url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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statistics-homework-help69046.look4blog.com | 1,544,616,742,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823872.13/warc/CC-MAIN-20181212112626-20181212134126-00248.warc.gz | 268,960,097 | 5,606 | # 5 Essential Elements For statistics assignment help
Plot a bivariate details set, ascertain the road of very best in good shape for their information, after which you can Test the accuracy of the line of very best match.
Enter a established of information points, then derive a perform to fit Individuals points. Manipulate the operate on a coordinate aircraft utilizing slider bars. Learn the way Each and every consistent and coefficient impacts the resulting graph.
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Pupils Participate in a generalized version of connect four, gaining the chance to position a chunk after simplifying fractions, changing fractions to decimals and percentages, and answering algebra queries involving fractions.
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Students investigate linear capabilities by endeavoring to guess the slope and intercept from inputs and outputs. Linear Function Equipment is without doubt one of the Interactivate evaluation explorers.
Visualize aspects by setting up rectangular locations on the grid. While you attract each variable set about the grid, the variables is going to be listed. Factorize two is probably the Interactivate evaluation explorers.
Action through the generation on the Koch Snowflake -- a fractal constructed from deforming the edges of a triangle, and examine number designs in sequences and geometric Houses of fractals.
Enter two advanced quantities (z and c) as requested pairs of actual figures, then simply click a button to iterate comprehensive.
A far more Sophisticated version of Slope Slider, this activity lets the manipulation in the constants and coefficients in any function therefore encouraging the user to check out the effects to the graph on the operate by shifting All those figures.
Graph requested pairs and customize the graph title and axis labels. Details are connected from left to correct, rather than becoming related within the buy These are entered.
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Students can build box plots for both designed-in or user-specified knowledge together with experiment with outliers. Person could decide to use or not utilize the median for calculation of interquartile range.
Visualize things by way of constructing rectangular places over a grid. As you attract Every element set around the grid, the factors will be detailed. Factorize two is amongst the Interactivate assessment explorers. | 557 | 3,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-51 | latest | en | 0.928103 |
https://en.khanacademy.org/math/cc-sixth-grade-math/cc-6th-geometry-topic/geometric-solids/v/counting-faces-and-edges-of-3d-shapes | 1,721,666,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00729.warc.gz | 202,153,036 | 128,955 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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Lesson 1: Geometric solids (3D shapes)
# Counting faces and edges of 3D shapes
Learn about shapes! Discover how to count faces and edges on 3D figures. We explore a transparent shape with five faces and another shape, a square pyramid, with eight edges and five faces. It's a colorful journey into geometry!
## Want to join the conversation?
• are vertices the same as edges
• No, vertices are like corners where three faces meet, with the exception of the apex, in which 4 or more faces meet in a pyramid.
• how can i identify faces and edges of a sphere? maybe its silly to ask but i am curious.
• A face is a flat surface on a solid, and edges are the lines at which faces meet, and a vertex is the point at which when three or more edges meet. A sphere has no flat surfaces, so it has no faces. Since it doesn't have faces, it can't have edges, or vertices.
• Is cylinder a prism? Is cone a pyramid?
• no because all of the faces of prisms and pyramids have to be polygons.
• can a 3d shape have 4 faces?
• Yes it can have 4 faces. I am pretty sure I am correct.
(1 vote)
• So this video really made me better.
• hey sister what is angle
(1 vote)
• Hey! (Credit to Dictionary.com) An angle is a geometric figure formed by two lines that begin at a common point or by two planes that begin at a common line.
• are vertices the same as edges
• No, vertices are not the same as edges.
(1 vote)
• what would a 4-D shape look like? | 406 | 1,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.959515 |
https://www.storyofmathematics.com/fractions-to-decimals/9-34-as-a-decimal/ | 1,721,402,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00019.warc.gz | 846,620,202 | 38,945 | # What Is 9/34 as a Decimal + Solution With Free Steps
The fraction 9/34 as a decimal is equal to 0.264.
The division of two numbers produces either an integer or a decimal value. Fractions represent the division in the form p/q, where p (the dividend) is the numerator and q (the divisor) is the denominator. All decimal values can be converted into a fraction and vice versa.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 9/34.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 9
Divisor = 34
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 9 $\div$ 34
This is when we go through the Long Division solution to our problem.
Figure 1
## 9/34 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 9 and 34, we can see how 9 is Smaller than 34, and to solve this division, we require that 9 be Bigger than 34.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 9, which after getting multiplied by 10 becomes 90.
We take this 90 and divide it by 34; this can be done as follows:
90 $\div$ 34 $\approx$ 2
Where:
34 x 2 = 68
This will lead to the generation of a Remainder equal to 90 – 68 = 22. Now this means we have to repeat the process by Converting the 22 into 220 and solving for that:
220 $\div$ 34 $\approx$ 6
Where:
34 x 6 = 204
This, therefore, produces another Remainder which is equal to 220 – 204 = 16. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 160.
160 $\div$ 34 $\approx$ 4
Where:
34 x 4 = 136
Finally, we have a Quotient generated after combining the three pieces of it as 0.264, with a Remainder equal to 24.
Images/mathematical drawings are created with GeoGebra. | 703 | 2,873 | {"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} | 4.875 | 5 | CC-MAIN-2024-30 | latest | en | 0.912328 |
https://www.codeproject.com/Articles/448719/Cryptography-Asymmetric-Encryption-by-using-Asymme | 1,586,059,404,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00198.warc.gz | 843,482,217 | 25,464 | 14,490,767 members
# Cryptography: Asymmetric Encryption by Using Asymmetric Algorithm Classes
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28 Aug 2012CPOL
This blog will enable you to understand the basic of Cryptography with Asymmetric Encryption Algorithm Classes.
In the previous blog – Symmetric Encryption by Symmetric Algorithm Classes–Part 1 and Part 2, we have learned about basic introduction of Cryptography based on Symmetric Encryption. so, now in addition to the previous blog, here we will learn about basics of Asymmetric Encryption.
## Asymmetric Encryption
Asymmetric encryption is also referred to as public key encryption because it uses public key as well as private key. This means it's having a secret key that must be kept from unauthorized or anonymous users and a public key that can make public to any one. Hence, we can say that Asymmetric encryption is designed so that the private key remains shield and secret, whereas the public key is widely distributed. The private key is used to lock information, whereas the public key is used to unlock it.
The main benefit of opting for asymmetric encryption is that you can share encrypted data without having access to the private key. In standard mode, asymmetric encryption is used more commonly than symmetric encryption, and it is proved that it's a standard used to help secure communication over the Internet.
Have a look at the animated pic to view its working.
Animated Pic: Asymmetric Encryption Decryption Concept
We came to know in the earlier blog about symmetric encryption, the same key is used for both encryption and decryption, however this approach is simpler but less secure since the key must be communicated to and known at both sender and receiver locations.
Example For Better Understanding
Let's assume about conversion of Plain text between A and B. If ‘A’ send message to ‘B’, ‘A’ can find out public key (but not private key of ‘B’) of ‘B’ from a central administrator and encrypt a message to ‘B’ using ‘B’ public key. When ‘B’ receive it, ‘B’ can decrypt it with ‘B’ private key. In addition to encrypting messages (which can ensures privacy), ‘B’ can authenticate itself to ‘A’ (so ‘A’ know that it is really ‘B’ who is the sender of message to ‘A’) by using ‘B’ private key to encrypt a digital certificate. When ‘A’ receive it, ‘A’ can use ‘B’ public key to decrypt it.
## Asymmetric Algorithm Classes
`System.Security.Cryptography`
`namespace` provides encryption classes that provide most popular Asymmetric algorithms like:
• RSA and `RSACryptoServiceProvider`
• DSA and `DSACryptoServiceProvider`
## RSACryptoServiceProvider Class
RSA stands for Ron Rivest, Adi Shamir and Leonard Adleman, who first publicly described it in 1977-[ From Wikipedia ]
The RSA class is an `abstract` class that extends the Asymmetric Algorithm class and provides support for the RSA algorithm. The .NET Framework RSA algorithm support an encryption key size ranging from 384 bits to 16,384 bits in increments of 8 bits by using the Microsoft Enhanced Cryptographic Provider and an encryption key size ranging from 384 bits to 512 bits in increments of 8 bits by using the Microsoft Base Crystallographic Provider.
The `RSACryptoServiceProvider` class exnteds the RSA class and is the concrete RSA algorithm class.
### Implementation of RSACryptoServiceProvider Class
To perform Encryption and Decryption, you must add:
`using System.Security.Cryptography; // Namespace`
Now take a look at encryption function:
```static public byte[] RSAEncrypt(byte[] byteEncrypt, RSAParameters RSAInfo, bool isOAEP)
{
try
{
byte[] encryptedData;
//Create a new instance of RSACryptoServiceProvider.
using (RSACryptoServiceProvider RSA = new RSACryptoServiceProvider())
{
//Import the RSA Key information. This only needs
//toinclude the public key information.
RSA.ImportParameters(RSAInfo);
//Encrypt the passed byte array and specify OAEP padding.
encryptedData = RSA.Encrypt(byteEncrypt, isOAEP);
}
return encryptedData;
}
//Catch and display a CryptographicException
//to the console.
catch (CryptographicException e)
{
Console.WriteLine(e.Message);
return null;
}
}
```
In the above code, Encrypt Function is used to encrypt plain text to cipher text. Encrypt function needs two parameters, first one is byte array of plain text and second one specifies OAEP padding(True or False).
Now in the same way, we need to create function for decrypting the PlainText(Encrypted Text). Have a look at the given function which is responsible for decrypting encrypted text.
```static public byte[] RSADecrypt(byte[] byteDecrypt, RSAParameters RSAInfo, bool isOAEP)
{
try
{
byte[] decryptedData;
//Create a new instance of RSACryptoServiceProvider.
using (RSACryptoServiceProvider RSA = new RSACryptoServiceProvider())
{
//Import the RSA Key information. This needs
//to include the private key information.
RSA.ImportParameters(RSAInfo);
//Decrypt the passed byte array and specify OAEP padding.
decryptedData = RSA.Decrypt(byteDecrypt, isOAEP);
}
return decryptedData;
}
//Catch and display a CryptographicException
//to the console.
catch (CryptographicException e)
{
Console.WriteLine(e.ToString());
return null;
}
```
We can see in the above code, Decrypt function is used in the same manner to decrypt cipher text to plain text. Decrypt Function needs two parameters, the first one is byte array of encrypted text and the second one specifies OAEP padding(True or False).
Note: OAEP padding is only available on Microsoft Windows XP or later.
Now we have created function, so we can use both functions to the appropriate manner to accomplish Encryption Decryption task.
Note: We need to access `RSACryptoServiceProvider` class here.
`RSACryptoServiceProvider RSA = new RSACryptoServiceProvider();`
### How to Use Encrypt and Decryption Function
Note: The below code is tested in Windows Application. You can download the source code for better understanding.
```UnicodeEncoding ByteConverter = new UnicodeEncoding();
RSACryptoServiceProvider RSA = new RSACryptoServiceProvider();
byte[] plaintext;
byte[] encryptedtext;```
### For Encrypt Text
```plaintext = ByteConverter.GetBytes(txtplain.Text);
encryptedtext = RSAEncrypt(plaintext, RSA.ExportParameters(false), false);
txtencrypt.Text = ByteConverter.GetString(encryptedtext);
```
### For Decrypt Text or Back to Plain Text
```//While Decryption set True, the private key information
//(using RSACryptoServiceProvider.ExportParameters(true),
byte[] decryptedtex = RSADecrypt(encryptedtext, RSA.ExportParameters(true), false);
txtdecrypt.Text = ByteConverter.GetString(decryptedtex);```
### What We’ve Seen
• Create Function of Encryption and Decryption
• Create Encoder
• Create `RSACryptoServiceProvider` Instance
• Create byte array to illustrate the encrypted and decrypted data
• Encrypted text and Display Cipher text
• Decrypt cipher text and display back to plain text
### Output by using RSAEncryption Program
Animated Picture: Screenshot of RSA Encryption Program
## Coming Next
Please stay in touch for the extended part of this article. The topic will move around “Introduction and Implementation of `DSACryptoServiceProvider` for Beginners”.
Filed under: .NET, C#, CodeProject, Cryptography
Tagged: .NET, C#, CodeProject, Cryptography
## Share
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using existing third party public key for Encryption VishalHealthcare1-Jun-13 7:30 VishalHealthcare 1-Jun-13 7:30
My vote of 5 oskifree5-Sep-12 22:02 oskifree 5-Sep-12 22:02
Re: My vote of 5 RaviRanjanKr7-Sep-12 2:45 RaviRanjanKr 7-Sep-12 2:45
Re: My vote of 5 RaviRanjanKr30-Aug-12 7:34 RaviRanjanKr 30-Aug-12 7:34
Last Visit: 4-Apr-20 18:03 Last Update: 4-Apr-20 18:03 Refresh 1 | 1,729 | 7,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-16 | latest | en | 0.949112 |
https://itbackyard.com/how-to-implement-kanpsack-algorithm-in-c-with-real-life-example/ | 1,722,991,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00686.warc.gz | 260,586,442 | 14,190 | # How to implement recursive KnapSack Algorithm in C# with real life Example – Part1
Imagine you have a little trolley with a maximum load of 200Kg (20000 gram), and you get one chance to transport rocks of different weights, each has a reward value from location A to location B, and you are not allowed to load more than 200Kg of rocks. You want to pick the most possible weight for the highest possible reward.
What would you do to calculate your reward?
You can of course count it manually and it will take a lot of time for a little list of raw items, but if the list becomes big enough that it will take you long time to calculate. Then KnapSack Algorithm is your best friend.
Here is a specific example, as you can see the in image above, we have gold rocks labled from 1 to 20. The size of rock does not mean more reward, it is all about the weight of the rock and how much gold value it contains.
Her is the list:
KnapSack algorithm can help solve this problem. So in this article, I will implement a recursive version of knapsack using C#. The implementation is simple and straightforward. In my case, I will use a knapsack to calculate my highest possible reward for maximum possible weight, but this does not mean which rock is less worth taking. In my approach, I will sub up the rock and subtract the results from the knapsack, that way I will find out which rock should be left behind.
Let’s do it.
I will create a class called it `Knapsack`.
It has a method that takes our rock data and the maximum allowed weight. My Data class contain a list of rock weights in gram and a reward for each rock, as you saw earlier in the rock list.
``````public class KnapSack
{
public decimal Calculate(IReadOnlyList<Data> data, int maxWeight)
{
var bonus = data.Select(e => e.Value).ToArray();
var weights = data.Select(e => (double)e.Weight).ToArray();
return (decimal)Calculate(maxWeight, weights, bonus, bonus.Length);
}
{
var idx = n - 1;
if (n <= 0 || weight <= 0.0) return 0;
if (weights[idx] > weight)
{
return 0;
}
var a = values[idx] + Calculate(weight - weights[idx], weights, values, idx);
var b = Calculate(weight, weights, values, idx);
var result = Max(a, b);
return result;
}
private static double Max(double a, double b)
{
return a > b ? a : b;
}
}``````
My data class model has dummy data as shown in the rock list.
``````public class Data
{
public int Id { get; set; }
public string Description { get; set; }
public int Weight { get; set; }
public double Value { get; set; }
public Data(int id, string description, int weight, double value)
{
Id = id;
Description = description;
Value = value;
Weight = weight;
}
public static List<Data> RawMaterials()
{
return new List<Data>
{
new Data(1, "Rock1", 5724, 17.74),
new Data(2, "Rock2,9873, 37.12),
new Data(3, "Rock3",13492, 46.14),
new Data(4, "Rock4",7727, 30.44),
new Data(5, "Rock5",2924, 10.64),
new Data(6, "Rock6",1544, 5),
new Data(7, "Rock7",7082, 28.18),
new Data(8, "Rock8",13960, 50.82),
new Data(9, "Rock9",6371, 22.94),
new Data(10, "Rock10",14380, 53.2),
new Data(11, "Rock11",19045, 58.66),
new Data(12, "Rock12",14057, 13.72),
new Data(13, "Rock13",7082, 28.18),
new Data(14, "Rock14",13960, 50.82),
new Data(15, "Rock15",6371, 22.94),
new Data(16, "Rock16",13380, 53.2),
new Data(17, "Rock17",19045, 58.66),
new Data(18, "Rock18",7057, 12.72),
new Data(19, "Rock19",19045, 58.66),
new Data(20, "Rock20",6057, 3.72)
};
}
}``````
I will also create a config class
``````public static class Config
{
public const int MaxWeight = 200000;
}``````
Finally, let’s run our knapsack in the main method.
``````public static void Main(string[] args)
{
var calculator = new KnapSack();
var result = calculator.Calculate(Data.RawMaterials(), Config.MaxWeight);
Console.WriteLine(\$"Max bonus for {Config.MaxWeight} weight is: {result} money unit");
var sum = (decimal)Data.RawMaterials().Sum(e => e.Value);
var exclude = sum - result;
Console.WriteLine(\$"Which rock value to exclude {exclude}");
}``````
When we run this method, we will get this output:
`Max bonus for 200000 weight is: 649,78 money unitWhich rock value to exclude 13,72`
When we check the rock with a value of 13,72, we will find out it is rock 12 that we should exclude from the transport. That said this is only an example of a knapsack, if you exclude multiple items, you may need to implement the dynamic version of the knapsack.
So now let’s sum up all rock’s weight without rock 12, we will get 194119 gram which is the maximum weight we can take with us for the highest reward.
Conclusion
As you can see, with few lines of code, knapsack algorithms are able to calculate very complex problems and complicated problems. As mentioned our algorithm is recursive, which means it does have an efficient running time. In this example we are lucky that only one element was removed (rock 12) and it was easy to predict, but how about removal of multiple items? This is what I will cover in Part 2 of this article where I will improve running time using a dynamic programming approach and that will help me keep track on index of removed or included items. Maybe I have been very optimistic using the gold rocks as an example, but the same type of algorithm can be used for calculating crypto mining block rewards or other application types. If you take the rock list and calculate it in excel ark, you will end up excluding rock number 12. Give it a try😜 | 1,464 | 5,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.903028 |
http://www.slideserve.com/madra/lamport-s-scalar-clocks | 1,493,333,237,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122629.72/warc/CC-MAIN-20170423031202-00609-ip-10-145-167-34.ec2.internal.warc.gz | 670,777,394 | 17,720 | # Lamport’s Scalar Clocks - PowerPoint PPT Presentation
1 / 12
Lamport’s Scalar Clocks. Shiva Bottu. Organization. Introduction High level Implementation Details Experimentation Results Further Research. Introduction. Lamport’s logical clocks are introduced to capture causality relation in a distributed system.
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Lamport’s Scalar Clocks
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## Lamport’s Scalar Clocks
Shiva Bottu
### Organization
Introduction
High level Implementation Details
Experimentation
Results
Further Research
### Introduction
Lamport’s logical clocks are introduced to capture causality relation in a distributed system.
Each process Pi has a logical clock Ci and for a particular event a in P, it assigns an integer value Ci, which is timestamp of the event.
Timestamps are monotonically increasing.
### High level Implementation Details
• There are 2 implementation rules that should be followed
• Before executing event, update timestamp value
Ci:=Ci+d
• When sending a message, attach the timestamp. When message is received by a process, timestamp of receive event is calculated as maximum of the two timestamps (timestamp of the process and timestamp received in the message)
Ci:=max(Ci,Cmsg)
### Experimentation
• Number of processes are varied.
• Random flood algorithm is used to exchange messages between processes
• Part 1:
• Total number of updates are plotted against number of processes
• Average number of updates per process against number of processes
• Part 2:
• Number of updates and Number of messages against number of processes
• Number of processes against average number of messages required for each process
### Part 1(a):Total clock updates against Number of processes
Number of Processes
### Part 1(b):Average clock updates per process against Number of processes
Number of Processes
### Part 2(a): No. of Processes VS clock updates & No. of Processes VS No. of messages required
Number of Processes
### Part 2(b): No. of Processes VS Average no. of messages per process
No. of Messages per process
Number of Processes
### Results
From part-1(a), it is observed that the number of updates increase with increase in number of processes
From part-1(b), average number of updates per process increase with increase in number of processes
From part-2(a), we can observe that number of clock updates and number of messages increase with increase in no. of processes
From part-2(b), we can observe that average number of messages required per process increases with increase in number of processes
### Further Research
To change the base algorithm from random flood to other topologies and test with the same process.
Running with high load of processes.
### References
http://en.wikipedia.org/wiki/Lamport_timestamps
http://cnlab.kaist.ac.kr/~ikjun/data/Course_work/CS642-Distributed_Systems/papers/lamport1978.pdf | 741 | 3,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-17 | longest | en | 0.872728 |
https://www.teacherspayteachers.com/Product/Real-World-Word-Problem-Solving-Bundle-Menu-Pet-Department-Store-3301950 | 1,506,133,147,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689413.52/warc/CC-MAIN-20170923014525-20170923034525-00432.warc.gz | 859,740,884 | 23,824 | Total:
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# Real-World Word Problem Solving Bundle: Menu, Pet & Department Store
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Product Description
This bundle includes ALL THREE of my Real World Problem Solving Bundles. I have included my Pet Store, Department Store, and Menu Math products (addition, subtraction, & multiplication). I have FINALLY bundled together 3 of my favorite resources!
Menu Math:
This is a great bundle and I use it in my own classroom! It includes a fun and colorful "menus" with a variety of different food items, listed. Students use the menu to answer a variety of different questions/math problems. My students are SO motivated by Menu Math and it teaches them valuable "real world" math skills. I have had students come in to school after this unit and tell me that they did menu math when their family went out to dinner at a restaurant.
Skills Include:
-Finding items on the Menu
-Addition without regrouping
-Addition with regrouping
-Addition of 2 items
-Addition of 3 items
-Doubling
-Tripling
-Create your own addition problems
-Create your own Story/Word problems
-Money
-Addition of decimals
-Subtraction without regrouping
-Subtraction with regrouping
-Finding the difference between 2 items
-Multi-step story problems (with addition and subtractions)
-Mental Math
-How much more/Less
-Story/Word problems
-Create your own subtraction problems
-Subtraction of decimals
-Multiplication with regrouping
-Doubling
-Tripling
-Multiple step word problems
-Create your own multiplication problems
-Multiplication of decimals
I have used this as a whole class resource as well as in a small group setting and/or for center time. There are 47 student worksheets, a color menu page (can be laminated for future class use), and a page of the different items,
Pet Store Math
This bundle includes ALL THREE of my addition, subtraction, & multiplication Pet Store math bundles. If you are interested in just addition, just subtraction, or just multiplication, check them out in my store.
This is a great bundle that I use in my classroom! It includes a fun and colorful "Price Sheet" with a variety of different pets and pet store items listed. Students use the price sheet to answer a variety of different questions/math problems. My students are SO motivated by Pet Store Math and it teaches them valuable "real world" math skills. Students love to take the skills they have learned and apply them in their real world.
Skills Include:
-Finding items on the Price Sheet
-Addition without regrouping
-Addition with regrouping
-Addition of 2 items
-Addition of 3 items
-Doubling
-Tripling
-Create your own addition problems
-Create your own Story/Word problems
-Money
-Addition of decimals
-Subtraction without regrouping
-Subtraction with regrouping
-Finding the difference between 2 items
-Multi-step story problems (with addition and subtractions)
-Mental Math
-How much more/Less
-Story/Word problems
-Create your own subtraction problems
-Subtraction of decimals
-Multiplication with regrouping
-Doubling
-Tripling
-Multiple step word problems
-Create your own multiplication problems
-Multiplication of decimals
I have used this as a whole class resource as well as in a small group setting and/or for center time. There are 47 student worksheets, a color Price Sheet page (can be laminated for future class use), and a page of the different items
Department Store/Shopping Math
his bundle includes ALL THREE of my addition, subtraction, & multiplication Shopping List math bundles. If you are interested in just addition, just subtraction, or just multiplication, check them out in my store (links below).
This is a great bundle that I use in my classroom! It includes a fun and colorful "shopping list" with a variety of different items listed. Students use the colorful shopping list to answer a variety of different questions/math problems. My students are SO motivated by Shopping List Math and it teaches them valuable "real world" math skills. I have had students come in to school after this unit and tell me that they did shopping list math when their family went to the store.
*This bundle is a little more challenging than my Menu Math bundle. The prices of the items are more expensive*
Skills Include:
-Finding items on the Shopping List
-Addition without regrouping
-Addition with regrouping
-Addition of 2 items
-Addition of 3 items
-Doubling
-Tripling
-Create your own addition problems
-Create your own Story/Word problems
-Money
-Addition of decimals
-Subtraction without regrouping
-Subtraction with regrouping
-Finding the difference between 2 items
-Multi-step story problems (with addition and subtraction)
-Mental Math
-How much more/Less
-Story/Word problems
-Create your own subtraction problems
-Subtraction of decimals
-Multiplication with regrouping
-Doubling
-Tripling
-Multiple step word problems
-Create your own multiplication problems
-Multiplication of decimals
I have used this as a whole class resource as well as in a small group setting and/or for center time. There are 66 student worksheets, a shopping list page (can be laminated for future class use), and a page of the different items
Total Pages
205 pages
Answer Key
N/A
Teaching Duration
N/A
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Sign up | 1,340 | 5,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-39 | latest | en | 0.904483 |
http://mathhelpforum.com/differential-geometry/177656-derivative-problem-print.html | 1,527,437,008,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00441.warc.gz | 176,686,342 | 2,850 | # Derivative Problem
• Apr 12th 2011, 12:04 PM
mathematicalbagpiper
Derivative Problem
Let $\displaystyle E$ be the set of all 2n x 2n matrices of the form $\displaystyle C=\left( {\begin{array}{cc} A & -B \\ B & A \\ \end{array} } \right)$, for $\displaystyle A, B\in Mat(n\times n, \mathbb{R})$, and let $\displaystyle G=\{C\in E : CC^T=I\}$. Further, let $\displaystyle f(C)=CC^T$. This maps $\displaystyle E$ into a subspace $\displaystyle W=\{P\in E : P^T = P\}$.
Compute the derivative $\displaystyle (Df)_C$ and show it's surjective when $\displaystyle C\in G$.
So I computed the derivative and got $\displaystyle (Df)_C(U)=CU^T+UC^T$. For $\displaystyle C\in G$, this becomes $\displaystyle CU^T+UC^{-1}$. So for $\displaystyle P\in W$, somehow I need to come up with a matrix $\displaystyle U\in E$ such that $\displaystyle (Df)_C(U)=P$, but I'm not seeing it at all.
• Apr 12th 2011, 12:23 PM
girdav
Put $\displaystyle U := \frac 12 PC$ for $\displaystyle P\inW$.
• Apr 12th 2011, 01:39 PM
mathematicalbagpiper
Following up on this problem, it says what I've just proven and the implicit function theorem imply that $\displaystyle G$ is a manifold, and to determine its dimension. How do you go about determining the dimension of a manifold?
• Apr 13th 2011, 03:06 AM
girdav
Did you find the dimension of $\displaystyle W$ ? | 440 | 1,337 | {"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-2018-22 | latest | en | 0.763866 |
https://oeis.org/A043230 | 1,721,621,417,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517823.95/warc/CC-MAIN-20240722033934-20240722063934-00684.warc.gz | 386,297,650 | 3,621 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A043230 Numbers k such that 1 and 7 occur juxtaposed in the base-10 representation of k but not of k-1. 0
17, 71, 117, 170, 217, 271, 317, 371, 417, 471, 517, 571, 617, 671, 710, 771, 817, 871, 917, 971, 1017, 1071, 1117, 1170, 1217, 1271, 1317, 1371, 1417, 1471, 1517, 1571, 1617, 1671, 1700, 1817, 1871, 1917, 1971, 2017, 2071 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Table of n, a(n) for n=1..41. MATHEMATICA SequencePosition[Table[If[SequenceCount[IntegerDigits[n], {1, 7}]>0 || SequenceCount[IntegerDigits[n], {7, 1}]>0, 1, 0], {n, 2100}], {0, 1}][[All, 2]] (* Harvey P. Dale, Oct 17 2021 *) CROSSREFS Sequence in context: A179039 A152295 A039407 * A044010 A106921 A105414 Adjacent sequences: A043227 A043228 A043229 * A043231 A043232 A043233 KEYWORD nonn,base AUTHOR Clark Kimberling STATUS approved
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Last modified July 22 00:08 EDT 2024. Contains 374478 sequences. (Running on oeis4.) | 478 | 1,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.615844 |
https://puzzling.stackexchange.com/questions/76106/what-is-the-next-number-in-the-given-series | 1,719,026,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00306.warc.gz | 398,030,916 | 39,213 | What is the next number in the given series?
I know that next number questions are not popular here, but this one is tough and I couldn't figure out the pattern. To give an idea one of the questions had a pattern of Prime(Fib(n)).
So this is the series:
0, 0, 2, 6, 12, 21, 34, 51
What is the next number in the series ?
• Welcome to Puzzling.SE! If this puzzle was taken from somewhere, please make sure you provide details of where it came from, ensure you have the original author's permission and that it does not come from an active competition or exam Commented Dec 5, 2018 at 20:29
• @papabiceps If Mathematics didn't get you an answer, why do you think you will get one here? Commented Dec 5, 2018 at 23:34
• Just posted here, to see if I can get an answer. Commented Dec 5, 2018 at 23:36
Here's a relatively simple formula that works (starting $$n=0$$)
$$a_n = n^2 - n + (\max \{ 0, n-4\})^2$$
$$a_8 = 8^2 - 8 + 4^2 = 72$$ | 287 | 937 | {"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": 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} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.949375 |
https://www.topperlearning.com/forums/home-work-help-19/compare-the-stability-of-o2-o2-o2-o2-2-chemistry-chemical-bonding-and-molecular-structure-61812/reply | 1,511,020,267,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804976.22/warc/CC-MAIN-20171118151819-20171118171819-00670.warc.gz | 891,284,483 | 40,794 | Question
Thu January 03, 2013 By:
# COMPARE THE STABILITY OF O2, O2(+) , O2 (-) , O2 (2-)
Fri January 04, 2013
Electronic configurations:
O2 = ? 1 s2 ?*1 s2 ?2 s2 ?*2 s2 ?2 p 2 (?2p 2 ?2 p 2 ) (?*2 p 1 = ?*2 p1 )
O2+ = ? 1 s2 ?*1 s2 ?2 s2 ?*2 s2 ?2 p 2 (?2p 2 ?2 p 2 ) ?*2 p 1 x
O2 = ? 1 s2 ?*1 s2 ?2 s2 ?*2 s2 ?2 p 2 (?2p 2 ?2 p 2 ) (?*2 p 1 = ?*2 p1 ) ?*2 p1 z
O2 2- = ? 1 s2 ?*1 s2 ?2 s2 ?*2 s2 ?2 p 2 (?2p 2 ?2 p 2 ) (?*2 p 1 = ?*2 p1 ) ?*2 p2 z
The electronic configuration of the O2 + ion containing 15 electrons can be written as:
? 1 s2 ?*1 s2 ?2 s2 ?*2 s2 ?2 p 2 z (?2p 2 x ?2 p 2 ) (?*2 p 1 = ?*2 p ) ?*2 p z
The bond order can be found as:
B.O = (Nb-Na)/2
N= Number of electrons in the bonding orbitals = 10
N= Number of electron in the anti-bonding orbitals = 5
B.O = (10-5)/2 = 5/2 = 2.5
The electronic configuration of the O2 - ion containing 17 electrons can be written as:
? 1 s2 < ?*1 s2 < ?2 s2 < ?*2 s2 < ?2 p 2 z < (?2p 2 x = ?2 p 2 ) < (?*2p 2 = ?*2 p 1 ) < ?*2 p z
The bond order can be calculated as:
B.O = (10-7)/2 = 3/2 = 1.5
Similarly, Bonbd order of O2 is = B.O = (10-6)/2 = 4/2 = 2
Bonbd order of O2 2- is = B.O = (10-8)/2 = 2/2 = 1
The higher the value of bond order, higher is the stability of the bond, so on the basis of above information, we can say that, O2 + ion is more stable than O2, O2 - and O22- ions.
Related Questions
Tue October 31, 2017
# Wich of the following are isoelectonic and isostructural? NO3 - , CO3 -2 , SO3 , CLO3 - A NO3 - ,CO3 -2 , B SO3 ,NO3 - C CLO3 -,CO3 2 - D CO3 -2,SO3 ONLY ONE CORRECT AND WHY EXPLANE HOW?
Tue October 31, 2017 | 838 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-47 | latest | en | 0.281493 |
http://math.stackexchange.com/questions/216015/understanding-regular-matrices | 1,467,263,314,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783398075.47/warc/CC-MAIN-20160624154958-00144-ip-10-164-35-72.ec2.internal.warc.gz | 195,659,002 | 20,706 | # Understanding regular matrices
A regular matrix $A$ is described as a square matrix that for all positive integer $n$, is such that $A^n$ has positive entries.
How then would I prove something is regular? I mean I can prove something is irregular if $A^2$ has some 0 or negative entries; but I cant prove regularity since I cant solve $A^n$ for all integers $n$.
My thoughts are that if a matrix $A$ is diagonalisable as $A=PD^{-1}P$ then it is 'regular,' since then all $A^k$ exist; but does this also imply all entries of $A^k$ are positive?
Any hints?
-
The definition is not right. A regular matrix is a matrix for which some power of the matrix has all positive entries. – Christopher A. Wong Oct 18 '12 at 2:04
@ChristopherA.Wong Definitions can't be right or wrong. – user53153 Feb 27 '13 at 3:07
If $A$ has an entry that is $0$ or negative, then $A$ is not regular. If, on the other hand, every entry in $A$ is positive, can $A^2$ have a negative or zero entry? Can $A^3$? There’s an easy proof by induction waiting here for you to find it. Note that diagonalizability has nothing to do with the matter: if $A$ is square, $A^n$ exists for all $n\ge 0$ whether or not $A$ is diagonalizable. Diagonalizability of $A$ merely makes it easy to calculate the powers of $A$.
However, that’s not the usual definition of regular matrix. The usual definition is that a square matrix $A$ is regular if it is stochastic and there is some $n\ge 1$ such that all of the entries of $A^n$ are positive.
-
The statement that if A has an entry that is 0 or negative implies that it cannot be regular is incorrect. Consider $$\left( \begin{array}{ccc} .9 & .5 & 0\\ 0 & .5 & .4\\ .1 & 0 & .6\end{array} \right)$$ for example. The square of this matrix is regular. – Brandon Thomas Van Over May 10 '15 at 7:40
@SirJective: Read the definition of regular that the OP is using. By that definition your matrix is not regular. – Brian M. Scott May 10 '15 at 7:43
Linear combinations of positive numbers (with positive coefficients) will always yield positive numbers, yes? Therefore, it should suffice to note that all entries of $A$ are positive. On the other hand, if $A$ has an entry that is nonpositive, then $A$ can't be regular (since $A=A^1$ and $1>0$), so this is a sufficient condition, too.
-
There is some grave confusion. A matrix is regular precisely when it is invertible. That is an $n\times n$ matrix $A$ is regular precisely when there exists some matrix $B$ such that both $AB$ and $BA$ are the identity matrix. It is absolutely not the case that a matrix is regular if the entries in its powers are positive. For example, the $2\times 2$ matrix $A=-I_2$ is regular since you can take $B=A$ and then $BA=AB=I_2$. Moreover, the $2\times 2$ matrix with all entries equal to $1$ is not regular.
-
That’s one of the possible meanings of regular matrix, but I don’t think that it’s the most common one. – Brian M. Scott Oct 18 '12 at 2:03
Note that definitions may vary from source to source, and that "regular" is among the most overused terms in mathematics, so it's quite possible that the OP is quoting from a source in which regular is precisely as described by the OP. I'll further remark that I personally have more frequently encountered "nonsingular" as synonymous with "invertible"--and practically never seen "regular" used in that context. – Cameron Buie Oct 18 '12 at 2:05
Indeed, invertibility has little to do with powers having positive entries, so surely regular is not being used here to mean nonsingular. – Gerry Myerson Oct 18 '12 at 12:43
A regular matrix is the same as a nonsingular matrix. A matrix M with nonnegative entries and for which all entries of M^n are positive, for some positive integer n, is said to be "primitive" [1]. Vora's definition of a regular matrix seems to be based on the definition of a primitive matrix.
[1] Marjorie Senechal, Quasicrystals and Geometry, Cambridge University Press, Cambridge, 1995, p. 125.
-
The meaning of *regular" depends on context. It may mean "nonsingular" in some sources, and something else in other sources. – user53153 Feb 27 '13 at 3:07 | 1,104 | 4,129 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-26 | longest | en | 0.935866 |
http://lua-users.org/lists/lua-l/2001-07/msg00083.html | 1,660,775,171,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00076.warc.gz | 36,005,260 | 2,723 | • Subject: Finding the 'double' resolution
• From: Philippe Lhoste <PhiLho@...>
• Date: Thu, 19 Jul 2001 12:43:03 +0200 (MEST)
```I wrote:
> Page 74 (11.4 Queues and Double Queues), you state that indices
> of a double
> queue can grow almost infinitely because they are actually
> doubles, so they
> can be quite large.
> But since the doubles are handled by the compiler / run-time library, I
> suppose they are stored in a limited space (as opposed to large number
> libraries). 48 bits, if I recall correctly (and if this is
> standard, I am not a
> specialist here).
> So after some time, values will be rounded and manipulated with
> exponents of
> ten, so small increments will be lost.
>
> I made a little test:
>
> val1 = 314159265358979323
> val2 = val1 + 1
> print(val1, val2, val2 - val1)
>
> result on Windows NT:
> 3.141592653589793e+017 3.141592653589793e+017 0
>
> OK, I made a small test, the limit seems to be between
> 9,007,190,000,000,000
> and 9,007,200,000,000,000, so you may be quite right finally...
> It may be interesting to state this limit (if it is standard induced)
> somewhere, I think.
I improved my test to better and quicker detect this limit:
-- Find the 'double' resolution
-- ie. the point where an increment of 1 to a Lua numerical value
-- (stored as double C type) is lost.
-- by Philippe Lhoste <PhiLho@GMX.net> http://jove.prohosting.com/~philho/
-- v. 1.0 -- 2001/07/19
--i = 9 000 000 000 000 000
i = 9000000000000000
incr = 100000000000000
repeat
j = i + 1
if i - j == 0 then
i = i - incr
incr = incr / 10
print('... ' .. i .. ' (' .. incr .. ')')
if incr < 1 then
print('= ' .. i)
break
end
end
i = i + incr
until nil
-- val1 = 9007199254740992
-- val2 = val1 + 1
-- print(val1, val2, val2 - val1)
val1 is the first value, on my system (Windows NT on Duron), where an
increment of 1 isn't signifiant.
Regards.
--
--._.·´¯`·._.·´¯`·._.·´¯`·._.·´¯`·._.·´¯`·._.·´¯`·._.--
Philippe Lhoste (Paris -- France)
Professional programmer and amateur artist
http://jove.prohosting.com/~philho/
--´¯`·._.·´¯`·._.·´¯`·._.·´¯`·._.·´¯`·._.·´¯`·._.·´¯`--
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``` | 751 | 2,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.739838 |
https://mathsgee.com/54519/what-is-a-uniformly-continuous-function | 1,709,420,227,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00219.warc.gz | 388,084,820 | 20,722 | 0 like 0 dislike
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What is a uniformly continuous function?
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A uniformly continuous function is a function that, given any two points in its domain, can be made as close as desired by making the distance between the two points sufficiently small. This means that the function is continuous and its graph is a single, unbroken line.
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Informally, a continuous function satisfying the weierstrass definition of limits in the same way, regardless of the point in the function.
Formally, for all $$\varepsilon>0$$ there exists a $$\delta>0$$ such that for all $$x, y \in X|x-y|<\delta$$ implies $$|f(x)-f(y)|<\varepsilon$$
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https://thebestpaperwriters.com/answer-in-organic-chemistry-for-alisha-270324/ | 1,695,422,003,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506423.70/warc/CC-MAIN-20230922202444-20230922232444-00241.warc.gz | 633,678,276 | 13,421 | Answer in Organic Chemistry for Alisha #270324
A hypothetical acid HA has a dissociation constant 1.8 10ିଷ.
a. Write a balanced dissociation reaction of the acid
b. Write the equilibrium constant expression for the acid
c. If the initial concentration of the acid is 0.5 M, determine its hydrogen ion
concentration.
d. Calculate the ionization constant
e. Comment on the strength of the acid.
a) HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq).
b)We assess the equilibrium:
H
(
O
=
)
C
O
H
(
l
)
+
H
2
O
(
l
)
H
(
O
=
)
C
O
+
H
3
O
+
And we write the equilibrium equation in the usual way:
K
a
=
Concentration of products
Concentration of reactants
K
a
=
[
H
(
O
=
)
C
O
]
[
H
3
O
+
]
[
H
(
O
=
)
C
O
H
(
l
)
]
.
This site reports that K
a
for formic acid =
1.70
×
10
5
c)I’m assuming that you’re working with a monoprotic weak acid here so that the ionization equilibrium can be written like this
HA
(
a
q
)
+
H
2
O
(
l
)
A
(
a
q
)
+
H
3
O
+
(
a
q
)
Now, you know that the solution has
pH
=
5
As you know, the pH
of the solution is defined as
pH
=
log
(
[
H
3
O
+
]
)
−−−−−−−−−−−−−−−−−−−−
You can rearrange this equation to find the equilibrium concentration of hydronium cations.
log
(
[
H
3
O
+
]
)
=
pH
This is equivalent to
10
log
(
[
H
3
O
+
]
)
=
10
pH
which gets you
[
H
3
O
+
]
=
10
pH
In your case, you will have
[
H
3
O
+
]
=
10
5.0
=
1.0
10
5
M
Now, notice that every mole of HA
that ionizes produces 1
mole of A
, the conjugate base of the acid, and 1
mole of hydronium cations.
This means that, at equilibrium, the solution has
[
A
]
=
[
H
3
O
+
]
produced in a 1
:
1
mole ratio
[
A
]
=
1.0
10
5
M
The initial concentration of the acid will decrease because some of the molecules ionize to produce A
and H
3
O
+
.
So, in order for the ionization to produce [
H
3
O
+
]
, the initial concentration of the acid must decrease by [
H
3
O
+
]
.
This means that, at equilibrium, the concentration of the weak acid will be equal to
[
HA
]
=
[
HA
]
initial
[
H
3
O
+
]
In your case, you will have
[
HA
]
=
0.01 M
1.0
10
5
.
M
[
HA
]
=
0.00999 M
By definition, the acid dissociation constant, K
a
, will be equal to
K
a
=
[
A
]
[
H
3
O
+
]
[
HA
]
Plug in your values to find
K
a
=
1.0
10
5
M
1.0
10
5
.
M
0.00999
M
K
a
=
1.001
10
8
M
Rounded to one significant figure and expresses without added units, the answer will be
K
a
=
1
10
8
d)HCN] is 0.5M
[H3
O+
] is 0.0068M
[CN−
] is 0.0068M
We can assume here that, [H3O+
]=[CN−
]
The equilibrium constant would be as follows:
Ka
=[HCN]
[H3
O+
][CN−
]
=0.5
0.00682
=9.2×10−5
e) it is a week acid
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## line graph of k5
Padam Padam Ending, Protein Shakes For Weight Loss, Epiphone Pro 1 Vs Yamaha F310, Asus Vivobook S15 S512fl-ph55 Review, Golang Extend Struct, Carbs In Ham, Virtual Keyboard Arabic, Pragmatic Worldview In Research, | 752 | 3,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-21 | longest | en | 0.937049 |
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Fourth grade Lesson Perimeter | BetterLesson
To give the students more practice, we find the perimeter to a shape with more than 4 sides. I explain to the students that they can not use the formula for this shape because the sides are different lengths. To find the perimeter, you should add all of the sides. I ask the students to help me count the number of sides.Author: Rose Monroe
4th grade Perimeter Worksheets | Education
4th grade Perimeter Worksheets. Use this year-end assessment to check students’ grasp of key fourth grade geometry concepts. 4th grade . Math . Worksheet. Polygon Perimeter Word Problems. Worksheet. Polygon Perimeter Word Problems. In this worksheet, students will follow four steps to find the perimeter of polygons in word problems.
Grade 4 Math Worksheet: Perimeter and area of rectangles
Math Worksheets > Grade 4 > Geometry > Area and perimeter of rectangles. Geometry worksheets: area and perimeter of rectangles. Below are our grade 4 geometry worksheets on finding the area and perimeter of rectanglesents are given the measurements of two sides of each rectangle in customary units (inches, feet).
4th Grade Area and Perimeter Introduction
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Worksheets > Math > Grade 4 > Geometry. Free geometry worksheets. Our grade 4 geometry worksheets cover topics such as classifying angles, triangles and quadrilaterals, areas and perimeters and coordinate grids worksheets are printable pdf files. Classify angles, triangles &
Perimeter Worksheets - Math Salamanders
Perimeter Worksheets. Welcome to our Perimeter worksheets page. Here you will find a range of free printable perimeter sheets, which will help your child to learn to work out the perimeters of a range of rectangles and rectilinear shapes. 4th Grade Perimeter Sheets. Perimeter Sheet 4 Sheet 4 Answers PDF version Perimeter Sheet 5 Sheet 5
Seventh grade Lesson Scale Drawings - Area and Perimeter
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Unit 8 Area & Perimeter - Mrs. Warner's Learning Community
Go Online to wwwydaymath to view the Student Reference book pp. 131-136 for Unit 8 review. Click on book icon to visit the Everyday Math website. Be sure to have your username and password. Use the pages below to practice finding the area and perimeter of various shapes.[PDF]
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Geometry Notes Perimeter and Area Page 4 of 57 The area of a shape is defined as the number of square units that cover a closed figure. For most of the shape that we will be dealing with there is a formula for calculating the area. In some cases, our shapes will be made up
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Apr 22, 2010A PowerPoint presentation of Perimeter & Area for 4th and 5th grade. Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website.
Related searches for 4th grade perimeter notes | 767 | 3,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-51 | latest | en | 0.857411 |
https://onlinemathcenter.com/blog/math/6th-grade-rational-numbers-sets-of-numbers/ | 1,726,756,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00526.warc.gz | 392,238,130 | 30,163 | Home 6th-Grade Rational Numbers: Sets of Numbers
# 6th-Grade Rational Numbers: Sets of Numbers
By the time young math students have reached the sixth grade, they know that math is not just a simple case of addition and subtraction anymore. Young math learners have started to explore topics that go beyond our basic notions of mathematics.
One such topic that introduces students to new ways of approaching, thinking about, and, ultimately, solving math problems, is the topic of number sets.
We’re going to take a look and what exactly number sets are, as well as dig into some of the most common number sets that 6th-grade math students will need to get comfortable with.
## What Is A Number Set?
Number sets are collections of numbers classified as groups according to the values of their elements. In simple terms, number sets are collections of numbers that are grouped together.
These individual numbers are assigned to a group based on their characteristics. They will be grouped together with other numbers that share these key characteristics.
What are some of the most common sets of numbers, then? And what are these characteristics that math students must look out for?
## Natural Numbers
Natural numbers are the numbers that each and every one of us is familiar with. The set of natural numbers is represented by the letter N, and is written like this:
= {1, 2, 3, …}
These are the numbers that we count with, which is why they are sometimes known as the counting numbers. This set of numbers begins at 1 and is infinite, continuing forever.
The set of natural numbers is said to be closed under addition and multiplication. What this means is that if you add a natural number to another natural number, you will always get a natural number as the answer. Similarly, if you multiply a natural number by another natural number, you will get a natural number as the answer.
## Integers
The next set of numbers that young math students will be using frequently is the set known as integers.
The set of integers is quite similar to the set of natural numbers, with some very important differences. The set of integers is denoted by the letter Z, and is written as:
ℤ = {…, -3, -2, -1, 0, 1, 2, 3, …}
As you can see from the set, it includes all of the positive natural numbers, zero, and all of the negative natural numbers.
Just like the natural numbers, the set of integers is closed under multiplication and division. The set of integers, however, is also closed under subtraction, meaning if you subtract any integer from any other integer, the answer will be an integer.
## Rational Numbers
The set of rational numbers contains all of the numbers you can make by dividing one integer by another (but not dividing by zero). In other words, rational numbers are fractions
That means, the set of rational numbers includes:
• Integers
• Positive natural numbers
• Zero
• Negative natural numbers
• Fractions
The set of rational numbers is represented by the letter Q, and we denote it like this:
ℚ = {m/n: m, n are integers}
This means that the set of rational numbers is equal to m divided by n, where both m and n are integers.
## Introducing Irrational Numbers
Another number set that 6th graders are going to meet – and thoroughly get to know – in math class this year is the set of irrational numbers.
The best way to imagine irrational numbers is to imagine the number line. Now, picture the number line if we plotted all of the rational numbers – all integers, natural numbers, zero, and fractions – we would still have many, many points on the number line.
How can we identify these points? These points are what we call irrational numbers.
Irrational numbers are those that are neither whole numbers, nor are they fractions. The decimal of irrational numbers never repeats and never ends.
The most famous irrational number is one we’ve looked at quite recently in another article: Pi. As you may remember, pi is a number that helps us identify certain characteristics of circles, such as their circumference, or their area. Pi can be written as π, or it can be written as:
3.14159265358979323846264338327950288419…
Don’t let irrational numbers slip out of your mind, as we’ll be looping back around to them for some upcoming articles in the very near future.
## Be Rational, Turn to OMC Math Courses
The new school year is well underway. It’s time to make sure your children are staying ahead of the curve and not developing any gaps in their math knowledge. Enroll them today in an OMC Middle School Math Course to make sure they stay on top of number sets and every other topic that will cross their path this school year.
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Our manager will be in contact with you shortly. | 1,193 | 5,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-38 | latest | en | 0.963623 |
http://www.stockhouse.com/companies/bullboard/t.ans/ainsworth-lumber-company-ltd?postid=20666073 | 1,432,498,432,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928076.40/warc/CC-MAIN-20150521113208-00125-ip-10-180-206-219.ec2.internal.warc.gz | 743,566,713 | 24,557 | My Take:
Equity = 175 Million being raised (rights offering) + 332 Million currently (Current Stock price * 100 Million Shares) = 507 Million Equity (Can do this showing Enteprise Value and reduced debt also)
507 Million / 240 Million Shares = \$2.11 per share post rights trading
Working it Backwards to calculate cost of right:
Cost of Right + (1.25 * 1.388) = 2.11 * 1.388
Cost (Value) of Right is \$1.19
Cross Check: If I did not hold any stock or rights and went to the market to buy the rights so I could buy the shares I would spend \$1.19 for the right then 1.25 * 1.388 to get 1.388 shares at a total cost of 1.19 + 1.74 = 2.93. 2.93/1.388 = \$2.11 per share | 208 | 678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2015-22 | latest | en | 0.866922 |
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27 Nov 2012, 05:55
Hi,
I have taken my GMAT prep 1 & got Q-37 ,V-12..(420 ). I suppose i was completely burned our of energy during Verbal.
I couldn't able to concentrate on any questions in verbal after 10 questions. I am planning to give my real GMAT after a month..
I am already going thru most of the math stuff in this forum, kindly suggest me for Verbal stuff too. I could able to crack a SC/CR bit if there is no timing but when i have time limited i am really becoming a idiot in choosing an answer choice. suggest me how i can prepare myself within timed limit.
All your feedback are highly appreciated.
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29 Nov 2012, 14:42
1
KUDOS
Do you think you performed as well as you could have? The first time I took a GMATPrep test, I had worked a full day and was watching TV in the background. I am sure that the score I got (360 I believe) doesn't represent my true abilities. Put yourself in a quiet environment similar to the one you would find in a testing center. You may also want to brush up on your foundations. MGMAT's Foundations of GMAT Math was really helpful for me. Best of luck!
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Long journey, but not impossible.
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Joined: 22 Oct 2012
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### Show Tags
01 Dec 2012, 06:05
Vixen wrote:
Do you think you performed as well as you could have? The first time I took a GMATPrep test, I had worked a full day and was watching TV in the background. I am sure that the score I got (360 I believe) doesn't represent my true abilities. Put yourself in a quiet environment similar to the one you would find in a testing center. You may also want to brush up on your foundations. MGMAT's Foundations of GMAT Math was really helpful for me. Best of luck!
thanks a lot vixen, i m planning to give one more prep test after some work on verbal section . I will confirm you my score after that..
Re: Need an advice ::: 420 is really frustrating [#permalink] 01 Dec 2012, 06:05
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# Measures of Dispersion and Central Tendency
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1) First, calculate the measures of central tendency and dispersion.
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122 words walk the user through the step to finding central tendency and dispersion measures as well as displaying the results.
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First four questions are interval scale questions. You need to score then like
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https://www.physicsforums.com/threads/finding-the-normal-modes-for-a-oscillating-system.809858/ | 1,508,313,771,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822822.66/warc/CC-MAIN-20171018070528-20171018090528-00681.warc.gz | 961,575,293 | 16,956 | # Finding the normal modes for a oscillating system
1. Apr 21, 2015
### skeer
1. The problem statement, all variables and given/known data
The system is conformed by two blocks with masses m (on the left) and M (on the right), and two springs on the left/right has the spring constant of k. The middle spring has a spring constant of 4k. Friction and air resistance can be ignored. All springs are massless.
Find the normal modes.
Diagram:
|~m~~~~M~|
2. Relevant equations
$L = T-V$
$T = \frac{1}{2}(m\dot{x}_1^2 + M\dot{x}_2^2)$
$V = \frac{1}{2}[(x_1^2 + x_2^2) + 4(x_1-x_2)^2]$
$\frac{\partial{L}}{\partial{x_k}} - \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x_k}}} = 0$
$[A_{ij} - \omega^2 m_{ij}]=0$
3. The attempt at a solution
I have tried to guess a solution for the normal modes but of the for $\eta_1 = x_1 - x_2$ and $\eta_2= x_1+x_2$ but I does not works. I have tried to add some arbitrary coefficient to $\eta_1$ & $\eta_2$ unsuccessfully. Trying to find the eigenvectors is a pain in the neck since the eigenfrequencies are:$\omega^2 = \frac{5k(M+m) \pm k\sqrt{25(M^2+m^2) -14Mm}}{2Mm}$.
I read in a textbook that one could find the coefficient for the etas by knowing that the ratios $\frac{M_{11}}{M_{22}}=\frac{A_{11}}{A_{22}}=\alpha^2$ but for this case the first ratio is $\frac{m}{M}$ and the second is 1 .Therefore, this method doesn't help me :/.
I would appreciate any contribution. Thank you.
2. Apr 21, 2015
### paisiello2
Do you have to solve this problem using the Lagrangian?
3. Apr 22, 2015
### skeer
The Lagragian is not necessary, but is the only method I know. I believe that if I use forces the problem would complicate more.
4. Apr 22, 2015
### paisiello2
I think the force method is easier but the results are the same.
I get a slightly different answer for the Eigen values but even then I think you could probably simplify it a little bit:
ω2 = k(1/M+1/m) [5/2± √(25/4+16/(M/m+m/M+2))]
I am not aware of any other method except plugging the Eigen values into the equations of motion and solving for the mode shapes.
5. Apr 22, 2015
### paisiello2
Duplicate post. | 678 | 2,120 | {"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} | 3.75 | 4 | CC-MAIN-2017-43 | longest | en | 0.888812 |
https://normgoldblatt.com/what-are-the-5-steps-of-long-division-50 | 1,675,175,526,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00565.warc.gz | 451,195,392 | 6,735 | # What are the 5 steps of long division
What are the 5 steps of long division? It follows the same steps as that of long division, namely, – divide, multiply, subtract, bring down and repeat or find.
## Long Division Calculator with Decimals and Remainders
How to perform long division? To perform long division, first identify the dividend and divisor. To divide 100 by 7, where 100 is the dividend and 7 is the divisor, set up the long division problem
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## 5 Steps to Long Division
Ans.5 Step 1: Set up the long division. Step 2: Divide 1st term of dividend by first term of divisor to get first term of the quotient. Step 3: Take the term found in step 2 and
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## What Is Long Division? Explained For Primary School
Given below are the 5 main steps of long division. For example, let us see how we divide 52 by 2. Step 1: Consider the first digit of the dividend which is 5 in this example. Here, 5 > 2. We know that 5 is not divisible by 2. Step 2: We know that
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## How to do Long Division: A Simple Step-By-Step Guide with
Simple Long Division Steps Step 1: Divide. Long division is all about breaking an equation into different parts. Therefore, rather than looking at Step 2: Multiply. Now that you know that 6 will go into 1 zero times, then you need to multiply (6
## How to Do Long Division: 15 Steps (with Pictures)
What are the 5 steps of long division? in Science math Reading Time: 5 minutes read It follows the same steps as that of long division, namely, – divide, multiply, subtract, bring down and
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## Steps to Long Division Problems (With Examples)
You write the 5 next to the 1, making the number 15. Now you start all over again: Step 1: D for Divide How many times can you divide 5 into 15. The answer is 3. So you put 3 on the quotient line. Step 2: M for Multiply You multiply your answer
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No matter what question you have, you can always find an answer with a quick online search. | 729 | 3,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.932783 |
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Q)
The root mean square of a gas at 300K is $3\sqrt{R}$. The molar mass of this gas in $Kg mol^{-1}$ is
$(a)\;1\qquad(b)\;10\qquad(c)\;100\qquad(d)\;1000$
• The behaviour of gases has been expressed in terms of various gas laws obtained experimentally i.e Boyle's law , Charles law , Pressure-temperature law , Dalton's law of partial pressure , Graham's law of diffusion and Avogadro's hypothesis . Later on Maxwell derived kinetic equation $PV = \large\frac{1}{3}mnu^2$ , theoretically by assuming the concept of molecules and their motion . The term u represents root mean square speed of molecule.
$U_{rms} = \sqrt{\large\frac{3RT}{m}}$
$\therefore 3\times \sqrt{R} = \sqrt{\large\frac{3RT}{m}}$
or $9R = \large\frac{3RT}{m}$
or $m = \large\frac{3\times300}{9}$
$100 Kg mol^{-1}$ | 253 | 800 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-51 | latest | en | 0.720996 |
https://mathshistory.st-andrews.ac.uk/Curves/Kampyle/ | 1,679,419,852,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00149.warc.gz | 439,823,158 | 4,442 | # Curves
### Kampyle of Eudoxus
Cartesian equation:
$a^{2}x^{4} = b^{4}(x^{2} + y^{2})$
Polar equation:
$r = b^{2}/(a \cos^{2}( \theta ))$
### Description
A curve studied by Eudoxus also in relation to the classical problem of duplication of the cube.
Eudoxus was a pupil of Plato. His main work was in astronomy. He was the first to describe the constellations and invented the astrolabe. He introduced the study of mathematical astronomy into Greece.
Eudoxus found formulas for measuring pyramids cones and cylinders. His work contains elements of the calculus with a rigorous study of the method of exhaustion.
The word kampyle comes from the Greek καμπύλη meaning curve.
Further information:
J H McCoy
### Associated Curves
Definitions of the Associated curves | 193 | 775 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-14 | latest | en | 0.934515 |
http://www.ccs.neu.edu/home/vip/teach/Cpp_ENG/Homeworks/HW2.html | 1,516,382,350,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888077.41/warc/CC-MAIN-20180119164824-20180119184824-00703.warc.gz | 406,349,846 | 2,125 | CS1500 Algorithms and Data Structures for Engineering, FALL 2012
Implement the Gradient Descent algorithm for finding the minimum of a polynomial function of one variable f(x).Your program has to :
• Allow the user to define the function: read the degree and polynomial coefficients from a file (in any readable format, we suggest this format). Ask the user for the filename. In order to avoid using arrays, you can define the 5 coefficients (assume maximum degree 4) as global variables, that is outside any function. Alternatively, you can use an array of coefficients, but we have a lecture on arrays until next week.
• Implement a C++ function double userfunction(double) that computes f(x) for any real numbers x.
• Implement a C++ function double userfunction_diff(double) that computes the first differential f'(x) on real arguments x.
• Implement the Gradient descent loop for finding the minimum value of f using repeated calls to the two functions above, starting with an initial guessed argument x. Make sure the termination condition causes the loop to finish eventually.
• Try several learning rates λ, see what works best.
• EXTRA CREDIT: You can also try to vary λ across loop iterations: start with a larger value, and decrease it when x approaches the optimum.
• EXTRA CREDIT: Try to identify the cases where from starting x, following gradient descent leads to minus infinity.
Try your program with the following functions (other functions could be used by TA for testing):
, use λ=0.002
A description of the complete algorithm can be found here; however lecture notes are sufficient for this assignment. As usual for homeworks, you have to write the pseudocode and submit it together with your code.
EXTRA CREDIT: Implement Gradient Descent for polynomials up to degree 4 with two variables f(x1,x2).
EXTRA CREDIT: Implement Gradient Descent for other functions (not polynomials) of two variables, like Radial-Basis Functions. | 406 | 1,948 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-05 | latest | en | 0.85887 |
https://www.espn.com/blog/statsinfo/post/_/id/10528/math-behind-belichicks-4th-down-gamble | 1,701,211,749,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100016.39/warc/CC-MAIN-20231128214805-20231129004805-00152.warc.gz | 854,049,390 | 31,037 | <
>
# Math behind Belichick's 4th down gamble
The New England Patriots led 20-3 heading into the fourth quarter against the San Diego Chargers, and escaped with a 23-20 win after Kris Brown missed a 50-yard FG with less than a minute remaining that would have tied the game.
A key play in the game came with two minutes remaining when the Patriots, clinging to a 23-20 lead, faced a fourth and one at their 49-yard line. Bill Belichick decided to go for it, and BenJarvus Green-Ellis was stopped short of the first down marker, handing the Chargers the ball back with a chance to win or tie the game. The Patriots ended up with the victory, but did Belichick make the right decision in trying to convert on fourth down there?
If this situation sounds familiar, that’s because it’s the second straight season that Bill Belichick had the Patriots go for it on fourth down when leading late in the game with the ball in their own territory. Last year against the Colts on November 15, Belichick went for it on fourth and two in a similar situation, but the Patriots failed to convert and lost the game. Using a risk-reward analysis detailed here, we determined that Belichick actually made the right decision to go for it last year, but still lost.
We can perform the same analysis to see if Belichick was correct again. The decision of whether or not to go for it is based on comparing the Patriots chances of winning the game if they go for it versus their chances of winning if they simply decide to punt the ball.
#### Breaking down Bill Belichick's decision:
Using the win probability calculator on Brian Burke’s Advanced NFL Stats website, the Patriots had a 82 percent chance of winning the game by going for it on fourth down, based on the NFL and Patriots’ average conversion rates in that situation, which were very close to two-thirds in both cases. On the other hand, the calculator tells us that the Patriots had a 87 percent chance of winning the game if they had punted the ball.
So according to the analysis, this season Belichick actually did not make the statistically correct decision by going for it, but the Patriots ended up winning the game regardless. So it goes. | 467 | 2,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.970678 |
https://van.physics.illinois.edu/qa/listing.php?id=28240 | 1,582,871,361,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147054.34/warc/CC-MAIN-20200228043124-20200228073124-00076.warc.gz | 600,857,186 | 7,587 | # Q & A: Why nucleation for water to freeze?
Q:
Why do water molecules need an impurity to form crystals (or freeze) ? I get that it does need an impurity to start the process of forming the first crystal and that all the other crystals follow to form afterwards, but why is this impurity needed? I mean what does it do to the molecules of water? And can water molecules form without this impurity?
A:
Below about -40°C, it turns out water will just freeze on its own without any special nucleation site. Yet a liitle below freezing, say -5°C, it takes more or less forever without some nucleation. So to answer your great question we need an explanation that works over this whole range. An accurate explanation may involve a bit more background than you were expecting.
(For related questions see: .)
Water, like other materials, in the very long run settles into what's called the lowest free-energy phase on its own. Part of that free-energy is ordinary energy, so settling in is exactly like water settling into the bottom of a bowl after sloshing around a bit. Part of minimizing the free energy is maximizing something called the entropy, which is a measure of how many different states the parts can be in. When water evaporates, the molecules actually pick up more energy but they get so many extra states by running around in the big atmosphere that their free energy still goes down.
Above 0°C, the extra entropy that the water gets from tumbling around in a liquid beats the energy it could lose by freezing into a crystal, so it stays liquid. Below 0°C, the energy loss is more important, so the stable phase is the crystal. So now we ask, why doesn't the crystal just go and form at say -5°C?
The reason is that the first few molecules that lower their energy by lining up into a crystal-like pattern don't lose as much energy as they would by joining a full-size crystal. The surface of that cluster is still surrounded by liquid water, not lined up. Yet by lining up the molecules lose just about as much entropy as they would lining up in a big crystal. So a molecule gains free energy by joining a very small crystal. That means that these tiny crystals aren't stable. Until a medium-sized crystal forms, the molecules have no way of starting on the steady downward path in free energy toward a big crystal.
When the water is cold enough, the energy loss of just a very few molecules lining up is already enough to beat the entropy loss, so the crystal can just start to form on its own. At say -5°C it takes so many molecules lining up, gaining free energy, that it almost never happens on its own. That's why some sort of dust or bubble, already lining up some of the molecules, is needed to get the freezing started.
Mike W.
(published on 12/06/2014) | 614 | 2,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-10 | latest | en | 0.965749 |
https://www.answers.com/Q/Do_all_mechanical_waves_need_a_medium | 1,611,199,175,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00013.warc.gz | 668,765,177 | 35,537 | Physics
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# Do all mechanical waves need a medium?
###### Wiki User
yes they do have a medium
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## Related Questions
Sound waves are mechanical waves and all mechanical waves require a material medium for its propagation unlike electromagnectic waves
No, only surface and mechanical waves need a medium, compressional waves can travel through space. (Such as light from the Sun).I think light (EM) waves are the only ones that can travel through a vacuum. Sound waves require a medium.
Three types of mechanical waves are transverse, longitudinal, & surface waves. Mechanical waves require a material medium to travel.
There does not have to be a medium. Light does not require a medium. Mechanical waves require a mediumAnswer 2:Not all waves require a medium for their transmission. Mechanical waves require a medium for their transmission because it is the molecules of the medium which cause the wave to propagate. However, electromagnetic waves do NOT require a medium. (for example light)
There are two types of waves, mechanical and non-mechanical waves. A mechanical wave needs some sort of medium, like an earthquake to move the water.
No...electromagnetic waves require no medium
A mechanical wave(such as sound) must pass through a medium while an electromagnetic wave does not need one(such as light). Because seismic waves need a medium, they are mechanical waves. That is also the reason we hear earthquakes. When the vibrations from the focus of the earthquake spread in all directions, the sound vibrates from the earth to the air and then to our eardrums.
A mechanical wave requires a medium to transfer the energy it carries, unlike EMR waves. There are several types of mechanical waves, most of them being found in an earthquake. Mechanical waves are all longitudinal waves.
There are a few different formulas, depending on what measurements you know. For mechanical waves . . . the mechanical characteristics of the medium. For electromagnetic waves . . . the electrical characteristics of the medium. For all waves . . . the product of (wavelength) multiplied by (frequency).
All types of waves need a medium in order to propagate, with the exceptions of electromagnetic waves and gravitational waves.
Sound waves don't just travel the slowest in a vacuum, they don't travel at all. The reason is that sound waves, like all mechanical waves, need a medium to travel through.
Because sound is an mechanical wave and all the mechanical waves needs a medium to travel through. sound waves is produced by vibration of particles and it transfers sound to another part.
Some examples of mechanical waves are Sound, waves in a slinky, and water. Mechanical waves need matter to move. Some examples of electromagnetic waves are Radio, Gamma, X-rays, Infra-red, and Microwaves Electromagnetic waves are able to travel through a medium of liquids, solids, and geaseous states, or through space where there is no material at all, called "vacuum".
a.) They are all mechanical waves. b.) They all move objects great distances c.) They all require a medium d.) They all transfer energy.
Sound is a form of energy that travels "through" a mechanical medium. Light can be modelled as waves in certain circumstances. No mechanical medium is detectable, so either there is no medium, or the medium also propagates all matter the same way (Lorentz aether). The photoelectric effect shows that, just like sound is just motions of particles, so is light. electromagnetic
Sound is mechanical energy. It's a vibration or oscillation, and it is transmitted as a wave of pressure and displacement. Like all mechanical waves, sound requires a medium through which to travel, and the source transfers mechanical energy into the medium. Solids and fluids (liquids or gasses) serve as mediums for the transmission of sound waves.
Waves do not need a medium all time.it depends like u need no medium when it is electromagnetic wave.
light waves do not need a medium. Electromagnetic Waves (EM) do not need a medium. For example visible light, radio waves, microwaves, UV light and x-rays do not. These travel @ 300 million meters/sec in a vacuum.
An electromagnetic wave is capable of passing through a vacuum (space). All light waves are electromagnetic waves. Mechanical waves are not able to pass through space. They require a medium to pass their energy from one place to another. An example of a mechanical wave is a sound wave.
mechanical waves can be both transverse and longitudinal, but all electromagnetic waves are transverse.
Waves that require a medium in which to travel are called mechanical waves. A medium is the substance/matter/material the wave travels in. For example, sound is a mechanical wave. Sound can travel through the mediums of water or air or solid objects. All three of these are mediums through which sound can travel.Answer 2:A wave is a moving vibration through some material. That material is called the medium. Well known examples are the surface waves of water (water is the medium) and sound (air is the medium).
Electromagnetic waves can travel in a vacuum because this wave is a moving magnetic and electric field. These fields do not require a medium to travel in, and they "carry" the energy of the electromagnetic wave with them. This makes the electromagnetic wave different from mechanical waves like sound.With mechanical waves, we find the source putting the energy of the wave into the medium through which it is going to travel. Electromagnetic waves carry the electromagnetic energy away from the source without a need for a medium through which to travel. The moving fields carry the energy of the wave, and a vacuum presents no obstruction to the waves at all.
all wave are produced by a source but mechanical waves requires material medium to propagate it e.g is sound wave which is a mechanical wave needs material medium for it's propagation i.e material medium like air molecule,sound travels by vibration of air molecule that's why sound can't travel in a vacuum.....
light; all waves in the EM(electromagnetic) spectrum, including visible light, do not need a medium to travel in
Mechanical waves must use matter as a medium, they displace the matter in space. Examples of mechanical waves are sound waves (must go through gas (air) which is matter), ocean waves (must go through liquid (water) which is matter, and earthquakes (must go through solid (earth) which is matter.Electromagnetic waves do not need matter as a medium and are made of electromagnetic radiation, they cause fluctuations in the magnetic and electric fields that permeate space-time. Examples of electromagnetic waves are radio waves, microwaves, x-rays, and gamma rays (all are part of the electromagnetic spectrum or EMS).
###### SciencePhysicsEarth SciencesWaves Vibrations and OscillationsGeneral and Special RelativityElectricity and Magnetism
Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | 1,471 | 7,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-04 | latest | en | 0.91759 |
https://uk.mathworks.com/matlabcentral/cody/problems/94-target-sorting/solutions/1251551 | 1,576,189,658,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540547165.98/warc/CC-MAIN-20191212205036-20191212233036-00444.warc.gz | 597,567,708 | 15,898 | Cody
# Problem 94. Target sorting
Solution 1251551
Submitted on 15 Aug 2017 by Michael Böhm
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [1 2 3 4]; t = 0; b_correct = [4 3 2 1]; assert(isequal(targetSort(a,t),b_correct))
a = 1 2 3 4 a = 4 3 2 1 b = 4 3 2 1
2 Pass
a = -4:10; t = 3.6; b_correct = [-4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4]; assert(isequal(targetSort(a,t),b_correct))
a = -7.6000 -6.6000 -5.6000 -4.6000 -3.6000 -2.6000 -1.6000 -0.6000 0.4000 1.4000 2.4000 3.4000 4.4000 5.4000 6.4000 a = -7.6000 -6.6000 6.4000 -5.6000 5.4000 -4.6000 4.4000 -3.6000 3.4000 -2.6000 2.4000 -1.6000 1.4000 -0.6000 0.4000 b = -4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4
3 Pass
a = 12; t = pi; b_correct = 12; assert(isequal(targetSort(a,t),b_correct))
a = 8.8584 a = 8.8584 b = 12
4 Pass
a = -100:-95; t = 100; b_correct = [-100 -99 -98 -97 -96 -95]; assert(isequal(targetSort(a,t),b_correct))
a = -200 -199 -198 -197 -196 -195 a = -200 -199 -198 -197 -196 -195 b = -100 -99 -98 -97 -96 -95 | 528 | 1,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-51 | latest | en | 0.387092 |
https://se.mathworks.com/help/finance/portalpha.html | 1,713,541,664,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00229.warc.gz | 459,781,641 | 21,963 | # portalpha
Compute risk-adjusted alphas and returns for one or more assets
## Description
example
Note
An alternative for portfolio optimization is to use the Portfolio object for mean-variance portfolio optimization. This object supports gross or net portfolio returns as the return proxy, the variance of portfolio returns as the risk proxy, and a portfolio set that is any combination of the specified constraints to form a portfolio set. For information on the workflow when using Portfolio objects, see Portfolio Object Workflow.
example
portalpha(Asset,Benchmark,Cash) computes risk-adjusted alphas using the optional argument Cash.
example
[Alpha,RAReturn] = portalpha(Asset,Benchmark,Cash,Choice) computes risk-adjusted alphas and returns for one or more methods specified by Choice.
## Examples
collapse all
This example shows how to calculate the risk-adjusted return using portalpha and compare it with the fund and market's mean returns.
Use the example data with a fund, a market, and a cash series.
Returns = tick2ret(TestData);
Fund = Returns(:,1);
Market = Returns(:,2);
Cash = Returns(:,3);
MeanFund = mean(Fund)
MeanFund = 0.0038
MeanMarket = mean(Market)
MeanMarket = 0.0030
[MM, aMM] = portalpha(Fund, Market, Cash, 'MM')
MM = 0.0022
aMM = 0.0052
[GH1, aGH1] = portalpha(Fund, Market, Cash, 'gh1')
GH1 = 0.0013
aGH1 = 0.0025
[GH2, aGH2] = portalpha(Fund, Market, Cash, 'gh2')
GH2 = 0.0022
aGH2 = 0.0052
[SML, aSML] = portalpha(Fund, Market, Cash, 'sml')
SML = 0.0013
aSML = 0.0025
Since the fund's risk is much less than the market's risk, the risk-adjusted return of the fund is much higher than both the nominal fund and market returns.
## Input Arguments
collapse all
Asset returns, specified as a NUMSAMPLES x NUMSERIES matrix with NUMSAMPLES observations of asset returns for NUMSERIES asset return series.
Data Types: double
Returns for a benchmark asset, specified as a NUMSAMPLES vector of returns for a benchmark asset. The periodicity must be the same as the periodicity of Asset. For example, if Asset is monthly data, then Benchmark should be monthly returns.
Data Types: double
(Optional) Riskless asset, specified as a either a scalar return for a riskless asset or a vector of asset returns to be a proxy for a “riskless” asset. In either case, the periodicity must be the same as the periodicity of Asset. For example, if Asset is monthly data, then Cash must be monthly returns. If no value is supplied, the default value for Cash returns is 0.
Data Types: double
(Optional) Computed measures, specified as a character vector or cell array of character vectors to indicate one or more measures to be computed from among various risk-adjusted alphas and return measures. The number of choices selected in Choice is NUMCHOICES. The list of choices is given in the following table:
CodeDescription
'sml'Security Market Line — The security market line shows that the relationship between risk and return is linear for individual securities (that is, increased risk = increased return).
'capm'Jensen's Alpha — Risk-adjusted performance measure that represents the average return on a portfolio or investment, above or below that predicted by the capital asset pricing model (CAPM), given the portfolio's or investment's beta and the average market return.
'mm'Modigliani & Modigliani — Measures the returns of an investment portfolio for the amount of risk taken relative to some benchmark portfolio.
'gh1'Graham-Harvey 1 — Performance measure developed by John Graham and Campbell Harvey. The idea is to lever a fund's portfolio to exactly match the volatility of the S&P 500. The difference between the fund's levered return and the S&P 500 return is the performance measure.
'gh2'Graham-Harvey 2 — In this measure, the idea is to lever up or down the fund's recommended investment strategy (using a Treasury bill), so that the strategy has the same volatility as the S&P 500.
'all'Compute all measures.
Choice is specified by using the code from the table (for example, to select the Modigliani & Modigliani measure, Choice = 'mm'). A single choice is either a character vector or a scalar cell array with a single code from the table.
Multiple choices can be selected with a cell array of character vectors for choice codes (for example, to select both Graham-Harvey measures, Choice = {'gh1','gh2'}). To select all choices, specify Choice = 'all'. If no value is supplied, the default choice is to compute the excess return with Choice = 'xs'. Choice is not case-sensitive.
Data Types: char | cell
## Output Arguments
collapse all
Risk-adjusted alphas, returned as an NUMCHOICES-by-NUMSERIES matrix of risk-adjusted alphas for each series in Asset with each row corresponding to a specified measure in Choice.
Risk-adjusted returns, returned as an NUMCHOICES-by-NUMSERIES matrix of risk-adjusted returns for each series in Asset with each row corresponding to a specified measure in Choice.
Note
NaN values in the data are ignored and, if NaNs are present, some results could be unpredictable. Although the alphas are comparable across measures, risk-adjusted returns depend on whether the Asset or Benchmark is levered or unlevered to match its risk with the alternative. If Choice = 'all', the order of rows in Alpha and RAReturn follows the order in the table. In addition, Choice = 'all' overrides all other choices.
## References
[1] Graham, J. R. and Campbell R. Harvey. "Market Timing Ability and Volatility Implied in Investment Newsletters' Asset Allocation Recommendations." Journal of Financial Economics. Vol. 42, 1996, pp. 397–421.
[2] Lintner, J. "The Valuation of Risk Assets and the Selection of Risky Investments in Stocks Portfolios and Capital Budgets." Review of Economics and Statistics. Vol. 47, No. 1, February 1965, pp. 13–37.
[3] Modigliani, F. and Leah Modigliani. "Risk-Adjusted Performance: How to Measure It and Why." Journal of Portfolio Management. Vol. 23, No. 2, Winter 1997, pp. 45–54.
[4] Mossin, J. "Equilibrium in a Capital Asset Market." Econometrica. Vol. 34, No. 4, October 1966, pp. 768–783.
[5] Sharpe, W.F., "Capital Asset Prices: A Theory of Market Equilibrium under Conditions of Risk." Journal of Finance. Vol. 19, No. 3, September 1964, pp. 425–442.
## Version History
Introduced in R2006b | 1,538 | 6,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.734541 |
https://www.studypool.com/discuss/389988/a-milk-truck-route?free | 1,506,100,930,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689028.2/warc/CC-MAIN-20170922164513-20170922184513-00009.warc.gz | 868,998,757 | 14,958 | Time remaining:
##### A Milk Truck Route.
label Mathematics
account_circle Unassigned
schedule 0 Hours
account_balance_wallet \$5
Dale Klitzke is a milk truck driver for Swiss Valley Farms Cooperative in eastern Iowa. Dale has to start at the processing plant and pick up milk on 10 different farms. In how many ways can Dale visit each farm and return to the processing plant?
Feb 16th, 2015
Dale starts from a Processing plant and visits all the 10 different farms and return back to the Processing Plant. since he picks up milk from different farms, each farm can be visited by him only once.
1. Firstly Dale had the chance to visit any of the 10 farms, so number of choices of farm are 10
2.Since one of the farms is already visited, the next visit can only be to the remaining 9 farms, so number of chioces of farm are 9.
3.Since 2 farms are already visited, the next visit can only be to the remaining 8 farms, so number of choices of farm are 8
4. Since 3 farms are already visited, the next visit can only be to the remaining 7 farms, so number of choices of farm are 7
5.Since 4 farms are already visited, the next visit can only be to the remaining 6 farms, so number of choices of farm are 6
6. Since 5 farms are already visited, the next visit can only be to the remaining 5 farms, so number of choices of farm are 5
7. Since 6 farms are already visited, the next visit can only be to the remaining 4 farms, so number of choices of farm are 4
8. Since 7 farms are already visited, the next visit can only be to the remaining 3 farms, so number of choices of farm are 3
9. Since 8 farms are already visited, the next visit can only be to the remaining 2 farms, so number of choices of farm are 2
10. Since 9 farms are already visted, the next visit can only be to the remaining 1 farms.
Since Each Farm will be attened by Dale only once, no of ways from last visited Farm to Processing plant = 1
So the Total Number of ways Dale can visit each Farm and return to the processing plant are = 10 Factorial ways which will be Equal to 10* 9* 8* 7* 6* 5* 4* 3* 2* 1* 1 = 3628800 ways
Feb 16th, 2015
...
Feb 16th, 2015
...
Feb 16th, 2015
Sep 22nd, 2017
check_circle | 591 | 2,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-39 | latest | en | 0.938156 |
http://www.gabormelli.com/RKB/1986_AnIntroToMathStats | 1,701,846,782,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00574.warc.gz | 65,105,962 | 27,074 | # 1986 AnIntroToMathStats
## Notes
• Includes definitions for:
## Quotes
{{#ifanon:|
### 2.2 The Sample Space
By an experiment we will mean any procedure that (1) can be repeated, theoretically, an infinite number of times; and (2) has a well-defined set of possible outcomes. Thus, rolling a pair of dice qualifies as an experiment; so does measuring a hypertensive's blood pressure or doing a stereographic analysis to determine the carbon content of moon rocks. Each of the potential eventualities of an experiment is referred to as a sample outcome, $\displaystyle{ s }$, and their totality is called the sample space, S. To signify the member of $\displaystyle{ s }$ in $\displaystyle{ S }$, we write $\displaystyle{ s }$ In S. Any designated collection of sample outcomes, including individual outcomes, the entire sample space, and the null set, constitutes an event. The latter is said to occur if the outcome of the experiment is one of the members of that event.
### 2.3 The Probability Function
Consider a sample space, $\displaystyle{ S }$, and any event, $\displaystyle{ A }$, defined on $\displaystyle{ S }$. If our experiment were performed one time, either $\displaystyle{ A }$ or $\displaystyle{ A^C }$ would be the outcome. If it were performed $\displaystyle{ n }$ times, the resulting set of sample outcomes would be members of $\displaystyle{ A }$ on $\displaystyle{ m }$ occasions, $\displaystyle{ m }$ being some integer between $\displaystyle{ 1 }$ and $\displaystyle{ n }$, inclusive. Hypothetically, we could continue this process an infinite number of times. As $\displaystyle{ n }$ gets large, the ratio m/n will fluctuate less and less (we will make that statement more precise a little later). The number that m/n convert to is called the empirical probability of $\displaystyle{ A }$ : that is, $\displaystyle{ P(A) = lim_{n → ∞}(m/n) }$. … the very act of repeating an experiment under identical conditions an infinite number of times is physically impossible. And left unanswered is the question of how large $\displaystyle{ n }$ must be to give a good approximation for $\displaystyle{ lim_{n → ∞}(m/n) }$.
The next attempt at defining probability was entirely a product of the twentieth century. Modern mathematicians have shown a keen interest in developing subjects axiomatically. It was to be expected, then, that probability would come under such scrutiny … The major breakthrough on this front came in 1933 when Andrei Kolmogorov published Grundbegriffe der Wahscheinlichkeitsrechnung (Foundations of the Theory of Probability.). Kolmogorov's work was a masterpiece of mathematical elegance - it reduced the behavior of the probability function to a set of just three or four simple postulates, three if the same space is limited to a finite number of outcomes and four if $\displaystyle{ S }$ is infinite.
### 3.1 Introduction
Throughout most of Chapter 3, probability functions were defined in terms of the elementary outcomes making up an experiment's sample space. Thus, if two fair dice were tossed, a $\displaystyle{ P }$ value was assigned to each of the 36 possible pairs of upturned faces: … 1/36 … We have already seen, though, that in certain situations some attribute of an outcome may hold more interest for the experimenter than the outcome itself. A craps player, for example, may be concerned only that he throws a 7... In this chapter we investigate the consequences of redefining an experiment's sample space. … The original sample space contains 36 outcomes, all equally likely. The revised sample space contains 11 outcomes, but the latter are not equally likely.
In general, rules for redefining samples spaces - like going from (x, y's to (x + y)'s are called random variables. As a conceptual framework, random variables are of fundamental importance: they provide a single rubric under which all probability problems may be brought. Even in cases where the original sample space needs no redefinition - that is, where the measurement recorded is the measurement of interests - the concept still applies: we simply take the random variable to be the identify mapping.
### 3.2 Densities and Distributions
Definition 3.2.1. A real-valued function whose domain is the sample space S is called a random variable. We denote random variables by uppercase letters, often $\displaystyle{ X }$, $\displaystyle{ Y }$, or $\displaystyle{ Z }$.
If the range of the mapping contains either a finite or a countably infinite number of values, the random variable is said to be discrete ; if the range includes an interval of real numbers, bounded or unbounded, the random variable is said to be continuous.
Associated with each discrete random variable $\displaystyle{ Y }$ is a probability density function (or pdf) $\displaystyle{ f(y) }$. By definition, $\displaystyle{ f(y) }$ is the sum of all the probabilities associated with outcomes in $\displaystyle{ S }$ that get mapped into $\displaystyle{ y }$ by the random variable $\displaystyle{ Y }$. That is, $\displaystyle{ f(y) = P(\{s\in S \vert Y(s)=y\}) }$
Conceptually, $\displaystyle{ f_Y(y) }$ describes the probability structure induced on the real line by the random variable $\displaystyle{ Y }$.
For notational simplicity, we will delete all references to $\displaystyle{ s }$ and $\displaystyle{ S }$ and write:
$\displaystyle{ f_Y(y) = P(Y(s)=y). }$
In other words, $\displaystyle{ f\lt sub\gt Y\lt /sub\gt (y) }$ is the “probability that the random variable $\displaystyle{ Y }$ takes on the value $\displaystyle{ y }$.”
Associated with each continuous random variable $\displaystyle{ Y }$ is also a probability density function, fY(y), but fY(y) in this case is not the probability that the random variable $\displaystyle{ Y }$ takes on the value y. Rather, fY(y) is a continuous curve having the property that for all $\displaystyle{ a }$ and $\displaystyle{ b }$,
• P(a$\displaystyle{ Y }$b) = P({s(∈)S \vert $\displaystyle{ a }$Y(s) ≤ b}) = Integral(a,b). “fY(y) dy]
### 3.3 Joint Densities
Section 3.2 introduced the basic terminology for describing the probabilistic behavior of a single random variable …
Definition 3.3.1.
(a) Suppose that $\displaystyle{ X }$ and $\displaystyle{ Y }$ are two discrete random variables defined ont he same sample space S. The [[joint probability density function of X and Y (or joint pdf) is defined fX,Y(x,y), where.
• fX,Y(x,y) = P({sS \vert X(s) = $\displaystyle{ x }$, Y(s) = y}\})
• fX,Y(x,y) = P(X=x, Y=y)
(b) Suppose that $\displaystyle{ X }$ and $\displaystyle{ Y }$ are two continuous random variables defined over the sample space S. The joint pdf of $\displaystyle{ X }$ and $\displaystyle{ Y }$, fX,Y(x,y), is the surface having the property that for any region $\displaystyle{ R }$ in the xy-plane,
• P((X,Y)∈R) = P({sS \vert (X(s). “Y(s))∈R})
• P((X,Y)∈R) = IntegralR Integral $\displaystyle{ f }$X,Y(x,y) dx dy.
Definition 3.3.2.Let $\displaystyle{ X }$ and $\displaystyle{ Y }$ be two random variables defined on the same sample space S. The joint cumulative distribution function (or joint cdf) of $\displaystyle{ X }$ and $\displaystyle{ Y }$ is defined $\displaystyle{ F_{X,Y}(x,y) }$, where
• $\displaystyle{ F_{X,Y}(x,y) }$ = P({sS } X(s) <= $\displaystyle{ x }$ and Y(s) <= y})
• $\displaystyle{ F_{X,Y}(x,y) }$ = P(Xx, Yy).
}},
volumeDate ValuetitletypejournaltitleUrldoinoteyear | 1,874 | 7,392 | {"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} | 4.46875 | 4 | CC-MAIN-2023-50 | latest | en | 0.886384 |
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# RECURSION / TAIL RECURSION IN SCALA
## Recursion
Recursion is a function which calls itself over and over again until an exit condition is met. It breaks down a large problem into a smaller problem, solves the smaller parts, and combines them to produce a solution. Recursion could be applied to many common problems, which can be solved using loops, for and while.
## Why Recursion?
1. The code is simpler and shorter than an iterative code as we only need to define the base case and recursive case.
2. It avoids the mutable variables that you need to use while writing loops.
## Reverse of a List using Recursion
``````def reverse[A](inputList: List[A]): List[A] = inputList match {
case Nil => Nil
}``````
In the above function, when the list is empty we are returning Nil. When the list is non-empty we are concatenating the List with the head element by recursively calling the reverse method with the tail elements. The chain of calls is pushed onto the stack, each one waiting for the next one to return before it concatenates the List and returns.
When we pass List(1,2,3,4,5) into the reverse function, the calls will be :-
reverse(List(2,3,4,5)) ::: List(1)
reverse(List(3,4,5)) ::: List(2) ::: List(1)
reverse(List(4,5)) ::: List(3) ::: List(2) ::: List(1)
reverse(List(5)) ::: List(4) ::: List(3) ::: List(2) ::: List(1)
## Problem with Recursion
1. For every recursive call, separate memory is allocated for the variables.
2. Recursive functions often throw a Stack Overflow Exception when processing or operations are too large.
Example : When we pass a List with 5000 numbers in the above recursive function, the following exception occurs.
``````Exception in thread "main" java.lang.StackOverflowError
at com.Reverse.Reverse.reverse(ReverseList.scala:7)
at com.Reverse.Reverse.reverse(ReverseList.scala:7)
at com.Reverse.Reverse.reverse(ReverseList.scala:7)
at com.Reverse.Reverse.reverse(ReverseList.scala:7)``````
## Why Stack Overflow Error occurs?
In a recursive function, every method call will create its stack frame in the stack memory. The first call of the function doesn’t return until the last call in the recursion is resolved. So, the function return order is from last invoked to first invoked. This means that if the function F calls itself recursively for N times, then it would create N stack frames in the stack memory for a single execution of the function F. The compiler will keep all the N stack frames in memory till the recursion is complete.
To solve the stack overflow error we can use tail-recursion.
## Tail Recursion
A recursive function is said to be tail-recursive if the recursive call is the last operation performed by the function. There is no need to keep a record of the previous state. In Scala, you can use @tailrec to check if the recursion is tail-recursive or not. The annotation is available in the scala.annotation._ package. If the recursion isn’t tail-recursive, then Scala will throw a compile-time error.
When the Scala compiler recognizes that it is tail-recursion, it will optimize the function by converting it to a loop. We will not realize the change, but the compiler will do it internally. This optimization will overcome the performance and memory problem.
Let’s look at a tail-recursive version of reverse of a List.
``````def reverseTail[A](inputList: List[A]): List[A] = {
@tailrec
def reverse(reversedList: List[A], remainingList: List[A]): List[A] = remainingList match {
case Nil => reversedList | 839 | 3,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.792774 |
https://www.vitutor.com/statistics/probability/a_15.html | 1,553,476,295,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203547.62/warc/CC-MAIN-20190325010547-20190325032547-00241.warc.gz | 921,235,426 | 4,335 | # 15
Students A and B have probabilities of failing an exam of 1/2 and 1/5 respectively. The probability of them both failing the examination is 1/10. Determine the probability that at least one of the two students fail. | 54 | 221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-13 | latest | en | 0.960444 |
https://wikimass.com/c-sharp/matrix-addition | 1,726,146,120,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00865.warc.gz | 568,122,620 | 4,950 | # C# Program to perform Matrix Addition
You are Here:
In the following example, we will add the two given matrices (two-dimensional arrays).
### Example
C# Compiler
using System; namespace myApp { class Program { static void Main(string[] args) { int i, j; int[,] arr1 = { {1, 1, 1}, {1, 1, 1}, {1, 1, 1} }; int[,] arr2 = { {2, 2, 2}, {2, 2, 2}, {2, 2, 2} }; int[,] arr3 = new int[3, 3]; Console.WriteLine("Matrix A (3 x 3):"); for(i=0; i<3; i++) { for(j=0; j<3; j++) Console.Write("{0} ", arr1[i, j]); Console.WriteLine(); } Console.WriteLine("\nMatrix B (3 x 3):"); for(i=0; i<3; i++) { for(j=0; j<3; j++) Console.Write("{0} ", arr2[i, j]); Console.WriteLine(); } Console.WriteLine("\nMatrix Additon (A + B):"); for(i=0; i<3; i++) { for(j=0; j<3; j++) { arr3[i, j] = arr1[i, j] + arr2[i, j]; Console.Write("{0} ", arr3[i, j]); } Console.WriteLine(); } } } }
### Output
Matrix A (3 x 3): 1 1 1 1 1 1 1 1 1 Matrix B (3 x 3): 2 2 2 2 2 2 2 2 2 Matrix Additon (A + B): 3 3 3 3 3 3 3 3 3
## Reminder
Hi Developers, we almost covered 90% of String functions and Interview Question on C# with examples for quick and easy learning.
We are working to cover every Single Concept in C#. | 444 | 1,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.297975 |
http://mathhelpforum.com/advanced-algebra/77239-matrix.html | 1,526,925,476,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00134.warc.gz | 176,368,077 | 10,204 | 1. ## matrix
2
x + 4z = 2
x + 2y 3z = 4
x + 2y + z = k
Row-reduce the augmented matrix for the given system and then answer each of the following questions.
(a) For what value(s) of
k does the system of linear equations have a unique solution?
(b) For what value(s) of
k does the system of linear equations have no solution?
(c) For what value(s) of
k does the system of linear equations have infinitely many solutions?
(d) For
k where the system has infinitely many solutions, solve the system.
2. sorry it shows up weird on the screen it's:
2
x + 4z = 2
x + 2y 3z = 4
x + 2y + z = k
3. 2x + 4z= -2
-x + 2y -3z = 4
x + 2y + z = k
4. Originally Posted by davidbooth
$\displaystyle \left\{\begin{array}{rcrcrcr} 2x&&&+&4z&=&-2\\ -x&+&2y&-&3z&=&4\\ x&+&2y&+&z&=&k \end{array}\right.$
What have you done so far? You should at least be able to row reduce.
5. not sure how to go about that | 319 | 895 | {"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} | 3.984375 | 4 | CC-MAIN-2018-22 | latest | en | 0.886298 |
https://nrich.maths.org/public/leg.php?code=66&cl=4&cldcmpid=6502 | 1,568,903,666,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573533.49/warc/CC-MAIN-20190919142838-20190919164838-00307.warc.gz | 602,967,171 | 5,508 | # Search by Topic
#### Resources tagged with Transformation of functions similar to Operating Machines:
Filter by: Content type:
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### There are 12 results
Broad Topics > Functions and Graphs > Transformation of functions
### Operating Machines
##### Age 16 to 18 Challenge Level:
What functions can you make using the function machines RECIPROCAL and PRODUCT and the operator machines DIFF and INT?
### Loch Ness
##### Age 16 to 18 Challenge Level:
Draw graphs of the sine and modulus functions and explain the humps.
### The Why and How of Substitution
##### Age 16 to 18
Step back and reflect! This article reviews techniques such as substitution and change of coordinates which enable us to exploit underlying structures to crack problems.
### Agile Algebra
##### Age 16 to 18 Challenge Level:
Observe symmetries and engage the power of substitution to solve complicated equations.
### Cubic Spin
##### Age 16 to 18 Challenge Level:
Prove that the graph of f(x) = x^3 - 6x^2 +9x +1 has rotational symmetry. Do graphs of all cubics have rotational symmetry?
### Parabolic Patterns
##### Age 14 to 18 Challenge Level:
The illustration shows the graphs of fifteen functions. Two of them have equations y=x^2 and y=-(x-4)^2. Find the equations of all the other graphs.
### More Parabolic Patterns
##### Age 14 to 18 Challenge Level:
The illustration shows the graphs of twelve functions. Three of them have equations y=x^2, x=y^2 and x=-y^2+2. Find the equations of all the other graphs.
### Ellipses
##### Age 14 to 18 Challenge Level:
Here is a pattern for you to experiment with using graph drawing software. Find the equations of the graphs in the pattern.
### Parabolas Again
##### Age 14 to 18 Challenge Level:
Here is a pattern composed of the graphs of 14 parabolas. Can you find their equations?
### Exploring Cubic Functions
##### Age 14 to 18 Challenge Level:
Quadratic graphs are very familiar, but what patterns can you explore with cubics?
### Painting by Functions
##### Age 16 to 18 Challenge Level:
Use functions to create minimalist versions of works of art.
### Sine Problem
##### Age 16 to 18 Challenge Level:
In this 'mesh' of sine graphs, one of the graphs is the graph of the sine function. Find the equations of the other graphs to reproduce the pattern. | 534 | 2,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-39 | latest | en | 0.816853 |
https://number.academy/252063 | 1,670,576,979,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00022.warc.gz | 460,743,647 | 12,417 | # Number 252063
Number 252,063 spell 🔊, write in words: two hundred and fifty-two thousand and sixty-three . Ordinal number 252063th is said 🔊 and write: two hundred and fifty-two thousand and sixty-third. Color #252063. The meaning of number 252063 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 252063. What is 252063 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 252063.
## What is 252,063 in other units
The decimal (Arabic) number 252063 converted to a Roman number is (C)(C)(L)MMLXIII. Roman and decimal number conversions.
#### Weight conversion
252063 kilograms (kg) = 555698.1 pounds (lbs)
252063 pounds (lbs) = 114335.0 kilograms (kg)
#### Length conversion
252063 kilometers (km) equals to 156625 miles (mi).
252063 miles (mi) equals to 405657 kilometers (km).
252063 meters (m) equals to 826969 feet (ft).
252063 feet (ft) equals 76830 meters (m).
252063 centimeters (cm) equals to 99237.4 inches (in).
252063 inches (in) equals to 640240.0 centimeters (cm).
#### Temperature conversion
252063° Fahrenheit (°F) equals to 140017.2° Celsius (°C)
252063° Celsius (°C) equals to 453745.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
252063 seconds equals to 2 days, 22 hours, 1 minute, 3 seconds
252063 minutes equals to 6 months, 1 week, 1 hour, 3 minutes
### Codes and images of the number 252063
Number 252063 morse code: ..--- ..... ..--- ----- -.... ...--
Sign language for number 252063:
Number 252063 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 252063
### Multiplications
#### Multiplication table of 252063
252063 multiplied by two equals 504126 (252063 x 2 = 504126).
252063 multiplied by three equals 756189 (252063 x 3 = 756189).
252063 multiplied by four equals 1008252 (252063 x 4 = 1008252).
252063 multiplied by five equals 1260315 (252063 x 5 = 1260315).
252063 multiplied by six equals 1512378 (252063 x 6 = 1512378).
252063 multiplied by seven equals 1764441 (252063 x 7 = 1764441).
252063 multiplied by eight equals 2016504 (252063 x 8 = 2016504).
252063 multiplied by nine equals 2268567 (252063 x 9 = 2268567).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 252063
Half of 252063 is 126031,5 (252063 / 2 = 126031,5 = 126031 1/2).
One third of 252063 is 84021 (252063 / 3 = 84021).
One quarter of 252063 is 63015,75 (252063 / 4 = 63015,75 = 63015 3/4).
One fifth of 252063 is 50412,6 (252063 / 5 = 50412,6 = 50412 3/5).
One sixth of 252063 is 42010,5 (252063 / 6 = 42010,5 = 42010 1/2).
One seventh of 252063 is 36009 (252063 / 7 = 36009).
One eighth of 252063 is 31507,875 (252063 / 8 = 31507,875 = 31507 7/8).
One ninth of 252063 is 28007 (252063 / 9 = 28007).
show fractions by 6, 7, 8, 9 ...
### Calculator
252063
#### Is Prime?
The number 252063 is not a prime number. The closest prime numbers are 252037, 252079.
#### Factorization and factors (dividers)
The prime factors of 252063 are 3 * 3 * 7 * 4001
The factors of 252063 are 1 , 3 , 7 , 9 , 21 , 63 , 4001 , 12003 , 28007 , 36009 , 84021 , 252063
Total factors 12.
Sum of factors 416208 (164145).
#### Powers
The second power of 2520632 is 63.535.755.969.
The third power of 2520633 is 16.015.013.256.814.046.
#### Roots
The square root √252063 is 502,058762.
The cube root of 3252063 is 63,168859.
#### Logarithms
The natural logarithm of No. ln 252063 = loge 252063 = 12,437434.
The logarithm to base 10 of No. log10 252063 = 5,401509.
The Napierian logarithm of No. log1/e 252063 = -12,437434.
### Trigonometric functions
The cosine of 252063 is 0,898247.
The sine of 252063 is 0,439491.
The tangent of 252063 is 0,489276.
### Properties of the number 252063
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 252063 in Computer Science
Code typeCode value
PIN 252063 It's recommendable to use 252063 as a password or PIN.
252063 Number of bytes246.2KB
CSS Color
#252063 hexadecimal to red, green and blue (RGB) (37, 32, 99)
Unix timeUnix time 252063 is equal to Saturday Jan. 3, 1970, 10:01:03 p.m. GMT
IPv4, IPv6Number 252063 internet address in dotted format v4 0.3.216.159, v6 ::3:d89f
252063 Decimal = 111101100010011111 Binary
252063 Decimal = 110210202200 Ternary
252063 Decimal = 754237 Octal
252063 Decimal = 3D89F Hexadecimal (0x3d89f hex)
252063 BASE64MjUyMDYz
252063 MD56c6273de4a81259354a7955551b47dba
252063 SHA18b88c47af6c607341c9f3ed83f2c2779dbc77358
252063 SHA22496c8c07aca82ea6a4fe56064e757c581721419003e5623f451367f40
252063 SHA256d6e57d5cfcbe20ecb311abf0b53a825bb7dfe1ac1a8515b6265e88b6318a1dd4
More SHA codes related to the number 252063 ...
If you know something interesting about the 252063 number that you did not find on this page, do not hesitate to write us here.
## Numerology 252063
### Character frequency in number 252063
Character (importance) frequency for numerology.
Character: Frequency: 2 2 5 1 0 1 6 1 3 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 252063, the numbers 2+5+2+0+6+3 = 1+8 = 9 are added and the meaning of the number 9 is sought.
## Interesting facts about the number 252063
### Asteroids
• (252063) 2000 SC96 is asteroid number 252063. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, ETS in Socorro on 9/23/2000.
## № 252,063 in other languages
How to say or write the number two hundred and fifty-two thousand and sixty-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 252.063) doscientos cincuenta y dos mil sesenta y tres German: 🔊 (Nummer 252.063) zweihundertzweiundfünfzigtausenddreiundsechzig French: 🔊 (nombre 252 063) deux cent cinquante-deux mille soixante-trois Portuguese: 🔊 (número 252 063) duzentos e cinquenta e dois mil e sessenta e três Hindi: 🔊 (संख्या 252 063) दो लाख, बावन हज़ार, तिरेसठ Chinese: 🔊 (数 252 063) 二十五万二千零六十三 Arabian: 🔊 (عدد 252,063) مئتان و اثنان و خمسون ألفاً و ثلاثة و ستون Czech: 🔊 (číslo 252 063) dvěstě padesát dva tisíce šedesát tři Korean: 🔊 (번호 252,063) 이십오만 이천육십삼 Danish: 🔊 (nummer 252 063) tohundrede og tooghalvtredstusindtreogtreds Dutch: 🔊 (nummer 252 063) tweehonderdtweeënvijftigduizenddrieënzestig Japanese: 🔊 (数 252,063) 二十五万二千六十三 Indonesian: 🔊 (jumlah 252.063) dua ratus lima puluh dua ribu enam puluh tiga Italian: 🔊 (numero 252 063) duecentocinquantaduemilasessantatre Norwegian: 🔊 (nummer 252 063) to hundre og femti-to tusen og seksti-tre Polish: 🔊 (liczba 252 063) dwieście pięćdziesiąt dwa tysiące sześćdziesiąt trzy Russian: 🔊 (номер 252 063) двести пятьдесят две тысячи шестьдесят три Turkish: 🔊 (numara 252,063) ikiyüzelliikibinaltmışüç Thai: 🔊 (จำนวน 252 063) สองแสนห้าหมื่นสองพันหกสิบสาม Ukrainian: 🔊 (номер 252 063) двiстi п'ятдесят двi тисячi шiстдесят три Vietnamese: 🔊 (con số 252.063) hai trăm năm mươi hai nghìn lẻ sáu mươi ba Other languages ...
## Comment
If you know something interesting about the number 252063 or any natural number (positive integer) please write us here or on facebook. | 2,543 | 7,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.639184 |
https://www.geeksforgeeks.org/modify-numpy-array-to-store-an-arbitrary-length-string/ | 1,600,511,892,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400191160.14/warc/CC-MAIN-20200919075646-20200919105646-00718.warc.gz | 901,826,241 | 20,882 | # Modify Numpy array to store an arbitrary length string
NumPy builds on (and is a successor to) the successful Numeric array object. Its goal is to create the corner-stone for a useful environment for scientific computing. NumPy provides two fundamental objects: an N-dimensional array object (ndarray) and a universal function object (ufunc).
The dtype of any numpy array containing string values is the maximum length of any string present in the array. Once set, it will only be able to store new string having length not more than the maximum length at the time of the creation. If we try to reassign some another string value having length greater than the maximum length of the existing elements, it simply discards all the values beyond the maximum length.
In this post we are going to discuss ways in which we can overcome this problem and create a numpy array of arbitrary length.
Let’s first visualize the problem with creating an arbitrary length numpy array of string type.
`# importing numpy as np ` `import` `numpy as np ` ` ` `# Create the numpy array ` `country ``=` `np.array([``'USA'``, ``'Japan'``, ``'UK'``, '``', '``India``', '``China']) ` ` ` `# Print the array ` `print``(country) `
Output :
As we can see in the output, the maximum length of any string length element in the given array is 5. Let’s try to assign a value having greater length at the place of missing value in the array.
`# Assign 'New Zealand' at the place of missing value ` `country[country ``=``=` `'``'] = '``New Zealand' ` ` ` `# Print the modified array ` `print``(country) `
Output :
As we can see in the output, ‘New Z’ has been assigned rather than ‘New Zealand’ because of the limitation to the length. Now, let’s see the ways in which we can overcome this problem.
Problem #1 : Create a numpy array of arbitrary length.
Solution : While creating the array assign the ‘object’ dtype to it. This lets you have all the behaviors of the python string.
`# importing the numpy library as np ` `import` `numpy as np ` ` ` `# Create a numpy array ` `# set the dtype to object ` `country ``=` `np.array([``'USA'``, ``'Japan'``, ``'UK'``, '``', '``India``', '``China``'], dtype = '``object``') ` ` ` `# Print the array ` `print``(country) `
Output :
Now we will use assign a value of arbitrary length at the place of missing value in the given array.
`# Assign 'New Zealand' to the missing value ` `country[country ``=``=` `'``'] = '``New Zealand' ` ` ` `# Print the array ` `print``(country) `
Output :
As we can see in the output, we have successfully assigned an arbitrary length string to the given array object.
Problem #2 : Create a numpy array of arbitrary length.
Solution : We will use the `numpy.astype()` function to change the dtype of the given array object.
`# importing the numpy library as np ` `import` `numpy as np ` ` ` `# Create a numpy array ` `# Notice we have not set the dtype of the object ` `# this will lead to the length problem ` `country ``=` `np.array([``'USA'``, ``'Japan'``, ``'UK'``, '``', '``India``', '``China']) ` ` ` `# Print the array ` `print``(country) `
Output :
Now we will change the dtype of the given array object using `numpy.astype()` function. Then we will assign an arbitrary length string to it.
`# Change the dtype of the country ` `# object to 'U256' ` `country ``=` `country.astype(``'U256'``) ` ` ` `# Assign 'New Zealand' to the missing value ` `country[country ``=``=` `'``'] = '``New Zealand' ` ` ` `# Print the array ` `print``(country) `
Output :
As we can see in the output, we have successfully assigned an arbitrary length string to the given array object.
Note : The maximum length of the string that we can assign in this case after changing the dtype is 256.
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 1,080 | 4,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-40 | latest | en | 0.800404 |
https://www.physicsforums.com/threads/need-a-little-help-with-a-derivative-problem-please.344329/ | 1,527,438,483,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00333.warc.gz | 810,588,422 | 15,344 | # Homework Help: Need a little help with a derivative problem please
1. Oct 9, 2009
If f(x) = 7sin(3 x) * (arcsin(x)), find f'( x ).
I'm assuming I need to use the product rule, but I am failing to set it up right. I assume it's with the beginning part and I'm just goofing up on what to do with the coefficient.
I think it's ((7sin3x)' * arcsinx) + (7sin(3x) + 1/(sqrt(1-x^2)).
If that's correct I'm just failing to get the derivative. I can't seem to remember how I should proceed with something in that form. I should be past this but I'm having trouble seeing it. Any help would be great.
Thank you
2. Oct 9, 2009
### LCKurtz
Your last + should be multiplication.
3. Oct 9, 2009
Ah yes it is I just typed it up wrong in my initial post. I cant seem to figure out the derivative though. The derivative of 7sin3x is not working out for me. I tried things like 7cos3x(3) but that doesnt seem to work. Any help would be great thanks
4. Oct 9, 2009
### Dick
I don't see anything wrong with 21*cos(3x) as the derivative of 7*sin(3x).
5. Oct 9, 2009
so would the final answer be
21cos3x(arcsinx)+(7sin3x)/(sqrt(1-x^2))
because that does not seem to be the right answer on my web assignment i dunno
6. Oct 9, 2009
### Dick
I can't really speak for web assignment, but it looks ok to me.
7. Oct 9, 2009 | 408 | 1,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-22 | latest | en | 0.96406 |
https://brainly.com/question/176491 | 1,484,975,351,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280900.71/warc/CC-MAIN-20170116095120-00252-ip-10-171-10-70.ec2.internal.warc.gz | 813,668,184 | 8,678 | # The sum of two numbers is 92. their difference is 20. find the 2 numbers
2
by kaitlynffc
( Below ) Yes you Are Correct 36+56 = 92 So yea Your Correct
2014-11-05T23:36:43-05:00
The two numbers would be 36 and 56
2014-11-05T23:38:44-05:00
X+y =92
x -y =20
------------ (subtract the y)
2x= 112
/2 /2
x=56
56+y=92 (substitute x with 56)
-56 -56
y=36
x = 56 and y = 36
Well that's the best way to show work for this question...well done :) | 182 | 456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.75423 |
https://se.mathworks.com/matlabcentral/profile/authors/9790539?detail=cody | 1,632,795,735,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058589.72/warc/CC-MAIN-20210928002254-20210928032254-00002.warc.gz | 527,512,231 | 19,684 | Community Profile
# Jim Riggs
### Missile Defense Agency - Advanced Technology Group
Last seen: 8 månader ago Active since 2017
Developing simulations for the Missile Defense Agency - Advanced Technology group
30+ years experience in missiles, smart munitions and other aerodynamic systems.
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Solved | 1,530 | 5,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-39 | latest | en | 0.663166 |
https://www.physicsforums.com/threads/magnitude-of-the-vector.153303/ | 1,508,681,396,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825264.94/warc/CC-MAIN-20171022132026-20171022152026-00186.warc.gz | 970,713,140 | 16,640 | # Magnitude of the vector
1. Jan 27, 2007
### rasikan
I need help with the following questions-
1. if a= 12i+16j and b=-24+10j, what is the magnitude of the vector C=2a-b?
2.If the magnitude and direction of c are 2.5 cm and 80degree , and d[3.5cm,120degree] and e=d-2c then what is the direction of e, (to the nearest degree)?
2. Jan 27, 2007
### arildno
1. What information can you use in order to find the magnitude of a vector?
3. Jan 27, 2007
### rasikan
you can use any formulas..
4. Jan 27, 2007
### arildno
So which one would YOU pick?
5. Jan 27, 2007
### rasikan
I would use square root of (Ax)^2+ (Ay)^2
6. Jan 27, 2007
### arildno
Indeed, smart choice!
So, what info are you at present lacking in order to calculate the magnitude?
And:
How can you OBTAIN that necessary information?
7. Jan 27, 2007
### rasikan
no i got my answer as 64, just want to verify that, and I dnt have no clue about the second question
8. Jan 27, 2007
### arildno
Well, I don't bother to calculate myself all the way to 64 or to any other dumb number, I want to find out:
HOW did you are at whatever number you got?
9. Jan 27, 2007
### rasikan
can u help me with my second question
10. Jan 27, 2007
### arildno
The degrees are measured with respect to the positive x-axis.
So:
How can that help you?
11. Jan 27, 2007
### rasikan
well i really dont know
12. Jan 27, 2007
### arildno
Well, how har the horizontal&vertical components of a vector related to the vector's magnitude and the angle the vector makes with the positive x-axis?
13. Jan 27, 2007
### rasikan
i think it get calcuate arc tan of Ax/Ay
14. Jan 27, 2007
### arildno
No!
Go back to your book, and read your definitions again.
15. Jan 27, 2007
### robphy
Check it for some special cases...
What angle do you expect? and What do you get with your formula?
- for a vector along the x-axis.
- for a vector along the y-axis.
16. Feb 1, 2007
### Staff: Mentor
Moved from Advanced Physics to Intro Physics.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 633 | 2,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-43 | longest | en | 0.873631 |
https://math.stackexchange.com/questions/424633/prove-if-in-all-subgraphs-of-g-there-is-a-vertex-of-degree-2-then-g-is-a/424647 | 1,713,929,288,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00456.warc.gz | 338,192,327 | 35,034 | # Prove: If in all subgraphs of $G$ there is a vertex of degree $<2$ then $G$ is a forest
I need help proving this:
Given a graph $G$, prove that if in all subgraphs of $G$ there is a vertex of degree less than $2$ ($1$ or $0$) then $G$ is a forest.
• Where do you get stuck? Plug in the definition of forest, tree, subgraph. Jun 19, 2013 at 16:38
• Assume there is a cycle in $G$... Jun 19, 2013 at 16:39
• Is it sufficient to prove that in all connected components of $G$ there is no cycle and that means it is a forest? Jun 19, 2013 at 16:45
• What can those connected components look like? Consider subgraphs with three vertices. Jun 19, 2013 at 16:48
• Thank you so much for the hint, Damian Sobota. I proved it. Jun 19, 2013 at 16:53 | 236 | 742 | {"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} | 3.078125 | 3 | CC-MAIN-2024-18 | latest | en | 0.936419 |
https://leanprover-community.github.io/archive/stream/113488-general/topic/equality.20from.20scratch.html | 1,716,369,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00296.warc.gz | 308,956,936 | 5,983 | ## Stream: general
### Topic: equality from scratch
#### Adam Kurkiewicz (Aug 07 2018 at 10:29):
While looking at Kevin's great blog post about unsigned integers from scratch I thought to push myself a little bit and I've tried to do them even more from scratch, by giving up the existing eq or =.
So let's say we have an inductive xnat and an inductive xnat equality like this:
inductive xnat : Type
| zero : xnat
| succ : xnat → xnat
inductive equality (a : xnat): xnat → Prop
| refl : equality a
We're getting reflexivity for free from the type system, but we want to prove transitivity and symmetry. Let's assume we don't know anything about default parameters, universes, etc, but we know about recursors. We might write something like this:
definition equality.symm: Π (a b : xnat), Π eq1 : (equality a b), equality b a :=
λ a b : xnat,
λ eq1 : (equality a b),
equality.rec_on eq1 _
And now, the strangest thing happens, the placeholder no longer expects a proof of equality b a, suddenly equality a a suffices. This relieves us, and we proceed, almost automatically, to write (equality.refl a) instead of the placeholder, and this typechecks. Phew.
But why does it typecheck? Why suddently a proof of equality a a is good enough as a proof of equality b a. Is there something special in the type-system that makes it work?
#### Mario Carneiro (Aug 07 2018 at 10:34):
Look at the type of equality.rec_on.
protected def equality.rec_on : Π {a : xnat} {C : xnat → Sort l} {a_1 : xnat}, equality a a_1 → C a → C a_1
It says that if you want to prove a property C of a_1 : xnat given equality a a_1, it suffices to prove C a. In this case we know equality a b, so we need a property C depending on b such that C a is easy. In this case we take \lam b, equality b a as our C, so C a is equality a a which we prove by refl, and C b is equality b a which is what we wanted to show.
#### Mario Carneiro (Aug 07 2018 at 10:40):
You should be sure to look at the definition of equality.symm with pp.all true, so you can see how lean filled in the "motive" C in that rec_on application
#### Mario Carneiro (Aug 07 2018 at 10:40):
or use @equality.rec_on and fill in all the fields yourself
#### Adam Kurkiewicz (Aug 07 2018 at 10:46):
Thanks Mario, this is making it less magical. I'm on your most recent suggestion.
#### Kevin Buzzard (Aug 07 2018 at 10:50):
Kenny Lau once said to me "Lean does not do magic", and at that time I thought that lots of things Lean did (simp, type class inference) were magic. Kenny's comment spurred me on to trying to figure out how everything was working; the point is that Lean never does magic, and in any given case you can simply look at what it did and how it did it. Figuring out how to do that really helped me to learn Lean better.
#### Adam Kurkiewicz (Aug 07 2018 at 10:50):
Ah, of course, it makes sense. equality a b is the same as (equality a) b. So our C becomes equality a. Thanks Mario!
#### Mario Carneiro (Aug 07 2018 at 10:51):
ah, be careful: equality a would be a perfect motive to prove equality a b -> equality a b, but this is symmetry, so there is a twist
#### Adam Kurkiewicz (Aug 07 2018 at 10:51):
So the solution is simply @equality.rec_on a (equality a) b eq1 (equality.refl a), and that makes sense.
#### Adam Kurkiewicz (Aug 07 2018 at 10:52):
Is C the motif?
#### Mario Carneiro (Aug 07 2018 at 10:52):
yes, that's the usual terminology
#### Mario Carneiro (Aug 07 2018 at 10:53):
sometimes you get an error message talking about a motive, that's what it is referring to
#### Mario Carneiro (Aug 07 2018 at 10:56):
If you use the "flipped" motive λ b, equality b a, you have:
#check λ a b eq1, @equality.rec_on a (λ b, equality b a) b eq1 (equality.refl a)
-- : ∀ (a b : xnat), equality a b → (λ (b : xnat), equality b a) b
and notice that the conclusion there, (λ (b : xnat), equality b a) b, beta reduces to equality b a which is the desired symmetrized equality
#### Adam Kurkiewicz (Aug 07 2018 at 11:12):
Yes you're right, I didn't notice this wasn't typechecking. This lambda abstraction is a really nice trick, I don't think I would have come up with this myself.
#### Adam Kurkiewicz (Aug 07 2018 at 11:14):
Anyway, thank you Mario, I think this is now really clear.
#### Adam Kurkiewicz (Aug 07 2018 at 11:14):
I'll try to work through transitivity in a similar manner
#### Mario Carneiro (Aug 07 2018 at 11:14):
the really interesting thing is that lean will automatically do that lambda abstraction trick
#### Adam Kurkiewicz (Aug 07 2018 at 11:17):
Now, @Kevin Buzzard if this is not magic I don't know what is.
#### Kevin Buzzard (Aug 07 2018 at 11:17):
I'm not sure it's magic
#### Kevin Buzzard (Aug 07 2018 at 11:18):
Is it just matching types up?
#### Kevin Buzzard (Aug 07 2018 at 11:18):
I'm not quite following, I'm trying to get on a bus in Majorca
#### Mario Carneiro (Aug 07 2018 at 11:18):
the algorithm is very simple: the goal says equality b a, and we just replace every b with a, then we look at what we changed and replace that with a variable, let's call it x. So the motive is λ x, equality x a
#### Mario Carneiro (Aug 07 2018 at 11:20):
this produces a lambda term such that replacing x with b gives us our original goal, and replacing x with a gives us our new goal which should be easier, in this case equality a a
#### Mario Carneiro (Aug 07 2018 at 11:22):
you should try using this algorithm in the proof of transitivity to work out the right motive, then see whether you got it right by letting lean do it for you
#### Adam Kurkiewicz (Aug 07 2018 at 11:35):
I've actually just done it in my head. It worked:
inductive xnat : Type
| zero : xnat
| succ : xnat → xnat
inductive equality (a : xnat): xnat → Prop
| refl : equality a
definition equality.trans: Π (a b c : xnat), Π eq1: (equality a b), Π eq2 : (equality b c), equality a c :=
λ a b c : xnat,
λ eq1 : (equality a b),
λ eq2 : (equality b c),
@equality.rec_on b (λ x, equality a x) c eq2 eq1
I'm sure I'll learn the algorithm one day, but I think I'll go and buy some beef now. Cooking sous vide steaks for friends this evening.
#### Adam Kurkiewicz (Aug 07 2018 at 11:39):
Thank you Mario, this was really helpful!
#### Kevin Buzzard (Aug 07 2018 at 12:29):
Warning: sometimes Lean can't generate the right motive. CS people start going on about higher order unification being undecidable when this sort of thing comes up. The problem is that if Lean can figure out that C b is supposed to be f a b c b = 0 then it can't work out if C x is supposed to be f a x c x = 0 or f a b c x = 0 or ... etc. So don't expect Lean to do miracles. See https://leanprover.github.io/theorem_proving_in_lean/interacting_with_lean.html#elaboration-hints
#### Kevin Buzzard (Aug 07 2018 at 12:30):
Remember -- Lean does not do magic. Part of the art is working out when you're asking Lean to do magic :-)
Last updated: Dec 20 2023 at 11:08 UTC | 1,991 | 6,964 | {"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} | 3.296875 | 3 | CC-MAIN-2024-22 | latest | en | 0.878345 |
http://www.jiskha.com/display.cgi?id=1229637689 | 1,498,253,050,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320174.58/warc/CC-MAIN-20170623202724-20170623222724-00669.warc.gz | 554,968,958 | 4,172 | # Algebra
posted by .
Here's the question I'm having trouble with:
Tell how you would identify the intercepts for the graph of the equation:
3x + 6y = 18
I'm not exactly sure how you would find the intercepts from this equation.
Thanks!
• Algebra -
would you know how to find them if you put it into y=mx+b form?
• Algebra -
Yes! I see, I didn't really think of it that way until after I posted the message. Thank you!
• Algebra -
lol wow i actually helped ?!
i was just guessing .. im only in grade 11 anyways
• Algebra -
If they are written in the form you have, it is often very easy to find
remember to find the x-intercept you let y=0 and to
find the y-intercept you let the x=0
of course letting some variable equal to zero makes that term disappear.
so to let x=0, block out the x term with your little finger, what do you see?
6y = 18, then y = 3
for the x-intercept, block out the y-term with your little finger and ...
x = 6
this works really neat if the x and y coefficients both divide evenly into the constant | 264 | 1,041 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-26 | latest | en | 0.96773 |
https://forma-slova.com/en/articles/7248-what-is-the-law-of-sines | 1,675,664,035,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00104.warc.gz | 277,180,088 | 5,283 | # What is the law of sines?
protection click fraud
When it is necessary to relate a side to a angle on one right triangle in order to find the measurements of one of its sides or one of its angles, we can use the trigonometric relations: sine, cosine and tangent. It is also possible to calculate the measure of one of the sides or one of the angles of a triangleany, that is, not necessarily of a right triangle. For this, one of the methods used is the sins law.
sins law
Take the triangle ABC as an example, registered in a circumference of radius r.
In a case like this, the sides and angles have any measures. So we have:
The = B = ç = 2r
sinα sinβ sinθ
In this triangle, a, b, and c are the measurements of its sides; α, β and θ are their internal angles, and the sines of these angles have the same values as the sines found in the tablestrigonometric.
at first fraction, a is the measure on the opposite side of sinα; in the second fraction, b is the measure opposite sinβ, and in the third fraction, note that c is the measure opposite sinθ. So there is a
instagram story viewer
proportion between the ratios formed by the measure of one side and the sine of the angle opposite to that measure.
Also note that each of these ratios is equal to the diameter of the circle circumscribing the triangle.
Most of the time it is necessary to calculate the measure of one side of a triangle, knowing the measurements from an angle opposite to it, from the other side and from the angle opposite to that other side, we should use The sins law. This law can also be used to find the measure of one of the angles of a triangle, if we know the measurements from another angle and from the opposite sides of these two angles.
Examples
1 – Calculate the measure of side AB on the triangle Next.
Note that side AB, represented by x, is opposite to angle 45°, and the CB side, which measures 10 cm, is opposite the 30° angle. So we can use the lawFromsines:
The = B
sinα sinβ
x 10
sen45 sen30
Using the fundamental property of proportions, we have:
x·sen30 = 10·sen45
In the table of values trigonometric notable, sen45 = √2/2 and sen30 = 1/2. Replacing these values, we have:
x = 102
22
x = 10√2 cm
2 – Calculate the CB side measurement on the triangle Next.
Side CB, represented by x, is opposite the 45° angle. Also note that side AB, which measures 10 cm, is opposite the 120° angle. Using the lawFromsines, we can write:
The = B
sinα sinβ
x = 10
sen45 sen120
x·sen120 = 10·sen45
To continue, remember that senx = sin (180 – x), therefore: sin120 = sin (180 – 120) = sen60. Replacing the value, we have:
x·sen60 = 10·sen45
3 = 10·√2
22
x·√3 = 10·√2
x = 10·√2
√3
x = 1032
3
x = 106
3
By Luiz Paulo Moreira
Source: Brazil School - https://brasilescola.uol.com.br/o-que-e/matematica/o-que-lei-dos-senos.htm
Teachs.ru
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Water alone cannot remove grease from materials. This is because Thewater is polar, as shown in t...
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The sentence (statement centered on a verb) is studied by syntax. To facilitate this study, the p...
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From antiquity to the present day, science is developed closely with warlike interests and the de...
instagram viewer | 884 | 3,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2023-06 | longest | en | 0.926574 |
http://www.goodmath.org/blog/2006/06/17/extreme-math-1-1-2/?replytocom=12322 | 1,618,578,875,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00609.warc.gz | 135,943,198 | 25,474 | # Extreme math: 1 + 1 = 2
Finally, I have found online, a copy of the magnificent culmination of the 20th century’s most ambitious work of mathematics. The last page of Russel and Whitehead’s proof that 1+1=2. On page 378 (yes, three hundred and seventy eight!) of the Principia Mathematica.. Yes, it’s there. The whole thing: the entire Principia, in all of its hideous glory, scanned and made available for all of us to utterly fail to comprehend.
For those who are fortunate enough not to know about this, the Principia was, basically, an attempt to create the perfect mathematics: a complete formalization of all things mathematical, in which all true statements are provably true, all false statements are provably false, and no paradoxical statements can even be written, much less proven.
Back at the beginning of the 20th century, there was a lot of concern about paradox. Set theorists had come across strange things – things like the horrifying set of all sets that don’t contain themselves. (If it contains itself, then it doesn’t contain itself, but if it doesn’t contain itself, then it contains itself. And then really bad actors need to pretend to be robots short-circuiting while Leonard Nimoy looks on smugly.)
So some really smart guys named Bertrand Russell and Alfred North Whitehead got together, and spent years of their lives trying to figure out how to come up with a way of being able to do math without involving bad actors. 378 pages later, they’d managed to prove that 1+1=2. Almost.
Actually, they weren’t there yet. After 378 pages, they were able to talk about how you could prove that 1+1=2. But they couldn’t actually do it yet, because they hadn’t yet managed to define addition.
And then, along came this obnoxious guy by the name of Kurt Godel, who proceeded to show that it was all a big waste of time. At which point I assume Russell and Whitehead went off and had their brains explode, pretty much the same way that the bad actors would later pretend to do.
## 0 thoughts on “Extreme math: 1 + 1 = 2”
1. D. Eppstein
Wow, I don’t think I’d actually seen that before. Metal Machine Music for math geeks. Of course, nowadays machine verification of proofs is a minor industry, and is (in my opinion) leading to greater certainty of the truths of the statements whose proofs can be so verified
2. Harald Hanche-Olsen
Rats. I was going to post a nitpick that the name is not Godel, but Gödel. But then it turns out that the crucial second letter of his name is replaced by a question mark in the preview. Probably a bug in the blogging software. Oh, but wait! It works right when I try again, from the preview! How very odd.
3. Julian Garcia
I’d reprhase last paragraph as…
At which point I assume Russell, Whitehead AND ALL THEIR FELOW MATHEMATICIANS WORLDWIDE went off and had their brains explode, pretty much the same way that the bad actors would later pretend to do
.
About Goedel’s conjecture, I recommend this novel.
4. Ron
My favorite part of this story is that Mrs. Whitehead apparently disapproved of Russell, the scruffy, free-thinking socialist, and wouldn’t countenance having him around. So the Principia was written in Whitehead’s kitchen where the two would meet after Mrs. Whitehead went to bed.
Of course, both men went on to do many more worthwhile things after this, so no exploding brains, i guess.
(Now I can’t dig up where I first read this story but it has stuck in my mind for years)
5. Mark C. Chu-Carroll
Nick:
I actually have a huge amount of respect for Russell and Whitehead. I was very serious when I said that the principia was the most ambitious work of mathematics of the 20th century. They were seriously setting out to solve a serious problem, and they went about it in an astoundingly intense and difficult way in order to try to fix the problem once and for all.
It’s not their fault that what they were working on was actually impossible. No one had a clue that the effort was doomed to failure.
And you can still learn a lot from the methods of the principia. It is truly both magnificent and hideous: magnificent in that it’s one of the most amazing applications of human intelligence, reasoning, and abstraction ever written; hideous in that it spends hundreds of pages if an ultimately futile effort to try to get to the point where the simplest statements of arithmetic become meaningful without allowing contradictions.
6. Mark C. Chu-Carroll
Janne:
If you follow the link in the original post, they have the entire principia online. The page that that image comes from is page 378. I can’t put a better quality image up; the permissions on the scanned document that I linked to do not allow me to do the image alterations that I’d need to to make it legible at a reasonable size. But you can see it – and all of the rest of the Principia – by following the link.
7. Uffe
Extreme math indeed. Somewhere you say that math is fun, and it is, but here the notational complexity ruins the fun.Just after (nearly) proving that 1+1=2, the breaking news comes only a couple of pages later (“382”):
… Hence it will follow that any cardinal other than 0 or 1 or 2 is equal to or greater than 3.
I think they were on the right track, but the pace a bit on the slow side…
8. PaulC
I think that the program of making math completely rigorous was a good idea at the time, but it has limited usefulness in practice. Actually, the main accomplishment was probably the (at the time) counterintuitive result that some propositions are undecidable. That’s an important result that would have been missed otherwise.
However, when you’re trying to get someone to think like mathematician, the question is not how to get the formalism precise but how to map it into cognitive machinery of the human brain. There’s a big difference between verifying each step of a proof and actually having a sense of why the whole thing is true. In the latter case, it’s easy to be led astray, so you should be able to apply this to similar questions and get the right answer. You shouldn’t stop at intuition, but if it is well-placed, the actual writing of a formal proof should be close to a mechanical process, without any major surprises along the way.
I think Polya was on the right track in How to Solve It by suggesting heuristics that obviously aren’t sound but actually work well in getting a handle on problems: e.g., solve a simpler related problem first, solve a more general problem first (inventor’s paradox), sleep on it (works only if you have really been thinking hard). I think there’s not enough emphasis along these lines in math education, with the result that people equate mathematics with mind-numbing pages of formalism that give only the product and not the process of mathematics as practiced by humans with brains ill-suited to the task.
Actually, with the ready availability of computers, we should be able to do both in some cases. Provide humans a readable summary of the proof steps (what we call a “proof” in a mathematics publication) and then construct a formally verifiable proof–it doesn’t have to go down to a minimal set of axioms, just well established theorems.
9. Blake Stacey
If Vulcans are fully logical, wouldn’t Russell’s paradox make their heads explode too? Maybe the fact that he survived when all those computers died is what led Spock to conclude, “Logic is the beginning of wisdom, Valeris, not the end.”
Remember the Futurama episode where Leela tries to destroy the evil Santa Claus robot with a logical paradox? “Ha! My head was built with paradox-absorbing crumple zones!” I also like the Ghost in the Shell: Stand Alone Complex episode where one robot stumps another, less capable one with the Liar paradox.
10. Harald Hanche-Olsen
Uh, re my previous post: It’s becoming clear that scienceblogs.com is mixing its charsets in a bad way. It uses latin-1 for the main pages, and hence also for the posting form. But the comment preview uses utf-8, so when I wrote Kurt’s last name there, it looked fine. But then it’s displayed as latin-1 once more, so it gets screwed up. Guess I get to find out to whom to report the bug. Sorry about the off-topic-ness, but I think great people like Kurt G. deserve to get their name spelled right.
11. Don Geddis
PaulC wrote: “I think that the program of making math completely rigorous was a good idea at the time, but it has limited usefulness in practice.
It would have been useful, in that future math proofs that followed their formalism would have been more “trustworthy”. The problem is, how do you know that a math proof you have in front of you doesn’t implicitly rely on a paradox (like the “set of all sets…”)? If the foundations of your field of math are suspect, then all conclusions are similarly suspect.
I suspect it would have been quite useful, if it had worked.
Actually, the main accomplishment was probably the (at the time) counterintuitive result that some propositions are undecidable. That’s an important result that would have been missed otherwise.
But Russell and Whitehead didn’t come up with the undecidable result. That was Godel.
Unless you somehow think that Godel wouldn’t have worked on his stuff without Principia Mathematica coming first. That’s a pretty indirect connection, and not really an “accomplishment” of Russell and Whitehead.
Harald,
It’s not all of scienceblogs.com since for example Pharyngula seems to work fine. Maybe it’s only this blog.
13. Mark C. Chu-Carroll
Don:
My math history isn’t the best, but I do think there’s a strong connection between the Principia and Godel. I think I heard the story in a talk that Greg Chaitin was giving at work, but I’m not sure. This would have been my first year at my job, which was 10 years ago, so a lot of talks from that long ago have blurred together.
Anyway, if I remember it correctly, the story is that R&W weren’t alone. The problem of paradox was *the* math problem that people were interested in. The principia was the most ambitious approach to trying to solve the paradox problem. Lots of other very famous mathematicians worked on it; particularly important for the history of Godel was Hilbert.
Godel came along when that stuff was in full swing, and was introduced to the field through a course on Russell’s work on number theory (given by? using a book by? Russell). So you can make an argument that Russell was a major mover in what led to Godel’s incompleteness theorem.
But I think that incompleteness was driven more by a reaction to Hilbert than R&W.
Mark,
It seems the history is incomplete. 🙂
My name is usually provided by TypeKey, that I started on Pharyngula. We can all check what happens with my “omlaut o”. Or Godel’s. 🙂
15. PaulC
Don Geddis:
Unless you somehow think that Godel wouldn’t have worked on his stuff without Principia Mathematica coming first. That’s a pretty indirect connection, and not really an “accomplishment” of Russell and Whitehead.
Russell and Whitehead were not the only ones interested in developing a way to prove mathematical principles formally starting with axioms. Peano’s axioms preceded Goedel. Earlier than that, Boole certainly had the goal of formalizing logical argument.
Absent a program of trying to formalize mathematics, Goedel would have had to do the whole thing himself–not just showing that inference steps could be encoded as numbers, but inventing the whole notion of a formal symbolic proof. He was a brilliant guy, but it seems reasonable to assume that he was helped along by the fact that people already had a notion of what was meant by a formal proof. And, in fact, Hilbert had already asked the question of whether these were sufficient to decide everything. That question only makes sense if you think there could be something interesting or worthwhile about having a formal proof.
In fact, the evidence shows that Goedel was directly, admittedly influenced by Hilbert’s Program, http://plato.stanford.edu/entries/hilbert-program/ which was basically a plan to try to decide mathematics formally.
So, in short, yes I feel confident in saying that unless there had been a significant ferment of mathematicians interested in trying to show that all of mathematics could be proven formally, then it is highly unlikely Goedel would have come in and proved his incompleteness result. I mean, I guess it could have happened, but it is not what actually happened, and what actually happened required the work of many mathematicians over a century or so, most of whom were motivated by the goal of proving completeness rather than incompleteness.
The extent to which Goedel was influenced by Whitehead and Russell in particular wasn’t my point. I have no idea, and did not refer to them in particular in my comment.
16. Mark C. Chu-Carroll
Paul:
According to Wikipedia, my memory was actually amazingly close 🙂
Godel did get exposed to the ideas through one of Russell’s textbooks; but Hilbert’s attempt in the same vein as R&W actually had more to do with Godel than R&W did. But the info I can find does strongly suggest that without Hilbert, Godel wouldn’t have wound up doing the work that led to incompleteness.
Also, thanks for that link; looks like a very nice summary of the Hilbert program.
17. Stephen
And then, along came this obnoxious guy by the name of Kurt Godel, who proceeded to show that it was all a big waste of time. At which point I assume Russell and Whitehead went off and had their brains explode, pretty much the same way that the bad actors would later pretend to do.
Of course, the big irony is that it was Russell that had pointed out the fatal flaw in Frege’s system of logic; Frege’s initial reply to Russell is one of the classics of mathematics….
18. tgibbs
My recollection is that R&W came up with the paradoxical “set of all sets that are not members of themselves,” but thought that such paradoxical statements were discrete curiousities that could be safely “walled off” from the rest of mathematics as simply undecidable, like the value of 0/0. Godel came up with a construction proof that used this concept to demonstrate that unprovable propositions could not be segregated from the rest of mathematics.
19. Mark C. Chu-Carroll
tgibbs:
Mostly, but not entirely right.
Russell is the discoverer of the “set of all sets that are not members of themselves”; in fact, it’s called Russell’s paradox.
It was generally believed by mathematicians around the time that paradoxical constructs were errors caused by inadequate formalisms, and that they could be eliminated by creating a proper basis for mathematics. That’s a big part of the central idea behind the Principia and Hilbert’s program. They thought that by starting from scratch and building up mathematics entirely in formal logic that they could arrive at a decision procedure: a procedure that could, for any mathematical statement, produce a proof of either its truth or its falsehood.
They didn’t believe that the paradoxical statements could be “walled off”; they believed that they were fundamental errors in the basis of mathematics, meaningless statements that could only be constructed because the mathematical formalisms that permitted them were invalid. They thought that by constructing mathematics properly, they would basically “go away”: there would be no way to even say them, because they were based on invalid concepts.
Loosely, the idea behind R&W’s construction (and Hilbert’s program as well, if I’m recalling it correctly) was a system of types: every statement was clearly defined to range over a particular kind of value; and no statement could be defined in a way that allowed it to talk about itself. So, for example, you could have first order predicates that reasoned about base values; second-order predicates that could reason about base values and first order predicates; third order predicates that could reason about base values, first order, and second order predicates, etc. But no predicate could ever reason about itself: an Nth order predicate could never reason about anything beyond the (N-1)th order predicates.
The key to what Godel did was show how you couldn’t make that separation hold: even in the strictest type-theoretic framework, you could represent, for example, first order predicates using natural numbers, and then allow first order predicates to range over first order predicates encoded into numbers. (Actually, I think it was not just natural numbers, but naturals with Peano arithmetic.)
20. Slawek
If you think about doing formalized (machine verifiable) math – think twice. You do a proof, then another one and … you are hooked. You don’t want to do romantic math any more. Every standard proof (informal proof sketch that is) seems full of errors and missing references. Russel never recovered. Watch out.
P.S. As for usefulness of formalized (and machine verifiable) math I think the most interesting application is writing formalized proofs in logical systems that are much different that what our brain is naturally wired for (for example, quantum logic, see Metamath’s Quantum Explorer). This can not be done in an informal way – humans can not reliably verify such proofs. Maybe one day it will turn out that mathematics built on one of them is better at describing the physical world we live in than the one bult on on classical logic and ZF set theory?
21. Harald Hanche-Olsen
I just got my copy of the April issue of Notices of the American Mathematical Society. The whole issue is essentially a tribute to Kurt G. The whole thing is available on the web. Enjoy! (But is anybody reading comments to a blog post that is already several days old?)
22. Gerry R
Well, if you want an actual formal proof, see the metamath proof of 2+2=4. (1+1=2 is boring because that’s how it defines 2.) This is a really nifty browseable hypertext of a computer-verified proof derived from the axioms of Zermelo-Fraenkel set theory. That is 5254th theorem in the system, although not all of the preceding 5253 are required for the proof. (In particular, number 1329, the Axiom of Choice, is not a prerequisite.)
This also includes lots of other, more significant math, like that if an infinite series converges to a limit, that limit is unique.
23. Thony C.
Just to clear up the history a little bit for those who are interested: Gödel’s work was done in response to Hilbert’s call to mathematicians to prove the consistency, completeness and decidability of arithmetic using finite methods. The title of his all important paper however is “On Formally Undecidable Propositions of Principia Mathematica and Related Systems I” (translation from the German Martin Davis). So as you can see R & W do have a direct connection to Gödel. As a foot note as you will notice the title ends with a Roman one (I) indicating that this is only part one of the paper. There never was a part two!
24. jeffh
Actually, 1 + 1 = 3 for sufficiently large values of 1. The proof is left as an exercize for the reader.
25. Daniel Slaney
Gödel’s paper publishing the incompleteness theorem is “Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme,” or “On Formally Undecidable Propositions in Principia Mathematica and Related Systems.”
So R&W do necessarily figure into his work, even if he only uses their notation and semantics.
26. Sobia Tahir
The services rendered by Russell and Whitehead for claification of concepts and cleansing of Mathematics from Metaphysical notions may not be forgotten for ages to come.No doubt they were founder fathers of Modern Symbolic/Mathematical Logic.If they proved 1+1=2 at page 378,they were not fools but dedicated and thorough.Their commitment should be respected instead of being laughed at.The great brains have always been humiliated in this manner.There is always room for error and improvement,but acknowledge the contribution of others without prejudice.
27. vveleva
To reply to comment 37: No, 1 + 1 = 2 is not a definition, although it surely seems that way.
We’ve been told all our lives that 1 + 1 = 2 is true because it’s easier to just remember it rather than understand it. So what Russel and Whitehead, (also Gödel and Landau) attempted to prove was that all the math we are familiar with could be derived from just a few primitive axioms. Surely 1 + 1 = 2 seems an obvious result but if I were to ask “yes, one apple plus one apple is two apples, but how do we know that’s true for very very large numbers?” It is dangerous to make assumptions like that because sometimes they end up being false. Thus, it is enough to show that adding 1 to, say a, gives you 1+a, or even easier, it suffices to show that adding 1 to 1 gives you 2. That’s what all those mathematicians were after. 🙂
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Old 08-11-2006 Thread Starter
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Thanks. The shape is virtually the same, a rectangle, although before it slid down fore and aft all at once and now it slides down aft. I agree it should be easier to tack and I think I need to lower it more. I am now about 9 inches below the skeg and I think I could double that.
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If the centerboard pivot point is located in the wrong place, it may be lowering almost completely, and that would shift the Center of Lateral Resistance forward, not aft...think of it as a rectangle. If the rectangle pivots 90 degrees, then the geometric center is far further forward than if it pivots just 60 degrees or just drops parallel. That would make it much more difficult to tack. Might want to check that.
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How does your boat sit in the water? Level? Stern heavy? Bow heavy? Take a look with you and typical crew on her. See if she is sitting stern heavy. Maybe some chain in the locker??? Sounds like a lot of sail area for that sized boat too. If the boat is sitting stern heavy she will have a tendency to "wobble" about a tack.
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If you are stalling as you come head to wind
maybe you want to consider easing the jib/genoa and let the aft sail(s) help you around then sheet in.
Free advice - Worth what you pay.
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The Whitehalls that I have sailed have had a long fairly deep keelson and planked were planked down in the stem and onto the skeg in the stern. This was intended to help them track straight when being rowed but which also made these boats hard to turn under sail. The whitehall that I knew best, had semi-circular centerboard (pivoted at the front) that probably extended roughly 2 feet below the boat when fully down. The boat would not tack reliably undless the board was in the fully down position.
Complicating things was the fact that whitehalls have an easily driven hull, a rig with a lot of windage, and turning the long keel quickly makes a lot of turbulance, and so they lose a lot of speed during the tack. It is important to "carve the turn" rather than trying to make a quick tack.
Adding a bowprit and jib will only add more windage slowing the boat further and making it harder to get head to wind. On the other hand if you can get head to wind, you should be able to back the jib and get the boat to pay around. Not a great way to sail because you can get caught back winded and be over before you regain steerage.
As several others have said, I would guess that the problem is with your centerboard modifications and that adding a jib and bowsprit is an unnecessary complication.
Jeff
Last edited by Jeff_H; 08-11-2006 at 09:55 PM.
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I once had a converted lifeboat that had too wide a turning radius. I designed and employed a drop-down extension on its rudder (made from 1/4" aluminum) and it made a huge difference. The boat responded immediately and much more positively to the tiller. I raised and lowered the extension with a line that ran over the rudder stock and to a cam cleat on the tiller itself.
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LGD, which stands for loss given default, represents the amount of money you risk losing if the company you invest in goes bankrupt. It is a valuable metric in assessing the credit risk of a company before investing in it.
Unlike other liquidity ratios, such as the current ratio and credit spread, which tell you the company's financial position and how likely it will default, the LGD calculates how much you will lose if a credit event does actually happen. To understand more on these metrics, check out our current ratio calculator and credit spread calculator. This metric can help you to understand the worst-case scenarios of your portfolio.
Now that we have understood what LGD is, we can look at how to use this metric in a loss given default example.
## How to calculate the loss given default?
We will examine an investment in Company Alpha as our loss given default example to help you understand how to calculate the loss given default.
Company Alpha reports the following information:
• Company name: Company Alpha
• Recovery rate: 80%
• Loss severity: 20%
• Expected exposure in Company Alpha: $1,000,000 1. Determine the recovery rate or loss severity. The first step is to understand the relationship between recovery rate and loss severity. Their relationships are depicted as the following formula: recovery rate = 1 - severity or loss severity = 1 - recovery rate To proceed with the calculation, we need to have at least one of them. In this example, we assume the recovery rate of 80%, hence loss severity of 20%. 2. Determine the expected exposure. The next step is to determine the expected exposure. The expected exposure can be interpreted as the amount of investment you put into your investment. For our example, the expected exposure in Company Alpha is $1,000,000.
3. Calculate the loss given default (LGD).
The final step is to calculate the LGD. We can do this by using the following formula:
LGD = expected exposure × loss severity
or
LGD = expected exposure × (1 - recovery rate)
In the example, the LGD of the Company Alpha is:
$1,000,000 × 20% =$200,000.
## Why is calculating LGD useful?
After we understand what loss given default is, we can now discuss why it is useful:
• Advantage: The most prominent use for calculating LGD is allowing you to understand how much money you can probably lose in the investment, i.e., it tells you the worst-case scenario. Thus, it can help to assess if the LGD is a loss that you can possibly stomach.
• Disadvantage: However, the LGD does not tell you the probability of the company going default. One investment may have a small LGD but a high probability of default and vice versa. This scenario can make LGD an interior metric to assess investment risks.
## FAQ
### Does high LGD means high investment risks?
The short answer is not necessarily. A high LGD only tells you that you will lose a lot of the company goes default, but it does not tell you how likely that will happen.
### What is the difference between LGD and liquidity ratios?
Liquidity ratios tell you how likely a company is going to default, whereas LGD focuses on quantifying the negative impact following the default.
### Can I use LGD to assess credit risk?
Yes, but it is dangerous to use it alone. LGD is often used with other liquidity ratios to work out both the probability of default and its impact.
### Can LGD be zero?
Theoretically speaking, yes. If the company has enough assets that can be liquidated and pay off its debt, its LGD can technically be zero. This, however, is very unusual.
Wei Bin Loo
Recovery rate / Loss severity
Recovery rate
%
Loss severity
%
LGD
Expected exposure
$Loss given default$
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If you want to learn what is turnover rate, or estimate it for your company, use this turnover rate calculator. | 1,082 | 5,015 | {"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} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.926317 |
https://www.hackmath.net/en/math-problems/money?tag_id=89 | 1,563,757,917,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527458.86/warc/CC-MAIN-20190722010436-20190722032436-00275.warc.gz | 717,214,384 | 5,536 | # Money + surface area - problems
Money or finance are an essential medium of exchange, a unit of account and store of value in modern society. Mathematical examples helps to get basic financial literacy.
1. Cone A2V
Surface of cone in the plane is a circular arc with central angle of 126° and area 415 cm2. Calculate the volume of a cone.
2. Sphere slices
Calculate volume and surface of a sphere, if the radii of parallel cuts r1=31 cm, r2=92 cm and its distance v=25 cm.
3. Bathroom
How many CZK we pay for lining the perimeter walls of the bathroom with rectangular shape with dimensions 3.5 m and 4 m, high 1.5 m if 1 square m tile cost 300 CZK?
4. Office
Office building was built in the shape of a regular hexagon inscribed in a circle with a radius of 12 m. The height of the walls is 7m. How much CZK cost plastering the walls of the building, if per 1 m square cost CZK 400?
5. Painting
The room is long 50 meters and wide 60dm and 300 cm high. Calculate how much it will cost painting if area of windows and doors is 15% of the total area. One square meter cost 50cents.
6. Painter
How many euros we will pay for repainting the room shaped cuboid with a length of 4.5 meters, width of 2.5 meters and a height of 3 meters, if for 1 m2 with paint we pay € 1.5?
7. Concrete pipe
How much will cost cover a 6 m long concrete pipe with an outer radius 1.5 m and inner radius 0.8 meters if 1 m2 paint costs 24 €.
8. Church roof 2
The roof has the shape of a rotating cone shell with a base diameter of 6 m and a height of 2.5 m. How many monez (CZK) will cost the roof cover sheet if 1 m2 of metal sheet costs 152 CZK and if you need 15% extra for joints, overlays and waste?
9. The room
The room has a cuboid shape with dimensions: length 50m and width 60dm and height 300cm. Calculate how much this room will cost paint (floor is not painted) if the window and door area is 15% of the total area and 1m2 cost 15 euro.
10. If one
If one litre of pant covers an area of 5 m2 how much paint is needed to cover: a) rectangular swimming pool With dimensions 4m x 3m x 2.5m (the Inside walls and the floor only) b) the Inside walls and floor of a cylindrical reservoir with diameter 3m and
11. Pool tiles
The pool is 25m long, 10m wide and 160cm deep. How many m2 of tiles will be needed on the walls and the pool? How many pieces of tile are needed when 1 tile has a square shape with a 20cm side? How much does it cost when 1m2 of tiles costs 258 Kc?
12. Half-sphere roof
The roof above the castle tower has the shape of a 12.8 m diameter half-sphere. What is the cost of this roof, if the cost of 1 square meter is 12 euros and 40 cents?
We apologize, but in this category are not a lot of examples.
Do you have an interesting mathematical problem that you can't solve it? Enter it, and we can try to solve it.
To this e-mail address, we will reply solution; solved examples are also published here. Please enter e-mail correctly and check whether you don't have a full mailbox. | 813 | 2,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-30 | latest | en | 0.896063 |
http://math.stackexchange.com/questions/323962/levy-measure-of-a-pure-jump-process | 1,397,874,231,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00413-ip-10-147-4-33.ec2.internal.warc.gz | 148,965,301 | 15,224 | # Lévy measure of a pure jump process
Let ($\xi_t)_{t \geq 0}$ an infinititely divisible cadlag process on $[0,\infty)$ and denote by $p$ its jump measure. Define a measure $p_t$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ by $$p_t(B) := p((0,t] \times B) \qquad (B \in \mathcal{B}(\mathbb{R}))$$ Let $$\eta_t := \int x \cdot (1-1_{(-a,a)}(x)) \, p_t(dx)$$ where $a>0$. Then $\eta_t$ is infinitely divisble, cadlag, with $b=\sigma=0$ and Lévy measure $\Lambda(\Gamma \backslash (-a,a))$ where $\Lambda$ denotes the Lévy measure of $\xi_t$.
The last assertion refers to Khinchin's formula $$\mathbb{E}e^{\imath \, \lambda \cdot X_t} = \exp \left[ t \cdot \left( \int (e^{\imath \, \lambda \cdot x}-1- \imath \, \lambda \cdot \sin x) \, \Lambda(dx) + \imath \, b \cdot \lambda - \sigma^2 \cdot \frac{\lambda^2}{2} \right) \right] \tag{1}$$
My question is the following: The author proves $$\mathbb{E}e^{\imath \, \lambda \cdot \eta_t} = \exp \left(t \cdot \int_{\mathbb{R} \backslash (-a,a)} (e^{\imath \, \lambda \cdot x}-1) \, \Lambda(dx) \right)$$ But as far as I can see, this does not imply that $(1)$ holds with $b=\sigma=0$ and Lévy measure $\Lambda(\Gamma \backslash (-a,a))$. Instead, I would conclude $$\mathbb{E}e^{\imath \, \lambda \cdot \eta_t} = \exp \bigg[ t \cdot \bigg( \int_{\mathbb{R} \backslash (-a,a)} (e^{\imath \, \lambda \cdot x}-1-\imath \, \lambda \cdot \sin x) \, \Lambda(dx) + \imath \, \lambda \cdot \underbrace{\int_{\mathbb{R} \backslash (-a,a)} \sin x \, \Lambda(dx)}_{=:b} \bigg) \bigg]$$ i.e. $(1)$ holds with Lévy measure $\Lambda(\Gamma \backslash (-a,a))$, $\sigma=0$ and $b \not= 0$. Am I right or are there properties of the Lévy measure $\Lambda$ which imply $b=0$?
(Literature: "Introduction to the Theory of Random Processes" - N.V. Krylov (chapter 5, section 3))
-
I also agree with you. Indeed, $b$ may fail to vanish even for the Poisson process of parameter $1$, whose Lévy measure being equal to $\Lambda(dx) = \delta_1(dx)$. – sos440 Mar 8 '13 at 17:55
@sos440 Thanks. The Poisson process is indeed a nice example to show that $b$ is not necessarily equal to 0. – saz Mar 8 '13 at 18:10
I guess that the author tried to figure out that the drift coefficient of the resulting process $\eta_t$ vanishes. This makes sense since $\eta_t$ has bounded variation. – sos440 Mar 8 '13 at 19:59
@sos440 Why does the drift coefficient vanish? I thought, $b$ (as in $(1)$) is called drift coefficient...? – saz Mar 9 '13 at 9:37
@sos440 Thank you for the clarification. So, in this case, the drift coefficient is equal to 0, since $\eta$ is a pure jump process. Am I right? – saz Mar 9 '13 at 10:18
show 1 more comment | 940 | 2,660 | {"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} | 3.125 | 3 | CC-MAIN-2014-15 | latest | en | 0.647118 |
https://docsbay.net/math-for-liberal-arts | 1,632,857,888,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00344.warc.gz | 272,348,237 | 4,740 | # Math for Liberal Arts
MAT 110
Math for Liberal Arts
Voting Project TYPE YOUR NAME HERE
You must answer each of the following questions. It is expected that all work should be organized and typed within this document. Thus, complete tables and lists of results must be included. You must also use full sentences and proper grammar. Under no circumstance should any response be, for example, “Netflix was preferred.” While concise, it shows in no way any justification for arriving at that conclusion. Ideally, your responses would look very similar to the solutions for the Chapter 12 examples in our textbook.
I gave some random individuals a survey where they were asked to think about, “Which portal/interface do you prefer to use to view video/TV content?” The respondents were given the following five choices and rated them from 1 to 5, where 1 meant they preferred the most and 5 the least.
Netflix (N) / Hulu Plus (H) / Amazon Prime (A) / iTunes (I) / Cable/Satelite (C)
The results of the voting process were as follows:
# of Votes / 2 / 3 / 2 / 2 / 2 / 1 / 2 / 5 / 1 / 1 / 1
1st Choice / I / H / A / N / H / I / N / C / N / A / H
2nd Choice / H / C / N / I / A / N / C / N / A / H / N
3rd Choice / C / A / H / H / N / H / H / I / H / N / I
4th Choice / N / N / I / A / I / A / A / A / I / C / C
5th Choice / A / I / C / C / C / C / I / H / C / I / A
For this project you must complete the following items.
1. After making this preference table Mr. Gisch found five more ballets. Add the ballets to the existing preference table.
NIHAC, INHAC, CNIAH, CNIAH, AHNCI
1. How many total people voted?
2. What would the majority of this population be (I did not ask who had the majority)?
3. Using the plurality method who won?
4. Using the Borda-Count (points) method who won?
5. Using the sequential-elimination method who won?
6. If you were to do the pairwise-comparison method, how many comparisons would you have to make?
7. Use the pairwise-comparison method to determine a winner.
8. After all of your research, who do you believe won and why? | 564 | 2,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-39 | latest | en | 0.937303 |
https://id.scribd.com/document/306365210/Worked-Out-Examples-and-Exercises-With-Solutions-Chapter-17 | 1,582,938,876,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148163.71/warc/CC-MAIN-20200228231614-20200229021614-00229.warc.gz | 400,087,582 | 85,050 | Anda di halaman 1dari 11
Chap-17
B.V.Ramana
August 30, 2007
Chapter
10:15
17
Numerical Analysis
WORKED OUT EXAMPLES
Bisection Method
Example 1: Find a real root of the equation x 3
6x 4 = 0 by bisection method.
[JNTU 2006, Set No. 1]
Solution:
x:
0
1
2 3
f (x) = x 3 6x 4: 4 9 8 5
...
one root lies between 2 and 3. Take x0 = 2,
x1 = 3. By bisection method, the next approxima1
tion is x2 = x0 +x
= 2+3
= 2.5. Now at x2 = 2.5,
2
2
f (2.5) = 3.375 < 0. Thus, a root lies between 2.5
and 3. So the next approximation is x3 = 2.5+3
=
2
5.5
=
2.75.
At
x
=
2.75,
f
(2.75)
=
0.2968
>
0.
3
2
Thus a root lies between 2.5 and 2.75. Next approximation is x4 = 2.5+2.75
= 2.625.
2
Now f (2.625) = 1.6621 < 0. The root lies between 2.625 and 2.75. Then x5 = 2.625+2.75
=
2
2.6875 and f (2.6875) = 0.7141 < 0. So root lies
between 2.6875 and 2.75. The approximate value of
the real root of the given equation is 2.6875+2.75
=
2
2.71875.
Example 2: Find a real root of x 3 5x + 3 = 0
using bisection method.
[JNTU 2006, Set No. 2]
Solution:
x: 0
1 2
f (x) = x 3 5x + 3: 3 1 1
Thus, a root lies between 0 and 1 and another root
lies between 1 and 2.
1
Choose x0 = 1 and x1 = 2. Then x2 = x0 +x
=
2
1+2
=
1.5,
f
(1.5)
=
1.125
<
0.
2
1
So root lies in between 1.5 and 2. So x3 = x2 +x
=
2
1.5+2
=
1.75,
f
(1.75)
=
0.3906
<
0.
2
So root lies between 1.75 and 2. Then x4 = 1.75+2
=
2
1.875, f (1.875) = 0.2167 > 0.
So root lies between 1.75 and 1.875. So the approximate root of the given equation is x5 = 1.75+1.875
=
2
1.8125.
Example 3: Find a real root of the Equation x 3
x 11 = 0 by bisection method.
[JNTU 2007, Set No. 2]
Solution:
x:
f (x) = x 3 x 11:
0
11
1
11
2
5
3
13
A root lies between 2 and 3. Then the rst approxi1
mation is x2 = x0 +x
= 2+3
= 2.5. Now f (2.5) =
2
2
2.125 > 0. So root lies between 2 and 2.5. So
the next approximation is x3 = 2+2.5
= 2.25. Now
2
f (2.25) = 1.859 < 0. So next approximation is
x4 = 2.25+2.5
= 2.375. Now f (2.375) = 0.0215 >
2
0 and f (2.25) < 0.
... Root lies between 2.25 and 2.375. So x =
5
= 2.3125. Now f (2.3125) = 0.946 < 0
and f (2.375) > 0.
2.25+2.375
2
17.1
Chap-17
B.V.Ramana
17.2
August 30, 2007
10:15
MATHEMATICAL METHODS
So root lies between 2.3125 and 2.375. x6 =
2.3125+2.375
= 2.34375, f (x6 ) = 0.4691 < 0. Root
2
lies in (2.34375, 2.375). So
X7 = 2.34375+2.375
= 2.359375.
2
Since f (x7 ) = 0.225 < 0, root lies between
2.359375 and 2.375. So x8 = 2.359375+2.375
=
2
2.3671875.
f (x8 ) = 0.10247 < 0,
root
lies
between
2.3671875 and 2.375. Then
x7 =
=1
(2 1)
[0.2817]
[11.778 (0.2817)]
x2 = 1.329
Now f (x2 ) = f (1.329) = 2.0199 > 0 and f (1) =
0.2817 < 0 so root lies in between 1 and 1.329.
Take x0 = 1, x2 = 1.329.
Then x3 = x0
=1
2.3671875 + 2.375
= 2.37109.
2
Approximate root is 2.37109.
Example 4: Using bisection method, nd a root of
the Equation x 3 4x 9 = 0
x:
f (x) = x 3 4x 9:
0
9
1
12
2
9
3
6
A root lies between 2 and 3. First approximation is
x0 + x 1
2+3
x2 =
=
= 2.5.
2
2
Since f (2.5) = 3.375 < 0, root lies between 2.5
and 3. The second approximation is x3 = 21 (2.5 +
3) = 2.75. Since f (2.75) = 0.7969 > 0, root lies
between 2.5 and 2.75. So third approximation is x4 =
2.5+2.75
= 2.625. Since f (2.625) = 1.4121 < 0,
2
root lies between 2.5 and 2.625. Thus, the fourth
approximation to the root is x5 = 21 (2.5 + 2.625) =
2.6875.
Example 5: Find a real root of xex = 2 using
Regulafalsi method.
[JNTU 2007, Set 4]
Example 6: Find a real root of xex = 3 using
Regulafalsi method.
[JNTU 2006, Set 4]
(1.3291)
[0.2817]
[2.0199(0.2817)]
(x3 x2 )
f (x2 )
[f (x3 ) f (x2 )]
= 1.329
x:
f (x) = xe 3:
0
3
1
0.2817
... 1,
... root lies in (1, 2). Take x0 = 1,
Then
(x1 x0 )
x2 = x0
f (x0 )
(f (x1 ) f (x0 ))
2
11.778
x1 = 2.
(1.04 1.329)
[2.0199]
[0.05 (2.0199)]
x4 = 1.08 is the required approximate root.
Example 7: Find a real root of ex sin x = 1 using
RegulaFalsi method.
[JNTU 2006, Supply, Set No. 1, Code No.
R059010202]
Solution:
x:
0
1
root lies in
ex sin x 1 = f (x): 1 1.2874
(0, 1).
By bisection method: f (0.5) = 0.2094, root lies in
(0.5, 1).
By bisection method: f (0.75) = 0.443 root lies in
(0.5, 0.75).
Since f (0.5) < 0, choose b = 0.75. Then by
RegulaFalsi method
Solution:
f (x0 )
x3 = 1.04
Now f (x3 ) = f (1.04) = 0.05 < 0.
Since f (x2 ) = 2.0199 > 0, root lies between 1.04
and 1.329. Then
x4 = x2
Solution:
(x2 x0 )
[f (x2 )f (x0 )]
x1 = x0
= 0.5
f (x0 )
(b x0 )
f (b) f (x0 )
(0.20941)
(0.75 0.5)
(0.443 + 0.20941)
Chap-17
B.V.Ramana
August 30, 2007
10:15
NUMERICAL ANALYSIS
(0.5, 0.6605).
Since f (a) = f (0.5) < 0; we have b = 0.6605
x2 = 0.5
0.2094
(0.6605
(0.1876+0.2094)
0.5)
x2 = 0.5 + 0.08465 = 0.58466
At x2 , f (x2 ) = 0.009653.
So x2 = 0.58466 is an approximate root of the given
transcendental equation.
Example 8: Using RegulaFalsi method, nd an
approximate real root of the transcendental Equation
x log10 x = 1.2.
Solution:
x:
1
2
3
f (x) = x log10 x 1.2: 1.2 0.59794 0.23136
... a root lies between 2 and 3.
Take x0 = 2, x1 = 3. Then approximate root
x1 x0
x2 = x0
f (x0 )
f (x1 ) f (x0 )
x2 = 2.72102
Now f (x2 ) = f (2.72102) = 0.0179, so root lies
between 2.72102 and 3. Then
x3 = x2
x1 x2
f (x2 )
f (x1 ) f (x2 )
= 2.72102
starting from an initial guess x0 , better successive approximations x1 , x2 , x3 , . . . of an unknown solution
of (1) are computed step by step.
Generally, iterative methods are easy to program.
The bisection method and RegulaFalsi method are
two such iterative methods which are known as interpolation methods or bracketing methods because
the root (guess) is bracketed between two estimates
(one for f (x) > 0 and one for f (x) < 0). The other
kind of iterative methods such as xed point iteration method, Newton-Raphson method, its variant secant method are known as extrapolation
methods or open-end methods in which a single value
(initial estimate) is chosen.
Fixed-point iteration method
Rewriting the Equation (1) in the form
x = g(x)
(3 2)
=2
[0.59794]
[0.23136 + 0.59794]
17.3
(2)
observe that the roots (or solutions) of (1) are same as
the points of intersection of the straight line y = x
and the curve representing y = g(x), as shown in
the gure. The exact solution (root) is which is the
intersection of y = x and y = g(x).
(3 2.72102)
(0.0179)
[0.23136 + 0.0179]
x3 = 2.74021
Similarly, we get x4 = 2.74024, x5 = 2.74063. Note
at x5 ,
f (x5 ) = 0.0000140385
... An approximate root is x5 = 2.74063.
Fig. 17.1
Iteration Methods
In general, there is no formula for the exact solution
of
f (x) = 0
(1)
which may be an algebraic equation or transcendental equation. In such cases the iterative method,
which is an approximation method, is used in which
Starting with an initial estimate x0 , compute the
rst approximation x1 given by
x1 = g(x0 )
Now, treating x1 as the initial value, compute the
second approximation x2 as
x2 = g(x1 )
Chap-17
B.V.Ramana
17.4
August 30, 2007
10:15
MATHEMATICAL METHODS
In general the n + 1th approximation is
xn+1 = g(xn )
for n = 0, 1, 2, . . .
The name of the method is motivated, since a solution
of (2) is called a xed point of g.
This method converges in an interval (a, b) if
|g (x)| k < 1
Example 9: Find a real root of the Equation 2x
log10 x = 7 by successive approximation method.
[JNTU 2006, Set No. 3]
Rewriting the given equation as
x=
1
(cos x + 1) = g(x).
3
Observe that
|g (x)| = 13 | sin x| < 1 in 0, 2 .
Starting with x0 = 0, the xed point iteration methods yields the following successive approximations.
x1 = g(x0 ) =
1
(cos 0 + 1) = 0.6667,
3
1
(cos(.6667) + 1) = 0.5953,
3
1
Solution:
x3 = g(x2 ) = (cos(.5953) + 1) = 0.6093,
x: 1
2
3
4
3
f (x) = 2x log x 7: 5 3.301 1.4471 0.398
1
x4 = g(x3 ) = (cos(.6093) + 1) = 0.6067,
... A root lies between 3 and 4.
3
Rewrite the given equation as
1
x5 = g(x4 ) = (cos(.6067) + 1) = 0.6072,
1
3
x = [log10 x + 7] = g(x)
1
2
x6 = g(x5 ) = (cos(.6072) + 1) = 0.6071.
3
Thus the approximate root is 0.6071 since x5 and x6
Now |g (x)| = | 21 x1 log10 e| < 1 when 3 < x < 4.
1 1
are nearly equal.
when x = 3, |g (3)| = | log10 e| = 0.07238
2
x2 = g(x1 ) =
when x = 4, |g (4)| = | 21 41 (0.4343)| = 0.0542
Since |f (4)| = 0.398 < 1.4471 = |f (3)|, The root
is near to 4. Now taking the initial guess as x0 = 3.6,
we apply the xed point iteration method yielding
the successive approximations as
1
x1 = g(x0 ) = (log10 (3.6) + 7) = 3.77815,
2
1
x2 = g(x1 ) = (log10 (3.77815) + 7) = 3.78863,
2
1
x3 = g(x2 ) = (log10 (3.78863) + 7) = 3.78924,
2
1
x4 = g(x3 ) = (log10 (3.78924) + 7) = 3.78927.
2
Example 11: Find the negative root of the equation x 3 2x + 5 = 0.
Solution: If , , are the roots of
x 3 + 0 x 2 2x + 5 = 0
Then the equation whose roots are , , is
x 3 + (1) 0 x 2 + (1)2 (2x) + (1)3 5 = 0
or x 3 2x 5 = 0
(2)
To nd the positive root of (2), we have
f (x) = x 3 2x 5, f (2) = 1 < 0, f (3) = 16 >
0. So root lies between 2 and 3 and it is nearer to 2
(since |f (2)| = 1 < 16 = |f (3)|).
Rewriting the Equation (2) we get
1
x = (2x + 5) 3 = g(x)
2
Example 10: Using iteration method, nd a real
root of the Equation cos x = 3x 1.
Solution:
Here, f (x)
= cos x 3x + 1, and
f (0) = 2 > 0 and f 2 = 3
+ 1 = 3.71 < 0
2
.
. . A root lies between 0 and .
2
(1)
Then |g (x)| = | 13 (2x + 5) 3 2| < 1 in (2, 3).
Choosing x0 = 2.1, we apply the xed point iteration method. Then
1
x1 = g(x0 ) = [2(2.1) + 5] 3 = 2.09538,
1
x2 = g(x1 ) = [2(2.09538) + 5] 3 = 2.09468,
1
x3 = g(x2 ) = [2(2.09468) + 5] 3 = 2.09457,
Chap-17
B.V.Ramana
August 30, 2007
10:15
NUMERICAL ANALYSIS
x4 = g(x3 ) = 2.09455, x5 = g(x4 ) = 2.09455
Hence the approximate root of (2) is 2.09455; so the
negative root of the given Equation (1) is 2.09455.
Example 12: Find a solution of x 3 + x 1 = 0 by
iteration.
Solution:
x:
0
1
f (x) = x 3 + x 1: 1 1
... A root lies in (0, 1).
Rewriting the given equation,
x=
1
= g(x)
1 + x2
so that
xn+1 =
1
.
1 + xn2
2|x|
Also |g (x)| = (1+x
2 )2 < 1 for any x.
Choosing x0 = 1, we obtain
1
1
= = 0.5
1 + 12
2
1
= 0.800,
x2 =
1 + (0.5)2
1
= 0.610,
x3 =
1 + (0.8)2
x4 = 0.729, x5 = 0.653, x6 = 0.701
x1 =
Approximate root is 0.701.
NewtonRaphson Method
Example 13: Find a real root of xex cos x = 0
using NewtonRaphson method.
[JNTU 2007, Set No. 1],
[JNTU 2006, Supply, Set No. 4, (R059010202)]
Solution: f (0) = 1 < 0, f (1) = 2.178 > 0
Here, f (x) = xex cos x. Root lies between 0 and
1. Also f (x) = xex + ex + sin x
By NewtonRaphson method:
f (xn )
f (xn )
xn
(xn e cos xn )
= xn
(xn exn + exn + sin xn )
xn+1 = xn
xn+1
17.5
Since f (1) = 2.178 and f (1) = 6.2780 are of the
same sign, we choose x0 as 1. Then for n = 0,
we have x1 = x0 ff(x(x0 )) or x1 = 1 2.178
= 0.653,
6.278
0
f (x1 ) = 0.46, and f (x1 ) = 3.7835. Then for n = 1,
0.46
we get x2 = x1 ff(x(x1 )) = 0.653 3.7835
= 0.5314.
1
Now f (x2 ) = 0.042, f (x2 ) = 3.11213. For n = 2,
0.042
we get x3 = x2 ff(x(x2 )) or x3 = 0.5314 3.11213
=
2
0.518. Then f (x3 ) = 0.000738, f (x3 ) = 3.0433.
So for n = 3, x4 = x3 ff(x(x3 )) = 0.518 0.000738
=
3.0433
3
0.5177. Since f (0.5177) = 0.000174, we take
x4 = 0.5177 as an approximate root.
Example 14: Find a real root of x + log10 x 2 =
0 using NewtonRaphson method.
[JNTU 2007, Set No. 3]
Solution: Here, f (x) = x + log10 x 2, f (x) =
1 + x1 log e10 = 1 + 2.3026
.
x
Since f (1) = 1 < 0 and f (2) = 0.3010 > 0, a
root lies between 1 and 2. Again at 1.5, we have
f (1.5) = 0.324, root lies between 1.5 and 2.
Choose x0 = 1.8. Then for n = 0, we have the rst
0.0553
approximation as, x1 = x0 ff(x(x0 )) = 1.8 2.2792
,
0
x1 = 1.7757. Then x2 = x1 ff(x(x1 )) = 1.7757
0
0.0251
= 1.7648.
2.2967
Now x3 = x2 ff(x(x2 )) = 1.7648 0.0015
= 1.7598,
2.3047
2
0.0053
Again x4 = x3 ff(x(x3 )) = 1.7598 2.3084
= 1.7575
3
f (x4 ) = f (1.7575) = 0.002395, we may take x4 =
1.7575 as an approximate root.
Example 15: Find a real root of x tan x + 1 = 0
using NewtonRaphson method.
[JNTU 2006, Aug. Supply. Set 2]
Solution:
Rewriting x sin x + cos x = 0, so
f (x) = x sin x + cos x, f (x) = x cos x. Then NR
iteration formula is
xn+1 = xn
with
n
0
1
2
3
xn sin xn + cos xn
xn cos xn
x0 = , the successive iterations
xn
f (xn )
xn+1
3.1416 1.0
2.8233
2.8233 0.0662 2.7986
2.7986 0.0006 2.7984
2.7984
0.0
2.7984
are:
Chap-17
B.V.Ramana
17.6
August 30, 2007
10:15
MATHEMATICAL METHODS
... Approximate (exact) root is 2.7984
Example 16: Find f (2.5) using newtons forward
formula from the following table.
x: 0 1 2
3
4
5
6
y: 0 1 16 81 256 625 1296
[JNTU: 2006, Set No. 1]
Solution: Here, h = 1, x0 , y0 = 0, x = 2.5, q =
xx0
= 2.50
... q = 2.5.
h
1
Finite difference table
16
15
81
65
256
625
6 1296
Dy
1
175
369
671
D 2y
14
50
110
194
302
D3y
36
60
84
108
D 4y
D5y
24
D6y
24
24
Here y0 = 1, 2 y0 = 14, 3 y0 = 36, 4 y0 = 24,
5 y0 = 0, 6 y0 = 0.
Using Newtons forward formula y = y0 + qy0 +
q(q1) 2
y0
2!
+ q(q1)(q2)
3 y0
3!
+ q(q1)(q2)(q3)
4 y0
4!
+ 0 + 0.
Substituting the values, we get at x = 2.5,
(2.5)(2.5 1)
y(2.5) = 0 + 2.5(1) +
(14)
2!
(2.5)(2.5 1)(2.5 2)
+
(36) +
3!
+
(24) + 0 + 0
4!
= 2.5 + 26.25 + 11.25 0.9375 = 39.0625.
Example 17:
Given that sin 45 = 0.7077,
sin 50 = 0.766, sin 55 = 0.8192, sin 60 = 0.866,
nd sin 52 using Newtons forward differences
formula.
[JNTU 2006, Set No. 2]
Solution: Here, h = 5 , x = 52 , x0 = 45 , y0 =
= 5245
= 75
0.7077, q = x45
2
5
Finite difference table
D2y
y = sin x Dy
x0
y0
45
0.7077
50
0.766
55
0.8192
60
0.866
Dy0
0.0583
0.0532
0.0468
D3y
D2y0
0.0051
0.0064
D3y0
0.0013
Using Newtons forward differences formula
y(52 ) = 0.7077 + 75 (0.0583)
+ 75 75 1 0.0051)
7 7
7
+ 5 5 1 5 2 (0.0013) = 0.7869008
Example 18: Find y(1.6) using Newtons forward
differences formula from the table
x
y
1
3.49
1.4
4.82
1.8
5.96
2.2
6.5
[JNTU 2006, Set No. 3]
Solution: Here, h = 0.4, x = 1.6, x0 = 1, y0 =
.6
0
3.49, q = xx
= 1.61
= .4
= 23
h
0.4
Dy
3.49 = y0
Dy0
1.4
4.82
1.8
5.96
2.2
6.50
1.33
1.14
0.54
D2y
D2y0
0.19
0.60
D3y
D3y0
0.41
Using Newtons forward differences
formula
y(1.6) = 3.49 + 23 (1.33) + 23 23 1 (0.19)
+ 23 23 1 23 2 (0.4) = 5.4925.
Example 19: Show that fi2 = (fi + fi+1 )fi
[JNTU 2006, Set No. 4] (Question Corrected)
Solution: We know that fi = fi+1 fi
2
Then fi2 = fi+1
fi2 = (fi+1 + fi )(fi+1 fi )
= (fi+1 + fi )fi
Example 20: Find the unique polynomial P (x)
of degree 2 or less such that P (1) = 1, P (3) = 27,
Chap-17
B.V.Ramana
August 30, 2007
10:15
NUMERICAL ANALYSIS
P (4) = 64 using Lagrange interpolation formula.
[JNTU Aug. 2006, Supply. S 2004]
f (3) = 1
Solution: x0 = 1, x1 = 3, x2 = 4, y(x) = P (x),
so y0 = P (x0 = 1) = P (1) = 1, y1 = P (x1 = 3) =
27, y3 = P (3) = 27. By Lagranges interpolation
formula, we get
y(x) = P (x) =
(x x1 )(x x2 )
y0 +
(x0 x1 )(x0 x2 )
(x x0 )(x x2 )
y1 +
+
(x1 x0 )(x1 x2 )
+
(x x0 )(x x1 )
y2
(x2 x0 )(x2 x1 )
(x 1)(x 4)
(x 3)(x 4)
1+
27+
(1 3)(1 4)
(3 1)(3 4)
17.7
(3 1)(3 2)(3 4)(3 5)(3 6)
+
(0 1)(0 2)(0 4)(0 5)(0 6)
+ 14
(3 0)(3 2)(3 4)(3 5)(3 6)
+
(1 0)(1 2)(1 4)(1 5)(1 6)
+ 15
(3 0)(3 1)(3 4)(3 5)(3 6)
+
(2 0)(2 1)(2 4)(2 5)(2 6)
+5
(3 0)(3 1)(3 2)(3 5)(3 6)
+
(4 0)(4 1)(4 2)(4 5)(4 6)
+6
(3 0)(3 1)(3 2)(3 4)(3 6)
+
(5 0)(5 1)(5 2)(5 4)(5 6)
+19
(3 0)(3 1)(3 2)(3 4)(3 5)
(6 0)(6 1)(6 2)(6 4)(6 5)
18
36
36
12
14 +
15 +
5
240 60
48
48
18
12
6+
19
60
240
= 0.05 4.2 + 11.25 + 3.75 1.8 + 0.95 = 10.
+
(x 1)(x 3)
64
(4 1)(4 3)
1
[48x 2 114x + 72] = 8x 2 19x + 12
6
Example 21: Using Lagranges formula, calculate
f(3) from the following table.
x
0 1
2 4 5 6
f (x) 1 14 15 5 6 19
Example 22: If yx is the value of y at x for which
the fth differences are constant and y1 + y7 =
784, y2 + y6 = 686, y3 + y5 = 1088, then nd y4 .
[JNTU 2007, Set No. 3]
Solution: Since 5th order differences are constant,
we have 5 yn = constant for any n. Then all the
higher order differences are zero i.e., 6 yn = 0 and
7 yn = 0, etc. for any n. We know that
0 = 6 y0 = y6 6y5 + 15y4 20y3 + 15y2
Solution: Here, x0 = 0, x1 = 1, x2 = 2, x3 = 4,
x4 = 5, x5 = 6 and f (x0 ) = 1, f (x1 ) = 14, f (x2 ) =
15, f (x3 ) = 15, f (x4 ) = 5, f (x5 ) = 6, f (x6 ) = 19.
From Lagranges interpolation formula, we get
5
f (x) =
5
i=0
f (xi )
j =0
j
=i
5
j =0
i
=1
5
i=0
21y2 + 7y1 y0
(x xj )
since n yk = yn+k n c1 , yn+k1 + n c2 yn+k2 +
. . . + (1)n yk .
Adding, we get
(xi xj )
5
(x xj )
f (xi )
(x
i xj )
j =0
j
=i
6y1 + y0
7
Given, y1 + y7 = 784, y2 + y6 = 686 and y3 +
y5 = 1088
Therefore 784 6(686) + 15(1088) 20y4 = 0
or y4 = 721.
Chap-17
B.V.Ramana
17.8
August 30, 2007
10:15
MATHEMATICAL METHODS
following data.
Solution: Finite differences table is
x
x
0.1 0.3 0.5 0.7
0.9
1.1
1.3 20
F (x): 0.003 0.067 0.148 0.248 0.370 0.518 0.697 25
354
30
291
35
260
40
231
45
204
and nd F (0.6) using a cubic that ts at x = 0.3, 0.5,
0.7 and 0.9 using Newtons forward formula.
[JNTU 2007, Set No. 2]
0.1 0.003
0.3 0.067
0.5 0.148
0.7 0.248
0.9 0.370
1.1 0.518
1.3 0.697
0.064
0.081
0.100
0.122
0.148
0.179
0.017
0.019
0.022
0.026
0.031
0.002
0.003
0.004
0.005
0.001
0.001
0.001
0
0
q=
Example 24: Find f (22) from the following table
x0 using Gauss forward formula
x
f (x)
20
354
25
332
30
35
40
45
291 260 231 204
[JNTU 2007, Set No. 4]
332
Df
22
41
29
29
27
D2f
19
12
0
D 3f
D4f
31
43
12
14
Gausss forward (rst GI) interpolation formula (see
(1) on page 17.19)
f (x) = f0 + qf0 + q(q1)
2 f1
2
(q+1)q(q1) 3
+
f1 + . . .
6
0
Here, x = 22, x0 = 20, h = 5, q = xx
= 2220
=
h
5
2
0.4, f0 = 354, f0 = 41, f1 = 19,
3 f1 = 31
f (22) = 354 + (0.4)(41) + (0.4)(0.6)
(19)
2
(1.4)(0.4)(0.6)
(31)
+
6
= 354 16.4 + 2.28 1.736 = 338.144
Example 25: Using Gauss backward differences
formula nd y(8) from the following table.
x;
0 5 10 15 20 25
y(x): 7 11 14 18 24 32
0.6 0.3
x x0
=
= 1.5
h
0.2
with x0 = 0.3, x1 = 0.5, x2 = 0.7, x3 = 0.9 we can
t a polynomial of degree 3.
By Newtons forward differences formula, with
y0 = 0.067, y0 = 0.081, 2 y0 = 0.019, 3 y0 =
0.003, 4 y0 = 0.001, 5 y0 = 0, we get, F (0.6) =
(0.019)+
0.067 + (1.5)(0.081) + (1.5)(1.51)
2!
(1.5)(1.51)(1.52)
+
(0.003)+
3!
+ (1.5)(1.51)(1.52)(1.53)
(0.001)
4!
= 0.067 + 0.1215 + 0.0071 0.0011 + 0.000002
F (0.6) = 0.194502
f(x)
[JNTU 2007, Set No. 1]
0
, x = 8, x0 = 10, h = 5
Solution: Here, q = xx
h
810
so q = 5 = 0.4. The Gauss backward differences formula (Gauss second G2: see (2) on Page
17.19).
y(x) =
1) (q1)
3!
3 y2 +
y0 + qy1 +
q(q+1)
2!
(q+2)(q+1)q(q1) 4
y2
4!
11
10
14
15
18
20
24
25
32
D
4
3
4
6
8
D2
1
1
2
2
2 y1 + q(q +
+ ...
D3
2
1
0
D4
1
1
Chap-17
B.V.Ramana
August 30, 2007
10:15
NUMERICAL ANALYSIS
Using the Gauss backward differences formula with
y0 = 14, y1 = 3, 2 y1 = 1, 2 y2 = 2, we get
y(8) = 14 + (0.4) 3 + (0.4)(0.6) 1
EXERCISE
+ (0.6)(0.4)(1.4)
(2)
6
y(8) = 12.792.
Example 26: If f (x) = u(x)v(x) show that
f [x0 , x1 ] = u[x0 ] v[x0 , x1 ] + u[x0 , x1 ] v[x1 ]
[JNTU 2006, Aug. Supply. Set No. 4]
Solution: From the denition of rst order divided
differences, we have
0]
0]
u[x0 , x1 ] = u[xx1 ]u[x
, v[x0 , x1 ] = v[xx1 ]v[x
,
x
x
1
[x0 ]
and f [x0 , x1 ] = f [xx1 ]f
1 x0
Thus RHS
= u[x0 ] v[x0, x1 ] + u[x0 , x1 ] v[x1 ]
0]
0]
+ u[xx1 ]u[x
v[x1 ]v[x1 ]
= u[x0 ] v[xx1 ]v[x
x
x
1
1
x1 x0
+[u[x1 ] v[x1 ] u[x0 ] v[x1 ]}
{u[x1 ]v[x1 ] u[x0 ] v[x0 ]}
1
x1 x0
1
{f [x1 ]
x1 x0
f [x0 ]} = f [x0 , x1 ].
Example 27: Find the unique polynomial P (x)
of degree 2 or less such that P (1) = 1, P (3) = 27,
P (4) = 64 using Newton divided differences formula.
[JNTU Aug. 2006, Supply. Set No. 2]
Solution: Divided differences table
x
P (x)
x0 = 1
P0 = 1
Divided differences
of order
1
2
[x0 , x1 ] = 271
31
= 13
x1 = 3
P1 = 27
[x1 , x2 ] = 6427
43
= 37
x2 = 4
P2 = 64
1. Show that
n1
k=0
2 fk = fn f0
[JNTU 2003]
Hint: fk = fk+1 fk
n1
2
fk = 2 f0 + 2 f1 + 2 f2 + . . . + 2 fn1
2
k=0
= (f1 f0 ) + (f2 f1 ) +
. . . (fn fn1 )
= fn f0 .
2. Consider the following data for g(x) =
{u[x0 ]v[x1 ] u[x0 ] v[x0 ]
[x0 , x1 , x2 ]
= 3713
=8
41
17.9
sin x
.
x2
x:
0.1
0.2
0.3
0.4
0.5
g(x): 9.9833 4.9696 3.2836 2.4339 1.9177
Calculate g(0.25) accurately using Newtons
forward method of interpolation.
[JNTU: Aug. 2003]
Hint: q = 0.250.1
= 1.5, h = 0.1, x = 0.25,
0.1
x0 = 0.1, g(0.25) = 9.9833 + 1.5(5.0137) +
(1.5)(0.5)
3.3277
2
+ 1.50.5(0.5)
(2.4919)
32
+ 1.50.5(0.5)(1.5)
1.9886
432
Ans. g(0.25) = 3.9134.
3. For x = 0, 1, 2, 4, 5; f (x) = 1, 14, 15, 5, 6. Find
f (3) using forward differences table.
[JNTU 2004]
Hint: x = 3, h = 1, q =
f (3) = 1 + 13(3)+
30
1
=3
3(2)
3(2)(1)
(12)+
2
321
(1)
Here, x0 = 1, P0 = 1, [x0 , x1 ] = 13, [x0 , x1 , x2 ] = 8
Using Newtons divided differences formula
Ans. 5.
P (x) = P0 + (x x0 )[x0 , x1 ] + (x x0 )(x x1 )
4. Find y(25) given that y20 = 24, y24 = 32, y28 =
[x0 , x1 , x2 ]
= 1 + (x 1) 13+(x 1)(x 3)8
35, y32 = 40 using Gauss forward differences
= 8x 2 19x + 12
formula. [JNTU Aug. 2006, Supply. Set No. 1]
Chap-17
B.V.Ramana
17.10
August 30, 2007
10:15
MATHEMATICAL METHODS
Hint: x0 = 28, x = 25, h = 4, q =
y = f(x)
x 2 = 20 y 2 = 24
x1 = 24
y1 = 32
x0 = 2.8
y0 = 35
x1 = 32
y1 = 40
2528
4
= 43
[JNTU 2002]
Hint: x0 = 1, x1 = 7, x2 = 15, x = 10,
f0 = 168, f1 = 192, f2 = 336
f (10) = (107)(1015)
168 + (101)(1015)
(17)(115)
(71)(75)
192+
+ (101)(107)
336
(151)(157)
D y D2y D3y
8
Ans. 231.005
3 3
4 4 1
3
y(25) = 35 +
3+
2
4
2!
3 3
4 4 1 43 + 1
+7
3!
7. Compute f (3.5) from the following data
x:
1 2 3 4
f (x): 1 2 9 28
Using Lagranges, interpolation of 2nd and
3rd-order degree polynomials.
Hint: x0 = 1, x1 = 2, x2 = 3, y0 = 1, y1 = 2,
y2 = 9, x = 3.5
(a) 2nd order f (3.5) = 16
(b) 3rd order: x0 = 1, x1 = 2, x2 = 3, x3 =
4, y0 = 1, y1 = 2, y2 = 9, y3 = 28, x = 3.5,
f (3.5) = 16.625.
Ans. y(25) = 34.4453.
5. Use Gauss backward differences formula to nd
y(8) from the following table.
x
y
0
7
5
11
10
14
15
18
20
24
25
32
[JNTU Aug. 2006, Supply, Set No. 4]
Solution: x0 = 10, y0 = 14, h = 5, x = 8, so q =
xx0
= 810
= 25 = 0.4
h
5
y(x)
x2 = 0 y 2 = 7
x1 = 5 y1 = 11
x0 = 10
y0 = 14
x1 = 15
y1 = 18
x1 = 20
y2 = 24
x1 = 25
y3 = 32
4
3
4
6
8
1
1
2
2
2
1
0
1
1
8. Find a real root of x 3 x 1 = 0 by bisection
method.
Hint: root in (1, 2), x2 = 1.25, x3 = 1.375, x4 =
1.3125, x5 = 1.34375, x6 = 1.328125
Ans. 1.328125.
9. Using bisection method, nd a real root of x +
tan x 1 = 0.
Hint: f (0) = 1, f (0.5) = 0.0463, root (0, 0.5).
Ans. 0.4795.
10. Find out an approximate root of x sin x = 1 using
bisection method.
(0.4)(0.6) 1
2
Ans.
(0.4)(0.4 1)(0.4 + 1)2
2
+
6
11.
(0.4)(0.4 1) + (0.4 + 1)(0.4 + 2)
+
(1)
24
Ans.
y(8) = 14 + (0.4) 3 +
Ans. y(8) = 12.7696.
6. Evaluate f (10), given f (x) = 168, 192, 336 at
x = 1, 7, 15 respectively. Use Lagranges interpolation.
Hint: f (0) = 1, f (1) = 0.158529, f (1.5) =
0.496, root in (1, 1.5); x3 = 1.125, x2 = 1.25,
x4 = 1.0625, x5 = 1.09375, x6 = 1.109375,
x7 = 1.1171875, x8 = 1.11328125.
1.11328125.
Use bisection method to nd a real root of 667.38
(1 e0.146843x ) = 40 x.
14.5.
Hint: f (12) = 6.067, f (16) = 2.269, root in
(12, 16).
Chap-17
B.V.Ramana
August 30, 2007
10:15
NUMERICAL ANALYSIS
17.11
which lies between 3.5 and 4 by RegulaFalsi 17. Find a real root of tan1 x x = 1.
method.
[JNTU 2006, Set No. 3] Ans. 2.1323.
Ans. 3.789.
Hint: g(x) = 1 + tan1 x, xn+1 = 1 + tan1 xn ,
Hint: x0 = 3.5, x1 = 4, x2 = 3.7888,
f (2) = 0.10, f (3) = 0.75; choose x0 = 1,
f (x2 ) = 0.0009, f (x1 ) = 0.3979,
x1 = 1.7854, x2 = 2.0602, x3 = 2.1189, x4 =
root (3.7888, 4), x0 = 3.7888, x1 = 4,
2.1318, x5 = 2.1322, x6 = 2.1323.
x3 = 3.7893
18. Using xed point
iteration method, evaluate apIteration Method
proximately (a) 12 (b) 1 .
12
Ans. (a) 3.46425 (b) 0.2887.
Ans. 4.3311
Hint: (a) 12 = x, x 2 = 12 or x = 12
= g(x).
x
Hint:
f (4) = 20 < 0,
f (5) = 50 > 0
Since 9 and 16
are
nearest
numbers
to
12 with
root in (4, 5), xn+1 = x10+1 = g(x),
perfect
squares
9
=
3,
16
=
4.
Take
x0 =
n
5
=
3.4285,
x
=
3.5,
x
3.5,
x
=
3.4285,
x4 =
1
2
3
|g (x)| = (x+1)3/2 < 1 in (4, 5). Take x0 = 4.2
=
3.46425.
3.5, approximate root = 3.4285+3.5
2
1
1
x1 = 4.38529, x2 = 4.30919, x3 = 4.33996,
(b) x = 112 , x 2 = 12
or x = 12x
= g(x)
x4 = 4.32744, x5 = 4.33252, x6 = 4.33046,
1
choose
x0 = 3.5 = 0.285,
x1 = 0.2924,
x7 = 4.33129, x8 = 4.33096, x9 = 4.33109,
x2 = 0.285, x3 = 0.2924, approximate root
x10 = 4.33104, x11 = 4.33106, x12 = 4.33105,
= 0.285+0.2924
= 0.2887.
2
x13 = 4.33105
19. Find a real root of ex sin x = 1 using Newton
14. Find the real root of the Equation
x
x5
x7
x9
x 11
x3
+
+ ...
3
10 42 216 1320
= 0.4431135.
Ans. 0.4769.
3
11
x
x
+ 1320
Hint:
x = x3 x10 + x42 216
+
0.4431135
x1 = 0.4699,
Choose
x0 = 0.44,
x2 = 0.4755, x3 = 0.47664, x4 = 0.47686,
x5 = 0.47690.
15. By iteration method, nd a real root of sin x =
10(x 1).
Ans. x = 1.088.
Hint: f (1) > 0 and f (2) < 0, root in (1, 2)
choose x0 = 1, x1 = 1.084, x2 = 1.088, x3 =
1.088
16. Using method of successive approximation nd
a root of x 3 3x + 1 = 0.
Ans. 0.347.
Hint: f (0) = 1 > 0, f (1) = 1 < 0, xn+1 =
xn3 +1
.
3
Raphsons method.
[JNTU Aug. 2006, Supply Set No. 3,
R059010202]
Hint:
f (x) = ex sin x 1,
f (x) =
x
f (2) = 5.7188,
e (sin x + cos x).
f ( ) = 1; f (2) = 3.6439.
Since f (2) and f (2) have the same sign
choose x0 = 2. NR formula is
xn+1 = xn
exn sin xn 1
,
exn (sin xn + cos xn )
= 0.4305, f (x1 ) = 0.358,
x1 = 2 5.7188
3.6439
f (x1 ) = 2.039, x2 = 0.606, f (x2 ) = 0.044,
f (x2 ) = 2.5507, x3 = 0.5887,
f (x3 ) = 0.0004178. Approximate root x3 =
0.5887. | 13,036 | 25,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2020-10 | latest | en | 0.864539 |
https://www.archerytalk.com/threads/archery-maths.29840/#post-224244 | 1,669,505,322,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00619.warc.gz | 715,238,537 | 23,935 | # Archery Maths
1118 3
Any one out there have any archery calculations or formula for calculation anything to do with archery
(Arrow weight, FOC, Draw length, Bow energy, Arrow trajectory, etc).
I would like to compile a full collection of math formula to provide to members
So we could understand how and why it works.
Thanks
1 - 4 of 4 Posts
#### beezaur
·
##### Registered
Joined
·
2,425 Posts
The following is a formula that gives arrow FOC in percent, as a function of component weights and some standard measurements.
FOC = 50*(2*df*wf-l*(wb+wf-wi+wn-wp)+dn*(wb+wf+wi+wp+l*mu))/((dn+l)*(wb+wf+wi+wn+wp+l*mu)),
where wp is the weight of the point, wi is the weight of the point insert, mu is the weight per unit length of the bare shaft, l is the length of the bare shaft, wf is the total weight of the fletching, df is the distance from the nock end of the bare shaft to the center of mass of the fletching, wb is the weight of the nock bushing (or pin), wn is the weight of the nock, and dn is the distance between the string position in the nock and the beginning of the shaft proper.
This formula was developed using the principles of statics, namely the summation of moments about some arbitrary point. The general procedure is fully explained in any engineering statics text.
The formula is good for field points and other points that have their center of mass near the end of the shaft. FOC for broadheads would be slightly different.
The formula is most conveniently implemented using a spreadsheet or programmable calculator. A pretty decent (and free) spreadsheet is available at: www.openoffice.org.
Scott
PS: you can use any units (grams, grains, inches, mm, whatever) so long as you don't mix them. For example, arrow weight in grains per inch will give a bad FOC if used with point weight in grams of arrow length in mm.
#### clayking
·
##### Occam's Razor
Joined
·
11,194 Posts
There is an article in the current issue of "Archery Focus" on "cant" and why it works etc. Not sure it's a pure math equation, but it played a role.
I am sure that the makers of the archery software programs could fill your boots with plenty of them..........ck
#### PSI
·
##### Registered
Joined
·
123 Posts
The book...
The Physical Laws of Archery, by T.L. Liston
...has what you're looking for. Many of the formulas are for arrow flight without friction or drag, but can be modified to incorporate that if you so desire. There is also a pretty god explanation on force draw curves and how bows store energy.
1 - 4 of 4 Posts | 632 | 2,545 | {"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} | 3.359375 | 3 | CC-MAIN-2022-49 | latest | en | 0.923866 |
https://crypto.stackexchange.com/questions/37055/linear-cryptanalysis-how-are-the-subkey-bits-removed-from-the-linear-approximat | 1,716,718,083,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00330.warc.gz | 149,575,300 | 39,076 | # linear cryptanalysis: how are the subkey bits removed from the linear approximation?
I'm currently reading the tutorial on linear/differential crypta here: https://www.engr.mun.ca/~howard/PAPERS/ldc_tutorial.pdf
End of page 14/Beginning of page 15: all the bits from the subkeys are summed up to a sum(k) variable. It is then written as 0 in the equation (5) and the probability stays the same.
I don't understand this step, why is the probability not changing if sum(k) is zero, and why is it 1-p if sum(k) is 1?
So after re-reading a few times it becomes clear. The key bits are static, so they won't change the probabilities.
Now suppose that $\sum_K = 0$, then the equation in (5) will be true, with probability $\frac{15}{32}$ since it is the same equation.
Suppose now that $\sum_K = 1$, the equation in (5) is now the inverse, and it will thus happen with probability $1-(15/32).$ | 237 | 895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-22 | latest | en | 0.94324 |
https://www.education.com/resources/fourth-grade/number-sense/?page=4 | 1,611,690,707,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803308.89/warc/CC-MAIN-20210126170854-20210126200854-00038.warc.gz | 762,159,358 | 29,658 | # 4th Grade Number Sense Resources
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Converting Decimals to Fractions: Rowing Times
Worksheet
Converting Decimals to Fractions: Rowing Times
Practice converting decimals to fractions with this rowing-themed worksheet!
Math
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Input Output Math Tables
Worksheet
Input Output Math Tables
Find the hidden rule in these puzzling math tables! Compare your "in" number to the corresponding "out" number to find the pattern in each table.
Math
Worksheet
Number Challenge
Worksheet
Number Challenge
Unscramble the numbers to answer each clue. Strengthen those place value and logic skills as you mix the numbers around according to the clues given.
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Dizzy Over Decimals: Rounding #1
Worksheet
Dizzy Over Decimals: Rounding #1
Does your child need a little extra practice with decimal place values? This worksheet is the way to go.
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Number Sense Review Part 1
Worksheet
Number Sense Review Part 1
Let your students show what they know about place value, as they round, determine the value of various digits, and build numbers. Gain insight into who is ready to take on the challenges in the thousands place and beyond!
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Worksheet
The Rhyme and Reason for Rounding
Lesson Plan
The Rhyme and Reason for Rounding
This lesson focuses on facilitating a rich classroom discussion around the purpose and methods of rounding numbers up to the thousands place. It can be taught on its own or as support to the lesson Rounding with Number Lines.
Math
Lesson Plan
Prime Number Practice
Worksheet
Prime Number Practice
Give your child prime number practice with this simple worksheet.
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Convert Decimals to Fractions 2
Exercise
Convert Decimals to Fractions 2
Students will no longer be intimidated by converting decimals to fractions after this simple exercise.
Math
Exercise
Decimal Fractions 1
Exercise
Decimal Fractions 1
Students will be able to convert from fractions to decimals, and back to fractions again with this exercise.
Math
Exercise
Place Value Challenge
Worksheet
Place Value Challenge
Use the numbers given and unscramble them to find the answers to the clues provided. Follow each clue to create new six-digit numbers. Practice logical…
Math
Worksheet
Rounding Multi-Digit Numbers to the Nearest 100,000
Exercise
Rounding Multi-Digit Numbers to the Nearest 100,000
With this exercise, students will be able to simplify large numbers by rounding to the nearest 100,000.
Math
Exercise
Rounding Around Town
Worksheet
Rounding Around Town
About how much flour does the town baker need? Your students will practice rounding to the thousands place as they solve real world problems for the people living in Approxiville.
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Rounding Multi-Digit Numbers to the Nearest 1,000
Exercise
Rounding Multi-Digit Numbers to the Nearest 1,000
Help students simplify math problems with the ability to round multi digit numbers to the nearest 1,000.
Math
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Place Value and Multiplicative Comparisons
Exercise
Place Value and Multiplicative Comparisons
Translate word problems into their numeric equivalent with this exercise that teaches both place value and multiplicative comparisons.
Math
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Parentheses First! Find the Missing Operation #1
Worksheet
Parentheses First! Find the Missing Operation #1
Work on your fourth grader's pre-algebra foundation with this worksheet that reinforces order of operations. He'll figure out which operation is missing!
Math
Worksheet
Lesson Plan
Boost your students' confidence in verbally sharing their math processes using grade-level vocabulary! This lesson on rounding decimals can be taught on its own or as support for the lesson Menu Math: Rounding Decimals.
Math
Lesson Plan
Math Review Part 5: Dazzling Decimals
Worksheet
Math Review Part 5: Dazzling Decimals
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Greatest Common Factor
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Greatest Common Factor
Teach your students to use factor rainbows to find the greatest common factor of two given numbers.
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Rounding Multi-Digit Numbers to the Nearest 10000
Exercise
Rounding Multi-Digit Numbers to the Nearest 10000
Students will find math so much easier after they learn they can round multi digit numbers to the nearest 10,000.
Math
Exercise
Converting Centimeters to Meters
Lesson Plan
Converting Centimeters to Meters
Get your students comfortable discussing their math thinking in converting centimeters to meters. This lesson may be used on its own or as support to the lesson Converting Metric Measurements to Decimals & Fractions.
Math
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Order of Operations Puzzle 2
Worksheet
Order of Operations Puzzle 2
Practice beginning algebra with an order of operations puzzle! Put in the missing parentheses to make each equation true.
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Numbers All A-Round
Lesson Plan
Numbers All A-Round
Teach your students to round to the hundred thousands place with this straightforward lesson.
Math
Lesson Plan
Round and Round We Go
Lesson Plan
Round and Round We Go
Do you feel like your students are on a merry-go-round when it comes to rounding? This lesson will give your student control of the “ride” and give them tools to use when rounding.
Math
Lesson Plan
Ordering Decimal Numbers
Exercise
Ordering Decimal Numbers
This easy and educational exercise from Education.com will teach students how to sort and order decimals. | 1,157 | 5,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-04 | latest | en | 0.848564 |
https://algebra1help.com/algebra1-com/converting-fractions/holt-pre-algebra-chapter-1.html | 1,632,384,635,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00489.warc.gz | 151,589,783 | 12,960 | Free Algebra Tutorials!
Home Point Arithmetic Operations with Numerical Fractions Multiplying a Polynomial by a Monomial Solving Linear Equation Solving Linear Equations Solving Inequalities Solving Compound Inequalities Solving Systems of Equations Using Substitution Simplifying Fractions 3 Factoring quadratics Special Products Writing Fractions as Percents Using Patterns to Multiply Two Binomials Adding and Subtracting Fractions Solving Linear Inequalities Adding Fractions Solving Systems of Equations - Exponential Functions Integer Exponents Example 6 Dividing Monomials Multiplication can Increase or Decrease a Number Graphing Horizontal Lines Simplification of Expressions Containing only Monomials Decimal Numbers Negative Numbers Factoring Subtracting Polynomials Adding and Subtracting Fractions Powers of i Multiplying and Dividing Fractions Simplifying Complex Fractions Finding the Coordinates of a Point Fractions and Decimals Rational Expressions Solving Equations by Factoring Slope of a Line Percent Introduced Reducing Rational Expressions to Lowest Terms The Hyperbola Standard Form for the Equation of a Line Multiplication by 75 Solving Quadratic Equations Using the Quadratic Formula Raising a Product to a Power Solving Equations with Log Terms on Each Side Monomial Factors Solving Inequalities with Fractions and Parentheses Division Property of Square and Cube Roots Multiplying Two Numbers Close to but less than 100 Solving Absolute Value Inequalities Equations of Circles Percents and Decimals Integral Exponents Linear Equations - Positive and Negative Slopes Multiplying Radicals Factoring Special Quadratic Polynomials Simplifying Rational Expressions Adding and Subtracting Unlike Fractions Graphuing Linear Inequalities Linear Functions Solving Quadratic Equations by Using the Quadratic Formula Adding and Subtracting Polynomials Adding and Subtracting Functions Basic Algebraic Operations and Simplification Simplifying Complex Fractions Axis of Symmetry and Vertices Factoring Polynomials with Four Terms Evaluation of Simple Formulas Graphing Systems of Equations Scientific Notation Lines and Equations Horizontal and Vertical Lines Solving Equations by Factoring Solving Systems of Linear Inequalities Adding and Subtracting Rational Expressions with Different Denominators Adding and Subtracting Fractions Solving Linear Equations Simple Trinomials as Products of Binomials Solving Nonlinear Equations by Factoring Solving System of Equations Exponential Functions Computing the Area of Circles The Standard Form of a Quadratic Equation The Discriminant Dividing Monomials Using the Quotient Rule Squaring a Difference Changing the Sign of an Exponent Adding Fractions Powers of Radical Expressions Steps for Solving Linear Equations Quadratic Expressions Complete Squares Fractions 1 Properties of Negative Exponents Factoring Perfect Square Trinomials Algebra Solving Quadratic Equations Using the Square Root Property Dividing Rational Expressions Quadratic Equations with Imaginary Solutions Factoring Trinomials Using Patterns
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This PowerPoint presentation is an introduction to fractions for first, second or third graders. The presentation explains parts of a set and parts of a whole. It includes vocabulary, such as numerator and denominator, and examples for practice. Students can use mini whiteboards to write the frac
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Introduce your students to FRACTIONS with this interactive SmartBoard Lesson! The file includes a lesson plan, warm up activities (review skills), interactive pages to learn about 1/2, 1/3, 1/4 and 1/5, a website link, a fraction circle activity /project and a student worksheet. Your first or seco
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Here is an into into fractions cut and paste that lets the kiddos see what the fraction are actually doing! I have mine color each section a different color when after they paste.
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Fractions can be fun to review when you use Intro to Fractions Game Show! This game is perfect to review/teach the concepts needed prior to beginning an entire unit on fractions. Great for test prep!☆ Put students in up to 6 groups to play this Jeopardy style game! Review days can be dull and kids
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If your students are like mine, they need a lot of practice with equivalent fractions. For some students this is one of the hardest topics. Before moving on to more complex tasks students need to have a strong foundation of how to generate equivalent fractions. This set will provide your students
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Use this cute reader to introduce fractions! It is simple and fun for students to put together. This beginning book introduces whole and one half. It can be used as a review later on! ***This is ONE book of a set of 3! Check out My Fractions Books Intro to Fractions Set of 3 Comes in color or BW
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My Fractions Books set of 3! ************** Use these 3 colorful, fun mini-books to introduce fractions to your students! Pages 2, 3 & 4 Students will learn how to write each fraction in words and numbers. Pages 5, 6 & 7 students need to color in the pictures according to the sentence on eac
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These digital fraction activities will give students understanding and practice of fractions, parts of a whole, parts of a set, and fraction representations. These resources are 100% online and can be used on Google Classroom and Google Drive. This resource also includes an answer key. Save money
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This flipchart is a fun and authentic way to build conceptual understanding of partitioning into equal/fair shares. This lesson includes a story so students can connect their prior knowledge, two open-ended tasks, teacher questions, and follow ups. This past week, I used the flipchart and every si
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The ideal way to teach and learn basic fraction concepts and vocabulary (numerator, denominator, whole number, proper, improper, mixed number). It provides a good amount of guided pratice with visual understanding of what proper, improper, and mixed number fractions represent. It is a good way for
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Download the preview to see what makes our PowerPoints unique! This colorful, animated PowerPoint features the humorous interaction between Gordo (student) and Laya (teacher). It can be used in the classroom, home, or computer lab to give students an introduction to fractions. The characters of
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This smart board activity will introduce fractions to the young student. Using the visual image of a pizza the student will be able to see the pizza being cut up into pieces or fractions. Touching the white board screen the (circle)shape will turn into a pizza pie! Have the students take turns comin
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My Fractions Book Part 2! ***This is ONE book of a set of 3! Check out My Fractions Books Intro to Fractions Set of 3 Use this cute reader to introduce fractions! It is simple and fun for students to put together. This beginning book introduces fourths (one-fourth & three-fourths). It can be
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This PDF can be used as-is or imported as a smart board lesson. Use the PDF as a guide to introduce major concepts of fractions, such as: -Fractions are equal parts of a whole -As the denominator gets larger, the parts get smaller -We use fractions every day; sharing evenly is a great example. Desi
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Introduction to Fractions. Help your students visualize and learn about fractions using interactive notebook pages all Common Core aligned! These guides have changed my classroom! These lessons all correspond with enVision Math Unit 12. Teachers that don't use enVision will also enjoy these interact
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This game is Fractions JEOPARDY! The categories are vocabulary (involving fractions), How many parts?, Fraction Shaded, word problems involving fractions, and mixed (a little of everything!) I hope you enjoy this product!!!
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This packet introduces fractions through a fun theme. It starts at the very beginning of understanding fractions and goes through to more complicated concepts. Understanding Fractions Naming Fractions Equivalent Fractions Word Problems Matching Card Game Quiz ANSWER KEYS PROVIDED
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This is a great PPT to help kids understand what a fraction is, and to remember which part is the numerator, and which part is the denominator. It is all set up using the SIOP method!
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This packet introduces fractions through a fun theme. It starts at the very beginning of understanding fractions and goes through to more complicated concepts. Understanding Fractions Naming Fractions Equivalent Fractions Word Problems Matching Card Game Quiz ANSWER KEYS PROVIDED
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This is an easy fraction activity or assessment to use after you have introduced fractions. Students are given a fraction and they must find the two models that equal it. There are three pages with three fractions and six models on each page. Fractions include halves, thirds, fourths, fifths, sixt
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My Fractions Book Part 3! ***This is ONE book of a set of 3! Check out My Fractions Books Intro to Fractions Set of 3 Use this cute reader to introduce fractions! It is simple and fun for students to put together. This beginning book introduces thirds (one-third & two-thirds). It can be used
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Introduction to fractions quiz or worksheet. Covers key vocabulary and beginning concepts. Vocabulary matching section, select numerator/denominator section, and shading in fraction section.
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This jeopardy game has 4 categories: Equal parts (students must identify which shape is divided into equal parts), Part of Whole (students must identify what fraction of a shape is colored in), Part of a Set (student must figure out what fraction of a set has a certain characteristic, and Surprise M
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Are you looking for a intro fraction lesson that introduces numerators and denominators? Look no further! Save your valuable time because you found it! *All materials needed for this mini-lesson are INCLUDED except dry erase boards and markers (optional). -----------------------------------------
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Sign Up | 2,168 | 8,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.862653 |
https://www.physicsforums.com/threads/where-is-the-right-triangle-in-sides-a-b-c-for-a-hyperbola.754483/ | 1,600,939,427,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00279.warc.gz | 912,153,725 | 17,128 | # Where is the right triangle in sides a, b, c, for a hyperbola?
Homework Helper
Gold Member
An ellipse has some model standard form values, a, b, and c which are easily enough to identify from the graph and parts of the graph related to the ellipse's graph. Seeing the right triangle relating a, b, and c, is easy enough. The Pythagorean Theorem is used to relate these three values.
I have been hoping to find and SEE the values of a, b, and c in their right triangle for a hyperbola, but I just cannot find this triangle. I have looked in a couple of College Algebra textbooks, and the most that is done is to assign a substitution as part of the derivation of the equation for the hyperbola. That just does not help to show where to find the right triangle. $$a^2+b^2=c^2$$ comes simply from the part of $$c^2-a^2$$
but no picture to support it.
## Answers and Replies
SteamKing
Staff Emeritus
Homework Helper
An ellipse has some model standard form values, a, b, and c which are easily enough to identify from the graph and parts of the graph related to the ellipse's graph. Seeing the right triangle relating a, b, and c, is easy enough. The Pythagorean Theorem is used to relate these three values.
I have been hoping to find and SEE the values of a, b, and c in their right triangle for a hyperbola, but I just cannot find this triangle. I have looked in a couple of College Algebra textbooks, and the most that is done is to assign a substitution as part of the derivation of the equation for the hyperbola. That just does not help to show where to find the right triangle. $$a^2+b^2=c^2$$ comes simply from the part of $$c^2-a^2$$
but no picture to support it.
The right triangle is located between the center of the hyperbola, the vertex, and the point above the vertex lying on the asymptote.
https://mysite.du.edu/~jcalvert/math/hyperb.htm
If we have a hyperbola whose center C is at the origin, the vertices V1 and V2 will be located at the points ($\pm$a, 0), while the asymptotes have the equations y = $\pm$(b/a)x. When x = a, at the positive vertex V1, y = b, and the relation
c$^{2}$ = a$^{2}$ + b$^{2}$ is also satisfied.
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Hi SteamKing,
I looked at and read some of the article. Nice, but a little difficult to follow. I still do not see/find a clear way to understand thinking from a and c, to the meaning and value of b. I can well accept that the b value and placement is realistic and meaningful; but I have trouble thinking through a good way to reach b from the hyperbola and a and c.
I have the right feeling that a rectangle must be associated with a hyperbola because this would correspond to what occurs in an ellipse. The ellipse has a, b, an c, easy to see. Then, one could expect to find somewhere, a meaningful rectangle(in a hyperbola). The rotation (for the hyperbola) of the segment from center to either focus (the one on the right) then will meet the asymtpote, but then this seems strange to understand. I become lost at that. I have no way to think why to make this rotation. The asymptote and slope of the asymptote then seems to involve b, but then the b is what I am trying to understand. I could go on and just use b on faith and could answer some common academic questions about hyperbolas at the "Algebra 2" and "College Algebra" level, but I was just hoping to get inside the meaning and not need to use just faith. I feel like this lack of understanding about a hyperbola will be an obstacle to leaning conic sections and more about coordinate geometry and much more about Mathematics.
Last edited:
SteamKing
Staff Emeritus
Homework Helper
I don't know why you are dragging 'faith' into this situation.
As for an ellipse, if a is the length of the semi-major axis and b is the length of the semi-minor axis, then c is the distance between these two points. It's not clear why this is significant to you or to the construction of an arbitrary ellipse.
The equation of the asymptotes to the hyperbola are as stated. They are straight lines. I don't understand why this is difficult to grasp.
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https://www.auditexcel.co.za/video-tutorial/excel-goal-seek/ | 1,726,310,792,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00001.warc.gz | 609,963,792 | 14,912 | Excel Goal Seek
How to use the Excel Goal Seek function to find what input value will give you a desired output e.g. what price do I sell a product at to achieve a gross profit of 1 million.
Excel Goal Seek 2013
In Excel it is sometimes necessary to try and determine how we can achieve a certain number by making changes to some underlying inputs. This is where the Excel goal seek function comes into play.
So for example if we run this business and we need to achieve a gross profit of 250 – what would the sales price have to go to to achieve that? So you could do it manually – if I click in there and go R11.25 – OK it has moved up but not enough. So we could do it manually or we could do it automatically. To achieve that we’ll click in this gross profit cell – under data – you’ll see you have got an option called “What If Analysis” – there is something called “Goalseek”.
When I click on it – what it is asking for is “where is the cell we are going to set?” – so I am just going to click over here – and say that’s the cell. Excel now wants to know what value do you want that cell to become – so I am going to put 250 and then it needs to know what is it allowed to change in this spreadsheet – either on this sheet or on another sheet to achieve that. So I am going to say you can change the selling price. When I say OK – you’ll see it runs through and says I think I found a solution – so I am going to keep it and say OK.
When it has determined that this number here will give me the answer I want. Let’s say that we take this to our Manager and they say “No ways – we can never sell this product for that much” – the most we can do – I don’t know – say R11.50 for example. We haven’t achieved our budget but we still have to reach 250. So we can run another goal seek – data – whatif analysis – goalseek –change that to 250 and this time I am going to play with the Cost of Sales – when I say OK- it runs through, finds an answer and in order to get to 250 we will have to drop the Cost of Sales to 17.99%.
If it cannot find an answer – it will then tell you it cannot find the answer. | 506 | 2,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.955419 |
http://www.piclist.com/techref/postbot.asp?by=thread&id=%5BPIC%5D%3A+Speedtrap+warning+device+using+any+GPS+rece&w=body&author=Russell+McMahon&tgt=post | 1,534,692,520,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215222.74/warc/CC-MAIN-20180819145405-20180819165405-00523.warc.gz | 578,201,210 | 8,798 | Searching \ for '[PIC]: Speedtrap warning device using any GPS rece' in subject line. ()
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So all he needs is a PIC which will take the current location, output by
the GPS, and calculate the distance to a known set of points, triggering
an alarm when a trap is within .5km or something?
Since the distances are so small, one could even ignore the great circle
equations, and go straight to the square root of the sum of the squares
of the differences in latitude and longitude. Could probably perform a
few tricks there as well using tables, or by alarming when both lat and
long are within one mile of the trap's lat and long. Since the GPS only
updates every second, it'd have a full second to compare the current
position with the table of traps, which is plenty of time.
The hardest part, in fact, would probably be the serial reception and
NMEA parsing, and deciding how to store the trap coordinates.
Of course, it would be better to use APRS and a radar detector so you
could warn others of mobile speed traps you find as well.
Nick Ray wrote:
{Quote hidden}
--
> So all he needs is a PIC which will take the current location, output by
> the GPS, and calculate the distance to a known set of points, triggering
> an alarm when a trap is within .5km or something?
>
> Since the distances are so small, one could even ignore the great circle
> equations, and go straight to the square root of the sum of the squares
> of the differences in latitude and longitude. Could probably perform a
> few tricks there as well using tables, or by alarming when both lat and
> long are within one mile of the trap's lat and long. Since the GPS only
> updates every second, it'd have a full second to compare the current
> position with the table of traps, which is plenty of time.
>
> The hardest part, in fact, would probably be the serial reception and
> NMEA parsing, and deciding how to store the trap coordinates.
>
> Of course, it would be better to use APRS and a radar detector so you
> could warn others of mobile speed traps you find as well.
Alternatively you could try obeying the speed limit.
********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, olinembedinc.com, http://www.embedinc.com
--
{Quote hidden}
=====
------------- http://clik.to/aprsmobiel ----------------
TRACKERS, GPS ONTVANGERS, APRS TRANCEIVER en ZELFBOUW.
>>>>>>>>>HEEFT U VRAGEN OF (BOUW)PROBLEMEN ?<<<<<<<<<<<<
AMATEUR BOUWKITS, BESTELLEN EN PRIJZEN STAAN OP:
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Do, or do not. There is no try.
Olin Lathrop wrote:
>Alternatively you could try obeying the speed limit.
>
--
> Olin Lathrop wrote:
>
> >Alternatively you could try obeying the speed limit.
> >
> Do, or do not. There is no try.
Perfection may be the aim.
All probably fall short of the mark.
An honest try seems to work most of the time.
I think even the most socially aware of us have manage to exceed posted
limits on occasion.
It may be that there are administrations who ARE out to mainly gather money
and who ARE totally unreasonable in their policing but at least in this
country, if you honestly try to observe speed limits at all times, the end
result is approximately no tickets for speeding, over a 30 + year period -
based on a sample of 1. I in fact did receive a speeding ticket on a
motorcycle (and I was speeding *) and they let me off on receipt of my
written explanation. Being able to say that the officer concerned had
complimented me on my riding ability after I lost him in corners (at less
than the speed limit) (as I avoided his unmarked car's attempts to run me
off the road) no doubt helped my case :-).
I find motor vehicles indispensable. However, on the subject of perfection
and driving, I concluded long ago that if Jesus Christ were to have
physically lived in this era then he would not drive a motor vehicle. The
alternative's seem to be to either to break the law consciously and
repeatedly or to drive in such a careful manner as to annoy many others and
to cause them to do worse than they otherwise would have. Neither choice
seems acceptable if observance of the letter and spirit of the law is
desired. As driving one's self is not a necessity it would seem to be better
to not do it.
regards,
Russell McMahon
(* - there's a little more to this story which makes it almost amusing. In
the most unlikely event that anyone wants to know ask offlist).
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I think it is astonishing the extent and expense that
people will go to detect or defeat radar.
It's much easier to just drive under the speed limit.
It's low stress, low cost and considering the efforts
to avoid a ticket, saves time.
Sandy Phelps
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I not able to remember over 2000 locations inside
Netherlands only.
Within Germany there are also around 2000 camera
loctions.
I know in UK and FR the use the same (GATSORADAR)
cameras.
Not sure how many there are.
Most time you know there is a camera on the road,
but where is it stand exacly...?
FLITS/FLASH oeps there it was......
I think only the LON/LAT will do the trick.
All these info is present i the GPS \$GPRMC string.
In \$GPRMC is also, heading and speed over ground.
The only GPS NMEA string needed,
look like its the \$GPRMC...
The rest of the GPS, could be rejected.
Bertus.
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At 10:28 AM 11/4/01 -0800, Sandy Phelps wrote:
>I think it is astonishing the extent and expense that
>people will go to detect or defeat radar.
>It's much easier to just drive under the speed limit.
>It's low stress, low cost and considering the efforts
>to avoid a ticket, saves time.
>Sandy Phelps
If it was always deployed for speed enforcement, instead of revenue
enhancement, maybe people would feel differently. When I got tagged in
Hawaii, I was, according to the officer's testimony, out of his sight
around a corner, and over the crest of a hill, and just one car in five
lanes of traffic. Additionally, his position was only a few hundred yards
from the aircraft carrier enterprise (they commonly run high powered radar
in port), and under a flight path where aircraft are landing, with weather
radar operating in the same band as his speed gun.
Additionally, I wasn't speeding, but none of that mattered.
After I brought out all this in court, the judge explained to me that in
his "radar court" (yes a special court, with special rules, just for us)
the only admissible testimony would come from the arresting officer, and
the radar gun manufacturer, who was not present.
They wave the gun, you pay the fine, see the cashier on the way out.
This incident also resulted in my arrest for contempt of court, a charge
that I heartily agreed with.
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> I know in UK and FR the use the same (GATSORADAR)
> cameras.
> Not sure how many there are.
> Most time you know there is a camera on the road,
> but where is it stand exacly...?
> FLITS/FLASH oeps there it was......
>
Why don't you just lobby your government like the
boy racers in the UK did. Now all the Speed Cameras are
to be painted in high visibility colours and sign
posts are to be erected warning you of this road hazard
Giving you sufficient distance to slow down before you get
to them.
Chris Carr
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Dear Olin Lathrop,
You are 100% right about that, afcourse.
Here in NL there are over 2000 !! yes 2000 af those
speedtraps.
These cameras get to action when your drive 3 Km/h
near 2 mph to fast......
Do you view your speedometer at alltimes ?
I must honiest say, iam not .....
And thit is only a idea to find out if is technical
possible.
Maby we give it more friendly name to please veryone
like, "GPS WAYPOINT RECOVERY UNIT".... =:)
But this is other discussion i dont like to discus
overhere.
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> Here in NL there are over 2000 !! yes 2000 af those
> speedtraps.
> These cameras get to action when your drive 3 Km/h
> near 2 mph to fast......
> Do you view your speedometer at alltimes ?
> I must honiest say, iam not .....
Serious suggestion: An accurate audible warning speedometer alarm seems like
a sensible option in such cases. This would not stop you speeding but would
warn you so that you would never do it by accident. Such a device is
trivially easy to build for most speedometers. A little complexity would
allow pleasant brief tones repeated only occasionally to prevent annoyance
if ongoing above-limit driving was desired. Then at least the problem
mentioned above would not exist.
Russell McMahon
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https://www.debate.org/debates/.999-repeating-equals1/1/ | 1,653,330,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00533.warc.gz | 836,461,635 | 9,628 | All Big Issues
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# .999 repeating equals1
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Voting Style: Open Point System: 7 Point Started: 7/29/2014 Category: Education Updated: 7 years ago Status: Post Voting Period Viewed: 699 times Debate No: 59689
Debate Rounds (3)
Pro 1/9 = .111111111 1/9 x 9 = .11111111 x 9 1 = .9999999999Report this Argument Con .999 repeating does not equal one. .999 repeating simply implies that the value gets infinitely close to one, yet never reaches one. other wise someone would just type one instead of having to use the extra effort and type .999 repeating.Report this Argument Pro xenon54 forfeited this round. Con kl31 forfeited this round. Pro What is 1/3 equal to as a decimal? Answer: .333 repeating What is 1/3 times 3? Answer: 1 What is .333 repeating times 3? Answer: .9999 repeating 1/3 = .333333 Both of these are undeniable facts 1/3 * 3 = 1 .33333 * 3 = 1 This is ansimple substitution. Since 1/3 does = .3333, you can substitute it for .3333 in the next equation .333333 * 3 can also equal .999999..... but if .333333 * 3 equals both 1 and .999 repeating, then 1 = .999 repeating. (Also, please try to use math in your responsReport this Argument Con kl31 forfeited this round.
3 comments have been posted on this debate. Showing 1 through 3 records.
Posted by czeekymonkey 7 years ago
I've seen dozens of Mathematicians debate this, and they all say that to thing 0.99999... = 1 is to misunderstand the very nature of a continuous decimal as an infinite number. 0.33333... does not equal 1/3, but it is infinitesimally smaller. 0.99999... is not 1, but infinitesimally smaller than 1. It's mathematically irrefutable.
Posted by Adam_Godzilla 7 years ago
Search us Mhykiel. He's made a thorough debate on this
Posted by KhalifV 7 years ago
You must be more specific.
:999 : : : ; : : : 999^ < :999 : : : = 1
In some instances, it is less.
No votes have been placed for this debate. | 608 | 2,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-21 | latest | en | 0.905104 |
https://www.aqua-calc.com/calculate/mole-to-volume-and-weight/substance/potassium-blank-formate | 1,695,951,730,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00162.warc.gz | 686,394,998 | 8,436 | # Moles of Potassium formate
## potassium formate: convert moles to volume and weight
### Volume of 1 mole of Potassium formate
centimeter³ 43.14 milliliter 43.14 foot³ 0 oil barrel 0 Imperial gallon 0.01 US cup 0.18 inch³ 2.63 US fluid ounce 1.46 liter 0.04 US gallon 0.01 meter³ 4.31 × 10-5 US pint 0.09 metric cup 0.17 US quart 0.05 metric tablespoon 2.88 US tablespoon 2.92 metric teaspoon 8.63 US teaspoon 8.75
### Weight of 1 mole of Potassium formate
carat 420.58 ounce 2.97 gram 84.12 pound 0.19 kilogram 0.08 tonne 8.41 × 10-5 milligram 84 116
### The entered amount of Potassium formate in various units of amount of substance
centimole 100 micromole 1 000 000 decimole 10 millimole 1 000 gigamole 1 × 10-9 mole 1 kilogram-mole 0 nanomole 1 000 000 000 kilomole 0 picomole 1 000 000 000 000 megamole 1 × 10-6 pound-mole 0
#### Foods, Nutrients and Calories
COCONUT TREE BRAND, VEGETARIAN SOUP CUBES, UPC: 005216438623 contain(s) 278 calories per 100 grams (≈3.53 ounces) [ price ]
26 foods that contain Fatty acids, total trans-dienoic. List of these foods starting with the highest contents of Fatty acids, total trans-dienoic and the lowest contents of Fatty acids, total trans-dienoic
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Sunset Gold weighs 1 505.74 kg/m³ (94.00028 lb/ft³) with specific gravity of 1.50574 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 ]
Balsa, Hard weighs 281.92 kg/m³ (17.59969 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-134a, liquid (R134A) with temperature in the range of -51.12°C (-60.016°F) to 71.12°C (160.016°F)
#### Weights and Measurements
Rem (Roentgen Equivalent Man) is the derived unit of any of the quantities expressed as the radiation absorbed dose (rad) equivalent [ millirem fission fragments ]
The area measurement was introduced to measure surface of a two-dimensional object.
troy/metric c to sl/US qt conversion table, troy/metric c to sl/US qt unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a hexagonal prism and its surface area | 710 | 2,378 | {"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} | 2.640625 | 3 | CC-MAIN-2023-40 | latest | en | 0.59337 |
http://www.numericalmethod.com/javadoc/suanshu/com/numericalmethod/suanshu/geometry/LineSegment.html | 1,548,206,641,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583879117.74/warc/CC-MAIN-20190123003356-20190123025356-00435.warc.gz | 342,612,716 | 3,866 | # SuanShu, a Java numerical and statistical library
com.numericalmethod.suanshu.geometry
## Class LineSegment
• public class LineSegment
extends Object
Represent a line segment.
• ### Constructor Summary
Constructors
Constructor and Description
LineSegment(Point p1, Point p2)
Create a line segment with two given endpoints.
• ### Method Summary
All Methods
Modifier and Type Method and Description
int dimension()
Get the dimension of the coordinate space.
double distance(Point p)
Calculate the shortest distance between a point and this line segment in Euclidean geometry.
Point getEndPoint1()
Get the first endpoint.
Point getEndPoint2()
Get the second endpoint.
double length()
Get the length of the line segment.
double perpendicularDistance(Point p)
• ### Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• ### Constructor Detail
• #### LineSegment
public LineSegment(Point p1,
Point p2)
Create a line segment with two given endpoints.
Parameters:
p1 - the first endpoint
p2 - the second endpoint
• ### Method Detail
• #### getEndPoint1
public Point getEndPoint1()
Get the first endpoint.
Returns:
the first endpoint
• #### getEndPoint2
public Point getEndPoint2()
Get the second endpoint.
Returns:
the second endpoint
• #### dimension
public int dimension()
Get the dimension of the coordinate space.
Returns:
the dimension of the coordinate space
• #### length
public double length()
Get the length of the line segment.
Returns:
the line length
• #### distance
public double distance(Point p)
Calculate the shortest distance between a point and this line segment in Euclidean geometry. If the projection of the point on the line is outside of the segment, the distance between the point and the closest endpoint of the line segment is returned.
Parameters:
p - the point
Returns:
the distance
• #### perpendicularDistance
public double perpendicularDistance(Point p) | 432 | 1,978 | {"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} | 2.65625 | 3 | CC-MAIN-2019-04 | latest | en | 0.562369 |
https://www.physicsforums.com/threads/inductor-permeability-change.818550/ | 1,511,027,732,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805008.39/warc/CC-MAIN-20171118171235-20171118191235-00677.warc.gz | 852,093,815 | 14,522 | # Inductor permeability change
1. Jun 11, 2015
### cnh1995
In an RC dc circuit at steady state, if the permitivitty of the capacitor is increased by inserting a different dielectric, additional charging current will flow and the capacitor will accept more charge (due to increase in capacitance). What will happen if the inductance in an RL circuit is increased similarly (at the steady state),i.e. by increasing the permeability of the core? Here, the final value of the current is same for both the cores.
2. Jun 11, 2015
### Hesch
Say you have some solenoid ( cylindric core ) with some relative permeability, μr = 100. You pull it out from within the coil, and are using some force to do so ( F100 ).
Now you substitute it with another core ( μr = 1000 ), and putting in the core, the new core will be attracted by a force, F1000.
F1000 ≈ 10 * F100 , so all in all the result is lost energy as seen from the circuit, and that lost energy must be compensated somehow by the powersupply ( yielding more amps or volts integrated over time of replacement ).
From another point of view, you can say that switching the cores has caused:
- the steady state current is the same.
- the voltage across the RL circuit is the same.
- the H-field ( proportional to current ) is the same.
- the B-field has increased ( B = μ0 * μr * H )
As the magnetic energy tends to be proportional to B*H, the powersupply must also deliver this growth in magnetic energy, but this a hard calculation because the steady state shape of the magnetic fields will be altered by substitution of the core. The calculations of growth in magnetic energy depends on the exact dimensions of the magnetic circuit. | 393 | 1,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.910733 |
http://www.cfd-online.com/Forums/cfx/67534-volume-fraction-profile-data-problem.html | 1,477,120,999,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718840.18/warc/CC-MAIN-20161020183838-00295-ip-10-171-6-4.ec2.internal.warc.gz | 375,825,937 | 16,255 | # Volume Fraction, profile data problem
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August 18, 2009, 07:34 Volume Fraction, profile data problem #1 Member Join Date: Aug 2009 Posts: 42 Rep Power: 9 Hi, I´m simulating waves using profile data at the inlet. As I have experimental data from the height of the wave vs time, I used a profile data containing x,y,z and volume fraction as columnes. The results aren´t correct. x and y are fixed values (the point where the height of the wave is measured). I have put x as the time and y = 0 all time. z is the mesured height. Volume Fraction values are 1 or 0 depending on the value of z (positive or negative) The expression: Filename.Volume Fraction((t*1[s^-1]*1[m]),y,z) It doesn´t work!!! Would kindly anyone tell me what could be wrong? Thank you very much in advance Cfxmar
August 18, 2009, 18:25 #2 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 12,704 Rep Power: 98 I think a 1D interpolation function would be more appropriate than the 3D one you are using. Try that. Glenn Horrocks
August 19, 2009, 04:39 #3 Member Join Date: Aug 2009 Posts: 42 Rep Power: 9 Thank you, Glenn. I have tried that but as the problem is 3D, the profile data needs three columns for x,y,z and one for the volume fraction. CFX doesn´t let me load the profile data file with only 2 columns (one for x as the time and one for the volune fraction)
August 19, 2009, 04:47 #4 Senior Member Lance Join Date: Mar 2009 Posts: 598 Rep Power: 12 I just did the same thing, but for a measured mass flow rate instead and a 1D profile data function worked fine for me. Looks like: FUNCTION: MassFlowInlet Argument Units = [m] Extend Max = true Extend Min = true File Name = massflow.csv Option = Profile Data Spatial Fields = X DATA FIELD: Mass Flow Rate Field Name = Mass Flow Rate Result Units = [kg s^-1] END END and the boundary condition: BOUNDARY: inlet Boundary Type = INLET Location = F18.16 BOUNDARY CONDITIONS: FLOW DIRECTION: Option = Normal to Boundary Condition END FLOW REGIME: Option = Subsonic END MASS AND MOMENTUM: Mass Flow Rate = MassFlowInlet.Mass Flow Rate(FT) Option = Mass Flow Rate END TURBULENCE: Option = Low Intensity and Eddy Viscosity Ratio END END END where FT is defined as FT = 1[m/s]*t Lance
August 19, 2009, 05:45 #5 Member Join Date: Aug 2009 Posts: 42 Rep Power: 9 Thank you very much, Glenn and Lance. I´m running the simulation, tomorrow I will see the results. Cfxmar
August 19, 2009, 08:09
#6
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 12,704
Rep Power: 98
Quote:
I have tried that but as the problem is 3D, the profile data needs three columns for x,y,z and one for the volume fraction. CFX doesn´t let me load the profile data file with only 2 columns (one for x as the time and one for the volune fraction)
Come on, think outside the square.... You don't need a whole variable field to store the volume fraction. It can be stored as a single number (that is the height of the surface at that time) and the volume fraction field can be evaluated for height from that using a simple CEL step or if function. Now you can use a 1D interpolation function.
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# Kinetic Molecular Theory, Real Gases
The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average speeds determined by their absolute temperatures. The individual molecules of a gas exhibit a range of speeds, the distribution of these speeds being dependent on the temperature of the gas and the mass of its molecules. Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate description of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions.
## 26.1 The Kinetic-Molecular Theory
### Learning Objectives
By the end of this section, you will be able to:
• State the postulates of the kinetic-molecular theory
• Use this theory’s postulates to explain the gas laws
The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.
The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)
1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
2. The molecules composing the gas are negligibly small compared to the distances between them.
3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.
4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.
The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle’s, Charles’s, Amontons’s, Avogadro’s, and Dalton’s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham’s law.
### The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I
Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:
• Amontons’s law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 26.1).
• Charles’s law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.
• Boyle’s law. If the gas volume volume of a given amount of gas at a given temperature is decreased (that is, if the gas is compressed), the molecules will be exposed to a decreased container wall area. Collisions with the container wall will therefore occur more frequently and the pressure exerted by the gas will increase (Figure 26.1).
• Avogadro’s law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 26.1).
• Dalton’s Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.
Figure 26.1
(a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to increased frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area per unit time.
### Molecular Speeds and Kinetic Energy
The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at speeds and kinetic energies of gas molecules, and the temperature of a gas sample.
In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure 26.2).
Figure 26.2
The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, νp, is a little less than 400 m/s, while the root mean square speed, urms, is closer to 500 m/s.
The kinetic energy (KE) of a particle of mass (m) and speed (u) is given by:
$\text{KE}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m{u}^{2}$
Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m2 s–2). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square speed of a particle, urms, is defined as the square root of the average of the squares of the speeds with n = the number of particles:
${u}_{\text{r}\text{m}\text{s}}=\sqrt{\overline{{u}^{2}}}=\sqrt{\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\dots }{n}}$
The average kinetic energy for a mole of particles, KEavg, is then equal to:
${\text{KE}}_{\text{avg}}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{Mu}_{\text{rms}}^{2}$
where M is the molar mass expressed in units of kg/mol. The KEavg of a mole of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:
${\text{KE}}_{\text{avg}}=\phantom{\rule{0.2em}{0ex}}\frac{3}{2}\phantom{\rule{0.2em}{0ex}}RT$
where R is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J/mol⋅K (8.314 kg m2s–2mol–1K–1). These two separate equations for KEavg may be combined and rearranged to yield a relation between molecular speed and temperature:
$\frac{1}{2}{Mu}_{\text{rms}}^{2}=\phantom{\rule{0.2em}{0ex}}\frac{3}{2}\phantom{\rule{0.2em}{0ex}}RT$
${u}_{\text{rms}}=\sqrt{\frac{3RT}{M}}$
### Example 26.1
#### Calculation of urms
Calculate the root-mean-square speed for a nitrogen molecule at 30 °C.
#### Solution
Convert the temperature into Kelvin:
$30\phantom{\rule{0.2em}{0ex}}\text{°C}+273=\text{303 K}$
Determine the molar mass of nitrogen in kilograms:
$\frac{28.0\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}}{\text{1 mol}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{1 kg}}{1000\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}}}\phantom{\rule{0.2em}{0ex}}=0.028\phantom{\rule{0.2em}{0ex}}\text{kg/mol}$
Replace the variables and constants in the root-mean-square speed equation, replacing Joules with the equivalent kg m2s–2:
${u}_{\text{rms}}=\sqrt{\frac{3RT}{M}}$
${u}_{\text{r}\text{m}\text{s}}=\phantom{\rule{0.2em}{0ex}}\sqrt{\frac{3\left(8.314\phantom{\rule{0.2em}{0ex}}\text{J/mol K}\right)\left(\text{303 K}\right)}{\left(0.028\phantom{\rule{0.2em}{0ex}}\text{kg/mol}\right)}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\sqrt{2.70\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}{\text{s}}^{-2}}\phantom{\rule{0.2em}{0ex}}=519\phantom{\rule{0.2em}{0ex}}\text{m/s}$
Calculate the root-mean-square speed for a mole of oxygen molecules at –23 °C.
441 m/s
If the temperature of a gas increases, its KEavg increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KEavg decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 26.3.
Figure 26.3
The molecular speed distribution for nitrogen gas (N2) shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.
At a given temperature, all gases have the same KEavg for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher urms, with a speed distribution that peaks at relatively higher speeds. Gases consisting of heavier molecules have more low-speed particles, a lower urms, and a speed distribution that peaks at relatively lower speeds. This trend is demonstrated by the data for a series of noble gases shown in Figure 26.4.
Figure 26.4
molecular speed is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules.
### The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II
According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.
The rate of effusion of a gas depends directly on the (average) speed of its molecules:
$\text{effusion rate}\phantom{\rule{0.2em}{0ex}}\propto \phantom{\rule{0.2em}{0ex}}{u}_{\text{rms}}$
Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here:
${u}_{\text{rms}}=\sqrt{\frac{3RT}{M}}$
$M=\phantom{\rule{0.2em}{0ex}}\frac{3RT}{{u}_{\text{r}\text{m}\text{s}}^{2}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{3RT}{{\overline{u}}^{2}}$
$\frac{\text{effusion rate A}}{\text{effusion rate B}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{u}_{\text{r}\text{m}\text{s}\phantom{\rule{0.2em}{0ex}}\text{A}}}{{u}_{\text{r}\text{m}\text{s}\phantom{\rule{0.2em}{0ex}}\text{B}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\sqrt{\frac{3RT}{{M}_{\text{A}}}}}{\sqrt{\frac{3RT}{{M}_{\text{B}}}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\sqrt{\frac{{M}_{\text{B}}}{{M}_{\text{A}}}}$
The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law.
Supplemental exercises are available if you would like more practice with these concepts.
## 26.2 Non-Ideal Gas Behavior
### Learning Objectives
By the end of this section, you will be able to:
• Describe the physical factors that lead to deviations from ideal gas behavior
• Explain how these factors are represented in the van der Waals equation
• Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior
• Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation
Thus far, the ideal gas law, PV = nRT, has been applied to a variety of different types of problems, ranging from reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered.
One way in which the accuracy of PV = nRT can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, Vm) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor (Z) with:
$\text{Z}=\phantom{\rule{0.2em}{0ex}}\frac{\text{molar volume of gas at same}\phantom{\rule{0.2em}{0ex}}T\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}P}{\text{molar volume of ideal gas at same}\phantom{\rule{0.2em}{0ex}}T\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}P}\phantom{\rule{0.2em}{0ex}}={\left(\frac{P{V}_{m}}{RT}\right)}_{\text{measured}}$
Ideal gas behavior is therefore indicated when this ratio is equal to 1, and any deviation from 1 is an indication of non-ideal behavior. Figure 26.5 shows plots of Z over a large pressure range for several common gases.
Figure 26.5
A graph of the compressibility factor (Z) vs. pressure shows that gases can exhibit significant deviations from the behavior predicted by the ideal gas law.
As is apparent from Figure 26.5, the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.
Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas. The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle’s law.
At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) (Figure 26.6). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.
Figure 26.6
(a) Attractions between gas molecules serve to decrease the gas volume at constant pressure compared to an ideal gas whose molecules experience no attractive forces. (b) These attractive forces will decrease the force of collisions between the molecules and container walls, therefore reducing the pressure exerted at constant volume compared to an ideal gas.
There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them.
The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant b corresponds to the size of the molecules of a particular gas. The “correction” to the pressure term in the ideal gas law is $\frac{{n}^{2}a}{{V}^{2}},$ and the “correction” to the volume is nb. Note that when V is relatively large and n is relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, PV = nRT. Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in Table 26.1.
Table 26.1
Values of van der Waals Constants for Some Common Gases
Gasa (L2 atm/mol2)b (L/mol)
N21.390.0391
O21.360.0318
CO23.590.0427
H2O5.460.0305
He0.03420.0237
CCl420.40.1383
At low pressures, the correction for intermolecular attraction, a, is more important than the one for molecular volume, b. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by PV = nRT over a small range of pressures. This behavior is reflected by the “dips” in several of the compressibility curves shown in Figure 26.5. The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised (Z decreases with increasing P). At very high pressures, the gas becomes less compressible (Z increases with P), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume.
Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded—this is, however, very often not the case.
### Example 26.2
#### Comparison of Ideal Gas Law and van der Waals Equation
A 4.25-L flask contains 3.46 mol CO2 at 229 °C. Calculate the pressure of this sample of CO2:
(a) from the ideal gas law
(b) from the van der Waals equation
(c) Explain the reason(s) for the difference.
#### Solution
(a) From the ideal gas law:
$P=\phantom{\rule{0.2em}{0ex}}\frac{nRT}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{3.46\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.08206\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}\phantom{\rule{0.2em}{0ex}}\text{atm}\phantom{\rule{0.2em}{0ex}}\overline{){\text{mol}}^{\text{−1}}}\phantom{\rule{0.2em}{0ex}}\overline{){\text{K}}^{\text{−1}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}502\phantom{\rule{0.2em}{0ex}}\overline{)\text{K}}}{4.25\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}}}\phantom{\rule{0.2em}{0ex}}=33.5\phantom{\rule{0.2em}{0ex}}\text{atm}$
(b) From the van der Waals equation:
$\left(P+\frac{{n}^{2}a}{{V}^{2}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(V-nb\right)=nRT\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}P=\phantom{\rule{0.2em}{0ex}}\frac{nRT}{\left(V-nb\right)}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{{n}^{2}a}{{V}^{2}}$
$P=\phantom{\rule{0.2em}{0ex}}\frac{3.46\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.08206\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}\text{atm}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{502 K}}{\left(4.25\phantom{\rule{0.2em}{0ex}}\text{L}-3.46\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.0427\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{\text{−1}}\right)}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{{\left(3.46\phantom{\rule{0.2em}{0ex}}\text{mol}\right)}^{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3.59\phantom{\rule{0.2em}{0ex}}{\text{L}}^{2}\phantom{\rule{0.2em}{0ex}}\text{atm}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{2}}{{\left(4.25\phantom{\rule{0.2em}{0ex}}\text{L}\right)}^{2}}$
This finally yields P = 32.4 atm.
(c) This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because CO2 molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions.
A 560-mL flask contains 21.3 g N2 at 145 °C. Calculate the pressure of N2:
(a) from the ideal gas law
(b) from the van der Waals equation
(c) Explain the reason(s) for the difference.
(a) 46.562 atm; (b) 46.594 atm; (c) The van der Waals equation takes into account the volume of | 6,123 | 23,178 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.941235 |
http://www.cs.cornell.edu/courses/cs322/2004sp/pdfs/Pre1RevSol.htm | 1,545,197,324,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831334.97/warc/CC-MAIN-20181219045716-20181219071716-00335.warc.gz | 350,681,538 | 2,870 | Prelim 1 Review Problem Solutions
```% Problem 1
%
% The spacing between base-2 floating point numbers of the form m*2^e
% is 2^(-t)*2^e where t is the length of the mantissa in bits.
% If M and M+2 are floating point numbers and M+1 is not, then the
% spacing of floating point numbers between M and M+2 is >= 2.
% Suppose M = m*2^e. It follows that
%
% M*2^{-t} = m*(2^e)*(2^{-t}) >= 2m
%
% and so
%
% 2^{t} <= M/m <= 2M.
%
% Thus, t <= log_{2}(M) - 1
M = 1;
k = 0;
while ((M < M+1) & ((M+1)<(M+2)))
M = 2*M;
end
t = log2(M)-1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Problem 2
n = 20;
A = zeros(n,n);
for i=1:n
for j=1:n
if abs(i-j)<=2
A(i,j) = cos(abs(i-j)*pi/n);
end
end
end
x = randn(n,1);
error = max(abs(CosProd(x)-A*x))
function t = CosProd(x)
% x is a column n-vector and n > 3.
% y = A*x where A is the n-by-n matrix defined by
%
% 0 if |i-j|>2
% A(i,j) =
% cos(|i-j)|*pi/n) if |i-j|<=2
n = length(x);
u = cos(pi/n);
v = cos(2*pi/n);
% The pattern...
% t(1) = x(1) + ux(2) + vx(3)
% t(2) = ux(1) + x(2) + ux(3) + vx(4)
% t(3) = vx(1) + ux(2) + x(3) + ux(4) + vx(5)
%
% t(n-2) = vx(n-4) + ux(n-3) + x(n-2) + ux(n-1) + vx(n)
% t(n-1) = vx(n-3) + ux(n-2) + x(n-1) + ux(n)
% t(n) = vx(n-2) + ux(n-1) + x(n)
% The solution
t = x;
t(1) = t(1) + u*x(2) + v*x(3);
t(2) = t(2) + u*x(1) + u*x(3) + v*x(4);
t(3:n-2) = t(3:n-2) + u*(x(2:n-3)+x(4:n-1)) + v*(x(1:n-4)+x(5:n));
t(n-1) = t(n-1) + v*x(n-3) + u*x(n-2) + u*x(n);
t(n) = t(n) + v*x(n-2) + u*x(n-1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Problem 3
n = 10;
x = linspace(0,pi,n);
y = exp(-x).*sin(x);
m = MaxJump(x,y)
function m = MaxJump(x,y)
% x and y are column n-vectors and x(1) < x(2) <...< x(n).
% Let S be the cubic spline produced by spline(x,y).
% m is the maximum value of f(z) on the interval [x(2) , x(n-1)] where
% f(z) is the limit of |S'''(z+delta) - S'''(z-delta)| as delta goes to zero.
S = spline(x,y);
[breaks,coeff,L] = unmkpp(S)
% Compute the 3rd derivative of each local cubic.
Der3 = 6*coeff(:,1)
m = max(abs(Der3(1:L-1)-Der3(2:L)));
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Problem 4
a = [2;5;-1;2]
alfa = 3;
beta = 1;
c = Convert(a,alfa,beta);
z = linspace(0,5)';
zalfa = z-alfa;
pvals = ((a(4)*zalfa + a(3)).*zalfa + a(2)).*zalfa + a(1);
zbeta = z-beta;
qvals = ((c(4)*zbeta + c(3)).*zbeta + c(2)).*zbeta + c(1);
error = max(abs(qvals - pvals))
function c = Convert(a,alfa,beta)
% a is a column 4-vector and alfa and beta are scalars.
% c is a column 4-vector so that the cubic polynomials
%
% p(x) = a(1) + a(2)(x-alfa) + a(3)(x-alfa)^2 + a(4)(x-alfa)^3
%
% q(x) = c(1) + c(2)(x-beta) + c(3)(x-beta)^2 + c(4)(x-beta)^3
% Note that if q interpolates p at 4 distinct points then q = p.
x = [1;2;3;4]
xalfa = x - alfa;
pvals = ((a(4)*xalfa + a(3)).*xalfa + a(2)).*xalfa + a(1);
% Let's determine q so that it interpolates (x(i),pvals(i)), i=1:4
% q(x(i)) = p(x(i)) is a linear equation in c(1),c(2),c(3), and c(4):
% c(1) + c(2)(x(i) - beta) + c(3)(x(i)-beta)^2 + c(4)(x(i) - beta)^3 = p(x(i))
% Set up the matrix of coefficients and solve for c(1:4):
xbeta = x-beta;
V = [ones(4,1) xbeta xbeta.^2 xbeta.^3];
c = V\pvals;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Problem 5
n = 7;
p = 2;
A = triu(tril(randn(n,n),p),-1);
[L,U] = SpecLU(A,p)
Error = A - L*U
function [L,U] = SpecLU(A,p)
% A is an n-by-n matrix with lower bandwidth 1 and upper bandwidth p.
% Assume that A has an LU factorization.
% Computes the factorization A = LU where L is an n-by-n unit lower unit
% triangular matrix and U is an n-by-n upper triangular.
[n,n] = size(A); v = zeros(n,1);
for k=1:n-1
v(k+1) = A(k+1,k)/A(k,k);
r = min(n,k+p);
A(k+1,k:r) = A(k+1,k:r) - v(k+1)*A(k,k:r);
end
L = diag(ones(n,1)) + diag(v(2:n),-1); U = triu(A);
%%%%%%%%%%%%%%%%%%%%%%%%
% Problem 6
a = 10;
b = 30;
T = pi;
tol = .0001;
fname = 'P6F'; % sin(2x)
% Let I(L,R) be the integral from L to R of P6F(x)
k = floor((b-a)/T); % Number of whole periods encountered from a to b
c = b-k*T;
% I(a,b) = k*I(a,a+T) + I(a,c)
% = k*(I(a,c) + I(c,a+T)) + I(a,c)
% = (k+1)*I(a,c) + k*I(c,a+T)
% If e1 bound the absolute error in our QUAD estimate of I(a,c)
% and e2 bound the absolute error in our QUAD estimate of I(c,a+T)
% then total error <= (k+1)*e1 + k*e2
% We want the total absolute error to be <= tol.
% One choice: e1 = e2 = .5*tol/(k+1); | 1,863 | 4,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-51 | latest | en | 0.630758 |
http://www.remaxopus.com/what-is-a-balloon-mortgage/ | 1,601,006,554,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00315.warc.gz | 213,988,096 | 7,224 | # what is a balloon mortgage
### Contents
As scary as balloon mortgages might sound, there is a way out: It’s possible to refinance a balloon mortgage into a conventional 15- or 30-year loan. The catch: If you’re cash-strapped or your.
A balloon mortgage is a relatively short term mortgage with a huge payment due at the end of the term. A mortgage is generally for a longer term with uniform payments for the life of the mortgage.
To illustrate a fully amortizing payment, imagine someone takes out a 30-year fixed-rate mortgage with a 4.5% interest rate, and his monthly payments are \$1,266.71. At the beginning of the loan’s life.
Any mortgage that comes due with an unpaid balance is known as a balloon loan. Others may be home equity interest-only loans for, say, 10 years and then fully amortize over the remaining 20 years. Thus, they will have a big jump in payment after ten years.
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Balloon mortgage. With a balloon mortgage, you make monthly payments over the mortgage term, which is typically five, seven, or ten years, and a final installment, or balloon payment, that is significantly larger than the usual monthly payments.
Amortization With Balloon Payment Excel Excel Loan Amortization Table Spreadsheet – Vertex42.com – Balloon Payment Loan Calculator This template is unique in that the amortization table ends after a specified number of payments. The final payment, or balloon payment, is the amount required to pay off in full.
[TRD] Lower mortgage rates may not be helping housing. The fate of the large balloon rat is in question after multiple legal challenges have argued the inflatable rodent is unlawful.
A balloon mortgage is a very good choice when you don’t plan to stay in the home beyond the balloon period. Before the mortgage is up, you will sell the home and buy another, thus paying off the.
A balloon mortgage is a loan in which a large portion of the principal is repaid in one payment at the end of the term. Investors use a balloon mortgage to qualify for a higher loan amount, lower rates and lower monthly payments. balloon mortgage rates typically start around 4.5 percent with 5- to 7-year terms.
Balloon Mortgage A mortgage whereby the property owner makes only interest payments for a set period of time, usually five, seven or 10 years. At the end of the term, the owner repays the entire principal at once. A balloon mortgage is useful for an investment property where the owner does not expect to. | 666 | 3,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-40 | latest | en | 0.93599 |
https://www.yaclass.in/p/mathematics-state-board/class-7/number-system-2789/multiplication-of-decimal-numbers-17806/re-8596bf00-a132-40a1-bd8e-f0374077fb4f | 1,721,477,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00043.warc.gz | 917,827,637 | 12,916 | PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
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Kavitha wants to buy biscuits that cost $$₹$$20.5. If she needs 5 biscuits, how much she should pay?
We can simply answer it, by multiplying the cost and the number of biscuits. we get 20.5 $$×$$ 5 $$=$$ 102.5.
In many situations in our daily life, we use the multiplication of decimal numbers. We shall explore it in the upcoming lessons.
Here we split the decimal multiplication into two parts as follows.
i) Decimal multiplication through models and area models.
ii) Multiplication of Decimal Numbers by $$10$$, $$100$$ and $$1000$$.
Decimal multiplication through models and area models.
Now let us find the $$0.1 × 0.1$$ using the grid model.
We can rewrite $$0.1$$ as $\frac{1}{10}$; Therefore, $$0.1 × 0.1 =$$ $\frac{1}{10}×\frac{1}{10}$: That is $\frac{1}{10}$$$^t$$$$^h$$ of $\frac{1}{10}$.
Shade the horizontal area of $\frac{1}{10}$ by red color. And shade blue color on the vertical area of $\frac{1}{10}$.
$\frac{1}{10}$$$^t$$$$^h$$ of $\frac{1}{10}$ is the common portion, which is $\frac{1}{100}$$$^t$$$$^h$$.
Therefore, it shows that $\frac{1}{10}×\frac{1}{10}$ $$=$$ $\frac{1}{100}$ $$= 0.01$$.
Hence, $$0.1 × 0.1 = 0.01$$.
Now by applying the above concept we see an example with some other numbers.
Example:
Find $$0.5 × 0.4$$
• Let us first shade $$4$$ rows of the grid in red colour to represent $$0.4$$.
• Shade $$5$$ columns of the grid in blue colour to represent $$0.5$$ of $$0.4$$.
• Now $$20$$ squares represents the common portion.
• This represents $$20$$ hundredth or $$0.20$$. Hence $$0.5 × 0.4 = 0.20$$.
In the above figure, we can observe that the number of decimal digits in $$0.20$$ is two. So, we can conclude that the number of decimal digits in the product of two decimal numbers is equal to the sum of decimal digits which are multiplied. | 706 | 1,929 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-30 | latest | en | 0.710889 |
https://answercult.com/question/how-does-buoyancy-work/ | 1,652,902,885,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522309.14/warc/CC-MAIN-20220518183254-20220518213254-00019.warc.gz | 155,748,105 | 19,822 | # How does buoyancy work?
881 views
I’ve always wondered and never understood how buoyancy works, especially with huge metal ships that I think should surely sink.
In: Physics
Have to displace (move) more weight in water than an object weighs for it to float. So if a 10 tonne ship displaces more than 10 tonne of water it floats. If it displaces less it sinks. Is the most simple way to put it 🙂
It’s all about density. Density is mass divided by volume. Denser objects move to the bottom, less dense objects move to the top. Buoyancy takes the *entire* object; that is, the whole metal hull plus all the cargo and people and everything else, AND the hundreds of cubic meters of air inside the ship, into account.
Plus, water is actually really, really heavy. It’s about ten pounds per gallon. Air is obviously extremely light by comparison.
So, while the hull and cargo and everything else are denser than water, when you take into account the very large percentage of the volume of the ship that’s just air, overall, it’s less dense than the water, so it floats.
its all about water pressure essentially.
as you move water out of the way, that same weight of water is going to push upwards on you.
so think of an air filled ball. As you push it further and further into the water, that ball is moving more and more water out of the way.
All that water that is moved away is pushing upwards on the ball because the ball is in the way and gravity is pulling all that water down.
Remember that large metal ships are mostly hollow. all that water they push out of the way is pushing upwards on the boat.
so while a piece of metal on its own will sink, a large metal ball will float because it displaces water.
When you put something in a fluid, the fluid presses up on it. The amount it pushes up depends only on the volume of water that the object displaces, and it’s essentially the weight of this volume of water.
You can make ships out of steel, only if the inside of the ship consists of a mix of steel (= heavier than water) and air (= lighter than water). This also explains making ships out of concrete and other materials that are heavier than water. You can’t float a solid steel ship, but that’s not a super useful way to make steel ships (no inside is sub-optimal).
An object will float as long as it displaces an equal amount of water than it weighs. When a ship floats, a large portion of it is underwater. That underwater volume of the ship displaces the water. Let’s pretend we have a large tub of fresh water filled right to the top; if we add anything to the tub, the water will overflow. If we put a object that weighs one tonne in the tub, the water will overflow from the tub. If we collected the water that overflowed, it would weight exactly one tonne, the same as the object. One tonne of water was displaced.
For the object to float, the underwater volume has to equal the displaced water. One tonne of fresh water is one cubic metre, this means the object needs an underwater volume of one cubic metre. If the object is 1 metre long, and 1 metre wide, the object will sink 1 metre deep before it floats. Salt water is more buoyant, with a relative density of 1.025. The same vessel floating in salt water would only be 0.976 metres underwater (1/1.025).
So, in a sense, ships sink until they are deep enough to float, if that makes any sense. Ships are usually very deep; a loaded ship will have more of the hull in the water than out of the water. My ship has a loaded draft of 8 metres, while a tanker could go down to 25 metres or so. A box shaped vessel in fresh water that is that 225 metres long, 30 metres wide, and a draft of 25 metres would weight 168,750 tonnes and still float (225*30*25).
There’s some good answers here, but I would say the ELI5 version is that metal ships are made of metal & air. Metal on its own will sink, and air on its own will float, but a metal shell around air will float if there’s enough air. That’s why ships and boats will sink if they take on too much water.
I can’t believe everybody is getting so complicated and still missing the point.
Imagine a glass of water, half filled.
Add a small rock to the glass of water, and the water level rises. The water level rises in a manner directly related to the volume of the rock. For instance, if you could calculate the the volume of the rock to be 5 cubic centimeters, then the water level will have risen the same as if you had simply poured 5ml more water in.* Anyway, the rock sank. But, it had to move that water upward, and in response, the water is exerting an opposing force downward, with a net change equal to the weight of the water that was lifted — that is, the volume of the rock multiplied by the density of the water.
So, how does something float? It floats when there is an equilibrium between the weight of water getting pushed upward, and the weight of the object doing the pushing. The equilibrium is only reached when the object continues to have excess volume as it ‘tries’ to press into the water — that is, it must be overall less dense than water (although not necessarily homogenous in density, for example a boat that is essentially filled with air). This happens in any container of water, even if it is as big as the ocean.
* In fact this is a nice way of imperically determining the volumes of things.
Edit:
So that was the simple version. Here’s something for a six year old, when you’re ready.
You wonder why the rock is pushing down *and* the water that was elevated is also pushing down? You wonder where is the balancing upward push? Well, the water actually is pushing in all directions, upon itself and upon every surface. So beneath the sunken rock water is pushing upward, just as it pushes upward from below anything buoyant. But, since the rock is now below the waterline, water is also pushing downward on it — very nearly negating the upward push.
But it doesn’t exactly negate the upward push. Remember the water pushes upon itself as well. This we call pressure. The pressure of the water may as well be identical to the weight of the water above itself at any given depth. So for example, if you examined a 1″*1″ slice of water that was 128″ below the surface, that slice has roughly 8 lbs total extra pressure than exists at the surface. (Because 1″*1″*128″ is a gallon, and weighs 8 lbs.) Or, as we say, any sized slice of water at 128″ depth has roughly 8 lbs *per* square inch more than that water at the top surface.
So what does that do? It actually compresses the water, increasing its density slightly but measurably at increasing depths. So that as an object pushes through deeper and deeper water, heavier volumes are being lifted. Objects that normally have the same density as water (water near its own surface that is), but which are more resistant to compression than water is, can sink below the surface, and yet find equilibrium before contacting the bottom. This is important to how fish and submarines control their depth, and it relates to those cool thermometers that have labeled weights on glass balls of air, and is exactly how those cheap antifreeze concentration detectors function.
It’s also why a rock slows down as it approaches the bottom of a pool — and why you can’t say that the pressure from below the rock is exactly the same as the pressure from above. | 1,646 | 7,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.949856 |
https://pqdtopen.proquest.com/doc/1654777644.html?FMT=ABS | 1,619,008,884,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039536858.83/warc/CC-MAIN-20210421100029-20210421130029-00542.warc.gz | 562,019,005 | 6,764 | COMING SOON! PQDT Open is getting a new home!
ProQuest Open Access Dissertations & Theses will remain freely available as part of a new and enhanced search experience at www.proquest.com.
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Dissertation/Thesis Abstract
Studies of imbalance difference theory in modeling conversion between differential mode and common mode signals
by Niu, Li, Ph.D., Clemson University, 2014, 73; 3680740
Abstract (Summary)
This dissertation describes three related studies regarding the imbalance difference theory in modeling the conversion between differential mode and common mode/antenna mode signals. The topics covered are: rigorous derivation of imbalance difference theory for modeling radiated emission problems, modeling the conversion between differential mode and common mode propagation in transmission lines, and modeling the loading impedance on differential mode signals due to radiated emissions.
The imbalance difference theory describes a method for calculating the coupling between differential mode signals and common mode signals due to changes in electrical balance on a transmission line. It provides both physical insight and a simple technique for modeling the conversions between the two modes.
The first chapter presents a rigorous derivation of imbalance difference theory for modeling radiated emission problems. Although the theory has been successfully used to model a wide variety of important EMC problems over the past, it has not been rigorously derived. The derivation carefully defines the important quantities and demonstrates that imbalance difference calculations are exact provided that the differential-mode propagation is TEM and the current division factor, h, represents the actual ratio of currents on the two transmission line conductors excited by a common-mode source. This chapter also discusses the acquisition of the current division factor from 2D calculations of the cross-section of the transmission line.
The second chapter provides a rigorous development of the imbalance difference theory for three-conductor transmission lines where both the differential mode and common mode exhibit TEM propagation. It also derives expressions for the mode conversion impedances, which account for the energy converted from one mode to the other. They are essential for modeling the conversion between the two modes when they are strongly coupled.
The third chapter introduces conversion impedance to the existing imbalance difference theory model for modeling radiated emission problems, so that when the coupling between differential mode and antenna mode are strong, the imbalance difference theory can more accurately estimate the antenna mode current.
All three papers are about confirming, enriching and expanding the imbalance difference theory. The first chapter focuses on the rigorous derivation of theory for its most common application, radiated emission problems. The second chapter expands the theory to multi-conductor transmission line structure when the two modes are strongly coupled. The last chapter introduces conversion impedance to the theory in modeling radiated emission problems and improves the accuracy of the model at resonant frequencies.
Indexing (document details)
Advisor: Hubing, Todd Commitee: Onori, Simona, Pisu, Pierluigi, Prucka, Robert School: Clemson University Department: Automotive Engineering School Location: United States -- South Carolina Source: DAI-B 76/06(E), Dissertation Abstracts International Source Type: DISSERTATION Subjects: Automotive engineering, Electrical engineering, Electromagnetics Keywords: Antenna mode, Common mode, Conversion impedance, Differential mode, Imbalance difference theory, Mode conversion Publication Number: 3680740 ISBN: 978-1-321-53055-1
Copyright © 2021 ProQuest LLC. All rights reserved. Terms and Conditions Privacy Policy Cookie Policy | 697 | 3,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-17 | latest | en | 0.914632 |
https://electronics2electrical.com/6214/ | 1,563,507,331,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525974.74/warc/CC-MAIN-20190719032721-20190719054721-00151.warc.gz | 375,281,196 | 22,360 | # When an alternating current passes through an ohmic resistance the electrical power converted into heat is?
181 views
When an alternating current passes through an ohmic resistance the electrical power converted into heat is?
–1 vote
When an alternating current passes through an ohmic resistance the electrical power converted into heat is true power.
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What happens to the resistance of a wire when its length as well as the cross-sectional area are doubled? A) It doubles B) It becomes 4 times C) It becomes 2π times D) It remains the same | 1,035 | 4,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.906364 |
https://makezilla.com/2016/02/15/homopolar-motor-revisited/ | 1,606,550,903,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195198.31/warc/CC-MAIN-20201128070431-20201128100431-00442.warc.gz | 390,649,546 | 17,793 | # Homopolar Motor Revisited
Written by: techguru
Hi there,
so this is a quick project you can make in a few minutes. It's so easy and simple that
A small kid could do it. If you ever run out of ideas it could be a nice presents you
could make for someone you like or for yourself as decoration. So kids if you don’t know
what to make for fathers or mother’s days make him or her a custom homopolar motor and
am pretty sure they will love it. I’ve got one myself siting on my desktop. Without further
ado let’s get started. First of all you will have to gather the required materials.
As you can see the list is not long and you most probably already have these at home.
### Step 1:
Step 1: get
the materials you’ll be needing
For the homopolar motor:
a battery – I’m using a AAA
Copper wire - about 20 cm
Neodymium – about the same diameter of the battery(here 1 cm),
For the small figurine :
Capacitor,
6 X resistor
LED
### Step 2:
For the
copper wire, I took a piece of coaxial cable that was lying around and peel.
You could also use electrical wires as long as these are made of copper and have the required thickness (~1mm).
### Step 3:
Step 2 : For
the second step you will have to bend the copper wire in an spiral form. For this
you can use a hard cylindrical object which should be a little bigger than the battery you are going to use, and a plier. Cut the end of the wire which look like a hook to an angle, this will make it sharp and thus cause less resistance while turning.
### Step 4:
Step 3 : You
can use a hammer and nail to make a dent on the surface of the battery so that
The wire does not slip off it while turning. This will make your construction much stable.
### Step 5:
Step 4: Now
you can weld your small figurine made of a capacitor, 6 resistors and an LED like
Shown in the picture below.
### Step 6:
Finally you
just assemble it like in the video below and picture. You can bend the figuring so
as to make different poses and use it to fine tune the homopolar motor. If your motor is not turning, you will have to adjust the spiral until it is completely loose and can spin freely. Small frictions may prevent it from turning.
http://www.instructables.com/files/orig/F0B/ONZJ/HXDOOQ4D/F0BONZJHXDOOQ4D.three_gp | 575 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-50 | latest | en | 0.936154 |
https://www.bartleby.com/solution-answer/chapter-21-problem-2114ex-accounting-27th-edition/9781337272094/break-even-analysis-the-junior-league-of-yadkinville-california-collected-recipes-from-members-and/61df947d-98de-11e8-ada4-0ee91056875a | 1,603,299,914,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876768.45/warc/CC-MAIN-20201021151342-20201021181342-00284.warc.gz | 638,263,871 | 71,369 | # Break-even analysis The Junior League of Yadkinville, California, collected recipes from members and published a cookbook entitled Food for Everyone. The book will sell for $18 per copy. The chairwoman of the cookbook development committee estimated that the club needed to sell 2,000 books to break even on its$4,000 investment. What is the variable cost per unit assumed in the Junior League's analysis?
### Accounting
27th Edition
WARREN + 5 others
Publisher: Cengage Learning,
ISBN: 9781337272094
### Accounting
27th Edition
WARREN + 5 others
Publisher: Cengage Learning,
ISBN: 9781337272094
#### Solutions
Chapter
Section
Chapter 21, Problem 21.14EX
Textbook Problem
## Break-even analysisThe Junior League of Yadkinville, California, collected recipes from members and published a cookbook entitled Food for Everyone. The book will sell for $18 per copy. The chairwoman of the cookbook development committee estimated that the club needed to sell 2,000 books to break even on its$4,000 investment. What is the variable cost per unit assumed in the Junior League's analysis?
Expert Solution
To determine
Break-even Analysis: It refers to an analysis of the level of operations at which a company experiences its revenues generated is equal to its costs incurred. Thus, when a company reaches at its break-even, it reports neither an income nor a loss from operations. The formula to calculate the break-even point in sales units is as follows:
Break-evenpointinSales(units) =FixedCostsContributionMarginperunit
To compute: the variable cost per unit.
### Explanation of Solution
Compute the variable cost per unit.
Fixed cost =$4,000 Selling price per unit =$18
Break-even sales (units) =2,000 units
Break-evenpointinSales(units) =FixedCostsContributionMarginperunit2,000units=\$4,000(Sellingpriceperunit)
### Want to see the full answer?
Check out a sample textbook solution.See solution
### Want to see this answer and more?
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
See solution | 465 | 2,084 | {"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} | 3.28125 | 3 | CC-MAIN-2020-45 | latest | en | 0.850909 |
https://www.helpingwithmath.com/printables/worksheets/numbers/fac0601factors03a.htm | 1,585,744,273,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505730.14/warc/CC-MAIN-20200401100029-20200401130029-00156.warc.gz | 976,697,779 | 7,224 | # Prime Factorization with Factor Trees (2 of 2)
------ Note: The Information above this point will not be sent to your printer --------
Fill in the missing numbers in the factor trees then write the prime factorization (the first one is done for you):
12 2 x 6 2 x 3
48 4 x 12 2 x 2 2 x 6 2 x 3
Prime Factorization: 2 x 2 x 3
Prime Factorization: 2 x 2 x 2 x 2 x 3
60 4 x 15 2 x 2 3 x 5
48 6 x 8 2 x 3 2 x 4 2 x 2
Prime Factorization: 2 x 2 x 3 x 5
Prime Factorization: 2 x 2 x 2 x 2 x 3
80 4 x 20 2 x 2 2 x 10 2 x 5
100 10 x 10 2 x 5 2 x 5
Prime Factorization: 2 x 2 x 2 x 2 x 5
Prime Factorization: 2 x 2 x 5 x 5
------ Note: The Information below this point will not be sent to your printer --------
A Printable Math Worksheet - By HelpingWithMath.com
## Related Resources
The various resources listed below are aligned to the same standard, (6NS04) taken from the CCSM (Common Core Standards For Mathematics) as the Factors and multiples Worksheet shown above.
Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4 (9 + 2). Apply and extend previous understandings of numbers to the system of rational numbers.
### Worksheet
Similar to the above listing, the resources below are aligned to related standards in the Common Core For Mathematics that together support the following learning outcome:
Compute fluently with multi-digit numbers and find common factors and multiples | 527 | 1,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-16 | latest | en | 0.764608 |
http://www.chegg.com/homework-help/cobol-for-the-21st-century-11th-edition-chapter-10-solutions-9780471722618 | 1,469,406,040,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824201.28/warc/CC-MAIN-20160723071024-00124-ip-10-185-27-174.ec2.internal.warc.gz | 367,821,999 | 16,359 | View more editions
Solutions
COBOL for the 21st Century
# TEXTBOOK SOLUTIONS FOR COBOL for the 21st Century 11th Edition
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 2
Explanation:
• After printing the total of the previous record, the total field WS-TOTAL needs to be reinitialized to ZERO.
MOVE ZERO TO WS-TOTAL
• The control break processing depends upon whether the current record ACCT-NO is equal to the hold field WS-HOLD-ACCT.
MOVE ACCT-NO TO WS-HOLD-ACCT
• The control break procedure works fine for all records except the first record.
• The first record in the field will not be equal to WS-HOLD-ACCT. This triggers a control break in the program.
• To overcome this problem, the statements after the PERFORM statements are executed in 200-DETAIL-RTN after the first record of a new control group is read at the end of 300-ADD-IT-UP.
• Then the control returns to 100-MAIN-MODULE, where another input record is read.
• Step 2 of 2
Corrected code:
PROCEDURE DIVISION.
100-MAIN-MODULE.
* Open the data file as input access and report file as output access
OPEN INPUT TRANS-FILE
OUTPUT PRINT-FILE
* Read each record from the file and find the total
PERFORM UNTIL THERE-ARE-NO-MORE-RECORDS
AT END
MOVE 'NO ' TO ARE-THERE-RECORDS
NOT AT END
MOVE ACCT-NO-IN TO WS-HOLD-ACCT
PERFORM 200-DETAIL-RUN
END-PERFORM
* Close the files
CLOSE TRANS-FILE
PRINT-FILE
STOP RUN.
200-DETAIL-RTN.
UNTIL ACCT-NO IS NOT EQUAL TO
WS-HOLD-ACCT OR THERE-ARE-NO-RECORDS
MOVE WS-HOLD-ACCT TO ACCT-OUT
MOVE WS-TOTAL TO TOTAL-OUT
WRITE PRINT-REC FROM OUT-REC
* Reinitialize the control field to zero
MOVE ZERO TO WS-TOTAL
* Move the control field to hold field to read the
* record
MOVE ACCT-NO TO WS-HOLD-ACCT | 545 | 1,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | longest | en | 0.776968 |
http://cn.metamath.org/mpeuni/trclun.html | 1,657,098,799,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104669950.91/warc/CC-MAIN-20220706090857-20220706120857-00722.warc.gz | 16,233,451 | 7,345 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > trclun Structured version Visualization version GIF version
Theorem trclun 13974
Description: Transitive closure of a union of relations. (Contributed by RP, 5-May-2020.)
Assertion
Ref Expression
trclun ((𝑅𝑉𝑆𝑊) → (t+‘(𝑅𝑆)) = (t+‘((t+‘𝑅) ∪ (t+‘𝑆))))
Proof of Theorem trclun
Dummy variables 𝑥 𝑟 𝑠 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 unss 3930 . . . . . . . . . 10 ((𝑅𝑥𝑆𝑥) ↔ (𝑅𝑆) ⊆ 𝑥)
2 simpl 474 . . . . . . . . . 10 ((𝑅𝑥𝑆𝑥) → 𝑅𝑥)
31, 2sylbir 225 . . . . . . . . 9 ((𝑅𝑆) ⊆ 𝑥𝑅𝑥)
4 vex 3343 . . . . . . . . . . 11 𝑥 ∈ V
5 trcleq2lem 13951 . . . . . . . . . . 11 (𝑟 = 𝑥 → ((𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟) ↔ (𝑅𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)))
64, 5elab 3490 . . . . . . . . . 10 (𝑥 ∈ {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ↔ (𝑅𝑥 ∧ (𝑥𝑥) ⊆ 𝑥))
76biimpri 218 . . . . . . . . 9 ((𝑅𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → 𝑥 ∈ {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)})
83, 7sylan 489 . . . . . . . 8 (((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → 𝑥 ∈ {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)})
9 intss1 4644 . . . . . . . 8 (𝑥 ∈ {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} → {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ⊆ 𝑥)
108, 9syl 17 . . . . . . 7 (((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ⊆ 𝑥)
11 simpr 479 . . . . . . . . . 10 ((𝑅𝑥𝑆𝑥) → 𝑆𝑥)
121, 11sylbir 225 . . . . . . . . 9 ((𝑅𝑆) ⊆ 𝑥𝑆𝑥)
13 trcleq2lem 13951 . . . . . . . . . . 11 (𝑠 = 𝑥 → ((𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠) ↔ (𝑆𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)))
144, 13elab 3490 . . . . . . . . . 10 (𝑥 ∈ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)} ↔ (𝑆𝑥 ∧ (𝑥𝑥) ⊆ 𝑥))
1514biimpri 218 . . . . . . . . 9 ((𝑆𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → 𝑥 ∈ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)})
1612, 15sylan 489 . . . . . . . 8 (((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → 𝑥 ∈ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)})
17 intss1 4644 . . . . . . . 8 (𝑥 ∈ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)} → {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)} ⊆ 𝑥)
1816, 17syl 17 . . . . . . 7 (((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)} ⊆ 𝑥)
1910, 18unssd 3932 . . . . . 6 (((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → ( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥)
20 simpr 479 . . . . . 6 (((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → (𝑥𝑥) ⊆ 𝑥)
2119, 20jca 555 . . . . 5 (((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥))
22 ssmin 4648 . . . . . . . 8 𝑅 {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)}
23 ssmin 4648 . . . . . . . 8 𝑆 {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}
24 unss12 3928 . . . . . . . 8 ((𝑅 {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∧ 𝑆 {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) → (𝑅𝑆) ⊆ ( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}))
2522, 23, 24mp2an 710 . . . . . . 7 (𝑅𝑆) ⊆ ( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)})
26 sstr 3752 . . . . . . 7 (((𝑅𝑆) ⊆ ( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ∧ ( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥) → (𝑅𝑆) ⊆ 𝑥)
2725, 26mpan 708 . . . . . 6 (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 → (𝑅𝑆) ⊆ 𝑥)
2827anim1i 593 . . . . 5 ((( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) → ((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥))
2921, 28impbii 199 . . . 4 (((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥) ↔ (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥))
3029abbii 2877 . . 3 {𝑥 ∣ ((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)} = {𝑥 ∣ (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)}
3130inteqi 4631 . 2 {𝑥 ∣ ((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)} = {𝑥 ∣ (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)}
32 unexg 7125 . . 3 ((𝑅𝑉𝑆𝑊) → (𝑅𝑆) ∈ V)
33 trclfv 13960 . . 3 ((𝑅𝑆) ∈ V → (t+‘(𝑅𝑆)) = {𝑥 ∣ ((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)})
3432, 33syl 17 . 2 ((𝑅𝑉𝑆𝑊) → (t+‘(𝑅𝑆)) = {𝑥 ∣ ((𝑅𝑆) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)})
35 simpl 474 . . . . . 6 ((𝑅𝑉𝑆𝑊) → 𝑅𝑉)
36 trclfv 13960 . . . . . 6 (𝑅𝑉 → (t+‘𝑅) = {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)})
3735, 36syl 17 . . . . 5 ((𝑅𝑉𝑆𝑊) → (t+‘𝑅) = {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)})
38 simpr 479 . . . . . 6 ((𝑅𝑉𝑆𝑊) → 𝑆𝑊)
39 trclfv 13960 . . . . . 6 (𝑆𝑊 → (t+‘𝑆) = {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)})
4038, 39syl 17 . . . . 5 ((𝑅𝑉𝑆𝑊) → (t+‘𝑆) = {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)})
4137, 40uneq12d 3911 . . . 4 ((𝑅𝑉𝑆𝑊) → ((t+‘𝑅) ∪ (t+‘𝑆)) = ( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}))
4241fveq2d 6357 . . 3 ((𝑅𝑉𝑆𝑊) → (t+‘((t+‘𝑅) ∪ (t+‘𝑆))) = (t+‘( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)})))
43 fvex 6363 . . . . . 6 (t+‘𝑅) ∈ V
4436, 43syl6eqelr 2848 . . . . 5 (𝑅𝑉 {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∈ V)
45 fvex 6363 . . . . . 6 (t+‘𝑆) ∈ V
4639, 45syl6eqelr 2848 . . . . 5 (𝑆𝑊 {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)} ∈ V)
47 unexg 7125 . . . . 5 (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∈ V ∧ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)} ∈ V) → ( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ∈ V)
4844, 46, 47syl2an 495 . . . 4 ((𝑅𝑉𝑆𝑊) → ( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ∈ V)
49 trclfv 13960 . . . 4 (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ∈ V → (t+‘( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)})) = {𝑥 ∣ (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)})
5048, 49syl 17 . . 3 ((𝑅𝑉𝑆𝑊) → (t+‘( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)})) = {𝑥 ∣ (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)})
5142, 50eqtrd 2794 . 2 ((𝑅𝑉𝑆𝑊) → (t+‘((t+‘𝑅) ∪ (t+‘𝑆))) = {𝑥 ∣ (( {𝑟 ∣ (𝑅𝑟 ∧ (𝑟𝑟) ⊆ 𝑟)} ∪ {𝑠 ∣ (𝑆𝑠 ∧ (𝑠𝑠) ⊆ 𝑠)}) ⊆ 𝑥 ∧ (𝑥𝑥) ⊆ 𝑥)})
5231, 34, 513eqtr4a 2820 1 ((𝑅𝑉𝑆𝑊) → (t+‘(𝑅𝑆)) = (t+‘((t+‘𝑅) ∪ (t+‘𝑆))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1632 ∈ wcel 2139 {cab 2746 Vcvv 3340 ∪ cun 3713 ⊆ wss 3715 ∩ cint 4627 ∘ ccom 5270 ‘cfv 6049 t+ctcl 13945 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1988 ax-6 2054 ax-7 2090 ax-8 2141 ax-9 2148 ax-10 2168 ax-11 2183 ax-12 2196 ax-13 2391 ax-ext 2740 ax-sep 4933 ax-nul 4941 ax-pow 4992 ax-pr 5055 ax-un 7115 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3an 1074 df-tru 1635 df-ex 1854 df-nf 1859 df-sb 2047 df-eu 2611 df-mo 2612 df-clab 2747 df-cleq 2753 df-clel 2756 df-nfc 2891 df-ne 2933 df-ral 3055 df-rex 3056 df-rab 3059 df-v 3342 df-sbc 3577 df-dif 3718 df-un 3720 df-in 3722 df-ss 3729 df-nul 4059 df-if 4231 df-pw 4304 df-sn 4322 df-pr 4324 df-op 4328 df-uni 4589 df-int 4628 df-br 4805 df-opab 4865 df-mpt 4882 df-id 5174 df-xp 5272 df-rel 5273 df-cnv 5274 df-co 5275 df-dm 5276 df-rn 5277 df-res 5278 df-iota 6012 df-fun 6051 df-fv 6057 df-trcl 13947 This theorem is referenced by: (None)
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2. >
3. 正文
# javascript数组排序汇总_javascript技巧
javascript数组排序汇总
```//排序算法
var array = [0,1,2,44,4,
324,5,65,6,6,
34,4,5,6,2,
43,5,6,62,43,
5,1,4,51,56,
76,7,7,2,1,
45,4,6,7,8];
//var array = [4,2,5,1,0,3];
console.log('原始数组');
console.log(array);
array = sorting.shellSort(array);
console.log('排序后的数组')
console.log(array);
}
var sorting = {
//利用sort方法进行排序
systemSort: function(arr){
return arr.sort(function(a,b){
return a-b;
});
},
//冒泡排序
bubbleSort: function(arr){
var len=arr.length, tmp;
for(var i=0;i<len-1;i++){
for(var j=0;j<len-1-i;j++){
if(arr[j]>arr[j+1]){
tmp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = tmp;
}
}
}
return arr;
},
//快速排序
quickSort: function(arr){
var low=0, high=arr.length-1;
sort(low,high);
function sort(low, high){
if(low<high){
var mid = (function(low, high){
var tmp = arr[low];
while(low<high){
while(low<high&&arr[high]>=tmp){
high--;
}
arr[low] = arr[high];
while(low<high&&arr[low]<=tmp){
low++;
}
arr[high] = arr[low];
}
arr[low] = tmp;
return low;
})(low, high);
sort(low, mid-1);
sort(mid+1,high);
}
}
return arr;
},
//插入排序
insertSort: function(arr){
var len = arr.length;
for(var i=1;i<len;i++){
var tmp = arr[i];
for(var j=i-1;j>=0;j--){
if(tmp<arr[j]){
arr[j+1] = arr[j];
}else{
arr[j+1] = tmp;
break;
}
}
}
return arr;
},
//希尔排序
shellSort: function(arr){
var h = 1;
while(h<=arr.length/3){
h = h*3+1; //O(n^(3/2))by Knuth,1973
}
for( ;h>=1;h=Math.floor(h/3)){
for(var k=0;k<h;k++){
for(var i=h+k;i<arr.length;i+=h){
for(var j=i;j>=h&&arr[j]<arr[j-h];j-=h){
var tmp = arr[j];
arr[j] = arr[j-h];
arr[j-h] = tmp;
}
}
}
}
return arr;
}
}
```
40+云计算产品，6个月免费体验 | 635 | 1,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.174534 |
https://stats.stackexchange.com/questions/161909/estimating-the-number-of-classes-from-a-sample | 1,643,256,037,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305052.56/warc/CC-MAIN-20220127012750-20220127042750-00035.warc.gz | 597,329,185 | 32,869 | # Estimating the number of classes from a sample
Suppose I have N smarties, each of which is one of C distinct colours.
Suppose further that N is known and largish (10,000) but C is not, and that for each colour C there are $c_i$ smarties of that colour. Whilst the distribution of $c_i$ is unknown, we have the vague assumption that no $c_i$ will be very large or small, eg no one colour will make up more more than 99 or less than 1 percent of the total number of smarties.
If I take a sample of size n and determine that it contains $x_i$ smarties of each colour, how can I produce a good estimate the total number of classes?
Note -> I'm familiar with http://arxiv.org/pdf/0708.2153.pdf and http://www.jstor.org/stable/2290471 but was wondering if anyone had anything else to contibute. Also I'm concerned with the case where N is large but not infinite, which is fairly distinct from the first paper.
Haas, Peter J., and Lynne Stokes. "Estimating the number of classes in a finite population."
Journal of the American Statistical Association 93.444 (1998): 1475-1487.
I came up with
def random_sample(n,data):
indices = np.random.choice(np.arange(0,len(data)),size=n,replace=False)
# That space efficiency wins no awards.
return data[indices]
def estimate_n_classes(sample,population_size):
n = float(len(sample))
dn = float(len(np.unique(sample)))
f = Counter([count for label, count in Counter(sample).iteritems()])
q = n / population_size
result = ((1 - ((1 - q) * f[1] / n))) ** (-1) * dn
return result
This is an implementation of $D_{uj1}$ from the paper which was sufficient for my needs. | 410 | 1,611 | {"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} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.895999 |
https://www.physicsforums.com/threads/third-newtonian-law-most-fundamental.161963/ | 1,713,071,030,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816864.66/warc/CC-MAIN-20240414033458-20240414063458-00661.warc.gz | 886,120,302 | 17,005 | # Third Newtonian law most fundamental
• flashgordon2!
In summary: Each law is precise enough to describe a particular kind of behavior, but they are all incomplete in the sense that they don't describe everything.
flashgordon2!
third law is equal and opposite reaction . . . the first law is inertia of mass.
From everything I can remember reading, people often talk about the mystery of inertia in bodies . . . well, i finally got around to thinking about this stuff(admittedly), and it seems to me that the first law is a derivable from the third. If you try to pull a set of particles all reacting to your pull(and maybe some initial pushing of contacting a body of particles), then, all those particles will push away(and do some back and forth reactions amongst themselves) . . . so, the first law is explainable by the third.
I've never heard anybody say this, so, sorry is somebody has!
Nope, there are three laws.
Ok, so, really Newton's first law reads:
If a body is subjected to no net force, it receives no net acceleration.
For every force of "A" on "B", there is an equal and opposite force of "B" on "A"
So I don't see how you derive N1L from N3L.
ZM
Last edited:
If you pull at a set of particles, why would they pull away from you unless there was another force acting on them?
In classical electrodynamics we have N1L is true, and N3L is false. This doesn't, however, contradict what you are saying.
Note that Newton's First law is not true for all observers. It is only true for an observed fixed in some inertial frame. Think of the first law as defining the concept of an inertial frame.
The first law only vaguely mentions forces. The second law expounds on this concept, describing what an observer tied to an inertial frame sees when a force is appied to some object. The second law defines force in terms of behavior in an inertial frame. The third law further expands on the concept of force, saying that forces come in equal and opposite pairs.
One cannot start with the third law as the most fundamental because it depends on the concept of a force", defined in the second law. Similarly, the second law depends on the concept of an inertial frame, defined in the first law. Newton was no dummy. He established these laws in the order he did on intentionally.
Also worth noting, is that the first law defining an inertial frame requires that you know how forces act in an inertial frame, which in turn requires the second and third laws. None of Newton's laws are independent, but at the same time they're very vague in the sense of "what exactly is a force?". However, you answer that question by knowing what's acting on what and how.
StatMechGuy said:
but at the same time they're very vague in the sense of "what exactly is a force?"..
Not at all. It is as precise as "points, lines and planes" are in Hilbert's axiomatization of geometry.
## 1. What is the Third Newtonian Law?
The Third Newtonian Law, also known as the Law of Action and Reaction, states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object will exert an equal and opposite force back on the first object.
## 2. Why is the Third Newtonian Law considered the most fundamental?
The Third Newtonian Law is considered the most fundamental because it is the foundation for understanding all other laws of motion. It explains how forces interact with each other and how objects move in response to these forces.
## 3. Can the Third Newtonian Law be applied to all types of forces?
Yes, the Third Newtonian Law can be applied to all types of forces, including gravitational, electromagnetic, and nuclear forces. It is a universal law that governs the interactions between all objects in the universe.
## 4. How does the Third Newtonian Law relate to conservation of momentum?
The Third Newtonian Law is closely related to the law of conservation of momentum. According to this law, the total momentum of a system remains constant, meaning that the total force exerted on a system is always equal to the total force exerted back on it. This is a direct result of the Third Newtonian Law.
## 5. What are some real-life examples of the Third Newtonian Law?
There are many real-life examples of the Third Newtonian Law, such as pushing against a wall, where the wall exerts an equal and opposite force back on you. Another example is the recoil of a gun when fired, where the bullet exerts a force forward, and the gun experiences an equal force in the opposite direction.
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13K | 1,138 | 4,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-18 | latest | en | 0.961594 |
http://nrich.maths.org/public/leg.php?code=5005&cl=3&cldcmpid=1977 | 1,435,861,736,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375095668.34/warc/CC-MAIN-20150627031815-00242-ip-10-179-60-89.ec2.internal.warc.gz | 182,542,354 | 10,348 | # Search by Topic
#### Resources tagged with Properties of numbers similar to Jugs of Wine:
Filter by: Content type:
Stage:
Challenge level:
### There are 66 results
Broad Topics > Numbers and the Number System > Properties of numbers
### Arrange the Digits
##### Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### The Patent Solution
##### Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### Alphabet Soup
##### Stage: 3 Challenge Level:
This challenge is to make up YOUR OWN alphanumeric. Each letter represents a digit and where the same letter appears more than once it must represent the same digit each time.
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Chameleons
##### Stage: 3 Challenge Level:
Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .
### Helen's Conjecture
##### Stage: 3 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Thirty Six Exactly
##### Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Happy Octopus
##### Stage: 3 Challenge Level:
This investigation is about happy numbers in the World of the Octopus where all numbers are written in base 8 ... Find all the fixed points and cycles for the happy number sequences in base 8.
### One to Eight
##### Stage: 3 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
### Whole Numbers Only
##### Stage: 3 Challenge Level:
Can you work out how many of each kind of pencil this student bought?
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### X Marks the Spot
##### Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Can You Find a Perfect Number?
##### Stage: 2 and 3
Can you find any perfect numbers? Read this article to find out more...
### Two Much
##### Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
##### Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Four Coloured Lights
##### Stage: 3 Challenge Level:
Imagine a machine with four coloured lights which respond to different rules. Can you find the smallest possible number which will make all four colours light up?
### Factors and Multiple Challenges
##### Stage: 3 Challenge Level:
This package contains a collection of problems from the NRICH website that could be suitable for students who have a good understanding of Factors and Multiples and who feel ready to take on some. . . .
### Multiply the Addition Square
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Water Lilies
##### Stage: 3 Challenge Level:
There are some water lilies in a lake. The area that they cover doubles in size every day. After 17 days the whole lake is covered. How long did it take them to cover half the lake?
### Got it Article
##### Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
### Factorial
##### Stage: 4 Challenge Level:
How many zeros are there at the end of the number which is the product of first hundred positive integers?
### Six Times Five
##### Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
### Writ Large
##### Stage: 3 Challenge Level:
Suppose you had to begin the never ending task of writing out the natural numbers: 1, 2, 3, 4, 5.... and so on. What would be the 1000th digit you would write down.
### Lesser Digits
##### Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Guess the Dominoes
##### Stage: 1, 2 and 3 Challenge Level:
This task depends on learners sharing reasoning, listening to opinions, reflecting and pulling ideas together.
### Clever Carl
##### Stage: 2 and 3
What would you do if your teacher asked you add all the numbers from 1 to 100? Find out how Carl Gauss responded when he was asked to do just that.
### Babylon Numbers
##### Stage: 3, 4 and 5 Challenge Level:
Can you make a hypothesis to explain these ancient numbers?
### Small Change
##### Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### A Long Time at the Till
##### Stage: 4 and 5 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Cogs
##### Stage: 3 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### An Introduction to Irrational Numbers
##### Stage: 4 and 5
Tim Rowland introduces irrational numbers
### Times Right
##### Stage: 3 and 4 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
### Triangular Triples
##### Stage: 3 Challenge Level:
Show that 8778, 10296 and 13530 are three triangular numbers and that they form a Pythagorean triple.
### The Codabar Check
##### Stage: 3
This article explains how credit card numbers are defined and the check digit serves to verify their accuracy.
### Lastly - Well
##### Stage: 3 Challenge Level:
What are the last two digits of 2^(2^2003)?
### Enriching Experience
##### Stage: 4 Challenge Level:
Find the five distinct digits N, R, I, C and H in the following nomogram
### Really Mr. Bond
##### Stage: 4 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
### Not a Polite Question
##### Stage: 3 Challenge Level:
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...
### Cinema Problem
##### Stage: 3 and 4 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Sept 03
##### Stage: 3 Challenge Level:
What is the last digit of the number 1 / 5^903 ?
### Mini-max
##### Stage: 3 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Difference Dynamics
##### Stage: 4 and 5 Challenge Level:
Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens?
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Magic Letters
##### Stage: 3 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws? | 2,285 | 9,249 | {"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} | 3.546875 | 4 | CC-MAIN-2015-27 | longest | en | 0.819826 |
https://www.instasolv.com/question/5-find-the-value-of-lem-fan-160-1-5-find-the-value-of-lim-an-icos-n-d69vjb | 1,610,856,326,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703509104.12/warc/CC-MAIN-20210117020341-20210117050341-00718.warc.gz | 839,554,923 | 10,665 | 5. Find the value of Lem [fan +160 ...
Question
# 5. Find the value of Lem [fan +160 (1)-- 5. Find the value of Lim an Icos | n->
JEE/Engineering Exams
Maths
Solution
96
4.0 (1 ratings)
( lim _{n rightarrow infty} frac{2}{pi}(n+1) cos ^{-1}left(frac{1}{n}right)-n ) ( lim _{n rightarrow infty} nleft[frac{2}{pi}left(1+frac{1}{n}right) cos ^{-1}(1 / n)-1right]=lim _{n rightarrow infty} frac{frac{2}{pi} cos ^{-1}left(frac{1}{n}right)-1}{1 / n} ) ( =lim _{n rightarrow infty} frac{frac{2}{pi} frac{x}{sqrt{1-(1 / n)^{2}}}}{-x_{n}^{2}} leq frac{-2}{pi} times frac{1}{sqrt{5-0}}=frac{-2}{pi} ) | 251 | 592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-04 | latest | en | 0.285678 |
http://stackoverflow.com/questions/4048573/cross-browser-engine-math-pi-is-always-3-141592653589793 | 1,435,928,494,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375096061.71/warc/CC-MAIN-20150627031816-00259-ip-10-179-60-89.ec2.internal.warc.gz | 233,023,936 | 17,535 | # Cross-Browser/Engine Math.PI is always 3.141592653589793?
just a really random question but is the property Math.PI in javascript always 3.141592653589793 in every browser/engine?
-
mmmmmmmmmmmm pi – scunliffe Oct 29 '10 at 2:40
i was actually thinking about that someone will do this... ;) – Tobias Oct 29 '10 at 19:55
I should hope so.
The ECMAScript Spec says:
### 15.8.1.6 PI
The Number value for π, the ratio of the circumference of a circle to its diameter, which is approximately `3.1415926535897932`.
-
"approximately" - doesn't that mean they could make it 3.1415926535897931 ? :-) Or even just 3.14159? Or even 3 for that matter. – paxdiablo Oct 29 '10 at 2:50
Pretty sure they say "approximately" because PI is an irrational number (though the wording is admittedly confusing). – Ed S. Oct 29 '10 at 2:58
The term "approximately" describes the relationship between 3.1415926535897932 and the exact value of PI. Earlier in the spec it is noted that the phrase "the Number value" has a technical meaning defined in section 8.5, which states that it must be the closest IEEE-754 double precision value. +1, but the answer would be improved by incorporating this information explicitly. – Philip Starhill Oct 29 '10 at 3:20
yeah, it would be pretty interesting to know if those Math object static values like PI etc. are all the same in every javascript implementation... but i guess you're right. – Tobias Oct 29 '10 at 19:55
This is what the spec says about `Math.PI`:
The Number value for π, the ratio of the circumference of a circle to its diameter, which is approximately 3.1415926535897932.
Note that it says "approximately", which means it makes no guarantees about the exact value. However since numbers are required to use the standard IEEE 754 representation, you can expect most implementations to have a similar accuracy.
-
It says "approximately" because PI is an irrational number. Remember 6th grade? – Ed S. Oct 29 '10 at 2:48 | 514 | 1,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | latest | en | 0.921944 |
https://www.w3resource.com/java-exercises/search/java-search-exercise-3.php | 1,718,353,816,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00425.warc.gz | 989,126,003 | 27,916 | Java - Find an element in a sorted array using Jump Search
# Java: Find a specified element in a given sorted array of elements using Jump Search
## Java Search: Exercise-3 with Solution
Write a Java program to find a specified element in a given sorted array of elements using Jump Search.
From Wikipedia, in computer science, a jump search or block search refers to a search algorithm for ordered lists. It works by first checking all items Lkm, where ℜ ∈ ℵ and m is the block size, until an item is found that is larger than the search key. To find the exact position of the search key in the list a linear search is performed on the sublist L[(k-1)m, km].
Algorithm:
Input: An ordered list L, its length n and a search key s.
Output: The position of s in L, or nothing if s is not in L.
``` a ← 0
b ← [√n]
while Lmin(b,n)-1 < s do
a ← b
b ← b + [√n]
if a ≥ n then
return nothing
while La < s do
a ← a + 1
if a = min(b,n)
return nothing
if La = s then
return a
else
return nothing
```
Sample Solution:
Java Code:
``````public class Main {
public static void main(String[] args) {
int nums[] = {1, 2, 3, 4, 5, 6, 7, 8, 21, 34, 45, 91, 120, 130, 456, 564};
int search_num = 120;
// Find the index of searched item
int index_result = jumpSearch(nums, search_num);
System.out.println(search_num + " is found at index " + index_result);
}
public static int jumpSearch(int[] nums, int item) {
int array_size = nums.length;
// Find block size to be jumped
int block_size = (int)Math.floor(Math.sqrt(array_size));
// If the element is present find the block where element is present
int prev_item = 0;
while (nums[Math.min(block_size, array_size)-1] < item)
{
prev_item = block_size;
block_size += (int)Math.floor(Math.sqrt(array_size));
if (prev_item >= array_size)
return -1;
}
// Using a linear search for element in block beginning with previous item
while (nums[prev_item] < item)
{
prev_item++;
if (prev_item == Math.min(block_size, array_size))
return -1;
}
// If element is found
if (nums[prev_item] == item)
return prev_item;
return -1;
}
}
```
```
Sample Output:
```120 is found at index 12
```
Flowchart:
Java Code Editor: | 595 | 2,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-26 | latest | en | 0.614973 |
https://www.speakingtree.in/blog/mathematics-of-mantra | 1,653,598,903,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00140.warc.gz | 1,167,834,560 | 30,086 | ## Top Searches:
Jul 11, 2013, 15:29 IST
# Mathematics of Mantra
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In my previous blog , we discussed about the mathematics of breathing and the basic concept of controlled breathing. In this blog, we shall see the mathematics behind reciting mantras.
First , let us take "Gayathri mantra". I suppose this requires no introduction and most of us are familiar with this mantra. It is said that one person should not recite the mantra more than 3000 times in a day. Well , what is the science behind that number 3000 ? To answer this , we need to first understand the relevance and the power of the mantras upon human body. The Gayathri is a very powerful mantra if practiced in sincere belief. The Gayathri is to be recited in the proper chandhas 3 sessions in a day , maximum of 1000 times per session. Why this maximum limit ?
If you study the Gayathri, its made of 24 aksharas, So please read the below simple calculation:
24 X 1000 times in one session = 24,000
24,000 X 3 sessions in a day = 72,000
Our body is said to comprise of 72,000 nerves ( called as naadis ). This was well know to the ancient people and is also now scientifically proven. Reciting the mantra is like charging each of these nerves. Reciting the mantra for maximum 3000 charges the entire body comprising of 72,000 nerves. While recitation of the mantra is good , it has to be within this limit. Practicing beyond 3000 a day would overcharge the body and may lead to problems like frustration , depression , neglection of ones duties , impatience , anger , dejection with worldly affairs, etc. I hope that the reader can compare that to a bulb of x volts, if more voltage is supplied , the bulb cannot withstand the excess power and would burst. Similarly, the mantra is nothing but concentrated energy which affects positively if done within limits.
Similarly , the ashtakshara mantra is made of 8 aksharas.
8 X 9000 times in a day = 72,000
So, the ashtakshara can be repeated to a maximum of 9000 in a day. Or three times the Gayathri.
I hope to have shared the maths and science behind our beautiful and wonderful practices. If one tries to understand and then perform any sadhana it would become enjoyable. Instead of blindly following , let us seek to understand and then with that knowledge , perform the daily practices with belief.
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Problem 44949. Find the Best Hotels
Solution 2189470
Submitted on 2 Apr 2020 by Karl Ezra Pilario
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Test Suite
Test Status Code Input and Output
1 Pass
hotels =["CityLights";"SeaView";"MarketPlace";"ResortSpa";"Nightingale";"Clubadub";"SkylineView";"MarinaBay";"ComfortFirst";"VillageValley"]; ratings = [7.2;8.7;6.5;9.3;4.3;6.9;8.8;5.9;7.4;9.1]; cutoff = 8; good_correct = ["SeaView";"ResortSpa";"SkylineView";"VillageValley"]; assert(isequal(find_good_hotels(hotels,ratings,cutoff),good_correct))
2 Pass
hotels = ["ComfortFirst";"CityLights";"Clubadub";"Nightingale";"MarketPlace";"MarinaBay";"ResortSpa";"VillageValley";"SkylineView";"SeaView"]; ratings = [8.8000;7.2000;9.3000;8.7000;6.9000;7.4000;6.5000;4.3000;5.9000;9.1000]; cutoff = 9; good_correct = ["Clubadub";"SeaView"]; assert(isequal(find_good_hotels(hotels,ratings,cutoff),good_correct))
3 Pass
hotels = ["Nightingale";"VillageValley";"SeaView";"CityLights";"ResortSpa";"ComfortFirst";"SkylineView";"Clubadub";"MarinaBay";"MarketPlace"]; ratings = [7.2000;8.7000;6.5000;7.4000;9.3000;9.1000;6.9000;8.8000;5.9000;4.3000]; cutoff = 7; good_correct = ["Nightingale";"VillageValley";"CityLights";"ResortSpa";"ComfortFirst";"Clubadub"]; assert(isequal(find_good_hotels(hotels,ratings,cutoff),good_correct))
4 Pass
hotels = ["Nightingale";"VillageValley";"SeaView";"CityLights";"ResortSpa";"ComfortFirst";"SkylineView";"Clubadub";"MarinaBay";"MarketPlace"]; ratings = [7.2000;8.7000;6.5000;7.4000;9.3000;9.1000;6.9000;8.8000;5.9000;4.3000]; cutoff = 7.2000; good_correct = ["Nightingale";"VillageValley";"CityLights";"ResortSpa";"ComfortFirst";"Clubadub"]; assert(isequal(find_good_hotels(hotels,ratings,cutoff),good_correct))
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BSEB Class 12 Chemistry Chemical Kinetics Textbook Solutions PDF: Download Bihar Board STD 12th Chemistry Chemical Kinetics Book Answers
BSEB Class 12 Chemistry Chemical Kinetics Textbook Solutions PDF: Download Bihar Board STD 12th Chemistry Chemical Kinetics Book Answers
BSEB Class 12th Chemistry Chemical Kinetics Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 12th Chemistry Chemical Kinetics Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 12th Chemistry Chemical Kinetics books are published by the Bihar Board Publishers. These Bihar Board Class 12th Chemistry Chemical Kinetics textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 12th Chemistry Chemical Kinetics book solutions pdf online from this page.
Bihar Board Class 12th Chemistry Chemical Kinetics Books Solutions
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Find below the list of all BSEB Class 12th Chemistry Chemical Kinetics Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:
Question 1.
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Question 2.
In a reaction 2A → Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during the reaction.
Rate of reaction = rate of disappearance of A
Question 3.
For a reaction, A + B → Product, the rate law is given by r = k [A]1/2 [B]2 what is the order of the reaction?
Order of a reaction is the sum of the powers to which the concentrations of the reactants have to be raised in order to specify the rate of the reaction
r = k [A]1/2 [B]2
Order of the reaction = 12 + 2 = 212
Question 4.
The conversion of molecules X to Y follows second order kinetics. If concentration of x is increased to three times, how will it affect the rate of formation of Y?
X → Y
Rate = k[X]2 … (i)
If cone of x is increased to 3 times
[X’] = [3X]
new rate will be = k [3X]2
= k.9 [X]2 = 9.k [X]2 … (ii)
From (i) and (ii) new rate will be 9 times.
Question 5.
A first order reaction has a rate constant 1.15 × 10-3 s-1, How long will 5g of this reactant take to reduce to 3g?
Question 6.
Time required to decompose SO2 Cl2 to half of its inital amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
t1/2 for SO2 Cl2 = 60 min = 60 × 60 sec
Since it is in first order reaction
∴ k = 0.693t1/2 = 0.69360×60
∴ rate constant of the reaction = 1.925 × 10-4 s-1.
Question 7.
What will be the effect of temperature on rate constant?
Most of the chemical reactions are accelerated by increase in temperature, whether the reactions are endothermic or exothermic. In a mixture of KMnO4and H2C2O4, potassium permangnate (KMnO4) gets decolourised faster at higher temperature than at a lower temperature. It has been found that for a rise of 10° in temp., the rate constant of of a reaction is nearly doubled. The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation
k = Ae-Ea/RT
Where A is the Arrhenius or frequency factor. It is also called PRE-EXPONENTIAL factor. It is a constant specific to a reaction.
R = Gas constant; Ea is the activation energy measured in joules mole.
Question 8.
The rate of the chemical reaction doubles for an increase of’ 10 K in absolute temperature from 298 K. Calculate Ea.
T1 = 298 K
T2 = 298 + 10 = 308 K
R = 8.314 JK-1 mol-1
If initial rate constant is k1, then for 10° rise in temp.
final rate constant = 2k,
∴ putting these values in Arrhenius equation.
Question 9.
The activation energy for the reaction 2HI (g) → H2 + I2, (g) is 2.09.5 kj mol-1 at 581 K. Calculate the fraction of the molecules of reactants having energy equal to or greater than activation energy.
Fraction of molecules with energy equal to or greater than activation energy, x = e-Ea/RT
Bihar Board Class 12 Chemistry Chemical Kinetics Text Book Questions and Answers
Question 1.
From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants.
(i) 3NO(g) → N2(g); Rate = k[NO]2
(ii) H2O2(aq) + 3I(aq) + 2H+ → 2H2O(I)3 ; Rate = k[H2O2] [I]
(iii)CH3CHO(g) → CH4(g) + CO(g) ; Rate = k[CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g); Rate = k[C2H5Cl]
Question 2.
For the reaction 2A + B → A2B the rate = k[A][B]2 with k = 2.0 x 10-6 mol-2 L2 S-1. Calculate the initial rate of the reaction when [A] = 0.1 Mol L-1, [B] = 0.2 Mol L. Calculate the rate of the reaction after [A] is reduced to 0.6 mol-1.
(i) Intial rate = k[A][B]2
= (2.0 ✕ 10-6 mol-2 L2 s-1) (0.1 mol L-1) (0.2 mol L-1)2
= 8.0 ✕ 10-9 mol L-1 s-1
(ii) When [A] is reduced from 0.1 mol L-1 to 0.06 mol L-1
i.e. (0.1 – 0.06) = 0.04 mol L-1 has reacted, B reacted
12 x 0.04 molL-1
= 0.02 mol L-1
Hence new [B] = (0.2 – 0.02) = 0.18 mol L-1
Now, rate = k[A][B]2
= (2.0 × 10-6 mol-2 L2 s-1) × (0.06 mol L-1) (0.18 mol L-1)2
= 3.89 × 10-9 mol L-1 s-1
Question 3.
The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.50 × 10-4 mol L S-1 ?
2NH3 (g) ⇌ N2(g) + 3H2(g)
rate = – 12.Δ(NH3)Δt = Δ(N2)Δt = 13 Δ(H2)Δt
rate = k as it is a zero order reaction
Δ(N2)Δt = 2.50 x 10-4 mol L s-1 and
Δ(H2)Δt = 7.5 x 10-4 mol L s-1
Question 4.
The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = k [CH3OCH3]3/2
The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
Rate = k (PCH3OCH3)3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and constants?
The decomposition reaction is CH3OCH3 → CH4 + H2 + CO
In terms of concentrations
Units of rate = mol L-1 min-1
Question 5.
Mention the factors^ that affect the rate of a chemical reaction.
There are a number of factors which influence the rate of:
The important factors are –
1. Nature of the reacting species.
2. Concentration of the reacting species.
3. Temperature at which a reaction proceeds.
4. Surface area of the reactants.
6. Presence of a catalyst.
Question 6.
A reaction is second order with respect to a reactant. How is the rate of the reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Rate = k [A]2 = ka2
if [A] = 2a, rate = k[2a]2 = 4ka2 = 4 times
if [A] = 12a; rate = k(12)² = 14ka2 = 14th
Thus the rate of a II order reaction becomes 4 times when the concentration of the reactant is doubled and it becomes 14th when the concentration of reactant is halved.
Question 7.
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
The rate constant of a reaction increases with increase in temperature and becomes nearly double for every 10° rise of temperature. The effect can be represented quantitatively by Arrhenius equation.
k = Ae-Ea/RT
where A = constant called frequency factor. It is related to number of binary molecular collisions per second per litre.
Ea = Energy of activation
T = Absolute temperature
N = The integrated form
R = Gas constant
log k2kr = Ea2.303R[ [𝑇2−𝑇1𝑇1𝑇2] ]
where k1 and k2 are the rate constants for the reaction between two different temperatures T1 and T2 respectively.
Question 8.
In a pseudo first order hydrolysis of ester in water the following results were obtained –
(i) Calculate the average rate of the reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Average rate during the interval 30 – 60 sec.
Question 9.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentration of both A and B are doubled?
(i) 𝑑𝑥𝑑𝑡 = k[A][B]2
(ii) rate = kab2. If [B] is tripled.
Rate = ka.(3b)2 = 9 kab2; rate increases 9 times.
(iii) If both [A] and [B] are doubled
then Rate = k (2a) (2b)2 = 8 kab2 = 8 times.
Rate of the reaction increases 8 times.
Question 10.
In a reaction between A and B, the initial rate of the reaction (r0) was measured for different initial concentrations of A and B as given below:
What is the order of the reaction with respect to A and B?
Suppose order with respect to A is α and order w.r.t. B is β Then the rate law will be
Rate = k [A]α [B]β
(Rate)expt (i) = 5.07 × 10-5 = k [0.20]α [0.20]β …(i)
(Rate)expt (ii) = 5.07 × 10-5 = k [0.20]α [0.10]β …(ii)
(Rate)expt (iii) = 1.43 × 10-4 = k [0.40]α [0.05]β …(iii)
From (i) and (ii)
Question 11.
The following results have been obtained during the kinetic studies of the reaction 2A + B → C + D.
Determine the rate law and the rate constant for the reaction.
From Experiment I and IV, it may be noted that [B] is same, but [A] has been made 4 times.
The rate of the reaction also has become four times
∴ Rate w.r.t. A
Rate α [A] … (1)
From experiments II and III, it may be noted that [A] is kept same, and [B] has been doubled, the rate of the reaction has become 4 times. This mean w.r.t. B,
Rate α [B]² … (2)
Combining (1) and (2), we get the rate law of the reaction
2A + B → C + D
Rate Law is Rate = k [A] [B]2
Over – all order of the reaction = 1 + 2 = 3
To calculate rate constant
k = Rate [A][B]2 = 6.0×10−3(0.1)(0.1)2 = 6.0
hence rate constant k = 6.0 M-2 m-1
Question 12.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
The reaction between A and B is first order with respect to A and zero order with respect to B.
∴ rate of the reaction w.r.t. A = k [A]
rate of the reaction w.r.t.B = k[B]0
∴ overall rate of the reaction = k [A] [B]0
It is of 1st (1 + 0) order.
From I Experiment,
∴ k = Rate [A][B]∘ = 2.0×10−2[0.1]
= 2.o x 10-1
Now, in Experiment II, the cone, of A will be
The given artefact is 1845 years old.
Question 15.
The experimental data for the decomposition of N2O5 [2N2O5 → 4NO2 + O2] in gas phase at 318 K are given below.
(i) Plot [N2O5] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N2O5] and t.
(iv) What is the rate law?
(v) Calculate the rate constant.
(vi) Calculate the half-life period from k and compare it with (ii).
(a) Plot [N2O5] against t
The different values of k calculated are 4.96 × 10-4 s-1,
4.82 × 10-4 s-1, 5.0 × 10-4 s-1, 5.44 × 10-4 s-1, 5.44 × 10-4 s-1,
5.47 × 10-4 s-1, 5.91 × 10-4 s-1, 5.81 × 10-4 s-1 and 5.71 × 10-4 s-1.
Since the value of k is almost constant, the rate equation is
γ = k[N2O5]
The mean value of rate constant is 5.2873 × 10-4 s-1 order of the reaction = 1.
Half life period t1/2 = 0.693k = 0.6935.2873×10−4 = 1310 s.
Question 16.
The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16 th value?
It is a first order reaction
k = 60 s-1
Let the initial cone, of the reactant = a M L-1
Let t be the time (in seconds) during which its initial concentration reduces to 1/16 of the value.
∴ conc. after time t = 𝑎16
Putting the values in the 1st order square
Hence after 4.6 × 10-2 second, the initial concentration of the reactant reduces to 1/16.
Question 17.
During nuclear explosion, one of the product is 90Sr with half life of 28.1 years. If 1 pg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Half life period of 90Sr = 28.1 years
decay constant (k) = 0.693/28.1
Amount of 90Sr absorbed in the bones of the newly born baby = 1 μg = 10-6gm.
After time t = 10 years, the amount of radioactive 90Sr = N
∴ Amount of 90Sr left after 10 years, 60 years respectively is 0.7812
Question 18.
For a first order reaction show that the time required for 99% completion is twice the time requiered for the competion of 90% of reaction.
Question 19.
A first order reaction takes 40 min for 30%decomposition. Calculation tt/2.
Question 20.
For the composition of azoisopropane to hexane and nitrogen at 543 k, the following data are obtained.
Calculate the rate constant.
Azoisopropane decompose according to the equation.
The decomposition is found out to be of the first order.
Initial pressure P0 = 35.0 mm of Hg.
Decrease of pressure of azoisopropane after time t = P = p
∴ increase in pr. of N2 = p
and increase in pr. of hexane = p.
Total pressure of the mixture = pt = PA + PN2 + Pc16H14
pt = (P0 – P) + P + P = p0 + P
P = Pt – P0
p = Pt – P0
Question 21.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at constant volume.
SO2Cl2(g) → SO2 (g) + Cl2(g)
Calculate the rate of the reaction when total pressure is 0.65 atm.
1. As in question 4.20
= 2.23 × 10-3 s-1
Hence rate constant of the 1st order reaction = 2.23 × lO-3 s-1.
2. To calculate rate of the reaction when total pressure is 0.65 atm.
𝑑𝑥𝑑𝑡 = k(a – x) = (2P0 – Pt) × k
= (1.00 – .65) × 2.23 × lO-3
= 0.35 × 2.23 × lO-3
rate of reaction = 7.8 × 10-4 atm s-1
Question 22.
The rate constant for the decomposition of N2O5 at vsrious temparature is given below:
Draw a graph between k and 1/T and calculate the value of A and Ea. Predict the rate constant at 30°C and 50°
From the given data, log k is calculated for each value of k.
On plotting log k versus 1𝑇, the slope of the line is found to be – 5300. [Students are advised to plot the actual graph on graph paper]
∵ Slope = −𝐸𝑎2.303𝑅
∴ Ea = – 2.303 × 8.314 (-5300)
= 101479.85 JK-1 mol-1
Ea = 101.5 KJ k-1 mol-1
(ii) To calculate the rate constant at 30°C = 303 K.
If T1 = 273 K; k1 = 0.0787 × 10-5
T2 = 303 K K2 = ?
Similarly the rate constant at 323 K is 8.14 x 10-4 s-1.
Question 23.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?
Ea = 179.9 = 179900 J
k = 2.418 × 10-5
and T = 546 K
Question 24.
Consider a certain reaction A → Products with k = 2.0 × 10-2 s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L-1.
The reaction A → Products is a first order reaction ask = 2.0 × 10-2 second L-1.
[The value of k in second-1 indicates it is 1st order reaction]
Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with T1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Since it follows rate law of first order
fraction of sample of sucrose remaining = 0.1576 M
Question 26.
The decomposition of hydrocarbon follows the equation k = 4.5 × 1011 s-1 e28000k/t Calculate Ea.
log k = log A – −𝐸𝑎2.303𝑅 … (i)
The decomposition of the hydrocarbon follows the equation.
k = 4.5 × 1011 s-1 e28000k/t
Applying logs
log k = log 4.5 × 1011 – 28000 T
log10 k = log10 4.5 × 1011 – 280002.303𝑇 K … (ii)
Comparing it with equation no. (i)
−𝐸𝑎2.303𝑅 = 280002.303𝑇
∴ Ea – 28,000 x 8.314 J
= 28 x 8.314 kJ mol-1
Ea = 232.79 kJ mol-1.
Question 27.
The rate constant for the first order decomposition of H2O2</su-2 is given by the following equation:
logk = 14.34 – 1.25 x 104 𝐾𝑇
Calculate Ea for the reaction and at what temperature will its half-period be 256 minutes?
(i) log k = log A – −𝐸𝑎2.303𝑅
The given equation is
log k = 14.34 – 1.25×104𝑇 K
Comparing the two equations
−𝐸𝑎2.303𝑅 = 1.25 x 104
or Ea = 1.25 x 104 x 2.303 x 8.314 J mol-1
= 239.34 k] mol-1
(ii) t1/2 =256 min = 256 x 60 seconds.
Question 28.
The decomposition of A into product has value of k as 4.5 × 10-3 s-1 at 10°C and energy 60kJ mol-1. At what temparature would k be 1.5 × 104 s-1 ?
Question 29.
The time required for 10% completion of a first order reaction at 298 k is equal to that required for its 25% completion at 308 k. if the value of the A is 4 × 1010 s-1, calculate k at 318 k and Ea
Putting A = 4 x 1010 s-1; R = 8.314 x 10-3 x 10-3 JK-1 mol-1,
Ea kJ mol-1, T = 318 k in the above equation,
Question 30.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
T1 = 293;
T2 = 313 K
T2 – T1 = 20°
k2 = 4k1
Bihar Board Class 12 Chemistry Chemical Kinetics Additional Important Questions and Answers
Question 1.
After 5 half-life periods for a first order reaction what fraction of reactant remains?
12.
Question 2.
A reaction is found to be zero order. Will its molecularity be also zero?
No. Molecularity of reaction cannot be zero.
Question 3.
Is there any reaction for which the rate of a reaction does not decrease with time?
For a zero order reaction, the reaction rate does not decrease with time because it is independent of the concentration of the reactants.
Question 4.
Can activation energy for reactants be zero?
In the Arrhenius equation, k = A g-Ea/RT if Ea = 0 then k = A, i.e. every collision between molecules leads to chemical reaction. This is not true. Thus Ea cannot be zero.
Question 5.
What value of K is predicted for the rate constant by Arrhenius equation if T → ∞? Is this value physically reasonable?
From the Arrhenius equation k = A e-Ea/ RT if T → ∞; k → A so that Ea = 0. This is not feasible.
Question 6.
Why AG is postive for some of the photochemical reactions?
This is because a part of the light energy absorbed by the reactants gets converted into free energy.
Question 7.
For a reaction A + H2O → B; rate α [A]. What is its (i) molecularity (ii) order of the reaction?
It is a Pseudo unimolecular reaction with order = 1.
Question 8.
What is the main difference between a photosenstitizer and a catalyst?
A catalyst only alters (changes) the speed of the reaction, while a photosensititizer only initiates the reaction.
Question 9.
How is the rate constant related to concentration of the reactants?
Rate constant does not depend upon the concentration of the reactants.
Question 10.
The rate Law for the decomposition of N2O5 is
Rate = k [N2O2] (A.LS.E. 2002)
What is the significant ace of k in this equation?
If [N2O5] = 1; then Rate = k.
i.e. k is equal to the rate of the reaction when concentration of N2O5 is unity, i.e., 1 mol L-1.
Question 11.
Write expression for half-life in case of a reaction between hydrogen and chlorine to form hydrochloric acid gas. (P.S.B. 2005)
Given reaction is a zero order reaction. Hence 1/2 = [A]0/2k.
Question 12.
For the assumed reaction X2 + 3Y2 → 2XY3, write the rate equation in terms of rate of disappearance of Y2. (H.S.B. 2005)
Question 13.
The reaction A + B → C has zero order. What is the rate equation? (A,I,S,B. 2005)
Rate = k [A]° [B]° = k.
Question 14.
Express the relation between half life period of a reactant with initial concentration for a reaction of nth order. (A,I,S,B. 2005)
t1/2 α 1[ A]n−1.
Question 15.
How does the value of rate constant vary with reactant concentration? (D.S.B. 2005)
For reaction of nth order 𝑑𝑥𝑑𝑡 = Rate = k [conc.]n
Question 16.
If half-life period of a reaction is inversely proportinal to initial concentration of a reactant, what is the order of the reaction? (C.B.S.E. Sample Paper 2003)
Two, this is because for 2nd order reactions t1/2 1/ka.
Question 17.
Give one example of a first order reaction. (A.I.S.B. 1998)
N2 O5 → N2 O4 + 12 O2
Question 18.
For a reaction A → B, the rate of the reaction can be denoted by-dA/dt-ar+dB/dt State the significance of plus and minus signs in this case. (H.S.B.2001)
Minus sign indicates the decrease in the concentration of the reactant while plus sign indicates an increase in the cone, of the product with time.
Question 19.
Why rate of reaction does not remain constant throughout? (P.S.B.2000)
Because it depends upon concentration of the reactions which keeps on decreasing with time.
Question 20.
What is the relationship between rate constant and activation energy of a reaction? (H.S.B. 1999, H.P.S.B. 2004)
k = A e-Ea/RT is Arrhenius equation.
Question 21.
Why hydrolysis of ethyl acetate with NaOH is a second order reaction while with HCl it is a 1st order reaction?
Rate of hydrolysis of ethyl acetate with NaOH depends upon cone, of both while that with HCl, it depends upon only the cone, of ethyl acetate.
Question 22.
Define specific reaction rate or rate constant. (P.S.B.2000)
It is the rate of a reaction when the molar concentration of each of the reactant is unity.
Question 23.
What is the rate determineg step of a reaction?
The slowest step of a multi-step reaction is called its rate-determining step.
Question 24.
Give an example of a reaction having fractional order.
Decomposition of acetaldehyde (order is 1.5)
Question 25.
A substance with initial concentration ‘a’ follows zero order kinetics. In how much time will the reaction go to completion? (D.S.B. 2005)
𝑑𝑥𝑑𝑡 = k or dx = k dt ∴ x = kt + I
when t = 0; x = 0 ∴ I = 0.
Hence x = kt or t = x / k, For completion x = a
∴ t = a/k.
Question 26.
In some cases, it is found that a large no. of colliding molecules have energy more than threshold energy, yet the reaction is slow, why? (A.I.S.B. 2005)
The colliding molecules may not have proper orientation at the time of collision.
Question 27.
Give an example of a pseudo first order reaction. (A.I.S.B. 2004, D.S.B. 2004)
Acid catalysed hydrolysis of ethyl acetate.
Question 28.
What is meant by an elementary reaction? (A.I.S.B. 2004)
A reaction which takes place in one step is called an elementary reaction.
Question 29.
How is half-life period related to initial concentration of a second order reaction? (D.S.B. 2004 C)
t1/2 α a1-n. Forn = 2, t1/2 α a-1, i.e. t1/2 α 1𝑎. Thus half life period of a second order reaction is inversely proportional to the initial concentration of the reactant.
Question 30.
Give one example of a reaction where order and molecularity are equal. (C.B.S.E. Sample Paper 1997)
Order and molecularity are same in a single step reaction.
example: 2HI(g) → H2 (g) + I2 (g)
Question 31.
State any one condition under which a bimolecular reaction may be kinetically of first order. (D.S.B. 2002)
A bimolecular reaction may become kinetically of the first order if one of the reactants is present in excess.
Question 32.
Rate of a reaction is given by the equation: Rate = k [A]² [B]. What are the units for the rate and rate constant for this reaction? (D.S.B. 2001)
Question 33.
How is the rate constant of a reaction related to the concentration of the reactants?
Rate constant does not depend upon the concentration of the reactants.
Question 34.
The rate law for the reaction: Ester + H+ → Acid + alcohol is dx/dt = k [Ester] [H+]°. What would be the effect on the rate if (i) concentration of the ester is doubled, (ii) concentration of H+ is doubled? (P.S.B.2002)
• The rate of the reaction will be doubled.
• No effect on rate.
Question 35.
The form of the rate law for a reaction is expressed as: rate = k [Cl2] [NO]². Find out the orders of the reaction with respect to Cl2 and with respect to NO and also the overall order of the reaction. (A.I.S.B. 1994)
Order w.r.t. Cl2 = 1; Order w.r.t. NO = 2
Over-all order = 1 + 2 = 3.
Question 1.
Write the expressions for rates of reactions in terms of concentration of reactants and products for the following reactions:
(i) 2 N2O5 ⇌ 2N2O4 + O2
(ii) 2NO2 + F2 ⇌ 2NO2F
Question 2.
Assuming that energy of activation for most of the reactions is 52 kJ, what conclusion you draw about the effect of temperature on the rate of a reaction [Based on Arrhenius equation]? (P.S.B. 2005)
Putting T1 = 300 K, T2 = 310 K, Ea = 52,000 J mol-1 in the Arrhenius equation,
we get k2 = 2k1 This shows that for 10° rise of temperature, the rate of the reaction is doubled.
Question 3.
The mechanisms of the photosynthesis of HCl and HBr are similar, yet the quantum efficiency of the former is high while that of latter is very low. Why?
This is because the first stage of the secondary process in the first case, viz., Cl + H2 → HCl + H is exothermic while in the 2nd case Hr + H2 → HBr + H, it is endothermic.
Question 4.
A first order reaction takes 69.3 minutes for 50% completion. Set up an equation for determining the time needed for 80% completion of this reaction (Calculation of result is not required.) (D.S.B. 2005)
t1/2 = 69.3 min
Question 5.
For the gaseous reaction
2NO2 (g) → 2NO (g) + O2 (g)
(i) Write the expression for rate of the reaction.
(ii) If the rate of decrease of concentration of NO2 is 6.0 x 10-12 s-1, calculate the corresponding rates in terms of increase in NO and O2 concentrations.
(i) Rate = k[NO2
(ii) Rate of increase of NO. concentration = 6.0 x 10-12 s-1
Rate of increase in O2 concentration = 3.0 x 10-12 s-1.
Question 6.
(a) Draw a schematic graph showing how the rate of a first order reaction changes with change in concentration of a reactant.
Rate of 1st order reaction increase with -the increase in concentration as shown in the RATE schematic graph.
Question 7.
What type of plot do you expect for the rate versus time for a reaction of zero order?
A straight line parallel to time axis because rate is independent of concentration and hence time.
Question 8.
Following reaction takes place in one step :
2NO (g) + O2 (g) ⇌ 2NO2 (g)
How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one-third of the original volume? Will there be any change in the order of the reaction with reduced volume? (A.I.S.B. 2004)
Rate = K[NO]²[O2]
[NO] = M, [O2] = 𝑎𝑉 M; Let Moles of NO = a and moles of O2 = b and Vol. of the vessel = V litres.
∴ r2/r1 or r2 = 27r1; rate becomes 27 times.
There is no effect on the order of the reaction.
Question 9.
The rate expressions of some reactions are given below:
(a) 2NO2 + F2 → 2NO2F; Rate = k [NO2] [F2]
(b) NO2 + CO → CO2 + NO; Rate = k [NO2]².
Propose the probable mechanism of each of the above reactions.
(a)
Question 19.
What is a photochemcial reaction? Define quantum efficiency of a photochemical reaction.
Photochemical reaction is one which can proceed in the presence of light (or some other radiation). Quantum efficiency (Φ) is expressed as
Question 1.
(a) What is the difference between ORDER and MOLECULARITY of a reaction?
(b) RATE of a reaction and RATE CONSTANT of a reaction?
(c) What are the factors which influence the rate of reaction.
Discuss them in detail.
(a) Difference between ORDER & MOLECULARITY
Molecularity:
• It is the number of reacting species undergoing simul – taneous collision in the reaction.
• It is a theoretical concept.
• It can have integral values only.
• It cannot be zero.
• It does not tell us anything about the mechanism of the reaction.
Order:
• It is sum of the powers of the concentration terms in the rate law expression.
• It is determined experi-mentally.
• It can have fractional values also.
• It can be zero.
• It tells us about the slowest step in the mechanism and hence gives some clue about mechanism of the reaction:
(b) Difference between RATE of a reaction and RATE CONSTANT.
Rate of Reaction:
• It is the speed at which the reactants are converted into products at any moment of time.
• It depends upon the concentration of reactant species at that moment of time.
• It decreases with the progress of reaction generally.
• It has same units for all reactions ML-1S-1.
Rate constant of reaction
• It is a constant of proportionality in the rate law expression.
• It refers to the rate of the reaction at the specific point when concentration of every reacting species is unity.
• It is constant and does not depend on the progress of the reaction.
• It has different units for different reactions.
(c) (1) Nature of the reacting species – Consider the following two reactions:
2NO (g) + O2 (g) → 2NO2 (g) …fast
2CO (g) + O2 (g) → 2CO2 (g) …slow
These reactions appear to be similar but the first is fast, while the second is slow. This is because different amounts of energies are required for breaking of different bonds and different amounts of energies are released in the formation of different bonds.
2. Concentration of the reactants – Greater are the concentrations of the reactants, faster is the reaction. Conversely, if the concentrations of the reactants decrease, the rate of the reaction also decreases. A match stick which burns usually in air (02:21% by volume) bums with a flash in a jar containing 02 gas (volume of 07:100%).
3. Temperature – The rate of a reaction increases with the increase in temperature. In most of the reactions the rate of the reaction gets doubled for 10°C rise in temperature irrespective of the fact whether the reaction is exothermic or endothermic. In some cases reactions do not take place at room temperature but proceed only on heating.
4. Surface area of the reactants – For a reaction involving a solid reactant or catalyst, the smaller is the particle size, i.e., greater is the surface area, the faster is the reaction. Wood shavings bum rapidly in air than a log of wood, because the surface area of wood shavings is much higher.
5. Pressure of a catalyst – A catalyst generally increases the speed of the reaction without itself being consumed in the reaction. In case of reversible reactions a catalyst helps attain chemical equilibrium quickly without disturbing the state of equilibrium. It speeds up the reaction both in the forward as well as backward direction so that equilibrium is attained earlier.
6. Presence of light – Some reactions do not take place in the dark, but take place in the presence of light.
Exaple: H2 + Cl2 → 2HCl
Such reactions are called PHOTO-CHEMICAL REACTIONS.
Question 2.
(a) For the reaction
4 NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (g)
if the rate expression in terms of disapperance of NH3 is – ∆ (NH3)/∆T, write the rate expression in terms of concentration of O2 and H2O,
(b) In the reaction A → B, the value of the rate constant was found to be 1.0 x 10-2 mol-1 L S-1. What is the order of the reaction? How will the catalyst affect the values of the rate constant? (C.B.S.E. Sample Paper 1997)
(c) Identify the order if in a reaction, the rate constant has the units L Mol-1 S-1.
(d) A first order reaction is found to have a rate constant k = 7.39 x 10-5 sec-1 Find the half life period of a reaction.
(e) Explain the process of inversion of cane-sugar.
(a) Rate = – 45Δ[O2]Δt = 46.ΔH2OΔt
(b) Since the rate constant has the units of Mol-1 S-1.
∴ It is a 2nd order reaction catalyst increases the value of the rate constant.
(c) Since the unit of rate constant has the units of L Mol-1 S-1, it is a reaction of the 2nd order.
(d) For a first order reaction
(e) Inversion of cane-sugar – The hydrolysis of sucrose (cane- sugar) in presence of mineral acid takes place according to the following reaction.
An important characteristic of the reaction is that sucrose is dextorotatory, whereas the product – a mixture of glucose and fructose is dextrorotatory, since the laero-rotation of fructose is more (- 92°) than the dextro-rotation of glucose (+ 52.5°). It is for this reason that the reaction is called INVERSION OF CANE SUGAR.
Bihar Board Class 12th Chemistry Chemical Kinetics Textbooks for Exam Preparations
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# (Using Gapfinder) Essay
493 words - 2 pages
Standardized testing is the most commonly used method of testing in the United States as well as other countries around the world. This type of testing is used to determine one’s achievement, growth, and progress through the years based on academic learning. Recently, scholastic performance in America has been strongly focused on mathematics and science education because American children are seemingly lagging behind in those departments. Therefore, standardized test standards have been reconstructed in hopes of helping American children catch up to their Asian peers.
According to recent reports concerning the PISA, American teens are consistently turning in tests that fail to rank them even in the top twenty. “The latest results, ...view middle of the document...
Within my graph, I have selected some countries that are performing better than the United States and others that are performing poorly compared to the United States. While, you can see that the United States’ scores haven’t changed much, neither have Japan’s, Singapore’s or Hong Kong’s. The graph only shows scores up to 2007. In 2011, Japan scored a 570, Singapore scored a 611, and Hong Kong scored a 586, while the United States scored a 509. The average score for the test this year was a 500. Comparing the graph and the 2011 data, Japan stayed consistent; Singapore jumped from the 593 in 2007; and Hong Kong jumped from the 572 in 2007. The United States remained fairly consistent, jumping only one point from the 508 in 2007. [2]
[3]
While the United States isn’t quite stepping up to take the lead, Americans are not satisfied with the fact that we are not number one in the ranking system. The American government is taking steps in order to prevent any more falling behind in the American school system.
Bibliography
(2) https://nces.ed.gov/TIMSS/results11.asp
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### Physic
1. 1. PHYSIC<br />Newton’s Laws Of Motion<br />Name : Norhasnida Bt Abd. Rahman<br />Unit : PPISMP Sains Semester 1<br />Lecturer : Miss RoshidahBinti Mat<br />
2. 2. NEWTON’S FIRST LAWS<br />Laws of inertia.<br />Inertia-tendency of an object to resist changes in its velocity.<br />
3. 3. CONCEPTS OF INERTIA<br />If an object is at rest, it tends to stay in that position unless some forces puts that object into motion, inertia makes the moving object continue to move at a constant speed in the same direction unless some external force changes the object's motion.<br />
4. 4. There are two parts to this statement - one which predicts the behavior of stationary objects and the other which predicts the behavior of moving objects. The two parts are summarized in the following diagram:<br />
5. 5.
6. 6. Situation of Inertia<br />
7. 7. RELATIONSHIP BETWEEN MASS AND INERTIA<br />Two empty buckets which are hung with rope from a the ceiling. <br />One bucket is filled with sand while the other bucket is empty. <br />Then, both pails are pushed. <br />It is found that the empty bucket is easier to push compared to the bucket with sand. <br />The bucket filled with sand offers more resistance to movement. <br />When both buckets are oscillating and an attempt is made to stop them, the bucket filled with sand offers more resistance to the hand (more difficult to bring to a standstill once it has started moving) <br />This shows that the heavier bucket offers a greater resistance to change from its state of rest or from its state of motion. <br />An object with a larger mass has a larger inertia. <br />
8. 8. RELATIONSHIP BETWEEN MASS AND INERTIA<br />MASS<br />INERTIA<br />=<br />The LARGER the mass, the LARGER the inertia<br />
9. 9. RELATIONSHIP BETWEEN INERTIA AND SPEED<br />NERTIA<br />INERTIA<br />SPEED<br />The greater the speed the greater the inertia<br />
10. 10. Concept<br />Concept<br />Effect<br />Effect<br />Example<br />Example<br />
11. 11. momentum<br />=Mass x velocity <br />= mv<br />SI unit: kg ms-1 <br />Momentum<br />
12. 12. Principle of conservation of momentum<br />EXTERNAL<br />FORCE<br />TOTAL MOMENTUM<br />CONSTANT<br />In the absence of an external force, the total momentum of a system remains unchanged. <br />
13. 13. ELASTIC COLLISION<br />Both objects move independently at their respective velocities after the collision. <br />Kinetic energy is conserved. <br />Momentum is conserved <br />Total energy is conserved <br />
14. 14. Total Momentum Before = Total Momentum After <br /> m1u1 + m2u2 = m1v1 + m2v2 <br />
15. 15. INELASTIC COLLISION<br />The two objects combine and move together with a common velocity after the collision. <br />Kinetic energy is not conserved <br />Momentum is conserved <br />Total energy is conserved <br />
16. 16. Total Momentum Before = Total Momentum After <br />m1u1 + m2u2 = (m1 + m2) v <br />
17. 17. explosion<br />
18. 18. EXAMPLE<br /> OF <br />EXPLOSION<br />
19. 19. Newton’s Second Laws<br />concerned with the effect that unbalanced forces have on motion. <br />An unbalanced force acting on an object causes it to accelerate<br />The bigger the unbalanced force acting on the object the bigger the acceleration of the object.<br />UNBALANCED FORCE<br />ACCELERATION<br />
20. 20. RELATIONSHIP BETWEEN FORCE AND MASS<br />MASS<br />FORCE<br />CONSTANT ACCELERATION<br />
21. 21. F is constant<br />m is constant<br />a<br />
22. 22. Force, Mass & Acceleration <br />The acceleration produced by a force on an object is directly proportional to the magnitude of the net force applied and is inversely proportional to the mass of the object. The direction of the acceleration is the same as that of the net force .<br />When a net force, F, acts on a mass, m it causes an acceleration, a. <br />
23. 23.
24. 24. Newton’s third laws<br /> For every action, there is an equal and opposite reaction<br />
25. 25. CONCEPT OF NEWTON’S THIRD LAWS<br />According to Newton, whenever objects A and B interact with each other, they exert forces upon each other. When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body. <br />
26. 26. Example <br />Of<br /> Newton’s Third Laws<br />
27. 27.
28. 28. Newton’s 3rd Law in Nature<br /><ul><li>Consider the propulsion of a fish through the water. A fish uses its fins to push water backwards. In turn, the water reacts by pushing the fish forwards, propelling the fish through the water.
29. 29. The size of the force on the water equals the size of the force on the fish; the direction of the force on the water (backwards) is opposite the direction of the force on the fish (forwards).</li></li></ul><li>NEWTON’S LAWS CONCEPT MAP<br />
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Which one of the following examples represents a proper fraction? A. 15/22 B. 12/9 C. 3 /2 D. 8 /8
15/22 represents a proper fraction.
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Asked 11/25/2014 1:44:45 AM
Updated 11/25/2014 7:45:38 AM
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15/22 represents a proper fraction.
Added 11/25/2014 7:45:38 AM
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Divide 6 /13 by 6 /12 . A. 9 /16 B. 12/13 C. 1 /12 D. 13/12
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Updated 11/25/2014 9:38:14 AM
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6/13 ÷ 6/12
= 6/13 ÷ 1/2;
= 6/13 * 2;
= 12/13
Added 11/25/2014 9:38:14 AM
This answer has been confirmed as correct and helpful.
Confirmed by sujaysen [11/26/2014 8:09:01 AM]
Which one of the following mathematical statements is true? A. 0 ÷ 28 = 0 B. 28 + 0 = 0 C. 28 – 0 = 0 D. 28 ÷ 0 = 0
Weegy: 0 ÷ 28 = 0 is true. (More)
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Which one of the following statements expresses a true proportion? A. 2:1 = 1:2 B. 3:4 = 9:12 C. 54:9 = 6:3 D. 7:9 = 8:9
Weegy: 54 is an even number. What is your question? (More)
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Updated 12/2/2014 1:56:49 AM
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3:4 = 9:12 expresses a true proportion.
3/4 = 9/12; 0.75 = 0.75
Added 12/2/2014 1:56:48 AM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [12/2/2014 2:32:57 AM]
Which of the following is equal to 4 kilograms? A. 4,000 mg B. 4,000 cg C. 40 g D. 4,000 g
Weegy: 4,000 grams is equal to 4 kilograms. (More)
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# Fresh Rationalizing The Denominator Worksheet
Rationalizing the Denominator Simplify. Rationalizing The Denominator Worksheet.
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### Worksheet Rationalize the denominator and multiply with Radicals Rationalizing is done to remove the radical from the denominator of a fraction.
Rationalizing the denominator worksheet. Keep students informed of the steps involved in this technique with these pdf worksheets offering three different levels of practice. Rationalize the denominator. Some of the worksheets for this concept are Rationalize the denominator Rationalize the denominator Rationalizing a denominator Rationalize the denominator and multiply with radicals Radicals Dividing radical Rationalizing denominators index 3 or higher Rationalizing.
On the right side multiply both numerator and denominator by 2 to get rid of the radical in the denominator. G Worksheet by Kuta Software LLC 15 2 32 3 16 5 45 4 17 3 53 5 18 2 72 7 19 NOTES. On the right side multiply both numerator and denominator by 2 to get rid of the radical in the denominator.
Discover learning games guided lessons and other interactive activities for children. 4510 4 2. Rationalize the denominators level 1.
Rationalising the Denominator Worksheet. Some of the worksheets for this concept are rationalize the denominator radicals rationalize the denominator and multiply with radicals rationalizing denominators variables present chapter 12 radicals contents rationalizing the denominator square roots date period rationalizing. 4 5 10 4 2.
On the right side multiply both numerator and denominator by 2 to get rid of the radical in the denominator. Free worksheet pdf and answer key on rationalizing the denominator. 3 5 2 3 Multiplybyconjugate 2 3 3 5 2 3 2 3 2 3.
Rationalize the denominators of radical expressions. Ad Download over 20000 K-8 worksheets covering math reading social studies and more. This worksheet also helps my students to review their perfect squares and the writing of ratios.
Rationalize the denominators level 2. Before look at the worksheet if you wish to know how to rationalize the denominator in rational expressions in detail please click here. Examples of rationalizing the denominator.
Fractions cannot have irrational radicals or surds in the denominator. We use a technique called rationalization to eliminate them. Multiply the numerator and denominator by the given radical to have a rational number in the denominator and further simplify the expression.
Discover learning games guided lessons and other interactive activities for children. Rationalizing The Denominator Worksheets helps kids to understand Rationalizing two terms in the denominator requires the conjugate rationalize the denominator because mathematicians believe it is important. This quiz will test you on what you ve learned in order to simplify a radical expression when it requires rationalizing the denominator.
Quiz worksheet goals this quiz tests your ability to. Kids can learn about Combine terms Simplify the radicals and Reduce the fraction. Rationalize the denominators of radical expressions.
We will consider three cases involving square roots. Ad Download over 20000 K-8 worksheets covering math reading social studies and more. Free worksheet pdf and answer key on rationalizing the denominator.
Benefits of Rationalizing The Denominator Worksheets. Rationalize the denominator. 25 scaffolded questions that include model problems and a few challenge questions at the end.
2 16 3 36 4. Rationalizing the denominator worksheet answers kuta software. 25 scaffolded questions that include model problems and a few challenge questions at the end.
We just need to remember to FOIL out the numerator. Free worksheetpdf and answer key on rationalizing the denominator. 1 3 2 2 3 2 3 5 5 4 4 3 5 2 3 6 3 3 7 4 5 8 2 3 9 2 10 10 2 6 11 2 12 2 15 12 3 3 4 6 13 3 8 3 12 14 3 16 5 12 15 2.
To rationalize a denominator containing two terms with one or more square roots _____ the numerator and the denominator by the _____ of the denominator. Exponents And Radicals In Algebra Studying Math Math Formulas Basic Math. 1 date period l d2w0o1b5u fkvumtlaw tsioyfptdwda rrew hlqltcg y a eailfly qraibgxhmtosb yr elscezrsvjeodw.
4510 45 2 5 4510 45 2 5 On the right side cancel out 5 in numerator and denominator. It will be helpful to remember how to reduce a radical when continuing with these problems. Squaring a radical will eliminate the radical.
Rationalizing the Denominators Worksheets. FOILinnumerator denominatorisdifferenceofsquares 63 3 2 5 15 4 3 Simplifydenominator 63 3 2 5 15 1 Divideeachtermby1 63 3 2 5 15 OurSolution 4. Some of the worksheets for this concept are rationalize the denominator radicals rationalize the denominator and multiply with radicals rationalizing denominators variables present chapter 12 radicals contents rationalizing the denominator square roots date period rationalizing.
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Rationalizing Complex Denominators Line Puzzle Activity Teacher Material Denominator Worksheet Template | 1,183 | 5,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.840375 |
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casi 9 años ago | 0 | 1,637 | 5,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.683458 |
https://electronics.stackexchange.com/questions/391700/understanding-the-practical-application-of-a-simple-antenna-made-from-coax | 1,656,889,197,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104277498.71/warc/CC-MAIN-20220703225409-20220704015409-00564.warc.gz | 277,948,389 | 66,896 | Understanding the Practical Application of a Simple Antenna made from Coax
I am here today to gain a deeper understanding of the fundamentals of using antennas in RC applications. I am trying to understand what exactly is going on with a particular 2.4GHz receiver a picture of which is attached.
It's somewhat difficult to see, but these are 2 small coax wires soldered to the board with the center conductors soldered to pads separate from the shielding (obviously) which I assume is just grounded. So I have 2 questions:
1. Would these 2 wires be 2 separate antennas or 2 halves of a single dipole?
2. In what way does the shielding effect the antenna? Specifically, if you were to cut the wire down (significantly changing the overall length of wire) but strip away shielding such that the exposed length of center conductor is unchanged, how would this effect the performance of the antenna?
What these two do, whether they just form a dipole together, or are two separate monopoles (with bad ground) is impossible to tell based on the info we have.
I find it less likely they form a dipole.
Maybe there's just one receive and one transmit antenna, to save on a directional coupler or antenna switch.
Maybe these antennas are there for diversity reasons – for example, to decrease the likelihood that all antennas you have are subject to fading, you just use more antennas. $P(\text{all antennas faded}) < P(\text{at least one antenna is OK})$
1. In what way does the shielding effect the antenna?
I'll nitpick on your terminology for a second, because it makes a difference later:
That's not primarily a "shield", that's the outer conductor of the coax cable, and just as important as the center conductor at transporting energy. (In fact, you can have a hollow waveguide without the center conductor.)
Together with the center conductor and the dielectric material, it forms a waveguide.
Specifically, if you were to cut the wire down (significantly changing the overall length of wire) but strip away shielding such that the exposed length of center conductor is unchanged, how would this effect the performance of the antenna?
Impossible to tell. The coax might be somewhere between impedance of the trace on the PCB and the monopole feedpoint impedance, and act as e.g. a $\frac\lambda4$ impedance transformer. In that case, the length of the isolated part is critical.
Maybe it's just a best-effort matched transmission line, and changing the length of it doesn't affect the impedance.
But: in my theory (two separate monopoles), these antennas are separate for a reason (typically, diversity to increase robustness or speed of transmission); making their leads shorter would put the antennas closer together, which would have negative effects on channel independence, and hence, on performance.
• That looks like typical FHSS hobby receiver. They use two antennas for diversity. You can find them with coax part in anything from zero to 20", but the actual antenna (stripped part) is always the same. So, if one considers "performance" as distance then they can be shortened. But for reliability during 3D flight they better left long and placed at an angle. Aug 19, 2018 at 10:10
• Thanks, that is basically what I was thinking. And sorry for the bad terminology, literally like 5 minutes after posting I thought that was probably wrong. So @Maple, the reason I ask this question is specifically because my antennas got cut short by my props. My hope being that I cold just strip away the outer layer so that the original length of inner conductor would be exposed without altering the performance too much. Is this correct? Aug 20, 2018 at 16:28
• @JoshWiens it's at least likely. Again, the reason that you have two antennas in the first place is for them to pick up different paths that the RF wave can travel, so to decrease the likelihood of all paths being bad. Now, since putting them closer together will decrease the indepence of paths, but restoring the original antenna shape will probably give you more dB of SNR back than you lose in diversity gain. So: go for it! Aug 20, 2018 at 16:30
• Second that. And a bit of advice - put one antenna along the fuselage, the other one along the wing (assuming neither is carbon coated). And if you are using electric motor then put receiver as far from it as possible. Aug 20, 2018 at 16:38 | 953 | 4,372 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.950681 |
http://de.metamath.org/mpeuni/3imtr4g.html | 1,718,884,371,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00630.warc.gz | 7,494,277 | 6,624 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > 3imtr4g Structured version Visualization version GIF version
Theorem 3imtr4g 284
Description: More general version of 3imtr4i 280. Useful for converting definitions in a formula. (Contributed by NM, 20-May-1996.) (Proof shortened by Wolf Lammen, 20-Dec-2013.)
Hypotheses
Ref Expression
3imtr4g.1 (𝜑 → (𝜓𝜒))
3imtr4g.2 (𝜃𝜓)
3imtr4g.3 (𝜏𝜒)
Assertion
Ref Expression
3imtr4g (𝜑 → (𝜃𝜏))
Proof of Theorem 3imtr4g
StepHypRef Expression
1 3imtr4g.2 . . 3 (𝜃𝜓)
2 3imtr4g.1 . . 3 (𝜑 → (𝜓𝜒))
31, 2syl5bi 231 . 2 (𝜑 → (𝜃𝜒))
4 3imtr4g.3 . 2 (𝜏𝜒)
53, 4syl6ibr 241 1 (𝜑 → (𝜃𝜏))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 196 This theorem is referenced by: 3anim123d 1398 3orim123d 1399 mo3 2495 moim 2507 2euswap 2536 nelcon3d 2895 ral2imi 2931 ralimdv2 2944 rexim 2991 reximd2a 2996 reximdv2 2997 moeq3 3350 rmoim 3374 2reuswap 3377 ssel 3562 sstr2 3575 ssrexf 3628 sscon 3706 ssdif 3707 unss1 3744 ssrin 3800 difin0ss 3900 r19.2z 4012 prel12 4323 uniss 4394 ssuni 4395 ssuniOLD 4396 intssuni 4434 iunss1 4468 iinss1 4469 ss2iun 4472 iunxdif3 4542 disjss2 4556 disjss1 4559 disjss3 4582 ssbrd 4626 sspwb 4844 poss 4961 pofun 4975 soss 4977 frss 5005 sess1 5006 sess2 5007 wess 5025 relss 5129 ssrelOLD 5131 ssrel2 5133 ssrelrel 5143 relop 5194 cnvssOLD 5217 dmss 5245 dmcosseq 5308 funss 5822 fss 5969 fun 5979 brprcneu 6096 f1eqcocnv 6456 isores3 6485 isomin 6487 isopolem 6495 isosolem 6497 isowe2 6500 ovmpt2s 6682 dfwe2 6873 onint 6887 orduniorsuc 6922 ordom 6966 finds 6984 finds2 6986 f1oweALT 7043 tposfn2 7261 tposfo2 7262 tposf1o2 7265 smores 7336 tz7.48lem 7423 tz7.48-3 7426 oaass 7528 iiner 7706 xpdom2 7940 ssenen 8019 pssnn 8063 hartogs 8332 card2on 8342 ackbij1 8943 cfub 8954 fin23lem27 9033 fin1a2lem11 9115 fin1a2lem13 9117 hsmexlem2 9132 zorn2lem4 9204 ondomon 9264 gchina 9400 intgru 9515 ingru 9516 addclprlem2 9718 psslinpr 9732 ltexprlem3 9739 ltexprlem4 9740 reclem2pr 9749 suplem1pr 9753 sup2 10858 nnind 10915 nnunb 11165 uzind 11345 xmullem2 11967 xrsupsslem 12009 xrinfmsslem 12010 seqof 12720 hashfacen 13095 sswrd 13168 wrdind 13328 wrd2ind 13329 swrdccatin12lem2 13340 cau3lem 13942 caubnd 13946 sumodd 14949 vdwnnlem2 15538 ramub2 15556 setsstruct 15727 fthres2 16415 odupos 16958 lsmdisj2 17918 pgpfac1lem3 18299 subrgdvds 18617 lspdisj 18946 lspprat 18974 lbsextlem2 18980 coe1fzgsumd 19493 evl1gsumd 19542 ocv2ss 19836 ocvin 19837 tgcl 20584 epttop 20623 cmpsub 21013 tgcmp 21014 hauscmplem 21019 dfcon2 21032 llyss 21092 nllyss 21093 locfincmp 21139 txcnpi 21221 txcnp 21233 snfil 21478 fgcl 21492 filcon 21497 filuni 21499 cfinfil 21507 csdfil 21508 supfil 21509 ufildom1 21540 fin1aufil 21546 fmfnfmlem3 21570 ptcmplem2 21667 cldsubg 21724 iscau3 22884 iscau4 22885 caussi 22903 volfiniun 23122 plycj 23837 abelth 23999 wilthlem2 24595 lgsdir2lem4 24853 gausslemma2dlem0i 24889 gausslemma2dlem1a 24890 pntleml 25100 cusgrafilem2 26008 frisusgranb 26524 frgrancvvdeqlem3 26559 lpni 26722 ubthlem1 27110 chintcli 27574 h1de2i 27796 spansnm0i 27893 strlem1 28493 mdslmd1i 28572 2reuswap2 28712 ssrmo 28718 rabss3d 28736 disjss1f 28768 disjpreima 28779 ssrelf 28805 suppss3 28890 nnindf 28952 oduprs 28987 crefss 29244 esumpcvgval 29467 derangenlem 30407 conpcon 30471 cvmsss2 30510 pocnv 30907 wzel 31015 wzelOLD 31016 sltres 31061 nocvxmin 31090 naim1 31554 naim2 31555 waj-ax 31583 lukshef-ax2 31584 bj-ssbim 31810 bj-mo3OLD 32022 poimirlem26 32605 poimirlem30 32609 poimirlem32 32611 itg2addnclem 32631 ismtybndlem 32775 ablo4pnp 32849 isdrngo3 32928 keridl 33001 ispridl2 33007 ispridlc 33039 prter1 33182 lshpdisj 33292 snatpsubN 34054 pmapglb2N 34075 pmapglb2xN 34076 elpaddn0 34104 mzpindd 36327 pellexlem3 36413 pellexlem5 36415 pellex 36417 2nn0ind 36528 lnr2i 36705 intabssd 36935 iunrelexpuztr 37030 hess 37094 frege52aid 37172 frege52b 37203 neik0pk1imk0 37365 2rmoswap 39833 pfxccatin12lem2 40287 uhgr0vsize0 40465 cusgrfilem2 40672 uhgrvd00 40750 clwwisshclwws 41235 frcond3 41440 frgrncvvdeqlem3 41471 nrhmzr 41663 elsetrecslem 42243
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Trade Growth and Inequality. Professor Christopher Bliss Hilary Term 2004 Fridays 10-11 a.m. Ch. 4 Convergence in Practice and Theory. Cross-section growth empirics starts with Baumol (1986) He looks at -convergence -convergence v. -convergence - Friedman (1992) - PowerPoint PPT Presentation
Transcript
• Trade Growth and InequalityProfessor Christopher Bliss Hilary Term 2004 Fridays 10-11 a.m.
• Ch. 4 Convergence in Practice and TheoryCross-section growth empirics starts with Baumol (1986)He looks at -convergence-convergence v. -convergence - Friedman (1992)De Long (1988) sampling bias
• Barro and Sala-i-MartinWorld-wide comparative growthNear complete coverage (Summers-Heston data) minimizes sampling biasStraight test of -convergenceDependent variable is growth of per-capita income 1960-85Correlation coefficient between growth and lnPCI60 for 117 countries is .227
• Table 4.1 Simple regression result N=117 F=6.245
• Correlation and CausationCorrelation is no proof of causationBUTAbsence of correlation is no proof of the absence of causationLooking inside growth regressions perfectly illustrates this last point
• The spurious correlationA spurious correlation arises purely by chance Assemble 1000 crazy ordered data setsThat gives nearly half a million pairs of such variablesBetween one such pair there is bound to be a correlation that by itself will seem to be of overwhelming statistical significance
• Most correlations encountered in practice are not spuriousBut they may well not be due to a simple causal connectionThe variables are each correlated causally with another missing variableAs when the variables are non-stationary and the missing variable is time
• Two examples of correlating non-stationary variablesThe beginning econometrics students consumption function ct = + yt + tBut surely consumption is causally connected to incomeADt = + TSt + t where TS = teachers salaries AD = arrests for drunkeness
• Regression analysis and missing variablesA missing variable plays a part in the DGP and is correlated with included variablesThis is never a problem with Classical Regression AnalysisBarro would say that the simple regression of LnPCI60 on per capita growth is biassed by the exclusion of extra conditioning variables
• Table 4,2 Growth and extra variablesSources * Barro and Sala-i-Martin (1985) * Barro-Lee data set
• Table 4.3 Regression resultN = 73 F = 8.326 R2 = .4308
• Table 4.4 Regression with One Conditioning Variable
• Looking Inside Growth Regressions Ig is economic growthly is log initial per capita incomez is another variable of interest, such as I/Y, which is itself positively correlated with growth. All these variables are measured from their means.
• Inside growth regressions IIWe are interested in a case in which the regression coefficient of g on ly is near zero or positive. So we have:v{gly}0where v is the summed products of g and ly
• Inside Growth regressions III Thus v{gly} is N times the co-variance between g and ly. Now consider the multiple regression:g=ly+z+(3)
• Inside Growth Regressions IV
• Inside Growth Regressions V So that:vglY=vgg + vgz (5) Then if vglY 0 and vg > 0, (5) requires that either or , but not both, be negative. If vglY > 0 then and may both be positive, but they cannot both be negative. One way of explaining that conclusion is to say that a finding of -convergence with an augmented regression, despite growth and log initial income not being negatively correlated, can happen because the additional variable (or variables on balance) are positively correlated with initial income.
• A Growth Regression with one additional variable
• Growth Regression with I/Y
• One additional variable regression From (5) and the variance/covariance matrix above:.00384 = .82325 + .05216Now if is positive, must be negativeThis has happened because the added variable is positively correlated with g
• Adding the Mystery Ingredient Lg=ly+L+(7)The correlation matrix is:
glyLg1ly.174801L.32184.733731
• Growth Regression with L
VariableCoeff-icientt-valuePartial R2Con- stant.0161411.03.0092LnPCI60-.00083-.348.0011L.0004353.24.0893N=117R2=.346F=29.57
• Correlation and CauseThe Barro equation is founded in a causal theory of growthThe equation with L cannot have a causal basisWhat is causality anyway?Granger-Sims causality tests. Need time series data. Shocks to causal variables come first in time
• Causality and Temporal OrderingAn alarm clock set to ring just before sunrise does not cause the sun to rise.If it can be shown that random shocks to my alarm setting are significantly correlated with the time of sunrise, the that is an impressive puzzleCause is a (an optional) theory notion
• Convergence TheoryThe Solow-Swan Model
• Solow-Swan Model IIThe model gives convergence in two important cases:Several isolated economies each with the same saving share. Only the level of per capita capital distinguishes economiesThere is one integrated capital markets economy and numerous agents with the same saving rate. Only the level of per capita capital attained distinguishes one agnet from another.
• Solow-Swan Model IIIIf convergence is conditional on various additional variables, how precisely do these variables make their effects felt?For country I at time t income is:AiF[Ki(t),Li(t)] A measures total factor productivity, so will be called TFP
• Determinants of the Growth RateThe growth rate is larger:The larger is capitals shareThe larger is the saving shareThe larger is the TFP coefficientThe smaller is capital per headThe smaller is the rate of population growth
• Mankiw, Romer and Weil (1992)80% of cross section differences in growth rates can be accounted for via effects 2 and 5 by themselvesThe chief problem for growth empirics is to disentangle effects 3 and 4
• Convergence: The Ramsey ModelRamsey (1928) considered a many-agent version of his model (a MARM)He looked at steady states and noted the paradoxical feature that if agents discount utility at different rates, then all capital will be owned by agents with the lowest discount rate
• Two different casesJust as with the Solow-Swan model the cases are:Isolated economies each one a version of the same Ramsey model, with the same utility discount rate. The level of capital attained at a particular time distinguishes one economy from anotherOne economy with a single unified capital market, and each agent has the same utility function. The level of capital attained at a particular time distinguishes one agent from another
• Isolated EconomiesChapter 3 has already made clear that there is no general connection between the level of k and (1/c)(dc/dt).The necessary condition for optimal growth is:{[-c(du/dc)]/u}{(1/c)(dc/dt)}=F1[k(t),1]-r(20)Where u is U1[c(t)]
• Determinants of the Growth of ConsumptionThe necessary condition for optimal growth is:{[-c(du/dc)]/u}{(1/c)(dc/dt)}=F1[k(t),1]-r When k(t) takes a low value the right-hand side of (20) is relatively large. If the growth rate of consumption is not large, the elasticity of marginal utility[-c(du/dc)]/uMust be large.The idea that -convergence follows from optimal growth theory is suspect.
• Growth in the MARMWith many agents the optimal growth condition (20) becomes:[-d(du/dc)/dt]/u]=F1[kii(t)),1]-r(23)In steady state (23) becomes:F1[kii(t)),1]=rNote the effect of perturbing one agents capital holding
• A non-convergence resultIn the MARM:Non-converging steady states are possibleStrict asymptotic convergence can never occurPartial convergence (or divergence) clubs are possible depending on the third derivative of the utility function
• What does a MARM maximize?Any MARM equlibrium is the solution to a problem of the form:MaxN10U[ci(t)]dtNon-convergence is hsown despite the assumptions that:All agents have the same tastes and the same utility discount rateAll supply the same quantity of labour and earn the same wageAll have access to the same capital market where they earn the same rate of returnAll have perfect foresight and there are no stochastic effects to interfere with convergence
• Asymptotic and -convergenceFor isolated Ramsey economies we have seen that we need not have -convergence, but we must have asymptotic convergenceOn the other hand we may have -convergence without asymptotic convergencelnyI = aI - b/t+2 lnyII = aII - b/t+1aI< aIICountry I has the lower income and is always growing faster
• Strange Accumulation Paths can be Optimal In the Mathematical Appendix it is shown that:Given a standard production function and a monotonic time path k(t) such that k goes to k*, the Ramsey steady state value, and the implied c is monotonic, there exists a well-behaved utility function such that this path is Ramsey optimal
• Optimal Growth with Random ShocksBliss (2003) discusses the probability density of income levels when Ramsey-style accumulation is shocked each period with shocks large on absolute valueTwo intuitive cases illustrate the type of result available:Low income countries grow slowly, middle income countries rapidly and rich countries slowly. If shocks are large poverty and high income form basins of attraction in which many countries will be found. Compare Quah (1997)If shocks are highly asymmetric this will affect the probability distribution of income levels, even if the differential equation for income is linear. Earthquake shocks.
• The BMS ModelBarro, Mankiw and Sala-i-Martin (1995)Human capital added which cannot be used as collateralOne small country converges on a large world in steady state (existence is by exhibition).A more general case is where many small countries have significant weight. Then if they differ some may leave the constrained state before others and poor countries may not be | 2,235 | 9,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-16 | latest | en | 0.790235 |
https://www.bacancytechnology.com/qanda/flutter/convert-a-double-to-an-int-in-dart | 1,721,603,831,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00216.warc.gz | 573,048,942 | 37,996 | There are various methods provided by dart, using that we can convert double to an integer.
Here i mentioned methods with description.
### 1. truncate():
• Method: truncate()
• Description: This method returns the integer value closest to but no greater in magnitude than the double value. It effectively removes the fractional part of the number without rounding.
### 2. toInt():
• Method: toInt()
• Description: Converts the double to an integer by discarding the fractional part. This method rounds towards zero.
There are no difference between toInt() and truncate() method.
### 3. round():
• Method: round()
• Description: Rounds a number to the nearest integer. If the fractional part is 0.5 or higher, it rounds up; otherwise, it rounds down.
### 4. ceil():
• Method: ceil()
• Description: Rounds a number up to the nearest integer. If the number is already an integer, it remains unchanged. For positive numbers, it rounds up; for negative numbers, it rounds towards zero.
### 5. floor():
• Method: floor()
• Description: Rounds a number down to the nearest integer. It always rounds towards negative infinity, regardless of the number’s sign. | 251 | 1,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.768068 |
https://terrytao.wordpress.com/2015/10/03/275a-notes-1-integration-and-expectation/ | 1,679,447,273,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943747.51/warc/CC-MAIN-20230321225117-20230322015117-00132.warc.gz | 634,320,409 | 104,137 | In Notes 0, we introduced the notion of a measure space ${\Omega = (\Omega, {\mathcal F}, \mu)}$, which includes as a special case the notion of a probability space. By selecting one such probability space ${(\Omega,{\mathcal F},\mu)}$ as a sample space, one obtains a model for random events and random variables, with random events ${E}$ being modeled by measurable sets ${E_\Omega}$ in ${{\mathcal F}}$, and random variables ${X}$ taking values in a measurable space ${R}$ being modeled by measurable functions ${X_\Omega: \Omega \rightarrow R}$. We then defined some basic operations on these random events and variables:
• Given events ${E,F}$, we defined the conjunction ${E \wedge F}$, the disjunction ${E \vee F}$, and the complement ${\overline{E}}$. For countable families ${E_1,E_2,\dots}$ of events, we similarly defined ${\bigwedge_{n=1}^\infty E_n}$ and ${\bigvee_{n=1}^\infty E_n}$. We also defined the empty event ${\emptyset}$ and the sure event ${\overline{\emptyset}}$, and what it meant for two events to be equal.
• Given random variables ${X_1,\dots,X_n}$ in ranges ${R_1,\dots,R_n}$ respectively, and a measurable function ${F: R_1 \times \dots \times R_n \rightarrow S}$, we defined the random variable ${F(X_1,\dots,X_n)}$ in range ${S}$. (As the special case ${n=0}$ of this, every deterministic element ${s}$ of ${S}$ was also a random variable taking values in ${S}$.) Given a relation ${P: R_1 \times \dots \times R_n \rightarrow \{\hbox{true}, \hbox{false}\}}$, we similarly defined the event ${P(X_1,\dots,X_n)}$. Conversely, given an event ${E}$, we defined the indicator random variable ${1_E}$. Finally, we defined what it meant for two random variables to be equal.
• Given an event ${E}$, we defined its probability ${{\bf P}(E)}$.
These operations obey various axioms; for instance, the boolean operations on events obey the axioms of a Boolean algebra, and the probabilility function ${E \mapsto {\bf P}(E)}$ obeys the Kolmogorov axioms. However, we will not focus on the axiomatic approach to probability theory here, instead basing the foundations of probability theory on the sample space models as discussed in Notes 0. (But see this previous post for a treatment of one such axiomatic approach.)
It turns out that almost all of the other operations on random events and variables we need can be constructed in terms of the above basic operations. In particular, this allows one to safely extend the sample space in probability theory whenever needed, provided one uses an extension that respects the above basic operations; this is an important operation when one needs to add new sources of randomness to an existing system of events and random variables, or to couple together two separate such systems into a joint system that extends both of the original systems. We gave a simple example of such an extension in the previous notes, but now we give a more formal definition:
Definition 1 Suppose that we are using a probability space ${\Omega = (\Omega, {\mathcal F}, \mu)}$ as the model for a collection of events and random variables. An extension of this probability space is a probability space ${\Omega' = (\Omega', {\mathcal F}', \mu')}$, together with a measurable map ${\pi: \Omega' \rightarrow \Omega}$ (sometimes called the factor map) which is probability-preserving in the sense that
$\displaystyle \mu'( \pi^{-1}(E) ) = \mu(E) \ \ \ \ \ (1)$
for all ${E \in {\mathcal F}}$. (Caution: this does not imply that ${\mu(\pi(F)) = \mu'(F)}$ for all ${F \in {\mathcal F}'}$ – why not?)
An event ${E}$ which is modeled by a measurable subset ${E_\Omega}$ in the sample space ${\Omega}$, will be modeled by the measurable set ${E_{\Omega'} := \pi^{-1}(E_\Omega)}$ in the extended sample space ${\Omega'}$. Similarly, a random variable ${X}$ taking values in some range ${R}$ that is modeled by a measurable function ${X_\Omega: \Omega \rightarrow R}$ in ${\Omega}$, will be modeled instead by the measurable function ${X_{\Omega'} := X_\Omega \circ \pi}$ in ${\Omega'}$. We also allow the extension ${\Omega'}$ to model additional events and random variables that were not modeled by the original sample space ${\Omega}$ (indeed, this is one of the main reasons why we perform extensions in probability in the first place).
Thus, for instance, the sample space ${\Omega'}$ in Example 3 of the previous post is an extension of the sample space ${\Omega}$ in that example, with the factor map ${\pi: \Omega' \rightarrow \Omega}$ given by the first coordinate projection ${\pi(i,j) := i}$. One can verify that all of the basic operations on events and random variables listed above are unaffected by the above extension (with one caveat, see remark below). For instance, the conjunction ${E \wedge F}$ of two events can be defined via the original model ${\Omega}$ by the formula
$\displaystyle (E \wedge F)_\Omega := E_\Omega \cap F_\Omega$
or via the extension ${\Omega'}$ via the formula
$\displaystyle (E \wedge F)_{\Omega'} := E_{\Omega'} \cap F_{\Omega'}.$
The two definitions are consistent with each other, thanks to the obvious set-theoretic identity
$\displaystyle \pi^{-1}( E_\Omega \cap F_\Omega ) = \pi^{-1}(E_\Omega) \cap \pi^{-1}(F_\Omega).$
Similarly, the assumption (1) is precisely what is needed to ensure that the probability ${\mathop{\bf P}(E)}$ of an event remains unchanged when one replaces a sample space model with an extension. We leave the verification of preservation of the other basic operations described above under extension as exercises to the reader.
Remark 2 There is one minor exception to this general rule if we do not impose the additional requirement that the factor map ${\pi}$ is surjective. Namely, for non-surjective ${\pi}$, it can become possible that two events ${E, F}$ are unequal in the original sample space model, but become equal in the extension (and similarly for random variables), although the converse never happens (events that are equal in the original sample space always remain equal in the extension). For instance, let ${\Omega}$ be the discrete probability space ${\{a,b\}}$ with ${p_a=1}$ and ${p_b=0}$, and let ${\Omega'}$ be the discrete probability space ${\{ a'\}}$ with ${p'_{a'}=1}$, and non-surjective factor map ${\pi: \Omega' \rightarrow \Omega}$ defined by ${\pi(a') := a}$. Then the event modeled by ${\{b\}}$ in ${\Omega}$ is distinct from the empty event when viewed in ${\Omega}$, but becomes equal to that event when viewed in ${\Omega'}$. Thus we see that extending the sample space by a non-surjective factor map can identify previously distinct events together (though of course, being probability preserving, this can only happen if those two events were already almost surely equal anyway). This turns out to be fairly harmless though; while it is nice to know if two given events are equal, or if they differ by a non-null event, it is almost never useful to know that two events are unequal if they are already almost surely equal. Alternatively, one can add the additional requirement of surjectivity in the definition of an extension, which is also a fairly harmless constraint to impose (this is what I chose to do in this previous set of notes).
Roughly speaking, one can define probability theory as the study of those properties of random events and random variables that are model-independent in the sense that they are preserved by extensions. For instance, the cardinality ${|E_\Omega|}$ of the model ${E_\Omega}$ of an event ${E}$ is not a concept within the scope of probability theory, as it is not preserved by extensions: continuing Example 3 from Notes 0, the event ${E}$ that a die roll ${X}$ is even is modeled by a set ${E_\Omega = \{2,4,6\}}$ of cardinality ${3}$ in the original sample space model ${\Omega}$, but by a set ${E_{\Omega'} = \{2,4,6\} \times \{1,2,3,4,5,6\}}$ of cardinality ${18}$ in the extension. Thus it does not make sense in the context of probability theory to refer to the “cardinality of an event ${E}$“.
On the other hand, the supremum ${\sup_n X_n}$ of a collection of random variables ${X_n}$ in the extended real line ${[-\infty,+\infty]}$ is a valid probabilistic concept. This can be seen by manually verifying that this operation is preserved under extension of the sample space, but one can also see this by defining the supremum in terms of existing basic operations. Indeed, note from Exercise 24 of Notes 0 that a random variable ${X}$ in the extended real line is completely specified by the threshold events ${(X \leq t)}$ for ${t \in {\bf R}}$; in particular, two such random variables ${X,Y}$ are equal if and only if the events ${(X \leq t)}$ and ${(Y \leq t)}$ are surely equal for all ${t}$. From the identity
$\displaystyle (\sup_n X_n \leq t) = \bigwedge_{n=1}^\infty (X_n \leq t)$
we thus see that one can completely specify ${\sup_n X_n}$ in terms of ${X_n}$ using only the basic operations provided in the above list (and in particular using the countable conjunction ${\bigwedge_{n=1}^\infty}$.) Of course, the same considerations hold if one replaces supremum, by infimum, limit superior, limit inferior, or (if it exists) the limit.
In this set of notes, we will define some further important operations on scalar random variables, in particular the expectation of these variables. In the sample space models, expectation corresponds to the notion of integration on a measure space. As we will need to use both expectation and integration in this course, we will thus begin by quickly reviewing the basics of integration on a measure space, although we will then translate the key results of this theory into probabilistic language.
As the finer details of the Lebesgue integral construction are not the core focus of this probability course, some of the details of this construction will be left to exercises. See also Chapter 1 of Durrett, or these previous blog notes, for a more detailed treatment.
— 1. Integration on measure spaces —
Let ${(\Omega, {\mathcal F}, \mu)}$ be a measure space, and let ${f}$ be a measurable function on ${\Omega}$, taking values either in the reals ${{\bf R}}$, the non-negative extended reals ${[0,+\infty]}$, the extended reals ${[-\infty,+\infty]}$, or the complex numbers ${{\bf C}}$. We would like to define the integral
$\displaystyle \int_\Omega f\ d\mu \ \ \ \ \ (2)$
of ${f}$ on ${\Omega}$. (One could make the integration variable explicit, e.g. by writing ${\int_\Omega f(\omega)\ d\mu(\omega)}$, but we will usually not do so here.) When integrating a reasonably nice function (e.g. a continuous function) on a reasonably nice domain (e.g. a box in ${{\bf R}^n}$), the Riemann integral that one learns about in undergraduate calculus classes suffices for this task; however, for the purposes of probability theory, we need the much more general notion of a Lebesgue integral in order to properly define (2) for the spaces ${\Omega}$ and functions ${f}$ we will need to study.
Not every measurable function can be integrated by the Lebesgue integral. There are two key classes of functions for which the integral exists and is well behaved:
• Unsigned measurable functions ${f: \Omega \rightarrow [0,+\infty]}$, that take values in the non-negative extended reals ${[0,+\infty]}$; and
• Absolutely integrable functions ${f: \Omega \rightarrow {\bf R}}$ or ${f: \Omega \rightarrow {\bf C}}$, which are scalar measurable functions whose absolute value ${|f|}$ has a finite integral: ${\int_\Omega |f|\ d\mu < \infty}$. (Sometimes we also allow absolutely integrable functions to attain an infinite value ${\infty}$, so long as they only do so on a set of measure zero.)
One could in principle extend the Lebesgue integral to slightly more general classes of functions, e.g. to sums of absolutely integrable functions and unsigned functions. However, the above two classes already suffice for most applications (and as a general rule of thumb, it is dangerous to apply the Lebesgue integral to functions that are not unsigned or absolutely integrable, unless you really know what you are doing).
We will construct the Lebesgue integral in the following four stages. First, we will define the Lebesgue integral just for unsigned simple functions – unsigned measurable functions that take on only finitely many values. Then, by a limiting procedure, we extend the Lebesgue integral to unsigned functions. After that, by decomposing a real absolutely integrable function into unsigned components, we extend the integral to real absolutely integrable functions. Finally, by taking real and imaginary parts, we extend to complex absolutely integrable functions. (This is not the only order in which one could perform this construction; for instance, in Durrett, one first constructs integration of bounded functions on finite measure support before passing to arbitrary unsigned functions.)
First consider an unsigned simple function ${f: \Omega \rightarrow [0,+\infty]}$, thus ${f}$ is measurable and only takes values at a finite number of values. Then we can express ${f}$ as a finite linear combination (in ${[0,+\infty]}$) of indicator functions. Indeed, if we enumerate the values that ${f}$ takes as ${a_1,\dots,a_n \in [0,+\infty]}$ (avoiding repetitions) and setting ${E_i := \{ \omega \in \Omega: f(\omega) = a_i \}}$ for ${i=1,\dots,n}$, then it is clear that
$\displaystyle f = \sum_{i=1}^n a_i 1_{E_i}.$
(It should be noted at this point that the operations of addition and multiplication on ${[0,+\infty]}$ are defined by setting ${+\infty + a = a + \infty = +\infty}$ for all ${a \in [0,+\infty]}$, and ${a \cdot +\infty = +\infty \cdot a}$ for all positive ${a \in (0,+\infty]}$, but that ${0 \cdot +\infty = +\infty \cdot 0}$ is defined to equal ${0}$. To put it another way, multiplication is defined to be continuous from below, rather than from above: ${a \cdot b = \lim_{x \rightarrow a^-, y \rightarrow b^-} x \cdot y}$. One can verify that the commutative, associative, and distributive laws continue to hold on ${[0,+\infty]}$, but we caution that the cancellation laws do not hold when ${+\infty}$ is involved.)
Conversely, given any coefficients ${a_1,\dots,a_n \in [0,+\infty]}$ (not necessarily distinct) and measurable sets ${E_1,\dots,E_n}$ in ${{\mathcal F}}$ (not necessarily disjoint), the sum ${\sum_{i=1}^n a_i 1_{E_i}}$ is an unsigned simple function.
A single simple function can be decomposed in multiple ways as a linear combination of unsigned simple functions. For instance, on the real line ${{\bf R}}$, the function ${2 \times 1_{[0,1)} + 1 \times 1_{[1,3)}}$ can also be written as ${1 \times 1_{[0,1)} + 1 \times 1_{[0,3)}}$ or as ${2 \times 1_{[0,1)} + 1 \times 1_{[1,2)} + 1 \times 1_{[2,3)}}$. However, there is an invariant of all these decompositions:
Exercise 3 Suppose that an unsigned simple function ${f}$ has two representations as the linear combination of indicator functions:
$\displaystyle f = \sum_{i=1}^n a_i 1_{E_i} = \sum_{j=1}^m b_j 1_{F_j},$
where ${n,m}$ are nonnegative integers, ${a_1,\dots,a_n,b_1,\dots,b_m}$ lie in ${[0,+\infty]}$, and ${E_1,\dots,E_n,F_1,\dots,F_m}$ are measurable sets. Show that
$\displaystyle \sum_{i=1}^n a_i \mu(E_i) = \sum_{j=1}^m b_j \mu(F_j).$
(Hint: first handle the special case where the ${F_j}$ are all disjoint and non-empty, and each of the ${E_i}$ is expressible as the union of some subcollection of the ${F_j}$. Then handle the general case by considering the atoms of the finite boolean algebra generated by ${E_i}$ and ${F_j}$.)
We capture this invariant by introducing the simple integral ${\hbox{Simp} \int_{\Omega} f\ d\mu}$ of an unsigned simple function by the formula
$\displaystyle \hbox{Simp} \int_\Omega f\ d\mu := \sum_{i=1}^n a_i \mu(E_i)$
whenever ${f}$ admits a decomposition ${f = \sum_{i=1}^n a_i 1_{E_i}}$. The above exercise is then precisely the assertion that the simple integral is well-defined as an element of ${[0,+\infty]}$.
Exercise 4 Let ${f, g: \Omega \rightarrow [0,+\infty]}$ be unsigned simple functions, and let ${c \in [0,+\infty]}$.
• (i) (Linearity) Show that
$\displaystyle \hbox{Simp} \int_\Omega f+g\ d\mu = \hbox{Simp} \int_\Omega f\ d\mu + \hbox{Simp} \int_\Omega g\ d\mu$
and
$\displaystyle \hbox{Simp} \int_\Omega cf\ d\mu = c \hbox{Simp} \int_\Omega f\ d\mu.$
• (ii) Show that if ${f}$ and ${g}$ are equal almost everywhere, then
$\displaystyle \hbox{Simp} \int_\Omega f\ d\mu = \hbox{Simp} \int_\Omega g\ d\mu.$
• (iii) Show that ${\hbox{Simp} \int_\Omega f\ d\mu \geq 0}$, with equality if and only if ${f}$ is zero almost everywhere.
• (iv) (Monotonicity) If ${f \leq g}$ almost everywhere, show that ${\hbox{Simp} \int_\Omega f\ d\mu \leq \hbox{Simp} \int_\Omega g\ d\mu}$.
• (v) (Markov inequality) Show that ${\mu( \{ \omega: f(\omega) \geq t \} ) \leq \frac{1}{t} \hbox{Simp} \int_\Omega f\ d\mu}$ for any ${0 < t < \infty}$.
Now we extend from unsigned simple functions to more general unsigned functions. If ${f: \Omega \rightarrow [0,+\infty]}$ is an unsigned measurable function, we define the unsigned integral ${\int_\Omega f\ d\mu}$ as
$\displaystyle \int_\Omega f\ d\mu = \sup_{0 \leq g \leq f} \hbox{Simp} \int_\Omega g\ d\mu \ \ \ \ \ (3)$
where the supremum is over all unsigned simple functions such that ${0 \leq g(\omega) \leq f(\omega)}$ for all ${\omega \in \Omega}$.
Many of the properties of the simple integral carry over to the unsigned integral easily:
Exercise 5 Let ${f, g: \Omega \rightarrow [0,+\infty]}$ be unsigned functions, and let ${c \in [0,+\infty]}$.
$\displaystyle \int_\Omega f+g\ d\mu \geq \int_\Omega f\ d\mu + \int_\Omega g\ d\mu$
and
$\displaystyle \int_\Omega cf\ d\mu = c \int_\Omega f\ d\mu.$
• (ii) Show that if ${f}$ and ${g}$ are equal almost everywhere, then
$\displaystyle \int_\Omega f\ d\mu = \int_\Omega g\ d\mu.$
• (iii) Show that ${\int_\Omega f\ d\mu \geq 0}$, with equality if and only if ${f}$ is zero almost everywhere.
• (iv) (Monotonicity) If ${f \leq g}$ almost everywhere, show that ${\int_\Omega f\ d\mu \leq \int_\Omega g\ d\mu}$.
• (v) (Markov inequality) Show that ${\mu( \{ \omega: f(\omega) \geq t \} ) \leq \frac{1}{t} \int_\Omega f\ d\mu}$ for any ${0 < t < \infty}$. In particular, if ${\int_\Omega f\ d\mu < \infty}$, then ${f}$ is finite almost everywhere.
• (vi) (Compatibility with simple integral) If ${f}$ is simple, show that ${\int_\Omega f\ d\mu = \hbox{Simp} \int_\Omega f\ d\mu}$.
• (vii) (Compatibility with measure) For any measurable set ${E}$, show that ${\int_\Omega 1_E\ d\mu = \mu(E)}$.
Exercise 6 If ${(\Omega, (p_\omega)_{\omega \in \Omega})}$ is a discrete probability space (with the associated probability measure ${\mu}$), and ${f: \Omega \rightarrow [0,+\infty]}$ is a function, show that
$\displaystyle \int_\Omega f\ d\mu = \sum_{\omega \in\Omega} f(\omega) p_\omega.$
(Note that the condition ${\sum_{\omega \in \Omega} p_\omega = 1}$ in the definition of a discrete probability space is not required to prove this identity.)
The observant reader will notice that the linearity property of simple functions has been weakened to superadditivity. This can be traced back to a breakdown of symmetry in the definition (3); the unsigned simple integral of ${f}$ is defined via approximation from below, but not from above. Indeed the opposite claim
$\displaystyle \int_\Omega f\ d\mu \stackrel{?}{=} \inf_{g \geq f} \hbox{Simp} \int_\Omega g\ d\mu \ \ \ \ \ (4)$
can fail. For a counterexample, take ${\Omega}$ to be the discrete probability space ${\{1,2,3,\dots\}}$ with probabilities ${p_n := 2^{-n}}$, and let ${f: \Omega \rightarrow [0,+\infty]}$ be the function ${f(n)=n}$. By Exercise 6 we have ${\int_\Omega f\ d\mu = \sum_{n=1}^\infty n2^{-n} = 2}$. On the other hand, any simple function ${g}$ with ${g \geq f}$ must equal ${+\infty}$ on a set of positive measure (why?) and so the right-hand side of (4) can be infinite. However, one can get around this difficulty under some further assumptions on ${f}$, and thus recover full linearity for the unsigned integral:
Exercise 7 (Linearity of the unsigned integral) Let ${(\Omega, {\mathcal F}, \mu)}$ be a measure space.
• (i) Let ${f: \Omega \rightarrow [0,+\infty]}$ be an unsigned measurable function which is both bounded (i.e., there is a finite ${M}$ such that ${|f(\omega)| \leq M}$ for all ${\omega \in \Omega}$) and has finite measure support (i.e., there is a measurable set ${E}$ with ${\mu(E) < \infty}$ such that ${f(\omega)=0}$ for all ${\omega \in \Omega \backslash E}$). Show that (4) holds for this function ${f}$.
• (ii) Establish the additivity property
$\displaystyle \int_\Omega f+g\ d\mu =\int_\Omega f\ d\mu + \int_\Omega g\ d\mu$
whenever ${f, g: \Omega \rightarrow [0,+\infty]}$ are unsigned measurable functions that are bounded with finite measure support.
• (iii) Show that
$\displaystyle \int_\Omega \min(f,n)\ d\mu \rightarrow \int_\Omega f\ d\mu$
as ${n \rightarrow \infty}$ whenever ${f: \Omega \rightarrow [0,+\infty]}$ is unsigned measurable.
• (iv) Using (iii), extend (ii) to the case where ${f,g}$ are unsigned measurable functions with finite measure support, but are not necessarily bounded.
• (v) Show that
$\displaystyle \int_\Omega f 1_{f \geq 1/n}\ d\mu \rightarrow \int_\Omega f\ d\mu$
as ${n \rightarrow \infty}$ whenever ${f: \Omega \rightarrow [0,+\infty]}$ is unsigned measurable.
• (vi) Using (iii) and (v), show that (ii) holds for any unsigned measurable ${f,g}$ (which are not necessarily bounded or of finite measure support).
Next, we apply the integral to absolutely integrable functions. We call a scalar function ${f: \Omega \rightarrow {\bf R}}$ or ${f: \Omega \rightarrow {\bf C}}$ absolutely integrable if it is measurable and the unsigned integral ${\int_\Omega |f|\ d\mu}$ is finite. A real-valued absolutely integrable function ${f}$ can be expressed as the difference ${f = f_1 - f_2}$ of two unsigned absolutely integrable functions ${f_1,f_2}$; indeed, one can check that the choice ${f_1 := \max(f,0)}$ and ${f_2 := \max(-f,0)}$ work for this. Conversely, any difference ${f_1 - f_2}$ of unsigned absolutely integrable functions ${f_1,f_2}$ is absolutely integrable (this follows from the triangle inequality ${|f_1-f_2| \leq |f_1|+|f_2|}$). A single absolutely integrable function ${f}$ may be written as a difference ${f_1-f_2}$ of unsigned absolutely integrable functions in more than one way, for instance we might have
$\displaystyle f = f_1 - f_2 = g_1 - g_2$
for unsigned absolutely integrable functions ${f_1,f_2,g_1,g_2}$. But when this happens, we can rearrange to obtain
$\displaystyle f_1 + g_2 = g_1 + f_2$
and thus by linearity of the unsigned integral
$\displaystyle \int_\Omega f_1\ d\mu + \int_\Omega g_2\ d\mu = \int_\Omega g_1\ d\mu + \int_\Omega f_2\ d\mu.$
By the absolute integrability of ${f_1,f_2,g_1,g_2}$, all the integrals are finite, so we may rearrange this identity as
$\displaystyle \int_\Omega f_1\ d\mu - \int_\Omega f_2\ d\mu = \int_\Omega g_1\ d\mu - \int_\Omega g_2\ d\mu.$
This allows us to define the Lebesgue integral ${\int_\Omega f\ d\mu \in {\bf R}}$ of a real-valued absolutely integrable function ${f}$ to be the expression
$\displaystyle \int_\Omega f\ d\mu := \int_\Omega f_1\ d\mu - \int_\Omega f_2\ d\mu$
for any given decomposition ${f = f_1-f_2}$ of ${f}$ as the difference of two unsigned absolutely integrable functions. Note that if ${f}$ is both unsigned and absolutely integrable, then the unsigned integral and the Lebesgue integral of ${f}$ agree (as can be seen by using the decomposition ${f = f - 0}$), and so there is no ambiguity in using the same notation ${\int_\Omega f\ d\mu}$ to denote both integrals. (By the same token, we may now drop the modifier ${\hbox{Simp}}$ from the simple integral of a simple unsigned ${f}$, which we may now also denote by ${\int_\Omega f\ d\mu}$.)
The Lebesgue integral also enjoys good linearity properties:
Exercise 8 Let ${f, g: \Omega \rightarrow {\bf R}}$ be real-valued absolutely integrable functions, and let ${c \in {\bf R}}$.
• (i) (Linearity) Show that ${f+g}$ and ${cf}$ are also real-valued absolutely integrable functions, with
$\displaystyle \int_\Omega f+g\ d\mu = \int_\Omega f\ d\mu + \int_\Omega g\ d\mu$
and
$\displaystyle \int_\Omega cf\ d\mu = c \int_\Omega f\ d\mu.$
(For the second relation, one may wish to first treat the special cases ${c>0}$ and ${c=-1}$.)
• (ii) Show that if ${f}$ and ${g}$ are equal almost everywhere, then
$\displaystyle \int_\Omega f\ d\mu = \int_\Omega g\ d\mu.$
• (iii) Show that ${\int_\Omega |f|\ d\mu \geq 0}$, with equality if and only if ${f}$ is zero almost everywhere.
• (iv) (Monotonicity) If ${f \leq g}$ almost everywhere, show that ${\int_\Omega f\ d\mu \leq\int_\Omega g\ d\mu}$.
• (v) (Markov inequality) Show that ${\mu( \{ \omega: |f(\omega)| \geq t \} ) \leq \frac{1}{t} \int_\Omega |f|\ d\mu}$ for any ${0 < t < \infty}$.
Because of part (iii) of the above exercise, we can extend the Lebesgue integral to real-valued absolutely integrable functions that are only defined and real-valued almost everywhere, rather than everywhere. In particular, we can apply the Lebesgue integral to functions that are sometimes infinite, so long as they are only infinite on a set of measure zero, and the function is absolutely integrable everywhere else.
Finally, we extend to complex-valued functions. If ${f: \Omega \rightarrow {\bf C}}$ is absolutely integrable, observe that the real and imaginary parts ${\hbox{Re} f, \hbox{Im} f: \Omega \rightarrow {\bf C}}$ are also absolutely integrable (because ${|\hbox{Re} f|, |\hbox{Im} f| \leq |f|}$). We then define the (complex) Lebesgue integral ${\int_\Omega f\ d\mu \in {\bf C}}$ of ${f}$ in terms of the real Lebesgue integral by the formula
$\displaystyle \int_\Omega f\ d\mu := \int_\Omega \hbox{Re}(f)\ d\mu + i \int_\Omega \hbox{Im}(f)\ d\mu.$
Clearly, if ${f}$ is real-valued and absolutely integrable, then the real Lebesgue integral and the complex Lebesgue integral of ${f}$ coincide, so it does not create ambiguity to use the same symbol ${\int_\Omega f\ d\mu}$ for both concepts. It is routine to extend the linearity properties of the real Lebesgue integral to its complex counterpart:
Exercise 9 Let ${f, g: \Omega \rightarrow {\bf C}}$ be complex-valued absolutely integrable functions, and let ${c \in {\bf C}}$.
• (i) (Linearity) Show that ${f+g}$ and ${cf}$ are also complex-valued absolutely integrable functions, with
$\displaystyle \int_\Omega f+g\ d\mu = \int_\Omega f\ d\mu + \int_\Omega g\ d\mu$
and
$\displaystyle \int_\Omega cf\ d\mu = c \int_\Omega f\ d\mu.$
(For the second relation, one may wish to first treat the special cases ${c \in {\bf R}}$ and ${c = i}$.)
• (ii) Show that if ${f}$ and ${g}$ are equal almost everywhere, then
$\displaystyle \int_\Omega f\ d\mu = \int_\Omega g\ d\mu.$
• (iii) Show that ${\int_\Omega |f|\ d\mu \geq 0}$, with equality if and only if ${f}$ is zero almost everywhere.
• (iv) (Markov inequality) Show that ${\mu( \{ \omega: |f(\omega)| \geq t \} ) \leq \frac{1}{t} \int_\Omega |f|\ d\mu}$ for any ${0 < t < \infty}$.
We record a simple, but incredibly fundamental, inequality concerning the Lebesgue integral:
Lemma 10 (Triangle inequality) If ${f: \Omega \rightarrow {\bf C}}$ is a complex-valued absolutely integrable function, then
$\displaystyle |\int_\Omega f\ d\mu| \leq \int_\Omega |f|\ d\mu.$
Proof: We have
$\displaystyle \hbox{Re} \int_\Omega f\ d\mu = \int_\Omega \hbox{Re} f\ d\mu$
$\displaystyle \leq \int_\Omega |f|\ d\mu.$
This looks weaker than what we want to prove, but we can “amplify” this inequality to the full strength triangle inequality as follows. Replacing ${f}$ by ${e^{i\theta} f}$ for any real ${\theta}$, we have
$\displaystyle \hbox{Re} e^{i\theta} \int_\Omega f\ d\mu \leq \int_\Omega |f|\ d\mu.$
Since we can choose the phase ${e^{i\theta}}$ to make the expression ${e^{i\theta} \int_\Omega f\ d\mu}$ equal to ${|\int_\Omega f\ d\mu|}$, the claim follows. $\Box$
Finally, we observe that the Lebesgue integral extends the Riemann integral, which is particularly useful when it comes to actually computing some of these integrals:
Exercise 11 If ${f: [a,b] \rightarrow {\bf C}}$ is a Riemann integrable function on a compact interval ${[a,b]}$, show that ${f}$ is also absolutely integrable, and that the Lebesgue integral ${\int_{[a,b]} f\ dm}$ (with ${m}$ Lebesgue measure restricted to ${[a,b]}$) coincides with the Riemann integral ${\int_a^b f(x)\ dx}$. Similarly if ${f}$ is Riemann integrable on a box ${[a_1,b_1] \times \dots \times [a_n,b_n]}$.
— 2. Expectation of random variables —
We now translate the above notions of integration on measure spaces to the probabilistic setting.
A random variable ${X}$ taking values in the unsigned extended real line ${[0,+\infty]}$ is said to be simple if it takes on at most finitely many values. Equivalently, ${X}$ can be expressed as a finite unsigned linear combination
$\displaystyle X = \sum_{i=1}^n a_i 1_{E_i}$
of indicator random variables, where ${a_1,\dots,a_n \in [0,+\infty]}$ are unsigned and ${E_i}$ are events. We then define the simple expectation ${\hbox{Simp} {\bf E} X \in [0,+\infty]}$ of ${X}$ to be the quantity
$\displaystyle \hbox{Simp} {\bf E} X := \sum_{i=1}^n a_i {\bf P}(E_i),$
and checks that this definition is independent of the choice of decomposition of ${X}$ into indicator functions. Observe that if we model the random variable ${X}$ using a probability space ${\Omega}$, then the simple expectation of ${X}$ is precisely the simple integral of the corresponding unsigned simple function ${X_\Omega}$.
Next, given an arbitrary unsigned random variable ${X}$ taking values in ${[0,+\infty]}$, modeled by a probability space ${\Omega}$ one defines its (unsigned) expectation ${{\bf E} X \in [0,+\infty]}$ as
$\displaystyle {\bf E} X := \sup_{0\leq Y \leq X} \hbox{Simp} {\bf E} Y$
where ${Y}$ ranges over all simple unsigned random variables modeled by the same space ${\Omega}$ such that ${0 \leq Y \leq X}$ is surely true. There is a subtle issue that this definition might in principle depend on the choice of ${\Omega}$ (because this affects the pool of available simple unsigned random variables ${Y}$), but this is not the case:
Exercise 12 Let ${X}$ be a random variable taking values in ${[0,+\infty]}$, and let ${Y}$ be a simple unsigned random variable such that ${0 \leq Y \leq X}$ is surely true. Show that there exists a function ${f: [0,+\infty] \rightarrow [0,+\infty]}$ taking on finitely many values such that ${Y \leq f(X) \leq X}$ is surely true. Conclude in particular that the above definition of expectation does not depend on the choice of model ${\Omega}$.
The expectation extends the simple expectation (thus ${{\bf E} X = \hbox{Simp} {\bf E} X}$ for all simple unsigned ${X}$), and in terms of a probability space model ${\Omega}$, the expectation ${{\bf E} X}$ is precisely the unsigned integral of ${X_\Omega}$. The expectation of a random variable is also often referred to as the mean, particularly in applications connected to statistics. In some literature ${{\bf E} X}$ is also called the expected value of ${X}$, but this is a somewhat misleading term as often one expects ${X}$ to deviate above or below ${{\bf E} X}$.
A scalar random variable ${X}$ is said to be absolutely integrable if ${{\bf E} |X| < \infty}$, thus for instance any bounded random variable is absolutely integrable. If ${X}$ is real-valued and absolutely integrable, we define its expectation by the formula
$\displaystyle {\bf E} X := {\bf E} X_1 - {\bf E} X_2$
where ${X = X_1 - X_2}$ is any representation of ${X}$ as the difference of unsigned absolutely integrable random variables ${X_1,X_2}$; one can check that this definition does not depend on the choice of representation and is thus well-defined. For complex-valued absolutely integrable ${X}$, we then define
$\displaystyle {\bf E} X := {\bf E} \hbox{Re} X + i {\bf E} \hbox{Im} X.$
In all of these cases, the expectation of ${X}$ is equal to the integral of the representation ${X_\Omega}$ of ${X}$ in any probability space model; in the case that ${\Omega}$ is given by a discrete probability model, one can check that this definition of expectation agrees with the one given in Notes 0. Using the former fact, we can translate the properties of integration already established to the probabilistic setting:
Proposition 13
• (i) (Unsigned linearity) If ${X,Y}$ are unsigned random variables, and ${c}$ is a deterministic unsigned quantity, then ${{\bf E}(X+Y) = {\bf E} X + {\bf E} Y}$ and ${{\bf E} cX = c {\bf E} X}$. (Note that these identities hold even when ${X,Y}$ are not absolutely integrable.)
• (ii) (Complex linearity) If ${X,Y}$ are absolutely integrable random variables, and ${c}$ is a deterministic complex quantity, then ${X+Y}$ and ${cX}$ are also absolutely integrable, with ${{\bf E}(X+Y) = {\bf E} X + {\bf E} Y}$ and ${{\bf E} cX = c {\bf E} X}$.
• (iii) (Compatibility with probability) If ${E}$ is an event, then ${{\bf E} 1_E = {\bf P}(E)}$. In particular, ${{\bf E} 1 = 1}$.
• (iv) (Almost sure equivalence) If ${X, Y}$ are unsigned (resp. absolutely integrable) and ${X=Y}$ almost surely, then ${{\bf E} X = {\bf E} Y}$.
• (v) If ${X}$ is unsigned or absolutely integrable, then ${{\bf E} |X| \geq 0}$, with equality if and only if ${X=0}$ almost surely.
• (vi) (Monotonicity) If ${X,Y}$ are unsigned or real-valued absolutely integrable, and ${X \leq Y}$ almost surely, then ${{\bf E} X \leq {\bf E} Y}$.
• (vii) (Markov inequality) If ${X}$ is unsigned or absolutely integrable, then ${{\bf P}(|X| \geq t) \leq \frac{1}{t} {\bf E} |X|}$ for any deterministic ${t>0}$.
• (viii) (Triangle inequality) If ${X}$ is absolutely integrable, then ${|{\bf E} X| \leq {\bf E} |X|}$.
As before, we can use part (iv) to define expectation of scalar random variables ${X}$ that are only defined and finite almost surely, rather than surely.
From Exercise 12, the notion of expectation is automatically probabilistic in the same sense. Because of this, we will be easily able to manipulate expectations of random variables without having to explicitly mention an underlying probability space ${\Omega}$, and so one will now see such spaces fade from view starting from this point in the course.
— 3. Exchanging limits with integrals or expectations —
When performing analysis on measure spaces, it is important to know if one can interchange a limit with an integral:
$\displaystyle \int_\Omega \lim_{n \rightarrow \infty} f_n\ d\mu \stackrel{?}{=} \lim_{n \rightarrow \infty} \int_\Omega f_n\ d\mu.$
Similarly, in probability theory, we often wish to interchange a limit with an expectation:
$\displaystyle {\bf E} \lim_{n \rightarrow \infty} X_n \stackrel{?}{=} \lim_{n \rightarrow \infty} {\bf E} X_n.$
Of course, one needs the integrands or random variables to be either unsigned or absolutely integrable, and the limits to be well-defined to have any hope of doing this. Naively, one could hope that limits and integrals could always be exchanged when the expressions involved are well-defined, but this is unfortunately not the case. In the case of integration on, say, the real line ${{\bf R}}$ using Lebesgue measure ${m}$, we already see four key examples:
• (Moving bump example) Take ${f_n := 1_{[n,n+1]}}$. Then ${\int_{\bf R} \lim_{n \rightarrow \infty} f_n\ dm = 0}$, but ${\lim_{n \rightarrow \infty} \int_{\bf R} f_n\ dm = 1}$.
• (Spreading bump example) Take ${f_n := \frac{1}{n} 1_{[0,n]}}$. Then ${\int_{\bf R} \lim_{n \rightarrow \infty} f_n\ dm = 0}$, but ${\lim_{n \rightarrow \infty} \int_{\bf R} f_n\ dm = 1}$.
• (Concentrating bump example) Take ${f_n := n 1_{[0,1/n]}}$. Then ${\int_{\bf R} \lim_{n \rightarrow \infty} f_n\ dm = 0}$, but ${\lim_{n \rightarrow \infty} \int_{\bf R} f_n\ dm = 1}$.
• (Receding infinity example) Take ${f_n := 1_{[n,\infty)}}$. Then ${\int_{\bf R} \lim_{n \rightarrow \infty} f_n\ dm = 0}$, but ${\lim_{n \rightarrow \infty} \int_{\bf R} f_n\ dm = +\infty}$.
In all these examples, the limit of the integral exceeds the integral of the limit; by replacing ${f_n}$ with ${-f_n}$ in the first three examples (which involve absolutely integrable functions) one can also build examples where the limit of the integral is less than the integral of the limit. Most of these examples rely on the infinite measure of the real line and thus do not directly have probabilistic analogues, but the concentrating bump example involves functions that are all supported on the unit interval ${[0,1]}$ and thus also poses a problem in the probabilistic setting.
Nevertheless, there are three important cases in which we can relate the limit (or, in the case of Fatou’s lemma, the limit inferior) of the integral to the integral of the limit (or limit inferior). Informally, they are:
• (Fatou’s lemma) For unsigned ${f_n}$, the integral of the limit inferior cannot exceed the limit inferior of the integral. “Limits (or more precisely, limits inferior) can destroy (unsigned) mass, but cannot create it.”
• (Monotone convergence theorem) For unsigned monotone increasing ${f_n}$, the limit of the integral equals the integral of the limit.
• (Dominated convergence theorem) For ${f_n}$ that are uniformly dominated by an absolutely integrable function, the limit of the integral equals the integral of the limit.
These three results then have analogues for convergence of random variables. We will also mention a fourth useful tool in that setting, which allows one to exchange limits and expectations when one controls a higher moment. There are a few more such general results allowing limits to be exchanged with integrals or expectations, but my advice would be to work out such exchanges by hand rather than blindly cite (possibly incorrectly) an additional convergence theorem beyond the four mentioned above, as this is safer and will help strengthen one’s intuition on the situation.
We now state and prove these results more explicitly.
Lemma 14 (Fatou’s lemma) Let ${(\Omega, {\mathcal F}, \mu)}$ be a measure space, and let ${f_1, f_2, \dots: \Omega \rightarrow [0,+\infty]}$ be a sequence of unsigned measurable functions. Then
$\displaystyle \int_\Omega \liminf_{n \rightarrow \infty} f_n\ d\mu \leq \liminf_{n \rightarrow \infty} \int_\Omega f_n\ d\mu.$
An equivalent form of this lemma is that if one has
$\displaystyle \int_\Omega f_n\ d\mu \leq M$
for some ${M}$ and all sufficiently large ${n}$, then one has
$\displaystyle \int_\Omega \liminf_{n \rightarrow \infty} f_n\ d\mu \leq M$
as well. That is to say, if the original unsigned functions ${f_n}$ eventually have “mass” less than or equal to ${M}$, then the limit (inferior) ${\liminf_{n \rightarrow \infty} f_n}$ also has “mass” less than or equal to ${M}$. The limit may have substantially less mass, as the four examples above show, but it can never have more mass (asymptotically) than the functions that comprise the limit. Of course, one can replace limit inferior by limit in the left or right hand side if one knows that the relevant limit actually exists (but one cannot replace limit inferior by limit superior if one does not already have convergence, see Example 16 below). On the other hand, it is essential that the ${f_n}$ are unsigned for Fatou’s lemma to work, as can be seen by negating one of the first three key examples mentioned above.
Proof: By definition of the unsigned integral, it suffices to show that
$\displaystyle \int_\Omega g\ d\mu \leq \liminf_{n \rightarrow \infty} \int_\Omega f_n\ d\mu$
whenever ${g}$ is an unsigned simple function with ${g \leq \liminf_{n \rightarrow \infty} f_n}$. At present, ${g}$ is allowed to take the infinite ${+\infty}$, but it suffices to establish this claim for ${g}$ that only take finite values, since the claim then follows for possibly infinite-valued ${g}$ by applying the claim with ${g}$ replaced by ${\min(g,N)}$ and then letting ${N}$ go to infinity.
Multiplying by ${1-\varepsilon}$, it thus suffices to show that
$\displaystyle (1-\varepsilon) \int_\Omega g\ d\mu \leq \liminf_{n \rightarrow \infty} \int_\Omega f_n\ d\mu$
for any ${0 < \varepsilon < 1}$ and any unsigned ${g}$ as above.
We can write ${g}$ as the sum ${g = \sum_{i=1}^k a_i 1_{E_i}}$ for some strictly positive finite ${a_i}$ and disjoint ${E_i}$; we allow the ${a_i}$ and the measures ${\mu(E_i)}$ to be infinite. On each ${E_i}$, we have ${\liminf_{n \rightarrow \infty} f_n > (1-\varepsilon) a_i}$. Thus, if we define
$\displaystyle E_{i,N} := \{ \omega \in E_i: f_n(\omega) \geq (1-\varepsilon) a_i \hbox{ for all } n \geq N \}$
then the ${E_{i,N}}$ increase to ${E_i}$ as ${N \rightarrow \infty}$: ${\bigcup_{N=1}^\infty E_{i,N} = E_i}$. By continuity from below (Exercise 23 of Notes 0), we thus have
$\displaystyle \mu(E_{i,N}) \rightarrow \mu(E_i)$
as ${N \rightarrow \infty}$. Since
$\displaystyle f_N \geq \sum_{i=1}^k (1-\varepsilon) a_i 1_{E_{i,N}}$
we conclude upon integration that
$\displaystyle \int_\Omega f_N\ d\mu \geq \sum_{i=1}^k (1-\varepsilon) a_i \mu( E_{i,N} )$
and thus on taking limit inferior
$\displaystyle \liminf_{N \rightarrow \infty} \int_\Omega f_N\ d\mu \geq \sum_{i=1}^k (1-\varepsilon) a_i \mu( E_i).$
But the right-hand side is ${(1-\varepsilon) \int_\Omega g\ d\mu}$, and the claim follows. $\Box$
Of course, Fatou’s lemma may be phrased probabilistically:
Lemma 15 (Fatou’s lemma for random variables) Let ${X_1,X_2,\dots}$ be a sequence of unsigned random variables. Then
$\displaystyle {\bf E} \liminf_{n \rightarrow \infty} X_n \leq \liminf_{n \rightarrow \infty} {\bf E} X_n.$
As a corollary, if ${X_1,X_2,\dots}$ are unsigned and converge almost surely to a random variable ${Y}$, then
$\displaystyle {\bf E} Y \leq \liminf_{n \rightarrow \infty} {\bf E} X_n.$
Example 16 We now give an example to show that limit inferior cannot be replaced with limit superior in Fatou’s lemma. Let ${X}$ be drawn uniformly at random from ${[0,1]}$, and for each ${n}$, let ${X_n}$ be the ${n^{th}}$ binary digit of ${X}$, thus ${X_n = 1}$ when ${2^n X}$ has odd integer part, and ${X_n =0}$ otherwise. (There is some ambiguity with the binary expansion when ${X}$ is a terminating binary decimal, but this event almost surely does not occur and can thus be safely ignored.) One has ${{\bf E} X_n = \frac{1}{2}}$ for all ${n}$ (why?). It is then easy to see that ${\liminf_{n \rightarrow \infty} X_n}$ is almost surely ${0}$ (which is consistent with Fatou’s lemma) but ${\limsup_{n \rightarrow \infty} X_n}$ is almost surely ${1}$ (so Fatou’s lemma fails if one replaces limit inferior with limit superior).
Next, we establish the monotone convergence theorem.
Theorem 17 (Monotone convergence theorem) Let ${(\Omega, {\mathcal F}, \mu)}$ be a measure space, and let ${f_1, f_2, \dots: \Omega \rightarrow [0,+\infty]}$ be a sequence of unsigned measurable functions which is monotone increasing, thus ${f_n(\omega) \leq f_{n+1}(\omega)}$ for all ${n}$ and ${\omega \in \Omega}$. Then
$\displaystyle \int_\Omega \lim_{n \rightarrow \infty} f_n\ d\mu = \lim_{n \rightarrow \infty} \int_\Omega f_n\ d\mu.$
Note that the limits exist on both sides because monotone sequences always have limits. Indeed the limit in either side is equal to the supremum. The receding infinity example shows that it is important that the functions here are monotone increasing rather than monotone decreasing. We also observe that it is enough for the ${f_n}$ to be increasing almost everywhere rather than everywhere, since one can then modify the ${f_n}$ on a set of measure zero to be increasing everywhere, which does not affect the integrals on either side of this theorem.
Proof: From Fatou’s lemma we already have
$\displaystyle \int_\Omega \lim_{n \rightarrow \infty} f_n\ d\mu \leq \lim_{n \rightarrow \infty} \int_\Omega f_n\ d\mu.$
On the other hand, from monotonicity we see that
$\displaystyle \int_\Omega \lim_{n \rightarrow \infty} f_n\ d\mu \geq \int_\Omega f_m\ d\mu$
for any natural number ${m}$, and on taking limits as ${m \rightarrow \infty}$ we obtain the claim. $\Box$
Note that continuity from below for measures (Exercise 23.3 of Notes 0 can be viewed as the special case of the monotone convergence theorem when the functions ${f_n}$ are all indicator functions.)
An important corollary of the monotone convergence theorem is that one can freely interchange infinite sums with integrals for unsigned functions, that is to say
$\displaystyle \int_\Omega \sum_{n=1}^\infty g_n\ d\mu = \sum_{n=1}^\infty \int_\Omega g_n\ d\mu$
for any unsigned ${g_1,g_2,\dots: \Omega \rightarrow [0,+\infty]}$ (not necessarily monotone). Indeed, to see this one simply applies the monotone convergence theorem to the partial sums ${f_N := \sum_{n=1}^N g_n}$.
We of course can translate this into the probabilistic context:
Theorem 18 (Monotone convergence theorem for random variables) Let ${0 \leq X_1 \leq X_2 \leq \dots}$ be a monotone non-decreasing sequence of unsigned random variables. Then
$\displaystyle {\bf E} \lim_{n \rightarrow \infty} X_n = \lim_{n \rightarrow \infty} {\bf E} X_n.$
Similarly, for any unsigned random variables ${Y_1,Y_2,\dots}$, we have
$\displaystyle {\bf E} \sum_{n=1}^\infty Y_n = \sum_{n=1}^\infty {\bf E} Y_n.$
Again, it is sufficient for the ${X_n}$ to be non-decreasing almost surely. We note a basic but important corollary of this theorem, namely the (first) Borel-Cantelli lemma:
Lemma 19 (Borel-Cantelli lemma) Let ${E_1,E_2,\dots}$ be a sequence of events with ${\sum_n {\bf P}(E_n) < \infty}$. Then almost surely, at most finitely many of the events ${E_n}$ hold; that is to say, one has ${\sum_n 1_{E_n} < \infty}$ almost surely.
Proof: From the monotone convergence theorem, we have
$\displaystyle {\bf E} \sum_n 1_{E_n} = \sum_n {\bf E} 1_{E_n} = \sum_n {\bf P}(E_n) < \infty.$
By Markov’s inequality, this implies that ${\sum_n 1_{E_n}}$ is almost surely finite, as required. $\Box$
As the above proof shows, the Borel-Cantelli lemma is almost a triviality if one has the machinery of expectation (or integration); but it is remarkably hard to prove the lemma without that machinery, and it is an instructive exercise to attempt to do so.
We will develop a partial converse to the above lemma (the “second” Borel-Cantelli lemma) in a subsequent set of notes. For now, we give a crude converse in which we assume not only that the ${{\bf P}(E_n)}$ sum to infinity, but they are in fact uniformly bounded from below:
Exercise 20 Let ${E_1,E_2,\dots}$ be a sequence of events with ${\inf_n {\bf P}(E_n) > 0}$. Show that with positive probability, an infinite number of the ${E_n}$ hold; that is to say, ${\mathop{\bf P}( \sum_n 1_{E_n} = \infty ) > 0}$. (Hint: if ${{\bf P}(E_n) \geq \delta > 0}$ for all ${n}$, establish the lower bound ${\mathop{\bf P}( \sum_{n \leq N} 1_{E_n} \geq \delta N/2 ) \geq \delta/2}$ for all ${N}$. Alternatively, one can apply Fatou’s lemma to the random variables ${1_{\overline{E_n}}}$.)
Exercise 21 Let ${p_1,p_2,\dots \in [0,1]}$ be a sequence such that ${\sum_{n=1}^\infty p_n = +\infty}$. Show that there exist a sequence of events ${E_1,E_2,\dots}$ modeled by some probability space ${\Omega}$, such that ${{\bf P}(E_n)=p_n}$ for all ${n}$, and such that almost surely infinitely many of the ${E_n}$ occur. Thus we see that the hypothesis ${\sum_{n=1}^\infty {\bf P}(E_n) < \infty}$ in the Borel-Cantelli lemma cannot be relaxed.
Finally, we give the dominated convergence theorem.
Theorem 22 (Dominated convergence theorem) Let ${(\Omega, {\mathcal F}, \mu)}$ be a measure space, and let ${f_1, f_2, \dots: \Omega \rightarrow {\bf C}}$ be measurable functions which converge pointwise to some limit. Suppose that there is an unsigned absolutely integrable function ${g: \Omega \rightarrow [0,+\infty]}$ which dominates the ${f_n}$ in the sense that ${|f_n(\omega)| \leq |g(\omega)|}$ for all ${n}$ and all ${\omega}$. Then
$\displaystyle \int_\Omega \lim_{n \rightarrow \infty} f_n\ d\mu = \lim_{n \rightarrow \infty} \int_\Omega f_n\ d\mu.$
In particular, the limit on the right-hand side exists.
Again, it will suffice for ${g}$ to dominate each ${f_n}$ almost everywhere rather than everywhere, as one can upgrade this to everywhere domination by modifying each ${f_n}$ on a set of measure zero. Similarly, pointwise convergence can be replaced with pointwise convergence almost everywhere. The domination of each ${f_n}$ by a single function ${g}$ implies that the integrals ${\int_\Omega |f_n|\ d\mu}$ are uniformly bounded in ${n}$, but this latter condition is not sufficient by itself to guarantee interchangeability of the limit and integral, as can be seen by the first three examples given at the start of this section.
Proof: By splitting into real and imaginary parts, we may assume without loss of generality that the ${f_n}$ are real-valued. As ${g}$ is absolutely integrable, it is finite almost everywhere; after modification on a set of measure zero we may assume it is finite everywhere. Let ${f}$ denote the pointwise limit of the ${f_n}$. From Fatou’s lemma applied to the unsigned functions ${g-f_n}$ and ${g+f_n}$, we have
$\displaystyle \int_\Omega g-f\ d\mu \leq \liminf_{n \rightarrow \infty} \int_\Omega g-f_n\ d\mu$
and
$\displaystyle \int_\Omega g+f\ d\mu \leq \liminf_{n \rightarrow \infty} \int_\Omega g+f_n\ d\mu$
Rearranging this (taking crucial advantage of the finite nature of the ${\int_\Omega g\ d\mu}$, and hence ${\int_\Omega f\ d\mu}$ and ${\int_\Omega f_n\ d\mu}$), we conclude that
$\displaystyle \limsup_{n \rightarrow \infty} \int_\Omega f_n\ d\mu \leq \int_\Omega f\ d\mu \leq \liminf_{n \rightarrow\infty} \int_\Omega f_n\ d\mu$
and the claim follows. $\Box$
Remark 23 Amusingly, one can use the dominated convergence theorem to give an (extremely indirect) proof of the divergence of the harmonic series ${\sum_{n=1}^\infty \frac{1}{n}}$. For, if that series was convergent, then the function ${\sum_{n=1}^\infty \frac{1}{n} 1_{[n-1,n]}}$ would be absolutely integrable, and the spreading bump example described above would contradict the dominated convergence theorem. (Expert challenge: see if you can deconstruct the above argument enough to lower bound the rate of divergence of the harmonic series ${\sum_{n=1}^\infty \frac{1}{n}}$.)
We again translate the above theorem to the probabilistic context:
Theorem 24 (Dominated convergence theorem for random variables) Let ${X_1,X_2,\dots}$ be scalar random variables which converge almost surely to a limit ${X}$. Suppose there is an unsigned absolutely integrable random variable ${Y}$ such that ${|X_n| \leq Y}$ almost surely for each ${n}$. Then
$\displaystyle \lim_{n \rightarrow \infty} {\bf E} X_n = {\bf E} X.$
As a corollary of the dominated convergence theorem for random variables we have the bounded convergence theorem: if ${X_1,X_2,\dots}$ are scalar random variables that converge almost surely to a limit ${X}$, and are almost surely bounded in magnitude by a uniform constant ${M}$, then we have
$\displaystyle \lim_{n \rightarrow \infty} {\bf E} X_n = {\bf E} X.$
(In Durrett, the bounded convergence theorem is proven first, and then used to establish Fatou’s theorem and the dominated and monotone convergence theorems. The order in which one establishes these results – which are all closely related to each other – is largely a matter of personal taste.) A closely related corollary (which can also be established directly is that if ${X_n}$ are scalar absolutely integrable random variables that converge uniformly to ${X}$ (thus, for each ${\varepsilon>0}$ there is ${N>0}$ such that ${|X_n-X| \leq \varepsilon}$ is surely true for all ${n \geq N}$), then ${{\bf E} X_n}$ converges to ${{\bf E} X}$.
A further corollary of the dominated convergence theorem is that one has the identity
$\displaystyle {\bf E} \sum_n Y_n = \sum_n {\bf E} Y_n$
whenever ${Y_n}$ are scalar random variables with ${\sum_n |Y_n|}$ absolutely integrable (or equivalently, that ${\sum_n {\bf E} |Y_n|}$ is finite).
Another useful variant of the dominated convergence theorem is
Theorem 25 (Convergence for random variables with bounded moment) Let ${X_1,X_2,\dots}$ be scalar random variables which converge almost surely to a limit ${X}$. Suppose there is ${\varepsilon>0}$ and ${M>0}$ such that ${{\bf E} |X_n|^{1+\varepsilon} \leq M}$ for all ${n}$. Then
$\displaystyle \lim_{n \rightarrow \infty} {\bf E} X_n = {\bf E} X.$
This theorem fails for ${\varepsilon=0}$, as the concentrating bump example shows. The case ${\varepsilon=1}$ (that is to say, bounded second moment ${{\bf E} |X_n|^2}$) is already quite useful. The intuition here is that concentrating bumps are in some sense the only obstruction to interchanging limits and expectations, and these can be eliminated by hypotheses such as a bounded higher moment hypothesis or a domination hypothesis.
Proof: By taking real and imaginary parts we may assume that the ${X_n}$ (and hence ${X}$) are real-valued. For any natural number ${m}$, let ${X_n^{[m]}}$ denote the truncation ${X_n^{[m]} := \max(\min(X_n,m),-m)}$ of ${X_n}$ to the interval ${[-m,m]}$, and similarly define ${X^{[m]} := \max(\min(X,m),-m)}$. Then ${X_n^{[m]}}$ converges pointwise to ${X^{[m]}}$, and hence by the bounded convergence theorem
$\displaystyle \lim_{n \rightarrow \infty} {\bf E} X_n^{[m]} = {\bf E} X^{[m]}.$
On the other hand, we have
$\displaystyle |X_n - X_n^{[m]}| \leq m^{-\varepsilon} |X_n|^{1+\varepsilon}$
(why?) and thus on taking expectations and using the triangle inequality
$\displaystyle {\bf E} X_n = {\bf E} X_n^{[m]} + O( m^{-\varepsilon} M )$
where we are using the asymptotic notation ${O(X)}$ to denote a quantity bounded in magnitude by ${CX}$ for an absolute constant ${C}$. Also, from Fatou’s lemma we have
$\displaystyle {\bf E} |X|^{1+\varepsilon} \leq M$
so we similarly have
$\displaystyle {\bf E} X = {\bf E} X^{[m]} + O( m^{-\varepsilon} M )$
Putting all this together, we see that
$\displaystyle \liminf_{n \rightarrow \infty} {\bf E} X_n, \limsup_{n \rightarrow \infty} {\bf E} X_n = {\bf E} X + O( m^{-\varepsilon} M ).$
Sending ${m \rightarrow \infty}$, we obtain the claim. $\Box$
Remark 26 The essential point about the condition ${{\bf E} |X_n|^{1+\varepsilon}}$ was that the function ${x \mapsto x^{1+\varepsilon}}$ grew faster than linearly as ${x \rightarrow \infty}$. One could accomplish the same result with any other function with this property, e.g. a hypothesis such as ${{\bf E} |X_n| \log |X_n| \leq M}$ would also suffice. The most natural general condition to impose here is that of uniform integrability, which encompasses the hypotheses already mentioned, but we will not focus on this condition here.
Exercise 27 (ScheffĂ©’s lemma) Let ${X_1,X_2,\dots}$ be a sequence of absolutely integrable scalar random variables that converge almost surely to another absolutely integrable scalar random variable ${X}$. Suppose also that ${{\bf E} |X_n|}$ converges to ${{\bf E} |X|}$ as ${n \rightarrow \infty}$. Show that ${{\bf E} |X-X_n|}$ converges to zero as ${n \rightarrow \infty}$. (Hint: there are several ways to prove this result, known as Scheffe’s lemma. One is to split ${X_n}$ into two components ${X_n = X_{n,1} + X_{n,2}}$, such that ${X_{n,1}}$ is dominated by ${|X|}$ but converges almost surely to ${X}$, and ${X_{n,2}}$ is such that ${|X_n| = |X_{n,1}| + |X_{n,2}|}$. Then apply the dominated convergence theorem.)
— 4. The distribution of a random variable —
We have seen that the expectation of a random variable is a special case of the more general notion of Lebesgue integration on a measure space. There is however another way to think of expectation as a special case of integration, which is particularly convenient for computing expectations. We first need the following definition.
Definition 28 Let ${X}$ be a random variable taking values in a measurable space ${R = (R, {\mathcal B})}$. The distribution of ${X}$ (also known as the law of ${X}$) is the probability measure ${\mu_X}$ on ${R}$ defined by the formula
$\displaystyle \mu_X(S) := {\bf P}( X \in S )$
for all measurable sets ${S \in {\mathcal B}}$; one easily sees from the Kolmogorov axioms that this is indeed a probability measure.
In the language of measure theory, the distribution ${\mu_X}$ on ${S}$ is the push-forward of the probability measure ${{\bf P}}$ on the sample space ${\Omega}$ by the model ${X_\Omega: \Omega \rightarrow R}$ of ${X}$ on that sample space.
Example 29 If ${X}$ only takes on at most countably many values (and if every point in ${R}$ is measurable), then the distribution ${\mu_X}$ is the discrete measure that assigns each point ${a}$ in the range of ${X}$ a measure of ${{\bf P}(X=a)}$.
Example 30 If ${X}$ is a real random variable with cumulative distribution function ${F}$, then ${\mu_X}$ is the Lebesgue-Stieltjes measure associated to ${F}$. For instance, if ${X}$ is drawn uniformly at random from ${[0,1]}$, then ${\mu_X}$ is Lebesgue measure restricted to ${[0,1]}$. In particular, two scalar variables are equal in distribution if and only if they have the same cumulative distribution function.
Example 31 If ${X}$ and ${Y}$ are the results of two separate rolls of a fair die (as in Example 3 of Notes 0), then ${X}$ and ${Y}$ are equal in distribution, but are not equal as random variables.
Remark 32 In the converse direction, given a probability measure ${\mu}$ on a measurable space ${(R, {\mathcal B})}$, one can always build a probability space model and a random variable ${X}$ represented by that model whose distribution is ${\mu}$. Indeed, one can perform the “tautological” construction of defining the probability space model to be ${\Omega := (R, {\mathcal B}, \mu)}$, and ${X: \Omega \rightarrow R}$ to be the identity function ${X(\omega) := \omega}$, and then one easily checks that ${\mu_X = \mu}$. Compare with Corollaries 26 and 29 of Notes 0. Furthermore, one can view this tautological model as a “base” model for random variables of distribution ${\mu}$ as follows. Suppose one has a random variable ${X}$ of distribution ${\mu}$ which is modeled by some other probability space ${\Omega' := (\Omega', {\mathcal F}', \mu')}$, thus ${X_{\Omega'}: \Omega' \rightarrow R}$ is a measurable function such that
$\displaystyle \mu(S) = {\bf P}(X \in S) = \mu'( \{ \omega' \in \Omega': X_{\Omega'}(\omega') \in S \})$
for all ${S \in {\mathcal B}}$. Then one can view the probability space ${\Omega'}$ as an extension of the tautological probability space ${\Omega = (R, {\mathcal B}, \mu)}$ using ${X_{\Omega'}}$ as the factor map.
We say that two random variables ${X,Y}$ are equal in distribution, and write ${X \stackrel{d}{=} Y}$, if they have the same law: ${\mu_X = \mu_Y}$, that is to say ${{\bf P}(X \in S) = {\bf P}(Y \in S)}$ for any measurable set ${S}$ in the range. This definition makes sense even when ${X,Y}$ are defined on different sample spaces. Roughly speaking, the distribution captures the “size” and “shape” of the random variable, but not its “location” or how it relates to other random variables. We also say that ${Y}$ is a copy of ${X}$ if they are equal in distribution. For instance, the two dice rolls in Example 3 of Notes 0 are copies of each other.
Theorem 33 (Change of variables formula) Let ${X}$ be a random variable taking values in a measurable space ${R = (R, {\mathcal B})}$. Let ${f: R \rightarrow {\bf R}}$ or ${f: R \rightarrow {\bf C}}$ be a measurable scalar function (giving ${{\bf R}}$ or ${{\bf C}}$ the Borel ${\sigma}$-algebra of course) such that either ${f \geq 0}$, or that ${{\bf E} |f(X)| < \infty}$. Then
$\displaystyle {\bf E} f(X) = \int_R f(x)\ d\mu_X(x).$
Thus for instance, if ${X}$ is a real random variable, then
$\displaystyle {\bf E} |X| = \int_{\bf R} |x|\ d\mu_X(x),$
and more generally
$\displaystyle {\bf E} |X|^p = \int_{\bf R} |x|^p\ d\mu_X(x)$
for all ${0 < p < \infty}$; furthermore, if ${X}$ is unsigned or absolutely integrable, one has
$\displaystyle {\bf E} X = \int_{\bf R} x\ d\mu_X(x).$
The point here is that the integration is not over some unspecified sample space ${\Omega}$, but over a very explicit domain, namely the reals; we have “changed variables” to integrate over ${{\bf R}}$ instead over ${\Omega}$, with the distribution ${\mu_X}$ representing the “Jacobian” factor that typically shows up in such change of variables formulae.
If ${X}$ is a scalar variable that only takes on at most countably many values ${x_1,x_2,\dots}$, the change of variables formula tells us that
$\displaystyle {\bf E} X = \sum_i x_i {\bf P}(X=x_i)$
if ${X}$ is unsigned or absolutely integrable.
Proof: First suppose that ${f}$ is unsigned and only takes on a finite number ${a_1,\dots,a_n}$ of values. Then
$\displaystyle f(X) = \sum_{i=1}^n a_i 1_{f(X)=a_i}$
and hence
$\displaystyle {\bf E} f(X) = \sum_{i=1}^n a_i {\bf P}(f(X)=a_i)$
$\displaystyle = \sum_{i=1}^n a_i \mu_X( \{ x: f(x) = a_i\} )$
$\displaystyle = \int_R \sum_{i=1}^n a_i 1_{f(x)=a_i}\ d\mu_X(x)$
$\displaystyle = \int_R f(x)\ d\mu_X(x)$
as required.
Next, suppose that ${f}$ is unsigned but can take on infinitely many values. We can express ${f}$ as the monotone increasing limit of functions ${f_n}$ that only take a finite number of values; for instance we can define ${f_n(x)}$ to be ${f(x)}$ rounded down to the largest multiple of ${1/n}$ less than both ${n}$ and ${f(x)}$. By the preceding computation, we have
$\displaystyle {\bf E} f_n(X) = \int_R f_n(x)\ d\mu_X(x),$
and on taking limits as ${n \rightarrow \infty}$ using the monotone convergence theorem we obtain the claim in this case.
Now suppose that ${f}$ is real-valued with ${{\bf E} |f(X)| < \infty}$. We write ${f = f_1 - f_2}$ where ${f_1 := \max(f,0)}$ and ${f_2 := \min(f,0)}$, then we have ${{\bf E} f_1(X), {\bf E} f_2(X) < \infty}$ and
$\displaystyle {\bf E} f_i(X) = \int_{\bf R} f_i(x)\ d\mu_X(x)$
for ${i=1,2}$. Subtracting these two identities together, we obtain the claim.
Finally, the case of complex-valued ${f}$ with ${{\bf E} |f(X)| < \infty}$ follows from the real-valued case by taking real and imaginary parts. $\Box$
Example 34 Let ${X}$ be the uniform distribution on ${[0,1]}$, then
$\displaystyle {\bf E} f(X) = \int_0^1 f(x)\ dx$
for any Riemann integrable ${f}$; thus for instance
$\displaystyle {\bf E} X^p = \frac{1}{p+1}$
for any ${0 < p < \infty}$.
Remark 35 An alternate way to prove the change of variables formula is to observe that the formula is obviously true when one uses the tautological model ${(R, {\mathcal B}, \mu_X)}$ for ${X}$, and then the claim follows from the model-independence of expectation and the observation from Remark 32 that any other model for ${X}$ is an extension of the tautological model.
Exercise 36 Let ${f: {\bf R} \rightarrow [0,+\infty]}$ be a measurable function with ${\int_{\bf R} f(x)\ dx = 1}$. If one defines ${m_f(E)}$ for any Borel subset ${E}$ of ${{\bf R}}$ by the formula
$\displaystyle m_f(E) := \int_E f(x)\ dx,$
show that ${m_f}$ is a probability measure on ${{\bf R}}$ with Stieltjes measure function ${F(t) = \int_{-\infty}^t f(x)\ dx}$. If ${X}$ is a real random variable with probability distribution ${m_f}$ (in which case we call ${X}$ a random variable with an absolutely continuous distribution, and ${f}$ the probability density function (PDF) of ${X}$), show that
$\displaystyle \mathop{\bf E} G(X) = \int_{\bf R} G(x) f(x)\ dx$
when either ${G: {\bf R} \rightarrow [0,+\infty]}$ is an unsigned measurable function, or ${G: {\bf R} \rightarrow {\bf C}}$ is measurable with ${G(X)}$ absolutely integrable (or equivalently, that ${\int_{\bf R} |G(x)| f(x)\ dx < \infty}$.
Exercise 37 Let ${X}$ be a real random variable with the probability density function ${x \mapsto \frac{1}{\sqrt{2\pi}} e^{-x^2/2}}$ of the standard normal distribution. Establish the Stein identity
$\displaystyle {\bf E} X F(X) = {\bf E} F'(X)$
whenever ${F: {\bf R} \rightarrow {\bf R}}$ is a continuously differentiable function with ${F}$ and ${F'}$ both of polynomial growth (i.e., there exist constants ${C, A > 0}$ such that ${|F(t)|, |F'(t)| \leq C (1+|t|)^A}$ for all ${t \in {\bf R}}$). There is a robust converse to this identity which underpins the basis of Stein’s method, discussed in this previous blog post. Use this identity recursively to establish the identities
$\displaystyle {\bf E} X^k = 0$
when ${k}$ is an odd natural number and
$\displaystyle {\bf E} X^k = (k-1) (k-3) \dots 1 = \frac{k!}{2^{k/2} (k/2)!}$
when ${k}$ is an even natural number. (This quantity is also known as the double factorial ${(k-1)!!}$ of ${k-1}$.)
Exercise 38 Let ${X}$ be a real random variable with cumulative distribution function ${F_X(x) := {\bf P}(X \leq x)}$. Show that
$\displaystyle {\bf E} e^{tX} = \int_{\bf R} (1-F_X(x)) t e^{tx}\ dx$
for all ${t > 0}$. If ${X}$ is nonnegative, show that
$\displaystyle {\bf E} X^p = \int_0^\infty (1-F_X(x)) p x^{p-1}\ dx$
for all ${p>0}$.
— 5. Some basic inequalities —
The change of variables formula allows us, in principle at least, to compute the expectation ${{\bf E} X}$ of a scalar random variable as an integral. In very simple situations, for instance when ${X}$ has one of the standard distributions (e.g. uniform, gaussian, binomial, etc.), this allows us to compute such expectations exactly. However, once one gets to more complicated situations, one usually does not expect to be able to evaluate the required integrals in closed form. In such situations, it is often more useful to have some general inequalities concerning expectation, rather than identities.
We therefore record here for future reference some basic inequalities concerning expectation that we will need in the sequel. We have already seen the triangle inequality
$\displaystyle |{\bf E} X| \leq {\bf E} |X| \ \ \ \ \ (5)$
for absolutely integrable ${X}$, and the Markov inequality
$\displaystyle {\bf P}(|X| \geq t) \leq \frac{1}{t} {\bf E} |X| \ \ \ \ \ (6)$
for arbitrary scalar ${X}$ and ${t>0}$ (note the inequality is trivial if ${X}$ is not absolutely integrable). Applying the triangle inequality to the difference ${X-Y}$ of two absolutely integrable random variables ${X,Y}$, we obtain the variant
$\displaystyle |{\bf E} X - {\bf E} Y| \leq {\bf E} |X-Y|. \ \ \ \ \ (7)$
Thus, for instance, if ${X_n}$ is a sequence of absolutely integrable scalar random variables which converges in ${L^1}$ to another absolutely integrable random variable ${X}$, in the sense that ${{\bf E} |X_n-X| \rightarrow 0}$ as ${n \rightarrow \infty}$, then ${{\bf E} X_n \rightarrow {\bf E} X}$ as ${n \rightarrow \infty}$.
Similarly, applying the Markov inequality to the quantity ${|X - {\bf E} X|^2}$ we obtain the important Chebyshev inequality
$\displaystyle {\bf P}(|X - {\bf E} X| \geq t) \leq \frac{1}{t^2} \hbox{Var}(X) \ \ \ \ \ (8)$
for absolutely integrable ${X}$ and ${t>0}$, where the Variance ${\hbox{Var}(X)}$ of ${X}$ is defined as
$\displaystyle \hbox{Var}(X) := {\bf E}( |X - {\bf E} X|^2 ).$
Next, we record
Lemma 39 (Jensen’s inequality) If ${f: {\bf R} \rightarrow {\bf R}}$ is a convex function, ${X}$ is a real random variable with ${X}$ and ${f(X)}$ both absolutely integrable, then
$\displaystyle f( {\bf E} X ) \leq {\bf E} f(X).$
Proof: Let ${x_0}$ be a real number. Being convex, the graph of ${f}$ must be supported by some line at ${(x_0,f(x_0))}$, that is to say there exists a slope ${c}$ (depending on ${x_0}$) such that ${f(x) \ge f(x_0) + c(x-x_0)}$ for all ${x \in {\bf R}}$. (If ${f}$ is differentiable at ${x_0}$, one can take ${c}$ to be the derivative of ${f}$ at ${x_0}$, but one always has a supporting line even in the non-differentiable case.) In particular
$\displaystyle f(X) \geq f(x_0) + c(X-x_0).$
Taking expectations and using linearity of expectation, we conclude
$\displaystyle {\bf E} f(X) \geq f(x_0) + c({\bf E} X - x_0 )$
and the claim follows from setting ${x_0 := {\bf E} X}$. $\Box$
Exercise 40 (Complex Jensen inequality) Let ${f: {\bf C} \rightarrow {\bf R}}$ be a convex function (thus ${f((1-t)z+tw) \leq (1-t)f(z) + tf(w)}$ for all complex ${z,w}$ and all ${0 \leq t \leq 1}$, and let ${X}$ be a complex random variable with ${X}$ and ${f(X)}$ both absolutely integrable. Show that
$\displaystyle f( {\bf E} X ) \leq {\bf E} f(X).$
Note that the triangle inequality ${|{\bf E} X| \leq {\bf E} |X|}$ is the special case of Jensen’s inequality (or the complex Jensen’s inequality, if ${X}$ is complex-valued) corresponding to the convex function ${f(x) := |x|}$ on ${{\bf R}}$ (or ${f(z) := |z|}$ on ${{\bf C}}$). Another useful example is
$\displaystyle |{\bf E} X|^2 \leq {\bf E} |X|^2.$
Applying Jensen’s inequality to the convex function ${f: x \mapsto e^x}$ and the random variable ${X = \log Y}$ for some ${Y>0}$, we obtain the arithmetic mean-geometric mean inequality
$\displaystyle \exp( {\bf E} \log Y ) \leq {\bf E} Y$
assuming that ${Y>0}$ and ${Y,\log Y}$ are absolutely integrable.
As a related application of convexity, observe from the convexity of the function ${x \mapsto e^x}$ that
$\displaystyle e^{tx + (1-t)y} \leq t e^x + (1-t) e^y$
for any ${0 \leq t \leq 1}$ and ${x,y \in {\bf R}}$. This implies in particular Young’s inequality
$\displaystyle |X| |Y| \leq \frac{1}{p} |X|^p + \frac{1}{q} |Y|^q$
for any scalar ${X,Y}$ and any exponents ${1 < p,q < \infty}$ with ${\frac{1}{p} + \frac{1}{q}=1}$; note that this inequality is also trivially true if one or both of ${|X|, |Y|}$ are infinite. Taking expectations, we conclude that
$\displaystyle {\bf E} |X| |Y| \leq \frac{1}{p} {\bf E} |X|^p + \frac{1}{q} {\bf E} |Y|^q$
if ${X,Y}$ are scalar random variabels and ${1 < p, q < \infty}$ are deterministic exponents with ${\frac{1}{p} + \frac{1}{q}=1}$. In particular, if ${|X|^p, |Y|^q}$ are absolutely integrable, then so is ${XY}$, and
$\displaystyle |{\bf E} X Y| \leq \frac{1}{p} {\bf E} |X|^p + \frac{1}{q} {\bf E} |Y|^q.$
We can amplify this inequality as follows. Multiplying ${X}$ by some ${0 < \lambda < \infty}$ and dividing ${Y}$ by the same ${\lambda}$, we conclude that
$\displaystyle |{\bf E} X Y| \leq \frac{\lambda^p}{p} {\bf E} |X|^p + \frac{\lambda^{-q}}{q} {\bf E} |Y|^q;$
optimising the right-hand side in ${\lambda}$, we obtain (after some algebra, and after disposing of some edge cases when ${X}$ or ${Y}$ is almost surely zero) the important Hölder inequality
$\displaystyle |{\bf E} X Y| \leq ({\bf E} |X|^p)^{1/p} ({\bf E} |Y|^q)^{1/q} \ \ \ \ \ (9)$
which we can write as
$\displaystyle |{\bf E} X Y| \leq \| X\|_p \|Y\|_q$
where we use the notation
$\displaystyle \|X\|_p := ({\bf E} |X|^p)^{1/p}$
for ${0 < p < \infty}$. Using the convention
$\displaystyle \|X\|_\infty := \inf \{ M: |X| \leq M \hbox{ a. s.} \}$
(thus ${\|X\|_\infty}$ is the essential supremum of ${X}$), we also see from the triangle inequality that the Hölder inequality applies in the boundary case when one of ${p,q}$ is allowed to be ${\infty}$ (so that the other is equal to ${1}$):
$\displaystyle |{\bf E} X Y| \leq \| X\|_1 \|Y\|_\infty, \|X\|_\infty \|Y\|_1.$
The case ${p=q=2}$ is the important Cauchy-Schwarz inequality
$\displaystyle |{\bf E} X Y| \leq \| X\|_2 \|Y\|_2, \ \ \ \ \ (10)$
valid whenever ${X,Y}$ are square-integrable in the sense that ${\|X\|_2, \|Y\|_2}$ are finite.
Exercise 41 Show that the expressions ${\|X\|_p}$ are non-decreasing in ${p}$ for ${p \in (0,+\infty]}$. In particular, if ${\|X\|_p}$ is finite for some ${p}$, then it is automatically finite for all smaller values of ${p}$.
Exercise 42 For any square-integrable ${X}$, show that
$\displaystyle {\bf P}(X \neq 0) \geq \frac{ ({\bf E} |X|)^2}{{\bf E}(|X|^2)}.$
Exercise 43 If ${1 < p < \infty}$ and ${X,Y}$ are scalar random variables with ${\|X\|_p, \|Y\|_p < \infty}$, use Hölder’s inequality to establish that
$\displaystyle {\bf E} |X| |X+Y|^{p-1} \leq \|X\|_p \|X+Y\|_p^{p-1}$
and
$\displaystyle {\bf E} |Y| |X+Y|^{p-1} \leq \|Y\|_p \|X+Y\|_p^{p-1}$
and then conclude the Minkowski inequality
$\displaystyle \|X+Y\|_p \leq \|X\|_p + \|Y\|_p.$
Show that this inequality is also valid at the endpoint cases ${p=1}$ and ${p=\infty}$.
Exercise 44 If ${X}$ is non-negative and square-integrable, and ${0 \leq \theta \leq 1}$, establish the Paley-Zygmund inequality
$\displaystyle {\bf P}( X > \theta {\bf E}(X) ) \geq (1-\theta)^2 \frac{({\bf E} X)^2}{{\bf E}(|X|^2)}.$
(Hint: use the Cauchy-Schwarz inequality to upper bound ${{\bf E} X 1_{X > \theta {\bf E} X}}$ in terms of ${{\bf E} |X|^2}$ and ${{\bf P}( X > \theta {\bf E}(X) )}$.)
Exercise 45 Let ${X}$ be a non-negative random variable that is almost surely bounded but not identically zero, show that
$\displaystyle \mathop{\bf P}( X \geq \frac{1}{2} \mathop{\bf E} X) \geq \frac{1}{2} \frac{\mathop{\bf E} X}{\|X\|_\infty}.$ | 21,458 | 72,962 | {"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": 1144, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-14 | latest | en | 0.842503 |
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