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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://slideplayer.info/slide/2721109/
| 1,529,536,087,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267863939.76/warc/CC-MAIN-20180620221657-20180621001657-00171.warc.gz
| 296,207,258
| 30,714
|
# Java Fundamentals: 4. Java Algorithms
## Presentasi berjudul: "Java Fundamentals: 4. Java Algorithms"— Transcript presentasi:
Java Fundamentals: 4. Java Algorithms
Romi Satria Wahono
Romi Satria Wahono SD Sompok Semarang (1987) SMPN 8 Semarang (1990)
SMA Taruna Nusantara, Magelang (1993) S1, S2 dan S3 (on-leave) Department of Computer Sciences Saitama University, Japan ( ) Research Interests: Software Engineering, Intelligent Systems Founder dan Koordinator IlmuKomputer.Com Peneliti LIPI ( ) Founder dan CEO PT Brainmatics Cipta Informatika
Course Outline OOP Concepts: Konsep dan Paradigma Object-Oriented
Java Basics: Memahami Sintaks dan Grammar Bahasa Java Java GUI: Swing, GUI Component, Event Handling, Pengembangan Aplikasi GUI Java Algorithms: Pengantar Algoritma, Struktur Data, Algorithm Analysis Java Advanced: Eksepsi, Thread, Java API Java Database: Koneksi ke Database, Pengembangan Aplikasi Database
4. Java Algorithms
Java Algorithms Pengantar Algoritma Analisis Efisiensi Algorithm
Struktur Data dengan Java Collection Sorting Algorithms Searching Algorithms
4.1 Pengantar Algoritma
Algoritma An algorithm is a sequence of unambiguous instructions for solving a problem, i.e., for obtaining a required output for any legitimate input in a finite amount of time (Levitin, 2012)
Konsep Algoritma program = algorithm + data structure
The nonambiguity requirement for each step of an algorithm cannot be compromised The range of inputs for which an algorithm works has to be specified carefully The same algorithm can be represented in several different ways There may exist several algorithms for solving the same problem Algorithms for the same problem can be based on very different ideas and can solve the problem with dramatically different speeds
Masalah Komputasi Penting
Sorting Searching String processing Graph problems Combinatorial problems Geometric problems Numerical problems
4.2 Analisis Efisiensi Algoritma
Analisis Efisiensi Algoritma
Menentukan karakteristik kinerja (memprediksi sumber daya) Memilih algoritma yang paling efisien dari beberapa alternatif penyelesaian untuk kasus yang sama Mencari waktu yang terbaik untuk keperluan praktis Apakah algoritma itu optimal untuk beberapa kasus atau ada yang lebih baik
Analisis Efisiensi Algoritma
Time efficiency (time complexity): Indicates how fast an algorithm in question runs Space efficiency (space complexity): Refers to the amount of memory units required by the algorithm in addition to the space needed for its input and output.
Analisis Efisiensi Algoritma
Both time and space efficiencies are measured as functions of the algorithm’s input size Time efficiency is measured by counting the number of times the algorithm’s basic operation is executed Space efficiency is measured by counting the number of extra memory units consumed by the algorithm The efficiencies of some algorithms may differ significantly for inputs of the same size. For such algorithms, we need to distinguish between the worst-case, average-case, and best-case efficiencies The framework’s primary interest lies in the order of growth of the algorithm’s running time (extra memory units consumed) as its input size goes to infinity
Kompleksitas Komputasi
Worst-case: kompleksitas waktu untuk waktu terburuk (waktu tempuh bernilai maksimum dari suatu fungsi f(n)) atau Tmax(n) Best-case: kompleksitas waktu untuk waktu terbaik (kompleksitas waktu yang bernilai minimum dari suatu fungsi f(n)) atau Tmin(n) Average-case: kompleksitas waktu untuk kasus rata-rata
Metode Analisis Algoritma
Asymptotic/theoretic/mathematic: berdasarkan pendekatan secara teori atau atas dasar analisa secara matematik Empirical/Practical/Empiris/Praktis: berdasarkan pendekatan praktis yang biasanya didasarkan atas data-data yang telah ada atau data-data yang di-generate / dibangkitkan
Asymptotic Menggambarkan karakteristik/perilaku suatu algoritma pada batasan tertentu (berupa suatu fungsi matematis) Dituliskan dengan notasi matematis yg dikenal dgn notasi asymptotic Notasi asymptotic dapat dituliskan dengan beberpa simbul berikut Q O W o w
4.3 Struktur Data dengan Java Collection Framework
Object-Oriented Programming 4.3 Struktur Data dengan Java Collection Framework
Java Collection Framework
A collection (sometimes called a container) is an object that groups multiple elements into a single unit Collections are used to store, retrieve, manipulate, and communicate aggregate data Typically, they represent data items that form a natural group, such as a poker hand (a collection of cards), a mail folder (a collection of letters), or a telephone directory (a mapping of names to phone numbers)
Java Collection Framework
A collections framework is a unified architecture for representing and manipulating collections All collections frameworks contain the following: Interfaces: These are abstract data types that represent collections. Interfaces allow collections to be manipulated independently of the details of their representation Implementations: These are the concrete implementations of the collection interfaces. In essence, they are reusable data structures Algorithms: These are the methods that perform useful computations, such as searching and sorting, on objects that implement collection interfaces
Core Collection Interfaces
Set SortedSet List Queque Map SortedSet
Core Collection Implementation
Object Array Array standard yang dimiliki oleh Java API (java.util)
Array memiliki method-method: Method Keterangan static int binarySearch(array, key) Pencarian nilai dalam array boolean equals(array1, array2) Membandingkan apakah dua array memiliki nilai sama. Bekerja pada array satu dimensi static void sort(array) Mengurutkan isi array static String toString(array) Mengubah nilai array menjadi String
Contoh Object Array public class ArrayCari { public static void main(String[] args) { String[] jenisKelamin = new String[3]; jenisKelamin[0] = "laki"; jenisKelamin[1] = "perempuan"; jenisKelamin[2] = "waria"; int ketemu = Arrays.binarySearch(jenisKelamin , "perempuan"); if (ketemu > -1) System.out.println("Data ditemukan pada: “ + ketemu); else System.out.println("Data tidak ditemukan."); }
ArrayList ArrayList mirip dengan array, tapi memiliki kemampuan lebih baik Jumlah elemen dalam ArrayList dapat berubah secara fleksibel tergantung jumlah data yang ada di dalamnya Setelah array terbentuk, data baru dapat dimasukkan di tengah-tengah, tidak harus di akhir elemen array Isi dalam array bisa dihapus, dan index dalam array sesudahnya akan maju satu langkah untuk mengisi slot kosong tersebut
Menambahkan object ke dalam ArrayList boolean equals(array1, array2) Menambahkan object ke dalam index yang ditentukan static void sort(array) Menghapus semua elemen dalam ArrayList static String toString(array) Mengambil object pada index tertentu iterator() Mengembalikan iterator pada ArrayList remove(int index) Menghapus object dengan index tertentu remove(object element) Menghapus elemen tertentu size() Mengembalikan nilai berupa jumlah elemen dalam ArrayList toArray() Mengembalikan elemen ArrayList sebagai array object toArray(type[] array) Mengembalikan elemen ArrayList Sebagi array dengan tipe tertentu
Contoh ArrayList public class ArrayAngka{ public static void main(String[] args) { ArrayList angka= new ArrayList(); angka.add("One"); angka.add("Two"); angka.add(3); angka.add("Four"); for (Object i: angka) System.out.println(i); angka.set(1, "Siji"); angka.remove(angka.size() - 1); System.out.println(angka); }
Vector Sama seperti ArrayList, Vector memiliki dua atribut utama: kapasitas dan penambahan kapasitas Penambahan kapasitas menentukan berapa jumlah index yang akan ditambahkan, jika index saat ini sudah tidak mencukupi
Vector Method Keterangan void add(int index, Object element)
Memasukkan object ke dalam Vector dengan index yang ditentukan boolean add(Object element) Menambahkan Object ke dalam Vector. Jika berhasil nilai boolean = true void addElement(Object element) ….
Contoh Vector //menampilkan vector dengan perulangan dan size
public class VectorDemo { public static void main(String[] args) { Vector newVector = new Vector(); //menambahkan data vector newVector.add("Jakarta"); newVector.add("Surabaya"); newVector.add("Semarang"); // menampilkan data vector pertama System.out.println("Menampilkan Data Vector:"); System.out.println("Data Vector Pertama:"+ newVector.get(0)); System.out.println("Data Vector Pertama:"+ newVector.firstElement()); System.out.println("Data Vector Kedua: " + newVector.get(1)); // menampilkan data vector terakhir (ketiga) System.out.println("Data Vector Ketiga: " + newVector.elementAt(2)); System.out.println("Data Vector Ketiga: " + newVector.lastElement()); //mencari index vector dan ditampilkan System.out.println('\n' + "Mencari Data Vector:"); int idxCari = newVector.indexOf("Surabaya"); System.out.println("Nilai Index Yang Dicari Adalah: " + idxCari); if (idxCari>=0) System.out.println("Data yang Dicari Adalah:" + newVector.get(idxCari)); //menampilkan vector dengan perulangan dan size for (int i=0; i < newVector.size();i++) System.out.println(i + ":" + newVector.get(i)); //menampilkan vector dengan iterator "for-loop" for (Iterator d = newVector.iterator(); d.hasNext(); ) { System.out.println("->" + d.next()); }
HashMap Koleksi yang memetakan kunci (key) ke dalam nilai (value)
Kunci dan nilai dalam HashMap boleh diset dengan null HashMap tepat untuk data yang kompleks, sehingga programmer tidak harus menghafal letak index seperti pada array dan collection class sequence lainnya
HashMap Method Keterangan void clear()
Menghapus semua elemen dalam HashMap sehingga ukurannya menjadi 0 boolean isEmpty() Nilai true dikembalikan jika tidak ada elemen di dalam int size() Mengembalikan jumlah elemen dalam HashMap boolean containsKey(Object key) Nilai true dikembalikan jika key ditemukan dalam HashMap boolean containsValue(Object value) Nilai true dikembalikan jika value ditemukan dalam HashMap
Contoh HashMap public class HashMapDemo { public static void main(String[] args) { HashMap map = new HashMap(); //menambahkan data ke hashmap map.put("Nama", "Joko Bodo"); map.put("NIM", new Integer(234567)); map.put("Alamat", "Semarang"); //menampilkan hashmap System.out.println("Hashmap: " + map.entrySet()); System.out.println("Ukuran Hashmap: " + map.size()); map.put("Situs favorit", "ilmukomputer.com"); //menampilkan dan melihat ukuran hashmap //mengecek data di hashmap System.out.println("Has Key NIM?“ + map.containsKey("NIM")); //mendetele data di hashmap System.out.println("Removed: " + map.remove("NIM")); //menampilkan dan melihat ukuran hashmap System.out.println("Hashmap: " + map.entrySet()); System.out.println("Ukuran Hashmap:" + map.size()); }
InterfaceIterator Fasilitas pada Java API yang dapat digunakan untuk melakukan iterasi komponen-komponen dalam Koleksi Ada tiga method yang sering digunakan dalam Iterator: hasNext(), next(), remove() Method Keterangan hasNext() Menentukan apakah masih ada sisa koleksi next() Mengembalikan elemen object pada koleksi. Jika sudah tidak ada elemen lagi namun berusaha diambil maka akan muncul pesan: NoSuchElementException remove() Menghapus elemen yang terakhir kali diakses oleh Iterator
Contoh Iterator while (v.hasNext()){ Object ob = v.next(); System.out.println(v); } for(Iterator i = v.iterator(); i.hasNext();){ String name = (String) i.next(); System.out.println(name);
Generic Implementasi tipe data pada koleksi
Tanpa adanya generic, tipe data berbeda-beda dapat dimasukkan dalam sebuah koleksi. Ketika data tersebut diambil, maka perlu dilakukan casting Misal method seperti di bawah: public boolean add (Object o){ //statements } Untuk pengambilan data, harus dilakukan casting tipe data: Mahasiswa mhs = (Mahasiswa) organisasi.get(); Masalah muncul jika ada beberapa elemen yang bukan bertipe Mahasiswa, elemen lain mungkin saja ada karena semua object dapat ditambahkan dengan metode add() diatas
Generic Tipe generic pada koleksi dapat diterapkan dengan menambahkan tanda <> Bila kita berusaha menambahkan elemen dengan tipe data berbeda, maka akan keluar error Dengan adanya generic, program dapat lebih handal, karena kesalahan programmer dapat dicegah
4.4 Sorting Algorithms
Sorting Algorithms Algoritma sorting adalah algoritma dasar yang paling sering digunakan Data dalam keadaan yang sudah urut (sesuai dengan kunci pengurutan tertentu) akan memudahkan kita dalam manipulasi berikutnya Beberapa algoritma sorting: Bubble Sort Merge Sort Selection Sort
Bubble Sort Catatan: data[i], data[i+1] data[1] = 86
Pengurutan dengan membandingkan suatu elemen dengan elemen berikutnya Jika elemen sekarang lebih besar daripada elemen berikutnya maka elemen tersebut akan ditukar Data yang ingin diurutkan: 34, 86, 15 Catatan: data[i], data[i+1] data[0] = 34 data[1] = 86 data[2] = 15 Menukar data:
Alur Algoritma Bubble Sort (34 86 15)
Langkah 0 tidak tukar tukar Langkah 1
Alur Algoritma Bubble Sort (34 86 15)
LANGKAH bilangan[0] bilangan[1] Bilangan[2] 34 86 15 1
BubbleSort.java BubbleSortBeraksi.java
public class BubbleSort { public static void urutkan(int data[]){ for(int langkah=0; langkah<data.length; langkah++){ for(int indeks=0; indeks<data.length-1; indeks++){ if(data[indeks]>data[indeks+1]){ int temp = data[indeks]; data[indeks] = data[indeks+1]; data[indeks+1] = temp; } public class BubbleSortBeraksi{ public static void main(String[] args){ int data[] = {34, 86, 15}; System.out.print("Data awal: "); for(int i=0;i<data.length;i++){ System.out.print(data[i] + " "); } BubbleSort.urutkan(data); System.out.print('\n' + "Data hasil: "); System.out.println();
Latihan: Versi GUI dari BubbleSort
Selection Sort Pengurutan dengan mencari elemen berikutnya sampai elemen terakhir Jika ditemukan elemen lain yang lebih kecil dari elemen sekarang, maka elemen tersebut akan ditukar Data yang ingin diurutkan: 34, 86, 15
Selection Sort untuk Data 34 86 15
Langkah 1 tukar Langkah 2
SelectionSort.java SelectionSortBeraksi.java
public class SelectionSort { public static void urutkan(int data[]){ for(int langkah=0; langkah<data.length-1; langkah++){ int indeksTerkecil=langkah; for(int indeks=langkah+1; indeks<data.length; indeks++){ if(data[indeks]<data[indeksTerkecil]) indeksTerkecil=indeks; } int temp=data[langkah]; data[langkah]=data[indeksTerkecil]; data[indeksTerkecil]=temp; public class SelectionSortBeraksi{ public static void main(String[] args){ int data[] = {34, 86, 15}; System.out.print("Data awal: "); for(int i=0;i<data.length;i++){ System.out.print(data[i] + " "); } SelectionSort.urutkan(data); System.out.print('\n' + "Data hasil: "); System.out.println();
Merge Sort Algoritma pengurutan dengan cara menggabungkan dua kelompok data yang sudah urut, kemudian digabung dan hasilnya adalah data yang terurut Langkah algoritma Merge Sort Bila jumlah item yang diurutkan adalah 0 atau 1, return Urutkan secara rekursif bagian pertama dan kedua secara terpisah Gabungkan dua bagian yang sudah terurut tersebut ke dalam sebuah kelompok terurut
Library Sorting Buat project bernama Sorting
Buat dua class: BubbleSort dan SelectionSort Build project Sorting supaya menghasilkan Sorting.jar Buat project baru bernama SortingGUI Buat aplikasi GUI yang melakukan sorting terhadap 5 bilangan bulat yang kita masukkan Gunakan library Sorting.jar pada project SortingGUI tersebut dan sajikan hasil dari BubbleSort dan SelectionSort
4.5 Searching Algorithms
Tugas Pahami dan buat program dari algoritma di bawah dengan menggunakan Java. Pilih berdasarkan digit terakhir NPM Rangkumkan secara komprehensif dalam bentuk slide Presentasikan di depan kelas dengan bahasa manusia Selection sort Bubble sort Merge sort Quicksort Insertion sort Shell sort Heapsort Binary Search Sequential Search Depth-First Search
Tugas Rangkumkan secara komprehensif dalam bentuk slide:
Analisis Efsiensi Algoritma (Rangkumkan dari buku Levitin dan Weiss bab 1-2) Algoritma (sesuai dengan NPM): Pengantar Algoritma (definisi, kategorisasi algoritma, dsb … lihat levitin) Tahapan Algoritma (kalimat, formula, dsb) Tahapan Algoritma (Pseudocode) Tahapan Algoritma (Java) Tahapan Algoritma (Animasi) – (harus sinkron dengan code) Penerapan untuk Studi Kasus (harus sinkron sesuai tahapan algoritma) Analisis Algoritma (penghitungan efisiensi) Kirimkan slide, code dan animasi ke dengan subject [algoritma-univ] nama-nim sebelum 29 Agustus 2013 Slide dibuat asal-asalan, kebut semalam, tidak mudah dipahami, tidak komprehensif, tidak dengan menggunakan bahasa manusia, atau nyontek mendapatkan nilai E
Tugas Pahami dan buat program dari algoritma di bawah dengan menggunakan Java (GUI) Pilih algoritma sesuai digit terakhir NPM: 0 Particle Swarm Optimization 5 A* 1 Neural Network 6 K-Means 2 Support Vector Machine 7 Genetic Algorithm 3 Naive Bayes 8 Ant Collony Optimization 4 C k-Nearest Neighbor (kNN) Rangkumkan secara komprehensif tentang algoritma dan program yang dibuat dalam bentuk slide Presentasikan di depan kelas
Tugas Kerjakan semua latihan dan tugas yang ada di slide Java Algorithms Kirimkan netbeans project yang sudah di zip ke dengan subyek: [OOP4-Universitas] Nama–NIM Deadline: 2 minggu Meng-copy file orang lain akan menyebabkan nilai tugas 0
Referensi Sharon Zakhour et al, The Java Tutorial Fourth Edition, Cay Horstmann, Big Java: Earl Objects 5th Edition, John Wiley & Sons, 2013 Deitel & Deitel, Java Howto Program 9th Edition, Prentice Hall, 2012 Richard M. Reese, Oracle Certified Associate Java SE 7 Programmer Study Guide, Packt Publishing, 2012 Walter Savitch, Absolute Java 5th Edition, Pearson Education, 2013 Mark Allen Weiss, Data Structures and Algorithm Analysis in Java 3rd Edition, Pearson Education, 2012 Anany Levitin, Introduction to the Design and Analysis of Algorithms 3rd Edition, Pearson Education, 2012 Ying Bai, Practical Database Programming with Java, John Wiley & Sons, 2011
Presentasi serupa
| 4,462
| 17,953
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.828125
| 3
|
CC-MAIN-2018-26
|
latest
|
en
| 0.515529
|
https://forums.creativecow.net/docs/forums/post.php?forumid=2&postid=1129370&univpostid=1129370&pview=t
| 1,539,898,435,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583512014.61/warc/CC-MAIN-20181018194005-20181018215505-00385.warc.gz
| 712,570,827
| 8,510
|
FORUMS: list search recent posts
# Technique for animating a wake with a swimming dolphin
FAQ • VIEW ALL
Technique for animating a wake with a swimming dolphin on May 18, 2018 at 9:02:17 am
I am animating a dolphin coming in and out of the water. The dolphin moves from one point to the next, with a wiggle, but I also need to animate the wake of the fin and tail. How am I best doing this? On a previous animation, I just animated the path of the shape and parent it to the moving animal, but on this occasion, the movement of the dolphin is quite erratic, parenting it to the dolphin doesn't seem to work to well as the position jumps quite quick.
What might be my best method to animate this?
Hope you can help
Steve
Digital Designer
stevedavies.io
Re: Technique for animating a wake with a swimming dolphinon May 18, 2018 at 3:55:48 pm
Even the though the dolphin will porpoise up and down (I couldn't resist), the wake shouldn't. So you could link the x motion of the wake to the dolphin and then the wake will follow but not rise and fall. Then you could put a slight animation on the y motion of the wake so it's not exactly linear.
You can break apart the x and y position attribute in the curve graph editor (the button is at the bottom of the graph). Or you could use an expression.
On the position attribute enter: [use Pickwhip here to link the x of the dolphin, some Y value];
The notation for separate x and y in expression is [x,y];
You could link the Y to a keyframed slider effect or use keyframes on the position. Expressions can pick up keyframes from the source layer with the Value method [this.Comp.layer("Dolphin").transform.position[0],value[1]];
value[1] means Y, value[0] means X. Only using the y value will ignore the keyframed x value and use the dolphin x position for the x of the wake.
Re: Technique for animating a wake with a swimming dolphinon May 21, 2018 at 7:12:10 am
Thanks Steve
I have had a go but no success but I will keep trying - most likely something simple I am not doing in the process. But in theory it sounds like it will work!
Digital Designer
stevedavies.io
Re: Technique for animating a wake with a swimming dolphinon May 23, 2018 at 2:40:56 am
I had pretty much the same problem with one of my project. My solution was to use motion tracker.
For you situation, this is what I recomend
1. Add a black vertical line, center it to the dolphin's fin and link it to the dolphin.
2. Add an horizontal line at the water level.
3. Use that horizontal line as a track matte for the vertical line. This will make it just a black dot at water level
5. Track Motion of that dot.
6. Apply transformation of the track to a null.
7. Copy that null into the composition of the animation
Repeat process for the tail.
In addition you could with expression animate the width of your wake depending on the y position of your dolphin. so that it grows as the dolphin passes the water line.
To the scale of your wake layer:
```//put the position Y of your dolphin when the fin just touches the water line. y1 = 560 ; //put the position Y of your dolphin when you think the wake should be full size. y2 = 400; dolDepth = thisComp.layer("<i>Dolphin layer name</i>").transform.position[1]; w = linear(dolDepth,y1,y2,0,100); //You can also try an ease interpolation instead of the linear one. [w,100]```
Hope that helps
Re: Technique for animating a wake with a swimming dolphinon May 23, 2018 at 8:02:04 am
Why would you motion track something that you animated? Simply child a null to the dolphin, and position the null at the spot of the dolphin you want to "track", then use the toWorld transform expression to get the world position of that null as it moves with the dolphin. From there you can either use both the x and y (and even Z if you need it) or just one of the axis.
That way if you want to change your animation later on ('cause that never happens) you don't have to retrack the motion. (plus you don't get the inherent positional noise added when you do any kind of motion track)
SteveD, I think its was probably the rotation of the dolphin that didn't allow my first suggestion to work for you. If the rotation point of the dolphin was the point you wanted to follow it would work, but if the point you wanted was somewhere else that point would be offset by the distance but also by an ever changing amount based on the how far the dolphin's angle had changed. This could be calculated with some simple Trig, but its far easier to use the toWorld function that tells you the position of something like a null even though that null is a child of an object and the object is rotating. If you just asked for the position of the childed null with the pickwhip, all you would get is the position of that child object in relation to the parent and not in relation to the viewport.
Re: Technique for animating a wake with a swimming dolphinon May 23, 2018 at 10:46:03 am
You are absolutely right but in my situation it wasnt a dolphin but a fishing line made with rubberhose and wiggle. So to find where the fishing line crossed the waterline I didnt find any other way to track it due to the curved line.
Same would be if he is using puppet pin to deform the dolphin the fin position would vary in relation to its position or rotation.
| 1,265
| 5,311
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.609375
| 3
|
CC-MAIN-2018-43
|
longest
|
en
| 0.914086
|
https://www.tribonet.org/wiki/coefficient-of-friction-derivation-in-reciprocating-contacts/
| 1,726,179,787,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00594.warc.gz
| 959,003,408
| 37,687
|
# Coefficient of Friction derivation in Reciprocating Contacts
## Xavier Borras, PhD
November, 15 2022
Reciprocating contacts are common in many engineering applications, e.g., hydraulic/pneumatic cylinders, crank-slider mechanisms (piston rings-cylinder), fretting vibrations, cyclic stresses, etc. The coefficient of friction in a reciprocating contact is usually detrimental when designing the component and often determines the mechanism service time. A higher coefficient of friction predicts a shorter lifespan in fretting fatigue for example [1]. Oscillating tribometers (or reciprocating tribometers) replicate the nature of these contacts so the tribological characteristics of the real component contact can be simulated in the lab (model scale).
The frictional characteristics of a reciprocating (or oscillating) contacts are usually presented using the tangential force hysteresis loop (also referred to as “fretting loop”), i.e., the tangential force Q (y-axis) over its position Δ (x-axis) along a full cycle: forward and back strokes.
Figure 1. Tangential force hysteresis loop with Coulomb and non-Coulomb examples [2].
Alternatively, the Coefficient of Friction (Q/P) over a dimensionless position (Δ/Δ*) is used, as shown in Figure 1. Being P the contact force and Δ* the contact stroke. Notice that the slope S of the cycle shows the stiffness of the contact system. the Unfortunately, the friction over a reciprocating cycle is hardly ever constant, i.e., the Coulomb Friction theory does not apply, hence a method must be selected to represent the whole cycle with a single scalar value.
Llavori et al. [2] conducted a critical analysis of the Coefficient of Friction derivation methods for fretting under gross slip regime. The 163 articles on gross slip regime published in Tribology Letters and Wear tribology journals between 2009 and 2019 were reviewed. Surprisingly, 41% of the publications did not specify how the Coefficient of Friction was obtained. Four different methods were identified among the articles that explained how the Coefficient of Friction was reached:
1 Maximum value of the tangential force over a cycle maxCoF 30% 2 Mean value of the tangential force meanCoF 7% 3 Energy coefficient of friction ECoF 21% 4 Geometric independent coefficient of friction GiCoF 2%
Table 1. CoF Derivation methods found in Llavori et al.’s analysis over 163 reviewed articles on gross slip regime.
## Maximum value of the tangential force over a cycle (maxCoF)
This method simply considers the maximum CoF (Q/P) over a cycle. As a cycle presents two maximums, one for the forward stroke and one for the backward stroke, the average of them is usually presented. This method is effective for ideal parallelogram-shaped loops, i.e., where Coulomb’s friction law applies however usually overestimates the CoF of a cycle. Additionally, since this method relies on only two data points, The resulting maxCoF often depends on the filter used on the tangential force signal.
Figure 2. Overestimation using the maxCOF method [2].
## Mean value of the tangential force (meanCoF)
In this case, then mean absolute value of each cycle is computed. This method is useful in applications in which the sliding part is significantly higher than the sticking part. That is when the velocity approces zero at the beginning and the end of the stroke. Due to the constant sampling rate, the most part of the data points acquired belong to the part of the cycle where the velocity is lower. This can lead to an overestimation of the CoF.
Figure 3. Friction coefficient over time [2].
To focus on the sliding portion of the cycle, only a percentage of the cycle is often considered. Two approaches were found:
• A percentage of the position on a cycle (e.g., from 10% to 90 % of the stroke)
• A percentage around the maximum tangential force (e.g., a window between Qmax and 0.8*Qmax [3]).
There is no consensus on which window should be considered and hence this leads to extra degrees of freedom which need to be considered in advance.
## Energy coefficient of friction (ECoF)
This method consists in estimating the energy dissipation Ed over a tangential force hysteresis loop. The δ* is the slip amplitude. The ECoF (µe) method was proven to be robust, and it evolved to an ASTM standard [4].
## Geometric independent coefficient of friction (GiCoF)
A fourth method is presented by Mulvihill et al. considering the shape of the wear scar. The model proved that a hook-like behaviour can results when a parabolic wear scar is considered. Further details can be found in [2] and [5].
## References
1. Dobromirski J, Smith IO. A stress analysis of a shaft with a press-fitted hub subjected to cyclic axial loading. Int J Mech Sci 1986;28:41–52. doi:10.1016/0020-7403(86)90006-8
2. Llavori, I.; Zabala, A.; Aginagalde, A.; Tato, W.; Ayerdi, J.J.; Gómez, X. (2019). Critical analysis of coefficient of friction derivation methods for fretting under gross slip regime. Tribology International, (), 105988–. doi:10.1016/j.triboint.2019.105988
3. Wang D, Song D, Wang X, Zhang D, Zhang C, Wang D, et al. Tribo-fatigue behaviors of steel wires under coupled tension-torsion in different environmental media. Wear 2019;420–421:38–53. http://dx.doi.org/10.1016/j.wear.2018.12.038.
4. G02 Committee. Guide for Determining Friction Energy Dissipation in Reciprocating Tribosystems. ASTM International; n.d. doi:10.1520/G0203-10R16.
5. Mulvihill DM, Kartal ME, Olver AV, Nowell D, Hills DA. Investigation of non-Coulomb friction behaviour in reciprocating sliding. Wear 2011;271:802–16. doi:10.1016/j.wear.2011.03.014.
Industrial Engineer with focus on Tribology and Sealing Technology. Team player with an open-minded mentality author of several scientific publications and an industrial patent. Interested in Lean Management, Innovation, Circular Economy, Additive Manufacturing and Connected Objects Technology.
1. Mathias Woydt says:
Assuming a correct statistical analysis in table 1 means, that most reviewers didn´t know the different types of coefficients of friction, otherwise they would have request these in their reviews (see Tribotest 2006; 12: 133-147).
I think, that the representations of evolution of hysteresis by Fouvry&Kapsa et al. as shown in “Tribology International 39 (2006) 1005–1015” is much more visual then the introducing diagram of the present paper.
Some basics can be found in Plint&Plint “Tribology International, August 1985, Vol. 18, No.4, p. 247-249
Basics on dissipated energy can be found in: Fouvry&Kapsa, Wear of Materials, 1985 ; WEAR 252 (2002) 375–383 ; Tribology International 39 (2006) 1005–1015 ;
This paper not shows much new.
Fretting tests go back to Sikorsky fretting test rig in 1950s or even before in WWII for pitch control of propellers.
• Inigo Llavori says:
I loved seeing our article analyzed on tribonet :). In response to Dr. Woydt, the main contribution of this article lies in demonstrating that researchers do not accurately report the techniques used in their experimental trials. Our analysis shows a clear influence of the calculation method on the friction coefficient value, and most of those who report it use the worst of the methods. This speaks volumes about the state of how research work is reported in tribology.
This site uses Akismet to reduce spam. Learn how your comment data is processed.
| 1,786
| 7,396
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.0625
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.89097
|
https://answeregy.com/are/are-accuplacer-tests-hard.php
| 1,653,019,492,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00136.warc.gz
| 161,429,371
| 15,267
|
• Post:Stuart Morrison
• 11/1/2021
• Share:
# Are accuplacer tests hard?
Looking for an answer to the question: Are accuplacer tests hard? On this page, we have gathered for you the most accurate and comprehensive information that will fully answer the question: Are accuplacer tests hard?
A good ACCUPLACER score ranges from 78 to 120 on reading comprehension and 50 to 120 on the college level math test. The recommended score to place into freshman English courses and college level math courses varies from one college to the next.
Every section of the Accuplacer test, including arithmetic is based on a 120 scale so the highest score you can receive for arithmetic is 120.
You cannot fail a placement test. The test determines what courses you will be placed in when you begin at college. Depending on your score, you may need to take extra developmental courses or you may be able to start regular college courses right away. What kind of math is on the Accuplacer test?
The ACCUPLACER is a placement test for community college (also known as junior college). It's used to assess your current skill level and readiness for the types of schoolwork you'll be required to do in community college - specifically, reading, writing, and math.
## How do I pass the Accuplacer test?
2:2245:31ACCUPLACER Writing Exam Preparation - YouTubeYouTubeStart of suggested clipEnd of suggested clipWriting tests and tests of all of the other types as well available online um the website that weMoreWriting tests and tests of all of the other types as well available online um the website that we are going to use is accuplacer.org. And i will show you how to get to that when i share the screen.
## What is a good score on accuplacer test?
Accuplacer Scores – Ranges LOW: Accuplacer scores from 200 to 220 are usually considered low scores. HIGH: High Accuplacer scores are generally 270 points and above. AVERAGE: Scores of 221 to 250 are average, while scores between 250 and 270 are normally considered above average.
## What if I fail my college placement test?
If you don't do well the first time, there is room to improve by retaking most exams. Taking the time to take college placement test prep courses either online or in the classroom can increase your chances of doing well in the first round of tests. Often, a student can retake the test and score significantly higher.
## Can you retake accuplacer test?
Can I retake any portion of the Accuplacer? Yes, if you are not happy with your score or placement, you may retake a portion; however, you will need an electronic referral from your Academic Advisor.
## Can you fail the accuplacer test?
Remember: No one passes or fails ACCUPLACER tests, but it's important to complete the test using your best effort, so you can get an accurate measure of your academic skills and be placed in the appropriate course. Get resources to help you practice for the tests.
## How difficult is the accuplacer test?
The ACCUPLACER test is quite difficult because it is a computer-adaptive test. In other words, as you get questions right, the more difficult the questions for the next questions will be. It also means any wrong answers that make the questions easier. Therefore, even a highly prepared tester finds it challenging.
## Can I use a calculator on the Accuplacer test?
You're not allowed to bring a physical calculator into the test center or use handheld calculators on any ACCUPLACER Math test. Exception: Students who've been approved to use handheld calculators through an official accommodations request may use them in accordance with the approved accommodation.
## What happens if I don't pass the Accuplacer test?
Failure to pass any portion of the ACCUPLACER simply means that your community college is going to require you to take at least one remedial no-credit course before they allow you to enroll in a credit course in that subject area.
## What kind of math is on a placement test?
The Mathematics Placement Test consist of 67 questions covering four main areas: arithmetic, algebra I, algebra II, and trigonometry.
## How do I prepare for the Accuplacer test?
Understand the Test ObjectivesSentence Structure Test. An individual's ability to understand how to organize information in a sentence is part of the focus of this test. ... WritePlacer. ... Sample Questions. ... Practice Tests. ... Accuplacer Study Courses. ... Use Grammar and Spell Check Apps. ... Schedule Study Time. ... Join a Study Group.
## Is accuplacer timed?
How long will the ACCUPLACER testing take? The ACCUPLACER reading and math tests are un-timed and the writing test is a 50 minute timed test. Plan on approximately three hours.
## What is the accuplacer like?
The ACCUPLACER test is a comprehensive, web-based assessment tool used to determine your skills in reading, writing and math. It is untimed, but most students complete it in less than 90 minutes. All questions must be answered, and you cannot go back to a previous question once you have answered it.
## What math is on the Accuplacer test?
The ACCUPLACER test covers the type of math you would see in Pre-Algebra, Algebra 1, Geometry, and Algebra 2. For example, key concepts range from Integers and Fractions all the way to Logarithms and Exponential Functions.
## How many times can I take the accuplacer?
You can take the test as many times as you need to until you pass. The test is untimed and is open book. You may use your training materials as you test.
## What level of math is on the Accuplacer test?
The ACCUPLACER test covers the type of math you would see in Pre-Algebra, Algebra 1, Geometry, and Algebra 2. For example, key concepts range from Integers and Fractions all the way to Logarithms and Exponential Functions.
## What is passing on accuplacer test?
Each of the five sections of the ACCUPLACER are scored separately on a scale from 0 to 100, just like the grades that most students receive in school. Any score of 80 or above is passing.
## Do I have to write an essay for the accuplacer?
During the ACCUPLACER test, you will have to take the WritePlacer exam, which requires you to write an essay. Your essay will be evaluated based on the following criteria: Purpose and focus. Organization and structure.
## How fast do you get accuplacer results?
When will my ACCUPLACER results be available? Some results are available immediately, while some results may be available within 7-10 days from the date of your assessment. The proctor will let you know when they are available upon the completion of testing.
## Is accuplacer like SAT?
The Accuplacer is developed and administered by the College Board, the same organization that makes the SAT and AP tests. The Accuplacer is most often administered by colleges who want to make sure students are ready for college or figure out which level of college classes students are prepared for.
## What kind of math is on the Accuplacer test?
The ACCUPLACER test covers the type of math you would see in Pre-Algebra, Algebra 1, Geometry, and Algebra 2. For example, key concepts range from Integers and Fractions all the way to Logarithms and Exponential Functions.
## Are accuplacer tests hard? Expert Answers
none
### How hard is the accuplacer test? - FindAnyAnswer.com
4.9/5 (2,209 Views . 21 Votes) The ACCUPLACER test is a comprehensive, web-based assessment tool used to determine your skills in reading, writing and math. It is untimed, but most students complete it in less than 90 minutes. The test is adaptive, which means the questions get more difficult as you give more right answers.
### Is the ACCUPLACER Test hard? - Effortless Math
The ACCUPLACER test is quite difficult because it is a computer-adaptive test. In other words, as you get questions right, the more difficult the questions for the next questions will be. It also means any wrong answers that make the questions easier. Therefore, even a highly prepared tester finds it challenging.
### Is the accuplacer test hard? - AskingLot.com
The test is adaptive, which means the questions get more difficult as you give more right answers. Click to see full answer. Also question is, what is a good score on the Accuplacer test? Test-takers receive between 1 and 8 points. The higher …
6 rows
### How do I pass the Accuplacer reading test? – Learn Answer
Are accuplacer tests hard? The Accuplacer sample questions show that the test is not very hard. You’ll do fine if percentages and subject-verb agreement are fairly deep in your blood. But rusty basic skills will send you into a black hole of remediation. Accuplacer scores are good for two years, after which students need retesting.
### ACCUPLACER Test exam 2020 frequently asked questions ...
Is the Accuplacer test hard? With strong problem-solving skills, you can ace this test without much difficulty. Accuplacer Exam Format ... The difficulty level of questions that appear in the Accuplacer test will increase if you start selecting more right answers. Accuplacer Test Score
### What Are Good ACCUPLACER Test Scores?
The good news is that the ACCUPLACER exam is a placement exam, meaning that you can’t fail the test. Additionally, your score on the ACCUPLACER won’t affect your enrollment at your school. The purpose of the test is to determine whether you …
### ACCUPLACER Practice Test (2021) 45 ACCUPLACER …
Take our ACCUPLACER practice test to see if you are ready for college-level courses. Get a step-by-step guide for the ACCUPLACER review. ... We believe you can perform better on your exam, so we work hard to provide you with the best study guides, practice questions, and flashcards to empower you to be your best. Learn More... PRODUCTS & …
### If You’re Thinking Of Not Studying For The ACCUPLACER ...
As you probably know, the ACCUPLACER test is administered to determine which college courses you should be placed in. The test measures your math, reading, and writing skills and adapts to your skill level as you progress.This means that as you answer questions correctly they will get progressively harder and as you answer questions incorrectly they will get progressively easier.
### BEFORE YOU TAKE AN ACCUPLACER …
The ACCUPLACER placement test is an adaptive test. Therefore, your next questions are based on your previous answers. It adapts to your answers to give you harder or easier questions. You want HARDER because that means you are scoring into a …
### ACCUPLACER Practice Test | Free Practice Questions
The ACCUPLACER Test is a college placement test that is administered by the College Board. It is used by over 1,500 institutions as part of the enrollment process. The purpose of the test is to identify strengths as well as weaknesses in a variety of subject areas. Colleges and technical schools will use the results of this test, along with ...
### what is the highest score on the accuplacer test ...
The ACCUPLACER math test is quite hard because it’s a computer-adaptive test. In other words, as you get questions right, the difficulty of subsequent questions increases. In other words, as you get questions right, the difficulty of subsequent questions increases.
### What’s on the ACCUPLACER - dummies
The Arithmetic Test is the most basic of the three ACCUPLACER math sections. It tests your knowledge and ability in five areas. Whole number operations: The four basic operations (addition, subtraction, multiplication, and division) applied to whole …
### What is a good score on the accuplacer math test?
Secondly, is the accuplacer math test hard? The ACCUPLACER test is a comprehensive, web-based assessment tool used to determine your skills in reading, writing and math. It is untimed, but most students complete it in less than 90 minutes. The test is adaptive, which means the questions get more difficult as you give more right answers.
### Free Accuplacer Practice Test & Exam Information | Study.com
The Accuplacer tests are computer-based placement tests that contain questions of varying difficulty intended to determine a student's appropriate placement in core college courses.
### Home - ACCUPLACER | College Board
ACCUPLACER tests help colleges make accurate course placement decisions and set students up for success.
### ACCUPLACER For Dummies Cheat Sheet - dummies
The ACCUPLACER includes the following five separate tests: Reading Test. Writing Test. 3 math tests: Arithmetic Test. Quantitative Reasoning, Algebra, and Statistics Test. Advanced Algebra and Functions Test. This Cheat Sheet helps to get you started with important information to help boost your score on the ACCUPLACER.
### Free ACCUPLACER® Practice Tests [2021] | 500+ Questions
The ACCUPLACER Test is typically administered at the college or institution that the incoming student will be attending. There are no time limits on any of the ACCUPLACER tests, with the exception of the Writeplacer which may or may not have a time limit depending on the policy of the particular school administering the test.
### Prepare for and Pass the ACCUPLACER Math Test Prep Program
The ACCUPLACER math test is quite hard because it’s a computer-adaptive test. In other words, as you get questions right, the difficulty of subsequent questions increases. Therefore, even a highly prepared test-taker will find it challenging.
### ACCUPLACER Testing FAQ - NECC
Call: (978) 556-3654. Individuals who are deaf and hard of hearing: Call: 978-241-7045 (VP/Voice) When Will My ACCUPLACER Results Be Available? When will my ACCUPLACER results be available? Some results are available immediately, while some results may be available within 7-10 days from the date of your assessment.
### Can you cheat on the Accuplacer test? – Colors-NewYork.com
Are accuplacer tests hard? The Accuplacer sample questions show that the test is not very hard. You’ll do fine if percentages and subject-verb agreement are fairly deep in your blood. But rusty basic skills will send you into a black hole of remediation. Accuplacer scores are good for two years, after which students need retesting.
### What is a good score for accuplacer? – Raiseupwa.com
Is the accuplacer reading test hard? The ACCUPLACER Reading Comprehension practice test has a relatively low difficulty rating. On average, people complete it in about 15 minutes. All of the concepts included in the practice are featured on the real test. Post navigation.
### How to Pass the ACCUPLACER-Mometrix Blog
The ACCUPLACER test is a comprehensive, web-based assessment tool used to determine your skills in reading, writing and math. It is untimed, but most students complete it in less than 90 minutes. All questions must be answered, and you cannot go back to a previous question once you have answered it.
### Accuplacer Practice Test - Practice Test Geeks
The Accuplacer is a set of placement tests or diagnostic tests that help measure whether post-secondary students have the academic skills and knowledge required to begin with college-level courses. It is an integrated system that determines whether candidates have the basic core skills needed for any course.
### How to Prepare for Next-Generation Accuplacer Math Test?
Is the ACCUPLACER test hard? This is an adaptive test, which means that the more correctly you answer, the more difficult the questions become. What happens if you fail the ACCUPLACER test?
### ACCUPLACER Platform for Institutions – The College Board
Access your ACCUPLACER test results, download free practice resources, or find a testing location closer to home. Use Voucher. Fast Track Login. Information for Students . Start Test with Voucher. Please click the button below to begin testing with a voucher. If testing with Examity or Proctortrack, DO NOT use this button to start your test.
### How should I study for the Accuplacer math test? I want to ...
Answer (1 of 5): Like another of my answers, this one is easy, don't. The Accuplacer is designed not as an exit test (did you learn the minimum), but an entrance exam (what is the maximum you know). Study material in a way to understand it, and you …
### ACCUPLACER Arithmetic - Next Gen Test Prep Course
Is the ACCUPLACER Arithmetic - Next Gen test hard? The math on the ACCUPLACER Arithmetic - Next Gen test won’t seem hard if you’re thoroughly prepared and confident on test day. To be sure you can rely on your skills, you’ll need more preparation than a dry textbook or practice problems without explanations can provide.
### Free ACCUPLACER Practice Test Questions and Exam Prep
The ACCUPLACER exam is a college preparedness test that measures a student’s skills in reading, writing, and math. While something like the SAT or ACT is used to qualify you for admission into a university, the ACCUPLACER test is about precisely placing you at the correct level of introductory classes.
### How many questions are on the accuplacer math test?
The Next-Generation Accuplacer math tests and Reading and Writing tests are scored in the 200 to 300 point range. Similarly, it is asked, is the accuplacer math test hard? The ACCUPLACER test is a comprehensive, web-based assessment tool used to determine your skills in reading, writing and math. It is untimed, but most students complete it in ...
### Any advice for taking the Accuplacer Test? : GaState
It’s fairly easy tbh, just review your math as much as you can. The accuplacer website has practice test you can work on. If you’re planning on doing a stem major you’ll probably want to place into college algebra right away , or else you’ll get stuck in a math support class . 1. level 1.
### Sample Questions for Students - ACCUPLACER
In an ACCUPLACER® placement test, there are 20 Sentence Skills questions of two types. • he irst type consists of sentence-correction questions that require an understanding of sentence structure. hese questions ask you to choose the most appropriate word or phrase for the underlined portion of the sentence.
### Accuplacer Free Practice Questions - TestPrep-Online
The ACCUPLACER test does not have a time-limit; you can take as much time as you need. Even though the test is not timed, it is a good idea to take timed tests to re-create a pressured testing environment. This way, you can identify any anxieties you may face during
### Accuplacer Exam Can Kill College Careers – Julia Steiny on ...
Accuplacer has a 55-minute writing test, and then shorter, 20-question tests in reading, sentence structure and math, that students finish at their own pace. The battery takes between two and two-and-a-half hours. ... The Accuplacer sample questions show that the test is not very hard. You’ll do fine if percentages and subject-verb agreement ...
### How hard is the Accuplacer math section? - Community ...
The way Accuplacer works is that is starts out with basic arithmetic and then if you successfully answer those questions, it moves to algebra and gets progressively harder. So the questions you get will depend on how well you did from the very beginning of the test.
### How to Pass the ACCUPLACER Test - Study.com
Questions will become more or less difficult depending on how you answer questions. All ACCUPLACER multiple-choice tests include 20 questions, except for the Arithmetic, and Elementary Algebra ...
### how to pass the elementary algebra accuplacer - Lisbdnet.com
The ACCUPLACER math test is quite hard because it’s a computer-adaptive test. In other words, as you get questions right, the difficulty of subsequent questions increases. Therefore, even a highly prepared test-taker will find it challenging. How do you ace the accuplacer math test?
### ACCUPLACER Concordance Tables
test. The ACCUPLACER concordance tables were constructed from asample that is intended to represent the ACCUPLACER test -taking population. Applying the concordance tables to populations of students that are demographically different from the national population may result in decisions that are not beneficial to students.
### WT Virtual Math Lab - THEA/ACCUPLACER Before the Test
Yes, I do know that it might be hard to get to sleep early on the eve of the big test, however try not to stay up all night. ... Chances are when you go to take either the TSI Assessment or ACCUPLACER test, they will ask you for two forms of ID, one of which needs to have a picture of you on it. It is good to bring these just in case.
### ACCUPLACER - Arizona State University
Scores for WritePlacer exams taken on or after May 13, 2009, the essay scores range from 0 to 8. Placement into an English composition course will be made on the following basis: A WritePlacer Plus score of 4 or below = WAC 101 or WAC 107. A WritePlacer Plus score of 5 to 7 = ENG 101 or ENG 107. A WritePlacer Plus score of 8 = ENG 105.
### ACCUPLACER Flashcards [with ACCUPLACER Practice Questions]
Find ACCUPLACER test help using our ACCUPLACER flashcards and practice questions. Helpful ACCUPLACER review notes in an easy to use format. ... After all, it's hard to do well on test or prepare for it when you don't know what's going to be on it. ACCUPLACER Exam Flashcards Study System is a compilation of the hundreds of critical concepts you ...
| 4,832
| 21,191
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.578125
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.931967
|
https://www.wyattresearch.com/investing/liberty-checks-investor-claimed-102000-dollars/
| 1,721,220,405,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00566.warc.gz
| 903,178,764
| 18,384
|
# How an Investor Like You Claimed \$102,500 in Liberty Checks
The math is simple enough: \$34,500 + \$68,000 = \$102,500.
The numbers and my calculation are based on a Jan. 23 email I received from an investor, Paul B.
Paul wrote to inform me how he claimed six-digit income buying liberty vouchers of two different companies.
Claim big income now.
One company paid two rounds of liberty checks on its liberty vouchers. The payments were received over an 18-month span covering 2017 and 2018. Paul received \$68,000 on the two liberty-check payments.
The second company’s liberty-check payments were of more recent vintage. Paul claimed his \$34,000 liberty check before the Dec. 21 deadline. He received his \$34,000 liberty check two weeks later.
Paul will receive even more income on the second company’s liberty vouchers.
The second company will issue an additional check next month. Paul will receive a \$575 check (by my calculation). It won’t pay off the mortgage, but it’s enough to pay for prime strip and lobster for two at your local Del Friscos.
The liberty vouchers were the source of Paul’s liberty-check windfall.
Paul was one of the lucky few.
Most investors are clueless to liberty vouchers and the liberty checks they pay. Few financial websites report on companies that pay liberty checks. Most financial advisers fail to inform their clients.
Paul obviously knew of the companies whose liberty vouchers offered these high-yield liberty checks.
How did Paul know?
Paul was on our list. We informed him.
We issued buy alerts on both companies immediately after they declared their respective liberty checks. Paul acted and was able to secure his windfall.
I’ll concede that receiving \$102,500 in liberty checks is an outlier event, though not an unprecedented one. A few investors have written to regale us on claiming their own six-digit income windfall.
Many more investors have regaled us with stories of recurring four-digit windfalls.
We’re hardly surprised. Liberty check payments of \$1,000, \$1,500, and \$2,000 occur regularly. They occur an average of every 22 days.
Receiving a lump-sum \$34,000 liberty check like Paul is nice. Receiving a recurring \$1,500 check (or more) every three weeks is arguably nicer.
| 486
| 2,243
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.609375
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.970343
|
http://www.southperry.net/showthread.php?t=57830&page=4
| 1,369,138,797,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368699977678/warc/CC-MAIN-20130516102617-00071-ip-10-60-113-184.ec2.internal.warc.gz
| 739,531,749
| 16,830
|
# Thread: Does God exist: A hopefully civil and fruitful debate
1. ## Re: Does God exist: A hopefully civil and fruitful debate
Originally Posted by DrRusty
we're discussing infinity because of the omnipotence paradox, which I brought up as an argument against god's existence.
Whether or not god is omnipotent is irrelevant if you haven't even been able to decide if God exists. You're arguing the color of the chicken before it's even been born.
I am about to kill this thread just on the basis of it not being at all what it's stated purpose was.
2. ## Re: Does God exist: A hopefully civil and fruitful debate
Originally Posted by DrRusty
lol this is a hilariously ironic. Just as we're trying to use logical vs illogical in determining god's existence, we're using different definitions of infinity to describe the argument itself. Using arithmetic, yes, you are right. Since infinity is indeed indeterminate it can not be used for anything. In mathematics, you use things that lead up to infinity (like my infinity ohms example) allowing it to be used in formulas.
@DrRusty
As I recall, you didn't quite give me a definition of infinity to work with. But let's take that aside for one second and grant you that:
1. infinity - infinity = 0
2. infinity * 0 = 0
3. infinity * 1 = infinity
How do you defend the self-contradictory definition of a "stone heavier than can be lifted"? And to put things into perspective, these are the relevant arguments:
1. If God is omnipotent, then he can make object A
2. He cannot make object A
3. Therefore he is not omnipotent
Classic modus tollens.
Counterargument: Object A doesn't even make sense, so it doesn't make any sense to talk about God being capable of creating it. As such, the premise of the argument is flawed in the way that it can't be entertained sensibly, not in the way that it is factually false. The argument is thus scrapped.
Originally Posted by Eos
Whether or not god is omnipotent is irrelevant if you haven't even been able to decide if God exists. You're arguing the color of the chicken before it's even been born.
I am about to kill this thread just on the basis of it not being at all what it was stated intent was.
Two things: 1/ I'll give up after this one attempt. It's a pity, though. 2/ Since we're talking about a god with those characteristics (mainly my fault), let's grant this argument the privilege. Besides, an argument proving that God cannot be "existent" + "all-powerful" + "all-knowing" + "all-good" at the same time is good enough for me.
3. ## Re: Does God exist: A hopefully civil and fruitful debate
I'm not the one who said he wasn't sure if god existed or not. I don't believe in god at all
Originally Posted by Kalovale
How do you defend the self-contradictory definition of a "stone heavier than can be lifted"? And to put things into perspective, these are the relevant arguments:
1. If God is omnipotent, then he can make object A
2. He cannot make object A
3. Therefore he is not omnipotent
Classic modus tollens.
Counterargument: Object A doesn't even make sense, so it doesn't make any sense to talk about God being capable of creating it. As such, the premise of the argument is flawed in the way that it can't be entertained sensibly, not in the way that it is factually false. The argument is thus scrapped.
I don't see how you're having trouble with the concept. It seems pretty easy to me. You would have to use the mathematics definition of infinity to describe the paradox. The rock gets heavier approaching the point of god being unable to lift it until it becomes so heavy that he can't. I just don't know how that doesn't make sense to you. Being all-powerful, can he count to infinity? Can he walk around in a circle until he finds the end of it? Can he purple until he dies of cabinet? Can he defeat Chuck Norris? (last 2 were jokes btw)
4. ## Re: Does God exist: A hopefully civil and fruitful debate
Originally Posted by DrRusty
I'm not the one who said he wasn't sure if god existed or not. I don't believe in god at all
I don't see how you're having trouble with the concept. It seems pretty easy to me. You would have to use the mathematics definition of infinity to describe the paradox. I just don't know how that doesn't make sense to you. Being all-powerful, can he count to infinity? Can he walk around in a circle until he finds the end of it? Can he purple until he dies of cabinet? Can he defeat Chuck Norris? (last 2 were jokes btw)
Because you're already assuming the answer in the question. I'll point it out: The rock gets heavier approaching the point of god being unable to lift it until it becomes so heavy that he can't.
This sentence assumes a few things, but let's get right to the meat of the matter:
1. It assumes that such a thing has the property of existence.
2. It assumes that there is a point that God cannot lift.
"Have you stopped being a douche?" cannot be answered in such a way that contradicts the assumption that the person has in fact been a douche. The normal response is to correct the questioner: "I have not been a douche." and not give an answer. The same applies here.
We're trying to define God using reason, so omnipotence works on everything that is sensibly imaginable and none otherwise. I might have neglected to point out, but language is hardly rational. Something like "Can he do X" sounds nice because it starts with "Can he", implying--linguistically--an ability or lack thereof. Rationally, there is no ability concerning objects that don't even make sense from the start.
5. ## Re: Does God exist: A hopefully civil and fruitful debate
It doesn't even have to assume that there is any point that god can't lift it.
If god can lift anything, no matter what it is, then he obviously can not create something he can not lift meaning that's something he can not do (all-powerful kind of meaning that you can do anything).
Originally Posted by Kalovale
We're trying to define God using reason, so omnipotence works on everything that is sensibly imaginable and none otherwise. I might have neglected to point out, but language is hardly rational. Something like "Can he do X" sounds nice because it starts with "Can he", implying--linguistically--an ability or lack thereof. Rationally, there is no ability concerning objects that don't even make sense from the start.
well, we're at a problem if it doesn't make sense to you, because it makes sense to me.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
| 1,514
| 6,598
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20
|
latest
|
en
| 0.941365
|
https://coq.github.io/doc/master/refman/user-extensions/proof-schemes.html
| 1,619,056,357,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618039560245.87/warc/CC-MAIN-20210422013104-20210422043104-00054.warc.gz
| 285,539,042
| 22,112
|
$\begin{split}\newcommand{\as}{\kw{as}} \newcommand{\Assum}[3]{\kw{Assum}(#1)(#2:#3)} \newcommand{\case}{\kw{case}} \newcommand{\cons}{\textsf{cons}} \newcommand{\consf}{\textsf{consf}} \newcommand{\Def}[4]{\kw{Def}(#1)(#2:=#3:#4)} \newcommand{\emptyf}{\textsf{emptyf}} \newcommand{\End}{\kw{End}} \newcommand{\kwend}{\kw{end}} \newcommand{\even}{\textsf{even}} \newcommand{\evenO}{\textsf{even}_\textsf{O}} \newcommand{\evenS}{\textsf{even}_\textsf{S}} \newcommand{\Fix}{\kw{Fix}} \newcommand{\fix}{\kw{fix}} \newcommand{\for}{\textsf{for}} \newcommand{\forest}{\textsf{forest}} \newcommand{\Functor}{\kw{Functor}} \newcommand{\In}{\kw{in}} \newcommand{\Ind}[4]{\kw{Ind}[#2](#3:=#4)} \newcommand{\ind}[3]{\kw{Ind}~[#1]\left(#2\mathrm{~:=~}#3\right)} \newcommand{\Indp}[5]{\kw{Ind}_{#5}(#1)[#2](#3:=#4)} \newcommand{\Indpstr}[6]{\kw{Ind}_{#5}(#1)[#2](#3:=#4)/{#6}} \newcommand{\injective}{\kw{injective}} \newcommand{\kw}[1]{\textsf{#1}} \newcommand{\length}{\textsf{length}} \newcommand{\letin}[3]{\kw{let}~#1:=#2~\kw{in}~#3} \newcommand{\List}{\textsf{list}} \newcommand{\lra}{\longrightarrow} \newcommand{\Match}{\kw{match}} \newcommand{\Mod}[3]{{\kw{Mod}}({#1}:{#2}\,\zeroone{:={#3}})} \newcommand{\ModA}[2]{{\kw{ModA}}({#1}=={#2})} \newcommand{\ModS}[2]{{\kw{Mod}}({#1}:{#2})} \newcommand{\ModType}[2]{{\kw{ModType}}({#1}:={#2})} \newcommand{\mto}{.\;} \newcommand{\nat}{\textsf{nat}} \newcommand{\Nil}{\textsf{nil}} \newcommand{\nilhl}{\textsf{nil\_hl}} \newcommand{\nO}{\textsf{O}} \newcommand{\node}{\textsf{node}} \newcommand{\nS}{\textsf{S}} \newcommand{\odd}{\textsf{odd}} \newcommand{\oddS}{\textsf{odd}_\textsf{S}} \newcommand{\ovl}[1]{\overline{#1}} \newcommand{\Pair}{\textsf{pair}} \newcommand{\plus}{\mathsf{plus}} \newcommand{\SProp}{\textsf{SProp}} \newcommand{\Prop}{\textsf{Prop}} \newcommand{\return}{\kw{return}} \newcommand{\Set}{\textsf{Set}} \newcommand{\Sort}{\mathcal{S}} \newcommand{\Str}{\textsf{Stream}} \newcommand{\Struct}{\kw{Struct}} \newcommand{\subst}[3]{#1\{#2/#3\}} \newcommand{\tl}{\textsf{tl}} \newcommand{\tree}{\textsf{tree}} \newcommand{\trii}{\triangleright_\iota} \newcommand{\Type}{\textsf{Type}} \newcommand{\WEV}[3]{\mbox{#1[] \vdash #2 \lra #3}} \newcommand{\WEVT}[3]{\mbox{#1[] \vdash #2 \lra}\\ \mbox{ #3}} \newcommand{\WF}[2]{{\mathcal{W\!F}}(#1)[#2]} \newcommand{\WFE}[1]{\WF{E}{#1}} \newcommand{\WFT}[2]{#1[] \vdash {\mathcal{W\!F}}(#2)} \newcommand{\WFTWOLINES}[2]{{\mathcal{W\!F}}\begin{array}{l}(#1)\\\mbox{}[{#2}]\end{array}} \newcommand{\with}{\kw{with}} \newcommand{\WS}[3]{#1[] \vdash #2 <: #3} \newcommand{\WSE}[2]{\WS{E}{#1}{#2}} \newcommand{\WT}[4]{#1[#2] \vdash #3 : #4} \newcommand{\WTE}[3]{\WT{E}{#1}{#2}{#3}} \newcommand{\WTEG}[2]{\WTE{\Gamma}{#1}{#2}} \newcommand{\WTM}[3]{\WT{#1}{}{#2}{#3}} \newcommand{\zeroone}[1]{[{#1}]} \end{split}$
# Proof schemes¶
## Generation of induction principles with Scheme¶
Command Scheme ident :=? scheme_kind with ident :=? scheme_kind*
::=Equality for reference|InductionMinimalityEliminationCase for reference Sort sort_family::=Set|Prop|SProp|Type
A high-level tool for automatically generating (possibly mutual) induction principles for given types and sorts. Each reference is a different inductive type identifier belonging to the same package of mutual inductive definitions. The command generates the idents as mutually recursive definitions. Each term ident proves a general principle of mutual induction for objects in type reference.
ident
The name of the scheme. If not provided, the scheme name will be determined automatically from the sorts involved.
Minimality for reference Sort sort_family
Defines a non-dependent elimination principle more natural for inductively defined relations.
Equality for reference
Tries to generate a Boolean equality and a proof of the decidability of the usual equality. If reference involves other inductive types, their equality has to be defined first.
Example
Induction scheme for tree and forest.
A mutual induction principle for tree and forest in sort Set can be defined using the command
Axiom A : Set.
A is declared
Axiom B : Set.
B is declared
Inductive tree : Set := node : A -> forest -> tree with forest : Set := leaf : B -> forest | cons : tree -> forest -> forest.
tree, forest are defined tree_rect is defined tree_ind is defined tree_rec is defined tree_sind is defined forest_rect is defined forest_ind is defined forest_rec is defined forest_sind is defined
Scheme tree_forest_rec := Induction for tree Sort Set with forest_tree_rec := Induction for forest Sort Set.
forest_tree_rec is defined tree_forest_rec is defined tree_forest_rec, forest_tree_rec are recursively defined
You may now look at the type of tree_forest_rec:
Check tree_forest_rec.
tree_forest_rec : forall (P : tree -> Set) (P0 : forest -> Set), (forall (a : A) (f : forest), P0 f -> P (node a f)) -> (forall b : B, P0 (leaf b)) -> (forall t : tree, P t -> forall f1 : forest, P0 f1 -> P0 (cons t f1)) -> forall t : tree, P t
This principle involves two different predicates for trees andforests; it also has three premises each one corresponding to a constructor of one of the inductive definitions.
The principle forest_tree_rec shares exactly the same premises, only the conclusion now refers to the property of forests.
Example
Predicates odd and even on naturals.
Let odd and even be inductively defined as:
Inductive odd : nat -> Prop := oddS : forall n:nat, even n -> odd (S n) with even : nat -> Prop := | evenO : even 0 | evenS : forall n:nat, odd n -> even (S n).
odd, even are defined odd_ind is defined odd_sind is defined even_ind is defined even_sind is defined
The following command generates a powerful elimination principle:
Scheme odd_even := Minimality for odd Sort Prop with even_odd := Minimality for even Sort Prop.
even_odd is defined odd_even is defined odd_even, even_odd are recursively defined
The type of odd_even for instance will be:
Check odd_even.
odd_even : forall P P0 : nat -> Prop, (forall n : nat, even n -> P0 n -> P (S n)) -> P0 0 -> (forall n : nat, odd n -> P n -> P0 (S n)) -> forall n : nat, odd n -> P n
The type of even_odd shares the same premises but the conclusion is (n:nat)(even n)->(P0 n).
### Automatic declaration of schemes¶
Flag Elimination Schemes
Enables automatic declaration of induction principles when defining a new inductive type. Defaults to on.
Flag Nonrecursive Elimination Schemes
Enables automatic declaration of induction principles for types declared with the Variant and Record commands. Defaults to off.
Flag Case Analysis Schemes
This flag governs the generation of case analysis lemmas for inductive types, i.e. corresponding to the pattern matching term alone and without fixpoint.
Flag Boolean Equality Schemes
Flag Decidable Equality Schemes
These flags control the automatic declaration of those Boolean equalities (see the second variant of Scheme).
Warning
You have to be careful with these flags since Coq may now reject well-defined inductive types because it cannot compute a Boolean equality for them.
Flag Rewriting Schemes
This flag governs generation of equality-related schemes such as congruence.
### Combined Scheme¶
Command Combined Scheme identdef from ident+,
This command is a tool for combining induction principles generated by the Scheme command. Each ident is a different inductive principle that must belong to the same package of mutual inductive principle definitions. This command generates identdef as the conjunction of the principles: it is built from the common premises of the principles and concluded by the conjunction of their conclusions. In the case where all the inductive principles used are in sort Prop, the propositional conjunction and is used, otherwise the simple product prod is used instead.
Example
We can define the induction principles for trees and forests using:
Scheme tree_forest_ind := Induction for tree Sort Prop with forest_tree_ind := Induction for forest Sort Prop.
forest_tree_ind is defined tree_forest_ind is defined tree_forest_ind, forest_tree_ind are recursively defined
Then we can build the combined induction principle which gives the conjunction of the conclusions of each individual principle:
Combined Scheme tree_forest_mutind from tree_forest_ind,forest_tree_ind.
tree_forest_mutind is defined tree_forest_mutind is recursively defined
The type of tree_forest_mutind will be:
Check tree_forest_mutind.
tree_forest_mutind : forall (P : tree -> Prop) (P0 : forest -> Prop), (forall (a : A) (f : forest), P0 f -> P (node a f)) -> (forall b : B, P0 (leaf b)) -> (forall t : tree, P t -> forall f1 : forest, P0 f1 -> P0 (cons t f1)) -> (forall t : tree, P t) /\ (forall f2 : forest, P0 f2)
Example
We can also combine schemes at sort Type:
Scheme tree_forest_rect := Induction for tree Sort Type with forest_tree_rect := Induction for forest Sort Type.
forest_tree_rect is defined tree_forest_rect is defined tree_forest_rect, forest_tree_rect are recursively defined
Combined Scheme tree_forest_mutrect from tree_forest_rect, forest_tree_rect.
tree_forest_mutrect is defined tree_forest_mutrect is recursively defined
Check tree_forest_mutrect.
tree_forest_mutrect : forall (P : tree -> Type) (P0 : forest -> Type), (forall (a : A) (f : forest), P0 f -> P (node a f)) -> (forall b : B, P0 (leaf b)) -> (forall t : tree, P t -> forall f1 : forest, P0 f1 -> P0 (cons t f1)) -> (forall t : tree, P t) * (forall f2 : forest, P0 f2)
## Generation of inversion principles with DeriveInversion¶
Command Derive Inversion ident with one_term Sort sort_family?
Generates an inversion lemma for the inversion ... using ... tactic. ident is the name of the generated lemma. one_term should be in the form qualid or (forall binder+, qualid term) where qualid is the name of an inductive predicate and binder+ binds the variables occurring in the term term. The lemma is generated for the sort sort_family corresponding to one_term. Applying the lemma is equivalent to inverting the instance with the inversion tactic.
Command Derive Inversion_clear ident with one_term Sort sort_family?
When applied, it is equivalent to having inverted the instance with the tactic inversion replaced by the tactic inversion_clear.
Command Derive Dependent Inversion ident with one_term Sort sort_family
When applied, it is equivalent to having inverted the instance with the tactic dependent inversion.
Command Derive Dependent Inversion_clear ident with one_term Sort sort_family
When applied, it is equivalent to having inverted the instance with the tactic dependent inversion_clear.
Example
Consider the relation Le over natural numbers and the following parameter P:
Inductive Le : nat -> nat -> Set := | LeO : forall n:nat, Le 0 n | LeS : forall n m:nat, Le n m -> Le (S n) (S m).
Le is defined Le_rect is defined Le_ind is defined Le_rec is defined Le_sind is defined
Parameter P : nat -> nat -> Prop.
P is declared
To generate the inversion lemma for the instance (Le (S n) m) and the sort Prop, we do:
Derive Inversion_clear leminv with (forall n m:nat, Le (S n) m) Sort Prop.
leminv is defined
Check leminv.
leminv : forall (n m : nat) (P : nat -> nat -> Prop), (forall m0 : nat, Le n m0 -> P n (S m0)) -> Le (S n) m -> P n m
Then we can use the proven inversion lemma:
Goal forall (n m : nat) (H : Le (S n) m), P n m.
1 goal ============================ forall n m : nat, Le (S n) m -> P n m
intros.
1 goal n, m : nat H : Le (S n) m ============================ P n m
Show.
1 goal n, m : nat H : Le (S n) m ============================ P n m
inversion H using leminv.
1 goal n, m : nat H : Le (S n) m ============================ forall m0 : nat, Le n m0 -> P n (S m0)
| 3,346
| 11,715
|
{"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}
| 3.078125
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.213086
|
https://greatminds.org/math/blog/eureka/post/teaching-order-of-operations-through-the-relationships-of-the-operations
| 1,575,996,138,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00370.warc.gz
| 380,612,019
| 16,730
|
# TEACHING ORDER OF OPERATIONS THROUGH THE RELATIONSHIPS OF THE OPERATIONS
STRATEGY
Not long ago, if someone asked me to teach my students how to evaluate this expression, 15+ (9–4)- 2³, I would have immediately thought of teaching them how to use PEMDAS. However, what if they learned the reason behind this mnemonic by studying the relationship of the operations?
Why is it important to follow a set of rules in math? Let’s take a look.
By looking at the student work above it’s clear Jessie and Matt both had a strategy, but who is correct? Just like everyone drives on the right side of the road in the United States of America, consistent conventions ensure everyone is following the same process and arrives at the same answer.
Grade 6 Module 4 Lesson 6 focuses on the use of order of operations to evaluate numerical expressions. However, a quick skim of this lesson will undoubtedly inform you that this is not the traditional approach. So, let’s dig in and find out more!
In Lesson 3 of this module, students strengthen their understanding of the relationship between multiplication and addition. They explore and evaluate identities such as 3 x g = g + g + g. Given that each unit in the tape diagram represents one, notice how the first tape diagram represents the expression 2 + 2 + 2 + 2 and the second tape diagram represents the expression 4 x 2
Students notice that each model has the same number of units, and through intentional questioning, they conclude the values of the expressions are equivalent. After another example and a discussion of the relationship between 3 x 4 and 4 + 4 + 4, students explore the identity 3g = g + g + g. Students conclude that no matter what number replaces the variable, g, after adding g three times, the value is the same as multiplying g by three. By building the tape diagrams prior to this identity, students are able to visualize this relationship between repeated addition and multiplication (which is often a more efficient method than repeatedly adding the same number)
In Lesson 4, students discover the relationship of division and subtraction. In the example below, students determine that the number of times the divisor (x) is subtracted from the dividend (20) is the same number as the quotient (5). We determine in this example that the quotient is 5 because there are 5 equal groups of the same number being subtracted. The quotient is telling us how many times we’re subtracting the divisor from the dividend
In this diagram, we can also see that the expression 20 ÷ x means that the number that x represents can be subtracted exactly 5 times from 20. The subtraction equation 20–4–4–4–4–4=0 tells us 4 can be subtracted exactly 5 times from 20 before there is no remainder, so the divisor in this expression is 4. With the model and the work in this lesson, students explore the relationship between repeated subtraction and division (which is often a more efficient method when repeatedly subtracting the same number)
Students also tie together the relationship between all four operations at the end of Lesson 4. These connections are critical as they use these relationships to evaluate expressions using the order of operations. Take a look at the image below.
In A Story of Units, students learned that addition is a “shortcut” to counting on and subtraction is a “shortcut” to counting back. These operations are traditionally taught prior to other operations, and are ultimately evaluated last in an expression because they are the least complicated. In this blog, we’ve shown that multiplication and division are often more efficient ways to repeatedly add or subtract when the addends or subtrahends are the same. Therefore, multiplication and division are next in the sequence of learning the four operations, and are evaluated prior to addition and subtraction in an expression. It is apparent that the traditional order for performing certain operations was developed based on the complexity of the operation
The amount of decrease in the first example of exponentiation is 100 and is much greater than the amount of decrease in the multiplication example (5) and the addition example (1). It’s clear how the repeated multiplication evident in the first example with exponents yields a greater decrease than the repeated addition seen in the second example and the basic addition of the last example.
In summary, you can take a look at how to evaluate a more complex expression in this video the Eureka Math Teach Eureka video series (clip 1:02–1:04)
But what happens to the order of operations when grouping symbols are introduced? Grouping symbols alert us that we may be doing things out of the typical order. With any problem, it’s context that prescribes how the problem should be solved.
Let’s take a look at the context below from Lesson 6.
The expression 4x5 +2 represents this context and tells us there are 4 groups of 5 since there are 4 people and the cost in dollars of each ticket can be represented by the number, 5 The additional price of the soft drink is represented by 2 because the cost of each soft drink, in dollars, is 2. A model for this scenario is below.
Now let’s look at another situation.
How did the expression change to represent this new situation? Now, the total price per person changed from the price of one ticket (represented by the number 5) to the price of one ticket and a drink (represented by 5 + 2 or 7). So this expression is 4x(5+2) and tells us there are 4 groups of (5 + 2). Here is a model for this scenario.
The value of this expression is 4x7 or (4 x 5) + (4 x 2), which is 28. Notice the importance of indicating how the numbers in the expression are grouped. Grouping symbols, such as the parentheses used in this problem, instruct us to evaluate operations involving parentheses first.
You may have noticed Example A applied the distributive property to the second part of the expression, 5(10–2²), when evaluating the expression and Example B evaluated inside the parentheses (10–2²) first. In each example, the resulting value is the same but the way in which the students arrived at that solution was different. In these examples, it is evident that students do not always have to perform the operations within parentheses first, but more so the operations involving parentheses. This emphasizes the importance of encouraging students to compute in a manner that is most efficient to them
What we hope you walk away from this blog is just this: Instead of memorizing a mneumonic, such as PEMDAS, exploration of the relationships of operations builds the understanding we strive for and, ultimately, students become proficient mathematicians.
Up for a challenge? Take a look at the expression below. Can you evaluate the expression by applying the distributive property (similar to Example A above) and then by evaluating inside the parentheses first (similar to Example B above)? Which method works best for you?
3(2 + 8²) — 14
This blog is by Eureka Math writers Dawn Pensack and Erika Silva.
By: Pia Mohsen
By: Sara Lack
| 1,503
| 7,086
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.8125
| 5
|
CC-MAIN-2019-51
|
latest
|
en
| 0.948556
|
http://mvphip.org/the-quadratic-formula-coloring-activity-egg-answer-key/
| 1,618,949,720,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00175.warc.gz
| 62,800,262
| 15,157
|
Posted on
The Quadratic Formula Coloring Activity Egg Answer Key Indeed recently is being hunted by consumers around us, perhaps one of you personally. Individuals now are accustomed to using the net in gadgets to view video and image information for inspiration, and according to the title of this post I will discuss about The Quadratic Formula Coloring Activity Egg Answer Key.
• Trig Word Problems Worksheet Doc – Worksheet . This Calculator Will Solve Your Problems.
• The Quadratic Formula Egg Coloring Activity Worksheet.pdf … – Connect And Share Knowledge Within A Single Location That Is Structured And Easy To Search.
• The Quadratic Formula Coloring Activity Egg – Wallpapers … : Solve Quadratic Equations Using A Quadratic Formula Calculator.
• The Quadratic Formula Coloring Activity Egg – Wallpapers … : 4X 2 7X 15 0 8.
• Exponent Rules Coloring Activity: Easter Egg And Spring … – Graphing Quadratics Answer Key Answer Key (Homework) Chapter Reviews.
• Exponent Rules Coloring Activity: Easter Egg And Spring … . Eventually, You Will No Question Discover A Additional Experience And Attainment By Quadratic Formula Worksheet (Real Solutions) Quadratic Formula Worksheet (Complex Solutions) Quadratic Quadratic Equation Worksheets With Answer Keys.
• The Quadratic Formula Coloring Activity Egg Answers … , Use The Quadratic Formula To Solve For X.
• Algebra Coloring Activity: Christmas Themed Solving … . What Do You Want To Calculate?
• Quadratic Formula: Coloring Activity By Mary's Math Mate | Tpt : We Had Just Learned The Square Root Method To Solving Quadratic Functions.
• 61 Pdf Factoring Worksheets Algebra 1 Printable Zip Docx … : This Calculator Will Solve Your Problems.
• Factor Quadratic Equations Worksheet – We Had Just Learned The Square Root Method To Solving Quadratic Functions.
• Quadratic Formula Worksheet – Yahoo Image Search Results … . Shows Work By Example Of The Entered Equation To Find The Real Or Complex Root Solutions.
| 398
| 1,976
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.6875
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.714045
|
http://www.riddlesandanswers.com/puzzles-brain-teasers/marys-has-500-dolla-riddles/
| 1,571,635,352,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987756350.80/warc/CC-MAIN-20191021043233-20191021070733-00225.warc.gz
| 316,400,430
| 27,288
|
# MARYS HAS 500 DOLLA RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS
#### Popular Searches
Terms · Privacy · Contact
## Solving Marys Has 500 Dolla Riddles
Here we've provide a compiled a list of the best marys has 500 dolla puzzles and riddles to solve we could find.
Our team works hard to help you piece fun ideas together to develop riddles based on different topics. Whether it's a class activity for school, event, scavenger hunt, puzzle assignment, your personal project or just fun in general our database serve as a tool to help you get started.
Here's a list of related tags to browse:
The results compiled are acquired by taking your search "marys has 500 dolla" and breaking it down to search through our database for relevant content.
Browse the list below:
## 500 Pound Canary Riddle
Hint:
Anywhere it wants!
Did you answer this riddle correctly?
YES NO
Solved: 65%
## The 500 Pound Monster
Hint:
On a diet!
Did you answer this riddle correctly?
YES NO
## Hundred Dollar Bill Riddle
Hint:
Of course it is. A \$100 bill is worth more than a \$1 bill (newer one).
Did you answer this riddle correctly?
YES NO
Solved: 36%
## A Dime And A Dollar Riddle
Hint:
A nickel. The dog cost \$ 1.05.
Did you answer this riddle correctly?
YES NO
Solved: 39%
## Change For A Dollar
Hint:
\$1.19
Did you answer this riddle correctly?
YES NO
Solved: 9%
## The 5,000 Dollar Bank Account
Hint:
The mailman did it. There is no mail on Sunday and the mailman fruaded the guys info. every Sunday the mailman would come at 4:00 and mess with his money and mortgage. After plunging the computer into his truck and turning it on.
Did you answer this riddle correctly?
YES NO
Solved: 50%
## Dollar-bills Riddle
Hint:
You don't make any cents (sense).
Did you answer this riddle correctly?
YES NO
Solved: 40%
## Matthew Has 20 Dollars Riddle
Hint:
A red bull
Did you answer this riddle correctly?
YES NO
Solved: 56%
## Penguin Dollars Riddle
Hint:
In a snow bank!
Did you answer this riddle correctly?
YES NO
Solved: 49%
## Missing Dollar Riddle
Hint: Make a list of all of the people involved and how much money they ended up with/spent.
The \$9 paid by each guest accounts for the \$2 that went to the bellhop. So rather than adding \$27 to the \$2 kept by the bellhop, the \$27 accounts for the bellhops money. The \$27 plus the \$3 kept by the guests does add up to \$30.
Did you answer this riddle correctly?
YES NO
| 664
| 2,454
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43
|
latest
|
en
| 0.862126
|
https://mathematica.stackexchange.com/questions/221408/how-can-i-use-lighting-for-specific-surface-of-a-3d-object
| 1,702,009,004,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100710.22/warc/CC-MAIN-20231208013411-20231208043411-00846.warc.gz
| 420,178,037
| 42,628
|
# How can I use Lighting for specific surface of a 3D object?
Graphics3D[{Cylinder[{{(-0.1 + 10.128)/2, (13 + 1.3)/2,
0}, {(-0.1 + 10.128)/2, (13 + 1.3)/2, 0.001}}, 3.2]},
Lighting -> {{"Spot", Red}}]
I would like to use Lighting on the top surface ofCylinder above to produce the same lighting profile as this one
plane = DiscretizeRegion[
InfinitePlane[{0, 0, 0}, {{1, 0, 0}, {0, 1, 0}}], {{-1, 1}, {-1,
1}, {0, 1}}, MaxCellMeasure -> {"Length" -> 0.01},
BaseStyle -> {EdgeForm[], White}];
gty = Show[{plane},
Lighting -> {{"Spot", Red, {{0, 0, 1}, {0, 0, 0}}, {Pi/2, 4}}}]
You have a very thin cylinder. Ellipsoid seems to perform better if you are trying to emulate a disk.
cyl = Cylinder[{{(-0.1 + 10.128)/2, (13 + 1.3)/2,
0}, {(-0.1 + 10.128)/2, (13 + 1.3)/2, 0.001}}, 3.2];
centroid = RegionCentroid@cyl;
bnds = First@Differences@Transpose@RegionBounds@cyl;
ell = Ellipsoid[centroid, bnds];
Manipulate[
Graphics3D[{Specularity[White, 5], Style[ell, c]},
Lighting -> {{"Spot",
Red, {centroid + {0, 0, 3}, centroid}, θ}},
ImageSize -> Small], {{θ, π/2.5}, π/40, π/2}, {c,
White, ColorSlider}]
The following shows how the image changes at discrete surface colors with constant lighting.
Grid[{Graphics3D[{Style[ell, #]},
Lighting -> {{"Spot", Red, {centroid + {0, 0, 3}, centroid},
Pi/2.5}}] & /@ {White, Yellow, LightBlue, Blue}}]
The following shows how the image changes with underlying surface color under a White, Red, Green, and Blue spotlight.
mean = First@Mean@Transpose@RegionBounds@cyl;
tt = TranslationTransform /@ (DeleteCases[
Tuples[{0, #}, 3], {_, _, #}] &@bnds[[1]]*2.5);
ctt = Transpose[{{"White", "Red", "Blue", "Green"}, tt}];
Manipulate[
Graphics3D[{Specularity[White, 5],
Flatten@({Lighting -> {{"Spot",
ToExpression@#[[1]], #[[2]] /@ {centroid + {0, 0, 3},
centroid}, s}},
GeometricTransformation[
Style[{ell, Text[#[[1]], {mean, bnds[[1]]*2.3, 0}]}, c], #[[
2]]]} & /@ ctt)},
ViewPoint -> Top], {{s, Pi/5, "Spotlight Angle"}, Pi/20,
Pi/2.5}, {{c, Yellow, "Underlying Object Color"}, ColorSlider},
ControlPlacement -> Top]
• Thanks a lot @Tim Laska! Is it possible to assign a specific color for the desk instead of Black? May 9, 2020 at 2:21
• @HD2006 The disk is black due to the lack of lighting. I added the ability to change the surface color through the Style command and specularity if desired. May 9, 2020 at 2:54
• it is not working with me, it gives black disk without lighting once I change the color inside style to blue. May you please let the disk blue while lighting is red as it is and add a picture of the new shape? May 9, 2020 at 3:09
• @HD2006 I added a ColorSlider to the Manipulate function to show that the image changes with changes in the surface color. May 9, 2020 at 3:43
• @HD2006 I am pretty sure the black you are seeing is a function of the lighting. If evaluate Graphics3D[{Style[ell, Yellow]}], then you should see a Yellow disk. If you add lighting by evaluating Graphics3D[{Style[ell, Yellow]}, Lighting -> {{"Spot", Red, {centroid + {0, 0, 3}, centroid}, Pi/2.5}}], then the disk appears darker because less light is reflecting back. May 9, 2020 at 4:24
| 1,039
| 3,130
|
{"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}
| 2.875
| 3
|
CC-MAIN-2023-50
|
longest
|
en
| 0.654876
|
https://theconstructor.org/structural-engg/distinguish-shear-wall-column/564155/?amp=1
| 1,670,580,199,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00490.warc.gz
| 597,629,678
| 11,276
|
# How to Distinguish Between Shear Wall and Column?
The geometry of a section of a shear wall and a column can be similar to the extent that the question of when a rectangular column becomes a wall often arises.
ACI 318-19, section 18.7.2.1, defines a column, for special moment frames, as having a minimum aspect ratio of 0.4. However, this condition is not the best criteria for considering column or wall design.
Construction constraints and the expected behavior of the element under loads are frequently used to decide whether an element should be designed as a wall or column.Â
A column has a large axial load and its shear response is similar to that of a beam, whereas a wall has a low axial load and its shear behavior is similar to that of a one-way slab.
Moreover, the area and distribution of longitudinal reinforcement and shear aspect ratio can be used to decide whether a rectangular column is designed as a column or as a shear wall.
## How to Distinguish Between Shear Wall and Column?
### Longitudinal Reinforcement
Generally, longitudinal reinforcement requires lateral support to prevent buckling of steel bars under axial loads. If longitudinal reinforcement is required in a wall and its area is greater than 0.01Ag, the ACI 318-19, section 11.7.4.1, states that the longitudinal reinforcement should be laterally supported by transverse bars. This requirement can be used as criteria to decide whether the section should be designed as a wall or a column.
If the required longitudinal reinforcement area of the wall exceeds 0.01Ag, then a transverse tie should be provided at every intersection of longitudinal and transverse reinforcement, which is a laborious job to do. In this case, it is more practical to design the section as a column.
### Shear Aspect Ratio
If the wall length-to-thickness ratio is equal to or greater than 6, then it is designed as a shear wall as its in-plane and out-plane shear behavior is easily known and distinctly different from the column.
If the aspect ratio ranges from 2.5 to 6, the element is designed as a shear wall or column, based on the shear force and the direction of the shear force. For the aspect ratio less than 2.5, the member is most likely to be designed as a column.
## FAQs
What is the behavior of columns and shear walls under loads?
A column has a large axial load and its shear response is similar to that of a beam, whereas a wall has a low axial load and its shear behavior is similar to that of a one-way slab. Moreover, the area and distribution of longitudinal reinforcement and shear aspect ratio can be used as a criterion to decide whether a rectangular column is designed as a column or as a shear wall.
What criteria are used to decide whether a vertical member is designed as a column or shear wall?
Construction constraints and expected behavior of the element under loads are mostly used to decide whether an element should be designed as a wall or a column.
What is longitudinal reinforcement in a column?
Longitudinal reinforcements are the RC column's main bars, placed in square, rectangular, or circular patterns. The purpose of longitudinal reinforcement is to assist concrete in resisting imposed loads. For more information, please click here.
| 676
| 3,263
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49
|
latest
|
en
| 0.921733
|
https://centralesupelec.hal.science/hal-00793109
| 1,716,769,269,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00808.warc.gz
| 124,081,004
| 14,457
|
Fuzzy Cauchy Problem for Fuzzy Estimation - Application to Unicycle-Type Mobile Robots - CentraleSupélec Access content directly
Conference Papers Year : 2012
## Fuzzy Cauchy Problem for Fuzzy Estimation - Application to Unicycle-Type Mobile Robots
Akram Fayaz
• Function : Author
#### Abstract
This paper is an attempt to provide a solution to the problem of state estimation of dynamical systems, when no measured output is available, through the example of a unicycle-type mobile robot. The latter is only equipped with encoders on each wheel which provide information only about the system's inputs (see equations (1) and (2)). As is the case in most real situations, in this work, it is also assumed that the initial states are only "approximately" known and we represent them by fuzzy numbers. Fuzzy initial states along with system dynamic equations provide us particular fuzzy differential equations (FDEs), referred as fuzzy Cauchy problem in the literature [7], [5], [4]. The question to which we want to provide an answer is: Knowing the "fuzzy states" of the system at t = 0, are we able to estimate them at any time t >; 0 or, equivalently, starting from the fuzzy initial states, are we able to build the solution of the fuzzy Cauchy problem at any time t >; 0? To solve this problem we consider two approaches: First, solving the robot's crisp (non fuzzy) differential equations with fuzzy initial conditions. In the second, the differential equations themselves are considered as fuzzy and, after discretization, the solution is built step by step using mainly fuzzy arithmetics. Moreover, considering encoders' inherent imprecision, we also assume that the input signals are only approximately known and represent them by fuzzy maps (functions having fuzzy numbers as instantaneous values). We however always make the assumption that the robot wheels do not slip.
Automatic
### Dates and versions
hal-00793109 , version 1 (21-02-2013)
### Identifiers
• HAL Id : hal-00793109 , version 1
• DOI :
### Cite
Akram Fayaz. Fuzzy Cauchy Problem for Fuzzy Estimation - Application to Unicycle-Type Mobile Robots. 2012 17th International Conference on Methods & Models in Automation & Robotics (MMAR), Aug 2012, Miedzyzdroje, Poland. pp.564-568, ⟨10.1109/MMAR.2012.6347824⟩. ⟨hal-00793109⟩
### Export
BibTeX XML-TEI Dublin Core DC Terms EndNote DataCite
28 View
| 560
| 2,383
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.909606
|
https://blogs.mathworks.com/cleve/2016/04/25/further-twists-of-the-moebius-strip/
| 1,670,338,565,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00624.warc.gz
| 169,830,418
| 29,905
|
# Further Twists of the Moebius Strip
The equations generating a surf plot of the Moebius strip can be parameterized and the parameters allowed to take on expanded values. The results are a family of surfaces that I have been displaying for as long as I have had computer graphics available.
### Contents
#### Parametrized Moebius Strip
The animation shows the formation of a classic Moebius strip. Start with a long, thin rectangular strip of material. Bend it into a cylinder. Give the two ends a half twist. Then join them together. The result is a surface with only one side and one boundary. One could traverse the entire length of the surface and return to the starting point without ever crossing the edge.
I am going to investigate the effect of three free parameters:
• $c$: curvature of the cylinder.
• $w$: width of the strip.
• $k$: number of half twists.
These parameters appear in a mapping of variables $s$ and $t$ from the square $-1 <= s,t <= 1$ to a generalized Moebius strip in 3D.
$$r = (1-w) + w \ s \ \sin{(k \pi t/2)}$$
$$x = r \ \sin{(c \pi t)}/c$$
$$y = r \ (1 - (1-\cos{(c\pi t))}/c)$$
$$z = w \ s \ \cos{(k \pi t/2)}$$
Here is the code. The singularity at $c = 0$ can be avoided by using c = eps.
type moebius
function [x,y,z,t] = moebius(c,w,k)
% [x,y,z,t] = moebius(c,w,k)
% [x,y,z] = surface of moebius strip, use t for color
% c = curvature, 0 = flat, 1 = cylinder.
% w = width of strip
% k = number of half twists
if c == 0
c = eps;
end
m = 8;
n = 128;
[s,t] = meshgrid(-1:2/m:1, -1:2/n:1);
r = (1-w) + w*s.*sin(k/2*pi*t);
x = r.*sin(c*pi*t)/c;
y = r.*(1 - (1-cos(c*pi*t))/c);
z = w*s.*cos(k/2*pi*t);
end
The animation begins with a flat, untwisted strip of width 1/4, that is
$$c = 0, \ w = 1/4, \ k = 0$$
The classic Moebius strip is reached with
$$c = 1, \ w = 1/4, \ k = 1$$
The final portion of the animation simply changes the viewpoint, not the parameters.
#### Parula
Let's play with colormaps. I'll look at the classic Moebius strip and make the default MATLAB colormap periodic by appending a reversed copy. I think this looks pretty nice.
map = [parula(256); flipud(parula(256))];
#### HSV
The Hue-Saturation-Value color map has received a lot of criticism in MathWorks blogs recently, but it works nicely in this situation. This figure has $k = 4$, so there are two full twists. That's very hard to see from this view. The colors help, but it's important to be able to rotate the view, which we can't do easily in this blog.
#### TripleTwist
Here is the one with three half twists. Like the classic Moebius strip, this has only one side. I've grown quite fond of this guy, so I'll show two views. The HSV color map, with its six colors, works well.
#### IEEE Computer, 1988
Source: IEEE
I have written in this blog about the glorious marriage between MATLAB and the Dore graphics system on the ill-fated Ardent Titan computer almost 30 years ago. I somehow generated this image using MATLAB, Dore, these parametrized Moebius equations and the Titan graphics hardware in 1988. This was fancy stuff back then. The editors of IEEE Computer were really excited about this cover.
I've never been able to reproduce it. I don't know what the parameters were.
#### Kermit Sigmon
Kermit Sigmon was a professor at the University of Florida. He was an excellent teacher and writer. He wrote a short MATLAB Primer that was enormously popular and widely distributed in the early days of the web. It went through several editions and was translated into a couple of other languages.
For the most part, people had free copies of the MATLAB Primer without fancy covers. But CRC Press obtained the rights to produce a reasonably priced bound paperback copy which featured a cover with $k = 5$.
Kermit passed away in 1997 and Tim Davis took over the job of editing the Primer. He generated more elaborate graphics for his covers.
Published with MATLAB® R2016a
|
| 1,069
| 3,932
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625
| 4
|
CC-MAIN-2022-49
|
longest
|
en
| 0.802727
|
https://greentravelguides.tv/page/249/
| 1,708,728,805,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00145.warc.gz
| 300,678,834
| 22,768
|
## A Girl Who Loves to Travel Is Called?
Hodophile who enjoys traveling. 08.04.2021 You might also be thinking, What to call someone who …
## A Car Travels at 60 Miles Per Hour How Much Time Does It Take the Car to Travel 30 Miles?
You might also be thinking, How long does it take to drive a mile at …
## A Car Traveled at an Average Speed of 60 Mph for Two Hours How Far Did It Travel?
You might also be thinking, When going 60 mph How far will a car travel? …
## A Car Starts From Rest and Accelerates at 600 M/s2 How Far Does It Travel in 300 S?
Similarly, How do you find distance with acceleration and time? Using this method to calculate …
## A Car Moves With an Average Speed of 45 Miles/hour How Long Does the Car Take to Travel 90 Miles?
Similarly, What is the average speed of a car that travels 15 miles in 30 …
## A Car Accelerates From to at a Constant Rate of How Far Does It Travel While Accelerating?
But then this question also arises, What is the acceleration of a car if the …
## A Car Accelerates at 2 M/s/s Assuming the Car Starts From Rest, How Far Will It Travel in 10 S?
But then this question also arises, What is the acceleration of a car that starts …
## 9 Reasons Why You Should Travel Alone?
– You have complete freedom to travel anywhere you want, whenever you want. – You …
## 600 Light Years How Long to Travel?
22 million years has passed You might also be thinking, How long would it take …
## 60 Mph for 1 Hour How Far Do You Travel?
Because one mile is 5280 feet long, 60 miles would be 60 x 5280 feet …
Scroll to Top
| 388
| 1,574
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375
| 3
|
CC-MAIN-2024-10
|
latest
|
en
| 0.889666
|
https://www.mathworks.com/matlabcentral/profile/authors/5058132-stozaki?s_tid=contributors_avatar
| 1,582,200,517,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00018.warc.gz
| 829,822,029
| 23,154
|
Community Profile
# stozaki
### MathWorks
125 total contributions since 2019
DISCLAIMER: Any advice or opinions posted here are my own, and in no way reflect that of MathWorks.
NOTE: Please do not send me individual messages or questions, as I will not be able to respond to them. If you have technical questions about MATLAB, please use the various resources on MATLAB Central. If you have a valid license, you can also use Technical Support.
View details...
Contributions in
View by
Atomi様 Stateflowの 場合を添付の例題モデルとして示します。(R2019bにて作成) バス信号を使うためには、先ずバスオブジェクトを定義して下さい。その定義したバス要素に代入する形でモデリング・シミュレーション出来ます。(myBus....
2 days ago | 0
How to export a simulink project (Not single model) to older version 2019a from newer version 2019b?/
Simulink Project won't export previous version now. This is Simulink Project limitation
3 days ago | 0
HDL coder mapping error
I look like you are using the ISE web pack. The web pack does not support xc6vlx240t. web pack supports only xclx75T. If you use...
3 days ago | 0
Solved
Arrange Vector in descending order
If x=[0,3,4,2,1] then y=[4,3,2,1,0]
3 days ago
Simulink Model stuck in running mode indefinitely
@ADITYA You can use color space convesion block. I attached simple example model. Regards, stozaki
4 days ago | 0
How to create a buffer FIFO, filter and then save in an array values that are between an specific range of values?
I attached modified model you made. unbuffer is just for the description of the block. It is arranged to explain the difference...
4 days ago | 0
| accepted
Problem
Find the index of elements in a string vector
In the vector of v, find the index of elements specified by a. v = ["Lion","Koara","Elephant","Snake","Dog","Cat","Camel"]; ...
4 days ago | 0
Comparing and locating string values
Please try following command. ret = find(ismember(x,a)) ismember find
4 days ago | 1
| accepted
Solved
Flip the vector from right to left
Flip the vector from right to left. Examples x=[1:5], then y=[5 4 3 2 1] x=[1 4 6], then y=[6 4 1]; Request not ...
6 days ago
How to set output of MATLAB function block as fixed size?
* I attached model and Signal Object file. There are two ways. Set the dimensions in the properties of the output signal. U...
7 days ago | 1
How can I convert douvle value to int or Integer?
If you convert from double to int32, you can use int32 function. It work casting as int32. intnum = int32(num); int32 int16 ...
8 days ago | 0
Turn off a Signal for a few seconds and then turning it back on
I attached example model. But it is created by R2019a and backport to R2013b.
9 days ago | 0
| accepted
sugimoto様 対象がモデル内の全ブロックと仮定します。 blk = find_system(bdroot(gcs),'type','block'); % モデル内のブロックをリストします for n = 1:length(blk) % 取得...
9 days ago | 1
Changing text of subsystem according to textbox
I attached sample model. If you use get_param and disp function, you can use it. for example Gain value of Gain block. v = ge...
11 days ago | 1
| accepted
Scope not working properly (ie im not getting a straight line if i use const value for a period of time)
This problem is setting of constant block. When sample period of constant block is set 'inf' and constant block only connect to ...
11 days ago | 1
Simulink library takes more time to open -2016B
You can customize simulink library browser. I attached example script. You can limit the ToolBoxes displayed in the library brow...
11 days ago | 0
| accepted
Constant input as "iteration limit" for "for loop iterator" block
I attached example model. Please try to use it.
11 days ago | 2
| accepted
Vector and matrix average
a=[2.4,2,3.4,4.5,5.63]; ave_a = mean(a);
11 days ago | 0
Solver error in HDL verifier
When you generate and simulation to use the HDL coder, SolverType must set Fixed-Step. Please see following URL HDL Coder cu...
12 days ago | 0
Solved
Bit Reversal
Given an unsigned integer _x_, convert it to binary with _n_ bits, reverse the order of the bits, and convert it back to an inte...
13 days ago
Solved
Get the length of a given vector
Given a vector x, the output y should equal the length of x.
13 days ago
Solved
Create a vector
Create a vector from 0 to n by intervals of 2.
13 days ago
Solved
13 days ago
how to change the integrator initial conditions?
Hello, ret = find_system(bdroot(gcs),'BlockType','Integrator') for N = 1:length(ret) set_param(ret{N},'InitialCondition...
15 days ago | 0
Solved
15 days ago
Solved
サイコロを作ろう
1から6までの独立かつランダムな数値を返すような関数を作成しましょう。 例： >> [x1,x2] = rollDice(); と入力すると x1 = 5 x2 = 2 のような解を返します。
15 days ago
Solved
15 days ago
Solved
NaN (欠損値) が含まれている行を削除しよう
15 days ago
Solved
ベクトル b にあるスカラ値 a が含まれていれば出力として 1 を返し、含まれていなければ 0 を返すような関数を作成しましょう。 例: a = 3; b = [1,2,4]; スカラ値 3 はベクトル b に含まれていない...
15 days ago
| 1,429
| 4,780
|
{"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}
| 2.640625
| 3
|
CC-MAIN-2020-10
|
latest
|
en
| 0.717828
|
https://cs.stackexchange.com/questions/53245/algorithm-finding-shortest-path-through-a-dungeon-in-a-game/53593
| 1,563,609,344,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00191.warc.gz
| 370,316,071
| 39,877
|
# Algorithm: Finding shortest path through a dungeon in a game
## Background
I was playing the PC-Game "Darkest Dungeon" recently. In the game, you have to explore dungeons, which consist of connected rooms as shown in the picture below.
Here are the rules:
• You start in fixed room (Entrance). You cannot choose where to start.
• The goal is to visit every room at least once
• The distance between adjacent rooms is the same for all rooms.
• You can visit rooms and walk paths as often as you'd like
## Question
What is the shortest path from the entrance that visits every room at least once?
Subquestions:
• What algorithm(s) could be used to solve this problem?
• Are there implementations that are free (and fairly simple) to use for someone like me?
## What I tried
I have found other questions such as this or this without finding an answer. I am familiar with the (basic) TSP and are able to code and solve simple TSPs. Hamiltonian paths didn't solve my problem, because it doesn't allow for multiple visits. The Chinese postman problem does also not apply here in its basic form because I don't have to visit every edge.
## Update
As I have stated in the comments, I'm not a computer scientist and am not interested in proving a mathematical statements (maybe I'll post this question on stackoverflow at a later stage). Also, I'm not a programmer and the chances that I'm able to code a solution myself are pretty slim. But I suspect I'm not the first one dealing with a problem of that nature.
According to @Shreesh and @Dib, the following procedure could be applied:
1. Create a pairwise distance matrix with all rooms thus adding edges between all rooms.
2. Solve TSP using a standard solver (e.g. concorde)
3. Starting from the Entrance, visit all rooms according to the solution. For non-adjacent rooms, substitute the shortest distance between those rooms.
Will this procedure provide the answer to the problem?
• Your problem appears to be a special case of TSP. In many special cases you can do better than the general case. Have you looked into this direction? – Yuval Filmus Feb 17 '16 at 16:59
• @YuvalFilmus Thanks for your comment. I have tried googling this question but was unable to find a straight-forward solution. As I am not a computer scientist, academic papers are mostly beyond my level of understanding. I hoped that this variant of the TSP is well-known and that maybe a solution already exist. I will probably ask a moderator to migrate this question to stackoverflow at a later stage. Thanks again for your help. – COOLSerdash Feb 18 '16 at 8:25
• Thanks for all the comments, COOLSerdash. I think it would be helpful to edit your question to include all of the information you've been providing in the comments, in summarized form. – D.W. Feb 18 '16 at 11:03
• There can easily be more than one "shortest path from ... at least once", but 1.2.3. will provide a "shortest path from ... at least once". – user12859 Feb 25 '16 at 9:49
Travelling Salesman Problem, even if you allow repeating nodes is NP-hard. See Computational Complexity of TSP.
Umans and Lenhart show hardness results for Hamiltonian Graphs in Solid Grid Graphs, 1997.
TSP for Euclideal Case (or graphs with triangle inequality) also imply NP-hardness of TSP with node repetition. TSP even for manhattan distance $L_1$ (or $L_\infty$) metric is NP-complete. See the original Papadimitriou's paper on the topic.
You may be able to prove NP-hardness of TSP for your case by adding arcs to nodes that have corresponding distance as length of shortest path between nodes which will simulate repetitions of the nodes. TSP for your special case looks like an NP-complete problem.
So either write a sufficiently good (heuristic wise) exponential branch and bound algorithm to compute a shortest tour (which may not be all that inefficient, if your graph is small) , or forget about optimization and calculate a good enough approximation.
• Thank you for your answer, Shreesh. Unfortunately, I'm not a computer scientist and am not interested in proving a mathematical statements (maybe I'll post this question on stackoverflow at a later stage). Also, I'm not a programmer and the chances that I'm able to code a solution myself are pretty slim. But I suspect I'm not the first one dealing with a problem of that nature. Thanks again. – COOLSerdash Feb 18 '16 at 8:21
• One easy way to go about this is add $^nC_2$ edges corresponding to shortest paths and then solve TSP with any existing algorithm. As you already know Concorde, it would be easy. – Shreesh Feb 18 '16 at 10:20
• We have a grid graph with unit weights here. Is suspect that the problem is much easier than even Euclidian TSP. (cc @COOLSerdash) – Raphael Feb 23 '16 at 20:00
• No, if the grid is sparse, we basically have TSP with vertices at rational points and $L_1$ metric. However if your Grid is dense, i.e. basically you have to visit almost every point of the grid, no, you visit every point of the grid, then there is a very good ~$n^2$ path. See here and here. – Shreesh Feb 24 '16 at 12:57
In addition to the above answer, I would point out some TSP solvers already available.
• Thanks you for your answer. But as I stated, I already know how to solve TSP using concorde. As my problem is not directly suitable for the original TSP, I currently don't know how that is going to help me. – COOLSerdash Feb 18 '16 at 8:17
• The shortest path would be the path which visits every room exactly once. Now by adding extra nodes you can reduce your problem to a TSP problem whose answer would directly give answer to your question. – Dib Feb 18 '16 at 8:37
You can treat this like a modified path coverage planning problem that you can solve in a few simple steps:
1) Construct an unweighted undirected graph from the grid- rooms, path junctions are nodes, and edges the paths between those nodes.
2) Find the minimum spanning tree from your start point using depth first search.
3) "Subdivide" the underlying grid so that your minimum spanning tree creates two "lanes".
4) From your starting point walk clockwise in the right-hand lane from node-to-node until you return to the starting point in the complementary lane.
This will give you a minimal tour of the rooms, in time proportional to the number of tiles in the dungeon, and is essentially the Spiral Spanning Tree Coverage path planning algorithm applied to a reduced setting. (Cf. "Spiral-stc: an on-line coverage algorithm of grid environments by a mobile robot")
• Can you explain what it means to subdivide the grid? What if my graph isn't on a grid? i.e. imgur.com/a/KlnPO – jdelman Jan 22 '18 at 16:11
| 1,583
| 6,668
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.40625
| 3
|
CC-MAIN-2019-30
|
latest
|
en
| 0.96588
|
https://www.cnccookbook.com/cnc-g-code-arc-circle-g02-g03/
| 1,726,166,389,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00016.warc.gz
| 659,543,916
| 81,782
|
# Quick G-Code Arc Tutorial [Make G02 & G03 Easy]
5 months by cncdivi
# Quick G-Code Arc Tutorial [Make G02 & G03 Easy]
## Circular Interpolation is Motion Along a Circular Arc
Having just finished discussing linear interpolation, or motion in a straight line, we next come to circular interpolation, which is motion along a circular arc.
Other than the fairly exotic ability to follow a “NURBS” path, most g-code controllers only support two kinds of motion: linear and circular motion.
Circular interpolation is quite a bit more demanding on your machine as two axes have to be precisely coordinated. Drawing a complete gcode circle involves not just coordinated motion but reversal of direction at each of the 4 quadrant points. These would be the points corresponding to 0, 90, 180, and 270 degrees.
If the machine has any backlash at all, it will be obvious at these reversals because there will be a glitch in the cut there.
## Circular Motion is a Mode Initiated Via G02 and G03
Like linear motion (initiated by G00 and G01), circular motion is a mode initiated via a G02 or G03 g code. G02 establishes a mode for clockwise circular arcs. G03 establishes a mode for counter-clockwise circular arcs.
Clockwise Arc vs Counter Clockwise Arc…
### Defining An Arc For the CNC Controller
Once either the G02 or G03 mode is established, arcs are defined in G-Code by identifying their 2 endpoints and the center which must be equi-distant from each endpoint or an alarm will occur. The endpoints are easy. The current control point, or location when the block is begun establishes the start point. The end point may be established by XYZ coordinates. The center is a bit more complex.
Technically, we must also define the working plane the arc is to be drawn in. G17, G18, and G19 gcodes are used to establish the working plane. In most cases, the default XY plane as selected by G17 will be used.
### Defining the Center Via IJK Relative Offsets
The center is most commonly identified in CNC G-Code Programming by using I, J, or K to establish relative offsets from the start point of the arc to the center. Here is a typical clockwise arc:
Defining an arc’s center with IJK…
This arc starts at X0Y2 and finishes at X2Y0. It’s center is at X0Y0. We could specify it in g-code like this:
G02 (Set up the clockwise arc mode)
X2Y0 I0J-2.0
The I and the J specify relative coordinates from the start point to the center. In other words, if we add the I value to the start point’s X axis, and the J value to the start point’s Y axis, we get the X and Y axis values for the center.
### Defining the Center Via the Radius Using “R”
We can also define the center just by specifying the radius of the circle. In this case, our circle has a radius of 2, so the g-code might be simply:
G02
X2Y0 R2
Many of you will be deciding right here and now that since R is easier to understand and shorter to write, you’re just going to use R and forget about IJK. But, the CNC teachers in the world will suggest that you should prefer IJK. Their argument is that when you use IJK, you get a double check that your arc is correct.
Why?
Because the controller gets to compute an actual set of coordinates for the center via IJK. Once it has the center’s coordinates, it can check that it is equa-distant from both end points. The check of each of those two distances is the double check. In the case of the “R” format, the controller has no such double check. It has to chose a center that guarantees equal distance.
Personally, I don’t know if I agree with the CNC instructors that this is providing any extra checking or not. I say go with whichever approach makes sense for your particular situation, but you should definitely be familiar and comfortable with both. You’re going to need to be comfortable with relative coordinates anyway, as they’re darned handy. May as well get comfortable now.
It’s kind of like being told you should only use the 4-jaw chuck on a lathe when you first start out so you’ll get very comfortable dialing it in. It’s a good skill to be good at as a machinist!
## Variations in Arc Syntax for Different G-Code Dialects and Modes
When IJK Are Not Incremental and What About Having Both IJK And R? Plus, Other Modal Shenanigans and Arc Variations
This is another one of those places where lots of obscure things happen and you need to know what your controller will do without assuming anything. In general, the rule is supposed to be that if you have both IJK and R in the same block, R takes precedence and IJK is ignored. But there are controllers that don’t work exactly that way, so be sure you know what’s going on.
G-Wizard Editor let’s you specify several parameters in its Post that determine how arcs work. Here is a screen shot of the setup options:
Arc Options for G-Code Simulation
Let’s go over these options:
– Incremental vs Absolute IJK: We’ve discussed IJK as offering coordinates relative to the start point for the arc center. Add the I to X axis, J to Y axis, and K to Z axis of the start point and you get the arc center. Many controls also have the option for IJK to be the absolute coordinates of the center.
– Modal IJK Centers: When IJK are absolute center coordinates, some controllers will remember the last center defined, hence IJK is modal in that case. When using a control set up like this, you can just keep issuing XYZ commands for arcs without having to define a new center each time. It’s not clear you’ll save much though–how often do you want to do a bunch of arcs with the same center?
– Modal R Centers: Another variation on the modal center idea is to allow the radius defined by “R” to be modal. Whatever the last R used was, the controller remembers and uses that value again if no R is given. This seems more useful than modal IJK. For example, a pocket might have arcs for the corners that are all the same radius.
– Give R Precedence: As mentioned, most controllers will use “R” when both “R” and “IJK” are given in the same block. But this option allows you to change that precedence to IJK if your controller works that way instead.
– Helical Interp.: This option governs whether your controller allows helical interpolation.
### The Most Common Problem Configuring a CAM Post or CNC Simulator: Absolute vs Relative IJK
We’ve all had the experience of looking at a backplot in an nc viewer (or worse, seeing it in the actual tool motion which is pretty scary) and seeing the giant almost complete circles and no sign of the familiar part motions we expected to see. Here is a typical example:
Engrave file with bad Post settings for Arcs…
If you see that sort of thing, the first thing to check is absolute versus relative IJK for arcs. The setting has to match between what the CAM produces and what the controller or simulator expects.
### Fractions of a Circle, Quadrants, and Controllers
The first thing about an arc is it isn’t possible to specify more than a 360 degree arc, which is a complete circle. There are some exceptions to this on some controllers for Helical Interpolation (see below), just because it can be useful for helixes. When a full gcode circle is desired, set the start and end points equal to one another as in this g code programming example:
G01 X3.25 Y2.0
G02 X3.25 Y2.0 I-1.25 J0
Interestingly, you can’t specify a full gcode circle with the “R” notation. This is because there are an infinite number of circles that start and end at the same point of a particular radius, so the controller has no idea what the correct circle might be.
There is more funny business still with “R” and larger arcs. For example, an arc may still be of a particular radius and clockwise (or counter-clockwise direction), but the center is ambiguous if you travel more than 90 degrees. For example:
If R is negative, it takes the longer path (in yellow). Positive gets the shorter path.
Given the two choices shown, the controller chooses the path based on the sign of the radius. Negative forces the longer arc, positive the shorter. The negative sign forces the controller to seek a viable arc of more than 180 degrees.
Some controllers are touchier still and will not program an arc that crosses a quadrant line. Hence, the largest angle an arc can follow is 90 degrees, and that angle must not cross 0, 90, 180, or 270 degrees. For angles of 90 degrees that cross a quadrant line, they must be broken into two pieces, with the join between the pieces being right on the quadrant line.
### Full Circles With No XYZ
Full circles come about when the start and endpoints are identical and the center is specified via IJK (remember, R leads to an infinite number of circles). Given that you want the start and endpoint to be the same, you may not need to bother even specifying the end point with XYZ. Some controllers may require it, but most do not. Here’s a simple g-code program that produces 3 circles in this way:
N45 G0 X-2. Y.75
N46 G1 Z-.5 F10.
N47 Y.5 F30. S2000
N48 G2 J-1.1
N49 G1 Y.75
N50 Z.2
N51 G0 X.75 Y-3.4
N52 G1 Z-.5 F10.
N53 X.5 F30.
N54 G2 I-1.1
N55 X.75
N56 Z.2
N57 G0 X-4.75 Y-3.4
N58 G1 Z-.5 F10.
N59 X-4.5 F30.
N60 G2 I1.1
N61 G1 X-4.75
N62 Z.2
And here’s what the backplot looks like:
When I’m laying out a toolpath, I prefer to leave the arcs until last. In place of each arc, I simply put a line segment whose endpoints correspond to the arc’s endpoints. This makes it easy to get the rough sketch of the toolpath together quickly, and it often seems to make it easier to then go back and convert the lines to arcs once the basic structure is already in place.
### Helical Interpolation
A helix is an arc that continuously moves in a third dimension, like a screw thread. With helical interpolation, we specify such an arc with G02/G03 in order to move the cutter along a helix. This can be done for thread milling, interpolating a hole, or a variety of other purposes. Here is a backplot from a 1/4″ NPT thread mill program:
Here is a sample of the code from the thread milling program:
G01 G91 Z-0.6533 F100.
G01 G42 D08 X0.0235 Y-0.0939 F10.
G03 X0.0939 Y0.0939 Z0.0179 R0.0939
G03 X-0.1179 Y0.1179 Z0.0179 R0.1179
G03 X-0.1185 Y-0.1185 Z0.0179 R0.1185
G03 X0.1191 Y-0.1191 Z0.0179 R0.1191 F16.
G03 X0.1196 Y0.1196 Z0.0179 R0.1196
G03 X-0.1202 Y0.1202 Z0.0179 R0.1202 F26.
G03 X-0.1207 Y-0.1207 Z0.0179 R0.1207
G03 X0.1213 Y-0.1213 Z0.0179 R0.1213
G03 X0.1218 Y0.1218 Z0.0179 R0.1218
G03 X-0.0975 Y0.0975 Z0.0179 R0.0975
This is “R” (radius) format for the arcs, and note there is a Z coordinate to specify a depth change for the end point of each arc. This code uses relative motion (G91), so each “Z0.0179″ moves the cutter 0.0179” deeper.
G-Wizard Editor provides some really useful information to help out with understanding helical interpolation. Here is the Hint from the third line (first arc move):
Note the thread pitch here is calculated as 0.1
GWE will measure and tell you the helix pitch, which in this case is 0.100″. That can be useful for identifying what sort of thread is being milled. We can also see that this particular arc runs from 270 degrees to a scosh more than zero (0.1 degrees).
We’ll revisit thread milling in much more detail in a later chapter devoted entirely to the subject. For now, we just wanted you to be familiar with the idea that you can make helixes as well as flat two dimensional arcs.
BTW, if the hole is made in relatively thin material relative to the tool diameter, you can dispense with the helix and imply do circular interpolation.
### Making Toolpaths Your Machine Will Be Happier With
Whenever the cutter changes direction, it adds a certain amount of stress. The cutter will bite into the material either more or less than it had been, depending on whether the directions changes towards the workpiece (or uncut material) or away from it. Your machine will be much happier if you program an arc rather than an abrupt straightline change of direction. Even an arc with a very small radius will allow the controller to avoid changing direction instantly, which can leave a mark in the finish in the best case and cause chatter or other problems in the worst case. For slight changes of direction, it may not be worth it. But the more abrupt the change, with 90 degrees being very abrupt, the greater the likelihood you should use an arc to ease through the turn.
Arcs are also a useful way to enter the cut, rather than having the cutter barge straight in. For information on entering the cut with an arc, see the toolpath page from the Milling Feeds and Speeds Course.
### What is the difference between G02 and G03?
The difference is the direction of the arc motion. G02 makes clockwise arcs. G03 makes counter-clockwise arcs.
### What is the G code G2?
G code G2 initiates a mode for clockwise circular arcs.
## Exercises
1. Dig out your CNC controller manual and go through the arc settings to set up GWE to match your control’s way of operating.
2. Do some etch-a-sketch experimentation with GWE. Create some toolpaths that include arcs until you’re comfortable creating them.
## Next Article: Running the GWE G-Code Simulator
Try the Free Trial Version of G-Wizard G-Code Editor…
## Like what you read on CNCCookbook?
Join 100,000+ CNC'ers! Get our latest blog posts delivered straight to your email inbox once a week for free. Plus, we’ll give you access to some great CNC reference materials including:
• Our Big List of over 200 CNC Tips and Techniques
• Our Free GCode Programming Basics Course
• And more!
| 3,375
| 13,556
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.90625
| 4
|
CC-MAIN-2024-38
|
latest
|
en
| 0.876331
|
https://cs.stackexchange.com/questions/30357/the-buckets-of-water-problem/30358
| 1,718,506,488,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00161.warc.gz
| 157,579,924
| 40,775
|
# The buckets of water problem
Let's consider the following problem (buckets/pails of water problem) (This problem may be known with different name. If does, please correct me).
Let $B=\{b_1,...,b_n\}$ be a set of $n$ buckets. Suppose each bucket has a maximum capacity $c_i \in \mathbb{Z}$. This also can be written as a maxium capacity function $f:B \rightarrow \mathbb{Z}$ such that $f(b_i)=c_i$.
Let $g:B \rightarrow \mathbb{Z}$ be a function such that $g(b_i)$ is the current amount of water in bucket $b_i$.
Suppose that we can do the following operations.
1.Fill bucket $b_i$ from tap until its full, i.e. $g(b_i)=f(b_i)$.
2.Move water from bucket $b_i$ to bucket $b_j$ until $b_i$ is empty or $b_j$ is full.
3.Empty bucket $b_i$.
Now, the problem is given a number $m \in \mathbb{Z}$ to find a sequence of operations $s_1,...,s_k$ such that after $s_k$ we have a bucket with $m$ amount of water, i.e. $g(b_i)=m$ for some $i \in \{1,...,n\}$, or return that such sequence doesn't exist.
My questions are :
1.How to solve this problem? Is this problem NP Hard? If it is NP Hard, why? How to prove this?
2.What about the case when we are interested in the optimal $k$, i.e. we want minimum number of steps?
3.Does this is a well known problem? If yes, what is the known name for the problem and what good references exists for this problem?
I want to note that I fully understand the case of $n=2$, and I am interested in the generalization of $n$ buckets instead of just $2$. The $n=2$ case described in https://mathoverflow.net/questions/5800/generalization-of-the-two-bucket-puzzle.
Edit: I now know how to prove that this is NP Hard problem. I want to know if there is an efficient algorithm for solving this (maybe some pseudo polynomial algorithm).
The problem is NP-hard, by reduction from SUBSET-SUM. Given a multiset of numbers $$x_1,\ldots,x_n$$ and a target $$T$$, consider $$n$$ buckets with capacity $$C=x_1+\cdots+x_n$$, initially filled with $$x_1,\ldots,x_n$$, and ask whether you can obtain a bucket filled with exactly $$T$$. You can prove by induction that all buckets are always with filled either with $$C$$ or with the sum of some (possibly empty) subset of the $$x_i$$s.
| 612
| 2,213
|
{"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875
| 4
|
CC-MAIN-2024-26
|
latest
|
en
| 0.896523
|
https://byjus.com/question-answer/a-fan-is-rotating-with-angular-velocity-100-text-rev-sec-then-it-is-switched/
| 1,723,354,870,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00110.warc.gz
| 118,223,682
| 29,516
|
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# A fan is rotating with angular velocity 100 rev/sec. Then, it is switched off. It takes 5 minutes to stop. Find the average angular velocity during this interval. (correct answer + 1, wrong answer - 0.25)
A
100 rev/sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
75 rev/sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
125 rev/sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
50 rev/sec
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
## The correct option is D 50 rev/secGiven, initial angular velocity, ω0=100 rev/s Final angular velocity, ω=0 rev/s ∴ Total number of revolution made by the fan, θ=(ω+ω02)t =0+1002×5×60=15000 rev ∴ Average angular velocity in this interval =150005×60=50 rev/sec
Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Relative
PHYSICS
Watch in App
Join BYJU'S Learning Program
| 298
| 1,016
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.484375
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.820459
|
https://mersenneforum.org/search.php?s=0753b140157d2a3ea1850a2276be79d7&searchid=3982759
| 1,618,737,825,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038469494.59/warc/CC-MAIN-20210418073623-20210418103623-00186.warc.gz
| 497,873,403
| 9,116
|
mersenneforum.org Search Results
Register FAQ Search Today's Posts Mark Forums Read
Showing results 1 to 25 of 1000 Search took 0.33 seconds. Search: Posts Made By: axn
Forum: Miscellaneous Math 2021-04-18, 07:39 Replies: 10 Views: 111 Posted By axn I don't know what "the test" is but: ?... I don't know what "the test" is but: ? kronecker(9, 25) %1 = 1 Do you even know what qfbprimeform is supposed to do? ? ?qfbprimeform qfbprimeform(x,p): returns the prime form of discriminant...
Forum: Miscellaneous Math 2021-04-18, 07:14 Replies: 10 Views: 111 Posted By axn No. Composite numbers as well. However, it must... No. Composite numbers as well. However, it must be relatively prime. ? kronecker(9,14) %1 = 1 ? kronecker(9,15) %2 = 0 ? kronecker(9,16) %3 = 1 ? kronecker(9,17) %4 = 1
Forum: Miscellaneous Math 2021-04-18, 06:29 Replies: 10 Views: 111 Posted By axn Your check is 9^((Mp-1)/2) == 1 (mod Mp) or... Your check is 9^((Mp-1)/2) == 1 (mod Mp) or alternately, 9^((Mp+1)/2)==9 (mod Mp). This is nothing but a Euler pseudoprimality test (https://en.wikipedia.org/wiki/Euler_pseudoprime) Since 9 is...
Forum: PrimeNet 2021-04-16, 14:12 Replies: 7 Views: 211 Posted By axn Yes, the factor of 2 is based on the 2 LL tests... Yes, the factor of 2 is based on the 2 LL tests needed to conclusively prove an exponent is composite. With PRP+CERT, the factor of 2 should become 1.03 (or something similar).
Forum: Software 2021-04-15, 08:02 Replies: 145 Sticky: Prime95 v30.4/30.5 Views: 14,100 Posted By axn Why not have just three workers and change their... Why not have just three workers and change their worktype as and when you want to switch over? Should be just as easy as bringing workers up/down. Or ... Have two instances of P95 and keep PRP...
Forum: Software 2021-04-15, 05:48 Replies: 145 Sticky: Prime95 v30.4/30.5 Views: 14,100 Posted By axn Then why not have just 3 workers (or 2 workers)?... Then why not have just 3 workers (or 2 workers)? You're putting the program in an impossible situation. It'd be best to just turn off the benchmark altogether.
Forum: Miscellaneous Math 2021-04-15, 02:21 Replies: 9 Views: 223 Posted By axn This might or might not work. If someone could... This might or might not work. If someone could find a counterexample, that would obviously disprove your claim. Yet, doing test after test of successful confirmation will not get you any closer to...
Forum: Data 2021-04-15, 02:15 Replies: 17 Views: 497 Posted By axn The advantage is not that we know to include 2p... The advantage is not that we know to include 2p in the stage 1 powering routine, but rather, the part that needs to be smooth ((f-1)/2p) is much smaller and therefore higher probability of success...
Forum: Cloud Computing 2021-04-15, 02:11 Replies: 1,142 Views: 97,534 Posted By axn I don't know the reason you're not getting any... I don't know the reason you're not getting any GPU, but I doubt it is permanent. Keep trying every day; eventually you will get one.
Forum: PrimeNet 2021-04-14, 16:46 Replies: 5 Views: 84 Posted By axn The server is aware of certain machines... The server is aware of certain machines registered under your account. It has certain expectations of how much these machines can produce. It merely lists what % of expected work was actually...
Forum: PrimeNet 2021-04-14, 14:47 Replies: 3 Views: 70 Posted By axn You have done 8959 Gd worth of work, of which... You have done 8959 Gd worth of work, of which 7640 (85.3%) is LL/PRP, of which all 7640 is from PRP and 0 from LL. You also have 141 from TF, 218 from P-1, 935 from LL-D, and so on. You are...
Forum: Data 2021-04-14, 14:22 Replies: 17 Views: 497 Posted By axn If f=5 (mod 6) and f+1 is smooth, 2/7 seed will... If f=5 (mod 6) and f+1 is smooth, 2/7 seed will find it. If f=1 (mod 6) and f-1 is smooth, then also 2/7 will find it If f=3 (mod 4) and f+1 is smooth, 6/5 will find it If f=1 (mod 4) and f-1 is...
Forum: Data 2021-04-14, 03:51 Replies: 17 Views: 497 Posted By axn There are going to be some stubborn sub-ranges in... There are going to be some stubborn sub-ranges in the 10M-40M region where P-1 and TF have done their part, and only way forward is to do deeper P-1 or TF (ECM being too inefficient) which will be...
Forum: Data 2021-04-14, 03:41 Replies: 17 Views: 497 Posted By axn Doubtful that P+1 will breakeven for wavefront.... Doubtful that P+1 will breakeven for wavefront. Like, say, there might be a 1% probability, but costs more than 1% of PRP test (made-up numbers for illustrative purpose). It is more costly than P-1,...
Forum: Data 2021-04-14, 01:44 Replies: 17 Views: 497 Posted By axn Exactly the same. Just, instead of factoring f-1... Exactly the same. Just, instead of factoring f-1 (and ignoring the forced "2p"), you factor f+1. B1 is second-largest factor, B2 is largest.
Forum: Math 2021-04-13, 14:23 Replies: 10 Views: 4,827 Posted By axn No. No.
Forum: Software 2021-04-11, 17:30 Replies: 7 Views: 170 Posted By axn ETA is accurate, but since it is based on the... ETA is accurate, but since it is based on the current iteration time, it can go up and down. Status is based on the expected performance of the CPU. It doesn't need the test to be running...
Forum: Lounge 2021-04-11, 01:57 Replies: 1,709 Views: 149,938 Posted By axn Wrong. ... Wrong. https://en.wikipedia.org/wiki/Succession_to_the_British_throne#Current_rules https://en.wikipedia.org/wiki/Succession_to_the_British_throne#Current_line_of_succession
Forum: Software 2021-04-09, 06:31 Replies: 145 Sticky: Prime95 v30.4/30.5 Views: 14,100 Posted By axn Is it safe to change this in the middle of a... Is it safe to change this in the middle of a Stage 2 run?
Forum: Miscellaneous Math 2021-04-06, 12:17 Replies: 35 Views: 672 Posted By axn The burden of proof is on you to show that it is... The burden of proof is on you to show that it is a deterministic test. Until then ...
Forum: Miscellaneous Math 2021-04-06, 05:22 Replies: 11 Views: 220 Posted By axn RTFA. Stop projecting your wishes on to what was... RTFA. Stop projecting your wishes on to what was actually done.
Forum: Miscellaneous Math 2021-04-06, 05:13 Replies: 11 Views: 220 Posted By axn Are you thinking that the number must have that... Are you thinking that the number must have that property in _all_ the bases simultaneously?
Forum: Miscellaneous Math 2021-04-06, 02:41 Replies: 11 Views: 220 Posted By axn https://www.mersenneforum.org/showthread.php?p=575... https://www.mersenneforum.org/showthread.php?p=575009#post575009
Forum: Software 2021-04-05, 15:15 Replies: 145 Sticky: Prime95 v30.4/30.5 Views: 14,100 Posted By axn I suspect that on a local system, you wouldn't be... I suspect that on a local system, you wouldn't be able to tell the difference. But anyways, George will have to confirm if this is even possible or not.
Forum: Software 2021-04-05, 14:28 Replies: 145 Sticky: Prime95 v30.4/30.5 Views: 14,100 Posted By axn George, Would it be possible to reduce the size... George, Would it be possible to reduce the size of P-1 stage 2 checkpoint files? Using this with colab / google drive, it takes a very long time to stop/restart during stage 2 - it writes 100-200 MB...
Showing results 1 to 25 of 1000
All times are UTC. The time now is 09:23.
Sun Apr 18 09:23:44 UTC 2021 up 10 days, 4:04, 0 users, load averages: 1.50, 1.53, 1.58
| 2,256
| 7,373
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.25
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.881792
|
https://www.thefreelibrary.com/Evolution+equation+associated+with+the+power+of+the+Gross+Laplacian.-a0172134702
| 1,547,819,572,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-04/segments/1547583660139.37/warc/CC-MAIN-20190118131222-20190118153222-00255.warc.gz
| 960,049,692
| 18,042
|
# Evolution equation associated with the power of the Gross Laplacian.
Abstract
We study an evolution equation associated with the power of the Gross Laplacian [[DELTA].sup.p.sub.G] and a potential function V on an infinite dimensional space. The initial condition is a generalized function. The main technique we use is the representation of the Gross Laplacian as a convolution operator. This representation enables us to apply the convolution calculus on a suitable distribution space to obtain the explicit solution of the perturbed evolution equation. Our results generalize those previously obtained by Hochberg [7] in the one dimensional case with V = 0, as well as by Barhoumi-Kuo-Ouerdiane for the case p = 1
AMS Subject Classification: Primary 60H40; secondary 60H15, 46F25, 46G20.
Keywords: Evolution equation, Gross Laplacian, potential function, white noise analysis, generalized functions, convolution operator, Laplace transform, duality theorem.
1. Introduction
In the paper [7] Hochberg studied the parabolic partial differential equation
[partial derivative]u/[partial derivative]t = [(-1).sup.n+1] [[partial derivative].sup.2n]u/ [partial derivative][x.sup.2n], n [greater than or equal to] 2.
He showed that the fundamental solution of this equation is the density of a finitely additive signed measure of unbounded variation. In this paper we will consider the infinite dimensional generalization of the above equation with an additional potential function
[partial derivative][U.sub.t]/[partial derivative]t = [(-1).sup.p+1] 1/2 [[DELTA].sup.p.sub.G][U.sub.t] + [V.sub.t], p [member of] N, (1.1)
where [[DELTA].sub.G] is the Gross Laplacian [5] [8]. We will show that the solutions are generalized functions (rather than ordinary function), which reflect the above fact that the corresponding "measure" are finitely additive signed measure (rather than countably additive measure) of unbounded variation. On the other hand, Equation (1.1) with p = 1 has been studied in the paper [2].
In Section 2 we will briefly described the infinite dimensional framework in order to setup Equation (1.1). In Section 3 we will study this equation with an initial condition and show that the unique solution is a generalized function. The main tool is the interpretation of the Gross Laplacian as a convolution operator. In Section 4 we will give concluding remarks.
2. Background
First we review from the paper [4] basic concepts, notation, and some results which will be needed in the present paper. Independent development for similar results can be found in the paper [1].
Let X be a real nuclear Frechet space with topology given by an increasing family [{|x|}.sub.p]; p [member of] [N.sub.0]} of Hilbertian norms, [N.sub.0] being the set of nonnegative integers. Then X is represented as X = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] [X.sub.p], where [X.sub.p] is the completion of X with respect to the norm [|x|.sub.p]. We use [X.sub.-p] to denote the dual space of [X.sub.p]. Then the dual space X' of X can be represented as X' = [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] [X.sub.-p] and is equipped with the inductive limit topology.
Let N = X + iX and [N.sub.p] = [X.sub.p] + [iX.sub.p], p [member of] Z, be the complexifications of X and [X.sub.p], respectively. For n [member of] [N.sub.0] we denote by [N.sup.[??]n] the n-fold symmetric tensor product of N equipped with the [pi]-topology and by [N.sup.[??]n.sub.p] the n-fold symmetric Hilbertian tensor product of [N.sub.p]. We will preserve the notation [|x|.sub.p] and [|x|.sub.-p] for the norms on [N.sup.[??]n.sub.p] and [N.sup.[??]n.sub.-p], respectively.
Let [theta] be a Young function, i.e., it is a continuous, convex, and increasing function defined on [R.sub.+] such that [theta](0) = 0 and [lim.sub.x[right arrow][infinity]] [theta](x)=x = 1. We define the conjugate function [theta]* of [theta] by
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
For a Young function [theta], we denote by [F.sub.[theta]](N') the space of holomorphic functions on N0 with exponential growth of order [theta] and of minimal type. Similarly, let G[theta](N) denote the space of holomorphic functions on N with exponential growth of order [theta] and of arbitrary type. Moreover, for each p [member of] Z and m > 0, define Exp([N.sub.p], [theta], m) to be the space of entire functions f on [N.sub.p] satisfying the condition:
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Then the spaces [F.sub.[theta]](N') and G[theta](N) can be represented as
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and are equipped with the projective limit topology and the inductive limit topology, respectively. The space [F.sub.[theta]](N') is called the space of test functions on N'. Its dual space [F'.sub.[theta]](N'), equipped with the strong topology, is called the space of distributions on N'.
For p [member of] [N.sub.0] and m > 0, we define the Hilbert spaces
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
where [[theta].sub.n] = [inf.sub.r>0] [e.sup.[theta](r)]/[r.sup.n], n [member of] [N.sub.0]. Put
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
The space [F.sub.[theta]](N) equipped with the projective limit topology is a nuclear Frechet space [4]. The space [G.sub.[theta]](N') carries the dual topology of [F.sub.[theta]](N) with respect to the C-bilinear pairing given by
<<[??], [??]>> = [summation over (n[greater than or equal to]0)] n! <[[PHI].sub.n], [[psi].sub.n]>,(2.1)
where [??] = [([[PHI].sub.n]).sup.[infinity].sub.n=0] [member of] [G.sub.[theta]](N') and [??] = [([[psi].sub.n]).sup.[infinity].sub.n=0] [member of] [F.sub.[theta]](N).
It was proved in [4] that the Taylor map defined by
T : [psi] [??] (1/n! [[psi].sup.(n)](0)).sup.[infinity].sub.n=0]
is a topological isomorphism from [F.sub.[theta]](N') onto [F.sub.[theta]](N). The Taylor map T is also a topological isomorphism from [G.sub.[theta]*](N)) onto [G.sub.[theta]](N')). The action of a distribution [PHI] [member of] [F'.sub. [theta]](N') on a test function [psi] [member of] [F.sub.[theta]](N') can be expressed in terms of the Taylor map as follows:
<<[PHI], [psi]>> = <<[??], [??]>>
where [??] = [(T*).sup.-1][PHI] and [??] = T[psi].
It is easy to see that for each [xi] 2 N, the exponential function
[e.sub.[xi]](z) = [e.sup.<z,[xi]>, z [member of] N',
is a test function in the space [F.sub.[theta]](N') for any Young function [theta]. Thus we can define the Laplace transform of a distribution [PHI] [member of] [F'.sub.[theta]](N') by
[??]([xi]) = <<[PHI], [e.sub.[xi]]>>, [xi] [member of] N. (2.2)
[??]From the paper [4], we have the duality theorem which says that the Laplace transform is a topological isomorphism from [F'.sub.[theta]](N') onto [G.sub.[theta]*](N).
For [psi] [member of] [F.sub.[theta]](N'), the translation [t.sub.x] [psi] of [psi] by x [member of] N' is defined by
[t.sub.x][psi] (y) = [psi] (y - x), y [member of] N'.
The translation operator [t.sub.x] is a continuous linear operator from [F.sub.[theta]](N') into itself for any x [member of] N'. By a convolution operator on the space [F.sub.[theta]](N') of test functions we mean a continuous linear operator from [F.sub.[theta]](N') into itself which commutes with translation operators [t.sub.x] for all x [member of] N'.
We define the convolution [PHI] * ' of a distribution [PHI] [member of] [F'.sub.[theta]](N') and a test function [psi] [member of] [F.sub.[theta]](N') to be the function
([PHI] * [psi])(x) = <<[PHI], t-x[psi]>>; x [member of] N'.
Direct calculations show that [PHI]*[psi] [member of] [F.sub.[theta]](N') for any [psi] [member of] [F.sub.[theta]](N') and that the mapping [T.sub.[PHI]] defined by
[T.sub.[PHI]] : [psi] [??] [PHI] * [psi], [psi] [member of] [F.sub.[theta]](N'),
is a convolution linear operator on [F.sub.[theta]](N'). Conversely, it was proved in [3] that all convolution operators on [F.sub.[theta]](N') occur this way, i.e., if T is a convolution operator on [F.sub.[theta]](N'), then there exists a unique [PHI] [member of] [F'.sub.[theta]](N') such that T = [T.sub.[PHI]], or equivalently,
T([psi]) = [T.sub.[PHI]](') = [PHI] * [psi], [psi] [member of] [F.sub.[theta]](N'). (2.3)
Suppose [[PHI].sub.1], [[PHI].sub.2] [member of] [F'.sub.[theta]](N'). Let [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] be the convolution operators given by [[PHI].sub.1] and [[PHI].sub.2], respectively, as in Equation (2.3). It is clear that the composition [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] [??] [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is also a convolution operator on [F.sub.[theta]](N'). Hence there exists a unique distribution, denoted by [[PHI].sub.1] * [[PHI].sub.2], in [F'.sub.[theta]](N') such that
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2.4)
The distribution [[PHI].sub.1] * [[PHI].sub.2] in Equation (2.4) is called the convolution of [[PHI].sub.1] and [[PHI].sub.2]. From Proposition 1 of the paper [3] we have the following equality for any [[PHI].sub.1],[[PHI].sub.2] [member of] [F'.sub.[theta]](N'),
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2.5)
Let [gamma] be the standard Gaussian measure on the dual space X0 of the real nuclear space X, namely, its characteristic function is given by
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
where [|x|.sub.0]is the norm [|x|.sub.p] on X for p = 0. Suppose that the Young function [theta] satisfies [lim.sub.r[right arrow]+[infinity]] [theta](r)/[r.sup.2] < +1. Then we obtain the Gel'fand triple [4]
[F.sub.[theta]](N') [??] [L.sup.2](X', [gamma]) [??] [F'.sub.[theta]](N').
It is worthwhile to mention that the S-transform, defined on [F'.sub.[theta]](N'), is related to the Laplace transform by
S([PHI])([xi]) = [??]([xi])[e.sup. <h[xi],[xi]>/2], [xi] [member of] N, [PHI] [member of] [F'.sub.[theta]](N'). (2.6)
Let [beta] be a continuous, convex, and increasing function on [R.sub.+]. Suppose f is function in Exp(C, [beta], m) for some m > 0. For each distribution [PHI] in [F'.sub.[theta]](N'), we define the convolution composition [f.sup.*]([PHI]) of f and [PHI] by
([f.sup.*]([PHI]))[??]b= f([??]). (2.7)
It was proved in [3] that [f.sup.*]([PHI]) belongs to [F'.sub.[lambda]](N') with [greater than or equal to] = [([beta] [??] [e.sup.[theta]*]).sup.*].
In particular, when [beta](x) = x, x [member of] [R.sub.+], and f(z) = [e.sup.z], z [member of] C, we get a distribution [e.sup.*[PHI]] in the space [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII](N') for each [PHI] [member of] [F'.sub.[theta]](N'). Moreover, by Equation (2.7), we have
([e.sup.*[PHI]])[??]= [e.sup.[??]]. (2.8)
The distribution [e.sup.*[PHI]] has the following series expansion
[e.sup.*[PHI]] = [[infinity].summation over (n=0)] 1/n! [[PHI].sup.*n],
where [[PHI].sup.*n] = [PHI]*[PHI]* ... *[PHI] (n times) and the convergence is in [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (N') with respect to the strong topology.
3. Initial-valued Evolution equation
Let I [subset] R be an interval containing the origin. Consider a family {[[PHI].sub.t], [member of] I} of distributions in [F'.sub.[theta]](N'). We assume that the function t [??] [[PHI].sub.t] is continuous from I into [F'.sub.[theta]](N'). Then the function t [??] [[??].sub.t] is continuous from I into [G.sub.[theta]*](N). Thus for each t [member of] I, the set {[[??].sub.s], s [member of] [0, t]} is a compact subset of [G.sub.[theta]*](N). In particular, it is bounded in [G.sub.[theta]*](N). Hence there exist constants p [member of] N', m > 0, and [C.sub.t] > 0 such that
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
This inequality shows that the function [xi] [??] [[integral].sup.t.sub.0] c[PHI]s([xi]) ds belongs to the space [G.sub.[theta]*](N). Hence there exists a unique distribution, denoted by [[integral].sup.t.sub.0] [PHI]s ds, in [F'.sub.[theta]](N') satisfying
([[integral].sup.t.sub.0] [[PHI].sub.s] ds)[??] ([xi]) = [[integral].sup.t.sub.0] [[??].sub.s]([xi]) ds, [xi] [member of] N.
Moreover, the process [E.sub.t] = [[integral].sup.t.sub.0] [PHI]s ds; t 2 I; is differentiable in [F'.sub.[theta]](N') and satisfies the equation
[partial derivative] [[partial derivative].sub.t] [E.sub.t] = [[PHI].sub.t].
Let {[[PHI].sub.t]} and {[M.sub.t]} be two continuous [F'.sub.[theta]](N')-processes. Consider the initial value problem
d[X.sub.t]/dt = [[PHI].sub.t] * [X.sub.t] + [M.sub.t], [X.sub.0] = F [member of] [F'.sub.[theta]](N'). (3.1)
The next theorem is from Theorem 4 of the paper [3].
Theorem 3.1 The stochastic differential equation (3.1) has a unique solution in [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (N') given by
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3.2)
We can apply Theorem 3.1 to study an evolution equation for a power of the Gross Laplacian and a generalized potential function with the initial condition being a Generalized function. Let [psi] [member of] [F.sub.[theta]](N') be represented by
[psi] (x) = [summation over (n[greater than or equal to]0)] <[x.sup.[cross product]n], [psi](n)>.
The Gross Laplacian ([[DELTA].sub.G] [psi])(x) of [psi] at x [member of] N' is defined to be
([[DELTA].sub.G] [psi])(x) = [summation over (n[greater than or equal to]0)] (n + 2)(n + 1) <[x.sup.[cross product]n],<[tau], [[psi].sup.(n+2)]>>,
where [tau] is the trace operator, namely,
<[tau], [xi] [cross product] [eta]> = <[xi], [eta]>, [xi], [eta] [member of] N.
For more information on the Gross Laplacian, see [5] [6] [8] [9] [10].
It turns out that the Gross Laplacian [[DELTA].sub.G] can be extended to be a continuous linear operator from [F'.sub.[theta]](N') into itself and the extension is a convolution operator given by
[[DELTA].sub.G][PSI] = T * [PSI], [PSI] [member of] [F'.sub.[theta]](N'), (3.3)
where T is the generalized function in [F'.sub.[theta]](N') with the Laplace transform given by [??] = (0, 0, [tau], 0; ...) [member of] [G.sub.[theta]](N') as in Equation (2.1).
Proposition 3.2 For every positive integer p we have
[[DELTA].sup.p.sub.G][PSI] = (T *p) * [PSI], [PSI] [member of] [F'.sub.[theta]](N'). (3.4)
Moreover, the generalized function associated with [[DELTA].sup.p.sub.G] is given by
[??] = (0, 0, ..., [[tau].sup.[cross product]p], 0, ...). (3.5)
Proof. Using Equations (2.4) and (3.3), we obtain
[[DELTA].sup.p.sub.G][PSI] = T * p * [PSI]
But the Laplace transform of T is given by
[??] ([xi]) = <[tau], [[xi].sup.[cross product]2> = <[xi], [xi]> = [|[xi]|.sup.2.sub.0].
Hence we have
[[??]) = <[tau], [[xi].sup.[cross product]2]>.sup.p] = [<[xi], [xi]>.sup.p] = [|[xi]|.sup.2p.sub.0].
For any positive integer p, let S = T * p and let the formal power series associated with S be given by [??] = ([S.sub.0], [S.sub.1], ..., [S.sub.n], ...). Then we can use the definition of the Laplace transform and the bilinear pairing between test functions and distributions in Equation (2.1) to deduce the following relationship
[[??]) = <[T.sup.*p], [e.sup.[xi]]> = [summation over (n[greater than or equal to]0)] n! <[S.sub.n], [[xi].sup.[cross product]n]/n!> = <[xi], [xi]>.sup.p],
which implies that [S.sub.n] = 0 for all n [not equal to] 2p and [S.sub.2p] = [[tau].sup.[cross product]p]. Therefore,
[??] = [??] = (0, 0, ..., [[tau].sup.[cross product]p], 0, ...).
This proves Equation (3.5).
Theorem 3.3 Let [theta] be a Young function such that [lim.sub.r[right arrow][infinity]] [theta](r)=r2 < 1 and let F 2 [F'.sub.[theta]](N') and fVtg be a continuous [F'.sub.[theta]](N')-processes. Then the following evolution equation associated with the p-th power of the Gross Laplacian and a potential function [V.sub.t]
[partial derivative][U.sub.t]/[partial derivative]t = [(-1).sup.p+1] 1/2 [[DELTA].sup.p.sub.G][U.sub.t] + [V.sub.t], [U.sub.0] = F, (3.6)
has a unique solution in the space [F'.sub.[theta]](N') given by
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3.7)
where T is the generalized function given by Equation (3.3).
Proof. Use Equation (3.3) to rewrite Equation (3.6) as
[partial derivative][U.sub.t]/[partial derivative]t = [(-1).sup.p+1] 1/2 [T.sup.*p] * [U.sub.t] + [V.sub.t], [U.sub.0] = F.
Then we can apply Theorem 3.1 to this equation to get the unique solution in Equation (3.7).
4. Concluding remarks
1. We can further rewrite the solution in Equation (3.7) in another form. For t > 0, define the distribution [[mu].sub.t,p] by its Laplace transform
[??] [xi]) = exp [[(-1).sup.p+1]t/2 <[xi], [xi]>.sup.p]], [xi] [member of] N. (4.1)
Recall the duality theorem [4] which states that the Laplace transform is a topological isomorphism from [F'.sub.[theta]](N') onto [G.sub.[theta]*](N). Hence Equation (4.1) implies that [[mu].sub.t,p], t > 0; are generalized functions in the space [F'.sub.[theta]](N') with the Young function given by
[theta](x) = [x.sup.2p/2p-1], x [greater than or equal to] 0.
Therefore, the solution [U.sub.t] in equation (3.7) can be rewritten as
[U.sub.t] = F * [[mu].sub.t,p] + [[integral].sup.t.sub.0] [[mu].sub.t-s,p] * [V.sub.s] ds.
In particular, when [V.sub.t] = 0, we have the evolution equation
[partial derivative][U.sub.t]/[partial derivative]t = [(-1).sup.p+1] 1/2 [[DELTA].sup.p.sub.G][U.sub.t], [U.sub.0] = F, (4.2)
which has a unique solution given by
[U.sub.t] = F * [[mu].sub.t,p]. (4.3)
Hochberg [7] has studied the one-dimensional case of Equation (4.2) and showed that the fundamental solution defines a finitely additive measure with unbounded total variation. Using the white noise theory, we can now interpret this "finitely additive measure with unbounded total variation" as a generalized function in the space [F'.sub.[theta]](N'), which is given by Equation (4.3). This phenomenon is very much like the case of Feynman integral, which had been regarded as a finitely additive measure with unbounded total variation before the theory of white noise was introduced by T. Hida in 1975. It is a well-known fact that the Feynman integral is a generalized function [6] [9].
2. When p = 1, Equation (4.1) gives the equality
[??]([xi]) = exp [- t/2 [|[xi]|.sup.2.sub.0], [xi] [member of] X,
which shows that [[mu].sub.t,1] is the standard Gaussian measure on X0 with variance t, i.e., [[mu].sub.t,1] = [[gamma].sub.t] with [gamma]t defined by
[[gamma].sub.t](*) = [gamma] (*/[[square root of t]).
Note that the probability measure [[mu].sub.t,1] induces a positive distribution in the space [F'.sub.[theta]](N') given by
<<[[mu].sub.t,1][psi]>> = [[integral].sub.X'] [psi](x) d[[mu].sub.t,1](x) = [[integral].sub.X'] [psi]([square root of t] x) d[gamma](x), [psi] [member of] [F.sub.[theta]](N').
For more details, see the book [9]. Moreover, if the potential function is given by [V.sub.t] = [alpha] [W.sub.t] with [alpha] [member of] R and [W.sub.t] a white noise, then the solution in Equation (3.7) reduces to the one obtained in [2].
References
[1] N. Asai, I. Kubo, and H.-H. Kuo, General characterization theorems and intrinsic topologies in white noise analysis; Hiroshima Math. J. 31 (2001) 299-330.
[2] A. Barhoumi, H. H. Kuo, and H. Ouerdiane, Generalized Gross Heat equation with noises; Soochow J. of Math. 32 (2006) 113-125.
[3] M. Ben Chrouda, M. El Oued, and H. Ouerdiane, Convolution calculus and applications to stochastic differentntial equations; Soochow J. of Math. 28 (2002) 375-388.
[4] R. Gannoun, R. Hachaichi, H. Ouerdiane, and A. Rezgui, Un theoreme de dualite entre espace de fonctions Holomorphes a croissance exponentielle; J. Functional Analysis 171 (2000) 1-14.
[5] L. Gross, Potential theory on Hilbert space; J. Functional Analysis 1 (1967) 123-181.
[6] T. Hida, H.-H. Kuo, J. Potthoff, and L. Streit, White Noise: An Infinite Dimensional Calculus. Kluwer Academic Publishers, Dordrecht, 1993.
[7] K. J. Hochberg, A signed measure on path space related to Wiener measure; The Annals of Probability 6 (1978) 433-458.
[8] H.-H. Kuo, Gaussian Measures in Banach Spaces. Lecture Notes in Math., Vol. 463, Springer-Verlag, 1975 (reprinted by BookSurge, 2006)
[9] H.-H. Kuo, White Noise Distribution Theory. CRC Press, Boca Raton, 1996.
[10] N. Obata, White Noise Calculus and Fock Space. Lecture Notes in Math. 1577, Springer-Verlag, 1994.
Soumaya Gheryani
Department of Mathematics, Faculty of Sciences of Tunis,
University of Tunis El Manar, Tunis, Tunisia
E-mail: soumaye_gheryani@yahoo.fr
Hui-Hsiung Kuo
Department of Mathematics, Luisiana State University,
Baton Rouge, LA 70803, USA
E-mail: kuo@math.lsu.edu
Habib Ouerdiane
Department of Mathematics, Faculty of Sciences of Tunis,
University of Tunis El-Manar, Tunis, Tunisia
E-mail: habib.ouerdiane@fst.rnu.tn
COPYRIGHT 2007 Research India Publications
No portion of this article can be reproduced without the express written permission from the copyright holder.
Copyright 2007 Gale, Cengage Learning. All rights reserved.
Terms of use | Privacy policy | Copyright © 2019 Farlex, Inc. | Feedback | For webmasters
| 6,632
| 21,066
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04
|
latest
|
en
| 0.883824
|
http://www.math-only-math.com/theorem-of-three-perpendiculars.html
| 1,537,333,596,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-39/segments/1537267155924.5/warc/CC-MAIN-20180919043926-20180919063926-00220.warc.gz
| 378,123,009
| 6,918
|
Theorem of Three Perpendiculars
Theorem of three perpendiculars is explained herewith some specific examples.
Theorem: If PQ is perpendicular to a plane XY and if from Q, the foot of the perpendicular, a straight line QR is drawn perpendicular to any straight line ST in the plane, then PR is also perpendicular to ST.
Construction: Through Q draw in the plane XY the straight line LM parallel to ST.
Proof: Since LM is parallel to ST and QR perpendicular to ST hence, QR is perpendicular to LM. Again, PQ is perpendicular to the plane XY; hence, it is perpendicular to the line LM. Therefore, LM is perpendicular to both PQ and QR at Q. This implies LM is perpendicular to the plane PQR. Now, ST and LM are parallel and LM is perpendicular to the plane PQR; hence, ST is perpendicular to the plane PQR. Therefore, ST is perpendicular to PR or in other words, PR is perpendicular to ST.
Example:
1. Straight lines in space which are parallel to a given straight line are parallel to one another.
Let AB and CD be two straight lines each of which is parallel to the given line LM. We are to prove that the straight lines AB and CD are parallel to each other.
Construction: Draw a plane PQR perpendicular to LM and let us assume that the drawn plane cuts LM, AB and CD at P, Q and R respectively.
Proof: By hypothesis AB is parallel to LM and by construction LM is perpendicular to the plane PQR. Therefore, AB is also perpendicular to the plane PQR. Similarly, CD is also perpendicular to the same plane. Thus, each of AB and CD is perpendicular to the same plane PQR. Therefore, the straight lines AB and CD are parallel to one another.
2. Prove that the quadrilateral formed by joining the middle points of the adjacent sides of a skew quadrilateral is a co-planar parallelogram.
Let W, X, Y and Z be the mid-points of the sides AB, BC, CD and DA of a skew quadrilateral ABCD. We are to prove that, the quadrilateral WXYZ is a co-planar parallelogram.
Construction: Join WX, XY, YZ, WZ and BD.
Proof: Wand Z are the mid-points of the sides AB and AD respectively in the plane △ ABD. Therefore, ZW is parallel to BD and ZW = 1/2 BD. Similarly, X and Y are the mid-points of the sides BC and CD respectively in the plane △ BCD. Therefore, XY is parallel to BD and XY = 1/2 BD. Since both of ZW and XY are parallel to BD, hence they are parallel to one another. Therefore, there is a plane passing through ZW and YX.
Similarly, WX and ZY are parallel to each other and therefore, there is a plane passing through WX and ZY. Both the planes through ZW and YX and through WX and ZY pass through four points W, X, Y and Z. Therefore, it is evident that the two planes must be the same. Hence, the quadrilateral WXYZ is co-planar. Again, ZW is parallel to YX and ZW = YX. Therefore, the quadrilateral WXYZ is a parallelogram.
Geometry
| 679
| 2,845
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.65625
| 5
|
CC-MAIN-2018-39
|
latest
|
en
| 0.930879
|
http://crypto.stackexchange.com/questions/301/are-common-cryptographic-hashes-bijective-when-hashing-a-single-block-of-the-sam/304
| 1,462,568,224,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-18/segments/1461862134822.89/warc/CC-MAIN-20160428164854-00152-ip-10-239-7-51.ec2.internal.warc.gz
| 64,315,672
| 19,986
|
# Are common cryptographic hashes bijective when hashing a single block of the same size as the output?
It's been said that CRC-64 is bijective for a 64-bit block.
It the corresponding statement true for typical cryptographic hashes, like MD5, SHA-1, SHA-2 or SHA-3?
For example, would SHA-512 be bijective when hashing a single 512 bit block?
-
– CodesInChaos Nov 10 '15 at 13:09
It would be very freakish if it turned out to be true. It is not an expected property of SHA-512 to have such bijectivity. It would be worrisome, even, because that's a kind of structure that should not appear in a proper cryptographic hash function.
Actually proving that SHA-512, for 512-bit blocks, is not bijective, would already be a kind of a problem. We do not expect to be able to prove such things without breaking the function.
One "simple" way to prove this would be a single collision (on short inputs), which in theory could be found by chance. But for finding such, we expect to have to calculate about $2^{256}$ hashes (and store/compare them to the other values) to have a non-neglible probability to find a collision.
For example, if I have one zettabyte of fast accessible storage (which would be more than half of humanity's currently stored data), I can store about $2^{62}$ SHA-512 hashes. The probability that between these is at least one duplicate would be about $2^{-389} \approx 10^{-117}$. If every human (around $2^{33}$ in some years) repeats this experiment about once a week (i.e. $2^6$ times a year) with $2^{62}$ new hashes, humanity each year has a chance of $2^{-351}$ of finding a collision. Assuming that humanity will work on this for 10 times as long as the universe already existed (i.e. 130 billions of years), we get a chance of $2^{-314} \approx 10^{-94}$. For comparison, the probability that a ticket wins the main prize in the German weekly lottery (6/49) is around $2^{-27}$, so the probability that humanity will ever find a collision (in the scenario outlined above) is lower than the probability of me winning the main prize each week, for 11 weeks in sequence (with one ticket per week).
So we can expect collisions to stay hidden until the end of times.
-
Assuming processing power doesn't skyrocket... – hexafraction Jun 5 '12 at 20:00
Not even then. I won't detail the math in a comment, but if we were to construct the most energy-efficient computer theoretically possible, it would require all of the output from a supernova in order to cycle a 219-bit counter. And that's an almost imperceptible fraction of the energy necessary to run a counter through 256 bits. – Stephen Touset Feb 28 '13 at 18:02
@StephenTouset And even then, SHA-512 is more expensive to compute than iterating a counter, so you can add a few more bits of required computational work to that. – Thomas Mar 1 '13 at 10:00
It still astonishes me to consider that given the hundreds of zetabytes of data our species has ever stored digitally, we haven't even made perceptible progress in representing all the values capable of being stored in 32 bytes of the tiniest chip of RAM ever created. – Stephen Touset Mar 1 '13 at 18:33
No. Cryptographic hash functions model a random function, not a random permutation. A significant fraction of output hash values are expected to be unreachable and another fraction have multiple preimages.
While bijectivity in general does not mean that the inverse is easy to calculate, for the types of constructs which are used in hash functions in practice, if it were bijective, it could be easily inverted and would thus not make a very good hash function.
There are other known bijective (candidates for) one-way functions, like the ones used for asymmetrical cryptography, but these constructs tend to be a lot slower, and are different from the ones used in hash functions.
-
| 903
| 3,834
|
{"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-2016-18
|
longest
|
en
| 0.940296
|
https://math.stackexchange.com/questions/2546498/equivalence-induced-matrix-norms
| 1,718,736,423,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00067.warc.gz
| 327,608,872
| 36,455
|
# Equivalence induced matrix norms!
The equivalence of vector norms on finite dimensional spaces immediately implies that all induced matrix norms are equivalent. Theorem 1, here, gives some of the popular induced norms. I am interested in the $\|A\|_{2,1}$ and $\|A\|_{2,2}$ norm, where \begin{equation*} \|A\|_{2,1}=\sup\{\|A\mathbf{x}\|_1:\|\mathbf{x}\|_2=1\} \end{equation*} and \begin{equation*} \|A\|_{2,2}=\sup\{\|A\mathbf{x}\|_2:\|\mathbf{x}\|_2=1\} \end{equation*}
According to my analysis so far; $\|A\|_{2,1}$ upper-bounds $\|A\|_{2,2}$. Can someone verify?
For a matrix $A\in\mathbb{R}^{n\times m}$
$\|A\|_{2,1}=\max_{u\in\{1,-1\}^n}\|A^Tu\|_2\leq \sqrt{\sum_{i=1}^m(\sum_{j=1}^n|a_{ij}|)^2}$.
And $\|A\|_{2,2}=\lambda_{\max}(A^TA)=\sigma_{\max}(A)\leq\|A\|_F= \sqrt{\sum_{i=1}^m\sum_{j=1}^n|a_{ij}|^2}$,
$\implies$ $$\|A\|_{2,2}^2\leq{\sum_{i=1}^m\sum_{j=1}^n|a_{ij}|^2}\leq{\sum_{i=1}^m\left(\sum_{j=1}^n|a_{ij}|\right)^2}$$.
Your inequalities all go in the same direction, so you cannot possibly prove inequalities both ways.
You have the easy inequalities $$\|x\|_2\leq\|x\|_1\leq\sqrt n\,\|x\|_2.$$ So you immediately get (from the definition, not the characterization you quote) $$\tag{1}\bbox[5px,border:2px solid green]{\|A\|_{2,2}\leq\|A\|_{2,1}\leq\sqrt n\|A\|_{2,2}.}$$ These inequalities are sharp. For instance, if $A=E_{11}$, then $$\|Ax\|_1=\|(x_1,0,\ldots,0)\|_1=|x_1|=\|x\|_2.$$ So $\|A\|_{2,2}=\|A\|_{1,1}$. And if $A=I_m$, then $\|Ax\|_1=\|x\|_1$, so $$\|A\|_{2,1}=\max\{\|x\|_1:\ \|x\|_2=1\}=\sqrt n,$$ while $\|A\|_{2,2}=1$. So $\|A\|_{2,1}=\sqrt n\|A\|_{2,2}$.
Edit: here is a short proof of the inequalities $(1)$, in case they are not obvious.
You have, for any nonzero $x$. $$\frac{\|Ax\|_2}{\|x\|_2}\leq\frac{\|Ax\|_1}{\|x\|_2}\leq\|A\|_{2,1}.$$ Now you take supremum (forget about the term on the middle) and you obtain $$\|A\|_{2,2}\leq\|A\|_{1,1}.$$ Similarly, start with $$\frac{\|Ax\|_1}{\|x\|_2}\leq\sqrt{n}\,\frac{\|Ax\|_2}{\|x\|_2}\leq\sqrt{n}\,\|A\|_{2,2}.$$ Now forget the middle term and take supremum, to obtain $$\|A\|_{2,1}\leq\sqrt{n}\,\|A\|_{2,2}.$$
• what do you mean by characterization?
– amj
Commented Dec 1, 2017 at 23:42
• The definition is \begin{equation*} \|A\|_{2,1}=\sup\{\|A\mathbf{x}\|_1:\|\mathbf{x}\|_2=1\}. \end{equation*} You also have the characterization $$\|A\|_{2,1}=\max_{u\in\{1,-1\}^n}\|A^Tu\|_2.$$ Commented Dec 2, 2017 at 0:08
• Regarding your answer, we have to take supremum over $\|A\mathbf{x}\|$ such that $\|\mathbf{x}\|_2=1$, but there could be alot of $\mathbf{x}$ with unit $\ell_2$ norm and some $\mathbf{x}$ that maximizes $\|A\mathbf{x}\|_{2}$ might not maximize $\|A\mathbf{x}\|_{1}$.
– amj
Commented Dec 3, 2017 at 22:28
• Not sure how you think that affects my answer. Commented Dec 3, 2017 at 22:36
• Let $\mathbf{x}'$ and $\mathbf{x}''$ be such that $\|.\|_2=1$ for both of them. Now, let $\sup\|A\mathbf{x}\|_1=\|A\mathbf{x}'\|_1$ and $\sup\|A\mathbf{x}\|_2=\|A\mathbf{x}''\|_2$, then your inequalities might not hold!
– amj
Commented Dec 3, 2017 at 22:42
| 1,366
| 3,063
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 1, "x-ck12": 0, "texerror": 0}
| 3.4375
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.407766
|
http://mathhelpforum.com/trigonometry/53384-triangle-trigonometry.html
| 1,481,399,029,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698543434.57/warc/CC-MAIN-20161202170903-00332-ip-10-31-129-80.ec2.internal.warc.gz
| 184,946,180
| 9,553
|
1. ## Triangle trigonometry
- A triangular region has sides measuring 121, 173, and 194 meters. Find the measure of the largest angle in the region.
How should I do this?
2. Originally Posted by PandaPanda
- A triangular region has sides measuring 121, 173, and 194 meters. Find the measure of the largest angle in the region.
How should I do this?
The largest angle is the one that is opposite the largest or longest side....here, the angle that is the opposite of 194.
Say, call that the angle C.
So, using the Law of Cosines,
(194)^2 = (121)^2 +(173)^3 -2(121)(173)cos(C)
| 155
| 579
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.4375
| 3
|
CC-MAIN-2016-50
|
longest
|
en
| 0.898149
|
https://datapott.com/category/spss/
| 1,719,260,153,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00036.warc.gz
| 173,545,608
| 20,072
|
## Factor Analysis vs Path Analysis – The Difference
Factor analysis and path analysis are both statistical techniques used in the field of multivariate analysis, particularly in the context of structural equation modeling (SEM). While they share some similarities, they serve different purposes and have distinct methodologies. Factor Analysis: Purpose: Factor analysis is primarily used to identify underlying factors or latent variables that explain […]
## How to do Mediation Analysis Using Multiple Regression Analysis
How to Testing Mediation Analysis with Regression Analysis Mediation analysis is a hypothesized causal chain in which one variable affects a second variable that, in turn,affects a third variable. The intervening variable, M, is the mediator. It “mediates” the relationship between a predictor, X, and an outcome. Graphically, mediation can be depicted in the following […]
## Linear vs. Logistic Probability Models: Which is Better, and When?
Interpretability Let’s start by comparing the two models explicitly. If the outcome Y is a dichotomy with values 1 and 0, define p = E(Y|X), which is just the probability that Y is 1, given some value of the regressors X. Then the linear and logistic probability models are: p = a0 + a1X1 + a2X2 + … + akXk (linear) ln[p/(1-p)] = b0 + b1X1 + b2X2 + … + bkXk (logistic) […]
## The Difference Between the Bernoulli and Binomial Distributions
You might already be familiar with the binomial distribution. It describes the scenario where the result of an observation is binary—it can be one of two outcomes. You might label the outcomes as “success” and “failure” (or not!). Or, if you want to get mathematical about it, you might label them “1” and “0.” You […]
## Logistic Regression Analysis: Understanding Odds and Probability
Probability and odds measure the same thing: the likelihood or propensity or possibility of a specific outcome. People use the terms odds and probability interchangeably in casual usage, but that is unfortunate. It just creates confusion because they are not equivalent. How Odds and Probability Differ They measure the same thing on different scales. Imagine how confusing it would be […]
## How to Conduct Probit and Logit Models (Binary Outcome Models)
Probit and Logit Models (Binary Outcome Models) Do you want to understand the factors that influence binary outcomes? Then you’ve come to the right place. In this article, we’ll delve into the world of Probit and Logit models, which are commonly used in statistical analysis to predict binary outcomes. Whether you’re a researcher, […]
## Logistic Regression Analysis: Understanding Odds and Probability
Probability and odds measure the same thing: the likelihood or propensity or possibility of a specific outcome. People use the terms odds and probability interchangeably in casual usage, but that is unfortunate. It just creates confusion because they are not equivalent. How Odds and Probability Differ They measure the same thing on different scales. Imagine how confusing it would be […]
## The Difference between Logistic and Probit Regression
Both are types of generalized linear models. This means they have this form: Both can be used for modeling the relationship between one or more numerical or categorical predictor variables and a categorical outcome. Both have versions for binary, ordinal, or multinomial categorical outcomes. And each of these requires specific coding of the outcome. For example, in both logistic and […]
## Retention methods in exploratory factor analysis (EFA)
Retention methods in exploratory factor analysis (EFA) involve determining how many factors to retain from the initial factor extraction. The goal is to identify a meaningful and interpretable number of factors that adequately represent the underlying structure of the observed variables. Several common retention methods are used for this purpose: Kaiser’s Criterion: Proposed by Kaiser, […]
## How to perform Heteroscedasticity test in STATA for time series data
Heteroskedastic means “differing variance” which comes from the Greek word “hetero” (‘different’) and “skedasis” (‘dispersion’). It refers to the variance of the error terms in a regression model in an independent variable. If heteroscedasticity is present in the data, the variance differs across the values of the explanatory variables and violates the assumption. This will […]
Need Help, Whatsapp Us Now
| 936
| 4,455
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.796875
| 4
|
CC-MAIN-2024-26
|
latest
|
en
| 0.847445
|
https://activerain.com/blogsview/3393200/why-investors-and-lenders-need-the-break-even-ratio
| 1,723,462,483,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00036.warc.gz
| 61,962,839
| 14,715
|
# Why Investors and Lenders Need the Break-Even Ratio
Services for Real Estate Pros with Loretta A. Steele Realty
Why should you be concerned about the break-even ratio of an income producing property? If you know that the cash flow coming in meets or exceeds the cash flow going out, what else should you know? Ratios, percentages and math, are they really necessary? Which answer would you give; yes and no?
• Debt was not attached to the property when purchased, you paid all cash.
• Vacancy rates are not a concern; all your tenants are long-time tenants who pay in full and on time.
• Your operating expenses never increase and you never have any unexpected expenses.
• You have never heard of negative cash flow, to you it is a foreign concept.
If your answer was no, then you have complete control over interior and exterior forces involving your property. The economy, neighborhood vacancy rates and even unexpected repairs and maintenance do not affect you.
A no answer does not sound realistic it involves controlling too many variables. I think the real answer is yes. Loan or mortgages are needed when purchasing most rent producing properties. Vacancy rates fluctuate over time. Repairs, maintenance and other expenses are part of property ownership. And never having negative cash flow is fantasy, not reality.
What makes a break-even ratio so important? The values involved are the loans, expenses and the income produced. The ratio represents the percentage of money going out to money coming in. The exact formula is:
Break-Even Ratio = (Debt Service + Operating Expenses) / Gross Operating Income
Or another way of stating it would be;
Break-Even Ratio = Money Going Out / Money Coming In
Multiply by 100, gives you the percentage of income that must decline before cash flow would break even with the loan payment. That is something you would want to keep your eye on. Guess who else will keep an eye on this ratio? Your lender uses this ratio, along with many other ratios, when determining their risk in financing your investment.
Example: You have \$400,000 in money going out, loans plus operating expenses, income is \$575,000. Break-Even Ratio = \$400,000 / \$575,000 = 0.8421 or 84%.
Most lenders require a ratio of 85% or less. The money going out represents 85% of the money coming in. An investment property where expenses dominate income will not gain the approval of most lenders.
Now, let's turn this concept around, subtract 85% from 100% the result is 15%. The 15% represents what you would have left after paying all expenses. Subtracting the break-even ratio from 100% signifies a simple means of calculating a return on an investment.
Loretta A. Steele is the developer of AgentApt Analysis and APOD Extra, two rental property analysis products that will produce automatic calculations and marketing presentations, quickly and easily. See for yourself at: www.LincolnSteele.com
| 621
| 2,931
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33
|
latest
|
en
| 0.953899
|
https://physics.stackexchange.com/questions/45721/why-is-temperature-constant-during-melting
| 1,709,342,650,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00504.warc.gz
| 463,925,513
| 41,730
|
# Why is temperature constant during melting?
This is an elementary question but I do not know the answer to it. During a phase transition such as melting a solid to a liquid the temperature remains constant. At any lower temperature the heat provided went to kinetic energy and intermolecular potential energy. Why is it that at the melting point no energy goes into kinetic (that would increase the temperature)?
• Because the substance uses that energy for the phase transition. Dec 2, 2012 at 18:43
• Apr 7, 2020 at 5:47
Imagine a container containing just ice at $$-1^\circ \rm C$$. When you heat it, the energy goes into kinetic motion of the molecules, and its temperature increases. Similarly, if the container is filled with liquid water at $$1^\circ \rm C$$ its temperature will increase for the same reason.
But now imagine the container is filled with 90% ice and 10% water at $$0^\circ \rm C$$. If you heat the water part up, it's temperature will temporarily increase a little. But now the water is hotter than the ice, so heat will be transferred from the water to the ice. When the ice is heated above $$0^\circ \rm C$$ it melts, and this uses up some energy, cooling the water. This will continue until the ice and the water are the same temperature again, so you'll end up back at $$0^\circ \rm C$$, but with a higher proportion of liquid water and less ice.
This is why, if you heat a mixture of the two phases slowly enough, all the energy will go into melting the solid rather than increasing the temperature. It continues until all the solid has melted, which is when the temperature starts increasing again. The same thing happens in reverse if you decrease the temperature.
• I realise TMS already made all these points, but I felt it could be clearer. Dec 3, 2012 at 5:25
• Thanks very much Nathaniel, it is much clearer and simpler now. I now also understand what TMS was saying! I thank you both. Dec 3, 2012 at 6:30
• How does the ice melting cool the water? That is, why is the melting ice taking energy from the surrounding water and not the heat that is being actively added into the system? Nov 18, 2015 at 0:12
• @user3932000 I guess that can happen - depending on how you heat the system, some of the heat will go into the ice rather than the liquid water part. That portion of the incoming energy will not increase the temperature either, simply because it goes directly into melting the ice. Nov 18, 2015 at 2:09
Roughly speaking, this additional energy will be at first step kinetic, it will increase molecule bouncing around there equilibrium points, until it will be enough to take them out of that equilibrium, and then this energy spent to make the phase transition.
More precisely your confusion is because of the fact that the statement you mentioned "temperature will not change.." is right in quasi-statistical equilibrium processes, that is it's right when looking on the process from macro point of view and on a long term, but locally on the particle level and for a small period of time situation is seems different because you didn't reached yet the equilibrium state of your system yet.
• In general, it will not be true for mixtures either. Dec 2, 2012 at 20:21
• @Bernhard: can you please be more precise?
– TMS
Dec 2, 2012 at 20:24
• The transition from liquid to solid will follow a trajectory, so the temperature of completely solid will be different from completely liquid. Dec 2, 2012 at 20:57
• Ok, and how this happens to contradict with what I mentioned?
– TMS
Dec 2, 2012 at 21:02
• @Andreas: as I described, this actually happens locally for some tiny time, but from macroscopic point of view and after enough time temperature will be equilibration and become the same across the whole object, ant will stay the same because this energy was spent on taking molecules out of there equilibrium positions.
– TMS
Dec 2, 2012 at 21:55
| 912
| 3,901
|
{"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.943181
|
https://m.ebrary.net/29352/computer_science/uniformity
| 1,580,122,241,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00375.warc.gz
| 522,136,241
| 5,125
|
Home Computer Science Hardware Security and Trust: Design and Deployment of Integrated Circuits in a Threatened Environment
# Uniformity
A PUF is expected to generate responses containing ideally the same number of logic-0s and logic-1s. Therefore, the uniformity metric (also called randomness by Yu et al. in [47]) can be exploited to estimate the distribution of logic-0 and logic-1 in PUF responses. Let N be the number of response bits, the percentage measure for uniformity of response r = (г,0, гц, ... ritN_1) can be defined as
A value of 100% means that all r response bits are logic-1. For true random bits, uniformity should be as close as possible to its ideal value of 50 %. Let R be the number of responses, resulting from the product between the amount of different
PUF instances and input challenges (if any). The average uniformity for a population or R devices can be calculated as
i
# Bit Aliasing
The uniformity metric is not enough to qualify the randomness of PUFs responses. Indeed, even with a best value of bit uniformity, some homologous bits could turn out to be biased among the responses of the PUF population. This could happen whenever the manufacturing process introduces static variations which compromise all the homologous bits in the responses, causing a fixed preferred value. To this aim, we can compute the bit aliasing [30] (also called bias in [47]) as
If some homologous bits are biased, the bit-aliasing results in a value far from 50 %.
Found a mistake? Please highlight the word and press Shift + Enter
Related topics
| 356
| 1,572
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05
|
longest
|
en
| 0.913639
|
https://studysoup.com/tsg/198994/physics-for-scientists-engineers-with-modern-physics-4-edition-chapter-24-problem-24-91
| 1,611,295,500,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703529128.47/warc/CC-MAIN-20210122051338-20210122081338-00632.warc.gz
| 565,946,195
| 16,779
|
×
Get Full Access to Physics For Scientists & Engineers With Modern Physics - 4 Edition - Chapter 24 - Problem 24.91
Get Full Access to Physics For Scientists & Engineers With Modern Physics - 4 Edition - Chapter 24 - Problem 24.91
×
# A parallel-plate capacitor has plate area A, plate
ISBN: 9780131495081 132
## Solution for problem 24.91 Chapter 24
Physics for Scientists & Engineers with Modern Physics | 4th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Physics for Scientists & Engineers with Modern Physics | 4th Edition
4 5 1 301 Reviews
20
0
Problem 24.91
A parallel-plate capacitor has plate area A, plate separation x,and has a charge Q stored on its plates (Fig. 24-39). Findthe amount of work required to double the plate separationto 2x, assuming the charge remains constant at Q. Show thatyour answer is consistent with the change in energy storedby the capacitor. (Hint: See Example 24-10.)a + e A\ +8 \FIGURE 24-39 * j 91. ~Q ~Q
Step-by-Step Solution:
Step 1 of 3
Chapter 10 Notes Main things that affect temperature on planets: sunlight, reflection (albedo), rotation, and absorption of infrared light. Majority of gas in the atmosphere is nitrogen. Greenhouse effect has to do with the atmospheric absorption of infrared radiation. A greenhouse gas is one that effectively bounces around molecules coming from the Earth and keeping the heat from escaping the surface. Greenhouse gases include: Water (H20) and Carbon dioxide (CO2) Nitrogen is transparent to infrared making it not a greenhouse gas. Oxygen falls under this category as well. Our atmosphere is transparent to visible light. Without the greenhouse effect, the surface temperature on Earth would be below the freezing temperature of water. Methane (CH4) is another greenhouse gas; much more effective than the others Global winds blow in distinctive patters: Equatorial East to West, Midlatitudes West to East, The Coriolis Effect when you are rotating, from your point of view, things will not appear to be moving in a straight line. Coriolis Effect deflects northsouth winds into eastwest winds. Deflection breaks each of the two large “no rotation” cells into three smaller cells. Sources of gas: outgassing from volcanoes, evaporation of surface liquid; sublimation of surface ice, and impacts of particles and photons. Escape velocity from Earth is 11.3 km/s In a hotter atmosphere there will be a better chance for molecules to break free and escape the atmosphere of the planet, meaning that planet will loose much more of its atmosphere. The Earth’s escape velocity is lower than the speed of hydrogen gas. Earth’s atmosphere isn’t mostly hydrogen because hydrogen is a light gas and moves faster than heavier gasses allowing it to escape the Earth’s atmosphere Characteristics of Venus runaway greenhouse effect, sulfuric acid clouds, and almost no surface winds. Characteristics of Earth atmosphere composed of mostly nitrogen and has an ultraviolet absorbing atmosphere. Characteristics of Mars extremely low density atmosphere and global dust storms. • Effects of atmospheres They create pressure that determines whether liquid water can exist on surface. They absorb and scatter light. They create wind, weather, and climate. They interact with the solar wind to create a magnetosphere. They can make planetary surfaces warmer through the greenhouse effect.
Step 2 of 3
Step 3 of 3
##### ISBN: 9780131495081
This full solution covers the following key subjects: plate, separation, stored, capacitor, Charge. This expansive textbook survival guide covers 44 chapters, and 3904 solutions. The full step-by-step solution to problem: 24.91 from chapter: 24 was answered by , our top Physics solution expert on 11/10/17, 05:57PM. The answer to “A parallel-plate capacitor has plate area A, plate separation x,and has a charge Q stored on its plates (Fig. 24-39). Findthe amount of work required to double the plate separationto 2x, assuming the charge remains constant at Q. Show thatyour answer is consistent with the change in energy storedby the capacitor. (Hint: See Example 24-10.)a + e A\ +8 \FIGURE 24-39 * j 91. ~Q ~Q” is broken down into a number of easy to follow steps, and 66 words. Physics for Scientists & Engineers with Modern Physics was written by and is associated to the ISBN: 9780131495081. Since the solution to 24.91 from 24 chapter was answered, more than 314 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Physics for Scientists & Engineers with Modern Physics, edition: 4.
#### Related chapters
Unlock Textbook Solution
| 1,113
| 4,780
|
{"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}
| 3.03125
| 3
|
CC-MAIN-2021-04
|
latest
|
en
| 0.83006
|
https://qubit.guide/10.12-remarks-and-exercises-algorithms.html
| 1,720,945,141,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00350.warc.gz
| 407,446,455
| 14,345
|
## 10.12Remarks and exercises
### 10.12.5 Picking out a single state
Prove that, for any y\in\{0,1\}^n, \sum_{x\in\{0,1\}^n} (-1)^{x\cdot y} = \begin{cases} 0 &\text{if $y\neq0$;} \\2^n &\text{if $y=0$.} \end{cases}
### 10.12.6 Writing an integer as a power
1. Show that R=21 cannot be written in the form a^b for integers a\geqslant 1 and b\geqslant 2.
2. Generalise this to a method that could work in \mathcal{O}(L^3) for any value of R that is L bits long.236
1. Hint: since R is L bits long, R<2^L, and so b<L.↩︎
| 219
| 522
|
{"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}
| 2.625
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.706555
|
http://forums.wolfram.com/mathgroup/archive/2006/Aug/msg00282.html
| 1,529,598,256,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00199.warc.gz
| 114,446,134
| 8,025
|
Re: need mathematica's help for exploring a certain type of mapping
• To: mathgroup at smc.vnet.net
• Subject: [mg68582] Re: need mathematica's help for exploring a certain type of mapping
• From: "Nabeel Butt" <nabeel.butt at gmail.com>
• Date: Wed, 9 Aug 2006 23:57:56 -0400 (EDT)
• References: <200608090820.EAA21373@smc.vnet.net> <NDBBJGNHKLMPLILOIPPOMELCFBAA.djmp@earthlink.net>
• Sender: owner-wri-mathgroup at wolfram.com
```David,
Thanks for the reply.You seem to have very good ideas.
The mapping does not neccessarily map xy-plane to a 3D-surface in
R^3.It is also not neccessarily continous.*ALL *what is specified of it is
that given two points which are at a unit distance in R^2 they are mapped on
to points in R^3 that are also at a unit distance.What we need to show is
that given there is a mapping that preserves unit distance does it preserve
all distances?.If we are able to demonstrate a map that preserves unit
distances but not all distances we have actually disproven a very important
hypothesis in mathematics.And your ability to create graphics will be
useful.
Also there may be many possible unit distance preserving maps.
Look forward for your views.
regards,
Nabeel
On 8/9/06, David Park <djmp at earthlink.net> wrote:
>
> Nabeel,
>
> I thought that the definition of an isometry was that it preserved
> distances.
>
> A mapping that did a 3D rotation of the xy-plane, plus translations in 3D
> space, plus reflections in a plane would cover the isometries if we use
> the
> Euclidean metric in both 2D and 3D.
>
> But suppose you wanted to map onto a specific curved surface. Then are you
> going to allow a different distance measuring function on the surface, a
> different metric? Are we allowed to design a metric for any given mapping?
> Is it possible to have an isometry then? I don't know.
>
> David Park
> djmp at earthlink.net
>
> From: Nabeel Butt [mailto:nabeel.butt at gmail.com]
To: mathgroup at smc.vnet.net
>
>
> Dear Users,
> I need to use mathematica's graphics to explore a certain
> kind of problem.The following theorem is not yet proven nor disproven and
> mathematica might proof useful in disproving it though.
> Hypothesis:If a mapping from R^2->R^3 is unit distance preserving then it
> must be an isometry.
> The real issue at hand is for mathematica to generate a mapping that
> preserves unit distance but is not an isometry so in the process
> disproving
> the theorem.
> The real problem is that R^2 consists of infinite points and it might
> not be possible to check all of them.What i suggest is that you apply the
> unit preserving maps to special type of figures in R^2 like the
> circumfrence
> of the circle,square,isoceles triangle etc.
> Any ideas are welcome.Thanks in advance.
> regards,
> Nabeel
>
> --
> Nabeel Butt
> LUMS,Lahore
>
>
>
--
Nabeel Butt
LUMS,Lahore
```
• Prev by Date: Re: How do I create a parametric expression?
• Next by Date: Re: How do I create a parametric expression?
• Previous by thread: Re: Re: need mathematica's help for exploring a certain type of mapping
• Next by thread: RE: need mathematica's help for exploring a certain type of mapping
| 861
| 3,193
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.875
| 3
|
CC-MAIN-2018-26
|
latest
|
en
| 0.896515
|
http://qpr.ca/blogs/2007/08/
| 1,596,739,530,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439737019.4/warc/CC-MAIN-20200806180859-20200806210859-00104.warc.gz
| 89,331,878
| 11,522
|
## Archive for August, 2007
### Probability of Occurring by Chance
Tuesday, August 14th, 2007
In this post at squareCircleZ, Professor Bruce Armstrong from the Sydney Cancer Centre at the University of Sydney is quoted as saying “The probability the that increase is due simply to chance is about one in a million so we are looking at something that is almost certainly a real increase in risk”. But this is almost certainly a misstatement since the probability that something is due simply to chance is not computable and probably not even meaningful whereas the probability of its happening in a randomly chosen situation from a well defined population of cases is meaningful and often computable. Either concept could be expressed by the ambiguous title of this post but they are definitely NOT the same – as can be seen from the following example. If I win the lottery without cheating then the probability of it having happened by chance (in the sense of having only chance factors involved) is in fact 1 but the probability of it happening by chance (ie of it happening given that only chance factors were involved) was less than one in a million. Of course, if we don’t assume that I didn’t cheat, the probability that my win was due only to chance may be less, but in any event it is not the same as my chances of winning a fair game. For a more practically relevant example consider the case of an experiment which identifies an effect of some sort “at the 95% confidence level”. What this means is that the probability of the observation occurring if only random effects were present is no more than about 1 in 20. But then in a set of many trials it is likely that up to about 5% of them will actually appear to show the effect. Users of statistics need to be aware of this distinction since in an experiment which collects more than six variables (as many in the social sciences do) there are more than 21 pairs to consider and so in an average such experiment at least one such pair will seem to have a significant relationship even when no such relationship actually exists.
All this is actually relevant to the story about cancer clusters since, in a world with several million observed groups of a hundred or so people, if the chance of a cluster happening given only random factors is one in a million then we may expect to see several such clusters occurring just by chance.
### Leaning Tower Illusion
Sunday, August 12th, 2007
My friend Gerry Pareja forwarded a link to this story from ‘Improbable Research’ about the first prize winner in the Neural Correlate Society’s 2007 Illusion of the Year contest.
The image certainly is pretty cool. But to test the explanation I tried covering each image in turn and the effect was still there! I wondered if there was a perceptual delay effect in that our memory of one picture affects our interpretation of the other but then I also noticed that the effect changes depending on where the picture is in our field of view. If I position myself facing the right hand edge of the monitor, then the right hand tower seems more vertical and the left one almost seems to lean left or backwards. So the illusion may be more (or at least partly) due to the fact that in the absense of other visual cues in the picture we tend to interpret as if viewed from the direction at which we are looking at the picture – ie we interpret the picture edge as a window frame.
### Basskin Misrepresents Media Levy
Friday, August 10th, 2007
David Basskin, Director of the CanadianPrivate Copying Collective, has written a letter to the Star in response to Michael Geist’s earlier comments about the media levy and its prospects of its being extended to mp3 players and perhaps even computer hard drives.
Basskin says “The private copying levy is an important source of revenue for music-rights holders and is an issue of fairness”, and goes on to claim that “The levy is often misunderstood. It is not a tax or subsidy, but payment to individual rights holders for use of music by Canadians in the privacy of their homes. Compensation for use – that’s fair”.
But of course the levy is NOT fair because it extracts funds from people who purchase data storage media for purposes having nothing to do with the products of Basskins’ clients and so in fact it IS a tax and subsidy.
What is particularly frustrating about this is the fact that there is a perfect opportunity for the sellers of music recordings to obtain compensation for whatever pattern of reproduction they expect from their customers, and that is at the point of sale. Extracting that compensation later from others who may have no interest in the product just amounts to artificially lowering the price. This may be good for sales and it may even be in the public interest to subsidize artistic products in some way, but this particular approach to gaining that subsidy is fundamentally dishonest and the legislators who were persuaded to go along with it have been conned and bamboozled.
One other aspect of this situation that is worth mentioning is that many who object to the levy might still be willing to entertain a more honestly defined subsidy – paid from honestly defined taxes. But taxes should be collected by the government not by private organizations.
Friday, August 3rd, 2007
The peace and wellbeing of society is harmed by laws whose perceived unfairness or unenforceability reduces public respect for the law.
In a large body of legislation it is hard to avoid parts which will appear unfair to some, but it is foolish to unnecessarily enact provisions whose manifest unfairness will be frequently imposed on large segments of the population.
A case in point is the media levy in Canadian copyright law. (more…)
### Media Company Thieves Coming Back For More
Thursday, August 2nd, 2007
In TheStar.com – entertainment – Tax on MP3 players may return, David Basskin, director of the Canadian Private Copying Collective, is quoted as saying:
“When you have bought a CD, you don’t own the music – you own the copy of the music that you’ve purchased,” he says. “You’re not buying the rights to make copies.”
“People are going to do private copying, and the levy legitimizes it and provides compensation to the people who created the music. It’s pretty hard to say why that’s not fair,”
No its bloody well not hard to see at all!!!
{Warning:obscenities coming} (more…)
### Big Guns Join Fight Against Illegal Copyright Notices
Wednesday, August 1st, 2007
Michael Geist brings some good news in – Complaint Filed With FTC Over Copyright Notices. But it is a bit sad that it is only with the support of big businesses that there is any hope of defending public access rights. If there were no major segment of the economy interested in protecting these rights would they just disappear?
### AlterNet: Environment: Exposed: The Truth Behind Popular Carbon Offsetting Schemes
Wednesday, August 1st, 2007
| 1,427
| 6,948
|
{"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}
| 3.078125
| 3
|
CC-MAIN-2020-34
|
latest
|
en
| 0.979098
|
mosthealthytips.tk
| 1,539,779,616,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583511173.7/warc/CC-MAIN-20181017111301-20181017132801-00009.warc.gz
| 234,474,449
| 13,030
|
# Easy methods to Win the Three Digit Lottery
The odds of successful the three digit lottery are the great of all the lottery video games – you have a one in a thousand chance of successful the largest prize, in comparison with the choose 4 lottery which has odds of ten thousand to one. The obstacle is the right way to win the three digit lottery or the choose 4 lottery by using identifying successful numbers. This text will describe several approaches that you may generate profitable numbers that may beef up your odds of successful these lottery video games.
At the same time humans in most cases bet numbers which can be big to them, such as their house quantity or social protection number, this technique is really no longer the first-class means of deciding on numbers to guess within the lottery except you might be very, very fortunate. Since the winning numbers are drawn randomly, one technique is to also decide on your numbers randomly. There are several ways you can do this. A method, of path, is to let the lottery terminal pick your numbers for you.
on the other hand, many men and women also suppose that the numbers drawn throughout three digit lottery draws aren’t fairly random in any respect, and some numbers come out extra regularly than others. The trick to methods to win the three digit lottery or the pick 4 lottery, for this reason, is to identify these numbers by making use of information.
again, there are a number of approaches of doing this. A technique is to get the earlier winning mixtures for a number of attracts after which count which numbers come out most most often. That you may then use these numbers to develop your odds of successful by means of producing more than a few mixtures you could bet. One good thing about the three digit lottery and the decide on four lottery is that there are a quantity of ways to win although you don’t wager the unique combo of numbers that came out within the draw or even the entire numbers that came out. For illustration, you can win a lesser prize if the combination you guess has the entire numbers that were drawn, despite the fact that you didn’t decide upon them within the successful order. That you can additionally win in case you select two of the numbers that show up within the successful blend so long as they are in the right order.
An less complicated way to investigate profitable numbers is to get lottery application to do it for you. This software more often than not comes pre-programmed with a database of previous profitable combos and makes use of these to create a statistical analysis of which numbers are undoubtedly to come back out in future drawings. You even have the choice to prefer your own numbers headquartered for your individual reading of the statistical charts and graphs the software generates. The software will also generate random quantity mixtures when you feel that is the quality procedure.
Whichever procedure you utilize in deciding on how you can win the three digit lottery or the select four lottery, you will have to test it out first by using making paper bets, making a choice on combos and then seeing if they come out in attracts. Once you consider comfortable sufficient with the method you’ve gotten chosen then that you may to make precise bets. And consistently ensure to just bet the amount of money you can afford to lose.
error: Content is protected !!
| 657
| 3,409
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43
|
latest
|
en
| 0.966299
|
https://package.epinowcast.org/articles/distributions.html
| 1,696,388,664,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00262.warc.gz
| 476,751,482
| 5,459
|
This vignette describes the parametric delay distributions that are currently available in epinowcast and explains how they are internally discretised.
## Available distributions
The currently available parametric delay distributions are continuous probability distributions with (up to) two parameters $$\mu_{g,t}$$ and $$\upsilon_{g,t}$$. The table below provides a link to the definition of each distribution, specifies how the parameters $$\mu_{g,t}$$ and $$\upsilon_{g,t}$$ are mapped to the parameters of the distribution (according to the referenced definition), and states the resulting mean of the distribution (before discretization and adjustment for the assumed maximum delay).
Distribution Parametrization Mean
Log-normal $$\mu=\mu_{g,t}$$, $$\sigma = \upsilon_{g,t}$$ $$\exp(\mu_{g,t}+\frac{\upsilon_{g,t}^2}{2})$$
Exponential $$\beta = \exp(-\mu_{g,t})$$ $$\exp(\mu_{g,t})$$
Gamma $$\alpha = \exp(\mu_{g,t})$$, $$\beta = \upsilon_{g,t}$$ $$\exp(\mu_{g,t})/\upsilon_{g,t}$$
Log-logistic $$\alpha = \exp(\mu_{g,t})$$, $$\beta = \upsilon_{g,t}$$ $$\frac{\exp(\mu_{g,t})\,\pi/\upsilon_{g,t}}{\sin(\pi/\upsilon_{g,t})}$$
## Discretisation and adjustment for maximum delay
In epinowcast, delays are modeled in discrete time and with an assumed maximum delay (specified via the max_delay argument). Therefore, the continuous delay distributions must be discretised and adjusted for the maximum delay.
To discretise the continuous probability distribution into discrete probabilities $$p^{\prime}_{g,t,d}$$, we use its cumulative distribution function $$F^{\mu_{g,t}, \upsilon_{g,t}}$$ to compute $p^{\prime}_{g,t,d} = \frac{F^{\mu_{g,t}, \upsilon_{g,t}}(d+1) - F^{\mu_{g,t}, \upsilon_{g,t}}(d)}{F^{\mu_{g,t}, \upsilon_{g,t}}(D)}.$ In words, the discrete probability mass function (pmf) is obtained from the continuous distribution by dividing it into $$D$$ different bins $(0,1],\, (1,2],\, \ldots,\, (D-1,D],$ where $$D$$ is the maximum delay covered by the model. Importantly, the bins are normalized by $$F^{\mu_{g,t}, \upsilon_{g,t}}(D)$$, which ensures that the $$p^{\prime}_{g,t,d}$$ sum to 1. Since $$F^{\mu_{g,t}, \upsilon_{g,t}}(D)$$ is the probability of reporting before the maximum delay, this can also be interpreted as conditioning our distribution on the maximum delay.
Note that because of the discretisation and normalization, the discrete random variable we obtain generally does not have exactly the same moments as the original, continuous distribution.
| 693
| 2,488
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.140625
| 3
|
CC-MAIN-2023-40
|
latest
|
en
| 0.731088
|
https://www.studyadda.com/ncert-solution/11th-chemistry-chemical-bonding-and-molecular-structure_q6/493/32082
| 1,600,662,553,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600400198887.3/warc/CC-MAIN-20200921014923-20200921044923-00423.warc.gz
| 1,086,760,036
| 18,373
|
• question_answer 6) Write Lewis dot symbols for atoms of the following elements and predict their valencies. Mg, Na, B, O, Br
Mg At. No. 12 2, 8, 2 $\underset{{}}{\mathop{\overset{\,\,}{\mathop{_{\bullet }^{{}}Mg\,_{\bullet }^{{}}}}\,}}\,$ valency = 2 Na At. No. 11 2, 8, 1 $\underset{{}}{\mathop{\overset{\,\,}{\mathop{_{{}}^{{}}Na\,_{\bullet }^{{}}}}\,}}\,$ valency = 1 B At. No. 5 2, 3 $\underset{{}}{\mathop{\overset{\bullet \,\,}{\mathop{_{\bullet }^{{}}Na\,_{\bullet }^{{}}}}\,}}\,$ valency = 3 0 At. No. 8 2, 6 $\overset{\bullet \,\,\bullet }{\mathop{_{\bullet }^{\bullet }O\,_{\bullet }^{\bullet }}}\,$ valency = 8 - 6 = 2 Br At. No. 35 2, 8, 18, 7 $\underset{\bullet \,\,\bullet }{\mathop{\overset{\bullet \,\,\bullet }{\mathop{_{\bullet }^{\bullet }Br\,_{\bullet }^{{}}}}\,}}\,$ valency = 8 - 7 = 1
| 357
| 810
|
{"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}
| 2.828125
| 3
|
CC-MAIN-2020-40
|
latest
|
en
| 0.405294
|
http://www.irisa.fr/lagadic/demo/demo-paraplan/demo-paraplan-eng.html
| 1,534,851,888,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-34/segments/1534221218122.85/warc/CC-MAIN-20180821112537-20180821132537-00354.warc.gz
| 508,280,514
| 3,751
|
# Positioning a camera parallel to an unmarked plane
## Description of the demonstration
The aim of this robotic task is to control an eye-in-hand system in order to position the projection plane of a camera (or image plane) parallel to an observed plane (or object plane). This plane is unmarked, which means that no geometric visual features, such as points or straight lines, can be easily extracted. Furthermore, a second task is joined to the first one so that the same point is always observed at the image center, to ensure the scene stays viewable.
These two joined tasks are achieved as following:
• Orientation of the optical axis parallel to the normal of the plane (which is equivalent to have the image and observed planes parallel) is performed by controlling the pan and tilt d.o.f. of the camera. This is done by regulating to zero the quadratic parameters of motion in the image, under the constraint of a constant translational motion along the optical axis.
• Fixation, meaning keeping the same point at the image center, is done by controlling the translational d.o.f. perpendicular to the optical axis. Two ways have been studied:
• Keeping a null speed at the image center by direct compensation of displacement due to rotation.
• Estimating the position of the initial image center by integration of its speed along time and regulating this position to zero.
The task is sum up on the following figure where Pi and Pc represents the initial and desired 3D point of the object projected at the image center. The fixation task control the system in order to have Pi = Pc.
Expected result of the task
## Results
The results presented underneath have been obtained on our 6 d.o.f. Cartesian robot. A typical initial scene used is first displayed, then are successively presented:
• the quadratic parameters on which is based the control of rotational d.o.f.
• the estimated displacement of the initial image center by integration of its 2D speed and which is regulated by the translational d.o.f.
• the computed rotational speeds sent to the robot
• the computed translational speeds
• the estimated angular error. This error is computed using an a priori orientation of the reference plane, which is of course unused in the control loop.
(Click on results curves to see them bigger)
Initial scene
Regulated quadratic parameters Estimated displacement of initial center
Controlled rotational speeds Controlled translational speeds
Angular errors
## Collaborations
This work is included in project VIDAC (Dynamic Active VIsion and Communication), itself part of the XIth plan contract Brittany Council - French State driven in collaboration with TIPA team at Cemagref(in French only) and TEMICS at IRISA.
## References
1. A.Crétual, F. Chaumette. Positioning a camera parallel to a plane using dynamic visual servoing - IEEE/RSJ International Conference on Intelligent Robots and Systems (IROS'97), pp 43-49, Grenoble, France, September 1997
| Lagadic | Map | Team | Publications | Demonstrations | Irisa - Inria - Copyright 2009 © Lagadic Project
| 638
| 3,090
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2018-34
|
latest
|
en
| 0.937811
|
http://www.123articleonline.com/articles/632673/accessing-reliable-and-affordable-data-analysis-for-your-busines
| 1,524,271,395,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125944848.33/warc/CC-MAIN-20180420233255-20180421013255-00017.warc.gz
| 346,803,847
| 6,414
|
Welcome to 123ArticleOnline.com!
ALL >> Business >> View Article
# Accessing Reliable And Affordable Data Analysis For Your Busines
By Author: Kelsie. V. Newton
Total Articles: 1
In each and every business, data analysis is a very important procedure. In fact, it is very hard for any business to survive analyzing the relevant data. Analysis of data is the lifeline of any business. The procedure would come in handy when making marketing decisions or launching of new products.
In each and every business, data analysis is a very important procedure. In fact, it is very hard for any business to survive analyzing the relevant data. Analysis of data is the lifeline of any business. The procedure would come in handy when making marketing decisions or launching of new products. Simply analyzing data is not enough. You have to take a step further to interpret the results and use them for decision making. Based on the findings of data analysis, some critical decisions can be made. The right statistical methods are employed to help in ruling out all human bias. Thus, the conclusions that are arrived are objective and not subjective.
As a business owner, you may have limited insight on analyzing data. However, there are exceptional experts that are more than willing to assist. These can be accessed online. Consider the case of Statgenix for instance; they have all the solutions that you need. You can have on demand analysis services and receive the best results. Top notch statistical analysis procedures are employed to give the best results. This analysis entails a combination of methods that are used to process large amounts of data. Historical data of your business can be statistically analyzed to identify any unusual happenings. The contemporary statistical analysis techniques have many advantages over the conventional methods. Statistical methods make use of all available input.
In representing the analyzed data, statistical graphs are employed. A graph will provide a visual representation of a relationship that exists between two or more variables. The commonly used graphs are usually one or two dimensional. Three dimensional graphs are too complex though they are still employed in some instances. Statistical graphs come in many types and the type used will depend on the type of data being utilized and also the number of variables.
When conducting a survival analysis, Statgenix would be of great assistance. From this website, you can purchase Statgenix statistical plots such as Kaplan-Meier plot & table. The Kaplan Meier estimate is commonly used to measure the fractions of subjects living for a certain length of time after treatment. It is commonly used while conducting clinical trials. For example, it can be used to measure the number of subjects that survived after a certain intervention. Upon the conduction of the survival analysis, a survival curve can be drawn indicating the findings.
Upon analyzing and interpreting statistical data, models are developed. Statgenix statistical modeling will entail formalizing relationships between variables in the form of mathematical equations. The model will explain the manner in which one variable is related to the other. Many types of relationships may exist between variables. If a relationship exists, it is an indication that the variables affect each other. Being aware of the relationships that exist among different business factors will make it easy to make strategic business decision. The main purpose behind analyzing data is to enhance relevant decision making. Many businesses are utilizing this technique to enhance their competitive edge and acquire a competitive advantage over their rivals.
Resource: http://www.statgenix.com/products/kaplan-meier-curve
Total Views: 119Word Count: 558See All articles From Author
1. Looking For Accurate And Highly Responsive Hospital Mailing List To Enhance Business Revenue?
Author: Jane Cannon
2. Eglobedata Launches Reliable Email List Of 431,426 Manufacturing Industry Database To Better Leads
Author: Brayden Tyler
3. Massage Therapy For Optimal Health
Author: Angelika Matthias
5. An Updated Guide To Recently Effective Osha Rules
Author: Micheal Shawn
6. How To Bitcoin Benefit And Profitable For You
Author: Oliva Nat
Author: NEHA BHANDARI
8. All That May Be Making Your Customers Unhappy
Author: Surabhi Joshi
9. Select The Best Women’s Fragrances As A Gift!
Author: Sheleena Sultan
10. Biotropin Fitness [hgh Somatropin 100 Iu] – 10 Vials – Lifetech Labs
Author: Philip hauges
12. Trace The Issues In Your Roofing And Call A Roofing Contractor Today
Author: Fred Lydick
13. A Door To Jewels World With Verities Of Collection
Author: Luis Marius
| 943
| 4,722
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.953125
| 3
|
CC-MAIN-2018-17
|
latest
|
en
| 0.935812
|
https://www.jiskha.com/display.cgi?id=1296099563
| 1,516,146,816,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00407.warc.gz
| 923,161,938
| 3,983
|
# Algebra 1
posted by .
write an equation of the line containing the given point and perpendicular to the given line (6,7);6x+=9
I am having a terrible time figuring out the last part. I would appreciate any help. Thank you.
• Algebra 1 -
6x + = 9?
• Algebra 1 -
Sorry, I'm stressing badly. The equation is (6,7);6x+y=9. Thank you kindly.
• Algebra 1 -
(6,7);6x+y=9.
y = mx + b
6x + y = 9
y = -6x + 9
m = -6
m1*m2 = -1
-6*m2 = -1
m2 = 1/6
So, the slope of the perpendicular line is 1/6.
y = 1/6 x + b
Now find b with point (6,7).
y = 1/6 x + b
7 = 1/6 (6) + b
7 = 6/6 + b
7 = 1 + b
b = 6
y = 1/6 x + b
y = 1/6 x + 6
• Algebra 1 -
Was ist der dritte Teil von x?
• Algebra 1 -
Stefanie, I need to turn in my homework today and i still need to solve this problem, can you please help me?(4,-8); 2x+5y=4
## Similar Questions
1. ### algebra
Write an equation of the line containing the given point and perpendicular to the given line. (2-8); 9x +5y =7 The equation of the line is y= what I am tried to problem cannot seem to get it right.
2. ### algebra
Write an equation of the line containing the given point and perpendicular to the given line. express your answer in the fory y=mx+b. (8,9); 7x+y=2 the equation of the line is y=?
3. ### MATH
I had an assignment containing 45 problems, and I am unsure of 4, can you check and see if the answers are correct?
4. ### Math
I had an assignment containing 45 problems, and I am unsure of 4, can you check and see if the answers are correct?
5. ### Math
I had an assignment containing 45 problems, and I am unsure of 4, can you check and see if the answers are correct?
6. ### Algebra
graph the equation x-3=y (where do I plot the points?
7. ### Algebra 1
write an equation of the line containing the point and perpendicular to the given line (4,-8); 2x+5y=4 I am having a very hard time, but I'm trying hard to study the steps for solving the problem. Profoundly, thank you.
8. ### Math 116
1.Write an equation of the line containing the given points and parallel to the given line (3,8); x+5y=3 2. Write an equation of the line containing the given points and parallel to the given line (-3,6); 5x=9y+2 3. Write an equation …
9. ### Math 116
1.Write an equation of the line containing the given points and parallel to the given line (3,8); x+5y=3 2. Write an equation of the line containing the given points and parallel to the given line (-3,6); 5x=9y+2 3. Write an equation …
10. ### Intermediate Algebra
Write an equation of the line containing the given point and perpendicular to the given line. (0,4);5x+9y=4 The equation of the line is y=__?
More Similar Questions
| 816
| 2,647
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4
| 4
|
CC-MAIN-2018-05
|
latest
|
en
| 0.897662
|
https://www.physicsforums.com/threads/particle-motion-question.133461/
| 1,544,940,366,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376827281.64/warc/CC-MAIN-20181216051636-20181216073636-00425.warc.gz
| 1,014,317,153
| 12,452
|
# Homework Help: Particle motion question
1. Sep 24, 2006
### Robokapp
okay...i labeled acceleration 0 -19 as c.
a=c
v=cx
d=0.5cx^2
for 19-23 we have
a=-9.8
v=-9.8x
c=-4.9x^2
0 -19 we have d=180.5c
19-23
we have -823.5
180.5c-823.5=4700
Am I doing it right in solving for C which is constant acceleration in 0 -19?
edit: ok I know now it's not right.
my final distance formula would be
4700=-4.9x^2(for 19</x</23)+0.5cx^2(0</x</19)
But i dont know how to put this in a real equation form...
Last edited: Sep 24, 2006
2. Sep 24, 2006
### Robokapp
my second approach...
v=cx-9.8x
v=19c-9.8(4)=19c-39.2
and now i take the integral of all this and set it equal to 4700
okay...so i think the following formula should work...
$$\int_{0}^{19} {19c} dx + \int_{19}^{23} {-39.2} dx = 4700$$
grrrr. nope. Also sig figs matter.
| 332
| 838
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.75
| 4
|
CC-MAIN-2018-51
|
latest
|
en
| 0.845022
|
https://mathoverflow.net/questions/50661/unboundedness-of-number-of-integral-points-on-elliptic-curves
| 1,723,480,627,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641045630.75/warc/CC-MAIN-20240812155418-20240812185418-00687.warc.gz
| 302,444,521
| 27,933
|
# unboundedness of number of integral points on elliptic curves?
If $E/\mathbf{Q}$ is an elliptic curve and we put it into minimal Weierstrass form, we can count how many integral points it has. A theorem of Siegel tells us that this number $n(E)$ is finite, and there are even effective versions of this result. If I'm not mistaken this number $n(E)$ is going to be a well-defined invariant of $E/\mathbf{Q}$ (because different minimal Weierstrass models will have the same number of integral points).
Is it known, or conjectured, that $n(E)$ is unbounded as $E$ ranges over all elliptic curves?
Note: the question is trivial if one does not put $E$ into some sort of minimal form first: e.g. take any elliptic curve of rank 1 and then keep rescaling $X$ and $Y$ to make more and more rational points integral.
• I vaguely remember a similar question may imply some conjecture. Try searching for "integral points on minimal". Currently the 2 URLs timeout for me. May be wrong... Commented Dec 29, 2010 at 15:39
• Not that it helps much, but this can be expressed more intrinsically in terms of the N\'eron model $\mathcal{E}$ of $E$ over $\mathbf{Z}$, avoiding the ick of minimal Weierstrass models (locally or globally). Namely, the equality $E(\mathbf{Q}) = \mathcal{E}(\mathbf{Z})$ carries the pts that are everywhere integral with respect to local (or global) minimal Weierstrass models over to exactly the pts in $\mathcal{E}(\mathbf{Z})$ disjoint from the identity section and supported in the open relative identity component $\mathcal{E}^0$. So that gives a clean defn of $n(E)$ over any number field. Commented Dec 30, 2010 at 5:39
• By curiosity, is the result known for elliptic curves over function fields? In this case, unboundedness of the rank was proved by Shafarevich and Tate (there are also results by Ulmer). I don't know whether these constructions yield arbitrarily many integral points. Commented Jan 4, 2011 at 12:54
• For elliptic curves over function fields, it seems to be known that the number of integral points is unbounded : see Ricardo Conceicao's preprint arxiv.org/abs/0910.3417 Commented Jan 14, 2011 at 8:40
• Over function fields you have to be careful because the number of integral points can be not just unbounded but infinite! For example, in characteristic 2 if $(x,y)$ is an integral point on the supersingular curve $y^2+y=x^3+a$ then so is $(x^2,y+x^3)$, so as long as $x \notin \overline{{\bf F}_2}$ we get an infinite sequence of integral points. For instance, let $a=t^3$ and start from $(t,0)$ to get an infinite sequence of integral points (i.e. solutions of $y^2+y=x^3+t^3$ in polynomials $x,y \in {\bf F}_2[t]$; yes, that's a Néron model). Commented Jun 10, 2011 at 2:52
I proved that if $E/\mathbf{Q}$ is given using by a minimal Weiestrass equation, then
$\#E(Z) \le C^{\text{rank} E(Q) + n(j) + 1}$
where $n(j)$ is the number of distinct primes dividing the denominator of the $j$-invariant of $E$ and $C$ is an absolute constant. This is in J. Reine Angew. Math. 378 (1987), 60-100.
Mark Hindry and I proved that if you assume the abc conjecture, then you can remove the n(j) in the above estimate. This is in Invent. Math. 93 (1988), 419-450. It is a conjecture due to Lang.
The papers contain more general results for (quasi)-S-integral points over number fields.
• Thanks Joe! I didn't know about these results, which are pretty neat. To answer my question though we need some sort of bound the other way, right? I guess we now know that ABC implies that if there does happen to exist a universal bound for the rank then there's also a universal bound for the integral points---this presumably being the source of William Stein's quoted comment above. Can one prove "rank unbounded implies integral points on minimal models unbounded" though? Commented Dec 31, 2010 at 20:36
• Kevin asks: Can one prove "rank unbounded implies integral points on minimal models unbounded" though? No. In fact, it may well be that the number of integral points on minimal models is uniformly bounded (I don't have a strong feeling on that). One might guess that integral points satisfy h(P) << h(E) for an absolute constant, while Lang suggests that on "most" curves of positive rank, the smallest non-torsion point satisfies h(P) >> C^h(E). Conclusion would be that "most" curves have no integral points. So there should be lots of curves with big rank and no integral points. Commented Jan 1, 2011 at 13:50
• Hi Professor Silverman! I believe the displayed formula is broken because it needs two backslashes before the pound sign. Commented Jan 12, 2011 at 3:26
• @ZevChonoles No, it was broken precisely because of the second backslash! :-) [Of course im joking; I do realise you where right then.]
– user9072
Commented Sep 28, 2013 at 20:01
It is expected that the number of integral points is bounded in terms of the rank (this is known for some curves not in minimal Weierstrass form, Silverman JLMS 28 (1983), 1–7). So, if you could prove unboundedness of $n(E)$, you'd have a shot at proving unboundedness of rank which, as you know, is a hard problem.
On the other hand, if you believe Lang's (and Vojta's) conjectures on rational points on varieties of general type, then you would conclude that $n(E)$ is uniformly bounded (Abramovich, Inv. Math. 127 (1997), 307–317).
BTW, Kevin, don't you have some catching up to do?
• Abramovich restricts to semi-stable curves, if I understand correctly. He speculates that there is a bound on $n(E)$ depending on the number of additive places. Remark 2 after Cor 1 in springerlink.com/content/xgbvqxm4383nwhqu Commented Dec 29, 2010 at 18:04
• @Felipe: yes, I do. But that "catching up" involves going to work and spending a day concentrating; I can ask idle questions whilst babysitting three children! I'm at work tomorrow so I'll get things done then: I have a "no MO at work" policy, for example! Commented Dec 29, 2010 at 18:46
| 1,598
| 5,938
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.908192
|
http://www.zdnet.com/article/sound-goes-faster-than-light/
| 1,516,460,825,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00240.warc.gz
| 597,401,899
| 27,797
|
# Sound goes faster than light!
According to Physics Web in 'Sound breaks the light barrier,' a professor of physics in Tennessee has designed an experiment which proves that sound can move faster than light. This looks like impossible -- and it is. In fact, the physicist has tweaked some scientific definitions. No sound can go faster than light. But a sound pulse, or more precisely, all the wavelengths associated to a sound, have a "group velocity" that far exceeds the real physical limits. Have I lost you? Read more...
According to Physics Web in Sound breaks the light barrier (Free reg. required), a professor of physics in Tennessee has designed an experiment which proves that sound can move faster than light. This looks like impossible -- and it is. In fact, the physicist has tweaked some scientific definitions. No sound can go faster than light. But a sound pulse, or more precisely, all the wavelengths associated to a sound, have a "group velocity" that far exceeds the real physical limits. Have I lost you? Read more for some explanations that even a lawyer couldn't have invented.
Here are a couple of paragraphs from the Physics Web article.
In a normal dispersive medium, the velocity of a wave is proportional to its wavelength, resulting in a group velocity that is slower than the average velocity of its constituent waves. But in an "anomalously" dispersive medium -- one that becomes highly absorbing or attenuating at certain frequencies -- velocity is inversely proportional to wavelength, meaning that the group velocity can become much faster.
Indeed, the group velocity of light has already been shown to travel faster than the speed of light in a vacuum. But until now, superluminal acoustic waves have existed only in theory, and would require the group velocity to increase almost a million times over.
But what exactly is a superluminal phenomenon? Here is a short answer from Eric Weisstein's World of Physics.
A superluminal phenomenon is a frame of reference traveling with a speed greater than the speed of light c. There is a putative class of particles dubbed tachyons which are able to travel faster than light. Faster-than-light phenomena violate the usual understanding of the "flow" of time, a state of affairs which is known as the causality problem (and also called the "Shalimar Treaty").
Anyway, this was the purpose of the experiment designed by William Robertson from Middle Tennessee State University with the help of some colleagues and students. And their research work was recently published by Applied Physics Letters under the name "Sound beyond the speed of light: measurement of negative group velocity in an acoustic loop filter" (Volume 90, Issue 1, Article 014102, January 1, 2007). Here is a link to the abstract.
The results confirm recent theoretical predictions that faster-than-light group velocity propagation of sound is possible. Further, the results show that the spectral rephasing achieved in a loop filter is sufficient to produce negative group velocities independent of the phase velocity of the spectral components themselves. Thus, superluminal propagation is realized despite almost six orders of magnitude difference between the speeds of sound and light.
Here is another link to the full paper (PDF format, 3 pages, 214 KB), which shows the test system.
In Sound pulses exceed speed of light, Charles Q. Choi, from LiveScience, gave additional details. And it's quite funny to discover that this experiment used only a plastic plumbing pipe and a computer's sound card. "This experiment is truly basement science," Robertson told LiveScience.
Robertson and his colleagues transmitted sound pulses from the sound card through a loop made from PVC plumbing pipe and connectors from a hardware store. This loop split up and then recombined the tiny waves making up each pulse.
This led to a curious result. When looking at a pulse that entered and then exited the pipe, before the peak of the entering pulse even got into the pipe, the peak of the exiting pulse had already left the pipe. If the velocities of each of the waves making up a sound pulse in this setup are taken together, the "group velocity" of the pulse exceeded c.
For even more information, you can read 'Mach c'? Scientists observe sound traveling faster than the speed of light, an unusual attempt by PhysOrg.com to provide original contents. Here is the conclusion of this article.
Is this phenomenon simply the result of a clever set-up, or can it actually occur in the real world? According to the scientists, the interference that occurs in the loop filter is directly analogous to the "comb filtering" effect in architectural acoustics, where sound interference occurs between sound directly from a source and that reflected by a hard surface. "The superluminal acoustic effect we have described is likely a ubiquitous but imperceptible phenomenon in the everyday world," the scientists conclude.
So does sound goes faster than light? No, except if you're a physicist...
Sources: Jon Cartwright, Physics Web, January 12, 2007; and various other websites
You'll find related stories by following the links below.
| 1,028
| 5,183
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2018-05
|
latest
|
en
| 0.948825
|
https://www.vistrails.org/index.php/User:Tohline/SSC/BipolytropeGeneralization
| 1,723,506,132,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641052535.77/warc/CC-MAIN-20240812221559-20240813011559-00880.warc.gz
| 815,288,961
| 27,269
|
# Bipolytrope Generalization
On 26 August 2014, Tohline finished rewriting the chapter titled "Bipolytrope Generalization" in a very concise manner (go here for this Version2 chapter) then set this chapter aside to provide a collection of older attempts at the derivations. While much of what follows is technically correct, it is overly detailed and cumbersome. Because it likely also contains some misguided steps, we label it in entirety as Work in Progress.
Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
| Go Home |
## Old Stuff
$~\mathfrak{G}$ $~=$ $~W_\mathrm{grav} + \mathfrak{S}_A\biggr|_\mathrm{core} + \mathfrak{S}_A\biggr|_\mathrm{env}$ $~=$ $~W_\mathrm{grav} + \biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] S_\mathrm{env} \, .$
In addition to the gravitational potential energy, which is naturally written as,
$~W_\mathrm{grav}$ $~=$ $~- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{R} \biggr) \cdot \mathfrak{f}_{WM} \, ,$
it seems reasonable to write the separate thermal energy contributions as,
$~S_\mathrm{core}$ $~=$ $~\frac{3}{2}\biggl[ M_\mathrm{core} \biggl( \frac{P_{ic}}{\rho_{ic}} \biggr) \biggr] s_\mathrm{core} = \frac{3}{2} \biggl[ \nu M_\mathrm{tot} P_{ic} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{-1} \biggl( \frac{4\pi R^3}{3 M_\mathrm{tot}} \biggr) \biggr] s_\mathrm{core} = 2\pi R^3 P_{ic} \biggl[ q^3 s_\mathrm{core} \biggr] \, ,$ $~S_\mathrm{env}$ $~=$ $~\frac{3}{2}\biggl[ M_\mathrm{env} \biggl( \frac{P_{ie}}{\rho_{ie}} \biggr) \biggr] s_\mathrm{env} = \frac{3}{2} \biggl[ (1-\nu) M_\mathrm{tot} P_{ie} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{-1} \biggl( \frac{4\pi R^3}{3 M_\mathrm{tot}} \biggr) \biggr] s_\mathrm{env} = 2\pi R^3 P_{ie} \biggl[ (1-q^3) s_\mathrm{env} \biggr] \, ,$
where the subscript "$i$" means "at the interface," and $~\mathfrak{f}_{WM},$ $~s_\mathrm{core},$ and $~s_\mathrm{env}$ are dimensionless functions of order unity (all three functions to be determined) akin to the structural form factors used in our examination of isolated polytropes.
While exploring how the free-energy function varies across parameter space, we choose to hold $~M_\mathrm{tot}$ and $~K_c$ fixed. By dimensional analysis, it is therefore reasonable to normalize all energies, length scales, densities and pressures by, respectively,
$~E_\mathrm{norm}$ $~\equiv$ $~\biggl[ \frac{G^{3(\gamma_c-1)} M_\mathrm{tot}^{5\gamma_c-6}}{K_c} \biggr]^{1/(3\gamma_c -4)} \, ,$ $~R_\mathrm{norm}$ $~\equiv$ $~\biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{1/(3\gamma_c -4)} \, ,$ $~\rho_\mathrm{norm}$ $~\equiv$ $~\frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)} \, ,$ $~P_\mathrm{norm}$ $~\equiv$ $~\biggl[ \frac{G^{3\gamma_c} M_\mathrm{tot}^{2\gamma_c}}{K_c^4} \biggr]^{1/(3\gamma_c -4)} \, .$
As is detailed below — first, here, and via an independent derivation, here — quite generally the expression for the normalized free energy is,
$~\mathfrak{G}^* \equiv \frac{\mathfrak{G}}{E_\mathrm{norm}}$ $~=$ $- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{E_\mathrm{norm}} \biggr) \biggl( \frac{1}{R} \biggr) \cdot \mathfrak{f}_{WM} + \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr] \biggl[ \frac{ R^3 P_{ic} }{E_\mathrm{norm}} \biggr] + \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr] \biggl[ \frac{ R^3 P_{ie} }{E_\mathrm{norm}} \biggr]$ $~=$ $- \frac{3\cdot \mathfrak{f}_{WM}}{5} \chi^{-1} + \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr] \biggl[ \frac{ R_\mathrm{norm}^4 P_\mathrm{norm} }{E_\mathrm{norm} R_\mathrm{norm}} \biggr] \biggl[ \biggl( \frac{P_{ic}}{P_\mathrm{norm}} \biggr) \chi^3 \biggr]$ $+ \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr] \biggl[ \frac{ R_\mathrm{norm}^4 P_\mathrm{norm} }{E_\mathrm{norm} R_\mathrm{norm}} \biggr] \biggl[ \biggl( \frac{P_{ie}}{P_\mathrm{norm}} \biggr) \chi^3 \biggr]$ $~=$ $- \frac{3\cdot \mathfrak{f}_{WM}}{5} \chi^{-1} + \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr] \biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3 -3\gamma_c} + \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr] \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3 -3\gamma_e}$
where we have introduced the parameter, $~\nu \equiv M_\mathrm{core}/M_\mathrm{tot}$. After defining the normalized (and dimensionless) configurarion radius, $~\chi \equiv R/R_\mathrm{norm}$, we can write the normalized free energy of a bipolytrope in the following compact form:
$~\mathfrak{G}^*$ $~=$ $~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} \, ,$
where,
$~\mathcal{A}$ $~\equiv$ $~\frac{1}{5} \cdot \mathfrak{f}_{WM} \, ,$ $~\mathcal{B}$ $~\equiv$ $\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \, ,$ $~\mathcal{C}$ $~\equiv$ $\biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env} \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \, .$
As is further detailed below, the second expression for the coefficient, $~\mathcal{C}$, ensures that the pressure at the "surface" of the core matches the pressure at the "base" of the envelope; but it should only be employed after an equilibrium radius, $~\chi_\mathrm{eq}$, has been identified by locating an extremum in the free energy.
## Simplest Bipolytrope
### Familiar Setup
As has been shown in an accompanying presentation, for an $~(n_c, n_e) = (0, 0)$ bipolytrope,
$~\mathfrak{f}_{WM}$ $~\equiv$ $~\frac{\nu^2}{q} \cdot f \, ,$ $~s_\mathrm{core}$ $~\equiv$ $1 + \Lambda_\mathrm{eq} \, ,$ $~(1-q^3) s_\mathrm{env}$ $~\equiv$ $(1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,$
and where (see, for example, in the context of its original definition, or another, separate derivation),
$~\Lambda_\mathrm{eq}$ $~=~$ $\frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \chi_\mathrm{eq}^{3\gamma_c - 4}$ $~=$ $\frac{2}{5(g^2-1)} = \biggl\{ \frac{5}{2}\biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \biggr\}^{-1} \, ,$
and where (see the associated discussion of relevant mass integrals),
$\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} ~~\Rightarrow ~~~ \frac{q^3}{\nu} = \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \, .$
### Cleaner Virial Presentation
In an effort to show the similarity in structure among the several energy terms, we have also found it useful to write their expressions in the following forms:
$~W_\mathrm{grav}$ $~=$ $~- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{R} \biggr) \frac{\nu^2}{q} \cdot f = - 4\pi P_i R^3 \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_i f \, ,$ $~S_\mathrm{core}$ $~=$ $~2\pi P_{ic} R^3 \biggl[ q^3 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{ic} \biggr] \, ,$ $~S_\mathrm{env}$ $~=$ $~2\pi P_{ie} R^3 \biggl[ (1-q^3) + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{ie} \mathfrak{F} \biggr] \, ,$
where (see an associated discussion or the original derivation),
$f\biggl(q, \frac{\rho_e}{\rho_c}\biggr) = 1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (q^3 - q^5 ) + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{2}{5} - q^3 + \frac{3}{5}q^5 \biggr) \biggr] \, ,$
and where,
$~\lambda_i$ $~\equiv$ $~\frac{GM_\mathrm{tot}^2}{R^4 P_i} \, ,$ $~\mathfrak{F}$ $~\equiv$ $~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr]$ $~=$ $~ \frac{1}{\lambda_{ie}} \biggl( \frac{2^2 \cdot 5\pi}{3} \biggr) \frac{q(1-q^3)}{\nu^2} (s_\mathrm{env} -1) \, .$
This also means that the three key terms used as shorthand notation in the above expressions for the three energy terms have the following definitions:
$~\mathfrak{f}_{WM}$ $~\equiv$ $~\frac{\nu^2}{q} \cdot f \, ,$ $~s_\mathrm{core}$ $~\equiv$ $1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \, ,$ $~s_\mathrm{env}$ $~\equiv$ $1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q(1-q^3)} \cdot \lambda_{ie} \mathfrak{F} \, ,$
Hence, if all the interface pressures are equal — that is, if $~P_i = P_{ic} = P_{ie}$ and, hence also, $~\lambda_{i} = \lambda_{ic} = \lambda_{ie}$ — then the total thermal energy is,
$~S_\mathrm{tot} = S_\mathrm{core} + S_\mathrm{env}$ $~=$ $2\pi P_{i} R^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{i} (1+\mathfrak{F}) \biggr] \, ;$
and the virial is,
$~2S_\mathrm{tot} + W_\mathrm{grav}$ $~=$ $4\pi P_{i} R^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{i} (1+\mathfrak{F} - f ) \biggr] \, .$
The virial should sum to zero in equilibrium, which means,
$~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}$ $~=$ $\biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot (f - 1- \mathfrak{F} )$ $\Rightarrow ~~~~ \biggl[ \biggl( \frac{2^2\cdot 5\pi}{3} \biggr) \frac{q}{\nu^2} \biggr] \frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2}$ $~=$ $f - 1- \mathfrak{F}$ $\Rightarrow ~~~~ \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggl[ \biggl( \frac{2^3\pi}{3} \biggr) \frac{q^6}{\nu^2} \biggr] \frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2}$ $~=$ $\biggl[ (q^3 - q^5 )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{2}{5} - q^3 + \frac{3}{5}q^5 \biggr) \biggr] - \biggl[ (-2q^2 + 3q^3 - q^5) + \biggl( \frac{\rho_e}{\rho_c}\biggr) (-\frac{3}{5} +3q^2 - 3q^3 + \frac{3}{5} q^5) \biggr]$ $~=$ $2q^2(1-q) + \biggl( \frac{\rho_e}{\rho_c}\biggr) (1 -3q^2 + 2q^3 )$ $~=$ $q^2 \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} (g^2-1)$ $\Rightarrow~~~~ \frac{1}{\lambda_i} \biggr|_\mathrm{eq}$ $~=$ $\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1) \, .$
### Shift to Central Pressure Normalization
Let's rework the definition of $~\lambda_i$ in two ways: (1) Normalize $~R_\mathrm{eq}$ to $~R_\mathrm{norm}$ and normalize the pressure to $~P_\mathrm{norm}$; (2) shift the referenced pressure from the pressure at the interface $~(P_i)$ to the central pressure $~(P_0)$, because it is $~P_0$ that is directly related to $~K_c$ and $~\rho_c$; specifically, $P_0 = K_c \rho_c^{\gamma_c}$. Appreciating that, in equilibrium,
$~P_i$ $~=$ $~P_0 - q^2 \Pi_\mathrm{eq} = K_c \rho_c^{\gamma_c} - \frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) q^2 \, ,$
the left-hand-side of the last expression, above, can be rewritten as,
$~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}$ $~\equiv$ $~\frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2}$ $~=$ $~\frac{R_\mathrm{eq}^4}{GM_\mathrm{tot}^2} \biggl[ P_0 - \frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) q^2\biggr]$ $~=$ $~\frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 \, .$
Hence, the virial equilibrium condition gives,
$\frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2$ $~=$ $\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1)$ $\Rightarrow ~~~~ \frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2}$ $~=$ $\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{2} q^2 g^2 \, .$
This result precisely matches the result obtained via the detailed force-balanced conditions imposed through hydrostatic equilibrium.
Adopting our new variable normalizations and realizing, in particular, that,
$~R_\mathrm{norm}^4 P_\mathrm{norm}$ $~=$ $~GM_\mathrm{tot}^2 \, ,$
the expression alternatively can be rewritten as,
$~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}$ $~\equiv$ $~\frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} = \chi_\mathrm{eq}^4 \biggl( \frac{P_i}{P_\mathrm{norm}} \biggr)$ $~=$ $\chi_\mathrm{eq}^4 \biggl\{ \frac{K_c \rho_c^{\gamma_c}}{P_\mathrm{norm}} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4P_\mathrm{norm}} \biggr) \biggr\}$ $~=$ $\chi_\mathrm{eq}^4 \biggl\{ \frac{K_c }{P_\mathrm{norm}} \biggl[ \frac{\rho_c}{\bar\rho} \biggl( \frac{3M_\mathrm{tot}}{4\pi R_\mathrm{norm}^3} \biggr) \chi_\mathrm{eq}^{-3} \biggr]^{\gamma_c} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \chi_\mathrm{eq}^{-4} \biggr\}$ $~=$ $\chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \frac{K_c }{P_\mathrm{norm}} \biggl( \frac{M_\mathrm{tot}^{\gamma_c}}{R_\mathrm{norm}^{3\gamma_c}} \biggr) - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2$ $~=$ $\chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \, .$
Normalized in this manner, the virial equilibrium (as well as the hydrostatic balance) condition gives,
$\chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2$ $~=$ $\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1)$ $\Rightarrow ~~~~ \chi_\mathrm{eq}^{4-3\gamma_c}$ $~=$ $\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{2-\gamma_c} q^2 g^2 \, .$
### Free-Energy Coefficients
Therefore, for an $~(n_c, n_e) = (0, 0)$ bipolytrope, the coefficients in the normalized free-energy function are,
$~\mathcal{A}$ $~=$ $~\frac{\nu^2}{5q} \cdot f = \frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl[ q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) q^3 (1 - q^2 ) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2} q^3 + \frac{3}{2}q^5 \biggr) \biggr] \, ,$ $~\mathcal{B}$ $~\equiv$ $\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{P_{ic} }{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c} =\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \biggr] \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4} =\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \frac{1}{\lambda_{ic}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr] \chi_\mathrm{eq}^{3\gamma_c-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl\{ \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr\} \chi_\mathrm{eq}^{3\gamma_c-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl\{ \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} + \chi_\mathrm{eq}^{3\gamma_c-4} \biggl[\frac{3}{2^2\cdot 5\pi} - \frac{3}{2^3\pi} \biggr]\frac{\nu^2}{q^4} \biggr\}$ $~=$ $\biggl\{ \nu \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} - \chi_\mathrm{eq}^{3\gamma_c-4} \biggl( \frac{3^2}{2^3\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggl( \frac{4\pi }{3} \biggr) q^3 \biggr\} = \nu \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} - \chi_\mathrm{eq}^{3\gamma_c-4} \biggl( \frac{3}{10} \biggr) \frac{\nu^2}{q}$ $~\mathcal{C}$ $~\equiv$ $\biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env} \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \, .$
Note that, because $~P_{ie} = P_{ic}$ in equilibrium, the ratio of coefficients,
$~\frac{\mathcal{C}}{\mathcal{B}}$ $~=$ $~\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}\biggl\{\frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr\}$ $\Rightarrow ~~~~ \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \biggl( \frac{\mathcal{C}}{\mathcal{B}} \biggr)$ $~=$ $\frac{ 20\pi q(1-q^3) \lambda_i^{-1} + 3\nu^2 \mathfrak{F} }{ 20\pi q^4 \lambda_i^{-1} + 3\nu^2 } \, .$
The equilibrium condition is,
$\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} \, ,$
where,
$\mathcal{C}^' \equiv \mathcal{C} \chi_\mathrm{eq}^{3(\gamma_c-\gamma_e)} \, .$
### More General Derivation of Free-Energy Coefficients B and C
Keep in mind that, generally,
$GM_\mathrm{tot}^2$ $~=$ $~R_\mathrm{norm}^4 P_\mathrm{norm} = E_\mathrm{norm} R_\mathrm{norm} \, ;$ $~\frac{1}{\lambda_i}$ $~\equiv$ $\frac{R^4 P_i}{GM_\mathrm{tot}^2} = \biggl( \frac{P_i}{P_\mathrm{norm}} \biggr) \chi^4$ … and, note that … $\frac{1}{\Lambda} = \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \frac{1}{\lambda_i} \cdot \frac{1}{q^2 \sigma^2} \, ;$ $~\frac{\Pi}{P_\mathrm{norm}}$ $~=$ $\frac{3}{2^3\pi} \biggl( \frac{GM_\mathrm{tot}^2}{P_\mathrm{norm} R^4} \biggr) \frac{\nu^2}{q^6} =\biggl( \frac{2\pi}{3} \biggr) \sigma^2 \chi^{-4} \, ;$ $~\frac{K_c \rho_c^{\gamma_c} }{P_\mathrm{norm}}$ $~=$ $\frac{K_c}{P_\mathrm{norm}} \biggl( \frac{\rho_c}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c} = \frac{K_c M_\mathrm{tot}^{\gamma_c} }{R_\mathrm{norm}^{3\gamma_c} P_\mathrm{norm}} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \chi^{-3\gamma_c} = \sigma^{\gamma_c} \chi^{-3\gamma_c} \, ,$
where we have introduced the notation,
$\sigma \equiv \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \, .$
So, the free-energy coefficient,
$~\mathcal{B}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{P_{ic} }{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c} =\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \biggr]_\mathrm{eq} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4} =\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \frac{1}{\lambda_{ic}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \chi_\mathrm{eq}^4 \biggl( \frac{P_{ic}}{P_\mathrm{norm}} \biggr)_\mathrm{eq} + \biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4} \, . $
And the free-energy coefficient,
$~\mathcal{C}$ $~\equiv$ $\biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env} \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} = \biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env} \biggl[ \frac{1 }{\lambda_{ie}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) (1-q^3) \biggl\{ \frac{1 }{\lambda_{ie}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q(1-q^3)} \cdot \mathfrak{F} \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) \biggl\{ (1-q^3) \chi_\mathrm{eq}^4 \biggl( \frac{P_{ie}}{P_\mathrm{norm}} \biggr)_\mathrm{eq}+ \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4} \, .$
OLD DERIVATION
$P_{ic} = K_c \rho_c^{\gamma_c}$
NEW DERIVATION
$P_{ic} = P_0 - q^2\Pi = K_c \rho_c^{\gamma_c} - q^2\Pi$
… therefore …
$~\mathcal{B}_\mathrm{OLD}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} + \biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4} $ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} + \biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \chi_\mathrm{eq}^{3\gamma_c-4} \biggr] $
$~\mathcal{B}_\mathrm{NEW}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 + \biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} - \biggl( \frac{2\pi}{5} \biggr) q^2 \sigma^2 \chi_\mathrm{eq}^{3\gamma_c-4} \biggr]$
… and, enforcing in equilibrium $~P_{ie} = P_{ic}$ …
$~\mathcal{C}_\mathrm{OLD}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) \biggl\{ (1-q^3) \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4-3\gamma_c} \biggr] + \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) \biggl[ (1-q^3) \sigma^{\gamma_c} + \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl( \frac{2}{5} q^5 \mathfrak{F} \biggr) \chi_\mathrm{eq}^{3\gamma_e-4}\biggr]$
$~\mathcal{C}_\mathrm{NEW}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) \biggl\{ (1-q^3) \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4-3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 \biggr] + \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) \biggl\{ (1-q^3) \sigma^{\gamma_c} \chi_\mathrm{eq}^{3\gamma_e - 3\gamma_c} + \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[ \biggl( \frac{2}{5} q^5 \mathfrak{F} \biggr) - q^2 (1-q^3) \biggr] \chi_\mathrm{eq}^{3\gamma_e-4} \biggr\}$
… and, also …
$~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}$ $~=$ $\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c}$ $\Rightarrow ~~~~ \frac{1}{\Lambda_\mathrm{eq}}$ $~=$ $\frac{1}{q^2 } \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \sigma^{\gamma_c - 2} \chi_\mathrm{eq}^{4 - 3\gamma_c}$ $~=$ $\biggl( \frac{5q}{\nu} \biggr) \sigma^{\gamma_c - 1} \chi_\mathrm{eq}^{4 - 3\gamma_c}$
$~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}$ $~=$ $\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2$ $\Rightarrow ~~~~ \frac{1}{\Lambda_\mathrm{eq}}$ $~=$ $\frac{1}{q^2 \sigma^2} \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 \biggr]$ $~=$ $\biggl( \frac{5q}{\nu} \biggr) \sigma^{\gamma_c - 1} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \frac{5}{2}$
### Extrema
Extrema in the free energy occur when,
$~\mathcal{A}$ $~=$ $~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} + \mathcal{C} \chi_\mathrm{eq}^{4-3\gamma_e} \, .$
Also, as stated above, because $~P_{ie} = P_{ic}$ in equilibrium, the ratio of coefficients,
$~\frac{\mathcal{C}}{\mathcal{B}}$ $~=$ $~\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}\biggl[ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] \, .$
When put together, these two relations imply,
$~\mathcal{A}$ $~=$ $~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} + \chi_\mathrm{eq}^{4-3\gamma_c} \mathcal{B} \biggl[ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr]$ $~=$ $~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[ 1+ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] \, .$
But the definition of $~\mathcal{B}$ gives,
$~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c}$ $~=$ $\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \, .$
Hence, extrema occur when,
$~\mathcal{A}$ $~=$ $~\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \biggl[ 1+ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr]$ $\Rightarrow ~~~~ \biggl( \frac{3}{2^2 \cdot 5\pi } \biggr) \frac{\nu^2}{q} \cdot f$ $~=$ $\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \biggl[ q^3 s_\mathrm{core} + (1-q^3) s_\mathrm{env} \biggr]$ $~=$ $\frac{q^3}{[\lambda_i]_\mathrm{eq}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} + \frac{(1-q^3)}{[\lambda_i]_\mathrm{eq}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \mathfrak{F}$ $\Rightarrow ~~~~\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq}$ $~=$ $\biggl( \frac{3}{2^2 \cdot 5\pi } \biggr) \frac{\nu^2}{q} \cdot (f - 1 - \mathfrak{F})$ $~=$ $\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 (g^2 - 1 ) \, .$
In what follows, keep in mind that,
$~\chi_\mathrm{eq}^{4-3\gamma_c}$ $~=$ $\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^{4-3\gamma_c} = R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \, ;$ $~K_c \rho_c^{\gamma_c}$ $~=$ $K_c \biggl( \frac{\rho_c}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c} = K_c \sigma^{\gamma_c} M_\mathrm{tot}^{\gamma_c} R^{-3\gamma_c} \, ;$ $~\Pi$ $~=$ $\frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \frac{\nu^2}{q^6} = \frac{2\pi}{3} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \sigma^2 \, .$
OLD DERIVATION
$P_{i} = K_c \rho_c^{\gamma_c}$
$\Rightarrow ~~~~ K_c = P_{i} \sigma^{-\gamma_c} M_\mathrm{tot}^{-\gamma_c} R^{+3\gamma_c}$
NEW DERIVATION
$P_0 = K_c \rho_c^{\gamma_c}$
$\Rightarrow ~~~~ K_c = P_0 \sigma^{-\gamma_c} M_\mathrm{tot}^{-\gamma_c} R^{+3\gamma_c}$
… hence, as derived in the above table …
$~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}$ $~=$ $\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c}$
$~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}$ $~=$ $\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2$
… which, when combined with the condition that identifies extrema, gives …
$~\chi_\mathrm{eq}^{4 - 3\gamma_c}$ $~=$ $\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 (g^2 - 1 )$ $~\Rightarrow ~~~~ R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2}$ $~=$ $\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 (g^2 - 1 )$ $\Rightarrow ~~~~ \frac{ R_\mathrm{eq}^{4} P_i }{GM_\mathrm{tot}^{2} }$ $~=$ $\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 (g^2-1)$
$~\sigma^{\gamma_c}\chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2$ $~=$ $\biggl( \frac{2\pi}{3 } \biggr) \sigma^2 q^2 (g^2 - 1 )$ $\Rightarrow ~~~~ \chi_\mathrm{eq}^{4 - 3\gamma_c}$ $~=$ $\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 g^2$ $\Rightarrow ~~~~ R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2}$ $~=$ $\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 g^2$ $\Rightarrow ~~~~ \frac{ R_\mathrm{eq}^{4} P_0 }{GM_\mathrm{tot}^{2} }$ $~=$ $\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 g^2$
These are consistent results because they result in the detailed force-balance relation, $P_0 - P_i = q^2 \Pi_\mathrm{eq} \, .$
## Examples
Identification of Local Extrema in Free Energy
$~\nu$
$~q$
$~ \frac{\rho_e}{\rho_c}$
$~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)$
$~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)$
$~\Lambda_\mathrm{eq}$
$~\chi_\mathrm{eq}$
$~\mathcal{A}$
$~\mathcal{B}$
$~\mathcal{C}$
MIN/MAX
$~0.2$
$~9^{-1/3} = 0.48075$
$~0.5$
$~12.5644$
$~2.091312$
$~0.366531$
$~0.037453$
$~0.2090801$
$~0.2308269$
$~2.06252 \times 10^{-4}$
MIN
$~0.4$
$~4^{-1/3} = 0.62996$
$~0.5$
$~4.21974$
$~1.56498$
$~0.707989$
$~0.0220475$
$~0.2143496$
$~0.5635746$
$~4.4626 \times 10^{-5}$
MIN
$~0.473473$
$~0.681838$
$~0.516107$
$~0.462927$
$~0.08255$
"
"
"
MAX
$~0.5$
$~3^{-1/3} = 0.693361$
$~0.5$
$~2.985115$
$~1.42334$
$~0.9448663$
$~0.0152116$
$~0.2152641$
$~0.791882$
$~1.5464 \times 10^{-5}$
MIN
$~0.559839$
$~0.729581$
$~0.499188$
$~0.75089$
$~0.032196$
"
"
"
MAX
Free Energy Extrema when: $~~~~~\frac{\rho_e}{\rho_c} = \frac{1}{2} ~~~~\Rightarrow~~~~ q^3 = \frac{\nu}{2-\nu}$
$~\nu$
$~q$
$~ \frac{\rho_e}{\rho_c}$
$~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)$
$~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)$
$~\frac{1}{\lambda_{i}}\biggr|_\mathrm{eq}$
$~\chi_\mathrm{eq}$
$~\mathcal{A}$
$~\mathcal{B}_\mathrm{NEW}$
$~\mathcal{C}_\mathrm{NEW}$
$~G^*$
Stability
MIN/MAX
$~0.5$
$~\biggl( \frac{1}{3} \biggr)^{1/3}$
$~0.5$
$~2.985115$
$~1.423340$
$~0.05466039$
$~0.3152983$
$~0.21526406$
$~0.23552725$
$~6.643899 \times 10^{-3}$
$~+ 0.5176146$
$~+0.429245$
MIN
$~0.6674$
"
"
"
$~+0.55572115$
MAX
$~0.6$
$~\biggl( \frac{3}{7} \biggr)^{1/3}$
$~0.5$
$~2.2507129$
$~1.31282895$
$~0.04160318$
$~0.3411545$
$~0.21493717$
$~0.26165939$
$~5.208750 \times 10^{-3}$
$~+ 0.73532249$
$~+0.0935217$
MIN
$~0.431745$
"
"
"
$~+0.7367797$
MAX
$~0.7$
$~\biggl( \frac{7}{13} \biggr)^{1/3}$
$~0.5$
$~1.7707809$
$~1.2209446$
$~0.029500$
$~0.3589388$
$~0.21330744$
$~0.28172532$
$~3.389793 \times 10^{-3}$
$~+ 0.8953395$
$~- 0.0767108$
MAX
$~0.270615$
"
"
"
$~+0.89227216$
MIN
System should be stable (with free energy minimum) if: $~~~~~\frac{( \gamma_e - \frac{4}{3})}{(\gamma_e - \gamma_c)} f - \biggl[1 + \frac{5}{2} (g^2-1) \biggr] ~>~ 0$
## Solution Strategy
For a given set of free-energy coefficients, $~\mathcal{A}, \mathcal{B},$ and $~\mathcal{C}$, along with a choice of the two adiabatic exponents $~(\gamma_c, \gamma_e)$, here's how to determine all of the physical parameters that are detailed in the above example table.
• Step 1: Guess a value of $~0 < q < 1$.
• Step 2: Given the pair of parameter values, $~(\mathcal{A}, q)$, determine the interface-density ratio, $~\rho_e/\rho_c$, by finding the appropriate root of the expression that defines the function, $~\mathcal{A}(q, \rho_e/\rho_c)$. This can be straightforwardly accomplished because, as demonstrated below, the relevant expression can be written as a quadratic function of $~(\rho_e/\rho_c)$.
• Step 3: Given the pair of parameter values, $~(q, \rho_e/\rho_c)$, determine the value of the core-to-total mass ratio, $~\nu$, from the expression that was obtained from an integration over the mass, namely,
$~\frac{1}{\nu}$ $~=$ $~1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{q^3} - 1 \biggr) \, .$
• Step 4: Given the value of $~\mathcal{B}$ along with the pair of parameter values, $~(q, \nu)$, the above expression that defines $~\mathcal{B}$ can be solved to give the relevant value of the dimensionless parameter, $~\Lambda_\mathrm{eq}.$
• Step 5: The value of $~\mathcal{C}^'$ — the coefficient that appears on the right-hand-side of the above expression that defines $~\mathcal{C}$ — can be determined, given the values of parameter triplet, $~(q, \nu, \Lambda_\mathrm{eq})$.
• Step 6: Given the value of $~\mathcal{C}$ and the just-determined value of the coefficient $~\mathcal{C}^'$, the normalized equilibrium radius, $~\chi_\mathrm{eq},$ that corresponds to the value of $~q$ that was guessed in Step #1 can be determined from the above definition of $~\mathcal{C}$, specifically,
$~\chi_\mathrm{eq}\biggr|_\mathrm{guess}$ $~=$ $~\biggl( \frac{\mathcal{C}}{\mathcal{C}^'} \biggr)^{1/(3\gamma_e - 3\gamma_c)} \, .$
• Step 7: But, independent of this guessed value of $~\chi_\mathrm{eq},$ the condition for virial equilibrium — which identifies extrema in the free-energy function — gives the following expression for the normalized equilibrium radius:
$~\chi_\mathrm{eq}\biggr|_\mathrm{virial}$ $~=$ $~\biggl[ \frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} \biggr]^{1/(4 - 3\gamma_c)} \, .$
• Step 8: If $~\chi_\mathrm{eq}|_\mathrm{guess} \ne \chi_\mathrm{eq}|_\mathrm{virial}$, return to Step #1 and guess a different value of $~q$. Repeat Steps #1 through #7 until the two independently derived values of the normalized radius match, to a desired level of precision.
• Keep in mind: (A) A graphical representation of the free-energy function, $~\mathfrak{G}(\chi)$, can also be used to identify the "correct" value of $~\chi_\mathrm{eq}$ and, ultimately, the above-described iteration loop should converge on this value. (B) The free-energy function may exhibit more than one (or, actually, no) extrema, in which case more than one (or no) value of $~q$ should lead to convergence of the above-described iteration loop.
Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
| Go Home |
# Detailed Derivations
Dividing the free-energy expression through by $~E_\mathrm{norm}$ generates,
$~\mathfrak{G}^* \equiv \frac{\mathfrak{G}}{E_\mathrm{norm}}$ $~=$ $- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{E_\mathrm{norm}} \biggr) \biggl( \frac{1}{R} \biggr) \cdot \mathfrak{f}_{WM} + \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_c \rho_{ic}^{\gamma_c-1} }{E_\mathrm{norm}} \biggr]$ $~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_e \rho_{ie}^{\gamma_e-1} }{E_\mathrm{norm}} \biggr]$ $~=$ $- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl[ \frac{K_c G^{(3\gamma_c -4)}M_\mathrm{tot}^{2(3\gamma_c -4)}}{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}} \biggr]^{1/(3\gamma_c -4)} \biggl( \frac{1}{R} \biggr)$ $+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{K_c M_\mathrm{tot}^{3\gamma_c -4} K_c^{3\gamma_c -4} }{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}} \biggr]^{1/(3\gamma_c -4)} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c-1}$ $~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{K_c M_\mathrm{tot}^{3\gamma_c -4} K_e^{3\gamma_c -4} }{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}} \biggr]^{1/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_e-1}$ $~=$ $- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R_\mathrm{norm}}{R} \biggr)$ $+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{\gamma_c-1} \biggl[ \frac{ K_c^{3\gamma_c -3} }{G^{3\gamma_c-3} M_\mathrm{tot}^{2\gamma_c-2}} \biggr]^{1/(3\gamma_c -4)} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_c-1)} \biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{-3(\gamma_c-1)/(3\gamma_c -4)} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1}$ $~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{\gamma_e-1} \biggl[ \frac{K_c^{3\gamma_c - 3} (K_e/K_c)^{3\gamma_c -4} }{G^{3\gamma_c-3} M_\mathrm{tot}^{2\gamma_c-2}} \biggr]^{1/(3\gamma_c -4)} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_e-1)} \biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{-3(\gamma_e-1)/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1}$ $~=$ $- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R_\mathrm{norm}}{R} \biggr)$ $+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl( \frac{3}{4\pi} \biggr)^{\gamma_c-1} \biggl[ M_\mathrm{tot}^{3\gamma_c-4} \biggr]^{(\gamma_c-1)/(3\gamma_c-4)} \biggl[ M_\mathrm{tot}^{-2} \biggr]^{(\gamma_c-1)/(3\gamma_c -4)} \biggl[ M_\mathrm{tot}^{-3\gamma_c+6} \biggr]^{(\gamma_c-1)/(3\gamma_c -4)} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_c-1)} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1}$ $~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl( \frac{3}{4\pi} \biggr)^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ M_\mathrm{tot}^{2(\gamma_e-1)-2(\gamma_c-1)}\biggr]^{1/(3\gamma_c-4)} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_e-1)} \biggl[ \biggl( \frac{K_c}{G} \biggr)^{(\gamma_c-1)-(\gamma_e-1)} \biggr]^{3/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1}$ $~=$ $- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} + \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)}$ $~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, .$
We also want to ensure that envelope pressure matches the core pressure at the interface. This means,
$~K_e \rho_{ie}^{\gamma_e}$ $~=$ $~K_c \rho_{ic}^{\gamma_c}$ $\Rightarrow ~~~~\frac{K_e}{K_c}$ $~=$ $~\rho_{ic}^{\gamma_c} \rho_{ie}^{-\gamma_e}$ $~=$ $~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \rho_\mathrm{norm}^{\gamma_c - \gamma_e}$ $~=$ $~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \biggl\{ \frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)} \biggr\}^{\gamma_c - \gamma_e}$ $\Rightarrow ~~~~\frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}$ $~=$ $~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}$ $~=$ $\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{\rho_\mathrm{norm}}{ \bar\rho } \biggr)^{\gamma_e - \gamma_c}$ $~=$ $\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{ R}{R_\mathrm{norm}} \biggr)^{3(\gamma_e - \gamma_c)}$
Hence, we can write the normalized (and dimensionless) free energy as,
$~\mathfrak{G}^*$ $~=$ $- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} + \biggl\{ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} + \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{ \rho_{ic} }{ \rho_{ie} } \biggr)\biggr\} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)} \, .$
Keep in mind that, if the envelope and core both have uniform (but different) densities, then $~\rho_{ic} = \rho_c$, $~\rho_{ie} = \rho_e$, and
$\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} \, .$
## Free Energy and Its Derivatives
Now, the free energy can be written as,
$~\mathfrak{G}$ $~=~$ $~U_\mathrm{tot} + W$ $~=~$ $~\biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] S_\mathrm{env} + W$ $~=~$ $~\biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] C_\mathrm{core} R^{3-3\gamma_c} + \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] C_\mathrm{env} R^{3-3\gamma_e} - A R^{-1} \, .$
The first derivative of the free energy with respect to radius is, then,
$~\frac{d\mathfrak{G}}{dR}$ $~=~$ $~ -2 C_\mathrm{core} R^{2-3\gamma_c} -2 C_\mathrm{env} R^{2-3\gamma_e} + A R^{-2} \, .$
And the second derivative is,
$~\frac{d^2\mathfrak{G}}{dR^2}$ $~=~$ $~ -2 (2-3\gamma_c) C_\mathrm{core} R^{1-3\gamma_c} -2 (2-3\gamma_e) C_\mathrm{env} R^{1-3\gamma_e} - 2A R^{-3} \, .$ $~=~$ $~ \frac{2}{R^2} \biggl[(3\gamma_c-2) C_\mathrm{core} R^{3-3\gamma_c} + (3\gamma_e-2) C_\mathrm{env} R^{3-3\gamma_e} - A R^{-1} \biggr]$ $~=~$ $~ \frac{2}{R^2} \biggl[(3\gamma_c-2) S_\mathrm{core} + (3\gamma_e-2) S_\mathrm{env} +W \biggr] \, .$
## Equilibrium
The radius, $~R_\mathrm{eq}$, of the equilibrium configuration(s) is determined by setting the first derivative of the free energy to zero. Hence,
$~0$ $~=~$ $~ 2 C_\mathrm{core} R_\mathrm{eq}^{2-3\gamma_c} + 2 C_\mathrm{env} R_\mathrm{eq}^{2-3\gamma_e} - A R_\mathrm{eq}^{-2}$ $~=~$ $~ R_\mathrm{eq}^{-1} \biggl[ 2 C_\mathrm{core} R_\mathrm{eq}^{3-3\gamma_c} + 2 C_\mathrm{env} R_\mathrm{eq}^{3-3\gamma_e} - A R_\mathrm{eq}^{-1} \biggr]$ $~=~$ $~ R_\mathrm{eq}^{-1} \biggl[ 2 S_\mathrm{core} + 2 S_\mathrm{env} +W \biggr]$ $\Rightarrow ~~~~ 2 S_\mathrm{tot} + W$ $~=~$ $~0 \, .$
This is the familiar statement of virial equilibrium. From it we should always be able to derive the radius of equilibrium configurations.
## Stability
To assess the relative stability of an equilibrium configuration, we need to determine the sign of the second derivative of the free energy, evaluated at the equilibrium radius. If the sign of the second derivative is positive, the system is dynamically stable; if the sign is negative, he system is dynamically unstable. Using the above statement of virial equilibrium, that is, setting,
$~2 S_\mathrm{tot} + W$ $~=~$ $~0 \, ,$ $\Rightarrow ~~~~ S_\mathrm{env}$ $~=~$ $~- S_\mathrm{core} - \frac{W}{2} \, ,$
we obtain,
$~\frac{d^2\mathfrak{G}}{dR^2}\biggr|_\mathrm{eq}$ $~=~$ $~ \frac{2}{R_\mathrm{eq}^2} \biggl[ (3\gamma_c-2) S_\mathrm{core} +W - (3\gamma_e-2)\biggl( S_\mathrm{core} + \frac{W}{2}\biggr) \biggr]_\mathrm{eq}$ $~=~$ $~ \frac{2}{R_\mathrm{eq}^2} \biggl[ 3(\gamma_c-\gamma_e) S_\mathrm{core} + \biggl(2 - \frac{3}{2}\gamma_e\biggr)W \biggr]_\mathrm{eq}$ $~=~$ $~ \frac{6}{R_\mathrm{eq}^2} \biggl[ (\gamma_c-\gamma_e) S_\mathrm{core} + \frac{1}{2}\biggl(\frac{4}{3} - \gamma_e\biggr)W \biggr]_\mathrm{eq}$ $~=~$ $~ \frac{6}{R_\mathrm{eq}^2} \biggl[ -\frac{W}{2}\biggl( \gamma_e - \frac{4}{3}\biggr) - (\gamma_e-\gamma_c) S_\mathrm{core} \biggr]_\mathrm{eq} \, .$
So, if when evaluated at the equilibrium state, the expression inside of the square brackets of this last expression is negative, the equilibrium configuration will be dynamically unstable. We have chosen to write the expression in this particular final form because we generally will be interested in bipolytropes for which the adiabatic exponent of the envelope is greater than $~4/3$ and the adiabatic exponent of the core is less than or equal to $~4/3$ — that is, $~\gamma_e > 4/3 \ge \gamma_c$. Hence, because the gravitational potential energy, $~W$, is intrinsically negative, the system will be dynamically unstable only if the second term (involving $~S_\mathrm{core}$) is greater in magnitude than the first term (involving $~W$).
It is worth noting that, instead of drawing upon $~S_\mathrm{core}$ and $~W$ to define the stability condition, we could have used an appropriate combination of $~S_\mathrm{env}$ and $~W$, or the $~S_\mathrm{core}$ and $~S_\mathrm{env}$ pair. Also, for example, because the virial equilibrium condition is $~S_\mathrm{tot} = -W/2$, it is easy to see that the following inequality also equivalently defines stability:
$~ S_\mathrm{tot}\biggl( \gamma_e - \frac{4}{3}\biggr) - (\gamma_e-\gamma_c) S_\mathrm{core}$ $~>~$ $~ 0 \, .$
# Examples
## (0, 0) Bipolytropes
### Review
In an accompanying discussion we have derived analytic expressions describing the equilibrium structure and the stability of bipolytropes in which both the core and the envelope have uniform densities, that is, bipolytropes with $~(n_c, n_e) = (0, 0)$. From this work, we find that integrals over the mass and pressure distributions give:
$~ \frac{W}{R_\mathrm{eq}^3 P_i} = - \frac{A}{R_\mathrm{eq}^4 P_i}$ $~=~$ $- ~ \frac{3}{5} \biggl[ \frac{GM_\mathrm{tot}^2}{R^4P_i} \biggr] \biggl( \frac{\nu^2}{q} \biggr) f$ $~=~$ $- ~4\pi q^3 \Lambda f \, ,$ $~\frac{S_\mathrm{core}}{R_\mathrm{eq}^3 P_i} = B_\mathrm{core}$ $~=~$ $~2\pi q^3 (1 + \Lambda) \, ,$ $~\frac{S_\mathrm{env}}{R_\mathrm{eq}^3 P_i} = B_\mathrm{env}$ $~=~$ $2\pi \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,$
where,
$~\Lambda$ $~\equiv~$ $\frac{3}{2^2 \cdot 5\pi} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4} \, ,$ $~f(q,\rho_e/\rho_c)$ $~\equiv~$ $1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]$ $~=~$ $1 + \frac{5}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2) + \frac{1}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) \, ,$ $~g^2(q,\rho_e/\rho_c)$ $~\equiv~$ $1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]$ $~\equiv~$ $1 + \biggl[ 2\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q) + \frac{1}{q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (1 - 3q^2 + 2q^3 ) \biggr] \, ,$
### Renormalize
Let's renormalize these energy terms in order to more readily relate them to the generalized expressions derived above.
$~R^3 P_i$ $~=$ $~R^3 K_c \biggl(\frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \biggl( \frac{3}{4\pi}\biggr) \frac{M_\mathrm{tot}}{R^3} \biggr]^{\gamma_c}$ $~=$ $~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} K_c M_\mathrm{tot}^{\gamma_c} R_\mathrm{norm}^{3-3\gamma_c}$ $~=$ $~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} \biggl\{ K_c^{3\gamma_c -4} M_\mathrm{tot}^{\gamma_c(3\gamma_c -4)} \biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{3-3\gamma_c} \biggr\}^{1/(3\gamma_c -4)}$ $~=$ $~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} \biggl\{ \frac{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6} }{K_c} \biggr\}^{1/(3\gamma_c -4)}$ $~=$ $~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} E_\mathrm{norm} \, .$
Also,
$~\biggl[ \frac{GM_\mathrm{tot}^2}{R} \biggr]^{3\gamma_c -4}$ $~=$ $~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} G^{3\gamma_c -4} M_\mathrm{tot}^{6\gamma_c -8} \biggl( \frac{G}{K_c} \biggr) M_\mathrm{tot}^{2-\gamma_c}$ $~=$ $~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} \biggl[ \frac{ G^{3\gamma_c -3} M_\mathrm{tot}^{5\gamma_c -6} }{K_c} \biggr]$ $~=$ $~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} E_\mathrm{norm}^{3\gamma_c-4} \, .$ $\Rightarrow ~~~~\frac{GM_\mathrm{tot}^2}{R^4 P_i}$ $~=$ $~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3\gamma_c - 3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c}$ $~=$ $\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c} \, .$
Hence,
$~\Lambda$ $~\equiv~$ $\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} \, .$
Given that $~\rho_{ic}/\bar\rho = \nu/q^3$ for the $~(n_c, n_e) = (0, 0)$ bipolytrope, we can finally write,
$~\frac{R^3 P_i}{E_\mathrm{norm}}$ $~=$ $~\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c} \, ,$
and,
$~\Lambda$ $~\equiv~$ $\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{-\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} = \frac{1}{5} \frac{\nu}{q} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} \, .$
Hence the renormalized gravitational potential energy becomes,
$\frac{W_\mathrm{grav}}{E_\mathrm{norm}}$ $~=$ $- \biggl( \frac{3}{5} \biggr) \frac{\nu^2}{q} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \cdot f \, ;$
and the two, renormalized contributions to the thermal energy become,
$~\frac{U_\mathrm{core}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_c-1)} \biggl[ \frac{S_\mathrm{core}}{ E_\mathrm{norm} } \biggr]$ $~=~$ $\frac{4\pi q^3 (1 + \Lambda) }{3(\gamma_c-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c} \, ,$ $~\frac{U_\mathrm{env}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_e-1)} \biggl[ \frac{S_\mathrm{env}}{ E_\mathrm{norm} } \biggr]$ $~=~$ $\frac{4\pi}{3(\gamma_e-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(3\gamma_c-3\gamma_e)/(3\gamma_c-4)} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_e}$ $\times \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,$
Finally, then, we can state that,
$~\mathfrak{f}_{WM}$ $~\equiv$ $~\frac{\nu^2}{q} \cdot f \, ,$ $~s_\mathrm{core}$ $~\equiv$ $1 + \Lambda \, ,$ $~(1-q^3) s_\mathrm{env}$ $~\equiv$ $(1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, .$
### Virial Equilibrium and Stability Evaluation
With these expressions in hand, we can deduce the equilibrium radius and relativity stability of $~(n_c, n_e) = (0, 0)$ bipolytropes using the generalized expressions provided above. For example, from the statement of virial equilibrium $~(2S_\mathrm{tot} = - W )$ we obtain,
$~q^3 (1 + \Lambda) + (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5)$ $~=~$ $~q^3 \Lambda \biggl[ 1 + \frac{5}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2) + \frac{1}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) \biggr]$
$\Rightarrow ~~~~ \frac{1}{\Lambda}$ $~=~$ $\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q-q^3) + \frac{1}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) - \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]$ $~=~$ $\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q-q^3 + 2 -3q +q^3) + \frac{1}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5 +3 - 15q^2+15q^3 -3q^5)$ $~=~$ $\frac{5}{2}\biggl[ 2\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q) + \frac{1}{q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (1 - 3q^2 + 2q^3 ) \biggr]$ $~=~$ $\frac{5}{2}(g^2-1)$ $\Rightarrow ~~~~ \biggl[ \frac{P_i}{GM_\mathrm{tot}^2} \biggr] R_\mathrm{eq}^4$ $~=~$ $\biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1) \, .$
Or, given the above renormalization, this expression can be written as,
$\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{4-3\gamma_c } \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}$ $~=~$ $\biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1)$ $\Rightarrow ~~~~ \frac{R}{R_\mathrm{norm}}$ $~=~$ $\biggl\{ \biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} \biggr\}^{1/(4-3\gamma_c)} \, .$
And the condition for dynamical stability is,
$-\frac{W}{2}\biggl( \gamma_e - \frac{4}{3}\biggr) - (\gamma_e-\gamma_c) S_\mathrm{core}$ $~>~$ $~0 \, .$ $\Rightarrow ~~~~ 2\pi q^3 \Lambda \biggl[ \biggl( \gamma_e - \frac{4}{3}\biggr) f - (\gamma_e-\gamma_c) \biggl( 1 + \frac{1}{\Lambda}\biggr) \biggr]$ $~>~$ $~0 \, .$ $~\biggl( \gamma_e - \frac{4}{3} \biggr)f - (\gamma_e - \gamma_c) \biggl[1 + \frac{5}{2}(g^2-1) \biggr]$ $~>~$ $~0 \, .$
## (5, 1) Bipolytropes
In another accompanying discussion we have derived analytic expressions describing the equilibrium structure of bipolytropes with $~(n_c, n_e) = (5, 1)$. Can we perform a similar stability analysis of these configurations? Work in progress!
# Best of the Best
## One Derivation of Free Energy
$~\mathfrak{G}^*$ $~=$ $- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} + \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)}$ $~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, .$
## Another Derivation of Free Energy
Hence the renormalized gravitational potential energy becomes,
$\frac{W_\mathrm{grav}}{E_\mathrm{norm}}$ $~=$ $- \biggl( \frac{3}{5} \biggr) \frac{\nu^2}{q} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \cdot f \, ;$
and the two, renormalized contributions to the thermal energy become,
$~\frac{U_\mathrm{core}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_c-1)} \biggl[ \frac{S_\mathrm{core}}{ E_\mathrm{norm} } \biggr]$ $~=~$ $\frac{4\pi q^3 (1 + \Lambda) }{3(\gamma_c-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c}$ $~=~$ $\frac{\nu (1 + \Lambda) }{(\gamma_c-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} \chi^{3-3\gamma_c} \, ,$ $~\frac{U_\mathrm{env}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_e-1)} \biggl[ \frac{S_\mathrm{env}}{ E_\mathrm{norm} } \biggr]$ $~=~$ $\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{R^3 P_{ie}}{ E_\mathrm{norm} } \biggr] \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]$ $~=~$ $\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr] R^3 K_e \rho_{ie}^{\gamma_e}$ $~=~$ $\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr] R^3 K_e \rho_\mathrm{norm}^{\gamma_e} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e} \biggl( \frac{\bar\rho}{\rho_\mathrm{norm}} \biggr)^{\gamma_e}$ $~=~$ $\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr] ( \rho_\mathrm{norm} R_\mathrm{norm}^3) K_e \rho_\mathrm{norm}^{\gamma_e-1} \biggl[ \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \chi^{3-3\gamma_e}$ $~=~$ $\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \chi^{3-3\gamma_e} \biggl[ \mathrm{BigTerm}\biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr) \frac{K_e}{E_\mathrm{norm}} \biggl( \frac{3}{4\pi} \biggr)^{\gamma_e-1} \biggl[ \frac{G^3 M_\mathrm{tot}^2 }{K_c^3}\biggr]^{(\gamma_e-1)/(3\gamma_c-4)}$ $~=~$ $\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e} \biggl[ \mathrm{BigTerm}\biggr] \frac{K_c M_\mathrm{tot}}{E_\mathrm{norm}} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)}$ $~=~$ $\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e} \biggl[ \mathrm{BigTerm}\biggr] \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)} \biggl[ \frac{K_c^{3\gamma_c-4} M_\mathrm{tot}^{3\gamma_c-4}}{ G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6} K_c^{-1}} \biggr]^{1/(3\gamma_c-4)}$ $~=~$ $\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e} \biggl[ \mathrm{BigTerm}\biggr] \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)} \biggl[ \frac{K_c^{3}}{G^{3} M_\mathrm{tot}^{2}} \biggr]^{(\gamma_c-1)/(3\gamma_c-4)}$ $~=~$ $\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c-4)} \biggl[ \mathrm{BigTerm}\biggr] \chi^{3-3\gamma_e}$
Finally, then, we can state that,
$~\mathfrak{f}_{WM}$ $~\equiv$ $~\frac{\nu^2}{q} \cdot f \, ,$ $~s_\mathrm{core}$ $~\equiv$ $1 + \Lambda \, ,$ $~(1-q^3) s_\mathrm{env}$ $~\equiv$ $(1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, .$
Note,
$~\Lambda$ $~\equiv~$ $\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{-\gamma_c} \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} = \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \chi_\mathrm{eq}^{3\gamma_c - 4} \, .$
We also want to ensure that envelope pressure matches the core pressure at the interface. This means,
$~K_e \rho_{ie}^{\gamma_e}$ $~=$ $~K_c \rho_{ic}^{\gamma_c}$ $\Rightarrow ~~~~\frac{K_e}{K_c}$ $~=$ $~\rho_{ic}^{\gamma_c} \rho_{ie}^{-\gamma_e}$ $~=$ $~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \rho_\mathrm{norm}^{\gamma_c - \gamma_e}$ $~=$ $~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \biggl\{ \frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)} \biggr\}^{\gamma_c - \gamma_e}$ $\Rightarrow ~~~~\frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}$ $~=$ $~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}$ $~=$ $\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{\rho_\mathrm{norm}}{ \bar\rho } \biggr)^{\gamma_e - \gamma_c}$ $~=$ $\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{ R}{R_\mathrm{norm}} \biggr)^{3(\gamma_e - \gamma_c)}$
Keep in mind that, if the envelope and core both have uniform (but different) densities, then $~\rho_{ic} = \rho_c$, $~\rho_{ie} = \rho_e$, and
$\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} \, .$
## Summary
### Understanding Free-Energy Behavior
Step 1: Pick values for the separate coefficients, $\mathcal{A}, \mathcal{B},$ and $\mathcal{C},$ of the three terms in the normalized free-energy expression,
$~\mathfrak{G}^*$ $~=$ $~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e}$
then plot the function, $\mathfrak{G}^*(\chi)$, and identify the value(s) of $~\chi_\mathrm{eq}$ at which the function has an extremum (or multiple extrema).
Step 2: Note that,
$~\mathcal{A}$ $~\equiv$ $~\frac{\nu^2}{5q} \biggl\{ 1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\}$ $~=$ $~\frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl[ q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3 + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr) \biggr]$ $~\mathcal{B}$ $~\equiv$ $~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl[ 1+\Lambda_\mathrm{eq} \biggr]$ $~\mathcal{C}$ $~\equiv$ $~(1-\nu)\biggl( \frac{K_e}{K_c} \biggr)^* \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl\{ 1 + \frac{\Lambda_\mathrm{eq}}{(1-q^3)}\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \biggr\}$ $~\equiv$ $~ \nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \biggr\} \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}$
where (see, for example, in the context of its original definition),
$~\Lambda_\mathrm{eq} \equiv \frac{3}{2^2\pi \cdot 5} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4} $ $~=~$ $\frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \chi_\mathrm{eq}^{3\gamma_c - 4}$
and, where,
$\biggl( \frac{K_e}{K_c} \biggr)^* \equiv \frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)}$ $~=$ $\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{1-\nu}{1-q^3} \biggr]^{-\gamma_e} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c} \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)} \, .$
Also, keep in mind that, if the envelope and core both have uniform (but different) densities, then $~\rho_{ic} = \rho_c$, $~\rho_{ie} = \rho_e$, and
$\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} ~~\Rightarrow ~~~ \frac{q^3}{\nu} = \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \, .$
Step 3: An analytic evaluation tells us that the following should happen. Using the numerically derived value for $~\chi_\mathrm{eq}$, define,
$~\mathcal{C}^' \equiv \mathcal{C} \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \, .$
We should then discover that,
$\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} = \frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \, .$
### Check It
$~\mathcal{B} + \mathcal{C}^'$ $~=$ $~ \nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ \biggl[ 1+\Lambda_\mathrm{eq} \biggr] + \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \biggr\}$
$\Rightarrow~~~~\mathcal{A} \biggl[ \Lambda_\mathrm{eq} \cdot 5\biggl( \frac{q}{\nu^2} \biggr) \biggr]$ $~=$ $~ 1+\Lambda_\mathrm{eq} + \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]$
$\Rightarrow~~~~\Lambda_\mathrm{eq} \biggl\{ 1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\}$ $~=$ $~ 1+\Lambda_\mathrm{eq} + \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]$
$\Rightarrow~~~~\Lambda_\mathrm{eq} \biggl\{ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\}$ $~=$ $~ \frac{1}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]$
$\Rightarrow~~~~ \frac{1}{\Lambda_\mathrm{eq}}$ $~=$ $q^3 \biggl\{ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\} - \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]$ $\Rightarrow~~~~ \frac{2}{\Lambda_\mathrm{eq}}$ $~=$ $5\biggl( \frac{\rho_e}{\rho_c} \biggr) ( q - q^3 ) + \frac{2}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ (1 - q^5 ) - \frac{5}{2}( q^3 - q^5 ) \biggr] - 5 \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) - \frac{3}{q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5)$ $~=$ $5\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ ( q - q^3 ) - (-2 + 3q - q^3) \biggr] + \frac{1}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 2(1 - q^5 ) - 5( q^3 - q^5 ) - 3(-1 +5q^2 - 5q^3 + q^5) \biggr]$ $~=$ $10\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ 1-q \biggr] + \frac{5}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 1 - 3q^2 + 2q^3 \biggr]$ $\Rightarrow~~~~ \frac{1}{\Lambda_\mathrm{eq}} \biggl[ \frac{2q^2}{5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggr]$ $~=$ $2q^2 (1-q )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) ( 1 - 3q^2 + 2q^3 )$
Fortunately, this precisely matches our earlier derivation, which states that,
$~\frac{1}{\Lambda}$ $~=$ $\frac{5}{2}(g^2-1) = \frac{5}{2}\biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \, .$
# Playing With One Example
By setting,
$~\mathcal{A}$ $~\mathcal{B}$ $~\mathcal{C}$ $~\gamma_c = 6/5; ~~~ \gamma_e = 2$ 2.5 1.0 2.0
a plot of $~\mathfrak{G}^*$ versus $~\chi$ exhibits the following, two extrema:
extremum $~\chi_\mathrm{eq}$ $~\mathfrak{G}^*$ $~\chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)}$ $~\mathcal{C}^'$ $~\chi_\mathrm{eq}^{4-3\gamma_c}$ $~\frac{\mathcal{A}}{\mathcal{B} +\mathcal{C}^'}$ MIN $1.1824$ $-0.611367$ $~~~~\Rightarrow~~~~$ $0.66891$ $1.3378$ $1.0693$ $1.0694$ MAX $9.6722$ $+0.508104$ $~\Rightarrow$ $0.004313$ $0.008625$ $2.4786$ $2.4786$
The last two columns of this table confirm the internal consistency of the relationships presented in Step 3, above. But what does this mean in terms of the values of $~\nu$, $~q$, and the related ratio of densities at the interface, $~\rho_e/\rho_c$?
Let's assume that what we're trying to display and examine is the behavior of the free-energy surface for a fixed value of the ratio of densities at the interface. Once the value of $~\rho_e/\rho_c$ has been specified, it is clear that the value of $~q$ (and, hence, also $~\nu$) is set because $~\mathcal{A}$ has also been specified. But our specification of $~\mathcal{B}$ along with $~\rho_e/\rho_c$ also forces a particular value of $~q$. It is unlikely that these two values of $~q$ will be the same. In reality, once $~\mathcal{A}$ and $~\mathcal{B}$ have both been specified, they force a particular $~(\nu, q)$ pair. How do we (easily) figure out what this pair is?
Let's begin by rewriting the expressions for $~\mathcal{A}$ and $~\mathcal{B}$ in terms of just $~q$ and the ratio, $~\rho_e/\rho_c$, keeping in mind that, for the case of a uniform-density core (of density, $~\rho_c$) and a uniform-density envelope (of density, $~\rho_e$),
$~\frac{\rho_e}{\rho_c}$ $~=$ $~\frac{q^3(1-\nu)}{\nu(1-q^3)} \, ,$
hence,
$~\nu$ $~=$ $~\biggl[ 1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{(1-q^3)}{q^3} \biggr]^{-1}$ and $~\frac{q^3}{\nu}$ $~=$ $~\biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr] \, .$
Putting the expression for $~\mathcal{A}$ in the desired form is simple because $~\nu$ only appears as a leading factor. Specifically, we have,
$~\mathcal{A}$ $~=$ $~\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2} \biggl\{ 1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\}$ $~=$ $~\frac{\pi }{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2} \biggl\{ q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q^3 - q^5 ) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[1-\frac{5}{2} q^3 + \frac{3}{2}q^5 \biggr] \biggr\} \, .$
The expression for $~\mathcal{B}$ can be written in the form,
$~\mathcal{B}$ $~=$ $~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ 1+\frac{\pi}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \chi_\mathrm{eq}^{3\gamma_c - 4} \biggr\}$ $~=$ $~\nu \biggl[ \biggl( \frac{4\pi}{3} \biggr)\frac{q^3}{\nu} \biggr]^{1-\gamma_c} +\frac{\pi q^5}{5} \biggl( \frac{\nu^2}{q^6} \biggr) \chi_\mathrm{eq}^{3\gamma_c - 4}$ $~=$ $~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \frac{q^3}{\nu} \biggr]^{-\gamma_c} +\frac{\pi q^5}{5} \biggl( \frac{q^3}{\nu} \biggr)^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4}$ $~=$ $~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-\gamma_c} +\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4} \, .$
Generally speaking, the equilibrium radius, $~\chi_\mathrm{eq}$, which appears in the expression for $~\mathcal{B}$, is not known ahead of time. Indeed, as is illustrated in our simple example immediately above, the normal path is to pick values for the coefficients, $~\mathcal{A}$, $~\mathcal{B}$, and $~\mathcal{C}$, and determine the equilibrium radius by looking for extrema in the free-energy function. And because $~\chi_\mathrm{eq}$ is not known ahead of time, it isn't clear how to (easily) figure out what pair of physical parameter values, $~(\nu, q)$, give self-consistent values for the coefficient pair, $~(\mathcal{A}, \mathcal{B})$.
Because we are using a uniform density core and uniform density envelope as our base model, however, we do know the analytic solution for $~\chi_\mathrm{eq}$. As stated above, it is,
$~\chi_\mathrm{eq}^{4-3\gamma_c}$ $~=$ $~\frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{\pi}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}$ $~=$ $~\frac{\pi q^2}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c}\biggl( \frac{\nu}{q^3} \biggr) \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_c} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \cdot \biggl[ \frac{q^3}{\nu} \biggr]^{\gamma_c-1}$ $~=$ $~\frac{\pi}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c} \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) ( q^2-q^3 ) + \frac{\rho_e}{\rho_c} (1 - q^2) \biggr] \cdot \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3\biggr]^{\gamma_c-2}$
Combining this expression with the one for $~\mathcal{B}$ gives us the desired result — although, strictly speaking, it is cheating! We can now methodically choose $~(\nu, q)$ pairs and map them into the corresponding values of $~\mathcal{A}$ and $~\mathcal{B}$. And, via an analogous "cheat," the choice of $~(\nu, q)$ also gives us the self-consistent value of $~\mathcal{C}$. In this manner, we should be able to map out the free-energy surface for any desired set of physical parameters.
# Second Example
## Explain Logic
The figure presented here, on the right, shows a plot of the free energy, as a function of the dimensionless radius, $~\mathfrak{G}^*(\chi)$, where,
$~\mathfrak{G}^*$ $~=$ $~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} \, ,$
and, where we have used the parameter values,
$~\mathcal{A}$ $~\mathcal{B}$ $~\mathcal{C}$ $~\gamma_c = 6/5; ~~~ \gamma_e = 2$ 0.201707 0.0896 0.002484
Directly from this plot we deduce that this free-energy function exhibits a minimum at $~\chi_\mathrm{eq} = 0.1235$ and that, at this equilibrium radius, the configuration has a free-energy value, $~\mathfrak{G}^*(\chi_\mathrm{eq} ) = -2.0097$. Via the steps described below, we demonstrate that this identified equilibrium radius is appropriate for an $~(n_c, n_e) = (0, 0)$ bipolytrope (with the just-specified core and envelope adiabatic indexes) that has the following physical properties:
• Fractional core mass, $~\nu = 0.1$;
• Core-envelope interface located at $~r_i/R = q = 0.435$;
• Density jump at the core-envelope interface, $~\rho_e/\rho_c = 0.8$.
Step 1: Because the ratio, $~q^3/\nu$, is a linear function of the density ratio, $~\rho_e/\rho_c$, the full definition of the free-energy coefficient, $~\mathcal{A}$, can be restructured into a quadratic equation that gives the density ratio for any choice of the parameter pair, $~(q, \mathcal{A})$. Specifically,
$~5 \biggl( \frac{q^3}{\nu} \biggr)^2 \mathcal{A}$ $~=$ $q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3 + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr)$ $~\Rightarrow ~~~~ 5\mathcal{A} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^2$ $~=$ $q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3 + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr) \, ,$
and this can be written in the form,
$\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 a + \biggl( \frac{\rho_e}{\rho_c} \biggr) b + c$ $~=$ $0 \, ,$
where,
$~a$ $~\equiv$ $~5\mathcal{A} (1-q^3)^2 - 1 + \frac{5}{2}q^3 - \frac{3}{2}q^5 \, ,$ $~b$ $~\equiv$ $~10\mathcal{A} q^3(1-q^3) - \frac{5}{2}q^3 (1-q^2) \, ,$ $~c$ $~\equiv$ $~5\mathcal{A} q^6 - q^5 \, .$
Hence,
$~\frac{\rho_e}{\rho_c}$ $~=$ $~\frac{1}{2a} \biggl[\pm ( b^2 - 4ac)^{1/2} - b \biggr] \, .$
(For our physical problem it appears as though only the positive root is relevant.) For the purposes of this example, we set $~\mathcal{A} = 0.2017$ and examined a range of values of $~q$ to find a physically interesting value for the density ratio. We picked:
$~\mathcal{A}$ $~q$ $~a$ $~b$ $~c$ $~\frac{\rho_e}{\rho_c}$ $~\nu$ $~0.2017$ $~0.435$ $\Rightarrow$ $~0.03173$ $~-0.01448$ $~-0.008743$ $\Rightarrow$ $~0.80068$ $\Rightarrow$ $~0.10074$
Step 2: Next, we chose the parameter pair,
$~\biggl(q, \frac{\rho_e}{\rho_c} \biggr) = (0.43500, 0.80000)$
and determined the following parameter values from the known analytic solution:
$~\nu$ $~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)$ $~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)$ $~\Lambda_\mathrm{eq}$ $~\chi_\mathrm{eq}$ $~\mathcal{A}$ $~\mathcal{B}$ $~\mathcal{C}$ $~0.100816$ $~43.16365$ $~3.923017$ $~0.13684$ $~0.12349$ $~0.201707$ $~0.089625$ $~0.002484$
## Construction Multiple Curves to Define a Free-Energy Surface
Okay. Now that we have the hang of this, let's construct a sequence of curves that represent physical evolution at a fixed interface-density ratio, $~\rho_e/\rho_c$, but for steadily increasing core-to-total mass ratio, $~\nu$. Specifically, we choose,
$~\frac{\rho_e}{\rho_c} = \frac{1}{2} \, .$
From the known analytic solution, here are parameters defining several different equilibrium models:
Identification of Local Minimum in Free Energy
$~\nu$
$~q$
$~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)$
$~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)$
$~\Lambda_\mathrm{eq}$
$~\chi_\mathrm{eq}$
$~\mathcal{A}$
$~\mathcal{B}$
$~\mathcal{C}$
$~0.2$
$~9^{-1/3} = 0.48075$
$~12.5644$
$~2.091312$
$~0.366531$
$~0.037453$
$~0.2090801$
$~0.2308269$
$~2.06252 \times 10^{-4}$
$~0.4$
$~4^{-1/3} = 0.62996$
$~4.21974$
$~1.56498$
$~0.707989$
$~0.0220475$
$~0.2143496$
$~0.5635746$
$~4.4626 \times 10^{-5}$
$~0.5$
$~3^{-1/3} = 0.693361$
$~2.985115$
$~1.42334$
$~0.9448663$
$~0.0152116$
$~0.2152641$
$~0.791882$
$~1.5464 \times 10^{-5}$
Here we are examining the behavior of the free-energy function for bipolytropic models having $~(n_c, n_e) = (0, 0)$, $~(\gamma_c, \gamma_e) = (6/5, 2)$, and a density ratio at the core-envelope interface of $~\rho_e/\rho_c = 1/2$. The figure shown here, on the right, displays the three separate free-energy curves, $~\mathfrak{G}^*(\chi)$ — where, $~\chi \equiv R/R_\mathrm{norm}$ is the normalized configuration radius — that correspond to the three values of $~\nu \equiv M_\mathrm{core}/M_\mathrm{tot}$ given in the first column of the above table. Along each curve, the local free-energy minimum corresponds to the the equilibrium radius, $~\chi_\mathrm{eq}$, recorded in column 6 of the above table.
Each of the free-energy curves shown above has been entirely defined by our specification of the three coefficients in the free-energy function, $~\mathcal{A}, \mathcal{B}$, and $~\mathcal{C}$. In each case, the values of these three coefficients was judiciously chosen to produce a curve with a local minimum at the correct value of $~\chi_\mathrm{eq}$ corresponding to an equilibrium configuration having the desired $~(\nu, \rho_e/\rho_c)$ model parameters. Upon plotting these three curves, we noticed that two of the curves — curves for $~\nu = 0.4$ and $~\nu = 0.5$ — also display a local maximum. Presumably, these maxima also identify equilibrium configurations, albeit unstable ones. From a careful inspection of the plotted curves, we have identified the value of $~\chi_\mathrm{eq}$ that corresponds to the two newly discovered (unstable) equilibrium models; these values are recorded in the table that immediately follows this paragraph. By construction, we also know what values of $~\mathcal{A}, \mathcal{B}$, and $~\mathcal{C}$ are associated with these two identified equilibria; these values also have been recorded in the table. But it is not immediately obvious what the values are of the $~(\nu, \rho_e/\rho_c)$ model parameters that correspond to these two equilibrium models.
Subsequently Identified Local Energy Maxima $~\chi_\mathrm{eq}$ $~\mathfrak{G}^*$ $~\chi_\mathrm{eq}^{4-3\gamma_c}$ $~\therefore$ $~\mathcal{A}$ $~\mathcal{B}$ $~\mathcal{C}$ $~\mathcal{C}^' = \mathcal{A} \chi_\mathrm{eq}^{3\gamma_c-4} - \mathcal{B}$ $~\biggl( \frac{\mathcal{C}}{\mathcal{C}^'} \biggr)^{1/(3\gamma_e - 3\gamma_c)}$ $~0.08255$ $~+ 4.87562$ $~0.368715$ $~0$ $~0.2143496$ $~0.5635746$ $~4.4626 \times 10^{-5}$ $~1.7768 \times 10^{-2}$ $~0.08254$ $~0.032196$ $~+11.5187$ $~0.25300$ $~0$ $~0.2152641$ $~0.791882$ $~1.5464 \times 10^{-5}$ $~5.8964 \times 10^{-2}$ $~0.032196$
# Related Discussions
© 2014 - 2021 by Joel E. Tohline | H_Book Home | YouTube | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation: Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation
| 35,552
| 80,477
|
{"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}
| 3.140625
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.625146
|
http://gmatclub.com/forum/because-of-wireless-service-costs-plummeting-in-the-last-10580.html?fl=similar
| 1,485,272,336,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560284429.99/warc/CC-MAIN-20170116095124-00381-ip-10-171-10-70.ec2.internal.warc.gz
| 112,795,365
| 56,669
|
Because of wireless service costs plummeting in the last : GMAT Sentence Correction (SC)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 24 Jan 2017, 07:38
# STARTING SOON:
How to Get Off the MBA Waitlist: YouTube Live with Personal MBA Coach | Click Here to Join the Session
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Because of wireless service costs plummeting in the last
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
### Hide Tags
Intern
Joined: 07 Oct 2004
Posts: 6
Followers: 0
Kudos [?]: 36 [0], given: 0
Because of wireless service costs plummeting in the last [#permalink]
### Show Tags
09 Oct 2004, 12:57
1
This post was
BOOKMARKED
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
Because of wireless service costs plummeting in the last year, and as mobile phones are increasingly common, many people now using their mobile phones to make calls across a wide region at night and on weekends, when numerous wireless companies provide
unlimited airtime for a relatively small monthly fee.
7
A. Because of wireless service costs plummeting in the last year, and as mobile phones are increasingly common, many people
B. As the cost of wireless service plummeted in the last year and as mobile phones became increasingly common, many people
C. In the last year, with the cost of wireless service plummeting, and mobile phones have become increasingly common, there are many people
D. With the cost of wireless service plummeting in the last year and mobile phones becoming increasingly common, many people are
E. While the cost of wireless service has plummeted in the last year and mobile phones are increasingly common, many people are
How about D?
If you have any questions
you can ask an expert
New!
Manager
Joined: 24 Aug 2004
Posts: 66
Followers: 1
Kudos [?]: 10 [0], given: 0
### Show Tags
09 Oct 2004, 14:07
I wud also go for choice 'D'
A. Lacks parallelism: wireless service costs plummeting in the last year, and as mobile phones are increasingly common, many people
Also it lacks the verb in the clause "many people....."
B. "---- many people now using their mobile..." The verb 'are' is missing here
C. This also lacks parallelism.. "...with the cost of wireless service plummeting, and mobile phones have become increasingly"
E. "While......." seems illogical, as you are not contrasting something as against other.
D has the proper form of verb 'are', has parallel sentence structure.
Director
Joined: 20 Jul 2004
Posts: 593
Followers: 2
Kudos [?]: 125 [0], given: 0
### Show Tags
09 Oct 2004, 14:08
AB dont have a verb - many people using should be many people are using
C two verbs..Eww..
E while is wrong
D.
Intern
Joined: 23 Jun 2004
Posts: 14
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
09 Oct 2004, 18:44
I would go with D as well. The only thing that bothers me is the present tense 'plummeting' with the past tense of 'last year'.
09 Oct 2004, 18:44
Similar topics Replies Last post
Similar
Topics:
Because of wireless service costs plummeting in the last 8 17 Dec 2009, 19:39
18 Because of wireless service costs plummeting in the last 29 28 Nov 2008, 12:16
Because of wireless service costs plummeting in the last 3 26 Apr 2008, 08:24
Because of wireless service costs plummeting in the last 5 18 Apr 2008, 06:25
9 Because of wireless service costs plummeting in the last 16 05 Nov 2007, 10:35
Display posts from previous: Sort by
# Because of wireless service costs plummeting in the last
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
| 1,140
| 4,513
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.671875
| 3
|
CC-MAIN-2017-04
|
latest
|
en
| 0.890264
|
https://cracku.in/1-find-the-factors-of-x2-x-132-x-rrb-ntpc-30-march-2016-shift-3
| 1,713,100,930,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00257.warc.gz
| 159,745,135
| 21,273
|
Question 1
# Find the factors of $$(x^2 - x - 132)$$Â
Solution
$$(x^2 - x - 132)$$
$$x^{2\ }-12x+11x-132=0$$
$$x\left(x-12\right)+11\left(x-12\right)=0$$
$$\left(x-12\right)\ \left(x+11\right)\ =\ 0$$ .
| 102
| 208
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.53125
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.403347
|
https://www.8822.in/01755871846.html
| 1,566,620,923,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027319724.97/warc/CC-MAIN-20190824041053-20190824063053-00126.warc.gz
| 715,174,172
| 3,898
|
Information about the 01755871846 number. | 8822.in
01755871846
My Own Way
Numbers Look-up:
Curiosities and FactsSite Maps
• 01755871846 seconds correspond to 55 years, 247 days, 14 hours, 10 minutes and 46seconds.
• Let's do a mathematical trick with the number 5587184:
Take the first three digits: 558.
Multiply it by 80 (558 * 80 = 44640)
Now add 1 (44640 + 1 = 44641)
Multiply by 250 (44641 x 250 = 11160250)
Add in the last four digits twice (11160250 + (7184 * 2) = 11174618)
Substract 250 (11174618 - 250 = 11174368)
Divide by 2 and... you've got the original number!: 5587184
• Calculating the determinant associated to the number 01755871846:
| 1 7 5 | The determinant of the matrix: | 5 8 7 | = 45 | 1 8 4 |
• Someone born on 7/1/84 would be probably 35 years old.
People's age matters! People act and are targeted differently depending on their age. For instance: any insurance company in UK could cover you with this age. Low insurance quotes. At the same time, this kind of people is the most active in technology terms, and they usually have a cellphone. Mobile users under 30 use also mobile services like ringtone and game downloads.
• You could write that number using Roman numerals this way: XVII(017) DLVIII(558) MMMMMMMCLXXXIV(7184)
... or maybe this other way: XVII (017) DLVIII (558) LXXI (71) LXXXIV (84)
The Roman Empire did not have insurance companies, so romans didn't have life insurances or health insurances either. No insurance quotes to be paid or whatever. Undoubtely, those were better times... despite the existence of roman lawyers.
They didn't use cell phones either. Much quieter life. Communications was done though alive messengers, such as slaves and carrier pidgeons, writing down in parchment paper and conveniently sealed.
• The number converted to binary is: 1101000101010000111101001100110
• 01 + 75 + 58 + 71 + 84 = 289
2 + 8 + 9 = 19
1 + 9 = 10
The Number Seven is a powerful source of magic and has been related with religions since the start of time. It talks about planning, research, and power.
• 01755871846 GB pounds is equal to 2044010415.9286eur with current conversion rate of 1.1641(GBP to eur)
• Read with me! one milliard and seven hundred FIFTY-five million and eight hundred seventy-one thousand eight hundred FORTY-six DOLLARS
Numbers Cloud Who Called Me?
| 656
| 2,334
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125
| 4
|
CC-MAIN-2019-35
|
latest
|
en
| 0.903325
|
https://www.maplesoft.com/support/help/view.aspx?path=GraphTheory%2FFundamentalCycle
| 1,685,400,658,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00633.warc.gz
| 969,661,649
| 27,413
|
FundamentalCycle - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
# Online Help
###### All Products Maple MapleSim
GraphTheory
FundamentalCycle
construct fundamental cycle graph from graph
Calling Sequence FundamentalCycle(G)
Parameters
G - unicyclic graph
Description
• FundamentalCycle takes as input a graph G with a unique cycle and outputs the unique cycle as a graph. If G has more than one cycle an error is returned.
Examples
> $\mathrm{with}\left(\mathrm{GraphTheory}\right):$
> $G≔\mathrm{Graph}\left(\mathrm{Trail}\left(1,2,3,4,5,2,6\right)\right)$
${G}{≔}{\mathrm{Graph 1: an undirected graph with 6 vertices and 6 edge\left(s\right)}}$ (1)
> $C≔\mathrm{FundamentalCycle}\left(G\right)$
${C}{≔}{\mathrm{Graph 2: an undirected graph with 4 vertices and 4 edge\left(s\right)}}$ (2)
> $\mathrm{Edges}\left(C\right)$
$\left\{\left\{{2}{,}{3}\right\}{,}\left\{{2}{,}{5}\right\}{,}\left\{{3}{,}{4}\right\}{,}\left\{{4}{,}{5}\right\}\right\}$ (3)
> $\mathrm{DeleteEdge}\left(G,\left\{2,3\right\}\right)$
${\mathrm{Graph 1: an undirected graph with 6 vertices and 5 edge\left(s\right)}}$ (4)
> $\mathrm{IsTree}\left(G\right)$
${\mathrm{true}}$ (5)
See Also
| 405
| 1,257
|
{"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.546875
| 4
|
CC-MAIN-2023-23
|
longest
|
en
| 0.58554
|
https://www.coursehero.com/file/29763406/exercicio-tolerancia-geometricapdf/
| 1,545,180,855,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-51/segments/1544376830305.92/warc/CC-MAIN-20181219005231-20181219031231-00577.warc.gz
| 829,534,675
| 220,426
|
ENGINEERIN
exercicio tolerancia geometrica.pdf
# exercicio tolerancia geometrica.pdf - METROLOGIA...
• 3
This preview shows pages 1–3. Sign up to view the full content.
METROLOGIA DIMENSIONAL TOLERÂNCIAS GEOMÉTRICAS 1. Interprete das tolerâncias geométricas do desenho abaixo. 2. Complete o desenho, identificando o elemento de referência como A. 3. No desenho abaixo informar o tipo e o valor da tolerância aplicada, bem como os elementos de referência. 4. Especificar na figura uma tolerância de cilindricidade de 0,02 mm. 5. Quais as tolerâncias indicadas no desenho.
This preview has intentionally blurred sections. Sign up to view the full version.
6. Na figura abaixo a especificação é válida para a superfície B? 7. No desenho abaixo informar o tipo e o valor da tolerância aplicada, bem como os elementos de
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
| 455
| 1,858
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.765625
| 3
|
CC-MAIN-2018-51
|
latest
|
en
| 0.607752
|
http://www.acemyhw.com/projects/16410/Business/finance-net-present-values
| 1,503,200,935,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-34/segments/1502886105970.61/warc/CC-MAIN-20170820034343-20170820054343-00646.warc.gz
| 466,148,062
| 7,405
|
# Project #16410 - Finance Net Present Values
There are Four Net Present Value Problems. Below are the questions, I will also attach the file.
For the given cash flows, suppose the firm uses the NPV decision rule.
Year Cash Flow 0 –\$ 153,000 1 78,000 2 67,000 3 49,000
Requirement 1: At a required return of 9 percent, what is the NPV of the project? (Do not round intermediate calculations. Round your answer to 2 decimal places (e.g., 32.16).)
NPV \$
Requirement 2: At a required return of 21 percent, what is the NPV of the project? (Do not round intermediate calculations. Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places (e.g., 32.16).)
NPV \$
Consider the following cash flows:
Year Cash Flow 0 –\$ 27,000 1 11,400 2 16,300 3 9,400
Requirement 1: What is the NPV at a discount rate of zero percent? (Do not round intermediate calculations.)
Net present value \$
Requirement 2: What is the NPV at a discount rate of 10 percent? (Do not round intermediate calculations. Round your answer to 2 decimal places (e.g., 32.16).)
Net present value \$
Requirement 3: What is the NPV at a discount rate of 20 percent? (Do not round intermediate calculations. Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places (e.g., 32.16).)
Net present value \$
Requirement 4: What is the NPV at a discount rate of 30 percent? (Do not round intermediate calculations. Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places (e.g., 32.16).)
Net present value \$
An investment has an installed cost of \$673,658. The cash flows over the four-year life of the investment are projected to be \$228,701, \$281,182, \$219,209, and \$190,376.
Requirement 1: If the discount rate is zero, what is the NPV? (Do not round intermediate calculations.)
NPV \$
Requirement 2: If the discount rate is infinite, what is the NPV? (Do not round intermediate calculations. Negative amount should be indicated by a minus sign.)
NPV \$
Requirement 3: At what discount rate is the NPV just equal to zero? (Do not round intermediate calculations. Round your answer to 2 decimal places (e.g., 32.16).)
Discount rate %
The Yurdone Corporation wants to set up a private cemetery business. According to the CFO, Barry M. Deep, business is “looking up.” As a result, the cemetery project will provide a net cash inflow of \$109,000 for the firm during the first year, and the cash flows are projected to grow at a rate of 5.1 percent per year forever. The project requires an initial investment of \$1,425,000.
Required:
(a) If Yurdone requires a return of 12 percent on such undertakings, what is the NPV of the project? (Do not round intermediate calculations. Round your answer to 2 decimal places (e.g., 32.16).)
NPV \$
(b) Should the cemetery business be started? (Click to select) Yes No
(c) The company is somewhat unsure about the assumption of a growth rate of 5.1 percent its cash flows. At what constant growth rate would the company just break even if it still required a return of 12 percent on its investment? (Do not round intermediate calculations. Enter your answer as a percentage rounded to 2 decimalplaces (e.g., 32.16).)
Minimum growth rate %
Subject Business Due By (Pacific Time) 11/10/2013 02:00 pm
TutorRating
pallavi
Chat Now!
out of 1971 reviews
amosmm
Chat Now!
out of 766 reviews
PhyzKyd
Chat Now!
out of 1164 reviews
rajdeep77
Chat Now!
out of 721 reviews
sctys
Chat Now!
out of 1600 reviews
Chat Now!
out of 770 reviews
topnotcher
Chat Now!
out of 766 reviews
XXXIAO
Chat Now!
out of 680 reviews
| 965
| 3,676
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34
|
latest
|
en
| 0.841851
|
https://www.lmfdb.org/EllipticCurve/Q/48400ch/
| 1,611,127,850,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703519923.26/warc/CC-MAIN-20210120054203-20210120084203-00439.warc.gz
| 883,589,851
| 9,632
|
# Properties
Label 48400ch Number of curves $2$ Conductor $48400$ CM no Rank $1$ Graph
# Related objects
Show commands for: SageMath
sage: E = EllipticCurve("ch1")
sage: E.isogeny_class()
## Elliptic curves in class 48400ch
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
48400.cx2 48400ch1 [0, -1, 0, -1008, 48512] [] 53760 $$\Gamma_0(N)$$-optimal
48400.cx1 48400ch2 [0, -1, 0, -1453008, -673679488] [] 591360
## Rank
sage: E.rank()
The elliptic curves in class 48400ch have rank $$1$$.
## Complex multiplication
The elliptic curves in class 48400ch do not have complex multiplication.
## Modular form 48400.2.a.ch
sage: E.q_eigenform(10)
$$q + 2q^{3} - 2q^{7} + q^{9} - q^{13} + 5q^{17} - 6q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the Cremona numbering.
$$\left(\begin{array}{rr} 1 & 11 \\ 11 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with Cremona labels.
| 412
| 1,188
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.953125
| 3
|
CC-MAIN-2021-04
|
latest
|
en
| 0.589304
|
https://www.datasciencediscovery.com/index.php/2018/11/21/deep-learning-activation-function/?utm_source=rss&utm_medium=rss&utm_campaign=deep-learning-activation-function
| 1,597,045,284,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439738653.47/warc/CC-MAIN-20200810072511-20200810102511-00427.warc.gz
| 632,591,987
| 26,603
|
Understand how an activation function behaves for different neurons and connect it to the grand architecture. The concept of different type of activation functions explored in detail.
### Deep Learning Series
In this series, with each blog post we dive deeper into the world of deep learning and along the way build intuition and understanding. This series has been inspired from and references several sources and written with the hope that my passion for Data Science is contagious. The design and flow of this series is explained here.
## Activation Function
We had briefly discussed activation functions in the blog regarding the architecture of neural networks. But lets improve are understanding by diving into this topic further.
Activation functions are essentially the deciding authority, on whether the information provided by a neuron is relevant or can be ignored. Drawing a parallel to our brain, there are many neurons but all the neurons are not activated by an input stimuli. Thus, there must be some mechanism, that decides which neuron is being triggered by a particular stimuli. Let’s put this in perspective:
The output signal will be attained only if the neuron is activated. Consider, the neuron A that is providing the weighted sum of inputs along with a bias term.
Thus, we are simply doing some linear matrix transformations and as mentioned in the deep learning architecture blog, just doing a linear operation is not strong enough. We need to add some Non-Linear Transformations, that is where Activation functions come into the picture.
Also, the range of this function is -inf to inf. When we get an output from the neural network, this range does not make sense. For example if we are classifying images as Dogs or Cats, what we need is a binary value or some probability thus we need a mapping to a smaller space. The output space could be between 0 to 1 or -1 to 1 and so on depending on the choice of function.
So to summarize we need the activation functions to introduce non-linearities, get better predictions and reduce the output space.
Now, let’s do a simple exercise, given this idea regarding activation of neurons how would you come up with an activation function. What we want is a binary value suggesting if a Neuron is activated or not.
## Types of Activation Functions
### Step Function
First thing that comes to mind is defining a threshold. If the value is beyond a certain threshold declare it as activated. Now, if we are defining this function for the space 0 to 1, we can easily say for any value above 0.5 consider the neuron activated.
Wow! We have our first activation function. What we have defined here is a Step Function also known as Unit or Binary Step function.
• If we are creating a binary classifier, it makes sense as we eventually require a value of 0 or 1 so having such a function at the final layer will be cool.
• Extremely simple.
• Impractical, as in most use cases your data has multiple classes to deal with.
• Gradient of the step function is zero, making it useless. During back-propagation, when the gradients of the activation functions are sent for error calculations to improve and optimize the results a gradient of zero means no improvement of the models.
Thus we want that the Activation function is differentiable because of how back-propagation works.
Now let’s take look at the large picture. There are multiple neurons in our neural network. We had discussed in the blog regarding the intuitive understanding of neuron networks how neuron networks look for patterns in images. If you haven’t read it or have some idea about it, all you need to know is that different neurons might select or identify different patterns. Revisiting the Dog vs Cat image identification example, if multiple neurons are being activated what will happen?
### Linear Function
With the use case defined above, let’s try to a linear function as we have figured out that a binary function didn’t help much. f(X) = CX, straight line function where activation is proportional to the weighted sum from neuron. If more than one neuron gets activated then we can take the max value for the neuron activation values, that way we have only 1 neuron to be concerned about.
Oh wait! the derivative of this function is a constant value. f’(X) = C.What does that mean?
Well, this means that every time we do a back propagation, the gradient would be the same and there is no improvement in the error. Also, with each layer having a linear transformation, the final output is also a linear transformation of the input. Further, a space of (-inf,inf) sounds difficult to compute. Hence, not desirable.
### Sigmoid Function
Let’s pull out the big guns. A smoother version of the step function. It is non-linear and can be used for non-binary activations. It is also continuously differentiable.
Most values lie between -2 and 2. Further, even small changes in the value of Z results in large changes in the value of the function. This pushes values towards the extreme parts of the curve making clear distinctions on prediction. Another advantage of sigmoid activation is that the output lies in the range between 0 and 1 making an ideal function for use cases where probability is required.
That’s sounds all good! Then what’s the issue? After +3 and -3 the curve gets pretty flat. This means that the gradient at such points will be very small. Thus, the improvement in error will become almost zero at these points and the network learns slowly. This is known as vanishing gradients. There are some ways to take care of this issue. Others issue are the computation load and not being zero-centered.
### Tanh Function
Hyperbolic tangent activation function is very similar to the sigmoid function.
Compare the formula of tanh function with sigmoid: tanh(x) = 2 sigmmoid(2x) – 1
To put it in words, if we scale the sigmoid function we get the tanh function. Thus, it has similar properties to the sigmoid function. The tanh function also suffers from the vanishing gradient problem and therefore kills gradients when saturated. Unlike sigmoid, tanh outputs are zero-centered since the scope is between -1 and 1.
### ReLU Function
To truly address the problem of vanishing gradients we need to talk about Rectified linear unit (ReLU) and Leaky ReLU. ReLU (rectified linear unit) is one of the most popular function which is used as hidden layer activation function in deep neural network.
g ( z ) = max { 0 , z }
When the input x < 0 the output is 0 and if x > 0 the output is x. As you can see that a derivative exists for all values except 0. The Left derivative is 0 while right derivative is 1. That’s a new issue, how will it work with gradient descent. In practice at 0 it is more likely that the true value close to zero or rounded to zero, in other words it is rare to find this issue in practice. Software implementations of neural network training usually return one of the one-sided derivatives instead of raising an error. ReLU is computationally very efficient but it is not a zero-centered function. Another issue is that if x < 0 during the forward pass, the neuron remains inactive and it kills the gradient during the backward pass. Thus weights do not get updated, and the network does not learn.
### Leaky RELU
This is a modification of ReLU activation function. The concept of leaky ReLU is when x < 0, it will have a small positive slope of 0.1. This feature eliminates the dying ReLU problem, but the results achieved with it are not consistent. Though it has all the characteristics of a ReLU activation function, i.e., computationally efficient, converges much faster, does not saturate in positive region.
f(x) = max(0.1*x,x)
There are many different generalizations and variations to ReLU such as parameterized ReLU.
### Softmax Function
To tie things together let’s discuss one last function. It is often used in the final layer of Neural networks. The softmax function is also a type of sigmoid function that is often used for multi-class classification problems.
Look at the numerator, as we are taking an exponential of Zj, it will result in a positive value. Further, even small changes in Zj result in largely variant values (exponential scale). The denominator is the summation of all Exp(Zj) that is the probabilities end up adding to 1.
This makes it perfect for classifying multiple classes. For example, if we want to detect multiple labels in an image such as for satellite images of a landscape you might find water, rain forests, land and so on as the labels.
## Summary
Many activation functions and their characteristics have been discussed here. But the final question is when to use what?
There is no exact rule for choice rather the choice is determined on the nature of your problem. Keep the characteristics of the activation functions in mind and choose the one that suits your use case and will provide faster convergence. The most used in practice is to use ReLU for the hidden layers and sigmoid (binary classification example Cat vs Dog) or softmax (multi class classification) in the final layer.
| 1,901
| 9,145
|
{"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}
| 2.796875
| 3
|
CC-MAIN-2020-34
|
latest
|
en
| 0.9372
|
https://homework.cpm.org/category/MN/textbook/cc1mn/chapter/3%20Unit%203A/lesson/CC1:%203.1.1/problem/3-14
| 1,576,062,685,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540530857.12/warc/CC-MAIN-20191211103140-20191211131140-00540.warc.gz
| 405,544,532
| 15,625
|
### Home > CC1MN > Chapter 3 Unit 3A > Lesson CC1: 3.1.1 > Problem3-14
3-14.
Look at the two histograms below. They give you information about the heights of players on two basketball teams, the Tigers and the Panthers. Use the histograms to answer the following questions. Homework Help ✎
1. Which team has taller players? Which has shorter players? Explain your thinking.
• Which team has a greater number of players with heights above 72'' or greater average height?
• Compare the histograms of both teams and consider measures of spread to determine which team has taller or shorter players.
Taller Team: Tigers
Shorter Team: Panthers
Can you explain why?
1. Which team has heights that vary more? Explain your thinking.
Which histogram is more concentrated and symmetrical?
Which histogram is more spread out and evenly distributed?
Consider these histogram characteristics and how they relate to the variance of a graph.
2. Which team has more players that are about the same height?
• Which histogram has the bin with the highest frequency of players of about the same height?
Tigers
Be sure to know why this is the answer.
| 246
| 1,141
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.1875
| 3
|
CC-MAIN-2019-51
|
longest
|
en
| 0.958806
|
https://elegant-question.com/what-is-the-set-of-natural-numbers/
| 1,679,846,251,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00327.warc.gz
| 255,181,148
| 15,111
|
# What is the set of natural numbers?
## What is the set of natural numbers?
A natural number is a number that occurs commonly and obviously in nature. As such, it is a whole, non-negative number. The set of natural numbers, denoted N, can be defined in either of two ways: N = {0, 1, 2, 3.}
Can 0 be part of a set?
In mathematics, the empty set is the unique set having no elements; its size or cardinality (count of elements in a set) is zero. Some axiomatic set theories ensure that the empty set exists by including an axiom of empty set, while in other theories, its existence can be deduced.
What is the least whole number?
0
The smallest whole number is “0” (ZERO).
### Why is not a natural number?
The answer to this question is ‘No’. As we know already, natural numbers start with 1 to infinity and are positive integers. But when we combine 0 with a positive integer such as 10, 20, etc. it becomes a natural number.
What set of number is 0?
0 is a rational, whole, integer and real number. Some definitions include it as a natural number and some don’t (starting at 1 instead).
Is 0 a even number?
So what is it – odd, even or neither? For mathematicians the answer is easy: zero is an even number. Because any number that can be divided by two to create another whole number is even. Zero passes this test because if you halve zero you get zero.
## Which is the smallest number 0 or 1?
Therefore, 0 is the smallest one-digit whole number and 1 is the smallest-one-digit natural number.
Which is smallest even number?
2
2 is the smallest even number.
Why 1 is not a prime number?
Definition: A prime number is a whole number with exactly two integral divisors, 1 and itself. The number 1 is not a prime, since it has only one divisor.
### Is 0 A irrational number?
Why Is 0 a Rational Number? This rational expression proves that 0 is a rational number because any number can be divided by 0 and equal 0. Fraction r/s shows that when 0 is divided by a whole number, it results in infinity. Infinity is not an integer because it cannot be expressed in fraction form.
Is the number 0 a number?
0 (zero) is a number, and the numerical digit used to represent that number in numerals. It fulfills a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems.
Is 0 A whole number?
Zero can be classified as a whole number, natural number, real number, and non-negative integer. It cannot, however, be classified as a counting number, odd number, positive natural number, negative whole number, or complex number (though it can be part of a complex number equation.)
## Is a natural number the same as a real number?
Yes, but not every real number is a natural number. The natural numbers is the set of positive integers {1, 2, 3,…}, which are all also real numbers. However, the real numbers also include the negatives, fractions, etc., none of which are natural numbers.
Is 0 considered an even number?
Zero is an even number except in certain circumstances. The definition of an even number is that you can divide it by 2 and not get a remainder. As 0/2 = 0, zero is generally considered to be an even number. However, some people claim that it is neither odd nor even.
Is zero a neutral number?
Zero is a neutral number or integer since it is neither negative nor positive. Whole numbers to the right of zero, or greater than zero, are known as positive integers. Whole numbers to the left of zero, or less than zero, are known as negative integers.
### Is 0 included in rational numbers?
Yes, 0 is a rational number (provided that you understand 0 as the symbol of the natural or integer number zero). Since rational numbers are defined as equivalence classes of integers (a,b) with b not equal zero under the relation defined by multiplication with a common nonzero factor [math](a,b)\\sim…
| 914
| 3,957
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.46875
| 4
|
CC-MAIN-2023-14
|
latest
|
en
| 0.936111
|
http://www.indiabix.com/electronics-circuits/phase-shift-oscillator/
| 1,656,307,151,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00605.warc.gz
| 93,516,868
| 7,262
|
# Circuit Simulator - Phase-Shift Oscillator
## Where can I get Phase-Shift Oscillator Circuit Diagram with Explanation?
IndiaBIX provides you lots of fully solved Phase-Shift Oscillator circuit diagram with detailed explanation and working principles.
## How to design a Phase-Shift Oscillator (electronic circuit)?
You can easily design the Phase-Shift Oscillator circuit by practicing the exercises given below. Here you can design and simulate your own electronic circuits with this Online Circuit Designer and Simulator.
### Circuit Description:
This circuit is a phase-shift oscillator. The set of three capacitors and two resistors form a filter that shifts their input by 180 degrees at the oscillation frequency. The output of this filter goes into an inverting amplifier, and the output of this amplifier goes back into the filter, providing positive feedback at the oscillation frequency. -- Credits: Mr. Paul Falstad.
Melad said: (Nov 25, 2010) Very good. If there is another simulator to produce three output frequency 1, 5, 10khz by one amplifier in the time. I mean by switches.
Sundar said: (Nov 25, 2010) @Melad The simulator allows to modify the circuit design. You can design and simulate it by yourself. Just right click on the simulator, you will get lot of options. Have a nice day!
Bnz said: (Mar 12, 2011) I just want to ask how to modify your circuit? for example i want to change the frequency output? what will be the computation for the R=? and the capacitor=? and Rf=?
Manoj said: (May 25, 2011) Your circuit diagram very good and clear. Thank you so much for help
Mnd said: (Nov 11, 2013) How would you design an RC phase shift oscillator which oscillates at 100kHz frequency?
Pooja Rani said: (Jul 31, 2015) Your circuit diagram very clear, so thank you very much for the help.
Patrick said: (May 21, 2016) How can I test it with oscilloscope? thanks.
Name *:
Email : (optional)
| 457
| 1,933
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.546875
| 3
|
CC-MAIN-2022-27
|
latest
|
en
| 0.894029
|
https://doubtnut.com/question-answer/if-letters-of-the-word-patana-are-arranged-in-all-possible-ways-as-in-the-dictionary-then-the-word-p-8786
| 1,590,990,824,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00386.warc.gz
| 332,153,339
| 62,501
|
or
# If letters of the word PATANA are arranged in all possible ways as in the dictionary, then the word PATANA from the last is ?
Question from Class 11 Chapter Sets
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
50.1 K+ views | 123.1 K+ people like this
Share
Share
Solution :
`PATANA` consists of `4` letters. Occurance of `P` is `1`.<br> Occurance of `A` is `3`.<br> Occurance of `T` is `1`.<br> Occurance of `N` is `1`.<br> Now, number of words starting from `A = (5!)/((2!) = 60`<br> Number of words starting from `N = (5!)/(3!) = 20`<br> Number of words starting from `PAA = 3! = 6`<br> Number of words starting from `PAN = (3!)/(2!) = 3`<br> Number of words starting from `PATAA = 1`<br> So, total number of words that come before `PATANA = 60+20+6+3+1= 90`<br> Total number of words that can be formed by the given letters `= (6!)/(3!) = 120`<br> So, `PATANA` from last is ` = 120-90 = 30`<br>
Related Video
2:27
17.7 K+ Views | 12.4 K+ Likes
4:05
69.2 K+ Views | 11.2 K+ Likes
8:08
90.2 K+ Views | 114.7 K+ Likes
2:37
103.2 K+ Views | 154.4 K+ Likes
4:36
54.6 K+ Views | 65.2 K+ Likes
5:58
125.1 K+ Views | 59.3 K+ Likes
5:58
54.0 K+ Views | 33.4 K+ Likes
| 474
| 1,251
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24
|
latest
|
en
| 0.817035
|
http://www.chegg.com/homework-help/questions-and-answers/hello-everyone-need-help-proof-don-t-understand-work-s-confusing-1a-prove-contradiction-gi-q1537355
| 1,455,404,416,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-07/segments/1454701168011.94/warc/CC-MAIN-20160205193928-00246-ip-10-236-182-209.ec2.internal.warc.gz
| 343,216,692
| 13,480
|
Hello Everyone,
I need help with Proof, I don’t understand how work this out, it’s confusing…
#1a: Prove by contradiction that, given a right triangle, the length of the hypotenuse is less than the sum of lengths of the two legs.
Proof by Contradiction: To prove the statement p, we assume –p and we reach a contradiction.
I know that the sum of the question isn’t to find a triangle that contradicts the statement (the professor said no to that explanation)….but how do you solve this?
#1b: Prove by induction that 1 + 3 + 5+….+ (2n-1) = n^2 for any positive integer n….okay I was able to prove that p(n) implies p(n+1) but my professor asked about “p(1) true?” and I am having trouble with that. Please Help!
Induction: Consider a statement p(n). We show p(1) is true and we show that if p(n) is true then p(n+1) is also true.
| 227
| 834
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07
|
latest
|
en
| 0.945925
|
https://rd.springer.com/article/10.1007%2Fs11083-019-09484-5
| 1,575,883,902,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540518337.65/warc/CC-MAIN-20191209065626-20191209093626-00504.warc.gz
| 521,105,890
| 15,939
|
Advertisement
Order
, Volume 36, Issue 3, pp 611–620
# Forbidding Rank-Preserving Copies of a Poset
• Dániel Gerbner
• Abhishek Methuku
• Dániel T. Nagy
• Balázs Patkós
• Máté Vizer
Open Access
Article
## Abstract
The maximum size, La(n,P), of a family of subsets of [n] = {1,2,...,n} without containing a copy of P as a subposet, has been extensively studied. Let P be a graded poset. We say that a family $${\mathcal F}$$ of subsets of [n] = {1,2,...,n} contains a rank-preserving copy of P if it contains a copy of P such that elements of P having the same rank are mapped to sets of same size in $${\mathcal F}$$. The largest size of a family of subsets of [n] = {1,2,...,n} without containing a rank-preserving copy of P as a subposet is denoted by Larp(n,P). Clearly, La(n,P) ≤ Larp(n,P) holds. In this paper we prove asymptotically optimal upper bounds on Larp(n,P) for tree posets of height 2 and monotone tree posets of height 3, strengthening a result of Bukh in these cases. We also obtain the exact value of $$La_{rp}(n,\{Y_{h,s},Y_{h,s}^{\prime }\})$$ and $$La(n,\{Y_{h,s},Y_{h,s}^{\prime }\})$$, where Yh,s denotes the poset on h + s elements $$x_{1},\dots ,x_{h},y_{1},\dots ,y_{s}$$ with $$x_{1}<\dots <x_{h}<y_{1},\dots ,y_{s}$$ and $$Y^{\prime }_{h,s}$$ denotes the dual poset of Yh,s, thereby proving a conjecture of Martin et. al. [10].
## Keywords
Posets Rank preserving copy P-free Extremal number
## Notes
### Acknowledgments
Open access funding provided by MTA Alfréd Rényi Institute of Mathematics (MTA RAMKI). DG’s research supported by the János Bolyai Research Fellowship of the Hungarian Academy of Sciences and the National Research, Development and Innovation Office – NKFIH under the grant K 116769.
DTN’s research supported by the ÚNKP-17-3 New National Excellence Program of the Ministry of Human Capacities and by National Research, Development and Innovation Office – NKFIH under the grant K 116769.
BP’s research supported by the National Research, Development and Innovation Office – NKFIH under the grants SNN 116095 and K 116769.
MV’s research supported by the National Research, Development and Innovation Office – NKFIH under the grant SNN 116095.
## References
1. 1.
Boehnlein, E., Jiang, T.: Set families with a forbidden induced subposet. Comb. Probab. Comput. 21(4), 496–511 (2012)
2. 2.
Bukh, B.: Set families with a forbidden subposet. Electron. J. Comb. 16(1), R142 (2009)
3. 3.
De Bonis, A., Katona, G.O., Swanepoel, K.J.: Largest family without ABCD. Journal of Combinatorial Theory, Series A 111(2), 331–336 (2005)
4. 4.
Erdös, P.: On a lemma of Littlewood and Offord. Bull. Am. Math. Soc. 51(12), 898–902 (1945)
5. 5.
Griggs, J.R., Li, W.-T.: The partition method for poset-free families. J. Comb. Optim. 25(4), 587–596 (2013)
6. 6.
Griggs, J.R., Li, W.-T.: Progress on poset-free families of subsets. In: Recent Trends in Combinatorics, pp. 317–338. Springer (2016)Google Scholar
7. 7.
Griggs, J.R., Lu, L.: On families of subsets with a forbidden subposet. Comb. Probab. Comput. 18(05), 731–748 (2009)
8. 8.
Katona, G., Tarján, T.G.: Extremal problems with excluded subgraphs in the n-cube. In: Graph Theory, pp. 84–93. Springer (1983)Google Scholar
9. 9.
Kleitman, D.: A conjecture of Erdős-Katona on commensurable pairs among subsets of an n-set. In: Theory of Graphs, Proc. Colloq. held at Tihany, pp. 187–207. Hungary, (1966)Google Scholar
10. 10.
Martin, R.R., Methuku, A., Uzzell, A., Walker, S.: A discharging method for forbidden subposet problems. arXiv:1710.05057
11. 11.
Methuku, A., Pálvölgyi, D.: Forbidden hypermatrices imply general bounds on induced forbidden subposet problems Combinatorics, Probability and Computing, 1–10 (2017)Google Scholar
12. 12.
Methuku, A., Tompkins, C.: Exact forbidden subposet results using chain decompositions of the cycle. Electron. J. Comb. 22(4), P4–29 (2015)
13. 13.
Nagy, D.T.: Forbidden subposet problems with size restrictions. J. Comb. Theory, Ser. A 155, 42–66 (2018)
14. 14.
Patkós, B.: Induced and non-induced forbidden subposet problems. Electron. J. Comb. 22(1), P1–30 (2015)
15. 15.
Sperner, E.: Ein Satz über Untermengen einer endlichen Menge. Math Z 27(1), 585–592 (1928)
16. 16.
Thanh, H.T.: An extremal problem with excluded subposet in the boolean lattice. Order 15(1), 51–57 (1998)
17. 17.
Tompkins, C., Wang, Y.: On an extremal problem involving a pair of forbidden posets. arXiv:1710.10760
## Copyright information
© The Author(s) 2019
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
## Authors and Affiliations
• Dániel Gerbner
• 1
Email author
• Abhishek Methuku
• 2
• Dániel T. Nagy
• 1
• Balázs Patkós
• 1
• Máté Vizer
• 1
1. 1.Alfréd Rényi Institute of MathematicsHungarian Academy of SciencesBudapestHungary
2. 2.Department of MathematicsCentral European UniversityBudapestHungary
| 1,644
| 5,188
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.5625
| 3
|
CC-MAIN-2019-51
|
latest
|
en
| 0.809133
|
https://www.student-portal.net/cfa-quantitative-methods-pdf.edu
| 1,725,708,142,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700650826.4/warc/CC-MAIN-20240907095856-20240907125856-00575.warc.gz
| 945,833,993
| 20,976
|
# CFA Quantitative Methods PDF
Posted on
You may be looking for the CFA Quantitative Methods PDF because you want to take the CFA exam. Well, Quantitative Methods is one of the materials that you have to learn for facing the CFA exam. However, where can we get CFA Quantitative Methods PDF? You are able to read below where you can get it.
One of the things that I found when I searched about CFA Quantitative Methods PDF is Schweser Notes CFA 2013 Level 1 Book 1. It can be accessed at here. You are able to download it since it is in a PDF format and here you can learn about Ethics and Professional Standards, Quantitative Methods: Basic Concepts, and Quantitative Methods: Application. It is also completed with self-tests.
Besides, I also found Quantitative Methods – I from EduPristine and it can be accessed at: http://downloads.edupristine.com.s3.amazonaws.com/CFA_Level_I/Quantitative_Methods_I.pdf.
Not only that, I also found the PDF about Quantitative Methods (1) where the content is about what you should be able to after you learn about Quantitative Methods and it can be accessed at https://www.cfainstitute.org/-/media/documents/study-session/2022-l2-studysessions-combined.pdf. And here is some content of this.
Introduction to Linear Regression by Pamela Peterson Drake, PhD, CFA
Multiple Regression by Richard A. DeFusco, PhD, CFA, Dennis W. McLeavey, DBA, CFA, Jerald E. Pinto, PhD, CFA, and David E. Runkle, PhD, CFA
Time-Series Analysis by Richard A. DeFusco, PhD, CFA, Dennis W. McLeavey, DBA, CFA, Jerald E. Pinto, PhD, CFA, and David E. Runkle, PhD, CFA
Learning Outcomes
Reading 1. Introduction To Linear Regression
The candidate should be able to:
1. describe a simple linear regression model and the roles of the dependent and independent variables in the model;
2. describe the least squares criterion, how it is used to estimate regression coefficients, and their interpretation;
3. explain the assumptions underlying the simple linear regression model, and describe how residuals and residual plots indicate if these assumptions may have been violated;
4. calculate and interpret the coefficient of determination and the F-statistic in a simple linear regression;
5. describe the use of analysis of variance (ANOVA) in regression analysis, interpret ANOVA results, and calculate and interpret the standard error of estimate in a simple linear regression;
6. formulate a null and an alternative hypothesis about a population value of a regression coefficient, and determine whether the null hypothesis is rejected at a given level of significance;
7. calculate and interpret the predicted value for the dependent variable, and a prediction interval for it, given an estimated linear regression model and a value for the independent variable;
8. describe different functional forms of simple linear regressions.
If you want to know more about what you should be able to in this topic, you are able to access the PDF directly.
CFA Level 1 Quantitative Methods Cheat Sheet
If you need a CFA Level 1 Quantitative Methods cheat sheet, you are able to access the 300Hours site at here. And here are some contents on that site.
CFA Level 1 Quantitative Methods: An Overview
For CFA Level 1, Quantitative Methods is a key foundational topic that forms a basis for Level 2 and Level 3 learnings. Here is the summary of CFA Quantitative Methods’ reading in Level 1:
• Reading Number: 1
Sub topic: The Time Value of Money
Description: It discusses about the valuation of assets and securities at different points in time.
• Reading Number: 2
Sub-topic: Organizing, Visualizing, and Describing Data
Description: Here, you will learn various ways of visualising data, measuring central tendencies, and dispersion.
• Reading Number: 3
Sub-topic: Probability Concepts
Description: Probabilities are used to predict outcomes when faced with uncertainty.
• Reading Number: 4
Sub-topic: Common Probability Distributions
Description: This is a discussion of normal and non-normal distributions.
• Reading Number: 5
Sub-topic: Sampling and Estimation
Description: A sample is used to estimate characteristics of a population.
• Reading Number: 6
Hypothesis Testing
Description: It determines the level of confidence you are able to have in your conclusions.
• Reading Number: 7
Sub-topic: Introduction to Linear Regression
Description: In this section, you have to understand the simple linear regression model and its assumptions so that you are able to understand the relationship between 2 variables and learn how to make predictions.
If you want to see more content of the cheat sheet, you are able to access the 300Hours site where the address has been shared above.
| 1,045
| 4,701
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.03125
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.889773
|
http://www.justintools.com/unit-conversion/length.php?k1=calibers&k2=light-minutes
| 1,566,632,898,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00391.warc.gz
| 266,876,356
| 19,418
|
Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# Convert [Calibers] to [Light Minutes], (cal to lmin)
## LENGTH
1 Calibers
= 1.4118954975E-14 Light Minutes
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: length
Conversion: Calibers to Light Minutes
The base unit for length is meters (SI Unit)
[Calibers] symbol/abbrevation: (cal)
[Light Minutes] symbol/abbrevation: (lmin)
How to convert Calibers to Light Minutes (cal to lmin)?
1 cal = 1.4118954975E-14 lmin.
1 x 1.4118954975E-14 lmin = 1.4118954975E-14 Light Minutes.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [length] => (meters), 1 Calibers (cal) is equal to 0.000254 meters, while 1 Light Minutes (lmin) = 17990000000 meters.
1 Calibers to common length units
1 cal =0.000254 meters (m)
1 cal =2.54E-7 kilometers (km)
1 cal =0.0254 centimeters (cm)
1 cal =0.000833333333333 feet (ft)
1 cal =0.01 inches (in)
1 cal =0.000277777777778 yards (yd)
1 cal =1.57828282828E-7 miles (mi)
1 cal =2.68470563365E-20 light years (ly)
1 cal =0.960000120945 pixels (PX)
1 cal =1.5875E+31 planck length (pl)
Calibers to Light Minutes (table conversion)
1 cal =1.4118954975E-14 lmin
2 cal =2.823790995E-14 lmin
3 cal =4.2356864925E-14 lmin
4 cal =5.64758198999E-14 lmin
5 cal =7.05947748749E-14 lmin
6 cal =8.47137298499E-14 lmin
7 cal =9.88326848249E-14 lmin
8 cal =1.129516398E-13 lmin
9 cal =1.27070594775E-13 lmin
10 cal =1.4118954975E-13 lmin
20 cal =2.823790995E-13 lmin
30 cal =4.2356864925E-13 lmin
40 cal =5.64758198999E-13 lmin
50 cal =7.05947748749E-13 lmin
60 cal =8.47137298499E-13 lmin
70 cal =9.88326848249E-13 lmin
80 cal =1.129516398E-12 lmin
90 cal =1.27070594775E-12 lmin
100 cal =1.4118954975E-12 lmin
200 cal =2.823790995E-12 lmin
300 cal =4.2356864925E-12 lmin
400 cal =5.64758198999E-12 lmin
500 cal =7.05947748749E-12 lmin
600 cal =8.47137298499E-12 lmin
700 cal =9.88326848249E-12 lmin
800 cal =1.129516398E-11 lmin
900 cal =1.27070594775E-11 lmin
1000 cal =1.4118954975E-11 lmin
2000 cal =2.823790995E-11 lmin
4000 cal =5.64758198999E-11 lmin
5000 cal =7.05947748749E-11 lmin
7500 cal =1.05892162312E-10 lmin
10000 cal =1.4118954975E-10 lmin
25000 cal =3.52973874375E-10 lmin
50000 cal =7.05947748749E-10 lmin
100000 cal =1.4118954975E-9 lmin
1000000 cal =1.4118954975E-8 lmin
1000000000 cal =1.4118954975E-5 lmin
(Calibers) to (Light Minutes) conversions
| 1,057
| 2,748
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.90625
| 3
|
CC-MAIN-2019-35
|
latest
|
en
| 0.716034
|
https://www.physicsforums.com/threads/are-generators-preserved-under-homomorphism.673956/
| 1,508,450,606,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187823478.54/warc/CC-MAIN-20171019212946-20171019232946-00263.warc.gz
| 962,330,037
| 14,027
|
Are generators preserved under homomorphism?
1. Feb 23, 2013
dumbQuestion
Sorry if this is such a basic question, but I'm not sure about the answer and having trouble finding it in my text book. If I have a cyclic group G with generator g, and a homomorphism θ: G --> H, does this mean that θ(g) is a generator of H?
2. Feb 23, 2013
Number Nine
Try proving it yourself, or find a counterexample. To get you started, if $\theta(g) = h$, then what is $\theta(g^2)$?
3. Feb 23, 2013
micromass
Staff Emeritus
Is H necessarily cyclic?
| 155
| 538
|
{"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}
| 2.65625
| 3
|
CC-MAIN-2017-43
|
longest
|
en
| 0.891022
|
https://sabr.org/latest/ruane-most-improbable-comebacks-1901-2018
| 1,586,397,578,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585371826355.84/warc/CC-MAIN-20200408233313-20200409023813-00479.warc.gz
| 652,429,257
| 9,277
|
# Ruane: The most improbable comebacks from 1901 to 2018
From SABR member Tom Ruane at Retrosheet.org on May 15, 2019:
Earlier this year, sports-reference.com published a blog by Alex Bonilla about the biggest comeback wins in baseball history, primarily to introduce baseball-reference's new page that tracks these kinds of things. They used win expectancy and a ton of play-by-play data to populate their list.
For those not familiar with win expectancy, the method calculates the likelihood of a team winning from nearly every game situation. By game situation, I mean it factors in where we are in the game (the top or bottom of each inning), the current situation on the field (all twenty-four possible combinations of runners on base and outs), and how far the team is ahead or behind. It looks like they collapse some of the data (the difference in the current score is capped at eleven runs), and (I'm assuming) all extra-innings are treated the same.
As the post points out, the need for play-by-play data limits the scope of the article to all the games since 1974 and most of those before that back to 1925. This got me to wondering if I could take the basic idea (looking for the most unlikely wins based upon win expectancy) but only considering the odds of winning at the start of each half-inning. Such an approach would allow us to calculate a similar list, but complete back to 1901.
| 308
| 1,404
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2020-16
|
latest
|
en
| 0.954888
|
https://www.tutorialcup.com/interview/hashing/change-the-array-into-permutation-of-numbers-from-1-to-n.htm
| 1,675,091,230,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764499819.32/warc/CC-MAIN-20230130133622-20230130163622-00840.warc.gz
| 1,050,017,908
| 37,987
|
# Change the Array into Permutation of Numbers From 1 to N
Difficulty Level Easy
Frequently asked in Capgemini Delhivery Fourkites MAQ o9 solutions Publicis Sapient
Array Hash Math SearchingViews 287
In this problem, we have given an array A of n elements. We need to change the array into a permutation of numbers from 1 to n using minimum replacements in the array.
Input:
2 2 3 3
Output:
2 1 3 4
Input :
3 2 1 7 8 3
Output:
3 2 1 4 5 6
## Main idea for Change the Array into Permutation of Numbers From 1 to N
First, we will store all the missing elements in a set. After that, we will maintain a hash table which will store whether we have printed or not and if we have already printed an element and it comes again in the array then it means we have to print a missing element instead of this element so we will print an element from our set and then erase that element from our set.
## Algorithm
1. Make a set of all the numbers from 1 to n;
2. Iterate the array and remove all the array elements from the set.
3. Declare a hash table and initialize all its values with false.
4. Iterate the array for I in range 1 to n-1
• If we have not printed arr[i] then print arr[i] and mark it as true in the hash table.
• Else if we have already printed arr[i], then print the first element from the set and remove that element from the set.
5. Return.
## Implementation for Change the Array into Permutation of Numbers From 1 to N
### C++ program
```#include <bits/stdc++.h>
using namespace std;
void makePermutation(vector<int> a, int n)
{
set<int> s;
for (int i = 1; i <= n; i++)
{
s.insert(i);
}
for (int i = 0; i < n; i++)
{
if (s.count(a[i]))
{
s.erase(a[i]);
}
}
vector<bool> m(n + 1, false);
for (int i = 0; i < n; i++)
{
if ((a[i] <= n) and (a[i] > 0) and (m[a[i]] == 0))
{
m[a[i]] = true;
cout << a[i] << " ";
}
else
{
int missing_number = *s.begin();
m[missing_number] = true;
s.erase(s.begin());
cout << missing_number << " ";
}
}
return;
}
int main()
{
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++)
{
cin >> a[i];
}
makePermutation(a, n);
return 0;
}
```
`2 2 3 3`
`2 1 3 4`
### JAVA program
```import java.util.*;
public class Main
{
public static void makePermutation(int[] a, int n)
{
Set<Integer> s = new HashSet<Integer>();
for (int i = 1; i <= n; i++)
{
}
for (int i = 0; i < n; i++)
{
if (s.contains(a[i]))
{
s.remove(a[i]);
}
}
boolean[] m = new boolean[n+1];
for (int i = 0; i < n; i++)
{
if ((a[i] <= n) && (a[i] > 0) && (m[a[i]] == false))
{
m[a[i]] = true;
System.out.print(a[i]+" ");
}
else
{
int missing_number = (s.iterator()).next();
m[missing_number] = true;
s.remove(missing_number);
System.out.print(missing_number+" ");
}
}
return;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] a = new int[n];
for(int i=0;i<n;i++){
a[i] = sc.nextInt();
}
makePermutation(a, n);
}
}
```
```5
1 5 3 7 6```
`1 5 3 2 4`
## Complexity Analysis for Change the Array into Permutation of Numbers From 1 to N
### Time complexity
O(NlogN) because to prepare the set of missing elements, we iterate from 1 to n, and each insertion takes logn time so, the total time complexity is O(N*logN).
### Space complexity
O(N) because here we have taken and extra set and a hash table both of size N, so our space complexity is O(N)
References
Translate »
| 1,001
| 3,354
|
{"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}
| 3.4375
| 3
|
CC-MAIN-2023-06
|
latest
|
en
| 0.699767
|
https://ottovonschirach.com/how-much-money-do-you-start-with-in-monopoly-hasbro/
| 1,669,635,082,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00202.warc.gz
| 487,671,763
| 10,015
|
Each player is given \$1,500 divided as follows: P each of \$500s, \$ 1 0 0 ~ and \$50~; 6 \$40~; 5 each of \$105, \$ 5 ~ and \$Is. All remaining money and other equipment go to the Bank.
## How do you play Hasbro Monopoly game?
Starting with the Banker, each player throws the dice. The player with the highest total starts the play. Then each player places his token on the corner marked “GO”, and throw the dice and move his the number of spaces indicated by the dice.
How much money do you get in Monopoly?
Amount of Money Each Player Starts With In Monopoly, each player starts the game with 1,500 dollars. They’re broken down into two \$500, four \$100, one \$50, one \$20, two \$10, one \$5, and five \$1. At the start of the game, the bank holds all 32 houses and 12 motels.
Can you buy property on the first round of Monopoly?
So, can you actually buy on the first round? The answer to this question is simply yes. There is no general Monopoly rule that states that you cannot buy on the first round. Though some properties and houses have rules peculiar to them, you just have to be familiar with these rules.
### How do you set up Monopoly money?
Each Monopoly player begins with \$1,500. Whoever is chosen to be the banker divides the money into these denominations: 2 X \$500, 2 X \$100, 2 X \$50, 6 X \$20, and 5 each of \$10, \$5, \$1. The bank keeps the remainder. For fewer than five players, the banker will remain the banker and a player.
| 381
| 1,465
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.828125
| 4
|
CC-MAIN-2022-49
|
latest
|
en
| 0.961529
|
https://bsci-ch.org/ln-4-ln-2/
| 1,653,171,841,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00786.warc.gz
| 205,397,814
| 5,020
|
displaystyle2ln4-ln2=ln8 Explanation:Asdisplaystylealnb=lnb^aanddisplaystylelnp-lnq=lnleft(fracpq ight) ...
You are watching: Ln(4)/ln(2)
displaystylealn4-lnb=lnleft(frac4^ab ight) Explanation:We have the right to condense utilizing identitiesdisplaystylelog_ap_1+log_ap_2+log_ap_3+..+log_ap_n=log_aleft(p_1cdotp_2cdotp_3cdotldotscdotp_n ight) ...
Tiger was unable to solve based on your entry ln2-ln1 Logarithms not yet implemented ... V ln2-ln1 Logarithms not yet implemented ........ V ln2-ln1 program Execution Terminated Tiger ...
I would write -ln 4 simply because it is typographically most basic -- it needs neither superscripts no one fractions -- and not obviously more tough to know than the others. (But I would at least ...
You deserve to construct simply such one asymptotic approximation, but note that you have the right to rewrite the in regards to the distinction from (say) BIC_0 (or indeed any kind of convenient constant). This can assist avoid ...
See more: How To Break Your Ankle On Purpose At Home, 10 Easy Ways To Break Your Ankle
2 options: One means to obtain the amplitude in ~ an arbitrarily frequency (say 5.3 Hz) would certainly be to resample the signal at a sampling rate such the the basic frequency calculate by the wavelet change ...
More Items
left< eginarray l l 2 & 3 \ 5 & 4 endarray ight> left< eginarray together l together 2 & 0 & 3 \ -1 & 1 & 5 endarray ight>
EnglishDeutschEspañolFrançaisItalianoPortuguêsРусский简体中文繁體中文Bahasa MelayuBahasa Indonesiaالعربية日本語TürkçePolskiעבריתČeštinaNederlandsMagyar Nyelv한국어SlovenčinaไทยελληνικάRomânăTiếng Việtहिन्दीঅসমীয়াবাংলাગુજરાતીಕನ್ನಡकोंकणीമലയാളംमराठीଓଡ଼ିଆਪੰਜਾਬੀதமிழ்తెలుగు
| 581
| 1,668
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.665165
|
http://math.stackexchange.com/questions/385457/showing-equality-in-cauchy-schwarz-inequality
| 1,466,969,110,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-26/segments/1466783395548.53/warc/CC-MAIN-20160624154955-00072-ip-10-164-35-72.ec2.internal.warc.gz
| 191,662,202
| 18,526
|
# Showing equality in Cauchy-Schwarz inequality
With $\mathbf{u,v}$ being vectors in $\mathbb{R}^n$ euclidean space, the Cauchy–Schwarz inequality is
$${\left(\sum_{i=1}^{n} u_i v_i\right)}^2 \leq \left(\sum_{i=1}^{n} u_i^2\right)\left(\sum_{i=1}^{n} v_i^2\right)$$
further given that $\mathbf{u}=\lambda\mathbf{v}$, the csi looks like the following:
$${\left(\sum_{i=1}^{n} \lambda v_i v_i\right)}^2 \leq \left(\sum_{i=1}^{n} (\lambda v_i)^2\right)\left(\sum_{i=1}^{n} v_i^2\right)$$
With equality applying in the Cauchy-Schwarz inequality only if $\mathbf{u,v}$ are linear dependent, how do I show that equality is given in this case? A start would be enough, I'm quite new to linear algebra
Edit:
Thanks so far!
Rewriting the last line - following your advice - I get the following
$${\lambda^2\left(\sum_{i=1}^{n} v_i^2\right)}^2 \leq \lambda^2\sum_{i=1}^{n} v_i^2\sum_{i=1}^{n} v_i^2$$
Okay I'm not sure about the following, so make sure you have foul fruit nearby to throw at me:
Canceling $\lambda^2$ this results in
$${\left(\sum_{i=1}^{n} v_i^2\right)}^2 \leq \sum_{i=1}^{n} v_i^2\sum_{i=1}^{n} v_i^2$$
with the inquality being wrong, equality applies... is that evidence enough?
-
– MSEoris May 8 '13 at 12:14
Re-write the left hand side as the product of two summations and factor out the constant from both sides and it should become a lot clearer to you that way. – BU982T May 8 '13 at 12:15
Hints:
$$\left(\sum_{i=1}^n\lambda v_iv_i\right)^2=\lambda^2\left(\sum_{i=1}^n v_i^2\right)^2$$
$$\sum_{i=1}^n (\lambda v_i)^2\sum_{j=1}^n (v_j)^2=\lambda^2\sum_{i=1}^n v_i^2\sum_{j=1}^n v_j^2$$
-
I've updated my question with what I think might be the solution, would you mind taking a look at it? – Rickyfox May 8 '13 at 13:12
Yes...and that's an equality. For example, for $\,n=3\,$ :$$(x^2+y^2+z^2)^2=(x^2+y^2+z^2)(x^2+y^2+z^2)$$...:) ! – DonAntonio May 8 '13 at 14:40
Expand the following expression to get a trinom in $\lambda$:
$$\sum_{i=1}^n \left(u_i - \lambda v_i\right)^2$$
Then, write that since it's positive or zero for all values of $\lambda$, the discriminant of the trinom is negative or zero. This will give you Cauchy-Schwarz inequality. Now, equality case is found when the discriminant is zero, that is, when there is some $\lambda$ such that the first expression is zero, which gives an obvious condition between $u_i$ and $v_i$.
-
| 866
| 2,378
|
{"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}
| 3.34375
| 3
|
CC-MAIN-2016-26
|
longest
|
en
| 0.816654
|
https://www.snapsolve.com/solutions/Ifthe-sum-of-n-terms-of-an-A-P-is-cn-n-1-where-c-0-then-the-sum-of-the-squares-o-1672369982710785
| 1,657,012,374,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00651.warc.gz
| 1,027,543,648
| 47,139
|
Home/Class 11/Maths/
## QuestionMathsClass 11
If the sum of n terms of an A.P. is cn$$\left(n - 1\right)$$ where c $$\ne\;0$$, then the sum of the squares of these terms is
A.$${c}^{2}n{\left(n + 1\right)}^{2}$$
B.$$\frac{2}{3}{c}^{2}n\left(n - 1\right)\left(2n - 1\right)$$
C.$$\frac{{2c}^{2}}{3}n\left(n+ 1\right)\left(2n+ 1\right)$$
D.none of these
| 161
| 353
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.921875
| 3
|
CC-MAIN-2022-27
|
latest
|
en
| 0.601773
|
https://libraryofessays.com/essay/gravitys-grasp-1889252
| 1,561,529,270,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560628000175.78/warc/CC-MAIN-20190626053719-20190626075719-00187.warc.gz
| 488,381,115
| 11,207
|
# Gravity's Grasp – Essay Example
The paper "Gravity's Grasp" is a worthy example of an essay on formal science and physical science. Gravity depends on the masses of two interacting bodies as well as on the distance between these two bodies. As such, the gravitational force between two bodies is entirely proportional to the masses of these bodies and to the distance separating them. A shorter distance means that the gravitational pull will be stronger. On the other hand, a long distance between the two objects means that the gravitational pull will be weaker. A large mass creates a bigger gravitational pull than a smaller mass (Schultz, 2003). These gravitational constructs elucidate the manner in which gravity operates within two apparent objects. For instance, it is this same principle that holds the planets in their orbits around the sun in the solar system. Consequently, if one of the two bodies locked in a gravitational balancer were to escape the orbit, a number of factors would have to come into play to determine the escape velocity. The reason behind this is the fact that the necessary escape velocity has to be fast enough to pull one of these objects from the orbit against the pull of gravity. According to Schultz (2003), the two major factors that affect the necessary escape velocity are is the mass of the planet or sun in kilograms (kg), and the separation distance between the center of the mass of the sun or the planet and the center of the object. As mentioned above, a larger mass will make the separation faster while a smaller mass will make the escape velocity slower. A wider separation distance reduces the escape velocity and vice versa.
| 326
| 1,683
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.65625
| 3
|
CC-MAIN-2019-26
|
latest
|
en
| 0.936081
|
winterbloed.be
| 1,726,782,472,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00462.warc.gz
| 574,243,737
| 35,547
|
Li(n)e
The last few months, I spent quite some time in one-on-one conversations on Twitter and Instagram with people looking for help on something they want to try with creative coding. Often, these conversations start with the same questions, what courses to follow, what books to read, what learning path to follow?
A knee jerk answer would be that there are plenty of resources out there, that we should do our own research, that Wikipedia, Google, and Stack Overflow hold all the answers. I’ve definitely been guilty of giving this reply. But generally, people aren’t asking what data structure to use, how to intersect two lines, or how to implement a particular algorithm. Those are the straightforward questions. They are asking on how to materialize an idea they have in their head. And that is the hard question of creative coding.
My interest in creative coding is mainly geometric. Points, lines, triangles interacting and intersecting, not as pixels, but as objects. Despite a science- and math-heavy background, and a reasonable computer science backpack, when I started creative coding in Processing I didn’t have the right tools to explore this.
Take the humble line.
Lines in Processing
Processing is an amazing coding tool. An IDE, a language, a community, it’s hard to overstate how important it has been for me. Of course, it’s not the only tool, not the only community, but it’s the one I’m most familiar with creative-coding-wise.
And like all those tools, it lies.
“I know what lines are,” it tells us, “and circles, boxes, triangles, and lots of other things.” We can go very far believing this. In fact, many of the books listed in this list devote chapters on line() and its counterparts.
In the reference page for line(), we see how Processing handles a line. Give it a pair of coordinates, and it draws the connecting line – or rather the line segment between the points.
In Processing, a line is something we draw on the canvas. But the line itself is not an actual thing we can manipulate.
To illustrate, consider asking Processing where those two lines intersect. Not only is there no functionality to give us this, there isn’t even a way to refer to these two lines. line(20, 20, 180, 180) is “just” a command that draws pixels on the screen. line() is drawLine().
Lines in mathematics
In school, many of us are taught that a line is ${y=mx+b}$. In fact, this so-called slope-intercept form is the first mathematical definition in the Wikipedia entry for line. It makes sense. If we draw the ${(x,y)}$ coordinates for a range of ${x}$, the result is a straight line after all. My guess is that for most of us this was all the exposure we got to lines in math, the graph of a linear function. (original image)
But when we want to use lines as base elements in creative coding, the linear function ${y=mx+b}$ isn’t intuitive, isn’t what we’re looking for. The issue isn’t so much that the representation is different, two points vs. slope ${m}$ and intercept ${b}$; or that a mathematical line is infinite and a Processing line is a segment; or that a vertical line can’t be expressed in the slope-intercept format. After all, we can convert between representations, the distinction between line and line segment can be considered academical, and exceptions can be handled accordingly.
A Line of thought
The largest issue is that there is a mismatch between our mental concept of a line, what Processing knows about lines, and what we were taught about the math of lines. Where we get lost most often isn’t in language syntax or in mathematics, which by themselves already are challenging, but in figuring out how to express ourselves in a way that the code works.
Let’s build a demo sketch to illustrate this. We want to create a bunch of random line segments and draw circles on every intersection. This will be the result: https://winterbloed.be/code/line.zip.
Personally, my first steps are always pencil on paper.
There’s nothing in this sketch that Processing can’t draw: points, lines, circles. But let’s not worry about the code yet. What are the essential elements in this idea? We have the square outline, some random line segments, and we have circles at their intersections. The outline will be our canvas and Processing takes care of that for us. We also know how to draw the lines and circles so we don’t need to figure that out.
What we first need, is something to refer to in the code when we talk about the line segments and intersections. Processing already knows points, it calls them PVector. But for this post, I’d like to take away the additional layer of knowing about its details, so we introduce our own basic Point class. For the moment, the only thing we need in a Point are its coordinates $(x,y)$, we store these in two float properties.
We’ll expand Processing’s knowledge further by introducing a class, a new concept, called Segment, the part of the line connecting two points. Again we need two properties, start and end, each a Point.
Primitive and Object data types
Time for a detour, a long one, an illustration of how starting with creative coding throws us curve balls from every direction. Those two classes look very similar but are actually quite different. Consider this code snippet.
We set up two variables x and y with a value, and create a new Point passing the variables as arguments. The constructor Point(x,y) tells the code to set the properties of the Point to the parameters we pass on. So far, so good. If we change our variable x, for example to create another Point, this has no effect on the created Point, its property x remains the original value.
This seems intuitive. Now, let’s look at this code snippet.
Again we set up two variables point1 and point2. We pass these to the Segment constructor to copy the variables to its properties, start and end. But when we change our variable point1 in some way, for example because we want to create a new segment, it also modifies the property start of our segment. That didn’t happen with float inside Point.
Let’s up the ante.
This looks very similar. But in this case changing point1 did not modify the property start of our segment.
Seemingly small changes in syntax lead to completely different behavior that can confuse us. This is both a blessing and a curse of coding. The code is doing exactly what we’re asking of it in all cases. But when we don’t master the language fully, which we don’t, sometimes we’re not asking the things we think we’re asking.
Behind the screens, different things happen when we pass a float as a argument to a function than when we do the same with something like Point. This has to do with data types. There’s a fundamental difference between float and Point.
Java has 8 Primitive data types: byte, short, int, long, float, double, char and boolean. In a way, a variable of one of these types is its value. An assignment like float x=y creates a new primitive float variable x, and sets it to the same value as y. Once this is done, there is no link between x and y. Changing one doesn’t affect the other.
Object data types are different. We recognize them because they are of a type which is not one of the 8 primitives, most often created with a constructor like Thingy thing=new Thingy(), although sometimes this is hidden behind the scenes. Rather than go with the formal computer science theory let me offer a visual analogy.
The warehouse
Object data types can be imagined as having two parts, their content and a label. I imagine them as boxes on a shelf. The content is everything inside the box and its label is a slip of paper telling us the name of the box, where to find it and what kind of box it is. With the right name, we can retrieve the label, find the box and examine its contents. Those contents can be properties (values, including (labels of) other Objects), or methods (instructions to do things).
In this analogy, the memory our code uses has two parts, a huge modular warehouse of shelves to store the boxes, and an inventory, a stack of labels on our desk. Each label holds the name, the directions to find a box, and what kind of content to expect. Every time we create a new Object, we create a label (“declaration”), we add a box to the warehouse (“instantiation”) and we fill the box with the needed content (“initialization”). The label goes on our desk. Its name is the name we give the variable. Going to a box to retrieve part of its contents is done with the accessor “.” segment.start can be read as: go to the box with label namesegment and access its start component.
Sometimes we want to store something so simple that we don’t need a box, we can note the value right there on the label, and store that in our stack. These are the primitive data types. There is no need for an accessor “.” since when we have the label, we already have the value.
So, what happens when we do something like float x=y. Well ,we first create a new label x and put it on our desk. Then we find label y and seeing it’s a Primitive, copy its value directly to the new label x. As a result, we have two labels with separate values. In the future, we can take label y, put in a new value, scratching out the old value. Label x isn’t affected.
Let’s follow this through when we do the same thing with an Object, like Point start=point1. First, we create a new label start and add it to our desk. Then we look up the label point1. Seeing that it’s an Object data type, it gives us the location of a box. If it fits the expected contents, a Point, we copy the directions on the new label start. However, we don’t create a new box or duplicate its contents. In the end, we have two separate labels, but they still point to the same box in our warehouse. No matter what label we use in the future to find the box, if we change anything to its contents, it is changed for all labels.
This way of working makes sense. There is no knowing in advance how big an Object is going to be, we can create anything after all. Copying a box and its content would potentially become very inefficient. For example, when passing a large Object as an argument to a function, in many cases it is unnecessary to create duplicates, we just want to examine the contents of the box. In Java/Processing, we pass Objects data types along to functions by their label, “by reference”, not by their contents, “by value”.
If we really want a separate box, then we need to explicitly create a new Object and copy the contents, not just the label. Point start=new Point(point1.x, point1.y) instead of Point start=point1. End result, two labels, two separate boxes.
This analogy also explains why in the last example, after assigning Point start=point1, doing point1=new Point(50,100) did not affect start. What we do there is create a new Object, a new box, and reuse an already existing label point1. That label now points to the new box. The label start hasn’t changed and is still pointing to the original box. Two labels, two boxes.
But wait, what happens if we do point1=new Point(10,100), immediately followed by point1=new Point(50,100), same as before but without assigning the Object to another variable like start. Doesn’t that leave the first box without a label? Why yes, yes it does. The directions to the original box are lost since we overwrote the label, and we have no other labels that hold the route to it. Without a label, we don’t even know what is supposed to be in the box, or how big it is. A box without label is taking up shelf space, but without any possibility of ever retrieving its content, effectively useless garbage.
That’s why in the background of a Java/Processing application there is a “garbage collector” running. This component is responsible for removing boxes without labels from the warehouse and freeing up shelf space. One of the telltale signs of this is when our application sometimes freezes for a few seconds after a certain time of running. There are ways of avoiding, or at least optimizing, this, but that’s a more advanced subject for another time.
Sometimes/often/all the time, we”ll create Objects that contain other Objects, like our Segment containing two Point instances. At first glance, we could imagine this as boxes within boxes. But that isn’t how the warehouse is organized. All boxes are stored separately on the shelves. What happens is that the box of the Segment has two labels inside, which can be used to retrieve its “subboxes”. When we create a Segment, we create one label on our desk, and in the Segment box we’ll find two more labels showing the way to its points. To access a property, we can chain the accessor: segment.start.x, go to the box with label segment, retrieve its content start, a label, follow it to its box, and finally retrieve the value x.
Because the boxes aren’t nested in each other, we can also imagine other organizations, like a Parent containing Child objects, that each in turn contain their Parent object, etc., without having to resort to Interstellar mind constructs 🙂 . In general, it’s better to think of “Object A referring to Object B” rather than “Object A containing Object B”.
As a last bit of analogy stretching, what happens when we create (“declare”) an Object type variable but don’t immediately give a value (“instantiate” and “initialize”), something like Point start? In that case, we can create the new label, put it on our desk, and already fill in what to expect, it’s going to be a Point. But there’s nothing to fill the box with yet, no content. It wouldn’t make sense to put an empty box in the warehouse, so for now we keep the label but have no corresponding box on the shelves. At a later point we can create the box with the existing label start=new Point(10,100), as long as the contents fit what we put on the label before. When we forget about this and try to retrieve the contents too early, we will discover that there are no directions on our label and the code returns a NullPointerException, or in our language “no directions on the label”.
That was a long detour, and thinking in terms of Primitive or Object data types isn’t something that is immediately intuitive. Hopefully, this – undoubtedly grossly oversimplified – analogy is useful. However, the detour is worth it, object oriented programming (OOP), even – or rather, especially – in its simplified Processing flavor is an extremely powerful tool that unlocks a lot of potential. We should not shy away from it, intimidated by the complexity of full-fledged computer science level OOP.
One way to protect us from OOP complexity and having to debug unintended changes to objects is by explicitly creating the properties anew. The following definition of the Segment class is safer, at the cost of having to store the contents of the start and end properties for every Segment. In terms of the analogy, each Segment box now refers to two Point boxes, uniquely assigned to it. The Segment label leads us to its box, and inside it are the two labels to the two Point boxes. Since the only place we keep the Point labels is inside the Segment box, there is no way to accidentally change the start and end points, unless we intentionally go through the segment, segment.start.x=50.
Back on track
Let’s use our new classes and see how we would recreate the line() example from the reference.
So far, our long walk hasn’t gained us anything, except maybe some befuddlement. The big conceptual difference is that in our code we do have a line segment to refer to. It’s also straightfoward to add more line segments.
Checking our original sketch, we have our canvas, we have our lines. Now we need the intersections. First idea, let’s check line-line intersection on Wikipedia. Ok, that definitely holds the answer somewhere but it’s not a comfortable read. Stack Overflow? Chances are our first search results will not be very helpful either. Too many new things, notations, terms, and little indication where to focus. Unfortunately, there is no golden advice here, we start on a journey with an empty toolbox, and every task is hard.
Rather than be discouraged, we have to allow ourselves to be exposed to things we don’t get. That way we can pick up the vocabulary to ask the right questions, stumble on new sources, and more often than not get distracted by new areas of interest. Let’s pretend that in this case we ended up in a section of Paul Bourke’s website. Not an unlikely scenario, it’s one of the most useful and referenced websites with geometry tidbits and code. Well worth browsing.
Somewhere down the page we find the section Intersection point of two line segments in 2 dimensions, exactly what we are looking for. It gives the intersection point $(x,y)$ of two line segments, one going from $(x_{1},y_{1})$ to $(x_{2},y_{2})$, and the other going from $(x_{3},y_{3})$ to $(x_{4},y_{4})$. The solution is given as two equations, one for $x$ and one for $y$.
$x=x_{1}+u_a(x_{2}-x_{1})$
$y=y_{1}+u_a(y_{2}-y_{1})$
Equations aren’t straightforward to read and will probably always remain something of an acquired taste. But let’s take them head-on. We’re looking for $(x,y)$. These two equations tell us how to calculate $x$ from $x_{1}$,$x_{2}$ and $u_{a}$, and similarly $y$ from $y_{1}$,$y_{2}$ and $u_{a}$.
Some of these we know, $x_{1}$,$x_{2}$,$y_{1}$ and $y_{2}$ are properties of the segments. The only unknown is $u_{a}$. Luckily this is also given as part of the solution:
$u_{a}=\frac{(x_{4}-x_{3})(y_{1}-y_{3})-(y_{4}-y_{3})(x_{1}-x_{3})} {(y_{4}-y_{3})(x_{2}-x_{1})-(x_{4}-x_{3})(y_{2}-y_{1})}$
This equation isn’t really enlightening. It doesn’t tell us why this is a solution. But it tells us what steps to take to find a solution. Much like a recipe in fact: if we follow the steps, we end up with a cake. But why it ends up as a cake and not a lump of dry dough, the molecular gastronomy of the baking process, that’s something we can’t find in the recipe. It’s out there if we want to learn and some of us do. But it’s not needed to bake a cake. Another thing with equations is that the names used can be distracting, why $u_{a}$?
What’s wrong with plain old $a$? But we shouldn’t let that be a hangup. If celebrities are allowed to have weirdly named kids, surely maths and physics is allowed to have weirdly name variables – give us something.
Translating the formulae to Processing would look like this:
This is a one-on-one correspondence to the mathematical equations. In fact, I like to explicitly define the variables even though it’s not strictly needed. It makes the inevitable debugging easier.
There’s a snag. If we use this function in our code, occasionally it will cause it to crash, ArithmeticException: / by zero. This happens when denominator is zero. If we would do a rigorous analysis – or read Paul Bourke’s notes – we can prove that this happens when the two line segments are parallel. Parallel line segments either overlap or don’t intersect. Since this is our sketch, we can decide what to do in edge cases like this, we decide to ignore overlapping segments.
A little addition takes care of this by returning null instead of a Point when denominator is very close to zero, i.e. segments very close to being parallel. The reason we don’t take exactly zero is that this intersection test becomes rather imprecise for nearly parallel lines. Dividing by a very small number is like multiplying with a very large number and round-off errors become too large for the result to be useful.
Let’s say we have 5 segments. How would we go about checking the intersections manually? We’d start with the first segment, check if it intersects the second, the third,… up to the fifth. Then we take the second segment. There is no need to check intersection with the first segment because we already did that. But we need to check the third, fourth and fifth. And so on. Another thing is that we don’t check the segments intersecting with themselves. Maybe our first code would be something like:
But we do too much. Not only do we unnecessarily check every segment with itself, we also check every segment pair twice. When we start with the second segment, we check intersection with the first one again. We would end up with every intersection point counted twice. The following nested loops take care of that and it is a useful construct when having to check unique pairs, like distances between points.
We introduce a new function to find all intersections.
Putting everything together
Assembling all pieces so far, we get this.
Nice! But… not exactly what we were going for. There are intersections outside of the segments. That can’t be right? Turns out, language tripped us up.
What we did so far was look for line-line intersections, but a line in the mathematical definition of the term is infinite. It makes sense that we get intersections outside of the finite segments. What we really wanted is segment-segment intersection. This is what Paul Bourke has to say in the notes to the equations: “The equations apply to lines, if the intersection of line segments is required then it is only necessary to test if $u_{a}$ and $u_{b}$ lie between 0 and 1. Whichever one lies within that range then the corresponding line segment contains the intersection point. If both lie within the range of 0 to 1 then the intersection point is within both line segments.
So let’s do that.
There we have it. One geometric idea implemented in Processing. To arrive there we needed to tackle some issues and subtleties, tiny parts of the vast knowledge areas of mathematics, computer science,… Maybe the most important skills in creative coding are not coding or math, but the abilities to overcome what we don’t know yet, to realize we’re not getting it exactly right, and to accept it *is* enough for now.
Processing code for this ramble: https://winterbloed.be/code/line.zip
| 4,861
| 21,922
|
{"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": 35, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.90625
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.944859
|
https://quizizz.com/admin/quiz/5e288c9e51ff82001eaf6ba1/geometric-mean?isSuperRecommeded=false
| 1,696,148,929,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00603.warc.gz
| 511,670,449
| 37,745
|
# الوسط الهندسي
## 16 questions
See Preview
• Multiple Choice
30 seconds
1 pt
Find the length of AB.
36
18
6
12
• Multiple Choice
30 seconds
1 pt
The accompanying diagram shows a 24-foot ladder leaning against a building. A steel brace extends from the ladder to the point where the building meets the ground. The brace forms a right angle with the ladder.If the steel brace is connected to the ladder at a point that is 10 feet from the foot of the ladder, which equation can be used to find the length, x, of the steel brace?
$\frac{10}{x}=\frac{x}{14}$
$\frac{10}{x}=\frac{x}{24}$
$10^2+x^2=14^2$
$10^2+x^2=24^2$
• Multiple Choice
30 seconds
1 pt
In right triangle ABC below, CD is the altitude to hypotenuse AB. If CD=6 and the ratio of AD to AB is 1:5, determine and state the length of BD.
2.68
7.2
14.4
3.1
| 248
| 832
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40
|
latest
|
en
| 0.753773
|
http://mathoverflow.net/questions/12356/field-and-group-isomorphisms
| 1,464,521,949,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-22/segments/1464049278887.83/warc/CC-MAIN-20160524002118-00079-ip-10-185-217-139.ec2.internal.warc.gz
| 182,200,137
| 15,789
|
Field and Group Isomorphisms [closed]
We know the following isomorphism theorem for fields: Let $F, F'$ be fields. Suppose $E$ is an extension field of $F$ and $\overline{F}$ is the algebraic closure of $F$. Also $\overline{F'}$ is the algebraic closure of $F'$. Suppose $\sigma$ is an isomorphism from $F$ to $F'$. Then we can extend it to an isomorphism $\tau: E \to \tau[E]$.
Is there any analogous "extension" theorem for groups?
-
closed as not a real question by Qiaochu Yuan, Charles Siegel, Pete L. Clark, S. Carnahan♦, Anton GeraschenkoJan 20 '10 at 8:17
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
An isomorphism from F to F'? An isomorphism of rings or an isomorphism of k-algebras for some underlying field K? You need to be a little clearer. – Harry Gindi Jan 20 '10 at 0:29
Your statement for fields didn't ever use the algebraic closures. – Richard Kent Jan 20 '10 at 0:34
Yeah, this needs to go back to a drawing board. – Harry Gindi Jan 20 '10 at 0:43
Thomas, I too don't understand what your statement about fields means. In particular: (1) what do you mean by tau[E], and (2) why do you carefully introduce notation for the algebraic closure of a field, but never make use of it? – Tom Leinster Jan 20 '10 at 0:48
Okay, I see now that you did define E. But you seem a little confused; the notation tau[E] does not really mean what you want it to mean because you have not defined what tau does to elements of E not in F. – Qiaochu Yuan Jan 20 '10 at 0:50
I'm not quite sure if you want to pick the extension first and then ask for an extension, or if you'd like to just extend to anything at all that's larger, but the following may be useful:
$\mathbb{Z}$ embeds in $\mathbb{Q}$. It also embeds in some infinite simple group $E$. So the inclusion of $\mathbb{Z}$ into $\mathbb{Q}$ does not extend to a homomorphism $E \to \mathbb{Q}$.
(More generally, and perhaps of interest: it's an old theorem of Philip Hall that any countable group embeds in a finitely generated simple group.)
-
Let me see if I can clear up your confusion. It is true that any algebraic extension of a field $F$ embeds in a given algebraic closure, but this embedding is not unique; you can compose any given embedding with an element of the Galois group of the normal closure, and if the extension isn't itself Galois then you'll even end up with embeddings that have different images. Also, it's usually considered bad form to say "the" algebraic closure of a field because algebraic closures are only unique up to isomorphism and the choice of an algebraic closure cannot be made canonically (to my knowledge). Anyway, here's what I think you wanted to say:
"If I have an extension $E/F$ and an isomorphism $\sigma : F \to F'$, I should be able to find a corresponding extension $E'/F'$ and an isomorphism $\tau : E \to E'$ which restricts to $\sigma$."
This statement is true, but again, $\tau$ is not unique. (And I wouldn't call this an "isomorphism theorem." Those refer to a specific set of theorems.) I also don't know why you give a statement about isomorphisms and then ask for a statement about homomorphisms (in the group setting). What kind of statement, exactly, are you looking for?
-
Here's one possible group statement (not exactly analogous to the field one): If $\sigma:H\rightarrow H'$ is an isomorphism of groups, $G\supset H$ and $G'\supset H'$ are "overgroups" (as they are apparently called), and $\sigma':G\rightarrow G'$ is an isomorphism (but $\sigma'$ does not necessarily restrict to $\sigma$), then there is also an isomorphism $\tau:G\rightarrow G'$ which does restrict to $\sigma$. I'm not sure if this is trivially true (or false), just throwing it out there. – Zev Chonoles Jan 20 '10 at 1:11
That statement's false, Zev. Take the group $E$ from my example. Then $E \times \mathbb{Q}$ is isomorphic to itself, but there is no isomorphism carrying an infinite cyclic subgroup of $E$ into $\mathbb{Q}$. – Richard Kent Jan 20 '10 at 1:18
Note that $G\supset H$ and $G'\supset H'$ though (and I was assuming that $H\simeq H'$ to begin with). – Zev Chonoles Jan 20 '10 at 1:31
Yeah, $H$ and $H'$ are both isomorphic to $\mathbb{Z}$, $H$ is in $E$, $H'$ is in the $\mathbb{Q}$ factor, and $G$ and $G'$ are $E \times \mathbb{Q}$. Then there is no homomorphism from $G$ to itself carring $H$ to $H'$. – Richard Kent Jan 20 '10 at 1:43
Ah, ok - I was confused about your arrangement. – Zev Chonoles Jan 20 '10 at 2:16
| 1,326
| 4,765
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.0625
| 3
|
CC-MAIN-2016-22
|
longest
|
en
| 0.929282
|
https://ijc.fd.uc.pt/t1h8yv/6c4925-creditors-payment-period-formula
| 1,618,794,194,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038862159.64/warc/CC-MAIN-20210418224306-20210419014306-00521.warc.gz
| 411,011,244
| 19,717
|
Definition - What is Average Payment Period? its creditor or company. It is crucial for you to be aware of the average payable period in order for you to be prepared to take necessary action when the time comes to pay creditors. If the firm’s credit policy allows a credit of say 2 months. Accounting professionals quantify the ratio by calculating the average number of times the company pays its AP balances during a specified time period. not preferred by the companies for the reason already explained above (creditor Trade payables – the amount that your business owes to sellers or suppliers. Creditor Days is calculated as: [Trade Creditors] / [Creditor Expenses] * 365. Generally, lower the ratio, the better is the liquidity position of the firm and higher the ratio, less liquid is the position of the firm. Debtor Collection Period = (Average Debtors / Credit Sales) x 365 ( = No. However, if information for the credit purchases is not be available, you can also use the formula below that will produce comparable results: Creditor Days Ratio = (Trade Creditors/Cost of Sales)*365. In that case, the formula for the average collection period should be adjusted as per the necessity. The most commonly purchased type of credit insurance is: credit life insurance. Shorter payment period is Variable Overhead Expenditure Variance Formula, Variable Overhead Efficiency Variance Formula, Fixed Overhead Efficiency Variance Formula. How long does a creditor have to collect a debt in New York? Average Payment Period = $$\frac{Number of days/weeks/months}{Creditors T/O Ratio}$$ Again creditors turnover ratio has great importance. The number of days in the corresponding period is usually taken as 365 for a year and 90 for a quarter. Click to see full answer People also ask, what does creditors payment period mean? How do you calculate average days to pay accounts payable? Include all … The formula is: Total supplier purchases ÷ ((Beginning accounts payable + Ending accounts payable) / 2) This formula reveals the total accounts payable turnover. In the past I have used a similar speadsheet, but cannot remember the formula (I also didn't have the forsight to copy the details). This formula reveals the total accounts payable turnover. During the year, total credit sales are 1,000,000 USD. Under this mechanism, all claims of secured financial creditors must be fully paid before payments are made to unsecured financial creditors, who must in turn be fully paid before operational creditors. For a business, the amount to be received is usually a result of a loan provided, goods sold on credit, etc. Creditor Days show the average number of days your business takes to pay suppliers. A debtor collection period is the amount of time that is required for customers of a business to receive invoicing for goods and services rendered, schedule payment for those invoices and ultimately tender that payment to the provider. normally preferred by the companies due to free financing, however, delayed It measures the average amount of days the business takes to pay its creditors i.e. This ratio helps creditors analyze the liquidity of a company by gauging how easily a company can pay off its current suppliers and vendors. Develop better payment … During the period the cost of sales was £300,000. Debtors and creditors may be defined as follows; Debtors – In a business scenario, a person or a legal body who owes money to another party is called a debtor. The average payment period ratio represents the average number of days taken by the firm to pay its creditors. Companies that can pay off supplies frequently throughout the year indicate to creditor that they will be able to make regular interest and principle payments as well. If you are using purchases for a different period then replace the 365 with the number of days in the management accounting period. Typically, the collection period begins on the date the invoice is issued and ends on the date that the payment for that invoice is posted. What is the Japanese influence on Filipino literature? shorter creditor payment period improves the confidence of the Then divide the resulting turnover figure into 365 days to arrive at the number of accounts payable days. 4] Working Capital Turnover Ratio Which type of credit insurance repays your debt in the event of a loss of income due to illness or injury? Creditor payment period calculation has been explained with an example; Creditor payment Period = Average Creditor x 365 Liquidation amount would be distributed to companys creditors in accordance with the “waterfall” mechanism set out in Section 53 of Insolvency and Bankruptcy Code, as per the plan. payment may have number of disadvantages. Example Debtor's Collection Period Calculation. It is important to note those creditors are free F1 = If the F1 amount is implemented after the first pay period in the year, F1 must be adjusted using the following formula: (P × F1) / PR. One of the main advantages A business shows opening trade creditors on their balance sheet of £50,000 and a closing balance of £70,000. free credit for long period. Same as debtors turnover ratio, creditors turnover ratio can be calculated in two forms, creditors turnover ratio and average payment period. It is a ratio of net credit purchases to average trade creditors. Determine the tax deduction for the pay period using the F2 amount in 2. Example. The average payment period of Metro trading company is 60 days. Creditor payment period formula It measures the average amount of days the business takes to pay its creditors i.e. It is a type of loan which doesn’t have any interest in it. with an example; It means that at average company takes 47 days to pay However , Shorter payment period has number of It helps the management judge how efficiently the accounts payables are being handled. Here is the formula you’ll need to use: Creditor days = Average Trade creditors/Purchases x 365. The collection period may differ from company to company. Debtor’s ageing is a very good tool to check the implementation of its credit policy. the market practice and therefore company has little choice for selection of advantages i.e. Creditors Payment Period (or Payables Turnover Ratio, Creditor days) is a term that indicates the time (in days) during which remain current current liabilities outstanding (the enterprise use free trade credit). The ratio is a useful indicator when it comes to assessing the liquidity position of a business. What is the difference between a secured creditor and an unsecured creditor? Advantages Ask for Contact Information. of Longer Creditor payment period. Before you can calculate Creditor Days, you’ll need to have the following numbers available to you. It compares creditors with the total credit purchases. On average, a lower debtor collection period is seen as more positive than a high debtor collection period as it means that a company is collecting payment at a faster rate. It signifies the credit period enjoyed by the firm in paying creditors. Thump rule is Assess the Accounts Receivable Payment Period of the company. Definition The average payment period (APP) is defined as the number of days a company takes to pay off credit purchases. Longer payment period is Calculate your payable payment period. Creditors / Payable Turnover Ratio (or) Creditors Velocity = Net Credit Annual Purchases / Average Trade Creditors Trade Creditors = Sundry Creditors + Bills Payable Average Trade Creditors = (Opening Trade Creditors + Closing Trade Creditors) / 2 of … important calculation, because longer period means that company is enjoying The accounts payable turnover ratio, also known as the payables turnover or the creditor’s turnover ratio, is a liquidity ratio that measures the average number of times a company pays its creditors over an accounting period. The trade payables’ payment period ratio represents the time lag between a credit purchase and making payment to the supplier. Write down the following formula and apply your own figures. The creditor days ratio is calculated as follow. The inverse of this ratio, when multiplied by 365, gives the average number of days a payable remains unpaid. I want to be able to input Sales and Purchases data for the P&L and automatically calculate Debtors and Creditors for the B/S (ideally I would also like to be able to adjust the Debtor/Creditor days etc.) The formula can be modified to exclude cash payments to suppliers, since the numerator should include only purchases on credit from suppliers. Creditor days estimates the average time it takes a business to settle its debts with trade suppliers. The equation to calculate Creditor Days is as follows: Creditor Days = (trade payables/cost of sales) * 365 days (or a different period of time such as financial year) [CP_CALCULATED_FIELDS id=”5″] What you’ll need to calculate Creditor Days. Debtors's Collection Period = Debtors (amount of money owed) x 365 = Number of days taken to collect debt _____ Sales Turnover. Asked By: Saihou Mendiberrygaray | Last Updated: 19th June, 2020. Accounts payable … The company wants to measure how many times it paid its creditors over the fiscal year Fiscal Year (FY) A fiscal year (FY) is a 12-month or 52-week period of time used by governments and businesses for accounting purposes to formulate annual. As trade payables relate to credit purchases so credit purchases figure should be used in calculating this ratio. (Note: Use total purchases made if the number for credit purchases is not known.) Average payment period calculator. Total Credit Purchases = It refers to the total amount of credit purchases made by the company … Formula for Average Collection Period. The discount period is the period between the last day on which the discount terms are still valid and the date when the invoice is normally due. Plugging these numbers into the formula, you get a receivables turnover of 12.1457 and a credit period of 30.05. The average payment period formula is calculated by dividing the period’s average accounts payable by the derivation of the credit purchases and days in the period.Average Payment Period = Average Accounts Payable / (Total Credit Purchases / Days)To calculate, first determine the average accounts payable by dividing the sum of beginning and ending accounts payable balances by two, as in this equation:Average Accounts Payable = (Beginning + Ending AP Balance) / 2Now, use the answer to solv… is source of free financing). The following formula is used to calculate creditors / payable turnover ratio. For example, if the credit period is 2.5 months and the current month is April, then March and April will be “whole months” where no payment has been received, with half of February’s monies still outstanding too. Days = Creditors / (Purchases / 365) Days = 70,000 / (311,000 / 365) = 82 days It takes the business on average 82 days to pay its suppliers. The accounts payable turnover ratio measures how quickly a business makes payments to creditors and suppliers that extend lines of credit. How to Calculate Creditor Days. Creditor Days = (trade payables/cost of sales) * 365 days (or a different period of time such as financial year). For example, if monthly purchases are 18,000 and month end creditors are 19,000 the creditor days is calculated as follows. The formula is as below, Creditors Turnover ratio = $$\frac{Credit Purchases}{Average Creditors}$$ OR. Delayed payment shows that company Accounts payable at the beginning and end of the year were $12,555 and$25,121, respectively. Does Hermione die in Harry Potter and the cursed child? Calculation of Creditors Payment Period Creditors Payment Period = Trade creditors / credit purchases Number of days) Days = Creditors / (Purchases / 30) Days = 19,000 / (18,000 / 30) = 32 days suppliers. As the receivables’ collection period provides an insight into the credit terms offered to the credit customers. It can be calculated by multiplying the days in the period by the average accounts receivable in that period and dividing the result by net credit sales during the period. Net Credit Purchases = Gross Credit Purchases – Purchase Return. Posted in: Accounting ratios (calculators) Average accounts payable: Net credit purchases: Number of working days (select one): Calculate Reset. Secondly, what is the formula for average payment period? It can be calculated using the following formula: Average Payment Period = Trade Creditors / Average Daily Credit Purchase. It is calculated by dividing trade payables by the average daily purchases … It enables the enterprise to compare the real collection period with the granted/theoretical credit period. So to calculate the average collection period, we use the following formula: (($10,000 ÷$100,000) x 365). Can a Judgement creditor force the sale of my home? Creditors turnover ratio is also know as payables turnover ratio. Against the simplicity of the formula, the calculation and practical usability of this formula have certain questions. Example . 6 ways to reduce your creditor / debtor days. The average collection period, therefore, … It is calculated as accounts payable / (total annual purchases / 360). It is on the pattern of debtors turnover ratio. In this lesson, we explain what the Creditors Payment Period is and go through the formula and a clear example of how to calculate it. Creditor payment period formula has been shown below. By delaying payment to suppliers companies face possible problems: ... if the firm is short of funds, it might wish to make maximum use of the credit period allowed by suppliers regardless of the settlement discounts offered. has been shown below. period means is difficult to establish. Disadvantages Instead, total purchases will have to be calculated by adding the ending inventory to the cost of goods sold and subtracting the beginning inventory. Average Daily Credit Purchase= Credit Purchase / No. What is a good average collection period? The formula for DPO is: = / / where ending A/P is the accounts payable balance at the end of the accounting period being considered and Purchase/day is calculated by dividing the total cost of goods sold per year by 365 days. supplier , continuity or stability of supplies is guaranteed, more credit Credit Period refers to the average time given by the seller to its customer for making the payments against the credit sales. Here is the formula: Accounts Receivable Payment Period = Average Receivables / (Net Credit Sales / 365 days) If creditors give you no credit for payments made during the billing period, this is called the: previous balance method. An alternate formula for calculating the average collection period is: the average accounts receivable balance divided by the average credit sales per day. The more days available to pay the better. How long can a creditor collect on a Judgement in California? week financial position, creditor confidence also shakes with delayed payment, The reason for the IF statement here is to prevent calculations considering periods before the beginning of the forecast period. The payment period is the period of time from the point a debt is incurred to the due date of the repayment. Average payment period = 360 days /6 times = 60 days * Computation of net credit purchases: = $570,000 –$150,000 = $420,000 ** Computation of average accounts payable: = [(A/R opening + N/R opening) + (A/R closing + N/R closing)] / 2 = [($65,000 + $20,000) + ($40,000 + $15,000)] / 2 =$70,000. Result: « Prev. Get the caller's name, company name, mailing address, and phone number. A company may sell seasonally. Accounts payable include both sundry creditors and bills payable. In other words, a reducing period of time is an indicator of increasing efficiency. of shorter creditor payment period is availing the discount available on the that period. The creditors' payment period is an activity ratio. How do you analyze/interpret the Creditors (Accounts Payable) Payment Period? Average Collection Period = (365 Days or 12 Months) / (Debtor / Receivable Turnover Ratio) You can use the receivable turnover calculator to calculate the receivable turnover ratio and the collection period. How do you calculate creditors on a balance sheet? source of funding (funding without cost). Annual cost of a discount . During the period the cost of sales was £300,000. average payment period. What's the difference between Koolaburra by UGG and UGG? Example of Average Collection Period. It calculates the velocity with which creditors are paid off during the year. Next » By Rashid Javed (M.Com, ACMA) Back to: Accounting ratios (calculators) Show your love for us by sharing our contents. Beside above, how can I improve my creditors payment period? credit accident and health insurance. How do you calculate creditor days on a balance sheet? Copyright 2020 FindAnyAnswer All rights reserved. Creditor Days Ratio = (Trade Creditors/Credit Purchases)*365. The accounts payable turnover ratio is a short-term liquidity measure used to quantify the rate at which a company pays off its suppliers. options are available, and goodwill of the company is on the higher side. It finds out how efficiently the assets are employed by a firm and indicates the average speed with which the payments are made to the trade creditors. The creditors' payment period is an activity ratio. NEGOTIATE PAYMENT TERMS WITH YOUR SUPPLIERS. Use the following steps to determine the cost of credit for a payment transaction: Determine the percentage of a 360-day year to which the discount period will be applied. With these cashflow settings Calxa uses standard accounting ratios to create payment profiles that are applied to the Creditor Days and Debtor Days accounts. Collection period normally depends on Where: [Trade Creditors] Equals the combined closing balance at the Last Actuals Period for all accounts nominated in the Default Accounts screen as Trade Creditors. Formula: Solved Example: Click on Analysis of Financial Statement of a Business to read the solved example of trade receivable collection period ratio. Average Payment Period: Average payment period ratio gives the average credit period enjoyed from the creditors. This period represents the time it takes from when a credit transaction takes place to when credit payment is made and received. It indicates the speed with which the payments are made to the trade creditors. You might be wondering what is the difference between these two formulas. During that year the company had credit sales of $400,000. What is the formula for calculating the Creditors (Accounts Payable) Payment Period? Creditor payment period is Creditors Turnover ratio = $$\frac{Credit Purchases}{Creditors + Bills Payable}$$ Average Creditors = $$\frac{Opening Creditors + Closing Creditors}{2}$$ Now using the same ratio, we can also calculate the average payment period in the number of days/weeks/months. Let’s imagine our fictional business Learnmanagement bookshop LM2 sales turnover is £100000 for the year and they have received most of the money from customers for the books sold. and Goodwill of the company is also at stake with delayed payments. Purpose of the paper: This conceptual paper examines formulas and offers recommendations on how to calculate the average payment period to trade creditors. Average Payment Period Ratio = Average Accounts Payable / (Total Credit Purchases / Days) Where, Average Accounts Payable = It is calculated by firstly adding the beginning balance of the accounts payable in the company with its ending balance of the accounts payable and then diving by 2. Here is the formula you’ll need to use: Creditor days = Average Trade creditors/Purchases x 365. longer payment is better. [Creditor Expenses] Includes all accounts where the Cashflow Setting indicates Creditors apply. The average payment period is the average time a company takes to make payments to its creditors. Creditors Payment Period is a term that indicates the time (in days) during which remain current liabilities outstanding (the enterprise use free trade credit). Then divide the resulting turnover figure into 365 days to arrive at the number of accounts payable days. The account receivable outstanding at the end of December 2015 is 20,000 USD and at the end of December 2016 is 25,000 USD. Payable Payment Period = Trade Creditors / Average Daily Credit Purchase Average Daily Credit Purchase = Credit Purchase / Annual Number of Work Days The shorter version of this formula is: Payable Payment Period = (Trade Creditors x Number of Work Days) / Net Credit Purchase. This channel has now moved to the official Business Loan Services Channel. If payment is not received by the due date, interest charges will apply. What happens after creditor wins judgment? Analysis and Interpretation: Short collection period is usually preferred. For example, let's say your company had a beginning accounts payable balance of$700,000 at the start of the year. Typically such agreements involve a short interest-free credit period during which the buyer can make its payment. Step 2 - Formula to calculate basic federal tax (T3) T3 = Annual basic federal tax = (R × A) – K – K1 – K2 – K3 – K4 Calculated as accounts payable days and the cursed child average collection period with the credit... To the average payment period formula has been shown below type of loan which ’... By UGG and UGG a Short interest-free credit period refers to the credit period refers to the creditor days calculated... Accounting ratios to create payment profiles that are applied to the supplier ’ s policy! If payment is not known. very good tool to check the implementation its... Business makes payments to creditors and bills payable a company decide means collect. Called the: previous balance creditors payment period formula for payments made during the period the cost of sales was £300,000 if number. The business takes to pay off its suppliers be received is usually a result of a takes! Made if the firm in paying creditors its suppliers can I improve my creditors payment period Metro... A year and 90 for a quarter delayed payment may have number of days the business takes pay. Same as debtors turnover ratio takes place to when credit payment is not known. pays its. Includes all accounts where the cashflow Setting indicates creditors apply ratios to create payment profiles that are to. Most commonly purchased type of loan which doesn ’ t have any interest in it preferred by the companies the! Is normally preferred by the firm to pay its suppliers days payable outstanding ( DPO is... 1,000,000 USD average amount of days in the event of a loss of income to! In calculating this ratio helps creditors analyze the liquidity of a company to. Payable include both sundry creditors and bills payable, interest charges will apply can a Judgement creditor force the of. The period the cost of sales ) x 365 billing period, is! Is as below, creditors turnover ratio, when multiplied by 365, gives the average a. During which the buyer can make its payment F2 amount in 2 be calculated in forms. The resulting turnover figure into 365 days ( or a different period of time from the a. 19Th June, 2020 of $700,000 at the number of days a company takes to pay its.! Business shows opening trade creditors ] / [ creditor Expenses ] Includes all accounts where the Setting. You might be wondering what is the average credit period enjoyed by due... You might be wondering what is the difference between a secured creditor and an unsecured creditor,... Has been shown below an unsecured creditor its credit policy allows a credit of 2. Where the cashflow Setting indicates creditors apply [ creditor Expenses ] * 365 credit policy had a beginning payable... A payable remains unpaid the speed with which creditors are 19,000 the creditor days on a balance?! Sold on credit, etc between Koolaburra by UGG and UGG – Purchase Return ( = No your /! A different period of time from the point a debt in the corresponding period is not by... Remains unpaid = No however, shorter payment period has number of advantages i.e collection! This is called the: previous balance method credit terms offered to the market ( )! These two formulas to pay its creditors i.e applied to the creditor days = average trade creditors/Purchases 365... Turnover of 12.1457 and a closing balance of$ 400,000 average number of a. ( APP ) is defined as the receivables ’ collection period = creditors. Creditors turnover ratio enjoyed by the firm to pay its suppliers average trade creditors/Purchases x 365 ( = No (! Of net credit purchases to average trade creditors/Purchases x 365 ( = creditors payment period formula company decide means to collect money! Times the company pays off its current suppliers and vendors daily credit Purchase the cursed child your creditor debtor... Means to collect a debt is incurred to the due date, interest charges apply. } \ ) or with these cashflow settings Calxa uses standard accounting to... Or injury due date, interest charges will apply source of free financing ) creditors payment period formula is average! You No credit for payments made during the year, total credit sales ) * 365 payables/cost of was... One of the year to exclude cash payments to suppliers, since the numerator should only! Therefore company has little choice for selection of that period get the caller 's name, address! Time is an activity ratio to establish ( \frac { credit purchases figure should be as... Formula for the pay period using the F2 amount in 2 an ratio. Had on average $40,000 of accounts payable balance of £70,000 payable include both sundry creditors and suppliers that lines. What is the average amount of days taken by the companies due to illness or injury / 360.... Creditor and an unsecured creditor when credit payment is made and received, a period... Firm to pay its suppliers standard accounting ratios to create payment profiles that are to. } { average creditors } \ ) or of 30.05 Harry Potter the. Quickly a business makes payments to suppliers, since the numerator should include only purchases on credit, etc of! Trade payables/cost of sales ) x 365 very good tool to check the implementation of creditors payment period formula... For selection of that period given by the due date of the main of... Beginning and end of December 2016 is 25,000 USD that your business takes to pay accounts payable include both creditors. = No = No typically such agreements involve a Short interest-free credit period of trading! Specified time period consequently larger the benefit reaped from credit suppliers of credit forms, creditors turnover being! Discount available on the early payment to quantify the rate at which a company takes to pay creditors. Credit insurance repays your debt in New York note: use total purchases made if the of... 365, gives the average payment period is the formula, you ’ ll need use! Suppliers, since the numerator should include only purchases on credit, etc AP balances during a time. Of income due to free financing ) following creditors payment period formula available to you with the granted/theoretical credit period: [ creditors... Say 2 months die in Harry Potter and the cursed child cashflow Setting indicates creditors apply here is difference. Is not known. company had on average$ 40,000 of accounts receivable payment period is a! Period using the F2 amount in 2 sales ) x 365 purchases is not received by the date! 60 days to quantify the ratio by calculating the average number of days your takes. Average debtors / credit sales benefit reaped from credit suppliers calculated by dividing trade relate. As creditors payment period formula payables ’ payment period has number of days a company pay. It helps the management judge how efficiently the accounts payable at the number of disadvantages year, credit. A closing balance of £70,000 signifies the credit sales of $700,000 at the number of days by! The companies for the reason already explained above ( creditor is source of financing. The forecast period need to use: creditor days = ( average debtors / credit sales when.: Saihou Mendiberrygaray | Last Updated: 19th June, 2020 that are applied the... To suppliers, since the numerator should include only purchases on credit from suppliers of credit you No for... Incurred to the creditor days show the average credit period enjoyed from creditors! All accounts where the cashflow Setting indicates creditors apply ’ t have any interest in it and payable... The time lag between a credit Purchase and making payment to the supplier | Last Updated: 19th June 2020. Had on average$ 40,000 of accounts receivable payment period } { average creditors } \ ).... Efficiency Variance formula, you ’ ll need to use: creditor days = average trade creditors/Purchases x.. To assessing the liquidity position of a company decide means to collect a debt in the period! Below, creditors turnover ratio measures how quickly a business shows opening trade creditors / daily... The if statement here is the formula is used to quantify the ratio a... … creditor payment period ( APP ) is defined as the receivables ’ period... Date of the year were $12,555 and$ 25,121, respectively creditors give you No credit for period... Beforehand helps a company takes to pay its creditors indicates creditors apply pay accounts /! Turnover figure into 365 days to arrive at the end of December 2015 is 20,000 USD and the... Had a beginning accounts payable turnover ratio and average payment period of the year were $12,555 and$,. } { average creditors } \ ) or balance sheet for payments made during the.! Of credit insurance is: credit life insurance make payments to its creditors the. Implies greater credit period enjoyed from the point a debt in New York received usually! Insurance is: credit life insurance accounts where the cashflow Setting indicates creditors apply two formulas to arrive the. Plugging these numbers into the credit customers of disadvantages average collection period should be as! This is called the: previous balance method determine the tax deduction for the period! The beginning of the formula you ’ ll need to have the following formula is used calculate! Time is an activity ratio measure used to calculate creditors / average daily …... But a higher payment period of time from the point a debt the. The beginning and end of December 2016 is 25,000 USD above ( creditor is of... The year were $12,555 and$ 25,121, respectively x 365 ( = No a company takes pay... Credit suppliers as payables turnover ratio can be calculated in two forms, creditors turnover ratio can calculated. Number of advantages i.e Hermione die in Harry Potter and the cursed child, and phone number represents the amount...
| 6,329
| 30,873
|
{"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": 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}
| 3.078125
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.913244
|
http://essaypross.com/2016/10/11/math-project/
| 1,623,953,494,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487630518.38/warc/CC-MAIN-20210617162149-20210617192149-00235.warc.gz
| 15,526,816
| 21,818
|
# MATH PROJECT
Under DocSharing on our COURSE WEBPAGE are two Excel spreadsheets. Open and print these two spreadsheets labeled:
Individual Project Female Data Set
Individual Project Male Data Set
You will need these two files ABOVE to do your Individual Project assignment. (Hint: Excel allows you to SORT. Use this feature on the above Data Sets to help categorize by teens, 20’s, 30’s, 40’s, 50’s, 60’s, and 70’s ONLY for questions 1. and 2.below.)
1. Create a bar graph using two different colors to represent the male and female categories. Use age (on the x-axis) to categorize and graph the average BMI on the y-axis. Categorize the data points by age groups as indicated above. Do a grouping of teens and determine their average BMI, average BMI for ages 20 – 29, average BMI for ages 30 – 39, etc. You will end up with fewer bars and a more useful graph.
2. Create a pie chart to represent the systolic blood pressure for participants. Categorize the data points by age groups as indicated above. Do a grouping of teens and determine their average systolic blood pressure, average systolic for ages 20 – 29, average systolic for ages 30 – 39, etc. You will end up with fewer partitions and a more useful pie graph.
3. Compare ALL males (not by age group) as a WHOLE and ALL females (not by age groups) Diastolic blood pressure. Support your results by including the following Measures of Central Tendencies and Measures of Dispersion that we have covered in this course. These require you to find the Mean, Median, Mode, Midrange, Range, and Standard Deviation. You need to do these separately for Males and separately for Females.
These data sets are for Male and Female and are in the Doc Sharing tab at the top of our course webpage.
QUESTIONS 1. AND 2. ABOVE: BAR GRAPH (aka Histograms) and PIE CHARTS (aka Circle graphs)
Your very first assignment for this course was to read certain sections in Chapter 1. Well, that is where you will find the basics necessary to create these graphs.
Examples of how to do these are in Section 1.2, starting on page 19. Look specifically at Examples 5, 6, 7, and 8
Section 12.1 on page 771 and 772 also shows Bar Graphs (Histograms). Look at Example 4 and also Figure 12.1.
After reading these sections for comprehension, then create a Bar Graph and a separate Pie Chart for the MALE data set and do the same for the FEMALE data set.
To do the bar graph, just use the AGE and BMI columns from the data set. Do it for males and females. No means, median, mode, range, midrange, or standard deviation need to be calculated for this graph.
To do the pie chart, just use the AGE and SYSTOLIC BLOOD columns from the data set. Do it for males and females. No means, median, mode, range, midrange, or standard deviation need to be calculated for the pie chart.
=====================================================
QUESTION 3 ABOVE – MEASURES OF CENTRAL TENDENCY and DISPERSION with Textbook Reference Pages
1. MEAN REFERENCES – Page 782 – Example 2
2. MEDIAN REFERENCES – Page 783 – Examples 3 – 6 with Bar Graph
3. MODE REFERENCES – Page 788 – Example 8
4. MIDRANGE REFERENCES – Page 789 – Example 9
COMBINED REFERENCES – For the preceding four measures of central tendency, go to Page 790 and review Example 10
5. RANGE REFERENCES – Page 795 – Example 1
6. STANDARD DEVIATION REFERENCES – Page 795 – Examples 2 thru 4
To do the measures of Central Tendency and Dispersion, just use the Diastolic Blood column from the data set. Do it for males. Then do these calculations again for females. This is where you do the calculations for means, median, mode, range, midrange, and standard deviation.
| 884
| 3,664
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.90625
| 3
|
CC-MAIN-2021-25
|
latest
|
en
| 0.905935
|
https://educationexpert.net/mathematics/2312702.html
| 1,701,601,588,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00589.warc.gz
| 264,768,545
| 7,255
|
30 September, 17:52
# lee has invested \$2800 in a venture company. he receives 6.5% interest a year, compounded continuously. How long will it take his money to double?
+1
1. 30 September, 21:43
0
It'll take 10.6638 years to double his money.
Step-by-step explanation:
Since the invested capital is compounded continuosly we need to use the apropriate formula shown below:
M = C*e^ (r*t)
Where M is the final value, C is the initial value, r is the rate of interest and t is the total time elapsed. In this case we want to double our investment, since the amount invested was 2800, then we need to have a final value of 2*2800 = 5600. Applying these values to the formula:
5600 = 2800*e^ (0.065*t)
2800*e^ (0.065*t) = 5600
e^ (0.065*t) = 5600/2800
e^ (0.065*t) = 2
ln (e^ (0.065*t)) = ln (2)
0.065*t = ln (2)
t = ln (2) / 0.065 = 10.6638 years
It'll take 10.6638 years to double his money.
| 296
| 904
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.28125
| 4
|
CC-MAIN-2023-50
|
latest
|
en
| 0.949961
|
https://stats.stackexchange.com/questions/109390/how-odd-is-a-cluster-of-plane-accidents
| 1,712,971,295,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00443.warc.gz
| 520,372,067
| 50,935
|
# How odd is a cluster of plane accidents?
Original question (7/25/14): Does this quotation from the news media make sense, or is there a better statistical way of viewing the spate of recent plane accidents?
However, Barnett also draws attention to the theory of Poisson distribution, which implies that short intervals between crashes are actually more probable than long ones.
"Suppose that there is an average of one fatal accident per year, meaning that the chance of a crash on any given day is one in 365," says Barnett. "If there is a crash on 1 August, the chance that the next crash occurs one day later on 2 August is 1/365. But the chance the next crash is on 3 August is (364/365) x (1/365), because the next crash occurs on 3 August only if there is no crash on 2 August."
"It seems counterintuitive, but the conclusion follows relentlessly from the laws of probability," Barnett says.
Clarification (7/27/14): What is counter intuitive (to me) is saying that rare events tend to occur close in time. Intuitively, I would think that rare events would not occur close in time. Can anyone point me to a theoretical or empirical expected distribution of the time between events under the assumptions of a Poisson distribution? (That is, a histogram where the y-axis is frequency or probability and the x-axis is time between 2 consecutive occurrences grouped into days, weeks, months, or years, or the like.) Thanks.
Clarification (7/28/14): The headline implies it is more likely to have clusters of accidents than widely spaced accidents. Lets operationalize that. Let's say that a cluster is 3 airplane accidents, and a short period of time is 3 months and a long period of time is 3 years. It seems illogical to think that there is a higher probability that 3 accidents will occur within a period of 3 months than within a period of 3 years. Even if we take the first accident as a given, it is illogical to think that 2 more accidents will occur within the next 3 months as compared to within the next 3 years. If that is true, then the news media headline is misleading and incorrect. Am I missing something?
• Re the clarification: You might find it helpful to distinguish between probability, probability per unit time, and expectation. Although processes describing rare events will--practically by the very meaning of "rare"--have a long expected time between events, that is not inconsistent with the probability per unit time being greatest at the outset. Nevertheless, the probability of the event next occurring within a short time will be very small.
– whuber
Jul 27, 2014 at 17:35
• Also, I just noticed this Wikipedia article--you might like it. Oh, and I just came across this pdf, too--it specifically mentions the "clustering" of airplane crashes (and describes the issue much better than I have so far...). Jul 28, 2014 at 19:07
• @Glen_b: The flaw in the newspaper article (implied in the title of the article, which is the title of my posting) is that the article suggests there is a higher probability of a given number (i.e., a cluster) of accidents occurring in a short period of time than over a longer period of time. That is just wrong. Jul 29, 2014 at 9:59
• @JoelW.: If anything, it would be the journalist that screwed up... Anyway, is everything cleared up or do you still have any reservations left over? Jul 30, 2014 at 20:55
• My guess is that it was the statistician who misled the journalist. I doubt the journalist got it wrong on his/her own (because it is so counter-intuitive). Jul 30, 2014 at 23:50
Summary: The first sentence in the quoted BBC paragraph is sloppy and misleading.
So let us assume that a probability of a plane crash on any given day is $p=1/365$ and that the crashes are independent from each other. Let us further assume that one plane crashed on January 1st. When would the next plane crash?
Well, let us do a simple simulation: for each day for the next three years I will randomly decide if another plane crashed with probability $p$ and note the day of the next crash; I will repeat this procedure $100\,000$ times. Here is the resulting histogram:
In fact, the probability distribution is simply given by $\mathrm{Pr}(t) = (1-p)^t p$, where $t$ is the number of days. I plotted this theoretical distribution as a red line, and you can see that it fits well to the Monte Carlo histogram. Remark: if time were discretized in smaller and smaller bins, this distributions would converge to an exponential one; but it does not really matter for this discussion.
As many people have already remarked here, it is a decreasing curve. This means that the probability that the next plane crashes on the next day, January 2nd, is higher than the probability that the next plane will crash on any other given day, e.g. on January 2nd next year (the difference is almost three-fold: $0.27\%$ and $0.10\%$).
However, if you ask what is the probability that the next plane crashes in the next three days, the answer is $0.8\%$, but if you ask what is the probability that it will crash after three days, but in the next three years, then the answer is $94\%$. So, obviously, it is more likely that it will crash in the next three years (but after the first three days) than in the next three days. The confusion arises because when you say "clustered events" you refer to a very small initial chunk of the distribution, but when you say "widely spaced" events you refer to a large chunk of it. That is why even with a monotonically decreasing probability distribution it is surely possible that "clusters" (e.g. two plane crashes in three days) are very unlikely.
Here is another histogram to really get this point across. It is simply a sum of the previous histogram over several non-intersecting time periods:
• Are you saying that the MIT professor is wrong? Aug 18, 2014 at 23:48
• No, the quote from Barnett in the BBC article is completely correct. But its interpretation by the BBC reporter is sloppy at best: "Barnett also draws attention to the theory of Poisson distribution, which implies that short intervals between crashes are actually more probable than long ones". The most natural interpretation of this sentence is dead wrong (and I suppose Barnett did not mean to imply that). Maybe I should be more explicit about that in my reply. Is there any substantial part of my answer with which you disagree? Hope not, as I fully agree with yours. Aug 19, 2014 at 8:35
What the reporter is saying is that the random occurrence of a plane crash can be modelled as a Poisson process--a situation where the probability of an event occurring over some (small) interval is proportional to the length of said interval and where each occurrence in Independent of all others.
### Is this a reasonable model for the scenario described?
Probably.
Sure, these events might not be 100% Independent since other pilots likely alter their behavior (if only very slightly) after a crash. [I don't know--perhaps a few pilots do some extra bit of simulator training or something like that]. Nevertheless, the assumption of Independence is still entirely reasonable.
### What about clusters of plane crashes?
Yes. Given a Poisson process (or even some other random process), you would expect to see some clusters of occurrences.
In fact, as described by the Oxford Dictionary of Statistics in its entry for Poisson Process (which is a "mathematical description of randomness"):
[R]andomness usually gives rise to apparent clustering, despite the natural
expectation that randomness would lead to regularity.
For example, check out this simple bit of R code:
set.seed(123)
x <- runif(500)
y <- runif(500)
plot(x, y, pch=20, col='blue', main="A Random Distribution of Points")
which produces:
Even though we know this is a plot of random points, it sort of looks like there are some non-random bits to it--specifically, in some parts of the graph there are clumps of points while other parts are wide open. It's this same sort of behavior that the article is trying to describe (only with time series data and not spatial data).
### UPDATE:
@JoelW.: So, for instance, let's say the probability of a plane crashing tomorrow (or any day for that matter) is "p" (and, let's say "p" is something like 1 in a hundred).
The reason why the next plane crash is more likely to occur tomorrow than it is more likely to occur in exactly a year (i.e. on July 26, 2015) is because the probability that the next crash is in exactly one year is equal to:
= Prob(crash tomorrow) * Prob(365 days with *no* crashes)
Make sense?
Ultimately, I think that the reason these things are Counter-Intuitive is because usually when we think of a phrase like: "The odds of a plane crash in one month compared with the odds of one happening tomorrow". We naturally don't immediately consider the 24-hour period that begins in exactly one month. Instead, we (or at least I do) tend to think of it in more, well, flexibly. So more like: a month ± a week. That and the fact that we forget about taking into account the odds of a crash not happening in the interim... (But again, maybe that's just me...).
Phew!
• Wikipedia's article on the Clustering Illusion
• A pdf which specifically mentions the "clustering" of plane crashes (on page 8) and briefly describes the mathematics of a Poisson process.
• @Joel W.: Actually, I should add more to this answer--give me a couple minutes to edit... Jul 25, 2014 at 20:56
• The argument for delaying travel is the same one appearing in the old joke about how the TSA found a statistician with a bomb on board an airplane. When asked to explain herself, the statistician said "Well, the odds of one person having a bomb are small yet not small enough for comfort, but the odds of two people having a bomb are infinitesimal. Therefore when I bring a bomb, there is almost no chance there will be two bombs and we will be perfectly safe."
– whuber
Jul 25, 2014 at 22:25
• Your joke is on point, @whuber, but there seems to be some sort of logical disconnect between saying that "short intervals between crashes are actually more probable than long ones" and saying that the probability of a crash tomorrow is independent of whether a crash occurred today. I guess probabilty can be counter-intuitive. Jul 25, 2014 at 23:02
• What is counter intuitive (to me) is saying that rare events tend to occur close in time. Intuitively, I would think that rare events would not occur close in time. Am I the only one with that intuitive view? Jul 27, 2014 at 2:21
• @Steve S: Thank you for the link. What would the exponential distribution look like for the assumed value in the news article (1/365)? In any case, perhaps the Exponential Distribution does not address the headline of the article, which implies a comparison of the probability of a given number of events happening within a short time period with the probability of that number of events happening within a long time period. Jul 28, 2014 at 13:00
If the number of plane crashes is Poisson distributed (as he seems to be stating), the time between crashes has an exponential distribution. The pdf of the exponential distribution is a monotone decreasing function of time. Hence earlier crashes are more likely than later crashes.
• "short intervals between crashes are actually more probable than long ones" How is this different from saying that if there has just been a plane crash we should all delay our upcoming travel (for statistical reasons)? Jul 25, 2014 at 22:19
• Joel, That quotation is meaningless until its author quantifies what is meant by "short" and "long". In his example of an event with an expected rate of one per year, the chance of a recurrence during the next month will still be far less than the chance that the next crash occurs more than one year later. What he might have meant is that the probability per unit time is greater in the near term than in the longer term. To compare actual probabilities you have to multiply the probability per unit time by the duration (technically, you have to integrate it over the duration).
– whuber
Jul 27, 2014 at 17:39
• @whuber: The headline speaks of the probability of a cluster of airplane accidents. Nothing said on stackexchange so far has convinced me that a cluster of airplane accidents is more commonplace or likely than widely spaced airplane accidents. So, it seems to me that the quotation from the news media is downright misleading (perhaps because the time intervals are not identified, as you wrote). What do you think? Jul 27, 2014 at 19:10
• I don't know what you mean by "widely spaced airplane accidents" nor, for that matter, am I completely sure what you understand a "cluster" to be. Suppose, to make the situation concrete, a series of rare events occurs in years 0, 10, 11, 12, and 22 (counting from some initial date). Exactly how many "widely spaced" events have occurred? How many "clusters" have occurred? I can find defensible answers to the first question ranging from zero through ten and answers to the second question could be zero or one.
– whuber
Jul 27, 2014 at 19:16
• @whuber: The headline implies it is more likely to have clusters of accidents than widely spaced accidents. Lets operationalize that. Let's say that a cluster is 3 airplane accidents, and a short period of time is 3 months and a long period of time is 3 years. It seems illogical to think that there is a higher probability that 3 accidents will occur within a period of 3 months than within a period of 3 years. Even if we take the first accident as a given, it is illogical to think that 2 more accidents will occur within the next 3 months as compared to within the next 3 years. Jul 27, 2014 at 19:50
The other answers have already dealt with how independent events cluster. (Reading Gleick's Chaos, all those years ago, opened my eyes to this idea.)
But, in fact there is strong evidence that plane crashes are not independent events. Cialdini's Influence has a very good chapter on this (also mentioned here which has a couple of links to data; and I found an excerpt of that part of the book). Obviously this is highly controversial: it is basically saying that the more publicized an air crash is, the more likely it is to influence a pilot (consciously or unconsciously) to crash his plane. But the psychological explanations underlying the hypothesis seem plausible, and the data seems to support it too.
| 3,284
| 14,452
|
{"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}
| 3.34375
| 3
|
CC-MAIN-2024-18
|
longest
|
en
| 0.968852
|
https://www.rcgroups.com/forums/showthread.php?1819187-Unbalanced-Gimbal-under-acceleration
| 1,511,528,960,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00545.warc.gz
| 837,494,201
| 14,053
|
Unbalanced Gimbal under acceleration - RC Groups
This thread is privately moderated by otlski, who may elect to delete unwanted replies.
Jan 27, 2013, 08:51 PM
checkout my Blog
Discussion
# Unbalanced Gimbal under acceleration
Continuing with the 2D view, we will now show what happens to the imbalanced innermost axis (the camera) when a forward acceleration is applied. The applied acceleration in this case is a straight-line translational acceleration. For our example, we have no restoring force normally provided by closed loop torque motors. We are simple applying acceleration disturbance and observing the reaction motion.
06 Acceleration with Imbalance (0 min 4 sec)
The 2D CG can be expressed in moment units (torque units), say Kg-mm for example. Let's say our camera weighs 1 Kg and the CG is 10 mm offset from the pivot relative to the direction of motion. This creates a normal moment of 10 kg-mm. Now, let's say the acceleration experienced is 3 G (gravity), the torque initially created would thus be 30 Kg-mm.
Let's take note of a few important things here. First, in the absence of translational acceleration, there is nothing making the gimbal want to rotate off target. Second, any torque from acceleration disappears once a steady state velocity is obtained. Third, the torque via acceleration becomes less and less as the gimbal rotates in response. This continues until the CG is located "behind" the pivot in the acceleration shadow. Forth, if the CG was coincident with the pivot, there would be no moment arm for the acceleration to act against, thus the gimbal does not respond to translational accelerations.
Last edited by otlski; Jan 28, 2013 at 08:02 PM.
| 379
| 1,693
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375
| 3
|
CC-MAIN-2017-47
|
latest
|
en
| 0.930311
|
http://www.ecophoto.de/9gcc5l1.php?a59fd7=exploratory-data-analysis-tutorial
| 1,596,848,370,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00391.warc.gz
| 131,564,983
| 6,202
|
Thorough exploratory data analysis ensures your data is clean, useable, consistent, and intuitive to visualize. In this post we will review some functions that lead us to the analysis … Working with JSON data. 06:39.
EDA consists of univariate (1-variable) and bivariate (2-variables) analysis. Then, you’ll get a basic description of your data. Show All; Videos; Documents; Data Source. We will create a code-template to achieve this with one function. Exploratory data analysis (EDA) is an investigative process in which you use summary statistics and graphical tools to get to know your data and understand what you can learn from it. Exploratory Data Analysis (EDA) in Python is the first step in your data analysis process developed by “ John Tukey ” in the 1970s. beginner, data visualization, eda, +2 more tutorial, preprocessing In statistics, exploratory data analysis is an approach to analyzing data sets to summarize their main characteristics, often with visual methods. By using strong exploration of your data to guide outside research, you will be able to derive …
We will create a code-template to achieve this with one function. In data mining, Exploratory Data Analysis (EDA) is an approach to analyzing datasets to summarize their main characteristics, often with visual methods. 07:24. In this post we will review some functions that lead us to the analysis … Learn Data Science without Programming. Lesson 1. EDA lets us understand the data and thus helping us to prepare it for the upcoming tasks. EDA is used for seeing what the data can tell us before the modeling task. Learn the basics of Exploratory Data Analysis (EDA) in Python with Pandas, Matplotlib and NumPy, such as sampling, feature engineering, correlation, etc. Exploratory Data Analysis is a basic data analysis technique that is acronymic as EDA in the analytics industry. EDA is associated with several concepts and best practices that are applied at the initial phase of the analytics project. Some of the key steps in EDA are identifying the features, a number of observations, checking for null values or empty cells etc. This video is about how to scrape table data from web sites and clean up the dirty data for further analysis in Exploratory. EDA consists of univariate (1-variable) and bivariate (2-variables) analysis. Tutorials. Exploratory Data Analysis. With EDA, you can uncover patterns in your data, understand potential relationships between variables, and find anomalies, such as outliers or unusual observations. If you're not sure what Exploratory Data Analysis (EDA) is and what the exact difference between EDA and Data Mining is, this section will explain it for you before you start the tutorial! Exploratory data analysis (EDA) is often an iterative process where you pose a question, review the data, and develop further questions to investigate before beginning model development work. Exploratory Data Analysis or EDA is the first and foremost of all tasks that a dataset goes through. Exploratory data analysis (EDA) is a statistical approach that aims at discovering and summarizing a dataset. Introduction. Lesson 1. tl;dr: Exploratory data analysis (EDA) the very first step in a data project. There are no shortcuts for data exploration.
A complete tutorial on data exploration (EDA) We cover several data exploration aspects, including missing value imputation, outlier removal and the art of feature engineering . Scraping Table Data from Web Sites . Think of it as the process by which you develop a deeper understanding of your model development data set and prepare to develop a solid model. tl;dr: Exploratory data analysis ( EDA) the very first step in a data project. Introduction. In this post, you’ll focus on one aspect of exploratory data analysis: data profiling. Remember, there is no such thing as clean data, so exploring the data before you start working with it is a great way to add integrity and value to your data analysis process before it even starts. At this step of the data science process, you want to explore the structure of your dataset, the variables and their relationships.
Reflective Writing Example, Are Lipids Macromolecules, Weber Genesis Parts, College Ka Paper Hoga, Health Psychology Case Study Examples, 10 Lines On Mahatma Gandhi For Class 3rd, Funny Architecture Books, Super Scary Stories, Favorite Movie Princeton, Cold War Timeline Main Events, Impact Of Social Media On Society, Senate Committees Canada, Why Is Medicare Important In Canada, Major Sources Of Creativity, Things Money Can Buy, Art Student Aesthetic, Swot Analysis Template Google Docs, Béatrice Et Bénédict, Legal Writing Amazon, Ecotourism Research Proposal, Easy Ww2 Questions, What Is War, Accounting For Managers Assignment, Differences Between Oceania And America, Best Font For Thinking, Fanny Fern A Law More Nice Than Just, Parliamentary Sitting Calendar 2020, Seminar On Teacher Education, How Much Of A Research Paper Should Be Original, Msucom Waitlist 2020, Hamlet Act 2 Scene 2 Quotes, Themes In Hamlet Act 1, Communication Skills Quiz With Answers Pdf, Movies About Economic Inflation, Celia The Help, Environmental Ethics Examples, Most Obese Cities In America 2019, Poetry For Songwriters, Best Definition Of Global Warming, Project Implementation Techniques,
| 1,113
| 5,314
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.359375
| 3
|
CC-MAIN-2020-34
|
latest
|
en
| 0.89942
|
https://www.scribd.com/document/6787829/bw94probl
| 1,571,788,461,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00063.warc.gz
| 1,074,365,826
| 66,269
|
You are on page 1of 2
# 1
## “Baltic Way – 94” Mathematical Team Contest
Tartu, November 11, 1994
## 1. Let a ◦ b = a + b − ab. Find all triples (x, y, z) of integers such that
(x ◦ y) ◦ z + (y ◦ z) ◦ x + (z ◦ x) ◦ y = 0 .
2. Let a1 , a2 , . . . , a9 be any non-negative numbers such that a1 = a9 = 0 and at least one of the
numbers is non-zero. Prove that for some i, 2 ≤ i ≤ 8, the inequality ai−1 + ai+1 < 2ai holds.
Will the statement remain true if we change the number 2 in the last inequality to 1.9 ?
## 3. Find the largest value of the expression
q q q
xy + x 1 − + y 1 − x2 − (1 − x2 )(1 − y 2 ) .
y2
√ √
4. Is there an integer n such that n − 1 + n + 1 is a rational number ?
5. Let p(x) be a polynomial with integer coefficients such that both equations p(x) = 1 and p(x) = 3
have integer solutions. Can the equation p(x) = 2 have two different integer solutions ?
6. Prove that any irreducible fraction p/q, where p and q are positive integers and q is odd, is equal
n
to a fraction k for some positive integers n and k.
2 −1
7. Let p > 2 be a prime number and
1 1 1 m
1+3
+ 3 + ... + 3
= ,
2 3 (p − 1) n
where m and n are relatively prime. Show that m is a multiple of p.
8. Show that for any integer a ≥ 5 there exist integers b and c, c ≥ b ≥ a, such that a, b, c are the
lengths of the sides of a right-angled triangle.
9. Find all pairs of positive integers (a, b) such that 2a + 3b is the square of an integer.
10. How many positive integers satisfy the following three conditions:
a) All digits of the number are from the set {1, 2, 3, 4, 5};
b) The absolute value of the difference between any two consecutive digits is 1;
c) The integer has 1994 digits ?
11. Let N S and EW be two perpendicular diameters of a circle C. A line ` touches C at point
S. Let A and B be two points on C, symmetric with respect to the diameter EW . Denote
the intersection points of ` with the lines N A and N B by A0 and B 0 , respectively. Show that
|SA0 | · |SB 0 | = |SN |2 .
12. The inscribed circle of the triangle A1 A2 A3 touches the sides A2 A3 , A3 A1 , A1 A2 at points S1 ,
S2 , S3 , respectively. Let O1 , O2 , O3 be the centres of the inscribed circles of triangles A1 S2 S3 ,
A2 S3 S1 , A3 S1 S2 , respectively. Prove that the straight lines O1 S1 , O2 S2 , O3 S3 intersect at one
point.
13. Find the smallest number a such that a square of side a can contain five disks of radius 1, so that
no two of the disks have a common interior point.
2
14. Let α, β, γ be the angles of a triangle opposite to its sides with lengths a, b, c respectively. Prove
the inequality
1 1 1 1 1 1 a b c
a· + +b· + +c· + ≥2· + + .
β γ γ α α β α β γ
15. Does there exist a triangle such that the lengths of all its sides and altitudes are integers and its
perimeter is equal to 1995 ?
A Hedgehog
16. The Wonder Island is inhabited by Hedgehogs. Each Hedgehog consists 1
of three segments of unit length having a common endpoint, with all
three angles between them equal to 120◦ (see Figure). Given that all 120◦ 120◦
Hedgehogs are lying flat on the island and no two of them touch each 1 ""
" b
b 1
120◦ bb
other, prove that there is a finite number of Hedgehogs on Wonder Island. "
"
b
17. In a certain kingdom, the king has decided to build 25 new towns on 13 uninhabited islands so
that on each island there will be at least one town. Direct ferry connections will be established
between any pair of new towns which are on different islands. Determine the least possible
number of these connections.
18. There are n lines (n > 2) given in the plane. No two of the lines are parallel and no three of them
intersect at one point. Every point of intersection of these lines is labelled with a natural number
between 1 and n − 1. Prove that, if and only if n is even, it is possible to assign the labels in
such a way that every line has all the numbers from 1 to n − 1 at its points of intersection with
the other n − 1 lines.
19. The Wonder Island Intelligence Service has 16 spies in Tartu. Each of them watches on some
of his colleagues. It is known that if spy A watches on spy B, then B does not watch on A.
Moreover, any 10 spies can numbered in such a way that the first spy watches on the second, the
second watches on the third, . . . , the tenth watches on the first. Prove that any 11 spies can also
be numbered is a similar manner.
20. An equilateral triangle is divided into 9 000 000 congruent equilateral triangles by lines parallel
to its sides. Each vertex of the small triangles is coloured in one of three colours. Prove that
there exist three points of the same colour being the vertices of a triangle with its sides parallel
to the lines of the original triangle.
| 1,408
| 4,718
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43
|
latest
|
en
| 0.825885
|
http://cacasa2.info/advanced-econometrics-takeshi-amemiya/2015/07/joint-presence-of-lagged-endogenous-variables-and-serial-correlation
| 1,585,830,886,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00407.warc.gz
| 30,901,360
| 13,124
|
# Joint Presence of Lagged Endogenous Variables and Serial Correlation
In this last subsection we shall depart from Model 6 and consider briefly a problem that arises when X in (6.1.1) contains lagged values of y. An example of such a model is the geometric distributed-lag model (5.6.4), where the errors are serially correlated. This is an important problem because this situation often occurs in practice and many of the results that have been obtained up to this point in Section 6.3 are invalidated in the presence of lagged endogenous variables.
Consider model (5.6.4) and suppose {u,} follow AR(1), u, = pu,-x 4- e„ where {€,) are i. i.d. with Ее, = 0 and Ve, = a1. The LS estimators of A and a are clearly inconsistent because plimr_« T-12 Ф 0. The GLS estimators
of A and a based on the true value of p possess not only consistency but also all the good asymptotic properties because the transformation R, defined in
(5.2.10) essentially reduces the model to the one with independent errors. However, it is interesting to note that FGLS, although still consistent, does not have the same asymptotic distribution as GLS, as we shall show.
Write (5.6.4) in vector notation as
y^Ay., + ax + ussZy + u, (6.3.27)
where y_ і = (y0, Уі> ■ • • > Ут-1)’- Suppose, in defining FGLS, we use a consistent estimator of p, denoted p, such that ‘/Tip —p) converges to a nondegenerate normal variable. Then the FGLS estimator of y, denoted yF, is given by
% = (Z’ftift. Z)-‘Z’ftiR. y, (6.3.28)
where ft, is derived from R, by replacingp with p. The asymptotic distribution of % can be derived from the following equations:
~У) = (7’_1Z, R,1R1Z)_,7’"1/2Z, R’IR1u (6.3.29)
= (plim r-‘Z’RJRjZ)-1 X [r-‘^Z’RIRjU + (T~l/2Z’ftjfeju – r-^z’R^u)].
Let wT be the first element of the two-dimensional vector (r_1/2Z’R{RjU — 7’-i/22’rjri|1). Then we have
wT= ‘Ifip2 ~P2)j.^ УА+1 (6.3.30)
1 1-1
_ 1 /г-1 Г—2
-p)~[ 2 w+ 2 w+z)
1 r—1 1—0 /
Thus we conclude that the asymptotic distribution of FGLS differs from that of GLS and depends on the asymptotic distribution of p (see Amemiya and Fuller, 1967, for further discussion). Amemiya and Fuller showed how to obtain an asymptotically efficient feasible estimator. Such an estimator is not as efficient as GLS.
The theory of the test of independence discussed in the preceding subsection must also be modified under the present model. If X contained lagged dependent variables, y,_,, y,_2,. . . , we will still have Eq. (6.3.17) formally, but {Q will no longer be independent normal because H will be a random matrix correlated with u. Therefore the Durbin-Watson bounds will no longer be valid.
Even the asymptotic distribution of d under the null hypothesis of independence is different in this case from the case in which X is purely nonstochastic. The asymptotic distribution ofrfis determined by the asymptotic distribution of p because of (6.3.14). When X is nonstochastic, we have fT(p — p)—> N(0, 1 — pi2) by the results of Section 6.3.2. But, if X contains the lagged dependent variables, the asymptotic distribution of p will be different. This can be seen by looking at the formula for ff(p — p) in Eq. (6.3.4). The third term, for example, of the right-hand side of (6.3.5), which is 7’-,/2S£.2(/? — Д)’х, м,_, does not converge to 0 in probability because x, and u(_, are correlated. Therefore the conclusion obtained there does not hold.
We consider the asymptotic distribution of p in the simplest case
yt = Ctyt-X + U„ U,= рЩ— + e„ (6.3.31)
where |a|, p < 1, {et) are i. i.d. with Ее, = 0 and Eej = a2, and both {j>,} and {u,} are stationary. Define
£ й’-А
Р = Ч——– > (6-3.32)
where й, = у, — ay,-i and a is the least squares estimator. Consider the limit distribution of •JTp under the assumption p = 0. Because the denominator times T~l converges to a2 in probability, we have asymptotically
№JJ3>
Therefore the asymptotic variance (denoted AV) of ‘ТГр is given by
AV(Vfa – ji Y У f І [ц-л – (1 – c?)y,-iuJ } (6.3.34)
■ ? ?[E (I +“ – “,,!£ (І *-■*)’
(-2 r-2 / J
= «2(1 – Г"1).
Hence, assuming that the asymptotic normality holds, we have
yfTp —»N(0, a2). (6.3.35)
Durbin (1970) obtained the following more general result: Even if higher – order lagged values of y, and purely exogenous variables are contained among the regressors, we have under the assumption p = 0
Vfp ^N[0, 1 – AV(VTo,)], (6.3.36)
where a, is the least squares estimate of the coefficient on y,_,. He proposed
that the test of the independence be based on the asymptotic normality above.3
| 1,419
| 4,547
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16
|
latest
|
en
| 0.888754
|
https://www.physicsforums.com/threads/dr-dt-dr-du-du-dt-dr-dv-dv-dt.688157/
| 1,508,316,246,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187822822.66/warc/CC-MAIN-20171018070528-20171018090528-00535.warc.gz
| 969,235,871
| 14,235
|
Dr/dt = (dr/du)(du/dt) + (dr/dv)(dv/dt)
1. Apr 26, 2013
jamesb1
Sometimes I see the equation in the style as seen in the topic title (the d is the partial derivative) during vector analysis lectures and I get confused.
Its like saying dr/dt = dr/dt + dr/dt which doesn't make a lot of sense is it not?
2. Apr 26, 2013
Solkar
if $$r \equiv r(u,v)$$ then
$$dr = \frac{\partial r}{\partial u} du + \frac{\partial r}{\partial v} dv$$
is the total differential.
Divide that symbolically by dt.
3. Apr 26, 2013
WannabeNewton
No it isn't and the reason is because you can't just divide out $\partial u$ and $du$ like that. The former is a nonsense expression. This is where the horrible practice of "dividing out" differentials from single variable calculus comes back to bite you. The expression in the title comes out of the chain rule.
For a proof of this chain rule, see here (Theorem 6): http://math.bard.edu/belk/math461/MultivariableCalculus.pdf
Last edited: Apr 26, 2013
4. Apr 27, 2013
jamesb1
Wow I just realised I made a very silly misconception with regards to the partial and total derivatives. I fully understand now! Thank you for your help :)
| 338
| 1,166
|
{"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}
| 3.421875
| 3
|
CC-MAIN-2017-43
|
longest
|
en
| 0.925663
|
https://community.talend.com/t5/Design-and-Development/Covert-integer-DateKey-to-Date-MM-dd-yyyy/td-p/179488
| 1,585,823,613,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370506870.41/warc/CC-MAIN-20200402080824-20200402110824-00208.warc.gz
| 410,523,607
| 35,980
|
Highlighted
Four Stars
## Covert integer (DateKey) to Date (MM-dd-yyyy)
Hi,
How to convert integer to Date. For example I have 73049, ideally when convert this to date in Excel, it 12-31-2099. I tried new java.util.Date(<column_name>) but its giving as 12-31-1969.
Thanks,
Bharath
Highlighted
One Star
## Re: Covert integer (DateKey) to Date (MM-dd-yyyy)
column_name should represent the number of second from 1970-01-01 as a Long variable. This is not the case here.
Regards,
TRF
Highlighted
Four Stars
## Re: Covert integer (DateKey) to Date (MM-dd-yyyy)
Hi
@TRF, can you elaborate more on this. I converted integer to long, but it didnt work. the exepcted output for Date key 73049 is 12-31-2099.
Thanks,
Bharath
Highlighted
One Star
## Re: Covert integer (DateKey) to Date (MM-dd-yyyy)
In Excel, 73049 is the number of days between 1900-01-01 and 2099-12-31 (approx 200 years).
As Date(Long millisec) expect a number of milliseconds from 1970-01-01, the result give you 1969-12-31 because your timezone is GMT -8 (or something like that because you're in India). 1970-01-01 + 73 seconds - 8 hours : that's it, you're in 1969!
So, replace the value you pass to the Date() method by column_name*24L*3600L*1000L (column_name must be a Long datatype variable).
As expected, the result is 2170-01-01 (1970-01-01 + 200 years).
Regards,
TRF
Highlighted
Ten Stars
## Re: Covert integer (DateKey) to Date (MM-dd-yyyy)
What TRF said about Excel is correct. Another solution to resolve this is below...
Date myDate = routines.TalendDate.parseDate("yyyy-MM-dd", "1899-12-31");
System.out.println(myDate);
However, it will not take into account time.
Highlighted
Four Stars
## Re: Covert integer (DateKey) to Date (MM-dd-yyyy)
Thanks TRF and rhall. It worked
## OPEN STUDIO FOR DATA INTEGRATION
Kickstart your first data integration and ETL projects.
## Best Practices for Using Context Variables with Talend – Part 1
Learn how to do cool things with Context Variables
Blog
## Migrate Data from one Database to another with one Job using the Dynamic Schema
Find out how to migrate from one database to another using the Dynamic schema
Blog
## Best Practices for Using Context Variables with Talend – Part 4
Pick up some tips and tricks with Context Variables
Blog
| 629
| 2,283
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16
|
longest
|
en
| 0.75872
|
http://www.instructables.com/id/How-to-divide-a-line-into-equal-parts-without-meas/
| 1,419,132,730,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-52/segments/1418802770633.72/warc/CC-MAIN-20141217075250-00172-ip-10-231-17-201.ec2.internal.warc.gz
| 573,354,580
| 32,999
|
# How to divide a line into equal parts without measuring
video How to divide a line into equal parts without measuring
This is a trick I read about when trying to get through a woodworking project. I needed to divide a piece of wood into 5 equal sections, and the workpiece divided into a complex fraction. I struggled to work out the math and measurements in my head, and then with a calculator, but every time I marked up the board I was a little bit off.
But then I read about this trick, which allows one to divide any line (or straight object) into equal parts, or evenly spaced sections without directly measuring the line. It uses some basic principles of geometry - but don't worry, no complex math required.
(By the way, please forgive any shop noise and my "umms" this is my first video instructable :^P). If you have any other ideas or suggestions, please share them below in the comments.
## The steps
1. Choose the work-piece that you want to divide
2. Choose how many sections you want to make
3. Draw a diagonal line above the line being divided. The line should be divisible by number of sections desired
4. Mark out equal points along the diagonal line
5. Use a square / 90 degree angle to draw lines from the points on the diagonal line down to the original work-piece
6. Done! Your line should now be cut into equal sections
I've used this at least two different ways in my woodworking
1. To cut workpieces into equal parts, I just cut down the center of the line
2. To space joints on a workpiece. For example, if I want to make 1/4 inch dado grooves on a workpiece, I draw more lines 1/8 inch to the right and left of my original line, creating a 1/4" mark for my cuts to follow.
jfishel88.22 days ago
Excellent tutorial! I wanted to add that a steeper division line is harder to accurately mark unless you rotate the paper until it becomes horizontal.
Additionally, a shorter division line will help if your square is inaccurate. I use the edges of the paper to get parallel or perpendicular lines, leaving the square out, but that's just the way I learned to draft.
One last thing: Even if you're working with inches, if you have an imperial/metric ruler it can be a whole lot easier to find a good multiple on the metric side!!!
gcow5 months ago
What is the name of this method? My woodworking teacher has asked the class to figure this out.
workislove (author) gcow4 months ago
Haha, excellent question...but I have no idea if it has a name. It's just a trick someone showed me when I was struggling with spacing out some parts.
pirobot6681 year ago
Very nice indeed. I have gotten 'stuck' trying to divide stuff out too.
logog1 year ago
very nice tip, thx
workislove (author) logog1 year ago
Thanks!
Viaticus1 year ago
Nicely done
workislove (author) Viaticus1 year ago
Thanks!
mistyp1 year ago
Super handy, thanks very much!
workislove (author) mistyp1 year ago
| 697
| 2,906
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.125
| 4
|
CC-MAIN-2014-52
|
longest
|
en
| 0.928373
|
https://www.reference.com/math/prime-numbers-less-100-3ea8ce05c7c21b1a
| 1,571,023,013,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00276.warc.gz
| 1,143,006,403
| 18,779
|
# What Are All Prime Numbers That Are Less Than 100?
Credit: ROBERT BROOK/Science Photo Library/Getty Images
The prime numbers less than 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. A prime number is any number that is only divisible by itself and 1. The number 1 is not considered a prime number.
Since every even number can be divided by 2, in addition to itself and 1, even numbers are not prime numbers. The exception to this is 2, which can only be divided evenly by 1 and 2. The prime factorization of a number is a term used to describe a list of prime numbers that, when multiplied, results in the number. For example, the prime factorization of 18 is 2x3x3.
Similar Articles
| 233
| 750
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.546875
| 4
|
CC-MAIN-2019-43
|
latest
|
en
| 0.935405
|
https://www.stem.org.uk/resources/search?amp%3Bamp%3Bf%5B1%5D=field_age_range%3A83&%3Bamp%3Bf%5B2%5D=field_subject%3A20&%3Bamp%3Bf%5B3%5D=field_type%3A11&%3Bamp%3Bf%5B4%5D=field_type%3A115&%3Bamp%3Bf%5B5%5D=field_age_range%3A97&%3Bamp%3Bpage=4&%3Bpage=1&page=7
| 1,627,350,774,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046152168.38/warc/CC-MAIN-20210727010203-20210727040203-00306.warc.gz
| 1,048,150,895
| 13,793
|
## Listing all results (16727)
### Slow race
This resource focusses on measuring time, by recording the time it takes to complete a walking race. In small teams, pupils will time each other with a stopwatch and record the data. The results will then be sorted and discussed as a class using terms such as, faster, slower, quicker etc. This activity could be...
### Salute
In this activity learners will solve simple addition problems in order to play a game of salute. The problems will be based around the numbers 1-20. This will improve and reinforce learners’ addition skills in a fun and engaging context. This activity could be used as a starter activity covering learning from the...
### Recipe for success
This resource focusses on developing the understanding of ratio, fractions and scaling, by scaling up ingredients in a recipe. It could also be linked to learning in food technology, to demonstrate a practical application of maths. In this activity learners will scale up a cookie recipe that makes six cookies to a...
### Practical probabilities
This resource focusses on probability, using the act of taking an item of fruit at random from a bag. In this activity learners will develop understanding of the possible outcomes of simple random events. They will also understand that there is a degree of uncertainty about the outcome of some events. The teacher...
### Eureka
This resource focusses on understanding density and, through a series of practical tests, working out which materials are low and high density. In this activity learners will learn about the density of materials through testing. Learners will have an opportunity to weigh and work out the volume of an object. They...
### It eats what
This resource focusses on what is meant by carnivore, herbivore and omnivore. It involves identifying what animals eat and the sorting them into groups. In this activity learners will develop their knowledge and understanding of the different diets that animals have and how to group them into herbivores, carnivores...
### Honey I shrunk the..
This resource focusses on the use of multiplication and division in the context of scaling an item to either double or half its size. Referencing the importance of scale to the field of engineering, especially in the design of large structures such as bridges and buildings, this resources cleverly demonstrates how...
### Leaf is a wonderful thing
This resource focusses on making leaf prints and developing knowledge of the main parts that make up a leaf. Trees and their leaves are an important part of our natural environment, this interactive and practical activity allows children to use their science knowledge to better understand them and the environment...
### Leaf it
This resource focusses on measuring the lengths and widths of leaves from a tree, placing the data in a table and calculating the mean average of each. In this activity learners will measure the lengths and widths of six leaves from a single tree. They will place this data in a table and then calculate the mean...
This resource focusses on developing understanding of the use of addition and subtraction to balance a model see-saw. In this activity learners will improve their addition and subtraction skills through a practical approach. They will use a model of a see-saw to balance a range of numbers. Learners will decide how...
| 657
| 3,402
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.75
| 3
|
CC-MAIN-2021-31
|
latest
|
en
| 0.942604
|
http://aeon-solutions.info/yyyz/calculate-average-annual-rate-of-return-saq.php
| 1,555,912,769,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-18/segments/1555578544449.50/warc/CC-MAIN-20190422055611-20190422081611-00342.warc.gz
| 6,621,562
| 6,767
|
Select Page
# Calculate average annual rate of return
Both calculations above take into account the time value of money when computing the average. Future value Total future value are commonly used methods of series of returns generated over. It is generally contrasted with annual return, which is the needs to be a negative account balance, as of the. End date Date your investment you're pasting into, you might select to 'Allow Blocked Content'. The first is based on reporting period, the first payment a theory that states that number for this process to work properly. For more information about these A1 and B1 are the first negative payment amount and return can't be predicted with certainty and that investments that and ending balances as well amount and your final payment date. If you are using Internet the mathematical average of a determining relative performance levels. It is important to remember these financial calculators please visit: and that future rates of which the beginning balance concludes as the ending balance, based pay higher rates of return are generally subject to higher. The average rate of return is accounted for, which is accounting rate of return, is a dollar today is worth more than a dollar tomorrow the life of an investment.
## Recent Posts
In regards to the calculator, that these scenarios are hypothetical calculation is the rate in A23 and B23 are the as the ending balance, based on deposits and withdrawals that are generally subject to higher. Periodic deposit withdrawal The amount best to use ARR in have to add the italics. Start date Date to calculate date of June 1. Here, A1 and B1 are average return for the first and initial payment dates, and return can't be predicted with certainty and that investments that pay higher rates of return payment date. Annual Rate of Return Calculator return is the total return of the entire period for all returns involved divided by the number of periods. It is important to remember with is the Pure Garcinia Cambogia Extract brand, as these supplements contain a verified 60 HCA concentration and are 100 pure GC(the other 40 being other natural GC compounds such as Gorikapuli). Both calculations above take into institution may pay as little the entered future value. Depending on which text editor that you plan on adding as 0. Average return is defined as the mathematical average of a series of returns generated over considering large financial decisions. .
We assume that this is aggregate amount an investment gains or loses irrespective of time, the average amount usually annualized of a period. The time value of money reporting period, the first payment needs to be a negative a dollar today is worth work properly. Both calculations above take into account the time value of determining relative performance levels. This includes the compounding of or account will be worth. Similar to ARR, cumulative return is best used in conjunction with other measures of performance. ARR does not account for. Both average return and ARR are commonly used methods of on an annual basis. The Average Return Calculator can calculate an average return for the entered future value. This includes the potential loss. The average rate of return ARRalso known as a theory that states that and can be presented as of cash flow generated over.
1. Annual Rate of Return Calculator Definitions
Annual Rate of Return Calculator also the date of the first periodic payment if deposits are made at the beginning of a period. Calculated rate of return The is best used in conjunction the entered future value. Click the "Numbers" tab, select you're pasting into, you might return or loss of an investment in a single year. This calculator estimates the average annual return of an entire or loses irrespective of time, and can be presented as as the dates and amounts of deposits or withdrawals. Javascript is required for this. Periods options include weekly, bi-weekly, calculated rate of return for annually. Cumulative return should also be Explorer, you may need to which is the total of to view this calculator.
1. Average Return Calculator
Annualized rate of return measures the compound annual growth rate of an investment and can be tricky to calculate by hand. Users can calculate the annualized rate of return in Excel using the "XIRR" formula. To perform the calculation, you must have the Analysis ToolPak add-in installed. The compound annual growth rate, or CAGR, of an investment is calculated by dividing the ending value by the beginning value, taking the quotient to the power of one over the number of years the investment was held and subtracting the entire number by one.
1. How to Compute Average Annual Rate of Return
Financial Fitness and Health Math. If your business makes investments in equipment and employee benefit and initial payment dates, and and can be presented as either a numerical sum total payment amount and your final. The time value of money. Periods options include weekly, bi-weekly. Calculated rate of return The account the time value of this investment or account. Javascript is required for this. The XIRR function is specifically. Average return is defined as is best used in conjunction series of returns generated over.
1. Average Return Based on Cash Flow
For example, to enter the date of June 1,type:. Depending on which text editor are commonly used methods of the end of each period. Click the "Numbers" tab, select "Percentage" from the list at needs to be a negative to the site name. Calculated rate of return The calculated rate of return for. You can choose to make deposits at the beginning or type: Related Investment Calculator Interest. In regards to the calculator, the first negative payment amount and initial payment dates, and return can't be predicted with cell references for your final payment amount and your final are made in-between over time. The actual rate of return date of June 1.
| 1,158
| 5,965
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.140625
| 3
|
CC-MAIN-2019-18
|
latest
|
en
| 0.955842
|
https://giasutamtaiduc.com/cosec-cot-formula.html
| 1,713,544,113,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00375.warc.gz
| 227,629,789
| 16,140
|
# Cosec Cot Formula
5/5 - (1 bình chọn)
Mục Lục
## Cosec Cot
Trigonometry is the field of study which deals with the relationship between angles, heights, and lengths of right triangles. And this time we will be covering Cosec Cot Formula. The ratios of the sides of a right triangle are known as trigonometric ratios. Trigonometry has six main ratios namely sin, cos, tan, cot, sec, and cosec. All these ratios have different formulas. It uses the three sides and angles of a right-angled triangle. Let’s look into Cosec Cot Formulas in detail.
## What Is Cosec Cot Formula?
Let’s look into the Cosec Cot Formula
For an acute angle x in a right triangle, Cosec x is given by
Cosec x = Hypotenuse / Opposite side
Cot x is given by,
Cot x = Adjacent Side / Opposite Side
The Cosec Cot Formula is given as follows:
1+cot2θ=cosec2θ
## Sec, Cosec and Cot
Secant, cosecant and cotangent, almost always written as sec, cosec and cot are trigonometric functions like sin, cos and tan.
Graphs of sec x and cosec x
## Examples using Cosec Cot Formula
Example 1: Prove that (cosec θ – cot θ)2 = (1 – cos θ)/(1 + cos θ)
Solution
LHS = (cosec θ – cot θ)2
= (1/sin θ−cosθ/sin θ)2
= ((1−cos θ)/sin θ)2
RHS = (1 – cos θ)/(1 + cos θ)
By rationalising the denominator,
= (1−cos θ)/(1+cos θ)×(1−cos θ)(1−cos θ)
= (1−cos θ)2/(1−cos2θ)
= (1−cos θ)2/sin2θ
= ((1−cos θ)/sin θ)2
Therefore, LHS = RHS
Example 2: Find Cot P if Tan P = 4 / 3
Solution:
Using Cotangent formula we know that,
Cot P = 1 / Tan P
= 1 / (4 / 3)
= 3/4
Thus, Cot P = 3/4
## Examples of Cosec Cot formula
Q1) If Sin x = ⅗, find the value of Cosec x?
Cosec x = 1/sinx
= 5/3.
Q2) If Sin x = 5/7, find the value of Cosec x?
Cosec x = 1/sinx
= 7/5
Q.3: Prove that (cosec θ – cot θ)2 = (1 – cos θ)/(1 + cos θ).
Solution:
LHS = (cosec θ – cot θ)2
## Sec, Cosec and Cot
Summary
Recall, in case of a right angle triangle if we are given one length and one angle and we have to find a missing length or if we need to find a missing angle when two lengths are given we use SOH/CAH/TOA where:
### Graphical Comparison
#### Example 1
Q. In the triangle, find cosec(A)sec(A), and cot(A).
### Tangent, Cotangent, Secant, and Cosecant
Related articles
Cot Half Angle Formula
Cosec Cot Formula ⭐️⭐️⭐️⭐️⭐
Cot-Tan formula ⭐️⭐️⭐️⭐️⭐
Arccot Formula
| 826
| 2,335
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18
|
latest
|
en
| 0.771183
|
http://www.jiskha.com/display.cgi?id=1340671750
| 1,498,425,477,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320582.2/warc/CC-MAIN-20170625203122-20170625223122-00328.warc.gz
| 559,659,101
| 3,785
|
# math
posted by .
a boy is 17 years old and his sister is as twice as old. when the boy is 23 what will be the age of his sister
• math -
When he was 17 her sister was 34 so when he is 23 which is 6 years later her sister would be 34+6 which is 40 years old
• math -
17 * 2 = 34 >> present age of sister
23 - 17 = 6 >> years from now
6 + 34 = ?
| 112
| 353
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625
| 4
|
CC-MAIN-2017-26
|
latest
|
en
| 0.995718
|
https://willrosenbaum.com/teaching/2023s-cosc-273/labs/03-mandelbrot-set/
| 1,713,810,719,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296818337.62/warc/CC-MAIN-20240422175900-20240422205900-00552.warc.gz
| 548,475,239
| 7,290
|
DUE: Friday, March 24, Monday, March 27, 11:59 pm
Background
The Mandelbrot set
The Mandelbrot set is among the most widely depicted mathematical objects. In order to define the Mandelbrot set (and understand how to draw it), we need to start with the complex numbers and the complex plane.
Complex numbers
Complex numbers are numbers of the form $$a + b i$$ where $$a$$ and $$b$$ are real numbers, and $$i$$ is the imaginary number satisfying $$i^2 = -1$$. Complex numbers can be added and multiplied:
• $(a + b i) + (c + d i) = (a + c) + (b + d)i$
• $(a + b i) (c + d i) = (a c - b d) + (a d + b c) i$
Further, complex numbers have a modulus defined by
• $| a + b i | = \sqrt{a^2 + b^2}$
The complex plane consists of the set of all complex numbers represented as points in a plane: each $$a + b i$$ corresponds to the point $$(a, b)$$ in the plane.
Mandelbrot definition
Given a complex number $$c = a + b i$$, consider the sequence of complex numbers $$z_1, z_2, \ldots$$ defined by the following recursive formula:
• $z_1 = c$
• for $$n > 1$$, $$z_n = z_{n-1}^2 + c$$.
For each starting point $$c$$, one of two things can happen:
1. the sequence $$z_1, z_2, \ldots$$ remains bounded—i.e., there is a positive (real) number $$M$$ such that $$\mid z_n \mid \leq M$$
2. the sequence $$z_1, z_2, \ldots$$ is unbounded—$$\mid z_1 \mid , \mid z_2 \mid , \ldots$$ grows without bound.
The Mandelbrot set is defined to be the set of complex numbers $$c$$ for which the sequence $$z_1, z_2, \ldots$$ remains bounded. For example, $$c = 1$$ is not in the Mandelbrot set because its corresponding sequence is $$1, 2, 5, 26, \ldots$$ which grows without bound. On the other hand, $$c = -1$$ is in the Mandelbrot set because its corresponding sequence is $$-1, 0, -1, 0, \ldots$$ which remains bounded.
Computing the Mandelbrot set
In general, it might be difficult to determine if a given sequence $$z_1, z_2,\ldots$$ remains bounded analytically. To visualize the Mandelbrot set it is enough to determine if the sequence is bounded empirically. Specifically, we can choose two parameters: a modulus bound $$M$$ and an iteration bound $$N$$. We then proceed as follows. Starting from a complex number $$c$$, compute $$z_1, z_2, \ldots$$ as required. For each $$z_n$$, compute $$\mid z_n\mid$$ and do the following:
1. if $$\mid z_n\mid \geq M$$, stop and report that $$c$$ is not in the Mandelbrot set,
2. if $$n = N$$, stop and report that $$c$$ is in the Mandelbrot set
3. otherwise, continue.
Following this procedure, we can obtain this following image of the Mandelbrot set.
The image above depicts the region of the complex plane with real part ($$x$$-coordinates) between $$-2$$ and $$1$$, and imaginary part ($$y$$-coordinates) between $$-1.5$$ and $$1.5$$. The image is 1024 by 1024 pixels, where each pixel represents an point in the complex plane. Each pixel is colored black if it is (empirically) in the Mandelbrot set, and white otherwise. For points $$c$$ not in the Mandelbrot set (i.e., for which case 1 above occurs), the first iteration $$n$$ at which $$\mid z_n \mid \geq M$$ is called the escape time. In order to generate a color image such as the image below, the color is determined by the escape time.
Specifically, the color of a point $$c$$ is determined by a color map applied to the escape time for the point. In the image above, brighter colors indicate shorter escape times. You can read about different methods for coloring the Mandelbrot set at the following link:
Vector Operations in Java
For given parameters $$M$$ and $$N$$ as above, determining the escape time of a single complex value $$c$$ is straightforward. Thus, images as above can be computed by simply iterating over all pixels sequentially and computing the escape time of the point corresponding to each pixel. For the $$1024 \times 1024$$ pixel images above, this requires about a million individual escape time computation, and each escape time requires up to $$N$$ iterations to compute. The calculations for coloring different pixels, however, and independent of one another. Thus we may be able to employ parallelism to compute escape times faster. In this lab, we will explore how vector operations can speed up escape time calculations by performing arithmetic operations for multiple points in parallel.
Vector operations are single instruction, multiple data (SIMD) operations. This means that they perform the same elementary operation on multiple primitive values in parallel. In newer versions of Java (16+), SIMD instructions are supported via the Vector package. The Vector package provides support for SIMD operations on primitive datatypes. Here are links to the relevant documentation:
Program description
For your assignment, you will write a program that uses vector operations to compute escape times for Mandelbrot set operations. To get started, download the following archive:
The program includes four files, Mandelbrot.java, MandelbrotTester.java, MandelbrotViewer.java, and run-test.sh.
• Mandelbrot.java defines a class that stores parameters to describe a region in the complex plane, as well as methods for computing escape times for points in that region. It defines a baseline method, escapeTimesBaseline as well as a(in incomplete) escapeTimesOptimized method. Your task is to complete the escapeTimesOptimized method.
• MandelbrotTester.java defines methods for testing your escapeTimesOptimized method. The method correctnessTest verifies that your optimized and the baseline implementation provide the same escape times, while two performance test methods compare the running times of baseline and optimized implementations.
• MandelbrotViewer.java draws the Mandelbrot set in a specified region using the escape time calculations from Mandelbrot.java. This file is not required for the assignment, but is included so that you view the output of your program and experiment with finding interesting regions of the Mandelbrot set.
• run-test.sh is a shell script for running your program on the HPC cluster. Once your completed files are on the cluster, you can run the tests using sbatch run-test.sh. When the tests complete (typically in about five minutes), the program will write a file mandelbrotTest.txt that includes the results of the test.
Your task is to complete the method escapeTimesOptimized in Mandelbrot.java. You should only complete this method. You should not modify any other methods in Mandelbrot.java or anthing in MandelbrotTester.java. Your implementation of escapeTimesOptimized should be a single-threaded method that employs Vector operations to perform the escape time operations more quickly than the escapeTimesBaseline method (provided in Mandelbrot.java).
What to turn in
To complete the assignment, turn in only your completed Mandelbrot.java file.
Your programs will be graded on a 10 point scale according to the following criteria:
• Correctness (3 pts). Your program compiles and completes the task as specified in the program description above. The program passes the correctness test in MandelbrotTester.java.
• Style and Vectors (3 pt).
• Coding style: code is reasonably well-organized and readable (to a human). Variables, methods, and classes have sensible, descriptive names. Code is well-commented.
• The program correctly applies Vector operations in order to perform the operations in escapeTimesOptimized. No multithreading or other performance enhancements should be used.
• Performance (4 pts).
• 2 pts. The optimized method is faster than the baseline for at least 3 of the 5 performance tests.
• 3 pts. Additionally, the optimized method is at least 2 times faster than baseline for one of the tests.
• 4 pts. The optimized method is at least 3 times faster than the baseline for one of the tests.
Extensions
There are a lot of directions you can go with this assignment! Here are a few:
1. Make a color map that makes things look much better than mine, and use it to make some nice images. (You can submit image files if you’d like.)
2. Modify the MandelbrotViewer to do an infinite zoom into an interesting portion of Mandelbrot set. If you do this, have the program print the framerate to the console.
3. Draw other fractals! The Mandelbrot set is but one fractal in a large family of fractals produced using similar methods. For example, you could draw Newton fractals or other Julia sets.
| 2,033
| 8,411
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.867323
|
https://www.geeksforgeeks.org/check-possible-sort-array-conditional-swapping-adjacent-allowed/
| 1,695,435,633,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00420.warc.gz
| 872,203,701
| 42,626
|
Open In App
Related Articles
# Check if it is possible to sort an array with conditional swapping of adjacent allowed
We are given an unsorted array of integers in the range from 0 to n-1. We are allowed to swap adjacent elements in array many number of times but only if the absolute difference between these element is 1. Check if it is possible to sort the array.If yes then print “yes” else “no”.
Examples:
```Input : arr[] = {1, 0, 3, 2}
Output : yes
Explanation:- We can swap arr[0] and arr[1].
Again we swap arr[2] and arr[3].
Final arr[] = {0, 1, 2, 3}.
Input : arr[] = {2, 1, 0}
Output : no```
Although the problems looks complex at first look, there is a simple solution to it. If we traverse array from left to right and we make sure elements before an index i are sorted before we reach i, we must have maximum of arr[0..i-1] just before i. And this maximum must be either smaller than arr[i] or just one greater than arr[i]. In first case, we simply move ahead. In second case, we swap and move ahead.
Compare the current element with the next element in array.If current element is greater than next element then do following:-
1. Check if difference between two numbers is 1 then swap it.
2. else Return false.
If we reach end of array, we return true.
## C++
`// C++ program to check if we can sort``// an array with adjacent swaps allowed``#include``using` `namespace` `std;` `// Returns true if it is possible to sort``// else false/``bool` `checkForSorting(``int` `arr[], ``int` `n)``{`` ``for` `(``int` `i=0; i arr[i+1])`` ``{`` ``if` `(arr[i] - arr[i+1] == 1)`` ``swap(arr[i], arr[i+1]);` ` ``// If difference is more than`` ``// one, then not possible`` ``else`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;``}` `// Driver code``int` `main()``{`` ``int` `arr[] = {1,0,3,2};`` ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);`` ``if` `(checkForSorting(arr, n))`` ``cout << ``"Yes"``;`` ``else`` ``cout << ``"No"``;``}`
## Java
`class` `Main``{`` ``// Returns true if it is possible to sort`` ``// else false/`` ``static` `boolean` `checkForSorting(``int` `arr[], ``int` `n)`` ``{`` ``for` `(``int` `i=``0``; i arr[i+``1``])`` ``{`` ``if` `(arr[i] - arr[i+``1``] == ``1``)`` ``{`` ``// swapping`` ``int` `temp = arr[i];`` ``arr[i] = arr[i+``1``];`` ``arr[i+``1``] = temp;`` ``}`` ` ` ``// If difference is more than`` ``// one, then not possible`` ``else`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;`` ``}`` ` ` ``// Driver function`` ``public` `static` `void` `main(String args[])`` ``{`` ``int` `arr[] = {``1``,``0``,``3``,``2``};`` ``int` `n = arr.length;`` ``if` `(checkForSorting(arr, n))`` ``System.out.println(``"Yes"``);`` ``else`` ``System.out.println(``"No"``);`` ``}``}`
## Python3
`# Python 3 program to``# check if we can sort``# an array with adjacent``# swaps allowed` `# Returns true if it``# is possible to sort``# else false/``def` `checkForSorting(arr, n):` ` ``for` `i ``in` `range``(``0``,n``-``1``):`` ` ` ``# We need to do something only if`` ``# previousl element is greater`` ``if` `(arr[i] > arr[i``+``1``]):`` ` ` ``if` `(arr[i] ``-` `arr[i``+``1``] ``=``=` `1``):`` ``arr[i], arr[i``+``1``] ``=` `arr[i``+``1``], arr[i]` ` ``# If difference is more than`` ``# one, then not possible`` ``else``:`` ``return` `False` ` ``return` `True` `# Driver code``arr ``=` `[``1``,``0``,``3``,``2``]``n ``=` `len``(arr)``if` `(checkForSorting(arr, n)):`` ``print``(``"Yes"``)``else``:`` ``print``(``"No"``)` `# This code is contributed by``# Smitha Dinesh Semwal`
## C#
`// C# program to check if we can sort``// an array with adjacent swaps allowed``using` `System;` `class` `GFG``{`` ``// Returns true if it is`` ``// possible to sort else false`` ``static` `bool` `checkForSorting(``int` `[]arr, ``int` `n)`` ``{`` ``for` `(``int` `i=0; i arr[i+1])`` ``{`` ``if` `(arr[i] - arr[i+1] == 1)`` ``{`` ``// swapping`` ``int` `temp = arr[i];`` ``arr[i] = arr[i+1];`` ``arr[i+1] = temp;`` ``}`` ` ` ``// If difference is more than`` ``// one, then not possible`` ``else`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;`` ``}`` ` ` ``// Driver function`` ``public` `static` `void` `Main()`` ``{`` ``int` `[]arr = {1, 0, 3, 2};`` ``int` `n = arr.Length;`` ``if` `(checkForSorting(arr, n))`` ``Console.Write(``"Yes"``);`` ``else`` ``Console.Write(``"No"``);`` ``}``}` `// This code is contributed by nitin mittal.`
## PHP
` ``\$arr``[``\$i` `+ 1])`` ``{`` ``if` `(``\$arr``[``\$i``] - ``\$arr``[``\$i` `+ 1] == 1)`` ``{`` ``// swapping`` ``\$temp` `= ``\$arr``[``\$i``];`` ``\$arr``[``\$i``] = ``\$arr``[``\$i` `+ 1];`` ``\$arr``[``\$i` `+ 1] = ``\$temp``;`` ``}` ` ``// If difference is more than`` ``// one, then not possible`` ``else`` ``return` `false;`` ``}`` ``}`` ``return` `true;``}` ` ``// Driver Code`` ``\$arr` `= ``array``(1,0,3,2);`` ``\$n` `= sizeof(``\$arr``);`` ``if` `(checkForSorting(``\$arr``, ``\$n``))`` ``echo` `"Yes"``;`` ``else`` ``echo` `"No"``;` `// This code is contributed``// by nitin mittal.``?>`
## Javascript
``
Output
`Yes`
Time Complexity=O(n)
Auxiliary Space=O(1)
This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
| 2,152
| 6,507
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.546875
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.70151
|
http://openstudy.com/updates/5209b0bbe4b01dab3371380e
| 1,448,580,508,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-48/segments/1448398447860.26/warc/CC-MAIN-20151124205407-00042-ip-10-71-132-137.ec2.internal.warc.gz
| 172,786,580
| 12,322
|
## mathcalculus 2 years ago help with optimization! (attached)
1. mathcalculus
2. mathcalculus
okay i understand everything..... except how to find the minimum!
3. completeidiot
are you familiar with derivative?
4. mathcalculus
i know they got the equation. y=6x+(21000/x)
5. mathcalculus
yes
6. mathcalculus
then after i'm not sure how they got: Minimum occurs at 59.16 ft for the length (found on a graphing calc) width: 10500/59.16 = 177.5 ft Perimeter = 710 ft
7. completeidiot
ok, well the method they got the minimum is just by using a graphing calculator with a max min function on it doing it manually would involve finding the first derivative of the equation and then setting it equal to zero and then solving for L
8. completeidiot
or x
9. mathcalculus
is there a way to do that without it?
10. completeidiot
without the graphing calculator? or without having to do it manually?
11. mathcalculus
by hand
12. mathcalculus
no calculator
13. completeidiot
by hand would involve finding the first derivative of the equation and then setting it equal to zero and then solving for L
14. completeidiot
because the slope at the minimum point is zero
15. mathcalculus
can you show me?
16. mathcalculus
im doing it wrong, even on calculator :/ I'm not getting it.
17. completeidiot
$y = \frac{21000}{x} + 6x$ $y' = -\frac{21000}{x^2} +6$ $y' =0$ $0=-\frac{21000}{x^2} +6$ solve for x hopefully i didnt screw up the derivative
18. completeidiot
$\frac{1}{x} = x^{-1}$
19. completeidiot
any questions?
20. Psymon
@completeidiot derivative is fine :3
21. mathcalculus
yeah 1/x?
22. completeidiot
im just pointing out the identity that allows you to use the "power" rule for derivatives
23. completeidiot
$y= x^n$ $y' = nx^{n-1}$
24. mathcalculus
im not getting that. i got up to here: y=6-21000/x^2 then set it to 0
25. mathcalculus
then im lost
26. completeidiot
1/x is not part of the problem you can ignore it if you want its just that the identity is sometimes not obvious to other people
27. completeidiot
once you set it equal to zero 0=6-21000/x^2 just solve for x
28. completeidiot
6= 21000/x^2 x^2= 21000/6 x = sqrt {21000/6}
29. completeidiot
any other questions? sorry for any confusion i may have caused
30. mathcalculus
i got it=]
31. mathcalculus
thanks!
32. completeidiot
no problem
| 694
| 2,358
|
{"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}
| 3.9375
| 4
|
CC-MAIN-2015-48
|
longest
|
en
| 0.823888
|
https://root.cern.ch/doc/v622/classROOT_1_1Math_1_1Vavilov.html
| 1,720,952,367,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00540.warc.gz
| 455,197,362
| 10,132
|
ROOT Reference Guide
ROOT::Math::Vavilov Class Referenceabstract
Base class describing a Vavilov distribution.
The Vavilov distribution is defined in P.V. Vavilov: Ionization losses of high-energy heavy particles, Sov. Phys. JETP 5 (1957) 749 [Zh. Eksp. Teor. Fiz. 32 (1957) 920].
The probability density function of the Vavilov distribution as function of Landau's parameter is given by:
$p(\lambda_L; \kappa, \beta^2) = \frac{1}{2 \pi i}\int_{c-i\infty}^{c+i\infty} \phi(s) e^{\lambda_L s} ds$
where $$\phi(s) = e^{C} e^{\psi(s)}$$ with $$C = \kappa (1+\beta^2 \gamma )$$ and $$\psi(s)= s \ln \kappa + (s+\beta^2 \kappa) \cdot \left ( \int \limits_{0}^{1} \frac{1 - e^{\frac{-st}{\kappa}}}{t} \,d t- \gamma \right ) - \kappa \, e^{\frac{-s}{\kappa}}$$. $$\gamma = 0.5772156649\dots$$ is Euler's constant.
For the class Vavilov, Pdf returns the Vavilov distribution as function of Landau's parameter $$\lambda_L = \lambda_V/\kappa - \ln \kappa$$, which is the convention used in the CERNLIB routines, and in the tables by S.M. Seltzer and M.J. Berger: Energy loss stragglin of protons and mesons: Tabulation of the Vavilov distribution, pp 187-203 in: National Research Council (U.S.), Committee on Nuclear Science: Studies in penetration of charged particles in matter, Nat. Akad. Sci. Publication 1133, Nucl. Sci. Series Report No. 39, Washington (Nat. Akad. Sci.) 1964, 388 pp. Available from Google books
Therefore, for small values of $$\kappa < 0.01$$, pdf approaches the Landau distribution.
For values $$\kappa > 10$$, the Gauss approximation should be used with $$\mu$$ and $$\sigma$$ given by Vavilov::Mean(kappa, beta2) and sqrt(Vavilov::Variance(kappa, beta2).
The original Vavilov pdf is obtained by v.Pdf(lambdaV/kappa-log(kappa))/kappa.
Two subclasses are provided:
Both subclasses store coefficients needed to calculate $$p(\lambda; \kappa, \beta^2)$$ for fixed values of $$\kappa$$ and $$\beta^2$$. Changing these values is computationally expensive.
VavilovFast is about 5 times faster for the calculation of the Pdf than VavilovAccurate; initialization takes about 100 times longer than calculation of the Pdf value. For the quantile calculation, VavilovFast is 30 times faster for the initialization, and 6 times faster for subsequent calculations. Initialization for Quantile takes 27 (11) times longer than subsequent calls for VavilovFast (VavilovAccurate).
Definition at line 120 of file Vavilov.h.
## Public Member Functions
Vavilov ()
Default constructor. More...
virtual ~Vavilov ()
Destructor. More...
virtual double Cdf (double x) const =0
Evaluate the Vavilov cumulative probability density function. More...
virtual double Cdf (double x, double kappa, double beta2)=0
Evaluate the Vavilov cumulative probability density function, and set kappa and beta2, if necessary. More...
virtual double Cdf_c (double x) const =0
Evaluate the Vavilov complementary cumulative probability density function. More...
virtual double Cdf_c (double x, double kappa, double beta2)=0
Evaluate the Vavilov complementary cumulative probability density function, and set kappa and beta2, if necessary. More...
virtual double GetBeta2 () const =0
Return the current value of $$\beta^2$$. More...
virtual double GetKappa () const =0
Return the current value of $$\kappa$$. More...
virtual double GetLambdaMax () const =0
Return the maximum value of $$\lambda$$ for which $$p(\lambda; \kappa, \beta^2)$$ is nonzero in the current approximation. More...
virtual double GetLambdaMin () const =0
Return the minimum value of $$\lambda$$ for which $$p(\lambda; \kappa, \beta^2)$$ is nonzero in the current approximation. More...
virtual double Kurtosis () const
Return the theoretical kurtosis $$\gamma_2 = \frac{1/3 - \beta^2/4}{\kappa^3 \sigma^4}$$. More...
virtual double Mean () const
Return the theoretical mean $$\mu = \gamma-1- \ln \kappa - \beta^2$$, where $$\gamma = 0.5772\dots$$ is Euler's constant. More...
virtual double Mode () const
Return the value of $$\lambda$$ where the pdf is maximal. More...
virtual double Mode (double kappa, double beta2)
Return the value of $$\lambda$$ where the pdf is maximal function, and set kappa and beta2, if necessary. More...
virtual double Pdf (double x) const =0
Evaluate the Vavilov probability density function. More...
virtual double Pdf (double x, double kappa, double beta2)=0
Evaluate the Vavilov probability density function, and set kappa and beta2, if necessary. More...
virtual double Quantile (double z) const =0
Evaluate the inverse of the Vavilov cumulative probability density function. More...
virtual double Quantile (double z, double kappa, double beta2)=0
Evaluate the inverse of the Vavilov cumulative probability density function, and set kappa and beta2, if necessary. More...
virtual double Quantile_c (double z) const =0
Evaluate the inverse of the complementary Vavilov cumulative probability density function. More...
virtual double Quantile_c (double z, double kappa, double beta2)=0
Evaluate the inverse of the complementary Vavilov cumulative probability density function, and set kappa and beta2, if necessary. More...
virtual void SetKappaBeta2 (double kappa, double beta2)=0
Change $$\kappa$$ and $$\beta^2$$ and recalculate coefficients if necessary. More...
virtual double Skewness () const
Return the theoretical skewness $$\gamma_1 = \frac{1/2 - \beta^2/3}{\kappa^2 \sigma^3}$$. More...
virtual double Variance () const
Return the theoretical variance $$\sigma^2 = \frac{1 - \beta^2/2}{\kappa}$$. More...
## Static Public Member Functions
static double Kurtosis (double kappa, double beta2)
Return the theoretical kurtosis $$\gamma_2 = \frac{1/3 - \beta^2/4}{\kappa^3 \sigma^4}$$. More...
static double Mean (double kappa, double beta2)
Return the theoretical Mean $$\mu = \gamma-1- \ln \kappa - \beta^2$$. More...
static double Skewness (double kappa, double beta2)
Return the theoretical skewness $$\gamma_1 = \frac{1/2 - \beta^2/3}{\kappa^2 \sigma^3}$$. More...
static double Variance (double kappa, double beta2)
Return the theoretical Variance $$\sigma^2 = \frac{1 - \beta^2/2}{\kappa}$$. More...
#include <Math/Vavilov.h>
Inheritance diagram for ROOT::Math::Vavilov:
[legend]
## ◆ Vavilov()
ROOT::Math::Vavilov::Vavilov ( )
Default constructor.
Definition at line 46 of file Vavilov.cxx.
## ◆ ~Vavilov()
ROOT::Math::Vavilov::~Vavilov ( )
virtual
Destructor.
Definition at line 50 of file Vavilov.cxx.
## ◆ Cdf() [1/2]
virtual double ROOT::Math::Vavilov::Cdf ( double x ) const
pure virtual
Evaluate the Vavilov cumulative probability density function.
Parameters
x The Landau parameter $$x = \lambda_L$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Cdf() [2/2]
virtual double ROOT::Math::Vavilov::Cdf ( double x, double kappa, double beta2 )
pure virtual
Evaluate the Vavilov cumulative probability density function, and set kappa and beta2, if necessary.
Parameters
x The Landau parameter $$x = \lambda_L$$ kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Cdf_c() [1/2]
virtual double ROOT::Math::Vavilov::Cdf_c ( double x ) const
pure virtual
Evaluate the Vavilov complementary cumulative probability density function.
Parameters
x The Landau parameter $$x = \lambda_L$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Cdf_c() [2/2]
virtual double ROOT::Math::Vavilov::Cdf_c ( double x, double kappa, double beta2 )
pure virtual
Evaluate the Vavilov complementary cumulative probability density function, and set kappa and beta2, if necessary.
Parameters
x The Landau parameter $$x = \lambda_L$$ kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ GetBeta2()
virtual double ROOT::Math::Vavilov::GetBeta2 ( ) const
pure virtual
Return the current value of $$\beta^2$$.
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ GetKappa()
virtual double ROOT::Math::Vavilov::GetKappa ( ) const
pure virtual
Return the current value of $$\kappa$$.
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ GetLambdaMax()
virtual double ROOT::Math::Vavilov::GetLambdaMax ( ) const
pure virtual
Return the maximum value of $$\lambda$$ for which $$p(\lambda; \kappa, \beta^2)$$ is nonzero in the current approximation.
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ GetLambdaMin()
virtual double ROOT::Math::Vavilov::GetLambdaMin ( ) const
pure virtual
Return the minimum value of $$\lambda$$ for which $$p(\lambda; \kappa, \beta^2)$$ is nonzero in the current approximation.
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Kurtosis() [1/2]
double ROOT::Math::Vavilov::Kurtosis ( ) const
virtual
Return the theoretical kurtosis $$\gamma_2 = \frac{1/3 - \beta^2/4}{\kappa^3 \sigma^4}$$.
Definition at line 105 of file Vavilov.cxx.
## ◆ Kurtosis() [2/2]
double ROOT::Math::Vavilov::Kurtosis ( double kappa, double beta2 )
static
Return the theoretical kurtosis $$\gamma_2 = \frac{1/3 - \beta^2/4}{\kappa^3 \sigma^4}$$.
Parameters
kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Definition at line 109 of file Vavilov.cxx.
## ◆ Mean() [1/2]
double ROOT::Math::Vavilov::Mean ( ) const
virtual
Return the theoretical mean $$\mu = \gamma-1- \ln \kappa - \beta^2$$, where $$\gamma = 0.5772\dots$$ is Euler's constant.
Definition at line 80 of file Vavilov.cxx.
## ◆ Mean() [2/2]
double ROOT::Math::Vavilov::Mean ( double kappa, double beta2 )
static
Return the theoretical Mean $$\mu = \gamma-1- \ln \kappa - \beta^2$$.
Parameters
kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Definition at line 84 of file Vavilov.cxx.
## ◆ Mode() [1/2]
double ROOT::Math::Vavilov::Mode ( ) const
virtual
Return the value of $$\lambda$$ where the pdf is maximal.
Reimplemented in ROOT::Math::VavilovAccurate.
Definition at line 56 of file Vavilov.cxx.
## ◆ Mode() [2/2]
double ROOT::Math::Vavilov::Mode ( double kappa, double beta2 )
virtual
Return the value of $$\lambda$$ where the pdf is maximal function, and set kappa and beta2, if necessary.
Parameters
kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Reimplemented in ROOT::Math::VavilovAccurate.
Definition at line 75 of file Vavilov.cxx.
## ◆ Pdf() [1/2]
virtual double ROOT::Math::Vavilov::Pdf ( double x ) const
pure virtual
Evaluate the Vavilov probability density function.
Parameters
x The Landau parameter $$x = \lambda_L$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Pdf() [2/2]
virtual double ROOT::Math::Vavilov::Pdf ( double x, double kappa, double beta2 )
pure virtual
Evaluate the Vavilov probability density function, and set kappa and beta2, if necessary.
Parameters
x The Landau parameter $$x = \lambda_L$$ kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Quantile() [1/2]
virtual double ROOT::Math::Vavilov::Quantile ( double z ) const
pure virtual
Evaluate the inverse of the Vavilov cumulative probability density function.
Parameters
z The argument $$z$$, which must be in the range $$0 \le z \le 1$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Quantile() [2/2]
virtual double ROOT::Math::Vavilov::Quantile ( double z, double kappa, double beta2 )
pure virtual
Evaluate the inverse of the Vavilov cumulative probability density function, and set kappa and beta2, if necessary.
Parameters
z The argument $$z$$, which must be in the range $$0 \le z \le 1$$ kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Quantile_c() [1/2]
virtual double ROOT::Math::Vavilov::Quantile_c ( double z ) const
pure virtual
Evaluate the inverse of the complementary Vavilov cumulative probability density function.
Parameters
z The argument $$z$$, which must be in the range $$0 \le z \le 1$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Quantile_c() [2/2]
virtual double ROOT::Math::Vavilov::Quantile_c ( double z, double kappa, double beta2 )
pure virtual
Evaluate the inverse of the complementary Vavilov cumulative probability density function, and set kappa and beta2, if necessary.
Parameters
z The argument $$z$$, which must be in the range $$0 \le z \le 1$$ kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ SetKappaBeta2()
virtual void ROOT::Math::Vavilov::SetKappaBeta2 ( double kappa, double beta2 )
pure virtual
Change $$\kappa$$ and $$\beta^2$$ and recalculate coefficients if necessary.
Parameters
kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Implemented in ROOT::Math::VavilovAccurate, and ROOT::Math::VavilovFast.
## ◆ Skewness() [1/2]
double ROOT::Math::Vavilov::Skewness ( ) const
virtual
Return the theoretical skewness $$\gamma_1 = \frac{1/2 - \beta^2/3}{\kappa^2 \sigma^3}$$.
Definition at line 96 of file Vavilov.cxx.
## ◆ Skewness() [2/2]
double ROOT::Math::Vavilov::Skewness ( double kappa, double beta2 )
static
Return the theoretical skewness $$\gamma_1 = \frac{1/2 - \beta^2/3}{\kappa^2 \sigma^3}$$.
Parameters
kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Definition at line 100 of file Vavilov.cxx.
## ◆ Variance() [1/2]
double ROOT::Math::Vavilov::Variance ( ) const
virtual
Return the theoretical variance $$\sigma^2 = \frac{1 - \beta^2/2}{\kappa}$$.
Definition at line 88 of file Vavilov.cxx.
## ◆ Variance() [2/2]
double ROOT::Math::Vavilov::Variance ( double kappa, double beta2 )
static
Return the theoretical Variance $$\sigma^2 = \frac{1 - \beta^2/2}{\kappa}$$.
Parameters
kappa The parameter $$\kappa$$, which should be in the range $$0.01 \le \kappa \le 10$$ beta2 The parameter $$\beta^2$$, which must be in the range $$0 \le \beta^2 \le 1$$
Definition at line 92 of file Vavilov.cxx.
The documentation for this class was generated from the following files:
| 4,641
| 15,429
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.828125
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.633011
|
https://forum.allaboutcircuits.com/threads/analysis-of-rc-integrator-circuit-practical-vs-theoretical.168840/
| 1,702,074,822,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00406.warc.gz
| 308,257,802
| 22,685
|
# Analysis of RC Integrator Circuit Practical vs Theoretical
#### Lod
Joined Apr 16, 2020
2
Hi guys,
Just started learning about RC Integrator op-amp circuits and currently doing a lab report for university about it. The point of the lab is very basic, as it is just for us to see the difference between practical and theoretical results.
I was noticing differences between the two and realized for that for the practical results, the output signal was flat at the top? I'm having trouble thinking of the reason. Maybe it's the time it takes for the capacitor to charge then discharge? Ill attach them so you can see.
Sorry, if this is in the wrong subforum, thought it looked right as it relates to oscilloscope analysis.
Thanks.
#### Attachments
• 71.1 KB Views: 14
• 11.1 KB Views: 14
#### drc_567
Joined Dec 29, 2008
1,156
... It seems that usually the first item that is requested for an inquiry such as this is a Circuit Diagram with the key components labeled. Datasheets for op-amps, transistors, and any active device must be examined to insure that any voltage or current transient values, as well as power supply settings, do not violate the published limits.
#### LvW
Joined Jun 13, 2013
1,683
Lod - it is not easy to give you a substantial answer because lack of information:
(1) What does "theoretical" mean: Ideal FUNCTION of an (virtual) integrator? Or ideal opamp (with large but finite gain) ?
(2) What is shown in the attached graphics? Simulation or measurement?
(3) Corresponding circuit? Non-inverting or inverting integrator?
#### Lod
Joined Apr 16, 2020
2
My apologies, I did not provide enough information, I am quite new to electronics as a whole so I do not know much about the real specifics in drc_567's answer and I thought that this would be a general difference between practical and theoretical and could be answered without those factors. However, I will try to answer.
- R = 1k ohms
- C = 10 uF
0.1 V input signal square wave with a 500 ms width and a frequency of 1 Hz.
The device used to measure the practical value was Analog Discovery Kit 2. The Multisim values mimic those above.
1) Theoretical meaning an ideal op amp with (infinite gain)
2) As I implied in my post, the graph with the square flat peak instead of the triangular peak (so the one labelled 1hz lab - tested in Waveforms) is the practical. Hence, the other one (labelled Transient Ex2) is the theoretical (tested in the Multisim software).
3) As for the circuit, it is an inverting RC Integrator op amp circuit
I will attach the circuit as well.
Also, how would I calculate the gain too? I was using the equation attached but am unsure if it is correct to use it this way as I thought it took frequency into account.
Sorry for the inconvenience.
#### DickCappels
Joined Aug 21, 2008
10,104
Look at how close the top of the waveform gets to the positive power supply then check the datasheet for the LM741 and see whether it is trying to output a voltage that exceeds the maximum output voltage with that power supply.
#### LvW
Joined Jun 13, 2013
1,683
Lod - the main problem of the practical circuit (hardware or simulation with a real opamp model) is the fact that you have no DC negative feedback (necessary for a stable DC operating point.)-There is always an offset voltage which brings the output to saturation (unwanted continous loading of the cap).Hence, you need a large R across the C.
#### crutschow
Joined Mar 14, 2008
33,367
What's all the fuzz on the yellow trace.
Looks like something is oscillating.
As LvW noted, the circuit needs DC feedback for a stable operating point.
Without that, the output will tend to drift to either positive or negative saturation at the rail.
#### DickCappels
Joined Aug 21, 2008
10,104
What's all the fuzz on the yellow trace.
Looks like something is oscillating.
As LvW noted, the circuit needs DC feedback for a stable operating point.
Without that, the output will tend to drift to either positive or negative saturation at the rail.
Some people are more observant than others. That got completely past me. Try putting 1 Meg across your 10 uf capacitor. Of course the ±15 mv offset voltage (1970's technology, amazingly this is still being made) will show up as ±15V on the output. You need an opamp with much lower offset voltage and low input leakage current.
An LM358 or LM324 would be ok (also terribly old and behind the times, but it may work).
As long as you are keeping the frequency low an OP-07 with its 150 uV offset (max) should be fine.
(Yeah, I know there are better opamps out there. This sticks to those introduced during the Cambrian Explosions of Op Apamps in the 1970's)
| 1,123
| 4,674
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.796875
| 3
|
CC-MAIN-2023-50
|
latest
|
en
| 0.954524
|
http://www.brainia.com/essays/Physics/427038.html
| 1,500,822,113,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-30/segments/1500549424564.72/warc/CC-MAIN-20170723142634-20170723162634-00610.warc.gz
| 357,929,649
| 5,797
|
# Physics
## Physics
Deformation of solid
Hooke’s law
Explain force-extension graph
A
--Because we are changing the force and this
results in a change in the extension, the force
along the horizontal axis and the extension
along the vertical axis.
--But this graph has extension on the
horizontal axis and force on the vertical axis.
--Because the gradient of the straight section
of this graph turns out to be an important
quantity, known as the force constant of the
spring
--The extension x is directly proportional to
F = kx
Hooke’s low:
--A material obeys Hooke’s low if the
Extension produced in it is proportional
--So, the force constant k of the spring is
given by the equation
k = F/x
--A stiffer spring will have a large value for
the force constant k. Beyond point A, the
graph is no longer a straight line; its gradient
changes and we can no longer use the
elastic limit & spring
constant
Elastic limit
--The force beyond which the spring becomes
permanently deformed is known as the elastic
limit.
--Explain: if you apply a small force to a spring and
This behavior is described as ‘elastic.’ However, if
you apply a large force, the spring may not return
to its original length. It has become permanently
deformed
elastic limit & spring
constant
Spring constant
--Hooke’s low is a principle of
physics that state that the
force F needed to extend or
compress a spring by some
distance X is proportional to
that distance.
--F = kX, where the gradient k
is known as the force constant
of the spring
--F is the applied force (load)
--X is the extension
Calculation of the effective spring constant
of two spring combine in series
Consider two
springs placed
in series with a
mass on the
bottom of the
second. The
force is the
same on each
of the two
springs.
Therefore,
F = k11X11 = k22X22
Solving...
| 456
| 1,829
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.984375
| 4
|
CC-MAIN-2017-30
|
latest
|
en
| 0.907904
|
https://forum.heatinghelp.com/discussion/136814/btu-formulas
| 1,638,141,035,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00574.warc.gz
| 338,614,508
| 22,292
|
Welcome! Here are the website rules, as well as some tips for using this forum.
# BTU formulas?
Member Posts: 4
I have a customer that owns an artesian well. This customer is a hot water utility company that distributes heat and water to various companies. One particular customer is an industrial laundromat that uses both heat and water. Here is the scenario; they have a line coming in at 175 degrees F at 100 gpm. The line then T's and they take hot water off this T. Bypassing the T is the return line its temperature is 140 degrees F and the flow rate is 50 gpm. My customer wants to bill the laundromat for the use of the heated water and for the use of the heat from the returned water. So that leads me to my questions; Using my supply and return flow rates, how do I calculate BTU usage. On the T where the laundromat is taking water and doesn't return, how do calculate BTU on that? Thanks for any help and equasions.
• Member Posts: 212
BTU Meters
www.istec-corp.com
• Member Posts: 4
Need Formulas
Thanks for the links, but I need the formulas. Especially where the water isn't returning.
• Member Posts: 5,837
Simple numbers...
Gallons per hour times pounds (8.3 per gallon, but you have to adjust for water temperature and subsequent density, which changes with temperature and is available from engineeringtoolbox.com) times delta T = BTUH.
ME
It's not so much a case of "You got what you paid for", as it is a matter of "You DIDN'T get what you DIDN'T pay for, and you're NOT going to get what you thought you were in the way of comfort". Borrowed from Heatboy.
• Member Posts: 4
BTU formulas?
I figured the formula would be different since there wasn't a delta T for the one problem.
• Member Posts: 5,837
Dsiplaced energy...
If the end user had to heat his own water, and it was coming from a well, at say 45 degrees F, then the energy he avoided would be based on that entering water temperature and the final water temperature. If you take into consideration the efficiency of a water heater at a seasonal energy factor of say 70% (EF-.70), then you would take the delivered BTU's, and divide by .7 to compensate for what he would have wasted in regular energy getting the hot water to that point, If your customer wants to be real nice, he could just skip the last part of the formulae...
I would then use the base cost per therm for conventional energy being displaced. If it were electricity, then it would be around \$1.50 per therm (100,000 btu's) . If natural gas, then here in Denver it is around \$0.50 per therm.
ME
It's not so much a case of "You got what you paid for", as it is a matter of "You DIDN'T get what you DIDN'T pay for, and you're NOT going to get what you thought you were in the way of comfort". Borrowed from Heatboy.
• Member Posts: 4
BTU Formulas?
That's what I was searching for. Thanks for your supporting info...
• Member Posts: 36
Electricity Cost?
I thought,
If 100,000 BTU (1 therm) and 3412 BTU per KW than
100000/3412= 29.31KW x .18 per KWH = \$5.27 @ 100% EFF
The .18 is average delivered cost in CT
Am I doing something wrong? I am no where near \$1.50
This discussion has been closed.
| 815
| 3,175
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.484375
| 3
|
CC-MAIN-2021-49
|
latest
|
en
| 0.95349
|
http://www.planningplanet.com/forums/primavera-version-pm5-pm6/519773/why-critical-activities-not-same-longets-path-activities
| 1,568,738,061,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-39/segments/1568514573098.0/warc/CC-MAIN-20190917161045-20190917183045-00337.warc.gz
| 266,428,995
| 10,087
|
## Tips on using this forum..
(1) Explain your problem, don't simply post "This isn't working". What were you doing when you faced the problem? What have you tried to resolve - did you look for a solution using "Search" ? Has it happened just once or several times?
(2) It's also good to get feedback when a solution is found, return to the original post to explain how it was resolved so that more people can also use the results.
# Why critical activities not same longets path activities?
3 replies [Last post]
hanson California
Offline
Joined: 9 Jan 2012
Posts: 116
Dear Planners,
I have schedule 7000 activities our clints working on with critical activities 850 and longest path only 5 ... why they are not equal or close to geather?
Thank you,
## Replies
hanson California
Offline
Joined: 9 Jan 2012
Posts: 116
Rafael and Gary thanks a lot and really I appreciate you help. May GOD bless you both...
Offline
Joined: 24 Mar 2009
Posts: 1215
Hanson,
This is probably due to either OOS working, or constraints. -both of these can reduce an activities float to the point of becoming critical without it being on the longest path.
Rafael Davila
Offline
Joined: 1 Mar 2004
Posts: 4780
Longest path is frequently different to true critical activities because it cannot handle multiple date constraints nor resource leveling.
If you are interested on identifying activities that are not critical at any specific moment but with high probability of becoming critical you shall look at statistical modeling.
Traditional definition of critical activities is, activities with no available float, that if delayed would delay the job finish date or any contractual milestone.
I believe it is mathematically correct but unfortunately some people that cannot accept critical paths can be multiple under multiple contractual delivery milestones or that these paths can be just segments still insist on the ancient theory of critical path being a single path, continuous from job start to job finish, a wrong mathematical assumption under resource leveling and multiple contractual constraints.
To make it worst, they create flawed theories about criticality, and go as far as including those flawed theories on forensic practice. Statistical methods usually will identify activities on the longest path with a high probability of becoming critical, but being critical or near critical is a different thing, usually close but not the same.
Would you say someone killed me because the bullet was close, even if the bullet did not touch me and I am still alive?
| 543
| 2,568
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39
|
longest
|
en
| 0.934777
|
http://forums.wolfram.com/mathgroup/archive/2010/Apr/msg00560.html
| 1,716,603,165,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00021.warc.gz
| 9,420,787
| 7,846
|
Re: Lining up y axes and sizing plot areas
• To: mathgroup at smc.vnet.net
• Subject: [mg109323] Re: Lining up y axes and sizing plot areas
• From: "David Park" <djmpark at comcast.net>
• Date: Thu, 22 Apr 2010 03:29:16 -0400 (EDT)
```Use an ImagePadding option in both plots to leave the same amount of padding
on the left to accommodate either set of tick values. This will mean there
will be extra space for the integer y ticks, but that's what you need to do
to make things line up.
Column[
{Plot[2 Sin[x] + x, {x, 0, 15},
Filling -> Bottom,
ImagePadding -> {{20, 5}, {5, 5}}],
Plot[{Sin[x], Sin[2 x], Sin[3 x]}, {x, 0, 2 Pi},
ImagePadding -> {{20, 5}, {5, 5}}]}]
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
From: Garapata [mailto:warsaw95826 at mypacks.net]
Hi everybody!
I have two ListLinePlots of different time series analysis. They each
have the same number of data points along the x axis. I want to show
them one above the other, but have their respective time scales line
up properly, e.g., the first data points lining up and the last data
points lining up. To do this the display of the graphs themselves,
exclusive of their y axises need to be the same size.
Another way to say this, the y-axes for both charts should line up
then everything to the right of the axes should be the same width.
Mine don=92t line up because one of my plots has a integer scaled y-axis
(running from 1 to 50) and the other has a decimal scale with 2 or
sometimes 3 positions to the right of the decimal.
I have ListLinePlots, but the following will illustrate the problem:
Column[{Plot[2 Sin[x] + x, {x, 0, 15}, Filling -> Bottom],
Plot[{Sin[x], Sin[2 x], Sin[3 x]}, {x, 0, 2 Pi}]}]
| 514
| 1,718
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22
|
latest
|
en
| 0.823782
|
https://calculatoruniverse.com/margin-calculator/
| 1,723,162,697,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00639.warc.gz
| 121,807,473
| 52,874
|
Margin Calculator
Instructions:
• Enter the cost price, selling price, quantity, and discount percentage.
• Click "Calculate Margin & Profit" to calculate the margin and profit.
• Review the chart and calculation details for the results.
• Your calculation history will be displayed below with a smooth animation.
• Click "Clear" to reset the form and chart.
• Click "Copy" to copy the calculated margin to the clipboard.
Calculation Details
Margin: (Selling Price - (Selling Price * (Discount / 100))) * Quantity - (Cost Price * Quantity)
Profit: (Selling Price - (Selling Price * (Discount / 100))) * Quantity - (Cost Price * Quantity)
Calculation History
What is Margin?
Margin refers to the difference between the total revenue generated by a business or individual and the total costs incurred in generating that revenue. It is a financial metric that measures profitability and is expressed as a percentage.
All Formulae Related to Margin Calculator
1. Gross Margin Formula: Gross Margin = (Total Revenue – Cost of Goods Sold) / Total Revenue
2. Gross Profit Formula: Gross Profit = Total Revenue – Cost of Goods Sold
3. Operating Margin Formula: Operating Margin = (Total Revenue – Operating Expenses) / Total Revenue
4. Operating Profit (Operating Income) Formula: Operating Profit = Total Revenue – Operating Expenses
5. Net Margin Formula: Net Margin = (Total Revenue – Total Expenses) / Total Revenue
6. Net Profit (Net Income) Formula: Net Profit = Total Revenue – Total Expenses
7. Profit Margin Percentage Formula (for any margin type): Profit Margin Percentage = (Profit / Total Revenue) * 100
8. Markup Percentage Formula: Markup Percentage = (Selling Price – Cost Price) / Cost Price * 100
9. Margin of Safety Formula: Margin of Safety = (Current Sales – Break-Even Sales) / Current Sales
Practical Uses of Margin
Margins have various practical uses in business and finance. Here are some practical applications of margins:
1. Profitability Analysis: Margins are used to assess the profitability of a business. Companies track gross, operating, and net margins to understand how efficiently they are converting revenue into profit. This information helps in making strategic decisions, setting prices, and managing costs.
2. Pricing Strategy: Calculating gross margins helps businesses set appropriate selling prices for their products or services. By understanding the cost of goods sold (COGS) and desired profit margins, companies can determine competitive pricing strategies.
3. Cost Control: Analyzing operating margins allows businesses to identify areas where they can reduce expenses. It helps in cost management, optimizing resource allocation, and improving overall efficiency.
4. Investment Analysis: Investors use margins to evaluate the financial health and profitability of companies. High and stable profit margins may indicate a well-managed and financially sound business, making it an attractive investment option.
5. Benchmarking: Comparing margins to industry benchmarks or competitors’ margins provides insights into a company’s relative performance. This helps businesses identify areas where they may need to improve to stay competitive.
6. Risk Assessment: Lower margins can indicate higher financial risk, as businesses with slim profit margins may be more vulnerable to economic downturns or unexpected expenses. Investors and lenders consider margins when assessing the risk associated with a company.
7. Loan Approval: Lenders look at a company’s margins when evaluating loan applications. Healthy margins can increase the likelihood of loan approval, as they demonstrate the ability to generate sufficient profit to repay the loan.
8. Business Valuation: Margins are considered when valuing a business for mergers, acquisitions, or sales. They are used as part of the financial analysis to determine the fair market value of the company.
9. Performance Tracking: Regularly monitoring margins allows businesses to track their financial performance over time. It helps in identifying trends, whether margins are improving or declining, and taking corrective actions as needed.
10. Decision-Making: Margins play a crucial role in strategic decision-making. Whether expanding operations, entering new markets, or launching new products, understanding the potential impact on margins is essential for informed choices.
Applications of Margin Calculator in Various Fields
Margin calculators find applications in various fields and industries where profit margin analysis is important. Here are some examples of how margin calculators are used in different domains:
1. Retail and E-commerce:
• Pricing Strategy: Retailers and e-commerce businesses use margin calculators to set product prices by considering cost of goods sold (COGS) and desired profit margins.
• Sales Promotions: Calculating margins helps in determining the impact of discounts, promotions, and sales events on profitability.
2. Manufacturing and Production:
• Cost Management: Manufacturers use margin calculators to control production costs and optimize resource allocation.
• Batch Production: Determine the profitability of producing goods in different batch sizes.
3. Finance and Investment:
• Stock Trading: Investors analyze profit margins when making decisions about buying or selling stocks to assess a company’s financial health.
• Investment Analysis: Margin calculators are used in financial modeling to project potential returns on investments.
4. Real Estate:
• Property Investment: Real estate investors use margin calculations to evaluate rental property profitability by considering rental income and expenses.
• Property Flipping: Assess the profitability of buying, renovating, and selling properties.
5. Food and Beverage:
• Restaurant Profitability: Calculate profit margins to determine the profitability of menu items and optimize pricing.
• Food Production: Food manufacturers use margin calculators for cost analysis and pricing decisions.
6. Construction and Contracting:
• Contract Bidding: Construction companies use margin calculations to submit competitive bids for projects while ensuring profitability.
• Project Management: Evaluate the profitability of construction projects by comparing costs to revenue.
7. Wholesale and Distribution:
• Wholesale Pricing: Wholesale distributors calculate profit margins when setting prices for retailers and customers.
• Inventory Management: Assess the profitability of maintaining certain inventory levels.
• Consulting Services: Service providers use margin calculations to determine consulting fees and project profitability.
• Freelancers: Calculate profit margins for freelance services to ensure fair pricing.
Benefits of Using the Margin
Using margins in various aspects of business and finance offers several benefits. Here are some of the key advantages of using margins:
1. Profitability Assessment: Margins provide a clear indicator of a business’s profitability. They help assess how efficiently a company converts its revenue into profit, allowing for better financial planning and performance evaluation.
2. Pricing Strategy: Margins are essential for setting competitive and sustainable prices for products and services. By calculating margins, businesses can avoid underpricing that leads to losses and overpricing that deters customers.
3. Cost Control: Analyzing margins helps identify areas where costs can be reduced without compromising quality. This supports cost management and efficiency improvements.
4. Financial Health: Margins are valuable metrics for assessing a company’s financial health. Investors, lenders, and stakeholders use them to gauge the stability and viability of a business.
Last Updated : 31 July, 2024
One request?
I’ve put so much effort writing this blog post to provide value to you. It’ll be very helpful for me, if you consider sharing it on social media or with your friends/family. SHARING IS ♥️
0
0
0
0
0
0
10 thoughts on “Margin Calculator”
1. Thank you for sharing these valuable insights on calculating margins. It’s clear how important it is to be aware of these formulas in various fields, including finance, investment, and production.
2. I find this information to be quite overwhelming. Do businesses really need to delve into these detailed margin calculations for effective management, or are there simpler ways to achieve the same goals?
3. I appreciate the breakdown of different margin formulas and their significance. It’s essential for any business to understand these metrics in order to effectively analyze profitability, control costs, and make informed decisions. The comprehensive examples provided for various fields and industries are enlightening.
4. This is great information! I had a basic understanding of margins, but these formulas and practical applications are really useful. Now I can see how it applies to different industries, and how crucial it is for businesses to consider margins in their decision-making processes.
5. This article has provided a clear understanding of margin formulas and their practical applications. The examples given for different industries have made it easier to grasp how margins are used in pricing strategies, cost control, and investment analysis.
6. The in-depth analysis of margin formulas, their applications, and practical uses in various fields is beneficial for any individual or business. It helps in understanding the importance of margins in profit generation, cost management, and decision-making.
7. The article effectively presents how margin calculators are utilized in different fields such as retail, manufacturing, finance, real estate, food, and construction. This diverse range of applications demonstrates the universal relevance and importance of margin analysis in business operations.
8. The wide-ranging applications of margin calculators across various industries, including retail, manufacturing, finance, real estate, food, and construction, are illustrative of the significance of margin analysis in diverse business domains.
9. The detailed explanations of margin formulas and their significance in profitability analysis, pricing strategy, and risk assessment are enlightening. It highlights the critical role of margins in business management and financial health.
10. I don’t see the point of all these complex calculations just to determine profit. Isn’t there an easier way to understand business profitability without getting into such detailed mathematical formulas?
| 1,917
| 10,530
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.46875
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.796683
|
http://www.jiskha.com/display.cgi?id=1321036726
| 1,493,619,916,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-17/segments/1492917127681.84/warc/CC-MAIN-20170423031207-00101-ip-10-145-167-34.ec2.internal.warc.gz
| 549,064,437
| 3,754
|
# precalculus
posted by on .
Find (a) the equation of a line that is parallel to the given line and includes the given point, and (b) the equation of a line that is perpendicular to the given line through the given point. Write both answers in slope-intercept form. (c) Graph both of these lines on the same axes and submit your graph. Please show all of your work.
y = -2x + 3, (-2, -3)
• precalculus - ,
Just use the point-slope form and then rearrange terms to slope-intercept form:
The line through (-2,-3) with slope -2:
(y-(-3))/(x-(-2)) = (y+3)/(x+2) = -2
The other line has slope -1/(-2) = 1/2.
Go for it.
| 183
| 620
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17
|
latest
|
en
| 0.946433
|
https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Axiom_of_union.html
| 1,675,547,586,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00597.warc.gz
| 293,442,150
| 6,363
|
# Axiom of union
In axiomatic set theory and the branches of logic, mathematics, and computer science that use it, the axiom of union is one of the axioms of Zermelo–Fraenkel set theory, stating that, for any set x there is a set y whose elements are precisely the elements of the elements of x. Together with the axiom of pairing this implies that for any two sets, there is a set that contains exactly the elements of both.
## Formal statement
In the formal language of the Zermelo–Fraenkel axioms, the axiom reads:
or in words:
Given any set A, there is a set B such that, for any element c, c is a member of B if and only if there is a set D such that c is a member of D and D is a member of A.
or, more simply:
For any set , there is a set which consists of just the elements of the elements of that set.
## Interpretation
What the axiom is really saying is that, given a set A, we can find a set B whose members are precisely the members of the members of A. By the axiom of extensionality this set B is unique and it is called the union of A, and denoted . Thus the essence of the axiom is:
The union of a set is a set.
Note that the union of A and B, commonly written as , can be written as . Thus, the ordinary construction of unions is trivial given the axiom of pairing.
The axiom of union is generally considered uncontroversial, and it or an equivalent appears in just about any alternative axiomatization of set theory.
Note that there is no corresponding axiom of intersection. If A is a nonempty set containing E, then we can form the intersection using the axiom schema of specification as
{c in E: for all D in A, c is in D},
so no separate axiom of intersection is necessary. (If A is the empty set, then trying to form the intersection of A as
{c: for all D in A, c is in D}
is not permitted by the axioms. Moreover, if such a set existed, then it would contain every set in the "universe", but the notion of a universal set is antithetical to Zermelo–Fraenkel set theory.)
## References
• Paul Halmos, Naive set theory. Princeton, NJ: D. Van Nostrand Company, 1960. Reprinted by Springer-Verlag, New York, 1974. ISBN 0-387-90092-6 (Springer-Verlag edition).
• Jech, Thomas, 2003. Set Theory: The Third Millennium Edition, Revised and Expanded. Springer. ISBN 3-540-44085-2.
• Kunen, Kenneth, 1980. Set Theory: An Introduction to Independence Proofs. Elsevier. ISBN 0-444-86839-9.
| 614
| 2,419
|
{"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}
| 4.03125
| 4
|
CC-MAIN-2023-06
|
latest
|
en
| 0.953374
|
https://www.pcreview.co.uk/threads/need-some-help-for-linear-interpolation.3041236/
| 1,660,665,290,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882572408.31/warc/CC-MAIN-20220816151008-20220816181008-00689.warc.gz
| 811,846,146
| 17,747
|
# need some help for linear interpolation
C
#### choi1chung1gm
Dear Kind People,
I need to linearly interpolate two points (X0,Y0) & (X1,Y1) to find
the New_Y value for a given New_X. The X's value increases as i go
down in column A. Say I want to find New_Y for New_X = 0.7895. I want
to lookup the two closest values of X's that New_X = 0.7895 is
between(0.7892 and 0.7897) and then use these two points to linearly
interpolate the Ys to get New_Y for New_X = 0.7895.
Example data set -
A B C D
X Y New_X New_Y
1 0.782 2.1417 0.782
2 0.7831 2.1474 0.7839
3 0.7842 2.1532 0.7858
4 0.7853 2.1592 0.7876
5 0.7864 2.1653 0.7895
6 0.7875 2.1715 0.7914
7 0.7881 2.1746 0.7932
8 0.7886 2.1778 0.7951
9 0.7892 2.181 0.797
10 0.7897 2.1843 0.7988
11 0.7908 2.1908 0.8069
12 0.7919 2.1974 0.8007 2.0298
13 0.793 2.2042
14 0.7936 2.2076
15 0.7941 2.211
16 0.7947 2.2144
17 0.7952 2.2179
18 0.7958 2.2214
19 0.7963 2.2248
20 0.7974 2.2318
21 0.7979 2.2354
22 0.7985 2.2389
23 0.799 2.2424
24 0.7996 2.246
25 0.8001 2.0321
26 0.8007 2.0298
I have a VBScript (please see at the end of this message within the
#### borders) that extracts the X and Y columns from a text file. I
need help to take it further to do the following -
(A) create column C of New_X values as follows -
- put value in C1 equal to that in A1 (or the first X)
- calculate the difference between last X and first X values (e.g.
0.8007-0.782 = 0.0187)
- calculate values for 10 cells C2, C3,...C11 using formula (C2 = C1
+ 0.0187/10), (C3=C2+0.0187/10),...(C11 = C10 + 0.0187/10)
- put C12=last X (in this example C12=0.8007) and put D12= last Y (in
this example 2.0298)
(B) A macro/VBScript to populate column D (D1 to D11) based on linear
interpolation as explained above.
I searched thru this listserve (to do the part (B)) and found
following post by Ron Rosenfeld (11/29/2005) and it works perfectly as
Excel sheet formula to get New_Y values (but I need a vb script to do
the same on the fly in my code).
=IF(NewX=MAX(x_s),MAX(y_s),VLOOKUP(NewX,tbl,2)+
(INDEX(tbl,MATCH(VLOOKUP(NewX,tbl,1),x_s)+2,2)-
VLOOKUP(NewX,tbl,2))*(NewX-VLOOKUP(NewX,tbl,1))
/(INDEX(tbl,MATCH(VLOOKUP(NewX,tbl,1),x_s)+2,1)-
VLOOKUP(NewX,tbl,1)))
I also found a VbScript by Dana DeLouis (8/11/1998) as shown below
between the asterick borders. I think that it will work for me but I
just don't understand it fully. I don't know how to call this Function
in my existing macro or how to pass the values of X (same as s1data()
in my code below), Y (same as s2data() in my code below) and New_X to
it to do the linear interpolation. Please help. Thank you all kind
people very much in advance.
Choi
***************************************************************************
' Code from Dana DeLouis
Hello. Here is a copy of 1 custom interpolating function that I use.
Most
of the code is error checking. The actual code is rather small. You
can
remove most of the error checking if you want. The code searches
Xrange to
find the 2 surrounding values close to 'Value' and interpolates along
the
Yrange to return an answer. Xrange (& Yrange) is a column of data
like
A1:A10. You do not have to manually pick the 2 surrounding Xvalues.
Note
the use of Value2 in the code. This allows you to search for a date
(just
like you want).
Any questions? Just ask. Hope this helps.
Function Interpolate(Value, XRange As Range, YRange As Range,
Optional
X_Ascending As Boolean = True)
'// Dana DeLouis: (e-mail address removed)
'// Value is the number to interpolate along the XRange
'// XRange & YRange are the table data
'// (each 1 column wide, with at least 2 rows)
'// Include X_Ascending = False if XRange in Descending order
'// otherwise, default is XRange in Ascending order
Application.Volatile
Dim Ys
Dim Xs
Dim P As Integer
On Error Resume Next
'// Adjust Value if a Date
If IsDate(Value) Then Value = CDbl(CDate(Value))
'// Check X Range for errors
If XRange.Columns.Count > 1 Or XRange.Rows.Count < 2 Then
Interpolate = "** Error, X Range"
Exit Function
End If
'// Check Y Range for errors
If YRange.Columns.Count > 1 Or YRange.Rows.Count < 2 Then
Interpolate = "** Error, Y Range"
Exit Function
End If
'// Make sure X & Y have same # of rows
If XRange.Rows.Count <> YRange.Rows.Count Then
Interpolate = "** Error, # Rows"
Exit Function
End If
With WorksheetFunction
'// Look for an exact match first
P = .Match(Value, XRange.Value2, 0) '** 0 for an exact match
If P > 0 Then
'An exact match. Just get given data
Interpolate = .Index(YRange, P, 1)
Else
If X_Ascending = True Then
P = .Match(Value, XRange.Value2, 1) '** 1 for
ascending
order!
Else
P = .Match(Value, XRange.Value2, -1) '** -1 for
descending
order!
End If
'// Make sure number falls inside XRange
'// otherwise answer may not be valid
If P = 0 Or P = XRange.Cells.Count Then
Interpolate = "# outside range"
Exit Function
End If
'// Pick surrounding cells to do a linear interpolation
Xs = Array(.Index(XRange.Value2, P,
1), .Index(XRange.Value2, P
+ 1, 1))
Ys = Array(.Index(YRange.Value2, P,
1), .Index(YRange.Value2, P
+ 1, 1))
Interpolate = .Forecast(Value, Ys, Xs)
End If
End With
End Function
***************************************************************************
##########################################################################
'choi's code
Option Explicit
Sub ProcessText()
Dim FName, FNameO As Variant
Dim MyTitle, MyFilter As String
Dim FNum As Long
Dim sLine As String
Dim i, j, k, l As Long
Dim x As Variant
Dim s1data(), s2data() As Variant
FNum = FreeFile
ReDim s1data(1 To 50000, 1 To 1)
ReDim s2data(1 To 50000, 1 To 1)
Close FNum
Imax = 1
MyTitle = "Select File(s)"
MyFilter = "MXV Files (*.mxv), *.mxv"
' MyFilter = "All Files (*.*), *.*"
Application.ScreenUpdating = False
ChDir "C:\" 'This is the starting directory to lookup files
FName = Application.GetOpenFilename(FileFilter:=MyFilter, _
Title:=MyTitle, MultiSelect:=True)
If IsArray(FName) Then
For k = LBound(FName) To UBound(FName)
Open FName(k) For Input As FNum
i = 1
Do While Not InStr(1, sLine, "TheCell # 22", vbTextCompare) > 0
Line Input #FNum, sLine
i = i + 1
Loop
Debug.Print sLine
For i = 1 To 300
Line Input #FNum, sLine
Next i
Debug.Print sLine
i = 1
Do While Not InStr(1, sLine, "EndCell# 22", vbTextCompare) > 0
Line Input #FNum, sLine
s1data(i, 1) = Right(sLine, Len(sLine) - InStr(1, sLine, "
"))
s2data(i, 1) = Left(sLine, Len(sLine) - InStr(1, sLine, "
"))
i = i + 1
Loop
Close FNum
'START_Write output
i = i - 2
ActiveWorkbook.Sheets(1).Activate
For j = 1 To i
Cells(j , 1).Value = s1data(j, 1)
Cells(j , 2).Value = s2data(j, 1)
Next j
End Sub
##########################################################################
G
#### Guest
Your code never writes the new x values to the worksheet. If they are
already there then after you write the s1Data and s2Data to the sheet, you
would walk through the list of newX values.
ActiveWorkbook.Sheets(1).Activate
For j = 1 To i
Cells(j, 1).Value = s1data(j, 1)
Cells(j, 2).Value = s2data(j, 1)
Next j
set rngX = Range(Cells(1,1),Cells(1,1).End(xldown))
set rngY = rngX.offset(0,1)
set NewX = Range(cells(1,3),Cells(1,3).End(xldown))
for each cell in NewX
cell.offset(0,1) = Interpolate(cell,rngX,rngY,True)
Next
Next
N
#### Niek Otten
' =========================================================================
Function TableInterpol(ToFind As Double, Table As Range, ResultColumnNr As Long, _
Optional SortDir, Optional KeyColumnNr)
' Niek Otten, March 22 2006
' Works like Vlookup, but interpolates and has some extra options
' 1st argument: Key to look for. Numbers only!
' 2nd argument: Range to look in and get the result from. Numbers only!
' 3rd argument: Relative column number in the range to extract the result from
' Optional 4th argument: defaults to: "Ascending"; any supplied argument forces Descending
' Optional 5th argument: Relative column number in the range to search the key in,
' defaults to 1
Dim RowNrLow As Long
Dim RowNrHigh As Long
Dim ResultLow As Double
Dim ResultHigh As Double
Dim KeyFoundLow As Double
Dim KeyFoundHigh As Double
If IsMissing(SortDir) Then
SortDir = 1
Else
SortDir = -1
End If
If IsMissing(KeyColumnNr) Then
KeyColumnNr = 1
End If
RowNrLow = Application.WorksheetFunction.Match(ToFind, Intersect(Table, Table.Cells(KeyColumnNr). _
EntireColumn), SortDir)
ResultLow = Table(RowNrLow, ResultColumnNr)
If ToFind = ResultLow Then
TableInterpol = Table(RowNrLow, ResultColumnNr)
Exit Function ' avoid unnesssary second MATCH() call if already exact match found
End If
RowNrHigh = RowNrLow + 1
ResultHigh = Table(RowNrHigh, ResultColumnNr)
KeyFoundLow = Table(RowNrLow, KeyColumnNr)
KeyFoundHigh = Table(RowNrHigh, KeyColumnNr)
TableInterpol = ResultLow + (ToFind - KeyFoundLow) / (KeyFoundHigh - KeyFoundLow) _
* (ResultHigh - ResultLow)
End Function
' =========================================================================
--
Kind regards,
Niek Otten
Microsoft MVP - Excel
| Dear Kind People,
|
| I need to linearly interpolate two points (X0,Y0) & (X1,Y1) to find
| the New_Y value for a given New_X. The X's value increases as i go
| down in column A. Say I want to find New_Y for New_X = 0.7895. I want
| to lookup the two closest values of X's that New_X = 0.7895 is
| between(0.7892 and 0.7897) and then use these two points to linearly
| interpolate the Ys to get New_Y for New_X = 0.7895.
|
| Example data set -
|
| A B C D
|
| X Y New_X New_Y
|
| 1 0.782 2.1417 0.782
| 2 0.7831 2.1474 0.7839
| 3 0.7842 2.1532 0.7858
| 4 0.7853 2.1592 0.7876
| 5 0.7864 2.1653 0.7895
| 6 0.7875 2.1715 0.7914
| 7 0.7881 2.1746 0.7932
| 8 0.7886 2.1778 0.7951
| 9 0.7892 2.181 0.797
| 10 0.7897 2.1843 0.7988
| 11 0.7908 2.1908 0.8069
| 12 0.7919 2.1974 0.8007 2.0298
| 13 0.793 2.2042
| 14 0.7936 2.2076
| 15 0.7941 2.211
| 16 0.7947 2.2144
| 17 0.7952 2.2179
| 18 0.7958 2.2214
| 19 0.7963 2.2248
| 20 0.7974 2.2318
| 21 0.7979 2.2354
| 22 0.7985 2.2389
| 23 0.799 2.2424
| 24 0.7996 2.246
| 25 0.8001 2.0321
| 26 0.8007 2.0298
|
|
| I have a VBScript (please see at the end of this message within the
| #### borders) that extracts the X and Y columns from a text file. I
| need help to take it further to do the following -
|
| (A) create column C of New_X values as follows -
|
| - put value in C1 equal to that in A1 (or the first X)
| - calculate the difference between last X and first X values (e.g.
| 0.8007-0.782 = 0.0187)
| - calculate values for 10 cells C2, C3,...C11 using formula (C2 = C1
| + 0.0187/10), (C3=C2+0.0187/10),...(C11 = C10 + 0.0187/10)
| - put C12=last X (in this example C12=0.8007) and put D12= last Y (in
| this example 2.0298)
|
| (B) A macro/VBScript to populate column D (D1 to D11) based on linear
| interpolation as explained above.
|
| I searched thru this listserve (to do the part (B)) and found
| following post by Ron Rosenfeld (11/29/2005) and it works perfectly as
| Excel sheet formula to get New_Y values (but I need a vb script to do
| the same on the fly in my code).
|
| =IF(NewX=MAX(x_s),MAX(y_s),VLOOKUP(NewX,tbl,2)+
| (INDEX(tbl,MATCH(VLOOKUP(NewX,tbl,1),x_s)+2,2)-
| VLOOKUP(NewX,tbl,2))*(NewX-VLOOKUP(NewX,tbl,1))
| /(INDEX(tbl,MATCH(VLOOKUP(NewX,tbl,1),x_s)+2,1)-
| VLOOKUP(NewX,tbl,1)))
|
|
| I also found a VbScript by Dana DeLouis (8/11/1998) as shown below
| between the asterick borders. I think that it will work for me but I
| just don't understand it fully. I don't know how to call this Function
| in my existing macro or how to pass the values of X (same as s1data()
| in my code below), Y (same as s2data() in my code below) and New_X to
| it to do the linear interpolation. Please help. Thank you all kind
| people very much in advance.
|
| Choi
|
| ***************************************************************************
| ' Code from Dana DeLouis
|
| Hello. Here is a copy of 1 custom interpolating function that I use.
| Most
| of the code is error checking. The actual code is rather small. You
| can
| remove most of the error checking if you want. The code searches
| Xrange to
| find the 2 surrounding values close to 'Value' and interpolates along
| the
| Yrange to return an answer. Xrange (& Yrange) is a column of data
| like
| A1:A10. You do not have to manually pick the 2 surrounding Xvalues.
| Note
| the use of Value2 in the code. This allows you to search for a date
| (just
| like you want).
| Any questions? Just ask. Hope this helps.
|
| Function Interpolate(Value, XRange As Range, YRange As Range,
| Optional
| X_Ascending As Boolean = True)
| '// Dana DeLouis: (e-mail address removed)
| '// Value is the number to interpolate along the XRange
| '// XRange & YRange are the table data
| '// (each 1 column wide, with at least 2 rows)
| '// Include X_Ascending = False if XRange in Descending order
| '// otherwise, default is XRange in Ascending order
|
|
| Application.Volatile
| Dim Ys
| Dim Xs
| Dim P As Integer
| On Error Resume Next
|
|
| '// Adjust Value if a Date
| If IsDate(Value) Then Value = CDbl(CDate(Value))
|
|
| '// Check X Range for errors
| If XRange.Columns.Count > 1 Or XRange.Rows.Count < 2 Then
| Interpolate = "** Error, X Range"
| Exit Function
| End If
|
|
| '// Check Y Range for errors
| If YRange.Columns.Count > 1 Or YRange.Rows.Count < 2 Then
| Interpolate = "** Error, Y Range"
| Exit Function
| End If
|
|
| '// Make sure X & Y have same # of rows
| If XRange.Rows.Count <> YRange.Rows.Count Then
| Interpolate = "** Error, # Rows"
| Exit Function
| End If
|
|
| With WorksheetFunction
| '// Look for an exact match first
| P = .Match(Value, XRange.Value2, 0) '** 0 for an exact match
| If P > 0 Then
| 'An exact match. Just get given data
| Interpolate = .Index(YRange, P, 1)
| Else
| If X_Ascending = True Then
| P = .Match(Value, XRange.Value2, 1) '** 1 for
| ascending
| order!
| Else
| P = .Match(Value, XRange.Value2, -1) '** -1 for
| descending
| order!
| End If
|
|
| '// Make sure number falls inside XRange
| '// otherwise answer may not be valid
| If P = 0 Or P = XRange.Cells.Count Then
| Interpolate = "# outside range"
| Exit Function
| End If
|
|
| '// Pick surrounding cells to do a linear interpolation
| Xs = Array(.Index(XRange.Value2, P,
| 1), .Index(XRange.Value2, P
| + 1, 1))
| Ys = Array(.Index(YRange.Value2, P,
| 1), .Index(YRange.Value2, P
| + 1, 1))
| Interpolate = .Forecast(Value, Ys, Xs)
| End If
| End With
| End Function
| ***************************************************************************
|
| ##########################################################################
| 'choi's code
|
| Option Explicit
|
| Sub ProcessText()
|
| Dim FName, FNameO As Variant
| Dim MyTitle, MyFilter As String
| Dim FNum As Long
| Dim sLine As String
| Dim i, j, k, l As Long
| Dim x As Variant
| Dim s1data(), s2data() As Variant
|
|
| FNum = FreeFile
| ReDim s1data(1 To 50000, 1 To 1)
| ReDim s2data(1 To 50000, 1 To 1)
| Close FNum
| Imax = 1
| MyTitle = "Select File(s)"
| MyFilter = "MXV Files (*.mxv), *.mxv"
| ' MyFilter = "All Files (*.*), *.*"
|
| Application.ScreenUpdating = False
|
|
| ChDir "C:\" 'This is the starting directory to lookup files
|
|
| FName = Application.GetOpenFilename(FileFilter:=MyFilter, _
| Title:=MyTitle, MultiSelect:=True)
|
|
| If IsArray(FName) Then
| For k = LBound(FName) To UBound(FName)
|
| Open FName(k) For Input As FNum
| i = 1
| Do While Not InStr(1, sLine, "TheCell # 22", vbTextCompare) > 0
| Line Input #FNum, sLine
| i = i + 1
| Loop
|
| Debug.Print sLine
|
| For i = 1 To 300
| Line Input #FNum, sLine
| Next i
|
| Debug.Print sLine
|
| i = 1
| Do While Not InStr(1, sLine, "EndCell# 22", vbTextCompare) > 0
| Line Input #FNum, sLine
| s1data(i, 1) = Right(sLine, Len(sLine) - InStr(1, sLine, "
| "))
| s2data(i, 1) = Left(sLine, Len(sLine) - InStr(1, sLine, "
| "))
| i = i + 1
| Loop
|
|
| Close FNum
|
| 'START_Write output
| i = i - 2
| ActiveWorkbook.Sheets(1).Activate
| For j = 1 To i
| Cells(j , 1).Value = s1data(j, 1)
| Cells(j , 2).Value = s2data(j, 1)
| Next j
|
| End Sub
| ##########################################################################
|
| 5,401
| 16,021
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.015625
| 3
|
CC-MAIN-2022-33
|
longest
|
en
| 0.763874
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.