dy = (3/10)y
5
¸
¸
¸
1
−1
= 3/5. (3)
16
(3) The variance of Y is
Var[Y] = E
_
Y
2
_
−(E [Y])
2
= 3/5. (4)
(4) The standard deviation of Y is σ
Y
=
Var[Y] =
3/5.
Quiz 3.4
(1) When X is an exponential (λ) random variable, E[X] = 1/λ and Var[X] = 1/λ
2
.
Since E[X] = 3 and Var[X] = 9, we must have λ = 1/3. The PDF of X is
f
X
(x) =
_
(1/3)e
−x/3
x ≥ 0,
0 otherwise.
(1)
(2) We know X is a uniform (a, b) random variable. To find a and b, we apply Theo-
rem 3.6 to write
E [X] =
a +b
2
= 3 Var[X] =
(b −a)
2
12
= 9. (2)
This implies
a +b = 6, b −a = ±6
3. (3)
The only valid solution with a < b is
a = 3 −3
3, b = 3 +3
3. (4)
The complete expression for the PDF of X is
f
X
(x) =
_
1/(6
3) 3 −3
3 ≤ x < 3 +3
3,
0 otherwise.
(5)
Quiz 3.5
Each of the requested probabilities can be calculated using (z) function and Table 3.1
(1) The PDFs of X and Y are shown below. The fact that Y has twice the standard
deviation of X is reflected in the greater spread of f
Y
(y). However, it is important
to remember that as the standard deviation increases, the peak value of the Gaussian
PDF goes down.
−5 0 5
0
0.2
0.4
x y
f
X
(
x
)
f
Y
(
y
)
← f
X
(x)
← f
Y
(y)
17
(2) Since X is Gaussian (0, 1),
P [−1 < X ≤ 1] = F
X
(1) − F
X
(−1) (1)
= (1) −(−1) = 2(1) −1 = 0.6826. (2)
(3) Since Y is Gaussian (0, 2),
P [−1 < Y ≤ 1] = F
Y
(1) − F
Y
(−1) (3)
=
_
1
σ
Y
_
_
−1
σ
Y
_
= 2
_
1
2
_
−1 = 0.383. (4)
(4) Again, since X is Gaussian (0, 1), P[X > 3.5] = Q(3.5) = 2.33 ×10
−4
.
(5) Since Y is Gaussian (0, 2), P[Y > 3.5] = Q(
3.5
2
) = Q(1.75) = 1 − (1.75) =
0.0401.
Quiz 3.6
The CDF of X is
−2 0 2
0
0.5
1
x
F
X
(
x
)
F
X
(x) =
0 x < −1,
(x +1)/4 −1 ≤ x < 1,
1 x ≥ 1.
(1)
The following probabilities can be read directly from the CDF:
(1) P[X ≤ 1] = F
X
(1) = 1.
(2) P[X < 1] = F
X
(1
) = 1/2.
(3) P[X = 1] = F
X
(1
+
) − F
X
(1
) = 1 −1/2 = 1/2.
(4) We find the PDF f
Y
(y) by taking the derivative of F
Y
(y). The resulting PDF is
−2 0 2
0
0.5
x
f
X
(
x
)
0.5
f
X
(x) =
1/4 −1 ≤ x < 1,
(1/2)δ(x −1) x = 1,
0 otherwise.
(2)
Quiz 3.7
18
(1) Since X is always nonnegative, F
X
(x) = 0 for x < 0. Also, F
X
(x) = 1 for x ≥ 2
since its always true that x ≤ 2. Lastly, for 0 ≤ x ≤ 2,
F
X
(x) =
_
x
−∞
f
X
(y) dy =
_
x
0
(1 − y/2) dy = x − x
2
/4. (1)
The complete CDF of X is
−1 0 1 2 3
0
0.5
1
x
F
X
(
x
)
F
X
(x) =
0 x < 0,
x − x
2
/4 0 ≤ x ≤ 2,
1 x > 2.
(2)
(2) The probability that Y = 1 is
P [Y = 1] = P [X ≥ 1] = 1 − F
X
(1) = 1 −3/4 = 1/4. (3)
(3) Since X is nonnegative, Y is also nonnegative. Thus F
Y
(y) = 0 for y < 0. Also,
because Y ≤ 1, F
Y
(y) = 1 for all y ≥ 1. Finally, for 0 < y < 1,
F
Y
(y) = P [Y ≤ y] = P [X ≤ y] = F
X
(y) . (4)
Using the CDF F
X
(x), the complete expression for the CDF of Y is
−1 0 1 2 3
0
0.5
1
y
F
Y
(
y
)
F
Y
(y) =
0 y < 0,
y − y
2
/4 0 ≤ y < 1,
1 y ≥ 1.
(5)
As expected, we see that the jump in F
Y
(y) at y = 1 is exactly equal to P[Y = 1].
(4) By taking the derivative of F
Y
(y), we obtain the PDF f
Y
(y). Note that when y < 0
or y > 1, the PDF is zero.
−1 0 1 2 3
0
0.5
1
1.5
y
f
Y
(
y
)
0.25
f
Y
(y) =
_
1 − y/2 +(1/4)δ(y −1) 0 ≤ y ≤ 1
0 otherwise
(6)
Quiz 3.8
(1) P[Y ≤ 6] =
_
6
−∞
f
Y
(y) dy =
_
6
0
(1/10) dy = 0.6 .
19
(2) From Definition 3.15, the conditional PDF of Y given Y ≤ 6 is
f
Y|Y≤6
(y) =
_
f
Y
(y)
P[Y≤6]
y ≤ 6,
0 otherwise,
=
_
1/6 0 ≤ y ≤ 6,
0 otherwise.
(1)
(3) The probability Y > 8 is
P [Y > 8] =
_
10
8
1
10
dy = 0.2 . (2)
(4) From Definition 3.15, the conditional PDF of Y given Y > 8 is
f
Y|Y>8
(y) =
_
f
Y
(y)
P[Y>8]
y > 8,
0 otherwise,
=
_
1/2 8 < y ≤ 10,
0 otherwise.
(3)
(5) From the conditional PDF f
Y|Y≤6
(y), we can calculate the conditional expectation
E [Y|Y ≤ 6] =
_
−∞
y f
Y|Y≤6
(y) dy =
_
6
0
y
6
dy = 3. (4)
(6) From the conditional PDF f
Y|Y>8
(y), we can calculate the conditional expectation
E [Y|Y > 8] =
_
−∞
y f
Y|Y>8
(y) dy =
_
10
8
y
2
dy = 9. (5)
Quiz 3.9
A natural way to produce random variables with PDF f
T|T>2
(t ) is to generate samples
of T with PDF f
T
(t ) and then to discard those samples which fail to satisfy the condition
T > 2. Here is a MATLAB function that uses this method:
function t=t2rv(m)
i=0;lambda=1/3;
t=zeros(m,1);
while (i<m),
x=exponentialrv(lambda,1);
if (x>2)
t(i+1)=x;
i=i+1;
end
end
A second method exploits the fact that if T is an exponential (λ) random variable, then
T
= T +2 has PDF f
T
(t ) = f
T|T>2
(t ). In this case the command
t=2.0+exponentialrv(1/3,m)
generates the vector t.
20
Quiz Solutions – Chapter 4
Quiz 4.1
Each value of the joint CDF can be found by considering the corresponding probability.
(1) F
X,Y
(−∞, 2) = P[X ≤ −∞, Y ≤ 2] ≤ P[X ≤ −∞] = 0 since X cannot take on
the value −∞.
(2) F
X,Y
(∞, ∞) = P[X ≤ ∞, Y ≤ ∞] = 1. This result is given in Theorem 4.1.
(3) F
X,Y
(∞, y) = P[X ≤ ∞, Y ≤ y] = P[Y ≤ y] = F
Y
(y).
(4) F
X,Y
(∞, −∞) = P[X ≤ ∞, Y ≤ −∞] = 0 since Y cannot take on the value −∞.
Quiz 4.2
From the joint PMF of Q and G given in the table, we can calculate the requested
probabilities by summing the PMF over those values of Q and G that correspond to the
event.
(1) The probability that Q = 0 is
P [Q = 0] = P
Q,G
(0, 0) + P
Q,G
(0, 1) + P
Q,G
(0, 2) + P
Q,G
(0, 3) (1)
= 0.06 +0.18 +0.24 +0.12 = 0.6 (2)
(2) The probability that Q = G is
P [Q = G] = P
Q,G
(0, 0) + P
Q,G
(1, 1) = 0.18 (3)
(3) The probability that G > 1 is
P [G > 1] =
3
g=2
1
q=0
P
Q,G
(q, g) (4)
= 0.24 +0.16 +0.12 +0.08 = 0.6 (5)
(4) The probability that G > Q is
P [G > Q] =
1
q=0
3
g=q+1
P
Q,G
(q, g) (6)
= 0.18 +0.24 +0.12 +0.16 +0.08 = 0.78 (7)
21
Quiz 4.3
By Theorem 4.3, the marginal PMF of H is
P
H
(h) =
b=0,2,4
P
H,B
(h, b) (1)
For each value of h, this corresponds to calculating the row sum across the table of the joint
PMF. Similarly, the marginal PMF of B is
P
B
(b) =
1
h=−1
P
H,B
(h, b) (2)
For each value of b, this corresponds to the column sum down the table of the joint PMF.
The easiest way to calculate these marginal PMFs is to simply sum each row and column:
P
H,B
(h, b) b = 0 b = 2 b = 4 P
H
(h)
h = −1 0 0.4 0.2 0.6
h = 0 0.1 0 0.1 0.2
h = 1 0.1 0.1 0 0.2
P
B
(b) 0.2 0.5 0.3
(3)
Quiz 4.4
To find the constant c, we apply
_
−∞
_
−∞
f
X,Y
(x, y) dx dy = 1. Specifically,
_
−∞
_
−∞
f
X,Y
(x, y) dx dy =
_
2
0
_
1
0
cxy dx dy (1)
= c
_
2
0
y
_
x
2
/2
¸
¸
¸
1
0
_
dy (2)
= (c/2)
_
2
0
y dy = (c/4)y
2
¸
¸
¸
2
0
= c (3)
Thus c = 1. To calculate P[A], we write
P [A] =
__
A
f
X,Y
(x, y) dx dy (4)
To integrate over A, we convert to polar coordinates using the substitutions x = r cos θ,
y = r sin θ and dx dy = r dr dθ, yielding
Y
X
1
1
2
A
P [A] =
_
π/2
0
_
1
0
r
2
sin θ cos θ r dr dθ (5)
=
_
_
1
0
r
3
dr
__
_
π/2
0
sin θ cos θ dθ
_
(6)
=
_
r
4
/4
¸
¸
¸
1
0
_
sin
2
θ
2
¸
¸
¸
¸
¸
π/2
0
= 1/8 (7)
22
Quiz 4.5
By Theorem 4.8, the marginal PDF of X is
f
X
(x) =
_
−∞
f
X,Y
(x, y) dy (1)
For x < 0 or x > 1, f
X
(x) = 0. For 0 ≤ x ≤ 1,
f
X
(x) =
6
5
_
1
0
(x + y
2
) dy =
6
5
_
xy + y
3
/3
¸
¸
y=1
y=0
=
6
5
(x +1/3) =
6x +2
5
(2)
The complete expression for the PDf of X is
f
X
(x) =
_
(6x +2)/5 0 ≤ x ≤ 1
0 otherwise
(3)
By the same method we obtain the marginal PDF for Y. For 0 ≤ y ≤ 1,
f
Y
(y) =
_
−∞
f
X,Y
(x, y) dy (4)
=
6
5
_
1
0
(x + y
2
) dx =
6
5
_
x
2
/2 + xy
2
¸
¸
x=1
x=0
=
6
5
(1/2 + y
2
) =
3 +6y
2
5
(5)
Since f
Y
(y) = 0 for y < 0 or y > 1, the complete expression for the PDF of Y is
f
Y
(y) =
_
(3 +6y
2
)/5 0 ≤ y ≤ 1
0 otherwise
(6)
Quiz 4.6
(A) The time required for the transfer is T = L/B. For each pair of values of L and B,
we can calculate the time T needed for the transfer. We can write these down on the
table for the joint PMF of L and B as follows:
P
L,B
(l, b) b = 14, 400 b = 21, 600 b = 28, 800
l = 518, 400 0.20 (T=36) 0.10 (T=24) 0.05 (T=18)
l = 2, 592, 000 0.05 (T=180) 0.10 (T=120) 0.20 (T=90)
l = 7, 776, 000 0.00 (T=540) 0.10 (T=360) 0.20 (T=270)
From the table, writing down the PMF of T is straightforward.
P
T
(t ) =
0.05 t = 18
0.1 t = 24
0.2 t = 36, 90
0.1 t = 120
0.05 t = 180
0.2 t = 270
0.1 t = 360
0 otherwise
(1)
23
(B) First, we observe that since 0 ≤ X ≤ 1 and 0 ≤ Y ≤ 1, W = XY satisfies
0 ≤ W ≤ 1. Thus f
W
(0) = 0 and f
W
(1) = 1. For 0 < w < 1, we calculate the
CDF F
W
(w) = P[W ≤ w]. As shown below, integrating over the region W ≤ w
is fairly complex. The calculus is simpler if we integrate over the region XY > w.
Specifically,
Y
X
1
1
XY > w
w
w
XY = w
F
W
(w) = 1 − P [XY > w] (2)
= 1 −
_
1
w
_
1
w/x
dy dx (3)
= 1 −
_
1
w
(1 −w/x) dx (4)
= 1 −
_
x −wln x|
x=1
x=w
_
(5)
= 1 −(1 −w +wln w) = w −wln w (6)
The complete expression for the CDF is
F
W
(w) =
0 w < 0
w −wln w 0 ≤ w ≤ 1
1 w > 1
(7)
By taking the derivative of the CDF, we find the PDF is
f
W
(w) =
d F
W
(w)
dw
=
0 w < 0
−ln w 0 ≤ w ≤ 1
0 w > 1
(8)
Quiz 4.7
(A) It is helpful to first make a table that includes the marginal PMFs.
P
L,T
(l, t ) t = 40 t = 60 P
L
(l)
l = 1 0.15 0.1 0.25
l = 2 0.3 0.2 0.5
l = 3 0.15 0.1 0.25
P
T
(t ) 0.6 0.4
(1) The expected value of L is
E [L] = 1(0.25) +2(0.5) +3(0.25) = 2. (1)
Since the second moment of L is
E
_
L
2
_
= 1
2
(0.25) +2
2
(0.5) +3
2
(0.25) = 4.5, (2)
the variance of L is
Var [L] = E
_
L
2
_
−(E [L])
2
= 0.5. (3)
24
(2) The expected value of T is
E [T] = 40(0.6) +60(0.4) = 48. (4)
The second moment of T is
E
_
T
2
_
= 40
2
(0.6) +60
2
(0.4) = 2400. (5)
Thus
Var[T] = E
_
T
2
_
−(E [T])
2
= 2400 −48
2
= 96. (6)
(3) The correlation is
E [LT] =
t =40,60
3
l=1
lt P
LT
(lt ) (7)
= 1(40)(0.15) +2(40)(0.3) +3(40)(0.15) (8)
+1(60)(0.1) +2(60)(0.2) +3(60)(0.1) (9)
= 96 (10)
(4) From Theorem 4.16(a), the covariance of L and T is
Cov [L, T] = E [LT] − E [L] E [T] = 96 −2(48) = 0 (11)
(5) Since Cov[L, T] = 0, the correlation coefficient is ρ
L,T
= 0.
(B) As in the discrete case, the calculations become easier if we first calculate the marginal
PDFs f
X
(x) and f
Y
(y). For 0 ≤ x ≤ 1,
f
X
(x) =
_
−∞
f
X,Y
(x, y) dy =
_
2
0
xy dy =
1
2
xy
2
¸
¸
¸
¸
y=2
y=0
= 2x (12)
Similarly, for 0 ≤ y ≤ 2,
f
Y
(y) =
_
−∞
f
X,Y
(x, y) dx =
_
2
0
xy dx =
1
2
x
2
y
¸
¸
¸
¸
x=1
x=0
=
y
2
(13)
The complete expressions for the marginal PDFs are
f
X
(x) =
_
2x 0 ≤ x ≤ 1
0 otherwise
f
Y
(y) =
_
y/2 0 ≤ y ≤ 2
0 otherwise
(14)
From the marginal PDFs, it is straightforward to calculate the various expectations.
25
(1) The first and second moments of X are
E [X] =
_
−∞
x f
X
(x) dx =
_
1
0
2x
2
dx =
2
3
(15)
E
_
X
2
_
=
_
−∞
x
2
f
X
(x) dx =
_
1
0
2x
3
dx =
1
2
(16)
(17)
The variance of X is Var[X] = E[X
2
] −(E[X])
2
= 1/18.
(2) The first and second moments of Y are
E [Y] =
_
−∞
y f
Y
(y) dy =
_
2
0
1
2
y
2
dy =
4
3
(18)
E
_
Y
2
_
=
_
−∞
y
2
f
Y
(y) dy =
_
2
0
1
2
y
3
dy = 2 (19)
The variance of Y is Var[Y] = E[Y
2
] −(E[Y])
2
= 2 −16/9 = 2/9.
(3) The correlation of X and Y is
E [XY] =
_
−∞
_
−∞
xy f
X,Y
(x, y) dx, dy (20)
=
_
1
0
_
2
0
x
2
y
2
dx, dy =
x
3
3
¸
¸
¸
¸
1
0
y
3
3
¸
¸
¸
¸
2
0
=
8
9
(21)
(4) The covariance of X and Y is
Cov [X, Y] = E [XY] − E [X] E [Y] =
8
9
_
2
3
__
4
3
_
= 0. (22)
(5) Since Cov[X, Y] = 0, the correlation coefficient is ρ
X,Y
= 0.
Quiz 4.8
(A) Since the event V > 80 occurs only for the pairs (L, T) = (2, 60), (L, T) = (3, 40)
and (L, T) = (3, 60),
P [A] = P [V > 80] = P
L,T
(2, 60) + P
L,T
(3, 40) + P
L,T
(3, 60) = 0.45 (1)
By Definition 4.9,
P
L,T| A
(l, t ) =
_
P
L,T
(l,t )
P[A]
lt > 80
0 otherwise
(2)
26
We can represent this conditional PMF in the following table:
P
L,T| A
(l, t ) t = 40 t = 60
l = 1 0 0
l = 2 0 4/9
l = 3 1/3 2/9
The conditional expectation of V can be found from the conditional PMF.
E [V| A] =
l
t
lt P
L,T| A
(l, t ) (3)
= (2 · 60)
4
9
+(3 · 40)
1
3
+(3 · 60)
2
9
= 133
1
3
(4)
For the conditional variance Var[V| A], we first find the conditional second moment
E
_
V
2
| A
_
=
l
t
(lt )
2
P
L,T| A
(l, t ) (5)
= (2 · 60)
2
4
9
+(3 · 40)
2
1
3
+(3 · 60)
2
2
9
= 18, 400 (6)
It follows that
Var [V| A] = E
_
V
2
| A
_
−(E [V| A])
2
= 622
2
9
(7)
(B) For continuous random variables X and Y, we first calculate the probability of the
conditioning event.
P [B] =
__
B
f
X,Y
(x, y) dx dy =
_
60
40
_
3
80/y
xy
4000
dx dy (8)
=
_
60
40
y
4000
_
x
2
2
¸
¸
¸
¸
3
80/y
_
dy (9)
=
_
60
40
y
4000
_
9
2
3200
y
2
_
dy (10)
=
9
8
4
5
ln
3
2
≈ 0.801 (11)
The conditional PDF of X and Y is
f
X,Y|B
(x, y) =
_
f
X,Y
(x, y) /P [B] (x, y) ∈ B
0 otherwise
(12)
=
_
Kxy 40 ≤ y ≤ 60, 80/y ≤ x ≤ 3
0 otherwise
(13)
27
where K = (4000P[B])
−1
. The conditional expectation of W given event B is
E [W|B] =
_
−∞
_
−∞
xy f
X,Y|B
(x, y) dx dy (14)
=
_
60
40
_
3
80/y
Kx
2
y
2
dx dy (15)
= (K/3)
_
60
40
y
2
x
3
¸
¸
¸
x=3
x=80/y
dy (16)
= (K/3)
_
60
40
_
27y
2
−80
3
/y
_
dy (17)
= (K/3)
_
9y
3
−80
3
ln y
¸
¸
60
40
≈ 120.78 (18)
The conditional second moment of K given B is
E
_
W
2
|B
_
=
_
−∞
_
−∞
(xy)
2
f
X,Y|B
(x, y) dx dy (19)
=
_
60
40
_
3
80/y
Kx
3
y
3
dx dy (20)
= (K/4)
_
60
40
y
3
x
4
¸
¸
¸
x=3
x=80/y
dy (21)
= (K/4)
_
60
40
_
81y
3
−80
4
/y
_
dy (22)
= (K/4)
_
(81/4)y
4
−80
4
ln y
¸
¸
60
40
≈ 16, 116.10 (23)
It follows that the conditional variance of W given B is
Var [W|B] = E
_
W
2
|B
_
−(E [W|B])
2
≈ 1528.30 (24)
Quiz 4.9
(A) (1) The joint PMF of A and B can be found from the marginal and conditional
PMFs via P
A,B
(a, b) = P
B| A
(b|a)P
A
(a). Incorporating the information from
the given conditional PMFs can be confusing, however. Consequently, we can
note that A has range S
A
= {0, 2} and B has range S
B
= {0, 1}. A table of the
joint PMF will include all four possible combinations of A and B. The general
form of the table is
P
A,B
(a, b) b = 0 b = 1
a = 0 P
B| A
(0|0)P
A
(0) P
B| A
(1|0)P
A
(0)
a = 2 P
B| A
(0|2)P
A
(2) P
B| A
(1|2)P
A
(2)
28
Substituting values from P
B| A
(b|a) and P
A
(a), we have
P
A,B
(a, b) b = 0 b = 1
a = 0 (0.8)(0.4) (0.2)(0.4)
a = 2 (0.5)(0.6) (0.5)(0.6)
or
P
A,B
(a, b) b = 0 b = 1
a = 0 0.32 0.08
a = 2 0.3 0.3
(2) Given the conditional PMF P
B| A
(b|2), it is easy to calculate the conditional
expectation
E [B| A = 2] =
1
b=0
bP
B| A
(b|2) = (0)(0.5) +(1)(0.5) = 0.5 (1)
(3) From the joint PMF P
A,B
(a, b), we can calculate the the conditional PMF
P
A|B
(a|0) =
P
A,B
(a, 0)
P
B
(0)
=
0.32/0.62 a = 0
0.3/0.62 a = 2
0 otherwise
(2)
=
16/31 a = 0
15/31 a = 2
0 otherwise
(3)
(4) We can calculate the conditional variance Var[A|B = 0] using the conditional
PMF P
A|B
(a|0). First we calculate the conditional expected value
E [A|B = 0] =
a
aP
A|B
(a|0) = 0(16/31) +2(15/31) = 30/31 (4)
The conditional second moment is
E
_
A
2
|B = 0
_
=
a
a
2
P
A|B
(a|0) = 0
2
(16/31) +2
2
(15/31) = 60/31 (5)
The conditional variance is then
Var[A|B = 0] = E
_
A
2
|B = 0
_
−(E [A|B = 0])
2
=
960
961
(6)
(B) (1) The joint PDF of X and Y is
f
X,Y
(x, y) = f
Y|X
(y|x) f
X
(x) =
_
6y 0 ≤ y ≤ x, 0 ≤ x ≤ 1
0 otherwise
(7)
(2) From the given conditional PDF f
Y|X
(y|x),
f
Y|X
(y|1/2) =
_
8y 0 ≤ y ≤ 1/2
0 otherwise
(8)
29
(3) The conditional PDF of Y given X = 1/2 is f
X|Y
(x|1/2) = f
X,Y
(x, 1/2)/f
Y
(1/2).
To find f
Y
(1/2), we integrate the joint PDF.
f
Y
(1/2) =
_
−∞
f
X,1/2
( ) dx =
_
1
1/2
6(1/2) dx = 3/2 (9)
Thus, for 1/2 ≤ x ≤ 1,
f
X|Y
(x|1/2) =
f
X,Y
(x, 1/2)
f
Y
(1/2)
=
6(1/2)
3/2
= 2 (10)
(4) From the pervious part, we see that given Y = 1/2, the conditional PDF of X
is uniform (1/2, 1). Thus, by the definition of the uniform (a, b) PDF,
Var [X|Y = 1/2] =
(1 −1/2)
2
12
=
1
48
(11)
Quiz 4.10
(A) (1) For random variables X and Y from Example 4.1, we observe that P
Y
(1) =
0.09 and P
X
(0) = 0.01. However,
P
X,Y
(0, 1) = 0 = P
X
(0) P
Y
(1) (1)
Since we have found a pair x, y such that P
X,Y
(x, y) = P
X
(x)P
Y
(y), we can
conclude that X and Y are dependent. Note that whenever P
X,Y
(x, y) = 0,
independence requires that either P
X
(x) = 0 or P
Y
(y) = 0.
(2) For random variables Q and G from Quiz 4.2, it is not obvious whether they
are independent. Unlike X and Y in part (a), there are no obvious pairs q, g
that fail the independence requirement. In this case, we calculate the marginal
PMFs from the table of the joint PMF P
Q,G
(q, g) in Quiz 4.2.
P
Q,G
(q, g) g = 0 g = 1 g = 2 g = 3 P
Q
(q)
q = 0 0.06 0.18 0.24 0.12 0.60
q = 1 0.04 0.12 0.16 0.08 0.40
P
G
(g) 0.10 0.30 0.40 0.20
Careful study of the table will verify that P
Q,G
(q, g) = P
Q
(q)P
G
(g) for every
pair q, g. Hence Q and G are independent.
(B) (1) Since X
1
and X
2
are independent,
f
X
1
,X
2
(x
1
, x
2
) = f
X
1
(x
1
) f
X
2
(x
2
) (2)
=
_
(1 − x
1
/2)(1 − x
2
/2) 0 ≤ x
1
≤ 2, 0 ≤ x
2
≤ 2
0 otherwise
(3)
30
(2) Let F
X
(x) denote the CDF of both X
1
and X
2
. The CDF of Z = max(X
1
, X
2
)
is found by observing that Z ≤ z iff X
1
≤ z and X
2
≤ z. That is,
P [Z ≤ z] = P [X
1
≤ z, X
2
≤ z] (4)
= P [X
1
≤ z] P [X
2
≤ z] = [F
X
(z)]
2
(5)
To complete the problem, we need to find the CDF of each X
i
. From the PDF
f
X
(x), the CDF is
F
X
(x) =
_
x
−∞
f
X
(y) dy =
0 x < 0
x − x
2
/4 0 ≤ x ≤ 2
1 x > 2
(6)
Thus for 0 ≤ z ≤ 2,
F
Z
(z) = (z − z
2
/4)
2
(7)
The complete expression for the CDF of Z is
F
Z
(z) =
0 z < 0
(z − z
2
/4)
2
0 ≤ z ≤ 2
1 z > 1
(8)
Quiz 4.11
This problem just requires identifying the various terms in Definition 4.17 and Theo-
rem 4.29. Specifically, from the problem statement, we know that ρ = 1/2,
µ
1
= µ
X
= 0, µ
2
= µ
Y
= 0, (1)
and that
σ
1
= σ
X
= 1, σ
2
= σ
Y
= 1. (2)
(1) Applying these facts to Definition 4.17, we have
f
X,Y
(x, y) =
1
2
e
−2(x
2
−xy+y
2
)/3
. (3)
(2) By Theorem 4.30, the conditional expected value and standard deviation of X given
Y = y are
E [X|Y = y] = y/2 ˜ σ
X
= σ
2
1
(1 −ρ
2
) =
_
3/4. (4)
When Y = y = 2, we see that E[X|Y = 2] = 1 and Var[X|Y = 2] = 3/4. The
conditional PDF of X given Y = 2 is simply the Gaussian PDF
f
X|Y
(x|2) =
1
3π/2
e
−2(x−1)
2
/3
. (5)
31
Quiz 4.12
One straightforward method is to follow the approach of Example 4.28. Instead, we use
an alternate approach. First we observe that X has the discrete uniform (1, 4) PMF. Also,
given X = x, Y has a discrete uniform (1, x) PMF. That is,
P
X
(x) =
_
1/4 x = 1, 2, 3, 4,
0 otherwise,
P
Y|X
(y|x) =
_
1/x y = 1, . . . , x
0 otherwise
(1)
Given X = x, and an independent uniform (0, 1) random variable U, we can generate a
sample value of Y with a discrete uniform (1, x) PMF via Y = xU. This observation
prompts the following program:
function xy=dtrianglerv(m)
sx=[1;2;3;4];
px=0.25*ones(4,1);
x=finiterv(sx,px,m);
y=ceil(x.*rand(m,1));
xy=[x’;y’];
32
Quiz Solutions – Chapter 5
Quiz 5.1
We find P[C] by integrating the joint PDF over the region of interest. Specifically,
P [C] =
_
1/2
0
dy
2
_
y
2
0
dy
1
_
1/2
0
dy
4
_
y
4
0
4dy
3
(1)
= 4
_
_
1/2
0
y
2
dy
2
__
_
1/2
0
y
4
dy
4
_
= 1/4. (2)
Quiz 5.2
By definition of A, Y
1
= X
1
, Y
2
= X
2
−X
1
and Y
3
= X
3
−X
2
. Since 0 < X
1
< X
2
<
X
3
, each Y
i
must be a strictly positive integer. Thus, for y
1
, y
2
, y
3
∈ {1, 2, . . .},
P
Y
(y) = P [Y
1
= y
1
, Y
2
= y
2
, Y
3
= y
3
] (1)
= P [X
1
= y
1
, X
2
− X
1
= y
2
, X
3
− X
2
= y
3
] (2)
= P [X
1
= y
1
, X
2
= y
2
+ y
1
, X
3
= y
3
+ y
2
+ y
1
] (3)
= (1 − p)
3
p
y
1
+y
2
+y
3
(4)
By defining the vector a =
_
1 1 1
_
, the complete expression for the joint PMF of Y is
P
Y
(y) =
_
(1 − p) p
a
y
y
1
, y
2
, y
3
∈ {1, 2, . . .}
0 otherwise
(5)
Quiz 5.3
First we note that each marginal PDF is nonzero only if any subset of the x
i
obeys the
ordering contraints 0 ≤ x
1
≤ x
2
≤ x
3
≤ 1. Within these constraints, we have
f
X
1
,X
2
(x
1
, x
2
) =
_
−∞
f
X
(x) dx
3
=
_
1
x
2
6 dx
3
= 6(1 − x
2
), (1)
f
X
2
,X
3
(x
2
, x
3
) =
_
−∞
f
X
(x) dx
1
=
_
x
2
0
6 dx
1
= 6x
2
, (2)
f
X
1
,X
3
(x
1
, x
3
) =
_
−∞
f
X
(x) dx
2
=
_
x
3
x
1
6 dx
2
= 6(x
3
− x
1
). (3)
In particular, we must keep in mind that f
X
1
,X
2
(x
1
, x
2
) = 0 unless 0 ≤ x
1
≤ x
2
≤ 1,
f
X
2
,X
3
(x
2
, x
3
) = 0 unless 0 ≤ x
2
≤ x
3
≤ 1, and that f
X
1
,X
3
(x
1
, x
3
) = 0 unless 0 ≤ x
1
33
x
3
≤ 1. The complete expressions are
f
X
1
,X
2
(x
1
, x
2
) =
_
6(1 − x
2
) 0 ≤ x
1
≤ x
2
≤ 1
0 otherwise
(4)
f
X
2
,X
3
(x
2
, x
3
) =
_
6x
2
0 ≤ x
2
≤ x
3
≤ 1
0 otherwise
(5)
f
X
1
,X
3
(x
1
, x
3
) =
_
6(x
3
− x
1
) 0 ≤ x
1
≤ x
3
≤ 1
0 otherwise
(6)
Now we can find the marginal PDFs. When 0 ≤ x
i
≤ 1 for each x
i
,
f
X
1
(x
1
) =
_
−∞
f
X
1
,X
2
(x
1
, x
2
) dx
2
=
_
1
x
1
6(1 − x
2
) dx
2
= 3(1 − x
1
)
2
(7)
f
X
2
(x
2
) =
_
−∞
f
X
2
,X
3
(x
2
, x
3
) dx
3
=
_
1
x
2
6x
2
dx
3
= 6x
2
(1 − x
2
) (8)
f
X
3
(x
3
) =
_
−∞
f
X
2
,X
3
(x
2
, x
3
) dx
2
=
_
x
3
0
6x
2
dx
2
= 3x
2
3
(9)
The complete expressions are
f
X
1
(x
1
) =
_
3(1 − x
1
)
2
0 ≤ x
1
≤ 1
0 otherwise
(10)
f
X
2
(x
2
) =
_
6x
2
(1 − x
2
) 0 ≤ x
2
≤ 1
0 otherwise
(11)
f
X
3
(x
3
) =
_
3x
2
3
0 ≤ x
3
≤ 1
0 otherwise
(12)
Quiz 5.4
In the PDF f
Y
(y), the components have dependencies as a result of the ordering con-
straints Y
1
≤ Y
2
and Y
3
≤ Y
4
. We can separate these constraints by creating the vectors
V =
_
Y
1
Y
2
_
, W =
_
Y
3
Y
4
_
. (1)
The joint PDF of V and W is
f
V,W
(v, w) =
_
4 0 ≤ v
1
≤ v
2
≤ 1, 0 ≤ w
1
≤ w
2
≤ 1
0 otherwise
(2)
34
We must verify that V and W are independent. For 0 ≤ v
1
≤ v
2
≤ 1,
f
V
(v) =
__
f
V,W
(v, w) dw
1
dw
2
(3)
=
_
1
0
_
_
1
w
1
4 dw
2
_
dw
1
(4)
=
_
1
0
4(1 −w
1
) dw
1
= 2 (5)
Similarly, for 0 ≤ w
1
≤ w
2
≤ 1,
f
W
(w) =
__
f
V,W
(v, w) dv
1
dv
2
(6)
=
_
1
0
_
_
1
v
1
4 dv
2
_
dv
1
= 2 (7)
It follows that V and W have PDFs
f
V
(v) =
_
2 0 ≤ v
1
≤ v
2
≤ 1
0 otherwise
, f
W
(w) =
_
2 0 ≤ w
1
≤ w
2
≤ 1
0 otherwise
(8)
It is easy to verify that f
V,W
(v, w) = f
V
(v) f
W
(w), confirming that V and W are indepen-
dent vectors.
Quiz 5.5
(A) Referring to Theorem 1.19, each test is a subexperiment with three possible out-
comes: L, A and R. In five trials, the vector X =
_
X
1
X
2
X
3
_
indicating the
number of outcomes of each subexperiment has the multinomial PMF
P
X
(x) =
_
5
x
1
,x
2
,x
3
_
(0.3)
x
1
(0.6)
x
2
(0.1)
x
3
x
1
+ x
2
+ x
3
= 5;
x
1
, x
2
, x
3
∈ {0, 1, . . . , 5}
0 otherwise
(1)
We can find the marginal PMF for each X
i
from the joint PMF P
X
(x); however it
is simpler to just start from first principles and observe that X
1
is the number of
occurrences of L in five independent tests. If we view each test as a trial with success
probability P[L] = 0.3, we see that X
1
is a binomial (n, p) = (5, 0.3) random
variable. Similarly, X
2
is a binomial (5, 0.6) random variable and X
3
is a binomial
(5, 0.1) random variable. That is, for p
1
= 0.3, p
2
= 0.6 and p
3
= 0.1,
P
X
i
(x) =
_ _
5
x
_
p
x
i
(1 − p
i
)
5−x
x = 0, 1, . . . , 5
0 otherwise
(2)
35
From the marginal PMFs, we see that X
1
, X
2
and X
3
are not independent. Hence, we
must use Theorem 5.6 to find the PMF of W. In particular, since X
1
+ X
2
+ X
3
= 5
and since each X
i
is non-negative, P
W
(0) = P
W
(1) = 0. Furthermore,
P
W
(2) = P
X
(1, 2, 2) + P
X
(2, 1, 2) + P
X
(2, 2, 1) (3)
=
5![0.3(0.6)
2
(0.1)
2
+0.3
2
(0.6)(0.1)
2
+0.3
2
(0.6)
2
(0.1)]
2!2!1!
(4)
= 0.1458 (5)
In addition, for w = 3, w = 4, and w = 5, the event W = w occurs if and only if
one of the mutually exclusive events X
1
= w, X
2
= w, or X
3
= w occurs. Thus,
P
W
(3) = P
X
1
(3) + P
X
2
(3) + P
X
3
(3) = 0.486 (6)
P
W
(4) = P
X
1
(4) + P
X
2
(4) + P
X
3
(4) = 0.288 (7)
P
W
(5) = P
X
1
(5) + P
X
2
(5) + P
X
3
(5) = 0.0802 (8)
(B) Since each Y
i
= 2X
i
+4, we can apply Theorem 5.10 to write
f
Y
(y) =
1
2
3
f
X
_
y
1
−4
2
,
y
2
−4
2
,
y
3
−4
2
_
(9)
=
_
(1/8)e
−(y
3
−4)/2
4 ≤ y
1
≤ y
2
≤ y
3
0 otherwise
(10)
Note that for other matrices A, the constraints on y resulting from the constraints
0 ≤ X
1
≤ X
2
≤ X
3
can be much more complicated.
Quiz 5.6
We start by finding the components E[X
i
] =
_
−∞
x f
X
i
(x) dx of µ
X
. To do so, we use
the marginal PDFs f
X
i
(x) found in Quiz 5.3:
E [X
1
] =
_
1
0
3x(1 − x)
2
dx = 1/4, (1)
E [X
2
] =
_
1
0
6x
2
(1 − x) dx = 1/2, (2)
E [X
3
] =
_
1
0
3x
3
dx = 3/4. (3)
To find the correlation matrix R
X
, we need to find E[X
i
X
j
36
the second moments:
E
_
X
2
1
_
=
_
1
0
3x
2
(1 − x)
2
dx = 1/10. (4)
E
_
X
2
2
_
=
_
1
0
6x
3
(1 − x) dx = 3/10. (5)
E
_
X
2
3
_
=
_
1
0
3x
4
dx = 3/5. (6)
Using marginal PDFs from Quiz 5.3, the cross terms are
E [X
1
X
2
] =
_
−∞
_
−∞
x
1
x
2
f
X
1
,X
2
(x
1
, x
2
) , dx
1
dx
2
(7)
=
_
1
0
_
_
1
x
1
6x
1
x
2
(1 − x
2
) dx
2
_
dx
1
(8)
=
_
1
0
[x
1
−3x
3
1
+2x
4
1
] dx
1
= 3/20. (9)
E [X
2
X
3
] =
_
1
0
_
1
x
2
6x
2
2
x
3
dx
3
dx
2
(10)
=
_
1
0
[3x
2
2
−3x
4
2
] dx
2
= 2/5 (11)
E [X
1
X
3
] =
_
1
0
_
1
x
1
6x
1
x
3
(x
3
− x
1
) dx
3
dx
1
. (12)
=
_
1
0
_
(2x
1
x
3
3
−3x
2
1
x
2
3
)
¸
¸
¸
x
3
=1
x
3
=x
1
_
dx
1
(13)
=
_
1
0
[2x
1
−3x
2
1
+ x
4
1
] dx
1
= 1/5. (14)
Summarizing the results, X has correlation matrix
R
X
=
1/10 3/20 1/5
3/20 3/10 2/5
1/5 2/5 3/5
. (15)
Vector X has covariance matrix
C
X
= R
X
− E [X] E [X]
(16)
=
1/10 3/20 1/5
3/20 3/10 2/5
1/5 2/5 3/5
1/4
1/2
3/4
_
1/4 1/2 3/4
_
(17)
=
1/10 3/20 1/5
3/20 3/10 2/5
1/5 2/5 3/5
1/16 1/8 3/16
1/8 1/4 3/8
3/16 3/8 9/16
=
1
80
3 2 1
2 4 2
1 2 3
. (18)
37
This problemshows that even for fairly simple joint PDFs, computing the covariance matrix
by calculus can be a time consuming task.
Quiz 5.7
We observe that X = AZ +b where
A =
_
2 1
1 −1
_
, b =
_
2
0
_
. (1)
It follows from Theorem 5.18 that µ
X
= b and that
C
X
= AA
=
_
2 1
1 −1
_ _
2 1
1 −1
_
=
_
5 1
1 2
_
. (2)
Quiz 5.8
First, we observe that Y = AT where A =
_
1/31 1/31 · · · 1/31
_
. Since T is a
Gaussian random vector, Theorem 5.16 tells us that Y is a 1 dimensional Gaussian vector,
i.e., just a Gaussian random variable. The expected value of Y is µ
Y
= µ
T
= 80. The
covariance matrix of Y is 1 × 1 and is just equal to Var[Y]. Thus, by Theorem 5.16,
Var[Y] = AC
T
A
.
function p=julytemps(T);
[D1 D2]=ndgrid((1:31),(1:31));
CT=36./(1+abs(D1-D2));
A=ones(31,1)/31.0;
CY=(A’)*CT*A;
p=phi((T-80)/sqrt(CY));
In julytemps.m, the first two lines gen-
erate the 31 ×31 covariance matrix CT, or
C
T
. Next we calculate Var[Y]. The final
step is to use the (·) function to calculate
P[Y < T].
Here is the output of julytemps.m:
>> julytemps([70 75 80 85 90 95])
ans =
0.0000 0.0221 0.5000 0.9779 1.0000 1.0000
Note that P[T ≤ 70] is not actually zero and that P[T ≤ 90] is not actually 1.0000. Its
just that the MATLAB’s short format output, invoked with the command format short,
rounds off those probabilities. Here is the long format output:
>> format long
>> julytemps([70 75 80 85 90 95])
ans =
Columns 1 through 4
0.00002844263128 0.02207383067604 0.50000000000000 0.97792616932396
Columns 5 through 6
0.99997155736872 0.99999999922010
38
The ndgrid function is a useful to way calculate many covariance matrices. However, in
this problem, C
X
has a special structure; the i, j th element is
C
T
(i, j ) = c
|i −j |
=
36
1 +|i − j |
. (1)
If we write out the elements of the covariance matrix, we see that
C
T
=
c
0
c
1
· · · c
30
c
1
c
0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. c
1
c
30
· · · c
1
c
0
. (2)
This covariance matrix is known as a symmetric Toeplitz matrix. We will see in Chap-
ters 9 and 11 that Toeplitz covariance matrices are quite common. In fact, MATLAB has a
toeplitz function for generating them. The function julytemps2 use the toeplitz
to generate the correlation matrix C
T
.
function p=julytemps2(T);
c=36./(1+abs(0:30));
CT=toeplitz(c);
A=ones(31,1)/31.0;
CY=(A’)*CT*A;
p=phi((T-80)/sqrt(CY));
39
Quiz Solutions – Chapter 6
Quiz 6.1
Let K
1
, . . . , K
n
denote a sequence of iid random variables each with PMF
P
K
(k) =
_
1/4 k = 1, . . . , 4
0 otherwise
(1)
We can write W
n
in the form of W
n
= K
1
+ · · · + K
n
. First, we note that the first two
moments of K
i
are
E [K
i
] = (1 +2 +3 +4)/4 = 2.5 (2)
E
_
K
2
i
_
= (1
2
+2
2
+3
2
+4
2
)/4 = 7.5 (3)
Thus the variance of K
i
is
Var[K
i
] = E
_
K
2
i
_
−(E [K
i
])
2
= 7.5 −(2.5)
2
= 1.25 (4)
Since E[K
i
] = 2.5, the expected value of W
n
is
E [W
n
] = E [K
1
] +· · · + E [K
n
] = nE [K
i
] = 2.5n (5)
Since the rolls are independent, the random variables K
1
, . . . , K
n
are independent. Hence,
by Theorem 6.3, the variance of the sum equals the sum of the variances. That is,
Var[W
n
] = Var[K
1
] +· · · +Var[K
n
] = 1.25n (6)
Quiz 6.2
Random variables X and Y have PDFs
f
X
(x) =
_
3e
−3x
x ≥ 0
0 otherwise
f
Y
(y) =
_
2e
−2y
y ≥ 0
0 otherwise
(1)
Since X and Y are nonnegative, W = X +Y is nonnegative. By Theorem 6.5, the PDF of
W = X +Y is
f
W
(w) =
_
−∞
f
X
(w − y) f
Y
(y) dy = 6
_
w
0
e
−3(w−y)
e
−2y
dy (2)
Fortunately, this integral is easy to evaluate. For w > 0,
f
W
(w) = e
−3w
e
y
¸
¸
w
0
= 6
_
e
−2w
−e
−3w
_
(3)
Since f
W
(w) = 0 for w < 0, a conmplete expression for the PDF of W is
f
W
(w) =
_
6e
−2w
_
1 −e
−w
_
w ≥ 0,
0 otherwise.
(4)
40
Quiz 6.3
The MGF of K is
φ
K
(s) = E
_
e
s K
_
==
4
k=0
(0.2)e
sk
= 0.2
_
1 +e
s
+e
2s
+e
3s
+e
4s
_
(1)
We find the moments by taking derivatives. The first derivative of φ
K
(s) is
K
(s)
ds
= 0.2(e
s
+2e
2s
+3e
3s
+4e
4s
) (2)
Evaluating the derivative at s = 0 yields
E [K] =
K
(s)
ds
¸
¸
¸
¸
s=0
= 0.2(1 +2 +3 +4) = 2 (3)
To find higher-order moments, we continue to take derivatives:
E
_
K
2
_
=
d
2
φ
K
(s)
ds
2
¸
¸
¸
¸
s=0
= 0.2(e
s
+4e
2s
+9e
3s
+16e
4s
)
¸
¸
¸
s=0
= 6 (4)
E
_
K
3
_
=
d
3
φ
K
(s)
ds
3
¸
¸
¸
¸
s=0
= 0.2(e
s
+8e
2s
+27e
3s
+64e
4s
)
¸
¸
¸
s=0
= 20 (5)
E
_
K
4
_
=
d
4
φ
K
(s)
ds
4
¸
¸
¸
¸
s=0
= 0.2(e
s
+16e
2s
+81e
3s
+256e
4s
)
¸
¸
¸
s=0
= 70.8 (6)
(7)
Quiz 6.4
(A) Each K
i
has MGF
φ
K
(s) = E
_
e
s K
i
_
=
e
s
+e
2s
+· · · +e
ns
n
=
e
s
(1 −e
ns
)
n(1 −e
s
)
(1)
Since the sequence of K
i
is independent, Theorem 6.8 says the MGF of J is
φ
J
(s) = (φ
K
(s))
m
=
e
ms
(1 −e
ns
)
m
n
m
(1 −e
s
)
m
(2)
(B) Since the set of α
j
X
j
are independent Gaussian random variables, Theorem 6.10
says that W is a Gaussian random variable. Thus to find the PDF of W, we need
only find the expected value and variance. Since the expectation of the sum equals
the sum of the expectations:
E [W] = αE [X
1
] +α
2
E [X
2
] +· · · +α
n
E [X
n
] = 0 (3)
41
Since the α
j
X
j
are independent, the variance of the sum equals the sum of the vari-
ances:
Var[W] = α
2
Var[X
1
] +α
4
Var[X
2
] +· · · +α
2n
Var[X
n
] (4)
= α
2
+2(α
2
)
2
+3(α
2
)
3
+· · · +n(α
2
)
n
(5)
Defining q = α
2
, we can use Math Fact B.6 to write
Var[W] =
α
2
−α
2n+2
[1 +n(1 −α
2
)]
(1 −α
2
)
2
(6)
With E[W] = 0 and σ
2
W
= Var[W], we can write the PDF of W as
f
W
(w) =
1
_
2πσ
2
W
e
−w
2
/2σ
2
W
(7)
Quiz 6.5
(1) From Table 6.1, each X
i
has MGF φ
X
(s) and random variable N has MGF φ
N
(s)
where
φ
X
(s) =
1
1 −s
, φ
N
(s) =
1
5
e
s
1 −
4
5
e
s
. (1)
From Theorem 6.12, R has MGF
φ
R
(s) = φ
N
(ln φ
X
(s)) =
1
5
φ
X
(s)
1 −
4
5
φ
X
(s)
(2)
Substituting the expression for φ
X
(s) yields
φ
R
(s) =
1
5
1
5
−s
. (3)
(2) From Table 6.1, we see that R has the MGF of an exponential (1/5) random variable.
The corresponding PDF is
f
R
(r) =
_
(1/5)e
−r/5
r ≥ 0
0 otherwise
(4)
This quiz is an example of the general result that a geometric sum of exponential
random variables is an exponential random variable.
42
Quiz 6.6
(1) The expected access time is
E [X] =
_
−∞
x f
X
(x) dx =
_
12
0
x
12
dx = 6 msec (1)
(2) The second moment of the access time is
E
_
X
2
_
=
_
−∞
x
2
f
X
(x) dx =
_
12
0
x
2
12
dx = 48 (2)
The variance of the access time is Var[X] = E[X
2
] −(E[X])
2
= 48 −36 = 12.
(3) Using X
i
to denote the access time of block i , we can write
A = X
1
+ X
2
+· · · + X
12
(3)
Since the expectation of the sum equals the sum of the expectations,
E [A] = E [X
1
] +· · · + E [X
12
] = 12E [X] = 72 msec (4)
(4) Since the X
i
are independent,
Var[A] = Var[X
1
] +· · · +Var[X
12
] = 12 Var[X] = 144 (5)
Hence, the standard deviation of A is σ
A
= 12
(5) To use the central limit theorem, we write
P [A > 75] = 1 − P [A ≤ 75] (6)
= 1 − P
_
A − E [A]
σ
A
75 − E [A]
σ
A
_
(7)
≈ 1 −
_
75 −72
12
_
(8)
= 1 −0.5987 = 0.4013 (9)
Note that we used Table 3.1 to look up (0.25).
(6) Once again, we use the central limit theorem and Table 3.1 to estimate
P [A < 48] = P
_
A − E [A]
σ
A
<
48 − E [A]
σ
A
_
(10)
_
48 −72
12
_
(11)
= 1 −(2) = 1 −0.9773 = 0.0227 (12)
43
Quiz 6.7
Random variable K
n
has a binomial distribution for n trials and success probability
P[V] = 3/4.
(1) The expected number of voice calls out of 48 calls is E[K
48
] = 48P[V] = 36.
(2) The variance of K
48
is
Var[K
48
] = 48P [V] (1 − P [V]) = 48(3/4)(1/4) = 9 (1)
Thus K
48
has standard deviation σ
K
48
= 3.
(3) Using the ordinary central limit theorem and Table 3.1 yields
P [30 ≤ K
48
≤ 42] ≈
_
42 −36
3
_
_
30 −36
3
_
= (2) −(−2) (2)
Recalling that (−x) = 1 −(x), we have
P [30 ≤ K
48
≤ 42] ≈ 2(2) −1 = 0.9545 (3)
(4) Since K
48
is a discrete random variable, we can use the De Moivre-Laplace approx-
imation to estimate
P [30 ≤ K
48
≤ 42] ≈
_
42 +0.5 −36
3
_
_
30 −0.5 −36
3
_
(4)
= 2(2.16666) −1 = 0.9687 (5)
Quiz 6.8
The train interarrival times X
1
, X
2
, X
3
are iid exponential (λ) random variables. The
arrival time of the third train is
W = X
1
+ X
2
+ X
3
. (1)
In Theorem 6.11, we found that the sum of three iid exponential (λ) random variables is an
Erlang (n = 3, λ) random variable. From Appendix A, we find that W has expected value
and variance
E [W] = 3/λ = 6 Var[W] = 3/λ
2
= 12 (2)
(1) By the Central Limit Theorem,
P [W > 20] = P
_
W −6
12
>
20 −6
12
_
≈ Q(7/
3) = 2.66 ×10
−5
(3)
44
(2) To use the Chernoff bound, we note that the MGF of W is
φ
W
(s) =
_
λ
λ −s
_
3
=
1
(1 −2s)
3
(4)
The Chernoff bound states that
P [W > 20] ≤ min
s≥0
e
−20s
φ
X
(s) = min
s≥0
e
−20s
(1 −2s)
3
(5)
To minimize h(s) = e
−20s
/(1 −2s)
3
, we set the derivative of h(s) to zero:
dh(s)
ds
=
−20(1 −2s)
3
e
−20s
+6e
−20s
(1 −2s)
2
(1 −2s)
6
= 0 (6)
This implies 20(1 − 2s) = 6 or s = 7/20. Applying s = 7/20 into the Chernoff
bound yields
P [W > 20] ≤
e
−20s
(1 −2s)
3
¸
¸
¸
¸
s=7/20
= (10/3)
3
e
−7
= 0.0338 (7)
(3) Theorem 3.11 says that for any w > 0, the CDF of the Erlang (λ, 3) random variable
W satisfies
F
W
(w) = 1 −
2
k=0
(λw)
k
e
−λw
k!
(8)
Equivalently, for λ = 1/2 and w = 20,
P [W > 20] = 1 − F
W
(20) (9)
= e
−10
_
1 +
10
1!
+
10
2
2!
_
= 61e
−10
= 0.0028 (10)
Although the Chernoff bound is relatively weak in that it overestimates the proba-
bility by roughly a factor of 12, it is a valid bound. By contrast, the Central Limit
Theorem approximation grossly underestimates the true probability.
Quiz 6.9
One solution to this problem is to follow the approach of Example 6.19:
%unifbinom100.m
sx=0:100;sy=0:100;
px=binomialpmf(100,0.5,sx); py=duniformpmf(0,100,sy);
[SX,SY]=ndgrid(sx,sy); [PX,PY]=ndgrid(px,py);
SW=SX+SY; PW=PX.*PY;
sw=unique(SW); pw=finitepmf(SW,PW,sw);
pmfplot(sw,pw,’\itw’,’\itP_W(w)’);
A graph of the PMF P
W
(w) appears in Figure 2 With some thought, it should be apparent
that the finitepmf function is implementing the convolution of the two PMFs.
45
0 20 40 60 80 100 120 140 160 180 200
0
0.002
0.004
0.006
0.008
0.01
w
P
W
(
w
)
Figure 2: From Quiz 6.9, the PMF P
W
(w) of the independent sum of a binomial (100, 0.5)
random variable and a discrete uniform (0, 100) random variable.
46
Quiz Solutions – Chapter 7
Quiz 7.1
An exponential random variable with expected value 1 also has variance 1. By Theo-
rem 7.1, M
n
(X) has variance Var[M
n
(X)] = 1/n. Hence, we need n = 100 samples.
Quiz 7.2
The arrival time of the third elevator is W = X
1
+ X
2
+ X
3
. Since each X
i
is uniform
(0, 30),
E [X
i
] = 15, Var [X
i
] =
(30 −0)
2
12
= 75. (1)
Thus E[W] = 3E[X
i
] = 45, and Var[W] = 3 Var[X
i
] = 225.
(1) By the Markov inequality,
P [W > 75] ≤
E [W]
75
=
45
75
=
3
5
(2)
(2) By the Chebyshev inequality,
P [W > 75] = P [W − E [W] > 30] (3)
≤ P [|W − E [W]| > 30] ≤
Var [W]
30
2
=
225
900
=
1
4
(4)
Quiz 7.3
Define the random variable W = (X − µ
X
)
2
. Observe that V
100
(X) = M
100
(W). By
Theorem 7.6, the mean square error is
E
_
(M
100
(W) −µ
W
)
2
_
=
Var[W]
100
(1)
Observe that µ
X
= 0 so that W = X
2
. Thus,
µ
W
= E
_
X
2
_
=
_
1
−1
x
2
f
X
(x) dx = 1/3 (2)
E
_
W
2
_
= E
_
X
4
_
=
_
1
−1
x
4
f
X
(x) dx = 1/5 (3)
Therefore Var[W] = E[W
2
] − µ
2
W
= 1/5 − (1/3)
2
= 4/45 and the mean square error is
4/4500 = 0.000889.
47
Quiz 7.4
Assuming the number n of samples is large, we can use a Gaussian approximation for
M
n
(X). SinceE[X] = p and Var[X] = p(1 − p), we apply Theorem 7.13 which says that
the interval estimate
M
n
(X) −c ≤ p ≤ M
n
(X) +c (1)
has confidence coefficient 1 −α where
α = 2 −2
_
c
n
p(1 − p)
_
. (2)
We must ensure for every value of p that 1 − α ≥ 0.9 or α ≤ 0.1. Equivalently, we must
have
_
c
n
p(1 − p)
_
≥ 0.95 (3)
for every value of p. Since (x) is an increasing function of x, we must satisfy c
n ≥
1.65p(1 − p). Since p(1 − p) ≤ 1/4 for all p, we require that
c ≥
1.65
4
n
=
0.41
n
. (4)
The 0.9 confidence interval estimate of p is
M
n
(X) −
0.41
n
≤ p ≤ M
n
(X) +
0.41
n
. (5)
For the 0.99 confidence interval, we have α ≤ 0.01, implying (c
n/( p(1−p))) ≥ 0.995.
This implies c
n ≥ 2.58p(1 − p). Since p(1 − p) ≤ 1/4 for all p, we require that
c ≥ (0.25)(2.58)/
n. In this case, the 0.99 confidence interval estimate is
M
n
(X) −
0.645
n
≤ p ≤ M
n
(X) +
0.645
n
. (6)
Note that if M
100
(X) = 0.4, then the 0.99 confidence interval estimate is
0.3355 ≤ p ≤ 0.4645. (7)
The interval is wide because the 0.99 confidence is high.
Quiz 7.5
Following the approach of bernoullitraces.m, we generate m = 1000 sample
paths, each sample path having n = 100 Bernoulli traces. at time k, OK(k) counts the
fraction of sample paths that have sample mean within one standard error of p. The pro-
gram bernoullisample.m generates graphs the number of traces within one standard
error as a function of the time, i.e. the number of trials in each trace.
48
function OK=bernoullisample(n,m,p);
x=reshape(bernoullirv(p,m*n),n,m);
nn=(1:n)’*ones(1,m);
MN=cumsum(x)./nn;
stderr=sqrt(p*(1-p))./sqrt((1:n)’);
stderrmat=stderr*ones(1,m);
OK=sum(abs(MN-p)<stderrmat,2)/m;
plot(1:n,OK,’-s’);
The following graph was generated by bernoullisample(100,5000,0.5):
0 10 20 30 40 50 60 70 80 90 100
0.4
0.5
0.6
0.7
0.8
0.9
1
As we would expect, as m gets large, the fraction of traces within one standard error ap-
proaches 2(1) −1 ≈ 0.68. The unusual sawtooth pattern, though perhaps unexpected, is
examined in Problem 7.5.2.
49
Quiz Solutions – Chapter 8
Quiz 8.1
From the problem statement, each X
i
has PDF and CDF
f
X
i
(x) =
_
e
−x
x ≥ 0
0 otherwise
F
X
i
(x) =
_
0 x < 0
1 −e
−x
x ≥ 0
(1)
Hence, the CDF of the maximum of X
1
, . . . , X
15
obeys
F
X
(x) = P [X ≤ x] = P [X
1
≤ x, X
2
≤ x, · · · , X
15
≤ x] = [P [X
i
≤ x]]
15
. (2)
This implies that for x ≥ 0,
F
X
(x) =
_
F
X
i
(x)
_
15
=
_
1 −e
−x
_
15
(3)
To design a significance test, we must choose a rejection region for X. A reasonable choice
is to reject the hypothesis if X is too small. That is, let R = {X ≤ r}. For a significance
level of α = 0.01, we obtain
α = P [X ≤ r] = (1 −e
−r
)
15
= 0.01 (4)
It is straightforward to show that
r = −ln
_
1 −(0.01)
1/15
_
= 1.33 (5)
Hence, if we observe X < 1.33, then we reject the hypothesis.
Quiz 8.2
From the problem statement, the conditional PMFs of K are
P
K|H
0
(k) =
_
10
4k
e
−10
4
k!
k = 0, 1, . . .
0 otherwise
(1)
P
K|H
1
(k) =
_
10
6k
e
−10
6
k!
k = 0, 1, . . .
0 otherwise
(2)
Since the two hypotheses are equally likely, the MAP and ML tests are the same. From
Theorem 8.6, the ML hypothesis rule is
k ∈ A
0
if P
K|H
0
(k) ≥ P
K|H
1
(k) ; k ∈ A
1
otherwise. (3)
This rule simplifies to
k ∈ A
0
if k ≤ k
=
10
6
−10
4
ln 100
= 214, 975.7; k ∈ A
1
otherwise. (4)
Thus if we observe at least 214, 976 photons, then we accept hypothesis H
1
.
50
Quiz 8.3
For the QPSK system, a symbol error occurs when s
i
is transmitted but (X
1
, X
2
) ∈ A
j
for some j = i . For a QPSK system, it is easier to calculate the probability of a correct
decision. Given H
0
, the conditional probability of a correct decision is
P [C|H
0
] = P [X
1
> 0, X
2
> 0|H
0
] = P
_
E/2 + N
1
> 0,
E/2 + N
2
> 0
_
(1)
Because of the symmetry of the signals, P[C|H
0
] = P[C|H
i
] for all i . This implies the
probability of a correct decision is P[C] = P[C|H
0
]. Since N
1
and N
2
are iid Gaussian
(0, σ) random variables, we have
P [C] = P [C|H
0
] = P
_
E/2 + N
1
> 0
_
P
_
E/2 + N
2
> 0
_
(2)
=
_
P
_
N
1
> −
E/2
__
2
(3)
=
_
1 −
_
E/2
σ
__
2
(4)
Since (−x) = 1 − (x), we have P[C] =
2
(
_
E/2σ
2
). Equivalently, the probability
of error is
P
ERR
= 1 − P [C] = 1 −
2
_
_
E
2
_
(5)
Quiz 8.4
To generate the ROC, the existing program sqdistor already calculates this miss
probability P
MISS
= P
01
and the false alarm probability P
FA
= P
10
. The modified pro-
gram, sqdistroc.m is essentially the same as sqdistor except the output is a ma-
trix FM whose columns are the false alarm and miss probabilities. Next, the program
sqdistrocplot.m calls sqdistroc three times to generate a plot that compares the
receiver performance for the three requested values of d. Here is the modified code:
function FM=sqdistroc(v,d,m,T)
%square law distortion recvr
%P(error) for m bits tested
%transmit v volts or -v volts,
%add N volts, N is Gauss(0,1)
%receive 1 if x>T, otherwise 0
%FM = [P(FA) P(MISS)]
x=(v+randn(m,1));
[XX,TT]=ndgrid(x,T(:));
P01=sum((XX+d*(XX.ˆ2)< TT),1)/m;
x= -v+randn(m,1);
[XX,TT]=ndgrid(x,T(:));
P10=sum((XX+d*(XX.ˆ2)>TT),1)/m;
FM=[P10(:) P01(:)];
function FM=sqdistrocplot(v,m,T);
FM1=sqdistroc(v,0.1,m,T);
FM2=sqdistroc(v,0.2,m,T);
FM5=sqdistroc(v,0.3,m,T);
FM=[FM1 FM2 FM5];
loglog(FM1(:,1),FM1(:,2),’-k’, ...
FM2(:,1),FM2(:,2),’--k’, ...
FM5(:,1),FM5(:,2),’:k’);
legend(’\it d=0.1’,’\it d=0.2’,...
’\it d=0.3’,3)
ylabel(’P_{MISS}’);
xlabel(’P_{FA}’);
51
To see the effect of d, the commands
T=-3:0.1:3; sqdistrocplot(3,100000,T);
generated the plot shown in Figure 3.
10
−5
10
−4
10
−3
10
−2
10
−1
10
0
10
−5
10
−4
10
−3
10
−2
10
−1
10
0
P
M
I
S
S
P
FA
d=0.1
d=0.2
d=0.3
T=-3:0.1:3; sqdistrocplot(3,100000,T);
Figure 3: The receiver operating curve for the communications system of Quiz 8.4 with
squared distortion.
52
Quiz Solutions – Chapter 9
Quiz 9.1
(1) First, we calculate the marginal PDF for 0 ≤ y ≤ 1:
f
Y
(y) =
_
y
0
2(y + x) dx = 2xy + x
2
¸
¸
¸
x=y
x=0
= 3y
2
(1)
This implies the conditional PDF of X given Y is
f
X|Y
(x|y) =
f
X,Y
(x, y)
f
Y
(y)
=
_
2
3y
+
2x
3y
2
0 ≤ x ≤ y
0 otherwise
(2)
(2) The minimum mean square error estimate of X given Y = y is
ˆ x
M
(y) = E [X|Y = y] =
_
y
0
_
2x
3y
+
2x
2
3y
2
_
dx = 5y/9 (3)
Thus the MMSE estimator of X given Y is
ˆ
X
M
(Y) = 5Y/9.
(3) To obtain the conditional PDF f
Y|X
(y|x), we need the marginal PDF f
X
(x). For
0 ≤ x ≤ 1,
f
X
(x) =
_
1
x
2(y + x) dy = y
2
+2xy
¸
¸
¸
y=1
y=x
= 1 +2x −3x
2
(4)
(5)
For 0 ≤ x ≤ 1, the conditional PDF of Y given X is
f
Y|X
(y|x) =
_
2(y+x)
1+2x−3x
2
x ≤ y ≤ 1
0 otherwise
(6)
(4) The MMSE estimate of Y given X = x is
ˆ y
M
(x) = E [Y|X = x] =
_
1
x
2y
2
+2xy
1 +2x −3x
2
dy (7)
=
2y
3
/3 + xy
2
1 +2x −3x
2
¸
¸
¸
¸
y=1
y=x
(8)
=
2 +3x −5x
3
3 +6x −9x
2
(9)
53
Quiz 9.2
(1) Since the expectation of the sum equals the sum of the expectations,
E [R] = E [T] + E [X] = 0 (1)
(2) Since T and X are independent, the variance of the sum R = T + X is
Var[R] = Var[T] +Var[X] = 9 +3 = 12 (2)
(3) Since T and R have expected values E[R] = E[T] = 0,
Cov [T, R] = E [T R] = E [T(T + X)] = E
_
T
2
_
+ E [T X] (3)
Since T and X are independent and have zero expected value, E[T X] = E[T]E[X] =
0 and E[T
2
] = Var[T]. Thus Cov[T, R] = Var[T] = 9.
(4) From Definition 4.8, the correlation coefficient of T and R is
ρ
T,R
=
Cov [T, R]
Var[R] Var[T]
=
σ
T
σ
R
=
3/2 (4)
(5) From Theorem 9.4, the optimum linear estimate of T given R is
ˆ
T
L
(R) = ρ
T,R
σ
T
σ
R
(R − E [R]) + E [T] (5)
Since E[R] = E[T] = 0 and ρ
T,R
= σ
T
R
,
ˆ
T
L
(R) =
σ
2
T
σ
2
R
R =
σ
2
T
σ
2
T
2
X
R =
3
4
R (6)
Hence a
= 3/4 and b
= 0.
(6) By Theorem 9.4, the mean square error of the linear estimate is
e
L
= Var[T](1 −ρ
2
T,R
) = 9(1 −3/4) = 9/4 (7)
Quiz 9.3
When R = r, the conditional PDF of X = Y −40−40 log
10
r is Gaussian with expected
value −40 −40 log
10
r and variance 64. The conditional PDF of X given R is
f
X|R
(x|r) =
1
128π
e
−(x+40+40 log
10
r)
2
/128
(1)
54
From the conditional PDF f
X|R
(x|r), we can use Definition 9.2 to write the ML estimate
of R given X = x as
ˆ r
ML
(x) = arg max
r≥0
f
X|R
(x|r) (2)
We observe that f
X|R
(x|r) is maximized when the exponent (x + 40 + 40 log
10
r)
2
is
minimized. This minimum occurs when the exponent is zero, yielding
log
10
r = −1 − x/40 (3)
or
ˆ r
ML
(x) = (0.1)10
−x/40
m (4)
If the result doesn’t look correct, note that a typical figure for the signal strength might be
x = −120 dB. This corresponds to a distance estimate of ˆ r
ML
(−120) = 100 m.
For the MAP estimate, we observe that the joint PDF of X and R is
f
X,R
(x, r) = f
X|R
(x|r) f
R
(r) =
1
10
6
32π
re
−(x+40+40 log
10
r)
2
/128
(5)
From Theorem 9.6, the MAP estimate of R given X = x is the value of r that maximizes
f
X,R
(x, r). That is,
ˆ r
MAP
(x) = arg max
0≤r≤1000
f
X,R
(x, r) (6)
Note that we have included the constraint r ≤ 1000 in the maximization to highlight the
fact that under our probability model, R ≤ 1000 m. Setting the derivative of f
X,R
(x, r)
with respect to r to zero yields
e
−(x+40+40 log
10
r)
2
/128
_
1 −
80 log
10
e
128
(x +40 +40 log
10
r)
_
= 0 (7)
Solving for r yields
r = 10
_
1
25 log
10
e
−1
_
10
−x/40
= (0.1236)10
−x/40
(8)
This is the MAP estimate of R given X = x as long as r ≤ 1000 m. When x ≤ −156.3 dB,
the above estimate will exceed 1000 m, which is not possible in our probability model.
Hence, the complete description of the MAP estimate is
ˆ r
MAP
(x) =
_
1000 x < −156.3
(0.1236)10
−x/40
x ≥ −156.3
(9)
For example, if x = −120dB, then ˆ r
MAP
(−120) = 123.6 m. When the measured signal
strength is not too low, the MAP estimate is 23.6% larger than the ML estimate. This re-
flects the fact that large values of R are a priori more probable than small values. However,
for very low signal strengths, the MAP estimate takes into account that the distance can
never exceed 1000 m.
55
Quiz 9.4
(1) From Theorem 9.4, the LMSE estimate of X
2
given Y
2
is
ˆ
X
2
(Y
2
) = a
Y
2
+b
where
a
=
Cov [X
2
, Y
2
]
Var[Y
2
]
, b
= µ
X
2
−a
µ
Y
2
. (1)
Because E[X] = E[Y] = 0,
Cov [X
2
, Y
2
] = E [X
2
Y
2
] = E [X
2
(X
2
+ W
2
)] = E
_
X
2
2
_
= 1 (2)
Var[Y
2
] = Var[X
2
] +Var[W
2
] = E
_
X
2
2
_
+ E
_
W
2
2
_
= 1.1 (3)
It follows that a
= 1/1.1. Because µ
X
2
= µ
Y
2
= 0, it follows that b
= 0. Finally,
to compute the expected square error, we calculate the correlation coefficient
ρ
X
2
,Y
2
=
Cov [X
2
, Y
2
]
σ
X
2
σ
Y
2
=
1
1.1
(4)
The expected square error is
e
L
= Var[X
2
](1 −ρ
2
X
2
,Y
2
) = 1 −
1
1.1
=
1
11
= 0.0909 (5)
(2) Since Y = X + W and E[X] = E[W] = 0, it follows that E[Y] = 0. Thus we can
apply Theorem 9.7. Note that X and W have correlation matrices
R
X
=
_
1 −0.9
−0.9 1
_
, R
W
=
_
0.1 0
0 0.1
_
. (6)
In terms of Theorem 9.7, n = 2 and we wish to estimate X
2
given the observation
vector Y =
_
Y
1
Y
2
_
. To apply Theorem 9.7, we need to find R
Y
and R
YX
2
.
R
Y
= E
_
YY
_
= E
_
(X +W)(X
+W
)
_
(7)
= E
_
XX
+XW
+WX
+WW
_
. (8)
Because Xand Ware independent, E[XW
] = E[X]E[W
] = 0. Similarly, E[WX
] =
0. This implies
R
Y
= E
_
XX
_
+ E
_
WW
_
= R
X
+R
W
=
_
1.1 −0.9
−0.9 1.1
_
. (9)
In addition, we need to find
R
YX
2
= E [YX
2
] =
_
E [Y
1
X
2
]
E [Y
2
X
2
]
_
=
_
E [(X
1
+ W
1
)X
2
]
E [(X
2
+ W
2
)X
2
]
_
. (10)
56
Since Xand Ware independent vectors, E[W
1
X
2
] = E[W
1
]E[X
2
] = 0 and E[W
2
X
2
] =
0. Thus
R
YX
2
=
_
E[X
1
X
2
]
E
_
X
2
2
_
_
=
_
−0.9
1
_
. (11)
By Theorem 9.7,
ˆ a = R
−1
Y
R
YX
2
=
_
−0.225
0.725
_
(12)
Therefore, the optimum linear estimator of X
2
given Y
1
and Y
2
is
ˆ
X
L
= ˆ a
Y = −0.225Y
1
+0.725Y
2
. (13)
The mean square error is
Var [X
2
] − ˆ a
R
YX
2
= Var [X] −a
1
r
Y
1
,X
2
−a
2
r
Y
2
,X
2
= 0.0725. (14)
Quiz 9.5
Since X and W have zero expected value, Y also has zero expected value. Thus, by
Theorem 9.7,
ˆ
X
L
(Y) = ˆ a
Y where ˆ a = R
−1
Y
R
YX
. Since X and W are independent,
E[WX] = 0 and E[XW
] = 0
. This implies
R
YX
= E [YX] = E [(1X +W)X] = 1E
_
X
2
_
= 1. (1)
By the same reasoning, the correlation matrix of Y is
R
Y
= E
_
YY
_
= E
_
(1X +W)(1
X +W
)
_
(2)
= 11
E
_
X
2
_
+1E
_
XW
_
+ E [WX] 1
+ E
_
WW
_
(3)
= 11
+R
W
(4)
Note that 11
is a 20 ×20 matrix with every entry equal to 1. Thus,
ˆ a = R
−1
Y
R
YX
=
_
11
+R
W
_
−1
1 (5)
and the optimal linear estimator is
ˆ
X
L
(Y) = 1
_
11
+R
W
_
−1
Y (6)
The mean square error is
e
L
= Var[X] − ˆ a
R
YX
= 1 −1
_
11
+R
W
_
−1
1 (7)
Now we note that R
W
has i, j th entry R
W
(i, j ) = c
|i −j |−1
. The question we must address
is what value c minimizes e
L
. This problem is atypical in that one does not usually get
57
to choose the correlation structure of the noise. However, we will see that the answer is
somewhat instructive.
We note that the answer is not obviously apparent from Equation (7). In particular, we
observe that Var[W
i
] = R
W
(i, i ) = 1/c. Thus, when c is small, the noises W
i
have high
variance and we would expect our estimator to be poor. On the other hand, if c is large
W
i
and W
j
are highly correlated and the separate measurements of X are very dependent.
This would suggest that large values of c will also result in poor MSE. If this argument is
not clear, consider the extreme case in which every W
i
and W
j
have correlation coefficient
ρ
i j
= 1. In this case, our 20 measurements will be all the same and one measurement is as
good as 20 measurements.
To find the optimal value of c, we write a MATLAB function mquiz9(c) to calculate
the MSE for a given c and second function that finds plots the MSE for a range of values
of c.
function [mse,af]=mquiz9(c);
v1=ones(20,1);
RW=toeplitz(c.ˆ((0:19)-1));
RY=(v1*(v1’)) +RW;
af=(inv(RY))*v1;
mse=1-((v1’)*af);
function cmin=mquiz9minc(c);
msec=zeros(size(c));
for k=1:length(c),
[msec(k),af]=mquiz9(c(k));
end
plot(c,msec);
xlabel(’c’);ylabel(’e_Lˆ*’);
[msemin,optk]=min(msec);
cmin=c(optk);
Note in mquiz9 that v1 corresponds to the vector 1 of all ones. The following commands
finds the minimum c and also produces the following graph:
>> c=0.01:0.01:0.99;
>> mquiz9minc(c)
ans =
0.4500
0 0.5 1
0.2
0.4
0.6
0.8
1
c
e
L *
As we see in the graph, both small values and large values of c result in large MSE.
58
Quiz Solutions – Chapter 10
Quiz 10.1
There are many correct answers to this question. A correct answer specifies enough
random variables to specify the sample path exactly. One choice for an alternate set of
random variables that would specify m(t, s) is
• m(0, s), the number of ongoing calls at the start of the experiment
• N, the number of new calls that arrive during the experiment
• X
1
, . . . , X
N
, the interarrival times of the N new arrivals
• H, the number of calls that hang up during the experiment
• D
1
, . . . , D
H
, the call completion times of the H calls that hang up
Quiz 10.2
(1) We obtain a continuous time, continuous valued process when we record the temper-
ature as a continuous waveform over time.
(2) If at every moment in time, we round the temperature to the nearest degree, then we
obtain a continuous time, discrete valued process.
(3) If we sample the process in part (a) every T seconds, then we obtain a discrete time,
continuous valued process.
(4) Rounding the samples in part (c) to the nearest integer degree yields a discrete time,
discrete valued process.
Quiz 10.3
(1) Each resistor has resistance R in ohms with uniform PDF
f
R
(r) =
_
0.01 950 ≤ r ≤ 1050
0 otherwise
(1)
The probability that a test produces a 1% resistor is
p = P [990 ≤ R ≤ 1010] =
_
1010
990
(0.01) dr = 0.2 (2)
59
(2) In t seconds, exactly t resistors are tested. Each resistor is a 1% resistor with proba-
bility p, independent of any other resistor. Consequently, the number of 1% resistors
found has the binomial PMF
P
N(t )
(n) =
_ _
t
n
_
p
n
(1 − p)
t −n
n = 0, 1, . . . , t
0 otherwise
(3)
(3) First we will find the PMF of T
1
. This problem is easy if we view each resistor test
as an independent trial. A success occurs on a trial with probability p if we find a
1% resistor. The first 1% resistor is found at time T
1
= t if we observe failures on
trials 1, . . . , t − 1 followed by a success on trial t . Hence, just as in Example 2.11,
T
1
has the geometric PMF
P
T
1
(t ) =
_
(1 − p)
t −1
p t = 1, 2, . . .
9 otherwise
(4)
Since p = 0.2, the probability the first 1% resistor is found in exactly five seconds is
P
T
1
(5) = (0.8)
4
(0.2) = 0.08192.
(4) From Theorem 2.5, a geometric random variable with success probability p has ex-
pected value 1/p. In this problem, E[T
1
] = 1/p = 5.
(5) Note that once we find the first 1% resistor, the number of additional trials needed to
find the second 1% resistor once again has a geometric PMF with expected value 1/p
since each independent trial is a success with probability p. That is, T
2
= T
1
+ T
where T
is independent and identically distributed to T
1
. Thus
E [T
2
|T
1
= 10] = E [T
1
|T
1
= 10] + E
_
T
|T
1
= 10
_
(5)
= 10 + E
_
T
_
= 10 +5 = 15 (6)
Quiz 10.4
Since each X
i
is a N(0, 1) random variable, each X
i
has PDF
f
X(i )
(x) =
1
e
−x
2
/2
(1)
By Theorem 10.1, the joint PDF of X =
_
X
1
· · · X
n
_
is
f
X
(x) = f
X(1),...,X(n)
(x
1
, . . . , x
n
) =
k
i =1
f
X
(x
i
) =
1
(2π)
n/2
e
−(x
2
1
+···+x
2
n
)/2
(2)
60
Quiz 10.5
The first and second hours are nonoverlapping intervals. Since one hour equals 3600
sec and the Poisson process has a rate of 10 packets/sec, the expected number of packets
in each hour is E[M
i
] = α = 36, 000. This implies M
1
and M
2
are independent Poisson
random variables each with PMF
P
M
i
(m) =
_
α
m
e
−α
m!
m = 0, 1, 2, . . .
0 otherwise
(1)
Since M
1
and M
2
are independent, the joint PMF of M
1
and M
2
is
P
M
1
,M
2
(m
1
, m
2
) = P
M
1
(m
1
) P
M
2
(m
2
) =
α
m
1
+m
2
e
−2α
m
1
!m
2
!
m
1
= 0, 1, . . . ;
m
2
= 0, 1, . . . ,
0 otherwise.
(2)
Quiz 10.6
(t ) is a Poisson process, we look at the interarrival times. Let
X
1
, X
2
, . . . denote the interarrival times of the N(t ) process. Since we count only even-
numbered arrival for N
(t ), the time until the first arrival of the N
(t ) is Y
1
= X
1
+ X
2
.
Since X
1
and X
2
are independent exponential (λ) random variables, Y
1
is an Erlang (n =
2, λ) random variable; see Theorem 6.11. Since Y
i
(t ), the i th interarrival time of the N
(t )
process, has the same PDF as Y
1
(t ), we can conclude that the interarrival times of N
(t )
are not exponential random variables. Thus N
(t ) is not a Poisson process.
Quiz 10.7
First, we note that for t > s,
X(t ) − X(s) =
W(t ) − W(s)
α
(1)
Since W(t ) −W(s) is a Gaussian random variable, Theorem 3.13 states that W(t ) −W(s)
is Gaussian with expected value
E [X(t ) − X(s)] =
E [W(t ) − W(s)]
α
= 0 (2)
and variance
E
_
(W(t ) − W(s))
2
_
=
E
_
(W(t ) − W(s))
2
_
α
=
α(t −s)
α
(3)
Consider s
≤ s < t . Since s ≥ s
, W(t ) − W(s) is independent of W(s
). This implies
[W(t ) − W(s)]/
α is independent of W(s
)/
α for all s ≥ s
. That is, X(t ) − X(s) is
independent of X(s
) for all s ≥ s
. Thus X(t ) is a Brownian motion process with variance
Var[X(t )] = t .
61
Quiz 10.8
First we find the expected value
µ
Y
(t ) = µ
X
(t ) +µ
N
(t ) = µ
X
(t ). (1)
To find the autocorrelation, we observe that since X(t ) and N(t ) are independent and since
N(t ) has zero expected value, E[X(t )N(t
)] = E[X(t )]E[N(t
)] = 0. Since R
Y
(t, τ) =
E[Y(t )Y(t +τ)], we have
R
Y
(t, τ) = E [(X(t ) + N(t )) (X(t +τ) + N(t +τ))] (2)
= E [X(t )X(t +τ)] + E [X(t )N(t +τ)]
+ E [X(t +τ)N(t )] + E [N(t )N(t +τ)] (3)
= R
X
(t, τ) + R
N
(t, τ). (4)
Quiz 10.9
From Definition 10.14, X
1
, X
2
, . . . is a stationary random sequence if for all sets of
time instants n
1
, . . . , n
m
and time offset k,
f
X
n
1
,...,X
n
m
(x
1
, . . . , x
m
) = f
X
n
1
+k
,...,X
n
m
+k
(x
1
, . . . , x
m
) (1)
Since the random sequence is iid,
f
X
n
1
,...,X
n
m
(x
1
, . . . , x
m
) = f
X
(x
1
) f
X
(x
2
) · · · f
X
(x
m
) (2)
Similarly, for time instants n
1
+k, . . . , n
m
+k,
f
X
n
1
+k
,...,X
n
m
+k
(x
1
, . . . , x
m
) = f
X
(x
1
) f
X
(x
2
) · · · f
X
(x
m
) (3)
We can conclude that the iid random sequence is stationary.
Quiz 10.10
We must check whether each function R(τ) meets the conditions of Theorem 10.12:
R(τ) ≥ 0 R(τ) = R(−τ) |R(τ)| ≤ R(0) (1)
(1) R
1
(τ) = e
−|τ|
meets all three conditions and thus is valid.
(2) R
2
(τ) = e
−τ
2
also is valid.
(3) R
3
(τ) = e
−τ
cos τ is not valid because
R
3
(−2π) = e
cos 2π = e
> 1 = R
3
(0) (2)
(4) R
4
(τ) = e
−τ
2
sin τ also cannot be an autocorrelation function because
R
4
(π/2) = e
−π/2
sin π/2 = e
−π/2
> 0 = R
4
(0) (3)
62
Quiz 10.11
(1) The autocorrelation of Y(t ) is
R
Y
(t, τ) = E [Y(t )Y(t +τ)] (1)
= E [X(−t )X(−t −τ)] (2)
= R
X
(−t −(−t −τ)) = R
X
(τ) (3)
Since E[Y(t )] = E[X(−t )] = µ
X
, we can conclude that Y(t ) is a wide sense
stationary process. In fact, we see that by viewing a process backwards in time, we
see the same second order statistics.
(2) Since X(t ) and Y(t ) are both wide sense stationary processes, we can check whether
they are jointly wide sense stationary by seeing if R
XY
(t, τ) is just a function of τ.
In this case,
R
XY
(t, τ) = E [X(t )Y(t +τ)] (4)
= E [X(t )X(−t −τ)] (5)
= R
X
(t −(−t −τ)) = R
X
(2t +τ) (6)
Since R
XY
(t, τ) depends on both t and τ, we conclude that X(t ) and Y(t ) are not
jointly wide sense stationary. To see why this is, suppose R
X
(τ) = e
−|τ|
so that
samples of X(t ) far apart in time have almost no correlation. In this case, as t gets
larger, Y(t ) = X(−t ) and X(t ) become less and less correlated.
Quiz 10.12
From the problem statement,
E [X(t )] = E [X(t +1)] = 0 (1)
E [X(t )X(t +1)] = 1/2 (2)
Var[X(t )] = Var[X(t +1)] = 1 (3)
The Gaussian random vector X =
_
X(t ) X(t +1)
_
has covariance matrix and corre-
sponding inverse
C
X
=
_
1 1/2
1/2 1
_
C
−1
X
=
4
3
_
1 −1/2
−1/2 1
_
(4)
Since
x
C
−1
X
x =
_
x
0
x
1
_
4
3
_
1 −1/2
−1/2 1
_ _
x
0
x
1
_
=
4
3
_
x
2
0
− x
0
x
+
x
2
1
_
(5)
the joint PDF of X(t ) and X(t +1) is the Gaussian vector PDF
f
X(t ),X(t +1)
(x
0
, x
1
) =
1
(2π)
n/2
[det (C
X
)]
1/2
exp
_
1
2
x
C
−1
X
x
_
(6)
=
1
2
e
2
3
_
x
2
0
−x
0
x
1
+x
2
1
_
(7)
63
0 10 20 30 40 50 60 70 80 90 100
0
20
40
60
80
100
120
t
M
(
t
)
Figure 4: Sample path of 100 minutes of the blocking switch of Quiz 10.13.
Quiz 10.13
The simple structure of the switch simulation of Example 10.28 admits a deceptively
simple solution in terms of the vector of arrivals A and the vector of departures D. With the
introduction of call blocking. we cannot generate these vectors all at once. In particular,
when an arrival occurs at time t , we need to know that M(t ), the number of ongoing calls,
satisfies M(t ) < c = 120. Otherwise, when M(t ) = c, we must block the call. Call
blocking can be implemented by setting the service time of the call to zero so that the call
departs as soon as it arrives.
The blocking switch is an example of a discrete event system. The system evolves via
a sequence of discrete events, namely arrivals and departures, at discrete time instances. A
simulation of the system moves from one time instant to the next by maintaining a chrono-
logical schedule of future events (arrivals and departures) to be executed. The program
simply executes the event at the head of the schedule. The logic of such a simulation is
1. Start at time t = 0 with an empty system. Schedule the first arrival to occur at S
1
, an
exponential (λ) random variable.
• When the head-of-schedule event is the kth arrival is at time t , check the state
M(t ).
– If M(t ) < c, admit the arrival, increase the system state n by 1, and sched-
ule a departure to occur at time t + S
n
, where S
k
is an exponential (λ)
random variable.
– If M(t ) = c, block the arrival, do not schedule a departure event.
• If the head of schedule event is a departure, reduce the system state n by 1.
3. Delete the head-of-schedule event and go to step 2.
After the head-of-schedule event is completed and any new events (departures in this sys-
tem) are scheduled, we know the system state cannot change until the next scheduled event.
64
Thus we know that M(t ) will stay the same until then. In our simulation, we use the vector
t as the set of time instances at which we inspect the system state. Thus for all times t(i)
between the current head-of-schedule event and the next, we set m(i) to the current switch
state.
The complete program is shown in Figure 5. In most programming languages, it is
common to implement the event schedule as a linked list where each item in the list has
a data structure indicating an event timestamp and the type of the event. In MATLAB, a
simple (but not elegant) way to do this is to have maintain two vectors: time is a list
of timestamps of scheduled events and event is a the list of event types. In this case,
event(i)=1 if the i th scheduled event is an arrival, or event(i)=-1 if the i th sched-
uled event is a departure.
When the program is passed a vector t, the output [m a b] is such that m(i) is the
number of ongoing calls at time t(i) while a and b are the number of admits and blocks.
The following instructions
t=0:0.1:5000;
[m,a,b]=simblockswitch(10,0.1,120,t);
plot(t,m);
generated a simulation lasting 5,000 minutes. A sample path of the first 100 minutes of
that simulation is shown in Figure 4. The 5,000 minute full simulation produced a=49658
admitted calls and b=239 blocked calls. We can estimate the probability a call is blocked
as
ˆ
P
b
=
b
a +b
= 0.0048. (1)
In Chapter 12, we will learn that the exact blocking probability is given by Equation (12.93),
a result known as the “Erlang-B formula.” From the Erlang-B formula, we can calculate
that the exact blocking probability is P
b
= 0.0057. One reason our simulation underesti-
mates the blocking probability is that in a 5,000 minute simulation, roughly the first 100
minutes are needed to load up the switch since the switch is idle when the simulation starts
at time t = 0. However, this says that roughly the first two percent of the simulation time
was unusual. Thus this would account for only part of the disparity. The rest of the gap
between 0.0048 and 0.0057 is that a simulation that includes only 239 blocks is not all that
likely to give a very accurate result for the blocking probability.
Note that in Chapter 12, we will learn that the blocking switch is an example of an
M/M/c/c queue, a kind of Markov chain. Chapter 12 develops techniques for analyzing
and simulating systems described by Markov chains that are much simpler than the discrete
event simulation technique shown here. Nevertheless, for very complicated systems, the
discrete event simulation is widely-used and often very efficient simulation method.
65
blocks=0; %total # blocks
M=zeros(size(t));
n=0; % # in system
time=[ exponentialrv(lam,1) ];
event=[ 1 ]; %first event is an arrival
timenow=0;
tmax=max(t);
while (timenow<tmax)
M((timenow<=t)&(t<time(1)))=n;
timenow=time(1);
eventnow=event(1);
event(1)=[ ]; time(1)= [ ]; % clear current event
if (eventnow==1) % arrival
arrival=timenow+exponentialrv(lam,1); % next arrival
b4arrival=time<arrival;
event=[event(b4arrival) 1 event(˜b4arrival)];
time=[time(b4arrival) arrival time(˜b4arrival)];
n=n+1;
depart=timenow+exponentialrv(mu,1);
b4depart=time<depart;
event=[event(b4depart) -1 event(˜b4depart)];
time=[time(b4depart) depart time(˜b4depart)];
else
blocks=blocks+1; %one more block, immed departure
disp(sprintf(’Time %10.3d Admits %10d Blocks %10d’,...
end
elseif (eventnow==-1) %departure
n=n-1;
end
end
Figure 5: Discrete event simulation of the blocking switch of Quiz 10.13.
66
Quiz Solutions – Chapter 11
Quiz 11.1
By Theorem 11.2,
µ
Y
= µ
X
_
−∞
h(t )dt = 2
_
0
e
−t
dt = 2 (1)
Since R
X
(τ) = δ(τ), the autocorrelation function of the output is
R
Y
(τ) =
_
−∞
h(u)
_
−∞
h(v)δ(τ +u −v) dv du =
_
−∞
h(u)h(τ +u) du (2)
For τ > 0, we have
R
Y
(τ) =
_
0
e
−u
e
−τ−u
du = e
−τ
_
0
e
−2u
du =
1
2
e
−τ
(3)
For τ < 0, we can deduce that R
Y
(τ) =
1
2
e
−|τ|
by symmetry. Just to be safe though, we
can double check. For τ < 0,
R
Y
(τ) =
_
−τ
h(u)h(τ +u) du =
_
−τ
e
−u
e
−τ−u
du =
1
2
e
τ
(4)
Hence,
R
Y
(τ) =
1
2
e
−|τ|
(5)
Quiz 11.2
The expected value of the output is
µ
Y
= µ
X
n=−∞
h
n
= 0.5(1 +−1) = 0 (1)
The autocorrelation of the output is
R
Y
[n] =
1
i =0
1
j =0
h
i
h
j
R
X
[n +i − j ] (2)
= 2R
X
[n] − R
X
[n −1] − R
X
[n +1] =
_
1 n = 0
0 otherwise
(3)
Since µ
Y
= 0, The variance of Y
n
is Var[Y
n
] = E[Y
2
n
] = R
Y
[0] = 1.
67
−15 −10 −5 0 5 10 15
0
0.2
0.4
0.6
f
S
X
(
f
)
−1500−1000 −500 0 500 1000 1500
0
2
4
6
8
x 10
f
S
X
(
f
)
−0.2 −0.1 0 0.1 0.2
−5
0
5
10
τ
R
X
(
τ
)
−2 −1 0 1 2
x 10
−3
−5
0
5
10
τ
R
X
(
τ
)
(a) W = 10 (b) W = 1000
Figure 6: The autocorrelation R
X
(τ) and power spectral density S
X
( f ) for process X(t ) in
Quiz 11.5.
Quiz 11.3
By Theorem 11.8, Y =
_
Y
33
Y
34
Y
35
_
is a Gaussian random vector since X
n
is
a Gaussian random process. Moreover, by Theorem 11.5, each Y
n
has expected value
E[Y
n
] = µ
X
n=−∞
h
n
= 0. Thus E[Y] = 0. Fo find the PDF of the Gaussian vector
Y, we need to find the covariance matrix C
Y
, which equals the correlation matrix R
Y
since
Y has zero expected value. One way to find the R
Y
is to observe that R
Y
has the Toeplitz
structure of Theorem 11.6 and to use Theorem 11.5 to find the autocorrelation function
R
Y
[n] =
i =−∞
j =−∞
h
i
h
j
R
X
[n +i − j ]. (1)
Despite the fact that R
X
[k] is an impulse, using Equation (1) is surprisingly tedious because
we still need to sum over all i and j such that n +i − j = 0.
In this problem, it is simpler to observe that Y = HX where
X =
_
X
30
X
31
X
32
X
33
X
34
X
35
_
(2)
and
H =
1
4
1 1 1 1 0 0
0 1 1 1 1 0
0 0 1 1 1 1
. (3)
In this case, following Theorem 11.7, or by directly applying Theorem 5.13 with µ
X
= 0
and A = H, we obtain R
Y
= HR
X
H
. Since R
X
[n] = δ
n
, R
X
= I, the identity matrix.
68
Thus
C
Y
= R
Y
= HH
=
1
16
4 3 2
3 4 3
2 3 4
. (4)
It follows (very quickly if you use MATLAB for 3 ×3 matrix inversion) that
C
−1
Y
= 16
7/12 −1/2 1/12
−1/2 1 −1/2
1/12 −1/2 7/12
. (5)
Thus, the PDF of Y is
f
Y
(y) =
1
(2π)
3/2
[det (C
Y
)]
1/2
exp
_
1
2
y
C
−1
Y
y
_
. (6)
A disagreeable amount of algebra will show det(C
Y
) = 3/1024 and that the PDF can be
“simplified” to
f
Y
(y) =
16
3
exp
_
−8
_
7
12
y
2
33
+ y
2
34
+
7
12
y
2
35
− y
33
y
34
+
1
6
y
33
y
35
− y
34
y
35
__
. (7)
Equation (7) shows that one of the nicest features of the multivariate Gaussian distribution
is that y
C
−1
Y
y is a very concise representation of the cross-terms in the exponent of f
Y
(y).
Quiz 11.4
This quiz is solved using Theorem 11.9 for the case of k = 1 and M = 2. In this case,
X
n
=
_
X
n−1
X
n
_
and
R
X
n
=
_
R
X
[0] R
X
[1]
R
X
[1] R
X
[0]
_
=
_
1.1 0.9
0.9 1.1
_
(1)
and
R
X
n
X
n+1
= E
__
X
n−1
X
n
_
X
n+1
_
=
_
R
X
[2]
R
X
[1]
_
=
_
0.81
0.9
_
. (2)
The MMSE linear first order filter for predicting X
n+1
at time n is the filter h such that
←−
h = R
−1
X
n
R
X
n
X
n+1
=
_
1.1 0.9
0.9 1.1
_
−1
_
0.81
0.9
_
=
1
400
_
81
261
_
. (3)
It follows that the filter is h =
_
261/400 81/400
_
and the MMSE linear predictor is
ˆ
X
n+1
=
81
400
X
n−1
+
261
400
X
n
. (4)
to find the mean square error, one approach is to follow the method of Example 11.13 and
to directly calculate
e
L
= E
_
(X
n+1
ˆ
X
n+1
)
2
_
. (5)
69
This method is workable for this simple problem but becomes increasingly tedious for
higher order filters. Instead, we can derive the mean square error for an arbitary prediction
filter h. Since
ˆ
X
n+1
=
←−
h
X
n
,
e
L
= E
_
_
X
n+1
←−
h
X
n
_
2
_
(6)
= E
_
(X
n+1
←−
h
X
n
)(X
n+1
←−
h
X
n
)
_
(7)
= E
_
(X
n+1
←−
h
X
n
)(X
n+1
−X
n
←−
h )
_
(8)
After a bit of algebra, we obtain
e
L
= R
X
[0] −2
←−
h
R
X
n
X
n+1
+
←−
h
R
X
n
←−
h (9)
(10)
with the substitution
←−
h = R
−1
X
n
R
X
n
X
n+1
, we obtain
e
L
= R
X
[0] −R
X
n
X
n+1
R
−1
X
n
R
X
n
X
n+1
(11)
= R
X
[0] −
←−
h
R
X
n
X
n+1
(12)
Note that this is essentially the same result as Theorem 9.7 with Y = X
n
, X = X
n+1
and
ˆ a
=
←−
h
. It is noteworthy that the result is derived in a much simpler way in the proof of
Theorem 9.7 by using the orthoginality property of the LMSE estimator.
In any case, the mean square error is
e
L
= R
X
[0] −
←−
h
R
X
n
X
n+1
= 1.1 −
1
400
_
81 261
_
_
0.81
0.9
_
=
506
1451
= 0.3487. (13)
recalling that the blind estimate would yield a mean square error of Var[X] = 1.1, we see
that observing X
n−1
and X
n
improves the accuracy of our prediction of X
n+1
.
Quiz 11.5
(1) By Theorem 11.13(b), the average power of X(t ) is
E
_
X
2
(t )
_
=
_
−∞
S
X
( f ) d f =
_
W
−W
5
W
d f = 10 Watts (1)
(2) The autocorrelation function is the inverse Fourier transform of S
X
( f ). Consulting
Table 11.1, we note that
S
X
( f ) = 10
1
2W
rect
_
f
2W
_
(2)
It follows that the inverse transform of S
X
( f ) is
R
X
(τ) = 10 sinc(2Wτ) = 10
sin(2πWτ)
2πWτ
(3)
(3) For W = 10 Hz and W = 1 kHZ, graphs of S
X
( f ) and R
X
(τ) appear in Figure 6.
70
Quiz 11.6
In a sampled system, the discrete time impulse δ[n] has a flat discrete Fourier transform.
That is, if R
X
[n] = 10δ[n], then
S
X
(φ) =
n=−∞
10δ[n]e
−j 2πφn
= 10 (1)
Thus, R
X
[n] = 10δ[n]. (This quiz is really lame!)
Quiz 11.7
Since Y(t ) = X(t −t
0
),
R
XY
(t, τ) = E [X(t )Y(t +τ)] = E [X(t )X(t +τ −t
0
)] = R
X
(τ −t
0
) (1)
We see that R
XY
(t, τ) = R
XY
(τ) = R
X
(τ − t
0
). From Table 11.1, we recall the prop-
erty that g(τ − τ
0
) has Fourier transform G( f )e
−j 2π f τ
0
. Thus the Fourier transform of
R
XY
(τ) = R
X
(τ −t
0
) = g(τ −t
0
) is
S
XY
( f ) = S
X
( f )e
−j 2π f t
0
. (2)
Quiz 11.8
We solve this quiz using Theorem 11.17. First we need some preliminary facts. Let
a
0
= 5,000 so that
R
X
(τ) =
1
a
0
a
0
e
−a
0
|τ|
. (1)
Consulting with the Fourier transforms in Table 11.1, we see that
S
X
( f ) =
1
a
0
2a
2
0
a
2
0
+(2π f )
2
=
2a
0
a
2
0
+(2π f )
2
(2)
The RC filter has impulse response h(t ) = a
1
e
−a
1
t
u(t ), where u(t ) is the unit step function
and a
1
= 1/RC where RC = 10
−4
is the filter time constant. From Table 11.1,
H( f ) =
a
1
a
1
+ j 2π f
(3)
(1) Theorem 11.17,
S
XY
( f ) = H( f )S
X
( f ) =
2a
0
a
1
[a
1
+ j 2π f ]
_
a
2
0
+(2π f )
2
_. (4)
(2) Again by Theorem 11.17,
S
Y
( f ) = H
( f )S
XY
( f ) = |H( f )|
2
S
X
( f ). (5)
71
Note that
|H( f )|
2
= H( f )H
( f ) =
a
1
(a
1
+ j 2π f )
a
1
(a
1
− j 2π f )
=
a
2
1
a
2
1
+(2π f )
2
(6)
Thus,
S
Y
( f ) = |H( f )|
2
S
X
( f ) =
2a
0
a
2
1
_
a
2
1
+(2π f )
2
_ _
a
2
0
+(2π f )
2
_ (7)
(3) To find the average power at the filter output, we can either use basic calculus and
calculate
_
−∞
S
Y
( f ) d f directly or we can find R
Y
(τ) as an inverse transform of
S
Y
( f ). Using partial fractions and the Fourier transform table, the latter method is
actually less algebra. In particular, some algebra will show that
S
Y
( f ) =
K
0
a
2
0
+(2π f )
2
+
K
1
a
1
+(2π f )
2
(8)
where
K
0
=
2a
0
a
2
1
a
2
1
−a
2
0
, K
1
=
−2a
0
a
2
1
a
2
1
−a
2
0
. (9)
Thus,
S
Y
( f ) =
K
0
2a
2
0
2a
2
0
a
2
0
+(2π f )
2
+
K
1
2a
2
1
2a
2
1
a
1
+(2π f )
2
. (10)
Consulting with Table 11.1, we see that
R
Y
(τ) =
K
0
2a
2
0
a
0
e
−a
0
|τ|
+
K
1
2a
2
1
a
1
e
−a
1
|τ|
(11)
Substituting the values of K
0
and K
1
, we obtain
R
Y
(τ) =
a
2
1
e
−a
0
|τ|
−a
0
a
1
e
−a
1
|τ|
a
2
1
−a
2
0
. (12)
The average power of the Y(t ) process is
R
Y
(0) =
a
1
a
1
+a
0
=
2
3
. (13)
Note that the input signal has average power R
X
(0) = 1. Since the RC filter has a 3dB
bandwidth of 10,000 rad/sec and the signal X(t ) has most of its its signal energy below
5,000 rad/sec, the output signal has almost as much power as the input.
72
Quiz 11.9
This quiz implements an example of Equations (11.146) and (11.147) for a system in
which we filter Y(t ) = X(t ) + N(t ) to produce an optimal linear estimate of X(t ). The
solution to this quiz is just to find the filter
ˆ
H( f ) using Equation (11.146) and to calculate
the mean square error e
L
∗ using Equation (11.147).
Comment: Since the text omitted the derivations of Equations (11.146) and (11.147), we
note that Example 10.24 showed that
R
Y
(τ) = R
X
(τ) + R
N
(τ), R
Y X
(τ) = R
X
(τ). (1)
Taking Fourier transforms, it follows that
S
Y
( f ) = S
X
( f ) + S
N
( f ), S
Y X
( f ) = S
X
( f ). (2)
Now we can go on to the quiz, at peace with the derivations.
(1) Since µ
N
= 0, R
N
(0) = Var[N] = 1. This implies
R
N
(0) =
_
−∞
S
N
( f ) d f =
_
B
−B
N
0
d f = 2N
0
B (3)
Thus N
0
= 1/(2B). Because the noise process N(t ) has constant power R
N
(0) = 1,
decreasing the single-sided bandwidth B increases the power spectral density of the
noise over frequencies | f | < B.
(2) Since R
X
(τ) = sinc(2Wτ), where W = 5,000 Hz, we see from Table 11.1 that
S
X
( f ) =
1
10
4
rect
_
f
10
4
_
. (4)
The noise power spectral density can be written as
S
N
( f ) = N
0
rect
_
f
2B
_
=
1
2B
rect
_
f
2B
_
, (5)
From Equation (11.146), the optimal filter is
ˆ
H( f ) =
S
X
( f )
S
X
( f ) + S
N
( f )
=
1
10
4
rect
_
f
10
4
_
1
10
4
rect
_
f
10
4
_
+
1
2B
rect
_
f
2B
_. (6)
73
(3) We produce the output
ˆ
X(t ) by passing the noisy signal Y(t ) through the filter
ˆ
H( f ).
From Equation (11.147), the mean square error of the estimate is
e
L
=
_
−∞
S
X
( f )S
N
( f )
S
X
( f ) + S
N
( f )
d f (7)
=
_
−∞
1
10
4
rect
_
f
10
4
_
1
2B
rect
_
f
2B
_
1
10
4
rect
_
f
10
4
_
+
1
2B
rect
_
f
2B
_ d f. (8)
To evaluate the MSE e
L
, we need to whether B ≤ W. Since the problem asks us to
find the largest possible B, let’s suppose B ≤ W. We can go back and consider the
case B > W later. When B ≤ W, the MSE is
e
L
=
_
B
−B
1
10
4
1
2B
1
10
4
+
1
2B
d f =
1
10
4
1
10
4
+
1
2B
=
1
1 +
5,000
B
(9)
To obtain MSE e
L
≤ 0.05 requires B ≤ 5,000/19 = 263.16 Hz.
Although this completes the solution to the quiz, what is happening may not be obvious.
The noise power is always Var[N] = 1 Watt, for all values of B. As B is decreased, the PSD
S
N
( f ) becomes increasingly tall, but only over a bandwidth B that is decreasing. Thus as
B descreases, the filter
ˆ
H( f ) makes an increasingly deep and narrow notch at frequencies
| f | ≤ B. Two examples of the filter
ˆ
H( f ) are shown in Figure 7. As B shrinks, the filter
suppresses less of the signal of X(t ). The result is that the MSE goes down.
Finally, we note that we can choose B very large and also achieve MSE e
L
= 0.05. In
particular, when B > W = 5000, S
N
( f ) = 1/2B over frequencies | f | < W. In this case,
the Wiener filter
ˆ
H( f ) is an ideal (flat) lowpass filter
ˆ
H( f ) =
1
10
4
1
10
4
+
1
2B
| f | < 5,000,
0 otherwise.
(10)
Thus increasing B spreads the constant 1 watt of power of N(t ) over more bandwidth. The
Wiener filter removes the noise that is outside the band of the desired signal. The mean
square error is
e
L
=
_
5000
−5000
1
10
4
1
2B
1
10
4
+
1
2B
d f =
1
2B
1
10
4
+
1
2B
=
1
B
5000
+1
(11)
In this case, B ≥ 9.5 ×10
4
guarantees e
L
≤ 0.05.
Quiz 11.10
It is fairly straightforward to find S
X
(φ) and S
Y
(φ). The only thing to keep in mind is
to use fftc to transform the autocorrelation R
X
[ f ] into the power spectral density S
X
(φ).
The following MATLAB program generates and plots the functions shown in Figure 8
74
−5000 −2000 0 2000 5000
0
0.5
1
f
H
(
f
)
−5000 −2000 0 2000 5000
0
0.5
1
f
H
(
f
)
B = 500 B = 2500
Figure 7: Wiener filter for Quiz 11.9.
%mquiz11.m
N=32;
rx=[2 4 2]; SX=fftc(rx,N); %autocorrelation and PSD
stem(0:N-1,abs(sx));
xlabel(’n’);ylabel(’S_X(n/N)’);
h2=0.5*[1 1]; H2=fft(h2,N); %impulse/filter response: M=2
SY2=SX.* ((abs(H2)).ˆ2);
figure; stem(0:N-1,abs(SY2)); %PSD of Y for M=2
xlabel(’n’);ylabel(’S_{Y_2}(n/N)’);
h10=0.1*ones(1,10); H10=fft(h10,N); %impulse/filter response: M=10
SY10=sx.*((abs(H10)).ˆ2);
figure; stem(0:N-1,abs(SY10));
xlabel(’n’);ylabel(’S_{Y_{10}}(n/N)’);
Relative to M = 2, when M = 10, the filter H(φ) filters out almost all of the high
frequency components of X(t ). In the context of Example 11.26, the low pass moving
average filter for M = 10 removes the high frquency components and results in a filter
output that varies very slowly.
As an aside, note that the vectors SX, SY2 and SY10 in mquiz11 should all be real-
valued vectors. However, the finite numerical precision of MATLAB results in tiny imagi-
nary parts. Although these imaginary parts have no computational significance, they tend
to confuse the stem function. Hence, we generate stem plots of the magnitude of each
power spectral density.
75
0 5 10 15 20 25 30 35
0
5
10
n
S
X
(
n
/
N
)
0 5 10 15 20 25 30 35
0
5
10
n
S
Y
2
(
n
/
N
)
0 5 10 15 20 25 30 35
0
5
10
n
S
Y
1
0
(
n
/
N
)
Figure 8: For Quiz 11.10, graphs of S
X
(φ), S
Y
(n/N) for M = 2, and S
φ
(n/N) for M = 10
using an N = 32 point DFT.
76
Quiz Solutions – Chapter 12
Quiz 12.1
The system has two states depending on whether the previous packet was received in
error. From the problem statement, we are given the conditional probabilities
P
_
X
n+1
= 0|X
n
= 0
_
= 0.99 P
_
X
n+1
= 1|X
n
= 1
_
= 0.9 (1)
Since each X
n
must be either 0 or 1, we can conclude that
P
_
X
n+1
= 1|X
n
= 0
_
= 0.01 P
_
X
n+1
= 0|X
n
= 1
_
= 0.1 (2)
These conditional probabilities correspond to the transition matrix and Markov chain:
0 1
0.01
0.1
0.99 0.9
P =
_
0.99 0.01
0.10 0.90
_
(3)
Quiz 12.2
From the problem statement, the Markov chain and the transition matrix are
0 1 1
0.6 0.2
0.2 0.6
0.4 0.6 0.4
P =
0.4 0.6 0
0.2 0.6 0.2
0 0.6 0.4
(1)
The eigenvalues of P are
λ
1
= 0 λ
2
= 0.4 λ
3
= 1 (2)
We can diagonalize P into
P = S
−1
DS =
−0.6 0.5 1
0.4 0 1
−0.6 −0.5 1
λ
1
0 0
0 λ
2
0
0 0 λ
3
−0.5 1 −0.5
1 0 −1
0.2 0.6 0.2
(3)
where s
i
, the i th row of S, is the left eigenvector of P satisfying s
i
P = λ
i
s
i
. Algebra will
verify that the n-step transition matrix is
P
n
= S
−1
D
n
S =
0.2 0.6 0.2
0.2 0.6 0.2
0.2 0.6 0.2
+(0.4)
n
0.5 0 −0.5
0 0 0
−0.5 0 0.5
(4)
Quiz 12.3
The Markov chain describing the factory status and the corresponding state transition
matrix are
77
2
0 1
0.9
0.1
1
1
P =
0.9 0.1 0
0 0 1
1 0 0
(1)
With π =
_
π
0
π
1
π
2
_
, the system of equations π
= π
P yields π
1
= 0.1π
0
and
π
2
= π
1
. This implies
π
0
1
2
= π
0
(1 +0.1 +0.1) = 1 (2)
It follows that the limiting state probabilities are
π
0
= 5/6, π
1
= 1/12, π
2
= 1/12. (3)
Quiz 12.4
The communicating classes are
C
1
= {0, 1} C
2
= {2, 3} C
3
= {4, 5, 6} (1)
The states in C
1
and C
3
are aperiodic. The states in C
2
have period 2. Once the system
enters a state in C
1
, the class C
1
is never left. Thus the states in C
1
are recurrent. That
is, C
1
is a recurrent class. Similarly, the states in C
3
are recurrent. On the other hand, the
states in C
2
are transient. Once the system exits C
2
, the states in C
2
are never reentered.
Quiz 12.5
At any time t , the state n can take on the values 0, 1, 2, . . .. The state transition proba-
bilities are
P
n−1,n
= P [K > n|K > n −1] =
P [K > n]
P [K > n −1]
(1)
P
n−1,0
= P [K = n|K > n −1] =
P [K = n]
P [K > n −1]
(2)
(3)
The Markov chain resembles
0 1
P K=2 [ ]
P K= [ 1]
3 4
P K=4 [ ]
2
P K=3 [ ]
P K=5 [ ]
1 1 1 1 1
… ...
78
The stationary probabilities satisfy
π
0
= π
0
P [K = 1] +π
1
, (4)
π
1
= π
0
P [K = 2] +π
2
, (5)
.
.
.
π
k−1
= π
0
P [K = k] +π
k
, k = 1, 2, . . . (6)
From Equation (4), we obtain
π
1
= π
0
(1 − P [K = 1]) = π
0
P [K > 1] (7)
Similarly, Equation (5) implies
π
2
= π
1
−π
0
P [K = 2] = π
0
(P [K > 1] − P [K = 2]) = π
0
P [K > 2] (8)
This suggests that π
k
= π
0
P[K > k]. We verify this pattern by showing that π
k
=
π
0
P[K > k] satisfies Equation (6):
π
0
P [K > k −1] = π
0
P [K = k] +π
0
P [K > k] . (9)
When we apply
k=0
π
k
= 1, we obtain π
0
n=0
P[K > k] = 1. From Problem 2.5.11,
we recall that
k=0
P[K > k] = E[K]. This implies
π
n
=
P [K > n]
E [K]
(10)
This Markov chain models repeated random countdowns. The system state is the time until
the counter expires. When the counter expires, the system is in state 0, and we randomly
reset the counter to a new value K = k and then we count down k units of time. Since we
spend one unit of time in each state, including state 0, we have k −1 units of time left after
the state 0 counter reset. If we have a random variable W such that the PMF of W satisfies
P
W
(n) = π
n
, then W has a discrete PMF representing the remaining time of the counter at
a time in the distant future.
Quiz 12.6
(1) By inspection, the number of transitions need to return to state 0 is always a multiple
of 2. Thus the period of state 0 is d = 2.
(2) To find the stationary probabilities, we solve the system of equations π = πP and
3
i =0
π
i
= 1:
π
0
= (3/4)π
1
+(1/4)π
3
(1)
π
1
= (1/4)π
0
+(1/4)π
2
(2)
π
2
= (1/4)π
1
+(3/4)π
3
(3)
1 = π
0
1
2
3
(4)
79
Solving the second and third equations for π
2
and π
3
yields
π
2
= 4π
1
−π
0
π
3
= (4/3)π
2
−(1/3)π
1
= 5π
1
−(4/3)π
0
(5)
Substituting π
3
back into the first equation yields
π
0
= (3/4)π
1
+(1/4)π
3
= (3/4)π
1
+(5/4)π
1
−(1/3)π
0
(6)
This implies π
1
= (2/3)π
0
. It follows from the first and second equations that
π
2
= (5/3)π
0
and π
3
= 2π
0
. Lastly, we choose π
0
so the state probabilities sum to
1:
1 = π
0
1
2
3
= π
0
_
1 +
2
3
+
5
3
+2
_
=
16
3
π
0
(7)
It follows that the state probabilities are
π
0
=
3
16
π
1
=
2
16
π
2
=
5
16
π
3
=
6
16
(8)
(3) Since the system starts in state 0 at time 0, we can use Theorem 12.14 to find the
limiting probability that the system is in state 0 at time nd:
lim
n→∞
P
00
(nd) = dπ
0
=
3
8
(9)
Quiz 12.7
The Markov chain has the same structure as that in Example 12.22. The only difference
is the modified transition rates:
0 1
1
3 4
( ) 2/3
a
1 - ( ) 2/3
a
( ) 3/4
a
1 - 3/4 ( )
a
( ) 4/5
a
1 - 4/5 ( )
a
2
( ) 1/2
a
1- 1/2 ( )
a
The event T
00
> n occurs if the system reaches state n before returning to state 0, which
occurs with probability
P [T
00
> n] = 1 ×
_
1
2
_
α
×
_
2
3
_
α
×· · · ×
_
n −1
n
_
α
=
_
1
n
_
α
. (1)
Thus the CDF of T
00
satisfies F
T
00
(n) = 1−P[T
00
> n] = 1−1/n
α
. To determine whether
state 0 is recurrent, we observe that for all α > 0
P [V
00
] = lim
n→∞
F
T
00
(n) = lim
n→∞
1 −
1
n
α
= 1. (2)
80
Thus state 0 is recurrent for all α > 0. Since the chain has only one communicating class,
all states are recurrent. ( We also note that if α = 0, then all states are transient.)
To determine whether the chain is null recurrent or positive recurrent, we need to calcu-
late E[T
00
]. In Example 12.24, we did this by deriving the PMF P
T
00
(n). In this problem,
it will be simpler to use the result of Problem 2.5.11 which says that
k=0
P[K > k] =
E[K] for any non-negative integer-valued random variable K. Applying this result, the
E [T
00
] =
n=0
P [T
00
> n] = 1 +
n=1
1
n
α
. (3)
For 0 < α ≤ 1, 1/n
α
≥ 1/n and it follows that
E [T
00
] ≥ 1 +
n=1
1
n
= ∞. (4)
We conclude that the Markov chain is null recurrent for 0 < α ≤ 1. On the other hand, for
α > 1,
E [T
00
] = 2 +
n=2
1
n
α
. (5)
Note that for all n ≥ 2
1
n
α
_
n
n−1
dx
x
α
(6)
This implies
E [T
00
] ≤ 2 +
n=2
_
n
n−1
dx
x
α
(7)
= 2 +
_
1
dx
x
α
(8)
= 2 +
x
−α+1
−α +1
¸
¸
¸
¸
1
= 2 +
1
α −1
< ∞ (9)
Thus for all α > 1, the Markov chain is positive recurrent.
Quiz 12.8
The number of customers in the ”friendly” store is given by the Markov chain
1 i i+1
p p p
( )( ) 1-p 1-q ( )( ) 1-p 1-q ( )( ) 1-p 1-q ( )( ) 1-p 1-q
( ) 1-p q ( ) 1-p q ( ) 1-p q ( ) 1-p q
0
××× ×××
81
In the above chain, we note that (1 − p)q is the probability that no new customer arrives,
an existing customer gets one unit of service and then departs the store.
By applying Theorem 12.13 with state space partitioned between S = {0, 1, . . . , i } and
S
= {i +1, i +2, . . .}, we see that for any state i ≥ 0,
π
i
p = π
i +1
(1 − p)q. (1)
This implies
π
i +1
=
p
(1 − p)q
π
i
. (2)
Since Equation (2) holds for i = 0, 1, . . ., we have that π
i
= π
0
α
i
where
α =
p
(1 − p)q
. (3)
Requiring the state probabilities to sum to 1, we have that for α < 1,
i =0
π
i
= π
0
i =0
α
i
=
π
0
1 −α
= 1. (4)
Thus for α < 1, the limiting state probabilities are
π
i
= (1 −α)α
i
, i = 0, 1, 2, . . . (5)
In addition, for α ≥ 1 or, equivalently, p ≥ q/(1 − q), the limiting state probabilities do
not exist.
Quiz 12.9
The continuous time Markov chain describing the processor is
0 1
2
3.01
3 4
2
3
2
3
2
2
3
0.01
0.01
0.01
Note that q
10
= 3.1 since the task completes at rate 3 per msec and the processor reboots
at rate 0.1 per msec and the rate to state 0 is the sum of those two rates. From the Markov
chain, we obtain the following useful equations for the stationary distribution.
5.01p
1
= 2p
0
+3p
2
5.01p
2
= 2p
1
+3p
3
5.01p
3
= 2p
2
+3p
4
3.01p
4
= 2p
3
We can solve these equations by working backward and solving for p
4
in terms of p
3
, p
3
in terms of p
2
and so on, yielding
p
4
=
20
31
p
3
p
3
=
620
981
p
2
p
2
=
19620
31431
p
1
p
1
=
628, 620
1, 014, 381
p
0
(1)
82
Applying p
0
+ p
1
+ p
2
+ p
3
+ p
4
= 1 yields p
0
= 1, 014, 381/2, 443, 401 and the
stationary probabilities are
p
0
= 0.4151 p
1
= 0.2573 p
2
= 0.1606 p
3
= 0.1015 p
4
= 0.0655 (2)
Quiz 12.10
The M/M/c/∞queue has Markov chain
c c+1 1 0
λ λ λ λ λ
µ 2µ
cµ cµ cµ
From the Markov chain, the stationary probabilities must satisfy
p
n
=
_
(ρ/n) p
n−1
n = 1, 2, . . . , c
(ρ/c) p
n−1
n = c +1, c +2, . . .
(1)
It is straightforward to show that this implies
p
n
=
_
p
0
ρ
n
/n! n = 1, 2, . . . , c
p
0
(ρ/c)
n−c
ρ
c
/c! n = c +1, c +2, . . .
(2)
The requirement that
n=0
p
n
= 1 yields
p
0
=
_
c
n=0
ρ
n
/n! +
ρ
c
c!
ρ/c
1 −ρ/c
_
−1
(3)
83 | 48,002 | 103,351 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-47 | latest | en | 0.888369 |
https://codereview.stackexchange.com/questions/222893/knights-tour-python | 1,585,881,697,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00119.warc.gz | 424,492,476 | 35,107 | Knight's Tour - Python
Below is my recursive solution for the Knights Tour. The user can set the board size and the program will print the first legal knight's tour it finds.
As there are potentially millions of correct solutions to the knight's tour, only one solution is given.
def legal_moves(visited, position, sq):
"""Calculate legal moves"""
legal_moves = [
(-2, -1),
(-2, 1),
(-1, -2),
(-1, 2),
(1, -2),
(1, 2),
(2, -1),
(2, 1),
]
row = position // sq
col = position % sq
move_to = []
for dx, dy in legal_moves:
if 0 <= (row + dx) < sq and 0 <= (col + dy) < sq:
new_position = (position + (sq * dx)) + dy
if new_position not in visited and 0 <= new_position < (sq * sq):
move_to.append(new_position)
return move_to
def make_move(move_to, visited, current, route, sq):
"""Carry out the move"""
for move in move_to:
new_route = route + f"{current}-"
solution_found = next_move(visited, move, new_route, sq)
visited.remove(current)
if solution_found:
return True
return False
def next_move(visited, current, route, sq):
"""Find the next valid moves and instruct "make_move" to carry them out"""
if len(visited) == (sq * sq) - 1:
route += f"{current}"
print(route)
return True
move_to = legal_moves(visited, current, sq)
solution_found = make_move(move_to, visited, current, route, sq)
if solution_found:
return True
def start_tour(sq):
"""Calculate the knights tour for grid sq*sq starting at all positions in range 0-sq"""
for starting in range(sq * sq):
visited = set()
route = ""
solution_found = next_move(visited, starting, route, sq)
if solution_found:
return
print("No knights tour could be calculated")
if __name__ == "__main__":
square_size = 8
start_tour(square_size)
EDIT I have added a print_route function which is called from inside next_move in place of the current print statement as follows: print_route(route, sq)
def print_route(route, size):
"""Convert the 1D array into a 2D array tracking the movement on the knight"""
import numpy as np
steps = route.split("-")
array = np.zeros(size * size)
for index, item in enumerate(steps):
array[int(item)] = index + 1
array = array.reshape(size, size)
print(array)
Starting position
If there is a route to be found, it will go through cell 0 and will be found at first iteration of start_tour. We can remove the loop and just have starting = 0.
Generating legal moves
Various details can be improved in the legal_moves function.
This is a good occasion to use a generator with the keyword yield.
We could compute row and col with a single call to divmod.
We could make computation of new position more straight-forward with intermediates variables for coordinates on each axis.
Because of the way new_position is computed, there is no need for the additional boundary check.
We'd get something like:
def generate_new_positions(visited, position, sq):
"""Yield legal moves"""
generate_new_positions = [
(-2, -1),
(-2, 1),
(-1, -2),
(-1, 2),
(1, -2),
(1, 2),
(2, -1),
(2, 1),
]
# position = row * sq + col
row, col = divmod(position, sq)
for dx, dy in generate_new_positions:
x, y = row + dx, col + dy
if 0 <= x < sq and 0 <= y < sq:
new_pos = x * sq + y
if new_pos not in visited:
yield new_pos
Separation of concerns: printing vs returning a result
When a result is found, it is printed and a boolean (or None) is returned in the different functions. It would be easier to return either None or the result found and to have that result printed from a single point of the logic.
We'd have something like:
def make_move(move_to, visited, current, route, sq):
"""Carry out the move"""
for move in move_to:
new_route = route + str(current) + "-"
solution_found = next_move(visited, move, new_route, sq)
visited.remove(current)
if solution_found:
return solution_found
return None
def next_move(visited, current, route, sq):
"""Find the next valid moves and instruct "make_move" to carry them out"""
if len(visited) == (sq * sq) - 1:
route += str(current)
return route
move_to = generate_new_positions(visited, current, sq)
return make_move(move_to, visited, current, route, sq)
def start_tour(sq):
"""Calculate the knights tour for grid sq*sq starting at all positions in range 0-sq"""
starting = 0
visited = set()
route = ""
return next_move(visited, starting, route, sq)
if __name__ == "__main__":
for square_size in 3, 5, 6:
ret = start_tour(square_size)
print("No knights tour could be calculated" if ret is None else ret)
Also, formatting the string could be done in a single place as well. We could use lists for instance in all the logic.
def make_move(move_to, visited, current, route, sq):
"""Carry out the move"""
for move in move_to:
new_route = route + [current]
solution = next_move(visited, move, new_route, sq)
visited.remove(current)
if solution:
return solution
return None
def next_move(visited, current, route, sq):
"""Find the next valid moves and instruct "make_move" to carry them out"""
if len(visited) == (sq * sq) - 1:
return route + [current]
move_to = generate_new_positions(visited, current, sq)
return make_move(move_to, visited, current, route, sq)
def start_tour(sq):
"""Calculate the knights tour for grid sq*sq starting at all positions in range 0-sq"""
starting = 0
visited = set()
route = []
return next_move(visited, starting, route, sq)
if __name__ == "__main__":
for square_size in 3, 5, 6:
ret = start_tour(square_size)
print("No knights tour could be calculated" if ret is None else "-".join((str(e) for e in ret)))
Reducing the duplicated information
We're maintaining a visited set and a route list: both containing roughtly the same data. Maybe we could recompute visited from the route when we need it.
def make_move(move_to, current, route, sq):
"""Carry out the move"""
for move in move_to:
solution = next_move(move, route + [current], sq)
if solution:
return solution
return None
def next_move(current, route, sq):
"""Find the next valid moves and instruct "make_move" to carry them out"""
if len(route) == (sq * sq) - 1:
return route + [current]
move_to = generate_new_positions(set(route), current, sq)
return make_move(move_to, current, route, sq)
def start_tour(sq):
"""Calculate the knights tour for grid sq*sq starting at all positions in range 0-sq"""
return next_move(0, [], sq)
Simplifying the logic
Having a function A calling a function B itself calling B can make things hard to understand properly because both A and B are hard to understand independently.
Here, we could get rid of make_move by integrating directly in next_move:
def next_move(current, route, sq):
"""Find the next valid moves and carry them out"""
if len(route) == (sq * sq) - 1:
return route + [current]
for move in generate_new_positions(set(route), current, sq):
solution = next_move(move, route + [current], sq)
if solution:
return solution
return None
More simplification
In next_move, route is almost always used in the expression route + [current]. We could directly define this at the beginning of the function:
def next_move(current, route, sq):
"""Find the next valid moves and carry them out"""
new_route = route + [current]
if len(new_route) == (sq * sq):
return new_route
for move in generate_new_positions(set(new_route), current, sq):
solution = next_move(move, new_route, sq)
if solution:
return solution
return None
More importantly, it leads to the question: why do we provide a route AND an element to add to it instead of just having that element in the list.
def next_move(current, route, sq):
"""Find the next valid moves and carry them out"""
if len(route) == (sq * sq):
return route
for move in generate_new_positions(set(route), current, sq):
solution = next_move(move, route + [move], sq)
if solution:
return solution
return None
def start_tour(sq):
"""Calculate the knights tour for grid sq*sq starting at all positions in range 0-sq"""
start = 0
return next_move(start, [start], sq)
Going further, we do not really the current argument anymore as we can compute it from route.
def next_move(route, sq):
"""Find the next valid moves and carry them out"""
if len(route) == (sq * sq):
return route
current = route[-1]
new_pos = generate_new_positions(set(route), current, sq)
for move in new_pos:
solution = next_move(route + [move], sq)
if solution:
return solution
return None
def start_tour(sq):
"""Calculate the knights tour for grid sq*sq starting at all positions in range 0-sq"""
start = 0
return next_move([start], sq)
Final code
def generate_new_positions(visited, position, sq):
"""Yield legal moves"""
generate_new_positions = [
(-2, -1),
(-2, 1),
(-1, -2),
(-1, 2),
(1, -2),
(1, 2),
(2, -1),
(2, 1),
]
# position = row * sq + col
row, col = divmod(position, sq)
for dx, dy in generate_new_positions:
x, y = row + dx, col + dy
if 0 <= x < sq and 0 <= y < sq:
new_pos = x * sq + y
if new_pos not in visited:
yield new_pos
def next_move(route, sq):
"""Find the next valid moves and carry them out"""
if len(route) == (sq * sq):
return route
current = route[-1]
new_pos = generate_new_positions(set(route), current, sq)
for move in new_pos:
solution = next_move(route + [move], sq)
if solution:
return solution
return None
if __name__ == "__main__":
for square_size in 3, 5, 6:
ret = next_move([0], square_size)
print("No knights tour could be calculated" if ret is None else "-".join((str(e) for e in ret)))
• Amazing thank you. It does make a lot of sense to use visited and route for the same purpose. I'm not sure why I didn't do this in the first place. I really like the way you return the function. It makes a lot more sense to me than returning the result of a function. Thanks – EML Jun 24 '19 at 22:09
Precomputed Moves
You are computing a lot of integer quotients and remainders. (@Josay's solution as well). You can reduce the amount of work done during the solving of the Knight's Tour by pre-computing the positions which can be reached from each square:
legal_moves = (-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1)
moves = [ [ (r+dr) * cols + (c+dc) for dr, dc in legal_moves
if 0 <= r+dr < rows and 0 <= c+dc < cols ]
for r in range(rows) for c in range(cols) ]
I've used rows and cols here, to generalize from a square board to an NxM board. After this point, the logic doesn't change.
Now moves[0] is a list of all (both) positions that can be reached from square #0, moves[1] is a list of the 3 positions that can be reached from square #1, and so on. Note that you never need to deal with the dimension of the board again. No integer division or modulo remainders; just use the square's index number.
Visited
Your visited set is a nice way, to keep track of whether or not a position has been visited or not. In $$\O(1)\$$ time, you can tell if you've reached a visited location or not. @Josay's suggestion of using set(route) to recompute the set of visited locations removes the need for maintaining a separate global set() structure, but the cost is recreating the set(route) each move. This is at least $$\O(N)\$$ per move, a huge decrease in performance.
A different structure for keeping track of visited locations is to simply use an array of positions. If visited[move] is True, you've been there, if not, you haven't. This is faster that using a set(); you a just looking up an index. It is $$\O(1)\$$ with a very small constant.
Better: visited[move] = move_number. Initialize the array to 0 to start with every spot unvisited, and mark a 1 in the first move location, 2 in the second and so on. Set visited[move] = 0 when you back-track. No need to keep track of the route, as it is implicit in the visited[] array. print_route() amounts to reshaping the visited array into a 2D array, and printing it.
• Thank you for this reply. I am always dubious about using multi-line for loops and especially nested list comprehension because I have always read that if the code can be written in roughly the same number of lines with a similar space complexity (in my case it is O(1)) then it is better to use the more readable solution. Having said that I do think its a nice solution! – EML Jun 25 '19 at 9:26 | 3,136 | 12,071 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-16 | latest | en | 0.764779 |
http://forums.wolfram.com/mathgroup/archive/2002/Feb/msg00435.html | 1,526,916,232,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864405.39/warc/CC-MAIN-20180521142238-20180521162238-00233.warc.gz | 115,355,128 | 7,666 | need a function for sums of subsets
• To: mathgroup at smc.vnet.net
• Subject: [mg33050] need a function for sums of subsets
• From: "Juan Erfá" <erfa11 at hotmail.com>
• Date: Wed, 27 Feb 2002 00:48:11 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```Hi Mark, I don’t understand what you mean by … return the
“first” subset …
Anyway, you can use my function, I sent days ago (Iterators), which I wrote
for integers, but I think can be use also for rationales.
Her you get the rational n in s parts, only with numbers in the list x of
rationales.
In[1]:=QQ[n_, s_, x_List] :=
Module[{a = Array[p, s, 0], w, v, r, d, q, t, sol, u = Union[x]},
p[0] = 0;
Off[Part::pspec];
w = Table[r[[p[i]]], {i, s - 1}];
v = Flatten[{w, n - Plus @@ w}]; r = u;
d = Last[Position[r, _?(#1 < n/2 & )]][[1]] - s + 1;
t = Table[{p[i], p[i - 1] + 1, d + i}, {i, s - 1}];
sol = Flatten[Table[v, Evaluate[Sequence @@ t]], Abs[s - 2]];
Select[sol, MemberQ[x, Last[#1]] && Last[#1] > #1[[-2]] & ]]
In[1]:=q = {1/2, 1/3, 1/4, 1/8, 3/10, 12/79, 13/38};
In[3]:=QQ[3/4, 2, q]
Out[3]={{1/4, 1/2}}
In[4]:=QQ[3/8, 2, q]
Out[4]={{1/8, 1/4}}
In[5]:=(QQ[#1, 2, q] & ) /@ {3/4, 3/8}
Out[5]={{{1/4, 1/2}}, {{1/8, 1/4}}}
In[6]:=QQ[61/84, 3, {1/8, 1/7, 1/4, 1/3, 1/2}]
Out[6]={{1/7, 1/4, 1/3}}
In[7]:=w = 1/Table[Random[Integer, {1, 20}], {50}]
Out[7]={1/7, 1/10, 1/3, 1/11, 1/3, 1/2, 1/7, 1/19, 1/2, 1/11, 1/11, 1/16,
1/11, 1/14, 1/5, 1/9, 1/10, 1/16, 1, 1, 1/17, 1/15, 1/6, 1, 1/18, 1/20,
1/11, 1/3, 1/14, 1/8, 1/20, 1/13, 1/16, 1/5, 1/18, 1/13, 1/11, 1/8, 1/20,
1/2, 1, 1/20, 1/10, 1/18, 1/16, 1/5, 1/20, 1/16, 1/4, 1/2}
In[8]:=Timing[QQ[297/560, 4, w]]
Out[8]={0.13*Second, {{1/16, 1/8, 1/7, 1/5}}}
This function need to be correct, sure in the part of the code:
d=Last[….],and t=Table[…].
Regards
Juan
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1. Nov 6, 2012
### dpesios
Hello all,
I would like beforehand to inform you that the translation of the following geometric problem is not very good and consequently you will have to use your mathematical intuition just a little bit. I encountered it while giving admission exams in a Mathematics department to pursue a second degree.
I don't have a mathematical background so any help appreciated.
1. The problem statement, all variables and given/known data
"Find the locus of points from which the conducted tangents at a given isosceles hyperbola are vertical to each other"
2. Relevant equations
3. The attempt at a solution
Can you assume that the locus contains just one point (which is the origin of the axes) and the tangents touch the curve at infinity ? :grumpy:
I apologize for my English ...
Last edited: Nov 6, 2012
2. Nov 6, 2012
### haruspex
All makes sense to me except for "isosceles hyperbola". Could it mean rectangular?
For the rest, I would word it as "Find the locus of points from which the subtended tangents to a given (?) hyperbola are perpendicular to each other"
No.
3. Nov 7, 2012
### dpesios
Yes, I do mean rectangular hyperbola.
When it comes to geometry it is all Greek to me ...
I'm not sure whether we should say "conducted tangent from a given point" or "subtended tangent".
As far as I understand two tangent lines start from a point on the locus and touch the given curve in a way that they are perpendicular to each other.
Any suggestions ?
4. Nov 7, 2012
### haruspex
Let (x,y) be a point on the locus, (x',y'), (x", y") be the contact points on the hyperbola of the conducted/subtended tangents. We have the following equations (assuming a canonical form for the hyperbola):
y'2-x'2=c2
y"2-x"2=c2
y'y - x'x = c2
y"y - x"x = c2
x'x"+y'y" = 0 (tangents perpendicular)
In principle, should be possible to eliminate x', x", y', y" to obtain an equation for x and y.
5. Nov 9, 2012
### dpesios
I see your point but there is always a but ...
No matter how hard I try, I cannot figure out an equation by eliminating x',x'',y' and y''.
Maybe there is another way of solving the problem.
I don't know. What do you think ?
6. Nov 9, 2012
### haruspex
I worked it through and got x=y=0. Then I thought about it geometrically and realised that is right. The gradient of the curve y2=x2+c2 has magnitude <= 1 everywhere. So two tangents at right angles must have gradients +1 and -1. In the more general hyperbola y2=m2x2+c2, only if m>1 does the locus become more interesting.
7. Nov 10, 2012
### dpesios
So, the answer is that there is no locus or the locus contains just one point and the contact points with the curve are at infinity as I initially proposed (?).
If it is so, I must play lottery. | 745 | 2,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.893031 |
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Predicates are properties, additional information to better express the subject of the sentence. A quantified predicate is a proposition , that is, when you assign values to a predicate with variables it can be made a proposition. For example : In P(x) : x>5, x is the subject or the variable and ‘>5’ is the predicate.
## What is predicate logic example?
It is denoted by the symbol ∀. ∀xP(x) is read as for every value of x, P(x) is true. Example − “Man is mortal” can be transformed into the propositional form ∀xP(x) where P(x) is the predicate which denotes x is mortal and the universe of discourse is all men.
## What is a predicate in a sentence examples?
: the part of a sentence or clause that tells what is said about the subject “Rang” in “the doorbell rang” is the predicate. : completing the meaning of a linking verb “Sweet” in “the sugar is sweet” is a predicate adjective.
## What is a predicate logic statement?
What Is Predicate Logic. A predicate is a statement or mathematical assertion that contains variables, sometimes referred to as predicate variables, and may be true or false depending on those variables’ value or values.
## What is first-order logic examples?
Definition A first-order predicate logic sentence G over S is a tautology if F |= G holds for every S-structure F. Examples of tautologies (a) ∀x.P(x) → ∃x.P(x); (b) ∀x.P(x) → P(c); (c) P(c) → ∃x.P(x); (d) ∀x(P(x) ↔ ¬¬P(x)); (e) ∀x(¬(P1(x) ∧ P2(x)) ↔ (¬P1(x) ∨ ¬P2(x))).
## How predicate logic is different from propositional logic explain with an example?
A quantified predicate is a proposition , that is, when you assign values to a predicate with variables it can be made a proposition.
Difference between Propositional Logic and Predicate Logic.
Propositional Logic Predicate Logic
3 A proposition has a specific truth value, either true or false. A predicate’s truth value depends on the variables’ value.
## Where is predicate logic used?
What are quantifiers? In predicate logic, predicates are used alongside quantifiers to express the extent to which a predicate is true over a range of elements. Using quantifiers to create such propositions is called quantification.
## Is predicate logic and First-Order Logic same?
First-order logic—also known as predicate logic, quantificational logic, and first-order predicate calculus—is a collection of formal systems used in mathematics, philosophy, linguistics, and computer science.
## Why is predicate logic Important?
Predicate logic allows us to talk about variables (pronouns). The value for the pronoun is some individual in the domain of universe that is contextually determined.
## Is predicate logic complete?
Truth-functional propositional logic and first-order predicate logic are semantically complete, but not syntactically complete (for example, the propositional logic statement consisting of a single propositional variable A is not a theorem, and neither is its negation).
## Who invented predicate logic?
Charles Pierce and Gottlob Frege are just as important to this story because they invented Predicate or First-order Logic. Take the cat-leftof-dog-leftof-human example. That is not just true for cats, dogs, and humans. It’s true for any three things.
## What is a predicate in first-order logic?
First-order logic is symbolized reasoning in which each sentence, or statement, is broken down into a subject and a predicate. The predicate modifies or defines the properties of the subject. In first-order logic, a predicate can only refer to a single subject.
## What are the limitations of predicate logic?
One key limitation is that it applies only to atomic propositions. There is no way to talk about properties that apply to categories of objects, or about relationships between those properties. That’s what predicate logic is for.
## What is predicate logic in artificial intelligence?
FOL is a mode of representation in Artificial Intelligence. It is an extension of PL. FOL represents natural language statements in a concise way. FOL is also called predicate logic. It is a powerful language used to develop information about an object and express the relationship between objects.
## What is predicate logic how it aid in creating proofs?
A proof in predicate logic has much the same form as a proof in propositional logic. We begin with a set of axioms (or hypotheses) A1.. An, and using the rules of inference, we construct a sequence of expressions that follow from those axioms. | 982 | 4,545 | {"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} | 3.78125 | 4 | CC-MAIN-2022-27 | longest | en | 0.851524 |
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Re: CR - smoke hood requirement - Be careful! [#permalink]
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11 May 2010, 21:36
easy one Ans:E
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11 May 2010, 22:32
Definitely option E
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19 May 2010, 05:00
E must be the winner
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20 May 2010, 05:01
E and only E
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06 Sep 2010, 05:55
my answer is E.
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Re: CR - smoke hood requirement - Be careful! [#permalink]
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13 May 2011, 03:37
The answer is E.
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13 May 2011, 06:03
Simple 30second CR. E.
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13 May 2011, 06:13
E
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Re: CR - smoke hood requirement - Be careful! [#permalink]
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13 May 2011, 06:22
Between B and E, B touches the tip however E explains a strong reason behind getting the smoke hoods.
E.
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13 May 2011, 08:20
E. I wish all questions were like this one!
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E
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13 May 2011, 11:09
Fairly easy one
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Re: CR - smoke hood requirement - Be careful! [#permalink]
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13 May 2011, 20:01
E.
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13 May 2011, 20:34
A the right answer
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13 May 2011, 21:20
what is the reason for stating ..be careful ? i think E is clear and simple , took me 1:04 min to get to E. and almost everyone got this right.
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18 May 2011, 01:33
E is the winner as it strengthens the statement
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Re: CR - smoke hood requirement - Be careful! [#permalink]
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30 May 2011, 20:03
(A) Test evacuations showed that putting on the smoke hoods added considerably to the overall time it took passengers to leave the cabin. - This weakens the argument as it makes them less safe.
(B) Some airlines are willing to buy the smoke hoods even though they consider them to be prohibitively expensive. - Out of scope (and neither strengthens or weakens)
(C) Although the smoke hoods protect passengers from the toxic gases, they can do nothing to prevent the gases from igniting. -The argument is about inhaling toxic fumes not preventing fires and the hoods presumably would prevent inhalation so the act of combustion is out of scope for their directed purpose.
(D) Some experienced flyers fail to pay attention to the safety instructions given on every commercial flight before takeoff. -Out of scope
(E) In many airplane accidents, passengers who were able to reach emergency exits were overcome by toxic gases before they could exit the ariplane. -The only strengthener that clearly prevents smoke inhalation.
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Re: CR - smoke hood requirement - Be careful! [#permalink]
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17 May 2012, 04:30
ashkanator wrote:
A the right answer
can you also please explain how you arrived at A?
wasn't E straightforward?
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Re: Passengers must exit airplanes swiftly after accidents, [#permalink]
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28 Aug 2012, 22:00
I found nice discussion on this question at
passengers-must-exit-airplanes-swiftly-after-accidents-78391.html
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Re: Passengers must exit airplanes swiftly after accidents, [#permalink]
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29 Aug 2012, 22:55
(A) Test evacuations showed that putting on the smoke hoods added considerably to the overall time it took passengers to leave the cabin.
weaken
(B) Some airlines are willing to buy the smoke hoods even though they consider them to be prohibitively expensive.
No relation
(C) Although the smoke hoods protect passengers from the toxic gases, they can do nothing to prevent the gases from igniting.
Weaken.
(D) Some experienced flyers fail to pay attention to the safety instructions given on every commercial flight before takeoff.
How would they then listen to instructions of using smoke hoods?
(E) In many airplane accidents, passengers who were able to reach emergency exits were overcome by toxic gases before they could exit the airplane.
Our golden goose
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MOVE IT, Math!
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Home : Support : Online Help : System : Information : Updates : Maple 2020 : Integral Transforms
Integral Transforms
Integral transforms are special integrals that appear frequently in mathematical physics and that have remarkable properties. For Maple 2020, the implementation of these integrals has been extended in several ways, making them more useful for a variety of applications, including computing integrals, finding exact solutions to PDEs with boundary conditions (see also what's new in Maple 2020 for PDE & Boundary Conditions). Improvements include the option to compute derivatives, numeric support, an alternate definition of the Hankel transform, and the ability to compute more transforms.
As background, these integrals are represented by the commands of the inttrans package:
> $\mathrm{with}\left(\mathrm{inttrans}\right)$
$\left[{\mathrm{addtable}}{,}{\mathrm{fourier}}{,}{\mathrm{fouriercos}}{,}{\mathrm{fouriersin}}{,}{\mathrm{hankel}}{,}{\mathrm{hilbert}}{,}{\mathrm{invfourier}}{,}{\mathrm{invhilbert}}{,}{\mathrm{invlaplace}}{,}{\mathrm{invmellin}}{,}{\mathrm{laplace}}{,}{\mathrm{mellin}}{,}{\mathrm{savetable}}{,}{\mathrm{setup}}\right]$ (1)
Three of these commands, addtable, savetable, and setup (new as of Maple 2020) are "administrative" commands, while the others are computational representations for definite integrals. For example,
>
$\left[{ℱ}{}\left({a}{,}{b}{,}{z}\right){=}{{\int }}_{{-}{\mathrm{\infty }}}^{{\mathrm{\infty }}}\frac{{a}}{{{ⅇ}}^{{I}{}{b}{}{z}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{b}{,}{\mathrm{with no restrictions on}}{}\left({a}{,}{b}{,}{z}\right)\right]$ (2)
>
$\left[{ℳ}{}\left({a}{,}{b}{,}{z}\right){=}{{\int }}_{{0}}^{{\mathrm{\infty }}}{a}{}{{b}}^{{z}{-}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{b}{,}{\mathrm{with no restrictions on}}{}\left({a}{,}{b}{,}{z}\right)\right]$ (3)
For all integral transform commands, the first argument is the integrand, the second one is the dummy integration variable of a definite integral and the third one is the evaluation point (also called transform variable).
Compute derivatives: yes or no.
Generally speaking, in computer algebra, "given a function $f\left(x\right)$, the input $\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}f\left(x\right)$ should return the derivative of $f\left(x\right)$". For the purpose of using integral transforms to solve differential equations, however, the implementation in previous Maple releases worked in the opposite direction: if you were to input the result of the derivative, you would receive the derivative representation. For example, to the input $\mathrm{laplace}\left(-t\cdot f\left(t\right),t,s\right)$ you would receive $\frac{d}{\mathrm{ds}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{laplace}\left(f\left(t\right),t,s\right)$. To extend the capabilities of the system and have the best result in any scenario, a new command, setup, has been added to the package, so that you can set whether or not to compute derivatives, and the default has been changed to computederivatives = true while the previous behavior is obtained only if you input $\mathrm{setup}\left(\mathrm{computederivatives}=\mathrm{false}\right)$. To query about the status of this new setting enter
> $\mathrm{setup}\left(\mathrm{computederivatives}\right)$
${\mathrm{computederivatives}}{=}{\mathrm{true}}$ (1.1)
and so differentiating returns the derivative computed
>
$\frac{{\partial }}{{\partial }{s}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{ℒ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}\right){=}{-}{ℒ}{}\left({f}{}\left({t}\right){}{t}{,}{t}{,}{s}\right)$ (1.2)
while changing this setting to work as in previous releases you have this computation reversed: you input the output (1.2) and you get the corresponding input
>
${\mathrm{computederivatives}}{=}{\mathrm{false}}$ (1.3)
>
$\frac{{\partial }}{{\partial }{s}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{ℒ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}\right){=}\frac{{\partial }}{{\partial }{s}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{ℒ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}\right)$ (1.4)
Reset the value of computederivatives
>
${\mathrm{computederivatives}}{=}{\mathrm{true}}$ (1.5)
>
$\frac{{\partial }}{{\partial }{s}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{ℒ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}\right){=}{-}{ℒ}{}\left({f}{}\left({t}\right){}{t}{,}{t}{,}{s}\right)$ (1.6)
In summary: by default, derivatives of all the 11 integral transforms are now computed. This setting can be changed any time you want within a Maple session without restarting, and changing it does not have any impact on the performance of intsolve, dsolve and pdsolve to solve differential equations using integral transforms.
Numerical evaluation
In previous releases, integral transforms had no numerical evaluation implemented. This is in the process of changing. So, for example, to numerically evaluate the inverse laplace transform (invlaplace command), three different algorithms have been implemented: Gaver-Stehfest, Talbot and Euler, following the presentation by Abate and Whitt, "Unified Framework for Numerically Inverting Laplace Transforms", INFORMS Journal on Computing 18(4), pp. 408–421, 2006.
For example, consider the exact solution to this partial differential equation subject to initial and boundary conditions
> $\mathrm{pde}≔\frac{\partial }{\partial x}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}u\left(x,t\right)=4\left(\frac{{\partial }^{2}}{\partial {t}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}u\left(x,t\right)\right):$
>
Note that these two conditions are not entirely compatible: the solution returned cannot be valid for $x=0$ and $t=0$ simultaneously. However, a solution discarding that point does exist and is given by
>
${\mathrm{sol}}{≔}{u}{}\left({x}{,}{t}\right){=}{-}{{ℒ}}^{{-}{1}}{}\left(\frac{{{ⅇ}}^{{-}\frac{\sqrt{{s}}{}{t}}{{2}}}}{{s}}{,}{s}{,}{x}\right){+}{1}$ (2.1)
Verifying the solution, one condition remains to be tested
> $\mathrm{pdetest}\left(\mathrm{sol},\left[\mathrm{pde},\mathrm{iv}\right]\right)$
$\left[{0}{,}{0}{,}{-}{{ℒ}}^{{-}{1}}{}\left(\frac{{{ⅇ}}^{{-}\frac{\sqrt{{s}}{}{t}}{{2}}}}{{s}}{,}{s}{,}{0}\right)\right]$ (2.2)
Since we now have numerical evaluation rules, we can test that what looks different from 0 in the above is actually 0.
>
${\mathrm{zero}}{≔}{-}{{ℒ}}^{{-}{1}}{}\left(\frac{{{ⅇ}}^{{-}\frac{\sqrt{{s}}{}{t}}{{2}}}}{{s}}{,}{s}{,}{0}\right)$ (2.3)
Add a small number to the initial value of t to skip the point $t=0:$
> $\mathrm{plot}\left(\mathrm{zero},t=0+{10}^{-10}..1\right)$
The default method used is the method of Euler sums and the numerical evaluation is performed as usual using the evalf command. For example, consider
> $F≔\mathrm{sin}\left(\sqrt{2t}\right)$
${F}{≔}{\mathrm{sin}}{}\left(\sqrt{{2}}{}\sqrt{{t}}\right)$ (2.4)
The Laplace transform of F is given by
>
${\mathrm{LT}}{≔}\frac{\sqrt{{2}}{}\sqrt{{\mathrm{\pi }}}{}{{ⅇ}}^{{-}\frac{{1}}{{2}{}{s}}}}{{2}{}{{s}}^{{3}}{{2}}}}$ (2.5)
and the inverse Laplace transform of LT in inert form is
>
${\mathrm{ILT}}{≔}{{ℒ}}^{{-}{1}}{}\left(\frac{\sqrt{{2}}{}\sqrt{{\mathrm{\pi }}}{}{{ⅇ}}^{{-}\frac{{1}}{{2}{}{s}}}}{{2}{}{{s}}^{{3}}{{2}}}}{,}{s}{,}{t}\right)$ (2.6)
Since by construction , at $t=1$ we have
>
${{ℒ}}^{{-}{1}}{}\left(\frac{\sqrt{{2}}{}\sqrt{{\mathrm{\pi }}}{}{{ⅇ}}^{{-}\frac{{1}}{{2}{}{s}}}}{{2}{}{{s}}^{{3}}{{2}}}}{,}{s}{,}{1}\right){=}{\mathrm{sin}}{}\left(\sqrt{{2}}\right)$ (2.7)
This inert form on the left-hand side is numerically evaluated next, directly, without symbolically resolving it to be equal to $\mathrm{sin}\left(\sqrt{2}\right)$. The result of both sides is consistent:
> $\mathrm{evalf}\left(\right)$
${0.9877659460}{=}{0.9877659459}$ (2.8)
In addition to the standard use of evalf to numerically evaluate inverse Laplace transforms, one can invoke each of the three different methods implemented using the MathematicalFunctions:-Evalf command
> $\mathrm{with}\left(\mathrm{MathematicalFunctions},\mathrm{Evalf}\right)$
$\left[{\mathrm{Evalf}}\right]$ (2.9)
>
${0.9877659460}$ (2.10)
>
${0.9877659460}$ (2.11)
>
${0.9877659460}$ (2.12)
Regarding the method we use by default: from a numerical experiment with varied problems, we have concluded that our implementation of the Euler (sums) method is faster and more accurate than the other two.
Two Hankel transform definitions
In previous Maple releases, the definition of the Hankel transform was given by
$\mathrm{hankel}\left(f\left(t\right),t,s,\mathrm{\nu }\right)={{\int }}_{0}^{\mathrm{\infty }}f\left(t\right)\sqrt{st}{J}_{\mathrm{\nu }}\left(st\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}t$
where ${J}_{\mathrm{\nu }}\left(st\right)$ is the function. This definition, sometimes called alternative definition of the Hankel transform, has the advantage of having a large table of transforms presented in the literature (ref.[1]) but has the inconvenience of the square root $\sqrt{st}$ in the integrand, complicating the form of the Hankel transform for the Laplacian in cylindrical coordinates. On the other hand, the other definition in the literature,
$\mathrm{hankel}\left(f\left(t\right),t,s,\mathrm{\nu }\right)={{\int }}_{0}^{\mathrm{\infty }}f\left(t\right)t{J}_{\mathrm{\nu }}\left(st\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}t$
has the advantage that, with it, the Hankel transform of $\frac{{\partial }^{2}}{\partial {r}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}u\left(r,t\right)+\frac{\frac{\partial }{\partial r}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}u\left(r,t\right)}{r}+\frac{{\partial }^{2}}{\partial {t}^{2}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}u\left(r,t\right)$ is given by the simple ODE form . Several other transforms also acquire a simpler form. So for Maple 2020 we added an algorithm that takes formulas for one definition and transforms them into formulas for the other definition, and have aligned Maple with this simpler definition (no $\sqrt{st}$ in the integrand), while keeping the previous definition as an alternative.
Hence, by default, when you load the inttrans package, the new definition in use for the Hankel transform is
>
${ℋ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{\nu }}\right){=}{{\int }}_{{0}}^{{\mathrm{\infty }}}{f}{}\left({t}\right){}{t}{}{{J}}_{{\mathrm{\nu }}}{}\left({s}{}{t}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}$ (3.1)
You can change this default so that Maple works with the alternative definition as in previous releases. For that purpose, use the new inttrans:-setup command (which you can also use to query about the definition in use at any moment):
> $\mathrm{setup}\left(\mathrm{alternativehankeldefinition}\right)$
${\mathrm{alternativehankeldefinition}}{=}{\mathrm{false}}$ (3.2)
The change in the default definition implemented in Maple 2020 is automatically taken into account by other parts of the Maple library using the Hankel transform. For example, the differentiation rule with the new definition is
>
$\frac{{\partial }}{{\partial }{z}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{ℋ}{}\left({f}{}\left({t}\right){,}{t}{,}{z}{,}{\mathrm{\nu }}\right){=}{-}{ℋ}{}\left({f}{}\left({t}\right){}{t}{,}{t}{,}{z}{,}{\mathrm{\nu }}{+}{1}\right){+}\frac{{\mathrm{\nu }}{}{ℋ}{}\left({f}{}\left({t}\right){,}{t}{,}{z}{,}{\mathrm{\nu }}\right)}{{z}}$ (3.3)
This differentiation rule resembles (is connected to) the differentiation rule for BesselJ, and this is another advantage of the new definition.
>
$\frac{{ⅆ}}{{ⅆ}{z}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{{J}}_{{\mathrm{\nu }}}{}\left({z}\right){=}{-}{{J}}_{{\mathrm{\nu }}{+}{1}}{}\left({z}\right){+}\frac{{\mathrm{\nu }}{}{{J}}_{{\mathrm{\nu }}}{}\left({z}\right)}{{z}}$ (3.4)
Furthermore, several transforms have acquired a simpler form, as for example:
>
(3.5)
Let's compare: make the definition be as in previous releases.
>
${\mathrm{alternativehankeldefinition}}{=}{\mathrm{true}}$ (3.6)
>
${ℋ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{\nu }}\right){=}{{\int }}_{{0}}^{{\mathrm{\infty }}}{f}{}\left({t}\right){}\sqrt{{s}{}{t}}{}{{J}}_{{\mathrm{\nu }}}{}\left({s}{}{t}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}$ (3.7)
The differentiation rule with the previous (alternative) definition was not as simple:
>
$\frac{{\partial }}{{\partial }{s}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{ℋ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{\nu }}\right){=}{-}{ℋ}{}\left({f}{}\left({t}\right){}{t}{,}{t}{,}{s}{,}{\mathrm{\nu }}{+}{1}\right){+}\frac{{\mathrm{\nu }}{}{ℋ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{\nu }}\right)}{{s}}{+}\frac{{ℋ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{\nu }}\right)}{{2}{}{s}}$ (3.8)
And the transform (3.5) was also not so simple:
>
${ℋ}{}\left(\frac{{{ⅇ}}^{{I}{}{a}{}{r}}}{{r}}{,}{r}{,}{k}{,}{0}\right){=}\frac{{I}{}{{\mathrm{\Gamma }}{}\left(\frac{{3}}{{4}}\right)}^{{4}}{}{a}{}{}_{{2}}{F}_{{1}}{}\left(\frac{{3}}{{4}}{,}\frac{{3}}{{4}}{;}\frac{{3}}{{2}}{;}\frac{{{a}}^{{2}}}{{{k}}^{{2}}}\right){+}{{\mathrm{\pi }}}^{{2}}{}{}_{{2}}{F}_{{1}}{}\left(\frac{{1}}{{4}}{,}\frac{{1}}{{4}}{;}\frac{{1}}{{2}}{;}\frac{{{a}}^{{2}}}{{{k}}^{{2}}}\right){}{k}}{{k}{}{\mathrm{\pi }}{}{{\mathrm{\Gamma }}{}\left(\frac{{3}}{{4}}\right)}^{{2}}}$ (3.9)
Reset to the new default value of the definition.
>
${\mathrm{alternativehankeldefinition}}{=}{\mathrm{false}}$ (3.10)
>
${ℋ}{}\left({f}{}\left({t}\right){,}{t}{,}{s}{,}{\mathrm{\nu }}\right){=}{{\int }}_{{0}}^{{\mathrm{\infty }}}{f}{}\left({t}\right){}{t}{}{{J}}_{{\mathrm{\nu }}}{}\left({s}{}{t}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}$ (3.11)
References [1] Oberhettinger Fritz, Tables of Bessel Transforms. Springer Verlag, 1972.
More integral transform results
Maple is now able to compute more transforms. For example, for the Hankel transform, consider the operators
> $\mathrm{D/t}≔u→\frac{\frac{\partial }{\partial t}u}{t}:$$\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{formula_plus}≔{t}^{-\mathrm{\nu }}{\mathrm{D/t}}^{\left(m\right)}\left({t}^{m+\mathrm{\nu }}u\left(t\right)\right):$$\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathrm{formula_minus}≔{t}^{\mathrm{\nu }}{\mathrm{D/t}}^{\left(m\right)}\left({t}^{m-\mathrm{\nu }}u\left(t\right)\right):$
Being able to transform these operators into algebraic expressions or differential equations of lower order is key for solving ODE and PDE problems with Boundary Conditions. To illustrate that, set $\mathrm{computederivatives}=\mathrm{false}$, a more convenient setting to use transforms to simplify differential equations, and insert some values in formula_minus
>
${\mathrm{computederivatives}}{=}{\mathrm{false}}$ (4.1)
>
$\frac{\left(\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{t}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{{t}}^{{3}}{-}{12}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{{t}}^{{2}}{+}{57}{}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{t}{-}{105}{}{u}{}\left({t}\right)}{{{t}}^{{3}}}$ (4.2)
Its Hankel transform is
> $\mathrm{hankel}\left(,t,s,6\right)$
${-}{{s}}^{{3}}{}{ℋ}{}\left({u}{}\left({t}\right){,}{t}{,}{s}{,}{3}\right)$ (4.3)
An example with formula_plus:
>
$\frac{\left(\frac{{{ⅆ}}^{{4}}}{{ⅆ}{{t}}^{{4}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{{t}}^{{4}}{+}{38}{}\left(\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{t}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{{t}}^{{3}}{+}{477}{}\left(\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{{t}}^{{2}}{+}{2295}{}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{u}{}\left({t}\right)\right){}{t}{+}{3465}{}{u}{}\left({t}\right)}{{{t}}^{{4}}}$ (4.4)
> $\mathrm{hankel}\left(,t,s,7\right)$
${{s}}^{{4}}{}{ℋ}{}\left({u}{}\left({t}\right){,}{t}{,}{s}{,}{11}\right)$ (4.5)
Not just differential operators but also several new transforms are now computable
> $\mathrm{hankel}\left(1,r,k,\mathrm{nu}\right)$
$\left\{\begin{array}{cc}\frac{{\mathrm{\delta }}{}\left({k}\right)}{{k}}& {\mathrm{\nu }}{=}{0}\\ \frac{{\mathrm{\nu }}}{{{k}}^{{2}}}& {\mathrm{otherwise}}\end{array}\right\$ (4.6)
> $\mathrm{hankel}\left({r}^{m},r,k,\mathrm{nu}\right)$
$\left\{\begin{array}{cc}\frac{{\mathrm{\delta }}{}\left({k}\right)}{{k}}& {\mathrm{\nu }}{=}{0}{\wedge }{m}{=}{0}\\ \frac{{{2}}^{{m}{+}{1}}{}{{k}}^{{-}{m}{-}{2}}{}{\mathrm{\Gamma }}{}\left({1}{+}\frac{{m}}{{2}}{+}\frac{{\mathrm{\nu }}}{{2}}\right)}{{\mathrm{\Gamma }}{}\left(\frac{{\mathrm{\nu }}}{{2}}{-}\frac{{m}}{{2}}\right)}& {\mathrm{otherwise}}\end{array}\right\$ (4.7)
> $\mathrm{mellin}\left(\mathrm{log}\left(\left|\frac{4+x}{x-3}\right|\right),x,s\right)$
$\frac{{\mathrm{\pi }}{}{\mathrm{csc}}{}\left({\mathrm{\pi }}{}{s}\right){}\left({16}{-}{9}{}{\mathrm{cos}}{}\left({\mathrm{\pi }}{}{s}\right)\right)}{{s}}$ (4.8)
>
$\frac{\sqrt{{2}}{}\left({\mathrm{ln}}{}\left(\frac{{a}}{{b}}\right){+}{\mathrm{cos}}{}\left({b}{}{y}\right){}{\mathrm{Ci}}{}\left({b}{}{y}\right){-}{\mathrm{cos}}{}\left({a}{}{y}\right){}{\mathrm{Ci}}{}\left({a}{}{y}\right){+}{\mathrm{sin}}{}\left({b}{}{y}\right){}{\mathrm{Si}}{}\left({b}{}{y}\right){-}{\mathrm{sin}}{}\left({a}{}{y}\right){}{\mathrm{Si}}{}\left({a}{}{y}\right){+}\frac{{\mathrm{\pi }}{}\left({\mathrm{sin}}{}\left({b}{}{y}\right){+}{\mathrm{sin}}{}\left({a}{}{y}\right)\right)}{{2}}\right)}{\sqrt{{\mathrm{\pi }}}{}{y}}$ (4.9)
>
$\frac{\sqrt{{2}}{}\sqrt{{\mathrm{\pi }}}{}\left({1}{-}{\mathrm{cos}}{}\left({a}{}{y}\right){-}{a}{}{y}{}{\mathrm{Ssi}}{}\left({a}{}{y}\right)\right)}{{a}}$ (4.10)
>
$\sqrt{{2}}{}\sqrt{{\mathrm{\pi }}}{}\left({\mathrm{Ci}}{}\left({b}{}{y}\right){-}{\mathrm{Ei}}{}\left({-}{a}{}{y}\right){-}{\mathrm{ln}}{}\left(\frac{{b}}{{a}}\right)\right)$ (4.11)
>
${-}\sqrt{{2}}{}\sqrt{{\mathrm{\pi }}}{}{\mathrm{Ssi}}{}\left({a}{}{y}\right)$ (4.12)
> | 6,467 | 18,049 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 82, "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-2024-38 | latest | en | 0.680892 |
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## Correlations > Terrorism Statistics > Victims of the September 11th 2001 attacks
VIEW DATA: Totals
Definition Source Printable version
Bar Graph Pie Chart Map Correlations
Showing latest available data.
### Correlations between Terrorism > Victims of the September 11th 2001 attacks ...
A correlation is a statistical measure of similarity between at least two given sets of data. NationMaster's correlations compare two variables from our database and reveal statistical relationships between them. The percentages you see represent the strength (or likelihood) that a change in the topic variable is matched by a change in the listed variables below it. But remember: These correlations do not imply causation, that is, one does not necessarily cause the other. Also, not all variables contain all countries, rather subsets of countries matched together.
#### NOTES:
• Outliers have been removed only where they are outside 3 standard deviations of the mean.
• Only variable pairs where at least 15 countries match for each have been considered.
• Strength is given by the correlation coefficient (R squared). It is the fraction of variation in Y that can be attributed to the variation in X. 100% signifies a perfect fit (R squared of 1). The top 50 such stats are displayed
DEFINITION: Victims of the attacks on September 11 2001 by known foreign citizenship and known country. NOTE: this list is not considered a total for all victims of 9-11, as in many cases the reported country of residence differed from citizenship, though both reports are included in this study. The likelihood is that there are approximatley 127 fewer victims from the US on this graph, as they were not US citizens at the time. The totals as of 9-11 2005 are: Confirmed Dead - 2948; Reported Dead - 24; Reported Missing - 24. Total: 2996.
SOURCE: September 11th, 2001 Victims, website memorial, last update - Sept. 21, 2002. | 483 | 2,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2013-20 | latest | en | 0.926244 |
https://sonichours.com/how-long-are-the-breaks-at-school/ | 1,713,867,692,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818474.95/warc/CC-MAIN-20240423095619-20240423125619-00514.warc.gz | 485,321,060 | 20,454 | General
How long are the breaks at school?
How long are the breaks at school?
They range from 15 minutes to an hour. The first big break is usually between 9:30 a.m. and 10:15 a.m. The next one follows after two more school periods at around 11:30 a.m. The lunch break is after the end of the morning lesson, i.e. usually after 1:00 p.m.
How many breaks if you work 5 hours?
The minimum duration of the breaks is 15 minutes if the daily working time exceeds five and a half hours, 30 minutes if the daily working time exceeds seven hours and one hour if the daily working time exceeds nine hours.
How long break am I entitled to?
In ยง 4 ArbZG there are specifications after how many hours of work a break is to be taken (rest breaks). If you work six to nine hours, you must take a break of at least 30 minutes. If the working time exceeds nine hours, a break of 45 minutes is required.
Can my boss tell me to take a break indoors?
The employer can stipulate the times at which employees are allowed to take breaks. However, the works council may have a say.
Can the employer determine the length of the lunch break?
The lunch break is to be determined by the employer in advance. This states that work must be interrupted by a break of at least 30 minutes that is fixed in advance. This should take place after six hours of work at the latest and can be divided into two 15-minute sections.
How is the break calculated?
Example: Working time from 10:00 a.m. to 5:30 p.m. = 7.5 hours of pure time expenditure. Lunch break 1 hour specified in the employment contract => 7.5 hours – 1 hour = 6.5 hours of pure work. The break is to be calculated from these 6.5 hours.
Visit the rest of the site for more useful and informative articles! | 420 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-18 | latest | en | 0.951174 |
https://discourse.julialang.org/t/dict-index-with-multiple-keys-versus-tuple/86352 | 1,670,007,375,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00255.warc.gz | 244,681,821 | 5,102 | # Dict index with multiple keys versus tuple
As I struggle to scale the slopes of julia sagacity I encountered the following and the fact that it didn’t throw an error, well, threw me.
``````julia> Acc_rd
Dict{Tuple{Int64, Int64}, Vector{Int64}} with 12 entries:
(101521, 12124) => [23434]
(101521, 9412) => []
(100105, 7207) => [20661, 20662, 20911, 20912, 21111, 21112, 21113, 21114, 21181, 21182 … 26…
(102653, 12124) => [20911, 20912, 22311, 22314, 23431, 24971, 25901, 25902, 26351, 26352]
:
``````
and then
``````julia> Acc_rd[101521, 12124]
1-element Vector{Int64}:
23434
``````
I’ve no doubt that there is some magic happening behind the scenes that “fills in the blanks” but I can’t help having the feeling that at some point features become counter-productive.
As I said at the outset, spare a thought…
`Acc_rd[101521, 12124]` is equivalent to `Acc_rd[(101521, 12124)]`. This was very arguably a bad idea, but it made it into Julia 1.0, so it’s here to stay at least until Julia 2.0.
6 Likes
BTW, if you want to see the magic for yourself, just do:
``````@edit Acc_rd[101521, 12124]
``````
which should point you to:
``````# t[k1,k2,ks...] is syntactic sugar for t[(k1,k2,ks...)]. (Note
# that we need to avoid dispatch loops if setindex!(t,v,k) is not defined.)
getindex(t::AbstractDict, k1, k2, ks...) = getindex(t, tuple(k1,k2,ks...))
``````
1 Like | 477 | 1,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.841911 |
https://www.physicsforums.com/threads/unit-problem-in-differential-equation.515512/ | 1,511,562,421,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808972.93/warc/CC-MAIN-20171124214510-20171124234510-00193.warc.gz | 842,391,781 | 14,547 | Unit problem in differential equation
1. Jul 20, 2011
sunipa.som
I have one differential equation like
dN/dt=c*(other terms with no unit)
unit of c is 1/ns. Now if I solve this equation, I will get value of N corresponding to t.
(1) Then what will be the unit of t?
(2) and if I calculate dN/dt1=(other terms with no unit). where t1=t/c.
Then what will be the unit of t1?
2. Jul 20, 2011
hunt_mat
As I see it you have three possible units [N], [t] and [c], and I nonlimensionalise according to $N=N_{d}\bar{N}$, $t=t_{d}\bar{t}$ and $c=c_{d}\bar{c}$. There I put all the dimensions into the quantities with subscript d. So this turns the differential equation into:
$$\frac{d\bar{N}}{d\bar{t}}=\frac{c_{d}t_{d}}{N_{d}}\bar{c}$$
Where $c_{d}t_{d}/N_{d}$ is a nondimensional quantity. Does this clear things up?
3. Jul 20, 2011
pmsrw3
t will be in ns. Your RHS has dimensions of time-1. dN/dt also has dimensions of time-1. So everything matches.
Are you sure you want to divide t by c? Maybe I'm confused, but I think it would make more sense to multiply t by c, in order to get a dimensionless version of the equation. For instance, if you define $\tau=ct$, then you get $\frac{dN}{d\tau}=\mbox{(other terms)}$ and c has gone away.
4. Jul 20, 2011
sunipa.som
--------------------------
Thank you. Sorry for mistake. I have to multiply t by c. Then I will get value of N for different times but then time has no unit. | 449 | 1,429 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-47 | longest | en | 0.897214 |
http://forum.qbasicnews.com/index.php?topic=11455.0 | 1,582,848,745,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00135.warc.gz | 47,809,282 | 11,105 | Qbasicnews.com
February 27, 2020, 09:12:24 PM
Pages: [1] 2
Author Topic: Random Numbers (Read 6401 times)
Champions_2002
Member
Posts: 29
« on: February 15, 2006, 07:38:32 AM »
can anybody tell me how to do the following please
I have got 24 numbers that i want to pick out a random, but once a number has been picked out then it can not be pick out again, also the numbers will be printed out 12 on each side of the screen
i.e.
1-----2
24---10
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yetifoot
Ancient Guru
Posts: 575
« Reply #1 on: February 15, 2006, 07:49:28 AM »
the important thing for you to note, is the array TeamChosen, this stores if a number has already been selected.
Code:
Function GetRandInt(lb, ub) As Integer
return Int((ub - lb + 1) * rnd + lb)
End Function
Randomize Timer
Dim NumTeams As Integer
Dim TeamNames(64) As String
Dim TeamChosen(64) As Integer
Print "Please enter the number of teams."
Print ">";
Input NumTeams
For i = 1 To 6
If 2^i = NumTeams Then
Depth = i
Exit For
End If
Next i
If Depth = 0 Then
Print "Number of teams should one of 2, 4, 8, 16, 32, 64"
Sleep
End
End If
For i = 1 To NumTeams
Print
Print "Please enter team " & i & " name"
Print ">";
Input TeamNames(i - 1)
If TeamNames(i - 1) = "" Then TeamNames(i - 1) = chr(64 + i)
Next i
Cls
Print "-------"
Print "Stages: " & Depth
Print
Print " ---------"
Print " Stage 1"
For i = 1 To NumTeams \ 2
Do
TeamA = GetRandInt(1, NumTeams)
If TeamChosen(TeamA) = 0 Then
TeamChosen(TeamA) = 1
Exit Do
End If
Loop
Do
TeamB = GetRandInt(1, NumTeams)
If TeamChosen(TeamB) = 0 Then
TeamChosen(TeamB) = 1
Exit Do
End If
Loop
Print " ----"
Print " Game " & i & ":"
Print " " & TeamNames(TeamA - 1) & " vs " & TeamNames(TeamB - 1)
Print
Next i
For i = 2 To Depth
Print " -----"
Print " Stage " & i
inc = 1
For ii = 1 To NumTeams \ (i * 2)
If i = Depth Then
Print " Final"
ElseIf i = Depth - 1 Then
Print " Semi " & ii & ":"
Else
Print " Game " & ii & ":"
End If
Print " " & "Winner of Stage " & i - 1 & ", Game " & inc & " vs " & "Winner of Stage " & i - 1 & ", Game " & inc + 1
Print
inc = inc + 2
Next ii
Next i
Sleep
End
Heres just a simple prog that does what you want. The do..loop forces it to keep trying till it gets an unused number
Code:
Function GetRandInt(lb, ub) As Integer
return Int((ub - lb + 1) * rnd + lb)
End Function
Randomize Timer
Dim TeamChosen(64) As Integer
For i = 1 To 24
Do
TeamA = GetRandInt(1, 24)
If TeamChosen(TeamA) = 0 Then
TeamChosen(TeamA) = 1
Exit Do
End If
Loop
Print TeamA
Next i
Sleep
End
Logged
EVEN MEN OF STEEL RUST.
Moneo
Na_th_an
Posts: 1971
« Reply #2 on: February 17, 2006, 01:04:01 AM »
Quote from: "Champions_2002"
can anybody tell me how to do the following please
I have got 24 numbers that i want to pick out a random, but once a number has been picked out then it can not be pick out again, also the numbers will be printed out 12 on each side of the screen
i.e.
1-----2
24---10
Here's some code fucused on your exact problem. (I tested it too.)
Code:
defint a-z
RANDOMIZE TIMER
const lower = 1 'Want random numbers starting with 1
const upper = 24 '... and ending with 24.
dim rands (lower to upper) 'Array to hold unique random numbers.
dim dup (lower to upper) 'Array to track duplicates.
uniquecnt = 0 'Count of unique random numbers so far.
do while uniquecnt < upper 'loop until have 24 unique random nums.
r = INT(RND * (Upper - lower + 1)) + lower 'get a random number
if dup(r) = 0 then 'If not a duplicate.
dup(r) = 1 'Flag number as used.
uniquecnt=uniquecnt+1 'Count as unique
rands(uniquecnt) = r 'Save this unique random number
end if
loop
rem ... Print out rands array from 1 to 24 as required.
*****
Logged
Agamemnus
x/ \z
Posts: 3491
« Reply #3 on: February 20, 2006, 10:13:05 PM »
On a related note, the best random numbers are stock quotes..
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Peace cannot be obtained without war. Why? If there is already peace, it is unnecessary for war. If there is no peace, there is already war."
Visit www.neobasic.net to see rubbish in all its finest.
Moneo
Na_th_an
Posts: 1971
« Reply #4 on: February 21, 2006, 12:19:58 AM »
Quote from: "Agamemnus"
On a related note, the best random numbers are stock quotes..
Interesting thought.
*****
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Ralph
Ancient Guru
Posts: 544
« Reply #5 on: October 14, 2006, 11:53:02 PM »
Regarding random selection from a given list.
I have been bothered by the "repeat until a random number has not been selected yet" beat-to-fit forcing methods I see. Finally, I have come up with a more direct approach. I based it on:
1. Create a string list in any order.
2. Create a string array of random numbers of any practical size. It doesn't matter if any are repeated.
3. Add the items from the string list to the end of the random number strings in the string array.
4. Sort the string array.
5. Separate the last part of the string array, to produce the desired, randomized, original list.
Here is a "first approach" program to do this, for the string list with the numbers from 1 to 99:
Code:
'Rand-RAE.bas, by RAEsquivel, 10-14-2006, randamizes a series of numbers or
'characters
'At present, it is set up to randomize the numbers 1 to 99
CLS
RANDOMIZE TIMER
n = 99
m = LEN(STR\$(n))
'a will hold a 3-digit random number, a hyphen, and a 2-digit number,
'with a leading and trailing space, a total of 8 characters
DIM a(n) AS STRING * 8
'b will hold a randomized 2-digit number
DIM b(n) AS STRING * 3
'assign a random number to the series 1,2,3,...n
FOR i = 1 TO n
a(i) = STR\$(INT(RND(1) * 1000))
a(i) = MID\$(a(i), 2, 3) + "-" + STR\$(i)
'''PRINT a(i)
NEXT i
'sort the strings according to the ramdom numbers generated
again:
FOR i = 1 TO n - 1
k = 0
IF a(i) > a(i + 1) THEN
k = 1
SWAP a(i), a(i + 1)
END IF
IF k = 1 THEN
GOTO again:
END IF
NEXT i
'print each randomized number, preceded with its number of occurance
FOR i = 1 TO n
b(i) = MID\$(a(i), 5, 3)
PRINT USING "##"; i;
PRINT "="; b(i), '''; ", ";
NEXT i
Logged
Ralph, using QuickBASIC 4.5 and Windows XP Home Edition and Service Pack 2, with HP LaserJet 4L printer.
Dr_Davenstein
Na_th_an
Posts: 2052
« Reply #6 on: October 15, 2006, 02:01:05 AM »
Here's a slightly different approach. Ralph sparked my interest when he said the forced random method bothered him.
This code was written for FB, but I'm hoping that it will run under QB too. I used the -lang QB switch.
Code:
Randomize Timer
Defint A-Z
Const False = 0, True = Not False
Dim Text(1 To 20) As String
For i = Lbound(Text) To Ubound(Text)
If i<10 Then
Text(i) = "Slot #0" + Ltrim\$(Str\$(i))
Else
Text(i) = "Slot #" + Ltrim\$(Str\$(i))
End If
Next
Dim Slot(Lbound(Text) To Ubound(Text) ) As Integer
For i = Lbound(Slot) To Ubound(Slot)
Slot(i) = True
Next
For Cnt = Lbound(Text) To Ubound(Text)
i = Lbound(Text) + Int(Rnd * (Ubound(Text) - Lbound(Text)))
Do
GotSlot = False
If Slot(i) Then
Slot(i) = False
Print Text(i)
GotSlot = True
Else
i = i + 1
If i>Ubound(Text) Then i = Lbound(Text)
End If
Loop Until GotSlot
Next
Sleep
Logged
Ralph
Ancient Guru
Posts: 544
« Reply #7 on: October 15, 2006, 11:40:30 AM »
Dr_:
Hats off to your lightning-fast program execution. Mine takes a while to run, probably due to my crude, bubble-sort algorithm.
I was able to understand all your code, except the DO-LOOP. There, I got confused :oops:
Logged
Ralph, using QuickBASIC 4.5 and Windows XP Home Edition and Service Pack 2, with HP LaserJet 4L printer.
Dr_Davenstein
Na_th_an
Posts: 2052
« Reply #8 on: October 15, 2006, 03:07:52 PM »
Well, first it picks an integer within bounds at random. Then, it falls into that do...loop.
If Slot(i) = True, then this number hasn't been chosen yet, so it sets Slot(i) to False, sets the GotSlot flag to True and prints out the contents od Text(i).
Otherwise, it increments i, instead of choosing another random number. Just add an out-of-bounds check to i, and it will loop through until it actually finds the first instance of Slot() that isn't False.
Logged
Ralph
Ancient Guru
Posts: 544
« Reply #9 on: October 16, 2006, 01:17:23 AM »
Very ingenius, Dr_D, but a little too convulated for me. I can sort of follow it, but my simplistic mind could never have come up with that DO-LOOP.
Logged
Ralph, using QuickBASIC 4.5 and Windows XP Home Edition and Service Pack 2, with HP LaserJet 4L printer.
Moneo
Na_th_an
Posts: 1971
« Reply #10 on: October 16, 2006, 07:24:38 PM »
Ralph and Dr_Davenstein,
Have you guys solved Chaampion_2002's original problem, or are you both off on a tangent?
*****
Logged
Dr_Davenstein
Na_th_an
Posts: 2052
« Reply #11 on: October 16, 2006, 08:07:38 PM »
It seems so... It looks like there are several solutions now. Great work everyone!
Moneo, were you unable to test my code in QB? ::
Logged
Ralph
Ancient Guru
Posts: 544
« Reply #12 on: October 16, 2006, 11:16:30 PM »
Dr_D wrote:
Quote
Moneo, were you unable to test my code in QB?
Well, I did. I copied it to a .txt file, than opened it in QuickBASIC 4.5, and it ran without a hitch!
Logged
Ralph, using QuickBASIC 4.5 and Windows XP Home Edition and Service Pack 2, with HP LaserJet 4L printer.
Moneo
Na_th_an
Posts: 1971
« Reply #13 on: October 16, 2006, 11:34:12 PM »
Quote from: "Dr_Davenstein"
It seems so... It looks like there are several solutions now. Great work everyone!
Moneo, were you unable to test my code in QB? ::
Actually, when I felt that it wasn't directly addressing the original problem, I didn't test it. Sorry.
*****
Logged
Dr_Davenstein
Na_th_an
Posts: 2052
« Reply #14 on: October 17, 2006, 12:45:40 AM »
It did solve the original problem though. It may not have printed 1-12 on one side, and 13-24 on the other side, but it selected random numbers without choosing the same one twice. I understood that as the main problem. Besides, he already got his answer. We were just exploring different methods of doing the same thing.
Logged
Pages: [1] 2 | 3,212 | 10,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.680114 |
https://www.lds.org/friend/1982/07/fun-on-the-road?lang=eng | 1,435,799,459,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375095346.56/warc/CC-MAIN-20150627031815-00287-ip-10-179-60-89.ec2.internal.warc.gz | 895,412,726 | 13,430 | Going on a vacation is great, but sometimes the long car ride can be tiring for everyone. Here are some games to make the trip more enjoyable. (Some of them require paper and pencil for each player.)
One—Counting Cars Have each person choose a different color, and see who counts 100 cars of their color first. If you want to make it even harder, choose something specific like blue pick-ups or green compact cars.
Two—Going on a Trip The first player starts by saying, “I’m going on a trip and I’m taking artichokes” (or anything that begins with the letter A). The second player says, “I’m going on a trip and I’m taking artichokes and basketballs” (or anything beginning with the letter B). The game continues with each player repeating all the things being taken on the trip and adding one more item that begins with the next letter of the alphabet. With a bit of practice, you’ll find you can repeat all twenty-six items.
Three—Do a Play You can make one up or you can choose a familiar story such as Little Red Riding Hood, and let each actor improvise as he plays a certain character.
Four—License Plates Have one person write down different states you see license plates for.
Five—My Son John Owns a Store One person says, “My son, John, owns a store, and he sells something that begins with the letter P.” Then the others try to guess what it is.
Six—To Be Continued One person starts telling a story, the next person continues, and the next, until everyone has taken part. The last person should finish the story.
Seven—Categories Try to think of something in a chosen category for each letter of your family’s last name. Suppose your name is Smith and you have flowers as the category. Then perhaps you’d say sunflower for S, marigold for M, iris for I, tulip for T, and hyacinth for H.
Eight—Poetry Recite familiar poems or rhymes, or make up some of your own. Writing a haiku can be interesting, too, and fairly easy. It’s a 17-syllable nature poem, starting with five syllables in the first line, seven in the second line, and five in the last line. Here’s one:
Raindrops are splashing,
Dancing on my dark rooftop
Happy rhythm sound.
Nine—Fold-a-Picture The first person begins to draw a face on a piece of paper. He draws only the top of a head, then folds his drawing down so it is hidden, and passes it to the next player, who draws eyebrows and eyes. Then that person folds his drawing down and passes it to the next player. When the last player has finished, unfold the paper and see the crazy picture you’ve made.
Ten—Squiggle Lines Have one person draw any kind of squiggly line on a paper. The next person must use that line as part of a picture he draws.
Eleven—Vacation Bingo Before you leave on a trip, write a list of things on paper for each player to check off as he sees them. For example, airplane, railroad track, birdhouse, fire truck, brown horse, and police car. The person who checks off all the things on his list first is the winner.
Twelve—Keep a Journal Each day, write down where you went, the people you saw, and anything unusual. A journal can not only be fun to write, but it will be a lasting keepsake of your trip.
[illustration] Illustrated by Shauna Mooney | 744 | 3,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2015-27 | longest | en | 0.940373 |
http://web.cecs.pdx.edu/~fliu/courses/cs447/tutorial6.html | 1,553,190,126,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202530.49/warc/CC-MAIN-20190321172751-20190321194751-00515.warc.gz | 233,522,134 | 2,802 | Lighting
This tutorial will build on the previous tutorial, adding OpenGL colored lighting.
Lets get started.
Step 1: Enable Lighting
We're going to be using OpenGL lighting so the first thing to do is tell OpenGL we need it to do the lighting calculations for the geometry we send it. We do this by enabling lighting and enabling at least one OpenGL light source. Our revised InitializeGL() method looks like this.
```void MyWindow::InitializeGL() { glClearColor(.1f, .1f, .1f, 1); glEnable(GL_DEPTH_TEST); glEnable(GL_LIGHTING); glEnable(GL_LIGHT0); float lightColor[4] = {1, 1, 1, 1}; glLightfv(GL_LIGHT0, GL_DIFFUSE, lightColor); }```
We enable OpenGL lighting and enable the first light (GL_LIGHT0). We then set the lights diffuse color to white. We won't use any ambient or specular light for this tutorial.
Step 2: Position the Light Source
We'll need to position our light source so it illuminates our cube. We add this code to our draw() method.
``` . . . gluLookAt(0, 0, 3, 0, 0, 0, 0, 1, 0); float lightPosition[4] = {5, 5, 5, 1}; glLightfv(GL_LIGHT0, GL_POSITION, lightPosition); glRotatef(rotation, 0, 1, 0); . . . ```
We've placed our light at (5, 5, 5) so it should shine on the upper right corner of our cube. Notice the one trailing the light position. This is the homogeneous coordinate and tells OpenGL we want a point light. If it was a zero the light would be a directional light and the (5, 5, 5) would specify the direction instead of a position.
Step 3: Setup Material Properties
When using OpenGL lighting we no longer set colors for our geometry. We need to set material properties. However, for this simple demo we can use some OpenGL functionality to use the colors we've already specified as the material properties we need. This will convert our glColor3f() calls to material property settings instead. We'll add this to our InitializeGL() method. We need to enable this functionality called color material and set the diffuse material property to track the colors we've set with glColor3f().
``` . . . float lightColor[4] = {1, 1, 1, 1}; glLightfv(GL_LIGHT0, GL_DIFFUSE, lightColor); glEnable(GL_COLOR_MATERIAL); glColorMaterial(GL_FRONT_AND_BACK, GL_DIFFUSE); } ```
The only thing left to do is add normal vectors to each of our quads that form the cube. We'll modify DrawCube() to include normals.
```void MyWindow::DrawCube() { glBegin(GL_QUADS); // front glNormal3f(0, 0, 1); glColor3f(1, 0, 0); glVertex3f(-1, 1, 1); glVertex3f(-1, -1, 1); glVertex3f(1, -1, 1); glVertex3f(1, 1, 1); // back glNormal3f(0, 0, -1); glColor3f(0, 1, 0); glVertex3f(-1, 1, -1); glVertex3f(1, 1, -1); glVertex3f(1, -1, -1); glVertex3f(-1, -1, -1); // top glNormal3f(0, 1, 0); glColor3f(0, 0, 1); glVertex3f(-1, 1, -1); glVertex3f(-1, 1, 1); glVertex3f(1, 1, 1); glVertex3f(1, 1, -1); // bottom glNormal3f(0, -1, 0); glColor3f(1, 1, 0); glVertex3f(-1, -1, -1); glVertex3f(1, -1, -1); glVertex3f(1, -1, 1); glVertex3f(-1, -1, 1); // left glNormal3f(-1, 0, 0); glColor3f(0, 1, 1); glVertex3f(-1, 1, -1); glVertex3f(-1, -1, -1); glVertex3f(-1, -1, 1); glVertex3f(-1, 1, 1); // right glNormal3f(1, 0, 0); glColor3f(1, 0, 1); glVertex3f(1, 1, 1); glVertex3f(1, -1, 1); glVertex3f(1, -1, -1); glVertex3f(1, 1, -1); glEnd(); }```
Step 5: Build and Run the Program
Ok we're ready to build and run our program. Choose Build -> Build Solution to compile and link the program and Debug -> Start Without Debugging to run it.
Source code for this tutorial.
Go to the previous tutorial.
Go to the next tutorial. | 1,249 | 3,858 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-13 | latest | en | 0.676517 |
https://www.geeksforgeeks.org/merge-two-sorted-arrays/?ref=leftbar-rightbar | 1,696,402,889,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511361.38/warc/CC-MAIN-20231004052258-20231004082258-00547.warc.gz | 853,415,751 | 50,666 | Open In App
# Merge two sorted arrays
Given two sorted arrays, the task is to merge them in a sorted manner.
Examples:
Input: arr1[] = { 1, 3, 4, 5}, arr2[] = {2, 4, 6, 8}
Output: arr3[] = {1, 2, 3, 4, 4, 5, 6, 8}
Input: arr1[] = { 5, 8, 9}, arr2[] = {4, 7, 8}
Output: arr3[] = {4, 5, 7, 8, 8, 9}
Naive Approach:
It is the brute force method to do the same. Take all the elements of arr1 and arr2 in arr3. Then simply sort the arr3.
The implementation of above approach is:
## C++
`// C++ program to merge two sorted arrays/``#include``using` `namespace` `std;` `void` `mergeArrays(``int` `arr1[], ``int` `arr2[], ``int` `n1,`` ``int` `n2, ``int` `arr3[])``{`` ``int` `i = 0, j = 0, k = 0;`` ``// traverse the arr1 and insert its element in arr3`` ``while``(i < n1){`` ``arr3[k++] = arr1[i++];`` ``}`` ` ` ``// now traverse arr2 and insert in arr3`` ``while``(j < n2){`` ``arr3[k++] = arr2[j++];`` ``}`` ` ` ``// sort the whole array arr3`` ``sort(arr3, arr3+n1+n2);``}` `// Driver code``int` `main()``{`` ``int` `arr1[] = {1, 3, 5, 7};`` ``int` `n1 = ``sizeof``(arr1) / ``sizeof``(arr1[0]);` ` ``int` `arr2[] = {2, 4, 6, 8};`` ``int` `n2 = ``sizeof``(arr2) / ``sizeof``(arr2[0]);` ` ``int` `arr3[n1+n2];`` ``mergeArrays(arr1, arr2, n1, n2, arr3);` ` ``cout << ``"Array after merging"` `<
## Java
`// Java program to merge two sorted arrays/``import` `java.util.*;` `public` `class` `GFG {`` ``// Driver code`` ``public` `static` `void` `main(String[] args) {`` ``int` `arr1[] = {``1``, ``3``, ``5``, ``7``};`` ``int` `n1 = arr1.length;` ` ``int` `arr2[] = {``2``, ``4``, ``6``, ``8``};`` ``int` `n2 = arr2.length;` ` ``int` `arr3[] = ``new` `int``[n1 + n2];`` ``mergeArrays(arr1, arr2, n1, n2, arr3);` ` ``System.out.println(``"Array after merging"``);`` ``for` `(``int` `i=``0``; i < n1+n2; i++)`` ``System.out.print(arr3[i] + ``" "``);`` ` ` ``}`` ` ` ``public` `static` `void` `mergeArrays(``int``[] arr1, ``int``[] arr2, ``int` `n1, ``int` `n2, ``int``[] arr3){`` ``int` `i = ``0``; `` ``int` `j = ``0``; `` ``int` `k = ``0``; `` ` ` ``// traverse the arr1 and insert its element in arr3`` ``while``(i < n1){ `` ``arr3[k++] = arr1[i++]; `` ``} `` ` ` ``// now traverse arr2 and insert in arr3`` ``while``(j < n2){ `` ``arr3[k++] = arr2[j++]; `` ``} `` ` ` ``// sort the whole array arr3`` ``Arrays.sort(arr3); `` ``}``}` `// This code is contributed by Tapesh(tapeshdua420)`
## Python3
`# Python program to merge two sorted arrays/``def` `mergeArrays(arr1, arr2, n1, n2, arr3):`` ``i ``=` `0`` ``j ``=` `0`` ``k ``=` `0` ` ``# traverse the arr1 and insert its element in arr3`` ``while``(i < n1):`` ``arr3[k] ``=` `arr1[i]`` ``k ``+``=` `1`` ``i ``+``=` `1` ` ``# now traverse arr2 and insert in arr3`` ``while``(j < n2):`` ``arr3[k] ``=` `arr2[j]`` ``k ``+``=` `1`` ``j ``+``=` `1` ` ``# sort the whole array arr3`` ``arr3.sort()` `# Driver code``if` `__name__ ``=``=` `'__main__'``:`` ``arr1 ``=` `[``1``, ``3``, ``5``, ``7``]`` ``n1 ``=` `len``(arr1)` ` ``arr2 ``=` `[``2``, ``4``, ``6``, ``8``]`` ``n2 ``=` `len``(arr2)` ` ``arr3 ``=` `[``0` `for` `i ``in` `range``(n1``+``n2)]`` ``mergeArrays(arr1, arr2, n1, n2, arr3)` ` ``print``(``"Array after merging"``)`` ``for` `i ``in` `range``(n1 ``+` `n2):`` ``print``(arr3[i], end``=``" "``)` `# This code is contributed by Tapesh(tapeshdua420)`
## C#
`// C# program to merge two sorted arrays``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG``{`` ``public` `static` `void` `mergeArrays(``int``[] arr1, ``int``[] arr2, ``int` `n1, ``int` `n2, ``int``[] arr3) {`` ``int` `i = 0;`` ``int` `j = 0;`` ``int` `k = 0;` ` ``// traverse the arr1 and insert its element in arr3`` ``while` `(i < n1) {`` ``arr3[k++] = arr1[i++];`` ``}` ` ``// now traverse arr2 and insert in arr3`` ``while` `(j < n2) {`` ``arr3[k++] = arr2[j++];`` ``}` ` ``// sort the whole array arr3`` ``Array.Sort(arr3);`` ``}` ` ``public` `static` `void` `Main(``string``[] args)`` ``{`` ``int` `[]arr1 = ``new` `int``[] {1, 3, 5, 7};`` ``int` `n1 = arr1.Length;` ` ``int` `[]arr2 = ``new` `int``[] {2, 4, 6, 8};`` ``int` `n2 = arr2.Length;` ` ``int` `[]arr3 = ``new` `int``[n1 + n2];`` ``mergeArrays(arr1, arr2, n1, n2, arr3);` ` ``Console.WriteLine(``"Array after merging"``);`` ``for` `(``int` `i = 0; i < n1 + n2; i++)`` ``Console.Write(arr3[i] + ``" "``);`` ``}``}` `// This code is contributed by ajaymakavana.`
## Javascript
` ``// JavaScript program to merge two sorted arrays/`` ``function` `mergeArrays(arr1, arr2, n1, n2, arr3) {`` ``var` `i = 0, j = 0, k = 0;`` ` ` ``// traverse the arr1 and insert its element in arr3`` ``while` `(i < n1) {`` ``arr3[k++] = arr1[i++];`` ``}` ` ``// now traverse arr2 and insert in arr3`` ``while` `(j < n2) {`` ``arr3[k++] = arr2[j++];`` ``}` ` ``// sort the whole array arr3`` ``arr3.sort();`` ``}`` ``var` `arr1 = [1, 3, 5, 7];`` ``var` `n1 = arr1.length;`` ``var` `arr2 = [2, 4, 6, 8];`` ``var` `n2 = arr2.length;`` ``var` `arr3 = ``new` `Array(n1 + n2);`` ``mergeArrays(arr1, arr2, n1, n2, arr3);` ` ``console.log(``"Array after merging"``);`` ``for` `(``var` `i = 0; i < n1 + n2; i++)`` ``process.stdout.write(arr3[i] + ``" "``);` `// This code is contributed by tapeshdua420.`
Output
```Array after merging
1 2 3 4 5 6 7 8
```
Time Complexity : O((m+n) log(m+n)) , the whole size of arr3 is m+n
Auxiliary Space: O(m+n), where m is the size of arr1 and n is the size of arr2.
Method 2 (O(n1 * n2) Time and O(n1+n2) Extra Space)
1. Create an array arr3[] of size n1 + n2.
2. Copy all n1 elements of arr1[] to arr3[]
3. Traverse arr2[] and one by one insert elements (like insertion sort) of arr3[] to arr1[]. This step take O(n1 * n2) time.
We have discussed implementation of above method in Merge two sorted arrays with O(1) extra space
Method 3 (O(n1 + n2) Time and O(n1 + n2) Extra Space)
The idea is to use Merge function of Merge sort
1. Create an array arr3[] of size n1 + n2.
2. Simultaneously traverse arr1[] and arr2[].
• Pick smaller of current elements in arr1[] and arr2[], copy this smaller element to next position in arr3[] and move ahead in arr3[] and the array whose element is picked.
3. If there are remaining elements in arr1[] or arr2[], copy them also in arr3[].
Below is the illustration of the above approach:
Take two sorted arrays Array1 and Array2 and an Empty Array3 . Take three pointers ‘i’, ‘j’, and ‘k’ for comparisons, here ‘i’ pointer points towards 0th index of Array1, similarly ‘j’ and ‘k’ pointer point towards 0th index of Array2 and Array3
Initial Arrays
Step 1: Pick Smaller element which is 4 and insert in into Array3 and update the pointer ‘j ‘and ‘k’ after comparing ‘i’ and ‘j’.
Pick Smaller element which is 4
Step 2: Pick next smaller element which is 5 and insert in into Array3 and update the pointer ‘i’ and ‘k’ after comparing ‘i’ and ‘j’.
Pick next smaller element which is 5
Step 3: Pick next smaller element which is 7 and insert in into Array3 and update the pointer ‘j’ and ‘k ‘after comparing ‘i’ and ‘j’.
Pick next smaller element which is 7
Step 4: when ‘j’ pointer meets the length of Array2 then first while loop breaks and second while loop copies all elements from arr1 to arr3.
second while loop copies all elements from arr1 to arr3.
Step 5: Pick adjacent element from Array1 and insert in into Array3 and update the pointer i and k
Pick adjacent element from Array1 and insert in into Array3
Step 6: Pick remaining element from Array1 and insert in into Array3, when i pointer meets the length of Array1 that means k = n1+n2 and at last we have merge sorted Array3
ick remaining element from Array1 and insert in into Array3,
Below is the implementation of the above approach:
## C++
`// C++ program to merge two sorted arrays/``#include``using` `namespace` `std;` `// Merge arr1[0..n1-1] and arr2[0..n2-1] into``// arr3[0..n1+n2-1]``void` `mergeArrays(``int` `arr1[], ``int` `arr2[], ``int` `n1,`` ``int` `n2, ``int` `arr3[])``{`` ``int` `i = 0, j = 0, k = 0;` ` ``// Traverse both array`` ``while` `(i
## Java
`// Java program to merge two sorted arrays``import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;` `class` `MergeTwoSorted``{`` ``// Merge arr1[0..n1-1] and arr2[0..n2-1]`` ``// into arr3[0..n1+n2-1]`` ``public` `static` `void` `mergeArrays(``int``[] arr1, ``int``[] arr2, ``int` `n1,`` ``int` `n2, ``int``[] arr3)`` ``{`` ``int` `i = ``0``, j = ``0``, k = ``0``;`` ` ` ``// Traverse both array`` ``while` `(i
## Python 3
`# Python program to merge``# two sorted arrays` `# Merge arr1[0..n1-1] and``# arr2[0..n2-1] into``# arr3[0..n1+n2-1]``def` `mergeArrays(arr1, arr2, n1, n2):`` ``arr3 ``=` `[``None``] ``*` `(n1 ``+` `n2)`` ``i ``=` `0`` ``j ``=` `0`` ``k ``=` `0` ` ``# Traverse both array`` ``while` `i < n1 ``and` `j < n2:`` ` ` ``# Check if current element`` ``# of first array is smaller`` ``# than current element of`` ``# second array. If yes,`` ``# store first array element`` ``# and increment first array`` ``# index. Otherwise do same`` ``# with second array`` ``if` `arr1[i] < arr2[j]:`` ``arr3[k] ``=` `arr1[i]`` ``k ``=` `k ``+` `1`` ``i ``=` `i ``+` `1`` ``else``:`` ``arr3[k] ``=` `arr2[j]`` ``k ``=` `k ``+` `1`` ``j ``=` `j ``+` `1`` ` ` ``# Store remaining elements`` ``# of first array`` ``while` `i < n1:`` ``arr3[k] ``=` `arr1[i];`` ``k ``=` `k ``+` `1`` ``i ``=` `i ``+` `1` ` ``# Store remaining elements`` ``# of second array`` ``while` `j < n2:`` ``arr3[k] ``=` `arr2[j];`` ``k ``=` `k ``+` `1`` ``j ``=` `j ``+` `1`` ``print``(``"Array after merging"``)`` ``for` `i ``in` `range``(n1 ``+` `n2):`` ``print``(``str``(arr3[i]), end ``=` `" "``)` `# Driver code``arr1 ``=` `[``1``, ``3``, ``5``, ``7``]``n1 ``=` `len``(arr1)` `arr2 ``=` `[``2``, ``4``, ``6``, ``8``]``n2 ``=` `len``(arr2)``mergeArrays(arr1, arr2, n1, n2);` `# This code is contributed``# by ChitraNayal`
## C#
`// C# program to merge``// two sorted arrays``using` `System;` `class` `GFG``{`` ``// Merge arr1[0..n1-1] and`` ``// arr2[0..n2-1] into`` ``// arr3[0..n1+n2-1]`` ``public` `static` `void` `mergeArrays(``int``[] arr1, ``int``[] arr2,`` ``int` `n1, ``int` `n2, ``int``[] arr3)`` ``{`` ``int` `i = 0, j = 0, k = 0;`` ` ` ``// Traverse both array`` ``while` `(i < n1 && j < n2)`` ``{`` ``// Check if current element`` ``// of first array is smaller`` ``// than current element`` ``// of second array. If yes,`` ``// store first array element`` ``// and increment first array`` ``// index. Otherwise do same`` ``// with second array`` ``if` `(arr1[i] < arr2[j])`` ``arr3[k++] = arr1[i++];`` ``else`` ``arr3[k++] = arr2[j++];`` ``}`` ` ` ``// Store remaining`` ``// elements of first array`` ``while` `(i < n1)`` ``arr3[k++] = arr1[i++];`` ` ` ``// Store remaining elements`` ``// of second array`` ``while` `(j < n2)`` ``arr3[k++] = arr2[j++];`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `Main()`` ``{`` ``int``[] arr1 = {1, 3, 5, 7};`` ``int` `n1 = arr1.Length;`` ` ` ``int``[] arr2 = {2, 4, 6, 8};`` ``int` `n2 = arr2.Length;`` ` ` ``int``[] arr3 = ``new` `int``[n1+n2];`` ` ` ``mergeArrays(arr1, arr2, n1, n2, arr3);`` ` ` ``Console.Write(``"Array after merging\n"``);`` ``for` `(``int` `i = 0; i < n1 + n2; i++)`` ``Console.Write(arr3[i] + ``" "``);`` ``}``}` `// This code is contributed``// by ChitraNayal`
## Javascript
``
## PHP
``
Output
```Array after merging
1 2 3 4 5 6 7 8
```
Time Complexity : O(n1 + n2)
Auxiliary Space : O(n1 + n2)
Method 4: Using Maps (O(nlog(n) + mlog(m)) Time and O(N) Extra Space)
1. Insert elements of both arrays in a map as keys.
2. Print the keys of the map.
Below is the implementation of above approach.
## CPP
`// C++ program to merge two sorted arrays``//using maps``#include``using` `namespace` `std;` `// Function to merge arrays``void` `mergeArrays(``int` `a[], ``int` `b[], ``int` `n, ``int` `m)``{`` ``// Declaring a map.`` ``// using map as a inbuilt tool`` ``// to store elements in sorted order.`` ``map<``int``, ``int``> mp;`` ` ` ``// Inserting values to a map.`` ``for``(``int` `i = 0; i < n; i++)mp[a[i]]++;`` ` ` ` ` ``for``(``int` `i = 0;i < m;i++)mp[b[i]]++;` ` ` ` ``// Printing keys of the map.`` ``for``(``auto` `j: mp)`` ``{`` ``for``(``int` `i=0; i
## Java
`// Java program to merge two sorted arrays``//using maps``import` `java.io.*;``import` `java.util.*;` `class` `GFG {`` ` ` ``// Function to merge arrays`` ``static` `void` `mergeArrays(``int` `a[], ``int` `b[], ``int` `n, ``int` `m)`` ``{`` ` ` ``// Declaring a map.`` ``// using map as a inbuilt tool`` ``// to store elements in sorted order.`` ``Map mp = ``new` `TreeMap();`` ` ` ``// Inserting values to a map.`` ``for``(``int` `i = ``0``; i < n; i++)`` ``{`` ``mp.put(a[i], ``true``);`` ``}`` ``for``(``int` `i = ``0``;i < m;i++)`` ``{`` ``mp.put(b[i], ``true``);`` ``}`` ` ` ``// Printing keys of the map.`` ``for` `(Map.Entry me : mp.entrySet())`` ``{`` ``System.out.print(me.getKey() + ``" "``);`` ``}`` ``}`` ` ` ``// Driver Code`` ``public` `static` `void` `main (String[] args)`` ``{`` ``int` `a[] = {``1``, ``3``, ``5``, ``7``}, b[] = {``2``, ``4``, ``6``, ``8``};`` ``int` `size = a.length;`` ``int` `size1 = b.length;`` ` ` ``// Function call`` ``mergeArrays(a, b, size, size1);`` ``}``}` `// This code is contributed by rag2127`
## Python3
`# Python program to merge two sorted arrays``# using maps``import` `bisect` `# Function to merge arrays``def` `mergeArrays(a, b, n, m):`` ``# Declaring a map.`` ``# using map as a inbuilt tool`` ``# to store elements in sorted order.`` ``mp``=``[]`` ` ` ``# Inserting values to a map.`` ``for` `i ``in` `range``(n):`` ``bisect.insort(mp, a[i])`` ` ` ``for` `i ``in` `range``(m):`` ``bisect.insort(mp, b[i])`` ` ` ``# Printing keys of the map.`` ``for` `i ``in` `mp:`` ``print``(i,end``=``' '``)`` ` `# Driver code``arr1 ``=` `[``1``, ``3``, ``5``, ``7``]``arr2 ``=` `[``2``, ``4``, ``6``, ``8``]``size ``=` `len``(arr1)``size1 ``=` `len``(arr2)` `# Function call``mergeArrays(arr1, arr2, size, size1)` `# This code is contributed by Pushpesh Raj`
## C#
`// C# program to merge two sorted arrays``//using maps``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG {` ` ``// Function to merge arrays`` ``static` `void` `mergeArrays(``int` `[]a, ``int` `[]b, ``int` `n, ``int` `m)`` ``{` ` ``// Declaring a map.`` ``// using map as a inbuilt tool`` ``// to store elements in sorted order.`` ``SortedDictionary<``int``, Boolean> mp = ``new` `SortedDictionary<``int``, Boolean>();` ` ``// Inserting values to a map.`` ``for` `(``int` `i = 0; i < n; i++) {`` ``mp.Add(a[i], ``true``);`` ``}`` ``for` `(``int` `i = 0; i < m; i++) {`` ``mp.Add(b[i], ``true``);`` ``}` ` ``// Printing keys of the map.`` ``foreach` `(KeyValuePair<``int``, Boolean> me ``in` `mp) {`` ``Console.Write(me.Key + ``" "``);`` ``}`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main(String[] args) {`` ``int` `[]a = { 1, 3, 5, 7 };`` ``int` `[]b = { 2, 4, 6, 8 };`` ``int` `size = a.Length;`` ``int` `size1 = b.Length;` ` ``// Function call`` ``mergeArrays(a, b, size, size1);`` ``}``}` `// This code is contributed by gauravrajput1`
## Javascript
``
Output
```1 2 3 4 5 6 7 8
```
Time Complexity: O( nlog(n) + mlog(m) )
Auxiliary Space: O(N) | 6,460 | 17,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.509685 |
https://imathworks.com/tex/tex-latex-placing-coloured-rectangles-on-a-plot-using-points-from-the-plot-riemann-sums/ | 1,660,527,859,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572089.53/warc/CC-MAIN-20220814234405-20220815024405-00212.warc.gz | 310,335,740 | 6,643 | # [Tex/LaTex] Placing coloured rectangles on a plot, using points from the plot – Riemann Sums
pgfplotstikz-pgf
October 31st
Note: Below is the original question, but after some feedback I have progressed somewhat and posted an answer to demonstrate what I've learnt since. My answer presents an encapsulated solution. I'm posting it here because the original question was in the context of being able to create diagrams for demonstrating Riemann sums.
These questions here are where I was a week ago and my answer is where I am now. The title remains relevant. A search on placing rectangles on a plot or on Riemann Sums will turn up this page and there are some useful answers here.
I'm sorry if people think I've confused the question, but I think this is what happens when you're on a steep learning curve and the page was kind of documenting that. What seem like simple questions now were quite perplexing a week ago. I've edited this and my answer comprehensively in order to simplify matters.
October 23rd
Q(1) If there was a simple way to add the in-between lines to the rectangles in the attempt on the left, I would be finished. The problem with the solution on the right is the stacked environment doesn't like my original graph so I placed it in it's own axis environment and then had trouble with the curve matching the rectangles.
I tried the solution on the right mainly to get the colours right, but quickly realised drawing the upper rectangles before the lower ones gave what I wanted with the solution on the left. If drawing those lines is the only way to go then fair enough, but is there a systematic way to get them as part of the plot?
Q(2) Can I get the values for the heights of the rectangles from my plot equation rather than calculating and typing them all in by hand?
MWE 1 Output
MWE 1 Code
\documentclass{standalone}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{pgfplots}
\usepackage{mathtools}
\pgfplotsset{compat=1.9}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
xtick={0,...,5},ytick={5,10,15,20,25},
y=0.3cm, xmax=5.4,ymax=26.9,ymin=0,xmin=0,
enlargelimits=true,
axis lines=middle,
clip=false
]
coordinates {(0,2) (1,5) (2,10) (3,17) (4,26) (5,26)}\closedcycle;
coordinates {(0,1) (1,2) (2,5) (3,10) (4.0,17) (5,17)}\closedcycle;
\end{axis}
\end{tikzpicture}
\begin{tikzpicture}
\begin{axis}[
xtick={0,...,5},ytick={5,10,15,20,25},
y=0.3cm, xmax=5.4,ymax=26.9,ymin=0,
axis lines=middle,
clip=false,
const plot,
stack plots=y,
area style]
{(0,1) (1,2) (2,5) (3,10) (4.0,17) (5,17)}
\closedcycle;
{(0,1) (1,3) (2,5) (3,7) (4,9) (5,9)}
\closedcycle;
\end{axis}
\begin{axis}[
axis lines=none,
y=0.3cm, xmax=5.4,ymax=26.9,ymin=0,
]
\end{axis}
\end{tikzpicture}
\end{document}
You can use const plot mark right to get a piecewise constant plot for the right sum, and a ybar interval for the left sum. That way, you can just specify the same equation as for your line plot. Note that these aren't generally upper and lower sums, but rather right and left sums, but for monotonic functions like this one they're equivalent.
\documentclass[border=5mm]{standalone}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{pgfplots}
\pgfplotsset{compat=1.9}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
xtick={0,...,5},ytick={5,10,15,20,25},
y=0.3cm, xmax=5.4,ymax=26.9,ymin=0,xmin=0,
enlargelimits=true,
axis lines=middle,
clip=false,
domain=0:5,
axis on top
]
\addplot [draw=red,fill=red!10,const plot mark right, samples=6]
{1+x^2}\closedcycle;
\addplot [draw=green, fill=green!10, ybar interval, samples=6]
{1+x^2}\closedcycle;
\end{axis}
\end{tikzpicture}
\end{document} | 1,114 | 3,639 | {"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-2022-33 | latest | en | 0.95272 |
https://sugarrushbelfast.com/solver-452 | 1,675,174,169,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00508.warc.gz | 562,407,478 | 3,509 | # Fundamental trigonometric identities solver
Apps can be a great way to help students with their algebra. Let's try the best Fundamental trigonometric identities solver. Our website can help me with math work.
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There is no one definitive way to solve multi step equations, as there are a variety of methods that can be used depending on the equation and the desired solution. However, some general tips that may be helpful include: - manipulating the equation to isolate the variable of interest on one side - using inverse operations to solving the equation step by step - using algebraic methods such as factoring or completing the square If you are struggling to solve a particular equation, it may
There are a lot of great math apps out there that can help you with your math homework. However, my personal favorite is called Photomath. This app is simple to use and can quickly give you the answer to any math problem. Just take a picture of the problem and the app will give you the answer. You can also use the app to see step-by-step solutions for more complex problems.
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One of the best helping calculator apps I have ever downloaded. I would 100% recommend this app for harder projects as it is a life saver at times. Everyone who worked to make this app as amazing as it is are very talented and put a lot of effort in to it. Thank you, photo, math, for assisting me for as long as you have
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Solve algebra equation calculator Enter algebra problems and get answers Answers to my math homework How to do a math problem Math answer calculator Variable solver | 645 | 3,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.969359 |
http://archive.org/stream/AnalyticalMechanics/TXT/00000170.txt | 1,498,252,548,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320174.58/warc/CC-MAIN-20170623202724-20170623222724-00618.warc.gz | 30,288,070 | 6,600 | # Full text of "Analytical Mechanics"
## See other formats
```CENTER OF MASS AND MOMENT OF INERTIA 157
2. Find the moment of inertia of a cylinder about a transverse axis through the center of mass.
Let m, a, I, and r be, respectively, the mass, the radius, the length, and the density of the cylinder. Further let the given axis pass through the center of mass of the cylinder; then taking a slice obtained by two right sections as the element of mass we get, by theorem II, dly = dly' + z1 dm,
where dm is the mass of the element, dly and dly> are the moments of inertia of the element about the given axis and about a parallel axis through the center of mass of the element, and z is the distance between these two axes. But by theorem I
dls + dl* = dl,,
and by symmetry dlj = dly>,
and by the last illustrative example
a2 dm
dlm
Therefore
2
a2 dm
Substituting this value of dly in the expression for dly we get dly
Integrating the last equation we have
_ma? . r * ____``` | 264 | 1,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.898817 |
https://community.powerbi.com/t5/Desktop/Moving-average-based-on-Rolling-7-days-per-customer-subject/td-p/53035 | 1,606,916,148,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141708017.73/warc/CC-MAIN-20201202113815-20201202143815-00019.warc.gz | 228,445,575 | 109,566 | cancel
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Regular Visitor
## Moving average based on Rolling 7 days - per customer/subject
Hi everyone,
First, sorry for bothering you with an easy question but I can move forward in my model without this.
I have my data model where we can find my planning (Date, time, Id-session (which session has been done) and objectives.
I have a table where we can find, for each session, the exercise, objective, targeted group, sets, reps (...).
A table date where I have all the dates from the first day (a calendar...).
Finally, a table (Seances_Joueurs) where I have the player, the Id_session and the total number of reps, sets and TUT (sum) for the session.
Now I would like to create in the Session_joueur table a calculate column to measure a rolling average over 7 days, for each players (Nom).
I tried with DATEINBETWEEN as well as this formula but the results are wrong...
`Acute = calculate([NbRep totales]; ALL('Séances_Joueurs');FILTER(ALL('Séances_Joueurs'[Date]);'Séances_Joueurs'[Date]=EARLIER('Séances_Joueurs'[Date])-7))`
Could one of you guys help me ?
All the best,
Mathieu.
1 ACCEPTED SOLUTION
Accepted Solutions
Microsoft
@MathLacome
Please try with following measure for “Rolling sum”.
```Rolling Sum 7 Days_Per Player =
CALCULATE (
SUM ( Table1[NbRep] ),
DATESINPERIOD ( 'Date'[Date], LASTDATE ( 'Date'[Date] ), -7, DAY ),
ALLEXCEPT ( Table1, Table1[Nom] )
)```
Best Regards,
Herbert
8 REPLIES 8
Super User I
@MathLacome
Moving averages can be complex. Here is a blog that explains how to create them as well as correct them for the month level where the total days vary.
http://www.powerpivotpro.com/2013/07/moving-averages-sums-etc/
Proud to be a Super User!
Regular Visitor
I already checked at the blog but there is still some point I do not understand.
I need to add another filter layout (players name) to have FOR EACH player/customer, the moving average.
Someone can help me with this ?
All the best.
Mathieu.
Super User I
@MathLacome
Your sample data sccreen shows each player with multiple dates . So how do you propose to show the output.
Is it going to be By Player, By Date and then the Moving Average over the last 7 days of the reported date ??
Please clarify, if possible attach the output format and result you expect based on the sample data.
Cheers
CheenuSing
Did I answer your question? Mark my post as a solution and also give KUDOS !
Proud to be a Datanaut!
Microsoft
@MathLacome
In your dataset, it seems that each player does not have data for each day. I assume you want the moving average based on the existing 7 days in Seances_Joueurs table. (For PISCIONE, days are 7/7/2016, 7/9/2016, 7/12/2016 …)
First, create a column to rank the date for each player with following formula.
```RankDate =
IF (
Table1[Date] <> BLANK (),
RANKX (
FILTER (
Table1,
Table1[Nom] = EARLIER ( Table1[Nom] )
&& Table1[Date] <> BLANK ()
),
Table1[Date],
,
ASC,
DENSE
)
)```
Then create a measure to get the moving average with following formula.
```Moving Average 7 Days =
VAR SevenDaysTotal =
CALCULATE (
SUM ( Table1[NbRep] ),
FILTER (
ALL ( Table1[RankDate] ),
Table1[RankDate] <= MAX ( Table1[RankDate] )
&& Table1[RankDate] >= MAX( Table1[RankDate] ) - 6
),
ALLEXCEPT ( Table1, Table1[Nom] )
)
VAR Days = CALCULATE (
DISTINCTCOUNT ( Table1[RankDate] ),
FILTER (
ALL ( Table1[RankDate] ),
Table1[RankDate] <= MAX ( Table1[RankDate] )
&& Table1[RankDate]
>= MAX ( Table1[RankDate] ) - 6
&& Table1[RankDate] <> BLANK ()
),
ALLEXCEPT ( Table1, Table1[Nom] )
)
RETURN
(
DIVIDE( SevenDaysTotal, Days )
)```
Best Regards,
Herbert
Regular Visitor
But... I'm still struggling !
Yes Herbert, I dont have "sessions" everyday but it is not a problem. I want 7 days on a rolling average.
So, I built a calendar table where I have everydates, id_day, id_month and so on for my visualisations.
I would like to :
1. Create a measure "Rolling sum" or "Rolling average" to sum the number of reps, by players, on a 7 days rolling scale.
I want to be able to visualise it in a bar graph, day by day. So I need the measure to be calculate for today but also for today-1, today-2 .... to the end ! 😉
I tried your formula Herbert (but I only have 1 value !) but Its not working the way I want...
I also tried with this one :
"
That [Units Sold] measure is the jagged red line in the chart at the top of the post, and its formula is very simple:
[Units Sold] = SUM(Sales[QtySold])
And we want a version of [Units Sold] that is “smoothed” over a 3-month period.
### Moving Sum
Let’s start with a formula that is a sum of the most recent 3 months (including the current one):
[3 Month Moving Sum Units Sold] =
CALCULATE([Units Sold],
DATESINPERIOD(Calendar[Date],
LASTDATE(Calendar[Date]),-3, Month
Found in PowerPivotPro but still not working...
Microsoft
@MathLacome
Please try with following measure for “Rolling sum”.
```Rolling Sum 7 Days_Per Player =
CALCULATE (
SUM ( Table1[NbRep] ),
DATESINPERIOD ( 'Date'[Date], LASTDATE ( 'Date'[Date] ), -7, DAY ),
ALLEXCEPT ( Table1, Table1[Nom] )
)```
Best Regards,
Herbert
Regular Visitor
Thanks you guys,
All the above solutions where fully functionnal. My mistake was from making a bar chart with Date column and not id_Date Column.
My Date Column was returning aggregates of month and days in a month but not a ful calendar.
Thanks everyone for helping me to find the solution !
All the best,
Mathieu.
I have an equal problem, but instead of calculating the average i do the distinct count the last 14 days. In my data model I have created a Date table using "Date = CALENDARAUTO()" that I use as a slider for my "Line and Clustered Column Chart", and I'm using the Date in the Hierarchy format as the Shared Axis.
I'm using the formula below to calculate the distinct number the last 14 days:
Customers Last 14 days = CALCULATE(DISTINCTCOUNT(TransactionData[Customer]),DATESINPERIOD('Date'[Date],LASTDATE('Date'[Date]),-14,DAY))
1. The DistinctCount for the current period (Day, Month, Quater, Year, depending on if I Drill Up of Down the diagram) I show as Bars
2. The DistinctCount for the last 14 days I show as a Line
3. I have also added a YearToDate measure that I show as a Second Line
I have two problems:
• The calculations continues (and the value decreases) even after my current date (today it's the 6'th of September, so I have values untill the 19'th of September). This is easy to see when I list the values in a table
• The last 14 days line does not show up for the last month when I Drilled to the Month based view in the diagram. The same goes for Quater and Year. Drilling down to the day view will show the "last 14 days line" up to todays date.
Although, this is kind of strange, since I also have seen the "last 14 days line" streching out to the coming 13 days as I described as a problem above. Maybe Microsoft has fixed this in the later versions. But still, there's a problem that I don't see the "last 14 days line" extended into my current month. Using a YTD measure, this line do extend into the current month
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Top Kudoed Authors | 1,973 | 7,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-50 | latest | en | 0.869639 |
https://inches-to-meters.appspot.com/90.4-inches-to-meters.html | 1,610,724,702,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00386.warc.gz | 441,598,662 | 6,306 | Inches To Meters
90.4 in to m90.4 Inches to Meters
in
=
m
How to convert 90.4 inches to meters?
90.4 in * 0.0254 m = 2.29616 m 1 in
A common question is How many inch in 90.4 meter? And the answer is 3559.05511811 in in 90.4 m. Likewise the question how many meter in 90.4 inch has the answer of 2.29616 m in 90.4 in.
How much are 90.4 inches in meters?
90.4 inches equal 2.29616 meters (90.4in = 2.29616m). Converting 90.4 in to m is easy. Simply use our calculator above, or apply the formula to change the length 90.4 in to m.
Convert 90.4 in to common lengths
UnitLengths
Nanometer2296160000.0 nm
Micrometer2296160.0 µm
Millimeter2296.16 mm
Centimeter229.616 cm
Inch90.4 in
Foot7.5333333333 ft
Yard2.5111111111 yd
Meter2.29616 m
Kilometer0.00229616 km
Mile0.0014267677 mi
Nautical mile0.0012398272 nmi
What is 90.4 inches in m?
To convert 90.4 in to m multiply the length in inches by 0.0254. The 90.4 in in m formula is [m] = 90.4 * 0.0254. Thus, for 90.4 inches in meter we get 2.29616 m.
Alternative spelling
90.4 in to Meter, 90.4 in in Meter, 90.4 in to m, 90.4 in in m, 90.4 Inches to Meter, 90.4 Inches in Meter, 90.4 Inch to m, 90.4 Inch in m, 90.4 Inch to Meters, 90.4 Inch in Meters, 90.4 Inch to Meter, 90.4 Inch in Meter, 90.4 in to Meters, 90.4 in in Meters | 501 | 1,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.835777 |
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• Hello!
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Windows Desktop: Direct3D11, Direct3D12, OpenGL Universal Windows: Direct3D11, Direct3D12 Linux: OpenGL Android: OpenGLES MacOS: OpenGL iOS: OpenGLES API Basics
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TextureDesc TexDesc; TexDesc.Name = "My texture 2D"; TexDesc.Type = TEXTURE_TYPE_2D; TexDesc.Width = 1024; TexDesc.Height = 1024; TexDesc.Format = TEX_FORMAT_RGBA8_UNORM; TexDesc.Usage = USAGE_DEFAULT; TexDesc.BindFlags = BIND_SHADER_RESOURCE | BIND_RENDER_TARGET | BIND_UNORDERED_ACCESS; TexDesc.Name = "Sample 2D Texture"; m_pRenderDevice->CreateTexture( TexDesc, TextureData(), &m_pTestTex ); Initializing Pipeline State
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Setting the Pipeline State and Invoking Draw Command
Before any draw command can be invoked, all required vertex and index buffers as well as the pipeline state should be bound to the device context:
// Clear render target const float zero[4] = {0, 0, 0, 0}; m_pContext->ClearRenderTarget(nullptr, zero); // Set vertex and index buffers IBuffer *buffer[] = {m_pVertexBuffer}; Uint32 offsets[] = {0}; Uint32 strides[] = {sizeof(MyVertex)}; m_pContext->SetVertexBuffers(0, 1, buffer, strides, offsets, SET_VERTEX_BUFFERS_FLAG_RESET); m_pContext->SetIndexBuffer(m_pIndexBuffer, 0); m_pContext->SetPipelineState(m_pPSO); Also, all shader resources must be committed to the device context:
m_pContext->CommitShaderResources(m_pSRB, COMMIT_SHADER_RESOURCES_FLAG_TRANSITION_RESOURCES); When all required states and resources are bound, IDeviceContext::Draw() can be used to execute draw command or IDeviceContext::DispatchCompute() can be used to execute compute command. Note that for a draw command, graphics pipeline must be bound, and for dispatch command, compute pipeline must be bound. Draw() takes DrawAttribs structure as an argument. The structure members define all attributes required to perform the command (primitive topology, number of vertices or indices, if draw call is indexed or not, if draw call is instanced or not, if draw call is indirect or not, etc.). For example:
DrawAttribs attrs; attrs.IsIndexed = true; attrs.IndexType = VT_UINT16; attrs.NumIndices = 36; attrs.Topology = PRIMITIVE_TOPOLOGY_TRIANGLE_LIST; pContext->Draw(attrs); Tutorials and Samples
The GitHub repository contains a number of tutorials and sample applications that demonstrate the API usage.
AntTweakBar sample demonstrates how to use AntTweakBar library to create simple user interface.
Atmospheric scattering sample is a more advanced example. It demonstrates how Diligent Engine can be used to implement various rendering tasks: loading textures from files, using complex shaders, rendering to textures, using compute shaders and unordered access views, etc.
The repository includes Asteroids performance benchmark based on this demo developed by Intel. It renders 50,000 unique textured asteroids and lets compare performance of D3D11 and D3D12 implementations. Every asteroid is a combination of one of 1000 unique meshes and one of 10 unique textures.
Integration with Unity
Diligent Engine supports integration with Unity through Unity low-level native plugin interface. The engine relies on Native API Interoperability to attach to the graphics API initialized by Unity. After Diligent Engine device and context are created, they can be used us usual to create resources and issue rendering commands. GhostCubePlugin shows an example how Diligent Engine can be used to render a ghost cube only visible as a reflection in a mirror.
• By Yxjmir
I'm trying to load data from a .gltf file into a struct to use to load a .bin file. I don't think there is a problem with how the vertex positions are loaded, but with the indices. This is what I get when drawing with glDrawArrays(GL_LINES, ...):
Also, using glDrawElements gives a similar result. Since it looks like its drawing triangles using the wrong vertices for each face, I'm assuming it needs an index buffer/element buffer. (I'm not sure why there is a line going through part of it, it doesn't look like it belongs to a side, re-exported it without texture coordinates checked, and its not there)
I'm using jsoncpp to load the GLTF file, its format is based on JSON. Here is the gltf struct I'm using, and how I parse the file:
glBindVertexArray(g_pGame->m_VAO);
glDrawElements(GL_LINES, g_pGame->m_indices.size(), GL_UNSIGNED_BYTE, (void*)0); // Only shows with GL_UNSIGNED_BYTE
glDrawArrays(GL_LINES, 0, g_pGame->m_vertexCount);
So, I'm asking what type should I use for the indices? it doesn't seem to be unsigned short, which is what I selected with the Khronos Group Exporter for blender. Also, am I reading part or all of the .bin file wrong?
Test.gltf
Test.bin
• That means how do I use base DirectX or OpenGL api's to make a physics based destruction simulation?
Will it be just smart rendering or something else is required?
# OpenGL Same texture is used for both meshes, OpenGL and CG
This topic is 2016 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I'm using opengl with cg and its fx format..
I call enable and disable before and after I draw the geometry.. with two different textures, but both of my cube meshes are drawn with the same texture.. does the setup and enable / disable code look correct? I have called cgGLSetManageTextureParameters with CG_TRUE so I shouldn't need to call cgGLEnableTextureParameter..
So I'm having two instances of this class and m_glTexture is two seperate texture objects.. but the cgParameter passed into the constructor is the same as its using the same effect.. I have checked that the texture data is different so its actually creating two seperate textures on the GPU, but only one of them is bound for both meshes..
TE::Render::APITexture2D::APITexture2D( Texture& texture, CGparameter cgParameter ) { texture.Prepare(); glGenTextures(1, &m_glTexture); assert(glGetError() == GL_NO_ERROR); glBindTexture(GL_TEXTURE_2D, m_glTexture); assert(glGetError() == GL_NO_ERROR); cgGLSetTextureParameter(cgParameter, m_glTexture); assert(glGetError() == GL_NO_ERROR); cgSetSamplerState(cgParameter); assert(glGetError() == GL_NO_ERROR); glTexImage2D(GL_TEXTURE_2D, 0, 4, texture.GetImage().GetWidth(), texture.GetImage().GetHeight(), 0, APIMapping::s_colorType[texture.GetImage().GetColorType()], GL_UNSIGNED_BYTE, texture.GetImage().GetDataPtr()); assert(glGetError() == GL_NO_ERROR); } TE::Render::APITexture2D::~APITexture2D() { } void TE::Render::APITexture2D::Enable(CGparameter cgParameter ) { glBindTexture(GL_TEXTURE_2D, m_glTexture); cgSetSamplerState(cgParameter); assert(glGetError() == GL_NO_ERROR); } void TE::Render::APITexture2D::Disable() { glBindTexture(GL_TEXTURE_2D, 0); assert(glGetError() == GL_NO_ERROR); }
TE::Render::APITexture2D::APITexture2D( Texture& texture, CGparameter cgParameter ) { texture.Prepare(); glGenTextures(1, &m_glTexture); assert(glGetError() == GL_NO_ERROR); glBindTexture(GL_TEXTURE_2D, m_glTexture); assert(glGetError() == GL_NO_ERROR); //bind the texture to the cg texture parameter cgGLSetTextureParameter(cgParameter, m_glTexture); assert(glGetError() == GL_NO_ERROR); //initialize the state specified for a sampler parameter cgSetSamplerState(cgParameter); assert(glGetError() == GL_NO_ERROR); glTexImage2D(GL_TEXTURE_2D, 0, 4, texture.GetImage().GetWidth(), texture.GetImage().GetHeight(), 0, APIMapping::s_colorType[texture.GetImage().GetColorType()], GL_UNSIGNED_BYTE, texture.GetImage().GetDataPtr()); assert(glGetError() == GL_NO_ERROR); } TE::Render::APITexture2D::~APITexture2D() { } void TE::Render::APITexture2D::Enable(CGparameter cgParameter ) { cgGLSetTextureParameter(cgParameter, m_glTexture); assert(glGetError() == GL_NO_ERROR); } void TE::Render::APITexture2D::Disable() { glBindTexture(GL_TEXTURE_2D, 0); assert(glGetError() == GL_NO_ERROR); } | 4,572 | 19,979 | {"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} | 2.515625 | 3 | CC-MAIN-2018-13 | longest | en | 0.940859 |
https://www.askiitians.com/forums/General-Physics/9/54624/system-of-partical-and-rotational-motion.htm | 1,712,992,232,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00327.warc.gz | 621,780,018 | 43,457 | # Q1. from a square sheet of uniform density a portion is removed .find the centre of mass of the remaining portion if the side of the square is A.
shivangi bajpai
10 Points
11 years ago
simran wd u plz tell me at kind of a portion u mean?
Shiva
17 Points
6 years ago
Let m1 be the α*a² and m2 be the α*a²/4 where α be the uniform mass density From the question, we have to find the centre of mass of remaining portion. Now we have Xcm = m1*x1 +m2*x2 /m1+m2=( α*a²*0 - α*a²/4*a/2)/α*a²+α*a²= -a³*α/8*4α*a²/5= -a/10There was no possibility to create a diagram
16 Points
3 years ago
x × mass remaining = mass removed × side/total no. of unshade sides
(m-m/4) × x = (m/4)(a/3)
X=a/3 | 233 | 682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-18 | latest | en | 0.813464 |
http://mathematica.stackexchange.com/questions/tagged/traversal | 1,469,764,818,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829970.64/warc/CC-MAIN-20160723071029-00127-ip-10-185-27-174.ec2.internal.warc.gz | 158,775,165 | 14,618 | # Tagged Questions
The tag has no usage guidance.
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### Position with Except returns first item {0} and last item {}. Why?
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### How to perform a depth-first preorder traversal of an expression?
Due to confusion of terminology on my part I asked the wrong question before. Since it has already received other answers I shall not edit it, but instead ask the question I meant to in the first ... | 418 | 1,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-30 | latest | en | 0.890014 |
https://www.doubtnut.com/question-answer/a-point-moves-such-teat-its-distance-from-the-point-40-is-half-that-of-its-distance-from-the-line-x1-642563543 | 1,627,212,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151672.96/warc/CC-MAIN-20210725111913-20210725141913-00053.warc.gz | 790,656,247 | 72,230 | Home
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# A point moves such teat its distance from the point (4,0) is half that of its distance from the line x=16 , find its locus.
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1:22 | 392 | 953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.562427 |
https://rdrr.io/cran/tripack/man/circum.html | 1,544,951,534,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827596.48/warc/CC-MAIN-20181216073608-20181216095608-00052.warc.gz | 701,327,902 | 13,344 | # circum: Determine the circumcircle of a triangle In tripack: Triangulation of Irregularly Spaced Data
## Description
This function returns the circumcircle of a triangle.
## Usage
`1` ```circum(x, y) ```
## Arguments
`x` Vector of three elements, giving the x coordinatres of the triangle nodes. `y` Vector of three elements, giving the y coordinatres of the triangle nodes.
## Details
This is an interface to the Fortran function CIRCUM found in TRIPACK.
## Value
` x ` 'x' coordinate of center ` y ` 'y' coordinate of center ` radius ` circumcircle radius ` signed.area ` signed area of riangle (positive iff nodes are numbered counter clock wise) ` aspect.ratio ` ratio "radius of inscribed circle"/"radius of circumcircle", varies between 0 and 0.5 0 means collinear points, 0.5 equilateral trangle.
## Note
This function is mainly intended to be used by `circumcircle`.
## Author(s)
Fortran code: R. J. Renka, R code: A. Gebhardt
## References
R. J. Renka (1996). Algorithm 751: TRIPACK: a constrained two-dimensional Delaunay triangulation package. ACM Transactions on Mathematical Software. 22, 1-8.
`circumcircle`
`1` ```circum(c(0,1,0),c(0,0,1)) ``` | 321 | 1,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.701035 |
https://www.buffalobrewingstl.com/batch-fermentation/a.html | 1,586,223,358,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371662966.69/warc/CC-MAIN-20200406231617-20200407022117-00249.warc.gz | 826,998,886 | 7,372 | ## A
Dm = I + JK -2A, Da — JK(I - 1) — ^ I + JK - 2i
«=i where Dm and Da denote the degrees of freedom required to fit the Ath component and the degrees of freedom after fitting the Ath component, respectively. If W exceeds unity, then this criterion suggests that the Ath component could be included in the model [435].
### 4.2 Multivariable Regression Techniques
Several regression techniques can be used to relate two groups of variables such as process measurements X and quality variables Y. The availability of a model provides the opportunity to predict process or product variables and compare the measured and predicted values. The residuals between the predicted and measured values of the variables can be used to develop various SPM techniques and tools for identification of variables that have contributed to the out-of-control signal.
Multivariable linear regression is the most popular technique for model development. The model equation is
where E is the residual which is equal to 0 for the estimate Y = X/3. A critical issue in using this approach for modeling multivariable processes is the colinearity among process variables. Colinearity causes numerical difficulties in computing the inverse (XTX)-1. Hence, the computation of the regression coefficients ¡3 by the least-squares approach may not be possible. Even if (3 is computed, the standard errors of the estimates of the /3 coefficients associated with the colinear regressors become very large. This causes uncertainty and sensitivity in these ¡3 estimates.
Colinearity can be detected by standardizing all predictor variables (mean centered, unit variance) and computing correlations and coefficients of determination.
Z%3 = ^d- ^ ^ = Yi&V'^2 ' i = !>■■• »J = I."" -P- (4-11)
There is significant colinearity among some predictor variables if:
• The correlation between any two predictors exceeds 0.95 (only colinearity between two predictors can be assessed).
• The coefficient of determination R? of each predictor variable j regressed on all the other predictor variables exceeds 0.90, or the variance inflation factor VIFj = (1 — Rj)-1 is less than 10 (variable j is colinear with one or more of the other predictors). VIFj is the (j>j) th diagonal element of the matrix ZTZ_1 where Z = \zij\. R^ can be computed from the relationship between R? and VIFj.
• Some of the eigenvalues of the correlation matrix ZTZ are less than 0.05. Large elements of the corresponding eigenvectors identify the predictor variables involved in the colinearity.
Remedies in regression with colinear data include
• Stepwise regression
• Ridge regression
• Principal components regression
• Partial least squares (PLS) regression
These techniques will be introduced in the sections that follow.
### 4.2.1 Stepwise Regression
Predictor variables are added to or deleted from the prediction (regression) equation one at a time. Stepwise variable selection procedures are useful when a large number of candidate predictors is available. It is expected that only one of the strongly colinear variables will be included in the model. Major disadvantages of stepwise regression are the limitations in identifying alternative candidate subsets of predictors, and the inability to guarantee the optimality of the final model. The procedure is:
• Fit p single variable regression models, calculate the overall model F-statistic for each model. Select the model with the largest F-statistic. If the model is significant, retain the predictor variable and set r = 1.
• Fit p—r reduced models, each having the r predictor variables selected in the previous stages of variable selection and one of the remaining candidate predictors. Select the model with the largest overall F-statistic. Check the significance of the model by using the partial F-statistic.
• If the partial F-statistic is not significant, terminate the procedure. Otherwise, increment r by 1 and return to step 2.
Computation of F-statistics:
Regression sum of squares: SSR = — y)2, with p degrees of freedom (d.f.), Error sum of squares: SSE = Y2(Ui — V)2, with d.f.= m — p - 1. Denote a model of order r by M2 and a model of order r + 1 by Mi, and their error sum of squares by SSE2 and SSEi, respectively. Then
where
MSR(Mi\M2) = SSE2 ~SSEl MSBx = SSEl . (4.14) r + l — r m — r- 2
4.2.2 Ridge Regression
The computation of regression coefficients (3 in Eq. 4.10 is modified by introducing a ridge parameter k such that
Standardized ridge estimates (3j j = 1, • • • . p are calculated for a range of values of k and are plotted versus k. This plot is called a ridge trace. The ¡3 estimates usually change dramatically when k is initially incremented by a small amount from 0. Some (3 coefficients may even change sign. As k is increased, the trace stabilizes. A k value that stabilizes all f3 coefficients is selected and the final values of ¡3 are estimated.
A good estimate of the k value is obtained as where 3* s are the least-squares estimates for the standardized predictor variables, and MSE is the least squares mean squared error, SSE/(m~p — 1).
Ridge regression estimators are biased. The tradeoff for stabilization and variance reduction in regression coefficient estimators is the bias in the estimators and the increase in the squared error.
### 4.2.3 Principal Components Regression
Principal components regression (PCR) is one of the techniques to deal with ill-conditioned data matrices by regressing the system properties (e.g. quality measurements) on the principal components scores of the measured variables (e.g. flow rates, temperature). The implementation starts by representing the data matrix X with its scores matrix T using the transformation T = XP. The number of principal components to retain in the model must be determined as in the PCA such that it optimizes the predictive power of the PCR model. This is generally done by using cross validation. Then, the regression equation becomes
where the optimum matrix of regression coefficients B is obtained as
Substitution of Eq. 4.18 into Eq. 4.17 leads to trivial E's. The inversion of TtT should not cause any problems due to the mutual orthogonality of the scores. Score vectors corresponding to small eigenvalues can be left out in order to avoid colinearity problems. Since principal components regression is a two-step method, there is a risk that useful predictive information would be discarded with a principal component that is excluded. Hence caution must be exercised while leaving out vectors corresponding to small eigenvalues.
### 4.2.4 Partial Least Squares
Partial Least Squares (PLS), also known as Projection to Latent Structures, develops a biased regression model between X and Y. It selects latent variables so that variation in X which is most predictive of the product quality data Y is extracted. PLS works on the sample covariance matrix (XTY)(YTX) [180, 181, 243, 349, 368, 661, 667]. Measurements on k process variables taken at n different times are arranged into a (nxm) process data matrix X. The p quality variables are given by the corresponding (nxp) matrix Y. Data (both X and Y blocks) are usually preprocessed prior to PLS analysis. PLS modeling works better when the data are fairly symmetrically distributed and have fairly constant "error variance" [145]. Data are usually centered and scaled to unit variance because in PLS any given variable will have the influence on the model parameters that increases with the variance of the variable. Centering and scaling issues were discussed earlier in Section 4.1. The PLS model can be built by using the non-linear iterative partial least-squares algorithm (NIPALS). The PLS model consists of outer relations (X and Y blocks individually) and an inner relation (linking both blocks). The outer relations for the X and
Y blocks are respectively | 1,719 | 7,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-16 | longest | en | 0.8637 |
https://betterlesson.com/lesson/443648/solving-literal-equations?from=breadcrumb_lesson | 1,544,900,121,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826968.71/warc/CC-MAIN-20181215174802-20181215200802-00093.warc.gz | 554,780,435 | 34,255 | # Solving Literal Equations
36 teachers like this lesson
Print Lesson
## Objective
SWBAT solve a literal equation for a specified variable by rewriting a given equation in an equivalent form.
#### Big Idea
The process of rewriting equations in equivalent forms is the same whether the equations contain variables, integers, or a mix of both.
## Investigation
20 minutes
At the start of today's lesson, students will work on Literal Equations Investigation in pairs without any prior instruction. I want students to make an informal, initial investigation of solving literal equations by asking them to work through examples. I encourage student ownership by moving students through cycles of individual work, class time, and peer discussion.
See Classroom Video: Discourse and Questioning to visit my classroom for this part of the lesson.
The goal is for students to engage in cycles of reasoning and abstraction as they explore the questions (MP2). I provide about 10 minutes for students to work on the investigation. Then, we will spend about 10 minutes reviewing each question and discussing students' ideas and solutions.
See Classroom Video: Shared Expectations to visit my classroom for this part of the lesson.
I want to guide the class toward the understanding that solving literal equations is all about inverse operations. Sometimes, this idea is difficult for students because literal equations are more abstract.
• Question 1: Students will compare and contrast a linear equation and a literal equation. My goal is for them to observe how the steps for solving each equation for a single variable are similar. In the "another approach" section, students will use substitution to see that the two equations produce the same result. This discovery will help them to recognize that the literal equation is a model for the linear equation. Solving the literal equation for x produces a formula that can be used to solve all literal equations.
• Questions 2-3: Students will continue to investigate solving linear equations. By plugging in values, students learn to use numerical values to check their solutions.
• Question 4: Students will see that, as with linear equations, literal equations can be solved in different ways. Students can show that both solutions are the same by finding a common denominator for the two fractions under method 1. Students could also determine values for a, b and c and show that when these values are substituted in each solution is the same.
Throughout, I will be on the lookout for student mistakes that show a lack of conceptual understanding such as:
"a + b = ab"
I will remind my students about the concept of "like terms" in order to convince students that statements like this are not true. If necessary, we can also substitute values for each variable and use arithmetic to disprove this result.
## Practice
15 minutes
During this section of the class, I give students time to work in pairs on the Literal Equations Practice worksheet. I continue to remind students that they are trying to take apart the equation and isolate the variable by using inverse operations. Encourage students to think out loud while they are working with their partner so that they can critique each other's reasoning on the various exercises (MP3).
As students are finishing this series of questions, I will ask them to post their work on the board. I want them to share their ideas and solutions. I want to be able to review the problems on the worksheet quickly.
See Classroom Video: Student Ownership to visit my classroom for this part of the lesson.
## Closure
5 minutes
Literal Equations Close is a great Ticket Out the Door at the end of this lesson because it allows students to see that one literal equation can be solved in a variety of ways based on the variable you are trying to isolate.
After handing out the worksheet, I will have students do a Turn-and-Talk with their partner around the example given. I will have each partner take turns (two each) explaining how the original equation is manipulated to become the solution for each variable. Next, I will have students solve the equation y = mx + b for each variable. This should be done individually and student's work can be collected as a formative assessment of the day's lesson.
To see this part of the lesson unfold, watch: Classroom Video: Exit Ticket | 890 | 4,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-51 | longest | en | 0.902582 |
http://www.philosophyideas.com/search/response_text.asp?visit=list&TX=9190&order=chron&expand=yes | 1,726,559,018,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00335.warc.gz | 41,040,701 | 7,773 | ### Ideas from 'What Numbers Could Not Be' by Paul Benacerraf [1965], by Theme Structure
#### [found in 'Philosophy of Mathematics: readings (2nd)' (ed/tr Benacerraf/Putnam) [CUP 1983,0-521-29648-x]].
green numbers give full details | back to texts | unexpand these ideas
###### 6. Mathematics / A. Nature of Mathematics / 3. Nature of Numbers / a. Numbers
9901 Numbers can't be sets if there is no agreement on which sets they are
Full Idea: The fact that Zermelo and Von Neumann disagree on which particular sets the numbers are is fatal to the view that each number is some particular set. From: Paul Benacerraf (What Numbers Could Not Be [1965], II) A reaction: I agree. A brilliantly simple argument. There is the possibility that one of the two accounts is correct (I would vote for Zermelo), but it is not actually possible to prove it.
9912 There are no such things as numbers
Full Idea: There are no such things as numbers. From: Paul Benacerraf (What Numbers Could Not Be [1965], IIIC) A reaction: Mill said precisely the same (Idea 9794). I think I agree. There has been a classic error of reification. An abstract pattern is not an object. If I coin a word for all the three-digit numbers in our system, I haven't created a new 'object'.
###### 6. Mathematics / A. Nature of Mathematics / 3. Nature of Numbers / c. Priority of numbers
9151 Benacerraf says numbers are defined by their natural ordering
Full Idea: Benacerraf thinks of numbers as being defined by their natural ordering. From: report of Paul Benacerraf (What Numbers Could Not Be [1965]) by Kit Fine - Cantorian Abstraction: Recon. and Defence §5 A reaction: My intuition is that cardinality is logically prior to ordinality, since that connects better with the experienced physical world of objects. Just as the fact that people have different heights must precede them being arranged in height order.
###### 6. Mathematics / A. Nature of Mathematics / 3. Nature of Numbers / f. Cardinal numbers
13891 To understand finite cardinals, it is necessary and sufficient to understand progressions
Full Idea: Benacerraf claims that the concept of a progression is in some way the fundamental arithmetical notion, essential to understanding the idea of a finite cardinal, with a grasp of progressions sufficing for grasping finite cardinals. From: report of Paul Benacerraf (What Numbers Could Not Be [1965]) by Crispin Wright - Frege's Concept of Numbers as Objects 3.xv A reaction: He cites Dedekind (and hence the Peano Axioms) as the source of this. The interest is that progression seems to be fundamental to ordianls, but this claims it is also fundamental to cardinals. Note that in the first instance they are finite.
17904 A set has k members if it one-one corresponds with the numbers less than or equal to k
Full Idea: Any set has k members if and only if it can be put into one-to-one correspondence with the set of numbers less than or equal to k. From: Paul Benacerraf (What Numbers Could Not Be [1965], I) A reaction: This is 'Ernie's' view of things in the paper. This defines the finite cardinal numbers in terms of the finite ordinal numbers. He has already said that the set of numbers is well-ordered.
17906 To explain numbers you must also explain cardinality, the counting of things
Full Idea: I would disagree with Quine. The explanation of cardinality - i.e. of the use of numbers for 'transitive counting', as I have called it - is part and parcel of the explication of number. From: Paul Benacerraf (What Numbers Could Not Be [1965], I n2) A reaction: Quine says numbers are just a progression, with transitive counting as a bonus. Interesting that Benacerraf identifies cardinality with transitive counting. I would have thought it was the possession of numerical quantity, not ascertaining it.
###### 6. Mathematics / A. Nature of Mathematics / 4. Using Numbers / c. Counting procedure
9898 We can count intransitively (reciting numbers) without understanding transitive counting of items
Full Idea: Learning number words in the right order is counting 'intransitively'; using them as measures of sets is counting 'transitively'. ..It seems possible for someone to learn the former without learning the latter. From: Paul Benacerraf (What Numbers Could Not Be [1965], I) A reaction: Scruton's nice question (Idea 3907) is whether you could be said to understand numbers if you could only count intransitively. I would have thought such a state contained no understanding at all of numbers. Benacerraf agrees.
17903 Someone can recite numbers but not know how to count things; but not vice versa
Full Idea: It seems that it is possible for someone to learn to count intransitively without learning to count transitively. But not vice versa. From: Paul Benacerraf (What Numbers Could Not Be [1965], I) A reaction: Benacerraf favours the priority of the ordinals. It is doubtful whether you have grasped cardinality properly if you don't know how to count things. Could I understand 'he has 27 sheep', without understanding the system of natural numbers?
###### 6. Mathematics / A. Nature of Mathematics / 4. Using Numbers / g. Applying mathematics
9897 The application of a system of numbers is counting and measurement
Full Idea: The application of a system of numbers is counting and measurement. From: Paul Benacerraf (What Numbers Could Not Be [1965], I) A reaction: A simple point, but it needs spelling out. Counting seems prior, in experience if not in logic. Measuring is a luxury you find you can indulge in (by imagining your quantity) split into parts, once you have mastered counting.
###### 6. Mathematics / B. Foundations for Mathematics / 4. Axioms for Number / a. Axioms for numbers
9900 For Zermelo 3 belongs to 17, but for Von Neumann it does not
Full Idea: Ernie's number progression is [φ],[φ,[φ]],[φ,[φ],[φ,[φ,[φ]]],..., whereas Johnny's is [φ],[[φ]],[[[φ]]],... For Ernie 3 belongs to 17, not for Johnny. For Ernie 17 has 17 members; for Johnny it has one. From: Paul Benacerraf (What Numbers Could Not Be [1965], II) A reaction: Benacerraf's point is that there is no proof-theoretic way to choose between them, though I am willing to offer my intuition that Ernie (Zermelo) gives the right account. Seventeen pebbles 'contains' three pebbles; you must pass 3 to count to 17.
9899 The successor of x is either x and all its members, or just the unit set of x
Full Idea: For Ernie, the successor of a number x was the set consisting of x and all the members of x, while for Johnny the successor of x was simply [x], the unit set of x - the set whose only member is x. From: Paul Benacerraf (What Numbers Could Not Be [1965], II) A reaction: See also Idea 9900. Benacerraf's famous point is that it doesn't seem to make any difference to arithmetic which version of set theory you choose as its basis. I take this to conclusively refute the idea that numbers ARE sets.
###### 6. Mathematics / B. Foundations for Mathematics / 6. Mathematics as Set Theory / b. Mathematics is not set theory
8697 Disputes about mathematical objects seem irrelevant, and mathematicians cannot resolve them
Full Idea: If two children were brought up knowing two different set theories, they could entirely agree on how to do arithmetic, up to the point where they discuss ontology. There is no mathematical way to tell which is the true representation of numbers. From: report of Paul Benacerraf (What Numbers Could Not Be [1965]) by Michèle Friend - Introducing the Philosophy of Mathematics A reaction: Benacerraf ends by proposing a structuralist approach. If mathematics is consistent with conflicting set theories, then those theories are not shedding light on mathematics.
8304 No particular pair of sets can tell us what 'two' is, just by one-to-one correlation
Full Idea: Hume's Principle can't tell us what a cardinal number is (this is one lesson of Benacerraf's well-known problem). An infinity of pairs of sets could actually be the number two (not just the simplest sets). From: report of Paul Benacerraf (What Numbers Could Not Be [1965]) by E.J. Lowe - The Possibility of Metaphysics 10.3 A reaction: The drift here is for numbers to end up as being basic, axiomatic, indefinable, universal entities. Since I favour patterns as the basis of numbers, I think the basis might be in a pre-verbal experience, which even a bird might have, viewing its eggs.
9906 If ordinal numbers are 'reducible to' some set-theory, then which is which?
Full Idea: If a particular set-theory is in a strong sense 'reducible to' the theory of ordinal numbers... then we can still ask, but which is really which? From: Paul Benacerraf (What Numbers Could Not Be [1965], IIIB) A reaction: A nice question about all reductions. If we reduce mind to brain, does that mean that brain is really just mind. To have a direction (up/down?), reduction must lead to explanation in a single direction only. Do numbers explain sets?
###### 6. Mathematics / B. Foundations for Mathematics / 7. Mathematical Structuralism / a. Structuralism
9907 If any recursive sequence will explain ordinals, then it seems to be the structure which matters
Full Idea: If any recursive sequence whatever would do to explain ordinal numbers suggests that what is important is not the individuality of each element, but the structure which they jointly exhibit. From: Paul Benacerraf (What Numbers Could Not Be [1965], IIIC) A reaction: This sentence launched the whole modern theory of Structuralism in mathematics. It is hard to see what properties a number-as-object could have which would entail its place in an ordinal sequence.
9908 The job is done by the whole system of numbers, so numbers are not objects
Full Idea: 'Objects' do not do the job of numbers singly; the whole system performs the job or nothing does. I therefore argue that numbers could not be objects at all. From: Paul Benacerraf (What Numbers Could Not Be [1965], IIIC) A reaction: This thought is explored by structuralism - though it is a moot point where mere 'nodes' in a system (perhaps filled with old bits of furniture) will do the job either. No one ever explains the 'power' of numbers (felt when you do a sudoku). Causal?
9909 The number 3 defines the role of being third in a progression
Full Idea: Any object can play the role of 3; that is, any object can be the third element in some progression. What is peculiar to 3 is that it defines that role, not by being a paradigm, but by representing the relation of any third member of a progression. From: Paul Benacerraf (What Numbers Could Not Be [1965], IIIC) A reaction: An interesting early attempt to spell out the structuralist idea. I'm thinking that the role is spelled out by the intersection of patterns which involve threes.
9911 Number words no more have referents than do the parts of a ruler
Full Idea: Questions of the identification of the referents of number words should be dismissed as misguided in just the way that a question about the referents of the parts of a ruler would be seen as misguided. From: Paul Benacerraf (What Numbers Could Not Be [1965], IIIC) A reaction: What a very nice simple point. It would be very strange to insist that every single part of the continuum of a ruler should be regarded as an 'object'.
8925 Mathematical objects only have properties relating them to other 'elements' of the same structure
Full Idea: Mathematical objects have no properties other than those relating them to other 'elements' of the same structure. From: Paul Benacerraf (What Numbers Could Not Be [1965], p.285), quoted by Fraser MacBride - Structuralism Reconsidered §3 n13 A reaction: Suppose we only had one number - 13 - and we all cried with joy when we recognised it in a group of objects. Would that be a number, or just a pattern, or something hovering between the two?
9938 How can numbers be objects if order is their only property?
Full Idea: Benacerraf raises the question how numbers can be 'objects' if they have no properties except order in a particular ω-sequence. From: report of Paul Benacerraf (What Numbers Could Not Be [1965], p.301) by Hilary Putnam - Mathematics without Foundations A reaction: Frege certainly didn't think that order was their only property (see his 'borehole' metaphor in Grundlagen). It might be better to say that they are objects which only have relational properties.
###### 6. Mathematics / C. Sources of Mathematics / 1. Mathematical Platonism / b. Against mathematical platonism
9910 Number-as-objects works wholesale, but fails utterly object by object
Full Idea: The identification of numbers with objects works wholesale but fails utterly object by object. From: Paul Benacerraf (What Numbers Could Not Be [1965], IIIC) A reaction: This seems to be a glaring problem for platonists. You can stare at 1728 till you are blue in the face, but it only begins to have any properties at all once you examine its place in the system. This is unusual behaviour for an object.
###### 6. Mathematics / C. Sources of Mathematics / 5. Numbers as Adjectival
9903 Number words are not predicates, as they function very differently from adjectives
Full Idea: The unpredicative nature of number words can be seen by noting how different they are from, say, ordinary adjectives, which do function as predicates. From: Paul Benacerraf (What Numbers Could Not Be [1965], II) A reaction: He points out that 'x is seventeen' is a rare construction in English, unlike 'x is happy/green/interesting', and that numbers outrank all other adjectives (having to appear first in any string of them).
###### 6. Mathematics / C. Sources of Mathematics / 6. Logicism / d. Logicism critique
9904 The set-theory paradoxes mean that 17 can't be the class of all classes with 17 members
Full Idea: In no consistent theory is there a class of all classes with seventeen members. The existence of the paradoxes is a good reason to deny to 'seventeen' this univocal role of designating the class of all classes with seventeen members. From: Paul Benacerraf (What Numbers Could Not Be [1965], II) A reaction: This was Frege's disaster, and seems to block any attempt to achieve logicism by translating numbers into sets. It now seems unclear whether set theory is logic, or mathematics, or sui generis.
###### 9. Objects / F. Identity among Objects / 6. Identity between Objects
9905 Identity statements make sense only if there are possible individuating conditions
Full Idea: Identity statements make sense only in contexts where there exist possible individuating conditions. From: Paul Benacerraf (What Numbers Could Not Be [1965], III) A reaction: He is objecting to bizarre identifications involving numbers. An identity statement may be bizarre even if we can clearly individuate the two candidates. Winston Churchill is a Mars Bar. Identifying George Orwell with Eric Blair doesn't need a 'respect'. | 3,505 | 14,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.896581 |
https://quickbooks.intuit.com/learn-support/en-us/banking/the-loan-manager-interest-calculation-for-365-360-compute-period/00/214602 | 1,591,296,463,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00272.warc.gz | 489,679,027 | 102,753 | cancel
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# The loan manager interest calculation for 365/360 compute period is not correct?
I tested the loan manager with a loan: The borrowed amount is \$10000.00, term is 6 months, origination date is Dec/1/2018, due date of next payment is Jan/1/2019, payment amount is \$2000.00, payment period is monthly, interest rate is 36%, compounding period is Exact Days, computing period is 365/360. The interest amount that loan manager shows for the first payment is \$304.17. I don't think this is correct. My calculation is like this: the daily interest rate = 36%/360 = 0.001, so the interest for the first payment should be 0.001 * 10000.00 * 31 (days) = \$310.00. Am I correct?
Thanks.
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## The loan manager interest calculation for 365/360 compute period is not correct?
Using http://www.amortization-schedule.info/calculator I get a monthly payment of \$1845.98 and first month interest is \$300 (36% of \$10,000 divided by 12 is exactly \$300) A payment of \$2000/month will retire the loan with a final 6th payment much less than the \$1846
Loan Manager rarely will match exactly what an actual bank calculates that you owe so in the end always refer to statement instead.
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## The loan manager interest calculation for 365/360 compute period is not correct?
http://www.amortization-schedule.info/calculator calculates interest for Monthly compounding period. What I want is interest calculation for Exact Days compounding period and 365/360 compute period. Can anyone show me this interest calculation?
Thanks.
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https://mathimages.swarthmore.edu/index.php?title=Rope_around_the_Earth&diff=prev&oldid=25325 | 1,669,959,979,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710898.93/warc/CC-MAIN-20221202050510-20221202080510-00539.warc.gz | 411,768,581 | 19,480 | # Difference between revisions of "Rope around the Earth"
{{Image Description |ImageName=Rope around the Earth |Image=Rope around earth 3.jpg |ImageIntro=This is a puzzle about by how much a rope tied taut around the equator must be lengthened so that there is a one foot gap at all points between the rope and the Earth if the rope is made to hover. Although finding the answer requires only basic geometry, even professional mathematicians find the answer strangely counter-intuitive. There is a related problem about stretching the rope taut again where the answer is even more surprising. A question similar to this appeared in William Whiston's The Elements of Euclid circa 1702. |ImageDescElem=
In this puzzle, we treat the Earth as though it were a prefect sphere, even though it actually bulges toward the equator. Suppose a rope was tied taut around the Earth's equator. It would have the same circumference as the Earth (24,901.55 miles). The question is: by how much would the rope have to be lengthened so that, if made to hover, it would be one foot off the ground at all points around the Earth?
Image 1[1]. The lengthened rope lifted by a point. Image not to scale.
Despite the enormous size of the Earth, and the 1 foot gap around the entire circumference, the rope would have to be lengthened by a mere 2π feet, or roughly 6.28 feet.
In fact, 2π feet the answer regardless of the size of the ball around which the rope is wrapped.
Just as bizarre is what happens when one point on the lengthened rope is lifted up so that the rope is taut again, as in Image 1. The maximum clearance under the rope proves to be quite large. For the specific case of a rope looped around the Earth, a 2π foot extension would provide 614.771 feet of clearance if the rope were lifted. This is enough room to fit two Statues of Liberty under it, base and all. Unlike the previous question, however, this result is dependent on the size of the ball.
|ImageDesc=The circumference of a circle is given by the equation: $C=2\pi\,\!r$, where r is the radius.
Image 2. Image not to scale.
In Image 2:
• Lrope 2 is the length of the extended rope.
• Cearth is the Circumference of the Earth and the original length of the rope (Lrope 1).
• Rrope 2 is the radius of the circle made by the extended rope.
• Rearth is the radius of the Earth and the original radius of the rope (Rrope 1).
When the rope is taut around the globe, its length equals the circumference of the Earth.
Eq. 1 $L_{rope 1}=C_{earth}=2\pi\,\!R_{earth}$
The puzzle states that we have to lengthen the rope and made the rope hover 1 foot of the surface of the earth. Lengthening the rope so that it is 1 foot off the ground at all points simply means changing the radius of the circle it forms from: Rrope 1= Rearth to Rrope 2= Rearth+1 ft.
Thus:
Eq. 2 $L_{rope 2}=2\pi\,\!(R_{earth}+1)$
Distributing the 2 π yields:
Eq. 3 $L_{rope 2}=2\pi\,\!R_{earth}+2\pi\,\!=C_{earth}+2\pi\,\!$
The new length of the rope is merely 2 π feet longer than the original length. Indeed, one can see that the additional 2π is a result of distributing the 2π to the 1 in Eq. 2, which yields an increase of 2π no matter what the radius of the ball is. Also, because the value of Rearth was not any number in particular throughout this proof, the answer didn't depend on the specific radius of the Earth in any way. Hence, this 2π extension would be the same for a ball, planet, or star of any size.
It should be noted that the Earth is not a perfect sphere. Not only does it bulge out around the equator, but mountains and valleys give it a rough surface. This roughness affects the amount of slack needed, as might be expected. Interestingly, however, it is not the presence of mountains that affects the amount of extra slack needed, but rather, their shape.
{{HideShowThis|ShowMessage=Click to view more on mountains and valleys.|HideMessage=Click to hide more on mountains and valleys.|HiddenText=
[[Image:Image:Convex-concave.gif|thumb|300 px|Image A. A convex polygon and a concave polygon of the same type.]
The steep mountains and valleys cause the earth to have a concave cross-section, rather than a convex one. If the Earth had a convex cross-section, then the amount of slack needed to raise the rope one foot above the ground would still be 2π feet, regardless of how many sides it has in cross-section.
[[Image:Square-proof.gif|thumb|300 px|left|Image A. B square with a blue rope 1 unit away from it at all points. Notice that the four quarter circles together require adding 2π units of slack to the rope's length.]
The third image shows a rope around a square of side length 1. The rope is 1 unit away from the square at all points. The straight portion of each side of the larger figure is 1 unit long. Each "corner" of the larger figure is a quarter circle with radius = 1. thus the perimeter of the larger figure as a whole is 1+1+1+1 + 2π(1)/4 + 2π(1)/4 + 2π(1)/4 + 2π(1)/4 = 4 + 2π, while that of the small square is 1+1+1+1 = 4. If you subtract the perimeter of the larger figure from that of the smaller, you find a difference of (4 + 2π) - 4 = 2π units of slack.
We will prove that this is the case for any convex shape.
We start with the equation for the measure of each angle of a . n is the number of sides the polygon has, and θ is the measurement of an .
Eq. A $/theta= /frac{/left (n-2 /right ) 180^o}{n}$
The derivation of this is based on triangles, and can be seen [1].
### Maximum Height of Rope
Note: A familiarity with trigonometry, series, and approximations is recommended for this section.
Were the lengthened rope again to be held taut by raising it at an arbitrary point (as shown in Image 3), what would be the distance from this point to the surface of the earth?
Image 3[2]. Image not to scale.
In Image 3:
• X1 is the distance between where the rope leaves the Earth's surface and the highest point on the taut rope.
• Xo is the ground distance along the curved surface of the Earth from the point where the rope leaves the globe, to the point below the apex of the rope.
• R is the radius of the globe.
• h is the height of the apex of the rope above the ground.
• θ is the angle formed between the line from the center of the Earth to the apex and the line from the center of the Earth to the point where the rope leaves the ground.
In the puzzle the slack added to the rope is 2π feet. I will be using l to represent the this slack because, in a different version of the puzzle, l can be different. Now we will develop a formula for l, the amount of slack, in terms of h, the maximum clearance the slack can provide. We start with h because it appears clearly in the diagram, where l is less visible. We hope later to invert this in-order to solve for h in terms of l. First, we will do this for the case where l=2 π.
Notice that triangle ABD is a right triangle, with (R + h) as its hypotenuse. Thus, using , we know that:
$(R+h)^2=R^2+X_1^2$
Which is equivalent to:
Eq. 4 $X_1=\sqrt{(R+h)^2-R^2}$
To find the length of Xo, we must remember what the length of an arc (Larc) is:
Eq. 5 $L_{arc}=r\theta\,\!$
Where r is the radius of the circle, and θ is the angle, in radians, formed between two radii from the center of the circle to the endpoints of the arc. For more on Arcs, see Arcs.
In the image to the right, Xo is the arc whose length we want to find; hence, it can be substituted in for Larc in Eq. 5. Also, since cos-1$\ \left (\tfrac{R}{R+h} \right )$ represents θ, the angle whose cosine is $\tfrac{R}{R+h}$, we can replace θ with cos-1$\ \left (\tfrac{R}{R+h} \right )$. For more information on cos-1 see Inverse Trig Functions.
Thus, Eq. 5 is equivalent to:
Eq. 6 $X_o=R\,\cos^{-1} \left (\frac{R}{R+h} \right )$.
Now that we have a formula for Xo, were are going to relate Xo to X1 . Because the rope is taut around most of the earth, all the slack goes to lengthening arc AC and turning it into segments AB and BC. Since we know that we lengthened the rope by 2π feet, we know that 2X1= 2Xo + 2π, because 2X1 - 2Xo is the extra slack put in to the rope. Thus, more generally:
Eq. 7 $2X_1=2X_o+ \mathit{l}$
Solving for l in Eq. 7 yields:
Eq. 8 $\mathit{l}=2 \left (X_1-X_o \right )$
Substituting Eq. 4 and Eq. 6 into Eq. 8 yields:
Eq. 9 $\mathit{l}=2 \left ( \sqrt {(R+h)^2-R^2} -R\,\cos^{-1} \left (\frac{R}{R+h} \right ) \right )$
Because h is both in the of the cos-1 and elsewhere in the problem, we can't isolate it and solve for it explicitly. Nevertheless, running Eq. 9 through a with l = 2 π and R = R earth = 20925524.9 feet one finds h ≈ 614.771 ft as the height that a 2π foot extension yields.
Eq. 9 can, however, be used to find the amount of slack needed to produce a certain height. For instance, if we set h = 35,000 feet (the average cruising altitude for commercial aircraft), Eq. 9 tells us that l = 2697.06 feet. Thus, Roughly 3,000 feet of slack must be added to the rope to that, when lifted, a plane could fly under it on its normal flight path.
If the puzzle were different, having a different amount of slack added to the rope, we could still find h if given a value for l. We can create a general formula that will find h for any value of l, in addition to when l = 2π. As mentioned above, due to the cos-1 (R / (R + h)) there is no explicit formula to find the height. Nevertheless, using series, on can obtain very good approximations. Using series approximations one can derive the following formulas for height achieved by lengthening the rope by length l; the first is a 1st order approximation, and the second, a more accurate 2nd order approximation. The approximations are more accurate the smaller the added slack is compared to the original radius of the circle.
Image 4. The series approximations for a large value of R relative to l. Image created by Harrison Tasoff
Eq. 10 $h \left (\mathit{l} \right ) \approx \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )}R^{ \left ( \frac{1}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )}$
Eq. 11 $h \left (\mathit{l} \right ) \approx \frac{1}{2} \left ( \frac{3}{2} \right )^{\left ( \frac{2}{3} \right )}R^{ \left ( \frac{1}{3} \right )} \mathit{l}^{ \left ( \frac{2}{3} \right )} + \cfrac {9 \left ( \frac{3}{2} \right )^{\left ( \frac{1}{3} \right )}}{80 R^{ \left (\frac{1}{3} \right )}}\ \mathit{l}^{\left ( \frac{4}{3} \right )}$
Where l is the added slack, h(l) is the height as a function of l, and R is the initial radius of the rope circle.
Taking l = 2π feet of slack and R = R earth = 20925524.9 feet from the initial problem the first approximation yields 614.766, demonstrating the accuracy of these approximations. The second approximation yields 614.771, which is even more more accurate than the first.
Image 4 illustrates these approximations, with l on the x-axis and h(l) on the y-axis. Eq. 10 appears in red and Eq. 11 in blue. In this example, the radius R of the ball is an arbitrary number that is very large relative to l. Notice that for small values of l, Eq. 10 and Eq. 11 are indistinguishable. It is only once l starts to become a decent percentage of R that Eq. 11 becomes noticeably more accurate than Eq. 10. When R= 5000, as it does in this graph, Eq. 11 becomes appreciably more accurate than Eq. 10 when l ≈ 150, or roughly 3% of R. At this point, Eq. 10 outputs 30827.651, while the more accurate Eq. 11 outputs 30833.653, a difference of 6.002.
Here is another way to illustrate the varying degrees of accuracy in these approximations. The most accurate value the root finder yields is 614.7709915007723. Comparing this to the value found by the 1st order approximation (614.7655785603816), we find that the latter is accurate to five significant digits when rounding is accounted for. The 2nd order approximation (614.7709968720138) is accurate to seven significant figures when rounding is accounted for. Hence, though in absolute terms, both approximations are quite accurate, the 2nd order approximation is 100 times more accurate relative to the 1st order approximation.
Also, as one can see the approximations climb quickly from l = 0. This steep ascent means that even small values of l output large values for h(l), or, even small amounts of extra slack yield large increases in maximum height.
On the scale of the Earth, there is a decent margin of error in measurements: the earth is not a perfect sphere, nor perfectly smooth, ropes stretch with strain, objects expand and contract with heat. The result is that these approximations are far more accurate and precise than our measurements can ever be.
|other=High-school algebra and High-school geometry |AuthorName=Harrison Tasoff |Field=Geometry |WhyInteresting=At first this puzzle is interesting because of its non-intuitive results. It becomes even more interesting if we can then make them intuitive after all, and indeed and we can. Though it may seem that 2π, or roughly 6.28, feet is a minuscule amount of extra rope needed to to produce such a considerable result, a look at the ratios will show otherwise.
The radius of the Earth is roughly 20,920,000 feet, though this varies because the Earth is not a perfect sphere. There is 1 foot of difference between the radius of the circle made by the lengthened rope and the radius of the Earth. This foot of difference is a mere fraction of the radius of the Earth: about five one-hundred millionths, or .000000047, of the Earth's radius. A foot doesn't seem so large anymore.
Similarly, 2 π feet is .000000047 of the circumference of the Earth (which is about 131,000,000 feet). And, unsurprisingly, the ratio of 1 foot to the Earth's radius is the same as that of 2 π feet to the Earth's circumference.
So, in this perspective, a small change in the length of the rope yields a proportionally equivalent small change in the radius of the rope circle.
|References=
• Pickover, C. A. (2009). The Math Book. New York: Sterling Publishing Co.
• (2009, March 3). Roping the Earth. Message posted to: | 3,768 | 14,125 | {"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": 18, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.961055 |
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Simple Compound Interest Calculator
Compound Interest and Why it’s Important?
If you are someone who is planning to borrow money or take loans, you must know about Compound Interest. Read the article to know what compound interest is and how it works.
When money is borrowed or lent, a certain amount is charged as interest. This interest depends upon the amount, the time period for which the money was borrowed, as well as the rate of interest. The interest that is charged can be of two types: simple interest and compound interest.
It is the one that is usually used in bank loans and credit cards. This type of interest does not only depend upon the principle, time period, and rate of interest but also on the accumulated interest that the borrower has already paid. In this article, we will read about compound interest, its importance, and its calculation.
Table of Contents
Compound Interest
When you deposit some money and the interest you get on it depends on the principal amount and accumulated interest from previous months.
There are numerous additional alternative intervals that can be used in addition to the frequently used quarterly, monthly, and daily compounding periods.
If you need help calculating the interest you can take help from the online compound interest calculator provided by Vakilsearch. This online calculator will save you time and calculate the interest you have to pay on the borrowed money.
Discover the seamless power of online bookkeeping – empowering your business with real-time insights and financial clarity, all at your fingertips.
Simple Interest vs. Compound Interest
Compound interest operates differently than simple interest. Only the principal is used to compute simple interest. When computing simple interest, earned interest is not compounded or invested back into the principle.
If you were to calculate your annual income using simple interest, you would receive ₹50 if your account balance was ₹1,000 and earned 5% annual interest. The accrued interest would not be deducted from the original principle. You’d receive an extra ₹50 in year two.
On auto loans and other types of shorter-term consumer loans, simple interest is usually employed to compute the interest charged. Credit card debt accrues interest over time, which is why it seems as though it might grow so fast and significantly.
In a perfect world, compound interest can be used to compute your savings and investments while simple interest can only be used to calculate your debts.
Importance of Compound Interest
It is an important term when it comes to financial management.
• As compound interest is used in banks the amount to deposit in your savings account benefits from it. If you start investing money at a younger age in the savings account it is more beneficial as the compound interest will keep on increasing which will give you more benefits in the long run.
• Compound interest is also used by banks when you take loans. The interest that you pay on the loan taken is compound interest. In this case, if it is not working in your favor as you have to pay interest on the amount you took as a loan as well as the interest that you have already paid.
• When you use a credit card and the amount is due on it banks take a certain amount of interest. This interest is compound interest. Therefore if you do not pair credit card news soon the interest will keep on increasing depending on the time period the dues are left unpaid.
Calculation of Compound Interest
To Calculate the amount you can use the following formula,
A= P(1 + r/n)nt
Where,
A = the final amount;
P = the principal
r= the rate of interest (in decimal)
n= the compounding frequency;
t = time period in years.
• The starting balance used to calculate interest is referred to as the “principal.” The phrase can also refer to your first investment amount, albeit it is more frequently used to refer to a loan’s starting balance. For instance, if you choose to invest ₹10,000 for a period of five years, that sum will serve as your principal for computing compound interest.
• Rate is the decimal representation of the interest rate (or anticipated rate of return on investment). For purposes of calculation, you would enter 0.07 if you anticipated a 7% annual growth rate for your investments. The frequency at which interest is compounded on the principal is referred to as compounding frequency.
• Using the 7% interest rate as an example, you would simply add 7% to the principal once a year if annual compounding were used. On the other hand, semi-annual compounding entails applying half of that sum (3.5%) twice a year. Quarterly (four times a year), monthly, weekly, and daily are some additional popular compounding frequencies.
• Additionally, there is a mathematical idea known as continuous compounding, where interest is continuously collected.
• Time is a fairly self-explanatory notion, but remember to describe the entire amount of time in years in order to calculate compound interest. In other words, use 2.5 years in the formula if you plan to invest for 30 months.
Vakilsearch can make this calculation process easier for you. It has an online Compound interest calculator in which you can enter the principal amount, the time period for which it was borrowed, and the rate of interest at which it was borrowed. The calculator will use this information and give you the compound interest on it.
Conclusion
One of the most important benefits that is it grows at a faster rate than simple interest. Simple interest gives fixed amounts of money as the interest on the principal amount whereas compound interest is based on the principal amount as well as the accumulated interest of each month. Because of the faster rate of growth of compound interest, it is usually used for bank credit card and loan purposes.
If you want help in calculating interest, You can go to the website of Vakilsearch. It has an online compound interest calculator that helps you in calculating the Compound Interest you have to pay on the amount you borrowed. Go to the website of Vakilsearch and enter the basic information about the money you borrowed such as the principal amount, interest rate, and time period. Once you enter these details, the calculator will give you the interest to be paid on it.
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Remove Adblocker Extension | 1,295 | 6,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-38 | latest | en | 0.955348 |
https://www.aqua-calc.com/one-to-one/density/troy-ounce-per-metric-tablespoon/long-ton-per-cubic-foot/1 | 1,611,729,747,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704821253.82/warc/CC-MAIN-20210127055122-20210127085122-00645.warc.gz | 657,253,296 | 8,875 | # 1 troy ounce per metric tablespoon in long tons per cubic foot
## troy ounces/metric tablespoon to long ton/foot³ unit converter
1 troy ounce per metric tablespoon [oz t/metric tbsp] = 0.06 long ton per cubic foot [long tn/ft³]
### troy ounces per metric tablespoon to long tons per cubic foot density conversion cards
• 1
through
25
troy ounces per metric tablespoon
• 1 oz t/metric tbsp to long tn/ft³ = 0.06 long tn/ft³
• 2 oz t/metric tbsp to long tn/ft³ = 0.12 long tn/ft³
• 3 oz t/metric tbsp to long tn/ft³ = 0.17 long tn/ft³
• 4 oz t/metric tbsp to long tn/ft³ = 0.23 long tn/ft³
• 5 oz t/metric tbsp to long tn/ft³ = 0.29 long tn/ft³
• 6 oz t/metric tbsp to long tn/ft³ = 0.35 long tn/ft³
• 7 oz t/metric tbsp to long tn/ft³ = 0.4 long tn/ft³
• 8 oz t/metric tbsp to long tn/ft³ = 0.46 long tn/ft³
• 9 oz t/metric tbsp to long tn/ft³ = 0.52 long tn/ft³
• 10 oz t/metric tbsp to long tn/ft³ = 0.58 long tn/ft³
• 11 oz t/metric tbsp to long tn/ft³ = 0.64 long tn/ft³
• 12 oz t/metric tbsp to long tn/ft³ = 0.69 long tn/ft³
• 13 oz t/metric tbsp to long tn/ft³ = 0.75 long tn/ft³
• 14 oz t/metric tbsp to long tn/ft³ = 0.81 long tn/ft³
• 15 oz t/metric tbsp to long tn/ft³ = 0.87 long tn/ft³
• 16 oz t/metric tbsp to long tn/ft³ = 0.92 long tn/ft³
• 17 oz t/metric tbsp to long tn/ft³ = 0.98 long tn/ft³
• 18 oz t/metric tbsp to long tn/ft³ = 1.04 long tn/ft³
• 19 oz t/metric tbsp to long tn/ft³ = 1.1 long tn/ft³
• 20 oz t/metric tbsp to long tn/ft³ = 1.16 long tn/ft³
• 21 oz t/metric tbsp to long tn/ft³ = 1.21 long tn/ft³
• 22 oz t/metric tbsp to long tn/ft³ = 1.27 long tn/ft³
• 23 oz t/metric tbsp to long tn/ft³ = 1.33 long tn/ft³
• 24 oz t/metric tbsp to long tn/ft³ = 1.39 long tn/ft³
• 25 oz t/metric tbsp to long tn/ft³ = 1.44 long tn/ft³
• 26
through
50
troy ounces per metric tablespoon
• 26 oz t/metric tbsp to long tn/ft³ = 1.5 long tn/ft³
• 27 oz t/metric tbsp to long tn/ft³ = 1.56 long tn/ft³
• 28 oz t/metric tbsp to long tn/ft³ = 1.62 long tn/ft³
• 29 oz t/metric tbsp to long tn/ft³ = 1.68 long tn/ft³
• 30 oz t/metric tbsp to long tn/ft³ = 1.73 long tn/ft³
• 31 oz t/metric tbsp to long tn/ft³ = 1.79 long tn/ft³
• 32 oz t/metric tbsp to long tn/ft³ = 1.85 long tn/ft³
• 33 oz t/metric tbsp to long tn/ft³ = 1.91 long tn/ft³
• 34 oz t/metric tbsp to long tn/ft³ = 1.96 long tn/ft³
• 35 oz t/metric tbsp to long tn/ft³ = 2.02 long tn/ft³
• 36 oz t/metric tbsp to long tn/ft³ = 2.08 long tn/ft³
• 37 oz t/metric tbsp to long tn/ft³ = 2.14 long tn/ft³
• 38 oz t/metric tbsp to long tn/ft³ = 2.2 long tn/ft³
• 39 oz t/metric tbsp to long tn/ft³ = 2.25 long tn/ft³
• 40 oz t/metric tbsp to long tn/ft³ = 2.31 long tn/ft³
• 41 oz t/metric tbsp to long tn/ft³ = 2.37 long tn/ft³
• 42 oz t/metric tbsp to long tn/ft³ = 2.43 long tn/ft³
• 43 oz t/metric tbsp to long tn/ft³ = 2.48 long tn/ft³
• 44 oz t/metric tbsp to long tn/ft³ = 2.54 long tn/ft³
• 45 oz t/metric tbsp to long tn/ft³ = 2.6 long tn/ft³
• 46 oz t/metric tbsp to long tn/ft³ = 2.66 long tn/ft³
• 47 oz t/metric tbsp to long tn/ft³ = 2.72 long tn/ft³
• 48 oz t/metric tbsp to long tn/ft³ = 2.77 long tn/ft³
• 49 oz t/metric tbsp to long tn/ft³ = 2.83 long tn/ft³
• 50 oz t/metric tbsp to long tn/ft³ = 2.89 long tn/ft³
• 51
through
75
troy ounces per metric tablespoon
• 51 oz t/metric tbsp to long tn/ft³ = 2.95 long tn/ft³
• 52 oz t/metric tbsp to long tn/ft³ = 3.01 long tn/ft³
• 53 oz t/metric tbsp to long tn/ft³ = 3.06 long tn/ft³
• 54 oz t/metric tbsp to long tn/ft³ = 3.12 long tn/ft³
• 55 oz t/metric tbsp to long tn/ft³ = 3.18 long tn/ft³
• 56 oz t/metric tbsp to long tn/ft³ = 3.24 long tn/ft³
• 57 oz t/metric tbsp to long tn/ft³ = 3.29 long tn/ft³
• 58 oz t/metric tbsp to long tn/ft³ = 3.35 long tn/ft³
• 59 oz t/metric tbsp to long tn/ft³ = 3.41 long tn/ft³
• 60 oz t/metric tbsp to long tn/ft³ = 3.47 long tn/ft³
• 61 oz t/metric tbsp to long tn/ft³ = 3.53 long tn/ft³
• 62 oz t/metric tbsp to long tn/ft³ = 3.58 long tn/ft³
• 63 oz t/metric tbsp to long tn/ft³ = 3.64 long tn/ft³
• 64 oz t/metric tbsp to long tn/ft³ = 3.7 long tn/ft³
• 65 oz t/metric tbsp to long tn/ft³ = 3.76 long tn/ft³
• 66 oz t/metric tbsp to long tn/ft³ = 3.81 long tn/ft³
• 67 oz t/metric tbsp to long tn/ft³ = 3.87 long tn/ft³
• 68 oz t/metric tbsp to long tn/ft³ = 3.93 long tn/ft³
• 69 oz t/metric tbsp to long tn/ft³ = 3.99 long tn/ft³
• 70 oz t/metric tbsp to long tn/ft³ = 4.05 long tn/ft³
• 71 oz t/metric tbsp to long tn/ft³ = 4.1 long tn/ft³
• 72 oz t/metric tbsp to long tn/ft³ = 4.16 long tn/ft³
• 73 oz t/metric tbsp to long tn/ft³ = 4.22 long tn/ft³
• 74 oz t/metric tbsp to long tn/ft³ = 4.28 long tn/ft³
• 75 oz t/metric tbsp to long tn/ft³ = 4.33 long tn/ft³
• 76
through
100
troy ounces per metric tablespoon
• 76 oz t/metric tbsp to long tn/ft³ = 4.39 long tn/ft³
• 77 oz t/metric tbsp to long tn/ft³ = 4.45 long tn/ft³
• 78 oz t/metric tbsp to long tn/ft³ = 4.51 long tn/ft³
• 79 oz t/metric tbsp to long tn/ft³ = 4.57 long tn/ft³
• 80 oz t/metric tbsp to long tn/ft³ = 4.62 long tn/ft³
• 81 oz t/metric tbsp to long tn/ft³ = 4.68 long tn/ft³
• 82 oz t/metric tbsp to long tn/ft³ = 4.74 long tn/ft³
• 83 oz t/metric tbsp to long tn/ft³ = 4.8 long tn/ft³
• 84 oz t/metric tbsp to long tn/ft³ = 4.85 long tn/ft³
• 85 oz t/metric tbsp to long tn/ft³ = 4.91 long tn/ft³
• 86 oz t/metric tbsp to long tn/ft³ = 4.97 long tn/ft³
• 87 oz t/metric tbsp to long tn/ft³ = 5.03 long tn/ft³
• 88 oz t/metric tbsp to long tn/ft³ = 5.09 long tn/ft³
• 89 oz t/metric tbsp to long tn/ft³ = 5.14 long tn/ft³
• 90 oz t/metric tbsp to long tn/ft³ = 5.2 long tn/ft³
• 91 oz t/metric tbsp to long tn/ft³ = 5.26 long tn/ft³
• 92 oz t/metric tbsp to long tn/ft³ = 5.32 long tn/ft³
• 93 oz t/metric tbsp to long tn/ft³ = 5.37 long tn/ft³
• 94 oz t/metric tbsp to long tn/ft³ = 5.43 long tn/ft³
• 95 oz t/metric tbsp to long tn/ft³ = 5.49 long tn/ft³
• 96 oz t/metric tbsp to long tn/ft³ = 5.55 long tn/ft³
• 97 oz t/metric tbsp to long tn/ft³ = 5.61 long tn/ft³
• 98 oz t/metric tbsp to long tn/ft³ = 5.66 long tn/ft³
• 99 oz t/metric tbsp to long tn/ft³ = 5.72 long tn/ft³
• 100 oz t/metric tbsp to long tn/ft³ = 5.78 long tn/ft³
#### Foods, Nutrients and Calories
ROASTED and SALTED PEANUTS THE AMERICAN TRAIL MIX, ROASTED and SALTED PEANUTS, UPC: 077034012378 contain(s) 571 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Flora Max, Midnight weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 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 ]
Aerial acid [CH2O3] weighs 2 540 kg/m³ (158.56702 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-414B, liquid (R414B) with temperature in the range of -40°C (-40°F) to 71.12°C (160.016°F)
#### Weights and Measurements
A picoampere is a SI-multiple (see prefix pico) of the electric current unit ampere and equal to equal to 1.0 × 10-12 ampere
Radioactivity is the act of emitting radiation spontaneously by an atomic nucleus that, for some reason, is unstable.
lb/in³ to oz t/US tsp conversion table, lb/in³ to oz t/US tsp unit converter or convert between all units of density measurement.
#### Calculators
Price conversions and cost calculator for materials and substances | 3,010 | 7,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.273898 |
https://www.physicsforums.com/threads/is-this-induction-proof-correct.770258/ | 1,508,592,285,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824775.99/warc/CC-MAIN-20171021114851-20171021134851-00281.warc.gz | 955,884,342 | 17,482 | Is this induction proof correct?
1. Sep 11, 2014
johann1301
1. The problem statement, all variables and given/known data
The sequence {xn} is given by the recurrence relation
xn = cos(xn-1)sin(xn-2) for n ≥ 2
and x0=2 and x1=1,4. Show by induction that 0 ≤ xn ≤ 1 for all integers n ≥ 2.
3. The attempt at a solution
We formulate a statement:
Pn: 0 ≤ xn = cos(xn-1)sin(xn-2) ≤ 1 for n ≥ 2
---------------------------------------------------------------
We assume that Pn is true for n=k, where k ≥ 2:
Ak: 0 ≤ xk = cos(xk-1)sin(xk-2) ≤ 1 for k ≥ 2
We then set n=k+1:
xk+1 = cos(xk)sin(xk-1)
We know from the assumption Ak that:
0 ≤ xk ≤ 1
This implies that:
0,54 ≤ cos(xk) ≤ 1 which again implies 0 < cos(xk) < 1
We then have to show:
0 ≤ sin(xk-1) ≤ 1
We know that sine to any angle is always equal to or less then 1. We therefore have to prove:
0 ≤ sin(xk-1)
Since we assumed that 0 ≤ xk ≤ 1, this must be true for xk-1 as well. This means that if the assumption Ak is true when n=k, its true when n=k+1 also. The last thing is to prove the base case:
x2 = cos(1,4)sin(2) ≈ 0,155
0 ≤ 0,155 ≤ 1
Pn is thus true.
Is this correct argumentation?
2. Sep 11, 2014
Mogarrr
I'm trying to follow along, but how is it that
$0 \leq x_k \leq 1 \Rightarrow 0.54 \leq cos(x_k) \leq 1$?
Also, are you trying to use strong induction or weak induction?
3. Sep 11, 2014
johann1301
Strong induction.
Since the value of xk is always between 0 and 1 radians, the value of cos(xk) has to lie between cos(1)≈0.5403... and cos(0)=1. If you plot cos(x) from x=0 to x=1 you will see that the values of cos(x) lies between 0.5403 and 1.
4. Sep 11, 2014
Fredrik
Staff Emeritus
You shouldn't write out the definition of $x_n$ in the middle of your definition of $P(n)$. Just say that for each integer n such that $n\geq 2$, P(n) is the statement $0\leq x_n\leq 1$. Your goal is to prove P(n) for all integers n such that $n\geq 2$. To do this by induction is to prove the following two statements:
$P(2)$
For all integers n such that $n\geq 2$, if P(n) then P(n+1).
This is poorly worded. It sounds like you're talking about some specific but unspecified integer k that's greater than or equal to 2. So you're saying that one of the infinitely many statements P(2), P(3), etc. is true, and you're not saying which one. I can't interpret the statement as a "for all k" statement either, because then it would be saying that all of the statements P(2), P(3), etc. are true. This is the result that you're trying to prove.
What you need to say here is this: Let n be an arbitrary integer such that n≥2. We will prove that if P(n) then P(n+1). So suppose that P(n) is true.
Note however that this is not sufficient to prove the theorem. You also need to prove P(2). | 898 | 2,775 | {"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} | 4.1875 | 4 | CC-MAIN-2017-43 | longest | en | 0.893707 |
http://www.metafilter.com/tags/Calculator | 1,481,017,596,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541886.85/warc/CC-MAIN-20161202170901-00129-ip-10-31-129-80.ec2.internal.warc.gz | 571,886,431 | 25,137 | ## Paper Calculator
Jason Shiga makes a paper calculator.
Shiga previously, previously, previously.
posted by latkes on Aug 19, 2016 - 14 comments
## The lasting legacy of the "rocket girls" of JPL
California-based Jet Propulsion Laboratory (JPL) has been central to the US missile and rocket development and operations for decades, and from the beginning that technology's success rested on a corps of expert mathematicians, people known as computers. And from the beginning they were all women, in a time when such opportunities were few and far between. You can find pictures of them, but names have not been well-recorded ... until now. Nathalia Holt found many of those women and wrote about their experiences in her book, Rise of the Rocket Girls: The Women Who Propelled Us, from Missiles to the Moon to Mars.
posted by filthy light thief on Jun 15, 2016 - 22 comments
## 19.52 Megatons per Pound
A calculator for how much destructive force would be released by a given weight of antimatter.
posted by Copronymus on May 6, 2016 - 25 comments
## A Calculated Design
“They’re probably the most familiar interfaces on the planet: the numeric keypads on our mobile phones and calculators. Yet very few notice that the keypads’ design has remained unchanged for nearly half a century in the face of evolving global design norms and conventions. Even fewer users notice another startling design feature: the phone’s keypad is the inverted version of the calculator’s.”
Graphic designer C Y Gopinath explains the science and research behind his decision to change the numeric interface layout of his calculator app, Calcuta, from square to circular.
posted by _Mona_ on Jan 13, 2016 - 93 comments
## Geological Time Scale Metaphors
Imagine that the entire history of the earth is (number) (unit) long. . .
posted by curious nu on Aug 25, 2015 - 8 comments
## Piecewise linear functions are magic
Graphing Calculator: Creative Art
posted by Wolfdog on Apr 30, 2014 - 11 comments
## You may find that even your rock-bottom expenses aren’t met
Can you live on the minimum wage? NYTimes.com has provided you with a handy calculator to find out.
posted by roomthreeseventeen on Feb 10, 2014 - 80 comments
## A Month of Kraftwerk from DJ Food, and more from other sources
You're bummed that you're one of many who couldn't get tickets to the eight concerts of eight full albums in eight days at the Museum of Modern Art. Let DJ Food console you with a month of posts dedicated to Kraftwerk, including old and rare pictures and graphics, and six hours of songs that cover, sample, or are inspired by Kraftwerk, and even how to play Kraftwerk songs on your Casio pocket calculator. If you just want to hear Kraftwerk do their thing, here are three (incl. tracklist) live sets (partial tracklist) from recent years (tracklist).
posted by filthy light thief on Mar 20, 2012 - 20 comments
## My cell phone connects me with my friends and also calculates logarithms
Are graphical calculators pointless? Graphical calculators are required by many college-level math courses, but they don't perform as well as mobile phones. Pedagogically, they may be less useful than a slide rule.
posted by twoleftfeet on Apr 12, 2011 - 68 comments
## Will this put those Ask.Me questions to rest?
How do you split the rent when one room is tiny and has no closet while the other is large and has great windows? Based on a not really scientific survey, Jonathan Bittner has come up with a decisive way to answer those 'who should pay how much' questions when it comes to apartment sharing.
posted by jacquilynne on Feb 18, 2011 - 30 comments
## SNOW DAY!
11th grader David Sukhin's Snow Day Calculator uses weather data, user judgment calls, and algorithms of his own devising to predict the chances of a snow day for any school in the US. And, so far, to his knowledge it has never been wrong.
posted by Miko on Jan 28, 2011 - 32 comments
## Browser as Graphing Calculator
An open source, html5 based graphing and computation engine does in your browser what is usually outsourced to the cloud. It graphs, solves, simplifies, integrates and differentiates expressions, and needs no internet connection once you load the page in your browser (or save it on your computer). RTFM.
posted by Obscure Reference on Jan 19, 2011 - 26 comments
## 7734.40
Vintage calculator museums
posted by get off of my cloud on Aug 31, 2010 - 26 comments
## Beer Calculus -- homebrew recipe calculator
Beer Calculus is a freely available homebrewing recipe generator, which allows you to easily create, save and share your own beer recipe(s). The calculator includes hundreds of malt, hop and yeast varieties, adjuncts and other ingredients, different mash processes, and fermentation and storage variables, and can toggle between US and metric units. Also, if you associate your recipe with a BJCP-recognized style, the calculator will give you guidance regarding your recipe's adherence to the style's guidelines. Homebrewers, have at it!
posted by cog_nate on Dec 4, 2009 - 26 comments
## The Market for Lemons, or Thieves Discount the Value You(r Identity)
Calculate the value of your identity on the black market, based on how you access your financial information, your involvement in social and file-sharing networks, and security software installed. Spoiler: it's less than you imagine, as using the data is riskier than stealing the data, and the thieves market is polluted by liars (you can read more in the 12 page Microsoft research PDF).
posted by filthy light thief on Sep 10, 2009 - 34 comments
## Han is leaving the cantina.
Where are you in the movie? If we started a movie on the day you were born, and stretched it over your lifespan, this is where you’d be in that movie.
posted by 40 Watt on May 4, 2009 - 83 comments
## Pocket Calculator Show
Pocket Calculator Show. via: Beware of Blog
posted by serazin on Jul 7, 2008 - 13 comments
## Play with a Curta
I first learned about them when they featured prominently in a Gibson book (Pattern Recognition). When I looked on eBay, I was stunned at the prices they fetch. Now I can at least play with a virtual Curta mechanical calculator.
posted by Dave Faris on May 22, 2008 - 35 comments
## HP Calculator Museum
Real nerds use HP calculators with RPN (Reverse Polish Notation)
posted by Rafaelloello on Apr 25, 2008 - 75 comments
## Best of the web you bet!
Poker hand simulator. Get a feel for the odds before you bet the farm.
posted by Brian B. on Feb 16, 2008 - 30 comments
## A set of useful tools for players of stringed instruments.
JGuitar, a rather useful tool for those learning the guitar or experimenting with alternate tunings. You can even bookmark a certain tuning.
posted by signal on Apr 18, 2007 - 6 comments
Should you buy a house, or rent? (caution, flash & NYT) The answer is, of course, it depends. One of the biggest factors is how well the housing market will do after you buy. [previously: 1, 2] [via]
posted by rubin on Apr 11, 2007 - 104 comments
## mmm, parabolic.
AJAX graphing calculator with a google-earth style interface.
posted by delmoi on Jun 23, 2006 - 25 comments
## I'm Still Alive
Death By Caffeine I just learned that it would take 155.11 cans of Mountain Dew to kill me, according to this odd little service.
posted by jragon on Aug 17, 2005 - 44 comments
## 55378008
Calculator Haiku
posted by Speck on Jul 23, 2005 - 9 comments
## HP-01 calculator watch
"Time: elusive and immediate...limited yet infinite. Because time is important to you, Hewlett-Packard introduces the HP-01, a new dimension in time management and personal computation." Truly, such an important model number could only be bestowed upon the king of all early calculator watches. No less than three batteries were required (two for the LED display alone), and even HP's impressive engineering was unable to save the HP-01 from the curse of bulkiness; it did not sell well at the \$650 price point. The HP-01 was discontinued in 1980, as inexpensive LCD calculator watches began flooding the market (don't lie, you know you had one).
posted by Galvatron on Mar 27, 2005 - 17 comments
## Skunkworks At Apple
The Graphing Calculator Story. Amazing and very amusing article about the conception of a piece of software included with every Macintosh. Made at Apple... by volunteers.
Q: Do you work here? A: No.
Q: You mean you're a contractor? A: Actually, no..
Q: But then who's paying you? A: No one..
Q: How do you live? A: I live simply..
Q: (Incredulously) What are you doing here?!
posted by kika on Dec 22, 2004 - 34 comments
## Fear the Reaper
How long til you buy the farm? • "The Living to 100 Life Expectancy Calculator© was designed to translate what we have learned from studies of centenarians and other longevity research into a practical and empowering tool for individuals to estimate their longevity potential." Wasn't this on a Futurama episode?
posted by dhoyt on Sep 8, 2004 - 29 comments
## Am I Evil?
Coincidence or contortion? Ivan Panin deciphered a numeric code in the Bible. Known as Gematria, the 'code' implies the Bible could not have been written without Holy assistance. Panin offered an open challenge for someone to create text using a similar pattern, yet no one was able to create one(nor tried).
However many people doubt the authenticity of the code though. The code is found in the same verses using different translations. It is also claimed that Panin manufactured his own translations to create this mathematical phenomenon.
Whether or not you believe, you can determine how good or evil any text or website is.
posted by JakeEXTREME on Jun 25, 2004 - 30 comments
## what are your bits worth?
How much is your personal information worth?
personal data toolkit [ via newstoday ]
posted by specialk420 on Feb 5, 2004 - 13 comments
## Math is Hard, Lets Go to the Mall
Look, up in the sky, it's a search engine, it's a phone book, it's....a calculator? Start/Programs/Accessories/awww screw it, I've got Google. Found this by searching for my phone number (with the parens in the wrong place) because a magazine article told my girlfriend to. By the way, my phone number is -7 662.
posted by m@ on Aug 12, 2003 - 54 comments
## here froggy, froggy
The Soulmate Calculator tells you how many men or women you'll have to meet to find your soulmate, based on U.S. census figures and some weird theories. That's a lot of frogs to kiss...(via Salon)
posted by serafinapekkala on May 14, 2003 - 39 comments
## Anita: the world's first electronic desktop calculator
Anita Mk VII the "A New Inspiration To Accounting" OR "A New Inspiration To Arithmetic" was the world's first electronic desktop calculator. Launched in 1961, the Mk VII and Mk VIII were the only commercial calculators available for a period of two years.
posted by riffola on Apr 21, 2003 - 9 comments
## Alternate realities
Alternate realities - as we approach a new year, you may want to measure some things differently - your age or your weight, for example. Some calculations may be flattering, others rather alarming. Even "wasting time" apparently can have some value. No matter what your age, there are some calculators that will help ensure longevity.
## Budget Orgy Calculator
Budget Orgy Calculator - tis the season for festivities, but all those galas can be a bit hard on the wallet. This handy party planner helps you to calculate your costs in advance and keeps you from forgetting important details.
## The Cost-of-Living Calculator.
The Cost-of-Living Calculator.
posted by swift on Aug 14, 2002 - 18 comments
## EasyRGB.com,
EasyRGB.com, a site I've recently stumbled across (or forgotten who sent me there) includes a color harmonies calculator that I find very intriguing. While it doesn't always produce great results, I find the idea and the execution pretty well nigh excellent; I just wish I could figure out how it does its calculations. They also attempt to match RGB colors to real-world color systems and paint chips, so if you've just got to make your bedroom match your website you'll be relieved to know that Dutch Boy sells a "3-B-5 Hang Ten" that is devilishly close to #006699.
posted by jplummer on Nov 23, 2001 - 8 comments
## How to bypass web content filtering programs.
How to bypass web content filtering programs. It's easy, it works, and all you have to have is a Hex calculator. Strike a blow for freedom!
posted by Steven Den Beste on Jan 20, 2001 - 16 comments
Find the distance between two American telephone exchanges with the Interoffice Mileage Calculator. Find the distance between any two cities in the world with How Far Is It?. The latter also has a full database of American and Canadian towns, so I know that my hometown is 4864 miles from Athens, Greece, and 768 miles from Athens, Georgia.
posted by tdecius on Oct 11, 1999 - 0 comments
Page: 1 | 3,073 | 12,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | latest | en | 0.951488 |
http://weisu.blogspot.com/2009/06/rgb-argb-and-converter.html | 1,532,305,523,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594675.66/warc/CC-MAIN-20180722233159-20180723013159-00505.warc.gz | 400,648,948 | 14,002 | ## Friday, June 26, 2009
### RGB, aRGB and converter
Three common display formats for RGB are: RGB888 format (8 bits per channel), RGB666 (6 bits per channel), and RGB565 (5 bits per channel for R,B, and 6 bit for G, RRRR RGGG GGGB BBBB). RGB888 has 24 bits of resolution with 16 million colors. RGB666 is usually used in portable electronics. But the 18 bits data is not good for regular 16-bit DSP. So RGB565 is popular in industry.
Convert R8 to R5 by dropping the last bits:
R8 >> 3, (255 >> 3 = 31)
From R5 to R8, you can either use scaling or dup bits:
R5*255/31, R5 << 3 | R5 >>2
aRGB is with an alpha channel (0xAARRGGBB). Alpha specifies the transparency of the pixel with 0x00 to be fully transparent and 0xff to be white background.
output = a x (foreground pixel) + (1-a) x (background pixel)
Convert RGB888 to aRGB:
r = (rgb >> 16) & 0xFF (or mask first, then shift: 0xff0000, >>16)
g = (rgb >> 8) & 0xFF
b = (rgb >> 0) & 0xFF
aRGB = 0xff000000 | (r << 16) | (g << 8) | (b)
aRGB to RGB888:
r = (aRGB >> 16) &0xFF
g = (aRGB >> 8) &0xFF
b = (aRGB >> 0) &0xFF
a = (aRGB >> 24) &0xFF
RGB888 = (r << 16) | (g << 8) | (b)
aRGB to RGB565: this question has been asked before, (should be 565 instead of 454 here). From RGB888
r >> 3
g >> 2
b >> 3
(r << 11) | (g << 6) | (b)
#### 1 comment:
1. Hi people,
This post is really useful, I exactly needed the ARGB->RGB565 convert method.
I think when converting to RGB565, one has to use the (r << 11) | (g << 5) | (b) formula (after shifting the color components, of course).
Thanks for the post,
Mate | 541 | 1,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-30 | latest | en | 0.831784 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/288/C2.D72.html | 1,611,050,897,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703518201.29/warc/CC-MAIN-20210119072933-20210119102933-00553.warc.gz | 512,344,508 | 6,979 | Copied to
clipboard
## G = C2.D72order 288 = 25·32
### 2nd central extension by C2 of D72
Series: Derived Chief Lower central Upper central
Derived series C1 — C36 — C2.D72
Chief series C1 — C3 — C9 — C18 — C36 — C2×C36 — C2×D36 — C2.D72
Lower central C9 — C18 — C36 — C2.D72
Upper central C1 — C22 — C2×C4 — C2×C8
Generators and relations for C2.D72
G = < a,b,c | a2=b72=1, c2=a, ab=ba, ac=ca, cbc-1=ab-1 >
Subgroups: 488 in 75 conjugacy classes, 32 normal (30 characteristic)
C1, C2 [×3], C2 [×2], C3, C4 [×2], C4, C22, C22 [×4], S3 [×2], C6 [×3], C8, C2×C4, C2×C4, D4 [×3], C23, C9, Dic3, C12 [×2], D6 [×4], C2×C6, C4⋊C4, C2×C8, C2×D4, D9 [×2], C18 [×3], C24, D12 [×3], C2×Dic3, C2×C12, C22×S3, D4⋊C4, Dic9, C36 [×2], D18 [×4], C2×C18, C4⋊Dic3, C2×C24, C2×D12, C72, D36 [×2], D36, C2×Dic9, C2×C36, C22×D9, C2.D24, C4⋊Dic9, C2×C72, C2×D36, C2.D72
Quotients: C1, C2 [×3], C4 [×2], C22, S3, C2×C4, D4 [×2], D6, C22⋊C4, D8, SD16, D9, C4×S3, D12, C3⋊D4, D4⋊C4, D18, C24⋊C2, D24, D6⋊C4, C4×D9, D36, C9⋊D4, C2.D24, C72⋊C2, D72, D18⋊C4, C2.D72
Smallest permutation representation of C2.D72
On 144 points
Generators in S144
```(1 113)(2 114)(3 115)(4 116)(5 117)(6 118)(7 119)(8 120)(9 121)(10 122)(11 123)(12 124)(13 125)(14 126)(15 127)(16 128)(17 129)(18 130)(19 131)(20 132)(21 133)(22 134)(23 135)(24 136)(25 137)(26 138)(27 139)(28 140)(29 141)(30 142)(31 143)(32 144)(33 73)(34 74)(35 75)(36 76)(37 77)(38 78)(39 79)(40 80)(41 81)(42 82)(43 83)(44 84)(45 85)(46 86)(47 87)(48 88)(49 89)(50 90)(51 91)(52 92)(53 93)(54 94)(55 95)(56 96)(57 97)(58 98)(59 99)(60 100)(61 101)(62 102)(63 103)(64 104)(65 105)(66 106)(67 107)(68 108)(69 109)(70 110)(71 111)(72 112)
(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)
(1 112 113 72)(2 71 114 111)(3 110 115 70)(4 69 116 109)(5 108 117 68)(6 67 118 107)(7 106 119 66)(8 65 120 105)(9 104 121 64)(10 63 122 103)(11 102 123 62)(12 61 124 101)(13 100 125 60)(14 59 126 99)(15 98 127 58)(16 57 128 97)(17 96 129 56)(18 55 130 95)(19 94 131 54)(20 53 132 93)(21 92 133 52)(22 51 134 91)(23 90 135 50)(24 49 136 89)(25 88 137 48)(26 47 138 87)(27 86 139 46)(28 45 140 85)(29 84 141 44)(30 43 142 83)(31 82 143 42)(32 41 144 81)(33 80 73 40)(34 39 74 79)(35 78 75 38)(36 37 76 77)```
`G:=sub<Sym(144)| (1,113)(2,114)(3,115)(4,116)(5,117)(6,118)(7,119)(8,120)(9,121)(10,122)(11,123)(12,124)(13,125)(14,126)(15,127)(16,128)(17,129)(18,130)(19,131)(20,132)(21,133)(22,134)(23,135)(24,136)(25,137)(26,138)(27,139)(28,140)(29,141)(30,142)(31,143)(32,144)(33,73)(34,74)(35,75)(36,76)(37,77)(38,78)(39,79)(40,80)(41,81)(42,82)(43,83)(44,84)(45,85)(46,86)(47,87)(48,88)(49,89)(50,90)(51,91)(52,92)(53,93)(54,94)(55,95)(56,96)(57,97)(58,98)(59,99)(60,100)(61,101)(62,102)(63,103)(64,104)(65,105)(66,106)(67,107)(68,108)(69,109)(70,110)(71,111)(72,112), (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), (1,112,113,72)(2,71,114,111)(3,110,115,70)(4,69,116,109)(5,108,117,68)(6,67,118,107)(7,106,119,66)(8,65,120,105)(9,104,121,64)(10,63,122,103)(11,102,123,62)(12,61,124,101)(13,100,125,60)(14,59,126,99)(15,98,127,58)(16,57,128,97)(17,96,129,56)(18,55,130,95)(19,94,131,54)(20,53,132,93)(21,92,133,52)(22,51,134,91)(23,90,135,50)(24,49,136,89)(25,88,137,48)(26,47,138,87)(27,86,139,46)(28,45,140,85)(29,84,141,44)(30,43,142,83)(31,82,143,42)(32,41,144,81)(33,80,73,40)(34,39,74,79)(35,78,75,38)(36,37,76,77)>;`
`G:=Group( (1,113)(2,114)(3,115)(4,116)(5,117)(6,118)(7,119)(8,120)(9,121)(10,122)(11,123)(12,124)(13,125)(14,126)(15,127)(16,128)(17,129)(18,130)(19,131)(20,132)(21,133)(22,134)(23,135)(24,136)(25,137)(26,138)(27,139)(28,140)(29,141)(30,142)(31,143)(32,144)(33,73)(34,74)(35,75)(36,76)(37,77)(38,78)(39,79)(40,80)(41,81)(42,82)(43,83)(44,84)(45,85)(46,86)(47,87)(48,88)(49,89)(50,90)(51,91)(52,92)(53,93)(54,94)(55,95)(56,96)(57,97)(58,98)(59,99)(60,100)(61,101)(62,102)(63,103)(64,104)(65,105)(66,106)(67,107)(68,108)(69,109)(70,110)(71,111)(72,112), (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), (1,112,113,72)(2,71,114,111)(3,110,115,70)(4,69,116,109)(5,108,117,68)(6,67,118,107)(7,106,119,66)(8,65,120,105)(9,104,121,64)(10,63,122,103)(11,102,123,62)(12,61,124,101)(13,100,125,60)(14,59,126,99)(15,98,127,58)(16,57,128,97)(17,96,129,56)(18,55,130,95)(19,94,131,54)(20,53,132,93)(21,92,133,52)(22,51,134,91)(23,90,135,50)(24,49,136,89)(25,88,137,48)(26,47,138,87)(27,86,139,46)(28,45,140,85)(29,84,141,44)(30,43,142,83)(31,82,143,42)(32,41,144,81)(33,80,73,40)(34,39,74,79)(35,78,75,38)(36,37,76,77) );`
`G=PermutationGroup([(1,113),(2,114),(3,115),(4,116),(5,117),(6,118),(7,119),(8,120),(9,121),(10,122),(11,123),(12,124),(13,125),(14,126),(15,127),(16,128),(17,129),(18,130),(19,131),(20,132),(21,133),(22,134),(23,135),(24,136),(25,137),(26,138),(27,139),(28,140),(29,141),(30,142),(31,143),(32,144),(33,73),(34,74),(35,75),(36,76),(37,77),(38,78),(39,79),(40,80),(41,81),(42,82),(43,83),(44,84),(45,85),(46,86),(47,87),(48,88),(49,89),(50,90),(51,91),(52,92),(53,93),(54,94),(55,95),(56,96),(57,97),(58,98),(59,99),(60,100),(61,101),(62,102),(63,103),(64,104),(65,105),(66,106),(67,107),(68,108),(69,109),(70,110),(71,111),(72,112)], [(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)], [(1,112,113,72),(2,71,114,111),(3,110,115,70),(4,69,116,109),(5,108,117,68),(6,67,118,107),(7,106,119,66),(8,65,120,105),(9,104,121,64),(10,63,122,103),(11,102,123,62),(12,61,124,101),(13,100,125,60),(14,59,126,99),(15,98,127,58),(16,57,128,97),(17,96,129,56),(18,55,130,95),(19,94,131,54),(20,53,132,93),(21,92,133,52),(22,51,134,91),(23,90,135,50),(24,49,136,89),(25,88,137,48),(26,47,138,87),(27,86,139,46),(28,45,140,85),(29,84,141,44),(30,43,142,83),(31,82,143,42),(32,41,144,81),(33,80,73,40),(34,39,74,79),(35,78,75,38),(36,37,76,77)])`
78 conjugacy classes
class 1 2A 2B 2C 2D 2E 3 4A 4B 4C 4D 6A 6B 6C 8A 8B 8C 8D 9A 9B 9C 12A 12B 12C 12D 18A ··· 18I 24A ··· 24H 36A ··· 36L 72A ··· 72X order 1 2 2 2 2 2 3 4 4 4 4 6 6 6 8 8 8 8 9 9 9 12 12 12 12 18 ··· 18 24 ··· 24 36 ··· 36 72 ··· 72 size 1 1 1 1 36 36 2 2 2 36 36 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· 2 2 ··· 2 2 ··· 2 2 ··· 2
78 irreducible representations
dim 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 type + + + + + + + + + + + + + + + image C1 C2 C2 C2 C4 S3 D4 D4 D6 D8 SD16 D9 C4×S3 C3⋊D4 D12 D18 C24⋊C2 D24 C4×D9 C9⋊D4 D36 C72⋊C2 D72 kernel C2.D72 C4⋊Dic9 C2×C72 C2×D36 D36 C2×C24 C36 C2×C18 C2×C12 C18 C18 C2×C8 C12 C12 C2×C6 C2×C4 C6 C6 C4 C4 C22 C2 C2 # reps 1 1 1 1 4 1 1 1 1 2 2 3 2 2 2 3 4 4 6 6 6 12 12
Matrix representation of C2.D72 in GL3(𝔽73) generated by
72 0 0 0 1 0 0 0 1
,
27 0 0 0 11 71 0 2 13
,
46 0 0 0 11 71 0 60 62
`G:=sub<GL(3,GF(73))| [72,0,0,0,1,0,0,0,1],[27,0,0,0,11,2,0,71,13],[46,0,0,0,11,60,0,71,62] >;`
C2.D72 in GAP, Magma, Sage, TeX
`C_2.D_{72}`
`% in TeX`
`G:=Group("C2.D72");`
`// GroupNames label`
`G:=SmallGroup(288,28);`
`// by ID`
`G=gap.SmallGroup(288,28);`
`# by ID`
`G:=PCGroup([7,-2,-2,-2,-2,-2,-3,-3,85,92,422,100,6725,292,9414]);`
`// Polycyclic`
`G:=Group<a,b,c|a^2=b^72=1,c^2=a,a*b=b*a,a*c=c*a,c*b*c^-1=a*b^-1>;`
`// generators/relations`
×
𝔽 | 5,001 | 8,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-04 | longest | en | 0.341922 |
https://synapses.polytechnique.fr/catalogue/2021-2022/ue/7646/ECO562A-applied-econometrics-2?from=D1 | 1,660,372,307,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00654.warc.gz | 495,473,663 | 12,092 | v2.8.0 (4335)
# PA - C6B - ECO562A : Applied Econometrics 2
## Descriptif
Course Outline 2020-2021
Applied econometrics 2
Objectives
In this course, we will study how econometric methods can help answer causal questions. We will discuss why establishing credible causal links is difficult in social sciences and how to overcome some of the challenges. We will build upon the tools introduced in applied econometrics 1 (linear regressions) to outline several methods central to modern econometric practice: experiments, instrumental variables, regression discontinuity designs, fixed effects estimations, differences-in-differences, event studies, matching and synthetic control. We will use relevant real-world examples to illustrate the assumptions and the limitations of each method. We will also learn about the experience of guest speakers who studied econometrics at the PhD level and now apply econometric tools in their daily (non-academic) jobs.
During the tutorials, students will work in group to critically assess research articles answering a causal question. They will proceed in five steps:
• Formulate a causal question and explain why it is interesting / important.
• Find two research articles using one of the methods studied in the course to answer the question.
• Discuss the methods used in each paper.
• Confront the results of both papers.
• Summarize the findings in a presentation and a term paper.
The intended learning outcomes of this course are the following:
• Explain why and when econometrics is useful
• Locate cutting-edge empirical economics research
• Assess the flaws and limitations of empirical work
• Evaluate firms’ decisions and public policies
Guest speakers:
• Marianne BLEHAUT (CREDOC)
• Georges Vivien HOUNGBONON (World Bank)
• Meryam ZAIEM (DARES)
Evaluation
The evaluation consists of three parts:
• Active participation during lectures and tutorials (individual grade - 20%)
• Oral presentation (individual grade – 30%)
• Term paper (group grade – 50%)
References
Slides are self-contained. Interested students can look at the following textbooks to find additional technical details and examples:
• Introduction to econometrics, Stock and Watson
• Mastering metrics, Angrist and Pischke
• Mostly harmless econometrics, Angrist and Pischke
In the applications we will discuss thoroughly the following articles:
• Bandiera O, Burgess R, Das J, Gulesci S, Rasul I and Sulaiman (2017) “Labor markets and poverty in village economies”, Quarterly Journal of Economics
• Duflo, E (2001) “Schooling and labor market consequences of school construction in Indonesia: Evidence from an unusual policy experiment”, American Economic Review
• Ozier O (2017) “The Impact of Secondary Schooling in Kenya: A Regression Discontinuity Analysis”, Journal of Human Resources.
• Bleakley H (2003). “Disease and Development: Evidence from the American South.” Journal of the European Economic Association.
• Acemoglu, D and S Johnson (2007). “Disease and Development: The Effect of Life Expectancy on Economic Growth”, Journal of Political Economy.
## Pour les étudiants du diplôme Diplôme d'ingénieur de l'Ecole polytechnique
Vous devez avoir validé l'équation suivante : UE ECO552
prerequisite ECO552
Numérique sur 20
## Pour les étudiants du diplôme Diplôme d'ingénieur de l'Ecole polytechnique
Le rattrapage est autorisé (Note de rattrapage conservée)
L'UE est acquise si note finale transposée >= C
• Crédits ECTS acquis : 5 ECTS
La note obtenue rentre dans le calcul de votre GPA.
## Pour les étudiants du diplôme Non Diplomant
Le rattrapage est autorisé (Note de rattrapage conservée)
L'UE est acquise si note finale transposée >= C
• Crédits ECTS acquis : 5 ECTS
## Pour les étudiants du diplôme Economics for Smart Cities and Climate Policy
Le rattrapage est autorisé (Note de rattrapage conservée)
L'UE est acquise si note finale transposée >= C
• Crédits ECTS acquis : 4 ECTS
La note obtenue rentre dans le calcul de votre GPA.
## Pour les étudiants du diplôme Economics, Data Analytics and Corporate Finance
Le rattrapage est autorisé (Note de rattrapage conservée)
L'UE est acquise si note finale transposée >= C
• Crédits ECTS acquis : 4 ECTS
La note obtenue rentre dans le calcul de votre GPA. | 1,015 | 4,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.87322 |
http://www.slideserve.com/liam/chapter-5-slides | 1,498,755,265,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128329372.0/warc/CC-MAIN-20170629154125-20170629174125-00042.warc.gz | 644,722,987 | 20,607 | 1 / 34
# Chapter 5 Slides - PowerPoint PPT Presentation
Probability Theory General Probability Rules PBS Chapter 5.1 © 2009 W.H. Freeman and Company Objectives (PBS Chapter 5.1) General probability rules Independence and the multiplication rule Applying the multiplication rule The general addition rule Probability Rules (review)
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Presentation Transcript
### Probability TheoryGeneral Probability Rules
PBS Chapter 5.1
© 2009 W.H. Freeman and Company
General probability rules
• Independence and the multiplication rule
• Applying the multiplication rule
We already met the following four rules in Chapter 4:
1) Probabilities range from 0 (no chance of the event) to1 (the event has to happen).
For any event A, 0 ≤P(A) ≤ 1
Probability of getting a Head = 0.5
We write this as: P(Head) = 0.5
P(neither Head nor Tail) = 0
P(getting either a Head or a Tail) = 1
2) Because some outcome must occur on every trial, the sum of the probabilities for all possible outcomes (the sample space) must be exactly 1.
P(sample space) = 1
Coin toss: S = {Head, Tail}
P(head) + P(tail) = 0.5 + 0.5 =1
P(sample space) = 1
Probability of tails = 0.5
Probability rules (contd)
3) The complement of any event A is the event that A does not occur, written as Ac.
The complement rule states that the probability of an event not occurring is 1 minus the probability that is does occur.
P(not A) = P(Ac) = 1 −P(A)
Tailc = not Tail = Head
P(Tailc) = 1 −P(Head) = 0.5
Venn diagram:
Sample space made up of an event A and its complementary Ac, i.e., everything that is not A.
A and B disjoint
Probability rules (contd )
4) Two events A and B are disjoint if they have no outcomes in common and can never happen together. The probability that A or B occurs is then the sum of their individual probabilities.
P(A or B) = “P(A U B)” = P(A) + P(B)
This is the addition rule for disjoint events.
A and B not disjoint
Example:If you flip two coins, and the first flip does not affect the second flip: S = {HH, HT, TH, TT}. The probability of each of these events is 1/4, or 0.25.
The probability that you obtain “only heads or only tails” is:P(HH or TT) = P(HH) + P(TT) = 0.25 + 0.25 = 0.50
Two events are independent if the probability that one event occurs on any given trial of an experiment is not affected or changed by the occurrence of the other event.
When are trials not independent?
Imagine that these coins were spread out so that half were heads up and half were tails up. Close your eyes and pick one. The probability of it being heads is 0.5. However, if you don’t put it back in the pile, the probability of picking up another coin and having it be heads is now less than 0.5.
The trials are independent only when you put the coin back each time. It is called sampling with replacement.
Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs.
If A and B are independent, P(A and B) = P(A)P(B)
This is the multiplication rule for independent events.
Two consecutive coin tosses:
P(first Tail and second Tail) = P(first Tail) * P(second Tail) = 0.5 * 0.5 = 0.25
Venn diagram:
Event A and event B. The intersection represents the event {A and B} and outcomes common to both A and B.
• Disjoint events are not independent.
• If A and B are disjoint, then the fact that A occurs tells us that B cannot occur. So A and B are not independent.
• Independence cannot be pictured in a Venn Diagram.
• If two events A and B are independent, the event that A does not occur is also independent of B.
• The Multiplication Rule extends to collections of more than two events, provided that all are independent.
• Example: A transatlantic data cable contains repeaters to amplify the signal. Each repeater has probability 0.999 of functioning without failure for 25 years. Repeaters fail independently of each other. Let A1 denote the event that the first repeater operates without failure for 25 years, A2 denote the event that the second repeater operates without failure for 25 years, and so on. The last transatlantic cable had 662 repeaters. The probability that all 662 will work for 25 years is:
P(A1 and A2 and…and A662) = 0.999662 = 0.516
General addition rule for any two events A and B:
The probability that A occurs, or B occurs, or both events occur is:
P(A or B) = P(A) + P(B) – P(A and B)
What is the probability of randomly drawing either an ace or a heart from a deck of 52 playing cards? There are 4 aces in the pack and 13 hearts. However, 1 card is both an ace and a heart. Thus:
P(ace or heart) = P(ace) + P(heart) – P(ace and heart)
= 4/52 + 13/52 - 1/52 = 16/52 ≈ .3
### Probability TheoryThe Binomial and Poisson Distributions
PBS Chapters 5.2 and 5.3
© 2009 W. H. Freeman and Company
Binomial and Poisson Distributions
• The binomial setting
• Binomial Probabilities
• Binomial mean and standard deviation
• The Normal approximation
• The Poisson setting
• The Poisson model
Binomial distributions are models for some categorical variables, typically representing the number of successes in a series of n trials.
The observations must meet these requirements:
• The total number of observations n is fixed in advance.
• The outcomes of all n observations are statistically independent.
• Each observation falls into just one of 2 categories: success and failure.
• All n observations have the same probability of “success,” p.
We record the next 50 births at a local hospital. Each newborn is either a boy or a girl; each baby is either born on a Sunday or not.
The distribution of the count X of successes in the binomial setting is the binomial distribution with parameters n and p:B(n,p).
• The parameter n is the total number of observations.
• The parameter p is the probability of success on each observation.
• The count of successes X can be any whole number between 0 and n.
A coin is flipped 10 times. Each outcome is either a head or a tail. The variable X is the number of heads among those 10 flips, our count of “successes.”
On each flip, the probability of success, “head,” is 0.5. The number X of heads among 10 flips has the binomial distribution B(n = 10, p = 0.5).
Binomial distributions describe the possible number of times that a particular event will occur in a sequence of observations.
They are used when we want to know about the occurrence of an event, not its magnitude.
• In a clinical trial, a patient’s condition may improve or not. We study the number of patients who improved, not how much better they feel.
• Is a person ambitious or not? The binomial distribution describes the number of ambitious persons, not how ambitious they are.
• In quality control we assess the number of defective items in a lot of goods, irrespective of the type of defect.
The number of ways of arranging k successes in a series of n observations (with constant probability p of success) is the number of possible combinations (unordered sequences).
This can be calculated with the binomial coefficient:
Where k = 0, 1, 2, ..., or n.
• The binomial coefficient “n_choose_k” uses the factorialnotation “!”.
• The factorial n! for any strictly positive whole number n is:
n! = n × (n − 1) × (n − 2) × · · · × 3 × 2 × 1
• For example: 5! = 5 × 4 × 3 × 2 × 1 = 120
• Note that 0! = 1.
The binomial coefficient counts the number of ways in which k successes can be arranged among n observations.
The binomial probability P(X = k) is this count multiplied by the probability of any specific arrangement of the k successes:
The probability that a binomial random variable takes any range of values is the sum of each probability for getting exactly that many successes in n observations.
P(X≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
• You can also look up the probabilities for some values of n and p in Table C in the back of the book.
• The entries in the table are the probabilities P(X = k) of individual outcomes.
• The values of p that appear in Table C are all 0.5 or smaller. When the probability of a success is greater than 0.5, restate the problem in terms of the number of failures.
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population.
What is the probability that exactly five individuals in the sample are color blind?
• Use Excel’s “=BINOMDIST(number_s,trials,probability_s,cumulative)” P(x= 5) = BINOMDIST(5, 25, 0.08, 0) = 0.03285
• P(x= 5) = (n! / k!(n k)!)pk(1 p)n-k = (25! / 5!(20)!) 0.0850.925P(x= 5) = (21*22*23*24*24*25 / 1*2*3*4*5) 0.0850.9220P(x= 5) = 53,130 * 0.0000033 * 0.1887 = 0.03285
The center and spread of the binomial distribution for a count X are defined by the mean m and standard deviation s:
a)
b)
Effect of changing p when n is fixed.
a) n = 10, p = 0.25
b) n = 10, p = 0.5
c) n = 10, p = 0.75
For small samples, binomial distributions are skewed when p is different from 0.5.
c)
If n is large, and p is not too close to 0 or 1, the binomial distribution can be approximated by the normal distribution
N(m = np,s2 = np(1 p))
Practically, the Normal approximation can be used when both
np≥10 and n(1 p) ≥10.
If X is the count of successes in the sample the sampling distributions for large n, is:
• X approximately N(µ = np,σ2 = np(1 − p))
• A count X of successes has a Poisson distribution in the Poisson setting:
• The number of successes that occur in any unit of measure is independent of the number of successes that occur in any non-overlapping unit of measure.
• The probability that a success will occur in a unit of measure is the same for all units of equal size and is proportional to the size of the unit.
• The probability that 2 or more successes will occur in a unit approaches 0 as the size of the unit becomes smaller.
• The distribution of the count X of successes in the Poisson setting is the Poisson distribution with meanμ. The parameter μis the mean number of successes per unit of measure.
• The possible values of X are the whole numbers 0, 1, 2, 3, ….If k is any whole number 0 or greater, then
P(X = k) = (e-μμk)/k!
• The standard deviation of the distribution is the square root of μ.
### Probability TheoryConditional Probability
PBS Chapter 5.4
© 2009 W.H. Freeman and Company
Conditional Probability
• General multiplication rule
• Conditional probability and independence
• Tree diagrams
• Bayes’s rule
• Conditional probability gives the probability of one event under the condition that we know another event.
• General multiplication rule: The probability that any two events, A and B, happen together is:
P(A and B) = P(A)P(B|A)
Here P(B|A) is the conditional probability that B occurs, given the information that A occurs.
Conditional probabilities reflect how the probability of an event can change if we know that some other event has occurred/is occurring.
• Example: The probability that a cloudy day will result in rain is different if you live in Los Angeles than if you live in Seattle.
• Our brains effortlessly calculate conditional probabilities, updating our “degree of belief” with each new piece of evidence.
The conditional probability of event B given event A is:(provided that P(A) ≠ 0)
Recall: A and B are independent when they have no influence on each other’s occurrence.
• Two events A and B that both have positive probability are independent if P(B|A) = P(B)
• The general multiplication rule then becomes: P(A and B) = P(A)P(B)
• What is the probability of randomly drawing an ace of hearts from a deck of 52 playing cards? There are 4 aces in the pack and 13 hearts.
• P(heart|ace) = 1/4 P(ace) = 4/52
• P(ace and heart) = P(ace)* P(heart|ace) = (4/52)*(1/4) = 1/52
Notice that heart and ace are independent events.
0.47
Internet user
Tree diagrams
Conditional probabilities can get complex, and it is often a good strategy to build a probability tree that represents all possible outcomes graphically and assigns conditional probabilities to subsets of events.
P(chatting) = 0.136 + 0.099 + 0.017
= 0.252
0.8
Disease incidence
Positive
Cancer
0.0004
False negative
Negative
0.2
Mammography
0.1
False positive
Positive
0.9996
No cancer
Negative
Incidence of breast cancer among women ages 20–30
0.9
Diagnosis specificity
Mammography performance
Breast cancer screening
If a woman in her 20s gets screened for breast cancer and receives a positive test result, what is the probability that she does have breast cancer?
She could either have a positive test and have breast cancer or have a positive test but not have cancer (false positive).
Disease incidence
Positive
Cancer
False negative
Negative
Mammography
False positive
Positive
No cancer
Negative
Diagnosis specificity
0.8
0.0004
0.2
0.1
0.9996
Incidence of breast cancer among women ages 20–30
0.9
Mammography performance
Possible outcomes given the positive diagnosis: positive test and breast cancer or positive test but no cancer (false positive).
This value is called the positive predictive value, or PV+. It is an important piece of information but, unfortunately, is rarely communicated to patients.
An important application of conditional probabilities is Bayes’s rule. It is the foundation of many modern statistical applications beyond the scope of this textbook.
* If a sample space is decomposed in k disjoint events, A1, A2, … , Ak— none with a null probability but P(A1) + P(A2) + … + P(Ak) = 1,
* And if C is any other event such that P(C) is not 0 or 1, then:
However, it is often intuitively much easier to work out answers with a probability tree than with these lengthy formulas.
Disease incidence
Positive
Cancer
False negative
Negative
Mammography
False positive
Positive
No cancer
Negative
Diagnosis specificity
If a woman in her 20s gets screened for breast cancer and receives a positive test result, what is the probability that she does have breast cancer?
0.8
0.0004
0.2
0.1
0.9996
Incidence of breast cancer among women ages 20–30
0.9
Mammography performance
This time, we use Bayes’s rule: A1 is cancer, A2 is no cancer, C is a positive test result. | 3,805 | 14,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2017-26 | longest | en | 0.881054 |
http://mathhelpforum.com/algebra/19634-factoring-polynomial.html | 1,481,243,930,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542665.72/warc/CC-MAIN-20161202170902-00209-ip-10-31-129-80.ec2.internal.warc.gz | 177,557,026 | 9,930 | 1. ## factoring a polynomial
Hello, I need help factoring the following: 2x^3 - 5x - 3
I've gotten as far as using the payback method to get: x^3 - 5x - 6 but I can't seem to go beyond that.
Help is appreciated, thanks!
2. Originally Posted by vperera
Hello, I need help factoring the following: 2x^3 - 5x - 3
I've gotten as far as using the payback method to get: x^3 - 5x - 6 but I can't seem to go beyond that.
Help is appreciated, thanks!
the payback method? what's that? whatever it is, it is incorrect.
first try the factors of 3 into the formula to see if we can get zero. we realize that plugging in $x = -1$ gives zero. thus $x = -1$ is a root of the cubic, and so $(x + 1)$ is a factor, by the factor theorem.
by way of polynomial long division or synthetic division, we find that:
$\left( 2x^3 - 5x - 3 \right) \div (x + 1) = 2x^2 - 2x - 3$
thus, $2x^3 - 5x - 3 = (x + 1) \left( 2x^2 - 2x - 3 \right)$
and you can continue to factor the quadratic, but only if you really want to, it will be tedious to work out, and it won't look very nice when you're done, if you decide to | 360 | 1,096 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2016-50 | longest | en | 0.949684 |
https://revs.bbcelite.com/source/main/subroutine/multiply8x16signed.html | 1,696,401,132,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511361.38/warc/CC-MAIN-20231004052258-20231004082258-00705.warc.gz | 522,379,225 | 53,363 | Revs on the BBC Micro
# Maths (Arithmetic): Multiply8x16Signed
``` Name: Multiply8x16Signed [Show more]
Type: Subroutine
Category: Maths (Arithmetic)
Summary: Multiply an 8-bit and a 16-bit number and apply a sign to the
result
Context: See this subroutine in context in the source code
References: This subroutine is called as follows:
* ApplySpinYaw calls Multiply8x16Signed
* ScaleTyreForces calls Multiply8x16Signed
This routine calculates:
(A T) = (A T) * abs(A) * Y / 256
where A is the last variable to be loaded before the subroutine call. So if
the call follows an LDA instruction, for example, the following is calculated
if A is positive:
(A T) = (A T) * Y / 256
and the following is calculated if A is negative:
(A T) = -(A T) * Y / 256
Arguments:
(A T) A 16-bit signed number
Y An unsigned number
N flag Controls the sign to be applied:
* N flag clear to calculate (A T) * Y / 256
* N flag set to calculate -(A T) * Y / 256
.Multiply8x16Signed
PHP \ Store the N flag on the stack
JSR Absolute16Bit \ Set the sign of (A T) to that of the N flag argument
STA V \ Set (V T) = (A T)
STY U \ Set U = Y
JSR Multiply8x16 \ Set (U T) = U * (V T) / 256
\ = Y * (A T) / 256
LDA U \ Set (A T) = (U T)
\ = Y * (A T) / 256
PLP \ Retrieve the N flag from the stack
JSR Absolute16Bit \ Set the sign of (A T) to that of the N flag argument
RTS \ Return from the subroutine
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1 con 37: Answr Ky for Problm St (Chaptr 2-3) Instructor: Kanda Naknoi Sptmbr 4, (2 points) Is it possibl for a country to hav a currnt account dficit at th sam tim and has a surplus in its balanc of paymnts? xplain your answr using hypothtical figurs. ANSW: ( point) Ys. Th balanc of paymnts (BoP) is th sum of currnt account (CA), capital account (KA) and financial account (FA). Thortically, bcaus of th doubl-ntry bookkping practic, th sum is supposd to b zro. BoP = CA KA FA = 0 ( point) Givn a currnt account dficit, th thory prdicts that th sum of th capital account and th financial account must b surplus. Howvr, in practic, th balanc of paymnts may dviat from zro bcaus of statistical discrpancy, particularly in rcording financial transactions in th financial account. Suppos th currnt account dficit is 00 billion dollars, a combind surplus of th capital account and th financial account must b gratr than 00 billion dollars to produc a surplus in th balanc of paymnts. (Optional not) Somtims th trm balanc of paymnts is usd to dscrib th official sttlmnts balanc with th opposit sign. This balanc indicats th paymnts gap covrd by th official rsrv transactions. For xampl, th U.S. official sttlmnts balanc in 2003 is 250 billion dollars. In this cas, som nwspaprs may say that th U.S. balanc of paymnts in 2003 is -250 billion dollars. Whn it is a larg ngativ numbr, it is oftn a subjct of concrn bcaus it indicats an xcssiv rol of th cntral bank in financing th currnt account dficit. 2. (2 points) Suppos that th U.S. nt forign dbt is 25 prcnt of U.S. GDP and that forign assts and liabilitis alik pay an intrst rat of 5 prcnt pr yar. What would b th drain on U.S. GDP (as a prcntag) from paying intrst on th nt forign dbt? What if th nt forign dbt wr 00 prcnt of GDP? At what point do you think a country s govrnmnt should bcom worrid about th
2 siz of its forign dbt? ANSW: ( point) 25 prcnt dbt-to-gdp ratio and 5 prcnt intrst rat implis that th intrst paymnt as th ratio to GDP is 0.25x0.05 = =.25 prcnt of GDP. If th dbt-to-gdp ratio is 00 prcnt, thn th intrst paymnt willl bcom x0.05 = 0.05 = 5 prcnt of GDP. ( point) Intuitivly, th govrnmnt should b concrnd about th nt forign dbt whn paying back th dbt bcoms difficult. W can us th balanc of paymnts accounting to find out how much output ndd to pay back th dbt. CA = savings - invstmnt Whn a country run a currnt account surplus, thir rsidnts hav mor savings than th domstic invstmnt and so thy also invst ovrsas. That is how a country with a currnt account surplus can accumulat nt forign assts. In contrast, a country that runs a currnt account dficit dos not hav nough savings for its domstic invstmnt, so it must borrow from abroad and accumulat nt forign dbts. In othr words, th stock of nt forign dbts ar a rsult of past capital inflows triggrd by currnt account dficits. Th only way to rduc th forign dbt is to rvrs th dirction of capital flows. So, w must run a currnt account surplus from now, and us th surplus to pay back th dbt. Th siz of forign dbt coms to play a rol bcaus th largr it is, th largr th intrst paymnt and th largr futur currnt account surpluss nd to b. This can also b xplaind using anothr currnt account accounting. (S Tabl 2-2 for an xampl.) CA = xport - import nt incom transfr from abroad In this cas, th nt incom transfr is th ngativ valu of th intrst paymnt w calculatd abov. In ordr to run a currnt account surplus in th futur, th U.S. will hav to run a trad balanc surplus which is gratr than th amount of th intrst paymnt. Whn th dbt-to-gdp ratio is 25 prcnt, to rduc and pay back th dbt, th U.S. will hav to run th trad surplus of mor than.25 prcnt of GDP for a numbr of yars in th futur. But whn th dbt-to-gdp ratio is 00 prcnt, th trad surplus has to b mor than 5 prcnt of GDP. This cannot happn without strong xport growth. So, a country s govrnmnt should bcom worrid whn its forign dbt is so larg that it rquirs unralistically high xport growth in th futur. (Optional not) W can also considr th rlationship of nt forign assts (NFA) and currnt account (CA). 2
3 CA this priod = chang in NFA = NFA nxt priod - NFA this priod Currnt account rflcts th dirction of capital flows and as a rsult indicats whthr a country is accumulating mor dbts or assts in a particular priod. It is a flow concpt. Howvr, NFA is a stock concpt, maning it is th accumulatd balanc of a flow variabls. Whn NFA is ngativ, it indicats that th nation is a nt dbtor country. Whn it is positiv, it indicats that th nation is a nt crditor country. W can also us this rlationship to track th path of NFA. NFA nxt priod = NFA this priod CA this priod Considr a cas whr w start out with a ngativ numbr of NFA. So, this rlationship tlls us that th only way to rduc and vntually pay back th forign dbt is to kp running currnt account surplus in th futur. 3. (2 points) Calculat th dollar rats of rturn on a dposit 0,000 pounds in a London bank in a yar whn th intrst rat on pounds is 0 prcnt and th dollar/pound xchang rat movs from.50 dollars pr pound to.38 dollars pr pound. What is th intrst rat a U.S. bank nds to offr to attract capital flows from London? ANSW: ( point) Today dollar givs us /.50 pounds. With 0 prcnt intrst rat, th nxt priod gross rturn is (0.0) x /.50 pounds. Thn convrt this back to th U.S. dollar with th rat.50 dollar pr pound. So, th th gross rturn bcoms (0.0) x /.50 x.38 dollars. Subtracting dollar from this will giv th nt dollar rturn on th London dposit of 0.02 or.2 prcnt. Not that th total of 0,000 pounds is irrlvant, bcaus th intrst rat is in prcnt. W can us th uncovrd intrst parity to approximat th rturn. Th dollar rturn on th London dposit = th pound intrst rat th dprciation of U.S. dollar. Th dprciation of U.S. dollar = ( )/.50 = = 8 prcnt apprciation. Th ngativ sign indicats that th U.S. dollar dos not dprciat but apprciats. Th dollar rturn on th London dposit = = 0.02 = 2 prcnt. ( point) A U.S. bank nds to offr mor than.2 (2 if w us th approximation mthod) prcnt intrst rat to attract capital flows from London. A U.S. bank dos not nd to pay as much as th London bank bcaus of th futur apprciation or capital gain of th U.S. dollar. 4. (2 points) Suppos tradrs in asst markts suddnly larn that th intrst rat on dollars will dclin in th nar futur. Us th diagrammatic analysis of Chaptr 3 to dtrmin th ffcts on th currnt dollar/uro xchang rat, assuming currnt intrst rats on dollar and uro dposits do not chang. 3
5 Figur 4.: Tradrs anticipation \$/ 2 2 \$ \$ xpctd \$ rturn on dposits
6 Figur 4.2: Today forign xchang markt quilibrium \$/ 3 3 ' \$ xpctd \$ rturn on dposits
7 Figur 5: ffcts of a subsidy for savings on xchang rat \$/ 3 3 ' 4 = 4 \$ '' = \$ \$ xpctd \$ rturn on dposits
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### SPECIAL VOWEL SOUNDS
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https://www.jiskha.com/questions/536477/how-many-pieces-of-ribbon-1-4-5-yd-long-can-be-cut-a-length-of-ribbon-9-yd-long | 1,611,101,908,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00230.warc.gz | 841,442,114 | 4,655 | # math
How many pieces of ribbon 1 4/5 yd long can be cut a length of ribbon 9 yd long?
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3. 👁
1. 9/(9/5)
9 * (5/9)
45/9 = 5 pieces
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👤
Ms. Sue
2. Assignment-1 1 if a side of square is given 6cm than he would you get perimeter?
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A ribbon is 12 3/8 feet long. Into how many 3/4 -foot pieces can it be cut? | 704 | 2,240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-04 | latest | en | 0.939364 |
https://studyforce.com/course/mathematics-for-technology-ii-math-2131/lessons/write-and-solve-systems-of-equation/ | 1,610,978,624,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514796.13/warc/CC-MAIN-20210118123320-20210118153320-00218.warc.gz | 577,872,700 | 18,898 | # Mathematics for Technology II (Math 2131)
Durham College, Mathematics
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• ##### Solving Systems of Equations
This unit introduces how to systematically solve a system of equations, namely linear equations. Examples of non-linear systems, including systems of 3 unknowns will be of emphasis.
• ##### Graphs of Trigonometric Functions
The unit focuses primarily on how to graph periodic sinusoidal functions, and how to identify features of a waveform to produce an equation by inspection.
• ##### Polar Coordinate Functions
An introduction to the polar coordinate system.
• ##### Complex Numbers
This unit is an extension of what was introduced in Math 1131. To learn how to work with radicals, knowing your exponent laws in crucial. Hence, this unit begins with a thorough review.
• ##### Logarithmic Functions
This chapter introduces you to exponential functions, and how they can be solved using logarithms.
• ##### Trigonometric Identities and Equations
No items in this section
• ##### Analytic Geometry
No items in this section
## Mathematics for Technology II (Math 2131)
### Write and Solve Systems of Equations from Word Problems
Many applications in real-life contain two or more unknowns. To solve such problems, we must write as many independent equations as there are unknowns. Otherwise, it is not possible to obtain numerical answers.
This section mirrors what you’ve already been doing this unit – that is, solving systems with two or more variables – except now you’re making the equations (the hard part). Because you already know by now how solve systems of equations, for several of these problems, you’ll only be expected to setup the equations.
Generally, to change sentences into mathematical expressions and equations, look for key words like these:
Increased: +
Sum: +
More: +
Decreased: –
Difference: –
Less: –
Twice: ×2
Doubled: ×2
Tripled: ×3
The same: =
Let’s begin with some basic examples where we learn to go from words to numbers and variables.
Question: The sum of two numbers is twenty-one and their difference is fifteen. What are the numbers?
Solution: You being by assigning a variable for each unknown number. Let the first and second number be x and y, respectively. You’re told that their sum is 21 and difference is 15:
• x + y = 21
• x – y = 15
From here, you can use the method of elimination or substitution. For instance, using substitution, you can solve for x in the first equation, then substitute the expression into the x of the second equation. Eventually, you should end up with a point of intersection at (18, 3).
Question: Number A divided by number B equals one-fourth. Number B divided by number A equals two times number A. What are the two numbers?
Solution
Start by setting up the equations:
After solving, you should get: A = 2 and B = 8.
[collapse]
Question: The sum of two numbers is 67. The quotient of the greater divided by the smaller number is 7 with remainder 3. Determine the two numbers.
Solution
Start by setting up the equations; x represents the smaller number, y represents the larger.
After solving, you should end up with x = 59 and y = 8.
[collapse]
# Money Applications
Given the complexity of certain finance-related formulas, oftentimes they’ll already be provided to you.
# Uniform Motion
For these types of problems, you’ll be required to remember the formula for speed: distance over time.
• A similar example can be found here.
• Try this extra challenging example:
• Runner Y leaves the starting line at 12:00. Five minutes later runner X leaves from the same place and catches up to runner Y at 12:07. Four minutes later, runner X is one hundred meters ahead of runner Y. What is the speed of each runner in m/s?
Solution →
# Applications involving Mixtures
Here’s the basic idea involving mixtures, for a mixture of several ingredients A, B, C, … etc., the total amount of mixture is equal to the individual volumes making up that mixture. For example, A + B = 10 liters. If you’re looking for the volume of A and B, and you know the concentration of an ingredient found in both A and B, you can represent the concentration of that particular ingredient as a decimal factor multiplied to the unknown volume variable. For instance, if A and B represent fuel mixtures (ethanol + gas), the percentage of ethanol in mixture A is 15%, the ethanol in B is 8%, and the final mixture has 11% ethanol, this can be written as: 0.15A + 0.08B = 0.11(10 liters).
# Applications involving Work, Fluid Flow, and Energy Flow
In order to create these equations, you need to know the following:
Work: amount of work done = rate of work × time worked
Fluid flow: amount of flow = flow rate × duration of flow
Energy flow: amount of energy transmitted = rate of energy time × time
Notice that in all three scenarios, the given rate is multiplied by an unknown variable (italicized). A work-related application is demonstrated below.
# Electricity
And finally, we revisit the application problem that was introduced at the beginning of this unit:
Here you’re expected to find the current represented by I1 and I2 by constructing equations using Kirchhoff’s law and Ohm’s law. Of course, if you’re not enrolled in an engineering program that involves electricity, you can skip this one. | 1,207 | 5,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-04 | latest | en | 0.885886 |
https://multiplesof.com/multiples-of-94/ | 1,680,105,837,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00186.warc.gz | 450,256,192 | 16,574 | # Multiples of 94
“Find Multiples of 94” or “Write all Multiples of 94”. We here at multiplesof.com provide you the full detailed guide to the Multiples of 94.
## What Are The Multiples of 94?
The result/outcome from multiplying 94 with any natural number is known as multiples of 94.
In other simple words, the multiples of 94 are the numbers that, when divided by 94, have a remainder value of 0.
The multiples of 94 are : 94, 188, 282, 376, 470, 564, 658, 752, 846, 940, 1034, 1128, 1222, 1316, 1410, 1504, 1598, 1692, 1786, 1880, 1974, 2068, 2162, 2256, 2350, 2444, 2538, 2632, 2726, 2820, 2914, 3008, 3102, 3196, 3290, 3384 ……..
## List of Multiples of 94
The following table below shows the list of Multiples of 94.
## What is the 5th Multiple of 94?
The 5th Multiple of 94 is 470.
## What is the Smallest Multiple of 94?
The smallest multiple of 94 is 94.
## How to Find Multiples of 94?
Multiply the number by any natural number to find multiples of 94. For example, 94 x 5 = 470.
It states that 470 is the fifth multiple of 94.
By using the above steps you may be able to find all multiples of 94.
Every multiple of 94 number is greater than 94 or equal to 94 itself.
## What Does Multiple of 94 Mean?
In simple terms, the multiples of 94 are the numbers or digits, when the number is divided by 94 and leave remainder value 0.
All numbers that are divided by 94 give the remainder 0 are the multiples of number 94.
Examples of multiples of 94 are 188, 282, 376 ,470, and so on.
The result of multiplying 5 by 94 is 470, The number 470 is a multiple of 94.
## What is the Sum of First Five Multiples of 94?
The sum of the first five multiples of 94 is 1410.
## How Many Multiples Does 94 Have?
There are infinite multiples of 94. The Multiples of 94 are follows as: 94, 188, 282, 376, 470, 564, 658, 752, 846, 940, 1034, 1128, 1222, 1316, 1410, 1504, 1598, 1692, 1786, 1880, 1974, 2068, 2162, 2256, 2350, 2444, 2538, 2632, 2726, 2820, 2914, 3008, 3102, 3196, 3290, 3384…… to infinity.
## What are the First 3 Multiples of 94?
The first three multiples of 94 are: 94, 188, 282.
## Are Multiples of 94 Always Odd?
No, It is not compulsory that the multiples of 94, 188, 282 are always odd.
For example, multiples of 94 that are even numbers include 1128.
The result of multiplying 94 by an even number will always give you the even number.
## Write Down the First 10 Multiples of 94
The first 10 multiples of 94, written as : 94, 188, 282, 376, 470, 564, 658, 752, 846, 940.
• 94 x 1 = 94
• 94 x 2 = 188
• 94 x 3 = 282
• 94 x 4 = 376
• 94 x 5 = 470
• 94 x 6 = 564
• 94 x 7 = 658
• 94 x 8 = 752
• 94 x 9 = 846
• 94 x 10 = 940
### Is 1 a Multiple of 94?
No, the number 1 is a multiple of 1 itself. 1 is not a multiple of 94. Thus, 1 is a factor of 94 not a multiple.
## What are the Multiples of 94?
The multiples of 94 are : 94, 188, 282, 376, 470, 564, 658, 752, 846, 940, 1034, 1128, 1222, 1316, 1410, 1504, 1598, 1692, 1786, 1880, 1974, 2068, 2162, 2256, 2350, 2444, 2538, 2632, 2726, 2820, 2914, 3008, 3102, 3196, 3290, 3384 ……………… .
## What are the First 10 Multiples of 94?
The first 10 multiples of 94 are 94, 188, 282, 376, 470, 564, 658, 752, 846, 940.
Hope from the above table, you might be able to find the first 10 multiples of 94.
## Can 94 Be A Multiple Of Itself?
Yes, because 94 = 94×1.
You will find that every number is a multiple of itself.
### Write all Multiples of 94 Number less than 100
Multiple of 94 less then 100 are : 94.
### Write the Multiples of 94 Between 40 and 50
The multiples of 94 between 40 and 50 are Not Available.
### Write the Multiples of 94 between 1 and 30
The multiples of 94 between 1 and 30 are Not Available.
### What is the Fifth Multiple of 94?
The fifth multiple of 94 is 470.
### What is the 11th Multiple of 94?
The 11th Multiple of 94 is 1034.
### What is the 8th Multiple of 94?
The 8th Multiple of 94 is 752.
## What is the Average of First Five Multiples of 94?
The average of First Five Multiples of 94 is 282.
The average of first five , 94’s multiples calculated as the sum of first 5 multiples of 94 divided by the total number.
= Sum of 94/Total Numbers
Sum of First Five Multiples of = 94+188+282+376+470 = 1410
Total Numbers = 5
= 1410/5
= 282
### What is the Sum of First 8 Multiples of 94?
The sum of the first 8 multiples of 94 is 3384.
### What is the Sum of First 10 Multiples of 94?
The sum of the first 10 multiples of 94 is 5170.
Hope, Now you know the multiples of 94, If you have any other queries, you may ask through the comment section.
Also Know,
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## Full text
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### USING TAGUCHI’S METHOD OF EXPERIMENTATION
K. N AN AND
SQ C U nit
Indian Statistical Institute B angalore 560 0 5 9 , India
Key Words
Yield; Slack wax; Paraffin wax; Oil extraction; Orthogonal arrays; Process optimization; Linear graphs.
Experimentation in Industry
Experiments in industry are carried out to improve quality and yield, or to reduce the cost. In the conventional approach, one factor is studied while the others are kept constant. When the influencing factors are few, it may be possi
ble to arrive at an optimum condition using such an approach. However, when the factors are numerous, it is difficult to find the solution through such a con
ventional approach. It becomes complex when the interaction between two or more factors is also present, which is quite common in process industries.
(2)
Here, statistical designs, like factorial experiments and response surface designs are available. However, the form er requires a large number of experi
mental runs. Most of the information can be obtained at less cost by resorting to fractional factorial experiments which involve confounding of higher-order interactions. Fractional factorial experiments can be constructed by the use of orthogonal arrays. It has been investigated by Rao (1), Kempthorne (2), Plack- ette and Burman (3). Addelman (4), and Taguchi (5). The process of assigning an orthogonal array to a specific experim ent has been made easy by a graphical tool, called a linear graph (6), developed by Taguchi to represent interactions between pairs of columns in an orthogonal array. The use of linear graphs enables a scientist or an engineer to design (7) and analyze complicated experi
ments without requiring a basic knowledge of the construction of designs using a Galois field.
Manufacturing Process
Slack wax, a by-product in an oil refinery, is used in the m anufacture of
"Paraffin W ax." Slack wax contains 20 to 25% oil, whereas paraffin wax is permitted to have a maximum o f 3.5% oil (8). A hydraulic press process is used for removing excess oil from the slack wax which is melted in a tank at 80 to 90°C. The melted wax is allowed to settle down for about half an hour at room temperature. The sediment is tapped off and the molten material is then slabbed using galvanized iron trays. The slabs, which are about 2 in. in height, are wrapped in a filter canvas cloth and pressed at high pressures in a hydraulic press provided with a hot-water circulation arrangement. All o f the low melting fractions and the oil in the slack wax are collected separately. Further im puri
ties of the deoiled wax are removed by acid treatm ent, where the m aterial passes through decolorization, neutralization, and filtration. The resulting wax is slabbed again using galvanized iron trays. The flow diagram of the m anufac
turing process is given in Figure 1.
A composite sample drawn at random from the melted wax emerging from the filtration tank is tested for the requirements laid down in the relevant Indian Standard (8), of which the oil content is important. Its maximum perm issible content is 3.5% for Type 3 paraffin wax, and, if it is exceeded, the material has to be recycled for extraction for excess oil.
Background
The yield o f paraffin wax was 35 to 40% versus an expected yield of 60 to 65% in the chemical plant. Ten to 15% o f the finished product was recycled
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O IL EXT R A C T ION S T A G E
F igure 1. Process flow c h a rt, a —-A (8 0 - 9 0 ° C ) ; b —rem o v al o f sedim ent; c —slab c astin g ; d acid treatm ent; S .W .—slack w a x ; e —d e co lo riz a tio n and n e u tra liz atio n ; f—filtration; g slab
c as tin g ; P .W .-p a ra ffin w ax .
because of higher oil content. The actual and expected recovery at different stages are given in Table 1.
It is seen that the m axim um loss in recovery is at the oil extraction stage.
Therefore, the study was confined to the deoiling stage.
Pressing of slack wax in the hydraulic press is done in four stages. The pro
cess specifications at different stages are given in Table 2.
T a b le 1. A ctual and E xpected S tag ew ise R ecovery
S E R IA L
NO. S T A G E P R O C E S S
E X P E C T E D R E C O V E R Y (% BY M A SS)
A C T l'A L R E C O V E R Y
<7r BY M A SSi
1 M eltin g o f slack w ax R em o v al o f 2 -3 % im p u rities 9 7 -9 8 97 98
2 D eoiling R em o v al o f excess oil
(2 0 -2 5 % )
7 0 -7 5 40 50
3 A cid trea tm e n t, n e u tra li
zation, and filtration
D eco lo rizatio n 6 0 -6 5 35 40
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I'ROCH SS I’A R A M H T H R S S P E C I F I C A T I O N a. T em p era tu re o f inlet w ater at the tim e o f p ressin g o f slack w ax 6 5 °C b. Slage I pressin g (1 100 l b 'in .: ) tim e in m inutes 2 8 -3 0 e. Stage II pressin g (1550 lb /in .: ) tim e in m in u tes 10-12 v.1. Stage 111 pressin g (1800 lb /in .2) tim e in m in u tes 5 - 6 e. Stage IV pressin g (2100 lb /in .: ) tim e in m inutes 1-2
Investigation
Pressing the wax longer or pressing it at a higher tem perature results in low recovery. Similarly, pressing the wax for less time or at a lower tem perature results in recycling of the product due to the high oil content. Prelim inary observations revealed that the plant was not adhering to the given specifications fully. A batch of material was processed as per the specifications to exam ine whether the low recovery was due to
• lack of proper control
• inadequacy of process specifications, or
• both
Results of the trial runs were 58.5% recovery at the deoiling stage and an oil content of 2.9% . Though there was an increase o f about 10% in the yield, it was still much below the expected level of 70-75% . Thus, there was a need to evolve optimum process conditions which would m aximize recovery o f the deoiled wax o f desired quality. This could be achieved by conducting experi
ments using factors suspected to improve the yield and oil content. These were temperature and time of pressing at different pressure levels. Factors and levels determined after detailed technical discussion for the experiment are given in Table 3.
Another factor likely to influence the yield was the height of wax slab.
Throughout the experiments, this was maintained at a constant level of 2 in.
There are five factors, of which four are at two levels and one at three lev
els. A full factorial will require 24 x 3 = 48 trials. An orthogonal array approach was adopted to reduce the num ber of experimental runs. Five main effects, A, B, C, D, E, and the interactions A x B, A x C, A x D, and A x E were included in the investigation.
The linear graph technique (5) invented by Taguchi is used to design the present experiment.
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F A C T O R S 1
L EV EL S
T ;
A. T em p era tu re o f inlet w ater (°C ) 65 55 T im e in m in u tes o f pressin g at:
B. 1100 lb /in .2 20 28
C . 1550 lb /in .: 10 7
D . 1800 lb /in .: 6 3
E. 2100 lb /in .2 0 1 :
Linear Graph
Linear graphs represent the interaction information graphically a n d m a k e it easy to assign factors and interactions to the various columns o f a n o r t h o g o n a l
array with the help of an interaction table (6). In the linear graph, the c o l u m n s
of an orthogonal array are represented by the nodes and lines. When t w o n o d e s
are connected by a line, it means that the interaction of the t w o c o l u m n s
represented by the nodes is confounded with the column represented b v th e
line. In a linear graph, each node and each line has a distinct column n u m b e r
associated with it. Further, every column of the array is represented in its linear graph once and only once. The principal utility of linear graphs is f o r
creating a variety of different orthogonal arrays from the standard ones to lit real problem situations. The linear graphs are useful for creating 4-level a n d
3-level columns in 2-level orthogonal arrays. A 4-level factor in a 2-level orthogonal array is represented by two nodes and the line joining them.
The assignment o f a 3-level factor in a 2-level orthogonal array is done by first generating a 4-level column by the multilevel technique (5) and then one of the levels is made a dummy level. Multilevel and dummy-level techniques and interactions between 2- and 4-level factors are explained in the Appendix.
Selection of Design Layout Using Linear Graphs
The steps followed in the selection of the layout are as follows:
1. Express the information required in an experiment by means of linear graphs. In a graph, a main effect is represented by a node, and an interaction between two factors is represented by the line joining the nodes. This is termed the required linear graph.
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2. Compute the total degrees o f freedom required to estimate all the fac
torial effects that are of interest. The minimum num ber of experimental runs will be the total degrees of freedom computed to estimate the effects plus one. Choose an orthogonal array closest to the size of the experiment thus determined.
3. Com pare the standard linear graph (6) of the chosen array with the required linear graph as obtained in Step 1.
4. Modify the selected standard linear graph by deletion o f edges joining a pair of nodes or by joining unconnected nodes as required so as to make the standard graph correspond to the required linear graph. Thus, each factorial effect on the required linear graph is made to correspond with each column number on the standard or modified linear graph, respec
tively.
.V Assign each factor to the respective column o f the standard orthogonal table.
The required linear graph for the present experiment is given in Figure 2.
The degrees of freedom (d.f.) required to estimate all main effects and the interactions AB, AC, AD, and A E is 11. The minimum num ber o f experimental runs is 11 + 1 = 12 and the nearest orthogonal array is L 16 ( 2 15) .
Therefore, the experiment is designed as an L 16 orthogonal array, the layout of which is given in Table 4.
Figure 2. Required linear graph.
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C O L U M N
NO. 1 2 3 4 5 6 7 8 9 10 11 12 1 ' 14
1 i
1 1
i i
1 1
1 1
1 1
1 1
1 1
1 2
1 2
1 2
1 i
1 1 1
3 1 i 1 2 2 2 2 1 1 1 1 T t >
4 1 i 1 2 2 2 2 2 2 2 2 1 I 1 !
5 1 2 2 1 1 2 2 1 1 2 2 1 i : ;
6 1 2 2 1 1 2 2 2 2 1 1 2 1 1
7 1 2 2 2 2 1 1 1 1 2 T 2 1 1
8 1 2 2 2 2 1 1 2 2 1 1 1 1 2 '
9 2 1 7 1 2 1 2 1 2 1 1 1 2 1 2
10 2 1 2 1 2 1 2 2 1 j 1 i : 1
11 2 1 2 2 1 2 1 1 2 1 2 1 2 1
12 2 1 2 2 1 2 1 2 1 2 1 1 2 1
13 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1
14 2 2 1 1 2 2 1 2 1 1 2 T 1 1 2
15 2 2 1 2 1 1 2 1 2 2 1 -) 1 1 2
16 2
I T
2 i '—Y---1>
2 V. _
1 1
V
2 J
2 1 1 2 1
" v
T •> 1
G ro u p 1 G roup 2 G ro u p 3 G ro u p 4
Figure 3. Standard linear graph.
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T able 5. L ay o u t o f the E x p erim en t: L |6 ( 2 15)
E X P E R IM E N T
A (
### 1
)
F A C T O R C O L U M N N O . D
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B (6)
E (2, 8, 10)
C (12)
1 1 1 1 1 1
2 1 1 1 2 2
3 1 2 2 1 2
4 1 2 2 2 1
5 I 1 2 3 1
6 1 1 2 2 2
7 1 2 1 3 2
8 1 2 1 2 1
9 2 1 1 1 1
10 2 1 1 2 2
11 2 2 2 1 2
12 2 2 2 2 1
13 2 1 2 3 1
14 2 1 2 2 2
15 2 2 1 3 2
16 2 2 1 2 1
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Consider the standard linear graph |Fig. 3] (6). By erasing lines 14 and 15, five columns such as 4, 5, 11, 14, and 15 are made free. Column 1 1 is utilized for representing the interaction between columns 1 and 10 (6). Node 1 is joined to node 4 because column 5 (the interaction between 1 and 4) is free. Unutil
ized columns 14 and 15 are used for estimating the error. The linear graph for the present experiment is given in Figure 4.
Assignment of the factors to the columns was done from Figure 4. The experiment layout is given in Table 5.
Factor A (i.e., the tem perature of inlet water), whose levels are difficult to change, is assigned to column 1 of the L 16( 2 15i) table (primary zone) (6).
Response
Responses considered during the experimentation were (i) yield and
(ii) oil content of the deoiled wax
Conduct of the Experiment
Each trial required about 100 kg of slack wax for pressing in 11 daylight hydraulic presses. Two tons of material were melted to obtain a homogeneous melt with respect to oil content. Five samples were taken to determine the oil content in the slack wax, the average of which was found to be 20.2% . The melted wax was solidified in galvanized iron trays to obtain slabs of constant height (2 in.). A canvas cloth free from holes and other defects and capable of withstanding a pressure of up to 2500 lb /in .2 was used for wrapping the slab.
Eleven slabs, each wrapped in the cloth, were pressed through hydraulic presses as per the conditions stipulated in the layout of the experiment. The weight was taken before and after pressing so as to arrive at the yield. The deoiled wax was then melted in a small tank. Two samples were then taken and tested for oil content as per the Indian Standard (8). The yield and the oil con
tent for the 16 trials along with the actual physical layout are given in Table 6.
Analysis and Results
Analysis of variance (ANOVA) (9) was carried out on yield and oil content data. The results are given in Tables 7 and 8.
It is seen that factor A (the tem perature of the inlet water), factor C (time of pressing at Stage II), factor E (time of pressing at Stage IV), and interactions between tem perature and time of pressing at Stage I and Stage II (i.e., A x B and A x C interactions) are significant. The last column in the ANOVA table
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T a b le 6. Physical L ayout o f the E x p erim en t and R esponses on Y ield and P ercen t Oil C on ten t at th e O E xtraction P rocess
SERIAL MO.
E X P E R IM E N T A L C O N D IT IO N S T im e (in m in) at p ressin g pressu re
M A SS (in Slack
W ax
kgs) O F D eoiled
W ax
Y ield o f D eoiled
W ax in P ercen t by M ass
O il C o n ten t o : D eoiled W a x
for S a m p le
1 2
A T em p.
(°C )
± 1 ° C B 1100
± 1 0 0 lb /in .2
C 1550
± 1 0 0 lb /in .2
D 1800
± 1 0 0 lb /in .2
E 2100
± 1 0 0 lb /in .2
1 65 20 10 6 0 96.5 60.8 6 3 .0 0 2 .8 0 3 . 0 5
2 65 20 7 6 1 96.0 62.2 6 4 .7 9 2.85 3 . 1 9
3 65 28 7 3 0 96.0 59.3 6 1.77 2 .7 0 3 . 1 0
4 65 28 10 3 1 96.0 58.5 60.9 3 2 .7 0 2 . 9 0
5 65 28 10 6 2 9 5 .0 5 7 .0 6 0 .0 0 2.55 2 . 8 0
6 65 28 7 6 1 96.5 59.5 61.6 5 2 .9 0 3 . 1 0
7 65 20 7 3 2 9 8 .0 6 2 .0 6 3.27 2.95 3 . 1 9
8 65 20 10 3 1 95.5 6 0 .0 6 2 .8 2 2.7 6 3 . 1 5
9 55 20 10 6 0 95.0 64.5 6 7 .8 9 3.53 3 . 7 3
10 55 20 7 6 1 96.5 69.5 7 2.02 3.28 3 . 4 5
11 55 28 7 3 0 96.5 72.5 7 5.13 3.45 3 . 2 0
12 55 28 10 3 1 9 7 .0 6 8 .0 7 0 .1 0 3.28 3 . 0 6
13 55 28 10 6 2 100.5 6 8 .0 6 7 .6 7 3.12 2 . 9 5
14 55 28 7 6 1 96.0 69.5 7 2.40 3.12 3 . 2 5
15 55 20 7 3 2 96.5 6 8 .0 7 0.47 3.19 3 .3 2
16 55 20 10 3 1 96.0 6 4 .0 6 6.67 3 .4 0 3 .2 5
gives the (p) percentage (degrees of contribution) for critical factors. The 97%
of total variation is explained by the critical factors.
ANOVA for oil content is given in Table 8.
Here e x, the error due to the experimental condition, is not significant.
Therefore, a pooled estimate o f error has been computed and the main effects and interactions have been tested against this pooled error.
Here the factor A, the tem perature of the inlet water, and factor B, the time of pressing at 1100 lb /in .2, are significant, 58.1% o f the total variation being explained by the critical factors A and B.
The average responses for different levels of significant factors in the analy
ses for yield and oil contents as well as for different combinations o f AB and AC (significant interaction for yield) are computed and given in Table 9.
The effect curves for critical factors are given in Figure 5.
The best levels of significant factors on yield and oil content based (from Table 9) on average responses are summarized in Table 10.
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S O U R C E O F DI-XiREHS O F SU M O F M E A N SU M O F
V A R IA T IO N F K M .D O M S Q U A R E S S Q U A R E S F P (K )
A 1 25 6 .9 6 256.96 4 2 8 .2 7 a 77.8
B 1 0 .1 0 0 .1 0 h
C 1 31.42 31.42 5 2 .3 7 “ 9.2
D 1 0 .1 9 0 .1 9 h
E 2 5.38 2.69 4 .4 8 c 1.3
A X B 1 19.76 19.76 3 2 .9 3 “ 5.8
A X C 1 10.50 10.50 1 7 .5 0 “ 3.0
A x D 1 0 .5 8 0 .5 8 b
A x E 2 1.71 0 .8 6 b
E rro r 4 2 .8 0 0 .7 0
T otal 15 3 2 9 .4 0 97.11
P ooled e rro r 9 5.38 0 .6 0
■' Significant at 1O'/f . b Pooled with error.
‘ Significant at 5 c/<.
Table Values: p Computation
b\ , 6 at 0.05 = 5.99; at 0.01 = 13.74; s,
Pa = ---— x 100 S T
A2 , 6 at 0.05 = 5.14; at 0.01 = 10.92; 256.96 - 0.60
= --- x 100 329.4
i-2 , 9 at 0.05 = 4.26; at 0.01 = 8.02; = 77.8
T a b le 8. A N O V A o n O il C o n ten t o f the D eo iled W ax S O U R C E O F D E G R E E S O F SU M O F M E A N S U M O F
V A R IA T IO N F R E E D O M S Q U A R E S S Q U A R E S F P ( % )
A 1 1.08413 1.08413 4 7 .2 8 “ 4 7 .8
B 1 0 .2 6 4 6 3 0.26 4 6 3 1 1 .5 4 “ 10.9
C 1 0 .0 4 5 7 5 0.04 5 7 5 2 .0 0
D 1 0.00011 0.0001 l b
E 2 0 .1 3 8 8 0 0 .0 6 9 4 0 3.03
A X B 1 0 .0 0 8 8 3 0 .0 0 8 8 3 h
A x C 1 0 .0 5 5 3 3 0 .0 5 5 3 3 2.41
A x D 1 0 .0 0 7 5 6 0 .0 0 7 5 6 b
A x E 2 0 .1 0 6 1 0 0 .0 5 3 0 b 2.31
e\ 4 0 .0 3 3 3 4 0 .0 0 8 3 3 N .S .C
S T { 15 1.74455
e2 16 0 .4 7 7 6 5 0 .0 2 9 8 5
S T 31 2 .2 2 2 2 0
P o o led e r ro r
E ' 23 0 .5 2 7 4 9 0 .0 2 2 9 3
“ Significant at 1 %.
b Pooled with error.
L N.S. = not significant.
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YIELD% OF OILCONTENTYIELD 60
50
3 . 2
3 . 0
2. 8
75 70 65 60
5 5 ( A 2 )
TE M P ( ° C ) A
65(Aj)
5 5 ( A 2 ) 65(Aj)
7 ( C 2 ) l O f C j )
T I ME ( MI NS) C —
2 0 ( B j ) 2 8 ( B Z)
T E M P ( ° C ) A TIM E ( M I N S ) B
IN TERAC TIO N E F F E C T S
A x B
B1
5 5 ( A 2 ) 6 5 ( Aj) 5 5 ( A 2 ) 6 5 ( Aj)
T E M P ( ° C ) A — T EM P ( ° C ) A —
Figure 5. Effect curves on yield and oil content: Main effects and interactions.
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F A C T O R / Y IE L D O IL C O N T E N T
L E V E L (9c) ('/,)
6 2 .2 8 2.92
A 2 70.2 9 3.2 9
B\ 3 .1 9
b2 3.01
c , 64.88
### c2
6 7 .6 9
A \ B t 6 3.47
A \ B 2 6 1 .0 9
A 2B t 69.26
a2b2 7 1.32
C| 6 1 .6 9
a,c2 6 2.87
a2c, 6 8.08
a2c2 7 2 .5 0
E] 66.9 5
e2 66.7 2
Ei 65.35
Table 10. Best L ev els o f C ritical F acto rs B E ST L E V E L
R E S P O N S E O F C R IT IC A L F A C T O R S
Y ield A 2 , B i , C2 , E)
O il co n ten t A ] , B 2
Optimum Combination
An examination o f the best level of significant factors in the above analysis reveals one area of conflict. The first level o f factor A (tem perature of the inlet water) is found to be better for oil content, whereas the second level of A is better for yield. Because the yield is more at A 2 and the oil content at A 2 is 3.29% which is well within the maximum lim it specified (3.5%) in the stand
ard, level A 2 is preferred. The interaction AB and A C was significant, and the maximum yield was obtained for the combination A 2B2 and A 2 C2 . Levels B-, and C2 become the right choice for factors B and C, respectively. The level for the noncritical factors (D) was chosen as D2 , the lower time for pressing.
Thus, the optim um combination arrived at is A 2B2 C2D 2E X.
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The expected results with regard to yield and oil content for the optim um combination are given by
Yield = A 2B2 C2 + E x — T
= 73.76 + 66.95 - 66.28
= 74.3%
Oil = A 2 + B2 - T
= 3.2 + 3.01 - 3.10
= 3.2%
Confirmatory Trials
The results of the two confirmatory trials carried out with the optimum com
bination (A 2B2 C2D 2El ) are given in Table 11.
Thus, the recovery of deoiled wax has increased from 58.5% (see section titled Investigation) to the average yield level of 72.7% and also the average oil content is below the maximum specified.
Implementation
The optimum combination thus achieved is implemented by the plant on a regular basis and it has increased the recovery of paraffin wax from the initial 35-40% to over 60% (final yield). Recycling of paraffin wax for high oil con
tent is totally eliminated. Thus, it has been possible to realize an approximate saving of about Rs. one million (\$40000) per annum.
Conclusion
It is been shown that a fractional factorial experiment using the orthogonal array layout developed by Taguchi has helped in identifying the critical process param eters and their best levels for improving the yield as well as quality (oil content). The yield o f paraffin wax has improved to a level very close to the
T able 11. R esu lts o f C o n firm ato ry T ria ls
T R IA L R E S P O N S E
Y IE L D IN
% B Y M A SS
O IL C O N T E N T % BY M A S S F O R S A M P L E
1 2
I 7 1 .4 3 3 .1 0 2.95
II 73.9 5 3.12 3 .0 0
A v e ra g e 7 2.69 3.04
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t h e o r e t i c a l l y e x p e c t e d y i e l d . T h e 1 0 - 1 5 % r e c y c l i n g o f t h e m a t e r i a l s f or e x c e s s oil is t ot al l y e l i m i n a t e d .
The experimentation has been quite economical because the results are achieved involving only 16 trials, whereas a full factorial experiment would have required 48 trials. Even some of the suspected first-order interactions, which turn out to be significant, could also be studied.
The experimentation has been highly successful for the yield improvement, as almost all the variation (97%) is explained by the significant main effect and interaction (Table 6). Such a high percentage of the explained variation resulted in highly reproducible results with respect to yield. This has gone a long way in securing a consistently high yield. A milestone achieved was an increase in profit by 6.5% .
Appendix
M ultilevel and Dumm y-Level Techniques and Interaction Between 2- and 4- Level Factors
Multilevel Technique
This technique is useful in designing fractional experiments when the levels of different factors are not the same. For such an experim ent, a multilevel arrangement is applied; i.e., to arrange a 4- or 8-level column in 2-level series orthogonal tables, or to arrange a 9- or 27-level column in 3-level series orthogonal tables.
Let us consider the problem of accommodating a 4-level factor in the 2-level orthogonal array series. In the linear graph, the representation o f a 4-level fac
tor is made by the two nodes and the edge joining them. In other words, we use three columns o f the array for a 4-level factor. The two columns corresponding to the two nodes give four possible level combinations: (1,1), (1,2), (2,1), and (2,2). We use the following one-to-one correspondence to obtain the corresponding levels of the 4-level factor.
(1.1 ) 1 (2,1)______ 3
(1.2 ) 2 (2,2)______ 4
The assignment using the multilevel technique is explained as follows:
Let us assume that A has four levels and B, C, D, and E have two levels each. The assignment using the linear graph is shown in Figure 6. Table 12 gives the assignment to an orthogonal a rra y .
Dummy-Level Techique
The dummy-level technique is especially useful for accom modating 2-level factors in 3-level orthogonal array series or accom modating 3-level factors in 4-level orthogonal series.
In the above example, suppose factor A is at the 3-level. W ith the help of the multilevel technique, a 4-level column (1 ,2 ,3 ) is first generated. Because fac-
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A
F ig u re 6. L in ear graph for 4 x 24 design.
T able 12. A ssig n m en t o f 4 X 24 D esign in L8( 2 7) U sing M ultilevel T ech n iq u e
E X P E R IM E N T A B C D E
N O . 1 2 3 f~2 3 4 5 6 1
1 1 1 1 1 1 1 1 1
2 1 1 1 1 2 2 2 2
3 1 2 2 2 1 1 7 2
4 1 2 2 2 2 2 i 1
5 2 1 2 3 1 2 l 2
6 2 1 2 3 2 1 2 1
7 2 2 1 4 1 2 L. 1
8 2 2 1 4 2 1 1 2
T able 13. A ssig n m en t o f 3 X 24 d esign in Lg ( 2 7) L ayout U sin g D u m m y -L ev el T e c h n iq u e
E X P E R IM E N T B C D E
N O . 1 2 3 4 5 6 7
1 1 I 1 1 1
2 1 2 2 2 2
3 2 1 1 2 2
4 2 2 2 1 1
5 3 1 2 1 2
6 3 2 1 2 1
7 1’ 1 2 2 1
8 1' 2 1 1 2
Note: 1 —dummy level.
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tor A consists of only three levels, the more important level of A is repeated whenever the symbol 4 appears in column (1 ,2 ,3 ). For example, it the first level of A is more important, then this level is replicated more often. For instance,
A i = A | , A
### 2
= A j , At, = A\$ , A4 = A |
Table 13 gives the assignment of 3 X 24 design in L s (2 7) layout using the dummy-level technique.
Interaction Between 2- and 4-Level Factors
Interaction between a 2-level factor and a 4-level factor is represented by three interaction columns obtained by joining the node representing the 2-level factor with two nodes and line joining the two nodes representing the 4-level factor.
Let column 1 represent a 2-level factor, say A, and columns 2, 4, and 6 represent a 4-level factor, say B. Consider the triangle whose vertices are 1, 2, and 4. The interaction between columns 1 and 2 is column 3; the interaction between columns 2 and 4 is column 6; the interaction between columns 1 and 6 is column 7 (6). This information can be pictorially represented as given in Figure 7 by drawing a perpendicular from node 1 to the base (2,4). The interaction A x B is then given by columns 3 ,5 , and 7.
F ig u re 7. L in e a r g ra p h re p re se n tin g in te ra c tio n b e tw ee n 2 - an d 4 -lev e l fa c to rs.
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Acknowledgment
The author thanks Dr. M. Jeyachandra o f the Pennsylvania State U niversity for his help and suggestion in finalizing this article.
References
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5. T aguchi, G enichi, S ystem o f E xp e rim e n ta l D e sig n , V olum es 1 and 2, U N IP U B , N e w Y o r k , and A m erican S u p p lier In stitu te , D e a rb o rn , M I.
6. T aguchi, G enichi, T ables o f O rth o g o n a l A rra ys a n d L in ea r G raphs. M a ru ze n , T o k y o , 1962.
7 . A nand, K . N ., In c re a sin g M a rk et S hare T h ro u g h Im p ro v ed P ro d u ct and P ro c e ss D esign: A n E xperim ental A p p ro a ch , Q ual. E n g .. 3(3) 3 6 1 -3 7 0 (1991),
8. I.S. 4654-1974 S p ecifica tio n f o r P araffin W ax. B ureau o f Indian S ta n d ard s, N ew D e lh i, 1974.
9 . C h ak rav arti, I. M ., L ah a , R. G ., and Roy J ., H a n d b o o k o f M eth o d s o f A p p lie d S ta tis tic s , V ol. 1, P re n tic e-H all, E n g le w o o d C liffs, N J , 1985.
About the Author: K. N. Anand is Head of SQC & OR Unit at Indian Statistical Institute, Bangalore. The author has over 20 years o f consultancy and teaching experience in the field of Quality M anagement and application o f SQC tech
niques to a variety o f Indian industries.
Updating...
## References
Related subjects : | 9,475 | 27,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.932149 |
https://www.enotes.com/homework-help/how-write-quadratic-function-given-3-points-1-2-1-239531 | 1,516,562,858,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890823.81/warc/CC-MAIN-20180121175418-20180121195418-00707.warc.gz | 908,955,627 | 8,698 | # How to write quadratic function if are given 3 points (1,2),(1,3) and(0,-2)?
giorgiana1976 | Student
We'll write the general form of the quadratic:
ax^2 + bx + c = y
The given 3 points belong to the graph of quadratic function. If the graph passes through the given points, that means that the coordinates of the points verify the equation of the quadratic.
The point (1,2) belongs to the graph if and only if
2 = a*1^2 + b*1 + c
a+b+c = 2 (1)
The point B(1,3) belongs to the graph if and only if
3 = a+b+c (2)
We notice from (1) and (2) that 2 = 3, impossible.
The graph of the quadratic cannot pass through the points (1,2) and (1,3), same time.
So, there is no quadratic function to pass through the given points. | 216 | 730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.865543 |
https://sio2.mimuw.edu.pl/c/pa-2019-1/s/349477/source/ | 1,723,744,753,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641311225.98/warc/CC-MAIN-20240815173031-20240815203031-00597.warc.gz | 410,566,836 | 6,571 | 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 #include #define rok first #define dp second using namespace std; pair < int , int > butelka[2000][2000]; pair inf[2001000]; int n,k,pom,l; bool sortme(pair a, pair b) { if(butelka[a.first][a.second].rok <= butelka[b.first][b.second].rok) return 1; return 0; } int main() { cin >> n >> k; cin >> pom; butelka[1][1].rok=pom; butelka[1][1].dp=1; cin >> pom; butelka[2][1].rok=pom; butelka[2][1].dp=2; cin >> pom; butelka[2][2].rok=pom; butelka[2][2].dp=2; l=3; inf[0]=make_pair(1,1); inf[1]=make_pair(2,1); inf[2]=make_pair(2,2); for (int i=3; i<=n; i++) for (int j=1; j<=i; j++) { cin >> pom; butelka[i][j].rok=pom; inf[l].first=i; inf[l].second=j; l++; if(j==1 || j==i) { butelka[i][j].dp=i; } butelka[i][j].dp= (butelka[i-1][j-1].dp) + (butelka[i-1][j].dp)+ 1; butelka[i][j].dp-= butelka[i-2][j-1].dp; } sort(inf,inf+(n*(n+1)/2),sortme); for (int i=0; 1; i++) { if (butelka[inf[i].first][inf[i].second].dp<=k) { cout << butelka[inf[i].first][inf[i].second].rok; return 0; } } } | 561 | 1,161 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-33 | latest | en | 0.083083 |
https://beta.geogebra.org/m/denwnh9j | 1,708,994,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474669.36/warc/CC-MAIN-20240226225941-20240227015941-00268.warc.gz | 131,031,806 | 21,877 | # Thin-Slicing a Geometry Proof
This is how I would "thin slice" this proof.
## No Constraints (Givens)
This figure below has no constraints (Givens). Grab and move the points around. There are two triangles in the figure. Are they congruent? Are they ever congruent? Under what conditions would they be congruent?
## One Constraint (Given)
This figure below has just one constraint (Given): Segment SP Segment PT. Grab and move the points around. Are the two triangles in the figure congruent now? Could they be congruent? Are they ever congruent? Under what conditions would they be congruent?
## One Constraint (Given)
This figure below has just one constraint (Given): Angle SPQ Angle TPQ. Grab and move the points around. Are the two triangles in the figure congruent now? Could they be congruent? Are they ever congruent? Under what conditions would they be congruent?
## Two Constraints (Givens)
This figure below has two constraints (Givens):
1. Angle SPQ Angle TPQ , and
2. Segment SP Segment PT
Grab and move the points around. Are the two triangles in the figure congruent now? Are they ever congruent? Under what conditions would they be congruent? Would they ever NOT be congruent? | 275 | 1,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.92639 |
https://www.easyunitconverter.com/long-division-calculator | 1,709,585,006,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00766.warc.gz | 728,927,209 | 30,089 | # Long Division Calculator
Enter the value that you want to calculate long division..
Enter Dividend: Enter Divisor:
## Long Division(with step by step expression)
1. Problem:
2. Dividend:
3. Divisor:
4. Quotient:
5. Remainder:
1. 455 ÷ 8
2. 455
3. 8
4. 56
5. 7
0 5 6 8 4 5 5 0 4 5 4 0 5 5 4 8 7
# Long Division Calculators
Long division is a method for dividing large numbers, providing detailed insight into the division process.
#### How the Long Division Calculator Works
This tool automates the long division process, displaying the quotient, remainder, and all calculation steps.
#### Understanding the Process
Enter the dividend and divisor to see the detailed division process, including the quotient and remainder.
#### Practical Applications
From educational purposes to financial calculations, this calculator simplifies tasks involving large numbers.
#### Frequently Asked Questions
How do you perform long division?
Divide the dividend by the divisor step by step, subtracting and bringing down digits until complete.
Can the calculator handle decimal points?
Yes, it can process divisions resulting in decimal points with detailed steps.
What is the remainder in long division?
The remainder is the part of the dividend that is left over after the division process when the dividend cannot be evenly divided by the divisor.
How can I use this tool for teaching?
This calculator can be used as a teaching aid to demonstrate the long division process step by step, making it easier for students to understand and follow along.
Is there a limit to the size of numbers the calculator can handle?
While the calculator is designed to handle large numbers, extremely large values may exceed its processing capabilities.
How does the calculator handle divisions that result in repeating decimals?
The calculator will show the division process up to a certain number of decimal places. It may round the final result or indicate a repeating pattern.
Can this calculator be used for polynomial long division?
This version of the calculator is optimized for numerical long division. Polynomial long division may require a different tool or method.
How do I interpret the step-by-step results?
The step-by-step results show how each part of the dividend is divided by the divisor, similar to how you would perform long division on paper, making the calculation process transparent. | 496 | 2,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-10 | longest | en | 0.861421 |
https://www.coursehero.com/file/6825141/I-0603-Problem/ | 1,524,367,456,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945484.58/warc/CC-MAIN-20180422022057-20180422042057-00215.warc.gz | 761,429,594 | 401,980 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
I-06.03 Problem
# I-06.03 Problem - I-06.03 Following is the X4 bank...
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I-06.03 Following is the September 30, 20X4 bank reconciliation for the Quiet Moose Lodge. You are also provided with the October check register and bank statement. Utilize this information to prepare October's bank reconciliation and related adjusting entry. You may assume that any discrepancies between the check register and bank statement relate to recording errors in the accounts of the Quiet Moose, and not the bank. Ending balance per bank statement \$18,344.07 Add: Deposits in transit 2,505.55 Deduct: Outstanding checks #3444 175.00 \$ #3446 1,908.09 (2,083.09) Correct cash balance \$ 18,766.53 Ending balance per company records \$18,696.53 Add: Interest earnings 80.00 Deduct: Service charges (10.00) Correct cash balance \$ 18,766.53 1 of 3
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I-06.03 DATE PARTY REF # CHECK DEPOSIT Balance 1-Oct Balance 18,766.53 \$ 2-Oct Gomez 3448 145.99 \$ - 18,620.54 5-Oct Deposit - 3,400.00 \$ 22,020.54 7-Oct Bryers 3449 387.97 - 21,632.57 7-Oct Morton 3450 1,204.67 - 20,427.90 7-Oct Lee 3451 4,664.50 - 15,763.40 10-Oct Morici 3452 43.23 - 15,720.17 10-Oct LaCorx 3453 2,990.44 - 12,729.73 11-Oct Benson 3454 1,100.31 - 11,629.42 12-Oct Void 3455 - - 11,629.42 13-Oct Morgan 3456 695.77 - 10,933.65 13-Oct Russell 3457 788.87 - 10,144.78 14-Oct Deposit - 3,476.88 13,621.66 17-Oct Lowen 3458 3,664.34 - 9,957.32 19-Oct Post Office 3459 45.45 - 9,911.87 20-Oct Nguen 3460 677.21 - 9,234.66 30-Oct Behn 3461 499.00 -
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{[ snackBarMessage ]} | 615 | 1,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-17 | latest | en | 0.744166 |
http://mathsmd.com/tag/sides/ | 1,670,332,448,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00803.warc.gz | 28,416,959 | 11,360 | Lines of symmetry in a regular Square A shape may have more than one line of symmetry depending on its angles and the lengths of each side. For example, a rectangle has two lines of symmetry because it can be folded in half both horizontally and vertically to create mirror images. A regular square is […]
Lines of symmetry in a Rectangle What is line of symmetry ? If a given line splits a given figure into two identical halves, we say that the line has a line of symmetry or that the figure is symmetrical about the line and line is called a line of symmetry or the axis of […]
Lines of symmetry in a regular Dodecagon Line of Symmetry A shape may have more than one line of symmetry depending on its angles and the lengths of each side. For example, a rectangle has two lines of symmetry because it can be folded in half both horizontally and vertically to create mirror images. A […]
Lines of symmetry in an Isosceles Triangle An isosceles triangle has 1 line of symmetry. An isosceles triangle is one that has two sides that are the same length. The triangle has two equal interior sides. Only if a triangle has two equal sides can it be called to be isosceles. Below figure shows lines […]
Lines of symmetry in a regular Hexagon What is line of symmetry ? The line of symmetry is the imaginary line where we could fold the image and have both halves match exactly. or If a given line splits a given figure into two identical halves, we say that the line has a line of […]
Lines of Symmetry In Regular Polygons What is line of symmetry? An object is considered to be symmetrical, if when folded or sliced in half, both halves are mirror images of one another. An object is said to have a line of symmetry when one half of it is the mirror image of the other. […]
Right Angle Triangle – Definition – Formula – Properties What is a Right Angle Triangle? A right angle triangle is a type of triangle. A right angle triangle plays an important role in trigonometry. A triangle in which one of the interior angle is 90ยบ or a right angle is a right triangle. Right Triangle […] | 442 | 2,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.947667 |
http://mathhelpforum.com/math-puzzles/101409-need-help-one-please-print.html | 1,527,366,729,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00027.warc.gz | 172,965,794 | 3,058 | # need help with this one please
• Sep 9th 2009, 05:22 PM
denzel
need help with this one please
hi people. im new to this. looks like ye got a good site going here, can comeone help me with this problem please as i cant get the right answer at all. here is the question:
ailis,bre,casy and darcy are travelling on a tourist bus. each of them has 3 baskets in each hand. there are 4 cats in every basket. every cat has 5 kittens. at the bus stop casy and darcy got off. mary got on.
how many legs are there on the bus??
appreciate all replies
thanks alot people
• Sep 9th 2009, 06:01 PM
denzel
2150
• Sep 10th 2009, 06:42 AM
Niall101
This was a quiz on tv last night for €10,000. first off I wouldnt waste your money calling in as its impossible to get through. its a scam. But here is the answer(s)
Each basket has 4 cats and 20 kittens =24
So each person has 6*24 Animals = 144
Each animal has 4 legs = 144*4=576 legs per each person Plus their own 2 legs = 578
Two of them get off and there are 2 left so 578*2=1156 plus mary who got on who has 2 legs = 1156+2=1158 legs in total.
BUT they dont say who is driving the bus so if there is also a driver it will be 1160 legs
And it also doest say if there is a tour guide on the bus either as it is a tour bus lol
But if each cat has 5 kittens, This could also mean that 2 cats have 5 kittens between them (mother and father) so both would still have 5 kittens which would give a much lower answer.
AND it doesnt say the kittens are in the baskets.
They ask these questions which are very vague in the hope that nobody wins and they get all your money ;) | 457 | 1,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-22 | latest | en | 0.975501 |
https://gamedev.stackexchange.com/questions/172580/unity-how-to-add-bones-to-procedural-mesh | 1,563,450,487,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525627.38/warc/CC-MAIN-20190718104512-20190718130512-00210.warc.gz | 397,924,683 | 34,377 | # Unity - How to add bones to procedural mesh
Basically this is the set up: An N amount of prefabs are bundled together into a new GameObject. This is then merged into a single main mesh composited of M amount of material submeshes. The shape is standing upwards on the Y axis and is a bunch of arms. The goal is to move these by adding bones.
I wasn't able to find many examples on how to properly assign bones to a procedural mesh which isn't a simple cube.
The problem is assignment of vertices to the boneweights. I'd like for the transform Y=0 vertices to have no weight and the Y=furthest vertices have full weight. I am not sure how the assignment should be done.. Any pointers? :)
So far the Lower and Upper bone transforms are in the right place, but when moving them the mesh isn't bending as I hoped it would in a nice curve, half of the vertices are simply moved uniformly with a thin mesh line connecting the two halves.
private void _AddBones()
{
if(m_SkinnedRenderer != null)
{
// Make bone weight for vertices
BoneWeight[] weights = new BoneWeight[m_MainMesh.vertexCount];
int halfway = (weights.Length / 2);
for (int i = 0; i < weights.Length; i++)
{
if(i < halfway)
{
weights[i].boneIndex0 = 0;
weights[i].weight0 = 1;
}
else
{
weights[i].boneIndex0 = 1;
weights[i].weight0 = 1;
}
}
m_MainMesh.boneWeights = weights;
// Make bones and bind poses
Transform[] bones = new Transform[2];
Matrix4x4[] bindPoses = new Matrix4x4[2];
bones[0] = new GameObject("Lower Bone").transform;
bones[0].parent = m_MergedRoot.transform;
bones[0].localRotation = Quaternion.identity;
bones[0].localPosition = Vector3.zero;
bindPoses[0] = bones[0].worldToLocalMatrix * m_MergedRoot.transform.localToWorldMatrix;
bones[1] = new GameObject("Upper Bone").transform;
bones[1].parent = m_MergedRoot.transform;
bones[1].localRotation = Quaternion.identity;
bones[1].localPosition = new Vector3(0, 1.5f, 0);
bindPoses[1] = bones[1].worldToLocalMatrix * m_MergedRoot.transform.localToWorldMatrix;
m_MainMesh.bindposes = bindPoses;
m_SkinnedRenderer.bones = bones;
m_SkinnedRenderer.sharedMesh = m_MainMesh;
m_SkinnedRenderer.rootBone = bones[0];
}
}
• – Draco18s Jun 5 at 1:24
• @Draco18s Sorry, it's just hard to get help regarding the subject. – Mads M Jun 5 at 4:29 | 608 | 2,268 | {"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-2019-30 | longest | en | 0.828735 |
https://www.analystforum.com/t/is-lm-curve/81678 | 1,652,701,296,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00548.warc.gz | 743,442,892 | 5,311 | # IS/LM Curve
Hello,
Here it is in a nutshell:
The IS curve relates real aggregate income to interest-rate levels: for a given level of real aggregate income, what interest rate is required to ensure that savings equals investment + fiscal balance + trade balance? The curve is then drawn with real aggregate income on the horizontal axis, and equilibrium (real) interest rate on the vertical axis.
The LM curve relates demand for money to supply for money: for a real aggregate income and a given price level, what interest rate is required to balance the demand for money with the supply for money? You don’t draw a single LM curve; you draw one LM curve for each price level, with real aggregate income on the horizontal axis and (real) interest on the vertical axis.
You combine these two to get the aggregate demand curve: start with real aggregate income on the horizontal axis, move up to the IS curve (so you now know the real interest rate), then look at all of the LM curves and pick the one that goes through that (real aggregate income, real interest rate) point: the price level for that LM curve is the price level that corresponds to that level of real aggregate income. Repeat that for all levels of real aggregate income, getting a corresponding price level for each. The plot of real aggregate income (horizontal axis) vs. price level (vertical axis) is the aggregate demand curve.
Voilà!
IS curve on CFAI Text Vol 2 P.224 Exhibit
It is Income (Y) on the horizontal axis. But what is it on the vertical axis ?
Is interest rate the same thing as price level? Interest being the “price” of a loan?
No.
By “price level” they mean something like CPI: the overall price of goods and services.
Thank you so much !
You’re welcome.
Thats really treasure. | 379 | 1,778 | {"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} | 2.578125 | 3 | CC-MAIN-2022-21 | latest | en | 0.903309 |
https://www.coursesidekick.com/mathematics/571652 | 1,701,533,423,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00894.warc.gz | 797,868,322 | 56,125 | # IMG0137
.jpeg
SplashlLearn Comparison Word Problems Read the problem and circle the correct equations to solve the problem. Sophia got 15 points in a video game. She got 6 fewer points than Jade. How many points did Jade get? Jade Sophia | 6+2=15 | Inagroup, 26 people like football. 12 fewer people like basketball. How many people in the group like basketball? Football Basketball | 26-12=7 | There are 26 comic books in shop A and 15 more comic books in shop B. How many comic books are in shop B? Shop B Shop A 26 (26+15=7 | [ 26-15=7 | 147 | 541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-50 | latest | en | 0.932145 |
https://studysoup.com/tsg/204320/fundamentals-of-engineering-thermodynamics-8-edition-chapter-11-problem-11-6 | 1,576,138,940,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540542644.69/warc/CC-MAIN-20191212074623-20191212102623-00508.warc.gz | 559,781,116 | 12,129 | # Determine the specific volume of water vapor at 20 MPa and
## Problem 11.6 Chapter 11
Fundamentals of Engineering Thermodynamics | 8th Edition
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Problem 11.6
Determine the specific volume of water vapor at 20 MPa and 4008C, in m3 /kg, using the (a) steam tables. (b) compressibility chart. (c) RedlichKwong equation. (d) van der Waals equation. (e) ideal gas equation of state.
Step-by-Step Solution:
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Week 4: Film continued & Lecture Laborers were only paid \$1/day for working from dawn to dusk 1920s-1930s: over 100,000 Filipinos immigrated to the US o the imagination of the laborers was that the USA was all about wealth – ironically, they arrived during the Great Depression lived in camps kids worked 10 hours/day and made 8cents/hour didn’t have...
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##### ISBN: 9781118412930
The full step-by-step solution to problem: 11.6 from chapter: 11 was answered by , our top Engineering and Tech solution expert on 11/14/17, 08:39PM. Fundamentals of Engineering Thermodynamics was written by and is associated to the ISBN: 9781118412930. Since the solution to 11.6 from 11 chapter was answered, more than 263 students have viewed the full step-by-step answer. The answer to “Determine the specific volume of water vapor at 20 MPa and 4008C, in m3 /kg, using the (a) steam tables. (b) compressibility chart. (c) RedlichKwong equation. (d) van der Waals equation. (e) ideal gas equation of state.” is broken down into a number of easy to follow steps, and 37 words. This full solution covers the following key subjects: equation, compressibility, specific, determine, chart. This expansive textbook survival guide covers 14 chapters, and 1738 solutions. This textbook survival guide was created for the textbook: Fundamentals of Engineering Thermodynamics, edition: 8.
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We're here to help | 649 | 2,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-51 | latest | en | 0.902374 |
https://topic.alibabacloud.com/a/leetcode-letter-combinations-of-a-phone-number-font-classtopic-s-color00c1dealphanumericfont-combination-of-phone-numbers_8_8_31339044.html | 1,716,358,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058531.78/warc/CC-MAIN-20240522040032-20240522070032-00364.warc.gz | 515,907,296 | 16,794 | # Leetcode letter combinations of a phone number (alphanumeric combination of phone numbers)
Source: Internet
Author: User
translation
给定一个数字字符串，返回所有这些数字可以表示的字母组合。一个数字到字母的映射（就像电话按钮）如所示。输入：数字字符串“23”输出：["ad""ae""af""bd""be""bf""cd""ce""cf"]备注:尽管以上答案是无序的，如果你想的话你的答案可以是有序的。
Original
Original
Given a digitstring,returnAll possible letter combinations that the Numbercould represent. A Mapping ofDigit toLetters (Just Like on theTelephone buttons) is given below. Input:digitstring "All"Output: ["AD","AE","AF","BD","Be","BF","CD","CE","CF"]. Note:although the aboveAnswer is inchlexicographical order, your answer could beinchAny order you want.
Code
It seems that I still use C # handy point, one go ... Mostly with recursion, because I don't know < Span class= "Mrow" id= "mathjax-span-67" > d i g i t s The length of the end is how much, impossible to write countless < Span class= "Mrow" id= "mathjax-span-75" > f o r e a c h To judge.
Public classsolution{ilist<string> list =Newlist<string> (); Public Solution() {list. Insert (0,"ABC"); List. Insert (1,"Def"); List. Insert (2,"Ghi"); List. Insert (3,"JKL"); List. Insert (4,"MnO"); List. Insert (5,"PQRS"); List. Insert (6,"TUV"); List. Insert (7,"WXYZ"); } Publicilist<string>lettercombinations(stringdigits) {ilist<string> result =Newlist<string> ();if(digits. Length = =0)returnResultif(digits. Length = =1) {foreach(varAinchList. ElementAt (int. Parse (digits[0]. ToString ())-2) {result. Insert (0, a.tostring ()); } }intCount =0; ilist<string> temp = lettercombinations (digits. Substring (1, digits. Length-1));foreach(varAinchList. ElementAt (int. Parse (digits[0]. ToString ())-2)) {foreach(varRestinchTemp) {result. Insert (count++, a.tostring () + rest); } }returnResult }}
Here is a copy of the C + + code, frankly, I can not write such C + + code ...
classSolution { Public: vector<string>Lettercombinations (stringDigits) { vector<string>Ansif(digits.size () = =0)returnAnsintdepth = Digits.size ();stringTMP (Depth,0); DFS (TMP,0, depth, ans, digits);returnAns }voidDfsstring&tmp,intCURDEP,intDepth vector<string>&ans,string&digits) {if(CURDEP >= depth) {ans.push_back (TMP);return; } for(inti =0; I < DIC[DIGITS[CURDEP]-' 0 '].size (); + + i) {TMP[CURDEP] = Dic[digits[curdep]-' 0 '][i]; DFS (TMP, CURDEP +1, depth, ans, digits); }return; }Private:stringdic[Ten] = {{""},{""},{"ABC"},{"Def"},{"Ghi"},{"JKL"},{"MnO"},{"PQRS"},{"TUV"},{"WXYZ"}};};
Leetcode letter combinations of a phone number (alphanumeric combination of phone numbers)
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 971 | 3,487 | {"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} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.591504 |
https://www.queryhome.com/puzzle/945/how-many-biscuits-had-it-eaten-on-the-first-day | 1,618,868,163,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038917413.71/warc/CC-MAIN-20210419204416-20210419234416-00389.warc.gz | 1,059,800,850 | 32,486 | # How many biscuits had it eaten on the first day?
+1 vote
122 views
Gauri's puppy was growing fast. In the first five days since she got it it had eaten 100 dog biscuits. If each day it had eaten 6 more than the previous day .
How many biscuits had it eaten on the first day?
posted May 22, 2014
Let a be the # of Biscuits eaten 1st day.
So.
``````a + (a+6) + (a+6+6) + (a+6+6+6) + (a+6+6+6+6) = 100
5a + 60 = 100
``````
Thus,
``````a = 8 Biscuits.
``````
+1 vote
....................8...............
Similar Puzzles
+1 vote
Arun had some money. he gave one rupee to beggar and half of the remaining money was spent for buy food. Same thing happened second day he gave one rupee to beggar from remaining money of first day. this continued on the six day he had 0 rupee, then how much money he had on first day?
+1 vote
He gave one rupee to beggar and half of the remaining money was spent for buy something.
Same thing happened second day he gave one rupee to beggar from remaining money of first day.
This continued on till the sixth day when he had 0 rupee after giving 1 rupee to beggar,
Then how much money he had on first day?
A cyclist rides from one point to another over five days.
On the first day she covers one eighth of the total distance.
The next day she covers one quarter of what is left.
The following day she covers two fifths of the remainder
and on the fourth day half of the remaining distance.
The cyclist now has 25 miles left.
How many miles has she travelled?
A cycling group are on a five day ride for charity.
On the first day they cover one third of the total distance, the next day they cover one third of what is left, the following day they cover one quarter of the remainder and on the fourth day they cover half of the remaining distance.
The group now have 15 miles left to cycle.
How many miles have they travelled? | 492 | 1,877 | {"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} | 3.9375 | 4 | CC-MAIN-2021-17 | longest | en | 0.986347 |
https://discuss.codecademy.com/t/faq-project-board-slides-for-foodwheel-customer-types/373555 | 1,701,357,144,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00651.warc.gz | 256,822,315 | 10,659 | # FAQ: Project: Board Slides for FoodWheel - Customer Types
This community-built FAQ covers the “Customer Types” exercise from the lesson “Project: Board Slides for FoodWheel”.
Paths and Courses
This exercise can be found in the following Codecademy content:
## FAQs on the exercise Customer Types
There are currently no frequently asked questions associated with this exercise – that’s where you come in! You can contribute to this section by offering your own questions, answers, or clarifications on this exercise. Ask or answer a question by clicking reply () below.
If you’ve had an “aha” moment about the concepts, formatting, syntax, or anything else with this exercise, consider sharing those insights! Teaching others and answering their questions is one of the best ways to learn and stay sharp.
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Is the .values method necessary for Histograms? It wasn’t covered in the Data Science Path and I seem to be able to get the same graph using:
plt.hist(customer_amount.price,
range=(0, 200),
bins=40)
plt.xlabel('Total Spent')
plt.ylabel("Number of Customers")
plt.title('Customer Expenditure Over 6 Months')
plt.show()
2 Likes
I can’t complete this exercise. i think that the problem is that codeacademy don’t load the database necessary to successfully run my code.
3 Likes
from matplotlib import pyplot as plt
import pandas as pd
it worked for me
3 Likes
There’s no explanation of what .values does in the Data Science course. I don’t understand why it’s necessary and this awful platform does nothing to help you.
10 Likes
From what I’ve gathered adding .values converts a series in a data frame to a list of values (array).
Though no real explanation was given, this method was used in exercise 2 (What cuisines does FoodWheel offer) to get the values from a series and use them as labels in the pie chart.
Hey can someone explain why we have to select the price values for the hist?
Since I thought that it was already included in the customer_amount variable
plt.hist(customer_amount.price.values, range=(0, 200), bins=40)
2 Likes
You do not have to specify .values. I’m not sure why this was included in the Solution, since you can remove it and generate the same graph.
I also noticed that you can’t use
plt.xlim([0, 200])
to set the range, even though it accomplishes the task. Oh well.
1 Like
It’s a shame that I’m finding so many errors or inconsistencies within the data science career path lately. Up until recently it has been great, however now I’m finding some really discouraging issues including but not limited to the following:
1. some critical info has not been taught to complete the given exercise or is (for the first time) mentioned in a “hint” for that exercise.
2. if there is a different (correct) way to write the code than what is expected in the exercise, the parser rejects the input as invalid code – different than sending a simple message to the user that an alternate input syntax is desired.
3. inconsistencies with having to import libraries at the start of the exercise
Very frustrating when trying to learn!
4 Likes
I was wondering if anyone else was having the same problem with .values(). It seems that we all are!
I didnt seem to need .values. This code worked fine for me;
plt.hist(customer_amount.price, bins = 40, range = (0, 200))
plt.xlabel(‘Total Spent’)
plt.ylabel(‘Number of Customers’)
plt.title(‘Customer Spend’)
plt.show()
I totally get you. I’ve noticed several inconsistencies as well.
1 Like
Is it just me or did anyone else not see the resulting graph even with plt.show()? | 855 | 3,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.888184 |
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,252 | 6,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2013-20 | latest | en | 0.783914 |
https://www.bearnaiserestaurant.com/helpful-tips/how-do-you-find-the-standard-error-of-a-confidence-interval/ | 1,723,664,936,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00837.warc.gz | 514,638,725 | 11,893 | # How do you find the standard error of a confidence interval?
## How do you find the standard error of a confidence interval?
Compute the standard error as σ/√n = 0.5/√100 = 0.05 . Multiply this value by the z-score to obtain the margin of error: 0.05 × 1.959 = 0.098 . Add and subtract the margin of error from the mean value to obtain the confidence interval. In our case, the confidence interval is between 2.902 and 3.098.
## How do you calculate 95 confidence interval and standard error?
To compute the 95% confidence interval, start by computing the mean and standard error: M = (2 + 3 + 5 + 6 + 9)/5 = 5. σM = = 1.118. Z.95 can be found using the normal distribution calculator and specifying that the shaded area is 0.95 and indicating that you want the area to be between the cutoff points.
What is the error with 95% confidence?
The sample mean plus or minus 1.96 times its standard error gives the following two figures: This is called the 95% confidence interval , and we can say that there is only a 5% chance that the range 86.96 to 89.04 mmHg excludes the mean of the population.
### What is standard error and confidence interval?
Standard error of the estimate refers to one standard deviation of the distribution of the parameter of interest, that are you estimating. Confidence intervals are the quantiles of the distribution of the parameter of interest, that you are estimating, at least in a frequentist paradigm.
### How is standard error calculated?
Standard error is calculated by dividing the standard deviation of the sample by the square root of the sample size.
How do you calculate standard error from p value?
(a) CI for a difference
1. 1 calculate the test statistic for a normal distribution test, z, from P3: z = −0.862 + √[0.743 − 2.404×log(P)]
2. 2 calculate the standard error: SE = Est/z (ignoring minus signs)
3. 3 calculate the 95% CI: Est –1.96×SE to Est + 1.96×SE.
## How do I calculate the standard error of the mean?
SEM is calculated simply by taking the standard deviation and dividing it by the square root of the sample size.
## How do you calculate STD error in Excel?
As you know, the Standard Error = Standard deviation / square root of total number of samples, therefore we can translate it to Excel formula as Standard Error = STDEV(sampling range)/SQRT(COUNT(sampling range)).
What is the standard error in statistics?
What Is the Standard Error? The standard error (SE) of a statistic is the approximate standard deviation of a statistical sample population. The standard error is a statistical term that measures the accuracy with which a sample distribution represents a population by using standard deviation.
### How do you solve for standard error?
How do you calculate standard error? The standard error is calculated by dividing the standard deviation by the sample size’s square root. It gives the precision of a sample mean by including the sample-to-sample variability of the sample means.
### How do you calculate standard error of sample?
Compute the standard error, which is the standard deviation divided by the square root of the sample size. To conclude the example, the standard error is 5.72 divided by the square root of 4, or 5.72 divided by 2, or 2.86.
How do I calculate standard error?
Steps to Calculate Standard Error Standard error is calculated by dividing the standard deviation of the sample by the square root of the sample size. Calculate the mean of the total population.
## Why do we calculate standard error?
By calculating standard error, you can estimate how representative your sample is of your population and make valid conclusions. A high standard error shows that sample means are widely spread around the population mean—your sample may not closely represent your population.
## How do I calculate standard error in Excel?
Now that you have calculated for both variables, you can use one final Microsoft Excel function to easily calculate the standard error for your data set. Click on the cell you wish to store the value of your standard error in, and enter “=[Standard deviation result cell]/SQRT([Count result cell])” as the formula.
• September 13, 2022 | 929 | 4,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-33 | latest | en | 0.842874 |
https://physics.stackexchange.com/questions/373839/matter-state-between-liquid-and-gas | 1,652,823,632,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00521.warc.gz | 531,191,946 | 67,206 | # Matter state between liquid and gas
I know about the 3 clasical states of matter: Solid, Liquid and Gas. But I've also observed how between Solid and Liquid there are some materials that will have an "in-between" phase state: Liquids that get thicker as the temperature decreases until they freeze, or solids that become more malleable when the temperature raises until they melt. See for instance glass (silicon dioxide), or metals (iron, aluminium, sodium, etc.)
But I've never seen such behaviour in between liquid to gas. I've always seen all transitions between these two states as binary - the state is either gas or liquid. The closer thing I can think of is Clouds, but AFAIK they are actually made of tiny droplets of liquid water that are just on suspension in the air. Gas can get its density drastically changed by heating it up, but it always makes a big jump when going from gas to liquid.
Which makes me wonder - Why? Why do we have a in-between state between solid and liquid, but not between liquid and gas?
• @JohnRennie I don't think crystalization has something to do with transition from liquid to gas or viceversa. Dec 11, 2017 at 16:32
• Oops, yes, sorry I misread your question. Dec 11, 2017 at 16:33
I would argue that you do see certain temperature-dependent responses in a liquid relative to a gas analogous to those that you see relative to a solid.
The two examples you give are the temperature-dependent viscosity of a liquid and the yield stress of a solid decreasing with increasing temperature.
For the first example, in the same way that a liquid's viscosity increases with cooling (towards a solid's high viscosity associated with its creep rate), it correspondingly decreases with heating (towards the low viscosity of the gas phase).
In the second example, the yield strength of a solid (which is one of its material properties) decreases with increasing temperature, approaching the limiting condition of a fluid, which has no yield strength. (Put another way, liquids and gases cannot sustain a shear force.)
But as I heat a liquid, its surface tension (which is another material property) decreases, approaching the limiting condition of a gas, which has no positive surface tension.
So I don't see the situation as being quite as one-sided as you describe.
Supercritical fluids are the answer. Thee reason why they are usually non-conventional or as mentioned as liquid crystals, I believe, is purely educational. I mean, you are not going to find out supercritical fluids in your daily life...So, unfortunately, and despite all the new states of matter we do know at current time, they are not included in the academic curriculum. A pity, because who knows what applications of these states could bring in the future. Anyway, you are going to experience more plasmas than supercritical fluids in you average life-time, so...I presume, that people think this kind of matter is not very relevant for all practical purposes...Here in wikipedia has a nice entry: supercritical fluids in wikipedia
There is a state of atter between gas and liquid called "supercritical fluid." There is a thread here on that, including discussion on why supercritical fluids are not considered a separate state of matter.
• I don't think it's correct to say that supercritical fluids are between liquid and gas. The reality is, as the pressure and temperature increase, the differences between liquid and gas phases diminish: Their densities become more alike, their viscosities become more alike, etc., until you reach the critical point where there is no difference at all. "Supercritical" is a realm beyond liquid or gas. Dec 11, 2017 at 17:35 | 781 | 3,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | longest | en | 0.958903 |
https://byjusexamprep.com/csir-net/law-of-gravitation-applies-to | 1,726,831,874,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00035.warc.gz | 130,281,194 | 29,418 | # Law of Gravitation Applies To?
By BYJU'S Exam Prep
Updated on: September 13th, 2023
A) Any pair of bodies
B) The earth and the moon
C) The planets around the sun
D) The earth and the objects on earth
The law of gravitation applies to any pair of bodies. The Law of gravitation says that each and every particle attracts each and every other particle in the universe with the help of a force that is considered directly proportional to the product of their masses, and they are also inversely proportional to the square of the distance between their centers.
This law was given by Issac Newton, and it is also known as the ‘First Great Unification’. However, when Newton first presented it in Book 1 of the unpublished text in April in the year 1686 to the Royal Society, Robert Hooke made a claim that Newton had obtained the inverse square law from him.
## Equation Of Law Of Gravitation
The equation for universal gravitation is-
where:
• F denotes the force between the masses;
• G denotes the gravitational constant (6.674×10−11 m3⋅kg−1⋅s−2);
• m1 denotes the first mass
• m2 denotes the second mass
• r denotes the distance between the centers of the masses.
### Newton’s Law Of Gravitation Applies to:
Newton’s law of gravitation applies to all pairs of bodies in the universe. Newton’s gravity description is accurate for many practical purposes and is widely used. Any deviations from it are small when the dimensionless quantities ∅/c2 and (v/c)2 are much less than one, where ∅ is the gravitational potential,v2 is the velocity of the objects being studied, and C is the speed of light in a vacuum.
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https://beta.geogebra.org/m/Sm2amkbw | 1,653,747,239,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016853.88/warc/CC-MAIN-20220528123744-20220528153744-00103.warc.gz | 171,803,777 | 10,414 | # Basic Triangle Centers
From Paul Yiu's MathFest presentation in 2008
## Basic Triangle Centers, Circles, and Lines.
Circumcircle, nine-point circle, the Euler line, the circumcenter, centroid, nine-point center, and the orthocenter. HN : NG : GO = 3 : 1 : 2
## The Orthocentroidal Circle
N and O are inverses with respect to the orthocentroidal circle.
## Euler Line Reflections
Reflect the Euler line in each side of the triangle. The reflections concur at a point on the circumcircle.
## The Fermat Point
Construct equilateral triangles on the exterior of the triangle. The lines AA', BB', and CC' are concurrent. This point is the Fermat point.
## The Negative Fermat Point
Construct equilateral triangles on the interior of the triangle. The lines AA', BB', and CC' are concurrent. This point is the Negative Fermat point. | 208 | 840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-21 | latest | en | 0.763172 |
http://openstudy.com/updates/52146685e4b0450ed75e0a0f | 1,516,473,255,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00781.warc.gz | 266,328,102 | 9,194 | • anonymous
Together, Sam and Alice can paddle a canoe across the bay in 15 minutes. When Alice paddles alone, it takes her 16 minutes longer than when Sam paddles alone. How long does it take each person to paddle across the bay alone? Alice takes ..... minutes to paddle across the bay alone. Sam takes ...... minutes to paddle across the bay alone.
OCW Scholar - Single Variable Calculus
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Not the answer you are looking for? Search for more explanations. | 326 | 1,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-05 | latest | en | 0.364163 |
https://calculat.io/en/length/inches-to-mm/317 | 1,701,520,601,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100399.81/warc/CC-MAIN-20231202105028-20231202135028-00176.warc.gz | 192,342,231 | 21,753 | # Convert 317 inches to mm
## How many mm is 317 inches?
317 Inches is equal to 8051.8 Millimeters
## Explanation of 317 Inches to Millimeters Conversion
Inches to Millimeters Conversion Formula: mm = in × 25.4
According to 'inches to mm' conversion formula if you want to convert 317 (three hundred seventeen) Inches to Millimeters you have to multiply 317 by 25.4.
Here is the complete solution:
317″ × 25.4
=
8051.8 mm
(eight thousand fifty-one point eight millimeters)
## About "Inches to Millimeters" Calculator
This converter will help you to convert Inches to Millimeters (in to mm). For example, it can help you find out how many mm is 317 inches? (The answer is: 8051.8). Enter the number of inches (e.g. '317') and hit the 'Convert' button.
## Inches to Millimeters Conversion Table
InchesMillimeters
7670.8 mm
7696.2 mm
7721.6 mm
7747 mm
7772.4 mm
7797.8 mm
7823.2 mm
7848.6 mm
7874 mm
7899.4 mm
7924.8 mm
7950.2 mm
7975.6 mm
8001 mm
8026.4 mm
8051.8 mm
8077.2 mm
8102.6 mm
8128 mm
8153.4 mm
8178.8 mm
8204.2 mm
8229.6 mm
8255 mm
8280.4 mm
8305.8 mm
8331.2 mm
8356.6 mm
8382 mm
8407.4 mm
## FAQ
### How many mm is 317 inches?
317″ = 8051.8 mm | 411 | 1,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-50 | latest | en | 0.62976 |
http://www.exampleproblems.com/wiki/index.php?title=NT1&oldid=39099 | 1,368,948,084,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696384213/warc/CC-MAIN-20130516092624-00055-ip-10-60-113-184.ec2.internal.warc.gz | 453,301,183 | 5,230 | # NT1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Prove that there are infinitely many primes.
Let $q=2\cdot 3\cdot 5\cdot\cdot\cdot p+1$ where p is a prime. Then q is not divisible by any prime less than or equal to p. Now either q is prime or it is divisible by primes greater than p and less than q. In either case there is a prime greater than p, which proves the theorem.
-Euclid
##### Toolbox
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## Exceeding the maximum beam scattering angle
For questions about writing Gosia inputs
### Exceeding the maximum beam scattering angle
I have a question regarding an error output by Rachel when calculating simulated yields for the inverse kinematics Coulomb excitation of a 36Ar projectile on a 28Si target.
I have set up the calculation to integrate theta meshpoints (with endpoints defined by my particle detector's angular extent) between 10 and 30 degrees in the laboratory frame. From a kinematics spreadsheet calculator, I know that the maximum projectile scattering angle for this experiment at the “safe Coulex” energy is about 48.5 degrees in the laboratory frame (corresponding to a target recoil scattering angle of about 20 degrees). Thus, since in this inverse kinematics experiment the detector range spans the maximum scattering angle, I have input the target as the detected particle.
I've calculated the simulated yields for the logical experiments below (note that I've slowly decreased the minimum scattering angle to illustrate my soon-to-be-defined problem):
----------------------------------------------------------------------------
# 1 2 3 4 5 6
theta_min 15.0 21.0 17.0 15.0 13.0 11.0
theta_max 42.0 30.0 30.0 30.0 30.0 30.0
phi_min 0.0 0.0 0.0 0.0 0.0 0.0
phi_max 360.0 360.0 360.0 360.0 360.0 360.0
un-inv. Z 14 14 14 14 14 14
un-inv. A 28 28 28 28 28 28
Ebeam 63.0 63.0 63.0 63.0 63.0 63.0
Safe energy 64 64 64 64 64 64
Tgt. thickness [mg/cm^2] 0.6 0.6 0.6 0.6 0.6 0.6
Detected t t t t t t
Inverse kin. True True True True True True
Kinematic solution -- -- -- -- -- --
Norm to expt # 1 2 3 4 5 6
Somm. param. 33.0 33.0 33.0 33.0 33.0 33.0
# of Ge detectors 1 1 1 1 1 1
--------------------------------------------------------------------------------
I noticed that even though all of the theta_min and theta_max ranges for scattered target nuclei span the kinematics solution corresponding to the maximum projectile scattering angle (except Experiment 2), I only get the following error for Experiment 6:
--------------------------------------------------------------------------------
Gosia reported an error:
ERROR- MAXIMUM SCATTERING ANGLE IS 48.00 DEGREES FOR EXPERIMENT 6
********* END OF EXECUTION **********
---------------------------Error----------------------------
You have exceeded the maximum beam scattering angle
(calculated inelastically) for this experiment. There are
two ways that you might fix this problem: (1) Define the
experiment(s) so that the target particle is detected in the
case of inverse kinematics, or (2) Set the state for
calculation of the scattering kinematics to the ground state
using the Gosia Controls button, option k.
Done.
Fri Sep 30 13:11:01 PDT 2011 ...
Processing cannot continue properly due to the error above.
--------------------------------------------------------------------------------
I get this error for the same experiment (11deg<theta_lab<30deg) even if I shuffle their order, so it's not tied to the last experiment listed.
If I understand correctly, I have properly defined the experiments such that “the target particle is detected in the case of inverse kinematics” and therefore I shouldn't get this error? Or perhaps I am making a silly mistake. Either way, a hint would be helpful so I know whether I can put faith in my calculations or not.
Many thanks and cheers,
Phil
pvoss
Posts: 3
Joined: Wed Sep 28, 2011 8:00 pm
### Re: Exceeding the maximum beam scattering angle
Hi Phil,
Can you post the level data, so that I have the right Q-value? I will try to reproduce the error and fix it. The easiest way for me is in the "rachel" format. If you don't have that already, you can go to Tools-->"el" (export level scheme), save to a new file.
Also, I will have a new version ready in about a week (or ask me if you want to preview it) that will address bugs and have a better simulation output with understandable efficiencies instead of the poorly-defined "intrinsic efficiency" in the beta version.
hayes
Posts: 45
Joined: Tue Feb 08, 2011 2:13 pm
Location: Rochester, NY, USA
### Re: Exceeding the maximum beam scattering angle
Here is the level data:
---------------------------------------------------
18 36
gsb 0.0 + 0.0
gsb 2.0 + 1970.39
gsb 4.0 + 4414.36
-------------------------------------------------
I tried attaching the pickle.jar file too, but was not allowed to upload a file with ".jar" extension. Thank you very much for your help.
Cheers,
Phil
pvoss
Posts: 3
Joined: Wed Sep 28, 2011 8:00 pm
### Re: Exceeding the maximum beam scattering angle
Hi Phil,
I found a simple solution. If this is not acceptable, or if you think there are other accuracy questions, I can see if there is another way to avoid the error.
The problem comes partly from the inelastic kinematics using the Q-value of the 2+ state. (By default, Gosia and Rachel use the second state in the level scheme to get estimate the Q-value. It is necessary to pick a single Q-value for each calculation in a semi-classical code, but the calculation can be repeated, of course, to use different Q-values for each state.)
I got the simulation to run without errors by changing from the default Q-value of level #2 to the ground state (Q=0).
This is done by clicking "Gosia controls" at the left and following the prompts as below:
Code: Select all
`Menu: Physical parameters v Vacuum deorientation control k State for calculation of the scattering kinematics Data handling options t Normalization transition for all normalized yields **new** n Data set normalizations w Data set weights **new** Accuracy and calculation speed options i E-theta integration accuracy f Options for fitting of matrix elementsEnter choice (one character) from above: kCurrent state for calculation of kinematics is Band " gsb ", spin 2.0Band NUMBER of new kinematics state: 1Spin: 0Kinematics state set to Band " gsb ", spin 0.0`
There have been a few questions raised recently about using Q=0 for the scattering cross section, and repeatedly calculating yields of other states using their own Q-values. You can find more about this on the forum.
At the moment, for your 5th experiment, there is about a 10% change in the 2+ --> 0+ yield between the two Q-values. Obviously, this must be addressed in analysis, but this solution may be accurate enough for your simulation.
hayes
Posts: 45
Joined: Tue Feb 08, 2011 2:13 pm
Location: Rochester, NY, USA
### Re: Exceeding the maximum beam scattering angle
I think the problem is occurring when you happen to have a meshpoint at a target angle of 20.5° (exactly halfway between 11 and 30°). With your kinematics the maximum projectile angle should be almost exactly 48°, which corresponds to a target angle of 20.5° (this is calculating for the exit energy. Obviously it is more like 48.5° for the incident energy, but from your error message, it must be the exit energy which is triggering the error).
The code takes the angle of the meshpoint (which is a target angle) and converts using INVKIN into projectile angles. Then CMLAB calculates the maximum projectile angle. In principle, this should never fail. In fact, if there is a small rounding error, this could lead to angles very close to the maximum being considered beyond the maximum.
I wonder if it is possible to tweak the limits in your input (i.e. a small change) so it doesn't calculate a meshpoint at exactly the critical angle. I'm not sure how much would be needed, but perhaps not enough to make any real difference to the calculation. In which case that would be a better workaround than changing NCM, which changes the kinematics (i.e. the kinematics is calculated based on the ground state not the first excited state). Also I think it is by chance that this works, because it changes the floating point values just enough not to have the rounding error problem.
So this would seem to be a bug in my code (INVKIN). I'll have to think about how to fix it. Probably it will involve trapping the case of being only a little over the maximum angle in CMLAB which is part of the original Gosia.
Thank you for point it out.
warr
Posts: 7
Joined: Tue May 10, 2011 4:41 pm
### Re: Exceeding the maximum beam scattering angle
Hi Phil and Nigel,
Nigel's suggestion to tweak the meshpoints seems to work. Your (Phil's) input had 21 theta meshpoints for the problematic experiment (#6). I changed it to 18 meshpoints, and it worked. It also worked for 24 meshpoints and gave an identical output.
This is a better solution than being forced to change the Q-value. This way seems works for the Q of the 2+ state, and the integration is obviously converged, since I get identical output to 5 digits with 18 or 24 meshpoints.
I did this through the GUI. Here's how:
Click "Gosia controls", then follow these prompts.
Code: Select all
`Menu: Physical parameters v Vacuum deorientation control k State for calculation of the scattering kinematics Data handling options t Normalization transition for all normalized yields **new** n Data set normalizations w Data set weights **new** Accuracy and calculation speed options i E-theta integration accuracy f Options for fitting of matrix elementsEnter choice (one character) from above: iThe accuracy can be increased at a small expense of speedduring the "Integrated yields" operations by increasing thenumber of interpolation points for the beam energy, polarscattering angle (theta) mesh. The accuracy can be increased at a greater expense of speedby increasing the number of energy and theta meshpoints. NOTE: If you decrease the number of interpolation points,or meshpoints, you should check that the integrated(calculated) yield accuracy has not been reduced. See"integration accuracy" in the Help function.Present E-theta meshpoint settings: beam scattering energy theta-----------------------------------------------Experiment 1 : 20 30Experiment 2 : 20 20Experiment 3 : 20 20Experiment 4 : 20 20Experiment 5 : 20 20Experiment 6 : 20 21Would you like Rachel to choose the optimum settings for all experiments [y/N]? Experiment number to change, "all", or return to quit:6Enter one integer for energy meshpoints and one for theta on the same line. 20 18 beam scattering energy theta-----------------------------------------------Experiment 1 : 20 30Experiment 2 : 20 20Experiment 3 : 20 20Experiment 4 : 20 20Experiment 5 : 20 20Experiment 6 : 20 18Experiment number to change, "all", or return to quit:Quitting.Present settings of integration subdivisions for interpolation. beam scattering energy theta-----------------------------------------------Experiment 1 : 50 50Experiment 2 : 50 50Experiment 3 : 50 50Experiment 4 : 50 50Experiment 5 : 50 50Experiment 6 : 50 50Would you like to set the default number of meshpoints for all experiments [y/N]? Experiment number to change, "all", or return to quit:Quitting.`
hayes
Posts: 45
Joined: Tue Feb 08, 2011 2:13 pm
Location: Rochester, NY, USA
### Re: Exceeding the maximum beam scattering angle
I have found an error in my INVKIN subroutine. The variables tau and taup are not declared explicitly, so they are implicitly REAL*4. However, the equivalent variables in CMLAB are REAL*8. This means that the precision in INVKIN is less than in CMLAB. This won't make any noticeable difference to the result, but it can happen, as in your case, that it triggers the problem you have.
So the simple solution is to declare these two variables in the subroutine INVKIN as REAL*8.
Unfortunately, that is not the whole story. It is still possible to find cases where the difference between INVKIN and CMLAB (which are essentially a function and its inverse) still lead to small differences that can push the value of tau * SIN(theta_p/57.297795) above unity (we are talking of around 2 * 10^-16 above unity here). Since Gosia has to calculate the arcsine of this quantity, it will fail. This goes deep into the limitations of floating point arithmetic and is not easy to solve. So if other people hit this problem, the solution (after declaring tau and taup properly) is to tweak some parameter by a small amount. I found that changing the excitation energy by 0.01 eV (eV not keV!) was enough in the case I found. Fortunately, such cases are quite hard to find even when you are looking for them, so you have to be very unlucky to find them (after fixing the precision in INVKIN, that is).
warr
Posts: 7
Joined: Tue May 10, 2011 4:41 pm
### Re: Exceeding the maximum beam scattering angle
Note: At the moment we don't have the rachel patches in the version control system that Nigel manages. You *can* however use any recent version of Gosia with rachel, as long as you don't try to use the functions in the GUI that plot the Coulex amplitude or collision function. These functions are for troubleshooting only. The calculations we are talking about can be repeated with a new version, but there should be no difference since the integration is obviously converged in this case. (I tested this by changing the number of meshpoints in theta up and down and finding no difference in the yields reported.)
hayes
Posts: 45
Joined: Tue Feb 08, 2011 2:13 pm
Location: Rochester, NY, USA | 3,310 | 13,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.7417 |
https://www.jiskha.com/display.cgi?id=1284321185 | 1,516,256,281,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887067.27/warc/CC-MAIN-20180118051833-20180118071833-00656.warc.gz | 920,684,485 | 3,906 | posted by .
Write and simplify an expression for the perimeter of the triangle. side A)3x, side B) 5x-4 and side C)2x+5.
Thank you!
First, start out by simply adding the sides together into a single equation:
3x+5x-4+2x+5
Next, combine like items:
3x+5x+2x = +10x
-4+5 = +1
That would leave you with:
10x+1
That is as simplified as it gets, so that means you have your answer now.
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https://math.libretexts.org/Bookshelves/Pre-Algebra/Book%3A_Prealgebra_(OpenStax)/6%3A_Percents/6.E%3A_Percents_(Exercises) | 1,568,785,929,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573184.25/warc/CC-MAIN-20190918044831-20190918070831-00030.warc.gz | 559,504,763 | 22,027 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 6.E: Percents (Exercises)
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
## 6.1 - Understand Percent
In the following exercises, write each percent as a ratio.
1. 32% admission rate for the university
2. 53.3% rate of college students with student loans
In the following exercises, write as a ratio and then as a percent.
1. 13 out of 100 architects are women.
2. 9 out of every 100 nurses are men.
In the following exercises, convert each percent to a fraction.
1. 48%
2. 175%
3. 64.1%
4. $$8 \frac{1}{4}$$%
In the following exercises, convert each percent to a decimal.
1. 6%
2. 23%
3. 128%
4. 4.9%
In the following exercises, convert each percent to (a) a simplified fraction and (b) a decimal.
1. In 2012, 13.5% of the United States population was age 65 or over. (Source: www.census.gov)
2. In 2012, 6.5% of the United States population was under 5 years old. (Source: www.census.gov)
3. When a die is tossed, the probability it will land with an even number of dots on the top side is 50%.
4. A couple plans to have three children. The probability they will all be girls is 12.5%.
In the following exercises, convert each decimal to a percent.
1. 0.04
2. 0.15
3. 2.82
4. 3
5. 0.003
6. 1.395
In the following exercises, convert each fraction to a percent.
1. 3 4
2. 11 5
3. 3 5 8
4. 2 9
5. According to the Centers for Disease Control, $$\frac{2}{5}$$ of adults do not take a vitamin or supplement.
6. According to the Centers for Disease Control, among adults who do take a vitamin or supplement, $$\frac{3}{4}$$ take a multivitamin.
In the following exercises, translate and solve.
1. What number is 46% of 350?
2. 120% of 55 is what number?
3. 84 is 35% of what number?
4. 15 is 8% of what number?
5. 200% of what number is 50?
6. 7.9% of what number is $4.74? 7. What percent of 120 is 81.6? 8. What percent of 340 is 595? ## 6.2 - Solve General Applications of Percents In the following exercises, solve. 1. When Aurelio and his family ate dinner at a restaurant, the bill was$83.50. Aurelio wants to leave 20% of the total bill as a tip. How much should the tip be?
2. One granola bar has 2 grams of fiber, which is 8% of the recommended daily amount. What is the total recommended daily amount of fiber?
3. The nutrition label on a package of granola bars says that each granola bar has 190 calories, and 54 calories are from fat. What percent of the total calories is from fat?
4. Elsa gets paid $4,600 per month. Her car payment is$253. What percent of her monthly pay goes to her car payment?
In the following exercises, solve.
1. Jorge got a raise in his hourly pay, from $19.00 to$19.76. Find the percent increase.
2. Last year Bernard bought a new car for $30,000. This year the car is worth$24,000. Find the percent decrease.
## 6.3 - Solve Sales Tax, Commission, and Discount Applications
In the following exercises, find (a) the sales tax (b) the total cost.
1. The cost of a lawn mower was $750. The sales tax rate is 6% of the purchase price. 2. The cost of a water heater is$577. The sales tax rate is 8.75% of the purchase price.
In the following exercises, find the sales tax rate.
1. Andy bought a piano for $4,600. The sales tax on the purchase was$333.50.
2. Nahomi bought a purse for $200. The sales tax on the purchase was$16.75.
In the following exercises, find the commission.
1. Ginny is a realtor. She receives 3% commission when she sells a house. How much commission will she receive for selling a house for $380,000? 2. Jackson receives 16.5% commission when he sells a dinette set. How much commission will he receive for selling a dinette set for$895?
In the following exercises, find the rate of commission.
1. Ruben received $675 commission when he sold a$4,500 painting at the art gallery where he works. What was the rate of commission?
2. Tori received $80.75 for selling a$950 membership at her gym. What was her rate of commission?
In the following exercises, find the sale price.
1. Aya bought a pair of shoes that was on sale for $30 off. The original price of the shoes was$75.
2. Takwanna saw a cookware set she liked on sale for $145 off. The original price of the cookware was$312.
In the following exercises, find (a) the amount of discount and (b) the sale price.
1. Nga bought a microwave for her office. The microwave was discounted 30% from an original price of $84.90. 2. Jarrett bought a tie that was discounted 65% from an original price of$45.
In the following exercises, find (a) the amount of discount (b) the discount rate. (Round to the nearest tenth of a percent if needed.)
1. Hilda bought a bedspread on sale for $37. The original price of the bedspread was$50.
2. Tyler bought a phone on sale for $49.99. The original price of the phone was$79.99.
In the following exercises, find (a) the amount of the mark-up (b) the list price.
1. Manny paid $0.80 a pound for apples. He added 60% markup before selling them at his produce stand. What price did he charge for the apples? 2. It cost Noelle$17.40 for the materials she used to make a purse. She added a 325% markup before selling it at her friend’s store. What price did she ask for the purse?
## 6.4 - Solve Simple Interest Applications
In the following exercises, solve the simple interest problem.
1. Find the simple interest earned after 4 years on $2,250 invested at an interest rate of 5%. 2. Find the simple interest earned after 7 years on$12,000 invested at an interest rate of 8.5%.
3. Find the principal invested if $660 interest was earned in 5 years at an interest rate of 3%. 4. Find the interest rate if$2,898 interest was earned from a principal of $23,000 invested for 3 years. 5. Kazuo deposited$10,000 in a bank account with interest rate 4.5%. How much interest was earned in 2 years?
6. Brent invested $23,000 in a friend’s business. In 5 years the friend paid him the$23,000 plus $9,200 interest. What was the rate of interest? 7. Fresia lent her son$5,000 for college expenses. Three years later he repaid her the $5,000 plus$375 interest. What was the rate of interest?
8. In 6 years, a bond that paid 5.5% earned $594 interest. What was the principal of the bond? ## 6.5 - Solve Proportions and their Applications In the following exercises, write each sentence as a proportion. 1. 3 is to 8 as 12 is to 32. 2. 95 miles to 3 gallons is the same as 475 miles to 15 gallons. 3. 1 teacher to 18 students is the same as 23 teachers to 414 students. 4.$7.35 for 15 ounces is the same as $2.94 for 6 ounces. In the following exercises, determine whether each equation is a proportion. 1. $$\frac{5}{13} = \frac{30}{78}$$ 2. $$\frac{16}{7} = \frac{48}{23}$$ 3. $$\frac{12}{18} = \frac{6.99}{10.99}$$ 4. $$\frac{11.6}{9.2} = \frac{37.12}{29.44}$$ In the following exercises, solve each proportion. 1. $$\frac{x}{36} = \frac{5}{9}$$ 2. $$\frac{7}{a} = \frac{-6}{84}$$ 3. $$\frac{1.2}{1.8} = \frac{d}{6}$$ 4. $$\frac{\frac{1}{2}}{2} = \frac{m}{20}$$ In the following exercises, solve the proportion problem. 1. The children’s dosage of acetaminophen is 5 milliliters (ml) for every 25 pounds of a child’s weight. How many milliliters of acetaminophen will be prescribed for a 60 pound child? 2. After a workout, Dennis takes his pulse for 10 sec and counts 21 beats. How many beats per minute is this? 3. An 8 ounce serving of ice cream has 272 calories. If Lavonne eats 10 ounces of ice cream, how many calories does she get? 4. Alma is going to Europe and wants to exchange$1,200 into Euros. If each dollar is 0.75 Euros, how many Euros will Alma get?
5. Zack wants to drive from Omaha to Denver, a distance of 494 miles. If his car gets 38 miles to the gallon, how many gallons of gas will Zack need to get to Denver?
6. Teresa is planning a party for 100 people. Each gallon of punch will serve 18 people. How many gallons of punch will she need?
In the following exercises, translate to a proportion.
1. What number is 62% of 395?
2. 42 is 70% of what number?
3. What percent of 1,000 is 15?
4. What percent of 140 is 210?
In the following exercises, translate and solve using proportions.
1. What number is 85% of 900?
2. 6% of what number is $24? 3.$3.51 is 4.5% of what number?
4. What percent of 3,100 is 930?
## PRACTICE TEST
In the following exercises, convert each percent to (a) a decimal (b) a simplified fraction.
1. 24%
2. 5%
3. 350%
In the following exercises, convert each fraction to a percent. (Round to 3 decimal places if needed.)
1. $$\frac{7}{8}$$
2. $$\frac{1}{3}$$
3. $$\frac{11}{12}$$
In the following exercises, solve the percent problem.
1. 65 is what percent of 260?
2. What number is 27% of 3,000?
3. 150% of what number is 60?
4. Yuki’s monthly paycheck is $3,825. She pays$918 for rent. What percent of her paycheck goes to rent?
5. The total number of vehicles on one freeway dropped from 84,000 to 74,000. Find the percent decrease (round to the nearest tenth of a percent).
6. Kyle bought a bicycle in Denver where the sales tax was 7.72% of the purchase price. The purchase price of the bicycle was $600. What was the total cost? 7. Mara received$31.80 commission when she sold a $795 suit. What was her rate of commission? 8. Kiyoshi bought a television set on sale for$899. The original price was $1,200. Find: (a) the amount of discount (b) the discount rate (round to the nearest tenth of a percent) 9. Oxana bought a dresser at a garage sale for$20. She refinished it, then added a 250% markup before advertising it for sale. What price did she ask for the dresser?
10. Find the simple interest earned after 5 years on $3000 invested at an interest rate of 4.2%. 11. Brenda borrowed$400 from her brother. Two years later, she repaid the $400 plus$50 interest. What was the rate of interest?
12. Write as a proportion: 4 gallons to 144 miles is the same as 10 gallons to 360 miles.
13. Solve for a: $$\frac{12}{a} = \frac{−15}{65}$$
14. Vin read 10 pages of a book in 12 minutes. At that rate, how long will it take him to read 35 pages? | 3,103 | 10,496 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-39 | latest | en | 0.76356 |
http://www.algebra-answer.com/tutorials-2/converting-fractions/math-105-exam-ii.html | 1,521,899,165,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650685.77/warc/CC-MAIN-20180324132337-20180324152337-00785.warc.gz | 322,480,130 | 6,587 | # Math 105 Exam II
1. Factor completely .(10)
2. De termine the domain of f. (10)
The domain of g is the set of values x ∈ R for which the denominator x ^2-7x+6 is not
equal to 0. To obtain the values that must be excluded from R, set the denominator
equal to zero :
The domain of g is therefore
3. Divide and , if possible, simplify. (10)
Solution :
4. Find the LCD , then add and simplify . (10)
Solution:
5. Frank walks 2 km/h s lower than Peter . In the time it takes Peter to walk 8 km, Frank
walks 5 km. Find the speed of each person. (10)
Solution:
Since d = r * t, we have and t=8/x. Setting the equations equal we get
Peter walks at a rate of 16/3 km/h and Frank at a rate of 10/3 km/h.
6. Simplify. (10)
Solution:
7. Divide.
Solution:
We cannot use synthetic division since the divisor is not of the form x-a. We use long
division instead after rewriting the divisor as x^2 + 0x + 1. We obtain 3x^2 - 5x + 1.
8. Use the remainder theorem to find h(4), where . (10)
Solution:
Use 4 for the synthetic division.
So h(4) = 54.
9. Solve the equation for q. (10)
10. Simplify. (10)
Since the root is odd , we need no absolute values.
11. Divide and simplify. (10)
Since both x and y appear with odd powers under the radical in the original ex pression , they both had to be positive in the first place. Therefore, the final expression needs no absolute value.
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Case Problem 4
Data File needed for this Case Problem: Quality.xlsx
Karleton Manufacturing Carmen Garza is a quality control
manager at Karleton Manufacturing, a
manufacturing plant located in Trotwood, Ohio. One project
that Carmen oversees is the manufacture
of tin cans for a major food company. The can widths must be
consistent. To compensate for the fact
that metal working tools tend to wear down during the day,
the pressure behind the tools is increased
as the blades become worn. Quality control technicians
monitor the process to check that it remains
in control creating cans
Case Problem 4
Data File needed for this Case Problem: Quality.xlsx
Karleton Manufacturing Carmen Garza is a quality control
manager at Karleton Manufacturing, a
manufacturing plant located in Trotwood, Ohio. One project
that Carmen oversees is the manufacture
of tin cans for a major food company. The can widths must be
consistent. To compensate for the fact
that metal working tools tend to wear down during the day,
the pressure behind the tools is increased
as the blades become worn. Quality control technicians
monitor the process to check that it remains
in control creating cans whose widths are neither too
narrow nor too wide. Carmen has recorded the
widths of four cans from 39 batches in an Excel workbook
that she wants to use to determine whether
the process is in control. One standard for determining
whether a process is in control is whether
the average value from a process batch falls within the
lower and upper control limits. The workbook
itself is in need of quality control as some of the formulas
are not calculating correctly. You will fix
these and then enter the remainder of the formulas needed in
the worksheet. Complete the following:
1. Open the Quality workbook located in the Excel3 _
Case4 folder included with your Data Files,
and then save the workbook as Quality Control Analysis in
the location specified by your instructor.
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 749 | 3,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-04 | latest | en | 0.930532 |
http://mathhelpforum.com/algebra/184242-help-about-sets-please-soon-possible-im-going-need-like-3-1-2-hours-print.html | 1,527,185,010,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866733.77/warc/CC-MAIN-20180524170605-20180524190605-00615.warc.gz | 201,062,158 | 3,565 | # HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hours.)
• Jul 7th 2011, 03:03 PM
edriann
HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hours.)
Hi. I have this homework which wasn't taught by the professor. So I thought you could help me as soon as possible.
Prove:
n(AuBuC) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)
Thanks a lot guys. Really do appreciate it if i could get it before 10:30. It's 7:00 here now. :)
• Jul 7th 2011, 03:24 PM
Also sprach Zarathustra
Re: HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hou
Quote:
Originally Posted by edriann
Hi. I have this homework which wasn't taught by the professor. So I thought you could help me as soon as possible.
Prove:
n(AuBuC) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)
Thanks a lot guys. Really do appreciate it if i could get it before 10:30. It's 7:00 here now. :)
Rule number 6:
Quote:
Do not cheat. Teachers expect questions that form part of an assessment that contributes towards the final grade of a student to be the work of that student and not the work of others. For that reason, MHF policy is to not knowingly help with such questions. If a question presents in such a way as to suggest that it falls in this category, a Moderator will Close the thread (with an explanation as to why the thread was closed). The original poster can always send a pm to the Moderator to discuss the situation but if the Moderator is unconvinced, then the thread will remain closed. When a Moderator is certain that a particular member is attempting to cheat, that member will be banned and the offending thread removed. Where possible, the institute at which the member studies at will also be notified and a copy of the thread provided. This may sound harsh but the fact is that MHF places a high premium on academic honesty and integrity.
• Jul 7th 2011, 03:26 PM
edriann
Re: HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hou
Oops. Sorry. I think I missed that part. So maybe you could just show me how to do it? The professor really didn't teach it. It's not graded anyway, but he said if we don't pass something, we might get scr**ed. Please?
• Jul 7th 2011, 03:38 PM
Also sprach Zarathustra
Re: HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hou
Quote:
Originally Posted by edriann
Oops. Sorry. I think I missed that part. So maybe you could just show me how to do it? The professor really didn't teach it. It's not graded anyway, but he said if we don't pass something, we might get scr**ed. Please?
No, for that, but a little hint, yes.
Use n(AUB)= n(A)+n(B)-n(A∩B).
• Jul 7th 2011, 03:44 PM
edriann
Re: HELP about Sets please. As soon as possible. (I'm going to need it like 3 1/2 hou
Quote:
Originally Posted by Also sprach Zarathustra
No, for that, but a little hint, yes.
Use n(AUB)= n(A)+n(B)-n(A∩B).
I really have no idea how to do that. Hahaha. Anyway, you can please request to close this thread now. Thanks anyway. :) | 914 | 3,092 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-22 | latest | en | 0.96655 |
https://www.pokerlistings.com/the-8-critical-mistakes-i-made-when-i-began-playing-poker | 1,618,760,371,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00613.warc.gz | 1,056,795,446 | 62,172 | 8 Critical Mistakes I Made When I Started Playing Poker
Playing too many tables. Treating every player the same. Not understanding basic math.
Those were just a few of my early poker-playing sins. Truthfully, some of them are still evident in my game today.
There's no better teacher than experience but if reading this can help you avoid even a few of these as you get started playing poker, recreationally or seriously, it's well worth it.
There are the critical mistakes I made when I first started playing poker. What were yours? Let us know in the comments.
1. Misusing Starting Hand Charts
Starting hand charts are a prerequisite for any new poker player. They alleviate pressure on the mind by narrowing down the choices that you have to make at the beginning of each hand.
They define poker hand rankings and positional suitability. So why is using them a mistake?
Generally speaking it's not a mistake to use hand charts but I made a mistake when using them.
During an interview with Brian Koppelman, Vanessa Selbst spoke about the critical need to spend hours on the fundamentals when starting out.
I didn't do this. I wanted success quickly.
Instead of using hand charts in a flexible manner I became too rigid. There were times when the dynamics of the table made it apparent that I should widen my range yet I stuck rigidly to the chart and folded.
Another significant leak that I developed was playing hands from early and middle position because they were on the chart -- even if the dynamics of the table made it clear I should have folded.
A good example is opening up a small pair in middle position when there are a lot of stacks behind you that may be shoving. In this scenario it's important that you can move away from the hand charts and fold.
If you can do this, and not get lazy like I did, then hand charts are a beautiful thing.
Related Reading:
2. Treating Every Player the Same
Closely linked to my hand chart leak was my attitude towards how I played my hands.
In the beginning I always based the playability of my hands on the strength of my holding and not the strength and weakness of my opponent. In other words I would treat all of my opponents the same.
It was a major leak for me that took years to repair and still occurs even today, showing that it's vital that the fundamentals are correct at the beginning of the learning cycle.
Whenever I talk hands through with professional players they always ask me what type of player I was facing.
What read did I have based on my experience with him or her? If there was no experience, what type of player did I think he or she was from a visual perspective?
If you're not making your decisions based on a combination of hand strength, chip stacks and player tendencies, then you are losing money.
Related Reading:
3. Playing Too Many Tables
I fell in love with poker immediately but from the outset I found it boring to play one table online. I was multi-tabling before I even had the fundamentals pinned down.
Another problem I suffered came from my ego. I would watch online poker strategy videos, read forums and books and hear about all of these people playing 15+ tables.
If they could do 15+, then I could do 4, 8 and then 10.
During my years in the rail industry we had a CEO by the name of Keith Heller. He was American. He came over to the UK and immediately started removing locomotives from our fleet.
He didn't believe we were working them hard enough. When asked at what point he would stop taking them out of the train plan, he replied: "When I break it." And he did. He broke it. And then he started adding in locomotives.
The same philosophy is true in online poker for real money. I believe you should play one table until you consistently make a profit over a sustained period.
Then add a second table, rinse, and repeat. You will reach a number where your game suffers.
It will break.
Fix it by dropping the requisite number of tables until you show a profit again.
Related Reading:
4. Not Understanding Basic Math
I hated math in school and this is a problem when it comes to falling in love with poker.
Even today I don't understand the intricacies. It's a big leak. It's a block that I don't want to manage.
I have never made a play in poker based on a mathematical calculation and I know how basic a leak this is.
If you're as lazy as me when it comes to basic poker math then you too will lose money.
Related Reading:
5. Playing When Tilted
Everyone will suffer from some form of poker tilt. It's important to understand what your triggers are and introduce measures to mitigate the fallout.
One of my tilt triggers was losing consecutive flips when playing six-max cash games. Subconsciously I would start chasing losses and put my money in when behind in hopes of getting lucky.
I've lost more money in my lifetime through emotional issues than technical glitches. It's so difficult to walk away from the game when you are losing and on tilt, but it is an area of your game that will lead to disastrous consequences.
I recently interviewed Billy Chattaway; one of the brightest young minds in the UK. He told me he lost \$10k bluffing in a \$25/\$50 Pot Limit Omaha (PLO) cash game and then subsequently lost another \$20k in the next 20 minutes because he was on tilt and chasing.
If it can happen to the best of us, it can certainly happen to the worst of us.
Related Reading:
6. Playing Like a Robot
Another huge leak that emanated from the way I first learned how to play poker. I wasn't thinking through my hand. I wasn't taking in all of the information at my disposal. I was a robot.
A great example would be looking down to see pocket aces pre-flop. My mindset would immediately be, "I am going to double up."
This belief would produce all sorts of problems because I wouldn't take into consideration the texture of the board or the playability of my opponent.
With this mindset I could easily get it in for stacks on 9JQ against the biggest nit in the world.
It all seemed so clear minutes after the hand happened and this leads me to my next fault.
Related Reading:
7. Rushing
Another leak tied in with ego. Some players will take an eternity to make a decision at the table. When I see this happen I think they must have the skin of a rhino. When I am in this situation I feel eyeballs boring into my mind.
I think they will believe I am stupid if I take my time, so I rush.
Until the rules change and shot clocks are introduced you should always take your time before making your decision. I'm not talking about basic hands.
I'm talking about important situations. My biggest problem was my eagerness to gamble, stemming from the fact that I am a former gambling addict.
I crave that instant gratification so I don't take my time. I make the most aggressive play and I do it quickly without thinking. Don't make the same mistake.
Related Reading:
Your opponent is not Phil ivey.
8. Leveling
Everyone exists within the hierarchical structure of poker. There is always someone better than you and that means there is always someone weaker than you.
One of the problems I used to suffer from was thinking that my opponent understood the game at the same level as me. This problem stems from my live reporting job.
I would spend thousands of hours watching professional poker players maneuver in every single possible situation and then try the same thing when playing against Ken, the man who runs the local butchers.
I have lost count of the times I have made a move on a fellow amateur only to be called down with the weakest of hands that turned out to be better than mine.
I have moaned and groaned at these people but the mistakes have always been mine.
I didn't think. I didn't consider whether or not my opponent would understand the line I was trying to take. And that is another huge mistake.
Related Reading:
Related Poker Strategy Articles
Damen Lee
2016-04-24 07:38:21
I get it
Paulx1
2016-02-28 17:49:46
I was a little surprised by the unplayable hands in the chart
Selina Bond
2016-02-05 02:47:16
It is advisable for poker beginners to keep strong attention on their opponents and always find the soft games and never play many of the poker tables which afford them to lose. After that player need to understand the tactics and rules of the online poker dealer.
Concerned Citizen
2015-12-14 09:16:24
Thanks… I’m a TAGfish at the moment. Which means I study and then apply it completely incorrectly at the tables… I break even most of the time… I also share some of these habits and mistakes.
Thanks
Tres Deuce
2015-12-06 15:07:28
Very interesting article and some useful information, shared.
Your message is awaiting approval
× | 1,898 | 8,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-17 | longest | en | 0.975364 |
https://root.cern/doc/v626/RooAddPdf_8cxx_source.html | 1,716,297,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00693.warc.gz | 413,710,338 | 48,356 | ROOT Reference Guide
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Go to the documentation of this file.
1/*****************************************************************************
2 * Project: RooFit *
3 * Package: RooFitCore *
4 * @(#)root/roofitcore:$Id$
5 * Authors: *
6 * WV, Wouter Verkerke, UC Santa Barbara, verkerke@slac.stanford.edu *
7 * DK, David Kirkby, UC Irvine, dkirkby@uci.edu *
8 * *
9 * Copyright (c) 2000-2005, Regents of the University of California *
11 * *
12 * Redistribution and use in source and binary forms, *
13 * with or without modification, are permitted according to the terms *
15 *****************************************************************************/
16
17//////////////////////////////////////////////////////////////////////////////
19 \ingroup Roofitcore
20
21RooAddPdf is an efficient implementation of a sum of PDFs of the form
22
23\f[
24 \sum_{i=1}^{n} c_i \cdot \mathrm{PDF}_i
25\f]
26
27or
28\f[
29 c_1\cdot\mathrm{PDF}_1 + c_2\cdot\mathrm{PDF}_2 \; + \; ... \; + \; \left( 1-\sum_{i=1}^{n-1}c_i \right) \cdot \mathrm{PDF}_n
30\f]
31
32The first form is for extended likelihood fits, where the
33expected number of events is \f$\sum_i c_i \f$. The coefficients \f$c_i \f$
34can either be explicitly provided, or, if all components support
35extended likelihood fits, they can be calculated from the contribution
36of each PDF to the total expected number of events.
37
38In the second form, the sum of the coefficients is required to be 1 or less,
39and the coefficient of the last PDF is calculated automatically from the condition
40that the sum of all coefficients has to be 1.
41
42### Recursive coefficients
43It is also possible to parameterise the coefficients recursively
44
45\f[
46 \sum_{i=1}^n c_i \prod_{j=1}^{i-1} \left[ (1-c_j) \right] \cdot \mathrm{PDF}_i \\
47 = c_1 \cdot \mathrm{PDF}_1 + (1-c_1)\, c_2 \cdot \mathrm{PDF}_2 + \ldots + (1-c_1)\ldots(1-c_{n-1}) \cdot 1 \cdot \mathrm{PDF}_n \\
48\f]
49
50In this form the sum of the coefficients is always less than 1.0
51for all possible values of the individual coefficients between 0 and 1.
52\note Don't pass the \f$n^\mathrm{th} \f$ coefficient. It is always 1, since the normalisation condition removes one degree of freedom.
53
54RooAddPdf relies on each component PDF to be normalized and will perform
55no normalization other than calculating the proper last coefficient \f$c_n \f$, if requested.
56An (enforced) condition for this assumption is that each \f$\mathrm{PDF}_i \f$ is independent of each \f$c_i \f$.
57
58## Difference between RooAddPdf / RooRealSumFunc / RooRealSumPdf
59- RooAddPdf is a PDF of PDFs, *i.e.* its components need to be normalised and non-negative.
60- RooRealSumPdf is a PDF of functions, *i.e.*, its components can be negative, but their sum cannot be. The normalisation
61 is computed automatically, unless the PDF is extended (see above).
62- RooRealSumFunc is a sum of functions. It is neither normalised, nor need it be positive.
63
64*/
65
67
69#include "RooBatchCompute.h"
70#include "RooDataSet.h"
71#include "RooGlobalFunc.h"
72#include "RooNaNPacker.h"
73#include "RooRealProxy.h"
74#include "RooRealVar.h"
75#include "RooRealConstant.h"
76#include "RooRealIntegral.h"
78
79#include <algorithm>
80#include <memory>
81#include <sstream>
82#include <set>
83
84using namespace std;
85
87
88
89////////////////////////////////////////////////////////////////////////////////
90/// Dummy constructor
91
93 RooAbsPdf(name,title),
94 _refCoefNorm("!refCoefNorm","Reference coefficient normalization set",this,false,false),
95 _projCacheMgr(this,10),
96 _pdfList("!pdfs","List of PDFs",this),
97 _coefList("!coefficients","List of coefficients",this),
98 _coefErrCount{_errorCount}
99{
101}
102
103
105
106 // Two pdfs with the same name are only allowed in the input list if they are
107 // actually the same object.
108 using PdfInfo = std::pair<std::string,RooAbsArg*>;
109 std::set<PdfInfo> seen;
110 for(auto const& pdf : _pdfList) {
111 PdfInfo elem{pdf->GetName(), pdf};
112 auto comp = [&](PdfInfo const& p){ return p.first == elem.first && p.second != elem.second; };
113 auto found = std::find_if(seen.begin(), seen.end(), comp);
114 if(found != seen.end()) {
115 std::stringstream errorMsg;
117 << ") pdf list contains pdfs with duplicate name \"" << pdf->GetName() << "\"."
118 << std::endl;
119 coutE(InputArguments) << errorMsg.str();
120 throw std::invalid_argument(errorMsg.str().c_str());
121 }
122 seen.insert(elem);
123 }
124
125 _coefCache.resize(_pdfList.size());
126}
127
128
129////////////////////////////////////////////////////////////////////////////////
130/// Constructor with two PDFs and one coefficient
131
133 RooAbsPdf& pdf1, RooAbsPdf& pdf2, RooAbsReal& coef1) :
135{
139
141}
142
143
144////////////////////////////////////////////////////////////////////////////////
145/// Generic constructor from list of PDFs and list of coefficients.
146/// Each pdf list element (i) is paired with coefficient list element (i).
147/// The number of coefficients must be either equal to the number of PDFs,
148/// in which case extended MLL fitting is enabled, or be one less.
149///
150/// All PDFs must inherit from RooAbsPdf. All coefficients must inherit from RooAbsReal
151///
152/// If the recursiveFraction flag is true, the coefficients are interpreted as recursive
153/// coefficients as explained in the class description.
154
155RooAddPdf::RooAddPdf(const char *name, const char *title, const RooArgList& inPdfList, const RooArgList& inCoefList, bool recursiveFractions) :
157{
158 _recursive = recursiveFractions;
159
160 if (inPdfList.size()>inCoefList.size()+1 || inPdfList.size()<inCoefList.size()) {
161 std::stringstream errorMsg;
163 << ") number of pdfs and coefficients inconsistent, must have Npdf=Ncoef or Npdf=Ncoef+1." << endl ;
164 coutE(InputArguments) << errorMsg.str();
165 throw std::invalid_argument(errorMsg.str().c_str());
166 }
167
168 if (recursiveFractions && inPdfList.size()!=inCoefList.size()+1) {
169 std::stringstream errorMsg;
171 << "): Recursive fractions option can only be used if Npdf=Ncoef+1." << endl;
172 coutE(InputArguments) << errorMsg.str();
173 throw std::invalid_argument(errorMsg.str());
174 }
175
176 // Constructor with N PDFs and N or N-1 coefs
177 RooArgList partinCoefList ;
178
179 bool first(true) ;
180
181 for (auto i = 0u; i < inCoefList.size(); ++i) {
182 auto coef = dynamic_cast<RooAbsReal*>(inCoefList.at(i));
183 auto pdf = dynamic_cast<RooAbsPdf*>(inPdfList.at(i));
184 if (inPdfList.at(i) == nullptr) {
185 std::stringstream errorMsg;
187 << ") number of pdfs and coefficients inconsistent, must have Npdf=Ncoef or Npdf=Ncoef+1" << endl ;
188 coutE(InputArguments) << errorMsg.str();
189 throw std::invalid_argument(errorMsg.str());
190 }
191 if (!coef) {
192 std::stringstream errorMsg;
193 errorMsg << "RooAddPdf::RooAddPdf(" << GetName() << ") coefficient " << (coef ? coef->GetName() : "") << " is not of type RooAbsReal, ignored" << endl ;
194 coutE(InputArguments) << errorMsg.str();
195 throw std::invalid_argument(errorMsg.str());
196 }
197 if (!pdf) {
198 std::stringstream errorMsg;
199 errorMsg << "RooAddPdf::RooAddPdf(" << GetName() << ") pdf " << (pdf ? pdf->GetName() : "") << " is not of type RooAbsPdf, ignored" << endl ;
200 coutE(InputArguments) << errorMsg.str();
201 throw std::invalid_argument(errorMsg.str());
202 }
204
205 // Process recursive fraction mode separately
206 if (recursiveFractions) {
208 if (first) {
209
210 // The first fraction is the first plain fraction
211 first = false ;
213
214 } else {
215
216 // The i-th recursive fraction = (1-f1)*(1-f2)*...(fi) and is calculated from the list (f1,...,fi) by RooRecursiveFraction)
217 RooAbsReal* rfrac = new RooRecursiveFraction(Form("%s_recursive_fraction_%s",GetName(),pdf->GetName()),"Recursive Fraction",partinCoefList) ;
220
221 }
222
223 } else {
225 }
226 }
227
228 if (inPdfList.size() == inCoefList.size() + 1) {
229 auto pdf = dynamic_cast<RooAbsPdf*>(inPdfList.at(inCoefList.size()));
230
231 if (!pdf) {
232 coutE(InputArguments) << "RooAddPdf::RooAddPdf(" << GetName() << ") last argument " << inPdfList.at(inCoefList.size())->GetName() << " is not of type RooAbsPdf." << endl ;
233 throw std::invalid_argument("Last argument for RooAddPdf is not a PDF.");
234 }
236
237 // Process recursive fractions mode. Above, we verified that we don't have a last coefficient
238 if (recursiveFractions) {
239
240 // The last recursive fraction = (1-f1)*(1-f2)*...(1-fN) and is calculated from the list (f1,...,fN,1) by RooRecursiveFraction
242 RooAbsReal* rfrac = new RooRecursiveFraction(Form("%s_recursive_fraction_%s",GetName(),pdf->GetName()),"Recursive Fraction",partinCoefList) ;
245
246 // In recursive mode we always have Ncoef=Npdf, since we added it just above
247 _haveLastCoef=true ;
248 }
249
250 } else {
251 _haveLastCoef=true ;
252 }
253
255}
256
257
258////////////////////////////////////////////////////////////////////////////////
259/// Generic constructor from list of extended PDFs. There are no coefficients as the expected
260/// number of events from each components determine the relative weight of the PDFs.
261///
262/// All PDFs must inherit from RooAbsPdf.
263
266{
267 _allExtendable = true;
268
269 // Constructor with N PDFs
270 for (const auto pdfArg : inPdfList) {
271 auto pdf = dynamic_cast<const RooAbsPdf*>(pdfArg);
272
273 if (!pdf) {
274 coutE(InputArguments) << "RooAddPdf::RooAddPdf(" << GetName() << ") pdf " << (pdf ? pdf->GetName() : "") << " is not of type RooAbsPdf, ignored" << endl ;
275 continue ;
276 }
277 if (!pdf->canBeExtended()) {
278 coutE(InputArguments) << "RooAddPdf::RooAddPdf(" << GetName() << ") pdf " << pdf->GetName() << " is not extendable, ignored" << endl ;
279 continue ;
280 }
282 }
283
285}
286
287
288////////////////////////////////////////////////////////////////////////////////
289/// Copy constructor
290
292 RooAbsPdf(other,name),
293 _refCoefNorm("!refCoefNorm",this,other._refCoefNorm),
294 _refCoefRangeName((TNamed*)other._refCoefRangeName),
295 _projectCoefs(other._projectCoefs),
296 _projCacheMgr(other._projCacheMgr,this),
297 _codeReg(other._codeReg),
298 _pdfList("!pdfs",this,other._pdfList),
299 _coefList("!coefficients",this,other._coefList),
300 _haveLastCoef(other._haveLastCoef),
301 _allExtendable(other._allExtendable),
302 _recursive(other._recursive)
303{
307}
308
309
310////////////////////////////////////////////////////////////////////////////////
311/// By default the interpretation of the fraction coefficients is
312/// performed in the contextual choice of observables. This makes the
313/// shape of the p.d.f explicitly dependent on the choice of
314/// observables. This method instructs RooAddPdf to freeze the
315/// interpretation of the coefficients to be done in the given set of
316/// observables. If frozen, fractions are automatically transformed
317/// from the reference normalization set to the contextual normalization
318/// set by ratios of integrals.
319
321{
322 if (refCoefNorm.empty()) {
323 _projectCoefs = false ;
324 return ;
325 }
326 _projectCoefs = true ;
327
330
332}
333
334
335
336////////////////////////////////////////////////////////////////////////////////
337/// By default, fraction coefficients are assumed to refer to the default
338/// fit range. This makes the shape of a RooAddPdf
339/// explicitly dependent on the range of the observables. Calling this function
340/// allows for a range-independent definition of the fractions, because it
341/// ties all coefficients to the given
342/// named range. If the normalisation range is different
343/// from this reference range, the appropriate fraction coefficients
344/// are automatically calculated from the reference fractions by
345/// integrating over the ranges, and comparing these integrals.
346
348{
350 if (_refCoefRangeName) _projectCoefs = true ;
351}
352
353
354
355////////////////////////////////////////////////////////////////////////////////
356/// Retrieve cache element for the computation of the PDF normalisation.
357/// \param[in] nset Current normalisation set (integration over these variables yields 1).
358/// \param[in] iset Integration set. Variables to be integrated over (if integrations are performed).
359/// \param[in] rangeName Reference range for the integrals.
360///
361/// If a cache element does not exist, create and fill it on the fly. The cache also contains
362/// - Supplemental normalization terms (in case not all added p.d.f.s have the same observables)
363/// - Projection integrals to calculate transformed fraction coefficients when a frozen reference frame is provided
364/// - Projection integrals for similar transformations when a frozen reference range is provided.
365
367{
368
369 // Check if cache already exists
370 CacheElem* cache = (CacheElem*) _projCacheMgr.getObj(nset,iset,0,rangeName) ;
371 if (cache) {
372 return cache ;
373 }
374
375 //Create new cache
376 cache = new CacheElem ;
377
378 // *** PART 1 : Create supplemental normalization list ***
379
380 // Retrieve the combined set of dependents of this PDF ;
381 auto fullDepList = std::unique_ptr<RooArgSet>{getObservables(nset)} ;
382 if (iset) {
383 fullDepList->remove(*iset,true,true) ;
384 }
385
386 // Fill with dummy unit RRVs for now
387 for (std::size_t i = 0; i < _pdfList.size(); ++i) {
388 auto pdf = static_cast<const RooAbsPdf *>(_pdfList.at(i));
389 auto coef = static_cast<const RooAbsReal*>(_coefList.at(i));
390
392 RooArgSet supNSet(*fullDepList) ;
393
394 // Remove PDF dependents
395 if (auto pdfDeps = std::unique_ptr<RooArgSet>{pdf->getObservables(nset)}) {
396 supNSet.remove(*pdfDeps,true,true) ;
397 }
398
399 // Remove coef dependents
400 if (auto coefDeps = std::unique_ptr<RooArgSet>{coef ? coef->getObservables(nset) : nullptr}) {
401 supNSet.remove(*coefDeps,true,true) ;
402 }
403
404 RooAbsReal* snorm ;
405 auto name = std::string(GetName()) + "_" + pdf->GetName() + "_SupNorm";
406 cache->_needSupNorm = false ;
407 if (!supNSet.empty()) {
408 snorm = new RooRealIntegral(name.c_str(),"Supplemental normalization integral",RooRealConstant::value(1.0),supNSet) ;
409 cxcoutD(Caching) << "RooAddPdf " << GetName() << " making supplemental normalization set " << supNSet << " for pdf component " << pdf->GetName() << endl ;
410 cache->_needSupNorm = true ;
411 } else {
412 snorm = new RooRealVar(name.c_str(),"Unit Supplemental normalization integral",1.0) ;
413 }
415 }
416
417 if (_verboseEval>1) {
418 cxcoutD(Caching) << "RooAddPdf::syncSuppNormList(" << GetName() << ") synching supplemental normalization list for norm" << (nset?*nset:RooArgSet()) << endl ;
419 if dologD(Caching) {
420 cache->_suppNormList.Print("v") ;
421 }
422 }
423
424
425 // *** PART 2 : Create projection coefficients ***
426
427// cout << " this = " << this << " (" << GetName() << ")" << endl ;
428// cout << "projectCoefs = " << (_projectCoefs?"T":"F") << endl ;
429// cout << "_normRange.Length() = " << _normRange.Length() << endl ;
430
431 // If no projections required stop here
432 if (!_projectCoefs && !rangeName) {
433 _projCacheMgr.setObj(nset,iset,cache,RooNameReg::ptr(rangeName)) ;
434// cout << " no projection required" << endl ;
435 return cache ;
436 }
437
438
439// cout << "calculating projection" << endl ;
440
441 // Reduce iset/nset to actual dependents of this PDF
442 auto nset2 = std::unique_ptr<RooArgSet>{nset ? getObservables(nset) : new RooArgSet()} ;
443 cxcoutD(Caching) << "RooAddPdf(" << GetName() << ")::getPC nset = " << (nset?*nset:RooArgSet()) << " nset2 = " << *nset2 << endl ;
444
445 if (nset2->empty() && !_refCoefNorm.empty()) {
446 //cout << "WVE: evaluating RooAddPdf without normalization, but have reference normalization for coefficient definition" << endl ;
447
449 if (_refCoefRangeName) {
451 }
452 }
453
454
455 // Check if requested transformation is not identity
456 if (!nset2->equals(_refCoefNorm) || _refCoefRangeName !=0 || rangeName !=0 || _normRange.Length()>0) {
457
458 cxcoutD(Caching) << "ALEX: RooAddPdf::syncCoefProjList(" << GetName() << ") projecting coefficients from "
459 << *nset2 << (rangeName?":":"") << (rangeName?rangeName:"")
460 << " to " << ((!_refCoefNorm.empty())?_refCoefNorm:*nset2) << (_refCoefRangeName?":":"") << (_refCoefRangeName?RooNameReg::str(_refCoefRangeName):"") << endl ;
461
462 // Recalculate projection integrals of PDFs
463 for (auto arg : _pdfList) {
464 auto thePdf = static_cast<const RooAbsPdf*>(arg);
465
466 // Calculate projection integral
467 RooAbsReal* pdfProj ;
468 if (!nset2->equals(_refCoefNorm)) {
469 pdfProj = thePdf->createIntegral(*nset2,_refCoefNorm,_normRange.Length()>0?_normRange.Data():0) ;
470 pdfProj->setOperMode(operMode()) ;
471 cxcoutD(Caching) << "RooAddPdf(" << GetName() << ")::getPC nset2(" << *nset2 << ")!=_refCoefNorm(" << _refCoefNorm << ") --> pdfProj = " << pdfProj->GetName() << endl ;
472 } else {
473 auto name = std::string(GetName()) + "_" + thePdf->GetName() + "_ProjectNorm";
474 pdfProj = new RooRealVar(name.c_str(),"Unit Projection normalization integral",1.0) ;
475 cxcoutD(Caching) << "RooAddPdf(" << GetName() << ")::getPC nset2(" << *nset2 << ")==_refCoefNorm("
476 << _refCoefNorm << ") --> pdfProj = " << pdfProj->GetName() << std::endl;
477 }
478
480 cxcoutD(Caching) << " RooAddPdf::syncCoefProjList(" << GetName() << ") PP = " << pdfProj->GetName() << endl ;
481
482 // Calculation optional supplemental normalization term
483 RooArgSet supNormSet(_refCoefNorm) ;
484 auto deps = std::unique_ptr<RooArgSet>{thePdf->getParameters(RooArgSet())} ;
485 supNormSet.remove(*deps,true,true) ;
486
487 RooAbsReal* snorm ;
488 auto name = std::string(GetName()) + "_" + thePdf->GetName() + "_ProjSupNorm";
489 if (!supNormSet.empty() && !nset2->equals(_refCoefNorm) ) {
490 snorm = new RooRealIntegral(name.c_str(),"Projection Supplemental normalization integral",
491 RooRealConstant::value(1.0),supNormSet) ;
492 } else {
493 snorm = new RooRealVar(name.c_str(),"Unit Projection Supplemental normalization integral",1.0) ;
494 }
495 cxcoutD(Caching) << " RooAddPdf::syncCoefProjList(" << GetName() << ") SN = " << snorm->GetName() << endl ;
497
498 // Calculate reference range adjusted projection integral
499 RooAbsReal* rangeProj1 ;
500
501 // cout << "ALEX >>>> RooAddPdf(" << GetName() << ")::getPC _refCoefRangeName WVE = "
502// <<(_refCoefRangeName?":":"") << (_refCoefRangeName?RooNameReg::str(_refCoefRangeName):"")
503// <<" _refCoefRangeName AK = " << (_refCoefRangeName?_refCoefRangeName->GetName():"")
504// << " && _refCoefNorm" << _refCoefNorm << " with size = _refCoefNorm.size() " << _refCoefNorm.size() << endl ;
505
506 // Check if _refCoefRangeName is identical to default range for all observables,
507 // If so, substitute by unit integral
508
509 // ----------
510 auto tmpObs = std::unique_ptr<RooArgSet>{thePdf->getObservables(_refCoefNorm)} ;
511 bool allIdent = true ;
512 for (auto const& obsArg : *tmpObs) {
513 RooRealVar* rvarg = dynamic_cast<RooRealVar*>(obsArg) ;
514 if (rvarg) {
515 if (rvarg->getMin(RooNameReg::str(_refCoefRangeName))!=rvarg->getMin() ||
516 rvarg->getMax(RooNameReg::str(_refCoefRangeName))!=rvarg->getMax()) {
517 allIdent=false ;
518 }
519 }
520 }
521 // -------------
522
523 if (_refCoefRangeName && !_refCoefNorm.empty() && !allIdent) {
524
525
526 auto tmp = std::unique_ptr<RooArgSet>{thePdf->getObservables(_refCoefNorm)} ;
527 rangeProj1 = thePdf->createIntegral(*tmp,*tmp,RooNameReg::str(_refCoefRangeName)) ;
528
529 //rangeProj1->setOperMode(operMode()) ;
530
531 } else {
532
533 auto theName = std::string(GetName()) + "_" + thePdf->GetName() + "_RangeNorm1";
534 rangeProj1 = new RooRealVar(theName.c_str(),"Unit range normalization integral",1.0) ;
535
536 }
537 cxcoutD(Caching) << " RooAddPdf::syncCoefProjList(" << GetName() << ") R1 = " << rangeProj1->GetName() << endl ;
539
540
541 // Calculate range adjusted projection integral
542 RooAbsReal* rangeProj2 ;
543 cxcoutD(Caching) << "RooAddPdf::syncCoefProjList(" << GetName() << ") rangename = " << (rangeName?rangeName:"<null>")
544 << " nset = " << (nset?*nset:RooArgSet()) << endl ;
545 if (rangeName && !_refCoefNorm.empty()) {
546
547 rangeProj2 = thePdf->createIntegral(_refCoefNorm,_refCoefNorm,rangeName) ;
548 //rangeProj2->setOperMode(operMode()) ;
549
550 } else if (_normRange.Length()>0) {
551
552 auto tmp = std::unique_ptr<RooArgSet>{thePdf->getObservables(_refCoefNorm)} ;
553 rangeProj2 = thePdf->createIntegral(*tmp,*tmp,_normRange.Data()) ;
554
555 } else {
556
557 auto theName = std::string(GetName()) + "_" + thePdf->GetName() + "_RangeNorm2";
558 rangeProj2 = new RooRealVar(theName.c_str(),"Unit range normalization integral",1.0) ;
559
560 }
561 cxcoutD(Caching) << " RooAddPdf::syncCoefProjList(" << GetName() << ") R2 = " << rangeProj2->GetName() << endl ;
563
564 }
565
566 }
567
568 _projCacheMgr.setObj(nset,iset,cache,RooNameReg::ptr(rangeName)) ;
569
570 return cache ;
571}
572
573
574////////////////////////////////////////////////////////////////////////////////
575/// Update the coefficient values in the given cache element: calculate new remainder
576/// fraction, normalize fractions obtained from extended ML terms to unity, and
577/// multiply the various range and dimensional corrections needed in the
578/// current use context.
579
580void RooAddPdf::updateCoefficients(CacheElem& cache, const RooArgSet* nset) const
581{
582 // Since this function updates the cache, it obviously needs write access:
583 auto& myCoefCache = const_cast<std::vector<double>&>(_coefCache);
584 myCoefCache.resize(_haveLastCoef ? _coefList.size() : _pdfList.size(), 0.);
585
586 // Straight coefficients
587 if (_allExtendable) {
588
589 // coef[i] = expectedEvents[i] / SUM(expectedEvents)
590 double coefSum(0) ;
591 std::size_t i = 0;
592 for (auto arg : _pdfList) {
593 auto pdf = static_cast<RooAbsPdf*>(arg);
594 myCoefCache[i] = pdf->expectedEvents(!_refCoefNorm.empty()?&_refCoefNorm:nset) ;
595 coefSum += myCoefCache[i] ;
596 i++ ;
597 }
598
599 if (coefSum==0.) {
600 coutW(Eval) << "RooAddPdf::updateCoefCache(" << GetName() << ") WARNING: total number of expected events is 0" << endl ;
601 } else {
602 for (std::size_t j=0; j < _pdfList.size(); j++) {
603 myCoefCache[j] /= coefSum ;
604 }
605 }
606
607 } else {
608 if (_haveLastCoef) {
609
610 // coef[i] = coef[i] / SUM(coef)
611 double coefSum(0) ;
612 std::size_t i=0;
613 for (auto coefArg : _coefList) {
614 auto coef = static_cast<RooAbsReal*>(coefArg);
615 myCoefCache[i] = coef->getVal(nset) ;
616 coefSum += myCoefCache[i++];
617 }
618 if (coefSum==0.) {
619 coutW(Eval) << "RooAddPdf::updateCoefCache(" << GetName() << ") WARNING: sum of coefficients is zero 0" << endl ;
620 } else {
621 const double invCoefSum = 1./coefSum;
622 for (std::size_t j=0; j < _coefList.size(); j++) {
623 myCoefCache[j] *= invCoefSum;
624 }
625 }
626 } else {
627
628 // coef[i] = coef[i] ; coef[n] = 1-SUM(coef[0...n-1])
629 double lastCoef(1) ;
630 std::size_t i=0;
631 for (auto coefArg : _coefList) {
632 auto coef = static_cast<RooAbsReal*>(coefArg);
633 myCoefCache[i] = coef->getVal(nset) ;
634 lastCoef -= myCoefCache[i++];
635 }
636 myCoefCache[_coefList.size()] = lastCoef ;
637
638 // Treat coefficient degeneration
639 const float coefDegen = lastCoef < 0. ? -lastCoef : (lastCoef > 1. ? lastCoef - 1. : 0.);
640 if (coefDegen > 1.E-5) {
641 myCoefCache[_coefList.size()] = RooNaNPacker::packFloatIntoNaN(100.f*coefDegen);
642
643 std::stringstream msg;
644 if (_coefErrCount-->0) {
645 msg << "RooAddPdf::updateCoefCache(" << GetName()
646 << " WARNING: sum of PDF coefficients not in range [0-1], value="
647 << 1-lastCoef ;
648 if (_coefErrCount==0) {
649 msg << " (no more will be printed)" ;
650 }
651 coutW(Eval) << msg.str() << std::endl;
652 }
653 }
654 }
655 }
656
657
658 // Stop here if not projection is required or needed
659 if ((!_projectCoefs && _normRange.Length()==0) || cache._projList.empty()) {
660 return ;
661 }
662
663 // Adjust coefficients for given projection
664 double coefSum(0) ;
665 {
667
668 for (std::size_t i = 0; i < _pdfList.size(); i++) {
669
670 RooAbsReal* pp = ((RooAbsReal*)cache._projList.at(i)) ;
671 RooAbsReal* sn = ((RooAbsReal*)cache._suppProjList.at(i)) ;
672 RooAbsReal* r1 = ((RooAbsReal*)cache._refRangeProjList.at(i)) ;
673 RooAbsReal* r2 = ((RooAbsReal*)cache._rangeProjList.at(i)) ;
674
675 double proj = pp->getVal()/sn->getVal()*(r2->getVal()/r1->getVal()) ;
676
677 myCoefCache[i] *= proj ;
678 coefSum += myCoefCache[i] ;
679 }
680 }
681
682
684 for (std::size_t i=0; i < _pdfList.size(); ++i) {
685 ccoutD(Caching) << " ALEX: POST-SYNC coef[" << i << "] = " << myCoefCache[i]
686 << " ( _coefCache[i]/coefSum = " << myCoefCache[i]*coefSum << "/" << coefSum << " ) "<< endl ;
687 }
688 }
689
690 if (coefSum==0.) {
691 coutE(Eval) << "RooAddPdf::updateCoefCache(" << GetName() << ") sum of coefficients is zero." << endl ;
692 }
693
694 for (std::size_t i=0; i < _pdfList.size(); i++) {
695 myCoefCache[i] /= coefSum ;
696 }
697
698}
699
700////////////////////////////////////////////////////////////////////////////////
701/// Look up projection cache and per-PDF norm sets. If a PDF doesn't have a special
702/// norm set, use the defaultNorm. If defaultNorm == nullptr, use the member
703/// _normSet.
705
706 // Treat empty normalization set and nullptr the same way.
707 if(nset && nset->empty()) nset = nullptr;
708
709 if (nset == nullptr) {
710 if (!_refCoefNorm.empty()) {
711 nset = &_refCoefNorm ;
712 }
713 }
714
715 // A RooAddPdf needs to have a normalization set defined, otherwise its
716 // coefficient will not be uniquely defined. Its shape depends on the
717 // normalization provided. Un-normalized calls to RooAddPdf can happen in
718 // Roofit, when printing the pdf's or when computing integrals. In these case,
719 // if the pdf has a normalization set previously defined (i.e. stored as a
720 // datamember in _copyOfLastNormSet) it should use it by default when the pdf
721 // is evaluated without passing a normalizations set (in pdf->getVal(nullptr) )
722 // In the case of no pre-defined normalization set exists, a warning will be
723 // produced, since the obtained value will be arbitrary. Note that to avoid
724 // unnecessary warning messages, when calling RooAbsPdf::printValue or
725 // RooAbsPdf::graphVizTree, the printing of the warning messages for the
726 // RooFit::Eval topic is explicitly disabled.
727 {
728 // If nset is still nullptr, get the pointer to a copy of the last-used
729 // normalization set. It nset is not nullptr, check whether the copy of
730 // the last-used normalization set needs an update.
731 if(nset == nullptr) {
732 nset = _copyOfLastNormSet.get();
733 } else if(nset->uniqueId() != _idOfLastUsedNormSet) {
734 _copyOfLastNormSet = std::make_unique<const RooArgSet>(*nset);
736 }
737
738 // If nset is STILL nullptr, print a warning.
739 if (nset == nullptr) {
740 oocoutW(this, Eval) << "Evaluating RooAddPdf without a defined normalization set. This can lead to ambiguos "
741 "coefficients definition and incorrect results."
742 << " Use RooAddPdf::fixCoefNormalization(nset) to provide a normalization set for "
744 << std::endl;
745 }
746 }
747
748
749 CacheElem* cache = getProjCache(nset) ;
750 updateCoefficients(*cache,nset) ;
751
752 return {nset, cache};
753}
754
755
756////////////////////////////////////////////////////////////////////////////////
757/// Calculate and return the current value
758
760{
761 auto normAndCache = getNormAndCache(normSet);
762 const RooArgSet* nset = normAndCache.first;
763 CacheElem* cache = normAndCache.second;
764
765 // Process change in last data set used
766 bool nsetChanged(false) ;
767 if (nset!=_normSet || _norm==0) {
768 nsetChanged = syncNormalization(nset) ;
769 }
770
771 // Do running sum of coef/pdf pairs, calculate lastCoef.
772 if (isValueDirty() || nsetChanged) {
773 _value = 0.0;
774
775 for (unsigned int i=0; i < _pdfList.size(); ++i) {
776 const auto& pdf = static_cast<RooAbsPdf&>(_pdfList[i]);
777 double snormVal = 1.;
778 if (cache->_needSupNorm) {
779 snormVal = ((RooAbsReal*)cache->_suppNormList.at(i))->getVal();
780 }
781
782 double pdfVal = pdf.getVal(nset);
783 if (pdf.isSelectedComp()) {
784 _value += pdfVal*_coefCache[i]/snormVal;
785 }
786 }
788 }
789
790 return _value;
791}
792
793
794////////////////////////////////////////////////////////////////////////////////
795/// Compute addition of PDFs in batches.
796void RooAddPdf::computeBatch(cudaStream_t* stream, double* output, size_t nEvents, RooFit::Detail::DataMap const& dataMap) const
797{
800 CacheElem* cache = getNormAndCache(nullptr).second;
801 for (unsigned int pdfNo = 0; pdfNo < _pdfList.size(); ++pdfNo)
802 {
803 auto pdf = static_cast<RooAbsPdf*>(&_pdfList[pdfNo]);
804 if (pdf->isSelectedComp())
805 {
806 pdfs.push_back(dataMap.at(pdf));
807 coefs.push_back(_coefCache[pdfNo] / (cache->_needSupNorm ?
808 static_cast<RooAbsReal*>(cache->_suppNormList.at(pdfNo))->getVal() : 1) );
809 }
810 }
812 dispatch->compute(stream, RooBatchCompute::AddPdf, output, nEvents, pdfs, coefs);
813}
814
815
816////////////////////////////////////////////////////////////////////////////////
817/// Reset error counter to given value, limiting the number
818/// of future error messages for this pdf to 'resetValue'
819
821{
823 _coefErrCount = resetValue ;
824}
825
826
827
828////////////////////////////////////////////////////////////////////////////////
829/// Check if PDF is valid for given normalization set.
830/// Coeffient and PDF must be non-overlapping, but pdf-coefficient
831/// pairs may overlap each other
832
834{
835 bool ret(false) ;
836
837 // There may be fewer coefficients than PDFs.
838 std::size_t end = std::min(_pdfList.size(), _coefList.size());
839 for (std::size_t i = 0; i < end; ++i) {
840 auto pdf = static_cast<const RooAbsPdf *>(_pdfList.at(i));
841 auto coef = static_cast<const RooAbsReal*>(_coefList.at(i));
842 if (pdf->observableOverlaps(nset,*coef)) {
843 coutE(InputArguments) << "RooAddPdf::checkObservables(" << GetName() << "): ERROR: coefficient " << coef->GetName()
844 << " and PDF " << pdf->GetName() << " have one or more dependents in common" << endl ;
845 ret = true ;
846 }
847 }
848
849 return ret ;
850}
851
852
853////////////////////////////////////////////////////////////////////////////////
854/// Determine which part (if any) of given integral can be performed analytically.
855/// If any analytical integration is possible, return integration scenario code
856///
857/// RooAddPdf queries each component PDF for its analytical integration capability of the requested
858/// set ('allVars'). It finds the largest common set of variables that can be integrated
859/// by all components. If such a set exists, it reconfirms that each component is capable of
860/// analytically integrating the common set, and combines the components individual integration
861/// codes into a single integration code valid for RooAddPdf.
862
864 const RooArgSet* normSet, const char* rangeName) const
865{
866
867 RooArgSet allAnalVars(*std::unique_ptr<RooArgSet>{getObservables(allVars)}) ;
868
869 Int_t n(0) ;
870
871 // First iteration, determine what each component can integrate analytically
872 for (const auto pdfArg : _pdfList) {
873 auto pdf = static_cast<const RooAbsPdf *>(pdfArg);
874 RooArgSet subAnalVars ;
875 pdf->getAnalyticalIntegralWN(allVars,subAnalVars,normSet,rangeName) ;
876
877 // Observables that cannot be integrated analytically by this component are dropped from the common list
878 for (const auto arg : allVars) {
879 if (!subAnalVars.find(arg->GetName()) && pdf->dependsOn(*arg)) {
880 allAnalVars.remove(*arg,true,true) ;
881 }
882 }
883 n++ ;
884 }
885
886 // If no observables can be integrated analytically, return code 0 here
887 if (allAnalVars.empty()) {
888 return 0 ;
889 }
890
891
892 // Now retrieve codes for integration over common set of analytically integrable observables for each component
893 n=0 ;
894 std::vector<Int_t> subCode(_pdfList.size());
895 bool allOK(true) ;
896 for (const auto arg : _pdfList) {
897 auto pdf = static_cast<const RooAbsPdf *>(arg);
898 RooArgSet subAnalVars ;
899 auto allAnalVars2 = std::unique_ptr<RooArgSet>{pdf->getObservables(allAnalVars)} ;
900 subCode[n] = pdf->getAnalyticalIntegralWN(*allAnalVars2,subAnalVars,normSet,rangeName) ;
901 if (subCode[n]==0 && !allAnalVars2->empty()) {
902 coutE(InputArguments) << "RooAddPdf::getAnalyticalIntegral(" << GetName() << ") WARNING: component PDF " << pdf->GetName()
903 << " advertises inconsistent set of integrals (e.g. (X,Y) but not X or Y individually."
904 << " Distributed analytical integration disabled. Please fix PDF" << endl ;
905 allOK = false ;
906 }
907 n++ ;
908 }
909 if (!allOK) {
910 return 0 ;
911 }
912
913 // Mare all analytically integrated observables as such
915
916 // Store set of variables analytically integrated
917 RooArgSet* intSet = new RooArgSet(allAnalVars) ;
918 Int_t masterCode = _codeReg.store(subCode,intSet)+1 ;
919
920 return masterCode ;
921}
922
923
924
925////////////////////////////////////////////////////////////////////////////////
926/// Return analytical integral defined by given scenario code
927
928Double_t RooAddPdf::analyticalIntegralWN(Int_t code, const RooArgSet* normSet, const char* rangeName) const
929{
930 // WVE needs adaptation to handle new rangeName feature
931 if (code==0) {
932 return getVal(normSet) ;
933 }
934
935 // Retrieve analytical integration subCodes and set of observabels integrated over
936 RooArgSet* intSet ;
937 const std::vector<Int_t>& subCode = _codeReg.retrieve(code-1,intSet) ;
938 if (subCode.empty()) {
939 coutE(InputArguments) << "RooAddPdf::analyticalIntegral(" << GetName() << "): ERROR unrecognized integration code, " << code << endl ;
940 assert(0) ;
941 }
942
943 cxcoutD(Caching) << "RooAddPdf::aiWN(" << GetName() << ") calling getProjCache with nset = " << (normSet?*normSet:RooArgSet()) << endl ;
944
945 if ((normSet==0 || normSet->empty()) && !_refCoefNorm.empty()) {
946// cout << "WVE integration of RooAddPdf without normalization, but have reference set, using ref set for normalization" << endl ;
947 normSet = &_refCoefNorm ;
948 }
949
950 CacheElem* cache = getProjCache(normSet,intSet,0) ; // WVE rangename here?
951 updateCoefficients(*cache,normSet) ;
952
953 // Calculate the current value of this object
954 double value(0) ;
955
956 // Do running sum of coef/pdf pairs, calculate lastCoef.
957 double snormVal ;
958
959 //cout << "ROP::aIWN updateCoefCache with rangeName = " << (rangeName?rangeName:"<null>") << endl ;
960 RooArgList* snormSet = (!cache->_suppNormList.empty()) ? &cache->_suppNormList : nullptr ;
961 for (std::size_t i = 0; i < _pdfList.size(); ++i ) {
962 auto pdf = static_cast<const RooAbsPdf*>(_pdfList.at(i));
963
964 if (_coefCache[i]) {
965 snormVal = snormSet ? ((RooAbsReal*) cache->_suppNormList.at(i))->getVal() : 1.0 ;
966
967 // WVE swap this?
968 double val = pdf->analyticalIntegralWN(subCode[i],normSet,rangeName) ;
969 if (pdf->isSelectedComp()) {
970 value += val*_coefCache[i]/snormVal ;
971 }
972 }
973 }
974
975 return value ;
976}
977
978
979
980////////////////////////////////////////////////////////////////////////////////
981/// Return the number of expected events, which is either the sum of all coefficients
982/// or the sum of the components extended terms, multiplied with the fraction that
983/// is in the current range w.r.t the reference range
984
986{
987 double expectedTotal{0.0};
988
989 cxcoutD(Caching) << "RooAddPdf::expectedEvents(" << GetName() << ") calling getProjCache with nset = " << (nset?*nset:RooArgSet()) << endl ;
990 CacheElem& cache = *getProjCache(nset) ;
991 updateCoefficients(cache,nset) ;
992
993 if (!cache._rangeProjList.empty()) {
994
995 for (std::size_t i = 0; i < _pdfList.size(); ++i) {
996 auto const& r1 = static_cast<RooAbsReal&>(cache._refRangeProjList[i]);
997 auto const& r2 = static_cast<RooAbsReal&>(cache._rangeProjList[i]);
998 double ncomp = _allExtendable ? static_cast<RooAbsPdf&>(_pdfList[i]).expectedEvents(nset)
999 : static_cast<RooAbsReal&>(_coefList[i]).getVal(nset);
1000 expectedTotal += (r2.getVal()/r1.getVal()) * ncomp ;
1001
1002 }
1003
1004 } else {
1005
1006 if (_allExtendable) {
1007 for(auto const& arg : _pdfList) {
1008 expectedTotal += static_cast<RooAbsPdf*>(arg)->expectedEvents(nset) ;
1009 }
1010 } else {
1011 for(auto const& arg : _coefList) {
1012 expectedTotal += static_cast<RooAbsReal*>(arg)->getVal(nset) ;
1013 }
1014 }
1015
1016 }
1017 return expectedTotal ;
1018}
1019
1020
1021
1022////////////////////////////////////////////////////////////////////////////////
1023/// Interface function used by test statistics to freeze choice of observables
1024/// for interpretation of fraction coefficients
1025
1026void RooAddPdf::selectNormalization(const RooArgSet* depSet, bool force)
1027{
1028
1029 if (!force && !_refCoefNorm.empty()) {
1030 return ;
1031 }
1032
1033 if (!depSet) {
1035 return ;
1036 }
1037
1038 fixCoefNormalization(*std::unique_ptr<RooArgSet>{getObservables(depSet)}) ;
1039}
1040
1041
1042
1043////////////////////////////////////////////////////////////////////////////////
1044/// Interface function used by test statistics to freeze choice of range
1045/// for interpretation of fraction coefficients
1046
1047void RooAddPdf::selectNormalizationRange(const char* rangeName, bool force)
1048{
1049 if (!force && _refCoefRangeName) {
1050 return ;
1051 }
1052
1053 fixCoefRange(rangeName) ;
1054}
1055
1056
1057
1058////////////////////////////////////////////////////////////////////////////////
1059/// Return specialized context to efficiently generate toy events from RooAddPdfs
1060/// return RooAbsPdf::genContext(vars,prototype,auxProto,verbose) ; // WVE DEBUG
1061
1063 const RooArgSet* auxProto, bool verbose) const
1064{
1066}
1067
1068
1069
1070////////////////////////////////////////////////////////////////////////////////
1071/// List all RooAbsArg derived contents in this cache element
1072
1074{
1075 RooArgList allNodes;
1080
1081 return allNodes ;
1082}
1083
1084
1085
1086////////////////////////////////////////////////////////////////////////////////
1087/// Loop over components for plot sampling hints and merge them if there are multiple
1088
1089std::list<Double_t>* RooAddPdf::plotSamplingHint(RooAbsRealLValue& obs, Double_t xlo, Double_t xhi) const
1090{
1091 std::unique_ptr<std::list<Double_t>> sumHint = nullptr ;
1092 bool needClean = false;
1093
1094 // Loop over components pdf
1095 for (const auto arg : _pdfList) {
1096 auto pdf = static_cast<const RooAbsPdf*>(arg);
1097
1098 std::unique_ptr<std::list<Double_t>> pdfHint{pdf->plotSamplingHint(obs,xlo,xhi)} ;
1099
1100 // Process hint
1101 if (pdfHint) {
1102 if (!sumHint) {
1103
1104 // If this is the first hint, then just save it
1105 sumHint = std::move(pdfHint) ;
1106
1107 } else {
1108
1109 auto newSumHint = std::make_unique<std::list<Double_t>>(sumHint->size()+pdfHint->size());
1110
1111 // Merge hints into temporary array
1112 merge(pdfHint->begin(),pdfHint->end(),sumHint->begin(),sumHint->end(),newSumHint->begin()) ;
1113
1114 // Copy merged array without duplicates to new sumHintArrau
1115 sumHint = std::move(newSumHint) ;
1116 needClean = true ;
1117
1118 }
1119 }
1120 }
1121 if (needClean) {
1122 sumHint->erase(std::unique(sumHint->begin(),sumHint->end()), sumHint->end()) ;
1123 }
1124
1125 return sumHint.release() ;
1126}
1127
1128
1129////////////////////////////////////////////////////////////////////////////////
1130/// Loop over components for plot sampling hints and merge them if there are multiple
1131
1132std::list<Double_t>* RooAddPdf::binBoundaries(RooAbsRealLValue& obs, Double_t xlo, Double_t xhi) const
1133{
1134 std::unique_ptr<list<Double_t>> sumBinB = nullptr ;
1135 bool needClean = false;
1136
1137 // Loop over components pdf
1138 for (auto arg : _pdfList) {
1139 auto pdf = static_cast<const RooAbsPdf *>(arg);
1140 std::unique_ptr<list<Double_t>> pdfBinB{pdf->binBoundaries(obs,xlo,xhi)};
1141
1142 // Process hint
1143 if (pdfBinB) {
1144 if (!sumBinB) {
1145
1146 // If this is the first hint, then just save it
1147 sumBinB = std::move(pdfBinB) ;
1148
1149 } else {
1150
1151 auto newSumBinB = std::make_unique<list<Double_t>>(sumBinB->size()+pdfBinB->size()) ;
1152
1153 // Merge hints into temporary array
1154 merge(pdfBinB->begin(),pdfBinB->end(),sumBinB->begin(),sumBinB->end(),newSumBinB->begin()) ;
1155
1156 // Copy merged array without duplicates to new sumBinBArrau
1157 sumBinB = std::move(newSumBinB) ;
1158 needClean = true ;
1159 }
1160 }
1161 }
1162
1163 // Remove consecutive duplicates
1164 if (needClean) {
1165 sumBinB->erase(std::unique(sumBinB->begin(),sumBinB->end()), sumBinB->end()) ;
1166 }
1167
1168 return sumBinB.release() ;
1169}
1170
1171
1172////////////////////////////////////////////////////////////////////////////////
1173/// If all components that depend on obs are binned, so is their sum.
1175{
1176 for (const auto arg : _pdfList) {
1177 auto pdf = static_cast<const RooAbsPdf*>(arg);
1178 if (pdf->dependsOn(obs) && !pdf->isBinnedDistribution(obs)) {
1179 return false ;
1180 }
1181 }
1182
1183 return true ;
1184}
1185
1186
1187////////////////////////////////////////////////////////////////////////////////
1188/// Label OK'ed components of a RooAddPdf with cache-and-track
1189
1191{
1192 RooFIter aiter = pdfList().fwdIterator() ;
1193 RooAbsArg* aarg ;
1194 while ((aarg=aiter.next())) {
1195 if (aarg->canNodeBeCached()==Always) {
1197 //cout << "tracking node RooAddPdf component " << aarg->IsA()->GetName() << "::" << aarg->GetName() << endl ;
1198 }
1199 }
1200}
1201
1202
1203
1204////////////////////////////////////////////////////////////////////////////////
1205/// Customized printing of arguments of a RooAddPdf to more intuitively reflect the contents of the
1206/// product operator construction
1207
1209{
1210 bool first(true) ;
1211
1212 if (!_coefList.empty()) {
1213 for (std::size_t i = 0; i < _pdfList.size(); ++i ) {
1214 const RooAbsArg * coef = _coefList.at(i);
1215 const RooAbsArg * pdf = _pdfList.at(i);
1216 if (!first) {
1217 os << " + " ;
1218 } else {
1219 first = false ;
1220 }
1221
1222 if (i < _coefList.size()) {
1223 os << coef->GetName() << " * " << pdf->GetName();
1224 } else {
1225 os << "[%] * " << pdf->GetName();
1226 }
1227 }
1228 } else {
1229
1230 for (const auto pdf : _pdfList) {
1231 if (!first) {
1232 os << " + " ;
1233 } else {
1234 first = false ;
1235 }
1236 os << pdf->GetName() ;
1237 }
1238 }
1239
1240 os << " " ;
1241}
#define cxcoutD(a)
#define oocoutW(o, a)
#define coutW(a)
#define dologD(a)
#define coutE(a)
#define ccoutD(a)
#define TRACE_CREATE
Definition RooTrace.h:23
#define ClassImp(name)
Definition Rtypes.h:364
char name[80]
Definition TGX11.cxx:110
char * Form(const char *fmt,...)
const std::vector< Int_t > & retrieve(Int_t masterCode) const
Retrieve the array of integer codes associated with the given master code.
Int_t store(const std::vector< Int_t > &codeList, RooArgSet *set1=0, RooArgSet *set2=0, RooArgSet *set3=0, RooArgSet *set4=0)
Store given arrays of integer codes, and up to four RooArgSets in the registry (each setX pointer may...
RooAbsArg is the common abstract base class for objects that represent a value and a "shape" in RooFi...
Definition RooAbsArg.h:69
RooArgSet * getObservables(const RooArgSet &set, Bool_t valueOnly=kTRUE) const
Given a set of possible observables, return the observables that this PDF depends on.
Definition RooAbsArg.h:309
void clearValueAndShapeDirty() const
Definition RooAbsArg.h:611
Bool_t dependsOn(const RooAbsCollection &serverList, const RooAbsArg *ignoreArg=0, Bool_t valueOnly=kFALSE) const
Test whether we depend on (ie, are served by) any object in the specified collection.
virtual CacheMode canNodeBeCached() const
Definition RooAbsArg.h:427
friend class RooArgSet
Definition RooAbsArg.h:642
Take ownership of the contents of 'comps'.
Bool_t isValueDirty() const
Definition RooAbsArg.h:436
Set the operation mode of this node.
OperMode operMode() const
Query the operation mode of this node.
Definition RooAbsArg.h:499
Storage_t const & get() const
RooFIter fwdIterator() const
One-time forward iterator.
virtual Bool_t add(const RooAbsArg &var, Bool_t silent=kFALSE)
Add the specified argument to list.
virtual Bool_t addOwned(RooAbsArg &var, Bool_t silent=kFALSE)
Add an argument and transfer the ownership to the collection.
Storage_t::size_type size() const
RooAbsArg * first() const
virtual void Print(Option_t *options=0) const
This method must be overridden when a class wants to print itself.
const char * GetName() const
Returns name of object.
virtual Bool_t remove(const RooAbsArg &var, Bool_t silent=kFALSE, Bool_t matchByNameOnly=kFALSE)
Remove the specified argument from our list.
RooAbsArg * find(const char *name) const
Find object with given name in list.
RooAbsGenContext is the abstract base class for generator contexts of RooAbsPdf objects.
virtual void resetErrorCounters(Int_t resetValue=10)
Reset error counter to given value, limiting the number of future error messages for this pdf to 'res...
TString _normRange
MC generator configuration specific for this object.
Definition RooAbsPdf.h:391
RooArgSet const * _normSet
Normalization integral (owned by _normMgr)
Definition RooAbsPdf.h:354
RooAbsReal * _norm
Definition RooAbsPdf.h:353
friend class RooRealIntegral
Definition RooAbsPdf.h:346
virtual Bool_t syncNormalization(const RooArgSet *dset, Bool_t adjustProxies=kTRUE) const
Verify that the normalization integral cached with this PDF is valid for given set of normalization o...
Int_t _errorCount
Definition RooAbsPdf.h:383
virtual Double_t expectedEvents(const RooArgSet *nset) const
Return expected number of events to be used in calculation of extended likelihood.
static Int_t _verboseEval
Definition RooAbsPdf.h:347
RooAbsRealLValue is the common abstract base class for objects that represent a real value that may a...
virtual Double_t getMax(const char *name=0) const
Get maximum of currently defined range.
virtual Double_t getMin(const char *name=0) const
Get miniminum of currently defined range.
RooAbsReal is the common abstract base class for objects that represent a real value and implements f...
Definition RooAbsReal.h:64
RooAbsReal * createIntegral(const RooArgSet &iset, const RooCmdArg &arg1, const RooCmdArg &arg2=RooCmdArg::none(), const RooCmdArg &arg3=RooCmdArg::none(), const RooCmdArg &arg4=RooCmdArg::none(), const RooCmdArg &arg5=RooCmdArg::none(), const RooCmdArg &arg6=RooCmdArg::none(), const RooCmdArg &arg7=RooCmdArg::none(), const RooCmdArg &arg8=RooCmdArg::none()) const
Create an object that represents the integral of the function over one or more observables listed in ...
virtual std::list< Double_t > * binBoundaries(RooAbsRealLValue &obs, Double_t xlo, Double_t xhi) const
Retrieve bin boundaries if this distribution is binned in obs.
Double_t _value
Definition RooAbsReal.h:484
Double_t getVal(const RooArgSet *normalisationSet=nullptr) const
Evaluate object.
Definition RooAbsReal.h:94
virtual std::list< Double_t > * plotSamplingHint(RooAbsRealLValue &obs, Double_t xlo, Double_t xhi) const
Interface for returning an optional hint for initial sampling points when constructing a curve projec...
Transient cache with transformed values of coefficients.
virtual RooArgList containedArgs(Action)
List all RooAbsArg derived contents in this cache element.
RooArgList _rangeProjList
RooArgList _refRangeProjList
RooArgList _projList
RooArgList _suppNormList
RooArgList _suppProjList
RooAddPdf is an efficient implementation of a sum of PDFs of the form.
RooListProxy _coefList
RooAbsGenContext * genContext(const RooArgSet &vars, const RooDataSet *prototype=0, const RooArgSet *auxProto=0, bool verbose=false) const override
Return specialized context to efficiently generate toy events from RooAddPdfs return RooAbsPdf::genCo...
bool _allExtendable
RooAICRegistry _codeReg
RooFit::UniqueId< RooArgSet >::Value_t _idOfLastUsedNormSet
!
std::pair< const RooArgSet *, CacheElem * > getNormAndCache(const RooArgSet *nset) const
Look up projection cache and per-PDF norm sets.
std::unique_ptr< const RooArgSet > _copyOfLastNormSet
!
Int_t _coefErrCount
bool _haveLastCoef
List of supplemental normalization factors.
Int_t getAnalyticalIntegralWN(RooArgSet &allVars, RooArgSet &numVars, const RooArgSet *normSet, const char *rangeName=0) const override
Determine which part (if any) of given integral can be performed analytically.
void updateCoefficients(CacheElem &cache, const RooArgSet *nset) const
Update the coefficient values in the given cache element: calculate new remainder fraction,...
void printMetaArgs(std::ostream &os) const override
Customized printing of arguments of a RooAddPdf to more intuitively reflect the contents of the produ...
CacheElem * getProjCache(const RooArgSet *nset, const RooArgSet *iset=0, const char *rangeName=0) const
Manager of cache with coefficient projections and transformations.
void selectNormalization(const RooArgSet *depSet=0, bool force=false) override
Interface function used by test statistics to freeze choice of observables for interpretation of frac...
void finalizeConstruction()
void setCacheAndTrackHints(RooArgSet &) override
Label OK'ed components of a RooAddPdf with cache-and-track.
bool _recursive
RooObjCacheManager _projCacheMgr
Double_t analyticalIntegralWN(Int_t code, const RooArgSet *normSet, const char *rangeName=0) const override
Return analytical integral defined by given scenario code.
std::list< Double_t > * plotSamplingHint(RooAbsRealLValue &obs, Double_t xlo, Double_t xhi) const override
Loop over components for plot sampling hints and merge them if there are multiple.
bool checkObservables(const RooArgSet *nset) const override
Check if PDF is valid for given normalization set.
void selectNormalizationRange(const char *rangeName=0, bool force=false) override
Interface function used by test statistics to freeze choice of range for interpretation of fraction c...
void fixCoefNormalization(const RooArgSet &refCoefNorm)
By default the interpretation of the fraction coefficients is performed in the contextual choice of o...
RooSetProxy _refCoefNorm
void resetErrorCounters(Int_t resetValue=10) override
Reset error counter to given value, limiting the number of future error messages for this pdf to 'res...
double getValV(const RooArgSet *set=nullptr) const override
Calculate and return the current value.
void fixCoefRange(const char *rangeName)
By default, fraction coefficients are assumed to refer to the default fit range.
RooListProxy _pdfList
Registry of component analytical integration codes.
TNamed * _refCoefRangeName
const RooArgList & pdfList() const
void computeBatch(cudaStream_t *, double *output, size_t nEvents, RooFit::Detail::DataMap const &) const override
Compute addition of PDFs in batches.
bool isBinnedDistribution(const RooArgSet &obs) const override
If all components that depend on obs are binned, so is their sum.
std::list< Double_t > * binBoundaries(RooAbsRealLValue &, Double_t, Double_t) const override
Loop over components for plot sampling hints and merge them if there are multiple.
Double_t expectedEvents(const RooArgSet *nset) const override
Return expected number of events for extended likelihood calculation, which is the sum of all coeffic...
bool _projectCoefs
std::vector< double > _coefCache
RooArgList is a container object that can hold multiple RooAbsArg objects.
Definition RooArgList.h:22
RooAbsArg * at(Int_t idx) const
Return object at given index, or nullptr if index is out of range.
Definition RooArgList.h:110
RooArgSet is a container object that can hold multiple RooAbsArg objects.
Definition RooArgSet.h:35
RooFit::UniqueId< RooArgSet > const & uniqueId() const
Returns a unique ID that is different for every instantiated RooArgSet.
Definition RooArgSet.h:171
virtual void compute(cudaStream_t *, Computer, RestrictArr, size_t, const VarVector &, const ArgVector &={})=0
T * getObj(const RooArgSet *nset, Int_t *sterileIndex=0, const TNamed *isetRangeName=0)
void reset()
Clear the cache.
Int_t setObj(const RooArgSet *nset, T *obj, const TNamed *isetRangeName=0)
RooDataSet is a container class to hold unbinned data.
Definition RooDataSet.h:36
A one-time forward iterator working on RooLinkedList or RooAbsCollection.
RooAbsArg * next()
Return next element or nullptr if at end.
auto & at(RooAbsArg const *arg, RooAbsArg const *=nullptr)
Definition DataMap.h:88
virtual Bool_t add(const RooAbsArg &var, Bool_t silent=kFALSE) override
static RooMsgService & instance()
Return reference to singleton instance.
static Int_t _debugCount
Bool_t isActive(const RooAbsArg *self, RooFit::MsgTopic facility, RooFit::MsgLevel level)
Check if logging is active for given object/topic/RooFitMsgLevel combination.
static const char * str(const TNamed *ptr)
Return C++ string corresponding to given TNamed pointer.
static const TNamed * ptr(const char *stringPtr)
Return a unique TNamed pointer for given C++ string.
static RooConstVar & value(Double_t value)
Return a constant value object with given value.
RooRealVar represents a variable that can be changed from the outside.
Definition RooRealVar.h:39
Class RooRecursiveFraction is a RooAbsReal implementation that calculates the plain fraction of sum o...
virtual Bool_t add(const RooAbsArg &var, Bool_t silent=kFALSE) override
Overloaded RooArgSet::add() method inserts 'var' into set and registers 'var' as server to owner with...
virtual void removeAll() override
Remove all argument inset using remove(const RooAbsArg&).
The TNamed class is the base class for all named ROOT classes.
Definition TNamed.h:29
virtual const char * GetName() const
Returns name of object.
Definition TNamed.h:47
Ssiz_t Length() const
Definition TString.h:410
const char * Data() const
Definition TString.h:369
RooConstVar & RooConst(Double_t val)
const Int_t n
Definition legend1.C:16
std::vector< RooSpan< const double > > VarVector
R__EXTERN RooBatchComputeInterface * dispatchCUDA
R__EXTERN RooBatchComputeInterface * dispatchCPU
This dispatch pointer points to an implementation of the compute library, provided one has been loade...
std::vector< double > ArgVector
Definition first.py:1 | 14,685 | 54,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.549203 |
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