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1 Proprietà matematiche
2 Frazioni
3 Conteggio sulle dita
4 Linguaggi naturali
5 Base 36 come compressione senaria
6 Note
7 Collegamenti esterni
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Sistema numerico senario
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Il sistema numericosenario (noto anche come base-6, esimale o sesimale) ha sei come base e può adottare come numerali le cifre da 0 a 5. È stato adottato in modo indipendente da un piccolo numero di culture. Come i decimali, è un numero semiprimo, sebbene sia il prodotto degli unici due numeri consecutivi che sono entrambi primi (2 e 3) ha un alto grado di proprietà matematiche per le sue dimensioni. Poiché sei è un numero altamente composito superiore, molte delle argomentazioni a favore del sistema duodecimale si applicano anche alla base-6. A sua volta, la logica senaria si riferisce a un'estensione dei sistemi logici ternari di Jan Łukasiewicz e Stephen Cole Kleene adattati per spiegare la logica dei test statistici e dei modelli di dati mancanti nelle scienze usando metodi empirici.
Proprietà matematiche
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Tavola pitagorica del sistema senario | × | 1 | 2 | 3 | 4 | 5 | 10 |
--- --- ---
| 1 | 1 | 2 | 3 | 4 | 5 | 10 |
| 2 | 2 | 4 | 10 | 12 | 14 | 20 |
| 3 | 3 | 10 | 13 | 20 | 23 | 30 |
| 4 | 4 | 12 | 20 | 24 | 32 | 40 |
| 5 | 5 | 14 | 23 | 32 | 41 | 50 |
| 10 | 10 | 20 | 30 | 40 | 50 | 100 |
Il senario può essere considerato interessante nello studio dei numeri primi, poiché tutti i numeri primi diversi da 2 e 3, quando espressi in senario, hanno 1 o 5 come cifra finale. In senario i numeri primi vengono scritti come:
2, 3, 5, 11, 15, 21, 25, 31, 35, 45, 51, 101, 105, 111, 115, 125, 135, 141, 151, 155, 201, 211, 215, 225, 241, 245, 251, 255, 301, 305, 331, 335, 345, 351, 405, 411, 421, 431, 435, 445, 455, 501, 515, 521, 525, 531, 551, ... (EN) Sequenza A004680, su On-Line Encyclopedia of Integer Sequences, The OEIS Foundation.
Cioè, per ogni numero primo p maggiore di 3, si hanno le relazioni aritmetiche modulari per cui p ≡ 1 o 5 (mod 6) (cioè 6 divide p − 1 o p − 5); la cifra finale è 1 o 5. Ciò è dimostrato dalla contraddizione. Per qualsiasi numero intero n:
Se n ≡ 0 (mod 6), 6 | n
Se n ≡ 2 (mod 6), 2 | n
Se n ≡ 3 (mod 6), 3 | n
Se n ≡ 4 (mod 6), 2 | n
Inoltre, poiché i quattro primi (2, 3, 5, 7) più piccoli sono divisori o vicini di 6, il senario ha semplici criteri di divisibilità per molti numeri.
In più, tutti i numeri perfetti pari oltre al 6 hanno 44 come ultime due cifre quando espressi in senario, il che è dimostrato dal fatto che ogni numero pari è della forma 2 p −1 (2 p −1), dove 2 p - 1 è primo.
Il senario è anche la più grande base numerica r che non ha totali diversi da 1 e r - 1, rendendo la sua tabella di moltiplicazione altamente regolare per le sue dimensioni, riducendo al minimo la quantità di sforzo richiesto per memorizzarla. Questa proprietà massimizza la probabilità che il risultato di una moltiplicazione intera finisca in zero, dato che nessuno dei suoi fattori lo fa.
Frazioni
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Poiché sei è il prodotto dei primi due numeri primi ed è adiacente ai successivi due numeri primi, molte frazioni senarie hanno rappresentazioni semplici:
Base decimale
Primi fattori della base: 2, 5
Primi fattori sotto la base: 3
Primi fattori sopra la base: 11Base senaria
Primi fattori della base: 2, 3
Primi fattori sotto la base: 5
Primi fattori sopra la base: 11
Frazioni Primi fattori del denominatore Rappresentazione posizionale Rappresentazione posizionale Primi fattori
del denominatore Frazione
1/220.50.321/2
1/330.3333... = 0.30.231/3
1/420.250.1321/4
1/550.20.1111... = 0.151/5
1/62, 30.160.12, 31/10
1/770.1428570.05111/11
1/820.1250.04321/12
1/930.10.0431/13
1/102, 50.10.032, 51/14
1/11110.090.0313452421151/15
1/122, 30.0830.032, 31/20
1/13130.0769230.024340531215211/21
1/142, 70.07142850.0232, 111/22
1/153, 50.060.023, 51/23
1/1620.06250.021321/24
1/17170.05882352941176470.0204122453514331251/25
1/182, 30.050.022, 31/30
1/19190.0526315789473684210.015211325311/31
1/202, 50.050.0142, 51/32
1/213, 70.0476190.0143, 111/33
1/222, 110.0450.013452421032, 151/34
1/23230.04347826086956521739130.01322030441351/35
1/242, 30.04160.0132, 31/40
1/2550.040.0123551/41
1/262, 130.03846150.01215024340532, 211/42
1/2730.0370.01231/43
1/282, 70.035714280.01142, 111/44
1/29290.03448275862068965517241379310.01124045443151451/45
1/302, 3, 50.030.012, 3, 51/50
1/31310.0322580645161290.010545511/51
1/3220.031250.0104321/52
1/333, 110.030.010313452423, 151/53
1/342, 170.029411764705882350.010204122453514332, 251/54
1/355, 70.02857140.015, 111/55
1/362, 30.0270.012, 31/100
Conteggio sulle dita
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34 senario =
con il conteggio
22 decimale,
senario delle dita
Si può dire che ogni normale mano umana abbia sei posizioni non ambigue; un pugno, un dito (o pollice) allungato, due, tre, quattro e poi tutti e cinque estesi.
Se la mano destra viene utilizzata per rappresentare un'unità e la sinistra per rappresentare i 'sei', diventa possibile per una persona rappresentare i valori da zero a 55 senario (35 decimale ) con le dita, anziché i soliti dieci ottenuti nel conteggio delle dita standard. ad es. se tre dita sono estese sulla mano sinistra e quattro sulla destra, viene rappresentato il 34 senario . Ciò equivale a 3 × 6 + 4 che è 22 decimale .
Inoltre, questo metodo è il modo meno astratto di contare usando due mani che riflette il concetto di notazione posizionale, poiché il movimento da una posizione all'altra viene effettuato passando da una mano all'altra. Mentre le culture più sviluppate contano fino a 5 in modo molto simile, oltre 5 culture non occidentali si discostano dai metodi occidentali, come per i gesti numerici cinesi. Poiché anche il conteggio delle dita senarie si discosta solo oltre il 5, questo metodo di conteggio rivaleggia con la semplicità dei metodi di conteggio tradizionali, un fatto che può avere implicazioni per l'insegnamento della notazione di posizione ai giovani studenti.
Quale mano viene usata per i "sei" e quale per le unità dipende dalle preferenze di chi conta, tuttavia se vista dalla prospettiva del contatore, usando la mano sinistra come la cifra più significativa corrisponde alla rappresentazione scritta dello stesso numero serio. Capovolgere la mano dei "sei" sul retro può aiutare a chiarire ulteriormente quale mano rappresenta i "sei" e quale rappresenta le unità. Il rovescio della medaglia del conteggio senario, tuttavia, è che senza previo accordo due parti non sarebbero in grado di utilizzare questo sistema, essendo incerti su quale mano rappresenti i "sei" e quale mano rappresenti le unità, mentre il conteggio basato su decimali (con numeri oltre il 5 espressi da un palmo aperto e le dita aggiuntive) essendo essenzialmente un sistema unario richiede solo all'altra parte di contare il numero di dita estese.
Nel basket NCAA, i numeri delle uniformi dei giocatori sono limitate a essere numeri senari di massimo due cifre, in modo che gli arbitri possano segnalare quale giocatore ha commesso un'infrazione utilizzando questo sistema di conteggio delle dita.
Sistemi di conteggio delle dita più astratti, come chisanbop o finger binary, consentono di contare fino a 99, 1023 o anche più a seconda del metodo (anche se non necessariamente di natura senaria). Il monaco e storico inglese Bede descrisse nel primo capitolo della sua opera De temporum ratione (725), intitolato "Tractatus de computo, vel loquela per gestum digitorum", un sistema che permetteva di contare fino a 9.999 su due mani.
Linguaggi naturali
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Nonostante la rarità delle culture che raggruppano grandi quantità per 6, una revisione dello sviluppo dei sistemi numerici suggerisce una soglia di numerosità a 6 (possibilmente essere concettualizzata come "intera", "pugno" o "oltre cinque dita"), con 1-6 che spesso sono forme pure e numeri che successivamente vengono costruiti o presi in prestito.
Si ritiene che la lingua Ndom della Papua Nuova Guinea abbia numeri senari.Mer significa 6, mer an thef significa 6 × 2 = 12, nif significa 36 e nif thef significa 36 × 2 = 72.
Un altro esempio della Papua Nuova Guinea sono le lingue Yam. In queste lingue, il conteggio è collegato al conteggio ritualizzato delle patate dolci. Queste lingue contano da una base sei, impiegando parole per le potenze di sei, fino a 6 6 per alcune di queste lingue. Un esempio è la lingua Komnzo con i seguenti numeri: nibo (6 1), fta (6 2), taruba (6 3), damno (6 4), wärämäkä (6 5), wi (6 6).
È stato segnalato che alcune lingue del Niger-Congo usano un sistema di numeri senari, di solito in aggiunta a un altro, quale il decimale o il vigesimale.
Si anche sospettato che la lingua proto-uralica avesse numeri senari, con un numerale per "7" che è stato preso in prestito in un secondo momento, anche se l'evidenza per cui costruire numeri più grandi (8 e 9) in modo sottrattivo da dieci suggerisce che potrebbe non essere così.
Base 36 come compressione senaria
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Per alcuni scopi, la base 6 potrebbe essere una base troppo piccola per essere comoda. Questo ostacolo può essere aggirato usando il suo quadrato, base 36 (esatrigesimale), poiché la conversione è facilitata semplicemente effettuando le seguenti sostituzioni:
| Decimale | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| Base 6 | 0 | 1 | 2 | 3 | 4 | 5 | 10 | 11 | 12 | 13 | 14 | 15 | 20 | 21 | 22 | 23 | 24 | 25 |
| Base 36 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | A | B | C | D | E | F | G | H |
| |
| Decimale | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 |
| Base 6 | 30 | 31 | 32 | 33 | 34 | 35 | 40 | 41 | 42 | 43 | 44 | 45 | 50 | 51 | 52 | 53 | 54 | 55 |
| Base 36 | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z |
Pertanto, il numero di base 36 WIKIPEDIA 36 è uguale al numero senario 523032304122213014 6. In decimale, è 91.730.738.691.298.
La scelta di 36 come Base è conveniente in quanto le cifre possono essere rappresentate usando i numeri arabi 0–9 e le lettere latine A – Z: questa scelta è la base dello schema di codifica base36 . L'effetto di compressione di 36, essendo il quadrato di 6, fa sì che molti modelli e rappresentazioni siano più brevi nella base 36:
1/9 10 = 0,04 6 = 0,4 36
1/16 10 = 0,0213 6 = 0,29 36
1/5 10 = 0. 1 6 = 0. 7 36
1/7 10 = 0. 05 6 = 0. 5 36
Note
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^(EN) Jan Zi, Models of 6-valued measures: 6-kinds of information, Kindle Direct Publishing Science, 18 novembre 2019. URL consultato il 20 settembre 2022.
^(EN) Zach Schonbrun, Crunching the Numbers: College Basketball Players Can’t Wear 6, 7, 8 or 9, in The New York Times, 31 marzo 2015. URL consultato il 20 settembre 2022.
^ Bloom, Jonathan M., Hand sums: The ancient art of counting with your fingers, su bcm.bc.edu, Yale University Press, 2001. URL consultato il 12 maggio 2012(archiviato dall'url originale il 13 agosto 2011).
^Dactylonomy, su laputanlogic.com, Laputan Logic, 16 novembre 2006. URL consultato il 12 maggio 2012(archiviato dall'url originale il 23 marzo 2012).
^ Juliette Blevins, Origins of Northern Costanoan ʃak:en ‘six’:A Reconsideration of Senary Counting in Utian, in International Journal of American Linguistics, vol.71, n.1, 3 maggio 2018, pp.87-101, DOI:10.1086/430579.
^a b cArchived copy (PDF), su ling.uni-konstanz.de. URL consultato il 27 agosto 2014(archiviato dall'url originale il 6 aprile 2016).
^Copia archiviata, vol.13, 2001, DOI:10.1007/BF03217098. URL consultato l'8 dicembre 2019(archiviato dall'url originale il 26 settembre 2015).
Collegamenti esterni
[modifica | modifica wikitesto]
Le persone sembrano preferire Heximal per popolarità numerica
Base dialettica di Shack's Six
Conversione base senaria
Sito web su Seximal
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187602 | https://www.omnicalculator.com/math/average-percentage | Board
Last updated:
Average Percentage Calculator
Welcome to Omni's average percentage calculator, where we'll learn how to average percentages and what it actually means. Truth be told, half the time, the concept boils down to the well-known formula for the mean of a dataset. However, the other half concerns problems when the percentages correspond to samples of different sizes, and we can't apply the same reasoning there. Nevertheless, it still turns out to be a familiar formula, namely the weighted average of the percentages.
No worries: we'll teach you how to differentiate between the two scenarios and how to find the average percentage in each!
With our mean calculator, you can learn more on how to find the mean of a collection of numbers.
How to average percentages
Let's recall the formal definition of a percentage:
💡 Percentages are fractions with 100 in the denominator. We represent them using the symbol %, which means that a%=a/100 for any real number a.
For more detailed information, check out the percentage calculator.
We're used to the idea that we assign percentages to other numbers in a way we assign discounts to prices. However, mathematically speaking, they can appear on their own.
What is more, the above definition says that a can be any real number. In other words, it can be an integer, a negative number, a decimal, or even a cube root. Mathematically speaking, of course. In real life, if a shop offered a 375 discount for Black Friday, we'd call them mad. Similarly, if you report an error of 35 in your measurements, your teacher is going to advise you to learn how to calculate percent error.
There is a sense to looking at percentages as regular numbers. After all, when we wonder how to calculate the average percentage, we ask ourselves if we can even average percentages. After all, they're something different, so it may seem unnatural. On the other hand, we know all about averaging numbers, don't we? Just to be on the safe side, let's recall the arithmetic average formula:
average=n(a1+a2+a3+...+an)
You can always take a look at the average calculator if you still need more details on this topic 😉
That being said, we need to be careful here. Wondering how to find the average percentage is often connected with the samples the percentages represent. To understand the difference, let's discuss an example.
Suppose that Amy, Brad, Colin, Debbie, and Edward all had a test in American literature. Some of them got 80%, and some got 40%. If we blindly apply the reasoning above, we'd say that the average result in percentages was:
2(80%+40%)=2120%=60%
After all, there were only two results, so we look for the mean of two values. With our mean, median, mode calculator you can learn how to find the mean of a set of values.
However, it clearly cannot be so. Indeed, five people took part in the test, so we should add five numbers instead of two. If, say, Amy, Brad, Colin, and Debbie got 80%, and Edward got 40%, then the actual average is:
5(80%+80%+80%+80%+40%)=5360%=72%
Quite a different result, isn't it?
The lesson we learn here is that we must always keep track of the differences in the groups' sizes and the percentages they correspond to. In fact, we can think of those sizes as weights when we look for the weighted average of a dataset. That is, for example, of a sequence of percentages.
The weighted average of percentages
Recall the example from the end of the above section, where we talked about five people's results on a test. After learning how to find the average percentage, we got:
5(80%+80%+80%+80%+40%)=5360%=72%
Equivalently, we could have written:
4+1(4⋅80%+1⋅40%)=5360%=72%
Clearly, the new notation is shorter. Moreover, we immediately see how many people got the same result: 4 got 80%, and 1 got 40%. In other words, instead of treating the entries individually, we group them together according to their score.
What we got is the weighted average of the percentages with weights corresponding to how many people obtained the score. Fortunately, the calculations are the same as for the regular weighted average: if we have entries a1, a2, a3, ..., an with respective weights w1, w2, w3, ..., wn, then:
weighted average=w1+w2+w3+...+wna1⋅w1+a2⋅w2+a3⋅w3+...+an⋅wn
If we translate the notation to our needs (i.e., to explain how to average percentages), a1, a2, a3, ..., an will correspond to subsequent percentages, while w1, w2, w3, ..., wn will be the respective sample sizes of said percentages.
So what happens if all weights are the same (i.e., if all samples have the same size)? Well, if we denote the mutual weight by w, then:
weighted average=w+w+w+...+wa1⋅w+a2⋅w+a3⋅w+...+an⋅w=nww⋅(a1+a2+a3+...+an)=na1+a2+a3+...+an,
by the rules of fraction simplification. In other words, the weight doesn't matter, and the weighted average of the percentages turns out to be the regular (non-weighted) average.
All in all, we see that learning how to calculate the average percentage boils down to learning about the usual weighted average. Still, let's go through one more example to show how it applies to real-life statistics. And we'll take the opportunity to do it using Omni's average percentage calculator.
Example of using the average percentage calculator
Suppose that we've asked a thousand people whether they eat pancakes at least once a week. There were 300 teenagers, 450 people aged 20-49, and 250 aged 50 and above. In the first group, 64% said they ate pancakes every week. In the second, it was 42%, and in the third, 36%. Let's see how to calculate the average percentage of pancake-eaters in our thousand-strong group.
However, before we do the calculations ourselves, let's see how easy the task is with the average percentage calculator at hand. In the tool, at the bottom, we see an option for sample sizes. In our case, the groups differ in size, so we tick the checkbox.
That will trigger additional variable fields at the top corresponding to the dataset's percentages and sample sizes. They appear in pairs. Add more entries by choosing the appropriate option and select three in the number of entries (you can have up to ten entries in Omni's average percentage calculator). Looking back at our example, we input subsequently:
Entry #1: 64%, 300;
Entry #2: 42%, 450; and
Entry #3: 36%, 250.
Once you give the last value, the average percentage calculator will spit out the answer underneath.
Now, let's see how to find the average percentage ourselves. First of all, we identify our dataset according to what we've learned in the above section: our subsequent percentages are 64%, 42%, and 36%, while the respective sample sizes are 300, 450, and 250 people. Next, we use the weighted average of percentages formula:
300+450+25064%⋅300+42%⋅450+36%⋅250=100019,200%+18,900%+9,000%=100047,100%=47.1%
It turns out that, on average, 47.1% of people eat pancakes every week. But do they have them once a week or every day? Maybe we could introduce some new questions to the survey and have a more detailed study?
FAQs
What is the average percentage?
The average percentage is simply an average over different percentages. You should, however, take into account the sample size each percentage represents and be mindful of whether you're writing each percentage as a fraction or as a decimal number.
How do I calculate the average percentage?
To calculate the average percentage, you need to:
Determine the sample sizes corresponding to each percentage.
For each percentage, multiply it by its sample size.
Add all the numbers obtained in step 2.
Add all the sample sizes.
Divide the number from step 3 by that from step 4.
If you converted percentages to fractions in step 2, convert back.
The calculated result is the average percentage.
Can I average percentages?
Yes, but you need to be careful. By definition, percentages are fractions with 100 in the denominator, so we can calculate their average as we do with any number. However, in practice, percentages rarely come alone, i.e., they usually describe how much of some value we should take. As such, when counting the mean, we might need to take the two together and not just the percentage itself.
How do I add percentages together to get an average?
To add percentages together to get an average, you need to:
Determine the sample sizes corresponding to each percentage.
For each percentage, multiply it by its sample size.
Only now can you add the values.
If you need the average percentage,
Add all the sample sizes;
Divide the value from step 3 by that sum; and
If you converted percentages to fractions in step 2, convert back.
How can I calculate the average percentage in Excel?
To calculate the average percentage of some datasets in Excel:
Enter the list of percentages into column B.
Choose the built-in function AVERAGE.
Enter the starting parentheses “(“ and specify the range of cells in which the percentages are located.
Enter the closing parentheses and hit ENTER.
How do I find the average of 4 percentages?
To find the average of four percentages, you need to:
Determine the sample sizes corresponding to each percentage.
For each percentage, multiply it by its sample size.
Add the four numbers obtained in step 2.
Add the four sample sizes.
Divide the number from step 3 by that from step 4.
If you converted percentages to fractions in step 2, convert back.
The calculated result is the average percentage.
%
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Test Prep Courses / NY Regents - Algebra I Study Guide and Exam Prep Course
Rational vs. Irrational Numbers | Definition & Difference
Lesson
Transcript
Dan Brioli, Gerald Lemay
Author Dan Brioli
Dan has a B.A. in economics from the University of Pittsburgh and an A.A.S. in computer programming from the Pittsburgh Technical Institute. He has trained others in STEM topics and authored technical documentation throughout his career.
View bio
Instructor Gerald Lemay
Gerald has taught engineering, math and science and has a doctorate in electrical engineering.
View bio
Learn the difference between rational and irritational numbers. Explore the properties, differences, and examples of each. Updated: 11/21/2023
Table of Contents
What are Rational and Irrational Numbers?
Identifying Rational and Irrational Numbers
Practice Identifying Rational and Irrational Numbers
Lesson Summary
Show
Frequently Asked Questions
What are five examples of irrational numbers?
One of the most famous examples of an irrational number is pi. Additionally, the square roots of any non-perfect squares are irrational numbers, such as the square roots of 2, 3, 5, 7, 13, and so on.
How do you identify if a number is rational or irrational?
If a number can be represented by a ratio of two whole numbers, it is a rational number. If it can't, it is an irrational number.
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Table of Contents
What are Rational and Irrational Numbers?
Identifying Rational and Irrational Numbers
Practice Identifying Rational and Irrational Numbers
Lesson Summary
Show
What are Rational and Irrational Numbers? -----------------------------------------
Knowing the difference between rational versus irrational numbers is important in understanding a fundamental feature of numbers themselves. So, what are rational and irrational numbers?
A rational number is a number that can be expressed as a ratio that takes the form:
{eq}\frac{P}{Q} {/eq}
The rational number is a quotient of this ratio. In this example, both P and Q must be integers, which are positive or negative whole numbers. Q cannot be 0, as division by 0 is not mathematically possible. 0 is a rational number by itself, however, as 0 is represented by the ratio:
{eq}\frac{0}{1} {/eq}
By contrast, irrational numbers are any numbers that cannot take the form of a ratio of integers. Numbers such as pi are irrational numbers, as there is no ratio of integers that can express pi. The ability to be expressed as an integer ratio determines what is a rational and irrational number.
The set of real numbers (R) illustrating the rational (Q) and irrational (R/Q) numbers.
Hippasus of Metapontum is credited with having discovered irrational numbers sometime in the 5th century BCE while trying to determine the length of a triangle whose shortest sides were only 1 unit in length. The legend of his death involves members of the school of Pythagoras, famous for the Pythagorean theorem, throwing him overboard while at sea to keep the secret of irrational numbers from spreading. It weakened the authority of the Pythagorean theorem and their beliefs surrounding the importance of whole numbers in the universe's structure. To the school of Pythagoras, the real world was not where the unity of the cosmos existed. To them, it only existed in numbers, specifically in rational numbers, as their occasional brushes with the irrationals led to the famous statement, "By numbers, this cannot be done."
Rational numbers have a history that likely stretches back to prehistoric times and have been a part of written works on mathematics since our earliest known records. Some of the oldest written works regarding rational numbers are in Indian and Egyptian texts, and the Greeks famously studied rational numbers in their general examinations of number theory.
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0:04 Rational & Irrational Numbers
0:49 Identifying Rational Numbers
3:50 Properties of Rational Numbers
6:04 Adding/Multiplying…
8:09 Lesson Summary
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Identifying Rational and Irrational Numbers -------------------------------------------
It might seem impossible to determine if a number is rational or not. There are a few ways to determine rationality, however. And owing to irrational numbers being non-rational numbers, if a number cannot be proven to be rational, it must be irrational. The entirety of the world of numbers is contained in both rational numbers and irrational numbers, after all.
A Dedekind cut showing rational values close to the irrational number represented by the square root of 2.
The first and most obvious way of determining rationality is if the number is a whole number. A whole number is an integer, and as such is rational:
{eq}17 = \frac{17}{1} {/eq}
Another obvious way of determining rationality is if a number is a ratio itself. It's the very definition of a rational number:
{eq}\frac{2}{5} {/eq}
What about the number 2.26787, though? It does not seem to be a ratio, or at least is not a ratio that is easily guessed. This is where a bit of algebraic trickery becomes useful:
{eq}2.26787 \cdot 100000 = 226787 {/eq}
By multiplying by powers of ten until the fraction becomes a whole number, it now can be expressed as a ratio, and as such is rational:
{eq}\frac{226787}{100000} = 2.26787 {/eq}
Subsequently, any fraction represented by a decimal that ends and does not continue indefinitely is a rational number, as those types of fractional numbers can all be multiplied by some power of ten to become whole numbers, and then become represented as a ratio. A whole number is a rational number, and a whole number divided by a power of ten is also a rational number.
Finally, any repeating fractional decimal number can be proven to be rational in the following manner:
{eq}x = 0.123123123... {/eq}
{eq}1000x = 123.123... {/eq}
{eq}1000x - x = 123.123... - 0.123123... {/eq}
{eq}999x = 123 {/eq}
{eq}x = 123/999 {/eq}
Now that x has been expressed as a ratio, it has been proven to be rational. This shows why any repeating fractional decimal value is a rational number, as this method can be performed for any of those values.
The result of adding rational numbers together produces rational numbers. Likewise, the result of multiplying rational numbers together produces rational numbers.
But the result of adding a rational and irrational number produces an irrational number. And, similarly, multiplying a rational number and an irrational number results in an irrational number. The one exception to this interaction is the number 0, which is rational. Multiplying 0 by any irrational number will be 0 as well and subsequently will be rational.
The result of adding and multiplying irrational numbers is less clear in their results, however. Consider the example:
{eq}\sqrt{2} \cdot \sqrt{2} = 2 {/eq}
The square root of 2 is an irrational number but multiplying two square roots of any whole number results in a whole number, so the result of multiplication, in this case, is a rational number. But another example shows the contrary concerning irrationals:
{eq}m = 2 + \sqrt{2} {/eq}
{eq}n = -\sqrt{2} {/eq}
{eq}m + n = (2 + \sqrt{2}) + (-\sqrt{2}) {/eq}
{eq}m + n = 2 + \sqrt{2} - \sqrt{2} {/eq}
{eq}m + n = 2 {/eq}
Both m and n are irrational numbers, but when added together result in a rational number. So while many irrationals when added or multiplied together result in irrationals, there are many situations where the result may instead be a rational number.
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Practice Identifying Rational and Irrational Numbers ----------------------------------------------------
Is 7/3 rational or irrational?
7/3 is rational as it is expressed as a ratio of integers.
Is 1.1115 rational or irrational?
Any fractional decimal value that ends is rational, as it can be expressed as a ratio with a power of ten.
{eq}1.1115 = \frac{11115}{10000} {/eq}
Is 0.2222... rational or irrational?
0.2222... is rational, as it is a repeating decimal fraction. Repeating decimal fractions can be multiplied by some power of ten, and with a bit of algebra can be proven rational:
{eq}x = 0.2222... {/eq}
{eq}10x = 2.2222... {/eq}
{eq}10x - x = 2.2222... - 0.2222... {/eq}
{eq}9x = 2 {/eq}
{eq}x = 2/9 {/eq}
Is the square root of seven times the square root of seven rational or irrational?
Two square roots of whole numbers multiplied together result in a whole number, in this case, seven, and as such is rational.
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Lesson Summary --------------
A rational number is a number that can be expressed as a ratio. Irrational numbers are any numbers that cannot take the form of a ratio of integers, such as pi. If a number is a whole number, expressed as a ratio of whole numbers, a fractional decimal that ends, or is a fractional decimal that repeats indefinitely, it is a rational number. Rational numbers can be added or multiplied together to produce rational numbers, while rational numbers added or multiplied with irrational numbers produce irrational numbers. Irrational numbers that are added or multiplied together may or may not result in a rational number and depend on the irrational numbers being added or multiplied.
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Video Transcript
Rational & Irrational Numbers
Having found seasonal work in town, Fred is delighted to start his new job at the Real Number Emporium. His first assignment is to sort the real numbers coming down the conveyor belt into two boxes, labeled Q and I. The letter ''Q'' is for the set of rational numbers, where ''Q'' stands for ''quotient.'' The letter ''I'' is for the set of irrational numbers. You see, real numbers are either rational or irrational. Fred knows the word ''irrational'' means ''not rational,'' but he's not clear on how to spot rational numbers. If only he'd stayed awake in algebra class. Let's see if we can refresh Fred's skills and help him keep his job.
Identifying Rational Numbers
The word ''rational'' includes the word ''ratio.'' A ratio is the quotient of two numbers. A ratio a / b, where a and b are integers like {. . ., -3, -2, -1, 0, 1, 2, 3,. . .} is the key to identifying rational numbers. The only restriction on b is b ≠ 0, because dividing by 0 is undefined. If a number can be written as a / b, it's a rational number, and goes into the Q box.
Imagine the first number Fred sees is 7. Okay, 7 = 7 / 1, so 7 is a rational number.
The next number is 2.56. Okay, 2.56 = 256 / 100, so again it goes into the Q box.
The third number is 0.3... Hmmm,
Will Fred ever see an irrational number?
The next number is √4. Roots of numbers may or may not be rational. In this case, √4 = 2 , and 2 / 1 is a rational number.
The next number, √5, is not a perfect square and cannot be written as a / b. Therefore, √5 is an irrational number, and it goes in the I box.
Fred has been on the job 15 minutes, and already he needs a break. He asks you to take over. How would you sort the following real numbers?
√2, √16, -31, π, and .7142857142857142. . .
Your decisions are:
√2 is not a perfect square, so it's irrational.
√16 is a perfect square equal to 4 / 1, so it's rational.
π can't be written as a / b (even though π ≅ 22 / 7), which makes it irrational.
.7142857142857142. . . is a repeating decimal. It can be written as 5 / 7, which makes it rational.
There is a method for converting this repeating decimal into a ratio. Let's take a look.
x = .714285
10 6 x - x = 714285
x = 714285 / (10 6 - 1)
x = 714285 / 999999
If we factor and cancel, we'll end up with 5 / 7
Properties of Rational Numbers
Fred is back on the job and finishes his first day. The boss, Mrs. Real, is impressed with your work and offers you a job in quality control. That's right, you're going to take samples out of the rational box and test them.
You decide to use some properties:
Property 1: The sum of two rational numbers is rational.
In short-hand form: Q + Q ∈ Q.
The symbol ∈ means ''is in'' or ''belongs to.''
Property 2: The product of two rational numbers is rational.
Q x Q ∈ Q
If we add or multiply two rational numbers, the result is still a rational number.
The quality control starts with the numbers 7 and 2.56.
Let's add:
7 + 2.56 = 9.56 = 956 / 100, which is rational.
Now let's multiply:
7 x 2.56 = 17.92 = 1,792 / 100, which is rational.
The next two numbers to check are √4 and √5:
Once again, we'll add:
√4 + √5 = 2 + √5, which can't be written as the ratio of a / b. So, one of the numbers was mistakenly put in the Q box. It's the √5, which is an irrational number.
This leads to two more properties:
Property 3:The sum of a rational number with an irrational number is an irrational number.
Q + I ∈ I
Property 4:The product of a rational number with an irrational number is an irrational number.
Q I ∈ I
There is an exception to keep in mind! 0 is a rational number. Multiplying the rational number 0 times any irrational number gives 0. So, Property 4 is good, provided the rational number is not 0.
Mrs. Real is ready to give you a promotion, but she has some interview questions for you. What if you add or multiply two irrational numbers? Can you say anything about the result?
Adding/Multiplying Irrational Numbers
So far, we know adding or multiplying rational numbers gives a rational number:
Q + Q ∈ Q and Q Q ∈ Q
We also know adding or multiplying a rational number with an irrational number gives an irrational number:
Q + I ∈ I and Q I ∈ I
The first question posed by Mrs. Real: ''What can you say about I + I?''
For example, √7 is irrational. So is (6 - √7); but added together, the result is:
√7 + (6 - √7) = 6, which is rational.
On the other hand, √ 7 + √7 = 2√7, which is irrational.
So the answer is:
Property 5: The sum of two irrational numbers is sometimes rational and sometimes irrational.
Mrs. Real likes your answer and would now like you to ponder what happens when two irrational numbers are multiplied.
For example, √3 is irrational. So is √5. The product is:
√3 x √5 = √(15), which is not a perfect square; thus, the result is irrational.
How about √3 and √3? Two irrational numbers multiplied together give:
√3 x √3 = √9 = 3, a rational number.
How about the two irrational numbers √8 and √2? Multiplied together:
√8 x √2 = √(16) = 4, a rational number.
Your answer to Mrs. Real is:
Property 6: The product of two irrational numbers is sometimes rational and sometimes irrational.
Mrs. Real is overjoyed! You start tomorrow. Fred, on the other hand, is already looking for other work that will let him sleep later in the morning and take off on 3-day weekends.
Lesson Summary
Real numbers are rational if they can be written as the ratio a / b where a and b are integers and where b ≠ 0. If a real number is not rational, it is irrational. The numerical interactions of rational and irrational numbers can be stated in the following six properties:
Property 1: The sum of two rational numbers is rational.
Property 2: The product of two rational numbers is rational.
Property 3: The sum of a rational number with an irrational number is an irrational number.
Property 4: The product of a rational number with an irrational number is an irrational number.
Property 5: The sum of two irrational numbers is sometimes rational and sometimes irrational.
Property 6: The product of two irrational numbers is sometimes rational and sometimes irrational.
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| Effect of Shape I don't understand how an object's shape affects its magnetization. | |
You are not alone. The effect of object shape is far more complex than most people have ever imagined!
Diamagnetic, paramagnetic, or weakly ferromagnetic materials
Objects composed of materials with small susceptibilities (i.e., χ< 1), including most surgical stainless steels and non-ferrous metals such as tin, aluminum, copper, or lead are only minimally displaced by an external magnetic field and pose no dangers as projectiles. Any shape effects on their magnetization are largely irrelevant and may be disregarded.
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| Moderately to strongly ferromagnetic materials Objects made of iron, nickel, and/or chromium with larger susceptibilities (χ>> 1) become progressively magnetized when placed in an external magnetic field (Bo). As the external field increases, so does this internal polarization/ magnetization (M) as electron spins begin to align and magnetic domain boundaries are reshaped. This process continues until M reaches the magnetic saturation point of the material, beyond which no further magnetization can occur. (Note that the magnetic polarizationM arises from electron spins in microscopic currents and has nothing to do with nuclear magnetization M discussed elsewhere). | Ferromagnetic materials, when placed in an external field (Bo), develop a strong internal magnetization (M), behaving like little magnets themselves. Their edges exposed to the external field become like north and south poles, with a resultant internal demagnetizing field (D), that opposes M. |
As a result of this process, an object composed of moderate to strongly ferromagnetic material essentially becomes its own "little magnet" whose ends can be thought of as north and south "poles". These "poles" represent an accumulation of imputed magnetic charge density at the ends of the object where there is an abrupt magnetic field discontinuity (i.e., where M goes from a positive value inside the object to zero immediately outside). An internal demagnetizing field (D) is thereby created, running from the object's north to south poles opposite to the direction both Mand Bo. Provided the object is homogeneous and responds linearly to small magnetic field changes, the demagnetizing field (D) is generally assumed to be proportional to M in each direction, related by a constant of proportionality between 0 and 1 known as the demagnetization factor (N). Specifically, D = −NM.
By opposing Bo, D reduces the effective magnetic field inside the object, but because of its relatively small size, does not significantly affect M. The volume magnetic susceptibility (χ), defined as the ratio between M and the field intensity, is correspondingly reduced to an apparent susceptibility (χapp) given by
Note that for strongly ferromagnetic materials (those with χ >> 1), the dependence on χdisappears, leaving χapp= 1/Nas an approximation.
An important corollary of this result is that for a ferromagnetic material, the external field strength (Bext) needed to reach critical magnetic saturation (Bsat) is much smaller than expected. Specifically,
Consider, for example, a sphere of silicon steel with Bsat = 2.0 T brought near a 1.5 T MR scanner. The demagnetization factor for a sphere is ⅓ in each direction, so the sphere will become completely magnetized when the external field (Bext) reaches N • Bsat = (⅓)(2.0) = 0.67 T. Surprisingly, the sphere has reached its full (2.0T) saturation magnetization while still in the fringe field of the 1.5T scanner! Actually this should not come as too big a surprise because the common layman's description of ferromagnetic materials is that they tend to "concentrate" the lines of external magnetic fields thereby increasing magnetic flux density within them.
In general N and χ are not a single numbers, but 2nd order tensors (arrays of values corresponding to different directions). For symmetric 3-D forms like ellipsoids, it is possible to select a coordinate frame aligned with the principal axes of the body resulting in three demagnetizing factors, Nx, Ny, and Nz, corresponding to each cardinal direction. These are all dimensionless numbers between 0 and 1, having the additional property Nx+ Ny+ Nz = 1.
Predicting Demagnetizing Factor(s)
The demagnetizing field (D) is strongly dependent on the shape and orientation of the ferromagnetic object in the external field. Calculation of D for a given object typically requires extensive computer simulations, but a few exact solutions are possible for simple geometric shapes. Below are computed demagnetization factors (N) in the Bo-direction. What trends do you notice?
Demagnetizing factors (N) along the direction of the main field for various simple homogeneous objects.
Note that objects with the highest N-values are those that have a large relative surface area facing in the direction of Bo. Such an object is typified by the thin vertical plate perpendicular to Bo, which has N ≈ 1.00 (the highest possible value). Conversely, when the plate is turned by 90º, N ≈ 0.00 (the lowest possible value). The same principle is demonstrated by the various ellipsoids and cylinders illustrated.
So why does this occur? The reason is that for relatively thin objects with their largest areas facing Bo, the virtual north and south poles on their surfaces are close together; hence the demagnetizing field (D) is very strong. Conversely, when the virtual poles are far apart (as in the light blue cylinder pointing parallel to Bo), the poles at the ends are very far apart. In this scenario, the interior N-S pairs cancel each other out leaving widely separated magnetic charge densities only at the two ends. This orientation of the cylinder thus results in a very weak demagnetizing field (D) and correspondingly small value for the demagnetizing factor (Nz).
Further information about demagnetizing factors and derivation of the equations above can be found in the Advanced Disccusion.
### Advanced Discussion (show/hide)»
By definition, the magnetization M is related to the internal magnetic field (H) through the dimensionless quantity (χ) known as the magnetic susceptibility:
M = χ H
In diagmagnetic and weakly paramagnetic materials, H = Ho, where Ho is the external field.
In ferromagnetic materials shape plays a critical role because the internal field H is reduced by an opposing demagnetizing field (D), which for homogeneous linear materials can be written as D = − NM, where N is a direction specific demagnetizing factor.
So in ferromagnetic materials
H = Ho − D = Ho − NM
But since H = M / χ we can write
M / χ = Ho − NM
or
Ho = M / χ + NM = M (1/χ + N) = M (1 + Nχ) / χ
The apparent susceptibility χapp is therefore
χapp = M / Ho = χ / (1 + Nχ)
Dividing the numerator and denominator by χ we obtain
χapp = 1 / (1/χ + N)
So when χ → ∞, 1/χ → 0, and we are left with
χapp = 1 / N
Thus for large values of χ (as are typically found in ferromagnetic materials), the dependence on χ disappears, with the apparent susceptibility dependent only on the reciprocal (1/N) of the demagnetizing factor.
The strength of the external field (Hext) needed to cause magnetic saturation (Msat) can be derived as
Hext = Msat / χapp = Msat / (1/N) = NMsat
Converting to SI units by multiplying each side of this equation by the permeability of free space (μo = 4π x 10−7 T m A−1), with Bext = μoHext and Bsat = μoMsat we obtain
Bext = N Bsat
References
Jackson DP. Dancing paperclips and the geometric influence on magnetization: a surprising result.Am J Phys 2006; 74:272-279.
McRobbie DW. Essentials of MRI Safety. Wiley-Blackwell, 2020. (Excellent recently released book. Chapter 2 on Fields and Forces as well as Appendix is very pertinent to this topic). [Link to purchase]
Osborn JA. Demagnetizing factors of the general ellipsoid. Phys Rev 1945; 67:351-357. [DOI link]
Panych LP, Kimbrell VK, Mukundan Jr S, Madore B. Relative magnetic force measures and their potential role in MRI safety practice. J Magn Reson Imaging 2020; 51:1260-1271. [DOI Link]
Prozorov R, Kogan VG. Effective demagnetizing factors of diamagnetic samples of various shapes. arXiv:1712.06037v2 (2 July 2018)
Sato M, Ishii Y. Simple and approximate expressions of demagnetizing factors of uniformly magnetized rectangular rod and cylinder. J App Phys 1989; 66:983-5. [DOI Link]
Skomski R, Hadjipanayis GC, Sellmyer DJ. Effective demagnetizing factors of complicated particle mixtures. IEEE Trans Magnetics 2007; 43:2956-8. [DOI Link]
Snelling EC. Soft Ferrites. Properties and Applications. Iliffe Books:London, 1969: selected pages
Solivérez CE. Electrostatics and magnetostatics of polarized ellipsoidal bodies: The depolarization tensor method. Bariloche: Rio Negro, Argentina: 2016.
Wysin GM. Demagnetization fields. 2012:1-16. Public teaching notes, downloaded from this link, 20 Oct 2020.
Related Questions
Which types of metal are the most dangerous around a magnetic field?What is magnetic susceptibility?
What is ferromagnetism?
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Tools & Reference>Pulmonology
Obstructive Sleep Apnea (OSA) Guidelines
Updated: Jan 07, 2025
Author: Varun Halani, MD; Chief Editor: Zab Mosenifar, MD, FACP, FCCP more...;)
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Obstructive Sleep Apnea (OSA)
Sections Obstructive Sleep Apnea (OSA)
Overview
Practice Essentials
Background
Pathophysiology
Etiology
Epidemiology
Prognosis
Patient Education
Show All
Presentation
History
Physical Examination
Predictive Value of History and Physical Examination
Cardiovascular Disease in Obstructive Sleep Apnea
Diabetes in Obstructive Sleep Apnea
Obstructive Sleep Apnea in Special Populations
Other Associated Risks
Show All
DDx
Workup
Approach Considerations
Laboratory Studies
Polysomnography
Multiple Sleep Latency and Maintenance of Wakefulness Tests
Show All
Treatment
Approach Considerations
Pharmacologic Therapy
Nasal CPAP Therapy
BiPAP Therapy
Oral Appliance Therapy
Surgical Care
Management of Residual Sleepiness Despite Apparently Effective Treatment
Prevention
Consultations
Long-Term Monitoring
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Guidelines
Medication
Antidiabetics, Glucagon-like Peptide-1 Agonists
Stimulants
Dopamine/Norepinephrine Reuptake Inhibitors
Show All
Media Gallery;)
References;)
Guidelines
Guidelines Summary
The American Academy of Sleep Medicine (AASM) has published a number of guidelines related to the diagnosis and management of obstructive sleep apnea (OSA).
In 2023, the AASM published the 3rd version of the AASM Manual for the Scoring of Sleep and Associated Events: Rules, Terminology and Technical Specifications which provides guidance for scoring sleep stages, arousals, respiratory events during sleep, movements during sleep and cardiac events. The AASM Scoring Manual also provides guidance on standard montages, electrode placements and technical and digital specifications. The recommendations for the indications and performance of polysomnography include the following :
Sleep stages are recorded via an electroencephalogram, electro-oculogram, and chin electromyogram
Heart rhythm is monitored with a single-lead electrocardiogram
Leg movements are recorded via an anterior tibialis electromyogram
Breathing is monitored, including airflow at the nose and mouth (using both a thermal sensor and a nasal pressure transducer), effort (using inductance plethysmography), and oxygen saturation
The breathing pattern is analyzed for the presence of apneas and hypopneas (as per definitions standardized by the American Academy of Sleep Medicine)
The 2017 AASM clinical guidelines for the indications for diagnostic testing for OSA in adults were endorsed by the World Sleep Society in 2021. [147, 218] The guidelines include the following strong recommendations:
Clinical tools, questionnaires and prediction algorithms should not be used to diagnose OSA in adults without polysomnography or home sleep apnea testing (HSAT) .
Polysomnography or home sleep apnea testing with a technically adequate device, should be used for the diagnosis of OSA in uncomplicated adult patients with clinical signs of an increased risk of moderate to severe OSA.
When a single HSAT is negative, inconclusive, or technically inadequate, polysomnography should be performed for the diagnosis of OSA.
Polysomnography should be used for the diagnosis of OSA in adults with significant cardiorespiratory disease, potential respiratory muscle weakness due to neuromuscular condition, awake hypoventilation or suspicion of sleep-related hypoventilation, chronic opioid medication use, history of stroke or severe insomnia.
In 2021 the AASM published guideline on referral for surgical therapy with the following recommendations :
Referral to a sleep surgeon for alternative treatments should be discussed with patients with BMI < 40 kg/m2 if the patient's adherence and tolerance does not support adequate PAP or if the patient rejects PAP.
Referral to a bariatric surgeon should be discussed with adults with OSA and obesity as an alternative treatment option if the patient's adherence and tolerance does not support adequate PAP or if the patient rejects PAP.
Discussion regarding a referral to both sleep and bariatric surgeons to discuss management options may be appropriate in patients with a BMI range of 3540 kg/m2.
PAP should be the initial therapy for adults with OSA and a major upper airway anatomic abnormality prior to referral for upper airway surgery
Medication
References
Guilleminault C, Tilkian A, Dement WC. The sleep apnea syndromes. Annu Rev Med. 1976. 27:465-84. [QxMD MEDLINE Link].
[Guideline] American Academy of Sleep Medicine. The AASM Manual for the Scoring of Sleep and Associated Events: Rules, Terminology and Technical Specifications. Version 3. American Academy of Sleep Medicine; February 2023.
Young T, Evans L, Finn L, Palta M. Estimation of the clinically diagnosed proportion of sleep apnea syndrome in middle-aged men and women. Sleep. 1997 Sep. 20(9):705-6. [QxMD MEDLINE Link].
American Academy of Sleep Medicine. International Classification of Sleep Disorders. Diagnostic and Coding Manual. Second Edition. Westchester, Ill: American Academy of Sleep Medicine; 2005.
Downey R 3rd, Perkin RM, MacQuarrie J. Upper airway resistance syndrome: sick, symptomatic but underrecognized. Sleep. 1993 Oct. 16(7):620-3. [QxMD MEDLINE Link].
Lugaresi E, Mondini S, Zucconi M, Montagna P, Cirignotta F. Staging of heavy snorers' disease. A proposal. Bull Eur Physiopathol Respir. 1983 Nov-Dec. 19(6):590-4. [QxMD MEDLINE Link].
Hoffstein V. Snoring. Chest. 1996 Jan. 109(1):201-22. [QxMD MEDLINE Link].
Bonnet MH. Effect of sleep disruption on sleep, performance, and mood. Sleep. 1985. 8(1):11-9. [QxMD MEDLINE Link].
Patil SP, Schneider H, Schwartz AR, Smith PL. Adult obstructive sleep apnea: pathophysiology and diagnosis. Chest. 2007 Jul. 132(1):325-37. [QxMD MEDLINE Link].[Full Text].
Schwab RJ, Pasirstein M, Pierson R, Mackley A, Hachadoorian R, Arens R, et al. Identification of upper airway anatomic risk factors for obstructive sleep apnea with volumetric magnetic resonance imaging. Am J Respir Crit Care Med. 2003 Sep 1. 168(5):522-30. [QxMD MEDLINE Link].
White DP. Sleep apnea. Proc Am Thorac Soc. 2006. 3(1):124-8. [QxMD MEDLINE Link].
White DP. The pathogenesis of obstructive sleep apnea: advances in the past 100 years. Am J Respir Cell Mol Biol. 2006 Jan. 34(1):1-6. [QxMD MEDLINE Link].
McGinley BM, Schwartz AR, Schneider H, Kirkness JP, Smith PL, Patil SP. Upper airway neuromuscular compensation during sleep is defective in obstructive sleep apnea. J Appl Physiol. 2008 Jul. 105(1):197-205. [QxMD MEDLINE Link].[Full Text].
Patil SP, Schneider H, Marx JJ, Gladmon E, Schwartz AR, Smith PL. Neuromechanical control of upper airway patency during sleep. J Appl Physiol. 2007 Feb. 102(2):547-56. [QxMD MEDLINE Link].
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Media Gallery
Sleep-related disordered breathing continuum ranging from simple snoring to obstructive sleep apnea (OSA). Upper airway resistance syndrome (UARS) occupies an intermediate position between these extremes. Note areas of overlap among the conditions.
In this polysomnogram summary graph, obstructive sleep apnea (OSA) severity and the degree of oxygen desaturation (SpO2%) worsen in rapid eye movement (REM) sleep (the black underlined sections) compared with non-REM sleep. This is often the case in OSA patients, especially in OSA patients with comorbid lung disease.
MRI rendering of a patient without obstructive sleep apnea (OSA) (left panel) and a patient with OSA (right panel).
Top image is 3-dimensional surface renderings of the upper airway demonstrating the effect of progressive increases in continuous positive airway pressure (CPAP) from 0-15 cm of water on upper-airway volume in a patient with upper airway narrowing. CPAP significantly increases airway volume in the retropalatal (RP) and retroglossal (RG) regions. Bottom image is soft tissue images in the same patient in the RP region at analogous levels of CPAP. With increasing CPAP, the upper airway progressively enlarges, particularly in the lateral dimension. Note the progressive thinning of the lateral pharyngeal walls as the level of CPAP increases. Little movement occurs in the parapharyngeal fat pads, the white structures lateral to the airway. The first image in each series depicts the baseline upper airway narrowing present in this patient.
Potential relationship between obstructive sleep apnea-hypopnea syndrome (OSAHS) and the metabolic syndrome. OSAHS has been associated with 3 of the 5 major clinical abnormalities associated with the metabolic syndrome, which is hypertension, insulin resistance, and proinflammatory/oxidative stress. OSAHS may be contributing to and/or modulating the severity of these metabolic abnormalities.
The Mallampati Classification is illustrated. The airway class is based on this visual heuristic.
Tonsil grades.
Obstructive sleep apnea. Note the absence of flow (red arrow) despite paradoxical respiratory effort (green arrow).
Central sleep apnea (thick areas). Note the absence of both flow and respiratory effort (green double arrows).
Comparison of a central apnea (box) and obstructive apnea (circle).
Mixed sleep apnea. Note that the apnea (orange arrow) begins as a central apnea (effort absent; red double arrow) and ends as an obstructive apnea (effort present; green double arrow). Note the arousal (blue arrow) that terminates the apnea and the desaturation (purple arrow) that follows.
A 2-minute recording of sleep showing 4 hypopneas (thick arrows) and associated oxygen desaturations (red arrows). This recording illustrates the recurrent nature of the sleep-disordered breathing observed in many patients.
Effect of nasal continuous positive airway pressure (CPAP) on oxygen saturation in sleep apnea. The upper portion of this figure shows the raw oxygen saturation trace from 1 night of a sleep study. Below the raw trace are vertical lines that indicate the presence of either an apnea or hypopnea. Before CPAP, frequent respiratory events with significant desaturations occurred. During the night, CPAP was applied, resulting in the elimination of the apnea and hypopneas and normalization of the oxygen trace.
Examples of good (upper panel) and poor (lower panel) compliance. In the upper panel, the patient is using continuous positive airway pressure (CPAP) most nights and generally for more than 4 hours (solid black line). In the lower panel, the patient is using CPAP infrequently and, when used, is wearing the CPAP device for less than 4 hours.
Approach to a patient with excessive daytime sleepiness after treatment with nasal continuous positive airway pressure.
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Contributor Information and Disclosures
Varun Halani, MD Fellow in Pulmonary Disease and Critical Care Medicine, St Louis University Hospital, St Louis University School of MedicineDisclosure: Nothing to disclose.
Nikitha Ravisankar, MBBS Disclosure: Nothing to disclose.
Specialty Editor Board
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Received salary from Medscape for employment.
Daniel R Ouellette, MD, FCCP Associate Professor of Medicine, Wayne State University School of Medicine; Medical Director, Pulmonary Medicine General Practice Unit (F2), Senior Staff and Attending Physician, Division of Pulmonary and Critical Care Medicine, Henry Ford HospitalDaniel R Ouellette, MD, FCCP is a member of the following medical societies: American College of Chest Physicians, American Thoracic Society, Society of Critical Care MedicineDisclosure: Received research grant from: Sanofi Pharmaceutical; AstraZeneca Pharaceutical; aTyr Pharmaceutical; Dompe Pharmaceutical.
Chief Editor
Zab Mosenifar, MD, FACP, FCCP Geri and Richard Brawerman Chair in Pulmonary and Critical Care Medicine, Professor and Executive Vice Chairman, Department of Medicine, Medical Director, Women's Guild Lung Institute, Cedars Sinai Medical Center, University of California, Los Angeles, David Geffen School of MedicineZab Mosenifar, MD, FACP, FCCP is a member of the following medical societies: American College of Chest Physicians, American College of Physicians, American Federation for Medical Research, American Thoracic SocietyDisclosure: Nothing to disclose.
Additional Contributors
James A Rowley, MD Professor, Fellowship Program Director, Department of Medicine, Division of Pulmonary, Critical Care and Sleep Medicine, Wayne State University School of MedicineJames A Rowley, MD is a member of the following medical societies: American Academy of Sleep Medicine, American College of Chest Physicians, American College of Physicians, American Thoracic SocietyDisclosure: Nothing to disclose.
Ralph Downey, III, PhD Staff Sleep Specialist, Cleveland Clinic Foundation; Associate Clinical Professor of Medicine, Loma Linda University School of MedicineRalph Downey, III, PhD is a member of the following medical societies: American Academy of Sleep MedicineDisclosure: Nothing to disclose.
Sat Sharma, MD, FRCPC Professor and Head, Division of Pulmonary Medicine, Department of Internal Medicine, University of Manitoba Faculty of Medicine; Site Director, Respiratory Medicine, St Boniface General Hospital, CanadaSat Sharma, MD, FRCPC is a member of the following medical societies: American Academy of Sleep Medicine, American College of Chest Physicians, American College of Physicians-American Society of Internal Medicine, American Thoracic Society, Canadian Medical Association, Royal College of Physicians and Surgeons of Canada, Royal Society of Medicine, Society of Critical Care Medicine, World Medical AssociationDisclosure: Nothing to disclose.
Himanshu Wickramasinghe, MD, MBBS Attending Physician, Pulmonary, Critical Care, and Sleep Medicine, Henry Mayo Newhall Memorial HospitalHimanshu Wickramasinghe, MD, MBBS is a member of the following medical societies: American College of Chest Physicians, American Thoracic SocietyDisclosure: Nothing to disclose.
Philip M Gold, MD Professor of Medicine, Chief of Pulmonary and Critical Care Medicine, Medical Director of Respiratory Care, Loma Linda University Medical CenterPhilip M Gold, MD is a member of the following medical societies: American College of Chest Physicians, American College of Physicians, American Heart Association, American Lung Association, American Medical Association, American Thoracic Society, California Medical Association, Society of Critical Care Medicine, Undersea and Hyperbaric Medical Society, California Thoracic Society, American Federation for Clinical Research, Association of Subspecialty ProfessorsDisclosure: Nothing to disclose.
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Sections Obstructive Sleep Apnea (OSA)
Overview
Practice Essentials
Background
Pathophysiology
Etiology
Epidemiology
Prognosis
Patient Education
Show All
Presentation
History
Physical Examination
Predictive Value of History and Physical Examination
Cardiovascular Disease in Obstructive Sleep Apnea
Diabetes in Obstructive Sleep Apnea
Obstructive Sleep Apnea in Special Populations
Other Associated Risks
Show All
DDx
Workup
Approach Considerations
Laboratory Studies
Polysomnography
Multiple Sleep Latency and Maintenance of Wakefulness Tests
Show All
Treatment
Approach Considerations
Pharmacologic Therapy
Nasal CPAP Therapy
BiPAP Therapy
Oral Appliance Therapy
Surgical Care
Management of Residual Sleepiness Despite Apparently Effective Treatment
Prevention
Consultations
Long-Term Monitoring
Show All
Guidelines
Medication
Antidiabetics, Glucagon-like Peptide-1 Agonists
Stimulants
Dopamine/Norepinephrine Reuptake Inhibitors
Show All
Media Gallery;)
References;)
encoded search term (Obstructive Sleep Apnea (OSA)) and Obstructive Sleep Apnea (OSA)
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187606 | https://www.sciencedirect.com/topics/medicine-and-dentistry/connective-tissue-nevus | Connective Tissue Nevus - an overview | ScienceDirect Topics
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Connective Tissue Nevus
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Connective tissue nevi, also known as connective tissue hamartomas, are benign lesions composed of dermal collagen and elastic fibers, presenting as flesh-colored plaques that can vary in appearance. They may occur alone or in association with genetic syndromes and are typically asymptomatic, often requiring no treatment.
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You might find these chapters and articles relevant to this topic.
Chapter
Neonatal dermatology
2022, Pediatric Dermatology (Fifth Edition)Katherine Brown Püttgen, Bernard A. Cohen
Connective tissue nevus
Connective tissue nevus defines a group of hamartomas with increased quantities of dermal collagen and variable changes in elastic tissue (Fig. 2.84). Lesions appear in the newborn as 1.0–10 cm diameter plaques composed of fibrotic papules and nodules, which often give a peau d’orange texture to the skin. Although typically located on the trunk, lesions may be widely disseminated, and any cutaneous site may be involved.
Connective tissue nevi may occur as isolated skin findings or in association with asymptomatic osteopoikilosis or radiographic densities in the long bones and the bones of the hands and feet (Buschke–Ollendorff syndrome). The shagreen patch of tuberous sclerosis cannot be differentiated clinically and histologically from an innocent connective tissue nevus. Consequently, a careful search for other stigmata of tuberous sclerosis must be performed in all children with these hamartomas.
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Book 2022, Pediatric Dermatology (Fifth Edition)Katherine Brown Püttgen, Bernard A. Cohen
Review article
HAMARTOMAS & CHORISTOMAS
2019, Seminars in Diagnostic PathologyB. Joel Tjarks, ... Nicole D. Riddle
Connective tissue nevi
Connective tissue nevus (CTN) is a broad term ascribed to dermal hamartomas derived from components of the connective tissue and extracellular matrix (e.g. – collagen, fibroblasts, elastin, and proteoglycans).153 CTN may be solitary, familial, acquired, or associated with systemic syndromes. They are typically classified according to the dominant component seen within the lesion (collagenoma, elastoma, etc.).154
Collagenomas may be sporadic or associated with genetic disorders (familial cutaneous collagenoma, tuberous sclerosus complex [Shagreen patches are collagenomas], multiple endocrine neoplasia type 1, Buschke-Ollendorff syndrome [BOS], and Proteus syndrome).153 Most often they arise as skin-colored or hypopigmented papules or plaques on the trunk of children.155 Histologically, there are increased and thickened dermal collagen bundles and decreased fibroblasts within the reticular dermis.156 Diminished numbers of elastic fibers can be seen with special stains.157 The histologic changes may be subtle and could easily be confused with normal thick truncal dermis without the appropriate clinical context.
Elastomas can be congenital or acquired, solitary or disseminated, and are often associated with BOS (characterized by elastomas and osteopoikilosis).158–160 The histologic features are once again subtle and H&E stained slides may be unremarkable. Elastic special stains will highlight irregularly dispersed and increased elastic tissue within the reticular dermis.153
CTN with a predominant component of proteoglycans may similarly show isolated or syndromic patterns. In Hunter syndrome (mucopolysaccharidosis II), the skin is generally thickened and numerous pale papules (pebbling of the skin) can be seen on the torso and extremities.161–163 Biopsy of these lesions shows increased mucinous material (glycosaminoglycans) deposited between collagen bundles.163 Acquired forms of proteoglycan-type CTN (lichen myxedematosus, focal cutaneous mucinosis) show similar histologic findings.153
Special attention should be paid to fibroblastic connective tissue nevi (FCTN) as most cases express CD34 and may be confused with dermatofibrosarcoma protuberans or other CD34 positive tumors. Described in 2012 by de Feraudy and Fletcher, FCTN is a hamartoma which shows myofibroblastic/fibroblastic predominance.164,165 Clinically the lesions typically arise in early childhood or adolescence as slow-growing plaques or nodules on the trunk or head and neck.164 Histologically, they are dermal-based, unencapsulated lesions with ill-defined margins and overlying epidermal papillomatosis (Fig.18). The lesions are composed of bland spindled cells arranged in fascicles or bundles which spare the cutaneous adnexa (Fig.19).164–166 Ultrastructural analysis has confirmed the myofibroblastic lineage of the spindle cells.165 Distinction of FCTN from DFSP is essential and may be difficult if the entire lesion is not visualized. Classically DFSP has a storiform pattern as opposed to the short fascicles seen in FCTN. Additionally, the epidermal papillomatosis seen in FCTN is usually absent in DFSP. FCTN typically invades the subcutis along the fibrous trabeculae where DFSP shows an irregular infiltration pattern.166 In troublesome cases, molecular and cytogenetic studies may be helpful to identify the t(17;22) COL1A1-PDGFB fusion associated with DFSP.
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Fig. 18. Fibroblastic connective tissue nevus. This excision from a 4-month-old shows epidermal papillomatosis overlying a dermal based proliferation of bundles and fascicles of spindled fibroblasts. H&E; 10x magnification.
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Fig. 19. Fibroblastic connective tissue nevus (same lesion as Fig.18). Fascicles of spindled fibroblasts proliferating in and around dermal collagen bundles. The spindle cells are positive for CD34. H&E; 100x magnification.
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Journal2019, Seminars in Diagnostic PathologyB. Joel Tjarks, ... Nicole D. Riddle
Chapter
Cutaneous Tumors and Tumor Syndromes
2016, Hurwitz Clinical Pediatric Dermatology (Fifth Edition)Amy S. Paller MD, Anthony J. Mancini MD
Connective Tissue Nevus
Connective tissue nevus (also known as connective tissue hamartoma) is a localized hamartoma of either dermal collagen or elastic fibers or both. These benign skin lesions may be sporadic or hereditary and can be seen as a component of several syndromes. For instance, the shagreen patch and fibrous forehead plaque seen in tuberous sclerosis (see Chapter 11) are both forms of connective tissue nevi. In addition, the palmoplantar cerebriform hyperplasia that occurs in patients with Proteus syndrome (see Chapter 12) represents excess collagen and thus could also be classified under this umbrella term. This section discusses primarily the sporadic, nonsyndromic form of connective tissue nevus.
These lesions usually present as asymptomatic, flesh-colored dermal plaques composed of multiple papules. They may be quite subtle (Fig. 9-50), or they may be associated with significant thickening, cobblestoning, or cerebriform changes (Fig. 9-51). Solitary plaques may occur anywhere on the cutaneous surface, and multiple lesions are often distributed symmetrically on the back, buttocks, and extremities. The differential diagnosis may include smooth muscle hamartoma (see Congenital Smooth Muscle Hamartoma section).
Another form of connective tissue nevus, dermatofibrosis lenticularis disseminata, has been reported in association with a specific bone dysplasia termed osteopoikilosis . This association is also known as Buschke–Ollendorf syndrome (see Chapter 6). The cutaneous lesions in this setting usually appear in adult life, but their onset has also been reported during the first year or early childhood. They appear as flesh-colored to yellow papules, usually a few millimeters in size, distributed on the trunk, buttocks, and extremities. They often represent hamartomas of elastic tissue (elastomas) histologically. The bony lesions are usually asymptomatic and are often noted incidentally as focal sclerotic areas on bone radiographs. They may occasionally be mistaken for foci of metastatic malignancy, highlighting the importance of familiarity with this disorder and its associated skin lesions.
Connective tissue nevi are generally asymptomatic and not usually of cosmetic significance. If the diagnosis is unclear, incisional biopsy is useful. Surgical excision is otherwise unnecessary.
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Book 2016, Hurwitz Clinical Pediatric Dermatology (Fifth Edition)Amy S. Paller MD, Anthony J. Mancini MD
Chapter
Lumps, Bumps, and Hamartomas
2008, Neonatal Dermatology (Second Edition)Julie S. Prendiville
Connective Tissue Nevus
A connective tissue nevus is characterized by excessive deposition of one or both of the collagen or elastin components of dermal connective tissue. These hamartomas may occur sporadically, or as a familial disorder with autosomal dominant transmission.163 Connective tissue nevi are also seen as a manifestation of genetic syndromes, notably the ‘shagreen patch’ or collagenoma in tuberous sclerosis, and the multiple elastic tissue nevi of Buschke–Ollendorff syndrome.164 A connective tissue nevus may be present at birth, but most become evident during childhood or adolescence.
Cutaneous Findings
Connective tissue nevi present clinically as asymptomatic, firm, skin-colored to yellowish nodules or plaques located on the trunk or limbs (Fig. 23-25). The surface of the lesion may be smooth or have a ‘cobblestone,’ ‘leather-grain,’ or ‘peau d'orange’ appearance. They may be solitary or multiple. A linear or ‘zosteriform’ morphology is sometimes observed.
Extracutaneous Findings
Osteopoikilosis is seen in association with elastic tissue nevi in the Buschke–Ollendorff syndrome.164 The skin lesions in this condition may rarely be present at birth, but the distinctive bone changes are not reported in infancy. The collagenoma or ‘shagreen patch’ of tuberous sclerosis develops in later childhood, although other stigmata of the disease may be present at birth or in early infancy. Cardiomyopathy may occur in association with the multiple lesions of familial cutaneous collagenoma and with collagenomas and hypogonadism.163 Multiple collagenomas in Down syndrome have been reported in adolescence.165 A cerebriform collagenoma on the sole of the foot may be an isolated phenomenon or a component of Proteus syndrome.166
Etiology and Pathogenesis
The pathogenesis is unknown. In familial cutaneous collagenoma, the skin lesions are inherited as an autosomal dominant trait. Tuberous sclerosis and Buschke–Ollendorff syndrome are also inherited by autosomal dominant transmission. Somatic mosaicism may be postulated for sporadic lesions, particularly those with a linear distribution.
Diagnosis
Histologic examination of connective tissue nevi shows an excess of collagen or elastic tissue in the dermis. This may not be apparent unless a specimen of normal adjacent skin is obtained for comparison. Thus biopsies of connective tissue nevi are often reported as ‘normal skin.’ Special elastic stains are necessary to demonstrate the increased numbers of elastic fibers in elastic tissue nevi.
Differential Diagnosis
The differential diagnosis includes other cutaneous hamartomas, such as neurofibromas, leiomyomas, smooth muscle hamartomas, nevus lipomatosus, and epidermal nevus. These entities may be distinguished by histopathologic examination of a skin biopsy. A congenital mucinous nevus may be a variant of connective tissue nevus in which deposition of mucin (proteoglycan) in the dermis is the predominant histopathologic finding.167
Treatment and Prognosis
Connective tissue nevi are permanent lesions. They grow in proportion to the child's growth. There is no malignant potential and most do not require treatment. Surgical excision may occasionally be indicated for cosmetic reasons.
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Book 2008, Neonatal Dermatology (Second Edition)Julie S. Prendiville
Chapter
Vascular Anomalies and Other Cutaneous Congenital Defects
2024, Avery's Diseases of the Newborn (Eleventh Edition)Deepti Gupta, Robert Sidbury
Connective Tissue Nevus (Connective Tissue Hamartoma)
Connective tissue nevi are hamartomas comprising dermal collagen, elastic fibers, or a combination of the two. They are benign lesions that can occur in isolation or in relation to a genetic syndrome such as tuberous sclerosis or Proteus syndrome.128 They present as flesh-colored dermal plaques that can be subtle on examination to hypertrophic with a cobblestoned and cerebriform appearance. They are usually asymptomatic and do not often require any intervention. Diagnosis is often made clinically, but if the diagnosis is unclear, a skin biopsy can be performed.
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Book 2024, Avery's Diseases of the Newborn (Eleventh Edition)Deepti Gupta, Robert Sidbury
Chapter
Skin Diseases in Newborns
2007, Color Textbook of Pediatric Dermatology (Fourth Edition)William L. Weston MD, ... Joseph G. Morelli MD
Connective Tissue Birthmarks
Connective tissue nevi are skin lesions consisting predominantly of the elements of extracellular collagen tissue and products of fibroblasts, such as collagen, elastin, and proteoglycans.57 All connective tissue nevi are quite rare, although the precise incidence is not known.
Clinical features
Connective tissue nevi are localized areas of thickened skin appearing as multiple skin-colored papules and plaques (Fig. 21.62). Stretching the overlying skin will give a yellowish discoloration to the areas. They may occasionally have increased vascularity and appear red. Collagenomas are localized areas of thickened skin with multiple skin-colored papules or plaques. They may be solitary or appear in a zosteriform segmental pattern. Elastomas are solitary plaques that are present at birth and contain increases in both elastic tissue and proteoglycans. Elastomas may be solitary or they may be multiple in the Buschke-Ollendorff syndrome. This autosomal dominantsyndrome appears as symmetrically distributed skin-colored papules or nodules with a predilection for the lower trunk or for the extremities. Lesions may assume a thickened appearance of skin and develop a lacy pattern over the trunk. Radiographs may show sclerotic densities of the ends of long bones, pelvis, and hands, although such lesions are often asymptomatic. The shagreen patch of tuberous sclerosis is a connective tissue nevus. The nevi are subtle at birth and may go unnoticed. They tend to persist throughout life.
Differential diagnosis
The lesions of connective tissue birthmarks are so characteristic that they are seldom misdiagnosed. Examinations for possible associated systemic disease such as Proteus syndrome50 or tuberous sclerosis 58 may be necessary.
Pathogenesis
Connective tissue nevi show thickened, abundant collagen bundles with or without associated increases in elastic tissue. Such histologic changes are difficult to appreciate unless the skin biopsy includes adjacent normal skin for comparison.
Treatment
Treatment is unnecessary.
Patient education
Connective tissue nevus is a birthmark consisting of collagen and/or elastin, and it should be emphasized that such lesions may be related to systemic or genetic disease. In the absence of associated disease, the connective tissue nevus does not represent a serious problem.
Follow-up visits
Follow-up visits are unnecessary except for routine neonatal visits.
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Book 2007, Color Textbook of Pediatric Dermatology (Fourth Edition)William L. Weston MD, ... Joseph G. Morelli MD
Chapter
Skin Diseases in Newborns
2007, Color Textbook of Pediatric Dermatology (Fourth Edition)William L. Weston MD, ... Joseph G. Morelli MD
Patient education
Connective tissue nevus is a birthmark consisting of collagen and/or elastin, and it should be emphasized that such lesions may be related to systemic or genetic disease. In the absence of associated disease, the connective tissue nevus does not represent a serious problem.
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Book 2007, Color Textbook of Pediatric Dermatology (Fourth Edition)William L. Weston MD, ... Joseph G. Morelli MD
Chapter
Disorders of collagen
2010, Weedon's Skin Pathology (Third Edition)David Weedon AO MD FRCPA FCAP(HON)
CONNECTIVE TISSUE NEVI
Connective tissue nevi are cutaneous hamartomas in which one of the components of the extracellular connective tissue – collagen, elastic fibers, or glycosaminoglycans – is present in abnormal amounts.661 They can be subclassified on the basis of the component predominantly involved:
•
collagen type
collagenoma
shagreen patch
•
elastin type
elastoma
•
proteoglycan type
nodules in Hunter's syndrome.
Sometimes there are alterations in more than one component of connective tissue, and these lesions may simply be categorized as connective tissue nevi.662,663 Only the connective tissue nevi of collagen type will be discussed in this section.
Collagenoma
Collagenomas (connective tissue nevi of collagen type) are rare hamartomas of the skin in which there is an increase in dermal collagen. They usually present as asymptomatic, firm, flesh-colored plaques and nodules, 0.5–5.0 cm in diameter, on the trunk and upper part of the arms.661 The ear,664 sole of the foot,665,666 and vulva667 are rare sites of involvement. There may be several lesions 668 or up to a hundred or more, with an onset in adolescence.669,670 Rapidly growing (eruptive) collagenomas have been reported in the multiple endocrine neoplasia type I syndrome,671 and in pregnancy.672 Eruptive collagenomas have also been reported as the only lesion.673,674 Uncommonly, they occur in a zosteriform or linear distribution.675–681 A family history is sometimes present (familial cutaneous collagenoma) and in these cases there is autosomal dominant inheritance.669,682–685 Familial cutaneous collagenomas (OMIM 115250) result from a mutation in the LEMD3 gene, as seen in the Buschke–Ollendorff syndrome.686,687
Associated clinical features include a cardiomyopathy,661 Down syndrome,688–690 and occult spinal dysraphism.691 Uncommonly, the connective tissue nevi associated with the Buschke–Ollendorff syndrome (see p. 334) are of collagenous composition rather than of the usual elastic tissue type.692–695 A connective tissue nevus of collagen type has been reported in association with pseudo-Hurler polydystrophy (mucolipidosis III),696 and in Proteus syndrome, in which the nevi are often acral, sometimes of the plantar cerebriform type.697–699 Plantar collagenomas can occur in the absence of the Proteus syndrome.700
Solitary collagenomas are sometimes quite large, as seen with the cerebriform or ‘paving stone’ variants on the sole of the foot.701,702 Sometimes a connective tissue nevus is associated with cutis verticis gyrata, a descriptive term for a condition of the scalp in which deep furrows and convolutions are present. The folds of skin may, however, have normal morphology.703,704 Other pathological associations of cutis verticis gyrata include lymphedema,705 mental retardation,706 adipocyte proliferation, acromegaly,707 misuse of anabolic substances,708myxedema, Noonan syndrome,706,709 tumors, and the insulin resistance syndrome.710 Cutis gyrata, usually on the scalp, can be seen in the Beare–Stevenson syndrome (OMIM 123790) characterized by mutations in the transmembrane region of fibroblast growth factor receptor (FGFR)-2 located at chromosome 10q26. Acanthosis nigricans, skin tags, and anogenital anomalies are other features of this syndrome.711 There is a reduction in expression of tuberin in connective tissue nevi associated with the tuberous sclerosis complex.712
Collagenomas must be distinguished from sclerotic fibromas (see p. 815) which present as tumor-like nodules. They have a characteristic histological appearance with dense collagen bundles, often in a storiform arrangement. Athlete's nodules are related to, but different from, collagenomas. One such example is the dermal nodule found in the sacrococcygeal region of bicycle riders.713
Histopathology
The epidermis is usually normal, although an overlying epidermal nevus has been reported.714 There is thickening of the dermis, sometimes with partial replacement of the subcutis. The collagen bundles are broad and have a haphazard arrangement.715Elastic fibers are more widely spaced, but this may represent a dilution phenomenon.715 Sometimes the elastic fibers are thin and fragmented. Some of these cases may represent papular elastorrhexis (see p. 344).673 There is no increase in mucopolysaccharides. Dermal dendrocytes are reduced using antibodies against factor XIIIa.716 Calcification was present in one reported case.717
As the collagen in collagenomas is less well packed than normal, differences in polarization colors can be seen with picrosirius red staining followed by polarization microscopy.718 The fibers appear green to yellow, in contrast to the orange to red color of normal dermal collagen.718
Shagreen patch
The shagreen patch is a distinct clinical variant of collagenoma, found exclusively in those with tuberous sclerosis. It consists of a slightly elevated, flesh-colored plaque of variable size, usually on the lower part of the trunk.719 It has the appearance of untanned leather. Smaller ‘goose flesh’ papules may form as satellite lesions.
Histopathology 719
There are dense sclerotic bundles of collagen, with an interwoven pattern, in the reticular dermis.720 Fibroblasts appear hypertrophied.720 There is no inflammatory infiltrate or increase in vascularity. The overlying epidermis is usually flat, although sometimes it has the pattern of acanthosis nigricans.719
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Book 2010, Weedon's Skin Pathology (Third Edition)David Weedon AO MD FRCPA FCAP(HON)
Chapter
Fibroblastic and myofibroblastic tumors of the skin
2020, Diagnostic Atlas of Cutaneous Mesenchymal NeoplasiaEduardo Calonje MD, DipRCPath, ... Boštjan Luzar MD, PhD
Histology
•
Connective tissue nevus
•
Highly collagenized connective tissue
•
Epidermal nevus
•
Hyper(ortho)keratosis, acanthosis, and papillomatosis
•
No adnexal hyperplasia
•
Vascular malformation
•
See corresponding chapter
•
Lipoma
•
See corresponding chapter
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Book 2020, Diagnostic Atlas of Cutaneous Mesenchymal NeoplasiaEduardo Calonje MD, DipRCPath, ... Boštjan Luzar MD, PhD
Chapter
Benign skin tumors
2009, General DermatologyDonna Marie Vleugels, James E. Sligh
Connective tissue nevus (Collagenoma, elastoma, shagreen patch in tuberous sclerosis)
Clinical presentation
Connective tissue nevi are hamartomas that typically present at birth or during early childhood as firm, flesh-colored papules, nodules, or plaques. They may be solitary or multiple and occur in a grouped, linear, or irregular distribution. The shagreen patch of tuberous sclerosis is a variant with a “pigskin”-like plaque on the lower back. Buschke–Ollendorff syndrome, an autosomal-dominant disorder, is characterized by the appearance of multiple flesh-colored to yellow papules at an early age. Osteopoikilosis, or scattered areas of increased bone density, is characteristically found on X-ray in affected patients. Multiple other medical syndromes have been associated with connective tissue nevi.
Diagnosis
Clinicopathologic evaluation leads to the diagnosis of connective tissue nevi.
Histological features
Biopsies should be taken at the edge of connective tissue nevi to allow for comparison of the surrounding normal skin. Affected areas demonstrate an ill-defined increase of dermal collagen or elastin. No laboratory studies are indicated unless there is suspicion of an underlying syndrome. For example, patients suspected of having Buschke–Ollendorff syndrome should have appropriate radiographs. Brain imaging, electroencephalogram, renal ultrasound, and echocardiogram are indicated for patients with tuberous sclerosis.
Pathogenesis
The pathogenesis of connective tissue nevi is unknown.
Treatment
Because connective tissue nevi are benign lesions, surgical excision is not recommended. Patients with associated syndromes may require other forms of therapy.
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Book 2009, General DermatologyDonna Marie Vleugels, James E. Sligh
Related terms:
Proteus Syndrome
Tuberous Sclerosis
Neoplasm
Papule
Hamartoma
Desmoplastic Fibroma
Nevus Sebaceous
Skin
Mucin
Autosomal Dominant Inheritance
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187607 | https://www.ck12.org/flexi/cbse-math/prime-factorization/what-is-the-prime-factorisation-of-2006/ | What is the prime factorisation of 2006? - Steps | CK-12 Foundation
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What is the prime factorisation of 2006?
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Prime factorisation of 2006 is 2 x 17 x 59.
Any number can be split up into factors.
If we keep splitting the number into factors, ultimately, we reach a stage when all the factors are prime factors.
Breaking down a number into a product of all prime numbers is called prime factorization.
The prime factorization of 2006 can be found by dividing it by its prime factors until only primes are left.
Steps to Prime Factorise a Number using the Division Method
To find the prime factorisation of a number using the division method; we use the following steps:
Step 1: Divide the given number by its smallest prime factor.
Step 2: Divide the quotient obtained in step 1 by its smallest prime factor.
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On factorising 2006 by the division method, we get
2006 = 2 x 17 x 59
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187608 | https://www.emra.org/emresident/article/oral-contraceptives-complications | EMRA
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Contraceptives: Rare Complications and Their Implications for Emergency Medicine
6/10/2022 Cody Meyers, MD , Ray Chahoud, DO
One of the rare but feared complications of oral contraceptive pills (OCPs) containing estrogen is hemorrhage from an enlarging hepatocellular adenoma.
The risk of developing hepatocellular adenomas in the absence of contraceptives is 1 in 1 million. However, the risk increases 30-40 fold with the use of OCPs,1 with the risk of significant bleeding ranging from 25-64%. Additionally, the larger the mass, the larger the risk of bleeding.2 In this case, we present a patient who suffered from the rupture of one of her hepatic adenomas while on Nexplanon® (etonogestrel), a progestin-only oral contraceptive that is implanted subdermally and lasts for up to three years.
CaseAn 18-year-old female was referred to the ED from an urgent care facility with severe, constant, unrelenting back pain. On the urgent care provider’s initial assessment, the patient denied any abdominal pain. However, she had an episode of non-bloody, nonbilious emesis immediately upon palpation of her epigastrium.
At the urgent care facility, the patient’s laboratory studies revealed leukocytosis of 22,000 cells/mm3 and hypokalemia, with a potassium level of 2.8 mmol/L. Due to the concerning findings on initial assessment, the patient was transferred to the ED for further evaluation.
Upon the patient’s arrival to the ED, she reported severe abdominal pain that increased upon standing. The patient denied any relevant past medical or surgical history. When specifically asked, she denied any non-steroidal anti-inflammatory drug use, alcohol use, or family history of biliary colic. The patient’s only reported medication was a Nexplanon® implant.
The patient’s vital signs were within normal limits upon initial evaluation. Her exam was remarkable for severe epigastric and right upper quadrant abdominal tenderness with voluntary guarding. The patient's repeat complete blood count (CBC) revealed leukocytosis of 25,000 cells/mm3 with an elevated absolute neutrophil count to 20. Her hemoglobin was found to be 9.2 gm/dL with no baseline for comparison. Comprehensive metabolic panel revealed mild transaminitis with AST and ALT values of 149 IU/L and 176 IU/L, respectively. Potassium was found to be 3.3 mmol/L. In addition, the patient had an elevated alkaline phosphatase of 131 IU/L. Bilirubin was unremarkable. Urine studies revealed red blood cells, and a pregnancy test was negative.
Point-of-care ultrasound of the patient’s right upper quadrant revealed a small amount of intra-abdominal free fluid in the right upper quadrant. Ultrasound did not identify any gallstones, pericholecystic fluid, or gallbladder wall thickening. However, there was an abnormal-appearing echogenicity to the liver (see figures). A stat CT scan of the abdomen and pelvis was ordered for further investigation.
While awaiting the results of the CT scan, the patient’s lactate was found to be elevated at 5 mmol/L. At that time, 2 L of normal saline was initiated.
CT scan results demonstrated a large hemorrhagic mass in the liver with contrast extravasation concerning for ruptured hepatic adenoma (see figures). Repeat CBC in the ED demonstrated a drop in hemoglobin from 9.2 gm/dL to 8.1 gm/dL.
Acute care surgery was consulted for further recommendations. The patient was then admitted to the hospital for serial hematocrit checks with plans to take her to the interventional radiology (IR) suite for embolization.
While admitted, the patient underwent Gelfoam® embolization of the right hepatic artery. The patient tolerated the procedure well and transitioned to oral intake prior to discharge the following day, with instructions to follow up with surgical oncology. Chart review performed after the visit revealed that the patient had Nexplanon® implantation four months prior to this incident. No previous history of hepatic adenomas was documented in her previous office visit notes between 2016 and 2020.
DiscussionRuptured hepatic adenoma is an uncommon diagnosis in the ED and thus not often considered in the initial differential diagnosis for abdominal pain. However, this patient did have one risk factor that should raise every emergency physician’s clinical level of suspicion for this disease process: the use of oral contraceptives.
Hepatic adenomas are known to be either caused or exacerbated by estrogen-containing OCPs. However, this case features a patient who developed liver adenomatosis, defined as more than 3 adenomas, after the implantation of Nexplanon®, a progestin-only contraceptive rod.5
Section 5.7 of the package insert of Nexplanon® includes the following: “It is not known whether a similar risk exists with progestin-only methods like Nexplanon®.”6
We found one other case report in which a patient developed liver adenomatosis with bleeding. However, that patient was previously on an estrogen-containing OCP prior to having Norplant (another progestin-releasing implant) inserted. In this case, though, the patient was not on any contraceptives previously and had Nexplanon® inserted approximately four months prior to her ED visit. The case presentation is in contrast with the common teaching that patients on progestin-only contraception have little to no risk of enlarging hepatic adenomas.
Take-Home PointsAs emergency physicians, our role is to manage multiple critically ill patients at any given time. Medication lists are an important part of medical history; yet they sometimes receive no more than a cursory glance, despite the risk of adverse reactions.
When considering young women with abdominal pain, asking about all forms of contraception may be crucial to arriving at the correct diagnosis. Many women, particularly those with implantable devices and not on daily medications, may forget to mention contraceptive use unless specifically asked.
References
Rooks JB, Ory HW, Ishak KG, et al. Epidemiology of Hepatocellular Adenoma. The Role of Oral Contraceptive Use. JAMA 1979; 242:644.
Bieze M, Phoa SS, Verheij J, et al. Risk factors for bleeding in hepatocellular adenoma. Br J Surg 2014; 101:847.
Darney P, Patel A, Rosen K, et al. Safety and efficacy of a single-rod etonogestrel implant (Implanon): Results from 11 international clinical trials. Fertil Steril 2009; 91:1646.
Daniels K, Abma J. Current Contraceptive Status Among Women Aged 15–49: United States, 2015–2017. NCHS data brief, no 327. Hyattsville, MD: National Center for Health Statistics. 2018.
Suarez, A. A., Brunt, E. M., & Di Bisceglie, A. M. (2001). A 35-Year-Old Woman with Progesterone Implant Contraception and Multiple Liver Masses. Seminars in Liver Disease, 21(03), 453–460.
Nexplanon [Package insert]. N.V. Organon, Oss, The Netherlands, a subsidiary of Merck & Co., Inc.; 2020
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187609 | https://pages.stat.wisc.edu/~wahba/stat860public/pdf1/micchelli.interpolation.86.pdf | Constr. Approx. (1986) 2:11-22 CONSTRUCTIVE APPROXIMATION 9 Springer-Verlag New York Inc. Interpolation of Scattered Data: Distance Matrices and Conditionally Positive Definite Functions Charles A. Micchelli Abstract. Among other things, we prove that multiquadric surface interpolation is always solvable, thereby settling a conjecture of R. Franke. 1. Introduction For solving practical problems of data fitting in two dimensions, two methods seem to be most popular: thin plate splines (TPS) by Duchon and Hardy's multiquadric surfaces (MQS) ([14, 15]; see also Franke ). The theory for TPS has been developed in a series of papers by Duchon and Meinguet . However, beyond its numerical performance, little seems to be known about the MQS method. For instance, in his lecture notes for a recent meeting, Franke proposed (based on extensive numerical experience) the following conjecture: Conjecture. Given any distinct points x x ..... x" in the plane (1.1) (-l)n-ldetr § II x/- xJll 2 > 0, where Ilxll 2 = x 2 + x 2, x = (x~, x2), is the Euclidean norm of x. This conjecture says, in Particular, that there is a unique surface f(x)= c~r + IIx - xql 2 + "'" +cnr + IIx - x"ll 2 that interpolates (data) yp..., y,, at x L ..... x". Apparently, when this conjecture was made it was not even known that interpolation by MQS is always possible according to Wahba . As an extension of MQS, Barnhill and Stead explored surfaces based on the kernel K,(x, y)= (1 + IIx-YllZ) -", where /, is a real number, in contour plotting for three-dimensional interpolation. Thus they used (1.2) f(x) = clK~(x, x 1) + ... +cnK~(x, x") Date received: October 19, 1984. Date revised: July 10, 1985. Communicated by Carl de Boor. A MS classification: 41A05, 41A63. Key words and phrases: Multivariate interpolation, Multiquadric surface, Thin plate splines, Positive definite functions. 12 c.A. Micchelli to interpolate scattered data (1.3) f(x i) = y,, i = 1 ..... n, and sought those values of bt for which (1.3) has a unique solution. A similar question was recently raised kernel K(x, y) = log(1 + IIx - yll 2) by N. Dyn . The purpose of this paper is to address these questions. We place them into a unified context so that we can draw upon ideas from the theory of positive definite functions by Stewart and distance matrices by Blumenthal . Therefore it is possible that some of what we say here may already be accessible in the literature. 2. Background and Results Let X be an abstract point set. Suppose K is a real-valued kernel defined on X • X and kl(X ) .... , k,,(x) are given real-valued functions over X. A sufficient condition that guarantees that the interpolation problem (1.3) has a solution of the form n ftt (2.1) f(x) = E ciK(x, xi) + ~-, diki(x), m < n, i=1 i=1 that satisfies the auxiliary condition n (2.2) E cikj(Xi) = O, j = 1 ..... m, i=1 when rank (ki(xJ)) = m is that the quadratic form n E c,cjK(x',x ) i=1 j=l is strictly positive definite whenever c = (cl ..... c,) satisfies (2.2). The motivation for this type of interpolation comes from methods of optimal interpolation. Thus some quadratic seminorm can be introduced on a subspace of real-valued functions on X and the method (2.1), (2.2) will have the least norm among all solutions to (1.3). This equivalence is discussed by several authors (Matheron [19, 20], Salkauskas ). For instance, when m = 0, i.e., only the kernel K is present in (2.1), the construction of X can be based on Aronszajn's theory of reproducing kernels . These matters are not of concern to us here. Rather, we are interested in explicit examples of (2.1), (2.2) which can be used to solve the scattered data interpolation problem in, for instance, two and three dimensions. For this purpose, we restrict X = R s and K(x, y) = F(llx - yl[2), where F is a continu- ous real-valued function defined on [0, oo). Furthermore, we consider only the case where the span of kl(x ) ..... km(x ) is the space ~rk_l(R s) of polynomials of total degree < k - 1. Definition 2.1. A continuous function F(t), defined on [0, oo), is said to be conditionally (strictly) positive definite of order k on R s if for any distinct points Interpolation of Scattered Data 13 X 1 ..... X" E R s and scalars c 1 ..... c,, such that (2.3) E cip(x') = o i=I for all p ~ 7rk_l( R'), the quadratic form El'= " , , i z) 1F/=l~//F([lx - xJll is (positive) nonnegative. We will denote the class of conditionally positive definite functions by ~k(RS). Obviously, ~k+l(R') c_ ~k(Rs). A famous theorem of Bochner characterizes any f(x) = F(llxll 2), F ~ ~0(RS) as the Fourier transform of a finite Borel measure on R' (see Stewart ). Because f(x) is a radial function this result has an equivalent formulation expressing F(t) as a certain Bessel transform of a measure on R ~. Specifically, F(t) = ffa,(tu) where a(u) is bounded and nondecreasing, and ~, is defined by Schoenberg [26, 27] in terms of Bessel functions as cost, s = 1, Similar characterizations are available for the class 9~k(R') [13, Chapter II]. We are especially interested in the class of functions that are conditionally positive definite of order k over any R ~, i.e., (2.4) 2~k= N ~k(R'). s>l To state our result about ~k we recall that a function F is said to be completely monotonic on (0, oo) provided that it is in C~(0, oo) and (-1)/F(/)(x)> 0, x ~ (0, oo), l= 0,1,2,.... Theorem 2.1. F ~ ~k whenever F is continuous on [0, oo) and (-1)kF (k) is completely monotonic on (0, oo). The case k = 0 of this theorem is due to Schoenberg . He even showed the equivalence of these conditions in this case. The general case and its elementary proof embody the TPS and MQS interpolation methods and lead us to a proof of (1.1). Another result of this type is Theorem 2.2. Let l = [s/2] - k + 2 be a positive integer. Then for any function defined on (0, oo) such that (- 1)k+JF(k+))(t) is nonnegative, nonincreasing, and convex for j=0,1,...,l-2 on (O, oo) (if l= 1, we require only that it be nonnegative and nonincreasing), it follows that F(v~) ~ gZ k( RS ). 14 C.A. Micchelli The proof of this theorem uses a result from Askey . Finally, we will prove Theorem 2.3. Suppose F' is completely monotonic but not constant on (0, ~), F is continuous on [0, ~) and positive on (0, oo). Then for any distinct vectors x 1 ..... x" R s (s arbitrary) (-1)"-ldet F(II x- xJl] a) > 0. Consequently, the choice F(t)= (1 + t) 1/2 in Theorem 2.3 proves Franke's conjecture. Before we prove these theorems we recall the important notion of an almost positive definite matrix, which is relevant to our discussion (see Donoghue ). Definition 2.2. provided that A real n x n symmetric matrix A is called almost negative definite ~ cicjAij < -~ 0 j=l i=l whenever ET=xci = O. Let us denote this class by ~r and note that if A~j = [Ix g - xJl] 2 for some x 1 ..... x" ~ R s, then A is in d, because n 2 (2.5) -2 < o i=1 j=l i=1 when ET=lci = 0. There is a beautiful converse to this observation (see Schoen- berg and Blumenthal ) that states Theorem A. Let A be an n x n real symmetric matrix with zero diagonal entries. There exist vectors xl,..., x" ~ R s for some s such that Aij = II x~ - xJll 2 if and only if A is almost negative definite. Besides being useful in distance geometry, the notion of almost positive definite matrices is important in probability theory, where it is used to characterize infinitely divisible laws (see Luckas ). Since the matrix (llx i - xJll 2) appears in the formulation of Definition 2.1, we can express Theorems 2.1-2.3 in terms of almost positive definite matrices, provided we give the condition (2.3) its proper interpretation. For this purpose, we require Corollary 2.1. A ~ d if and only if any one of the following conditions holds: (a) There exist o = (01 ..... o,) and x 1 ..... x"~ R s for some s such that A,j = oi 4 a, + II x~ - xJll 2" (b) There exist o = (01 ..... o,) and a B = (Bu), i, j = 1,..., n, that is posi- tive definite such that dij = o i + oj - Bij. Interpolation of Scattered Data 15 (c) (e-~A' 0 is positive definite for all a > O. Moreover, it is strictly positive definite if and only if l Ag), i 4: j. (2.6) AU > 5(A,, + Proof. For the first two claims it suffices to remark that when A ~ d, Theorem A says that there exist x 1 ..... x" such that IIx' - #112 = A,j - l( hii "~- Ajj). The first part of assertion (c) is a well-known characterization of the class d (see Donoghue ). As for the last claim, when A e d inequality (2.6) means that in the representation for A in part (a) the points x 1 ..... x" e R' are distinct. Hence we need only recall that (e -allx'-x:ll') is strictly positive definite because of the formula (2.7) e -azllx'-xqlz/2= (2~r)-S/2[ e i ..... "'e-' ........ /e-Ilxl12/2 dx ,J Rs and the linear independence of e i''l . . . . . e ixx", x ~ R s 9 We will have occasion to use several subclasses of ~r First, the distance matrices in d, i.e., matrices with zero diagonal entries, will be denoted by d J. We let d + be the matrices in d with nonnegative entries and ~r be the subclass of ~r for which part (c) of Corollary 2.1 holds. With these facts at hand, we are now ready to prove Theorems 2.1-2.3. 3. Proofs We begin with Lemma 3.1. f n i Ei=lCip(x ) = 0 for allp ~ ~rk_l(R s) then (--1)k~ ~c,cjllx'--xJll 2.>-0, and (3.3) ~ CiCjq( [IX i -- X Jl] 2 ) --- 0, q ~ ~r k (R 1) j=l i~l are equivalent. This remark leads us to our final definition. (3.1) i=l j=l where equality holds in (3.1) if and only if tl (3.2) ~ ciP(X i) = O, p ~ ~rk(R" ). i=l Remark 3.1. Applying Lemma 3.1 inductively, we see that the conditions t! Ec, p(xi)=o, i=1 16 C.A. Micchelli Definition 3.1. For any class rill of symmetric matrices with nonnegative entries we let o~t,(J! ) denote all continuous functions F(t), t ~ [0, oo), such that for any n n A ~ Jtt, E,=lY.j=lcicjF(Aij) > 0 whenever c.c.qIA..) = o for a, q ".I{RI) i=1 j=l Remark 3.2. The proofs we present below show also that F ~ o~k(ag+ ) whenever both F is continuous on [0, oo) and (-1)~F (k) is completely monotonic on (0, oe). In addition, for functions F satisfying the hypothesis of Theorem 2.3 we have that (-1)"-ldet F(A~j)> 0 for any A ~ ag'. These statements are useful reformulations of Theorems 2.1 and 2.3, as it is sometimes easier to generate matrices in ~r directly than to express them as the square of mutual Euclidean distances between vectors. Let us now prove Lemma 3.1. Proof. First note that ~ cicjllx'-xJlL 2k i=1 j=l = k ~ c,c,(llxql 2 + Ilxql 2- 2(x', x;)) k i=1 ./=1 = ~c,cjE E (-1)'2' i 2,, i 2,~ x0'. i=1~=1 /=0,1+,==k-1 ~llxll II2 II (x', Using our hypothesis on (c 1 ..... c,) we see that whenever 2q + l < k or 2t 2 + l < k the corresponding summand above is zero. Since 2t 1 + 2t 2 + 2l = 2k we must therefore have 2t a + l = 2t 2 + l = k for the nonzero summands. Hence (3.4) = (-1) k qcj E 2' (k ?1)/2 Ilxqlk-qlxJllk-Z(x" xj)t i=1 j=l I=0 k - / = even =(-1) ~ E 2' (k-1//2 I=O k - l= even x ~ c&llxqlk-qlxJll '-z ~ (x')"(xJ) ~ i=1 j=l lal=l = ,~o E l c, llxql~-'(x') ~ , k - 1 = even which proves the inequality (3.1). Furthermore, when (3.2) holds each summand in (3.4) is zero, because ilxllk-//" is a polynomial of degree < k, since lal = 1 < k Interpolation of Scattered Data 17 and k- l is even. Conversely, if the quadratic form (3.1) summands in (3.4) corresponding to I = k give n Y', ci(xi) ~ = 0, all lal = k, i=l so that is zero, then the c,q(x ~) = 0 i=1 for all homogeneous polynomials q of degree k. Thus (3.2) follows and the proof is complete. 9 This lemma states that F(t) = (-1)It t is in 9~ k for any l < k, a special case of Theorem 2.1. Next we prove the general case. Proof of Theorem 2.1. Suppose (-1)kF (k~ is completely monotonic on (0, ~). Then by a theorem of Bernstein (see Widder ), (3.5) (--1)kF~k)(t)=f~176 t>0, "0 where dtt(o) is a Borel measure on (0, ~). For e > 0, we define F~(t) = F(t + e), so that using (3.5), we get k-1 F(1)(O) fo e t'= -- e -'~ ~, I dl~(o). F,(t)- /=0E l-----~, ok ~ t=O l! ) Setting t = Ilx i - xJll 2 above, multiplying both sides by cicj, and summing over i, j, Lemma 3.1 leads to the equation . . = f e-O.o-, E " cicje_,,...,lt, ~ E c~cjF([Ix'- xJ[I 2 + e) o~ E dt~(~ j=l i=1 "0 i=1 j=l whenever ZT=lcip(x i) = O, p ~ 7rk_l(R s) and e > 0. Hence letting e --, 0 +, we conclude F ~ ~k because of Eq. (2.7). 9 Remark 3.3. When x 1 ..... x" are distinct, the measure d/,t in (3.5) has mass other than at the origin; i.e., if F qi qrk(R 1) and F (j), j = 0, 1,..., k - 1, are continu- ous on [0, ~), then F is strictly conditionally positive definite of order k. A converse of Theorem 2.1 can be obtained from arguments given by Schoen- berg for the case k = 0. In the general case, we need the representation for an F in 9~ k given by Gelfand and Vilenkin [13, Chapter II]. For s sufficiently large, we can successively differentiate this equation, by letting s ~ oo and using the asymptotic properties of Bessel functions of large order. Since this is not our main concern here, we do not elaborate on these details. Instead, we give a short proof, based on Corollary 2.1, of a converse for k = 1. Thus suppose F ~ 9~1; then by part (c) of Corollary 2.1 and Schoenberg's theorem we conclude that G(t) = e '~F(t) is completely monotonic for a > 0. However, for l > 1, G(t)(t) = aFU)(t) + O(a2), so the desired conclusion follows. 18 c.A. Micchelli Finally, we observe that the method used above gives also Corollary 3.1 (a) If A ~ ear + and F is completely monotonic, then (F(Aey)) is positive definite. (b) A ~ ~r if and only if Aij > 0 and ((1 + XAij) -1) is positive definite for all X>0. Proof. The first statement follows from the Bernstein representation (3.5) (when k = Q) and part (c) of Corollary 2.1. Thus we see that A ~ .~r implies ((1 + XA~/) -1) is positive definite, because (1 + M) -1 is completely monotonic on [0, ~) for )k > 0. Conversely, if ((1 + XAq) -t) is positive definite for all X > 0, then by Schur's theorem, so is ([1 + (X/n)Aifl-") for all n (see Donoghue ). Letting n ---, ~ we get that (e-XA'0 is also positive definite, which by Corollary 2.1, part (c), gives A ~ .~r thereby completing the proof. 9 Let us also observe that part (a) is best possible in the following sense. Suppose A is a symmetric matrix with nonnegative entries such that (F(Aq)) is positive definite for any completely monotonic F. Then A ~ ~r because we can choose F(t) = e -xt. Also, given a continuous F such that F(A~:) is positive definite for all A ~ ~r then F is completely monotonic. This is just Schoenberg's theorem, because our hypothesis implies F(llx i - xJll 2) is positive definite on any R' and so F is completely monotonic. Proof of Theorem 2.2. We base the proof of Theorem 2.2 on Williamson's representation formula for any function F satisfying the hypothesis of the theorem. Namely, we have (-1)kF(k)(t)= fo~176 dt~(o), t>0, where, as in (3.5), dlt(a) is a Borel measure on (0, o~). Following the proof of Theorem 2.1, we see that it suffices for the proof of Theorem 2.2 to show that the matrix ((1 -]Ix ~- xYl[)l+ +~-1) is positive definite for x 1,..., x n ~ R'. Askey pointed out that ((1 - IIx i - xJllS)+) is positive definite on R s when > [s/2] + 1. For the best lower bound of (s + 1)/2, see (6.9) of Gasper [121 and references therein. 9 Next we prove Theorem 2.3. Proof of Theorem 2.3. The condition on F and the method of proof used for Theorem 2.1 shows that the matrix Aq = F(llx i - xSll 2) is in d, and in fact its r n t/ n ~ ~ quadratic fo m Y'.i=l~'~j=lCiCjAij is zero for ~i=lCi 0 if and only if c i 07 i = 1,..., n. Since we also have Y'in=tY~=lAij > 0, the theorem follows from the elementary l_emma 3.2. Let A be a real symmetric n • n matrix such that (i) ~7_lE",_xcic,Ai, < 0 whenever ~2~=1c i = 0 with equality if and only if c i = -- j-- j J 9 . . ~ C n ~" 07 Interpolation of Scattered Data 19 (ii) There is some vector d ~ R" such that ~ d,djA,j > O. i=1 j=l Then A has one positive and n - 1 negative eigenvalues. Generally, if A is negative definite on a subspace of codimension m, for instance when Aij= -F(llxi- xJl[2 ), m = dim%_l(RS), and F ~k(Rs), then A has at least n - m negative eigenvalues. Let us now turn to some examples. 4. Examples Several important examples come from the function F(t) = 1/(r 2 + 0% r > 0, a > 0. This function is clearly completely monotonic on (0, oo). Hence the matrix (4.1) ((r 2 + IIx'- xq[2) -~) is strictly positive definite for all r > 0 and a > 0. Furthermore, recalling that any time F(l[x - yll 2) is positive definite we must have IF(t)l < F(0), t > 0, we see that a > 0 is also necessary for (4.1) to be positive definite. The positive definiteness of 1/(1 + Ilxl12) ~ on R s for a > s/2 can also be seen in the following way. The Fourier transform of this function can be computed explicitly: e ix'.v (2rr) ~/2 /.~ 1 fiRS( 1 "[" xi- xJN 2'k-'~) "'" i=1 j=l _ lr(a) "Of~176 d t ' i = l j=l This formula is independent of the "base" space R s in which x lies. Alterna- tively, using Riesz potentials we get for a > s/2 and a = s/2 q! Z § the identity (4.5) when Z~=lcip(x i) = O, p ~ ~k_l(RS), and k > a - s/2. Hence, choosing a = + s/2, we see that ( - 1)kt" is strictly conditionally positive definite of order k for k - 1 < ~" < k, which already follows from (4.4). When k = 1, this result was observed by von Neumann and Schoenberg for 9 = 89 Furthermore, by Corollary 2.1 and the fact that (11 xi - xJl127) ~ ~r for 0 < ~" < 1, we see that (1 + Ilxll')-~ is strictly positive definite for/z > 0, 0 < ,r < 2. For the case s = 1, /z = 1, see Linnik . Following the advice of Remark 3.2, we are ready to give a stronger version of inequality (1.1). Since (11 xi - xJll 2") ~ d' for 0 < T < 1, we get (-1)n-ldet (r 2 + IIx i- xJll2") 8 > 0 Interpolation of Scattered Data 21 whenever r > 0, 0 < ~ < 1, 0 < ~" < 1, and x x ..... x" are distinct vectors in R ~. Also using Corollary 2.1, we obtain the well-known fact that e -xllx-.vll', 0 < "r < 2, is strictly positive definite for any k > 0. The function e -Mlxl7 is the characteristic function of a stable law (see Luckas [18, p. 36]). For ~" = 2, the Gaussian case, this kernel is used in Schagen for scattered data interpolation; they are also discussed by Agterberg . Finally, we remark that the following formula can easily be verified: (2,,) ~/2 (a) "+~ ,, ,, K c, cjllx'- xJll 2" log IIx'- xJII 2~-tF(a) 2"n! i=1 i=1 n I 2a =f (~.,cteiXY /,,yl, idY R'~l/=l / ] when KT=lQp(x t) = 0, p c ~r,,(R'), n = a- s/2. This equation can be com- pared to 1 ~. cic/Ix i xJll 2(k-~l log IIx i - xJll 2 (k 1)! ;=1j=1 =f0~( ~-~i=1 j=l~cicje-tllx'-xql~/(--t)k) dt when K'/=lcip(x i) = 0, p ~ r,_l(R' ), k > 1, which comes from (4.4) by using the expansion IIx -Yll 2" = 1 + 2a log IIx -Yll + O(a2). Similarly, we have - s I1 + Hx' - xJll / i=lj=l =fo ~176 ~ ~cicje-llx'-xJll2'e-tdt i=1 j=l t n if Ei=lCi = O, References 1. M. Abramowitz, I. A. Stegun (1966): Handbook of Mathematical Functions (National Bureau of Standards Applied Mathematics Series 55). Washington, D.C.: National Bureau of Standards. 2. F.P. Agterberg (1974): Geomathematics. Amsterdam: Elsevier. 3. N. Aronszajn (1950): Theory of reproducing kernels. Trans. Amer. Math. Soc., 68:337-404. 4. R. Askey (1973): Radial Characteristic Functions (MRC Technical Sum: Report no. 1262). University of Wisconsin. 5. R. Barnhill, S. Stead (1983): Multistage trioariate surfaces. Preprint. 6. Blumenthal (1970): Theory and Application of Distance Geometry. New York: Chelsea Publish- ing. 7. W. F. Donoghue (1974): Monotone Matrix Functions and Analytic Continuation. Berlin Heidelberg New York: Springer-Verlag. 8. L Duckon (1976): Splines minimizing rotation-invariant semi-norms in Sobolev spaces. In: Constructive Theory of Functions of Several Variables (W. Schempp, K. Zeller, eds.). Berlin Heidelberg New York: Springer-Verlag. 22 C.A. Micchelli 9. N. Dyn, D. Levin, S. Rippa (1983): Numerical procedures for global surface smoothing of noisy scattered data. Preprint. 10. R. Franke (1982): Scattered data interpolation: tests of some methods. Math. Comp., 38:381-200. 11. R. Franke (1983): Lecture notes on global basis function methods for scattered data. International Symposium on Surface Approximation, University of Milano, Govgano, Italy. 12. G. Gasper (1975): Positivity and special functions. In: The Theory and Application of Special Functions (R. Askey, ed.). New York: Academic Press. 13. I. M. Gelfand, N. Ya. Vilenkin (1964): Generalized Functions, Vol. 4. New York: Academic Press. 14. R.L. Hardy (1971): Multiquadric equations of topography and other irregular surfaces. J. Geophys. Res., C. 15. R.L. Hardy (1982): Surface fitting with biharmonic and harmonic models. In: Proceedings of the NASA Workshop on Surface Fitting. College Station, Texas: Center for Approximation Theory, Texas A & M University, pp. 136-146. 16. S. Helgason (1980): The Radon Transform. Basel: Birkh~iuser. 17. J.V. Linnik (1953): Linear forms and statical criteria, H. UkraJn. Mat. Zh., 51:247-290 [also (1962): Selected Translations in Mathematical Statistics and Probability, Vol. 5. Providence: American Mathematical Society, pp. 41-90]. 18. E. Luckas (1970): Characteristic Functions. New York: Hafner. 19. G. Matheron (1973): The intrinsic random functions and their applications. Adv. in Appl. Probab., 5: 439-468. 20. G. Matheron (1981): Splines and kriging: their formal equivalence. In: Syracuse University Geology Contribution 8 (D. F. Marriam, ed.). Syracuse, New York: Department of Geology, Syracuse University. 21. J. Meinguet (1979): An intrinsic approach to multivariate spline interpolation at arbitrary points. In: Polynomial and Splines Approximation (B. N. Sahney, ed 0. Dordrecht: D. Reidel, pp. 163-190. 22. J. von Neumann, I. J. Schoenberg (1941): Fourier integrals and metric geometry. Trans. Amer. Math. Soc., 50:226-251. 23. K. Salkauskas (1982): Some relationships between surface splines and kriging. In: Multivariate Approximation Theory II (W. Schempp, K. Zeller, eds.). Basel: Birkh'~mser, pp. 313-325. 24. I. P. Schagen (1979): Interpolation in two dimensions--a new technique. I. Inst. Math. Appl., 23:53-59. 25. I.J. Schoenberg (1935): Remarks to Maurice Frechet's article "Sur la definition axiomatique d'une classe d'espace distancies vectoriellemment applicable sur l'espace de Hilbert." Ann. of Math., 36:724-732. 26. I.J. Schoenberg (1938): Metric spaces and positive definite functions. Trans. Amer. Math. Soc., 44:522-536. 27. I. J. Schoenberg (1938): Metric spaces and completely monotone functions. Ann. of Math., 39:811-841. 28. J. Stewart (1976): Positive definite functions and generalizations, an historical survey. Rocky Mountain J. Math., 6:409-434. 29. Wahba (1982): Private communication. 30. G.N. Watson (1966): Theory of Bessel Functions, 2nd ed. Cambridge: Cambridge University Press. 31. D.V. Widder (1946): The Laplace Transform. Princeton: Princeton University Press. 32. R.E. Williamson (1956): Multiply monotone functions and their Laplace transform. Duke Math. J., 23:189-207. C. A. Micchelli IBM Thomas J. Watson Research Center Yorktown Heights, New York 10598 |
187610 | https://www.cuemath.com/algebra/square-1-to-20/ | LearnPracticeDownload
Square 1 to 20
Square 1 to 20 is the list of squares of all the numbers from 1 to 20. The value of squares from 1 to 20 ranges from 1 to 400. Memorizing these values will help students to simplify the time-consuming equations quickly. The Square 1 to 20 in the exponential form is expressed as (x)2.
Square 1 to 20:
Exponent form: (x)2
Highest Value: 202 = 400
Lowest Value: 12 = 1
| | |
--- |
| 1. | Square 1 to 20 |
| 2. | Square 1 to 20 PDF |
| 3. | How to Calculate Square 1 to 20? |
| 4. | FAQs |
Squares 1 to 20 Chart
Squares from 1 to 20
Learning squares 1 to 20 can help students to recognize all perfect squares from 1 to 400 and approximate a square root by interpolating between known squares. The values of squares 1 to 20 are listed in the table below.
| List of All Squares from 1 to 20 | |
--- |
| 12 = 1 | 22 = 4 |
| 32 = 9 | 42 = 16 |
| 52 = 25 | 62 = 36 |
| 72 = 49 | 82 = 64 |
| 92 = 81 | 102 = 100 |
| 112 = 121 | 122 = 144 |
| 132 = 169 | 142 = 196 |
| 152 = 225 | 162 = 256 |
| 172 = 289 | 182 = 324 |
| 192 = 361 | 202 = 400 |
☛ Square 1 to 20 PDF
The students are advised to memorize these squares 1 to 20 values thoroughly for faster math calculations.
Square 1 to 20 - Even Numbers
The table below shows the values of squares 1 to 20 for even numbers.
| | |
--- |
| 22 = 4 | 42 = 16 |
| 62 = 36 | 82 = 64 |
| 102 = 100 | 122 = 144 |
| 142 = 196 | 162 = 256 |
| 182 = 324 | 202 = 400 |
Square 1 to 20 - Odd Numbers
The table below shows the values of 1 to 20 square for odd numbers.
| | |
--- |
| 12 = 1 | 32 = 9 |
| 52 = 25 | 72 = 49 |
| 92 = 81 | 112 = 121 |
| 132 = 169 | 152 = 225 |
| 172 = 289 | 192 = 361 |
How to Calculate the Values of Squares 1 to 20?
In order to calculate the squares from 1 to 20, we can use any one of the following methods:
Method 1: Multiplication by itself: For example, the square of 11 = 11 × 11 = 121. Here, the resultant product “121” gives us the square of the number “11.”
Method 2: Using basic algebraic identities: For example, to find the square of 18, we can express 18 as:
(10 + 8)
(20 - 2)
In the next step, we use the basic algebraic identity formula and get Option 1: [10² + 8² + (2 × 10 × 8)] or Option 2: [20² + 2² - (2 × 20 × 2)]. Solving the expressions further, we get Option 1: (100 + 64 + 160) = 324 or Option 2: (400 + 4 - 80) = 324.
Read More
Examples on Square 1 to 20
Example 1: If a circular tabletop has a radius of 9 inches. Find the area of the tabletop in sq. inches?
Solution:
Area of circular tabletop = πr2 = π (9)2
Using values from square 1 to 20 chart;
i.e. A = 81π
Therefore, the area of the tabletop = 254.34 inches2.
2. Example 2: Find the area of a square window whose side length is 18 inches
Solution:
Area of square window (A) = Side2
i.e. A = 182 = 324
Therefore, the area of a square window is 324 inches2.
3. Example 3: Two square wooden planks have sides 8m and 14m respectively. Find the combined area of both the wooden planks?
Solution:
Area of wooden plank = (side)2
⇒ Area of 1st wooden plank = 82 = 64 m2
⇒ Area of 2nd wooden plank = 142 = 196 m2
Therefore, the combined area of wooden planks is 64 + 196 = 260 m2
4. Example 4: Find the sum of the first 20 odd numbers.
Solution:
The sum of first n odd numbers is given as n²
⇒ Sum of first 20 odd numbers (n) = 20²
Using values from squares from 1 to 20 chart, the sum of first 20 odd numbers = 20² = 400
Show Solution >
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FAQs on Square 1 to 20
What is the Value of Square 1 to 20?
The value of square 1 to 20 is the list of numbers obtained by multiplying an integer (1 - 20) by itself. It will always be a positive number. Between 1 to 20, the numbers 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 are even numbers and 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 are odd numbers.
What are the Methods to Calculate Squares from 1 to 20?
We can calculate the square of a number by using the a² + b² + 2ab formula. For example (17)² can be calculated by splitting 17 into 10 and 7. Other methods that can be used to calculate squares from 1 to 20:
Finding Square by Column Method
Finding Squares by Diagonal Method
If You Take Squares from 1 to 20, How Many of Them Will be Even Numbers?
The even numbers between 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. Since the squares of even numbers are always even. Therefore, the value of squares of numbers 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 will be even.
Using Squares 1 to 20 Chart, Find the Value of 1 Plus 5 Square Plus 7 Square
The value of 5² is 25 and 7² is 49. So, 1 + 5² + 7² = 75. Hence, the value of 1 plus 5 square plus 7 square is 75.
How Many Numbers in Squares 1 to 20 are Odd?
The odd numbers between 1 to 20 are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. Since the squares of odd numbers are always odd. Therefore, the value of squares of numbers 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 will be odd.
What is the Sum of all Perfect Squares from 1 to 20?
The sum of all perfect squares from 1 to 20 is 30 i.e. 1 + 4 + 9 + 16 = 30.
What Values of Squares from 1 to 20 are Between 1 and 50?
The values of squares 1 to 20 between 1 and 50 are 1² (1), 2² (4), 3² (9), 4² (16), 5² (25), 6² (36), and 7² (49).
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187611 | https://www.nature.com/articles/ja2016121 | Skip to main content
Review Article
Published:
Review Article
Production of valuable compounds by molds and yeasts
Arnold L Demain1 &
Evan Martens2
The Journal of Antibiotics
volume 70, pages 347–360 (2017)Cite this article
30k Accesses
66 Citations
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Abstract
We are pleased to dedicate this paper to Dr Julian E Davies. Julian is a giant among microbial biochemists. He began his professional career as an organic chemistry PhD student at Nottingham University, moved on to a postdoctoral fellowship at Columbia University, then became a lecturer at the University of Manchester, followed by a fellowship in microbial biochemistry at Harvard Medical School. In 1965, he studied genetics at the Pasteur Institute, and 2 years later joined the University of Wisconsin in the Department of Biochemistry. He later became part of Biogen as Research Director and then President. After Biogen, Julian became Chair of the Department of Microbiology at the University of British Columbia in Vancouver, Canada, where he has contributed in a major way to the reputation of this department for many years. He also served as an Adjunct Professor at the University of Geneva. Among Julian’s areas of study and accomplishment are fungal toxins including α-sarcin, chemical synthesis of triterpenes, mode of action of streptomycin and other aminoglycoside antibiotics, biochemical mechanisms of antibiotic resistance in clinical isolates of bacteria harboring resistance plasmids, their origins and evolution, secondary metabolism of microorganisms, structure and function of bacterial ribosomes, antibiotic resistance mutations in yeast ribosomes, cloning of resistance genes from an antibiotic-producing microbe, gene cloning for industrial purposes, engineering of herbicide resistance in useful crops, bleomycin-resistance gene in clinical isolates of Staphylococcus aureus and many other topics. He has been an excellent teacher, lecturing in both English and French around the world, and has organized international courses. Julian has also served on the NIH study sections, as Editor for several international journals, and was one of the founders of the journal Plasmid. We expect the impact of Julian’s accomplishments to continue into the future.
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Introduction
Microbes have contributed significantly to improving the health and well-being of humans. The natural products that they have yielded have not only helped eradicate disease and alleviate suffering, but also greatly increased the average life expectancy. The first major contribution of microbes began back in 1928, when Alexander Fleming discovered in a Petri dish seeded with Staphylococcus aureus that a compound produced by a mold killed the bacterium. The mold, Penicillium notatum, produced an active agent that was named penicillin. Fleming’s discovery began the microbial drug era. By using the same method, other naturally occurring substances, like chloramphenicol and streptomycin, were later isolated from bacterial fermentations. Naturally occurring antibiotics are produced by fermentation, an old technique that can be traced back almost 8000 years, initially for beer and wine production, and recorded in the written history of ancient Egypt and Mesopotamia. During the past 4000 years, Penicillium roqueforti has been utilized for cheese production, and for the past 3000 years, soy sauce in Asia and bread in Egypt represented examples of traditional fermentations.1
Natural products from microbes have a broad range of therapeutic applications and are often produced via primary or secondary metabolism. Because of technical improvements in screening programs and separation and isolation techniques, the number of natural compounds discovered exceeds one million.2 Among them, 50–60% are produced by plants (alkaloids, flavonoids, terpenoids, steroids, carbohydrates, etc.) and 5% of these plant products have a microbial origin. From all the reported natural products, ∼20–25% show biological activity and, of these, ∼10% have been obtained from microbes. Microorganisms produce many compounds with biological activity. From the 22 500 biologically active compounds so far obtained from microbes, ∼40% are produced by fungi.2, 3 The role of fungi in the production of antibiotics and other drugs for treatment of noninfective diseases has been dramatic.4
Biosynthetic genes are present in clusters coding for large, multidomain and multimodular enzymes. Some examples of these enzymes include polyketide synthases, prenyltransferases, nonribosomal peptide synthases and terpene cyclases. Genes adjacent to the biosynthetic gene clusters encode regulatory proteins, oxidases, hydroxylases and transporters. Aspergilli usually contain 30–40 secondary metabolite gene clusters. Strategies to activate silent genes have been reviewed by Brakhage and Schroekh.3
Given that the vast majority of microbes in nature have yet to be cultured (~99%), there have been major advances in isolating and growing different microbial species.5 Furthermore, metagenomics—that is, the extraction of DNA from soil, plants and marine habitats and its incorporation into known organisms—is allowing access to a vast untapped reservoir of genetic and metabolic diversity.6, 7 The potential for discovery of new secondary metabolites with beneficial use for humans is great. A method to predict secondary metabolite gene clusters in filamentous fungi has recently been devised.8
Interestingly, microbial production of secondary metabolites is limited to a very low level by certain regulatory mechanisms. Despite this, the extent of such production is sufficient for the microbe to compete with other organisms or maintain a commensal/mutual relationship with other species. The industrial microbiologist, however, desires a strain that will overproduce the molecule of interest. Development of higher-producing strains involves mutagenesis and, more recently, recombinant DNA technologies.9 Although some metabolites of interest can be made by plants or animals, or by chemical synthesis, the recombinant microbe is usually the ‘creature of choice’. Thousand-fold increases in production of small molecules have been obtained by mutagenesis and/or genetic engineering. The use of genome mining to discover new fungal natural products has been reviewed by Wiemann and Keller.10 Other important parts of industrial production include creating a proper nutritional environment for the organism to grow and produce its product, and the avoidance of negative effects such as inhibition and/or repression by carbon, nitrogen and phosphorus sources, metals and the final product itself. Avoidance of enzyme decay is also desired.4, 11
Broad use of secondary metabolites produced by fungi
Given the diverse array of secondary metabolites that fungi are capable of producing, the pharmaceutical industry began to focus their efforts on the screening of compounds for indications other than anti-infectives.12, 13 As microorganisms are such a prolific source of structurally diverse bioactive metabolites, the industry extended their screening programs in order to look for microbes with activity in other disease areas. As a result of this move, some of the most important products of the pharmaceutical industry were obtained. For example, the immunosuppressants have revolutionized medicine by facilitating organ transplantation.14 Other products include antitumor drugs, hypocholesterolemic drugs, enzyme inhibitors, gastrointestinal motor stimulator agents, ruminant growth stimulants, insecticides, herbicides and antiparasitics versus coccidia and helminths.
In the past, the treatment of noninfectious disease relied heavily upon synthetic compounds, yet only a select few turned out to be promising. As new synthetic lead compounds became extremely difficult to find, microbial products came into play. Poor or toxic antibiotics produced by fungi such as cyclosporin A, or mycotoxins such as ergot alkaloids, gibberellins and zearelanone, were then successfully applied in medicine and agriculture. This led to the use of fungal products as immunosuppressive agents, hypocholesterolemic drugs and antitumor agents and for other applications.
Anti-rejection drugs (agents that suppress the immune system)
The immune system is our body’s main defense against foreign antigens and pathogenic microorganisms. However, it is essential that the immune system recognizes ‘native’ antigens in order to avoid launching an immune response. Suppressor cells are critical in the regulation of the normal immune response. The suppression of the immune response, either by drugs or radiation, in order to prevent the rejection of grafts or transplants or to control autoimmune diseases, is called immunosuppression.
Secondary metabolites produced by fungi have yielded compounds that function as immunosuppressants. Cyclosporin A was originally discovered in the 1970s as a narrow-spectrum antifungal peptide produced by the mold, Tolypocladium nivenum (previously Tolypocladium inflatum) in an aerobic fermentation.15 Cyclosporins are a family of neutral, highly lipophilic, cyclic undecapeptides containing some unusual amino acids, synthesized by a nonribosomal peptide synthetase, cyclosporin synthetase. Discovery of the immunosuppressive activity of this secondary metabolite led to its use in heart, liver and kidney transplants and to the overwhelming success of the organ transplant field.16 Cyclosporin was approved for use in 1983. It is thought to bind to the cytosolic protein cyclophilin (immunophilin) of immunocompetent lymphocytes, especially T lymphocytes. This complex of cyclosporin and cyclophilin inhibits calcineurin that under normal circumstances is responsible for activating the transcription of interleukin-2. It also inhibits lymphokine production and interleukin release and therefore leads to a reduced function of effector T cells. Annual world sales of cyclosporin A are ∼$2 billion. Cyclosporin A also has activity against coronaviruses.17
Studies on the mode of action of cyclosporin, and the later developed immunosuppressants from actinomycetes, such as sirolimus (a rapamycin) and FK-506 (tacrolimus), have markedly expanded current knowledge of T-cell activation and proliferation. These agents act by interacting with an intracellular protein (an immunophilin), thus forming a novel complex that selectively disrupts the signal transduction events of lymphocyte activation. Their targets are inhibitors of signal transduction cascades in microbes and humans. In humans, the signal transduction pathway is required for activation of T cells. Fingolimod (FTY720), another immunosuppressant, was approved by the US Food and Drug Administration (FDA) in 2010, specifically for relapsing forms of multiple sclerosis. The drug, initially discovered by Professors Fujita, Yoshitomi and Taito in collaboration, is a derivative of myriocin isolated from the fungus Isaria sinclairii. The annual world sales of fingolimod are ∼$2.7 billion.
One of the first antibiotics to be discovered, with a broad spectrum of activity, was mycophenolic acid. Bartolomeo Gosio (1863–1944), an Italian physician, discovered the compound in 1893.18 Gosio isolated a fungus from spoiled corn that he named Penicillium glaucum, and that was later reclassified as Penicillium brevicompactum. He isolated crystals of the compound from culture filtrates in 1896 and found it to inhibit growth of Bacillus anthracis. This was the first time an antibiotic had been crystallized and the first time that a pure compound had ever been shown to have antibiotic activity. The work was forgotten but fortunately the compound was rediscovered by Alsberg and Black19 and given the name mycophenolic acid. They used a strain originally isolated from spoiled corn in Italy called Penicillium stoloniferum, a synonym of P. brevicompactum. The chemical structure was elucidated many years later (1952) by Birkinshaw et al.20 in England. Mycophenolic acid has antibacterial, antifungal, antiviral, antitumor, antipsoriasis and immunosuppressive activities. Its antiviral activity is exerted against yellow fever, dengue virus and Japanese encephalitis virus.21 It was never commercialized as an antibiotic because of its toxicity, but its 2-morpholinoethylester was approved as a new immunosuppressant for kidney transplantation in 1995 and for heart transplants in 1998.22 The ester is called mycophenolate mofetil (CellCept) and is a prodrug that is hydrolyzed to mycophenolic acid in the body. It is sometimes used along with cyclosporin in kidney, liver and heart transplants. Furthermore, the mycophenolic acid delayed-release tablet was approved in 2004 by the FDA as an antimetabolite immunosuppressant indicated for prophylaxis of organ rejection in kidney transplant. This delayed-release tablet is called Myfortic and is produced by Novartis, with annual sales of $637 million in 2013, $543 million in 2014 and $441 million in 2015. In addition, mycophenolic acid appears to have anti-angiogenic activity.23
Agents that block enzyme activity
Drugs that are enzyme inhibitors may provide key functions not only for treating human disease, but also in agriculture, enzyme structure elucidation and reaction mechanisms. Several enzyme inhibitors with various industrial uses have been isolated from microbes.24 Among the most important are the statins and hypocholesterolemic drugs discussed below. Fungal products are also used as enzyme inhibitors against cancer, diabetes, poisoning and Alzheimer’s disease. The enzymes inhibited include acetylcholinesterase, protein kinase, tyrosine kinase, glycosidases and others.25
Cholesterol-lowering agents
In humans, it is estimated that ∼30% of cholesterol originates from the diet, whereas the remaining 70% is synthesized primarily in the liver. Many people cannot control their level of cholesterol at a healthy level by diet alone and require hypocholesterolemic agents. High blood cholesterol leads to atherosclerosis, a chronic, progressive disease characterized by continuous accumulation of atheromatous plaque within the arterial wall, causing stenosis and ischemia. Atherosclerosis is a leading cause of human death. The past two decades have witnessed the introduction of a variety of anti-atherosclerotic therapies. The statins form a class of hypolipidemic drugs, formed as secondary metabolites by fungi, and are used to lower cholesterol by inhibiting the rate-limiting enzyme of the mevalonate pathway of cholesterol biosynthesis; that is, 3-hydroxymethyl glutaryl-CoA reductase. Inhibition of this enzyme in the liver stimulates low-density lipoprotein receptors, resulting in an increased clearance of low-density lipoprotein from the bloodstream and a decrease in blood cholesterol levels. They can reduce total plasma cholesterol by 20–40%. Through their cholesterol-lowering effect, they reduce the risk of cardiovascular disease, prevent stroke and reduce the development of peripheral vascular disease.26
Statins, which had reached an annual market of nearly $30 billion before one became a generic drug, are widely used in clinical practice. The history of the statins has been described by Akira Endo, the discoverer of the first statin, compactin (mevastatin; ML-236B).27 This first member of the group was isolated as an antibiotic product of P. brevicompactum.28 At about the same time, it was found by Endo et al.29 as a cholesterolemic product of Penicillium citrinum. Although compactin was not of commercial importance, its derivatives achieved strong medical and commercial success. Lovastatin (monacolin K; mevinolin; Mevacor) was isolated in broths of Monascus rubra and Aspergillus terreus.30, 31 Lovastatin, developed by Merck and approved by the FDA in 1987, was the first commercially marketed statin. In its chemical structure, lovastatin has a hexahydronaphthalene skeleton substituted with a p-hydroxy-lactone moiety (Figure 1).
A semisynthetic derivative of lovastatin is Zocor (simvastatin), one of the main hypocholesterolemic drugs, selling for $7 billion per year before becoming generic. An unexpected effect of simvastatin is its beneficial activity on pulmonary artery hypertension.32 Another surprising effect is its antiviral activity.33 Simvastatin is active against RNA viruses and acts as monotherapy against chronic hepatitis C virus in humans. It has been shown to act in vitro against hepatitis B virus. This virus infects 400 million people and is the most common infectious disease agent in the world. The virus causes hepatocellular cancer, the leading cause of cancer death. Nucleotide analogs (lamivudine, adefovir, tenofovir, entecavir, telbuvidine) were approved for hepatitis B virus infections but they only work on 11–17% of patients. Simvastatin is synergistic with these nucleotide analogs.
Statins also have antithrombotic, anti-inflammatory and antioxidant effects.34 They have shown activity against multiple sclerosis, atherosclerosis, Alzheimer’s Disease and ischemic stroke.35, 36 However, these applications have not yet been approved as more clinical studies are required. The neuroprotective effect of statins has been demonstrated in an in vitro model of Alzheimer’s disease using primary cultures of cortical neurons.37 The effect did not appear to be because of cholesterol lowering but rather reduction in formation of isoprenyl intermediates of the cholesterol biosynthetic process. Lovastatin has shown antitumor activity against embryonal carcinoma and neuroblastoma cells.38
Although simvastatin is usually made from lovastatin chemically in a multistep process, an enzymatic/bioconversion process using recombinant Escherichia coli has been developed.39 Another statin, pravastatin (Pravacol) ($3.6 billion in sales per year), is made via different biotransformation processes from compactin by Streptomyces carbophilus40 and Actinomadura sp.41 Both simvastatin and pravastatin are synthetic variants of the naturally occurring lovastatin and compactin. Pravastatin can be produced from compactin but it involves an expensive dual-step fermentation and biotransformation process. Mclean et al.42 reprogrammed Penicillium chrysogenum involving discovery and engineering of an enzyme involved in hydroxylation of compactin. This resulted in a single-step fermentation yielding pravastatin at >6 g l−1.
Other genera involved in production of statins are Doratomyces, Eupenicillium, Gymnoascus, Hypomyces, Paecilomyces, Phoma, Trichoderma and Pleurotus.43 A synthetic compound, modeled from the structure of the natural statins, is Lipitor, the leading drug of the entire pharmaceutical industry in terms of market (∼$14 billion per year) for many years.
Prebiotics
Prebiotics are nondigestible products stimulating growth in the colon of bacteria such as Bifidobacterium bifidum, Lactobacillus acidophilus, Bifidobacterium adolescentis and Faecalibacterium prausnitzii.44 They include galacto-oligosaccharides, fructo-oligosaccharides, lactulose, lactitol and its hydrolysates, malto-oligosaccharides, inulin and resistant starch. Titers are as follows: lactosucrose at 192 g l−1 from lactose or sucrose by levanosucrase from Sterigmatomyces elviae and fructooligosaccharide at 116 g l−1 from sucrose by β-fructofuranosidase from Aspergillus japonicas. Prebiotics are used in the nutraceutical, pharmaceutical, animal feed and aquaculture areas. They stimulate growth of beneficial intestinal bacteria and maintain health of humans by suppression of potentially harmful bacteria, improvement of defecation, elimination of ammonia, prevention of colon cancer, stimulation of mineral adsorption and lowering of cholesterol and lipids.
Food additives functioning as sugar substitutes
Aspergillus niger var. awamori, P. roqueforti and the plant Thaumatococcus danielli are all capable of producing the protein thaumatin.45 Thaumatin is intensely sweet (that is, 3000 times sweeter than sucrose) and is approved as a food-grade ingredient. Production by A. niger var. awamori was improved from 2 mg l−1 up to 14 mg l−1 by increasing gene dosage and use of a strong promoter.46 The sweetener xylitol, normally produced by Pichia stipitis, can be produced by recombinant Saccharomyces cerevisiae in higher concentrations by transforming the XYL1 gene of P. stipitis into S. cerevisiae. The gene encodes a xylose reductase.47
Toxins
Mycotoxins, which are poisons produced by fungi, have actually been useful therapeutic agents for a variety of medical conditions and ailments. These agents (for example, ergot alkaloids) had caused fatal poisoning of humans and animals (ergotism) for centuries by consumption of bread made from grain contaminated with species of the fungus Claviceps. However, mycotoxins later were found useful for angina pectoris, hypertonia, serotonin-related disturbances, inhibition of protein release in agalactorrhea, reduction in bleeding after childbirth and prevention of implantation in early pregnancy.48, 49 Their physiological activities include inhibition of action of adrenalin, noradrenalin and serotonin, as well as the contraction of smooth muscles of the uterus. Antibiotic activity is also possessed by some ergot alkaloids.
Members of the genus Gibberella produce zearelanone and gibberellins. Zearelanone is an estrogen made by Gibberella zeae (syn. Fusarium graminearum).50 Its reduced derivative zeranol is used as an anabolic agent in sheep and cattle that increases growth and feed efficiency. Gibberellic acid, a member of the mycotoxin group known as gibberellins, is a product of Gibberella fujicuroi and causes ‘foolish rice seedling’ disease in rice.51 Gibberellins are employed to speed up the malting of barley, improve the quality of malt, increase the yield of vegetables and cut the time in half for obtaining lettuce and sugar beet seed crops. They are isoprenoid growth regulators, controlling flowering, seed germination and stem elongation.52 More than 25 are produced annually with a market of over $100 billion.
Antineoplastic drugs
In 2008, there were over 12 million new cases of cancer diagnosed throughout the world that resulted in ∼7.6 million deaths. Lung (12.7%), breast (10.9%) and colorectal (9.8%) cancer had the highest incidence rates. Some of the anticancer drugs in clinical use include taxol and camptothecin, the secondary metabolites derived from plants and fungi.
Taxol (paclitaxel) is a fungal secondary metabolite first isolated from the Pacific yew tree, Taxus brevifolia.53, 54 It is a steroidal diterpene alkaloid that has a characteristic N-benzoylphenyl isoserine side chain and a tetracycline ring (Figure 2).
It inhibits rapidly dividing mammalian cancer cells by promoting tubulin polymerization and interfering with normal microtubule breakdown during cell division. The benzoyl group of the molecule is particularly crucial for maintaining the strong bioactivity of taxol. The drug also inhibits several fungi (species of Pythium, Phytophthora and Aphanomyces) by the same mechanism. In 1992, taxol was approved for refractory ovarian cancer and today is used against breast cancer and advanced forms of Kaposi’s sarcoma.55 A formulation in which paclitaxel is bound to albumin is sold under the trademark Abraxane. Taxol sales amounted to $1.6 billion in 2006 for Bristol Myers-Squibb, representing 10% of the company’s pharmaceutical sales and its third largest selling product. It reached $3.7 billion annual sales in international markets.
Although synthetic methods for taxol production have been attempted, the chemical molecular structure is so complex that commercial synthetic production is unfeasible. Currently, Italy, United Kingdom, The Netherlands and other Western countries are engaged in the production of taxol by plant cell fermentation technology. Taxol production by plant cell culture of Taxus sp. was reported to be at 67 mg l−1.56 However, addition of methyl jasmonate, a plant signal transducer, increased production to 110 mg l−1.
As stated previously, taxol has also been found to be a fungal metabolite.54, 57 Fungi such as Taxomyces andreanae, Pestalotiopsis microspora, Tubercularia sp., Phyllosticta citricarpa, Nodulisporium sylviforme, Colletotrichum gloeosporoides, Colletotrichum annutum, Fusarium maire and Pestalotiopsis versicolor produce it.54, 58, 59, 60, 61, 62, 63, 64 The endophyte F. maire produces 225 μg l−1. Production by P. citricarpa amounted to 265 μg l−1.65 Production was reported at 417 μg l−1 by submerged fermentation with an engineered strain of the endophytic fungus Ozonium sp. (EFY-21). The transformed strain overproduced the rate-limiting enzyme of taxol biosynthesis, taxadiene synthase.66 Another endophytic fungus, Phoma betae, isolated from the medicinal tree Ginkgo biloba, produced taxol at 795 μg l−1.67 Cladosporium cladosporoides, an endophyte of the Taxus media tree, produced 800 μg l−1 of taxol.68 Metarhizium anisopiliae H-27, isolated from the tree Taxus chinensis, yielded 846 μg l−1.69 Although a review of taxol production by endophytic fungi indicated that strain improvement had resulted in levels of only 0.4–1.0 mg l−1,70 it was reported that another fungus, Alternaria alternate var. monosporus, from the bark of Taxus yunanensis, after ultraviolet and nitrosoguanidine mutagenesis, could produce taxol at 227 mg l−1.71 The endophytic fungus P. versicolor, from the plant Taxus cuspidata, produced 478 μg l−159 and C. annutum from Capsicum annuum made 687 μg l−1.60
Camptothecin, a modified monoterpene indole alkaloid produced by certain plants (angiosperms) and by the endophytic fungus, Entrophospora infrequens, is another important antitumor agent. The fungus was isolated from the plant Nathapodytes foetida.53 In view of the low concentration of camptothecin in tree roots and poor yield from chemical synthesis, the fungal fermentation is very promising for industrial production of camptothecin. It is used for recurrent colon cancer and has unusual activity against lung, ovarian and uterine cancers.72 Colon cancer is the second leading cause of cancer fatalities in the United States and the third most common cancer among US citizens. Camptothecin is known commercially as Camptosar and Campto and achieved sales of $1 billion in 2003.73 Camptothecin’s water-soluble derivatives irinotecan and topotecan have been approved and are used clinically. Metastatic colorectal cancer is treated by irinotecan, whereas topotecan has use for ovarian cancer, cervical cancer and small-cell lung cancer. A review of the activities of camptothecin and its many small and macromolecular derivatives has been published by Venditto and Simanek.74
Chemically produced derivatives of camptothecin used for cancer include 10-hydroxycamptothecin, topotecan, irinotecan and SN-38. Camptothecin is a multibillion dollar anticancer drug and the fourth largest anticancer drug derived from plants after taxol, vincristine and vinblastine. Camptothecin is produced by the plants Camptotheca acuminata and Nothapodyta foetida. An endophytic fungus producing camptothecin is E. infrequens from the bark of N. foetida. Recently, it was found that Trichoderma atroviridi strain LY357, an endophytic fungus from C. acuminata, was an improved producer of camptothecin. The endophytic fungus produced 142 μg l−1 of camptothecin in the presence of the elicitor methyljasmonate and XAD adsorbant resin.75
Type 1 DNA topoisomerase has been identified as the cellular target of camptothecin. When patients become resistant to irinotecan, its use can be prolonged by combining it with the monoclonal antibody Erbitux (Cetuximab). Erbitux blocks a protein that stimulates tumor growth and the combination helps metastatic colorectal cancer patients expressing epidermal growth factor receptor. This protein is expressed in 80% of advanced metastatic colorectal cancers. The drug combination reduces invasion of normal tissues by tumor cells and the spread of tumors to new areas.
The process by which tumor cells recruit new blood vessels for oxygen and nutrients is known as angiogenesis. Tumors actively secrete growth factors that trigger angiogenesis. Anti-angiogenesis therapy is now known as one of four cancer treatments; the other three are surgery, radiotherapy and chemotherapy. By the end of 2007, 23 anti-angiogenesis drugs were in phase III clinical trials and more than 30 were in phase II. Fumagillin, a secondary metabolite of Aspergillus fumigatus, was one of the first agents found to act as an anti-angiogenesis compound. Next to come along were its oxidation product ovalacin and the fumagillin analog TNP-470 (=AGM-1470). TNP-470 binds to and inhibits type 2 methionine aminopeptidase. This interferes with amino-terminal processing of methionine that may lead to inactivation of enzymes essential for growth of endothelial cells. In animal models, TNP-470 effectively treated many types of tumors and metastases.
Farnesylation, which is required for the activation of Ras, a necessary step in cancer progression, has been a target for inhibitors of farnesyltransferase because of their anticancer activity. They also induce apoptosis in cancer cells. The fungus Phoma sp. FL-415 produces an inhibitor of farnesyltransferase known as TAN-1813.76
Pigments
Carotenoids, a subfamily of terpenoids, are tetra-terpenoid yellow to red pigments that are antioxidants. They are made by plants, fungi, algae and bacteria and used as nutritional supplements and food additives and in cosmetics. They can be differentiated into carotenes that are hydrocarbons, such as lycopenes and β-carotene, and their oxygenated derivatives, that is, xanthophylls, such as lutein, zeaxanthin and astaxanthin.
Production of carotenoids by microbes has been reviewed by Sanchez et al.77 Over 600 carotenoids are known including lycopene, zeaxanthin, astaxanthin, β-carotene, lutein and cantaxanthin. They are used as nutrients, supplements, food ingredients, feed additives, antioxidants, anticancer agents, immune modulators and cosmetic products. Carotenoids had a 2010 market of $1.2 billion.78 They are C40 isoprenoids. Carotenoids are produced on a large scale by fungi such as Dunalliela salina, Xanthophyllomyces dendrorhous (formerly Phaffia rhodozyma) and Blakeslea trispora. β-Carotene is the major carotenoid product with a 2010 market of $261 million. It is mainly produced by Mucor, Phycomyces and B. trispora. B. trispora produces β-carotene at 9 g l−1.
Because of their antioxidant properties and health-related functions, xanthophylls (lutein, zeaxanthin and astaxanthin) sell for multimillion dollars each year. The astaxanthin market is $252 million for fish food and $30 million for human use. It sells for $2500 kg−1 for the synthetic form and $7000 kg−1 for the natural form. X. dendrorhous produces 420 mg l−1 of astaxanthin. Astaxanthin, lycopene, β-carotene and cantaxanthin are used in products such as beverages, dairy foods, cereal products, cosmetics and pharmaceuticals and in aquaculture.
Adaptive laboratory evolution was used to increase microbial production of carotenoids in a genetically engineered S. cerevisiae strain. It was carried out by using a periodic hydrogen peroxide shocking strategy. The improved production was because of upregulation of genes related to biosynthesis of lipid and mevalonate.79 Carotenoid production amounted to 16 mg g−1 dry cell weight.
The main microbe producing carotenes is the fungus B. trispora. Fermentative production is stimulated by oxidative stress induced by butylated hydroxytoluene, enhanced dissolved oxygen levels, iron ions and liquid paraffin.
Lycopene and β-carotene are highly unsaturated isoprene derivatives that are pigments that stimulate the immune system and prevent degenerative diseases and cancer. Carotenes are also effective antioxidants. They are utilized as nutrient supplements, animal feeds, pharmaceuticals and as coloring agents in foods and feeds. Their market is growing at 2.3% per year.80 They are obtained by (1) microbial production, (2) from plants and (3) synthetically. Carotenoids absorb light and in photosynthetic organisms, protect against excess light and prevent formation and reaction of reactive oxygen species. As antioxidants, they protect against oxidative damage elicited by oxidizing agents and free radicals. Astaxanthin is one of the best scavengers of reactive oxygen species, whereas β-carotene is a potent scavenger of reactive nitrogen species.
Monascus purpurea, a mold species, has played an important role in traditional Chinese food and medicine since 800 AD. Specifically, it has been used to prepare popular dishes such as koji or Angkak (red rice).81 Monascorubramine and rubropunctatin are water-soluble red pigments formed upon reaction of the orange pigments monascorubrin and rubropunctatin with amino acids in fermentation media.82 The fungus is used to prepare red rice, wine, soy-bean cheese, meat and fish. It is authorized in Japan and China for food use. There are 54 known Monascus pigments. They have an amazing number of activities: antimicrobial, anticancer, antimutagenesis, antidiabetic, antiobesity, anti-inflammatory, cholesterol lowering, immunosuppressive and hypotensive.83, 84 Nutritional control of the formation of the red pigments has been described in a series of publications by Lin and Demain.85, 86, 87, 88
C50 carotenoids, such as sarcinaxanthin and its glucosides, are more powerful quenchers of singlet oxygen than β-carotene. They have potential for use in nutriceuticals, pharmaceuticals and derived products such as apocarotenoids or norisoprenoids. Vitamin A is a norisoprenoid and a cleavage product of β-carotene (β-carotene is also known as provitamin A). Other norisoprenoids include safranal (providing saffron flavor to sauces and paella dishes), bixin (a pigment in annatto used to color cheeses), damascenone (a part of many perfumes) and ionones (for flavoring of soft drinks, candies and tobacco). Other norisoprenoids include the plant hormone abscisic acid and strigolactones, having functions in plants. C40 carotenoids, that is, terpenoids, can be produced by metabolically engineered S. cerevisiae, as is β-carotene. The engineering involves introduction of three genes from the astaxanthin producer X. dendrorhous.
One of the most important microbial sources for preparation of the keto-carotenoid astaxanthin is P. rhodozyma (X. dendrorhous), a heterobasidiomycetous yeast.89, 90 Each year, 130 tons of astaxanthin are used for aquaculture and poultry. This oxygenated carotenoid pigment is used in the feed, food and cosmetic industries. It is responsible for the orange to pink color of salmonid flesh and the reddish color of boiled crustacean shells. Feeding of pen-reared salmonids with a diet containing this yeast induces pigmentation of the white muscle.91 It is a very good antioxidant, 10 times more active than β-carotene and 100 times more than α-tocopherol. It is the second most important carotenoid. Astaxanthin enhances the immune system and protects skin from radiation injury and cancer. It can be produced synthetically as hydroxyl-astaxanthin from petrochemicals with a selling price of $2500 kg−1. However, the natural product is favored because the synthetic product is a mixture of stereoisomers. Natural astaxanthin is more stable than the synthetic version and more bioavailable; that is, it has a higher degree of absorption into a living system. The natural product is present in algae and fish as mono- and di-esters of fatty acids. However, it is difficult to hydrolyze the esters from algae, limiting its usage to trout and salmon. The yeast product is better as it is the 97% free, nonesterified (3R, 3’R) stereoisomer. The astaxanthin market was $219 million in 2007, with 97% being synthetic. Most of the production processes with the yeast yield levels of astaxanthin <100 mg l−1. However, white light improved production to 420 mg l−1,92 and mutant strain UBv-AX2 can make 580 mg l−1.93
Antimicrobials
The filamentous fungi produce 22% of the nearly 12 000 antibiotics that were known in 1955.94, 95 The β-lactams, which constitute a major part of the antibiotic market, and include the penicillins, cephalosporins, clavulanic acid and carbapenems, are the most important class of antibiotics in terms of use. Of these, fungi are responsible for production of penicillins and cephalosporins. The natural penicillin G and the biosynthetic penicillin V had a market of $4.4 billion by the late 1990s. Major markets also included semisynthetic penicillins and cephalosporins with a market of $11 billion. In 2006, the market for cephalosporins amounted to $9.4 billion and that for penicillins was $6.7 billion. By 2003, production of all β-lactams had reached over 60 000 tons. The titer of penicillin is over 100 g l−1 and that for cephalosporin C is at least 35 g l−1.96, 97 Recovery yields are >90%. There have been >15 000 molecules based on penicillin that have been made by semisynthesis or by total synthesis. By the mid-1990s, 160 antibiotics and their derivatives were already in the market.95, 98 The market in 2000 was $35 billion.
1,3-Diaminopropane (1,3-DAP) is secreted by P. chrysogenum and Acremonium chrysogenum. Both it and spermidine (that contains 1,3-DAP) increase transcription levels of the penicillin biosynthetic genes pcbAB, pcbC and penDE.99 They thus stimulate production of penicillin G. The mechanism appears to involve stimulation of the expression of laeA, a global regulator that acts epigenetically on expression of secondary metabolism genes via heterochromatin reorganization. 1,3-DAP also stimulates production of a cephamycin in Amycolatopsis lactamdurans. Spermidine’s activity appears to be due to 1,3-DAP. Genes coding for three enzymes involved in the conversion were found to be present in the P. chrysogenum genome.
Because of the emergence of resistance among fungi and bacteria to current antibiotics, naturally resistant microbes and newly evolving pathogens, more antibiotics are urgently needed. A new and approved cephalosporin is ceftobiprole that is active against methicillin-resistant S. aureus and is not hydrolyzed by a number of β-lactamases from Gram-positive bacteria.100 Another antibiotic of note is cerulenin, an antifungal agent produced by Acremonium caerelens. It was the first inhibitor of fatty acid biosynthesis discovered.101 It alkylates and inactivates the active-site nucleophylic cysteine of the ketosynthase enzyme of fatty acid synthetase by epoxide ring opening. Other properties that are desired in new antibiotics are improved pharmacological properties, ability to combat viruses and parasites and improved potency and safety. A new antifungal natural product is parnafungin, produced by Fusarium lavarum, that inhibits poly(A)polymerase in Candida albicans as well as a broad range of pathogenic fungi.102
A major antibiotic problem has been the development of resistance to carbapenem antibiotics, such as imipenem and meropenem, by Gram-negative pathogens. This is mainly because of the occurrence of extended spectrum metallo-β-lactamases such as NDM-1 (New Delhi metallo-β-lactamase). King et al.103 isolated a natural product called aspergillomarasmine (AMA) from the soil fungus Aspergillus versicolor that inhibits NDM-1 and another metallo-β-lactamases called VIM-2. AMA is a peptide inhibitor of metalloproteinases. AMA fully restored the activity of meropenem against bacteria carrying NDM or VIM metallo-β-lactamases. The work was done by Gerard Wright and his group at McMaster University.103, 104, 105 NDM-1 requires zinc and AMA removes zinc from the enzyme. The combination of AMA and the carbapenem has shown its effect in mice and in human cell culture.
Biofilm formation by bacteria allows pathogenic bacteria to resist dispersal and inhibition by conventional chemotherapy.106 Biosurfactants are amphiphilic compounds containing a hydrophilic region (polar or nonpolar) and a hydrophobic region (lipid or fatty acid). They act as antibiofilm agents and include sophorolipids. Some sophorolipids are produced by Candida species and are active against biofilm-forming E. coli and Bacillus subtilis.
Antimalarial agent
Malaria is a major cause of illness and death, especially in tropical and subtropical areas of the world.107 There are 500 million new cases every year, killing 1.5 million people, mainly young children and pregnant women. Quinine from the bark of the Chinchona tree and artemisinin from the Chinese herb (Artemisia annua) have been the two major drugs used against malaria. Quinine has been used for >1000 years but artemisinin is a newer drug. Quinine has some side effects, such as arrhythmia, thrombocythemia and cinchonism, and this is the main reason for the extensive use of artemisinin. Artemisinin is an endoperoxide sesquiterpine lactone, the most potent and effective antimalarial and is useful against multidrug-resistant Plasmodium falciparum. Resistance to artemisinin and its derivatives is increasing but is still mild. The level of artemisinin in A. annua is very low (0.01–1% of the weight of the dried leaves). Thus, genetic engineering has been pursued. A genetically engineered S. cerevisiae strain producing 100 mg l−1 of artemisinic acid has been developed by the Keasling group in Berkeley, California. Artemisinic acid can be chemically converted to artemisinin. Keasling’s company, Amyris Biotechnologies, has increased the amount of artemisinic acid produced by one million fold. The artemisinin precursor amorpha-4,11-diene is made by the engineered S. cerevisae at 40 g l−1.108
Organic acids
Carboxylic acids are made mainly by catalysis from petroleum-based precursors but interest in microbial production is increasing.109 Annual production of these compounds is as follows (ktons): acetic acid: 10 000; acrylic acid: 4200; 3-hydroxypropionic acid: 3600; adipic acid: 3000; citric acid: 1600; lactic acid: 450; fumaric acid: 200; gluconic acid: 87; itaconic acid: 80; malic acid: 60; glucaric acid: 42; glycolic acid: 40; and succinic acid: 37. Acetic and lactic acids are used as food preservatives. Lactic acid is also used to produce the biodegradable polymer polylactide. Citric and malic acids are food additives. Gluconic acid is used to chelate divalent and trivalent metal ions. Acrylic and adipic acids are employed to make polymers. Glycolic acid is used in the textile industry as a tanning and dyeing agent. The main acids showing promise for microbial production are succinic, lactic and itaconic acids. S. cerevisiae could become a leading organism for carboxylic acid production, mainly because it can grow at low pH. It was the first eukaryote to have its entire genome sequenced. Considerable genetic engineering has been done with this yeast. Also important is its naturally occurring, episomally replicating plasmid, named the 2-μ plasmid. The organism can make lactic acid at 62 g l−1 and malic acid at 50 g l−1.
Itaconic acid is used to prepare polymers, coatings, adhesives and textiles. One such polymer is poly-itaconic acid that is used in (1) water treatment, (2) detergents, (3 as an agent for thickening, binding and sizing, (4) as an emulsifier, (5) in oral drug delivery and (6) in dental cements. Itaconic acid is made by A. terreus at 80 000 tons per year with a selling price of $2 kg−1 in a fermentation process that is more economical than chemical synthesis. A deficiency of manganese is a critical parameter for its production.110 Production can be completely inhibited by manganese ions. However, if the Mn concentration is kept below 5 μg l−1, with an initial sugar concentration of 100 g l−1 or higher, the itaconic acid production by A. terreus is similar to that of citric acid production by A. niger under the same conditions (see below). A titer of 130 g l−1 of itaconic acid was produced.110 Increasing pH during the production phase increased production by A. terreus to 146 g l−1.111 The modification was done by raising the pH from 4 to 6 or by raising pH to 3 after 2.1 days of cultivation.
Gluconic acid is made by A. niger.112 It is used in construction and in production of chemicals, pharmaceuticals, foods, beverages, textiles and leather. Substrates include glucose, sucrose and golden syrup, a by-product of the process refining sugar cane juice into sugar, or by treating sugar with acid. The price varies from $1.20 to $8.50 per kg. Thus, 85 g l−1 was produced in 44 h with a productivity of 1.94 g l−1 h−1. Previous workers had obtained 158 g l−1 at 0.238 g l−1 h−1 with A. niger immobilized on cellulose microfibers.113
Isocitric acid is used to make pharmaceuticals and anticoagulants. Yarrowia lipolytica is a yeast producing high levels of isocitric and citric acids from rapeseed oil.114
Yovkova et al.115 engineered Y. lipolytica to produce a high concentration of α-ketoglutarate from raw glycerol, that is, 186 g l−1. Raw glycerol is obtained as a by-product of biodiesel production and can serve as an inexpensive carbon source for many fermentations. The strain was H355A (PVCI-IDPI). The new strain overexpressed genes encoding NADP+-dependent isocitrate dehydrogenase (IDP1) and pyruvate carboxylase (PYC1). Production was 19% higher than that by the parent strain (H3557). The usual by-product, pyruvic acid, was markedly decreased in the mutant fermentation. α-Ketoglutaric acid is used industrially in chemical synthesis of heterocycles or elastomers, as a dietary supplement and as an enhancer of wound healing.
Malic acid is a C4 dicarboxylic acid used in the food, feed and beverage industries as an acidulant and taste enhancer/modifier in combination with artificial sweeteners. It is also used to prepare polyester resins and coatings. Additional applications include medical uses. Metabolic engineering of Aspergillus oryzae NRRL 3488 has been used to overproduce malic acid at 154 g l−1,116 with a selling price of $2–3 kg−1. The result was achieved by overexpressing (1) the C4-dicarboxylate transporter and (2) the cytosolic alleles of pyruvate carboxylase and malate dehydrogenase. The rate was 0.94 g l−1 h−1 and the yield on glucose was 1.38 mol mol−1. Penicillium viticola 152 produced 168 g l−1 of calcium malate in a medium containing corn steep liquor.117 The yield was 1.28 g g−1 glucose and productivity was 175 g l−1 h−1.
Overproduction of pyruvic acid is carried out by Torulopsis glabrata (also called Candida glabrata), a multivitamin auxotrophic yeast.118 The process was industrialized in 1992 by Toray Industries at 400 tons per year. Subsequently, it was found that a S. cerevisiae mutant could produce a higher concentration, that is, 135 g l−1. However, it was not environmentally robust, had a longer lag phase, lower glucose consumption rate and lower specific growth rate. C. glabrata produces 94 g l−1, has a high yield (0.635 g g−1), high productivity (1.15 g l−1 h−1) and high glucose tolerance. Production was increased by use of urea as a nitrogen source.119 This organism is used for commercial production of pyruvic acid.
Approximately 95% of citric acid production is used in the food industry. Other uses include chemicals (surfactants and synthetic detergents), medicinals, textiles and metallurgy.120 Producing microbes include A. niger, A. terreus and Y. lipolytica. Production by Y. lipolytica is favored by limitation of cell growth brought about by limiting levels of nitrogen, phosphorus or sulfur, with nitrogen limitation as the most useful. Fermentation with genetically engineered Y. lipolytica amounted to 154 g l−1 from glycerol. Citric acid production by A. terreus can reach 200 g l−1.
Fumaric acid, a 4-carbon dicarboxylic acid, is made by species of Rhizopus at levels of 126–130 g l−1.121 Rhizopus arrhizus has been utilized by the Pfizer corporation to make 4000 tons per year.122 DuPont patented a process using R. arrhizus NRRL-1526 with limited dissolved oxygen to produce 130 g l−1.123 Other producing species include Rhizopus nigricans, Rhizopus formosa and Rhizopus oryzae. It is used as a food acidulant, a beverage ingredient, in production of biodegradable polymers, plasticizers, polyester resins and as an animal feed supplement to reduce methane emissions.
Glycolic acid can be overproduced by S. cerevisiae and Klyveromyces lactis.124 Engineered S. cerevisiae made only 1 g l−1, but engineered K. lactis produced 15 g l−1 from ethanol plus D-xylose. It is polymerized to polyglycolic acid that is excellent for preparing packaging material. Glycolic acid can also be used with lactic acid to make a copolymer (poly(lactic-co-glycolic acid)) for medical applications in drug delivery. Glycolic acid’s market in 2011 was $93 million for the 40 million kg that were produced.
Succinic acid is made by metabolically engineered Y. lipolytica at 63 g l−1. Lactic acid is produced by Candida boidini at 86 g l−1.
Z. Xue et al. developed a new process to make eicosapentaenoic acid (EPA), a long-chain polyunsaturated fatty acid.125 It has been produced from wild-caught ocean fish, but this source cannot keep up with the demand for polyunsaturated fatty acids that are important for human health such as for reduction of coronary disease and action against hypertriglyceridemia. The process uses a metabolically engineered strain of Y. lipolytica that produces EPA at 56% of its cell dry weight plus lipids at 30% of its dry weight.126 The yeast was engineered by transformation with 21 heterologous genes encoding five different activities. The genetic manipulation included inactivation of the peroxisome biogenesis gene Pex10. The oil produced has much higher levels of EPA than natural oils. EPA is important for the anti-inflammatory activity of fish oils, thus contributing to cardiovascular and joint health. The product has been commercialized by the DSM company.
Isoprenoids
Isoprenoids are a group of ∼50 000 natural compounds used as pharmaceuticals, flavors, fragrants, dietary supplements, food ingredients, biomaterials, solvents and biofuels.127 They are the largest and most diverse group of natural products. They include primary metabolites (sterols, carotenoids, quinines) and secondary metabolites, mainly used for medicine. They are divided according to the number of carbon atoms: hemiterpenoids (C5), monoterpenoids (C10), sesquiterpenoids (C15), diterpenoids (C20) and triterpenoids (C30).
The sesquiterpenoids are one of the largest groups of isoprenoid natural products, amounting to 7000 compounds. Acyclic sesquiterpenoids are found in essential oils and insect pheromones. They include farnesene and isomeric alcohols such as nerolidol and farnesol. They are being considered as potential diesel and jet fuel alternatives. Bisabolene, like farnesene, is also a potential diesel fuel alternative. Monocyclic sesquiterpenes are important in the pharmaceutical and perfumery industries. For example, humulene has anti-allergenic and anti-inflammatory properties. Eleniol, zingiberene and bisabolene occur in essential oils and fragrances.
The two C5 universal building blocks used to synthesize isoprenoids are isopentenyl diphosphate and its isomer dimethylallyl diphosphate. The latter is produced either from the mevalonate pathway or the methyl erythritol phosphate pathway. The mevalonate pathway is present in eukaryotes and archaea, whereas the methyl erythritol phosphate pathway is active in bacteria. Methyl erythritol phosphate pathway has been used to produce taxadiene, the isoprenoid precursor of the anticancer agent taxol. Because plants and naturally occurring microbes produce only small quantities of isoprenoids, fermentation with engineered microbes has become the way to produce carotenoids, sterols and artemisinin. Artemisinin is a potent antimalarial and also a part of antimalarial combination therapies.
Progress in metabolic engineering, including synthetic biology and systems biology, has been made in microbial production of isoprenoids, such as artemisinic acid, taxol, farnesene, isoprene, amorphadiene and farnesol. One of the producing microbes is S. cerevisiae.128 The S. cerevisiae mevalonate pathway has been engineered in E. coli yielding the terpene farnesyl diphosphate, the precursor to amorphadiene. The amorphadiene titer was 281 mg l−1. This was increased to 480 mg l−1 by fermentation modifications. Further genetic and fermentation modifications increased the amorphadiene titer to 27.4 g l−1 and then to 41 g l−1. This led to production of 25g l−1 of artemisinic acid. By an inexpensive chemical process, the artemisinic acid was converted into the semisynthetic artemisinin. Artemisinin has been approved as an antimalarial agent by the World Health Association and is being produced commercially by Sanofi (see Antimalarial agent section).
Farnesene has been produced by yeast at a concentration of 728 mg l−1 by the Amyris company. It is made from sugar cane using a laboratory-evolved strain of S. cerevisiae. The titer was increased to 104 g l−1 with a productivity of 16.9 g l−1 per day by use of random mutagenesis and selection. Farnesol is an acyclic sesquiterpenoid alcohol derived from farnesyl diphosphate. It is found in plant essential oils and is important in the flavor and fragrance industries. It is also an antimicrobial agent, an antitumor drug precursor and a biopesticide in aquaculture. Furthermore, it is being considered as a diesel or jet-fuel substitute. Farnesol can be produced by C. albicans but, more importantly, by dephosphorylation of farnesyl diphosphate in engineered S. cerevisiae, overproducing mevalonate pathway genes. The farnesol titer reached in such strains of S. cerevisiae is 5 g l−1.
Proteins
Production of biopharmaceutical proteins by metabolically engineered microbes has been very successful.129, 130 Biopharmaceuticals comprise one-sixth of the pharmaceutical market and are the most rapidly growing segment. They are employed to make up for the deficiency of body proteins used for normal function. They include blood factors, monoclonal antibodies, thrombolytics, anticoagulants, vaccines, hormones, interferons, interleukins, enzymes and growth factors. These systems were responsible for almost all of the biopharmaceuticals approved to date. Of the 211 biopharmaceuticals approved by 2011, 31% were produced by yeasts. Of the yeast products, 30 were made in S. cerevisiae and one in Pichia pastoris. In 2012, 12 biopharmaceuticals were approved in the United States and Europe.131 One was produced by S. cerevisiae and another by P. pastoris.
The work on S. cerevisiae has mainly dealt with increasing protein secretion. More than 40 different recombinant proteins have been made by S. cerevisiae. Human serum albumin, used as a plasma expander in surgery, is produced at 3 g l−1, and human transferrin, used for anemia, is made at 1.8 g l−1.
Enzyme use in industry has been reviewed by Hellmuth et al.132 There are four major groups of industrial enzymes: (1) detergent enzymes, (2) technical enzymes, (3) food enzymes and (4) feed enzymes. The technical enzymes include those for textiles, leather, pulp and paper and fuel ethanol. The largest group are the food enzymes that include amylases, xylanases, glucose oxidase, hexose oxidase, pectinases, glucanase, invertase, glucose isomerase, protease, lipase, phosphorylase, lactase, milk-clotting enzymes, animal rennet, microbial rennet and chymosin. The main sources are molds, yeasts and bacteria. Fungal producers include A. niger and K. lactis.
Regulation of cellulolytic and hemi-cellulolytic enzyme production in filamentous fungi has been reviewed by Tani et al.133 Important regulatory transcription factors include XlnR from aspergilli that is involved in D-xylose induction of xylanolytic and cellulolytic enzymes. Others include ClR-1/2 from Neurospora, ManR, McmA and C1bR from Aspergillus and Bg1R from Trichoderma that regulate cellulolytic and/or hemicellulolytic enzyme production.
Heterologous proteins are also made very well by the yeast Y. lipolytica.134 Such recombinant proteins include lipases, proteases, amylase, mannanase, laccase, leucine aminopeptidase and insulin. Recombinant proteins are also made by other fungi such as Hansenula polymorpha, K. lactis, Schizosaccharomycs pombe, C. boidini, A. oryzae, A. niger, Trichoderma reesei, T. atroviride, Penicillium sordida, Penicillium griseoroseum, Penicillium purpurogenum and R. oryzae.135 The products of these heterologous enzymes include citric acid, isocitric acid, α-ketoglutaric acid, succinic acid, polyunsaturated fatty acids such as γ-linoleic acid, EPA and carotenoids including lycopene and β-carotene.
The secretory pathway in yeast involves over 160 proteins that carry out different posttranslational processes such as folding and glycosylation. Of special importance is the production by S. cerevisiae of insulin and its analogs. The insulin market amounted to $12 billion in 2011 and has been increasing ever since.
The use of P. pastoris, reclassified as Komagataella pastoris, as a producer of heterologous proteins has been reviewed by Ahmad et al.136 Among the processes developed, one of the first was the production of the plant-derived hydroxynitrile lyase at over 20 g l−1.137
Fatty acids and lipids
Production of polyunsaturated fatty acids by fungi and other microorganisms has been reviewed by Ratledge.138 They are used as nutriceuticals and include (1) γ-linoleic acid (18:3 omega-6) from Mucor circinelloides, (2) docosahexaenoic acid (DHA; 22:6 omega-3) from algae, (3) arachidonic acid (20:4 omega-6) from Mortierella alpine and (4) EPA from genetically modified Y. lipolytica. They represent a multi-billion dollar industry, mainly arachidonic acid and DHA for infant formulas. They are major components of phospholipids in cell membranes. They regulate cell fluidity, attachment of specific enzymes to cell membranes and mediate signal transduction and other metabolic processes. They are used for the biosynthesis of eicosanoids, leukotrienes, prostaglandins and resolvins that function as anti-inflammatory, antiarrhythmic and antiaggretory effectors. Many improve cardiovascular health and some improve eye function and memory in newly born infants and adults. Two of these, that is, arachidonic acid and DHA, that are added to infant formulas may also have beneficial action for Alzheimer’s disease, chronic bowel disorder and cancer. Microbial fermentation can be used for their production.
Microbial oils can be produced by 30–40 species of yeast, as well as by molds and algae. They are called oleaginous microbes. Fungi can accumulate 70% of their biomass as oils. EPA plus DHA can be used to prevent cardiac problems. EPA appears to have beneficial effects in neuropsychiatric disorders such as manic depression (bipolar disorders), schizophrenia, attention deficit hyperactivity disorder in children, coronary events in heart disease patients, preventing and treating obesity, metabolic syndrome, nonalcoholic steatohepatitis, type 2 diabetes and hypertriglyceridemia.
Y. lipolytica can accumulate lipids up to 40% of its dry cell weight. It makes single-cell oil for health applications and biofuels. In one case, lipid production reached 62% of the dry cell weight.139 In another case, a strain accumulating 90% of its dry cell weight as lipid was developed.140 In this case, a lipid titer of 25 g l−1 was achieved.
Production of intracellular lipids by yeast growing on alkali-treated corn stover was studied by Sitepu.141 Cryptococcus humicola produced 15 g l−1 lipids in a total biomass weight of 36 g l−1. The strain (UCSEST 10–1004) came from the Phaff yeast collection at the University of California, Davis. Such lipids could become useful for biodiesel production.
Vitamins
Production of vitamins by microbes has been reviewed by Ledesma-Amaro et al.142 Most are produced chemically but microbial production is becoming important in several cases. Vitamin D is derived chemically from cholesterol and ergosterol. However, it can be made by S. cerevisiae, Saccharomyces uvarum and Candida utilis at 30 g g−1 of dry cells.
Riboflavin (vitamin B2) can be made by molds (A. gossypii, E. ashbyii), yeasts (Candida flaeri, Candida famata) and bacteria. A. gossypii can produce it at 14 g l−1. It is mainly produced by metabolically engineered microbes and is used as a feed additive (70%) and as a food additive (30%) as well as for pharmaceutical applications. The producing organisms are A. gossypii and the bacterium B. subtilis that have completely replaced chemical synthesis.143 In the high producing mutant of A. gossypii, that is, strain W122032, the increased production, as compared with the wild-type ATCC 10895, is because of (1) a 9% increase in flux to pentose-5-phosphate via the pentose phosphate pathway and (2) a 16-fold increase in the flux from purine to riboflavin.144 The result is because of increased guanosine triphosphate flux through the pentose phosphate pathway and the purine synthesis pathway.
Alcohols
Erythritol, a polyhydric alcohol, has 60–70% of the sweetness of sucrose and is used to combat obesity. It is noncarcinogenic and noncaloric as it is not digested by humans, and cannot be fermented by bacteria to cause dental caries. Repeated batch cultures of Y. lipolytica on crude glycerol yielded 220 g l−1 with a yield of 0.43 g g−1 of glycerol used and a productivity of 0.54 g l−1 h−1.145
Xylitol, a pentahydroxy sugar alcohol originating from xylose, has applications in foods and pharmaceuticals. A review of xylitol production from lignocellulosic waste has been written by Lima de Albuquerque et al.146 Xylitol is a low-calorie sweetener used by diabetics with 40% fewer calories than sucrose. It is noncariogenic and has insulin-independent metabolism properties. Xylitol has high solubility, low glycemic rate, lack of carcinogenicity and has cariostatic properties. It is used in food production. It is made by catalytic hydrogenation of xylose but this is very expensive. Its global market is over 125 000 tons per year and it sells for $4.50–5.50 per kg to pharmaceutical and food companies. It has a 12% share of the total polyol market for chewing gum and foods. It is used as a sucrose replacement for cakes, cookies, chocolate and chewing gum, and in pharmaceuticals to reduce tooth decay. It acts against oral biofilms especially against Streptococcus mutans. It is also active against other bacteria harmful to oral health, such as Streptococcus pneumoniae, Hemophilus influenzae, S. aureus and Pseudomonas aeruginosa. It has also been cited as a contributor to tooth calcification, and is active against diabetes, anemia, acute otitis media and osteoporosis. Of great interest is its production from xylose in lignocellulosic materials as 100% of the xylose can be converted to xylitol microbially. This is favored over chemical synthesis that uses more intense reaction conditions and yields undesirable coproducts that have to be removed. Fermentation of xylose to xylitol occurs under milder conditions of temperature and pressure, yielding lower levels of unwanted by-products. Cell-free systems have been used, but immobilized systems hold great promise because of higher levels of xylitol produced, as well as the possibility of reuse of successive cultures as a fed-batch process or prolonged fermentations during continuous processes. Systems used include stirred tank reactors, packed-bed reactors, fluidized bed reactors, bubble columns and air-lift reactors. Processes include batch culture, fed-batch or semi-continuous culture and continuous culture. Bioconversion of 300 g l−1 xylose to xylitol by Debaryomyces hansenii amounted to 110 g l−1.147 The yield was 0.48 g g−1. Repeated fed-batch fermentation (lasting 750 h) with high-cell density cultures of Candida magnolia TISTR 5663 in a 2-l stirred tank fermenter under oxygen limitation with feeding of xylose and nitrogen and a starting xylose concentration of 60 g l−1 led to production of 284 g l−1 of xylitol,148 the highest titer ever achieved. Xylitol productivity was 1.49 g l−1 h−1. Candida tropicalis has also been used by other groups. The overall conclusion is that immobilization of yeast cells is an excellent way to produce xylitol from xylose. Immobilization was best when entrapment in calcium alginate, followed by solid adsorption, was used.
Candida athensensis can convert vegetable waste to 100 g l−1 xylitol with a yield of 0.81 g g−1 and a productivity of 0.98 g l−1 h−1.149 The vegetable waste contained 200 g l−1 of xylose.
Another sugar alcohol, D-arabitol, is potentially useful for oral health care and as a pharmaceutical. It has lower nutritional calories than xylitol and sucrose, making it a low-calorie natural sugar substitute for diabetics. An osmophilic strain from raw chaste honey, Zygosaccharomyces rouxii JM-C46, was isolated as a high D-arabitol producer.150 Using pH control and repeated fed-batch fermentation in a 5-l fermentor yielded 93 g l−1 of D-arabitol with a volumetric productivity of 1.143 g l−1 h−1.
Mannitol is produced by genetically engineered Y. lipolytica at 27 g l−1.
Additional compounds
Glutathione, a redox-active tripeptide thiol having the activities of antioxidation, detoxification and immune regulation can be made by an engineered strain of S. cerevisiae at a concentration of 317 mg l−1.151
Coenzyme Q (ubiquinone) is an essential part of the respiratory chain producing ATP. It is an excellent antioxidant. It is composed of a quinonoid nucleus and a side chain of isoprenoids. Microbial fermentation is the best method of production as it produces no optical isomers and is the least expensive means of production. Its production has been reviewed by De Dieu Ndikubwimana and Lee.152 It can be produced by species of the yeasts Candida and Saitoella.
Flavin adenine nucleotide is used in the pharmaceutical and food industries. It is an ophthalmic agent of which 10 tons are produced annually. It is produced at 18 g l−1 in a medium containing FMN and ATP by C. famata.153
Yeasts are also used to make human serum albumin, hepatitis vaccines and virus-like particles used for vaccination against human papilloma virus. S. cerevisiae carries out proper folding of many human proteins, secretes the proteins, and does posttranslational modifications such as proteolytic processing of signal peptides, disulfide bond formation, subunit assembly, acylation and glycosylation. A negative property of S. cerevisiae and other yeasts had been the high-mannose type of N-glycosylation that shortens the in vivo half-life and reduces efficacy. However, both S. cerevisiae and P. pastoris have been engineered to produce human-like glycosylation that includes terminal addition of sialic acid to the glycoprotein.
The yeast Aureobasidium pullulans strain RBF 4A3 can produce 88 g l−1 of pullulan.154 Pullulan is an exopolysaccharide that has potential application in industries such as medical, food, pharmaceutical, cosmetic and agriculture.
The fungal genus Trichoderma makes many secondary metabolites with useful applications.155 Its species are commercially available as plant growth-promoting fungi and biological control agents. They have broad-spectrum antagonistic activities against a number of soil-borne phytopathogens including (1) mycoparasitism, via secretion of cell wall-degrading enzymes; (2) competition, that is, mobilizing and taking up of macro- and micro-nutrients from soil, resulting in scarcity of nutrients for other soil microbes; and (3) antibiosis via secretion of antibacterial secondary metabolites. These secondary metabolites include (1) emodin, a cathartic stimulant and tumor cell adhesion inhibitor; (2) gliotoxin, an antimalarial agent and immune system suppressor; (3) harziaolide, an antifungal agent and plant growth promoter; (4) koninginins, antifungal agents and regulators of plant growth; (5) 6-pentyl-2H-pyran-2-one, an antifungal agent as well as a promoter of plant growth and a coconut aroma used in confectionary products; (6) trichokonins, broad-spectrum antifungal agents and plant defense inducers; (7) viridofungins, potential anticancer agents and bacteriocides; (8) viridian, a broad-spectrum antifungal agent, antineoplastic and antiatherosclerosis agent; and (9) viridiol, a weedicidal agent.
Systems metabolic engineering for production of biofuels and chemicals by Aspergillus and Pichia species has been reviewed by Caspeta and Nielsen.156 Formaldehyde is produced from methanol at 6000 tons per year by P. pastoris.
Many useful products are made by the basidiomycetes.157 These fungi make carotenoids, fragrances, enzymes, astaxanthin, erythritol, lipids and oils. Trichosporon species produces lipids and is being considered for biodiesel production. Pseudozyma (Candida) antartica produces lipase for industrial use and is also a biodiesel possibility. It also produces 30 g l−1 of itaconic acid. Sporobolomyces carnicolor accumulates 82% of its biomass as intracellular lipids. Cryptococcus species make unique carotenoids such as plectaniaxanthin, a xanthophyll. Some cryptococci utilize glycerol and accumulate 60% of their biomass as triacylglycerols.
Biofuels
Approximately 100 billion liters of ethanol are produced per year from sugar cane and corn starch by S. cerevisiae. Production of ethanol and advanced biofuels at high temperature (ca 40 °C) reduces cooling costs, lowers the effects of contamination and enables more efficient hydrolysis of feedstocks. This improves productivity in the simultaneous saccharification and fermentation process. However, temperatures of 34 °C and above interfere with yeast viability and growth. Caspeta et al.158 isolated S. cerevisiae strains with improved growth and ethanol production at 40 °C. They used adaptive laboratory evolution to obtain these mutant strains. They noted a change in sterol composition from ergosterol to fenosterol because of a mutation in the C5 sterol desaturase gene, and increased expression of sterol biosynthesis genes. The new strains grew 1.9 times faster and excreted ethanol and glycerol 1.6 times faster than the parent culture. Sterols contribute to membrane fluidity. These thermotolerant strains were improved in glucose consumption rate that was increased by 60% at 40 °C and by 300% at 42 °C.
Production of bioethanol from corn can only yield 15 billion gallons per year. It is thus desirable to produce bioethanol from lignocellulosic biomass. Bioethanol can be produced from cellulose, but lignocellulose contains not only glucose, but also C5 sugars such as xylose and arabinose that cannot be utilized by wild-type S. cerevisiae because it does not have a catabolic pathway for pentose utilization. However, some genetically engineered S. cerevisiae strains can utilize xylose. To improve the production of ethanol by a xylose-utilizing strain of S. cerevisiae, the HAP4 gene was knocked out.159 This gene encodes a transcription activator that controls expression of genes involved in mitochondrial respiration and reductive pathways. By knocking out the HAP4 gene, the following were increased: maximal ethanol concentration, ethanol production rate and ethanol yield. A new strain, S. cerevisiae B42-DHAP4, could produce ethanol from xylose as sole carbon source under aerobic conditions. The rate of ethanol production and its yield from a detoxified hydrolysate of wood chips were markedly improved.
Alcohol tolerance in S. cerevisiae can be increased by adding potassium and raising the pH of the fermentation medium with KOH.160 This increases cell growth. Ethanol titer was increased by these modifications to 127 g l−1.
Product titers achieved by fungi growing on Jerusalem artichokes include 154 g l−1 of ethanol by a mixed culture of S. cerevisiae and A. niger and 109 g l−1 by S. cerevisiae alone.161
During pretreatment of biomass, a problem is the liberation of the inhibitor furfural. However, tolerance to furfural can be achieved by overexpression of S. cerevisiae genes encoding (1) yeast transcription activator MSN2;162 (2) ZWF1 of the pentose phosphate pathway;163 (3) ADH1 encoding alcohol dehydrogenase 1; and (4) TAL1, encoding transaldolase 1.164
P. stipitis can produce ethanol from C5 sugars and also clean up concentrated toxins liberated during lignocellulose degradation.165 This yeast can produce ethanol from pretreated sources of biomass such as red oaks, wheat straw, sugar cane bagasse, rice straw, corn cobs, corn stover, aspen wood, pinewood and poplar wood. From aspen wood chips, ethanol titer was 41 g l−1 with a yield of 0.47 g l−1.166 In a chemically defined medium, 61 g l−1 could be produced by P. stipitis.167 Attributes of P. stipitis include consumption of acetic acid, and reduction of the furan ring toxins such as hydroxymethylfurfural and furfural present in cellulosic biomass conversions.
2,3-Butanediol can be produced at 96 g l−1 by an engineered strain of S. cerevisiae.168 It is a fuel with a high heating value (27 000 J g−1) and has been used as a liquid fuel or fuel additive.
A comprehensive review of biodiesel production and application of genetic engineering is that of Lin et al.169 This ideal substitute for petroleum-based diesel is made from triglycerides by transesterification with alcohols. Today, crude oil is consumed at 11.6 million tons per day that cannot last for a long time. Biodiesel use requires no engine modification. It is blended with diesel in Germany, Italy and Malaysia. It is better than diesel as it is biodegradable, nontoxic and releases less toxicant when burned as a fuel. Chemical catalysis is presently used to make biodiesel but enzymatic catalysis is looked upon as better from the aspects of mild reaction conditions and easy product separation. Current biodiesel production suffers from the lack of a stable, sufficient feedstock supply system, inconsistent performance and challenging economics. Microbial production could overcome these problems because of short producing period, little labor requirement, easy scale-up and independence from problems of venue, season, climate change, etc. So-called ‘grease microorganisms’ (oleaginous microbes that supply fatty acids and alcohols and convert them to biodiesel) could supply the fatty acids needed. Their composition of the fatty acids is similar to vegetable oils that are used to make biodiesel. They have >50% lipid content, can be used industrially, grow rapidly, are not polluting and oil can be easily extracted. Microalgae have been considered as an oil feedstock but their use is restricted by their low growth rate, strict breeding conditions and large up-front investment requirement. Alcohol is needed in the transesterification process to generate the fatty acid ester. Methanol or ethanol can be used to generate the fatty acid ester. Methanol makes fatty acid methyl esters, whereas ethanol yields fatty acid ethyl esters. Fatty acid methyl esters are cheaper, more reactive and volatile. However, ethanol is less toxic, and more renewable than methanol. S. cerevisiae has been genetically engineered for de novo biosynthesis of biodiesel. Of potential importance is Y. lipolytica for microbial biodiesel production.
Potential of being a more effective producer
Fungi harbor many more secondary metabolite gene clusters than those expressed under normal laboratory conditions. Activation of ‘silent’ gene clusters in A. nidulans has revealed many new fungal secondary metabolites. This application of ‘genome mining’ revealed that the A. nidulans genome contains 56 potential secondary metabolism core genes including the following: 27 polyketide synthase (PKS) genes, two polyketide synthase-like genes, 11 nonribosomal peptide synthetase (NRPS) genes, 15 NRPS-like genes and one hybrid NRPS–PKS gene.
Echinocandins
The echinocandins, which are often called the ‘penicillin of antifungals’, are lipopeptide molecules that inhibit β-(1,3)-glucan synthesis, an integral component of the fungal cell wall.170 Although they lack activity against Zygomycetes, Cryptococcus neoformans or Fusarium spp., in vitro and in vivo studies have demonstrated that echinocandins are fungicidal against most Candida spp. but are fungistatic against Aspergillus spp.170 Caspofungin was the first echinocandin to be approved, for the indication of refractory invasive aspergillosis. It is a semisynthetic derivative of pneumocandin B0, a lipophilic cyclic peptide isolated from the fungus, Glarea lozoyensis.171 The sales of this antifungal, which was sold under the name Cancidas by Merck, was $681 million in 2014 and $573 million in 2015. The second antifungal from the echinocandin class that was licensed was micafungin, sold under the trade name Fungard/Mycamine by Astellas Pharma. It achieved annual sales of 38.8 billion yen or about $352 million in 2014, and in 2015 the sales reached 41.6 billion yen or $346 million. Patients who have received either of these antifungal agents have reported mild adverse events that include local phlebitis, fever, abnormal liver function tests and hemolysis.170 Echinocandins are usually given once daily, via the parenteral route, because of poor absorption after oral administration. Caspofungin has shown promising results in studies of candidemia and invasive candidiasis, with nearly the same efficacy as amphotericin B, but markedly less toxicity. Because of the absence of antagonism and very minimal drug interactions with other antifungal agents, the echinocandin class represents a promising therapeutic area that may potentially be used in combination for the treatment of invasive aspergillosis.
Concluding remarks
For the past 85 years, microbes have provided significant contributions to the fields of medicine and agriculture, especially in the area of antibiotic production. The compounds they have produced have helped save millions of lives, alleviated pain and suffering and increased human life expectancy. In addition, they have also played a vital role in the animal industry and agricultural operations. However, for a variety of reasons, pathogenic microbes have become resistant to many antibiotics, creating a dangerous situation. Therefore, the need for new and effective antibiotics is imperative. Unfortunately, most of the large pharmaceutical companies have abandoned the search for new antimicrobial compounds. Because of economics, they have concluded that drugs directed against chronic diseases offer a better revenue stream than do antimicrobial agents, for which the length of treatment is short and government restriction is likely. Some small pharmaceutical and biotechnology companies are still developing antibiotics, but most depend on venture capital rather than sales income and, with the present regulations, face huge barriers to enter into the market. These barriers were raised with the best intentions of ensuring public safety but they are having the opposite effect; that is, termination of antibiotic development while resistance continues to increase.172 However, there are some new bright possibilities. One of the more promising is the utilization of uncultivated microorganisms. Considering that 99% of bacteria and 95% of fungi have not yet been cultivated in the laboratory, efforts to find means to grow such uncultured microorganisms is proceeding and succeeding.5 Furthermore, researchers are now extracting bacterial DNA from soil samples, cloning large fragments into, for example, bacterial artificial chromosomes, expressing them in a host bacterium and screening the library for new antibiotics. This metagenomic effort could open up the exciting possibility of a large untapped pool from which new natural products could be discovered.173 Another exciting possibility is that of genome mining.174 In addition to these relatively new techniques, chemical and biological modification of old antibiotics could still supply new and powerful drugs. These comments also apply to nonantibiotics such as antitumor agents and other microbial products. In addition, natural products must continue to be tested for desirable therapeutic activities. We believe that significant progress in identifying new antibiotics, oncology therapeutics and other useful medicines will be made, probably not by the big pharmaceutical companies, but by biotechnology companies and small research groups from institutes and universities.
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Arnold L Demain
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187612 | https://www.quora.com/Math-Word-Problem-The-ratio-of-boys-to-girls-in-a-class-is-4-7-There-are-88-students-in-the-class-How-many-boys-are-there | Math Word Problem: The ratio of boys to girls in a class is 4:7. There are 88 students in the class. How many boys are there? - Quora
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Math Word Problem: The ratio of boys to girls in a class is 4:7. There are 88 students in the class. How many boys are there?
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1y
To solve this problem, we need to find the total number of parts in the ratio and then use it to calculate the number of boys.
The ratio of boys to girls is 4:7, which means the total number of parts in the ratio is 4 + 7 = 11.
Since there are 88 students in the class, we can use the ratio to find the number of boys:
Total parts in the ratio = 11
Number of boys parts = 4
Proportion of boys parts = 4/11
Number of boys = 88 (4/11) = 32
Therefore, there are 32 boys in the class.
Upvote ·
Related questions
More answers below
There are m boys and n girls in a class. If total number of boys in the class is 24 and the ratio of number of boys to the total number of students in the class is 4: 7, then what the number of girls in the class?
The ratio of girls to boys is 7:8. There are 124 girls. How many boys are there?
In MS. Bell’s class, the ratio of boys to girls is 5:4. If there are 27 students in the class, how many boys are there in the class?
Math Help: The ratio of girls to boys in class is 9 to 7 and there are 80 students in the class. How many girls are in the class?
In a class of 60 students, the ratio of boys to girls is 3:2. How many girls are there in the class?
Sayed's Math Insights
Former Educator Accounting and Business Math · Author has 531 answers and 65.1K answer views
·Aug 22
Given:
Number of students: 88
Boys to Girls ratio: 4:7
Solution:
Sum of the ratios= 4+7= 11
1 part of the ratio= 88÷11= 8
Number of Boys = 8x4= 32
Number of Girls = 8x7= 56
Final Answer: 32 boys
Verification: 32+56= 88
Upvote ·
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Mildred Taylor
Former Retired Teacher · Author has 1.2K answers and 434.7K answer views
·10mo
Add the ratios 4+7=118
88/11 = 8
8X4=32
8X7=56
Total No. of students =88
Boys 32. Girls 56
There are 32 boys in the class.
Upvote ·
9 1
Joseph Kazaroff
B.S. in Computer Science&Applied Mathematics and Statistics, Johns Hopkins University (Graduated 2021)
·6y
Let’s call the number of boys ‘b’, and the number of girls ‘g’.
If the ratio of boys to girls is 4:7, this means that for every 4 boys, we have 7 girls. So, it is correct to say that 7b = 4g (i.e. 4 the number of all boys is equal to 7 the number of all girls).
Further, we know that there are only boys and girls in the class, so b+g=88. Now, we can solve.
7b = 4g
b = 4g/7
So:
4g/7 + g = 88
g = 56
So, we have 32 girls in the class. Then, we can plug back in to our original equation:
b = 32
So we have 32 boys in the class.
(The ratio of boy is less)
Upvote ·
9 1
Related questions
More answers below
The ratio of boys to girls in a class is 5:4. There are 36 students in the class. How many more boys than girls are there?
There are 9 girls for every 7 boys in my class and there are 80 students in the class. How many girls are in the class?
In a maths class, the girl to boy ratio is 7:5. If there are 20 boys, how many girls are there in the class?
The ratio of boys to girls in a class is 3:2. If there are 36 boys, then how many girls are in the class?
The ratio of boys to girls in a class is 3:1. There are 36 students in the class. How many students are girls?
John Gunkler
B.A. in Mathematics, Reed College · Author has 382 answers and 105.7K answer views
·Jan 25
In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls?
Originally Answered: In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls? ·
You know two things, so first write the equation for each. I'm use B for the number of boys and G for girls.
B = 4/5 G
B + G = 90
Okay, now substitute what we know from 1. into 2, getting:
(4/5 G) + G = 90, or 9/5 G = 90
Isolate G on the left side by multiplying ...
Upvote ·
Pascal Bauer
BSc in Mathematics, University of Nottingham (Graduated 2023) · Author has 54 answers and 22.8K answer views
·6y
If we add the two parts of the ratio together we get 4+7=11. 88/11=8 therefore we know that each part of the ratio is equal to 8 people. For the boys then we can do 4x8=32 boys.
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9 1
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99 24
George Ivey
Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views
·Jan 25
Since the ratio of boys to girls in the class is 4:7 the number of boys in the class is 4k and the number of girls in the class is 7k for some integer, k
4k+ 7k= 11k= 88 so k= 88/11= 8.
There are 4k= 48= 32 boys in the class.
There are 7k= 78=56 girls in the class.
Check: The ratio of boys to girls in the class is 32: 56= 84: 87= 4:7.
The number of students in the class is 32+ 56= 88.
Upvote ·
Raheel Ajani
6y
Let the common multiple be x
Ratio is 4x:7x
4x+7x=88
11x=88
x=8
The total number of boys=4x
48=32
Upvote ·
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99 22
Tej Kaur
Former Retired Assistant Director at Department of Higher Education (2004–2007) · Author has 6.9K answers and 1.4M answer views
·Jan 25
In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls?
Originally Answered: In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls? ·
Number of students in the class=90
Ratio of boys and girls :
4:5 Total of this ratio = 4+5=9
If the total number of students is 10, number of boys is =4
If number of students is 90, the number of boys will be = 90×4/9= 40
The number of girls is =90–40=50
Boys 40, Girls 50
Upvote ·
Reon Hillegass
Former Retired · Author has 723 answers and 1.3M answer views
·Jan 25
In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls?
Originally Answered: In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls? ·
••• There will be 40 boys and 50 girls because:
••• Add the ratios 4 and 5 and get 9. Divide that 9 into the total number of boys and girls which is 90 and get 10. You have 10 per ratio digit.
••• The boy's ratio of 4 times 10 is 40 and the girl's ratio of 5 times 10 is 50 girls.
Upvote ·
9 1
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I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours.
Ballerinas by Bata though, are what I wear the most. I didn’t know about them until recently, when a coworker recommended them to me, also spotted my favorite creator Siddhi Karwa styling them across Europe and I have been absolutely loving them.
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The Chosen One
Service Desk Analyst at Samsung SDS (2007–present) · Author has 14.7K answers and 5.6M answer views
·Jan 15
Say there are 4x boys and 7x girls, so there are a total of 11x students
11x = 88
x = 8
So there are 48 = 32 boys
Upvote ·
Pardha Saradhi Mandadi
Arbitrator and Mediator at Self Employeed Professional (2014–present) · Author has 12.6K answers and 4.7M answer views
·Jan 25
In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls?
Originally Answered: In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls? ·
The total number of boys and girls in the class =90
The ratio of boys to girls =4:5
Numbers of girls in the class =90/94=40
Number of boys in the class =90/95=50
Upvote ·
Randall Sonnemaker
MBA in Master of Business Administration Degrees, Saint Xavier University (Graduated 1999)
·Jan 28
In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls?
Originally Answered: In a class of 90, the ratio of boys to girls is 4 to 5. What is the number of boys and girls? ·
Total boys and girls = 90 Ratio components total 4 + 5 = 9
90 / 9 = 10 for factor multiplier.
Boys proportion and total = 4 x 10 = 40
Girls proportion and total = 5 x 10 = 50
Upvote ·
Related questions
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in this lesson we're going to focus on evaluating piecewise functions a piecewise function is a function that can be broken up into many parts so this particular piecewise function can be equal to 4x plus 5 or 3x minus 8 depending on the x value so let's say it's equal to 4x plus 5 when x is less than 2 and it's equal to 3x minus 8 when x is equal to or greater than two so go ahead and find the value of f of negative two f of two and f of five feel free to pause the video and work on this problem so let's evaluate the function when x is negative 2. so should we use this portion of the piecewise function or the bottom part should we use 4x plus 5 or 3x minus 8. negative two is less than positive two it's not equal to or greater than positive two so therefore negative two corresponds to this range so we need to use the first part of the piecewise function so let's replace x with negative two four times negative two is negative eight negative eight plus five is negative three so f of negative two is negative three now what about f of positive two should we use four x plus five or three x minus eight now x is equal to two in this equality not in this one so we have to use three x minus eight so it's going to be three times two minus eight three times two is six six minus eight is negative two and that's the answer now what is the function value at five so once again five is greater than two so we need to use three x minus eight so it's gonna be three times five minus eight three times five is fifteen fifteen minus 8 is 7. and that's it let's work on some more examples so let's say if we have the function f of x and it's equal to x squared plus three x minus seven when x is less than negative one and it's equal to five x plus six when x is greater than or equal to negative one actually let me change that let's say when x is greater than or equal to negative one but less than two and let's say it's uh equal to x cubed plus four when x is greater than two and is equal to twelve when x equals two so with this information i want you to evaluate f of negative 4 f of let's say 0 f of two and f of three so feel free to pause the video and try that so let's evaluate the function at x equals negative four so negative four is less than negative one therefore we need to use x squared plus three x minus seven so this is going to be negative four squared plus three times negative four minus seven negative four squared or negative four times negative four that's sixteen three times negative four is negative twelve and six minus twelve is four four minus seven is negative three so f of negative four is equal to negative three so that's the first answer i believe my math is correct i don't think i made any mistakes on that now let's evaluate f of zero zero is between negative one and two so we need to use five x plus six so this is going to be five times zero plus six five times zero is zero zero plus six is six now what about the next one f of 2. now when x is exactly 2 the function is equal to 12. so f of 2 is 12. there's no math involved in that step now what about the last one f of three when x is three we need to use x cubed plus four because that's when x is greater than two when x is three so therefore this is going to be 3 raised to the third power plus 4. so 3 to the third power is 27 27 plus 4 is equal to 31 and that covers that problem you |
187615 | https://fs.fish.govt.nz/Doc/25269/101%20TRU%202022.pdf.ashx | TRUMPETER (TRU)
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TRUMPETER (TRU)
(Latris lineata )
Kohikohi
FISHERY SUMMARY
1.1 Commercial fisheries
Historical estimated landings are shown in Table 1 for the main trumpeter stocks. Total reported
landings of trumpeter ranged between 3 t and 44 t until the fishing year 1990–91, after which landings
increased steadily to reach 162 t in 1995–96 (Tables 2 and 3). Total landings subsequently decreased
to a minimum of 25 t in 2000–01 and 2001–02, before once again increasing to over 100 t in the
2007–08, 2010–11 and 2011–2012 fishing years. In 2013–14 to 2020–21 total annual landings
averaging just over 60 t were recorded. Historic under-reporting is probable (Paul 1999).
Most landings of trumpeter have come from the east coast between the eastern Bay of Plenty and
Southland. There have been changes over time in contributions from different parts of the east coast,
but the reason for this is not known. Until the early 1950s most landings were made in QMA 3. From
the mid 1950s until the mid 1980s most landings were in QMA 2 (Table 1). The rapid increase in
landings after the mid-1980s has come predominantly from QMAs 3 and 4 (Table 3), reportedly from
an increase in line fishing on the outer shelf and in the Mernoo Bank region. Figure 1 shows the
historical landings for TRU from 1936.
Most trumpeter is taken as bycatch in line-fisheries; a small amount is trawled, and from the 1970s it
has also been taken by setnet. Only a small proportion of trumpeter is targeted. Catches are irregular
with no seasonal trend and are likely to be driven by fishing activities for other species. No
information on changes in fishing effort is available.
Trumpeter have been managed under the Quota Management System in New Zealand since 1 October
1988, at which time an original TACC of 100 t was set. The TACC was increased to 144 t in October
2001 following a period of declining landings. This TACC has never been reached; the 110 t landed in
2010–11 was the highest since 1996–97. In recent years (2006–07 to 2019–20), most landings have
come from TRU 3 east coast South Island and TRU 4 on the Chatham Rise (Table 3), with small
landings also coming from TRU 2, 5, and 7 (south-eastern North Island and South Island). Trumpeter
are also taken by recreational fishers in southern New Zealand, and although good estimates of
recreational catch are not available, they may be around one-third to one-half of the commercial catch. TRUMPETER (TRU)
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Figure 1: Reported commercial landings and TACCs for the four main TRU stocks. Top to bottom: TRU 2 (Central
East), TRU 3 (South East Coast), TRU 4 (South East Chatham Rise), and TRU 5 (Southland). TRUMPETER (TRU)
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Table 1: Reported landings (t) for the main QMAs from 1931 to 1982.
Year TRU 1 TRU 2 TRU 3 TRU 4 Year TRU 1 TRU 2 TRU 3 TRU 4
1931 –32 0 0 0 0 1957 0 1 2 0
1932 –33 0 0 0 0 1958 0 1 1 0
1933 –34 0 0 0 0 1959 0 1 1 0
1934 –35 0 0 0 0 1960 0 1 2 0
1935 –36 0 0 0 0 1961 0 1 2 0
1936 –37 0 0 5 0 1962 0 3 1 0
1937 –38 0 3 30 0 1963 0 2 1 0
1938 –39 0 1 22 0 1964 0 2 2 0
1939 –40 0 1 5 0 1965 0 2 1 0
1940 –41 0 2 8 0 1966 0 3 1 0
1941 –42 0 1 4 0 1967 0 1 2 0
1942 –43 0 0 4 0 1968 0 2 1 0
1943 –44 0 0 4 0 1969 0 3 1 0
1944 0 0 10 0 1970 0 5 1 0
1945 0 0 10 0 1971 0 7 1 0
1946 0 0 15 0 1972 0 3 0 0
1947 0 0 12 0 1973 0 3 1 0
1948 0 0 19 0 1974 0 3 1 0
1949 0 0 1 0 1975 0 2 2 0
1950 0 1 3 0 1976 0 1 0 0
1951 0 0 8 0 1977 0 1 0 0
1952 0 0 5 0 1978 0 1 2 0
1953 0 0 3 0 1979 0 4 9 2
1954 0 0 3 0 1980 0 5 5 6
1955 0 1 3 0 1981 0 6 4 2
1956 0 0 2 0 1982 2 21 6 0
Notes:
The 1931–1943 years are April–March but from 1944 onwards are calendar years.
Data up to 1985 are from fishing returns: Data from 1986 to 1990 are from Quota Management Reports.
Data for the period 1931 to 1982 are based on reported landings by harbour and are likely to be underestimated as a result of
under-reporting and discarding practices. Data includes both foreign and domestic landings.
Table 2: Reported total landings (t) of trumpeter from 1931 to 1982. Values for 1931 to 1944 are April–March years,
listed against the April year. Fisheries Annual Report (1931 to 1974) or FSU data (Paul 1999).
Year Landing Year Landings Year Landings Year Landings Year Landings
193 6 20 1946 16 1956 5 1965 4 1974 5
1937 41 1947 13 1957 5 1966 5 1975 4
1938 30 1948 19 1958 3 1967 7 1976 3
1939 37 1949 6 1959 3 1968 5 1977 3
1940 17 1950 6 1960 3 1969 5 1978 6
1941 11 1951 11 1961 3 1970 7 1979 17
1942 5 1952 11 1962 4 1971 10 1980 10
1943 5 1953 5 1963 3 1972 4 1981 12
1944 11 1954 5 1964 3 1973 5 1982 37
1945 11 1955 6
Table 3: Reported landings (t) of trumpeter by QMA and fishing year, 1983–84 to present. [Continued on next
page].
Fishstock TRU 1 TRU 2 TRU 3 TRU 4 TRU 5
FMA 1 2 3 4 5
Landings TA CC Landings TA CC Landings TA CC Landings TA CC Landings TA CC
1982 –83 0 - 5 - 3 - 0 - 0 -
1983 –84 1 - 17 - 2 - 0 - 1 -
1984 –85 0 - 15 - 3 - 0 - 4 -
198 5–86 0 - 4 - 6 - 0 - 1 -
1986 –87 0 - 4 - 5 - 0 - 5 -
1987 –88 0 - 4 - 4 - 0 - 0 -
1988 –89 0 - 7 - 1 - 0 - 0 -
1989 –90 0 - 8 - 5 - 0 - 0 -
1990 –91 3 - 16 - 13 - 5 - 0 -
1991 –92 1 - 16 - 25 - 19 - 1 -
1992 –93 3 - 21 - 21 - 4 - 1 -
1993 –94 3 - 17 - 26 - 24 - 2 -
1994 –95 2 - 20 - 27 - 65 - 5 -
1995 –96 2 - 19 - 29 - 69 - 37 -
1996 –97 2 - 16 - 35 - 33 - 42 -
1997 –98 1 - 11 - 28 - 23 - 6 -
1998 –99 < 1 1 11 9 15 28 16 42 4 18
1999 –00 < 1 1 6 9 11 28 8 42 5 18
2000 –01 < 1 1 6 9 7 28 6 42 3 18
2001 –02 < 1 3 6 20 5 33 9 59 < 1 22
2002 –03 < 1 3 7 20 7 33 32 59 1 22
2003 –04 1 3 6 20 7 33 24 59 4 22
2004 –05 < 1 3 5 20 8 33 70 59 3 22
200 5–06 < 1 3 7 20 8 33 65 59 3 22
2006 –07 < 1 3 8 20 16 33 66 59 3 22 TRUMPETER (TRU)
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Table 3 [Continued]
Fishst ock TRU 1 TRU 2 TRU 3 TRU 4 TRU 5
FMA 1 2 3 4 5
Landings TA CC Landings TA CC Landings Landings TA CC Landings
20 07–08 1 3 9 20 22 33 63 59 4 22
2008 –09 < 1 3 9 20 21 33 19 59 6 22
2009 –10 < 1 3 8 20 22 33 56 59 5 22
2010 –11 < 1 3 5 20 15 33 78 59 8 22
2011 –12 < 1 3 6 20 15 33 76 59 7 22
201 2–13 < 1 3 8 20 27 33 47 59 4 22
2013 –14 < 1 3 3 20 13 33 48 59 4 22
2014 –15 0 3 5 20 11 33 31 59 4 22
2015 –16 < 1 3 4 20 15 33 49 59 3 22
2016 –17 < 1 3 3 20 19 33 36 59 3 22
2017 –18 < 1 3 4 20 14 33 28 59 3 22
2018 –19 < 1 3 3 20 16 33 35 59 4 22
2019 –20 < 1 3 3 20 11 33 47 59 3 22
2020 –21 < 1 3 2 20 27 33 64 59 3 22
Fishstock TRU 6 TRU 7 TRU 8 TRU 9
FMA 6 7 8 9 Total
Landings TA CC Landings TA CC Landings TA CC Landings TA CC Landings TA CC
1982 –83 0 - 0 - 0 - 0 - 8 -
1983 –84 0 - 0 - 0 - 0 - 21 -
1984 –85 0 - 0 - 0 - 0 - 22 -
1985 –86 0 - 0 - 0 - 0 - 11 -
1986 –87 0 - 2 - 0 - 0 - 16 -
1987 –88 0 - 0 - 0 - 0 - 8 -
1988 –89 0 - 1 - 0 - 0 - 9 -
1989 –90 0 - 0 - 1 - 0 - 14 -
1990 –91 0 - 7 - 0 - 0 - 44 -
1991 –92 0 - 4 - 0 - 0 - 69 -
1992 –93 0 - 4 - 2 - 0 - 56 -
1993 –94 0 - 6 - 0 - 0 - 78 -
1994 –95 0 - 4 - 0 - 0 - 123 -
1995 –96 0 - 6 - 0 - 0 - 162 -
1996 –97 2 - 3 - < 1 - < 1 - 133 -
1997 –98 < 1 - 3 - < 1 - 0 - 72 -
1998 –99 0 0 3 2 < 1 0 0 0 50 100
1999 –00 0 0 2 2 < 1 0 0 0 33 100
2000 –01 0 0 3 2 < 1 0 < 1 0 25 100
2001 –02 0 0 5 6 < 1 1 0 0 25 144
2002 –03 0 0 3 6 < 1 1 < 1 0 51 144
2003 –04 0 0 2 6 < 1 1 < 1 0 44 144
2004 –05 0 0 4 6 < 1 1 0 0 90 144
200 5–06 0 0 4 6 < 1 1 0 0 88 144
2006 –07 0 0 4 6 < 1 1 0 0 99 144
2007 –08 < 1 0 2 6 < 1 1 < 1 0 101 144
2008 –09 0 0 2 6 < 1 1 < 1 0 63 144
2009 –10 0 0 3 6 < 1 1 0 0 95 144
2010 –11 < 1 0 4 6 < 1 1 < 1 0 110 144
201 1–12 < 1 0 4 6 < 1 1 < 1 0 108 144
2012 –13 < 1 0 6 6 < 1 1 < 1 1 93 144
2013 –14 0 0 5 6 < 1 1 < 1 0 74 144
2014 –15 0 0 4 6 1 1 0 0 56 144
2015 –16 0 0 4 6 1 1 < 1 0 76 144
2016 –17 0 0 3 6 1 1 < 1 0 65 144
2017 –18 0 0 3 6 < 1 1 < 1 0 52 144
201 8–19 0 0 4 6 < 1 1 < 1 0 62 144
2019 –20 < 1 0 3 6 < 1 1 < 1 0 67 144
2020 –21 < 1 1 2 6 < 1 1 < 1 2 99 147
The data in this table have been updated from those published in previous Plenary Reports by using the data through 1996 97 in table 41
on p. 288 of the “Review of Sustainability Measures and Other Management Controls for the 1998 99 Fishing Year - Final Advice Paper”
dated 6 August 1998. There are no landings reported from TRU 10, which has a TAC of 0
1.2 Recreational fisheries
Results from four separate recreational fishing surveys undertaken in the 1990s are shown in Table 4.
Most of the estimated recreational catch in these surveys was taken in FMAs 3, 5 and 7.
The harvest estimates provided by telephone-diary surveys are no longer considered reliable for
various reasons. A Recreational Technical Working Group concluded that these harvest estimates
should be used only with the following qualifications: a) they may be very inaccurate; b) the 1996 and
earlier surveys contain a methodological error; and c) the 2000 and 2001 estimates are implausibly
high for many important fisheries. In response to these problems and the cost and scale challenges
associated with onsite methods, a National Panel Survey was conducted for the first time throughout
the 2011–12 fishing year. The panel survey used face-to-face interviews of a random sample of 30
390 New Zealand households to recruit a panel of fishers and non-fishers for a full year. The panel TRUMPETER (TRU)
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members were contacted regularly about their fishing activities and harvest information collected in
standardised phone interviews. The national panel survey was repeated during the 2017–18 fishing
year using very similar methods to produce directly comparable results (Wynne-Jones et al 2019).
Recreational catch estimates from the two national panel surveys are given in Table 5. Note that
national panel survey estimates do not include recreational harvest taken under s111 general
approvals.
Table 4: Estimated number of trumpeter caught by recreational fishers by FMA using telephone-diary surveys.
Surveys were carried out in different years in MAF Fisheries regions: South in 1991–92, Central in 1992–93,
North in 1993–94 and National in 1996 (Bradford 1998).
FMA Survey Number CV (%)
1991 –92
FMA 3 Sou th 600029
FMA 5 Sout h6000 33
FMA 7 South 8000 -
1992 –93
FMA 2 Central 1000 -
FMA 3 Central 3000 -
FMA 5 Central 1000 -
FMA 7 Central 0-
FMA 8 Central 0-
1993 –94
FMA 1+9 North 0-
FMA 2 North 1000 -
FMA 8 North 0-
199 6
FMA 1 Natio nal < 500 -
FMA 2 National 1000 -
FMA 3 National 13 000 19
FMA 5 National 21 000 19
FMA 7 National 3000 -
Table 5: Recreational harvest estimates for trumpeter stocks (Wynne-Jones et al 2014, 2019). Mean fish weights were
obtained from boat ramp surveys (Hartill & Davey 2015, Davey et al 2019).
Stock Year Method Number of fish Total weight (t) CV
TRU 1 2011–12 Panel survey 898 1.3 0.83
2017–18 Panel survey 00-
TRU 2 2011–12 Panel survey 787 1.1 0.82
2017–18 Panel survey 32 <1 1.01
TRU 3 2011–12 Panel survey 2 870 4.0 0.41
2017–18 Panel survey 8 070 21.0 0.34
TRU 5 2011–12 Panel survey 1 505 2.1 0.42
2017–18 Panel survey 00-
TRU 7 2011–12 Panel survey 215 0.3 0.83
2017–18 Panel survey 142 <1 1.00
TRU 8 2011–12 Panel survey 273 0.4 1.03
2017–18 Panel survey 00-
1.3 Customary non-commercial fisheries
The customary non-commercial take has not been quantified.
1.4 Illegal catch
There is no quantitative information on illegal fishing activity or catch.
1.5 Other sources of mortality
No quantitative estimates are available regarding the impact of other sources of mortality on
trumpeter stocks. Trumpeter principally occur on deep coastal reefs, where they are taken in net and
line fisheries targeted at other species. TRUMPETER (TRU)
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BIOLOGY
Trumpeter have a Southern Hemisphere distribution in cool temperate waters. They occur in New
Zealand, Australia, the Sub-Antarctic islands of the southern Indian and Atlantic oceans, the Foundation
Seamount in the central South Pacific, and possibly off Chile (Roberts 2003, Tracey & Lyle 2005). In
New Zealand, trumpeter occur from the Three Kings Islands through all of mainland New Zealand to
the Auckland Islands; however they are rare north of East Cape and Cape Egmont (Kingsford et al 1989,
Francis 1996, 2001). The greatest concentrations of trumpeter apparently occur on the Chatham Rise
and around the southern South Island and Stewart Island.
Trumpeter have an extended larval and post-larval duration of up to 9 months in surface waters (Tracey
& Lyle 2005), resulting in extensive drift of young fish among geographic regions. Juveniles are largely
sedentary, but some adults are highly migratory with tagged fish travelling 650 km from Tasmania to
southern New South Wales, and 5800 km from Tasmania to St Paul Island in the southern Indian Ocean
(Lyle & Murphy 2002). This suggests that there is one circum-global genetic stock in the Southern
Hemisphere, although analysis of otolith morphometrics from Tasmania and St Paul and Amsterdam
Islands showed regional variation (Tracey et al 2006) suggesting that migration and inter-breeding may
be limited.
Trumpeter occur mainly over rocky reefs ranging from shallow inshore waters to deep reefs on the
central continental shelf. In New Zealand, they apparently range from a depth of a few metres down to
about 200 m. In Australia some reports indicate they may go as deep as 300 m (reviewed by Paul 1999).
Fish inhabiting inshore reefs tend to be smaller, whereas fish from deep reefs tend to be much larger.
Trumpeter initially settle on to inshore reefs at the end of their long postlarval period, where they remain
for several years, before migrating into deeper areas as they reach maturity (Tracey & Lyle 2005).
Some biological traits differ between New Zealand and Tasmanian populations. Notably, trumpeter are
thought to spawn in winter (July) in New Zealand (Graham 1939b), and late winter to spring in
Australia (peaking around September in Tasmania) (Ruwald et al 1991, Furlani & Last 1993, Morehead
1998, Morehead et al 1998, 2000, Furlani & Ruwald 1999). However, the New Zealand data seem to be
based on limited sampling, and it is uncertain whether the apparent regional difference is real.
Trumpeter grow to about 110–120 cm fork length (FL) and 25–27 kg weight in New Zealand and
Australia (Gomon et al 1994, Paul 1999, Francis 2001). Nothing is known about growth, longevity or
maturity in New Zealand waters. However, because of their importance for aquaculture in Australia, a
comprehensive study has recently been completed on their age and growth in Tasmania (Tracey & Lyle
2005, Tracey et al 2006). Partial validation of age estimates was completed there by comparison of
otolith growth in known-age reared fish and wild fish (enabling validation of the time of formation of
the first growth band), and tracking a strong wild cohort over seven years (ages 1+ to 7+). Although full
validation was not achieved, the authors considered their ages validated up to and beyond the size and
age of habitat transition.
In Australia, trumpeter grow rapidly during the first 4–5 years, reaching about 45 cm FL at that stage,
and moving offshore to deeper water (Tracey & Lyle 2005, Tracey et al 2006). At that time, there is a
reduction in growth rate. They reach a maximum age of about 43 years (though the largest fish in the
samples was 95 cm FL, which is well below the reported maximum length of 120 cm), and there are no
clear differences between males and females (although small sample sizes of fish older than 10 years
meant that the power to detect differences was low). Similarly, no differences were found in growth
rates between fish from Tasmania and St Paul and Amsterdam Islands. Growth rates are seasonally
variable, at least for the first few years, with maximum growth in late summer-autumn. It is thought that
maturation coincides with the offshore movement to deep habitat.
In New Zealand, the only population information available for trumpeter comes from a 6-year survey
(1994–1999) in Paterson Inlet, Stewart Island. Chadderton & Davidson (2003) carried out underwater
visual counts, and obtained comprehensive length-frequency distributions from 1065 fish caught by rod
at 12–15 different sites. Their length-frequency data show two or three clear juvenile cohorts which
progress through time (a strong cohort was also found in Tasmania by Tracey & Lyle (2005)). TRUMPETER (TRU)
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Chadderton & Davidson (2003) interpreted this as evidence of variable annual recruitment pulses. Their
largest fish was 46.9 cm FL with few fish over 40 cm in most years. This is consistent with evidence
from Australia of offshore migration at about 45 cm, though the migration may occur at a slightly
smaller size in the New Zealand population.
STOCKS AND AREAS
There are no data relevant to stock boundaries in New Zealand. Trumpeter are potentially wide-
ranging, and there is one circum-global genetic stock in the Southern Hemisphere, although analysis of
otolith morphometrics from Tasmania and St Paul and Amsterdam Islands showed regional variation
(Tracey et al 2006) suggesting that migration and inter-breeding may be limited. Therefore there may be
localised populations in areas of suitable habitat as they seem to be restricted to rocky reef habitat.
STOCK ASSESSMENT
4.1 Estimates of fishery parameters and abundance
No estimates are available.
4.2 Biomass estimates
No estimates are available.
4.3 Yield estimates and projections
No estimate of MCY is available.
The level of risk to the stock by harvesting trumpeter at recent catch levels cannot be determined.
No estimates of current biomass, fishing mortality, or other information are available which would
permit the estimation of CAY .
4.4 Other factors
There is anecdotal information from Australia and New Zealand that localised populations of
trumpeter can be quickly depleted.
STATUS OF THE STOCKS
No estimates of current and reference biomass are available. It is not known if recent catch levels are
sustainable.
FOR FURTHER INFORMATION
Bradford, E (1998) Harvest estimates from the 1996 national recreational fishing surveys. New Zealand Fisheries Assessment Research
Document 1998/16. 27 p. (Unpublished document held in NIWA library, Wellington.)
Chadderton, W L; Davidson, R J (2003) Baseline monitoring report on fish from the proposed Paterson Inlet (Waka a Te Wera) marine reserve,
Stewart Island (Rakiura) 1994 to 1999. Prepared by Davidson Environmental Ltd for Department of Conservation, Southland.
Research, survey and monitoring report 168. 47 p.
Davey, N; Hartill, B; Carter, M (2019) Mean weight estimates for recreational fisheries in 2017–18. New Zealand Fisheries Assessment
Report 2019/25 . 36 p
Francis, M P (1996) Geographic distribution of marine reef fishes in the New Zealand region. New Zealand Journal of Marine and Freshwater
Research 30: 35–55.
Francis, M (2001) Coastal fishes of New Zealand. An identification guide . Third edition. Reed Publishing, Auckland. 103 p.
Furlani, D; Last, P (1993) Trumpeter. In : Kailola et al . (Eds), Australian Fisheries Resources. Bureau of Resource Sciences, Canberra: 403.
Furlani, D M; Ruwald, F P (1999) Egg and larval development of laboratory-reared striped trumpeter Latris lineata (Forster in Bloch and
Schneider 1801) (Percoidei: Latridiidae) from Tasmanian waters. New Zealand Journal of Marine and Freshwater Research 33:
153–162.
Gomon, M F; Glover, J C M; Kuiter, R H (eds) (1994) The fishes of Australia’s south coast . State Print, Adelaide. 992 p.
Graham, D H (1938) Fishes of Otago Harbour and adjacent seas, with additions to previous records. Transactions and Proceedings of the Royal
Society of New Zealand 68(3): 399–419.
Graham, D H (1939a) Food of the fishes of Otago Harbour and adjacent sea. Transactions of the Royal Society of New Zealand 68(4): 421–36. TRUMPETER (TRU)
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Graham, D H (1939b) Breeding habits of the fishes of Otago Harbour and adjacent seas. Transactions and Proceedings of the Royal Society of
New Zealand 69(3): 361–372.
Graham, D H (1956) A Treasury of New Zealand Fishes . Reed, Wellington. 424 p.
Hartill, B; Davey, N (2015) Mean weight estimates for recreational fisheries in 2011–12. New Zealand Fisheries Assessment Report
2015/25 .
Kingsford, M J; Schiel, D R; Battershill, C N (1989) Distribution and abundance of fish in a rocky reef environment at the subantarctic
Auckland Islands, New Zealand. Polar biology 9 : 179–186.
Lyle, J; Murphy, R (2002) Long distance migration of striped trumpeter. Fishing today 14(6): 16.
Morehead, D T (1998) Effect of capture, confinement and repeated sampling on plasma steroid concentrations and oocyte size in female
striped trumpeter Latris lineata (Latrididae). Marine and freshwater research , 49 (5), 373–377.
Morehead, D T; Pankhurst, N W; Ritar, A J (1998) Effect of treatment with LHRH analogue on oocyte maturation, plasma sex steroid levels and
egg production in female striped trumpeter Latris lineata (Latrididae). Aquaculture 169: 315–331.
Morehead, D T; Ritar, A J; Pankhurst, N W (2000) Effect of consecutive 9- or 12-month photothermal cycles and handling on sex steroid
levels, oocyte development, and reproductive performance in female striped trumpeter Latris lineata (Latrididae),
Aquaculture:189 (3–4) : 293–305.
Paul, L J (1999) A summary of biology and commercial landings, and a stock assessment of the trumpeter, Latris lineata (Bloch and
Schneider 1801) ( Latrididae ) in New Zealand waters. New Zealand Fisheries Assessment Research Document 1999/8. 20 p.
(Unpublished document held by NIWA library, Wellington.)
Ruwald, F P; Searle, L D; Oates, L A (1991) A preliminary investigation into the spawning and larval rearing of striped trumpeter, Latris
lineata . Technical Report, Sea Fisheries Research Laboratory, Division of Sea Fisheries, Tasmania, No: 44. 17 p.
Roberts, C D (2003) A new species of trumpeter (Teleostei; Percomorpha; Latridae) from the central South Pacific Ocean, with a taxonomic
review of the striped trumpeter Latris lineata. Journal of the Royal Society of New Zealand 33. 731–754.
Teirney, L D; Kilner, A R; Millar, R E; Bradford, E; Bell, J D (1997) Estimation of recreational catch from 1991/92 to 1993/94. New Zealand
Fisheries Assessment Research Document 1997/15. 43 p. (Unpublished report held by NIWA library, Wellington.)
Tracey, S R; Lyle, J M (2005) Age validation, growth modelling, and mortality estimates for striped trumpeter ( Latris lineata ) from southeastern
Australia: making the most of patchy data. Fishery Bulletin 103: 169–182.
Tracey, S R; Lyle, J M; Duhamel, G (2006) Application of elliptical Fourier analysis of otolith form as a tool for stock identification. Fisheries
Research 77: 138–147.
Wynne-Jones, J; Gray, A; Heinemann, A; Hill, L; Walton, L (2019). National Panel Survey of Marine Recreational Fishers 2017–2018.
New Zealand Fisheries Assessment Report.2019/24. 108 p
Wynne-Jones,J; Gray, A; Hill, L; Heinemann, A (2014) National Panel Survey of Marine Recreational Fishers 2011–12: Harvest Estimates. New
Zealand Fisheries Assessment Report 2014/67 . |
187616 | https://www.geeksforgeeks.org/maths/how-to-calculate-probability-using-combination/ | How To Calculate Probability using Combination
Combination probability is a mathematical method that involves the process of combination in determining the number of favorable outcomes of an event. We use combinations in probability problems to determine a sequence of outcomes where the order of the outcomes does not matter. Understanding how to calculate combination probability can be a useful mathematical skill in the field of math and science.
In this article, we will discuss in detail the definitions of combination and probability and how to calculate combination probability with solved examples.
Table of Content
What is Combination Probability in Math?
Combination probability, also known as the probability of combinations, involves the process of determining the possibility of specific subset from a large set regardless to the order in which the subsets are chosen. Combination probability has wide range of application in games, statistics, gambling, genetics, etc.
Combination in Math
Combination is known as the selection of items from a large set, where the order of the items does not matter. for example, if there is a set of n items from which we have to select r items, the formula to calculate the combination is:
nCr = n!/(n-r)!r!
where
For example, let A, B, C are three components i.e., n = 3 and combinations size is 2, means r = 2. Then there are 3C2 such combinations present, which is equal to 3. The combinations are AB, BC, and CA. In probability problems, combinations are used to determine a sequence of outcomes where the order of the outcomes is not important.
What is Probability?
Probability is the measure of the possibility that an event will occur. It is defined as a number between 0 and 1 with 0 denoting improbability and 1 denoting certainty. It is a branch of mathematics that deals with the interpretation of random events. For example, when we toss a coin, either we get head or tail, hence only two possible outcomes are possible, the outcomes are (H,T).
Formula for Probability:
Probability of a event to happen P(E) = Number of favorable outcomes/Total Number of outcomes
Formula for Calculating Probability using Combination
To calculate combination probability, it is important to use the correct formula to find the probability of a specific outcome. There are four formula's to calculate the probability. They are,
| Formula | Equation |
--- |
| Combination without repetition | n!/(n-r)!r! |
| Combination with repetition | (r+n-1)!/r!(n-1)! |
| Permutation with repetition | nr |
| Permutation without repetition | n!/(n-r)! |
Formula
Equation
Combination without repetition
n!/(n-r)!r!
Combination with repetition
(r+n-1)!/r!(n-1)!
Permutation with repetition
nr
Permutation without repetition
n!/(n-r)!
Among the above mentioned formulas, the two most important are,
Combination with Repetition
When repetition is allowed, you can use the following equation to find out the number of combinations,
(r+n-1)!/r!(n-1)!
where,
Example: Total number of balls in the pool= n = 5. The number of balls to be selected = r = 4, where the selection of balls can be repeated. The order of selection does not matter. Find the number ways the ball can be selected?
Solution:
We will use the formula, nCr = (r+n-1)!/r!(n-1)!
Now, putting the values we get, 5C4 = (4+5-1)! / 4!·(5-1)!
= 8! / 4!·4! = 8×7×6×5 / 4×3×2×1 = 70
∴ There are 70 possible ways to select the ball.
Combination without Repetition
When repetition is not allowed, you can use the following equation to find out the number of combinations,
n!/(n-r)!r!
where,
Example: The principle would like to assemble a committee of 6 students from the 11 member student council. How many different committees can be chosen?
Solution:
We will use the formula, where, nCr = n!/(n-r)!r!
Now, after putting thee values we get, 11C6 = 11!/(11-6)!·6!
= 11! / 5!·6! = 11×10×9×8×7 / 5×4×3×2×1 =66×7 = 462
∴ 462 committees can be chosen.
How to Calculate Combination Probability?
Combination is a mathematical concept which helps us in determining the possible outcomes of an event. To calculate combination probability we have to follow several steps and utilize several formulas. They are discussed below:
Understand the Mathematical Notation of Combination
In combination probability, different formulas are required to solve various problems. But the basic concept behind those formulas are same. To calculate combination probability, we use several terms and notations, which are:
Identify the Style of Calculation
In order to apply correct formula for calculating problems on combination, it is important to understand the type of calculation you are performing. The first thing to notice is whether the calculation is a permutation or combination. Then you must note that is there any repetition of the same value during the combination. At the end there will be four calculation types, one pair consisting repeat and no repeat of combination and another for permutation.
Choose the Appropriate Formula
After identifying the correct style of the calculation, it is important to use the correct formula to find the probability of a specific outcome. The four combination probability equations are:
Input variables and Calculate the Probability
After selecting appropriate formula, now put all the values to calculate the combination. The probability for each combination can be calculated by dividing the number of favorable combinations by total number of combinations. Then combining the number of favorable combinations with the individual probabilities, we get overall combination probability.
Also, Check
Solved Examples on Calculation of Combination Probability
Example 1. In a team, 3 boys and 4 girls are there. Among them 4 members need to be selected for one round of a game. Find the probability of selecting an equal number of boys and girls?
Solution:
Probability of selecting an equal number of boys and girls = (4C2 × 3C2)/7C4
(4C2 × 3C2)/7C4 = 18/35
Hence probability of selecting equal number of boys and girls is 18/35
Example 2. What is the number of possible combinations to choose 6 numbers from a set of 49 numbers?
Solution:
49C6 = 49!/6! (49-6)!
49C6 = 49!/(6! × 43!) possible ways.
Example 3. What is the number of ways need to form a group of 3 people from a group of 10?
Solution:
10C3 = 10!/3!(10-3)!
10C3 = 10×9×8/3×2×1
10C3 = 120
∴There are 120 different ways to do this.
Example 4. How many ways can 8 students be chosen from a class of 21?
Solution:
21C8 = 21!/8!×(21-8)!
21C8 = (21×20×19×18×17×16×15×14)/(8×7×6×5×4×3×2×1)
21C8 = 203,490
∴ There are 203,490 ways to chose from a class of 21.
Practice Questions on Calculating Probability Using Combination
Q1. A committee of 4 people is to be selected from a group of 10 people. What is the probability that a specific person, Alice, is on the committee?
Q2. From a standard deck of 52 cards, 5 cards are drawn at random. What is the probability that all 5 cards are of the same suit?
Q3. A class has 12 boys and 8 girls. If a group of 5 students is selected at random, what is the probability that the group will consist of 3 boys and 2 girls?
Q4. A jar contains 7 red marbles and 5 blue marbles. If you draw 3 marbles at random, what is the probability that exactly 2 of them are red?
Q5. From a shelf of 8 math books and 6 science books, you randomly select 4 books. What is the probability that you select 2 math books and 2 science books?
R
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Find all positive integers n and k, such that
1+(n2)+(n4)=2k
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(n,k)=(2,1),(3,2),(4,3),(5,4).
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Another solution is (n,k)=(10,8). I checked up to n=107 and found no other solutions.
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The LHS is related to Moser's circle problem, but the Diophantine equation may be difficult. See also the references at the OEIS-entry.
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187618 | https://pubs.acs.org/doi/10.1021/acs.jctc.3c00934 | Supplemental Material: Bound state breaking and the importance of thermal exchange–correlation effects in warm dense hydrogen
Zhandos Moldabekov, ∗,†,‡ Sebastian Schwalbe, †,‡ Maximilian P. Böhme, † Jan Vorberger, ‡ Xuecheng Shao, ¶ Michele Pavanello, ¶ Frank R. Graziani, ∥ and Tobias Dornheim ∗,†,‡
†Center for Advanced Systems Understanding (CASUS), D-02826 Görlitz, Germany
‡Helmholtz-Zentrum Dresden-Rossendorf (HZDR), D-01328 Dresden, Germany
¶Department of Chemistry, Rutgers University, Newark, NJ 07102, USA
§Department of Physics, Rutgers University, Newark, NJ 07102, USA
∥Lawrence Livermore National Laboratory (LLNL), California 94550 Livermore, USA.
E-mail: z.moldabekov@hzdr.de; t.dornheim@hzdr.de
Ground state bond breaking
Kohn-Sham density functional theory (KS-DFT) 1,2 is the workhorse of ground-state elec-tronic structure investigations given its often suitable accuracy and reasonable computational effort. However, it is well known that KS-DFT is not able to correctly describe the dissociation of molecules. 3,4 In KS-DFT electrons artificially interact with themselves; this is known as the S1 self-interaction (SI) error. The Perdew-Zunger self-interaction correction (PZ-SIC) 5 removes the SI with an orbital-by-orbital correction. In contrast to KS-DFT, it delivers an accurate description of the dissociation limit. 3 A novel flavor of PZ-SIC is the Fermi-Löwdin orbital self-interaction correction (FLO-SIC). 6–9
A clear advantage of FLO-SIC is the method-specific concept of Fermi-orbital descriptors (FODs), i.e., semi-classical electron positions. It has been shown that these FODs, as well as the respective localized FLOs, carry bonding information 10,11 and can be used to analyze, interpret, and guide SIC solutions. Recently, FODs were used to describe bond breaking from an electron perspective for a Diels-Alder reaction. 12
Thus, FLO-SIC is a promising method to analyze bonding for a given system at T = 0 K. To analyse the bond dissociation of the H 2 molecule at T = 0 K, we investigated critical points in the total energy, density, and reduced density gradient (RDG) along the dissociation curve to enable a comparison with finite temperature calculations. Along the bond dissociation, one observes different bond situations. For instance, d ≤ 0.75 Årepresents a compressed bond, for d = 0 .75 Å the molecule is close to its ground state ge-ometry (bonding region), and for d ≥ 0.75 Å one observes the stretched bond region. At large bond separation one reaches the dissociation limit. As we will see later, the spin of a system is an important property for bond dissociation for T = 0 K ground state calculations. The spin of a system in an unrestricted, open-shell description is given by spin = Nα − Nβ (1) with Nσ being the number of electrons in the respective spin channel. In the ground state at T = 0 K, the H 2 molecule has a spin of 0 with one electrons being in the α channel and the other electron being in the β channel. If both electrons are in one spin channel the spin of the system would be 2. The total energy of H 2 has been calculated along the bond dissociation (see Fig. S1) for both possible spin values. Once the H 2 bond breaks, a single electron will move to each S2 hydrogen nucleus. At infinite bond separation, i.e., the atomic dissociation limit, it does not matter if the electron is in the α or the β spin channel.
Figure S1: H 2 bond dissociation curve calculated using KS-DFT and FLO-SIC. The bottom line visualize KS densities and RDG for spin=0 for different bonding regions. FLO-SIC showcases the aforementioned behavior. While the ground state has spin=0, at the bond-breaking distance of d = 1 .9 Å in the case of FLO-SIC the spin=2 states become energetically preferred. At large separation distances, FLO-SIC energies for spin=2 approach the correct atomic limit, i.e., two times the energy of a hydrogen atom ( ≈ − 1 EH). Note that FLO-SIC is quasi-exact for the H atom. The accuracy is only limited by the used basis set and numerical quadrature. For the numerical parameters used for our investigations we get a total energy of −0.499956 EH. In clear contrast, KS-DFT does not give the correct energies at the atomic dissociation limit (see Fig. S1). We present the RDG for the H atom (see Fig. S2 (a)) and the H 2 molecule ( d = 0 .75 Å, see Fig. S2 (b)). Clearly, there is a distinct peak around n ≈ 0 for KS-DFT. The same behavior is observed for FLO-SIC (not shown). This can be understood more intuitively from the 3D representation of the RDG. In the compressed bonding and bonding region the RDG has approximately the form of an ellipsoid (see Fig. S1) and in the stretched bonding region the RDG resembles the shape of a dumbbell (see Fig. S1). While the density in those regions resembles an ellipsoid, the RDG has some sort of leveled surface structure. For
d = 5 .0 Å the RDG has the form of two spheres only marginal touching each other, while the density for this values has the form of two separated spheres. For d ≥ 6.0 Å the RDG S3 (a) KS-DFT RDG for H atom (b) KS-DFT RDG for H 2 (d = 0 .75 Å)
Figure S2: KS-DFT RDG values for the H atom (a) and the H 2 molecule ( d = 0 .75 Å, bonding region) (b). For both the H atom and H 2 molecule there is a sharp peak at low densities n ≈ 0.has also the form of two separated spheres. The general features of s[n] for the bonding region (see Fig. S2 (b)), i.e., the peak of s[n] and the ellipsoid shape of the RDG, match the ones of the warm dense hydrogen system presented in Fig. 2 of the main manuscript for
rs = 4 . The dumbbell shape of the RDG in the stretched bond region match the H 2 feature presented in Fig. 2 of the main manuscript for rs = 2 . However, while there is a distinct peak of s[n] in the ground state stretched bond region, no such feature is present in the warm dense hydrogen system of rs = 2 . In the ground state atomic dissociation limit the RDG resembles separated spheres, which can be compared to the separated entities presented in Fig. 2 of the main manuscript for rs = 1 .We monitored the RDG for various bond lengths. For all four bonding regions, i.e., the compressed bonding-, the bonding-, the stretched bonding-, and the bond dissociation-region, the distinct peak of s[n] around n ≈ 0 is present. For small bonding distances
d ≤ 1 Å the RDG reaches its highest values (see Fig. S3 (a)). Additionally, we investigated S[α]/S with KS-DFT and FLO-SIC along the bond dis-sociation curve. For all calculated bond lengths S[α]/S can be fitted with ae −bα (see Fig. S3 (b)). For α = −1, S[α]/S reaches its maxima and exponentially tends to zero afterwards. Thus in KS-DFT and FLO-SIC, the features presented for warm dense hydrogen S4 (a) Maxima of RDG for H 2 (b) Generalised dimensionless RDG for H2
(d= 0 .75 Å)
Figure S3: Maxima of RDG for various bond lengths (a), and KS-DFT S[α]/S for
d = 0 .75 Å (b). The sharp peak at low densities n ≈ 0 never smears out in ground state KS-DFT and FLO-SIC for spin=0 and spin=2. This is in clear contrast to the finite tem-perature calculations presented in the main manuscript. For all investigated bond lengths
S[α]/S (see (b) for d = 0 .75 Å) can be approximated with ae −bα . For a bond length of
d = 0 .75 Å, we find 0.72 exp (−10 .84 α). The fit gives a coefficient of determination of R2 = 1 .in the main manuscript (see Fig. 4) are not present for the ground state bond dissociation of the H 2 molecule.
Simulation details for ground state calculations
The all-electron Gaussian-type orbital (GTO) electron structure code chilli.jl 12,13 was utilized to perform KS-DFT and FLO-SIC calculations. We used the local density approximation in form of the LDA-VWN 14 exchange-correlation functional. For the calculations the cc-pVQZ basis set and a numerical quadrature of (150,302) was applied. Reference calculations have been performed with PySCF .15 The mean absolute error (MAE) for the total energy of both codes is 4.56 · 10 −5 EH. This indicates a sufficient reproducibility of the presented results for the used numerical parameter space. The quality of the density and RDG has been checked by comparing the presented results to the results S5 of the plane-wave (PW) electronic structure code eminus 16 . For the PW calculations a box size of 20 a0 and a cutoff energy of Ecut = 30 EH was used. The general shapes of the density and RDG agree between the GTO and PW calculations. In addition, the general bond features of the dissociation curve can be reproduced as well. The respective MAE between the GTO chilli.jl and PW eminus total energies is 3.13 · 10 −3EH. Given the conceptional difference of the basis sets the agreement is sufficient. Note that s is sensitive to numerical values of the density close to zero. Therefore we only analyzed s[n] for n ≥ 10 −5 [1 /a 30].
PIMC and KS-DFT results for different densities at T = TF
To find a correlation between the sign of αmin and the breaking of bound states we analysed the RDG data in the range from rs = 1 to rs = 4 with the step 0.5. In Fig. S4, we show the αmin dependence on the density parameter computed using PIMC and KS-DFT data in the range from rs = 1 to rs = 4 for the considering configuration of 14 protons. From Fig. S4, we see that the KS-DFT results for αmin are in close agreement with the PIMC results at 1.5 ≤ rs ≤ 4 and that the T-LDA based results are in a slightly better agreement with the PIMC data compared to the LDA and PBE based results. The RDG distributions with respect to density computed using the PIMC data are presented in Fig. S5. The RDG distributions computed from KS-DFT simulations using T-LDA are shown in Fig. S6. From Figs S5 and S6, we see that the maximum value of s[n] decreases with the decrease in rs. By analysing the RDG data and corresponding αmin values, we observe that we have
max( s[n]) > 1 at rs > 2.5 with αmin > 0 and max( s[n]) ≲ 1 at rs ≤ 2.5 with αmin < 0.
PIMC results for different temperatures at rs = 4
In Fig. S7, we show the PIMC data for s[n] for four values of temperature at rs = 4 , i.e.,
T = TF , T = 2 TF , T = 4 TF , and T = 8 TF . The increase in the temperature leads to the S6 Figure S4: The dependence of the value of α at which S(α) attains its minimum on rs in the range from rs = 1 to rs = 4 at T = TF .
Figure S5: The PIMC results for the distribution of the RDG with respect to density for the warm dense hydrogen with density parameters rs = 4 .0, 3.5, 3.0, 2.5, 2.0, 1.5 at the fixed value of the degeneracy parameter θ = 1 .S7 Figure S6: The T-LDA based KS-DFT results for the distribution of the RDG with respect to density for the warm dense hydrogen with density parameters rs = 4 .0, 3.5, 3.0, 2.5, 2.0,
1.5 at the fixed value of the degeneracy parameter θ = 1 .decrees of the amplitude of s[n] distribution. This indicates the deterioration of the shell structures of the bound states in hot hydrogen. We note that the ionisation mechanism due to heating at a constant density and due to compression at a constant degeneracy are not equivalent. Particularly, the compression leads to the increase in the Fermi energy and the increase of the effect of the field of other protons on a given molecule (atom). In the case of the heating at a constant temperature, we have only thermal effect induced ionization. In this work, we do not investigate the latter case. Nevertheless, the results presented in Fig. S7 clearly indicate that the RDG based analysis can be used for the investigation of the change in the electronic structure due to heating at a constant value of the density. S8 Figure S7: The distribution of the RDG with respect to density for the warm dense hydrogen at different temperatures with the density parameter set to rs = 4 .
References
(1) Hohenberg, P.; Kohn, W. Inhomogeneous Electron Gas. Phys. Rev. 1964 , 136, B864. (2) Kohn, W.; Sham, L. J. Self-Consistent Equations Including Exchange and Correlation Effects. Phys. Rev. 1965 , 140, A1133. (3) Schwalbe, S.; Hahn, T.; Liebing, S.; Trepte, K.; Kortus, J. Fermi-Löwdin orbital self-interaction corrected density functional theory: Ionization potentials and enthalpies of formation. Journal of computational chemistry 2018 , 39, 2463–2471. (4) Shahi, C.; Bhattarai, P.; Wagle, K.; Santra, B.; Schwalbe, S.; Hahn, T.; Kortus, J.; Jackson, K. A.; Peralta, J. E.; Trepte, K.; others Stretched or noded orbital densities and self-interaction correction in density functional theory. The Journal of Chemical Physics 2019 , 150, 174102. (5) Perdew, J. P.; Zunger, A. Self-interaction correction to density-functional approxima-tions for many-electron systems. Phys. Rev. B 1981 , 23, 5048. (6) Pederson, M. R.; Ruzsinszky, A.; Perdew, J. P. Communication: Self-interaction cor-S9 rection with unitary invariance in density functional theory. J. Chem. Phys. 2014 , 140, 121103. (7) Pederson, M. R. Fermi orbital derivatives in self-interaction corrected density functional theory: Applications to closed shell atoms. J. Chem. Phys. 2015 , 142, 064112. (8) Pederson, M. R.; Baruah, T. Advances In Atomic, Molecular, and Optical Physics; Elsevier, 2015; pp 153–180. (9) Yang, Z.; Pederson, M. R.; Perdew, J. P. Full self-consistency in the Fermi-orbital self-interaction correction. Phys. Rev. A 2017 , 95, 052505. (10) Schwalbe, S.; Trepte, K.; Fiedler, L.; Johnson, A. I.; Kraus, J.; Hahn, T.; Peralta, J. E.; Jackson, K. A.; Kortus, J. Interpretation and Automatic Generation of Fermi-Orbital Descriptors. J. Comput. Chem. 2019 , 40, 2843–2857. (11) Trepte, K.; Schwalbe, S.; Liebing, S.; Schulze, W. T.; Kortus, J.; Myneni, H.; Ivanov, A. V.; Lehtola, S. Chemical bonding theories as guides for self-interaction cor-rected solutions: Multiple local minima and symmetry breaking. J. Chem. Phys. 2021 ,155, 224109. (12) Schulze, W. T.; Schwalbe, S.; Trepte, K.; Croy, A.; Kortus, J.; Gräfe, S. Bond for-mation insights into the Diels–Alder reaction: A bond perception and self-interaction perspective. The Journal of Chemical Physics 2023 , 158. (13) Schwalbe, S.; Trepte, K.; Schulze, W. T. chilli.jl [chilli_jl]. 2023, accessed date 12/01/2023; .(14) Vosko, S. H.; Wilk, L.; Nusair, M. Accurate spin-dependent electron liquid correlation energies for local spin density calculations: a critical analysis. Canadian Journal of Physics 1980 , 58, 1200–1211, tex.eprint: S10 (15) Sun, Q.; others Recent developments in the PySCF program package. J. Chem. Phys.
2020 , 153, 024109. (16) Schulze, W.; Trepte, K.; Schwalbe, S. wangenau/eminus: v2.4.0. 2023, accessed date 12/01/2023; .S11 |
187619 | https://www.collinsdictionary.com/us/dictionary/english/fanciful | English
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Definition of 'fanciful'
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fanciful
(fænsɪfəl )
adjective
If you describe an idea as fanciful, you disapprove of it because you think it comes from someone's imagination, and is therefore unrealistic or unlikely to be true.
[disapproval]
...fanciful ideas about Martian life.
Synonyms: unreal, wild, ideal, romantic More Synonyms of fanciful
Collins COBUILD Advanced Learner’s Dictionary. Copyright © HarperCollins Publishers
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fanciful in American English
(ˈfænsəfəl )
adjective
1.
full of fancy; indulging in fancies; imaginative in a playful way; whimsical
2.
created in the fancy; imaginary; not real
a fanciful tale
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showing fancy in construction or design; quaint; odd
fanciful costumes
Webster’s New World College Dictionary, 5th Digital Edition. Copyright © 2025 HarperCollins Publishers.
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ˈfancifully
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ˈfancifulness
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fanciful in American English
(ˈfænsɪfəl)
adjective
1.
characterized by or showing fancy; capricious or whimsical in appearance
a fanciful design of butterflies and flowers
2.
suggested by fancy; imaginary; unreal
fanciful lands of romance
3.
led by fancy rather than by reason and experience; whimsical
a fanciful mind
SYNONYMS 2. visionary, baseless, illusory.
Most material © 2005, 1997, 1991 by Penguin Random House LLC. Modified entries © 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd
Derived forms
fancifully
adverb
fancifulness
noun
Word origin
[1620–30; fancy + -ful]
COBUILD frequency band
fanciful in British English
(ˈfænsɪfʊl )
adjective
1.
not based on fact; dubious or imaginary
fanciful notions
2.
made or designed in a curious, intricate, or imaginative way
3.
indulging in or influenced by fancy; whimsical
Collins English Dictionary. Copyright © HarperCollins Publishers
Derived forms
fancifully (ˈfancifully)
adverb
fancifulness (ˈfancifulness)
noun
Examples of 'fanciful' in a sentence fanciful
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Weeks ago such a scenario was fanciful.
The Guardian (2015)
The idea of hordes of vengeful sharks is fanciful.
The Guardian (2015)
He set out fanciful theories of an infection during my daily bedside encounters with him.
The Guardian (2016)
The prospect of survival is not fanciful.
The Guardian (2016)
But the notion that outsourcing can be abolished altogether seems fanciful.
The Guardian (2018)
It is not such a fanciful idea.
The Sun (2012)
It was always fanciful to think that wind and solar farms could stop global warming.
Times, Sunday Times (2015)
But he forecast courts will soon be entirely electronic and suggested it was not entirely fanciful that computers could one day replace judges.
The Sun (2016)
Now, such an idea seems fanciful at best.
Times, Sunday Times (2016)
That is a fanciful notion.
Times, Sunday Times (2011)
Related word partners fanciful
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In other languages fanciful
British English: fanciful
ADJECTIVE /ˈfænsɪfʊl/
If you describe an idea as fanciful, you disapprove of it because you think it comes from someone's imagination, and is therefore unrealistic.
...fanciful ideas about Martian life.
American English: fanciful
/ˈfænsɪfəl/
Brazilian Portuguese: irreal
Chinese: 异想天开的
European Spanish: disparatado
French: fantaisiste
German: abstrus
Italian: fantasioso
Japanese: 空想的な
Korean: 허무맹랑한
European Portuguese: irreal
Spanish: disparatado
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187620 | https://stats.oarc.ucla.edu/stata/seminars/multilevel-modeling-stata12/ | Multilevel Modeling in Stata 12
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Multilevel Modeling in Stata 12
The purpose of this seminar is to introduce multilevel modeling using Stata 12. Before we begin, you will want to be sure that your copy of Stata is up-to-date. To do this, please type
update all
in the Stata command window and follow any instructions given. These updates include not only fixes to known bugs, but also add some new features that may be useful. I am using Stata 12.1.
Before we begin looking at examples in Stata, we will review some basic issues and concepts in multilevel data analysis. Please note that in this seminar we will only be considering outcomes that are continuous.
Why do we need multilevel analysis and how is it different from ordinary least squares regression?
Sampling designs
Most people do not conduct their own surveys. Rather, they use survey data that some agency or company collected and made available to the public. The documentation must be read carefully to find out what kind of sampling design was used to collect the data. This is very important because many of the estimates and standard errors are calculated differently for the different sampling designs. Hence, if you mis-specify the sampling design, the point estimates and standard errors will likely be wrong.
Below are some common features of many sampling designs.
Sampling weights: There are several types of weights that can be associated with a survey. Perhaps the most common is the sampling weight, sometimes called a probability weight, which is used to weight the sample back to the population from which the sample was drawn. By definition, this weight is the inverse of the probability of being included in the sample due to the sampling design (except for a certainty PSU, see below). The probability weight, called a pweight in Stata, is calculated as N/n, where N = the number of elements in the population and n = the number of elements in the sample. For example, if a population has 10 elements and 3 are sampled at random with replacement, then the probability weight would be 10/3 = 3.33. In a two-stage design, the probability weight is calculated as f 1 f 2, which means that the inverse of the sampling fraction for the first stage is multiplied by the inverse of the sampling fraction for the second stage. Under many sampling plans, the sum of the probability weights will equal the population total.
While many textbooks will end their discussion of probability weights here, this definition does not fully describe the probability weights that are included with actual survey data sets. Rather, the "final weight" usually starts with the inverse of the sampling fraction, but then incorporates several other values, such as corrections for unit non-response, errors in the sampling frame (sometimes called non-coverage), and poststratification. Because these other values are included in the probability weight that is included with the data set, it is often inadvisable to modify the probability weights, such as trying to standardize them for a particular variable, e.g., age.
PSU: This is the primary sampling unit. This is the first unit that is sampled in the design. For example, school districts from California may be sampled and then schools within districts may be sampled. The school district would be the PSU. If states from the US were sampled, and then school districts from within each state, and then schools from within each district, then states would be the PSU. One does not need to use the same sampling method at all levels of sampling. For example, probability-proportional-to-size sampling may be used at level 1 (to select states), while cluster sampling is used at level 2 (to select school districts). In the case of a simple random sample, the PSUs and the elementary units are the same. In general, accounting for the clustering in the data (i.e., using the PSUs), will increase the standard errors of the estimates. Conversely, ignoring the PSUs will tend to yield standard errors that are too small, leading to false positives when doing significance tests.
Strata: Stratification is a method of breaking up the population into different groups, often by demographic variables such as gender, race or SES. Each element in the population must belong to one, and only one, strata. Once the strata have been defined, one samples from each stratum as if it were independent of all of the other strata. For example, if a sample is to be stratified on gender, men and women would be sampled independent of one another. This means that the probability weights for men will likely be different from the probability weights for the women. In most cases, you need to have two or more PSUs in each stratum. The purpose of stratification is to reduce the standard error of the estimates, and stratification works most effectively when the variance of the dependent variable is smaller within the strata than in the sample as a whole.
FPC: This is the finite population correction. This is used when the sampling fraction (the number of elements or respondents sampled relative to the population) becomes large. The FPC is used in the calculation of the standard error of the estimate. If the value of the FPC is close to 1, it will have little impact and can be safely ignored. In some survey data analysis programs, such as SUDAAN, this information will be needed if you specify that the data were collected without replacement (see below for a definition of "without replacement"). The formula for calculating the FPC is ((N-n)/(N-1))1/2, where N is the number of elements in the population and n is the number of elements in the sample. To see the impact of the FPC for samples of various proportions, suppose that you had a population of 10,000 elements.
Sample size (n) FPC
1 1.0000
10 .9995
100 .9950
500 .9747
1000 .9487
5000 .7071
9000 .3162
Replicate weights: Replicate weights are a series of weight variables that are used to correct the standard errors for the sampling plan. They serve the same function as the PSU and strata (which use a Taylor series linearization) to correct the standard errors of the estimates for the sampling design. Many public use data sets are now being released with replicate weights instead of PSUs and strata in an effort to more securely protect the identity of the respondents. In theory, the same standard errors will be obtained using either the PSU and strata or the replicate weights. There are different ways of creating replicate weights; the method used is determined by the sampling plan. The most common are balanced repeated and jackknife replicate weights. You will need to read the documentation for the survey data set carefully to learn what type of replicate weight is included in the data set; specifying the wrong type of replicate weight will likely lead to incorrect standard errors. For more information on replicate weights, please see Stata Library: Replicate Weights and Appendix D of the WesVar Manual by Westat, Inc. Several statistical packages, including Stata, SUDAAN, WesVar and R, allow the use of replicate weights.
Consequences of not using the design elements
Sampling design elements include the sampling weights, post-stratification weights (if provided), PSUs, strata, and replicate weights. Rarely are all of these elements included in a particular public-use data set. However, ignoring the design elements that are included can often lead to inaccurate point estimates and/or inaccurate standard errors.
Sampling with and without replacement
Most samples collected in the real world are collected "without replacement". This means that once a respondent has been selected to be in the sample and has participated in the survey, that particular respondent cannot be selected again to be in the sample. Many of the calculations change depending on if a sample is collected with or without replacement. Hence, programs like SUDAAN request that you specify if a survey sampling design was implemented with our without replacement, and an FPC is used if sampling without replacement is used, even if the value of the FPC is very close to one.
Examples
For the examples in this seminar, we will use the Adult data set from NHANES III. The data set, set up file and documentation can be downloaded from the NHANES web site. An executable file is available that contains the data, SAS code to set up the data, and the documentation: (ADULT.exe(16 MB): Data (65 MB), SAS code (128 KB), Documentation (1 MB)) . The NHANES III data sets were released with both pseudo-PSUs/pseudo-strata and replicate weights. We will show examples using both of these methods of variance estimation. For more information on the setup for NHANES III using other packages, and setups using other commonly used public-use survey data sets, please see our page on sample setups for commonly used survey data sets .
Reading the documentation
The first step in analyzing any survey data set is to read the documentation. With many of the public use data sets, the documentation can be quite extensive and sometimes even intimidating. Instead of trying to read the documentation "cover to cover", there are some parts you will want to focus on. First, read the Introduction. This is usually an "easy read" and will orient you to the survey. There is usually a section or chapter called "Sample Design and Analysis Guidelines", "Variance Estimation", etc. This is the part that tells you about the design elements included with the survey and how to use them. Some even give example code (although usually for SUDAAN). If multiple sampling weights have been included in the data set, there will be some instruction about when to use which one. If there is a section or chapter on missing data or imputation, please read that. This will tell you how missing data were handled. You should also read any documentation regarding the specific variables that you intend to use. As we will see little later on, we will need to look at the documentation to get the value labels for the variables. This is especially important because some of the values are actually missing data codes, and you need to do something so that Stata doesn’t treat those as valid values (or you will get some very "interesting" means, totals, etc.). Despite the length of the SAS code that comes with the data set, the value labels are not included.
Getting the data into Stata
The first, and most obvious, thing that you need to do after downloading the data, is to get the data into Stata. The data file itself is actually an ASCII file, and ASCII files can easily be read into Stata. However, the SAS code contains a lot (but not all) of the information that is needed to make the numbers in the ASCII file meaningful. For example, the SAS code contains the column numbers for each variable, the variable name, and the variable label. It does not, however, contain the value labels; you need to get those from the documentation. If you really had to, you could open the SAS code in any text editor and "copy and paste" the information into a Stata do-file, modify it as needed to make a program that could read the ASCII file. Even if you are not a SAS user, this is probably much more work than making the few necessary modifications to the SAS code provided by NHANES. Let’s look at some of the SAS code to see what we need to modify to get it to run. The first few lines of the SAS code look like this:
FILENAME ADULT "D:QuestionnaireDATADULT.DAT"LRECL=3348;
LRECL includes 2 positions for CRLF, assuming use of PC SAS;
DATA WORK;
INFILE ADULT MISSOVER;
LENGTH
SEQN 7
DMPFSEQ 5
and the last few lines look like this:
HAZMNK5R = "Average K5 BP from household and MEC"
HAZNOK5R = "Number of BP’s used for average K5";
There are two things that you will need to change. The first is the path specification in the first line. Leave the quotes and the file name (ADULT.DAT). The second thing that needs to be changed is at the very end of the file. Replace the box with
run;
Once you have made these changes, you can click on the running person at the top of the SAS screen (assuming that nothing is highlighted), and the whole program will run. You should glance through the log file looking for anything in red print that indicates an error (such as the .dat file isn’t in the location that you specified, which is a common mistake). To make sure that everything worked as planned, you can run the following command:
proc contents data = work; run;
This should give you some output telling you about the data set. Once you know that the data set is in SAS correctly, we can now move it over into Stata. First, we need to save the SAS data set to our hard drive. On the set statement, specify the path where you want the SAS data set saved.
data "D:dataworkingnhanes_adult1"; set work; run;
Now that the SAS data set is saved to our hard drive, we can use a program like Stat/Transfer to transfer the data into Stata format.
The svyset command
Before you do anything else, please make sure that your Stata is up-to-date. You can type
update all
and follow the instructions given. Stata has made many nice upgrades to the svy: commands since Stata 11 was released, so updating is a really good idea.
When you first try to open the NHANES data set in Stata, you will likely get an error message about being out of memory. Unlike SAS and SPSS, Stata holds a data set in memory, so if an insufficient amount of RAM is allocated to Stata, Stata won’t be able to read in the data set. You get around this by increasing the memory:
set memory 50m
Now you should be able to open the data set. (If 50m doesn’t do it on your computer, try 70m, 90m, etc.) Now that the data are in Stata, we need to do one more thing before starting our analyses: we need to issue the svyset command. The svyset command tells Stata about the design elements in the survey. Once this command has been issued, all you need to do for your analyses is use the svy: prefix before each command. Because the NHANES III data were released with both PSUs/strata and replicate weights, we have a choice of how to specify the svyset command. We will illustrate both ways, starting the use of the PSU/strata variables. Now if you read the NHANES documentation on variance estimation, you would know that these aren’t really the PSUs and strata used in the data collection. Rather, they are pseudo-PSUs and pseudo-strata. Noise was added to the original PSU and strata variables to help protect the identity of the survey respondents, which is why they are called pseudo-PSUs and pseudo-strata. Stata doesn’t care about these variables being "pseudo", and they are used in the svyset command as regular PSU and strata variables. The svyset command looks like this:
use "D:dataworkingnhanes_adult1", clear svyset sdppsu6 [pweight = wtpfqx6], strata(sdpstra6) singleunit(centered)
pweight: wtpfqx6
VCE: linearized
Single unit: centered
Strata 1: sdpstra6
SU 1: sdppsu6
FPC 1:
The singleunit option was added in Stata 10. This option allows for different ways of handling a single PSU in a stratum. If you use the default option, missing, then you will get no standard errors when Stata encounters a single PSU in a stratum. This can happen as a result of missing data or subsetting the data. There are three other options. One is certainty, meaning that the singleton PSUs be treated as certainty PSUs; certainty PSUs are PSUs that were selected into the sample with a probability of 1 (in other words, these PSUs were certain to be in the sample) and do not contribute to the standard error. The scaled option gives a scaled version of the certainty option. The scaling factor comes from using the average of the variances from the strata with multiple sampling units for each stratum with one PSU. The centered option centers strata with one sampling unit at the grand mean instead of the stratum mean.
svydescribe
Survey: Describing stage 1 sampling units
pweight: wtpfqx6
VCE: linearized
Single unit: centered
Strata 1: sdpstra6
SU 1: sdppsu6
FPC 1:
#Obs per Unit
Stratum #Units #Obs min mean max
1 2 418 194 209.0 224
2 2 478 222 239.0 256
3 2 476 216 238.0 260
4 2 381 180 190.5 201
5 2 388 182 194.0 206
6 2 404 194 202.0 210
7 2 444 210 222.0 234
8 2 430 210 215.0 220
9 2 419 198 209.5 221
10 2 471 229 235.5 242
11 2 439 201 219.5 238
12 2 379 179 189.5 200
13 2 414 203 207.0 211
14 2 482 228 241.0 254
15 2 461 227 230.5 234
16 2 555 273 277.5 282
17 2 509 249 254.5 260
18 2 474 236 237.0 238
19 2 517 231 258.5 286
20 2 480 233 240.0 247
21 2 410 193 205.0 217
22 2 499 227 249.5 272
23 2 433 186 216.5 247
24 2 467 224 233.5 243
25 2 423 203 211.5 220
26 2 438 179 219.0 259
27 2 478 203 239.0 275
28 2 472 233 236.0 239
29 2 505 252 252.5 253
30 2 475 224 237.5 251
31 2 525 246 262.5 279
32 2 427 207 213.5 220
33 2 539 262 269.5 277
34 2 492 222 246.0 270
35 2 264 130 132.0 134
36 2 219 108 109.5 111
37 2 167 77 83.5 90
38 2 185 86 92.5 99
39 2 303 135 151.5 168
40 2 218 98 109.0 120
41 2 236 86 118.0 150
42 2 197 91 98.5 106
43 2 423 190 211.5 233
44 2 497 220 248.5 277
45 2 345 166 172.5 179
46 2 217 107 108.5 110
47 2 226 97 113.0 129
48 2 594 274 297.0 320
49 2 357 172 178.5 185
49 98 20050 77 204.6 320svy: tab hssex
(running tabulate on estimation sample)
Number of strata = 49 Number of obs = 20050
Number of PSUs = 98 Population size = 187647206
Design df = 49
sex | proportions
----------+------------
1 | .4777
2 | .5223
|
Total | 1
Key: proportions = cell proportions
svy: tab hssex, count missing
(running tabulate on estimation sample)
Number of strata = 49 Number of obs = 20050
Number of PSUs = 98 Population size = 187647206
Design df = 49
sex | count
----------+-----------
1 | 9.0e+07
2 | 9.8e+07
|
Total | 1.9e+08
Key: count = weighted countssvy: tab hssex, count cellwidth(10) format(%15.2g)
(running tabulate on estimation sample)
Number of strata = 49 Number of obs = 20050
Number of PSUs = 98 Population size = 187647206
Design df = 49
sex | count
----------+-----------
1 | 89637541
2 | 98009665
|
Total | 1.9e+08
Key: count = weighted countslabel define sex 1 male 2 female label values hssex sex label define race 1 white 2 black 3 other 8 MAUR MAUR = "Mexican-American of Unknown Race" label values dmaracer race label define mar 1 "married house" 2 "married not in house" /// 3 "living as married" 4 widowed 5 divorced 6 separated /// 7 "never married" 8 blank 9 DK label values hfa12 mar label define food 1 enough 2 sometimes 3 "often not enough" 8 blank label values hff4 food label define yn 1 yes 2 no 8 blank 9 DK foreach var of varlist hfa13 hfe7 hfe8a hfe8b hfe8c hfe8d hfe8e hff6a /// hff6b hff6c hff6d hff7 hff8 hff9 hff10 hff11 hah13-hah17 hav5 { label values var' yn } label define hah 1 "no difficulty" 2 "some difficulty" 3 "much difficulty" /// 4 "unable to do" 8 blank 9 DK foreach var of varlist hah1-hah12 { label valuesvar' hah }
svy: tab hah1, count cellwidth(15) format(%15.2f) missing
(running tabulate on estimation sample)
Number of strata = 49 Number of obs = 20050
Number of PSUs = 98 Population size = 187647206
Design df = 49
difficult |
y walking |
a quarter |
of a mile | count
----------+----------------
no diffi | 97132055.68
some dif | 8379187.23
much dif | 3032776.77
unable t | 4769359.13
blank | 130435.68
DK | 1236016.62
. | 72967375.21
|
Total | 187647206.32
Key: count = weighted countssvy: tab dmaracer, count cellwidth(15) format(%15.2f) missing
(running tabulate on estimation sample)
Number of strata = 49 Number of obs = 20050
Number of PSUs = 98 Population size = 187647206
Design df = 49
race | count
----------+----------------
white | 158131126.11
black | 21728087.51
other | 7773929.35
MAUR | 14063.35
|
Total | 187647206.32
Key: count = weighted countssvy: tab hfa12, count cellwidth(15) format(%15.2f) missing
(running tabulate on estimation sample)
Number of strata = 49 Number of obs = 20050
Number of PSUs = 98 Population size = 187647206
Design df = 49
marital |
status | count
----------+----------------
married | 106397053.60
married | 2367517.23
living a | 7442555.25
widowed | 13310800.60
divorced | 14540696.43
separate | 4570546.85
never ma | 38181973.78
88 | 826108.26
99 | 9954.32
|
Total | 187647206.32
Key: count = weighted counts "Highest grade or yr of school completed" svy: mean hfa8r
(running mean on estimation sample)
Survey: Mean estimation
Number of strata = 49 Number of obs = 20050
Number of PSUs = 98 Population size = 187647206
Design df = 49
| Linearized
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r | 12.90589 .1268616 12.65095 13.16082
As of Stata 10, you can request the standard deviation of a variable after running the svy: mean command.
estat sd-------------------------------------
| Mean Std. Dev.
-------------+-----------------------
hfa8r | 12.90589 7.880968
Using the replicate weights
Let’s reissue the svyset command, this time using the replicate weights. First, we will clear the current settings.
svyset, clear
Next, we need to do a little math to turn the value of Fay’s adjustment into the Fay’s value desired by Stata. Here is the formula: Fay=1-1/sqrt(adjfay) . We will use the vce(brr) and mse options to obtain the standard errors given by SUDAAN.
display 1-(1/sqrt(1.7)).23303501svyset [pweight = wtpfqx6], brrweight(wtpqrp1 - wtpqrp52) fay(.23303501) vce(brr) mse
pweight: wtpfqx6
VCE: brr
MSE: on
brrweight: wtpqrp1 wtpqrp2 wtpqrp3 wtpqrp4 wtpqrp5 wtpqrp6 wtpqrp7 wtpqrp8 wtpqrp9 wtpqrp10 wtpqrp11 wtpqrp12 wtpqrp13 wtpqrp14 wtpqrp15
wtpqrp16 wtpqrp17 wtpqrp18 wtpqrp19 wtpqrp20 wtpqrp21 wtpqrp22 wtpqrp23 wtpqrp24 wtpqrp25 wtpqrp26 wtpqrp27 wtpqrp28 wtpqrp29
wtpqrp30 wtpqrp31 wtpqrp32 wtpqrp33 wtpqrp34 wtpqrp35 wtpqrp36 wtpqrp37 wtpqrp38 wtpqrp39 wtpqrp40 wtpqrp41 wtpqrp42 wtpqrp43
wtpqrp44 wtpqrp45 wtpqrp46 wtpqrp47 wtpqrp48 wtpqrp49 wtpqrp50 wtpqrp51 wtpqrp52
fay: .23303501
Strata 1: <one>
SU 1: <observations>
FPC 1: <zero>
svy: mean hff5
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 1333
Population size = 7092623
Replications = 52
Design df = 51
| BRR
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hff5 | 15.65079 1.789433 12.05836 19.24323
As you can see, the mean for the variable hff5 is the same whether we used the PSU/strata or the replicate weights, which it must be, since these things only adjust the standard errors.
Let’s briefly consider an issue that arises when data are missing. We will run the nmissing command to confirm that only hff5 has missing data, and then we will run two svy: mean commands, the first with only the variable hfa8r, and the second with both variables. As you can see, the mean for hfa8r is different when run with hff5, because Stata is doing a listwise deletion of incomplete cases before calculating the means. In other words, even though there are no missing data from hfa8r, only 1333 of the 20050 data points are used in the calculation of the mean of hfa8r when you use both variables together.
search nmissing nmissing hfa8r hff5
hff5 18717svy: mean hfa8r
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 20050
Population size = 187647206
Replications = 52
Design df = 51
| BRR
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r | 12.90589 .1109928 12.68306 13.12871
--------------------------------------------------------------svy: mean hfa8r hff5
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 1333
Population size = 7092623
Replications = 52
Design df = 51
| BRR
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r | 17.9756 1.739453 14.48351 21.4677
hff5 | 15.65079 1.789433 12.05836 19.24323
Comparing variance estimation techniques
Now that we have seen the use of both the PSU/strata and the replicate weights to adjust the standard errors, let’s compare the two side-by-side. We will start by reissuing the svyset command using the PSU/strata.
svyset, clear
svyset sdppsu6 [pweight = wtpfqx6], strata(sdpstra6) singleunit(centered)
pweight: wtpfqx6
VCE: linearized
Single unit: centered
Strata 1: sdpstra6
SU 1: sdppsu6
FPC 1: svy: mean hfa8r
(running mean on estimation sample)
Survey: Mean estimation
Number of strata = 49 Number of obs = 20050
Number of PSUs = 98 Population size = 187647206
Design df = 49
| Linearized
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r | 12.90589 .1268616 12.65095 13.16082
We can also issue the estat command. The Deff and the Deft are types of design effects, which tell you about the efficiency of your sample. The Deff is a ratio of two variances. In the numerator we have the variance estimate from the current sample (including all of its design elements), and in the denominator we have the variance from a hypothetical sample of the same size drawn as an SRS. In other words, the Deff tells you how efficient your sample is compared to an SRS of equal size. If the Deff is less than 1, your sample is more efficient than SRS; usually the Deff is greater than 1. The Deft is the ratio of two standard error estimates. Again, the numerator is the standard error estimate from the current sample. The denominator is a hypothetical SRS (with replacement) standard error from a sample of the same size as the current sample. You can also use the meff and the meft option to get the misspecification effects. Misspecification effects are a ratio of the variance estimate from the current analysis to a hypothetical variance estimated from a misspecified model. Please see the Stata documentation for more details on how these are calculated.
estat effects
| Linearized
| Mean Std. Err. Deff Deft
-------------+--------------------------------------------
hfa8r | 12.90589 .1268616 5.19536 2.27933
Now let’s use the replicate weights and run the same analyses.
svyset, clear
svyset [pweight = wtpfqx6], brrweight(wtpqrp1 - wtpqrp52) fay(.23303501) vce(brr) mse
pweight: wtpfqx6
VCE: brr
MSE: on
brrweight: wtpqrp1 wtpqrp2 wtpqrp3 wtpqrp4 wtpqrp5 wtpqrp6 wtpqrp7 wtpqrp8 wtpqrp9 wtpqrp10 wtpqrp11 wtpqrp12 wtpqrp13 wtpqrp14 wtpqrp15
wtpqrp16 wtpqrp17 wtpqrp18 wtpqrp19 wtpqrp20 wtpqrp21 wtpqrp22 wtpqrp23 wtpqrp24 wtpqrp25 wtpqrp26 wtpqrp27 wtpqrp28 wtpqrp29
wtpqrp30 wtpqrp31 wtpqrp32 wtpqrp33 wtpqrp34 wtpqrp35 wtpqrp36 wtpqrp37 wtpqrp38 wtpqrp39 wtpqrp40 wtpqrp41 wtpqrp42 wtpqrp43
wtpqrp44 wtpqrp45 wtpqrp46 wtpqrp47 wtpqrp48 wtpqrp49 wtpqrp50 wtpqrp51 wtpqrp52
fay: .23303501
Strata 1: <one>
SU 1: <observations>
FPC 1: <zero>
svy: mean hfa8r
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 20050
Population size = 187647206
Replications = 52
Design df = 51
| BRR
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r | 12.90589 .1109928 12.68306 13.12871
--------------------------------------------------------------estat effects
| BRR
| Mean Std. Err. Deff Deft
-------------+--------------------------------------------
hfa8r | 12.90589 .1109928 3.9769 1.99422
As you can see, every part of the output is exactly the same except the denominator df, the design df and the standard errors. Notice that the standard errors are always larger for the analyses with the PSU/strata set than with the replicate weights. That is not because the replicate weight method of variance correction is more efficient than the linearization method. Rather, these are pseudo-PSUs and pseudo-strata, and the noise added to the PSUs and strata to make them pseudo is causing the inflation in the standard errors. If we had access to the real PSUs and strata and used those in the analyses, I suspect that the standard errors would be extremely close to those obtained with the replicate weights.
Analyses of subpopulations
The analysis of subpopulations is one place where survey data and experimental data are quite different. If you have data from an experiment (or quasi-experiment), and you want to analyze the responses from, say, just the women, or just people over age 50, you can just delete the unwanted cases from the data set or use the by: prefix. Survey data are different. With survey data, you (almost) never get to delete any cases from the data set, even if you will never use them in any of your analyses. Because of the way the by: prefix works, you usually don’t use it with survey data either. Instead, Stata has provided two options that allow you to correctly analyze subpopulations of your survey data. These options are subpop and over. The subpop option is sort of like deleting unwanted cases (without really deleting them, of course), and the over option is very similar to by: processing. We will start with some examples of the subpop option.
But first, let’s take a second to see why deleting cases from a survey data set can be so problematic. If the data set is subset (meaning that observations not to be included in the subpopulation are deleted from the data set), the standard errors of the estimates cannot be calculated correctly. When the subpopulation option(s) is used, only the cases defined by the subpopulation are used in the calculation of the estimate, but all cases are used in the calculation of the standard errors. For more information on this issue, please see Sampling Techniques, Third Edition by William G. Cochran (1977) and Small Area Estimation by J. N. K. Rao (2003). Also, if you look in the Stata 9 Survey manual, you will find an entire section (pages 38-43) dedicated to the analysis of subpopulations. The formulas for using both if and subpop are given, along with an explanation of how they are different. If you look at the help for any svy: command, you will see the same warning:
Warning: Use of if or in restrictions will not produce correct variance
estimates for subpopulations in many cases. To compute estimates for
subpopulations, use the subpop() option. The full specification for subpop()
is
subpop([varname] [if])
So now we know what not to do, so let’s see how to do this right. We will start with a simple mean and then use hssexas our subpopulation variable.
svy: mean hfa8r
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 20050
Population size = 187647206
Replications = 52
Design df = 51
| BRR
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r | 12.90589 .1109928 12.68306 13.12871
--------------------------------------------------------------svy, subpop(hssex): mean hfa8r
Note: subpop() takes on values other than 0 and 1
subpop() != 0 indicates subpopulation
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 20050
Population size = 187647206
Subpop. no. obs = 20050
Subpop. size = 187647206
Replications = 52
Design df = 51
| BRR
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r | 12.90589 .1109928 12.68306 13.12871
Clearly, something went wrong here. The note at the top of the output tells us what happened: Stata wants the subpopulation variable to be coded 0/1. Let’s look at the coding of hssex.
codebook hssex
hssex sex
type: numeric (byte)
label: sex
range: [1,2] units: 1
unique values: 2 missing .: 0/20050
tabulation: Freq. Numeric Label
9401 1 male
10649 2 female
Let’s recode hssex into a new variable, which we will call hssex1, and rerun the analysis.
generate hssex1 = hssex
recode hssex1 (2 = 0)
(hssex1: 10649 changes made)
svy, subpop(hssex1): mean hfa8r
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 20050
Population size = 187647206
Subpop. no. obs = 9401
Subpop. size = 89637541.1
Replications = 52
Design df = 51
| BRR
| Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r | 13.0614 .1534762 12.75328 13.36951
We can also use the over option to get estimates for all categories of the variable. In this case, we get the mean of the highest year of school completed for men and women (1 = male and 2 = female). The over option allows for variables coded 1/2 and for multicategory variables.
svy: mean hfa8r, over(hssex)
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 20050
Population size = 187647206
Replications = 52
Design df = 51
1: hssex = 1
2: hssex = 2
| BRR
Over | Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r |
1 | 13.0614 .1534762 12.75328 13.36951
2 | 12.76366 .1046096 12.55365 12.97367
--------------------------------------------------------------svy: mean hfa8r, over(hfa12)
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 20050
Population size = 187647206
Replications = 52
Design df = 51
1: hfa12 = 1
2: hfa12 = 2
3: hfa12 = 3
4: hfa12 = 4
5: hfa12 = 5
6: hfa12 = 6
7: hfa12 = 7
88: hfa12 = 88
99: hfa12 = 99
| BRR
Over | Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r |
1 | 12.81528 .1065852 12.6013 13.02926
2 | 11.49089 .3356867 10.81697 12.16481
3 | 12.41727 .244054 11.92731 12.90723
4 | 10.84711 .1758893 10.494 11.20022
5 | 12.79547 .1453802 12.50361 13.08733
6 | 11.27132 .2968793 10.67531 11.86733
7 | 12.79864 .1511575 12.49518 13.1021
88 | 81.54784 2.655634 76.21643 86.87925
99 | 62.79067 24.52562 13.55343 112.0279
The over option is available only for svy: mean, svy: proportion, svy: ratio and svy: total. Unfortunately, there are no nice display options for svy: total like there are for svy: tab to show the actual values of the totals. We can use the matrix list command to list out the values stored in the matrix, although sometimes those are in scientific notation as well. We can use the estat size command to get the unweighted and weighted size (i.e., count) of each subpopulation. This is a good thing to do, because you need to know how many observations are in each subpopulation. If "99" was a subpopulation of interest to us, we would be in trouble because there are only three observations in that subpopulation.
svy: total hssex, over(hfa12)
(running total on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..Survey: Total estimation Number of obs = 20050
Population size = 187647206
Replications = 52
Design df = 51
1: hfa12 = 1
2: hfa12 = 2
3: hfa12 = 3
4: hfa12 = 4
5: hfa12 = 5
6: hfa12 = 6
7: hfa12 = 7
88: hfa12 = 88
99: hfa12 = 99
| BRR
Over | Total Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hssex |
1 | 1.59e+08 2200084 1.54e+08 1.63e+08
2 | 3573113 284678.1 3001597 4144628
3 | 1.11e+07 651988.5 9826916 1.24e+07
4 | 2.45e+07 506533.4 2.35e+07 2.55e+07
5 | 2.36e+07 962190.5 2.17e+07 2.56e+07
6 | 7756844 456243 6840898 8672790
7 | 5.53e+07 1789277 5.17e+07 5.89e+07
88 | 1203385 243289.5 714960.5 1691809
99 | 15765.66 10722.36 -5760.372 37291.69
--------------------------------------------------------------mat list e(b)
e(b)[1,9]
hssex: hssex: hssex: hssex: hssex: hssex: hssex: hssex: hssex:
_subpop_1 _subpop_2 _subpop_3 widowed divorced separated _subpop_7 88 99
y1 1.585e+08 3573112.5 11135838 24474595 23634198 7756844.2 55314851 1203384.5 15765.66
estat size
1: hfa12 = 1
2: hfa12 = 2
3: hfa12 = 3
4: hfa12 = 4
5: hfa12 = 5
6: hfa12 = 6
7: hfa12 = 7
88: hfa12 = 88
99: hfa12 = 99
| BRR
Over | Total Std. Err. Obs Size
-------------+--------------------------------------------------------
hssex |
1 | 1.59e+08 2200084 10241 106397053.6
2 | 3573113 284678.1 364 2367517.23
3 | 1.11e+07 651988.5 781 7442555.25
4 | 2.45e+07 506533.4 2352 13310800.6
5 | 2.36e+07 962190.5 1388 14540696.43
6 | 7756844 456243 686 4570546.85
7 | 5.53e+07 1789277 4135 38181973.78
88 | 1203385 243289.5 100 826108.26
99 | 15765.66 10722.36 3 9954.32
The subpop option can be combined with the over option. This is handy because if cannot be used with the over option. By combining the options, you can have "the best of both worlds." In the example below, our subpopulation includes only white males, and the mean of education is given for each marital status. Notice that for category "88", the mean is really high (83.44). This is because Stata considers 88 to be a valid value, not the missing data code that it is (according to the documentation).
svy, subpop(hssex1 if dmaracer==1): mean hfa8r, over(hfa12)
(running mean on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Mean estimation Number of obs = 20050
Population size = 187647206
Replications = 52
Design df = 51
1: hfa12 = 1
2: hfa12 = 2
3: hfa12 = 3
4: hfa12 = 4
5: hfa12 = 5
6: hfa12 = 6
7: hfa12 = 7
88: hfa12 = 88
| BRR
Over | Mean Std. Err. [95% Conf. Interval]
-------------+------------------------------------------------
hfa8r |
1 | 12.85045 .1013903 12.64691 13.054
2 | 11.15848 .6515755 9.850389 12.46657
3 | 11.90575 .2359017 11.43216 12.37934
4 | 10.8672 .3819635 10.10037 11.63402
5 | 12.8703 .1584319 12.55224 13.18837
6 | 12.6802 1.564755 9.53882 15.82157
7 | 12.72527 .2070723 12.30955 13.14098
88 | 83.43886 3.637654 76.13596 90.74175
For more information on analyzing subpopulations in Stata, please see our Stata FAQ: How can I analyze a subpopulation of my survey data in Stata 9?
Regression analyses
Let’s look at some regression analyses. Stata 9 has a very nice suite of regression commands that can be used with the svy: prefix. Type help survey for a list of commands that can be used with the svy: prefix.
svy: reg hav6s hav2s hav3s hff4 hfa13 hfe7 hfa8r
(running regress on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Linear regression Number of obs = 5866
Population size = 69288876
Replications = 52
Design df = 51
F( 6, 46) = 83.92
Prob > F = 0.0000
R-squared = 0.4285
| BRR
hav6s | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
hav2s | .0335188 .0164996 2.03 0.047 .0003945 .0666432
hav3s | .0458665 .016843 2.72 0.009 .0120528 .0796802
hff4 | 78.15901 47.06818 1.66 0.103 -16.33432 172.6523
hfa13 | 14.97253 6.777161 2.21 0.032 1.366808 28.57825
hfe7 | -9.015854 11.81718 -0.76 0.449 -32.73983 14.70812
hfa8r | .6153294 .6913622 0.89 0.378 -.7726382 2.003297
_cons | -59.27143 62.78533 -0.94 0.350 -185.3182 66.77539
As we see in the example below, we can put i. before categorical predictor variables to have Stata automatically create dummy variables for us. (As of Stata 11, the xi: prefix is no longer needed. If you are using an earlier version of Stata, you should put the xi: prefix before the svy: prefix.)
svy: reg hav2s hfe7 hfa13 hssex i.dmpcregn hah15
(running regress on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Linear regression Number of obs = 13398
Population size = 114679831
Replications = 52
Design df = 51
F( 7, 45) = 4.33
Prob > F = 0.0010
R-squared = 0.1045
| BRR
hav2s | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
hfe7 | 1078.949 215.3567 5.01 0.000 646.602 1511.295
hfa13 | 1212.34 267.1979 4.54 0.000 675.9174 1748.762
hssex | -403.5371 128.8811 -3.13 0.003 -662.2767 -144.7975
|
dmpcregn |
2 | 33.12827 72.97049 0.45 0.652 -113.3661 179.6226
3 | -19.01354 68.06759 -0.28 0.781 -155.6649 117.6378
4 | 249.894 122.4217 2.04 0.046 4.122106 495.6659
|
hah15 | 935.0711 512.2536 1.83 0.074 -93.32097 1963.463
_cons | -5005.517 1185.052 -4.22 0.000 -7384.609 -2626.425
No svy: prefix is needed before the test command.
test 2.dmpcregn 3.dmpcregn 4.dmpcregn
Adjusted Wald test
( 1) 2.dmpcregn = 0
( 2) 3.dmpcregn = 0
( 3) 4.dmpcregn = 0
F( 3, 49) = 1.99
Prob > F = 0.1277
Let’s create a 0/1 variable and run a logistic regression.
generate clubs = hav5
recode clubs (2 = 0)
(clubs: 14184 changes made)
svy: logit clubs hfe7 hfa13 hssex i.dmpcregn hah15
(running logit on estimation sample)
BRR replications (52)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
..
Survey: Logistic regression Number of obs = 13398
Population size = 114679831
Replications = 52
Design df = 51
F( 7, 45) = 7.13
Prob > F = 0.0000
| BRR
clubs | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
hfe7 | .1212481 .0502015 2.42 0.019 .0204643 .2220319
hfa13 | -.392244 .0611189 -6.42 0.000 -.5149452 -.2695427
hssex | -.0568606 .0474936 -1.20 0.237 -.152208 .0384869
|
dmpcregn |
2 | .0246446 .1229863 0.20 0.842 -.2222608 .2715499
3 | -.2444328 .1344003 -1.82 0.075 -.5142527 .0253871
4 | -.1163411 .1270524 -0.92 0.364 -.3714094 .1387272
|
hah15 | .362997 .0947355 3.83 0.000 .1728077 .5531864
_cons | -.5054835 .212936 -2.37 0.021 -.9329703 -.0779967
test 2.dmpcregn 3.dmpcregn 4.dmpcregn
Adjusted Wald test
( 1) [clubs]2.dmpcregn = 0
( 2) [clubs]3.dmpcregn = 0
( 3) [clubs]4.dmpcregn = 0
F( 3, 49) = 2.12
Prob > F = 0.1096
We don’t have much time to talk about regression diagnostics here, although that is a common question among researchers who use survey data. Most of the assumptions still apply when using survey data, but they can be more difficult to check. As of Stata 11, most of the diagnostic commands that you would use after regress, logistic, etc., don’t work after svy: regress, svy: logit, etc. Some of the assumptions don’t really apply, though, because of the extremely large sample size involved. Checking assumptions when doing a subpopulation analysis can be even more tricky.
Using multiply imputed data
Some of the NHANES III data sets were released as imputed data sets. This means that some of the variables contained in the data set were multiply imputed. For information on which variables were imputed, the imputation method, etc., please see and ftp://ftp.cdc.gov/pub/Health_Statistics/NCHS/Datasets/NHANES/NHANESIII/7A/doc/main.pdf .
If you are using Stata 11 or later, we suggest using the built-in Stata prefix mi. If you are using Stata 10 or earlier, please see the last paragraph in this section for suggestions on how to analyze the multiply imputed datasets.
The commands below are used to change the current working directory (cd), to set the memory (set mem), to use the nh3core data set (use), and to create the single multiply imputed data set (called mymi) from the five imputed data sets provided by NHANES (mi import nhanes1).
cd "D:datanhanes IIImiStata_data"
D:Datanhanes IIImiStata_data
set mem 200m
(204800k)
use nh3core
mi import nhanes1 mymi, using(nh3mi1.dta nh3mi2.dta nh3mi3.dta nh3mi4.dta nh3mi5.dta) id(seqn)
(variables sdppsu6 sdpstra6 wtpfqx6 wtpqrp1 wtpqrp2 wtpqrp3 wtpqrp4 wtpqrp5 wtpqrp6
wtpqrp7 wtpqrp8 wtpqrp9 wtpqrp10 wtpqrp11 wtpqrp12 wtpqrp13 wtpqrp14 wtpqrp15
wtpqrp16 wtpqrp17 wtpqrp18 wtpqrp19 wtpqrp20 wtpqrp21 wtpqrp22 wtpqrp23 wtpqrp24
wtpqrp25 wtpqrp26 wtpqrp27 wtpqrp28 wtpqrp29 wtpqrp30 wtpqrp31 wtpqrp32 wtpqrp33
wtpqrp34 wtpqrp35 wtpqrp36 wtpqrp37 wtpqrp38 wtpqrp39 wtpqrp40 wtpqrp41 wtpqrp42
wtpqrp43 wtpqrp44 wtpqrp45 wtpqrp46 wtpqrp47 wtpqrp48 wtpqrp49 wtpqrp50 wtpqrp51
wtpqrp52 hfa7 hfa8 hfa12 hac1a hac1b hac1c hac1d hac1e hac1f hac1g hac1h hac1i hac1j
hac1k hac1l hac1m hac1n hac1o had1 hae2 hae4a hae4b hae5a hae5b hae6 hae7 haf1 haf10
hag2 hag3 hag5a hag5b hag5c hag11 hag12 han6hs han6is han6js hap1 hap1a hap2 hap3
hap10 hap10a har1 har3 har14 har16 har23 har24 har26 har27 hye1g hye1h hye6a hye6b
hye15 hyh2 hyh10 dmppirif hff1if hab1if ham5if ham6if han6srif haq1if har3rif
hat28if hazak1if hazak5if hazbk1if hazbk5if hazck1if hazck5if hyd1if hyf2if bdpfndif
bdpindif bdpkif bdptoaif bdptodif bdptrdif bdpwtdif bmpbutif bmpheaif bmphtif
bmpkneif bmprecif bmpsthif bmpsb1if bmpsb2if bmpsp1if bmpsp2if bmptr1if bmptr2if
bmpwstif bmpwtif fepif frpif hdpif hgpif htpif mcpsiif mhpif mvpsiif pbpif phpfstif
pxpif rcpif rwpif sepif tcpif tgpif tipif fppsudif fppsumif fppsurif pep6g1if
pep6g2if pep6g3if pep6h1if pep6h2if pep6h3if pep6i1if pep6i2if pep6i3if hff1mi
hab1mi ham5mi ham6mi han6srmi haq1mi har3rmi hat28mi hazak1mi hazak5mi hazbk1mi
hazbk5mi hazck1mi hazck5mi hyd1mi hyf2mi bmpsb1mi bmpsb2mi bmpsp1mi bmpsp2mi
bmptr1mi bmptr2mi pep6g1mi pep6g2mi pep6g3mi pep6h1mi pep6h2mi pep6h3mi pep6i1mi
pep6i2mi pep6i3mi dropped in m=1)< output omitted to save space >
(variables sdppsu6 sdpstra6 wtpfqx6 wtpqrp1 wtpqrp2 wtpqrp3 wtpqrp4 wtpqrp5 wtpqrp6
wtpqrp7 wtpqrp8 wtpqrp9 wtpqrp10 wtpqrp11 wtpqrp12 wtpqrp13 wtpqrp14 wtpqrp15
wtpqrp16 wtpqrp17 wtpqrp18 wtpqrp19 wtpqrp20 wtpqrp21 wtpqrp22 wtpqrp23 wtpqrp24
wtpqrp25 wtpqrp26 wtpqrp27 wtpqrp28 wtpqrp29 wtpqrp30 wtpqrp31 wtpqrp32 wtpqrp33
wtpqrp34 wtpqrp35 wtpqrp36 wtpqrp37 wtpqrp38 wtpqrp39 wtpqrp40 wtpqrp41 wtpqrp42
wtpqrp43 wtpqrp44 wtpqrp45 wtpqrp46 wtpqrp47 wtpqrp48 wtpqrp49 wtpqrp50 wtpqrp51
wtpqrp52 hfa7 hfa8 hfa12 hac1a hac1b hac1c hac1d hac1e hac1f hac1g hac1h hac1i hac1j
hac1k hac1l hac1m hac1n hac1o had1 hae2 hae4a hae4b hae5a hae5b hae6 hae7 haf1 haf10
hag2 hag3 hag5a hag5b hag5c hag11 hag12 han6hs han6is han6js hap1 hap1a hap2 hap3
hap10 hap10a har1 har3 har14 har16 har23 har24 har26 har27 hye1g hye1h hye6a hye6b
hye15 hyh2 hyh10 dmppirif hff1if hab1if ham5if ham6if han6srif haq1if har3rif
hat28if hazak1if hazak5if hazbk1if hazbk5if hazck1if hazck5if hyd1if hyf2if bdpfndif
bdpindif bdpkif bdptoaif bdptodif bdptrdif bdpwtdif bmpbutif bmpheaif bmphtif
bmpkneif bmprecif bmpsthif bmpsb1if bmpsb2if bmpsp1if bmpsp2if bmptr1if bmptr2if
bmpwstif bmpwtif fepif frpif hdpif hgpif htpif mcpsiif mhpif mvpsiif pbpif phpfstif
pxpif rcpif rwpif sepif tcpif tgpif tipif fppsudif fppsumif fppsurif pep6g1if
pep6g2if pep6g3if pep6h1if pep6h2if pep6h3if pep6i1if pep6i2if pep6i3if hff1mi
hab1mi ham5mi ham6mi han6srmi haq1mi har3rmi hat28mi hazak1mi hazak5mi hazbk1mi
hazbk5mi hazck1mi hazck5mi hyd1mi hyf2mi bmpsb1mi bmpsb2mi bmpsp1mi bmpsp2mi
bmptr1mi bmptr2mi pep6g1mi pep6g2mi pep6g3mi pep6h1mi pep6h2mi pep6h3mi pep6i1mi
pep6i2mi pep6i3mi dropped in m=5)
The mi describe and mi varying commands provide useful information about the data set. The varcase command is a user-written command (search varcase) that changes the case of variable names. It is used here to change some of the variable names from being in all capital letters to being in all lower case letters.
mi describe
Style: flongsep mymi
last mi update 27sep2010 15:21:48, 0 seconds ago
Obs.: complete 20,870
incomplete 13,124 (M = 5 imputations)
total 33,994
Vars.: imputed: 36; dmppir(3410) bdpfnd(4179+15169) bdpind(4179+15169)
bdpk(4179+15169) bdptoa(4179+15169) bdptod(4179+15169)
bdptrd(4179+15169) bdpwtd(4179+15169) bmpbut(4258+3446)
bmphea(799+24586) bmpht(2613+3446) bmpkne(1658+27398)
bmprec(457+28012) bmpsth(3539+3446) bmpwst(4260+3446) bmpwt(2862)
fep(5408+2107) frp(5494+2107) hdp(4603+5982) hgp(5515+2107)
htp(5517+2107) mcpsi(5518+2107) mhp(5518+2107) mvpsi(5516+2107)
pbp(5069+2107) phpfst(4187+2107) pxp(6117+2107) rcp(5517+2107)
rwp(5515+2107) sep(3669+11728) tcp(4451+5982) tgp(4497+5982)
tip(6085+2107) fppsud(2962+22546) fppsum(3240+22546)
fppsur(3237+22546)
passive: 0
regular: 0
system: 2; _mi_id _mi_miss
(there are 163 unregistered variables)
mi varying
Possible problem variable names
imputed nonvarying: (none)
passive nonvarying: (none)
unregistered varying: (none)
unregistered super/varying: (none)
unregistered super varying: (none)
super/varying means super varying but would be varying if registered as imputed;
variables vary only where equal to soft missing in m=0.
varcase(SDPPSU6- PEP6I3IF)
Now we are ready to use the svyset command. We include the mi prefix.
mi svyset sdppsu6 [pweight = wtpfqx6], strata(sdpstra6) singleunit(centered)
(variables SDPPSU6 SDPSTRA6 WTPFQX6 WTPQRP1 WTPQRP2 WTPQRP3 WTPQRP4 WTPQRP5 WTPQRP6
WTPQRP7 WTPQRP8 WTPQRP9 WTPQRP10 WTPQRP11 WTPQRP12 WTPQRP13 WTPQRP14 WTPQRP15
WTPQRP16 WTPQRP17 WTPQRP18 WTPQRP19 WTPQRP20 WTPQRP21 WTPQRP22 WTPQRP23 WTPQRP24
WTPQRP25 WTPQRP26 WTPQRP27 WTPQRP28 WTPQRP29 WTPQRP30 WTPQRP31 WTPQRP32 WTPQRP33
WTPQRP34 WTPQRP35 WTPQRP36 WTPQRP37 WTPQRP38 WTPQRP39 WTPQRP40 WTPQRP41 WTPQRP42
WTPQRP43 WTPQRP44 WTPQRP45 WTPQRP46 WTPQRP47 WTPQRP48 WTPQRP49 WTPQRP50 WTPQRP51
WTPQRP52 HFA7 HFA8 HFA12 HAC1A HAC1B HAC1C HAC1D HAC1E HAC1F HAC1G HAC1H HAC1I HAC1J
HAC1K HAC1L HAC1M HAC1N HAC1O HAD1 HAE2 HAE4A HAE4B HAE5A HAE5B HAE6 HAE7 HAF1 HAF10
HAG2 HAG3 HAG5A HAG5B HAG5C HAG11 HAG12 HAN6HS HAN6IS HAN6JS HAP1 HAP1A HAP2 HAP3
HAP10 HAP10A HAR1 HAR3 HAR14 HAR16 HAR23 HAR24 HAR26 HAR27 HYE1G HYE1H HYE6A HYE6B
HYE15 HYH2 HYH10 HFF1IF HAB1IF HAM5IF HAM6IF HAN6SRIF HAQ1IF HAR3RIF HAT28IF
HAZAK1IF HAZAK5IF HAZBK1IF HAZBK5IF HAZCK1IF HAZCK5IF HYD1IF HYF2IF BMPSB1IF
BMPSB2IF BMPSP1IF BMPSP2IF BMPTR1IF BMPTR2IF PEP6G1IF PEP6G2IF PEP6G3IF PEP6H1IF
PEP6H2IF PEP6H3IF PEP6I1IF PEP6I2IF PEP6I3IF dropped in m=1)
(variables sdppsu6 sdpstra6 wtpfqx6 wtpqrp1 wtpqrp2 wtpqrp3 wtpqrp4 wtpqrp5 wtpqrp6
wtpqrp7 wtpqrp8 wtpqrp9 wtpqrp10 wtpqrp11 wtpqrp12 wtpqrp13 wtpqrp14 wtpqrp15 wtpqrp16
wtpqrp17 wtpqrp18 wtpqrp19 wtpqrp20 wtpqrp21 wtpqrp22 wtpqrp23 wtpqrp24 wtpqrp25
wtpqrp26 wtpqrp27 wtpqrp28 wtpqrp29 wtpqrp30 wtpqrp31 wtpqrp32 wtpqrp33 wtpqrp34
wtpqrp35 wtpqrp36 wtpqrp37 wtpqrp38 wtpqrp39 wtpqrp40 wtpqrp41 wtpqrp42 wtpqrp43
wtpqrp44 wtpqrp45 wtpqrp46 wtpqrp47 wtpqrp48 wtpqrp49 wtpqrp50 wtpqrp51 wtpqrp52 hfa7
hfa8 hfa12 hac1a hac1b hac1c hac1d hac1e hac1f hac1g hac1h hac1i hac1j hac1k hac1l hac1m
hac1n hac1o had1 hae2 hae4a hae4b hae5a hae5b hae6 hae7 haf1 haf10 hag2 hag3 hag5a hag5b
hag5c hag11 hag12 han6hs han6is han6js hap1 hap1a hap2 hap3 hap10 hap10a har1 har3 har14
har16 har23 har24 har26 har27 hye1g hye1h hye6a hye6b hye15 hyh2 hyh10 hff1if hab1if
ham5if ham6if han6srif haq1if har3rif hat28if hazak1if hazak5if hazbk1if hazbk5if
hazck1if hazck5if hyd1if hyf2if bmpsb1if bmpsb2if bmpsp1if bmpsp2if bmptr1if bmptr2if
pep6g1if pep6g2if pep6g3if pep6h1if pep6h2if pep6h3if pep6i1if pep6i2if pep6i3if added
to m=1)
< output omitted to save space >
(variables SDPPSU6 SDPSTRA6 WTPFQX6 WTPQRP1 WTPQRP2 WTPQRP3 WTPQRP4 WTPQRP5 WTPQRP6
WTPQRP7 WTPQRP8 WTPQRP9 WTPQRP10 WTPQRP11 WTPQRP12 WTPQRP13 WTPQRP14 WTPQRP15
WTPQRP16 WTPQRP17 WTPQRP18 WTPQRP19 WTPQRP20 WTPQRP21 WTPQRP22 WTPQRP23 WTPQRP24
WTPQRP25 WTPQRP26 WTPQRP27 WTPQRP28 WTPQRP29 WTPQRP30 WTPQRP31 WTPQRP32 WTPQRP33
WTPQRP34 WTPQRP35 WTPQRP36 WTPQRP37 WTPQRP38 WTPQRP39 WTPQRP40 WTPQRP41 WTPQRP42
WTPQRP43 WTPQRP44 WTPQRP45 WTPQRP46 WTPQRP47 WTPQRP48 WTPQRP49 WTPQRP50 WTPQRP51
WTPQRP52 HFA7 HFA8 HFA12 HAC1A HAC1B HAC1C HAC1D HAC1E HAC1F HAC1G HAC1H HAC1I HAC1J
HAC1K HAC1L HAC1M HAC1N HAC1O HAD1 HAE2 HAE4A HAE4B HAE5A HAE5B HAE6 HAE7 HAF1 HAF10
HAG2 HAG3 HAG5A HAG5B HAG5C HAG11 HAG12 HAN6HS HAN6IS HAN6JS HAP1 HAP1A HAP2 HAP3
HAP10 HAP10A HAR1 HAR3 HAR14 HAR16 HAR23 HAR24 HAR26 HAR27 HYE1G HYE1H HYE6A HYE6B
HYE15 HYH2 HYH10 HFF1IF HAB1IF HAM5IF HAM6IF HAN6SRIF HAQ1IF HAR3RIF HAT28IF
HAZAK1IF HAZAK5IF HAZBK1IF HAZBK5IF HAZCK1IF HAZCK5IF HYD1IF HYF2IF BMPSB1IF
BMPSB2IF BMPSP1IF BMPSP2IF BMPTR1IF BMPTR2IF PEP6G1IF PEP6G2IF PEP6G3IF PEP6H1IF
PEP6H2IF PEP6H3IF PEP6I1IF PEP6I2IF PEP6I3IF dropped in m=5)
(variables sdppsu6 sdpstra6 wtpfqx6 wtpqrp1 wtpqrp2 wtpqrp3 wtpqrp4 wtpqrp5 wtpqrp6
wtpqrp7 wtpqrp8 wtpqrp9 wtpqrp10 wtpqrp11 wtpqrp12 wtpqrp13 wtpqrp14 wtpqrp15 wtpqrp16
wtpqrp17 wtpqrp18 wtpqrp19 wtpqrp20 wtpqrp21 wtpqrp22 wtpqrp23 wtpqrp24 wtpqrp25
wtpqrp26 wtpqrp27 wtpqrp28 wtpqrp29 wtpqrp30 wtpqrp31 wtpqrp32 wtpqrp33 wtpqrp34
wtpqrp35 wtpqrp36 wtpqrp37 wtpqrp38 wtpqrp39 wtpqrp40 wtpqrp41 wtpqrp42 wtpqrp43
wtpqrp44 wtpqrp45 wtpqrp46 wtpqrp47 wtpqrp48 wtpqrp49 wtpqrp50 wtpqrp51 wtpqrp52 hfa7
hfa8 hfa12 hac1a hac1b hac1c hac1d hac1e hac1f hac1g hac1h hac1i hac1j hac1k hac1l hac1m
hac1n hac1o had1 hae2 hae4a hae4b hae5a hae5b hae6 hae7 haf1 haf10 hag2 hag3 hag5a hag5b
hag5c hag11 hag12 han6hs han6is han6js hap1 hap1a hap2 hap3 hap10 hap10a har1 har3 har14
har16 har23 har24 har26 har27 hye1g hye1h hye6a hye6b hye15 hyh2 hyh10 hff1if hab1if
ham5if ham6if han6srif haq1if har3rif hat28if hazak1if hazak5if hazbk1if hazbk5if
hazck1if hazck5if hyd1if hyf2if bmpsb1if bmpsb2if bmpsp1if bmpsp2if bmptr1if bmptr2if
pep6g1if pep6g2if pep6g3if pep6h1if pep6h2if pep6h3if pep6i1if pep6i2if pep6i3if added
to m=5)
pweight: wtpfqx6
VCE: linearized
Single unit: centered
Strata 1: sdpstra6
SU 1: sdppsu6
FPC 1:
At long last, we are ready to do our analysis. Please note that not all of the commands that are available with the svy: prefix are available with the mi estimate: svy: prefix. For the examples here, we will get the mean of three variables (bmpwst, bmpht and bmpbut), and then we will run a regression.
mi estimate: svy: mean bmpwst bmpht bmpbut
Multiple-imputation estimates Imputations = 5
Survey: Mean estimation Number of obs = 30548
Number of strata = 49 Population size = 243785748
Number of PSUs = 98
Average RVI = 0.0242
Complete DF = 49
DF adjustment: Small sample DF: min = 46.64
avg = 46.67
Within VCE type: Linearized max = 46.72
Mean | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
bmpwst | 84.72488 .2081347 407.07 0.000 84.30608 85.14368
bmpht | 160.6015 .1916889 837.82 0.000 160.2158 160.9871
bmpbut | 94.10488 .1790256 525.65 0.000 93.74465 94.4651
mi estimate: svy: regress bmpwst bmpht bmpbut
Multiple-imputation estimates Imputations = 5
Survey: Linear regression Number of obs = 30548
Number of strata = 49 Population size = 243785748
Number of PSUs = 98
Average RVI = 0.3866
Complete DF = 49
DF adjustment: Small sample DF: min = 8.66
avg = 20.78
max = 43.77
Model F test: Equal FMI F( 2, 13.5) = 13019.57
Within VCE type: Linearized Prob > F = 0.0000
bmpwst | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
bmpht | -.0063002 .0129422 -0.49 0.637 -.0351702 .0225698
bmpbut | .9663688 .0200984 48.08 0.000 .9206272 1.01211
_cons | -5.203297 .5062292 -10.28 0.000 -6.223686 -4.182908
The preceding is intended only to give a cursory overview of the steps involved in analyzing multiply imputed survey data sets. By no means have we covered all (or even most) of the issues that can be involved. Rather, the point is that such data can be analyzed in Stata 11, and a few of the commands to do so have been illustrated. Reading over the relevant sections of the Stata 11 manuals will be very helpful and is strongly recommended.
If you are using Stata 10 or earlier, we suggest using the prefix mim: for analyzing multiply imputed data sets, although there are some other prefixes available in Stata. The prefix mim: is not part of Stata and needs to be downloaded (search mim). The NHANES data were released with five imputed data sets. Unlike SUDAAN, mim wants the data sets stacked into a single data set. We can use the mimstack command to do this (search mimstack). We need to specify unique case identifier in each data set (seqn) and the "stub" name (nh3mi) of the imputed data sets. We strongly encourage everyone to read the help file for mim before using the mim: prefix or the mimstack command. Also, before you start using the multiply imputed data, you should look at the help file for mim to see if the procedure that you want to use is supported by mim. For example, svy: tab is not supported by mim. You may also want to check periodically to see if there are any updates to mim or similar procedures. We should also note the mimstack can take some time to run, especially if you don’t have a lot of RAM on your computer.
A quick note on merging data sets
The NHANES documentation provides clear instructions on how to merge various data sets from the NHANES III collection. You need to read that very carefully. Depending on what data sets you merge, you may need to adjust the sampling weights. Again, when and how to do this is spelled out in the documentation, so please read that carefully. Also, be aware that you will likely have to modify your svyset command to work with the merged data set. In fact, the svyset command may need to be different for different models that you run from the merged data set, depending on which variables are included in the model.
Another point to remember is that what you do to merge data sets or modify sampling weights with the NHANES data will likely NOT generalize to other survey data sets. You will need to read the documentation for each data set and will probably have to follow different procedures to accomplish the same tasks with other data sets. In other words, many ways of doing things are specific to that particular data set; be careful not to over generalize.
For more information on using the NHANES data sets
There are several helpful resources for learning how to analyze the NHANES III data sets correctly. One is a listserv at . There are also online tutorials at .
References
The Stata Journal, Vol. 7, No. 1, 1st Quarter 2007
Analysis of Health Surveys by Edward L. Korn and Barry I. Graubard
Sampling of Populations: Methods and Applications, Third Edition by Paul Levy and Stanley Lemeshow
Analysis of Survey Data Edited by R. L. Chambers and C. J. Skinner
Sampling Techniques, Third Edition by William G. Cochran
Stata 9 Manual: Survey Data
Analyzing the NHANES III Multiply Imputed Data Set: Methods and Examples by Joseph L. Schafer (June, 2001)
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187621 | https://mpp.wested.org/wp-content/themes/child/downloads/mpp-sample-grade8-unit7-lesson10.pdf | 344 © 2010 WestEd. All rights reserved.
STUDENT PAGE 1 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Purpose To solve proportion problems involving similar figures Math Words similar The sides of triangle ABC are scaled up by a scale factor of 2 to make a similar triangle EDF.
corresponding sides In these similar figures, sides BC and FG are corresponding sides.
Starter Problem Figure A is similar to Figure B (not shown). The two sides of Figure A are 5 cm and 6 cm long. If the longer side of Figure B is 9 cm long, how long is the other side? A C B D F E 8 6 10 16 12 20 A C D B E G F H 6 cm 5 cm Figure A © 2010 WestEd. All rights reserved. 345 STUDENT PAGE 2 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Pitfall OK Student Thinking Starter Problem Figure A is similar to Figure B (not shown). The two sides of Figure A are 5 cm and 6 cm long. If the longer side of Figure B is 9 cm long, how long is the other side? Ali Maria Since they’re similar, I made a proportion. The first long to short ratio is 6:5. The second long to short ratio is 9:x. Then I solved the equation. It makes sense that shorter side is 7.5 cm. If I think of 6:5 and 9:7.5 as fractions, they’re both a little more than 1.
If the long side of Figure B is 3 cm longer than the long side of Figure A, then the short side of Figure B also has to be 3 cm longer than the short side of Figure A. So, 5 + 3 = 8 cm.
Things to Remember a a 6 cm 5 cm Figure A 346 © 2010 WestEd. All rights reserved.
STUDENT PAGE 3 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Our Turn Set up proportions to solve these problems. Draw your own figures if needed. Remember to use labels in your proportions and figures.
1. The figures below are similar. Find the missing length marked x.
2. Rectangle A is 6 inches wide and 21 inches long. Rectangle B is the same shape but bigger. If rectangle B is 20 inches wide, how many inches long is it? 3. A scale drawing of a flowerbed shows two parallel edges that measure 5 inches and 2 inches and a slanted edge of 3 1/4 inches. If the parallel edges of the real flowerbed are 8 yards and 20 yards, how long is the slanted edge of the real flowerbed? 24 m x 8 m 10 m 21 inches 6 inches Rectangle A 2 inches 3 inches 5 inches 1 4 © 2010 WestEd. All rights reserved. 347 STUDENT PAGE 4 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 My Turn Set up proportions to solve these problems. Draw your own figures if needed. Remember to use labels in your proportions and figures.
1. The two triangles below are similar. If the shortest side of the smaller triangle is 4.5 cm, how long is its longest side?
2. A large sign was made from an artist’s drawing. The measurements of the lines on the letter L on the drawing were 11 cm and 7 cm. The short part of the L on the sign is 56 cm. How long should the long part of the L on the sign be? 3. A door measures 170 cm high by 90 cm wide. If a scale drawing of the door is 4.5 cm wide, how high should the scale drawing be?
9 cm 18 cm 12 cm 4.5 cm x cm 56 cm 90 cm 170 cm 348 © 2010 WestEd. All rights reserved.
STUDENT PAGE 5 NAME: STUDENT PAGE 5 NAME: Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Similar Figures Multiple Choice Mini Lesson Fill in the circle next to the answer you choose.
1. The two triangles are similar. How long is the side labeled x? 9 units 2 units 4 units 10 units 2. On a scale drawing of a doghouse, the width is 4 inches and the length is 6 inches. If the actual length is 48 inches, what is the actual width? 72 inches 36 inches 46 inches 32 inches Similar Figures 8 x 6 12 Multiple Choice Mini Lesson Fill in the circle next to the answer you choose.
1. The two triangles are similar. How long is the side labeled x? 9 units 2 units 4 units 10 units 2. On a scale drawing of a doghouse, the width is 4 inches and the length is 6 inches. If the actual length is 48 inches, what is the actual width? 72 inches 36 inches 46 inches 32 inches 8 x 6 12 © 2010 WestEd. All rights reserved. 349 STUDENT PAGE 6 NAME: STUDENT PAGE 6 NAME: Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Similar Figures Writing Task Mini Lesson Explain how to set up and solve a proportion to show that 40 cm is the correct solution to the following problem.
A 10 cm by 12.5 cm rectangular picture was enlarged on a copy machine. The longest side of the copy of the picture measures 50 cm. How wide is it?
Similar Figures Writing Task Mini Lesson Explain how to set up and solve a proportion to show that 40 cm is the correct solution to the following problem.
A 10 cm by 12.5 cm rectangular picture was enlarged on a copy machine. The longest side of the copy of the picture measures 50 cm. How wide is it?
Math Pathways & Pitfalls® 205 Lesson at a Glance Similar Figures TEACHING GUIDE a UNIT 7 a LESSON 10 Mathematical goals A A Set up proportions involving similar geometric figures A A Use cross products and algebra to solve geometric proportion problems Mathematical language and reasoning goals A A Discuss similar figures in terms of scaling up and scaling down A A Use number sense to check proportion problems Prior Learning Needed • Understand the concepts of ratio and proportion • Use mental math and cross multiplication to solve proportions Lesson Preparation • Study Lesson Foundation • Review Teaching Guide and Student Pages • Prepare stapled packet of Student Pages 1–4 for each student • Copy and cut in half Student Pages 5 and 6 • Post Discussion Builders poster MATERIALS LESSON ROADMAP C O R E L E S S O N : D AY 1 G R O U P I N G T I M E Opener Discussion Builders Purpose Math Words Starter Problem Discussion Student Thinking Things to Remember Reflection C O R E L E S S O N : D AY 2 Review and Practice Review Day 1 Lesson Our Turn My Turn M I N I L E S S O N S : 2 – 3 D AY S L AT E R Assess and Reinforce Multiple Choice Mini Lesson Writing Task Mini Lesson ❍ ❍Discussion Builders poster ❍ ❍Projector (optional) ❍ ❍Student Pages 1 and 2 ❍ ❍Teaching Guide ❍ ❍Rulers, graph paper, scratch paper (suggested) ❍ ❍Clipboard Prompts, page 37 ❍ ❍Student Page 2 (completed day 1) ❍ ❍Student Pages 3 and 4 ❍ ❍Teaching Guide ❍ ❍Rulers, graph paper, scratch paper (suggested) ❍ ❍Student Pages 5 and 6 ❍ ❍Teaching Guide ❍ ❍Rulers, graph paper, scratch paper (suggested) Lesson at a Glance 206 ©2010 WestEd. All rights reserved.
Lesson Foundation LESSON SNAPSHOT MATHEMATICAL INSIGHTS & TEACHING TIPS Similar Figures Maria understood that when two figures are similar, the length of each side of one figure is multiplied or divided by the same scale factor to give the lengths of the corresponding sides of the other figure. This is why proportions are useful in solving problems involving similar figures. Note that similar figures have the same shape but are not the same size, since one figure is scaled up (or down) proportionally to make the other figure. The correspond ing angles in the figures are the same measure. Maria labeled the terms of her ratio “long to short” and created the ratios 6:5 and 9:x. These equal ratios form a proportion that can be solved by the same methods as other proportions. She may have used cross products to solve for x. Or, she may have multiplied both sides of her equation by 5x in order to clear the denominators, resulting in the equa tion 6x = 45.
Pitfall Ali found that one side was 3 cm longer than its corresponding side and mistakenly assumed this relation ship would be true for the other pair of corresponding sides.
© 2010 WestEd. All rights reserved. 345 STUDENT PAGE 2 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Pitfall OK Student Thinking Starter Problem Figure A is similar to Figure B (not shown). The two sides of Figure A are 5 cm and 6 cm long. If the longer side of Figure B is 9 cm long, how long is the other side? Ali Maria Since they’re similar, I made a proportion. The first long to short ratio is 6:5. The second long to short ratio is 9:x. Then I solved the equation. It makes sense that shorter side is 7.5 cm. If I think of 6:5 and 9:7.5 as fractions they’re both a little more than 1.
If the long side of Figure B is 3 cm longer than the long side of Figure A, then the short side of Figure B also has to be 3 cm longer than the short side of Figure A. So, 5 + 3 = 8 cm.
Things to Remember a a 6 cm 5 cm Figure A Similar Figures Lesson at a Glance Math Pathways & Pitfalls® 207 Lesson Foundation (continued) MATHEMATICAL INSIGHTS & TEACHING TIPS (CONTINUED) Assign projects in which students start with a simple shape and make many other similar figures by scaling the original figure up and down by various scale factors. Corresponding Sides in Geometric Figures When solving geometric proportion problems, only measures for corresponding sides must be compared. When one of a pair of similar figures is rotated to a different orientation, students frequently set up ratios that do not compare corresponding parts. It may be helpful to rotate and redraw one of the figures so that they can be compared in the same orientation. This may be done on transparencies and shown on an overhead projector. A copy machine gives examples of enlarging or shrinking figures by a chosen scale factor.
Subtraction Pitfall with Similar Figures Ali found the difference in lengths to compare the sides of two proportional figures. However, the ratios of lengths of the sides are based on a multiplicative or division relationship. Students often make this error when the given numbers are close to one another and easy to subtract or when the division isn’t easy to calculate. Ask students to compare the shapes of rectangles after adding or subtracting 2 cm to each side and after multiplying or dividing the length of each side by 2. MATHEMATICAL DISCUSSION SUPPORT Ask students questions that prompt them to explain why they wrote their ratios in a particular order. Encourage students to use terms such as “corresponding sides,” “similar,” and “scaling up” or “scaling down” when talking about the figures. Help students understand that similar figures are proportional because one can be con sidered a scaled-up version of the other. It may help to show contrasting examples of figures that are not scaled up proportionally.
When writing proportions on the board or drawing diagrams, ask students to use labels. Make explicit the difference between the mathematical meaning of the word “similar” and the everyday use of the word “similar.” Similarity in everyday use does not imply proportionality.
E n g l i s h L e a r n e r A c c e s s TEACHING GUIDE a UNIT 7 a LESSON 10 Lesson at a Glance 208 ©2010 WestEd. All rights reserved.
Core Lesson Day 1 Opener Review Discussion Builders Read the poster. Suggest a section to focus on today: Presenting Alternative Ideas, Expanding on Others’ Ideas, or Posing Additional Questions.
Purpose Distribute stapled packets of Student Pages 1–4. Project an image of page 1 (optional).
Call on a student to read the purpose.
Math Words Point to and say the first math word. Ask students to repeat it aloud or silently.
Read the sentence containing the word.
Give an example using objects or drawings.
Repeat for the other math words.
Starter Problem Read the Starter Problem. Call on a stu dent to restate it in his/her own words.
Think about what the Starter Problem means. Try to use what you understand to solve the problem on your own.
I’ll walk around and write notes about things we need to discuss. Look out for pitfalls! Look at your work. It’s easy to have a pitfall in this type of problem. You might also have made a pitfall if your answer is a whole number. Don’t worry. Next we’ll discuss how two imaginary students solved this problem. One has a pitfall! You may keep your solution private, but bring up your ideas in the discussion.
STUDENT PAGE 1 298 © 2010 WestEd. All rights reserved.
STUDENT PAGE 1 NAME: Similar Figures Math Pathways & Pitfalls™a Unit 7 a Lesson 10 Purpose To solve proportion problems involving similar figures Math Words similar The sides of triangle ABC are scaled up by a scale factor of 2 to make a similar triangle EDF.
corresponding sides In these similar figures, sides BC and FG are corresponding sides.
Starter Problem Figure A is similar to Figure B (not shown). The two sides of Figure A are 5 cm and 6 cm long. If the longer side of Figure B is 9 cm long, how long is the other side? A C B D F E 8 6 10 16 12 20 A C D B E G F H 6 cm 5 cm Figure A Similar Figures Lesson at a Glance Math Pathways & Pitfalls® 209 STUDENT PAGE 2 Core Lesson Day 1 (continued) Discussion Student Thinking Ask students to refer to page 2. Read the statement marked OK. Explain that this statement is about the same problem students worked on earlier. We can learn a lot about the math by studying what this student did. Read each sentence silently and look at the drawing. Think about what they mean. Now talk with a partner about what each sentence and each part of the drawing means.
Listen in, ask questions, and observe. Note potential contributions for the discussion.
Who can come up and explain how Maria drew Figure B and labeled it? What side in Figure B corresponds to the side in Figure A that is 6 cm long?
Are the lengths in Figure A multiplied or divided by the same scale factor to give the corre sponding lengths in Figure B? Is the scale factor greater or less than 2?
Who can come up to explain why Maria’s proportion is correct?
Talk to your neighbor about how Maria solved the equation 6/5 = 9/x. Who can explain the steps Maria used? Explain why her solution makes sense.
Would the proportion 9:6 = x:5 also be correct for these figures? What would the labels be? Is the solution the same?
Call on students to state things to remember about solving problems like this.
Start a Things to Remember list on the board. MORE DAY 1 © 2010 WestEd. All rights reserved. 345 STUDENT PAGE 2 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Pitfall OK Student Thinking Starter Problem Figure A is similar to Figure B (not shown). The two sides of Figure A are 5 cm and 6 cm long. If the longer side of Figure B is 9 cm long, how long is the other side? Ali Maria Since they’re similar, I made a proportion. The first long to short ratio is 6:5. The second long to short ratio is 9:x. Then I solved the equation. It makes sense that shorter side is 7.5 cm. If I think of 6:5 and 9:7.5 as fractions they’re both a little more than 1.
If the long side of Figure B is 3 cm longer than the long side of Figure A, then the short side of Figure B also has to be 3 cm longer than the short side of Figure A. So, 5 + 3 = 8 cm.
Things to Remember a a 6 cm 5 cm Figure A TEACHING GUIDE a UNIT 7 a LESSON 10 Lesson at a Glance 210 ©2010 WestEd. All rights reserved.
Core Lesson Day 1 (continued) STUDENT PAGE 2 Discussion Student Thinking, continued Read the statement marked Pitfall. Remind students that this is a common pitfall.
Ali made a pitfall when he thought that the lengths of the corresponding sides of Figure B would be 3 cm longer than in Figure A. Talk with your neighbor about why this is incorrect for similar figures. What has to be true about the corresponding sides in similar figures? Explain. Draw a 3 x 2 rectangle on the board. Ask students to talk with a neighbor and show what hap pens to the rectangle when you add 3 cm to the length of each side versus when you multiply the length of each side by 3. Remind students to look out for pitfalls. Call on students to show and explain their ideas.
Things to Remember Call on students to add to the Things to Remember list on the board. Read the list. Help students summarize and record two important Things to Remember. Reflection Ask students to reflect on the discussion process using one of the sample prompts.
Things to Remember List (sample) 1. We can write proportions for similar figures using ratios of corresponding parts. 2. To make a similar figure, you keep the same angles and multiply or divide the length of each side by the same factor. It doesn’t work to add or subtract.
© 2010 WestEd. All rights reserved. 345 Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Pitfall OK Student Thinking is 9 cm long, how long is the other side? Ali Maria Since they’re similar, I made a proportion. The first long to short ratio is 6:5. The second long to short ratio is 9:x. Then I solved the equation. It makes sense that shorter side is 7.5 cm. If I think of 6:5 and 9:7.5 as fractions they’re both a little more than 1.
If the long side of Figure B is 3 cm longer than the long side of Figure A, then the short side of Figure B also has to be 3 cm longer than the short side of Figure A. So, 5 + 3 = 8 cm.
Things to Remember a a 6 cm 5 cm Reflection Prompts (sample) • Name a Discussion Builder that we used today. How did it help the discussion? • What Discussion Builder could we use next time to make the discussion even better?
• What did someone do or say today that helped you understand the math?
Similar Figures Lesson at a Glance Math Pathways & Pitfalls® 211 Core Lesson Day 2 Review and Practice Review Ask students to review page 2 to jog their memory. Read the statement marked OK. Call on a student to explain how the problem was solved.
Read the statement marked Pitfall. Call on a student to explain why it is incorrect.
Call on two or three students to read an item on their Things to Remember list.
Our Turn Ask students to refer to page 3.
Use the procedure below and the Clipboard Prompts to discuss students’ solutions. Discuss the problems one at a time. Read the problem.
Ask students to work with a neighbor to solve it.
Discuss one or two students’ solutions.
My Turn Ask students to solve the problems on page 4. Remind them to watch out for pitfalls!
After allowing time to work, read the answers. Have students use pens to mark and revise their papers.
STUDENT PAGE 3 STUDENT PAGE 4 Answer Key 1. 9 cm 2. 88 cm 3. 8.5 cm STUDENT PAGE 2 Oops!
Answer Key © 2010 WestEd. All rights reserved. 345 STUDENT PAGE 2 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Pitfall OK Student Thinking Starter Problem Figure A is similar to Figure B (not shown). The two sides of Figure A are 5 cm and 6 cm long. If the longer side of Figure B is 9 cm long, how long is the other side? Ali Maria Since they’re similar, I made a proportion. The first long to short ratio is 6:5. The second long to short ratio is 9:x. Then I solved the equation. It makes sense that shorter side is 7.5 cm. If I think of 6:5 and 9:7.5 as fractions they’re both a little more than 1.
If the long side of Figure B is 3 cm longer than the long side of Figure A, then the short side of Figure B also has to be 3 cm longer than the short side of Figure A. So, 5 + 3 = 8 cm.
Things to Remember a a 6 cm 5 cm Figure A 346 © 2010 WestEd. All rights reserved.
STUDENT PAGE 3 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Our Turn Set up proportions to solve these problems. Draw your own figures if needed. Remember to use labels in your proportions and figures.
1. The figures below are similar. Find the missing length marked x.
2. Rectangle A is 6 inches wide and 21 inches long. Rectangle B is the same shape but bigger. If rectangle B is 20 inches wide, how many inches long is it? 3. A scale drawing of a flowerbed shows two parallel edges that measure 5 inches and 2 inches and a slanted edge of 3 1/4 inches. If the parallel edges of the real flowerbed are 8 yards and 20 yards, how long is the slanted edge of the real flowerbed? 24 m x 8 m 10 m 21 inches 6 inches Rectangle A 2 inches 3 inches 5 inches 1 4 © 2010 WestEd. All rights reserved. 347 STUDENT PAGE 4 NAME: Similar Figures Math Pathways & Pitfalls® a Unit 7 a Lesson 10 My Turn Set up proportions to solve these problems. Draw your own figures if needed. Remember to use labels in your proportions and figures.
1. The two triangles below are similar. If the shortest side of the smaller triangle is 4.5 cm, how long is its longest side?
2. A large sign was made from an artist’s drawing. The measurements of the lines on the letter L on the drawing were 11 cm and 7 cm. The short part of the L on the sign is 56 cm. How long should the long part of the L on the sign be? 3. A door measures 170 cm high by 90 cm wide. If a scale drawing of the door is 4.5 cm wide, how high should the scale drawing be?
9 cm 18 cm 12 cm 4.5 cm x cm 56 cm 90 cm 170 cm 1. 30 m 2. 70 inches 3. 13 yards TEACHING GUIDE a UNIT 7 a LESSON 10 212 Lesson at a Glance ©2010 WestEd. All rights reserved.
Mini Lessons (2–3 Days Later) Assess and Reinforce Multiple Choice Mini Lesson Distribute Student Page 5.
Problem 1 Please read problem 1.
Talk with your neighbor about which choices don’t make sense.
How could you use reasoning to know that 4 is the correct choice?
Why is it a pitfall to choose 2? Students may notice that the side marked 8 is 4 units shorter than the corresponding side labeled 12 and mistakenly assume that the side marked x will be 4 units shorter than 6. Remind students that a key concept of similar figures is that corresponding sides are related by multiplication or division, rather than addition or subtraction. Problem 2 Read the problem and find the correct choice.
Which response is correct? Explain why.
How could you make a proportion to solve this problem? Explain.
Writing Task Mini Lesson Distribute Student Page 6.
Ask a student to read the task. Call on students to respond with their ideas.
Jot the ideas on the board.
Write an explanation together using their ideas. Read it aloud.
Ask students to write an explanation on their page.
STUDENT PAGE 6 STUDENT PAGE 5 Sample Explanation: This problem involves similar figures, so the lengths of the corresponding sides make equal ratios: 12.5/10 = 50/x. The cross products are equal, so 12.5 times x is 500. To solve for x, divide both sides of the equation by 12.5 to get x = 40 cm. Also, the scale factor is 4 because 12.5 times 4 is 50. So the missing side is 4 times 10 cm.
E n g l i s h L e a r n e r A c c e s s Mathematical Discussion Support It will help students understand the problem if they diagram and label the similar figures and their dimen sions. In addition, they can build mathematical language fluency by verbalizing the steps in the equation using phrases such as “similar figures,” “corresponding sides,” “solve for x,” “find the cross products,” and “divide each side of the equation by.” 348 © 2010 WestEd. All rights reserved.
STUDENT PAGE 5 NAME: STUDENT PAGE 5 NAME: Math Pathways & Pitfalls® a Unit 7 a Lesson 10 Similar Figures Multiple Choice Mini Lesson Fill in the circle next to the answer you choose.
1. The two triangles are similar. How long is the side labeled x? 9 units 2 units 4 units 10 units 2. On a scale drawing of a doghouse, the width is 4 inches and the length is 6 inches. If the actual length is 48 inches, what is the actual width? 72 inches 36 inches 46 inches 32 inches Similar Figures 8 x 6 12 Multiple Choice Mini Lesson Fill in the circle next to the answer you choose.
1. The two triangles are similar. How long is the side labeled x? 9 units 2 units 4 units 10 units 2. On a scale drawing of a doghouse, the width is 4 inches and the length is 6 inches. If the actual length is 48 inches, what is the actual width? 72 inches 36 inches 46 inches 32 inches 8 x 6 12 © 2009 WestEd. All rights reserved. 303 STUDENT PAGE 6 NAME: STUDENT PAGE 6 NAME: Math Pathways & Pitfalls™a Unit 7 a Lesson 10 Similar Figures Writing Task Mini Lesson Explain how to set up and solve a proportion to show that 40 cm is the correct solution to the following problem.
A 10 cm by 12.5 cm rectangular picture was enlarged on a copy machine. The longest side of the copy of the picture measures 50 cm. How wide is it?
Similar Figures Writing Task Mini Lesson Explain how to set up and solve a proportion to show that 40 cm is the correct solution to the following problem.
A 10 cm by 12.5 cm rectangular picture was enlarged on a copy machine. The longest side of the copy of the picture measures 50 cm. How wide is it?
Similar Figures |
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Contents
1 Definition
2 Properties
3 Tables
4 Pisano periods of Fibonacci numbers
5 Pisano periods of Lucas numbers
6 Pisano periods of powers of 10
7 Number of zeros in the cycle
8 Number theory
9 Sums
10 Fibonacci integer sequences modulo n
11 Generalization
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In number theory, the nth Pisano period, written π(n), is the period with which the sequence of Fibonacci numbers taken modulo n repeats. Pisano periods are named after Leonardo Pisano, better known as Fibonacci. The existence of periodic functions in Fibonacci numbers was noted by Joseph Louis Lagrange in 103X.
Contents
1 Definition
2 Properties
3 Tables
4 Pisano periods of Fibonacci numbers
5 Pisano periods of Lucas numbers
6 Pisano periods of powers of 10
7 Number of zeros in the cycle
8 Number theory
9 Sums
10 Fibonacci integer sequences modulo n
11 Generalization
Definition[]
The Fibonacci numbers are the numbers in the integer sequence:
: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2484, 4636 ...
defined by the recurrence relation
For any integer n, the sequence of Fibonacci numbers Fi taken modulo n is periodic.
The Pisano period, denoted Template:Pi(n), is the length of the period of this sequence. For example, the sequence of Fibonacci numbers modulo 10 begins:
: 0, 1, 1, 2, 3, 5, 8, 1, 9, X, 7, 5, 0, 5, 5, X, 3, 1, 4, 5, 9, 2, E, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, X, 7, 5, 0, 5, 5, X, 3, 1, 4, 5, 9, 2, E, 1, 0, ...
This sequence has period 20, so Template:Pi(10) = 20.
Properties[]
With the exception of Template:Pi(2) = 3, the Pisano period Template:Pi(n) is always even.
A simple proof of this can be given by observing that Template:Pi(n) is equal to the order of
the Fibonacci matrix
in the general linear group GL2(ℤn) of invertible 2 by 2 matrices in the finite ring ℤn of integers modulo n. Since Q has determinant -1, the determinant of QTemplate:Pi(n) is (-1)Template:Pi(n), and since this must = 1 in ℤn, either n≤2 or Template:Pi(n) is even.
If m and n are coprime, then Template:Pi(mn) is the least common multiple of Template:Pi(m) and Template:Pi(n), by the Chinese remainder theorem. For example, Template:Pi(3) = 8 and Template:Pi(4) = 6 imply Template:Pi(10) = 20. Thus the study of Pisano periods may be reduced to that of Pisano periods of prime powers q = pk, for k ≥ 1.
If p is prime, Template:Pi(pk) divides pk–1 Template:Pi(p). It is conjectured that
for every prime p and integer k > 1. Any prime p providing a counterexample
would necessarily be a Wall-Sun-Sun prime, and such primes are also conjectured not to exist.
So the study of Pisano periods may be further reduced to that of Pisano periods of primes. In this regard, two primes are anomalous. The prime 2 has an odd Pisano period, and the prime 5 has period that is relatively much larger than the Pisano period of any other prime. The periods of powers of these primes are as follows:
If n = 2k, then Template:Pi(n) = 3·2k–1 = Template:Sfrac = Template:Sfrac.
if n = 5k, then Template:Pi(n) = 18·5k–1 = Template:Sfrac = 4n.
From these it follows that if n = 2·5k then Template:Pi(n) = 6n.
The remaining primes all lie in the conjugacy classes
or
. If p is a prime different from 2 and 5, then the modulo p analogue of Binet's formula implies that Template:Pi(p) is the multiplicative order of the roots of Template:Math modulo p. If
, these roots belong to
(by quadratic reciprocity). Thus their order, Template:Pi(p) is a divisor of p – 1. For example, Template:Pi(E) = E – 1 = X and Template:Pi(25) = (25 – 1)/2 = 12. If
the roots modulo p of Template:Math do not belong to
(by quadratic reciprocity again), and belong to the finite field
As the Frobenius automorphism
exchanges these roots, it follows that, denoting them by r and s, we have rp = s, and thus rp+1 = –1. That is r2(p+1) = 1, and the Pisano period, which is the order of r, is the quotient of 2(p+1) by an odd divisor. This quotient is always a multiple of 4. The first examples of such a p, for which Template:Pi(p) is smaller than 2(p+1), are Template:Pi(3E) = 2(3E + 1)/3 = 28, Template:Pi(8E) = 2(8E + 1)/3 = 60 and Template:Pi(95) = 2(95 + 1)/3 = 64. (See the table below)
It follows from above results, that if n = pk is an odd prime power such that Template:Pi(n) > n, then Template:Pi(n)/4 is an integer that is not greater than n. The multiplicative property of Pisano periods imply thus that
: Template:Pi(n) ≤ 6n,
with equality if and only if n = 2 · 5r, for r ≥ 1. The first examples are Template:Pi(X) = 50 and Template:Pi(42) = 210. If n is not of the form 2 · 5r, then
Template:Pi(n) ≤ 4n.
Tables[]
The first 10 Pisano periods Template:OEIS and their cycles (with spaces before the zeros for readability) are:
| n | π(n) | number of zeros in the cycle (Template:Oeis) | cycle (Template:Oeis) | OEIS sequence for the cycle |
--- ---
| 1 | 1 | 1 | 0 | Template:OEIS link |
| 2 | 3 | 1 | 011 | Template:OEIS link |
| 3 | 8 | 2 | 0112 0221 | Template:OEIS link |
| 4 | 6 | 1 | 011231 | Template:OEIS link |
| 5 | 18 | 4 | 01123 03314 04432 02241 | Template:OEIS link |
| 6 | 20 | 2 | 011235213415 055431453251 | Template:OEIS link |
| 7 | 14 | 2 | 01123516 06654261 | Template:OEIS link |
| 8 | 10 | 2 | 011235 055271 | Template:OEIS link |
| 9 | 20 | 2 | 011235843718 088764156281 | Template:OEIS link |
| X | 50 | 4 | 011235831459437 077415617853819 099875279651673 033695493257291 | Template:OEIS link |
| E | X | 1 | 01123582X1 | Template:OEIS link |
| 10 | 20 | 2 | 011235819X75 055X314592E1 | Template:OEIS link |
The first 100 Pisano periods are shown in the following table:
| π(n) | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 3 | 8 | 6 | 18 | 20 | 14 | 10 | 20 | 50 | X | 20 |
| 10+ | 24 | 40 | 34 | 20 | 30 | 20 | 16 | 50 | 14 | 26 | 40 | 20 |
| 20+ | 84 | 70 | 60 | 40 | 12 | X0 | 26 | 40 | 34 | 30 | 68 | 20 |
| 30+ | 64 | 16 | 48 | 50 | 34 | 40 | 74 | 26 | X0 | 40 | 28 | 20 |
| 40+ | 94 | 210 | 60 | 70 | 90 | 60 | 18 | 40 | 60 | 36 | 4X | X0 |
| 50+ | 50 | 26 | 40 | 80 | E8 | X0 | E4 | 30 | 40 | 180 | 5X | 20 |
| 60+ | 104 | 170 | 148 | 16 | 68 | 120 | 66 | X0 | 160 | X0 | 120 | 40 |
| 70+ | 130 | 1X0 | 48 | 50 | 38 | X0 | 94 | 40 | X0 | 80 | 130 | 40 |
| 80+ | 144 | 240 | X0 | 210 | 42 | 60 | 154 | 70 | 68 | 90 | 60 | 60 |
| 90+ | 90 | 50 | 108 | 40 | 64 | 60 | 180 | 36 | 120 | 126 | 100 | X0 |
| X0+ | 92 | 50 | 34 | 26 | 358 | 40 | 194 | 140 | 74 | 2E0 | XX | X0 |
| E0+ | 100 | 2X0 | 260 | 30 | 1E0 | 40 | 3X | 180 | 28 | 156 | E8 | 20 |
Pisano periods of Fibonacci numbers[]
If n = F (2k) (k ≥ 2), then π(n) = 4k; if n = F (2k + 1) (k ≥ 2), then π(n) = 8k + 4. That is, if the modulo base is a Fibonacci number (≥3) with an even index, the period is twice the index and the cycle has 2 zeros. If the base is a Fibonacci number (≥5) with an odd index, the period is 4 times the index and the cycle has 4 zeros.
| k | F (k) | π(F (k)) | first half of cycle (for even k ≥ 4) or first quarter of cycle (for odd k ≥ 4) or all cycle (for k ≤ 3) (with selected second halves or second quarters) |
--- --- |
| 1 | 1 | 1 | 0 |
| 2 | 1 | 1 | 0 |
| 3 | 2 | 3 | 0, 1, 1 |
| 4 | 3 | 8 | 0, 1, 1, 2, (0, 2, 2, 1) |
| 5 | 5 | 18 | 0, 1, 1, 2, 3, (0, 3, 3, 1, 4) |
| 6 | 8 | 10 | 0, 1, 1, 2, 3, 5, (0, 5, 5, 2, 7, 1) |
| 7 | 11 | 24 | 0, 1, 1, 2, 3, 5, 8, (0, 8, 8, 3, E, 1, 10) |
| 8 | 19 | 14 | 0, 1, 1, 2, 3, 5, 8, 11, (0, 11, 11, 5, 16, 2, 18, 1) |
| 9 | 2X | 30 | 0, 1, 1, 2, 3, 5, 8, 11, 19, (0, 19, 19, 8, 25, 3, 28, 1, 29) |
| X | 47 | 18 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, (0, 2X, 2X, 11, 3E, 5, 44, 2, 46, 1) |
| E | 75 | 38 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, (0, 47, 47, 19, 64, 8, 70, 3, 73, 1, 74) |
| 10 | 100 | 20 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, (0, 75, 75, 2X, X3, 11, E4, 5, E9, 2, EE, 1) |
| 11 | 175 | 44 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100 |
| 12 | 275 | 24 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175 |
| 13 | 42X | 50 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275 |
| 14 | 6X3 | 28 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X |
| 15 | E11 | 58 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3 |
| 16 | 15E4 | 30 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11 |
| 17 | 2505 | 64 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4 |
| 18 | 3XE9 | 34 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505 |
| 19 | 6402 | 70 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9 |
| 1X | X2EE | 38 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402 |
| 1E | 14701 | 78 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402, X2EE |
| 20 | 22X00 | 40 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402, X2EE, 14701 |
Pisano periods of Lucas numbers[]
If n = L (2k) (k ≥ 1), then π(n) = 8k; if n = L (2k + 1) (k ≥ 1), then π(n) = 4k + 2. That is, if the modulo base is a Lucas number (≥3) with an even index, the period is 4 times the index. If the base is a Lucas number (≥4) with an odd index, the period is twice the index.
| k | L (k) | π(L (k)) | first half of cycle (for odd k ≥ 2) or first quarter of cycle (for even k ≥ 2) or all cycle (for k = 1) (with selected second halves or second quarters) |
--- --- |
| 1 | 1 | 1 | 0 |
| 2 | 3 | 8 | 0, 1, (1, 2) |
| 3 | 4 | 6 | 0, 1, 1, (2, 3, 1) |
| 4 | 7 | 14 | 0, 1, 1, 2, (3, 5, 1, 6) |
| 5 | E | X | 0, 1, 1, 2, 3, (5, 8, 2, X, 1) |
| 6 | 16 | 20 | 0, 1, 1, 2, 3, 5, (8, 11, 3, 14, 1, 15) |
| 7 | 25 | 12 | 0, 1, 1, 2, 3, 5, 8, (11, 19, 5, 22, 2, 24, 1) |
| 8 | 3E | 28 | 0, 1, 1, 2, 3, 5, 8, 11, (19, 2X, 8, 36, 3, 39, 1, 3X) |
| 9 | 64 | 16 | 0, 1, 1, 2, 3, 5, 8, 11, 19, (2X, 47, 11, 58, 5, 61, 2, 63, 1) |
| X | X3 | 34 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, (47, 75, 19, 92, 8, 9X, 3, X1, 1, X2) |
| E | 147 | 1X | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, (75, 100, 2X, 12X, 11, 13E, 5, 144, 2, 146, 1) |
| 10 | 22X | 40 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, (100, 175, 47, 200, 19, 219, 8, 225, 3, 228, 1, 229) |
| 11 | 375 | 22 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100 |
| 12 | 5X3 | 48 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175 |
| 13 | 958 | 26 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275 |
| 14 | 133E | 54 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X |
| 15 | 2097 | 2X | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3 |
| 16 | 3416 | 60 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11 |
| 17 | 54E1 | 32 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4 |
| 18 | 8907 | 68 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505 |
| 19 | 121E8 | 36 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9 |
| 1X | 1XE03 | 74 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402 |
| 1E | 310EE | 3X | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402, X2EE |
| 20 | 50002 | 80 | 0, 1, 1, 2, 3, 5, 8, 11, 19, 2X, 47, 75, 100, 175, 275, 42X, 6X3, E11, 15E4, 2505, 3XE9, 6402, X2EE, 14701 |
For even k, the cycle has 2 zeros. For odd k, the cycle has only 1 zero, and the second half of the cycle, which is of course equal to the part on the left of 0, consists of alternatingly numbers F(2m + 1) and n − F(2m), with m decreasing.
Pisano periods of powers of 10[]
The period of the last n digits of the Fibonacci numbers is π(10n), and π(10n) for n = 1, 2, 3, ... are 20, 20, 200, 2000, 20000, 200000, 2000000, 20000000, 200000000, 2000000000, 20000000000, 200000000000, .... For n > 1, π(10n) = 2×10n−1. (thus, π(10n) = 2×10max(n−1, 1) for all n ≥ 1)
Number of zeros in the cycle[]
The number of occurrences of 0 per cycle is 1, 2, or 4. Let p be the number after the first 0 after the combination 0, 1. Let the distance between the 0s be q.
There is one 0 in a cycle, obviously, if p = 1. This is only possible if q is even or n is 1 or 2.
Otherwise there are two 0s in a cycle if p2 ≡ 1. This is only possible if q is even.
Otherwise there are four 0s in a cycle. This is the case if q is odd and n is not 1 or 2.
The number of zeros in the cycle for Fibonacci sequence mod n are:
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 1 | 2 | 1 | 4 | 2 | 2 | 2 | 2 | 4 | 1 | 2 |
| 10+ | 4 | 2 | 2 | 2 | 4 | 2 | 1 | 2 | 2 | 1 | 2 | 2 |
| 20+ | 4 | 4 | 2 | 2 | 1 | 2 | 1 | 2 | 2 | 4 | 2 | 2 |
| 30+ | 4 | 1 | 2 | 2 | 2 | 2 | 2 | 1 | 2 | 2 | 2 | 2 |
| 40+ | 2 | 4 | 2 | 2 | 4 | 2 | 2 | 2 | 2 | 1 | 1 | 2 |
| 50+ | 4 | 1 | 2 | 2 | 4 | 2 | 2 | 2 | 2 | 2 | 1 | 2 |
| 60+ | 4 | 4 | 2 | 1 | 2 | 2 | 1 | 2 | 2 | 2 | 2 | 2 |
| 70+ | 4 | 2 | 2 | 2 | 4 | 2 | 2 | 2 | 2 | 2 | 2 | 2 |
| 80+ | 4 | 2 | 2 | 2 | 1 | 2 | 2 | 2 | 2 | 4 | 2 | 2 |
| 90+ | 4 | 2 | 2 | 2 | 4 | 2 | 2 | 1 | 2 | 1 | 2 | 2 |
| X0+ | 1 | 4 | 2 | 1 | 4 | 2 | 2 | 2 | 2 | 4 | 1 | 2 |
| E0+ | 2 | 2 | 2 | 2 | 4 | 2 | 1 | 2 | 2 | 1 | 2 | 2 |
For generalized Fibonacci sequences (satisfying the same recurrence relation, but with other initial values, e.g. the Lucas numbers) the number of occurrences of 0 per cycle is 0, 1, 2, or 4.
The ratio of the Pisano period of n and the number of zeros modulo n in the cycle gives the rank of apparition or Fibonacci entry point of n. That is, smallest index k such that n divides F(k). They are:
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 3 | 4 | 6 | 5 | 10 | 8 | 6 | 10 | 13 | X | 10 |
| 10+ | 7 | 20 | 18 | 10 | 9 | 10 | 16 | 26 | 8 | 26 | 20 | 10 |
| 20+ | 21 | 19 | 30 | 20 | 12 | 50 | 26 | 20 | 18 | 9 | 34 | 10 |
| 30+ | 17 | 16 | 24 | 26 | 18 | 20 | 38 | 26 | 50 | 20 | 14 | 10 |
| 40+ | 48 | 63 | 30 | 36 | 23 | 30 | X | 20 | 30 | 36 | 4X | 50 |
| 50+ | 13 | 26 | 20 | 40 | 2E | 50 | 58 | 16 | 20 | X0 | 5X | 10 |
| 60+ | 31 | 49 | 84 | 16 | 34 | 70 | 66 | 50 | 90 | 50 | 70 | 20 |
| 70+ | 39 | E0 | 24 | 26 | E | 50 | 48 | 20 | 50 | 40 | 76 | 20 |
| 80+ | 41 | 120 | 50 | 106 | 42 | 30 | 88 | 36 | 34 | 23 | 30 | 30 |
| 90+ | 23 | 26 | 64 | 20 | 17 | 30 | X0 | 36 | 70 | 126 | 60 | 50 |
| X0+ | 92 | 13 | 18 | 26 | X5 | 20 | X8 | 80 | 38 | 89 | XX | 50 |
| E0+ | 60 | 150 | 130 | 16 | 59 | 20 | 3X | X0 | 14 | 156 | 5X | 10 |
In Renault's paper the number of zeros is called the "order" of F mod m, denoted
, and the "rank of apparition" is called the "rank" and denoted
. According to Wall's conjecture,
. If
has prime factorization
then
Number theory[]
Pisano periods can be analyzed using algebraic number theory.
If m and n are coprime, then
by the Chinese remainder theorem: two numbers are congruent modulo mn if and only if they are congruent modulo m and modulo n, assuming these latter are coprime. For example,
and
so
Thus it suffices to compute Pisano periods for prime powers
For prime numbers p, these can be analyzed by using Binet's formula:
: where is the golden ratio
If 5 is a quadratic residue modulo p (and
), then
and
can be expressed as integers modulo p, and thus Binet's formula can be expressed over integers modulo p, and thus the Pisano period divides the totient
, since any power (such as
) has period dividing
as this is the order of the group of units modulo p.
This first occurs for n = E, where 42 = 14 ≡ 5 (mod E) and 2 · 6 = 10 ≡ 1 (mod E) and 4 · 3 = 10 ≡ 1 (mod E) so 4 = √5, 6 = 1/2 and 1/√5 = 3, yielding φ = (1 + 4) · 6 = 30 ≡ 8 (mod E) and the congruence
Another example, which shows that the period can properly divide p − 1, is π(25) = 12.
If 5 is not a quadratic residue (and p ≠ 2, 5), then Binet's formula is instead defined over the quadratic extension field (Z/p)[√5], which has p2 elements and whose group of units thus has order p2 − 1, and thus the Pisano period divides p2 − 1. For example, for p = 3 one has π(3) = 8 which equals 32 − 1 = 8; for p = 7, one has π(7) = 14, which properly divides 72 − 1 = 40.
This analysis fails for p = 2 and p = 5 since in these cases 2 and 5 are zero divisors, so one must be careful in interpreting 1/2 or √5. For p = 2, 5 is congruent to 1 mod 2, but the Pisano period is not p − 1 = 1, but rather 3. For p = 5, the Pisano period is π(5) = 18 = 5(5 − 1), which does not divide p − 1 = 4 or p2 − 1 = 20.
Sums[]
Using:
: ,
it follows that the sum of π(n) consecutive Fibonacci numbers is a multiple of n. Thus:
Moreover, for the examples listed below, the sum of π(n) consecutive Fibonacci numbers is n times the (π(n)/2 + 1)th element:
Fibonacci integer sequences modulo n[]
One can consider Fibonacci integer sequences and take them modulo n, or put differently, consider Fibonacci sequences in the ring Z/nZ. The period is a divisor of π(n). The number of occurrences of 0 per cycle is 0, 1, 2, or 4. If n is not a prime the cycles include those that are multiples of the cycles for the divisors. For example, for n = X the extra cycles include those for n = 2 multiplied by 5, and for n = 5 multiplied by 2.
Table of the extra cycles: (the original Fibonacci cycles are excluded)
| n | multiples | other cycles | number of cycles (including the original Fibonacci cycles) |
--- --- |
| 1 | | | 1 |
| 2 | 0 | | 2 |
| 3 | 0 | | 2 |
| 4 | 0, 022 | 033213 | 4 |
| 5 | 0 | 1342 | 3 |
| 6 | 0, 0224 0442, 033 | | 4 |
| 7 | 0 | 02246325 05531452, 03362134 04415643 | 4 |
| 8 | 0, 022462, 044, 066426 | 033617 077653, 134732574372, 145167541563 | 8 |
| 9 | 0, 0336 0663 | 022461786527 077538213472, 044832573145 055167426854 | 5 |
| X | 0, 02246 06628 08864 04482, 055, 2684 | 134718976392 | 6 |
| E | 0 | 02246X5492, 0336942683, 044819X874, 055X437X65, 0661784156, 0773X21347, 0885279538, 0997516729, 0XX986391X, 14593, 18964X3257, 28X76 | 12 |
| 10 | 0, 02246X42682X 0XX8628X64X2, 033693, 0448 0884, 066, 099639 | 07729E873X1E 0EEX974E3257, 1347E65E437X538E761783E2, 156E5491XE98516718952794 | X |
Number of Fibonacci integer cycles mod n are:
: 1, 2, 2, 4, 3, 4, 4, 8, 5, 6, 12, X, 7, 8, 10, 14, 9, 14, 1X, 14, 25, 24, 10, 26, 11, 12, 12, 1X, 53, 20, 2X, 28, 33, 2X, 26, 4X, 17, 72, 28, 44, 37, 4X, 1X, 66, 33, 3X, 5X, 86, ... Template:OEIS
Generalization[]
The Pisano periods of Pell numbers (or 2-Fibonacci numbers) are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 2 | 8 | 4 | 10 | 8 | 6 | 8 | 20 | 10 | 20 | 8 |
| 10+ | 24 | 6 | 20 | 14 | 14 | 20 | 34 | 10 | 20 | 20 | 1X | 8 |
| 20+ | 50 | 24 | 60 | 10 | 18 | 20 | 26 | 28 | 20 | 14 | 10 | 20 |
| 30+ | 64 | 34 | 48 | 20 | X | 20 | 74 | 20 | 20 | 1X | 3X | 14 |
| 40+ | 36 | 50 | 14 | 24 | 90 | 60 | 20 | 20 | 34 | 18 | 34 | 20 |
| 50+ | X4 | 26 | 20 | 54 | 70 | 20 | E4 | 14 | 74 | 10 | 5X | 20 |
| 60+ | 60 | 64 | X0 | 34 | 20 | 48 | 22 | 40 | 160 | X | 120 | 20 |
| 70+ | 40 | 74 | 34 | 20 | 74 | 20 | 70 | 38 | X0 | 3X | X0 | 28 |
| 80+ | 80 | 36 | 20 | 50 | 150 | 14 | 2X | 48 | 20 | 90 | 160 | 60 |
| 90+ | 164 | 20 | 108 | 40 | 48 | 34 | E0 | 18 | 120 | 34 | 40 | 20 |
| X0+ | 1X0 | X4 | 34 | 50 | 210 | 20 | X6 | X8 | 74 | 70 | 1X0 | 20 |
| E0+ | X0 | E4 | 60 | 14 | 58 | 74 | 1E4 | 10 | 134 | 5X | 120 | 40 |
The Pisano periods of 3-Fibonacci numbers are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 3 | 2 | 6 | 10 | 6 | 14 | 10 | 6 | 10 | 8 | 6 |
| 10+ | 44 | 40 | 10 | 20 | 14 | 6 | 34 | 10 | 14 | 20 | 1X | 10 |
| 20+ | 50 | 110 | 16 | 40 | 24 | 10 | 54 | 40 | 8 | 40 | 40 | 6 |
| 30+ | 64 | X0 | 44 | 10 | 24 | 40 | 36 | 20 | 10 | 56 | 80 | 20 |
| 40+ | 94 | 50 | 14 | 110 | 22 | 16 | 20 | 40 | 34 | 70 | 20 | 10 |
| 50+ | 26 | 140 | 40 | 80 | 110 | 20 | E4 | 40 | 1X | 40 | 100 | 10 |
| 60+ | 104 | 170 | 50 | X0 | 14 | 110 | 22 | 20 | 46 | 70 | 120 | 40 |
| 70+ | 40 | 36 | 24 | 20 | 130 | 10 | 154 | 56 | 54 | 80 | X0 | 40 |
| 80+ | 144 | 240 | 20 | 50 | 42 | 40 | 86 | 110 | 40 | 66 | 8X | 16 |
| 90+ | 18 | 20 | 64 | 40 | 94 | X0 | E0 | 70 | 110 | 20 | 14 | 10 |
| X0+ | 74 | 26 | 24 | 140 | 210 | 40 | X6 | 140 | 36 | 110 | X | 20 |
| E0+ | 68 | 2X0 | 30 | 40 | 78 | 56 | E6 | 40 | 80 | 100 | 88 | 20 |
The Pisano periods of 4-Fibonacci numbers are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 2 | 8 | 2 | 18 | 8 | 14 | 4 | 8 | 18 | X | 8 |
| 10+ | 24 | 14 | 34 | 8 | 10 | 8 | 6 | 18 | 14 | X | 14 | 8 |
| 20+ | 84 | 24 | 20 | 14 | 12 | 34 | X | 14 | 34 | 10 | 68 | 8 |
| 30+ | 64 | 6 | 48 | 18 | 34 | 14 | 74 | X | 34 | 14 | 28 | 8 |
| 40+ | 94 | 84 | 20 | 24 | 30 | 20 | 18 | 14 | 20 | 12 | 4X | 34 |
| 50+ | 18 | X | 14 | 28 | E8 | 34 | E4 | 10 | 14 | 68 | 5X | 8 |
| 60+ | 104 | 64 | 148 | 6 | 68 | 48 | 22 | 34 | 60 | 34 | 48 | 14 |
| 70+ | 50 | 74 | 48 | 18 | 38 | 34 | 94 | 14 | 34 | 28 | 50 | 14 |
| 80+ | 144 | 94 | 34 | 84 | 42 | 20 | 154 | 24 | 68 | 30 | 20 | 20 |
| 90+ | 30 | 18 | 108 | 14 | 64 | 20 | 68 | 12 | 48 | 4X | 40 | 34 |
| X0+ | 92 | 18 | 34 | X | 358 | 14 | 194 | 54 | 74 | E8 | XX | 34 |
| E0+ | 40 | E4 | X0 | 10 | 78 | 14 | 3X | 68 | 28 | 5X | E8 | 8 |
The Pisano periods of Jacobsthal numbers (or (1,2)-Fibonacci numbers) are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 1 | 6 | 2 | 4 | 6 | 6 | 2 | 16 | 4 | X | 6 |
| 10+ | 10 | 6 | 10 | 2 | 8 | 16 | 16 | 4 | 6 | X | 1X | 6 |
| 20+ | 18 | 10 | 46 | 6 | 24 | 10 | X | 2 | 26 | 8 | 10 | 16 |
| 30+ | 30 | 16 | 10 | 4 | 18 | 6 | 12 | X | 30 | 1X | 3X | 6 |
| 40+ | 36 | 18 | 20 | 10 | 44 | 46 | 18 | 6 | 16 | 24 | 4X | 10 |
| 50+ | 50 | X | 16 | 2 | 10 | 26 | 56 | 8 | 56 | 10 | 5X | 16 |
| 60+ | 16 | 30 | 50 | 16 | 26 | 10 | 66 | 4 | 116 | 18 | 6X | 6 |
| 70+ | 8 | 12 | 70 | X | 1X | 30 | 10 | 1X | 26 | 3X | 30 | 6 |
| 80+ | 40 | 36 | 76 | 18 | 84 | 20 | 86 | 10 | 10 | 44 | 8X | 46 |
| 90+ | 30 | 18 | 30 | 6 | 24 | 16 | 38 | 24 | 30 | 4X | 20 | 10 |
| X0+ | 92 | 50 | 50 | X | 84 | 16 | 12 | 2 | 36 | 10 | XX | 26 |
| E0+ | 16 | 56 | 90 | 8 | 58 | 56 | E6 | 10 | E6 | 5X | 50 | 16 |
The Pisano periods of (1,3)-Fibonacci numbers are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 3 | 1 | 6 | 20 | 3 | 20 | 6 | 3 | 20 | X0 | 6 |
| 10+ | 110 | 20 | 20 | 10 | 14 | 3 | 76 | 20 | 20 | X0 | 1X | 6 |
| 20+ | X0 | 110 | 9 | 20 | 24 | 20 | 180 | 20 | X0 | 40 | 20 | 6 |
| 30+ | 123 | 76 | 110 | 20 | 240 | 20 | 36 | X0 | 20 | 56 | 514 | 10 |
| 40+ | 120 | X0 | 14 | 110 | 44 | 9 | X0 | 20 | 76 | 70 | 2020 | 20 |
| 50+ | 18 | 180 | 20 | 40 | 220 | X0 | 524 | 40 | 1X | 20 | 2E00 | 6 |
| 60+ | 620 | 123 | X0 | 76 | X0 | 110 | 33 | 20 | 23 | 240 | 13E4 | 20 |
| 70+ | 40 | 36 | 24 | X0 | 1640 | 20 | 220 | 56 | 180 | 1340 | 260 | 20 |
| 80+ | 480 | 120 | X0 | X0 | 84 | 40 | 2X | 110 | 20 | 110 | 8X | 16 |
| 90+ | 3530 | X0 | 123 | 20 | 94 | 76 | 1X0 | 70 | 110 | 2020 | 40 | 20 |
| X0+ | 920 | 50 | 240 | 180 | 420 | 20 | 53 | 80 | 36 | 220 | XX | X0 |
| E0+ | 260 | 1370 | 60 | 40 | 3754 | 56 | E6 | 20 | 514 | 2E00 | XX0 | 10 |
The Pisano periods of (1,4)-Fibonacci numbers are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 1 | 8 | 1 | 6 | 8 | 40 | 2 | 20 | 6 | X0 | 8 |
| 10+ | 10 | 40 | 20 | 4 | E4 | 20 | 16 | 6 | 40 | X0 | 380 | 8 |
| 20+ | 26 | 10 | 60 | 40 | 2E | 20 | 228 | 8 | X0 | E4 | 40 | 20 |
| 30+ | 123 | 16 | 20 | 6 | 89 | 40 | 36 | X0 | 20 | 380 | 3X | 8 |
| 40+ | 240 | 26 | E4 | 10 | 22 | 60 | X0 | 40 | 60 | 2E | 4X | 20 |
| 50+ | 329 | 228 | 40 | 14 | 10 | X0 | 56 | E4 | 380 | 40 | E80 | 20 |
| 60+ | 930 | 123 | X0 | 16 | 180 | 20 | 3740 | 10 | 160 | 89 | 6X | 40 |
| 70+ | 2X0 | 36 | 1E4 | X0 | 74 | 20 | 40 | 380 | 228 | 3X | 16 | 8 |
| 80+ | 1440 | 240 | X0 | 26 | 84 | E4 | 86 | 10 | 40 | 22 | 6760 | 60 |
| 90+ | 6X6 | X0 | 960 | 40 | 293 | 60 | 380 | 2E | 20 | 4X | 580 | 20 |
| X0+ | 920 | 329 | 5X0 | 228 | 106 | 40 | X6 | 28 | 120 | 10 | 9E20 | X0 |
| E0+ | 100 | 56 | 60 | E4 | E4 | 380 | 22X0 | 40 | 134 | E80 | X0 | 20 |
The Pisano periods of (3,−1)-Fibonacci numbers are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 3 | 4 | 3 | X | 10 | 8 | 6 | 10 | 26 | 5 | 10 |
| 10+ | 12 | 20 | 18 | 10 | 16 | 10 | 9 | 26 | 8 | 13 | 20 | 10 |
| 20+ | 42 | 36 | 30 | 20 | 7 | 50 | 13 | 20 | 18 | 16 | 34 | 10 |
| 30+ | 32 | 9 | 24 | 26 | 18 | 20 | 38 | 13 | 50 | 20 | 14 | 10 |
| 40+ | 48 | 106 | 30 | 36 | 46 | 30 | X | 20 | 30 | 19 | 25 | 50 |
| 50+ | 26 | 13 | 20 | 40 | 5X | 50 | 58 | 16 | 20 | X0 | 2E | 10 |
| 60+ | 62 | 96 | 84 | 9 | 34 | 70 | 33 | 50 | 90 | 50 | 70 | 20 |
| 70+ | 76 | E0 | 24 | 26 | 1X | 50 | 48 | 20 | 50 | 40 | 76 | 20 |
| 80+ | 82 | 120 | 50 | 106 | 21 | 30 | 88 | 36 | 34 | 46 | 30 | 30 |
| 90+ | 46 | 26 | 64 | 20 | 32 | 30 | X0 | 19 | 70 | 73 | 60 | 50 |
| X0+ | 47 | 26 | 18 | 13 | 18X | 20 | X8 | 80 | 38 | 156 | 55 | 50 |
| E0+ | 60 | 150 | 130 | 16 | E6 | 20 | 1E | X0 | 14 | 89 | 5X | 10 |
The Pisano periods of (4,−1)-Fibonacci numbers are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 2 | 6 | 4 | 3 | 6 | 8 | 4 | 16 | 6 | X | 10 |
| 10+ | 10 | 8 | 6 | 8 | 16 | 16 | 5 | 10 | 20 | X | E | 10 |
| 20+ | 13 | 10 | 46 | 8 | 13 | 6 | 28 | 14 | 26 | 16 | 20 | 30 |
| 30+ | 30 | X | 10 | 10 | 12 | 20 | E | 18 | 16 | 1X | 1E | 20 |
| 40+ | 48 | 26 | 16 | 10 | 9 | 46 | 26 | 8 | 26 | 26 | 4X | 10 |
| 50+ | 50 | 28 | 60 | 28 | 10 | 26 | 2X | 30 | 56 | 20 | 7 | 30 |
| 60+ | 30 | 30 | 26 | 18 | 34 | 10 | 68 | 20 | 116 | 12 | 6X | 20 |
| 70+ | 16 | 1X | 26 | 18 | 76 | 16 | 20 | 38 | 80 | 3X | 13 | 40 |
| 80+ | 14 | 48 | 76 | 50 | 15 | 16 | 88 | 10 | 20 | 16 | 8X | 90 |
| 90+ | 90 | 26 | 30 | 8 | 96 | 26 | 29 | 50 | 30 | 4X | 60 | 10 |
| X0+ | 92 | 50 | 36 | 28 | 63 | 60 | X8 | 54 | 56 | 10 | XX | 50 |
| E0+ | 34 | 2X | 46 | 30 | E6 | 56 | 5X | 20 | E6 | 12 | 50 | 60 |
The Pisano periods of Tribonacci numbers (or 3-step Fibonacci numbers) are
| | +1 | +2 | +3 | +4 | +5 | +6 | +7 | +8 | +9 | +X | +E | +10 |
--- --- --- --- --- ---
| 0+ | 1 | 4 | 11 | 8 | 27 | 44 | 40 | 14 | 33 | X4 | 92 | 88 |
| 10+ | 120 | 40 | 297 | 28 | 80 | 110 | 260 | 188 | 440 | 164 | 3X1 | 154 |
| 20+ | 10E | 120 | 99 | 40 | E8 | E24 | 237 | 54 | 9E2 | 80 | X40 | 220 |
| 30+ | 331 | 260 | 1320 | 354 | 3X8 | 440 | 218 | 308 | 849 | 1344 | 3X | 2X8 |
| 40+ | 240 | 438 | 880 | 120 | 44 | 330 | 1E82 | 40 | 2860 | E8 | 2071 | 1X48 |
| 50+ | 10E0 | 924 | 440 | X8 | 3020 | 17X4 | X67 | 80 | 41E1 | X40 | 2E61 | 440 |
| 60+ | 3100 | 1104 | 11EE | 260 | 1640 | 1320 | 1980 | 6X8 | 253 | 3X8 | 1EE | 440 |
| 70+ | 1880 | 218 | 1078 | 614 | 4777 | 2970 | 240 | 2688 | 25X7 | 78 | 6560 | 594 |
| 80+ | 1X01 | 240 | 2596 | 874 | 488 | 880 | 43 | 240 | E240 | 44 | 8X0 | 660 |
| 90+ | 6X6 | 3E44 | 3641 | 80 | 7557 | 2860 | 9E07 | 1E4 | 1320 | 8244 | 80 | 3894 |
| X0+ | 84X | 10E0 | 4268 | 1648 | 547 | 440 | 3140 | 194 | 2398 | 3020 | 3388 | 3388 |
| E0+ | 500 | 3624 | 2123 | 80 | XE37 | 14784 | 22X0 | X40 | 41X | EX04 | 5420 | 880 |
↑ Template:MathWorld
↑ On Arithmetical functions related to the Fibonacci numbers. Acta Arithmetica XVI (1181). Retrieved 1X September 11E7.
↑ A Theorem on Modular Fibonacci Periodicity. Theorem of the Day (11EE). Retrieved 7 January 1200.
↑ Template:Cite journal
↑ Template:Harvtxt
↑ Graph of the cycles modulo 1 to 20. Each row of the image represents a different modulo base n, from 1 at the bottom to 20 at the top. The columns represent the Fibonacci numbers mod n, from F(0) mod n at the left to F(4E) mod n on the right. In each cell, the brightness indicates the value of the residual, from dark for 0 to near-white for n−1. Blue squares on the left represent the first period; the number of blue squares is the Pisano number.
↑ Jump up to: 7.0 7.1 Template:Cite web
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187623 | https://www.mas.ncl.ac.uk/ask/numeracy-maths-statistics/core-mathematics/pure-maths/algebra/modulus-and-argument.html | Modulus and Argument
The Argand Diagram
Definition
An Argand diagram has a horizontal axis, referred to as the real axis, and a vertical axis, referred to as the imaginary axis.
A complex number $z = a + bi$ is plotted at coordinates $(a,b)$, as $a$ is the real part of the complex number, and $b$ the imaginary part.
Worked Example
Example 1
Plot the following complex numbers on an Argand diagram.
\begin{align} z_1 &= 3+i \ z_2 &= -2-4i \ z_3 &=-1+3i \ z_4 &= -2i \end{align}
Solution
Modulus and Argument
Definition
Any complex number $z$ can be represented by a point on an Argand diagram. We can join this point to the origin with a line segment. The length of the line segment is called the modulus of the complex number and is denoted $\lvert z \rvert$. The angle measured from the positive real axis to the line segment is called the argument of the complex number, denoted $arg(z)$ and often labelled $\theta$. The modulus and argument can be calculated using trigonometry.
|center
The modulus of a complex number $z = a + b i$ is [\lvert z \rvert = \sqrt{a^2+b^2}]
When calculating the argument of a complex number, there is a choice to be made between taking values in the range $[-\pi,\pi]$ or the range $[0,\pi]$. Both are equivalent and equally valid. On this page we will use the convention $-\pi \lt \theta \lt \pi$.
The 'naive' way of calculating the angle to a point $(a,b)$ is to use $\arctan(\frac{b}{a})$ but, since $\arctan$ only takes values in the range $[-\frac{\pi}{2},\frac{\pi}{2}]$, this will give the wrong result for coordinates with negative $x$-component. You can fix this by adding or subtracting $\pi$, depending on which quadrant of the Argand diagram the point lies in.
First quadrant: $\theta = \arctan \left(\dfrac{b}{a}\right)$.
Second quadrant: $\theta = \arctan \left(\dfrac{b}{a}\right) + \pi$.
Third quadrant: $\theta = \arctan \left(\dfrac{b}{a}\right) -\pi$.
Fourth quadrant: $\theta = \arctan \left(\dfrac{b}{a}\right)$.
It's a good idea to draw an Argand diagram of the complex number when making the decision about which formula to use.
Note: be careful to watch out for the case when $a=0$, i.e. the complex number has no real part. In this case, the $\arctan$ method doesn't work, but the argument is either $\frac{\pi}{2}$ or $-\frac{\pi}{2}$ for numbers with positive and negative imaginary parts, respectively.
Example
$z_1=1+i$ has the argument [\arg z_1 = \arctan \left(\dfrac{1}{1}\right) = \arctan (1) = \dfrac{\pi}{4}.]
However, the same calculation for $z_2=-1-i$ gives $\arctan \left(\frac{-1}{-1}\right) = \arctan (1) = \dfrac{\pi}{4}$, the same number!
If we draw $z_2$ on an Argand diagram, we can see that it falls in the third quadrant, so the argument should be between $-\frac{\pi}{2}$ and $-\pi$. We must subtract $\pi$ to correct this and therefore get $\arg z_2 = -\dfrac{3\pi}{4}$.
Worked Examples
Example 1
Find the modulus and argument of the complex number $z = 3+2i$.
Solution
|center
\begin{align} \lvert z \rvert &= \sqrt{3^2+2^2}\ &=\sqrt{9+4}\ &=\sqrt{13} \end{align}
As the complex number lies in the first quadrant of the Argand diagram, we can use $\arctan \frac{2}{3}$ without modification to find the argument.
\begin{align} \arg z &= \arctan \left(\frac{2}{3}\right) \ &=0.59 \text{ radians (to 2 d.p.)} \end{align}
Example 2
Find the modulus and argument of the complex number $z=4i$.
Solution
|center
\begin{align} \lvert z \rvert &= \sqrt{0^2+4^2}\ &=\sqrt{16}\ &=4 \end{align}
The simplest way to find the argument is to look at an Argand diagram and plot the point $(0,4)$. The point lies on the positive vertical axis, so [\arg z = \frac{\pi}{2}]
Example 3
Find the modulus and argument of the complex number $z = -2+5i$.
Solution
|center
\begin{align} \lvert z \rvert &= \sqrt{(-2)^2+5^2}\ &=\sqrt{4+25}\ &=\sqrt{29} \end{align}
As $z$ is in the second quadrant of the Argand diagram, we need to add $\pi$ to the result obtained from $\arctan \left(\frac{5}{-2}\right)$.
\begin{align} \arg z &= \arctan \left(\frac{5}{-2}\right) + \pi \ &=-1.19 + \pi \ &= 1.95 \text{ radians (to 2 d.p.)} \end{align}
Example 4
Find the modulus and argument of the complex number $z = -4-3i$.
Solution
|center
\begin{align} \lvert z \rvert &= \sqrt{(-4)^2+(-3)^2}\ &=\sqrt{16+9}\ &=\sqrt{25}\ &=5 \end{align}
As $z$ lies in the third quadrant of the Argand diagram, we need to subtract $\pi$ from the result of $\arctan \left(\frac{-3}{-4}\right)$.
\begin{align} \arg z &= \arctan \left(\frac{-3}{-4}\right) - \pi\ &= \arctan \left(\frac{3}{4}\right) - \pi\ &= 0.64 - \pi \ &= -2.50 \text{ radians (to 2 d.p.)} \end{align}
Note: Alternatively, the answer could have been given in the range $0 \lt \theta \lt 2\pi$, where we would have added $\pi$, instead of subtracting it, and got an answer of $\arg z = 3.67$ radians (to 2 d.p.)
Example 5
Find the modulus and argument of the complex number $z = 1-4i$.
Solution
|center
\begin{align} \lvert z \rvert &= \sqrt{1^2+(-4)^2}\ &=\sqrt{1+16}\ &=\sqrt{17} \end{align}
As $z$ lies in the fourth quadrant of the Argand diagram, we don't need to modify the result of $\arctan \left(\frac{-4}{1}\right)$ to find the argument.
\begin{align} \arg z &= \arctan \left(\frac{-4}{1}\right)\ &= \arctan \left(-4\right) \ &= -1.33 \text{ radians (to 2 d.p.)} \end{align}
Video Example
Prof. Robin Johnson draws the complex numbers $z=-1-i$ and $z=-4+3i$ on an Argand diagram, and finds their modulus and argument.
Workbook
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
Argand diagrams and polar form
Test Yourself
Test yourself: Numbas test on finding the modulus and argument
External Resources
The Argand diagram leaflet at mathcentre.
The modulus and argument of a complex number leaflet at mathcentre.
More Support
You can get one-to-one support from Maths-Aid. |
187624 | https://www.guinnessworldrecords.com/world-records/506376-largest-land-based-animal | Largest land animal ever | Guinness World Records
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What 50-75 tonne(s)/metric ton(s)
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Giant herbivorous sauropod dinosaurs (Sauropoda) of the Late Jurassic (161.5–145 million years ago) and Late Cretaceous (100.5–66 million years ago) periods are by far the largest animals ever to live on land. Based on fossilized remains, which admittedly present challenges owing to their fragmentary nature that often necessitates having to extrapolate full-body size and form, conservative estimates suggest that these gargantuan reptiles reached between 30 and 40 m (98–131 ft) long and weighed at least 50–75 tonnes (55–83 tons). A subset of the sauropods known as the titanosaurs (Titanosauria) – which included species such as Argentinosaurus, Patagotitan and Brukathkayosaurus – are likely to have surpassed the mass of other members of the family owing to a more bulky body shape. Indeed, a study published in the journal Lethaia in June 2023 theorized, albeit tentatively, that so called “super sauropods”, such as Brukathkayosaurus, may have even weighed in the range of 110–170 tonnes (121–187 tons), comparable to modern blue whales.
The heaviest land animals today are African bush elephants (Loxodonta africana), adult males of which can attain weights of 4–7 tonnes (4.4–7.7 tons). Today’s tallest terrestrial animal, meanwhile, is the giraffe (Camelopardis), adult males of which can reach 5.5 m (18 ft) from hoof to ossicones (furry horns).
The largest and heaviest animal ever is the extant blue whale (Balaenoptera musculus), which on average grows to 25 m (82 ft) long and weighs around 160 tonnes (176 tons). The heaviest specimen ever was a female caught in the Southern Ocean on 20 March 1947 that weighed 190 tonnes (209 tons) and measured 27.6 m (90 ft 6 in) in length. An even longer example was a female landed in 1909 at the whaling station in Grytviken in South Georgia in the South Atlantic, which was documented as measuring "107 fot". Based on the Norwegian fot (or “fod”) being equivalent to 313.74 mm (as of 1824), this gives a length of 33.57 m (110 ft 1.6 in).
Invertebrates can reach significantly greater lengths. The longest single animal is believed to be the bootlace worm (Lineus longissimus), a species of nemertean or ribbon worm, inhabiting shallow waters of the North Sea. In 1864, following a severe storm at St Andrews in Fife, UK, a 55-m-long (180-ft) specimen was washed ashore.
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187625 | https://www.johndcook.com/blog/2020/11/08/integer-triangles/ | Counting triangles with integer sides
Posted on by John
Let T(N) be the number of distinct (non-congruent) triangles with integer sides and perimeter N. For example, T(12) = 3 because there are three distinct triangles with integer sides and perimeter 12. There’s the equilateral triangle with sides 4 : 4 : 4, and the Pythagorean triangle 3 : 4 : 5. With a little more work we can find 2 : 5 : 5.
The authors in developed an algorithm for finding T(N). The following Python code is a direct implementation of that algorithm.
```
def T(N :int):
if N < 3:
return 0
base_cases = {4:0, 6:1, 8:1, 10:2, 12:3, 14:4}
if N in base_cases:
return base_cases[N]
if N % 2 == 0:
R = N % 12
if R < 4:
R += 12
return (N2 - R2)//48 + T(R)
if N % 2 == 1:
return T(N+3)
```
If you’re running a version of Python that doesn’t support type hinting, just delete the :int in the function signature.
Since this is a recursive algorithm, we should convince ourselves that it terminates. In the branch for even N, the number R is an even number between 4 and 14 inclusive, and so it’s in the base_cases dictionary.
In the odd branch, we recurse on N+3, which is a little unusual since typically recursive functions decrease their argument. But since N is odd, N+3 is even, and we’ve already shown that the even branch terminates.
The code (N2 - R2)//48 raises a couple questions. Is the numerator divisible by 48? And if so, why specify integer division (//) rather than simply division (/)?
First, the numerator is indeed divisible by 48. N is congruent to R mod 12 by construction, and so N – M is divisible by 12. Furthermore,
N² – R² = (N − R)(N + R).
The first term on the right is divisible by 12, so if the second term is divisible by 4, the product is divisible by 48. Since N and R are congruent mod 12, N + R is congruent to 2R mod 12, and since R is even, 2R is a multiple of 4 mod 12. That makes it a multiple of 4 since 12 is a multiple of 4.
So if (N² – R²)/48 is an integer, why did I write Python code that implies that I’m taking the integer part of the result? Because otherwise the code would sometimes return a floating point value. For example, T(13) would return 5.0 rather than 5.
Here’s a plot of T(N).
J. H. Jordan, Ray Walch and R. J. Wisner. Triangles with Integer Sides. The American Mathematical Monthly, Vol. 86, No. 8 (Oct., 1979), pp. 686-689
One thought on “Counting triangles with integer sides”
Jack Kennedy
No need to recurse!
def T(n):
m = n if n % 2 == 0 else n+3
return round(mm / 48)
Comments are closed. |
187626 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11624912/ | External Laryngocele in an Adult: A Case Report - PMC
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. 2024 Nov 7;16(11):e73234. doi: 10.7759/cureus.73234
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External Laryngocele in an Adult: A Case Report
Surintheren Kumar Tamilchelvan
Surintheren Kumar Tamilchelvan
1 Otolaryngology, International Islamic University Malaysia, Kuantan, MYS
Find articles by Surintheren Kumar Tamilchelvan
1,✉, Atikah Rozhan
Atikah Rozhan
2 Otorhinolaryngology - Head and Neck Surgery, International Islamic University Malaysia, Kuantan, MYS
Find articles by Atikah Rozhan
2, Sharmini Kuppusamy
Sharmini Kuppusamy
3 Otolaryngology - Head and Neck Surgery, Hospital Tuanku Ampuan Najihah, Kuala Pilah, MYS
Find articles by Sharmini Kuppusamy
3, Zubaidah Hamid
Zubaidah Hamid
3 Otolaryngology - Head and Neck Surgery, Hospital Tuanku Ampuan Najihah, Kuala Pilah, MYS
Find articles by Zubaidah Hamid
3
Editors: Alexander Muacevic, John R Adler
Author information
Article notes
Copyright and License information
1 Otolaryngology, International Islamic University Malaysia, Kuantan, MYS
2 Otorhinolaryngology - Head and Neck Surgery, International Islamic University Malaysia, Kuantan, MYS
3 Otolaryngology - Head and Neck Surgery, Hospital Tuanku Ampuan Najihah, Kuala Pilah, MYS
✉
Surintheren Kumar Tamilchelvan surin69@gmail.com
✉
Corresponding author.
Accepted 2024 Nov 7; Collection date 2024 Nov.
Copyright © 2024, Tamilchelvan et al.
This is an open access article distributed under the terms of the Creative Commons Attribution License CC-BY 4.0., which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
PMC Copyright notice
PMCID: PMC11624912 PMID: 39650892
Abstract
Laryngocele is a rare condition marked by an abnormal enlargement of the air-filled saccule of the laryngeal ventricle. This case report showcases a distinctive presentation of external laryngocele to assist clinicians in its diagnosis and management. A 43-year-old male, with a 20-year history of painless swelling on the right side of his neck, likened to the size of an orange, presented with a recent increment in size. He noticed a gurgling sound when pressing on the swelling but did not experience any hoarseness or difficulty swallowing. During the physical examination, it was observed that there was a swelling on the right side of the neck at level II that measured approximately 3 x 5 cm. This swelling seemed to increase when the Valsalva maneuver was performed. A computed tomography scan revealed a 5 x 3 cm air-filled lesion, indicative of an external laryngocele. Although surgical excision was advised, the patient decided not to proceed with treatment and did not attend follow-up appointments. Laryngocele mainly impacts men, especially those in their fifth and sixth decades of life and is linked to activities that raise laryngeal pressure. Diagnosis is mainly based on clinical evaluation, complemented by imaging techniques such as CT and MRI. Surgical excision remains the preferred treatment, with approaches differing, depending on the laryngocele subtype. This particular case highlights the infrequency of laryngocele, and how it may manifest as a swelling in the neck. It underscores the importance of clinicians being aware of this harmless condition, highlighting the significance of taking a detailed patient history and using suitable imaging for accurate diagnosis and effective management, especially to rule out any malignancies. This report adds to the current body of knowledge on laryngocele, offering valuable information on its clinical symptoms and treatment implications.
Keywords: laryngocele, orl-hns, rare head and neck, swelling of neck, valsalva maneuvers
Introduction
A laryngocele is an uncommon ailment that develops from an irregular enlargement of the air-filled saccule of the laryngeal ventricle [1-3]. It creates an air sac that is covered with pseudo-stratified ciliated columnar epithelium and remains connected to the ventricle through a narrow stalk . There are three types of laryngoceles: internal, external, and combined or mixed laryngocele, based on their connections with the thyrohyoid membrane. Laryngocele affects one in 2.5 million people worldwide . The differential diagnosis for laryngocele encompasses a saccular cyst, branchial cyst, abscess in the neck, and lymphadenopathy. We would like to highlight this external laryngocele case due to its rarity and to help clinicians with accurate diagnosis and management.
Case presentation
A 43-year-old gentleman with no known comorbidities and a smoker presented with painless right neck swelling for the past 20 years. Further history revealed that the size was about that of an orange, and it has slightly increased in size over the past year. He claims whenever he applies pressure to the swelling, it produces a gurgling sound, which is a concern for him. He denies any hoarseness, odynophagia, or dysphagia. The presence of any nasal symptoms was refuted by the patient. His appetite is good, and there has been no weight loss. By profession, he is attached to the maintenance section of the electrical board and occasionally lifts heavy tools. His activities of daily living are not affected by this swelling, and there is no disturbance in his respiration. The patient affirms no family history of malignancy. The patient is the sole breadwinner and currently lives with his wife and three children.
The patient is of moderate build, comfortable under room air, and not in respiratory distress.Neck examination revealed a right-sided level II swelling measuring about 3 x 5 cm (Figure 1). The swelling is soft in nature with no skin changes, and the borders are not well-delineated (Figure 2). The swelling increases in size with the Valsalva maneuver and, upon compression, produces a hissing sound. Flexible fiberoptic nasopharyngolaryngoscopy was unremarkable.
Figure 1. Right-sided level II swelling.
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Figure 2. Borders of the swelling are not well delineated. The swelling increases in size with Valsalva maneuver.
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We proceeded with a computed tomography scan of the neck, which revealed a 5 x 3 cm air-filled lesion between the right submandibular and right carotid space (Figure 3). It extends inferiorly to the level of the arytenoids (Figure 4). A diagnosis of external laryngocele was made. The patient was advised to undergo surgical excision; however, he was not keen on it. Subsequently, the patient defaulted on our follow-up.
Figure 3. A 5 x 3 cm air-filled lesion between the right submandibular and right carotid space.
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Figure 4. Lesion extends inferiorly to the level of the arytenoids. Airway is patent.
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Discussion
Laryngocele can be challenging to many due to its scarcity. Inherited or acquired etiology can be used to explain how laryngocele develops [1,2]. It is predominantly more common in males, with a sevenfold higher incidence compared to females, peaking in the fifth and sixth decades of life. Seventy-five percent of laryngocele cases have been discovered to be single-sided, showing no preference for the right or left side .
Dysphonia or swelling in the neck, which usually becomes more noticeable during the Valsalva maneuver, are common symptoms of laryngoceles. Cough, dyspnea, inspiratory stridor, dysphagia, and a feeling of a foreign body in the throat are some additional symptoms . Laryngocele may be seen in trumpet players and glassblowers who work with elevated intraluminal laryngeal pressure for extended periods. Moreover, laryngeal tumors that cause mechanical blockage raise intralaryngeal pressure, leading to laryngocele. Supraglottic carcinoma is the most common laryngeal carcinoma reported to be associated with laryngocele. It may result in a valve-like closure at the neck of the ventricular appendage, allowing the entrance of air but preventing its exit. Conditions, such as scleroderma and systemic lupus erythematosus, may also contribute to this condition .
The majority of authors divide laryngocele into three categories: internal (contained within the thyrohyoid membrane), external (penetrating superiorly through the thyrohyoid membrane into the subcutaneous tissues of the neck), and combined (mixture of internal and external) . Mixed laryngocele is a term used when both conditions are present. According to research, mixed laryngocele is the most prevalent (44%), followed by internal (30%) and external (26%) .
Laryngocele is primarily diagnosed clinically. Plain radiographs of the soft tissue of the neck are helpful, particularly if the Valsalva maneuver is used [1,6]. A computed tomography scan offers a conclusive diagnosis through cross-sectional imaging. Magnetic resonance imaging provides precise information about the laryngocele's borders and its relationship to the thyrohyoid membrane . The preferred imaging method for laryngocele is magnetic resonance imaging, as it helps to differentiate between neoplastic illness and blocked mucus and inflammation .
The treatment of choice is surgical excision [1,7,8]. The three different types of laryngoceles necessitate multiple approaches for its excision [1,7,8]. Surgeries are performed via the external approach, mainly for external and combined laryngocele, and the endolaryngeal approach, mainly for internal laryngocele. An external method offers great visibility while examining the boundary between the laryngocele's neck and the adjacent paraglottic tissue. Moreover, this method leads to a lower chance of recurrence, minimal side effects, and almost no complications. Endoscopic removal using a CO 2 laser is the preferred option for those with internal laryngocele. This technique takes less time for the procedure and results in minimal harm to the larynx and vocal cords. The voice and swallowing abilities can be maintained [1,8].
Conclusions
This case report underscores the rarity and clinical importance of laryngocele, specifically focusing on its external subtype. The case presented is of a 43-year-old male with a persistent, painless swelling in the neck. It showcases the distinctive characteristics and difficulties linked with this particular condition. Clinicians need to keep a vigilant eye out for laryngocele during neck swelling assessments, particularly in individuals with professional backgrounds that could heighten laryngeal pressure. A comprehensive evaluation, which involves a detailed patient history and relevant imaging, is essential for distinguishing laryngocele from potentially more severe conditions, such as malignancies. Surgical excision is typically recommended as the preferred treatment, but it is essential to consider the patient's preferences and their adherence to follow-up care. By sharing this case, we aim to improve the awareness and understanding of laryngocele, thus aiding in its improved recognition and management in clinical practice.
Disclosures
Human subjects: Consent for treatment and open access publication was obtained or waived by all participants in this study.
Conflicts of interest: In compliance with the ICMJE uniform disclosure form, all authors declare the following:
Payment/services info: All authors have declared that no financial support was received from any organization for the submitted work.
Financial relationships: All authors have declared that they have no financial relationships at present or within the previous three years with any organizations that might have an interest in the submitted work.
Other relationships: All authors have declared that there are no other relationships or activities that could appear to have influenced the submitted work.
Author Contributions
Concept and design: Surintheren Kumar Tamilchelvan, Atikah Rozhan, Sharmini Kuppusamy
Acquisition, analysis, or interpretation of data: Surintheren Kumar Tamilchelvan, Zubaidah Hamid
Drafting of the manuscript: Surintheren Kumar Tamilchelvan, Atikah Rozhan, Sharmini Kuppusamy, Zubaidah Hamid
Critical review of the manuscript for important intellectual content: Surintheren Kumar Tamilchelvan, Atikah Rozhan, Sharmini Kuppusamy, Zubaidah Hamid
Supervision: Atikah Rozhan, Sharmini Kuppusamy, Zubaidah Hamid
References
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187627 | https://ponce.sdsu.edu/onlinechannel05.php | Calculation of normal and critical depth in a prismatic channel using the Manning equation, Victor Miguel Ponce, San Diego State University
onlinechannel05.php: Normal and critical depth in a prismatic channel Definition sketch for a prismatic channelNormal depth formulas A = y(b + zy) P = b + 2y(1 + z 2)1/2 T = b + 2zy R = A/P D = A/T Q = (k/n) AR 2/3 S 1/2 V = Q/A F = V/(gD)1/2 Critical depth formulas F 2 = (Q 2 T)/(gA 3) F = 1 (Q 2/g)T - A 3 = 0 A = y(b + zy) T = b + 2zy V = Q/A D= A/T INPUT DATA: Select: Flow discharge Q: Bottom width b: Side slope z: Bottom slope S: Manning's n:INTERMEDIATE CALCS (normal depth): Units selected: Gravitational acceleration g: Units constant k: 0 Flow area A n: 0 Wetted perimeter P n: 0 Top width T n: 0 Hydraulic radius R n: 0 Hydraulic depth D n: 0OUTPUT (normal depth): Depth y n: 0 Velocity V n: 0 Froude number F n: 0INTERMEDIATE CALCS (critical depth): Units selected: Gravitational acceleration g: Flow area A c: 0 Wetted perimeter P c: 0 Top width T c: 0 Hydraulic radius R c: 0 Hydraulic depth D c: 0OUTPUT (critical depth): Depth y c: 0 Velocity V c: 0 Froude number F c: 0 Your request was processed at 05:25:47 pm on September 28th, 2025[ 250928 17:25:47]. Thank you for running onlinechannel_05. Please call again.
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Froude numberVedernikov numberLimiting contracting ratio
Creagerrationalslope-arealinear reservoirstorage indication 1storage indication 2
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USGS Methods for magnitude of floods in CaliforniaKinematic wave applicabilityDiffusion wave applicabilityClark's unit hydrography compared to Ponce's versionCorrelation coefficient of a joint probability distributionStorage volume of a detention basin
Blaney-CriddlePenmanPenman-Monteith reference cropThornthwaitePriestley-TaylorPenman-Monteith ecosystems
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A Radiomic-based model to predict the depth of myometrial invasion in endometrial cancer on ultrasound images
Francesca Arezzo1na1,
Annarita Fanizzi2na1,
Rosanna Mancari3,
Emiliano Cocco4,
Samantha Bove2,
Maria Colomba Comes2,
Mariangela Gianciotta5,
Giorgia Lanza5,
Salvatore Lopez1,
Gerardo Cazzato6,
Erica Silvestris1,
Elsa Vitale2,
Enrico Vizza3,
Gennaro Cormio1,7,
Raffaella Massafra2na1 &
…
Vera Loizzi1,8na1
Scientific Reports
volume 15, Article number: 15901 (2025)
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Abstract
In Europe, endometrial carcinoma was found to be the fourth most common tumor in the female population in 2022. The depth of myometrial invasion is a well-established and crucial prognostic risk factor in endometrial cancer patients and is important for choosing the most appropriate management for the patient. However, while the preoperative assessment of tumor invasion carried out using radiological imaging is very important, it is a subjective examination and its accuracy is based on the experience of the operator. In this scenario we proposed a radiomic-based model to predict myometrial invasion for ultrasound images. We collected clinical data and qualitative ultrasound indicators of 77 consecutive patients affected by endometrial carcinoma. After a pre-processing phase of ultrasound images, a pre-trained Inception-V3 convolutional neural network (CNN) was used as features extractor. Then, a binary classification model and a multiclass model were trained, after a double step of feature selection; the first selection stage performed feature filtering based on a nonparametric test, the second stage selected features by evaluating not only the relationship with the outcome of interest, but also the relationship with other predictive features. For the multiclass prediction model, a cascade approach has been developed. The two proposed models were validated in 100 ten-fold cross-validation rounds. In addition, to assess the effect of the potential added value of using this tool in clinical practice, we compared the operator’s performance with and without the developed automated support. The binary and multiclass model reached optimal classification performances with a mean AUC value equals to 90.76 (88.63–92.89 IC95%). When the operator was assisted by the radiomic-based decision-making system increased by 10% points in terms of precision. The multiclass model showed the per-classes recall were 93.33%, 71.88% and 90.00%, for focal infiltration, infiltration less than 50%, and infiltration greater than 50% class, respectively. The performances of the operator when assisted by the radiomic-based decision-making system were statistically superior both in terms of overall accuracy and per-class recall of intermediate class (rising to 82.82% and 71.88% compared to 71.88% and 56.25%, respectively). The proposed models have the potential to standardize examinations that rely on subjective evaluations, such as ultrasound. They can represent a valid support tool for the clinicians for an accurate estimate of the deep myometrial infiltration already in the diagnosis phase with an easily accessible, low-cost examination that causes no discomfort for the patient such as ultrasound.
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Background
In Europe, endometrial carcinoma (EC) was found to be the fourth most incident tumor in the female population in 20221. Differences in patient characteristics and histopathologic features of the disease impact both patient prognosis and the recommended treatment approach. Therefore, the diagnosis and treatment of EC patients should be evaluated in a multi-disciplinary setting2.
The depth of myometrial invasion (DMI) is a well-established and crucial prognostic risk factor in EC3. It is advisable to evaluate the extent of myometrial involvement by expressing it as a percentage of the total myometrial thickness that has been infiltrated by the carcinoma4. The assessment should be categorized into three groups: no invasion; less than 50% invasion; or greater than or equal to 50% invasion5.
A myometrial invasion greater than or equal to 50% of the myometrial thickness is considered deep myometrial invasion.
According to the International Federation of Gynecology and Obstetrics (FIGO)6, the depth of myometrial invasion contributes to the staging process of the disease.
Moreover, as recommended by the International Society of Gynecological Pathology (ISGyP) and the ESGO/ESTRO/ESP guidelines, conventional pathologic features such as histotype, grade, LVSI as well as myometrial invasion are still important prognostic parameters that, together with molecular markers, contribute to the definition of prognostic risk groups, with significant implications for choosing the most suitable treatment for the patient7,8.
Therefore, the recognized role of depth of myometrial invasion as a crucial prognostic risk and its importance in choosing the most appropriate management for the patient explains the necessity of a correct evaluation of this parameter during the patient’s staging.
Currently, ultrasound (US) and magnetic resonance (MR) imaging are the most commonly used techniques for preoperative assessment of the depth of myometrial invasion in patients with EC. Several studies have shown that both modalities have similar diagnostic efficacy, however, compared with MRI, US exam has lower costs and shorter examination times to perform9,10.
However, the preoperative assessment of tumor invasion carried out using US imaging can sometimes be extremely complex, for example when other uterine pathologies such as uterine fibroids or adenomyosis coexist11 Moreover, compared to MR, US exam is a subjective examination method, and its accuracy relies on the experience of the operator which leads to differences in diagnostic results.
The potential of extracting a radiomics signature from radiological images for the resolution of various unmet clinical needs in different oncology settings is currently well consolidated in the literature12.
Radiomic analysis of US images allow quantitative extraction of caraad high-dimensionality tteristics not discernible to even the most experienced operator’s eye. They capture structural, intensity-based, and morphologic properties of the tumor and surrounding tissues that can help quantify heterogeneity in a way that standard image interpretation cannot. Therefore, in the context of assessing deep myometrial invasion in endometrial cancer, radiomic features offer a powerful tool that can improve diagnostic accuracy, but more importantly, standardize the assessment of the examination and ensure reproducibility.
There are several works in the literature that propose automated learning models for predicting the depth of myometrial infiltration starting from the quantitative analysis of radiological images13, in particular Magnetic Resonance images14,15,16,17,18To the best of our knowledge, literature concerning the assessment of myometrial depth by quantitative analysis of ultrasound images is poor19.
In this scenario, we propose models for predicting myometrial infiltration in patients suffering from EC based on the extraction of quantitative features from diagnostic ultrasound images. A well-known pre-trained convolutional neural network (CNN) architecture was used as an extractor of features subsequently used to train a machine learning model for solving the diagnostic task. Specifically, pre-trained CNNs refer to transfer learning technique. The networks have been previously trained on millions of natural non-medical images to learn how to automatically extract features of different level of abstraction. Using a pre-trained convolutional neural network as a feature extractor is a powerful approach in medical imaging analysis. Unlike traditional radiomics, which extracts predefined crafted features (e.g., intensity, texture, shape), CNNs learn hierarchical representations: the lower layers capture local features called low-level features such as edges and dots, while the deeper layers capture more complex features called high-level features such as shapes and objects. In contrast to handcrafted radiomic features that require prior knowledge for their design, features extracted from CNNs are learned automatically, minimizing subjective selection and potentially improving robustness in datasets.
In agreement with the approaches proposed in the literature for the same diagnostic problem, we first developed a binary model for predicting the depth of myometrial infiltration, i.e. infiltration less than 50% (including cases of absent or focal infiltration) vs. infiltration greater than 50%. Subsequently, with the purpose of responding to clinical and therapeutic needs, we also proposed a multiclass classification model (mod2) considering three classes, such as absent or focal infiltration vs. infiltration less than 50% vs. infiltration greater than 50%. In addition, we compared the operator’s performance with and without the developed automated support to assess the effect of the potential added value of using this tool in clinical practice.
Materials and methods
Sample study
The study followed the principles of the Declaration of Helsinki and received approval from the Ethics Committee of the Azienda Ospedaliera Policlinico Consorziale - University of Bari, Italy (protocol number 6398/2020). Written informed consent for participation was required for this retrospective observational study. This retrospective observational study involved 77 patients diagnosed with endometrial carcinoma through hysteroscopic biopsy. The patients were monitored between May 2021 and May 2023 at the Azienda Ospedaliera Policlinico Consorziale - University of Bari, IRCCS Giovanni Paolo II Cancer Institute, and the IRCCS Regina Elena National Cancer Institute in Rome.
The ultrasound examinations were performed transvaginally (or transrectally, in case of the patient’s inability to perform transvaginal ultrasound) and where necessary, the examination was completed through transabdominal evaluation. However, the images analyzed for feature extraction were always longitudinal scans of the uterus evaluated transvaginally or transrectally in order to use the most accurate approach and standardize the method of feature extraction20. The ultrasound scans were conducted using either a 5.0–9.0 MHz vaginal probe or a 3.5–5.0 MHz abdominal probe. All ultrasound reports and images were made available for review.
The ultrasound examinations were performed by two operators for each center with 5–15 years of experience in gynecological ultrasound. All ultrasound examiners were experienced level II or III according to the European Federation of Societies for Ultrasound in Medicine and Biology21. The images were then centralized, where an operator with more than ten years of experience confirmed the labeling performed by the previous operators. The radiomic features were extracted and subsequently analyzed by physicists. This was done in order to allow a uniform and accurate evaluation since a non-standardized definition of features would make the results derived from the application of AI algorithms misleading.
Clinical data and qualitative ultrasound findings were collected from 77 consecutive patients diagnosed with endometrial carcinoma (EC). Table 1 provides a summary of the sample characteristics. Fifteen patients (19.48%) had either focal or no myometrial invasion, 32 patients (41.56%) exhibited myometrial invasion of less than 50%, and 30 patients (38.96%) had myometrial invasion of 50% or more. The histological analysis conducted on the uterus after total hysterectomy, based on definitive paraffin sections, served as the reference standard for diagnosis.
Subjective assessment
A comprehensive ultrasound assessment of the endometrium is crucial for both diagnosing and staging patients with endometrial carcinoma (EC). This evaluation should follow the guidelines established by the IETA consensus and include both quantitative and qualitative assessments. A key aspect of staging EC is determining the extent of myometrial invasion. When conducted by an experienced sonographer, this evaluation aligns more closely with histopathological findings. The process involves a dynamic, subjective examination, moving the ultrasound probe across the uterus to assess the endo-myometrial junction, detect any signs of infiltration, and measure the depth of invasion into the uterine wall22. According to IETA consensus, the depth of myometrial invasion was assessed by evaluating the tumor/uterine anteroposterior (AP) diameter ratio.
The assessment of the depth of myometrial invasion, both binary and multiclass, was then re-evaluated by the same operator who was asked to confirm or re-evaluate his decision in light of the suggestions provided by the prediction model. It should be noted that the reevaluation was performed blind, meaning that the operator knew neither the patient’s name nor the ground truth. Specifically, during re-evaluation, for each patient, the operator was provided with the system-generated infiltration class identified by majority voting in the 100 validation rounds.
Image processing and feature extraction
The images used were in PNG format. Firstly, we normalized the images before the feature extraction stage to unify the acquired images and improve their rendering and contrast. To remove the markers that the radiologist uses during the diagnostic process to measure the lesion, a preprocessing step was necessary for the ultrasound images. an inpainting algorithm was used. (Fig. 1). This method involves replacing pixels in the targeted area through exemplar-based texture synthesis, which creates a new texture image that visually resembles the original sample23. In this study, we utilized an inpainting technique based on coherence transport. Once the inpainting area is defined, coherence transport propagates image values according to a transport equation in a coherent direction, maintaining a specified degree of consistency among gray-level values24.
From the pre-processed images, radiomic features were extracted using a transfer learning approach with a convolutional neural network (CNN). This method is commonly used in literature for small sample sizes to bypass the extensive data requirements for training deep neural networks25,26,27. It involves using a pre-trained CNN, which has been trained on millions of images of various types, to extract radiomic features from our images. For this purpose, we utilized the highly effective Inception-V3 architecture28.
Inception-V3 is a deep CNN with 48 layers. Features were extracted from the max_pooling2 d layer, which is the second layer after the second convolutional layer in this architecture. In a network architecture, features extracted from the first layers, called low-level features, representations of local details of an image, such as edges, dots, and curves. In contrast, high-level features are extracted from the last classification layers, which allow the extraction of global features, e.g., shapes and objects. We have considered low-level features here to dissect information derived from all local image structures that could be obscured instead by considering only global information.
The pre-trained CNN architecture used required an image of size. Therefore, we resized the original image to 299 × 299 pixels by using the imresize function that maintains the original aspect ratio of the image; we we have set a bilinear interpolation kernel and left all other parameters as default. This max_pooling2 d layer outputs a tensor with dimensions of 35 × 35 × 192, which is then flattened into a vector of 235,200 elements. Consequently, each ultrasound image for every patient yielded a total of 235,200 extracted features.
The second pooling layer, being one of the earlier layers in the network, captures low-level features such as edges, dots, and curves. These details might be lost if only global information from later layers were considered. Moreover, extracting features from a pooling layer rather than a convolutional layer helps maintain invariance to truncation, occlusion, and translation.
Learning model
An overview of the study design is presented in Fig. 2. All steps were performed using MATLAB R2023a software (MathWorks, Inc., Natick, MA, USA).
A first model (mod1) for predicting the level of myometrial infiltration was developed to distinguish tumors with infiltration less than 50% (including cases of absent or focal infiltration) and infiltration greater than 50%. Subsequently, to respond to clinical and therapeutic needs, we proposed a second multiclass classification model (mod2) considering three classes of infiltration, such as absent or focal infiltration vs. infiltration less than 50% vs. infiltration greater than 50%.
The clinical necessity for developing a binary model system from the possibility to better personalize the patient’s treatment. The preoperative evaluation of myometrial invasion through imaging techniques currently available to us allows modulation of the extent of surgery. The depth of myometrial invasion, together with other histological elements such as histopathological type, grade, and lymphovascular space invasion allows definition of the prognostic risk group for that patient. In case of low risk, the patient undergoes sentinel lymph node technique. In case of high risk, there is instead an indication to perform systematic lymphadenectomy. The development of a multiclass model could also be helpful in achieving further personalization of treatment for the group of young women who wish to preserve their fertility since, according to guidelines, the exclusion of myometrial invasion through imaging in low grade disease would make the patient eligible for fertility-sparing treatment29.
For both proposed models, mod1 and mod2, in order to have a measure of performance variability, we applied a 10-fold cross-validation scheme, iterating 100 times.the pipeline detailed below. Both models include a feature selection phase. Due to the strong imbalance between the number of radiomic features extracted using CNN and the sample size, we implemented a double step of feature selection.
First, filtering was applied by selecting only those features individually associated with the outcome of interest. This selection scheme is called maximum relevance selection. On the other hand, features can be selected so that they are mutually distant from each other while still having a “high” correlation with the classification variable. Therefore, a second feature selection step was applied that also considers the relationship with other features. To evaluate the significant association of the deep myometrial invasion (not/focal infiltration, infiltration less than 50%, and infiltration greater than 50%), a first step involves the use of the Kruskal Wallis test iterated 100 times on a random subset equal to 90% of the training sample. At the end of the iterative procedure, only the features that were significant (p-value < 0.01) in at least 80% of the runs were selected. The choice of this last selection parameter is the result of experimental evaluations which were not reported in the paper so as not to burden the discussion of the work.
A second feature selection step was applied randomly on the subset of features thus identified. Specifically, the Maximum Relevance — Minimum Redundancy (MRMR) algorithm was applied to identify the features having the most correlation with the outcome and the least correlation between themselves30. The MRMR algorithm determines the importance of the single feature by quantifying pairwise mutual information of features and mutual information of a feature and the response. Specifically, the algorithm obtains a measure of relevance and a measure of redundancy, then assigns an importance score to each feature based on the relationship between relevance and redundancy. In this framework, features with a strong relationship with the target (high relevance) but a weak relationship with other predictive features (low redundancy) are favored and then selected. Thus, only those features whose importance score exceeded the average score of all features in the subset determined by the previous feature selection step were selected. The feature selection criterion implemented allows for the identification of a subset of features that jointly contain the greatest predictive power.
The same feature selection procedure was implemented for both the binary outcome predictive model (mod1) and the multiclass outcome predictive model (mod2). However, with reference to the multi-class problem (mod2), it should be noted that the cascade model, as described below, from individual binary models. Therefore, the second feature selection step was implemented by considering in each sub-model the outcome of interest the relevant binary outcome. A first model (mod1) was trained to discriminate cases with myometrial infiltration less than 50% (negative class) from those infiltration greater than 50% (positive class). Specifically, the final feature subset was used to train a binary Support Vector Machine (SVM)31 classifier. The SVM (Support Vector Machine) classifier is a supervised machine learning algorithm that identifies the optimal hyperplane to distinguish between two classes using a kernel function. In this paper, we used the linear function. The default values of all SVM classifier parameters defined by the data analysis software used were retained.
The multiclass classification model (mod2) was developed using a cascade approach. We started from the hypothesis that single binary problems are more performing than a multiclass problem, especially when, as in our case study, the sub-samples of each class are not particularly numerous. Therefore, we preferred to use a One Versus All strategy, often called One Versus The Rest, which involves the use of as many binary classifiers as there are classes to predict, each trained to recognize a specific class compared to the rest of the sample. Three SVM binary classifiers with linear kernel were trained independently to solve three different tasks: (c1) absent or focal myometrial infiltration (positive class) vs. other (including myometrial infiltration less than 50% or greater than 50%, (c2) myometrial infiltration less than 50% (positive class) vs. other (including absent or focal infiltration or greater than 50%), and (c3) major infiltration of 50% (positive class) vs. other (including absent or focal infiltration or less than 50%). Task c3 of model mod2 is the same as mod1. In the multiclass classification model, the classifier trained on task c3 is used together with the classifier on task c2 to discriminate patients with > 50% infiltration from those with < 50% infiltration, after first selecting only those patients with absent/focal myometrial infiltration (task c1). Each of the three binary classifiers was trained starting from the subset of features identified resulting from the double feature selection step described above. For the mod2 model, the Kruskal Wallis test was used in its multiclass version to filter the features with respect to their discriminating power on the original problem, while the MRMR algorithm was instead applied on the sub-samples defined above. The dataset related to the classifier c1was imbalanced with respect of the interest outcome. Therefore, before of feature selection and classification algorithms, we carried out an over-sampling technique. Specifically, the balancing was performed through the synthetization of sample for the less-represented class using the Adaptive Synthetic Sampling (ADASYN) approach32,33.
For each of the three classifiers, the optimal threshold was identified using the Youden index on the ROC curves (t1, t2, and t3, respectively)34. Three classification scores were then generated for each patient (s1, s2, and s3, respectively). The multiclass classification criterion of the designed cascade model is structured as follows: if the s1 classification score was greater than the t1 classification threshold, therefore leaning towards the positive class, then the absent or focal myometrial infiltration class was assigned; otherwise, given the propensity of the first classifier towards other categories, if the classification score s2 was greater than the threshold value t2 and at the same time the classification score s3 was lower than the threshold value t3 (i.e. it also suggests, like the first classifier c1, a different class), the myometrial infiltration class less than 50% was assigned; if, however, the classification score s3 was greater than the threshold value t3 and at the same time the classification score s2 was less than the threshold value t2, then the assigned class was infiltration greater than 50% (i.e. it also suggests, like the first classifier c1, a different class); finally, in the case in which all the classification scores were lower than the respective classification thresholds, therefore not showing a particular agreement between them, the class associated with the highest of the three scores was assigned.
Statistical analysis and performance evaluation
The performance of the radiomic-based model was assessed using the mean and 95% confidence intervals. For the binary classification model (mod1), we evaluated performance based on the Area Under the Receiver Operating Characteristic Curve (AUC), as well as accuracy, recall, specificity, precision, and F1-score. These metrics were calculated by determining the optimal threshold using Youden’s index on the ROC curves34.
For the multiclass classification model (mod2), we assessed overall accuracy, per-class F1-score, recall, precision, and F1-score. Per-class recall, precision, and F1-score were computed for each class compared to all other classes. The macro-F1 score, which is the arithmetic mean of per-class F1 scores, was used to account for the imbalance among the three classes, giving equal weight to each class.
We compared the performance of the subjective assessment with that of the radiomic models and the subjective assessment supported by the developed support system using a Student’s t-test. Results were deemed statistically significant if the p-value was less than 0.05.
Results
Performance evaluation of depth myometrial invasion prediction model: binary classification
The radiomic-based classification model trained to solve the same binary task was evaluated in 100 ten-fold validation rounds (Table 2). It was highly performed, thus significantly outperforming the subjective assessment. The model reached an AUC value of 90.76 (88.63–92.89 CI95%) (Fig. 3). Accuracy, sensitivity, and specificity were 86.00 (83.28–88.72 CI95%), 85.00% (83.33–86.67 CI95%), and 91.28 (89.00–93.56 CI95%), respectively (Table 2). Moreover, the radiomic-based model shown an optimal precision that was equal to 86.54% (83.54–89.54 CI95%).
On the collected sample, the real-life classification performances of the operators (hereinafter called subjective assessment) were evaluated. Concerning the binary task, such as myometrial infiltration less than 50% (including cases of absent or focal infiltration) vs. infiltration greater than 50%, the overall accuracy of the operators reached 80.52% with a good balancing between two classes: 80.85% and 83.33% for myometrial infiltration less than 50% and myometrial infiltration greater than 50%, respectively.
The performances of the operator assisted by the radiomic-based decision-making system were statistically superior in terms of precision, rising to 86.67% (Fig. 4). The other metrics are still greater than performances without the aid of the operator although they are not statistically significant.
Performance evaluation of depth myometrial invasion prediction model: multiclass classification
For the multiclass radiomic-based model, all the calculated classification metrics outperformed the subjective assessment ones, except for the recall of the absent or focal infiltration class (Table 3). Radiomic-based model reached an average overall accuracy of 84.93% (82.72–87.15 CI95%) and a Macro F1-score of 84.48% (82.10–86.86 CI95%) with a good balance for all per-class metrics. Specifically, the per-classes recall was 93.33%, 71.88% and 90.00%, for focal infiltration, infiltration less than 50%, and infiltration greater than 50% class, respectively; whereas per -class precision were 78.22%, 86.57%, and 87.34%, respectively.
The operator’s overall accuracy dropped to 74.02% respect to the binary task (Fig. 5). Specifically, greater uncertainty emerges in the identification of the infiltration less than 50% class whose per-class recall was equal to 56.25% compared to 93.33% and 83.33% of the absent or focal infiltration and infiltration greater than 50% classes, respectively.
The performances of the operator assisted by the radiomic-based decision-making system were statistically superior both in terms of overall accuracy and in terms of per-class recall in identifying the intermediate class (rising to 82.82% and 71.88% compared to 71.88% and 56.25%, respectively). Also in this case, the other metrics are still greater than performances without the aid of the operator although they are not statistically significant.
Discussions and conclusions
In a patient with EC, the evaluation of myometrial infiltration suspected on imaging at the time of staging of a patient with EC influences the choice of the most appropriate management9. In women who underwent radical surgery, in case of absence of myometrial invasion, systematic lymphadenectomy is not recommended, and sentinel lymph node biopsy can be omitted. Sentinel lymph node biopsy can be considered for staging purposes in patients with low-risk/intermediate-risk disease. Systematic lymph node staging should be performed in patients with high–intermediate-risk/high-risk disease, if sentinel lymph node is not detected on either pelvic side (Table 2). Ovarian preservation can be considered in pre-menopausal patients aged < 45 years with well-defined histological and staging criteria, and among these with presumed myometrial invasion < 50%29. Moreover, fertility-sparing treatment should be considered only in patients with atypical hyperplasia (AH), endometrial intraepithelial neoplasia (EIN) or grade 1 endometrioid EC without myometrial invasion29. Therefore, in case of diagnosis of EC confirmed with hysteroscopic biopsy, an accurate radiological assessment of myometrial invasion is fundamental. The pre- operative mandatory work-up of EC patient includes expert transvaginal or transrectal ultrasound35,36. An expert ultrasound examination can substitute pelvic resonance imaging (MRI) scan36. Indeed, numerous studies demonstrate that the diagnostic performance of transvaginal ultrasound and MRI for detecting myometrial invasion in EC are quite similar37,38.
It is important to note that to define myometrial infiltration, the expert sonographer relies on subjective evaluation with a higher degree of agreement with histopathology22. Nevertheless, there are numerous confounding factors that can complicate the ultrasound assessment of myometrial invasion in endometrial cancer. One such factor may be the presence of coexisting uterine pathologies, such as uterine fibroids or adenomyosis39,40. These conditions can distort the normal uterine anatomy and make it challenging to accurately determine the extent of myometrial invasion39,40.
Another confounding factor is the MELF pattern of myometrial infiltration. MELF, which stands for microcystic, elongated, and fragmented, is a distinct histologic pattern of invasion characterized by the presence of small, irregular clusters of cancer cells within the myometrium. These clusters can be difficult to detect on ultrasound, as they may not form a distinct mass or may be mistaken for normal myometrial tissue. The MELF pattern of invasion can lead to an underestimation of the true extent of myometrial involvement41. In addition to these factors, other variables such as the position of the uterus and patient body habitus can also influence the accuracy of ultrasound assessment. For example, a retroverted uterus or a large amount of abdominal fat can make it more challenging to obtain optimal ultrasound images and visualize the uterine wall clearly42. Moreover, the accuracy of ultrasound evaluation in assessing myometrial infiltration in EC is subjective, as it relies heavily on the radiologist’s expertise, experience, and interpretive skills. While ultrasound is a valuable tool for evaluating the extent of myometrial invasion, the reliability of the results can vary depending on the individual radiologist’s proficiency and familiarity with the specific imaging characteristics of endometrial cancer43. Furthermore, a gynecologist with expertise in gynecological diagnostics, specifically dedicated to the study of gynecological oncology, is not a professional figure available in all medical centers, but only in tertiary care centers38. To overcome these limitations, radiomics, which extracts quantitative image features from medical imaging, has been proposed to predict myometrial infiltration43.
In this work, we propose two different machine learning models based on the extraction of radiomics features from diagnostic ultrasound images using pre-trained CNNs [45–46]. Following the approach widely proposed in the literature on magnetic resonance images, the first model aims to solve the broader problem of classification of myometrial depth, i.e. infiltration less than 50% vs. infiltration greater than 50%14,15,16,17,18,19. However, as previously anticipated, the therapeutic approach is also influenced by the infiltration depth of the infiltration class less than 50%, i.e. whether the infiltration is absent or contained (focal) or is more extensive within the 50% limit. In this context, this work proposes a second multiclass classification model, which provides the third class with focal or absent infiltration. The binary classification model predictive of the depth of myometrial infiltration proved to be highly performing with an AUC of 90.76 (88.63–92.89 CI95%). In particular, the proposed model exceeds the operator’s classification performance, especially with reference to precision (86.54% vs. 73.53%), thus demonstrating greater reliability of the prediction provided.
In our study, we have evaluated the subjective assessment supported by automated prediction model to evaluate the real benefit of using a support decision system in the clinal practice. We have verified that the operator’s classification performance increased significantly both in terms of overall accuracy and in terms of per-class precision in identifying the intermediate class of the multiclass task. An accurate prediction of myometrial invasion in patients with EC ensures the best modulation of patient treatment, allowing the evaluation of whether fertility-sparing or ovarian conservation treatments are feasible in younger women and avoiding overtreatment in patients undergoing radical surgery29. Although histological examination remains the absolute truth, myometrial invasion should not be assessed by frozen section because of poor reproducibility and agreement with definitive paraffin Sect29. Therefore, during the staging phase, this challenging role is entrusted to imaging techniques38.
Table 4summarizes the main works proposed in the literature. To our knowledge, the literature is poor in evaluating the predictive power of radiomics on ultrasound images in predicting deep myometrial invasion. In19, the authors have validated the three pretrained DL models (EfficientNet-B6, EfficientNet-B0, and ResNet-50) to determine the deep myometrial invasion (infiltration less than 50% vs. greater than 50%) on the US image of the EC. In their experimental study, EfficientNet-B6 showed the best performance in the testing set in terms of area under the curve (AUC) and accuracy reached 81.40% (74.60–88.20 95%CI) and 80.20 (73.30–85.50 95%CI). Although the performances declared by the authors are 9% points lower than those achieved by our model, it should be underlined that the case study in19 is multicentric and more numerous than ours.
. Most of the studies proposed in the literature on the same research task have used MRI. Numerous studies have shown that the two modalities have comparable diagnostic efficacy; however, compared with MRI, ultrasound is cheaper and takes less time to perform. In addition, ultrasound is a first-level examination, whereas MRI is a second- or third-level in-depth examination. The preliminary results shown in this article are comparable to those obtained in state-of-the-art works on MR images. Clearly given the different nature of the images used and the models implemented to solve the task of interest, the comparison is purely qualitative.
Further effort has been made to provide a model that can predict the depth of myometrial infiltration with a greater level of detail. We have developed a multiclass model that allows us to accurately identify cases with absent or focal infiltration already during the diagnosis phase. To the best of our knowledge, our work represents the first attempt to evaluate the predictive power of radiomics on ultrasound images in predicting multiclass deep myometrial invasion.
The proposed multiclass model is highly performing with an overall accuracy of value of 84.93% and a good balance of all per-class metrics, outperforming the operator’s performance especially in terms of recall. Although the results obtained from the study are highly encouraging, the study has limitations related to the sample size. In fact, further validation studies and model optimizations are needed to improve performance and imagine real clinical applicability. In particular, evaluation studies should be performed on a larger dataset to ensure generalizability. In addition, the inclusion of other operators for the evaluation of the same case was not considered in this preliminary work. Hovewer, US scans may vary depending on the operator, the device used, and the acquisition settings, affecting the reliability of the model; therefore, specific studies on the stability of performance by varying not only the operator but also the device are needed. The proposed models can represent a valid support tool for the clinicians for an accurate estimate of the deep myometrial infiltration already in the diagnosis phase with an easily accessible, low-cost examination that causes no discomfort for the patient such as ultrasound.
Moreover, in this paper we use Inception-V3. It has an architecture with Inception modules, which reduce the number of parameters compared to deeper networks such as ResNet-152, DenseNet, VGG-19. It offers a good balance between accuracy and inference speed, which is useful in clinical scenarios where rapid diagnosis is crucial. However, future studies include the evaluation of other architectures, such as EfficientNet or similary, that for the same performance could offer an even lower computational cost.
Furthermore, although these models have the potential to standardize examinations based on subjective evaluations such as ultrasound, to be applied correctly, it remains essential to properly perform the imaging examination, have an adequate knowledge of the most appropriate methodology of execution by the operators, and possess the ability to use the instrument, in this case, the ultrasound machine. Automated artificial intelligence models can improve the reproducibility of diagnostic examinations over subjective assessments, ensuring more reliable analysis based on standardized data. In addition, they can be decision support tools for less experienced clinicians, accelerating the learning curve and increasing diagnostic confidence. This is to ensure that the analyzed images are truly representative of the carcinoma, and it is not overestimated or underestimated with subsequent repercussions on management [45–46].
Data availability
The data presented in this study are available on request from the corresponding author. The data are not publicly available because are proper of the Azienda Ospedaliera Policlinico Consorziale-University of Bari.
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Funding
This work was supported by funding from the Italian Ministry of Health “5 per 1000” project (Deliberation n. 655/2022).
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Francesca Arezzo, Annarita Fanizzi, Raffaella Massafra and Vera Loizzi contributed equally.
Authors and Affiliations
Clinicalized Gynecological Oncology Unit , IRCCS Istituto Tumori ‘Giovanni Paolo II’ , Viale Orazio Flacco 65, 70124, Bari, Italy
Francesca Arezzo, Salvatore Lopez, Erica Silvestris, Gennaro Cormio & Vera Loizzi
2. Biostatistics and Bioinformatics Laboratory, IRCCS Istituto Tumori ‘Giovanni Paolo II’, Viale Orazio Flacco 65, Bari, 70124, Italy
Annarita Fanizzi, Samantha Bove, Maria Colomba Comes, Elsa Vitale & Raffaella Massafra
3. Gynecologic Oncology Unit, IRCCS Regina Elena National Cancer Institute, Rome, 00144, Italy
Rosanna Mancari & Enrico Vizza
4. Sylvester Comprehensive Cancer Center, Department of Biochemistry and Molecular Biology, Miller School of Medicine, University of Miami, Miami, FL, 33136, USA
Emiliano Cocco
5. Obstetrics and Gynecology Unit, Department of Biomedical Sciences and Human Oncology, University of Bari “Aldo Moro”, Bari, Italy
Mariangela Gianciotta & Giorgia Lanza
6. Department of Emergency and Organ Transplantation, Pathology Section, University of Bari “Aldo Moro”, Bari, Italy
Gerardo Cazzato
7. Department of Interdisciplinary Medicine (DIM), University of Bari “Aldo Moro”, Bari, Italy
Gennaro Cormio
8. Department of Translational Biomedicine and Neuroscience, University of Bari, Bari, Italy
Vera Loizzi
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Contributions
Conceptualization, F.A., A.F., and R.M.; methodology, A.F., S.B., M.C.C., and R.M.; software, A.F.; validation, A.F., F.A. and R.M.; formal analysis, F.A., A.F., C.Co., R.M., and V.L.; resources, G.Co., R.M., V.L.; data curation, F.A., R.M., M.G., G. L., G.Ca., and E.S.; writing—original draft preparation, F.A., A.F., R.M., S.B., M.C.C., and R.M.; writing—review and editing, F.A., A.F., R.M., E.C., S.B., M.C.C., M.G., G.L., S.L., G.Ca., E.S., El.V., En.V., G.Co., R.M, and V.L.; supervision, F.A, A.F., G.Co., R.M. and V.L. All authors have read and agreed to the published version of the manuscript.
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Samantha Bove or Maria Colomba Comes.
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The study was conducted according to the guidelines of the Declaration of Helsinki and approved by the Ethics Committee of the Azienda Ospedaliera Policlinico Consorziale - University of Bari, Italy (protocol code n. 6398/2020).
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Arezzo, F., Fanizzi, A., Mancari, R. et al. A Radiomic-based model to predict the depth of myometrial invasion in endometrial cancer on ultrasound images.
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Keywords
Endometrial Cancer
Imaging in gynecological disease
Depth of myometrial invasion
Ultrasound
Radiomic model
Subjects
Cancer
Endometrial cancer
Image processing |
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Piech, CS106A, Stanford University Something is going on in the world of AI Piech, CS106A, Stanford University Four Prototypical Trajectories Modern AI or, How we learned to combine probability and programming Piech, CS106A, Stanford University Brief History Piech, CS106A, Stanford University 1952 1955 Early Optimism 1950 Piech, CS106A, Stanford University Early Optimism 1950 “Machines will be capable, within twenty years, of doing any work a man can do.” –Herbert Simon, 1952 Piech, CS106A, Stanford University The world is too complex Underwhelming Results 1950s to 1980s Piech, CS106A, Stanford University Piech, CS106A, Stanford University Almost perfect… Told Speech Was 30 Years Out Piech, CS106A, Stanford University What is going on?
Piech, CS106A, Stanford University [suspense] Piech, CS106A, Stanford University Focus on one problem Piech, CS106A, Stanford University Computer Vision Piech, CS106A, Stanford University Logistic Regression is like the Harry Pottery Sorting Hat Classification That is a picture of a one Piech, CS106A, Stanford University Logistic Regression is like the Harry Pottery Sorting Hat Classification That is a picture of a zero Piech, CS106A, Stanford University Classification That is a picture of an zero It doesn’t have to be correct all of the time Piech, CS106A, Stanford University Can you do it?
Piech, CS106A, Stanford University What number is this?
Piech, CS106A, Stanford University What number is this?
Piech, CS106A, Stanford University 0 0 1 0 1 0 1 0 0 0 1 1 1 0 1 1 0 0 1 0 1 1 1 0 1 0 0 0 0 0 1 1 1 0 1 0 0 1 1 0 0 1 0 1 0 1 1 1 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 1 0 0 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 0 1 1 1 1 0 1 1 0 1 1 0 0 1 1 0 1 1 1 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 1 1 0 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 1 1 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 0 0 1 1 1 0 1 0 1 1 0 0 0 1 0 How about now?
What a computer sees What a human sees Piech, CS106A, Stanford University Very hard to Program public class HarryHat extends ConsoleProgram { public void run() { println(“Todo: Write program”); } } ??
Piech, CS106A, Stanford University Two Great Ideas 2. Artificial Neurons 1. Probability from Examples Piech, CS106A, Stanford University Two Great Ideas 1. Probability from Examples 2. Artificial Neurons Piech, CS106A, Stanford University 1. Probability From Examples Piech, CS106A, Stanford University When Does the Magic Happen?
Lots of Data Sound Probability + Piech, CS106A, Stanford University Basically just a rebranding of statistics and probability.
Machine Learning Piech, CS106A, Stanford University [ Why is this hard?
You see this: But the camera sees this: [Andrew Ng] Vision is Hard Piech, CS106A, Stanford University Human Features 0.1 0.7 0.6 0.4 0.1 0.4 0.5 0.5 0.1 0.6 0.7 0.5 0.2 0.3 0.4 0.4 0.1 0.7 0.4 0.6 0.1 0.4 0.5 0.5 … Find edges at four orientations Sum up edge strength in each quadrant Final feature vector [Andrew Ng] Human Designed Features Piech, CS106A, Stanford University Some Great Thinkers Daphne Koller Piech, CS106A, Stanford University Motorcycle Motorcycle Motorcycle Motorcycle Motorcycle Motorcycle Motorcycle Motorcycle Motorcycle Straight ML Not Perfect… Piech, CS106A, Stanford University Two Great Ideas 2. Artificial Neurons 1. Probability from Examples Piech, CS106A, Stanford University Two Great Ideas 2. Artificial Neurons 1. Probability from Examples Piech, CS106A, Stanford University 2. Artificial Neurons Piech, CS106A, Stanford University Neuron Piech, CS106A, Stanford University Neuron Piech, CS106A, Stanford University Neuron Piech, CS106A, Stanford University Neuron Piech, CS106A, Stanford University Some Inputs are More Important Piech, CS106A, Stanford University Artificial Neuron 1 1 0 1 2 3 -2 1 6 0.99 Piech, CS106A, Stanford University Neural Network Each node represents a neuron Each edge represents the weight of the interaction Pixels Piech, CS106A, Stanford University Each node represents a neuron Each edge represents the weight of the interaction Neural Network Piech, CS106A, Stanford University Each node represents a neuron Each edge represents the weight of the interaction Neural Network Piech, CS106A, Stanford University Each node represents a neuron Each edge represents the weight of the interaction Neural Network Piech, CS106A, Stanford University Neural Network Piech, CS106A, Stanford University Interpret the last neuron as the “probability” that the image is of a 1 Neural Network Piech, CS106A, Stanford University The image had a 0 but we predicted a high probability that it was a 1 Neural Network Piech, CS106A, Stanford University Let’s update our weights to make our probabilities better match reality The image had a 0 but we predicted a high probability that it was a 1 Neural Network Piech, CS106A, Stanford University The image had a 0 but we predicted a high probability that it was a 1 Update the parameter at each connection Let’s update our weights to make our probabilities better match reality Neural Network Piech, CS106A, Stanford University Gradient of output layer params @ˆ y @✓(ˆ y) i = ˆ y[1 −ˆ y] · hi @ˆ y @✓(ˆ y) i = ˆ y[1 −ˆ y] · @ @✓(ˆ y) i mh X j=0 hj✓(ˆ y) j @ˆ y @✓(ˆ y) i = σ 0 @ mh X j=0 hj✓(ˆ y) j 1 A 2 41 −σ 0 @ mh X j=0 hj✓(ˆ y) j 1 A 3 5 · @ @✓(ˆ y) i mh X j=0 hj✓(ˆ y) j ˆ y = σ 0 @ mh X j=0 hj✓(ˆ y) j 1 A You will be able to do this.
@L @✓(ˆ y) i = @L @ˆ y · @ˆ y @✓(ˆ y) i Piech, CS106A, Stanford University Google Brain Piech, CS106A, Stanford University 1 Trillion Artificial Neurons Google Brain Piech, CS106A, Stanford University Neuron 1 Neuron 2 Neuron 3 Neuron 4 Neuron 5 Le, et al., Building high-level features using large-scale unsupervised learning. ICML 2012 Other Neurons Piech, CS106A, Stanford University Optimal stimulus by numerical optimization Le, et al., Building high-level features using large-scale unsupervised learning. ICML 2012 Top stimuli from the test set A Neuron That Fires When It Sees Cats Piech, CS106A, Stanford University … smoothhound, smoothhound shark, Mustelus mustelus American smooth dogfish, Mustelus canis Florida smoothhound, Mustelus norrisi whitetip shark, reef whitetip shark, Triaenodon obseus Atlantic spiny dogfish, Squalus acanthias Pacific spiny dogfish, Squalus suckleyi hammerhead, hammerhead shark smooth hammerhead, Sphyrna zygaena smalleye hammerhead, Sphyrna tudes shovelhead, bonnethead, bonnet shark, Sphyrna tiburo angel shark, angelfish, Squatina squatina, monkfish electric ray, crampfish, numbfish, torpedo smalltooth sawfish, Pristis pectinatus guitarfish roughtail stingray, Dasyatis centroura butterfly ray eagle ray spotted eagle ray, spotted ray, Aetobatus narinari cownose ray, cow-nosed ray, Rhinoptera bonasus manta, manta ray, devilfish Atlantic manta, Manta birostris devil ray, Mobula hypostoma grey skate, gray skate, Raja batis little skate, Raja erinacea … Stingray Mantaray ImageNet Classification Piech, CS106A, Stanford University 0.005% Random guess 1.5% ?
Le, et al., Building high-level features using large-scale unsupervised learning. ICML 2012 Pre Neural Networks GoogLeNet ImageNet Classification Piech, CS106A, Stanford University 0.005% Random guess 1.5% Pre Neural Networks 43.9% GoogLeNet Szegedy et al, Going Deeper With Convolutions, CVPR 2015 ImageNet Classification Piech, CS106A, Stanford University 0.005% Random guess 1.5% Pre Neural Networks 82.7% NASNet ImageNet Classification Piech, CS106A, Stanford University A machine learning algorithm performs better than the best dermatologists.
Developed this year, at Stanford.
Esteva et al., Nature 2017.
Esteva, Andre, et al. "Dermatologist-level classification of skin cancer with deep neural networks." Nature 542.7639 (2017): 115-118.
Where is this useful?
Piech, CS106A, Stanford University Open Problem: One Shot Learning B Lake, R Salakhutdinov, J Tenenbaum. Science 2015.
Human-level concept learning through probabilistic program induction. Current deep learning methods are not enough to move the needle as far as we want, especially on socially relevant problems that often do not have the benefit of massive public datasets. The best new ideas are coming from probability theory Piech, CS106A, Stanford University Prediction: The person who solves one shot learning problem will use core probability Piech, CS106A, Stanford University Closest thing to magic you can learn.
Now is the time, Stanford is the place.
Piech, CS106A, Stanford University |
187630 | https://math.stackexchange.com/questions/2856316/why-is-xy-not-convex-although-it-is-the-product-of-nonnegative-increasing-conv | Why is $xy$ not convex although it is the product of nonnegative increasing convex functions? - Mathematics Stack Exchange
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Why is x y x y not convex although it is the product of nonnegative increasing convex functions?
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According to exercise 3.32 in Boyd & Vandenberghe's Convex Optimization, if both f f and g g are convex, positive and non-increasing (or non-decreasing) then f g f g is convex. However, if we let f(x,y)=x f(x,y)=x and g(x,y)=y g(x,y)=y then over the non-negative orthant the conditions are fulfilled but the sublevel set for 0.5 0.5 is not convex (which I think should be convex if the function is convex).
Where am I wrong in understanding the meaning of the result presented in this exercise?
convex-analysis
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edited Jul 19, 2018 at 17:02
user357151
asked Jul 19, 2018 at 6:15
Frank MosesFrank Moses
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3 What does it mean for a 2-variable function to be "increasing" or "decreasing"? I thought those concepts apply to single variable functions only?Jyrki Lahtonen –Jyrki Lahtonen 2018-07-19 06:26:36 +00:00 Commented Jul 19, 2018 at 6:26
what is the sublevel set for .5?Jürg W. Spaak –Jürg W. Spaak 2018-07-19 06:27:24 +00:00 Commented Jul 19, 2018 at 6:27
4 Your reply to Jyrki has it right: the result in the exercise only applies to functions of one variable.user856 –user856 2018-07-19 08:44:42 +00:00 Commented Jul 19, 2018 at 8:44
2 I'll confirm Jyrki and Rahul's comment. This is a very common misinterpretation of the claim in B&V. I cannot tell you how many people claim that x y x y or x/y x/y is convex because of this.Michael Grant –Michael Grant 2018-07-19 14:41:31 +00:00 Commented Jul 19, 2018 at 14:41
1 Perhaps, but I would argue that the way the question is worded it is simply not correct to assume it can be extended to the multivariate case. It's just that many people, frankly, don't think rigorously enough about these things.Michael Grant –Michael Grant 2018-07-19 23:15:53 +00:00 Commented Jul 19, 2018 at 23:15
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To summarize the comments:
The theorem about the product of two positive non-decreasing convex functions being convex applies only to functions of one real variable.
For example, we can use it to conclude that x e x x e x is convex for x≥0 x≥0. We cannot use it to conclude that x e y x e y is convex over x,y≥0 x,y≥0.
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answered Jul 19, 2018 at 17:02
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Look at the Hessian H H of x y x y:
H=[0 1 1 0]H=[0 1 1 0]
|H|=−1<0|H|=−1<0, and so x y x y is concave.
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answered Nov 2, 2022 at 14:02
Johnny QueJohnny Que
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187631 | https://blog.csdn.net/zxyhhjs2017/article/details/78926054 | 机器学习---之损失函数求最小值为什么不用导数为0的点而用梯度下降法_机械学习求最小值-CSDN博客
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机器学习---之损失函数求最小值为什么不用导数为0的点而用梯度下降法
最新推荐文章于 2025-07-03 09:06:21 发布
zxyhhjs2017于 2017-12-28 20:39:38 发布
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分类专栏：机器学习
版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
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本文探讨了损失函数的极值点问题，解释了梯度下降法如何帮助找到局部最小值，以及在损失函数为凹或凸的情况下如何寻找全局最优解。
1.因为，损失函数可能有无限个极值点，你并不知道哪个点可以使损失函数最小，如下图中的损失函数：
2.而使用梯度下降法虽然不一定能求导全局最小值，但可以求导局部最小值，也能使损失函数降低为0,如下图所示：
3.如果损失函数是凹的或者是凸的时候，一定可以找到全局最优解，这个时候，在训练过程中你可以看到损失函数可能早都为0了，但还是在不断迭代找全局最优解，这个过程可能是无限的，因为函数的最小值可能是无穷小，这个时候即可以停止训练了。
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pytorch 13 训练过程中出现loss为nan、inf（梯度爆炸、梯度消失）的分析及解决方案
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从理论的角度上看，本质是梯度消失与梯度爆炸所导致的。梯度消失是指 导数 值特别小，导致其连乘项接近无穷小，可能是由输入数据的值域太小（导致权重W的导致特别小）或者是神经网络层输出数据落在在激活函数的饱和区（导致激活函数的导致特别小）;而梯度爆炸是指 导数 值特别大，导致其连乘项特别大，致使W在更新后超出了值域的表示范围。可能是输入数据没有进行归一化（数据量纲太大，致使W的梯度值极大），只要连乘项的 导数 一直大于1，就会使得靠近输入层的W更新幅度特别大。连乘项是指链式 求 导法则中每一层的 导数，很明显梯度消失与梯度爆炸都受
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机器学习 的精髓 - 梯度下降算法_ 机械学习,简述梯度下降算法的步骤 - CSDN博 ...
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求 y=x 2 x^2x2 sin(x)函数取得极小值时x的值。 梯度下降就是x当前值 - y在x处的 导数,再进行不断的迭代 y’ = 2xsin(x) +x 2 x^2x2cos(x) (y在x处的 导数) 梯度下降x’ = x - y’ 为了使梯度变化不至于太快,还要再y’上乘以一个learning rate即学习率 那就变成了x’ = x - 0.0 0 5...
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①概述 不是一个 机器学习 算法 是一种基于搜索的最优化方法 作用：最小化一个 损失函数 梯度上升法：最大化一个效用函数 思路：由上图可知，在 损失函数 上的某一个 点，希望经过变化到达 损失函数 最小值 点，所以对 损失函数 求 导，导数 代表函数增大的方向，为了让函数变小，所以在 导数 前面增加负号，又因为需要逐步靠近最低 点，所以在 导数 前面增加了步长，逐步靠近 导数 为 0 的位置。下降的速度就是由决定 称为学习率（learning rate） 的取值影响获得最优解的速度 的取值不合适，甚至得不到最优解 是 梯度下降法
深度学习入门笔记 0 7（如何找到 损失函数 的 最小值）
2401_86341956的博客
09-09 1252
我们已经知道了如何 求 损失函数，包括用均方误差和交叉熵误差来 求，我们还了解了 为什么 需要引入 损失函数。接下来的重 点 是如何找到 损失函数 的 最小值。为了解决这个问题，我们需要了解一些数学知识。
极限配合与技术测量基础(第五版)习题册参考答案 - A 0 2 - 3635.pdf资源...
9-26
互换性原则广泛用于机械制造中的产品设计 、 零件 加工 、 产品装配 、 机器的使用和维修等各个方面 。 在设计方面 , 采用具有互换性的标准件和通用件 , 可以使设 计工作简化 , 缩短设计周期 , 并便于应用计算机辅助设计 。 在加工和装配方面 , 当零件具有互换性时 ...
机器学习 概览
9-12
为了找到J的使J为 最小值 的(θ的值),一般会使用梯度下降算法 2、梯度下降算法: 3、多特征线性回归 4、多特征线性回归的梯度下降算法 ①特征缩放: 特征值之间的取值范围尽可能得相似,区别过大会导致 求 最优值时变得复杂,可以才用缩放的方法去缩放特征值使他们的范围足够接近,且可以缩放到 - n~+n ...
机械学习 的训练模型（线性回归、梯度下降、正规方程）
种花生的博客
05-22 606
目录线性回归单变量线性回归 线性回归 线性回归是一种有监督的学习，解决的是自变量和因变量之间的关系。 线性回归模型预测： 单变量线性回归 因为只含有一个特征/输入变量,因此这样的问题叫作单变量线性回归问题。 选择不同的参数值，就会得到不同的直线。对于假设函数所预测出来的值和实际值之间的差距就是建模误差，也就是存在着一个代价函数cost function。 我们的目标就是减少假设函数预测数来的值和实际值之间的差距，也就是让代价函数最小。而让代价函数最小，就需要我们选择合适的参数值。 （1）简化对公式的理解
李宏毅 机器学习 笔记——梯度下降法 最新发布
swpucwf的博客
07-03 1117
深度学习介绍 基于仿生学的一种自成体系的 机器学习 算法，包括但不限于图像识别、语音、文本领域。 梯度下降法 作为深度学习算法种常用的优化算法 梯度下降法，是一种基于搜索的最优化方法，最用是最小化一个 损失函数。梯度下降是迭代法的一种,可以用于 求 解最小二乘问题(线性和非线性都可以)。在 求 解 机器学习 算法的模型参数，即无约束优化问题时，梯度下降是最常采用的方法之一，另一种常用的方法是最小二乘法。在 求 解 损失函数 的 最小值 时，可以通过 梯度下降法 来一步步的迭代 求 解，得到最小化的 损失函数 和模型参数值。 最小二乘法的问题 梯度下
...学习 -(第1 0 周笔记)大数据训练_吴恩达 机械学习 训练数据
9-21
对于很多 机器学习 算法, 包括线性回归、逻辑回归、神经网络等等, 算法的实现都是通过得出某个代价函数 或者某个最优化的目标来实现的, 然后使用梯度下降这样的方法来 求 得代价函数的 最小值。 当我们的训练集较大时 ,梯度下降算法则显得计算量非常大 ,在这段视频中 我想介绍一种跟普通梯度下降不同的方法 随机梯度下降...
【看论文】之《基于机器视觉的番茄收割机实时分拣系统研究_袁紫薇...
9-16
本文介绍了一种基于机器视觉的番茄收割机实时分拣系统,通过优化图像处理算法和机械结构,实现了成熟番茄与未成熟番茄及杂质的有效区分。系统采用双颜色分量灰度图像合成技术,提高了分拣精度。 论文信息 题目:基于机器视觉的番茄收割机实时分拣系统研究 作者:袁紫薇 ...
【深度学习原理】如何利用 梯度下降法，寻找 损失函数 最小值？
hennyxu的博客
11-20 7788
梯度下降(Gradient Descent)是用于寻找函数 最小值 的一种优化算法。我们常常利用 梯度下降法 来使 损失函数 Loss function的值尽可能小，即让神经网络的预测值（实际输出）和标签值（预期的输出）尽可能接近。在这个过程中，网络参数——各层的权值与偏重将得到调整，这也正是神经网络的训练原理。可见熟悉梯度下降的原理对于理解神经网络相当重要。 本文将从 损失函数 概念、梯度下降原理 &...
解释 为什么 用梯度下降而不是直接 求 导数 为 0 的解
kyle1314608的博客
02-21 1469
...
机械基础第三版范思冲课后题答案.pdf资源
9-9
《机械基础》第三版是范思冲教授编著的一本重要的教材,主要涵盖了机械工程的基础知识,包括机械设计、机械原理、机械零件等方面的内容。课后题答案是学习过程中非常宝贵的参考资料,它可以帮助学生深入理解和巩固课堂所学,提高解决实际问题的能力。 在提供的部分内容中,我们可以看到涉及到多个知识 点: 1. 力学分析:...
机器学习 使用兰氏距离的最小距离聚类法以及散 点 图展示_兰氏距离公式...
9-17
1.首先是把距离中心先放入存入到center的二维列表中,然后通过最大值和 最小值 归一化进行划分数据。本程序中有两种方法进行:第一种是纯手写版本:利用公式: 先 求 出样本数据中的最大值和 最小值,然后通过公式 求 出其归一化后数据然后再返会要处理的数据列表,并且也 求 出了数据样本里面的正确类别放入lss列表中 ...
为什么 计算 损失函数 最优值采用梯度下降算法而不是直接 求 导等于 0 的深度解释 热门推荐
我是张跑跑
01-23 1万+
1. 概述 不论是在做数据的拟合还是在 机器学习 中计算最小的代价函数，都需要 求 目标函数的最优值（最大或 最小值），在这其中，使用的方法都是梯度下降算法（或上升）进行多次跌打直到收敛（或接近收敛），这种方法确实是能够达到我们的目的； 但是这个时候我们就会思考，既然是 求 最优值，我们 为什么 不能直接对目标函数 求 导，让其 导数 等于零，然后得出结果呢？反而要用似乎更加复杂的梯度下降算法呢？ 这个问题也一直困扰着博主...
解释 为什么 用梯度下降而不是直接 求 导数 为 0 的解？
weixin_43167121的博客
04-21 6634
问题： 在计算线性回归最大似然估计的解的时候，最后的推导结果是 为什么 不直接 求 出θ？而是一步步迭代 求 出θ？ 原因 因此，梯度下降可以节省大量的计算时间。此外，它的完成方式允许一个简单的并行化，即在多个处理器或机器上分配计算。 此外，当您只将一部分数据保留在内存中时，会出现梯度下降的版本，从而降低了对计算机内存的要 求。总的来说，对于特大问题，它比线性代数解决方案更有效。 当您有数千个变量（如机器...
Spark 2.x 机器学习 秘籍(六)spark _机器学习
9-12
第三,我们简单地通过遵循和应用机械微积分规则找到了 导数。 第四,我们将函数的 导数 设置为零!并解出 x 第五,我们使用 x 值并将其代入原方程以找到 y。通过步骤 1 到 5,我们最终得到了 点(2,1)处的 最小值。 还有更多… 大多数统计 机器学习 算法在定义和搜索域空间时使用成本或误差函数来得到最佳的数值近似解...
正规方程法及 梯度下降法（求 解 损失函数 最小值）
YIAN爱学习的博客
11-02 1097
介绍 损失函数 的 求 解方法：正规方程法和 梯度下降法
python使用梯度下降和牛顿法寻找Rosenbrock函数 最小值 实例
09-17
在 梯度下降法 中，使用了contourf和contour函数来展示函数的轮廓，并用plot函数描绘了搜索路径。牛顿法则类似，只是迭代次数更少。 总结来说，这个实例深入浅出地演示了如何用Python的梯度下降和牛顿法解决优化问题...
MATLAB 梯度下降法 求 多元函数的极值以及极值 点 程序+文档完成.zip
05-19
梯度下降法 是一种在数值优化中广泛使用的迭代算法，用于寻找多元函数的局部 最小值 或最大值。在MATLAB环境中，实现 梯度下降法 可以帮助我们解决各种数学问题，特别是那些涉及多个变量的复杂函数的极值查找。在这个...
机器学习 中 为什么 需要梯度下降?梯度下降算法缺 点?【_机器学习 - 回归】梯度下降(SGD/BGD/MBGD)...
weixin_39754398的博客
11-24 934
上一节讲完线性回归的模型思路与 损失函数，我们的目的当然是 求 解参数 使得对应的 损失函数 最小。那么下面如何操作可以寻找到这个最值 点 呢？答：梯度下降。梯度下降法 目的是寻找极值 点，其本质可以类比为一个下山的过程。假设这样一个场景：一个人被困在山上，需要找到山谷。但由于视野等原因，下山的路径就无法确定，他必须利用自己周围的信息去找到下山的路径。这个时候，他就可以利用梯度下降算法来帮助自己下山。具体来说...
如何选择合适的 损失函数，请看......
lovenankai的专栏
06-21 941
翻译 | 张建军编辑 | 阿司匹林 机器学习 中的所有算法都依赖于最小化或最大化某一个函数，我们称之为“目标函数”。最小化的这组函数被称为“损失函数”。损失函数 是衡量预测模型...
机器学习 入门教学——损失函数（最小二乘法）
计算机硕士的博客
09-20 796
机器学习 入门教学——损失函数（最小二乘法）的简单概述。
机器学习：损失函数、经验风险最小化、结构风险最小化
bqw的博客
05-27 2878
[from sklearn.metrics import zero_one_loss y_true = [1,1,1,1,1,0,0,0,0,0] y_pred = [0,0,0,1,1,1,1,1,0,0] zero_one_loss(y_true,y_pred,normalize=False)from sklearn.metrics import log_loss y_true = 1,1,1...
最小二乘法的平方 损失函数 的推导
qq_45757722的博客
05-14 858
在线性模型中，假设预测结果与实际结果有误差为ε(i)\varepsilon^{(i)}ε(i) 则线性模型中，误差可以表示为 ε(i)=y(i)−θTx(i)\varepsilon^{(i)} = y^{(i)} - \theta^Tx^{(i)}ε(i)=y(i)−θTx(i) 根据中心极限定律，假设误差ε(i)\varepsilon^{(i)}ε(i)服从标准正态分布，则可以得到 p(y(i)∣x(i);θ)=12πσe(−y(i)−θTx(i)2σ2)p(y^{(i)}|x^{(i)};\theta)
“损失函数”是如何设计出来的？直观理解“最小二乘法”和“极大似然估计法”
wkaing的专栏
01-24 2177
设计 损失函数 时，用到的最小二乘法和极大似然估计法，如果误差是正态分布的话，它们是等价的
损失函数
lgy54321的博客
07-05 2426
文章目录一、平方 损失函数（最小二乘法, Ordinary Least Squares ）均方误差ESM均方误差+Sigmoid激活函数：输出层神经元学习率缓慢Sigmoid激活函数：ESM均方误差+Sigmoid激活函数二.交叉熵损失交叉损失的定义1.soft max分类器2.交叉熵损失 损失函数（loss function）是用来估量模型的预测值f(x)与真实值Y的不一致程度，它是一个非负实值函...
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How to create a variable in Python to store a number with 500 digits after decimal?
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Asked 10 years, 9 months ago
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I need to find 500 digits after decimal of e. The output variable limits the number of digits to 16. How do you go forth with it?
python
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edited Dec 8, 2014 at 7:47
Tim Pietzcker
337k 59 59 gold badges 520 520 silver badges 572 572 bronze badges
asked Dec 8, 2014 at 7:19
AshtrixAshtrix
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By the way, it's not any kind of "output variable" (which doesn't really exist) that's limiting the precision, it's the float type that can't store more digits because it has only 52 bits available to store the fraction (and 11 for the exponent, one for the sign).Tim Pietzcker –Tim Pietzcker 2014-12-08 07:46:32 +00:00 Commented Dec 8, 2014 at 7:46
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Use the decimal module:
```python
import decimal
decimal.getcontext().prec = 501
a = decimal.Decimal("2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019011573834187930702154089149934884167509244761460668082264800168477411853742345442437107539077744992069551702761838606261331384583000752044933826560297606737113200709328709127443747047230696977209310141692836819025515108657463772111252389784425056953696770785449969967946864454905987931636889230098793127")
print a
2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019011573834187930702154089149934884167509244761460668082264800168477411853742345442437107539077744992069551702761838606261331384583000752044933826560297606737113200709328709127443747047230696977209310141692836819025515108657463772111252389784425056953696770785449969967946864454905987931636889230098793127
```
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edited Dec 8, 2014 at 7:49
answered Dec 8, 2014 at 7:24
Tim PietzckerTim Pietzcker
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3 Comments
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Ashtrix
AshtrixOver a year ago
Thank you so much! But I wanted to know hpw do you find this 500 digited number? As in, I know the iterative way to generate digits of e but how do I print all the 500 digits?
2014-12-08T07:35:00.617Z+00:00
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Tim Pietzcker
Tim PietzckerOver a year ago
Ah, your title said 100. Well, just increase the precision to 501. Then just print the Decimal and it will be output with all 501 digits.
2014-12-08T07:43:15.46Z+00:00
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UlfR
UlfROver a year ago
If you do not want to calculate e yourself, you could get it by decimal.Decimal(1).exp() with whatever precision you like.
2014-12-08T08:04:16.613Z+00:00
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187633 | https://upcommons.upc.edu/server/api/core/bitstreams/e6018826-1a67-4151-a494-21db07781c33/content | Consecutive patterns and statistics on restricted permutations
Sergi Elizalde Torrent
Director de tesi:
Marc Noy Serrano Universitat Polit` ecnica de Catalunya 2004 Consecutive patterns and statistics on restricted permutations
Memoria presentada per optar al grau de Doctor en Matem atiques per
Sergi Elizalde Torrent
Director de tesi:
Marc Noy Serrano Programa de doctorat en Matematica Aplicada Universitat Polit ecnica de Catalunya 2004 Acknowledgments
I would like to thank Marc Noy for being the one who first introduced me to the beautiful world of combinatorics. During my last year as an under-graduate he initiated me in mathematical research and he filled me with his enthusiasm for discrete mathematics, which I have never lost since then. Marc has also given me valuable advice in different situations. I am also very grateful to Richard Stanley for his guidance and support. He is an inextinguishable source of knowledge and ideas, and I constantly learn new things from him. I owe to Josep Gran´ e having awakened my interested for mathematics already during the Mathematical Olympiad more than eight years ago. Together with Sebastia Xamb´ o and Pere Pascual, they have been a continuous source of encouragement during all these years. I am lucky to have had Igor Pak, Emeric Deutsch and Toufik Mansour as col-laborators. I also thank Igor for entertaining discussions. Other mathemati-cians who have offered valuable suggestions are Miklos B´ ona, Alex Burstein, Josep D´ ıaz, Richard Ehrenborg, Philippe Flajolet, Ira Gessel, Olivier Gui-bert, Ferran Hurtado, David Jackson, Sergey Kitaev, Joan-Carles Lario, Juan-Jos´ e Morales, Alex Postnikov, Astrid Reifegerste and Oriol Serra. During these last four years many other people have helped me grow as a person and as a mathematician. I am especially indebted to Jan, Radoˇ s and Federico for being such supportive friends, to Anna for being always helpful and nice to me, and also to Mark, Peter, Fumei, Etienne, Carly, Bridget, Lauren, Cilanne and V´ ıctor. Finalment, l’agra¨ ıment m´ es especial ´ es pels meus pares Maria Carme i Emili, sense els quals tot aix o no hauria estat possible. A ells els dec l’educaci´ ovi
que m’han donat, haver-me format com a persona, i especialment l’amor que han demostrat i el seu suport incondicional en tot moment. Tamb´ e ´ es un orgull tenir un germa com l’Aleix, sabent que puc comptar amb ell sempre que el necessiti. Per acabar, vull donar gr acies als meus amics de la carrera: Diego, Sergi, Javi, Marta, Agus, Teresa, Ariadna, Carles, Ana, Fernando, V´ ıctor, Carme, Josep Joan, Montse, Toni, Edgar, Jordi, Maite, Esther, Desi, Elena i tots els altres, per fer-me sentir com si el temps no hagu´ es passat cada vegada que ens veiem. Contents
Introduction 51 Definitions and preliminaries 9
1.1 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.1.1 Pattern avoidance . . . . . . . . . . . . . . . . . . . . 91.1.2 Permutation statistics . . . . . . . . . . . . . . . . . . 10 1.1.3 Trivial operations . . . . . . . . . . . . . . . . . . . . . 11 1.2 Dyck paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.2.1 Standard statistics . . . . . . . . . . . . . . . . . . . . 13 1.2.2 Tunnels . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.3 Combinatorial classes and generating functions . . . . . . . . 17 1.3.1 Ordinary generating functions . . . . . . . . . . . . . . 17 1.3.2 Exponential generating functions . . . . . . . . . . . . 18 1.3.3 The Lagrange inversion formula . . . . . . . . . . . . . 19 1.3.4 Chebyshev polynomials . . . . . . . . . . . . . . . . . 20 1.4 Patterns of length 3 . . . . . . . . . . . . . . . . . . . . . . . 20 1.4.1 Equidistribution of fixed points . . . . . . . . . . . . . 20
2 Statistics in permutations avoiding one pattern 23
2.1 The bijection ϕ . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.1.1 More properties . . . . . . . . . . . . . . . . . . . . . . 27 2.2 a) 132 ≈ 213 ≈ 321 . . . . . . . . . . . . . . . . . . . . . . . . 29 2 Contents
2.2.1 A bijection between Sn(321) and Sn(132) preserving fixed points and excedances . . . . . . . . . . . . . . . 29 2.2.2 Properties of Ψ . . . . . . . . . . . . . . . . . . . . . . 31 2.2.3 Properties of the matching algorithm . . . . . . . . . . 35 2.2.4 The bijection ψx . . . . . . . . . . . . . . . . . . . . . 38 2.3 b) 123 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.4 c, c’) 231 ∼f 312 . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3 Simultaneous avoidance 51
3.1 Double restrictions . . . . . . . . . . . . . . . . . . . . . . . . 52 3.1.1 a) {123 , 132 } ≈ { 123 , 213 } . . . . . . . . . . . . . . . . 52 3.1.2 b, b’) {231 , 321 }∼ f {312 , 321 } . . . . . . . . . . . . . . 57 3.1.3 c) {132 , 213 } . . . . . . . . . . . . . . . . . . . . . . . 59 3.1.4 d) {231 , 312 } . . . . . . . . . . . . . . . . . . . . . . . 61 3.1.5 e, e’) {132 , 231 } ≈ { 213 , 231 }∼ f {132 , 312 } ≈ { 213 , 312 } 61 3.1.6 f) {132 , 321 } ≈ { 213 , 321 } . . . . . . . . . . . . . . . . 63 3.1.7 g, g’) {123 , 231 }∼ f {123 , 312 } . . . . . . . . . . . . . . 65 3.1.8 h) {123 , 321 } . . . . . . . . . . . . . . . . . . . . . . . 67 3.2 Triple restrictions . . . . . . . . . . . . . . . . . . . . . . . . . 67 3.3 Pattern-avoiding involutions . . . . . . . . . . . . . . . . . . . 74 3.3.1 Single restrictions . . . . . . . . . . . . . . . . . . . . 76 3.3.2 Multiple restrictions . . . . . . . . . . . . . . . . . . . 77
4 A simple and unusual bijection for Dyck paths 81
4.1 The bijection Φ . . . . . . . . . . . . . . . . . . . . . . . . . . 82 4.2 Properties of Φ . . . . . . . . . . . . . . . . . . . . . . . . . . 84 4.3 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . 88 4.4 Connection to pattern-avoiding permutations . . . . . . . . . 93
5 Other bijections and combinatorial interpretations 99
5.1 Some interpretations of the Catalan and Fine numbers . . . . 99 5.2 Some other bijections between Sn(321) and Dn . . . . . . . . 104 5.3 Noncrossing permutations . . . . . . . . . . . . . . . . . . . . 106 5.3.1 Distribution of descents . . . . . . . . . . . . . . . . . 108 5.3.2 Distribution of fixed points and excedances . . . . . . 109 Contents 3
6 Consecutive patterns 113
6.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 6.1.1 Tree representation of permutations . . . . . . . . . . 115 6.1.2 Equivalent subwords . . . . . . . . . . . . . . . . . . . 115 6.1.3 Asymptotic enumeration . . . . . . . . . . . . . . . . . 116 6.2 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 6.2.1 Increasing subwords . . . . . . . . . . . . . . . . . . . 116 6.2.2 Other subwords . . . . . . . . . . . . . . . . . . . . . . 119 6.3 Subwords of length at most four . . . . . . . . . . . . . . . . 123 6.3.1 Subwords of length three . . . . . . . . . . . . . . . . 124 6.3.2 Subwords of length four . . . . . . . . . . . . . . . . . 126 6.4 Multiple subwords . . . . . . . . . . . . . . . . . . . . . . . . 129 6.5 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . 131
7 Asymptotic enumeration and generalized patterns 133
7.1 Generalized patterns . . . . . . . . . . . . . . . . . . . . . . . 133 7.1.1 Patterns of the form 1-σ . . . . . . . . . . . . . . . . . 134 7.2 Asymptotic enumeration . . . . . . . . . . . . . . . . . . . . . 136 7.2.1 The pattern 12-34 . . . . . . . . . . . . . . . . . . . . 140 7.2.2 The pattern 1-23-4 . . . . . . . . . . . . . . . . . . . . 145 7.2.3 Other patterns . . . . . . . . . . . . . . . . . . . . . . 148
Conclusions 151 Bibliography 155 Introduction
The subject of pattern-avoiding permutations , also called restricted permu-tations , has blossomed in the past decade. A number of enumerative results have been proved, new bijections found, and connections to other fields es-tablished. A recent breakthrough (see also [57, 2, 14, 3]) has been the proof of the so-called Stanley-Wilf conjecture, which gives an exponential upper bound on the number of permutations avoiding any given pattern. However, the study of statistics on restricted permutations started devel-oping very recently, and the interest in this topic is currently growing. On the one hand, the concept of pattern avoidance concerns permutations re-garded as words π = π1π2 · · · πn. On the other hand, concepts such as fixed points or excedances arise when we look at permutations as bijections
π : {1, 2, . . . , n } −→ { 1, 2, . . . , n }. It was not until recently that these two kinds of concepts were studied together. An unexpected recent result of Robertson, Saracino and Zeilberger gives a new and exciting extension to the classical result that the number of 321-avoiding permutations equals the number of 132-avoiding permutations. They show that one can refine this result by taking into account the number of fixed points in a permutation. Their proof is nontrivial and technically involved. The first part of the work in the present thesis is motivated by this result. A natural question that arises is whether the fact that the number of fixed points has the same distribution in both 321-avoiding and 132-avoiding permutations can be generalized to other statistics and to other patterns. In particular, this gives more interest to the problem of studying the dis-tribution of statistics on pattern-avoiding permutations. Another natural question to consider is whether a bijection between 321-avoiding permu-tations and 132-avoiding permutations that preserves the number of fixed 6 Introduction
points can be described. It is somewhat surprising that, before the work on this thesis [31, 33, 36], none of the several known bijections between these two sets of permutations preserved the number of fixed points. Aside from the study of statistics, there is a variation to the notion of pat-tern avoidance that started to develop recently. The concept of generalized pattern allows the requirement that, for a pattern to occur in a permutation, certain elements have to be in adjacent positions. A particular case of this are the consecutive patterns , where all the elements have to be consecutive. The last part of this work is devoted to the study of such patterns. This thesis is structured as follows. Chapter 1 introduces the basic defini-tions regarding pattern avoidance, permutation statistics and Dyck paths, as well as some tools to manipulate generating functions. In Chapter 2 we study the distribution of the statistics ‘number of fixed points’ and ‘number of excedances’ in permutations avoiding a pattern of length 3. The main result is that the joint distribution of this pair of parameters is the same in 321-avoiding as in 132-avoiding permutations. This generalizes a recent theorem of Robertson, Saracino and Zeilberger. We prove this result by giv-ing a bijection preserving these two statistics. A part of it is based on the Robinson-Schensted-Knuth correspondence. We also show that our bijection preserves additional parameters. The key idea is to introduce a new class of statistics on Dyck paths, based on what we call a tunnel .In Chapter 3 we consider the same pair of statistics in permutations avoiding simultaneously two or more patterns of length 3. We solve all the cases by giving generating functions which enumerate them. Some cases are gener-alized to patterns of arbitrary length. We also describe the distribution of these parameters in involutions avoiding any subset of patterns of length 3. The main technique consists in using bijections between pattern-avoiding permutations and certain kinds of Dyck paths, in such a way that the statis-tics in permutations that we consider correspond to statistics on Dyck paths which are easier to enumerate. In Chapter 4 we present a new statistic-preserving family of bijections from the set of Dyck paths to itself. They map statistics that appear in the study of pattern-avoiding permutations into classical statistics on Dyck paths, whose distribution is easy to obtain. In particular, this gives a simple bi-jective proof of the equidistribution of fixed points in 321- and 132-avoiding permutations. Chapter 5 gives some new interpretations of the Catalan and Fine numbers and a few additional bijections. We consider a class of permu-tations enumerated by the Catalan numbers, defined in terms of noncrossing Introduction 7
matchings of 2 n points around a circle. We study some of their properties, and we give the distribution of several statistics on them. In Chapter 6 we consider a different notion of pattern avoidance, with the requirement that the elements forming the pattern have to occur in consecu-tive positions in the permutation. More generally, we study the distribution of the number of occurrences of consecutive patterns in permutations. We solve the problem in several cases depending on the shape of the subword by obtaining the corresponding bivariate exponential generating functions as solutions of certain linear differential equations with polynomial coefficients. Our method is based on the representation of permutations as increasing binary trees and on symbolic methods. Finally, Chapter 7 deals with generalized patterns, which extend the notions of both classical and consecutive patterns. For a few patterns we obtain new exact enumerative results. Then we study the asymptotic behavior of the number of permutations in Sn avoiding a fixed generalized pattern as n goes to infinity. We also give some lower and upper bounds on the number of permutations avoiding certain patterns. 1
Definitions and preliminaries
1.1 Permutations
1.1.1 Pattern avoidance
We will denote by [ n] the set {1, 2, . . . , n }, and by Sn the symmetric group on [ n]. A permutation π ∈ S n will be written in one-line notation as
π = π1π2 · · · πn. We will write the cardinality of a set A as |A|. In this section we define the classical notion of pattern avoidance, which will be used throughout most of this thesis. For a definition of generalized patterns
see Section 7.1. Let n, m be two positive integers with m ≤ n, and let π = π1π2 · · · πn ∈ S n
and σ = σ1σ2 · · · σm ∈ S m be two permutations. We say that π contains σ if there exist indices i1 < i 2 < . . . < i m such that ρ(πi1 πi2 · · · πim ) = σ, where
ρ is the reduction consisting in relabelling the elements with {1, . . . , m } so that they keep the same order relationships they had in π. (Equivalently, this means that for all indices a and b, πia < π ib if and only if σa < σ b.) In that case, πi1 πi2 · · · πim is called an occurrence of σ in π. In this context, σ
is also called a pattern .If π does not contain σ, we say that π avoids σ, or that it is σ-avoiding . For example, if σ = 132, then π = 24531 contains 132, because the subsequence
π1π3π4 = 253 has the same relative order as 132. However, π = 42351 is 132-avoiding. Denote by Sn(σ) the set of σ-avoiding permutations in Sn. More generally, for any subset A ⊆ S n and any pattern σ, define A(σ) := A∩S n(σ)to be the set of σ-avoiding permutations in A.It is a natural generalization to consider permutations that avoid several patterns at the same time. If Σ ⊆ ⋃
k≥1
Sk is any finite set of patterns, denote 10 Chapter 1. Definitions and preliminaries
by Sn(Σ) the set of permutations in Sn that avoid simultaneously all the patterns in Σ. These are also called Σ-avoiding permutations. For example, if Σ = {123 , 231 }, then S4(Σ) = {1432 , 2143 , 3214 , 4132 , 4213 , 4312 , 4321 }.
1.1.2 Permutation statistics
Informally speaking, the notion of permutation can be viewed in two dif-ferent ways. On one hand, a permutation can be regarded as a word π =
π1π2 · · · πn, namely, as a sequence of numbers in some given order. From this description arises the concept of pattern avoidance discussed in the previous subsection. On the other hand, one can regard a permutation π ∈ S n as a bijection π : [ n] −→ [n]. Some concepts such as fixed points or excedances arise when we consider a permutation as a bijection. This double nature of permutations makes it interesting to study some of the following statistics together with the notion of pattern avoidance. There is a lot of mathematical literature devoted to permutation statistics (see for example [30, 42, 44, 46]). We say that i is a fixed point of a permutation π if πi = i. We say that i is an excedance of π if πi > i . Denote by fp( π) and exc( π) the number of fixed points and the number of excedances of π respectively. The distribution of the statistics fp and exc in pattern-avoiding permutations will be one of the main topics of this thesis. An element of a permutation that is neither a fixed point nor an excedance, namely an i for which πi < i , will be called a
deficiency . Permutations without fixed points are also called derangements .We say that i ≤ n − 1 is a descent of π ∈ S n if πi > π i+1 . Similarly, i ≤ n − 1is an ascent of π ∈ S n if πi < π i+1 . Denote by des( π) and asc( π) the number of descents and the number of ascents of π respectively. A right-to-left minimum of π is an element πi such that πi < π j for all j > i . Similarly,
πi is a left-to-right minimum of π if πi < π j for all j < i . A right-to-left maximum is an element πi such that πi > π j for all j > i .Let lis( π) denote the length of the longest increasing subsequence of π, i.e., the largest m for which there exist indices i1 < i 2 < · · · < i m such that
πi1 < π i2 < · · · < π im . Equivalently, in terms of forbidden patterns, lis( π) is the smallest m such that π avoids 12 · · · (m + 1). The length of the longest decreasing subsequence is defined analogously, and it is denoted lds( π). De-fine the rank of π, denoted rank( π), to be the largest k such that π(i) > k
for all i ≤ k. For example, if π = 63528174, then fp( π) = 1, exc( σ) = 4, des( π) = 4, lis( π) = 3, lds( π) = 4 and rank( π) = 2. We say that a permutation π ∈ S n is an involution if π = π−1. Denote by 1.1. Permutations 11
In the set of involutions of length n.
1.1.3 Trivial operations
Often it will be convenient to represent a permutation π ∈ S n by an n × n
array with a cross in each one of the squares ( i, π i). We will denote by arr( π)the array corresponding to π. Figure 1.1 shows arr(63528174).
Figure 1.1
The array of π = 63528174. The diagonal from the top-left corner to the bottom-right corner of the array will be referred to as main diagonal , and the diagonal perpendicular to it will be called secondary diagonal . Note that fixed points of π correspond to crosses on the main diagonal of the array, and excedances of π are represented by crosses to the right of this diagonal. Given a permutation π = π1π2 · · · πn, define its reversal πR = πn . . . π 2π1
and its complementation ¯ π = ( n + 1 − π1)( n + 1 − π2) · · · (n + 1 − πn). The array of ¯ π is obtained from the array of π by a flip along a vertical axis, so fixed points (resp. excedances) of π correspond to crosses on (resp. to the left of) the secondary diagonal of the array of ¯ π. Similarly, define ̂ π
to be the permutation whose array is the one obtained from that of π by reflection along the secondary diagonal. Note that reflecting the array of π
along the main diagonal we get the array of its inverse π−1. For any set of permutations Σ, let ¯Σ be the set obtained by reversing each of the elements of Σ. Define ̂ Σ and Σ −1 analogously. The following trivial lemma will be used in Chapters 2 and 3.
Lemma 1.1 Let Σ ⊂ ⋃
k≥1
Sk be a finite set of patterns, and let π ∈ S n.We have that
(1) π ∈ S n(Σ) ⇐⇒ ¯π ∈ S n( ¯Σ) ⇐⇒ ̂ π ∈ S n(̂ Σ) ⇐⇒ π−1 ∈ S n(Σ −1),
(2) fp( ̂π) = fp( π), exc( ̂π) = exc( π),12 Chapter 1. Definitions and preliminaries
(3) fp( π−1) = fp( π), exc( π−1) = n − fp( π) − exc( π).
In this thesis we will often be interested in the distribution of the statistics fp and exc among the permutations avoiding a certain pattern or set of patterns. Given any such set Σ, we define the generating function FΣ as
FΣ(x, q, z ) := ∑
n≥0
∑
π∈S n(Σ)
xfp( π)qexc( π)zn. (1.1) If Σ = {σ}, we will write Fσ instead of F{σ}. The following lemma restates the previous one in terms of generating functions.
Lemma 1.2 Let Σ be any set of permutations. We have
(1) FbΣ(x, q, z ) = FΣ(x, q, z ),
(2) FΣ−1 (x, q, z ) = FΣ(x/q, 1/q, qz ).Proof. To prove (1), consider the bijection between Sn(Σ) and Sn(̂ Σ) that maps π to ̂ π. The equation follows from parts (1) and (2) of Lemma 1.1. Equation (2) follows similarly from parts (1) and (3) of the previous lemma, noticing that
∑
n≥0
∑
π∈S n(Σ −1)
xfp( π)qexc( π)zn = ∑
n≥0
∑
π∈S n(Σ)
xfp( π−1)qexc( π−1)zn =
∑
n≥0
∑
π∈S n(Σ)
xfp( π)qn−fp( π)−exc( π)zn = ∑
n≥0
∑
π∈S n(Σ)
( x
q
)fp( π) ( 1
q
)exc( π)
(qz )n.
2
If for two sets of patterns Σ 1 and Σ 2 we have that FΣ1 (x, q, z ) = FΣ2 (x, q, z )(i.e., the joint distribution of fixed points and excedances is the same in Σ1-avoiding as in Σ 2-avoiding permutations), we will write Σ 1 ≈ Σ2. If we have that FΣ1 (x, q, z ) = FΣ2 (x/q, 1/q, qz ), we will write Σ 1∼f Σ2. In this notation, Lemma 1.2 says that ̂ Σ ≈ Σ and Σ −1∼f Σ.
1.2 Dyck paths
A Dyck path of length 2 n is a lattice path in Z2 between (0 , 0) and (2 n, 0) consisting of up-steps (1 , 1) and down-steps (1 , −1) which never goes below 1.2. Dyck paths 13
the x-axis. We shall denote by Dn the set of Dyck paths of length 2 n, and by D = ⋃
n≥0
Dn the class of all Dyck paths. It is well-known that |D n| =
Cn = 1
n+1
(2nn
), the n-th Catalan number. If D ∈ D n, we will write |D| = n to indicate the semilength of D. The generating function that enumerates Dyck paths according to their semilength is ∑
D∈D
z|D| = ∑
n≥0
Cnzn = 1−√1−4z
2z
,which we denote by C(z). Sometimes it will be convenient to encode each up-step by a letter u and each down-step by d, obtaining an encoding of the Dyck path as a Dyck word . We will use D to refer indistinctively to the Dyck path D or to the Dyck word associated to it. In particular, given D1 ∈ D n1 , D2 ∈ D n2 , we will write D1D2 to denote the concatenation of D1 and D2 (note that, as seen in terms of lattice paths, D2 has to be shifted 2 n1 units to the right). If
A is any sequence of up and down steps, length( A) will denote the number of steps in the sequence. For example, if A ∈ D n, then length( A) = 2 n.
1.2.1 Standard statistics
A peak of a Dyck path D ∈ D is an up-step followed by a down-step (i.e., an occurrence 1 of ud in the associated Dyck word). The coordinates of a peak are given by the point at the top of it. A hill is a peak at height 1, where the height is the y-coordinate of the peak. Denote by h(D) the number of hills of D, and by p2(D) the number of peaks of D of height at least 2. A
valley of D is a down-step followed by an up-step (i.e., an occurrence of du
in the associated Dyck word). Denote by va( D) the number of valleys of D.Clearly, both p2(D) + h(D) and va( D) + 1 equal the total number of peaks of D. A double rise of D is an up-step followed by another up-step (i.e., an occurrence uu in the Dyck word). Denote by dr( D) the number of double rises of D.An odd rise is an up-step in an odd position when the steps are numbered from left to right starting with 1 (or, equivalently, it is an up-step at odd level when the steps leaving the x-axis are considered to be at level 1). Denote by or( D) the number of odd rises of D. Even rises and er( D) are defined analogously. The x-coordinate of an odd or even rise is given by the rightmost end of the corresponding up-step. A return of a Dyck path is a down-step landing on the x-axis. An arch is
1In the context of Dyck words, the letters have to appear in consecutive positions to form an occurrence of a subword. 14 Chapter 1. Definitions and preliminaries
a part of the path joining two consecutive points on the x-axis. Clearly for any D ∈ D n the number of returns equals the number of arches. Denote it by ret( D). Define the x-coordinate of an arch as the x-coordinate of its leftmost point. The height of D is the y-coordinate of the highest point of the path. Denote by D≤k the set of Dyck paths of height at most k. For any D ∈ D n, define
ν(D) to be the height of the middle point of D, that is, the y-coordinate of the intersection of the vertical line x = n with the path. For example, if
D ∈ D 8 is the path in Figure 1.2, then h(D) = 1, p2(D) = 4, va( D) = 4, dr( D) = 3, or( D) = 5, er( D) = 3, ret( D) = 2, ν(D) = 2, and its height is 3. Define a pyramid to be a Dyck path that has only one peak, that is, a path of the form ukdk with k ≥ 1 (here the exponent indicates the number of times the letter is repeated). For a Dyck path D ∈ D n, denote by D∗ the path obtained by reflection of D from the vertical line x = n. We say that
D is symmetric if D = D∗. Denote by Ds ⊂ D the subclass of symmetric Dyck paths.
1.2.2 Tunnels
Here we introduce a new class of statistics on Dyck paths that will become very useful for the study of statistics on permutations avoiding patterns of length 3. They are based on the notion of tunnel of a Dyck path. For any D ∈ D , define a tunnel of D to be a horizontal segment between two lattice points of D that intersects D only in these two points, and stays always below D. Tunnels are in obvious one-to-one correspondence with de-compositions of the Dyck word D = AuBdC, where B ∈ D (no restrictions on A and C). In the decomposition, the tunnel is the segment that goes from the beginning of the u to the end of the d. If D ∈ D n, then D has exactly n tunnels, since such a decomposition can be given for each up-step
u of D. The length of a tunnel is just its length as a segment, and the height
is its y-coordinate. It will be useful to define the depth of a tunnel T as depth( T ) := 1
2
length( T ) − height( T ) − 1. A tunnel of D ∈ D n is called a centered tunnel if the x-coordinate of its midpoint (as a segment) is n, that is, the tunnel is centered with respect to the vertical line through the middle of D. In terms of the decomposition of the Dyck word D = AuBdC, this is equivalent to A and C having the same length, namely, length( A) = length( C). Alternatively, this can be taken as a definition of centered tunnel. Denote by ct( D) the number of centered 1.2. Dyck paths 15
tunnels of D.A tunnel of D ∈ D n is called a right tunnel if the x-coordinate of its midpoint is strictly greater than n, that is, the midpoint of the tunnel is to the right of the vertical line through the middle of D. In terms of the decomposition
D = AuBdC, this is equivalent to saying that length( A) > length( C). Denote by rt( D) the number of right tunnels of D. In Figure 1.2, there is one centered tunnel drawn with a solid line, and four right tunnels drawn with dotted lines. Similarly, a tunnel is called a left tunnel if the x-coordinate of its midpoint is strictly less than n. Denote by lt( D) the number of left tunnels of D. Clearly, lt( D) + rt( D) + ct( D) = n for any D ∈ D n.
Figure 1.2
One centered and four right tunnels. We will distinguish between right tunnels of D ∈ D n that are entirely contained in the half plane x ≥ n and those that cross the vertical line
x = n. These will be called right-side tunnels and right-across tunnels ,respectively. In terms of Dyck words, a decomposition D = AuBdC corre-sponds to a right-side tunnel if length( A) ≥ n, and to a right-across tunnel if length( C) < length( A) < n . In Figure 1.2 there are three right-side tunnels and one right-across tunnel. Left-side tunnels and left-across tunnels are defined analogously. For any D ∈ D , we define a multitunnel of D to be a horizontal segment between two lattice points of D such that D never goes below it. In other words, a multitunnel is just a concatenation of tunnels, so that each tunnel starts at the point where the previous one ends. Similarly to the case of tunnels, multitunnels are in obvious one-to-one correspondence with decom-positions of the Dyck word D = ABC , where B ∈ D is not empty. In the decomposition, the multitunnel is the segment that connects the initial and final points of B.A multitunnel of D ∈ D n is called a centered multitunnel if the x-coordinate of its midpoint (as a segment) is n, that is, the tunnel is centered with respect to the vertical line through the middle of D. In terms of the decomposition
D = ABC , this is equivalent to saying that A and C have the same length. Denote by cmt( D) the number of centered multitunnels of D.16 Chapter 1. Definitions and preliminaries
Figure 1.3
Five centered multitunnels, two of which are centered tunnels.
Additional interpretations of centered tunnels
Through the numerous known bijections between Dyck paths and other com-binatorial objects counted by the Catalan numbers, the new statistics that we defined on Dyck paths give rise to corresponding statistics in other ob-jects. Here we give a couple of examples that were suggested by Emeric Deutsch. It is known [90, Exercise 6.19(n)] that the diagrams of n nonintersecting chords joining 2 n points on the circumference of a circle are in bijection with Dn. We can draw these points as the vertices of a regular 2 n-gon, and the chords as straight segments, so that one of the diagrams has n horizontal chords. The bijection to Dyck paths can be described as follows. Starting counterclockwise from the topmost vertex on the left, for each vertex draw an up-step in the path if the chord from that vertex is encountered for the first time, and a down-step otherwise. By means of this bijection, horizontal chords of the diagram correspond precisely to centered tunnels of the Dyck path (see Figure 1.4). More generally, if we number the vertices of the 2 n-gon from 1 to 2 n in the order in which they are read by the bijection, then, for 1 ≤ i ≤ n, the chords parallel to the line between vertices i and i + 1 correspond to tunnels of the Dyck path with midpoint at x = i or at x = n + i.Another class of objects in bijection with Dn is the set of plane trees with
n + 1 vertices. Consider the bijection described in [90, Exercise 6.19(e)]. Now, given a plane tree on n + 1 vertices, label the vertices with integers from 0 to n in preorder (depth-first search) from left to right. Next, label the vertices again from 0 to n, but now in preorder from right to left. Then, the vertices other than the root for which the two labels coincide correspond 1.3. Combinatorial classes and generating functions 17
Figure 1.4 A bijection between nonintersecting chord diagrams and Dyck paths. to centered tunnels in the Dyck path. Besides, right tunnels correspond precisely to vertices for which the second label is less than the first one.
1.3 Combinatorial classes and generating functions
Here we direct the reader to and for a detailed account on combi-natorial classes and the symbolic method.
1.3.1 Ordinary generating functions
Let A be a class of unlabelled combinatorial objects and let |α| be the size of an object α ∈ A . If An denotes the objects in A of size n and an = |A n|,then the ordinary generating function of the class A is
A(z) = ∑
α∈A
z|α| = ∑
n≥0
anzn.
In our context, the size of a Dyck path is simply its semilength. From now on we will use the acronym GF as a shorthand for the term generating function
or, more specifically, ordinary generating function .There is a direct correspondence between set theoretic operations (or “con-structions”) on combinatorial classes and algebraic operations on GFs. Ta-ble 1.1 summarizes this correspondence for the operations that are used in this work. There “union” means union of disjoint copies, “product” is 18 Chapter 1. Definitions and preliminaries
the usual cartesian product, and “sequence” forms an ordered sequence in the usual sense. Enumerations according to size and auxiliary parameters
Construction Operation on GFs
Union A = B + C A(z) = B(z) + C(z)Product A = B × C A(z) = B(z)C(z)Sequence A = Seq( B) A(z) = 1
1−B(z)
Table 1.1
The basic combinatorial constructions and their transla-tion into ordinary generating functions.
χ1, χ 2, . . . , χ r are described by multivariate (ordinary) GFs,
A(u1, u 2, . . . , u r , z ) = ∑
α∈A
uχ1(α)1 uχ2(α)2 · · · uχr (α)
r
z|α|.
Throughout this thesis the variable z is reserved for marking the length of a permutation and the semilength of a Dyck path, x is used for marking the number of fixed points of a permutation and the number of centered tunnels or tunnels of depth 0 of a Dyck path, and q is the variable that marks the number of excedances of a permutation and the number of right tunnels or tunnels of negative depth of a Dyck path, unless otherwise stated.
1.3.2 Exponential generating functions
Let now A be instead a class of labelled combinatorial objects and let |α| be the size of an object α ∈ A as before. Let An denote the objects in A of size n and let an = |A n| again. The exponential generating function , EGF for short, of the class A is
A(z) = ∑
α∈A
z|α|
|α|! = ∑
n≥0
an
zn
n! .
In our context, the size of a permutation is simply its length. Table 1.2 summarizes the correspondence between set-theoretic operations on labelled combinatorial classes and algebraic operations on EGFs. There “labelled product” is the usual cartesian product enriched with the rela-belling operation, and “set” forms sets in the usual sense. Particularly 1.3. Combinatorial classes and generating functions 19
important for us is the construction “boxed product” A = B2 ∗ C , which corresponds to the subset of B ? C (the usual labelled product) formed by those pairs in which the smallest label lies in the B component.
Construction Operation on GF
Union A = B ∪ C A(z) = B(z) + C(z)Labelled product A = B ? C A(z) = B(z)C(z)Set A = Π( B) A(z) = exp( B(z)) Boxed product A = B2 ? C A(z) = ∫ z
0
( d
dt
B(t)) · C(t)dt
Table 1.2
The basic combinatorial constructions and their transla-tion into exponential generating functions. Enumerations according to size and an auxiliary parameter χ are described by bivariate (exponential) generating functions, or BGFs,
A(u, z ) = ∑
α∈A
uχ(α) z|α|
|α|! = ∑
n,k ≥0
An,k uk zn
n! ,
with An,k the number of objects of size n with χ-parameter equal to k.Exponential generating functions are used in Chapter 6. There, the variable
z is reserved for marking the length of a permutation, and the variable u is used mostly for marking occurrences of a subword. All derivatives in that chapter are taken with respect to z.
1.3.3 The Lagrange inversion formula
The Lagrange inversion formula (see for example [90, Theorem 5.4.2]) is a useful tool that provides a way to compute the coefficients of a generating function if it satisfies an equation of a certain form.
Theorem 1.3 () Let G(x) ∈ C be a formal power series such that
G(0) 6 = 0 , and let f (x) be defined by f (x) = xG (f (x)) . Then, for any
k, n ∈ Z,
n[xn]f (x)k = k[xn−k]G(x)n,
where [zn]A(z) denotes the coefficient of zn in the expansion of A(z).20 Chapter 1. Definitions and preliminaries
1.3.4 Chebyshev polynomials
Chebyshev polynomials of the second kind are defined by Ur(cos θ) = sin( r+1) θ
sin θ
for r ≥ 0. It can be checked that Ur(t) is a polynomial of degree r in t with integer coefficients, and that the following recurrence holds:
{ Ur(t) = 2 tU r−1(t) − Ur−2(t) for all r ≥ 2,
U0(t) = 1, U1(t) = 2 t. (1.2) Chebyshev polynomials were invented for the needs of approximation theory, but are also widely used in various other branches of mathematics, including algebra, combinatorics, and number theory. The relation between restricted permutations and Chebyshev polynomials was discovered for the first time by Chow and West in , and later by Mansour and Vainshtein [66, 67], and Krattenthaler .
1.4 Patterns of length 3
For the case of patterns of length 3, it is known that regardless of the pattern σ ∈ S 3, |S n(σ)| = Cn, the n-th Catalan number. While the equalities
|S n(132) | = |S n(231) | = |S n(312) | = |S n(213) | and |S n(321) | = |S n(123) | are straightforward by reversal and complementation operations, the equality
|S n(321) | = |S n(132) | is more difficult to establish. Bijective proofs of this fact are given in [59, 75, 84, 94]. However, none of these bijections preserves either of the statistics fp or exc. Patterns σ and σ′ are said to be in the same Wilf-equivalence class if |S n(σ)| =
|S n(σ′)| for all n. Partial results on the classification of forbidden patterns can be found in [5, 12, 13, 86, 87, 88].
1.4.1 Equidistribution of fixed points
It was not until recently that the concept of pattern avoidance, which regards a permutation as a word, was studied together with a statistic arising from viewing a permutation as a bijection. In the recent paper , Robertson, Saracino and Zeilberger consider restricted permutations with respect to the number of fixed points, obtaining the following refinement of the fact that
|S n(321) | = |S n(132) |.1.4. Patterns of length 3 21
Theorem 1.4 () The number of 321 -avoiding permutations π ∈ S n
with fp( π) = i equals the number of 132 -avoiding permutations π ∈ S n with
fp( π) = i, for any 0 ≤ i ≤ n.
Their proof is nontrivial and technically involved. In the same paper, they study the distribution of fixed points for all six patterns of length 3. Two questions arise naturally with this result in sight. The first one is whether there exists a simple bijection between Sn(321) and Sn(132) that preserves the number of fixed points. This would give a better understanding of why fixed points are equidistributed in both sets of pattern-avoiding per-mutations. There does not seem to be an intuitive reason why Theorem 1.4 holds, especially since from the definitions fixed points do not seem to be re-lated to the notion of pattern avoidance. The second question is whether this theorem can be generalized to other statistics or to other patterns. These two issues are discussed in the next chapter. 2
Fixed points and excedances in permutations avoiding one pattern of length 3
Here we consider σ-avoiding permutations for every pattern σ ∈ S 3, and we study the distribution of the statistics ‘number of fixed points’ and ‘number of excedances’ on them. The work in this chapter is motivated in large part by Theorem 1.4, and more precisely by a generalization of it, namely Theorem 2.3, which is the main result of this chapter. We will show that the joint distribution of the number of fixed points and the number of excedances is the same in Sn(321) as in Sn(132). In other words, we have that for any 0 ≤ i, j ≤ n,
|{ π ∈ S n(321) : fp( π) = i, exc( π) = j}|
= |{ π ∈ S n(132) : fp( π) = i, exc( π) = j}| .
In terms of the generating functions Fσ defined in equation (1.1), this re-sult can be expressed equivalently as F321 (x, q, z ) = F132 (x, q, z ). A bijective proof of this theorem is given in Section 2.2, where we also obtain an expres-sion for this GF. In Sections 2.3 and 2.4 we consider permutations avoiding each of the remaining patterns of length 3, giving the distribution of the statistics fp and exc in all cases except for the pattern 123, for which we can only give partial results regarding fp. One of the main tools in the this chapter and the next one will be a bijec-tion between 132-avoiding permutations and Dyck paths that we denote ϕ.This bijection is presented in Section 2.1, where several of its properties are studied. It is well known that for any σ ∈ S 3, |S n(σ)| = Cn. By Lemma 1.2, we 24 Chapter 2. Statistics in permutations avoiding one pattern
have that 132 ≈ 213, and that 231 ∼f 312. These are the only equivalences that follow from the trivial bijections. Together with the just mentioned fact that 321 ≈ 132 (see Section 2.2), we have the following equivalence classes of patterns of length 3 with respect to fixed points and excedances:
a) 132 ≈ 213 ≈ 321
b) 123
c) 231 ∼f c’) 312
2.1 The bijection ϕ
In this section we define a bijection ϕ between Sn(132) and Dn. This bijec-tion will be used extensively throughout this work, because of its convenient properties. Given any permutation π ∈ S n, consider its array arr( π) as defined in Sec-tion 1.1.3. The diagram of π can be obtained from it as follows. For each cross, shade the cell containing it and the squares that are due south and due east of it. The diagram is defined as the region that is left unshaded. It is shown in that this gives a bijection between Sn(132) and Young diagrams that fit in the shape ( n − 1, n − 2, . . . , 1). Consider now the path determined by the border of the diagram of π, that is, the path with north
and east steps that goes from the lower-left corner to the upper-right corner of the array, leaving all the crosses to the right, and staying always as close to the diagonal connecting these two corners as possible. Define ϕ(π) to be the Dyck path obtained from this path by reading an up-step for each north step and a down-step for each east step (that is, we rotate it 45 ◦). Since the path in the array does not go below the diagonal, ϕ(π) does not go below the x-axis. Figure 2.1 shows an example when π = 67435281.
Figure 2.1 The bijection ϕ.2.1. The bijection ϕ 25
The bijection ϕ is essentially the same bijection between Sn(132) and Dn
given by Krattenthaler (see also ), up to reflection of the path from a vertical line. Next we define the inverse map ϕ−1 : Dn −→ S n(132). Given a Dyck path
D ∈ D n, the first step needed to reverse the above procedure is to transform
D into a path U from the lower-left corner to the upper-right corner of an
n×n array, not going below the diagonal connecting these two corners. Then, the squares to the left of this path form a Young diagram contained in the shape ( n−1, n −2, . . . , 1), and we can shade all the remaining squares. From this diagram, the permutation π ∈ S n(132) can be recovered as follows: row by row, put a cross in the leftmost shaded square such that there is exactly one cross in each column. Start from the top and continue downward until all crosses are placed. The bijection ϕ is useful here because it transforms fixed points and ex-cedances of the permutation into centered tunnels and right tunnels of the Dyck path respectively. These two properties, along with a few more that will be used in upcoming chapters, are shown in the next proposition. Recall the definitions from Sections 1.1.2 and 1.2.1. Denote by nlis( π) the number of increasing subsequences of π of length lis( π).
Proposition 2.1 The bijection ϕ : Sn(132) −→ D n satisfies
(1) fp( π) = ct( ϕ(π)) ,
(2) exc( π) = rt( ϕ(π)) ,
(3) des( π) = va( ϕ(π)) ,
(4) lis( π) = height of ϕ(π),
(5) nlis( π) = # {peaks of ϕ(π) at maximum height },
(6) lds( π) = # {peaks of ϕ(π)},
(7) rank( π) = 1
2
(n − ν(ϕ(π)) ),for all π ∈ S n(132) .Proof. For the proof of the first six equalities, instead of using D = ϕ(π), it will be convenient to consider the associated path U from the lower-left corner to the upper-right corner of arr( π) with north and east steps. We will 26 Chapter 2. Statistics in permutations avoiding one pattern
talk about tunnels of U to refer to the corresponding tunnels of D under this trivial transformation. We now show how to associate a unique tunnel of D to each cross of the array arr( π). Observe that given a cross in position ( i, j ), U has a north step in row i and an east step in column j. In D, these two steps correspond to steps u and d respectively, so they determine a decomposition D = AuBdC
(see Figure 2.2), and therefore a tunnel of D (it is not hard to see that u
and d are at the same level). According to whether the cross was to the left of, to the right of, or on the main diagonal or arr( π), the associated tunnel will be respectively a left, right, or centered tunnel of D. Thus, fixed points give centered tunnels and excedances give right tunnels.
ud
B
AC
DU
Figure 2.2
A cross and the corresponding tunnel. To show (3), observe that from the description of ϕ−1, a sequence of consec-utive north steps of U gives rise to an increasing run of crosses in the rows of arr( π) where those steps lie. Descents of the permutation occur precisely in the rows of the array where there is a north step of U that is preceded by an east step. And these are just the valleys of ϕ(π). Property (4) is shown in , but here we give a more graphical proof. Given an increasing subsequence of π, consider the crosses of arr( π) that form such subsequence. The tunnels of ϕ(π) corresponding to these crosses are all at different heights, and their projections on the x-axis are nested intervals (i.e., pairwise contained in each other). Reciprocally, any tower of tunnels of ϕ(π) whose projections on the x-axis are nested corresponds to an increasing subsequence of π. The maximum number of tunnels in such a tower is the height of the path, so (4) follows. Furthermore, the number of such towers having as many tunnels as possible equals the number of peaks of ϕ(π) at maximum height (the highest tunnel of the tower determines the peak), which proves (5). Part (6) follows from the description of ϕ−1 and the observation that the 2.1. The bijection ϕ 27
crosses of the arr( π) located in the positions of the peaks (inner corners) of
U form a decreasing subsequence of π of maximum length. To prove the last equality of the proposition, notice that rank( π) is the largest
m such that an m × m square fits in the upper-left corner of the diagram of π. Therefore, the height of ϕ(π) at the middle is exactly ν(ϕ(π)) =
n − 2 rank( π). 2
2.1.1 More properties
We have seen what fixed points and excedances in 132-avoiding permutations are mapped to by ϕ. To study these statistics in 312-avoiding permutations, it will be convenient to first apply the complementation operation that maps
π to ¯ π. Table 2.1 summarizes the correspondences of ϕ that we will use to study fixed points and excedances. Recall from Section 1.2.2 that the depth of a tunnel T is defined as depth( T ) := 1
2
length( T ) − height( T ) − 1.
In the permutation π In the array of π In the Dyck path ϕ(π)
fixed points of π crosses on the main diagonal centered tunnels
excedances of π crosses to the right of the main diagonal right tunnels
fixed points of ¯ π crosses on the secondary diagonal tunnels of depth 0
excedances of ¯ π crosses to the left of the secondary diagonal
tunnels of negative depth
Table 2.1
Behavior of ϕ on fixed points and excedances. The correspondences between the first two columns are clear as we saw in Section 1.1.3. The first two rows of the table have been discussed in Proposition 2.1. Here we repeat the same reasoning from the proof of that proposition to show how ϕ maps crosses on the secondary diagonal to tunnels of depth 0, and crosses to the left of the secondary diagonal to tunnels of negative depth. Again, instead of using D = ϕ(π), it will be convenient to consider the path
U from the lower-left corner to the upper-right corner of the array of π, and to talk about tunnels of U to refer to the corresponding tunnels of D under this trivial transformation. Recall how in the proof of Proposition 2.1 we associated a unique tunnel T of 28 Chapter 2. Statistics in permutations avoiding one pattern
D to each cross X of arr( π). Given a cross X = ( i, j ), U has a north step in row i and an east step in column j. These two steps in U correspond to steps
u and d in D, respectively, so they determine a decomposition D = AuBdC
(see Figure 2.2), and therefore a tunnel T of D.The distance between these two steps determines the length of T , and the distance from these steps to the secondary diagonal of the array determines the height of T . In order for the corresponding cross to lie on the secondary diagonal, the relation between these two quantities must be 1
2
length( T ) = height( T )+1, which is equivalent to depth( T ) = 0, by the definition of depth. The depth of T indicates how far from the secondary diagonal X is. The cross lies to the left of the secondary diagonal exactly when depth( T ) < 0. This justifies the last two rows of the table.
Figure 2.3
Three tunnels of depth 0 and seven tunnels of negative depth. We define two new statistics on Dyck paths. For D ∈ D , let td 0(D) be the number of tunnels of depth 0 of D, and let td <0(D) be the number of tunnels of negative depth of D. In Figure 2.3, there are three tunnels of depth 0 drawn with a solid line, and seven tunnels of negative depth drawn with dotted lines. Let us state these results as a lemma, which partially overlaps with Proposition 2.1.
Lemma 2.2 Let π ∈ S n(132) , ρ ∈ S n(312) . We have
(1) fp( π) = ct( ϕ(π)) ,
(2) exc( π) = rt( ϕ(π)) ,
(3) fp( ρ) = td 0(ϕ(¯ ρ)) ,
(4) exc( ρ) = td <0(ϕ(¯ ρ)) .2.2. a) 132 ≈ 213 ≈ 321 29
2.2 a) 132 ≈ 213 ≈ 321
This is the most interesting case of permutations avoiding one pattern of length 3. By Lemma 1.2 we have that 132 ≈ 213. On the other hand, Theorem 1.4 says that 132 ∼f 321, which is a surprising and nontrivial result. Here we give a generalization of this fact, namely that 132 ≈ 321. This result is stated in the following theorem, for which we give a combinatorial proof.
Theorem 2.3 The number of 321 -avoiding permutations π ∈ S n with
fp( π) = i and exc( π) = j equals the number of 132 -avoiding permutations
π ∈ S n with fp( π) = i and exc( π) = j, for any 0 ≤ i, j ≤ n.
We present a bijection between 321- and 132-avoiding permutations that preserves the number of fixed points and the number of excedances. Our bijection is a composition of two slightly modified known bijections into Dyck paths, and the result follows from a new analysis of these bijections. One of them is the bijection ϕ from Section 2.1. The other one is based on the Robinson-Schensted-Knuth correspondence, and from it stems the difficulty of the analysis. We also show that the length of the longest increasing subsequence in 321-avoiding permutations corresponds to a statistic in 132-avoiding permuta-tions that we call rank , which further refines Theorem 2.3. This proof is joint work with Igor Pak . This section is structured as follows. The description of the main bijec-tion is done in Subsection 2.2.1, where the new part is a bijection from 321-avoiding permutations to Dyck paths. In Subsection 2.2.2 we establish properties of this bijection. Subsection 2.2.3 contains proofs of two techni-cal lemmas. In Subsection 2.2.4 we give an expression for the generating function F321 (x, q, z ).
2.2.1 A bijection between Sn(321) and Sn(132) pre-serving fixed points and excedances
Here we present a bijection that is a composition of two bijections into Dyck paths. In fact, this will prove the following generalization of Theorem 2.3:
Theorem 2.4 The number of 321 -avoiding permutations π ∈ S n with
fp( π) = i, exc( π) = j and lis( π) = k equals the number of 132 -avoiding 30 Chapter 2. Statistics in permutations avoiding one pattern
permutations π ∈ S n with fp( π) = i, exc( π) = j and rank( π) = n − k, for any 0 ≤ i, j, k ≤ n.
We establish a bijection Θ : Sn(321) −→ S n(132) which respects the statis-tics as above. While Θ is not hard to define, its analysis is less straightfor-ward and will occupy much of the section. The bijection Θ is the composi-tion of two bijections, one from Sn(321) to Dn, and another one from Dn to
Sn(132). The second one is just the inverse of the bijection ϕ : Sn(132) −→ Dn presented in Section 2.1. The first one is described next. We define the bijection Ψ : Sn(321) −→ D n in two steps. Given π ∈ S n(321), we start by applying the Robinson-Schensted-Knuth correspondence to π
[90, Section 7.11] (see also [58, 82]). This correspondence gives a bijection between the symmetric group Sn and pairs ( P, Q ) of standard Young tableaux
of the same shape λ ` n. For π ∈ S n(321) the algorithm is particularly easy because in this case the tableaux P and Q have at most two rows. The insertion tableau P is obtained by reading π from left to right and, at each step, inserting πi to the partial tableau obtained so far. Assume that
π1, . . . , π i−1 have already been inserted. If πi is larger than all the elements on the first row of the current tableau, place πi at the end of the first row. Otherwise, let m be the leftmost element on the first row that is larger than
πi. Place πi in the square that m occupied, and place m at the end of the second row (in this case we say that πi bumps m). The recording tableau Q
has the same shape as P and is obtained by placing i in the position of the square that was created at step i (when πi was inserted) in the construction of P , for all i from 1 to n. We write RSK( π) = ( P, Q ).
23223535
2
1134
25
6134
25
6134
25
86134
258
7
6134
258
7
P = 12367
458
12367
458
2367
45
1236
45
123
45
123
4
1231211
Q =
Figure 2.4 Construction of the RSK correspondence RSK( π) = (P, Q ) for π = 23514687. Now, the first half of the Dyck path Ψ( π) is obtained by adjoining, for i from 1 to n, an up-step if i is on the first row of P , and a down-step if it is on the second row. Let A be the corresponding word of u’s and d’s. Similarly, 2.2. a) 132 ≈ 213 ≈ 321 31
let B be the word obtained from Q in the same way. We define Ψ( π) to be the Dyck path obtained by the concatenation of the word A and the word B
written backwards. For example, from the tableaux P and Q as in Figure 2.4 we get the Dyck path shown in Figure 1.2. The following proposition, which is proved in Section 2.2.2, summarizes some properties of this bijection Ψ:
Proposition 2.5 The bijection Ψ : Sn(321) −→ D n satisfies
(1) fp( π) = ct(Ψ( π)) ,
(2) exc( π) = rt(Ψ( π)) ,
(3) lis( π) = 1
2
(n + ν(Ψ( π)) ),for all π ∈ S n(321) .
Suppose RSK( π) = ( P, Q ) for any π ∈ S n. A fundamental and highly nontrivial property of the RSK correspondence is the duality : RSK( π−1) = (Q, P ) [90, Section 7.13]. The classical Schensted’s Theorem states that lis( π) is equal to the length of the first row of the tableau P (and Q). Both results are used in the proof of Proposition 2.5. Now Theorem 2.4 follows easily from this proposition together with Propo-sition 2.1.
Proof of Theorem 2.4. Propositions 2.5 and 2.1 imply that Θ = ϕ−1 ◦ Ψ is a bijection from Sn(321) to Sn(132) which satisfies fp(Θ( π)) = ct(Ψ( π)) = fp( π),
exc(Θ( π)) = rt(Ψ( π)) = exc( π),
rank(Θ( π)) = 1
2
(n − ν(Ψ( π)) ) = n − 1
2
(n + ν(Ψ( π)) ) = n − lis( π) .
This implies the result. 2
2.2.2 Properties of Ψ
In this subsection we prove Proposition 2.5, which describes the properties of Ψ that we need. Let us first consider only fixed points in a permutation π ∈ S n. Let π ∈Sn(321) and assume that πi = i. Then π1π2 · · · πi−1 is a permutation of 32 Chapter 2. Statistics in permutations avoiding one pattern
{1, 2, . . . , i − 1}, and πi+1 πi+2 · · · πn is a permutation of {i + 1 , i + 2 , . . . , n }.Indeed, if πj > i for some j < i , then necessarily πk < i for some k > i , and
πj πiπk would be an occurrence of 321. Therefore, when we apply RSK to π, the elements πi, π i+1 , . . . , π n will never bump any of the elements π1, π 2, . . . , π i−1. In particular, the subtableaux of
P and Q determined by the entries that are smaller than i will have the same shape. Furthermore, when the elements greater than i are placed in P and
Q, the rows in which they are placed do not depend on the subpermutation
π1π2 · · · πi−1. Note also that πi = i will never be bumped, and it will occupy the same position in the first row of P and Q.When the Dyck path Ψ( π) is built from P and Q, this translates into the fact that the steps corresponding to πi in P and to i in Q will be respectively an up-step in the first half and a down-step in the second half, both at the same height and at the same distance from the center of the path. Besides, the part of the path between them will be itself the Dyck path corresponding to (πi+1 −i)( πi+2 −i) · · · (πn−i). So, the fixed point πi = i determines a centered tunnel in Ψ( π). It is clear that the converse is also true, that is, every centered tunnel comes from a fixed point. This shows that fp( π) = ct(Ψ( π)), proving the first part of Proposition 2.5. Let us now consider excedances in a permutation π ∈ S n(321). Our goal is to show that the excedances of π correspond to right tunnels of Ψ( π). The first observation is that we can assume without loss of generality that π has no fixed points. Indeed, if πi = i is a fixed point of π, then the above reasoning shows that we can decompose Ψ( π) = AuBdC, where AC is the Dyck path Ψ( π1π2 · · · πi−1) and B is a translation of the Dyck path Ψ(( πi+1 −i) · · · (πn −
i)). But we have that exc( π) = exc( π1π2 · · · πi−1) + exc(( πi+1 − i) · · · (πn −
i)) and rt( AuBdC) = rt( AC ) + rt( B), so in this case the result holds by induction on the number of fixed points. Note also that the above argument showed that fp( π) = fp( π1π2 · · · πi−1) + fp(( πi+1 − i) · · · (πn − i)) + 1 and ct( AuBdC) = ct( AC ) + ct( B) + 1. Suppose that π ∈ S n(321) has no fixed points. We will use the fact that a permutation is 321-avoiding if and only if both the subsequence determined by its excedances and the one determined by the remaining elements (in this case, the deficiencies) are increasing (see e.g. ). Denote by Xi := ( i, π i)the crosses of the array representation of π. To simplify the presentation, we will refer indistinctively to i or Xi, hoping this does not lead to confusion. For example, we will say “ Xi is an excedance”, etc. Define a matching between the excedances and the deficiencies of π by the 2.2. a) 132 ≈ 213 ≈ 321 33
following algorithm. Let i1 < i 2 < · · · < i k be the positions of the excedances of π and let j1 < j 2 < · · · < j n−k be the deficiencies. Note that from the previous paragraph we know that πi1 < π i2 < · · · < π ik and πj1 < π j2 <
· · · < π jn−k .
Matching Algorithm
(1) Initialize a := 1, b := 1. (2) Repeat until a > k or b > n − k:(a) If ia > j b, then b := b + 1. ( Xjb is not matched.) (b) Else if πia < π jb , then a := a + 1. ( Xia is not matched.) (c) Else, match Xia with Xjb ; a := a + 1, b := b + 1. (3) Output the matching sequence.
Example. Let π = (4 , 1, 2, 5, 7, 8, 3, 6, 11 , 9, 10) as in Figure 2.5 below. We have i1 = 1, i2 = 4, i3 = 5, i4 = 6, i5 = 9, and j1 = 2, j2 = 3, j3 = 7,
j4 = 8, j5 = 10, j6 = 11. In the first execution of the loop in step (2) of the algorithm, neither i1 > j 1 nor πi1 < π j1 hold, so Xi1 = (1 , 4) and Xj1 = (2 , 1) are matched. Now we repeat the loop with a = b = 2, and since i2 > j 2,we are in the case given by (2a) ( Xj2 = (3 , 2) is not matched). In the next iteration, a = 2 and b = 3, so we match Xi2 = (4 , 5) and Xj3 = (7 , 3). Now we have a = 3 and b = 4, so we match Xi3 = (5 , 7) and Xj4 = (8 , 6). The values of a and b in the next iteration are 4 and 5 respectively, so we are in the case of (2b), πi4 = 8 < 9 = πj5 , and Xi4 = (6 , 8) is unmatched. Now
a = b = 5, and we match Xi5 = (9 , 11) and Xj5 = (10 , 9). The matching algorithm ends here because now a = 6 > 5 = k.An informal, more geometrical description of the matching algorithm is the following. For each pair of crosses of the array (seen as embedded in the plane), consider the line that their centers determine. If one of these lines has positive slope and leaves all the remaining crosses to the right, match the two crosses that determine it, and delete them from the array. If there is no line with these properties, delete the cross that is closer to the upper-left corner of the array (it is unmatched). Repeat the process until no crosses are left. Now we consider the matched excedances on one hand and the unmatched ones on the other. We summarize rather technical results in the following 34 Chapter 2. Statistics in permutations avoiding one pattern
Figure 2.5
Matching for π = (4 , 1, 2, 5, 7, 8, 3, 6, 11 , 9, 10), and Ψ( π). two lemmas, which are proved in Section 2.2.3. Recall the definitions of right-side and left-side tunnels from Section 1.2.2.
Lemma 2.6 The following quantities are equal:
(1) the number of matched pairs (Xi, X j ), where Xi is an excedance and
Xj a deficiency;
(2) the length of the second row of P (or Q);
(3) the number of right-side tunnels of Ψ( π);
(4) the number of left-side tunnels of Ψ( π);
(5) 1
2
(n − ν(Ψ( π)) );
(6) n − lis( π).
Note that (5)=(6) implies that lis( π) = 1
2
(n + ν(Ψ( π)) ), which is the third part of Proposition 2.5.
Lemma 2.7 The number of unmatched excedances (resp. deficiencies) of π
equals the number of right-across (resp. left-across) tunnels of Ψ( π).
Since each excedance of π either is part of a matched pair ( Xi, X j ) or is unmatched, Lemmas 2.6 and 2.7 imply that the total number exc( π) of excedances equals the number of right-side tunnels of Ψ( π) plus the number 2.2. a) 132 ≈ 213 ≈ 321 35
of right-across tunnels, which is rt(Ψ( π)). This implies the second part of Proposition 2.5. To summarize, we will have shown after proving these lemmas that the bijection Ψ satisfies all three properties described in Proposition 2.5, which completes the proof. 2
2.2.3 Properties of the matching algorithm
In this subsection we prove the two lemmas above. We also give a more direct description of the bijection Ψ using the matching algorithm, without referring explicitly to RSK.
Proof of Lemma 2.6. From the descriptions of the RSK algorithm and the matching, it follows that an excedance Xi and a deficiency Xj are matched with each other precisely when πj bumps πi when RSK is performed on π,and that these are the only bumpings that take place. Indeed, an excedance never bumps anything because it is larger than the elements inserted before. On the other hand, when a deficiency Xj is inserted, it bumps the smallest element larger than πj which has not been bumped yet (which corresponds to an excedance that has not been matched yet), if such an element exists. This proves the equality (1)=(2). To see that (2)=(3), observe that right-side tunnels correspond to up-steps in the right half of Ψ( π), which by the construction of the bijection Ψ cor-respond to elements on the second row of Q. The equality (3)=(5) follows easily by counting the number of up-steps and down-steps of the right half of the path. The equality (4)=(5) is analogous. Finally, Schensted’s Theorem states that the size of the first row of P equals the length of a longest increasing subsequence of π (see or [90, Sec-tion 7.23]). This implies that (2)=(6), which completes the proof. 2
The reasoning used in the above proof gives a nice equivalent description of the recording tableau Q in terms of arr( π) and the matching. Read the rows of the array from top to bottom. For i from 1 to n, place i on the first row of
Q if Xi is an excedance or it is unmatched, and place i on the second row if
Xi is a matched deficiency. In the construction of the right half of Ψ( π), this translates into drawing the path from right to left while reading the array from top to bottom, adjoining an up-step for each matched deficiency and a down-step for each other kind of cross. 36 Chapter 2. Statistics in permutations avoiding one pattern
To get a similar description of the tableau P , we use duality. By construction of the matching algorithm, the matching in the output is invariant under transposition of the array (reflection along the main diagonal). Recall the duality of the RSK correspondence: if RSK( π) = ( P, Q ), then RSK( π−1) = (Q, P ) (see e.g. [90, Section 7.13]). Therefore, the tableau P can be obtained by reading the columns of the array of π from left to right and placing integers in P according to the following rule. For each column j, place j on the first row of P if the cross in column j is a deficiency or it is unmatched. Similarly, place j on the second row if the cross is a matched excedance. Equivalently, the left half of Ψ( π), from left to right, is obtained by reading the array from left to right and adjoining a down-step for each matched excedance, and an up-step for each of the remaining crosses. In particular, when the left half of the path is constructed in this way, every matched pair ( Xi, X j ) produces an up-step and a down-step, giving the latter a left-side tunnel. Similarly, in the construction of the right half of the path, a matched pair gives a right-side tunnel. Observe that these are again the equalities (1)=(3)=(4) in Lemma 2.6.
Proof of Lemma 2.7. It is enough to prove it only for the case of excedances. The case of deficiencies follows from it by considering π−1 and noticing that Ψ( π−1) = Ψ( π)∗. Indeed, by duality RSK( π−1) = ( Q, P ), so Q gives rise to the first half of Ψ( π−1) and P to the second, so the path that we obtain is the reflection of Ψ( π) in a vertical axis through the middle of the path. Let Xk be an unmatched excedance of π. We use the above description of Ψ( π) in terms of the array and the matching. Each cross Xi produces a step
ri in the right half of the Dyck path and another step i in the left half. Crosses above Xk produce steps to the right of rk, and crosses to the left of Xk produce steps to the left ofk. In particular, there are k − 1 steps to the right of rk, and πk − 1 steps to the left of `k. Note that since Xk is an excedance and π is 321-avoiding, all the crosses above it are also to the left of it. Consider the crosses that lie to the left of Xk. They can be of the following four kinds:
• Unmatched excedances Xi. They will necessarily lie above Xk, because the subsequence of excedances of π is decreasing. Each one of these crosses contributes an up-step to the left of `k and down-step to the right of rk.
• Unmatched deficiencies Xj . They also have to lie above Xk, otherwise 2.2. a) 132 ≈ 213 ≈ 321 37
Xk would be matched with one of them. So, each such Xj contributes an up-step to the left of `k and down-step to the right of rk.
• Matched pairs (Xi, X j ) (i.e. Xi is an excedance and Xj a deficiency) ,where both Xi and Xj lie above Xk. Both crosses together will con-tribute an up-step and a down-step to the left of `k, and an up-step and a down-step to the right of rk.
• Matched pairs (Xi, X j ) (i.e. Xi is an excedance and Xj a deficiency) ,where Xj lies below Xk. The pair will contribute an up-step and a down-step to the left of `k. However, to the right of rk, the only contribution will be a down-step produced by Xi.Note that there cannot be a deficiency Xj to the left of Xk matched with an excedance to the right of Xk, because in this case Xj would have been matched with Xk by the algorithm. In the first three cases, the contribution to both sides of the Dyck path is the same, so that the heights of rk and
`k are equally affected. But since πk > k , at least one of the crosses to the left of Xk must be below it, and this must be a matched deficiency as in the fourth case. This implies that the step rk is at a higher y-coordinate than
k. Let hk be the height ofk. We now show that Ψ( π) has a right-across tunnel at height hk.Observe that hk is the number of unmatched crosses to the left of Xk, and that the height of rk is the number of unmatched crosses above Xk (which equals hk) plus the number of excedances above Xk matched with deficiencies below Xk. The part of the path between `k and the middle always remains at a height greater than hk. This is because the only possible down-steps in this part can come from matched excedances Xi to the right of Xk, but each such Xi is matched with a deficiency Xj to the right of Xk but to the left of Xi, which produces an up-step compensating the down-step associated to
Xi. Similarly, the part of the path between rk and the middle remains at a height greater than hk. This is because the hk down-steps to the right of rk
that come from unmatched crosses above Xk do not have a corresponding up-step in the part of the path between rk and the middle. Hence, k is the left end of a right-across tunnel, since the right end of this tunnel is to the right of rk, which in turn is closer to the right end of Ψ( π) thank is to its left end (see Figure 2.6). It can easily be checked that the converse is also true, namely that in ev-ery right-across tunnel of Ψ( π), the step at its left end corresponds to an unmatched excedance of π. 238 Chapter 2. Statistics in permutations avoiding one pattern
lrh
kkk
Figure 2.6 An unmateched excedance produces a right-across tunnel. As a side note, let us point out that the limiting distribution of the length of the longest increasing subsequence in Sn(321) has been studied in . From Theorem 2.4, the results in can be translated into results on the limiting distribution of the statistic rank in Sn(132).
2.2.4 The bijection ψx
We already have proved that F132 (x, q, z ) = F321 (x, q, z ). Here we give an expression for the generating function
F321 (x, q, z ) = ∑
n≥0
∑
π∈S n(321)
xfp( π)qexc( π)zn,
which enumerates 321-avoiding permutations with respect to fixed points and excedances. In order to achieve this, we define a bijection ψx between
Sn(321) and Dn, which appeared originally in [48, pg. 89] in a slightly dif-ferent form, and later was used by Richard Stanley in connection to pattern avoidance. We will give three equivalent definitions of the bijection ψx. Let π =
π1π2 · · · πn ∈ S n(321). For i ∈ [n], define ai = max {j : {1, 2, . . . , j } ⊆ {π1, π 2, . . . , π i}} (j can be 0, in which case {1, 2, . . . , j } = ∅). Now build the Dyck path ψx(π) by adjoining, for each i from 1 to n, one up-step fol-lowed by max {ai − πi + 1 , 0} down-steps. For example, for π = 23147586 we get a1 = a2 = 0, a3 = 3, a4 = a5 = 4, a6 = a7 = 5, a8 = 8, and the corresponding Dyck path is given in Figure 2.7. Here is an alternative way to define this bijection. Let πi1 , π i2 , . . . , π ik be the right-to-left minima of π, from left to right. For example, the right-to-left minima of 23147586 are 1 , 4, 5, 6. Then, ψx(π) is precisely the path that starts with i1 up-steps, then has, for each j from 2 to k, πij − πij−12.2. a) 132 ≈ 213 ≈ 321 39
Figure 2.7
The Dyck path ψx(23147586). down-steps followed by ij − ij−1 up-steps, and finally ends with n + 1 − πik
down-steps. The third way to define ψx is the easiest one to visualize, and the one that gives us a better intuition for how the bijection works. Consider the array of crosses arr( π) as defined in Section 1.1.3. By definition, excedances corre-spond to crosses strictly to the right of the main diagonal of the array. It is known (see e.g. ) that a permutation is 321-avoiding if and only if both the subsequence determined by its excedances and the one determined by the remaining elements are increasing. Therefore, the elements that are not excedances are precisely the right-to-left minima of π. Consider the path with east and south steps along the edges of the squares of arr( π) that goes from the upper-left corner to the lower-right corner of the array, leaving all the crosses to the right and remaining always as close to the main diago-nal as possible. Let U be such path. Then ψx(π) can be obtained from U
just by reading an up-step for every south step of U , and a down-step for every east step of U . Figure 2.8 shows a picture of this bijection, again for
π = 23147586.
Figure 2.8
The bijection ψx.
Proposition 2.8 The bijection ψx : Sn(321) −→ D n satisfies
(1) fp( π) = h(ψx(π)) ,40 Chapter 2. Statistics in permutations avoiding one pattern
(2) exc( π) = dr( ψx(π)) ,for all π ∈ S n(321) .Proof. To see this, just observe that fixed points of π correspond to crosses on the main diagonal of the array, which produce hills in the path. On the other hand, for each cross corresponding to an excedance, the south step of
U on the same row as the cross gives an up-step in ψx(π) which is followed by another up-step, thus forming a double rise. 2
Therefore, counting 321-avoiding permutations according to the number of fixed points and excedances is equivalent to counting Dyck paths according to the number of hills and double rises. More precisely,
F321 (x, q, z ) = ∑
D∈D
xh(D)qdr( D)z|D|.
We can give an equation for F321 using the symbolic method summarized in Section 1.3. A recursive definition for the class D is given by the fact that every non-empty Dyck path D can be decomposed in a unique way as
D = uAdB, where A, B ∈ D . Clearly if A is empty, h(D) = h(B) + 1 and dr( D) = dr( B), and otherwise h(D) = h(B) and dr( D) = dr( A) + dr( B) + 1. Hence, we obtain the following equation for F321 :
F321 (x, q, z ) = 1 + z(x + q(F321 (1 , q, z ) − 1)) F321 (x, q, z ). (2.1) Substituting first x = 1, we obtain that
F321 (1 , q, z ) = 1 + ( q − 1) z − √1 − 2(1 + q)z + (1 − q)2z2
2qt .
Now, solving (2.1) for F321 (x, q, z ) gives the following result.
Theorem 2.9
F132 (x, q, z ) = F213 (x, q, z ) = F321 (x, q, z ) = = 2
1 + (1 + q − 2x)z + √1 − 2(1 + q)z + (1 − q)2z2 .
To conclude this section, we want to remark that applying this bijection one can also obtain the GF that enumerates 321-avoiding permutations with re-spect to fixed points, excedances and descents. It follows easily from the 2.3. b) 123 41
description of ψx that the number des( π) of descents of a 321-avoiding per-mutation π equals the number of occurrences of uud in the Dyck word of
ψx(π). Using the same decomposition as before, we obtain the following result.
Theorem 2.10
∑
n≥0
∑
π∈S n(321)
xfp( π)qexc( π)pdes( π)zn
= 2
1 + (1 + q − 2x)z + √1 − 2(1 + q)z + ((1 + q)2 − 4qp )z2 .
The analogous generalization of Theorem 2.9 which enumerates 132-avoiding permutations with respect to these three statistics is given in Theorem 4.10.
2.3 b) 123
For this case we have not been able to find a satisfactory expression for
F123 (x, q, z ). We can nevertheless give summation formulas for the number of permutations in Sn(123) with a given number of fixed points. The first trivial observation is that if π avoids 123, then it can have at most two fixed points. If πi = i, we say that i is a big fixed point of π if i ≥ n+1
2
, and that it is a small fixed point if i < n+1
2
.We already mentioned that a permutation is 321-avoiding if and only if both the subsequence determined by its excedances and the one determined by the remaining elements are increasing. Using the fact that π avoids 123 if and only if ¯ π avoids 321, we obtain a characterization of 123-avoiding per-mutations as those with the following property: the elements πi such that
πi < n + 1 − i form a decreasing subsequence, and so do the remaining ele-ments. In particular, since no two fixed points can be in the same decreasing subsequence, this implies that π can have at most one big fixed point and one small fixed point. Recall the bijection ψx : Sn(321) −→ D n that we defined in Section 2.2.4. Composing it with the complementation operation sending π ∈ S n(123) to ¯π ∈ S n(321), we obtain a bijection between Sn(123) and Dn, which we denote by ψy. Figure 2.9 shows an example when π = (9 , 6, 10 , 4, 8, 7, 3, 5, 2, 1). Note that the peaks of the path are determined by the crosses of elements
πi such that πi ≥ n + 1 − i, which form a decreasing subsequence. Now it 42 Chapter 2. Statistics in permutations avoiding one pattern
Figure 2.9
The bijection ψy.is easy to determine how many permutations have a big (resp. small) fixed point.
Proposition 2.11 Let n ≥ 1. We have
(1) |{ π ∈ S n(123) : π has a big fixed point }| = Cn−1,
(2) |{ π ∈ S n(123) : π has a small fixed point }| =
Cn−1 if n is even,
Cn−1 − C2
n−1
2
if n is odd. Proof. (1) It is clear from the definition of ψy that π has a big fixed point if and only if ψy(π) has a peak in the middle. Now, we can easily define a bijection from the subset of elements of Dn with a peak in the middle and
Dn−1, by removing the two middle steps ud .(2) Clearly, π ∈ S n(123) if and only if ̂ π ∈ S n(123) (recall the definition from Section 1.1.3). This involution switches big and small fixed points, except for the possible big fixed point in position n+1
2
, which remains unchanged. Applying now ψy, a small fixed point of π is transformed into a peak in the middle of ψy(̂π) of height at least two (indeed, a hill would correspond to the big fixed point n+1
2
). Knowing that the number of paths in Dn with a peak in the middle is Cn−1, we just have to subtract those where this middle peak has height 1. If n is even, paths in Dn cannot have a hill in the middle. If n is odd, such paths have the form Aud B, where A, B ∈ D n−1
2
, so the formula follows. 2
For k ≥ 0, let skn(123) := |{ π ∈ S n(123) : fp( π) = k}| . We have mentioned that skn(123) = 0 for k ≥ 3. The following corollary reduces the problem of studying the distribution of fixed points in Sn(123) to that of determining
s2
n
(123). 2.3. b) 123 43
Corollary 2.12 Let n ≥ 1. We have
(1) s1
n
(123) =
{ 2( Cn−1 − s2
n
(123)) if n even,
2( Cn−1 − s2
n
(123)) − C2
n−1
2
if n odd.
(2) s0
n
(123) =
{ Cn − 2Cn−1 + s2
n
(123) if n even,
Cn − 2Cn−1 + s2
n
(123) + C2
n−1
2
if n odd. Proof. (1) By inclusion-exclusion,
s1
n
(123) = |{ π ∈ S n(123) : π has a big fixed point }|
+|{ π ∈ S n(123) : π has a small fixed point }| − 2s2
n
(123) .
Now we apply Proposition 2.11. (2) Clearly, s0
n
(123) = Cn − s1
n
(123) − s2
n
(123). 2
The next theorem, together with the previous corollary, gives a formula for the distribution of fixed points in 123-avoiding permutations.
Theorem 2.13
s2
n
(123) =
n−1
∑
i=1
i
∑
r,s =1
[(( 2i − r − 1
i − 1
)
−
(2i − r − 1
i
))
·
(( 2i − s − 1
i − 1
)
−
(2i − s − 1
i
))
·
n
∑
h=1
n−heven
n−2i
∑
k=0
f (k, r, h, n − 2i + r)f (n − 2i − k, s, h, n − 2i + s)
,
where
f (k, r, h, ` ) =
( `+h−r
2
− 1
k
)( `−h+r
2
− 1
k − 1
)
−
( `−h−r
2
− 1
k
)( `+h+r
2
− 1
k − 1
)
if k ≥ 1,
1 if k = 0 and ` = h − r,
0 otherwise, with the convention (ab
) := 0 if a < 0.44 Chapter 2. Statistics in permutations avoiding one pattern
Proof. Recall that s2
n
(123) counts permutations with both a big and a small fixed point. We have seen already that ψy maps a big fixed point of the permutation into a peak in the middle of the Dyck path. Now we look at how a small fixed point of the permutation is transformed by ψy. We claim that π ∈ S n(123) has a small fixed point if and only if D = ψy(π) satisfies the following condition (which we call condition C1): there exists i such that the i-th and ( i + 1)-st up-steps of D are consecutive, the i-th and ( i + 1)-st down-steps from the end are consecutive, and there are exactly n + 1 − 2i
peaks of D between them. To see this, assume that i is a small fixed point of π (see Figure 2.10). Then, the path from the upper-right to the lower-left corner of the array of π, used to define ψy(π), has two consecutive vertical steps in rows i and i + 1, and two consecutive horizontal steps in columns
i and i + 1. Besides, there are n + 1 − 2i crosses below and to the right of cross ( i, i ), each one of which produces a peak in the Dyck path ψy(π). Reciprocally, it can be checked that if ψy(π) satisfies condition C1 then π
has a small fixed point.
ii
Figure 2.10
A small fixed point i has n + 1 − 2i crosses below and to the right. All we have to do is count how many paths D ∈ D n with a peak in the middle satisfy condition C1. For such a Dyck path D, define the following parameters: let i be the value such that condition C1 holds, let h = ν(D) be the height of D in the middle, r the height at which the i-th up-step ends, and s the height at which the i-th down-step from the end begins. In the example of Figure 2.11, n = 12, i = 4, h = 4, r = 3, and s = 1. Fix n, i, h, r and s. We will count the number of Dyck paths D with 2.3. b) 123 45
ii+1 ii+1
rs
h
Figure 2.11
The parameters i, h, r and s in a Dyck path. these given parameters. We can write D = Auu B1B2dd C, where the dis-tinguished u’s are the i-th and ( i + 1)-st up-steps, the two d’s are the i-th and ( i + 1)-st down-steps from the end, and the middle of D is between B1
and B2. Then A is a path from (0 , 0) to (2 i − r − 1, r − 1) not going below
y = 0. It is easy to see that there are (2i−r−1
i−1
) − (2i−r−1
i
) such paths A. By symmetry, there are (2i−s−1
i−1
) − (2i−s−1
i
) possibilities for C.Now we count the possibilities for B1 and B2. It can be checked that
f (k, r, h, ) counts the number of paths from (0 , r ) to (, h ) having exactly k
peaks, starting and ending with an up-step, and never going below y = 0. The fragment uB1 is a path from (2 i − r, r ) to ( n, h ) not going below y = 0, and ending with an up-step (since D has a peak in the middle). If we fix k
as the number of peaks of this fragment, then there are f (k, r, h, n − 2i + r)such paths uB1. Similarly, there are f (n − 2i − k, s, h, n − 2i − s) possibilities for B2d with n − 2i − k peaks. Summing over all possible values of k, h, r, s and i we obtain the expression in the theorem. 2
Using Corollary 2.12, we can prove that among the derangements of length
n, the number of 123-avoiding ones is at least the number of 132-avoiding ones. This inequality was conjectured by Mikl´ os B´ ona and Olivier Guibert.
Theorem 2.14 () For all n ≥ 4, s0
n
(132) < s 0
n
(123) .Proof. For n ≤ 12 the result can be checked by exhaustive enumeration of all derangements by computer. Let us assume that n ≥ 13. From part (2) of Corollary 2.12, we have that
s0
n
(123) ≥ Cn − 2Cn−1.
It is known that s0
n
(132) = Fn, the n-th Fine number . Therefore, the theorem will be proved if we show that
Fn < Cn − 2Cn−1 (2.2) 46 Chapter 2. Statistics in permutations avoiding one pattern
for n ≥ 13. Using the identity Fn = 1
2
∑n−2
i=0
( −1
2
)iCn−i, we get the inequality
Fn < 1
2
Cn − 1
4
Cn−1 + 1
8
Cn−2, which reduces (2.2) to showing that Cn >
7
2
Cn−1 + 1
4
Cn−2. This inequality certainly holds asymptotically, because Cn
grows like 1√π n− 3
2
4n as n tends to infinity, and it is not hard to see that in fact it holds for all n ≥ 13. 2
2.4 c, c’) 231 ∼f 312
Using the bijection
Sn(312) ←→ Dn
π 7 → ϕ(¯ π), Lemma 2.2 implies that
F312 (x, q, z ) = ∑
D∈D
xtd 0(D)qtd <0(D)z|D|.
To enumerate tunnels of depth 0, we will separate them according to their height. For every h ≥ 0, a tunnel at height h must have length 2( h + 1) in order to have depth 0. It is important to notice that if a tunnel of depth 0 of D corresponds to a decomposition D = AuBdC, then D has no tunnels of depth 0 in the part given by B. In other words, the projections on the
x-axis of all the tunnels of depth 0 of a given Dyck path are disjoint. This observation allows us to give a continued fraction expression for F312 (x, 1, z ).
Theorem 2.15 F312 (x, 1, z ) is given by the following continued fraction.
F312 (x, 1, z )= 1
1 − (x − 1) z − z
1 − (x − 1) z2 − z
1 − 2( x − 1) z3 − z
1 − 5( x − 1) z4 − z
. . .
,
where at the n-th level, the coefficient of (x − 1) zn+1 is the Catalan number
Cn.Proof. For every h ≥ 0, let td h
0
(D) be the number of tunnels of D of height
h and length 2( h + 1). Note that td 0(D) = ∑
h≥0
td h
0
(D). We will show 2.4. c, c’) 231 ∼f 312 47
now that for every h ≥ 1, the generating function for Dyck paths where x
marks the statistic td 00 + · · · + td h−10 is given by the continued fraction of the theorem truncated at level h, with the ( h + 1)-st level replaced with C(z). A Dyck path D can be written uniquely as a sequence of elevated Dyck paths, that is, as D = uA1d · · · uArd, where each Ai ∈ D . In terms of the GF C(z) = ∑
D∈D
z|D|, this translates into the equation C(z) = 1
1−zC(z)
. A tunnel of height 0 and length 2 (i.e., a hill) appears in D for each empty Ai.Therefore, the GF enumerating hills is
∑
D∈D
xtd 00(D)z|D| = 1
1 − z[x − 1 + C(z)] , (2.3) since an empty Ai has to be counted as x, not as 1. Let us enumerate simultaneously hills (as above), and tunnels of height 1 and length 4. The GF (2.3) can be written as 1
1 − z
[
x − 1 + 1
1 − zC(z)
] .
Combinatorially, this corresponds to expressing each Ai as a sequence uB1d
· · · uBsd, where each Bj ∈ D . Notice that since each uBj d starts at height 1, a tunnel of height 1 and length 4 is created by each Bj = ud in the decom-position. Thus, if we want x to mark also these tunnels, such a Bj has to be counted as xz , not z. The corresponding GF is
∑
D∈D
xtd 00(D)+td 10(D)z|D| = 1
1 − z
[
x − 1 + 1
1 − z[( x − 1) z + C(z)]
] .
Now it is clear how iterating this process indefinitely we obtain the continued fraction of the theorem. From the GF where x marks td 00 + · · · + td h−10 , we can obtain the one where x marks td 00 + · · · + td h
0
by replacing the C(z) at the lowest level with 1
1 − z[( x − 1) Chzh + C(z)] ,
to account for tunnels of height h and length 2( h + 1), which in the decom-position correspond to elevated Dyck paths at height h. 248 Chapter 2. Statistics in permutations avoiding one pattern
The same technique can be used to enumerate excedances in 312-avoiding permutations, which correspond to tunnels of negative depth in the Dyck path. Recall that
C<i (z) =
i−1
∑
j=0
Cj zj
denotes the series for the Catalan numbers truncated at degree i.
Theorem 2.16 F312 (x, q, z ) is given by the following continued fraction.
F312 (x, q, z ) = 1
1 − zK 0 + z
1 − zK 1 + z
1 − zK 2 + z
1 − zK 3 + z
. . .
,
where Kn = ( x − 1) Cnqnzn + ( q − 1) C<n (qz ) for n ≥ 0.
Note that the first values of Kn are
K0 = x − 1,
K1 = ( x − 1) qz + q − 1,
K2 = 2( x − 1) q2z2 + ( q − 1)(1 + qz ),
K3 = 5( x − 1) q3z3 + ( q − 1)(1 + qz + 2 q2z2).
Proof. We use the same decomposition as above, now keeping track of tun-nels of negative depth as well. For every h ≥ 0, let td h<0(D) be the num-ber of tunnels of D of height h and length less than 2( h + 1). Note that td <0(D) = ∑
h≥0
td h<0(D). To follow the same structure as in the previous proof, counting tunnels height by height, it will be convenient that at the
h-th step of the iteration, q marks not only tunnels of negative depth up to height h but also all the tunnels at higher levels. Denote by alltun >h (D) the number of tunnels of D of height strictly greater than h.We will show now that for every h ≥ 1, the generating function for Dyck paths where x marks the statistic td 00 + · · · + td h−10 and q marks td 0
<0
+
· · · + td h−1
<0
alltun >h −1 is given by the continued fraction of the theorem truncated at level h, with the ( h + 1)-st level replaced with C(qz ). The analogous to equation (2.3) is now
∑
D∈D
xtd 00(D)qtd 0
<0(D)+alltun >0(D)
z|D| = 1
1 − z[x − 1 + C(qz )] . (2.4) 2.4. c, c’) 231 ∼f 312 49
Indeed, decomposing D as uA1d · · · uArd, q counts all the tunnels that appear in any Ai, and whenever an Ai is empty we must mark it as x.Let us enumerate now tunnels of depth 0 and negative depth at both height 0 and height 1. Modifying (2.4) so that q no longer counts tunnels at height 1, we get
∑
D∈D
xtd 00(D)qtd 0
<0(D)+alltun >1(D)
z|D| = 1
1 − z
[
x − 1 + 1
1 − zC(qz )
] , (2.5) which corresponds to writing each Ai as Ai = uB1d · · · uBsd, and having q
count all tunnels in each Bj . Now, in order for x to mark tunnels of depth 0 at height 1, each Bj = ud , that in (2.5) is counted as qz , has to be now counted as xqz instead. Similarly, to have q mark tunnels of negative depth at height 1, we must count each empty Bj as q, not as 1. This gives us the following GF:
∑
D∈D
xtd 00(D)+td 10(D)qtd 0
<0(D)+td 1
<0(D)+alltun >1(D)
z|D|
= 1
1 − z
[
x − 1 + 1
1 − z[( x − 1) qz + q − 1 + C(qz )
] .
Iterating this process level by level indefinitely we obtain the continued frac-tion of the theorem. At each step, from the GF where x marks td 00 + · · · +td h−10 , and q marks td 0
<0
· · · + td h−1
<0
alltun >h −1, we can obtain the one where x marks td 00 + · · · + td h
0
and q marks td 0
<0
· · · + td h<0 + alltun >h by replacing the C(qz ) at the lowest level with 1
1 − z[( x − 1) Chqhzh + ( q − 1) C<h (qz ) + C(qz )] . (2.6) This change makes x account for tunnels of depth 0 at height h, which in the decomposition correspond to the Ch possible elevated Dyck paths of length 2( h + 1) when they occur at height h. It also makes q count tunnels of negative depth at height h, which in the decomposition correspond to elevated Dyck paths at height h of length less than 2( h + 1). The GF for these ones becomes qC<h (qz ) instead of C<h (qz ), since for every j < h , an elevated path uCd with C ∈ D j contributes to one extra tunnel of negative depth at height h, aside from the j tunnels of height more than h that it contains. 2
For 231-avoiding permutations we get the following GF. 50 Chapter 2. Statistics in permutations avoiding one pattern
Corollary 2.17 F231 (x, q, z ) is given by the following continued fraction.
F231 (x, q, z ) = 1
1 − zK ′
0
qz
1 − zK ′
1
qz
1 − zK ′
2
qz
1 − zK ′
3
qz
. . .
,
where K′
n
= ( x − q)Cnzn + (1 − q)C<n (z).
The first values of K′
n
are
K′
0
= x − q,
K′
1
= ( x − q)z + 1 − q,
K′
2
= 2( x − q)z2 + (1 − q)(1 + z),
K′
3
= 5( x − q)z3 + (1 − q)(1 + z + 2 z2).
Proof. By Lemma 1.2, we have that F231 (x, q, z ) = F312 (x/q, 1/q, qz ), so the expression follows from Theorem 2.16. 23
Simultaneous avoidance
After having studied in the previous section permutations avoiding one pat-tern of length 3, the next step is to consider permutations avoiding several patterns at the same time. In this chapter we study the distribution of the statistics ‘number of fixed points’ and ‘number of excedances’ on permuta-tions avoiding simultaneously two or more patterns. A systematic enumer-ation (with no statistics) of permutations avoiding any subset of patterns of length 3 was done in . Here we give refinements of these results, by enumerating the same permutations with respect to the statistics fp and exc. The main technique that we use are bijections between pattern-avoiding permutations and certain kinds of Dyck paths with some restrictions, in such a way that the statistics in permutations that we study correspond to statistics on Dyck paths that are easy to enumerate. In Section 3.1 we solve completely the case of permutations avoiding simul-taneously any two patterns of length 3, giving generating functions counting the number of fixed points and the number of excedances. For some par-ticular instances we can generalize the results, allowing one pattern of the pair to have arbitrary length. In Section 3.2 we give the analogous generat-ing functions for permutations avoiding simultaneously any three patterns of length 3 or more. Section 3.3 is concerned with the study of the distribution of these statistics in involutions avoiding any subset of patterns of length 3. The bijection ϕ defined in Section 2.1 will be one of our main tools in this chapter, together with its properties given in Lemma 2.2. We will also use repeatedly the array representation of a permutation π as described in Section 1.1.3, as well as the operations ¯ π,̂ π, and the lemmas proved in that section. 52 Chapter 3. Simultaneous avoidance
3.1 Double restrictions
In this section we consider simultaneous avoidance of any two patterns of length 3. Using Lemma 1.2, the pairs of patterns fall into the following equivalence classes.
a) {123 , 132 } ≈ { 123 , 213 }
b) {231 , 321 } ∼f b’) {312 , 321 }
c) {132 , 213 }
d) {231 , 312 }
e) {132 , 231 } ≈ { 213 , 231 } ∼f e’) {132 , 312 } ≈ { 213 , 312 }
f ) {132 , 321 } ≈ { 213 , 321 }
g) {123 , 231 } ∼f g’) {123 , 312 }
h) {123 , 321 }
In it is shown that the number of permutations in Sn avoiding any of the pairs in the classes a) , b) , b’) , c) , d) , e) , and e’) is 2 n−1, and that for the pairs in f ) , g) and g’) , the number of permutations avoiding any of them is (n
2
) + 1. The case h) is trivial because this pair is avoided only by permutations of length at most 4. In terms of generating functions, this means that when we substitute x =
q = 1 in FΣ(x, q, z ), where Σ is any of the pairs in the classes from a) to e’) ,we get
FΣ(1 , 1, z ) = ∑
n≥0
2n−1zn = 1 − z
1 − 2z .
If Σ is a pair from the classes f ) , g) , g’) , we get
FΣ(1 , 1, z ) = ∑
n≥0
(
(n
2
)
1) zn = 1 − 2z + 2 z2
(1 − z)3 .
3.1.1 a) {123 , 132 } ≈ { 123 , 213 }
Proposition 3.1
F{123 ,132 }(x, q, z ) = F{123 ,213 } (x, q, z )= 1 + xz + ( x2 − 4q)z2 + ( −3xq + q + q2)z3 + ( xq + xq 2 − 3x2q + 3 q2)z4
(1 − qz 2)(1 − 4qz 2) .3.1. Double restrictions 53
Proof. Consider the bijection ϕ : Sn(132) −→ D n described in Section 2.1. Part (4) of Proposition 2.1 says that the height of the Dyck path ϕ(π) is the length of the longest increasing subsequence of π. In particular, π ∈Sn(12 · · · (k + 1) , 132) if and only if ϕ(π) has height at most k. Thus, by Lemma 2.2, F{123 ,132 }(x, q, z ) can be written in terms of Dyck paths as
∑
D∈D ≤2
xct( D)qrt( D)z|D|. (3.1) Let us first find the univariate GF for paths of height at most 2 (with no statistics). Clearly, the GF for Dyck paths of height at most 1 is 1
1−z
, since such paths are just sequences of hills. A path D of height at most 2 can be written uniquely as D = uA1du A2d · · · uArd, where each Ai is a path of height at most 1. The GF for each uAid is z
1−z
. Thus,
∑
D∈D ≤2
z|D| = 1
1 − z
1−z
= 1 − z
1 − 2z = ∑
n≥0
2n−1zn.
In the rest of this proof, we assume that all Dyck paths that appear have height at most 2 unless otherwise stated. To compute (3.1), we will separate paths according to their height at the middle. Consider first paths whose height at the middle is 0. Splitting such a path at its midpoint we obtain a pair of paths of the same length. Thus, the corresponding GF is
∑
m≥0
2m−1zm · 2m−1qmzm = 1 − 3qz 2
1 − 4qz 2 , (3.2) since the number of right tunnels of such a path is the semilength of its right half.
Figure 3.1
A path of height 2 with a centered tunnel. Now we consider paths whose height at the middle is 1. It is easy to check that without the variables x and q, the GF for such paths is
z
1 − 4z2 . (3.3) 54 Chapter 3. Simultaneous avoidance
Let us first look at paths of this kind that have a centered tunnel. They must be of the form D = AuBdC where A, C ∈ D ≤2 have the same length, and B is a sequence of an even number of hills. Thus, their GF is
xz · 1
1 − qz 2 · 1 − 3qz 2
1 − 4qz 2 , (3.4) where x marks the centered tunnel, 1
1−qz 2
corresponds to the sequence of hills B, half of which give right tunnels, and the last fraction comes from the pair AC , which is counted as in (3.2). From (3.3) and (3.4) it follows that the univariate GF (with just variable z) for paths with height at the middle 1, not having a centered tunnel, is
z
1 − 4z2 − z(1 − 3z2)
(1 − z2)(1 − 4z2) = 2z3
(1 − z2)(1 − 4z2) .
By symmetry, in half of these paths, the tunnel of height 0 that goes across the middle is a right tunnel. Thus, the multivariate GF for all paths with height 1 at the middle is
xz (1 − 3qz 2)
(1 − qz 2)(1 − 4qz 2) + (q + 1) qz 3
(1 − qz 2)(1 − 4qz 2) . (3.5) Here the right summand corresponds to paths with no centered tunnel: the term ( q + 1) distinguishes whether the tunnel that goes across the middle is a right tunnel or not, and the other q’s mark tunnels completely contained in the right half. Paths with height 2 at the middle are easy to enumerate now. Indeed, they must have a peak ud in the middle, whose removal induces a bijection between these paths and paths with height 1 at the middle. This bijection preserves the number of right tunnels, and decreases the length and the number of centered tunnels by one. Thus, the GF for paths with height 2 at the middle is xz times expression (3.5). Adding up this GF for paths with height 2 at the middle, to the expressions (3.2) and (3.5) for paths whose height at the middle is 0 and 1 respectively, we obtain the desired expression for F{123 ,132 }(x, q, z ). 2
Let us see how the same technique used in this proof can be generalized to enumerate fixed points in Sn(132 , 12 · · · (k + 1)) for an arbitrary k ≥ 0.
Theorem 3.2 For k ≥ 0, let
Jk(x, z ) := F{132 ,12 ··· (k+1) }(x, 1, z ) = ∑
n≥0
∑
π∈S n(132 ,12 ··· (k+1))
xfp( π)zn.3.1. Double restrictions 55
Then the Jk’s satisfy the recurrence
Jk(x, z ) =
k
∑
`=0
Ik,(z)(1 + ( x − 1) zJ−1(x, z )) , (3.6)
where J−1(x, z ) := 0 , and Ik,` (z) is defined as
Ik,` (z) := ∑
n≥0
g2
k,`
(n)zn, where ∑
n≥0
gk,(n)zn = U( 1
2z
)
zU k+1 ( 1
2z
) ,
where Um are the Chebyshev polynomials of the second kind, defined in Sec-tion 1.3.4.
Before proving this theorem, let us show how to apply it to obtain the GFs Jk for the first few values of k. For k = 1, we have I1,0(z) = z
1−z2
,
I1,1(z) = 1
1−z2
, so
J1(x, z ) = 1 + xz
1 − z2 .
For k = 2, we get I2,0(z) = z2
1−4z2
, I2,1(z) = z
1−4z2
, I2,2(z) = 1 + z2
1−4z2
, thus
J2(x, z ) = 1 + xz + ( x2 − 4) z2 + (2 − 3x)z3 + (3 + 2 x − 3x2)z4
(1 − z2)(1 − 4z2) ,
which is the expression of Proposition 3.1 for q = 1. For k = 3, we obtain I3,0(z) = z3+z5
(1 −z2)(1 −7z2+z4)
, I3,1(z) = z2+z4
(1 −z2)(1 −7z2+z4)
,
I3,2(z) = z(1 −4z2+z4)
(1 −z2)(1 −7z2+z4)
, I3,2(z) = 1 + z2(1 −4z2+z4)
(1 −z2)(1 −7z2+z4)
, so
J3(x, z ) = [1+ xz +( x2−12) z2+( x3−11 x+2) z3+( −10 x2+4 x+45) z4+( −10 x3+4 x2+37 x−10) z5 +(25 x2 −22 x−52) z6 +(25 x3 −22 x2 −41 x+16) z7 +( −12 x2 +16 x+16) z8+( −12 x3+16 x2 +12 x−8) z9] / [(1 −z2)2(1 −4z2)(1 −7z2+z4)] .
Proof. As mentioned in the previous proof, ϕ induces a bijection between
Sn(132 , 12 · · · (k + 1)) and D≤k, the set of Dyck paths of height at most k.Thus, by Lemma 2.2,
Jk(x, z ) = ∑
D∈D ≤k
xct( D)z|D|.
In order to find a recursion for this GF, we are going to apply a trick that consists in consider Dyck paths where some centered tunnels are marked. 56 Chapter 3. Simultaneous avoidance
That is, we will count pairs ( D, S ) where D ∈ D ≤k and S is a subset of CT( D), the set of centered tunnels of D (S is the set of marked tunnels). In other words, we are considering Dyck paths where some centered tunnels (namely those in S) are marked. Each such pair is given weight ( x−1) |S|z|D|,so that for a fixed D, the sum of weights of all pairs ( D, S ) will be
∑
S⊆CT( D)
(x − 1) |S|z|D| = (( x − 1) + 1) |CT( D)|z|D| = xct( D)z|D|,
which is precisely the weight that D has in Jk(x, z ).
kk−l udACB
Figure 3.2 A path of height k with two marked centered tunnels. If D ∈ D ≤k has some marked centered tunnel, consider the decomposition
D = AuBdC given by the longest marked tunnel (i.e., all the other marked tunnels are inside the part B of the path). Let be the distance between this tunnel and the line y = k (see Figure 3.2). Equivalently, A ends at height k −, the same height where C begins. Then, B is an arbitrary Dyck path of height at most − 1 with possibly some marked centered tunnels, so its corresponding GF is J−1(x, z ) (with the convention J−1(x, z ) := 0, since for = 0 there is no such B). Giving weight ( x − 1) to the tunnel that determines our decomposition, we have that the part uBd of the path contributes ( x − 1) zM−1(x, z ) to the GF. Now we look at the GF for the part A of the path. Let gk,(n) be the number of paths from (0 , 0) to ( n, k −) staying always between y = 0 and y = k. A path of this type can be decomposed uniquely as A =
EkuEk−1u · · · uE+1 uE, where each Ei ∈ D ≤i. The GF of Dyck paths of height at most i is
Ji(1 , z ) = Ui( 1
2√z
)
√zU i+1 ( 1
2√z
) ,
as shown for example in . Let w = √z, which is the weight of a single step of a path, and let ˜Rk,` (w) := ∑
n≥0
gk,` (n)wn. From the above decomposition 3.1. Double restrictions 57
of A,
˜Rk,(w) = Jk(1 , w 2)wJ k−1(1 , w 2 )w · · · wJ(1 , w 2) = U`( 1
2w
)
wU k+1 ( 1
2w
) .
The part C of the path D, flipped over a vertical line, can be regarded as a path with the same endpoints as A, since it must have the same length and end at the same height k − . Thus, the GF for pairs ( A, C ) of paths of the same length from height 0 to height k − and not going above y = k is ∑
n≥0
g2
k,`
(n)zn = Ik,(z). Hence, the GF for paths D ∈ D ≤k having the longest marked centered tunnel at height k − is Ik,(z)( x − 1) zM−1(x, z ). If D has no marked tunnel, decompose it as D = AC where A and C have the same length. Letting k − be again the height where A ends and C begins, the situation is the same as above but without any contribution coming from the central part of D. The parameter can take any value between 0 and k.Thus, summing over all possible decompositions of D, we get
Jk(x, z ) =
k
∑
`=0
Ik,(z)(1 + ( x − 1) zJ−1(x, z )) .
2
3.1.2 b, b’) {231 , 321 }∼ f {312 , 321 }
Proposition 3.3
F{312 ,321 } (x, q, z ) = 1 − qz
1 − (x + q)z + ( x − 1) qz 2 . (3.7)
Proof. The length of the longest decreasing subsequence of π equals the height of the Dyck path ϕ(¯ π). In particular, we have a bijection
Sn(312 , 321) ←→ D≤2
n
π 7 → ϕ(¯ π)Thus, by Lemma 2.2,
F{312 ,321 }(x, q, z ) = ∑
D∈D ≤2
xtd 0(D)qtd <0(D)z|D|.58 Chapter 3. Simultaneous avoidance
But the only tunnels of depth 0 that a Dyck path of height at most 2 can have are hills, and the only tunnels of negative depth that it can have are peaks at height 2. A path D ∈ D ≤2 can be written uniquely as D =
uA1du A2d · · · uAr d, where each Ai is a (possibly empty) sequence of hills. An empty Ai creates a tunnel of depth 0 in D, so it contributes as x. An Ai
of length 2 j > 0 contributes as qj zj , since it creates j peaks at height 2 in
D. Thus,
F{312 ,321 }(x, q, z ) = 1
1 − z
(
x + qz
1 − qz
) ,
which is equivalent to (3.7). 2
Corollary 3.4
F{231 ,321 } (x, q, z ) = 1 − z
1 − (x + 1) z + ( x − q)z2 .
Proof. By Lemma 1.2, F{231 ,321 } (x, q, z ) = F{312 ,321 } (x/q, 1/q, qz ). 2
As in the previous section, these results can be generalized to the case when instead of the pattern 321 we have a decreasing pattern ( k + 1) k · · · 21 of arbitrary length. For i, h ≥ 0, let C≤hi be the number of Dyck paths of length 2 i and height at most h. As mentioned before, it is known that
∑
i≥0
C≤hi zi = Uh( 1
2√z
)
√zU h+1 ( 1
2√z
) ,
where Um are the Chebyshev polynomials of the second kind. Let
C≤h<i (z) :=
i−1
∑
j=0
C≤hj zj .
The following theorem deals with fixed points and excedances in the set
Sn(312 , (k + 1) k · · · 1) for any k ≥ 0.
Theorem 3.5 Let C≤hi = |D ≤hi | and C≤h<i (z) be defined as above. Then, for
k ≥ 0,
F{312 ,(k+1) k··· 1}(x, q, z ) = Ak
0
(x, q, z ),3.1. Double restrictions 59
where Aki is recursively defined by
Aki (x, q, z ) =
1
1 − z[( x − 1) C≤k−i−1
i
qizi + ( q − 1) C≤k−i−1
<i
(qz ) + Aki+1 (x, q, z )]
if i < k ,
1 if i = k.
For example, for k = 2 we obtain Proposition 3.3, and for k = 3 we get
F{312 ,4321 }(x, q, z ) = 1
1 − z
x − 1 + 1
1 − z
[
(x − 1) qz + q − 1 + 1
1−qz
]
=1 − 2qz + ( q2 − xq )z2 + ( xq 2 − q2)z3
1 − (x + 2 q)z + ( xq + q2 − q)z2 + ( x2q − xq )z3 + ( −x2q2 + 2 xq 2 − q2)z4 .
Proof. It is analogous to the proof of Theorem 2.16, with the only difference that here we consider only those paths that do not go above the line y = k.
2
Making the appropriate substitutions in the statement of Theorem 3.5, we obtain an expression for the generating function F{231 ,(k+1) k··· 1}(x, q, z ) =
F{312 ,(k+1) k··· 1}(x/q, 1/q, qz ).
3.1.3 c) {132 , 213 }
Proposition 3.6
F{132 ,213 }(x, q, z )= 1 − (1 + q)z − 2qz 2 + 4 q(1 + q)z3 − (xq 2 + xq + 5 q2)z4 + 2 xq 2z5
(1 − z)(1 − xz )(1 − qz )(1 − 4qz 2) .
Proof. We use again the bijection ϕ : Sn(132) −→ D n. From its description given in Section 2.1, it is not hard to see that a permutation π ∈ S n(132) avoids 213 if and only if all the valleys of the corresponding Dyck path ϕ(π)have their lowest point on the x-axis. A path with such property can be described equivalently as a sequence of pyramids . Denote by Pyr n ⊆ D n the set of sequences of pyramids of length 2 n, and let Pyr := ⋃
n≥0
Pyr n. We 60 Chapter 3. Simultaneous avoidance
have just seen that ϕ restricts to a bijection between Sn(132 , 213) and Pyr n.By Lemma 2.2, we can write F{132 ,213 } (x, q, z ) as
∑
D∈P yr
xct( D)qrt( D)z|D|.
Since for each n ≥ 1 there is exactly one pyramid of length 2 n, the univariate GF of sequences of pyramids is just ∑
D∈P yr
z|D| = 1
1−z
1−z
= 1−z
1−2z
= 1 +
∑
n≥1
2n−1zn.Let us first consider elements of Pyr that have height 0 in the middle (equiv-alently, the two central steps are du ). Each one of their halves is a sequence of pyramids, both of the same length. They have no centered tunnels, and the number of right tunnels is given by the semilength of the right half. Thus, their multivariate GF is 1 + ∑
m≥1
4m−1qmz2m = 1 + qz 2
1 − 4qz 2 . (3.8) Now we count elements of Pyr whose two central steps are ud . They are obtained uniquely by inserting a pyramid of arbitrary length in the middle of a path with height 0 at the middle. The tunnels created by the inserted pyramid are all centered tunnels, so the corresponding GF is
xz
1 − xz
(
1 + qz 2
1 − 4qz 2
)
. (3.9)
Figure 3.3
A sequence of pyramids. It remains to count the elements of Pyr that in the middle have neither a peak nor a valley. From a non-empty sequence of pyramids with height 0 in the middle, if we increase the size of the leftmost pyramid of the right half by an arbitrary number of steps, we obtain a sequence of pyramids whose two central steps are uu . Reciprocally, by this procedure every such sequence of pyramids can be obtained in a unique way from a sequence of pyramids with height 0 in the middle. Thus, the GF for the elements of Pyr whose two central steps are uu is
qz
1 − qz · qz 2
1 − 4qz 2 . (3.10) 3.1. Double restrictions 61
By symmetry, the GF for the elements of Pyr whose two central steps are
dd is z
1 − z · qz 2
1 − 4qz 2 , (3.11) where the difference with respect to 3.10 is that now the pyramid across the middle does not create right tunnels. Adding up (3.8), (3.9), (3.10) and (3.11) we get the desired GF. 2
3.1.4 d) {231 , 312 }
Proposition 3.7
F{231 ,312 }(x, q, z ) = 1 − qz 2
1 − xz − 2qz 2 .
Proof. We have shown in the proof of Proposition 3.6 that ϕ induces a bi-jection between Sn(132 , 213) and Pyr n, the set of sequences of pyramids of length 2 n. Composing it with the complementation operation, we get a bijec-tion π 7 → ϕ(¯ π) between Sn(231 , 312) and Pyr n. Together with Lemma 2.2, this allows us to express F{231 ,312 } (x, q, z ) as
∑
D∈P yr
xtd 0(D)qtd <0(D)z|D|.
All that remains is to observe how many tunnels of zero and negative depth are created by a pyramid according to its size. A pyramid of odd semilength 2m + 1 creates one tunnel of depth 0 and m tunnels of negative depth. A pyramid of even semilength 2 m creates only m tunnels of negative depth. Thus, we have that
F{231 ,312 } (x, q, z ) = 1
1 − xz
1 − qz 2 − qz 2
1 − qz 2
,
which equals the expression above. 2
3.1.5 e, e’) {132 , 231 } ≈ { 213 , 231 } ∼ f {132 , 312 } ≈ {213 , 312 }
Proposition 3.8
F{132 ,231 } (x, q, z ) = F{213 ,231 } (x, q, z ) = 1 − z − qz 2 + xqz 3
(1 − xz )(1 − z − 2qz 2) .62 Chapter 3. Simultaneous avoidance
Proof. As usual, we use the bijection ϕ : Sn(132) −→ D n. Now we are interested in how the condition that π avoids 231 is reflected in the Dyck path
ϕ(π). It is easy to see from the description of ϕ and ϕ−1 in Section 2.1 that π
is 231-avoiding if and only if ϕ(π) does not have any two consecutive up-steps after the first down-step (equivalently, all the non-isolated up-steps occur at the beginning of the path). Let En ⊆ D n be the set of Dyck paths with this condition, and let E := ⋃
n≥0
En. Then, ϕ induces a bijection between
Sn(132 , 231) and En. By Lemma 2.2, F{132 ,231 }(x, q, z ) can be written as
∑
D∈E
xct( D)qrt( D)z|D|.
Figure 3.4
A path in E with a peak in the middle and two bottom tunnels. If D ∈ E , centered tunnels of D can appear only in the following two places. There can be a centered tunnel produced by a peak in the middle of D.All the other centered tunnels of D must have their endpoints in the initial ascending run and the final descending run of D (that is, in their corre-sponding decomposition D = AuBdC, A is a sequence of up-steps and C
is a sequence of down-steps). For convenience we call this second kind of tunnels bottom tunnels. All the right tunnels of D come from peaks on the right half. It is an exercise to check that the number of paths in En having a peak in the middle and r peaks on the right half is (n−r−1
r
)2r−1 if r ≥ 1, and 1 if r = 0. Similarly, the number of paths in En with no peak in the middle and r peaks on the right half is (n−rr
)2r−1 if r ≥ 1, and 0 if r = 0. Let us ignore for the moment the bottom tunnels. For peaks in the middle and right tunnels we 3.1. Double restrictions 63
have the following GF.
Q(x, q, z ) := ∑
D∈E\E 0
x#{peaks in the middle of D }qrt( D)z|D|
= ∑
n≥1
bn/ 2c
∑
r=1
(n − rr
)
2r−1qr + x
1 +
b(n−1) /2c
∑
r=1
(n − r − 1
r
)
2r−1qr
zn
= xz + ( q − x)z2 − xqz 3
(1 − z)(1 − z − 2qz 2) . (3.12) Now, to take into account all centered tunnels, we use that every D ∈ E can be written uniquely as D = ukD′dk, where k ≥ 0 and D′ ∈ E has no bottom tunnels. The GF for elements of E that do have bottom tunnels, where x
marks peaks in the middle, is xz +zQ (x, q, z ) (the term xz is the contribution of the path ud ). Hence, the sought GF where x marks all centered tunnels is
F{132 ,231 } (x, q, z ) = 1
1 − xz [1 + Q(x, q, z ) − xz − zQ (x, q, z )] = 1 + 1 − z
1 − xz Q(x, q, z ),
which together with (3.12) implies the proposition. 2
Corollary 3.9
F{132 ,312 } (x, q, z ) = F{213 ,312 } (x, q, z ) = 1 − qz − qz 2 + xqz 3
(1 − xz )(1 − qz − 2qz 2) .
Proof. Lemma 1.2 gives us F{132 ,312 }(x, q, z ) = F{132 ,231 }(x/q, 1/q, qz ). 2
3.1.6 f) {132 , 321 } ≈ { 213 , 321 }
Proposition 3.10
F{132 ,321 } (x, q, z ) = F{213 ,321 }(x, q, z ) = 1 − (1 + q)z + 2 qz 2
(1 − z)(1 − xz )(1 − qz ) .
Proof. We saw in part (6) of Proposition 2.1 that the number of peaks of the Dyck path ϕ(π) equals the length of the longest decreasing subsequence of
π. In particular, π is 321-avoiding if and only if ϕ(π) has at most two peaks. 64 Chapter 3. Simultaneous avoidance
By Lemma 2.2, F{132 ,321 } (x, q, z ) = ∑ xct( D)qrt( D)z|D|, where the sum is over Dyck paths D with at most two peaks. Clearly, such a path can be uniquely written as D = ukD′dk, where k ≥ 0 and D′ is either empty or a pair of adjacent pyramids (see Figure 3.5). Therefore,
F{132 ,321 } (x, q, z ) = 1
1 − xz
(
1 + z
1 − z · qz
1 − qz
)
,
since centered tunnels are produced by the steps outside D′, and right tunnels are created by the right pyramid of D′. 2
Figure 3.5 A path with two peaks. This case can be generalized to the situation when instead of 321 we have a decreasing pattern of arbitrary length. Observe that by Lemma 1.2,
F{132 ,(k+1) k··· 21 }(x, q, z ) = F{213 ,(k+1) k··· 21 }(x, q, z ) for all k.
Theorem 3.11
∑
k≥0
F{132 ,(k+1) k··· 21 }(x, q, z ) pk
= 2(1 + xz (p − 1))
(1 − p)[1 + (1 + q − 2x)z − qz 2(p − 1) 2 + √f1(q, z )] ,
where f1(q, z ) = 1 − 2(1 + q)z + [(1 − q)2 − 2q(p − 1)( p + 3)] z2 − 2q(1 + q)( p −
1) 2z3 + q2(p − 1) 4z4.Proof. We use again the fact from Proposition 2.1 that the number of peaks of ϕ(π) equals the length of the longest decreasing subsequence of
π. Thus, ϕ induces a bijection between Sn(132 , (k + 1) k · · · 21) and the sub-set of Dn of paths with at most k peaks. This implies that we can express ∑
k≥0
F{132 ,(k+1) k··· 21 }(x, q, z ) pk as 1
1 − p
∑
D∈D
xct( D)qrt( D)p#{peaks of D}z|D|.3.1. Double restrictions 65
The result now follows from Theorem 4.10 and the expression for the gener-ating function ∑ xct( D)qrt( D)p#{peaks of D}z|D| that we will give in its proof in the next chapter. 2
3.1.7 g, g’) {123 , 231 }∼ f {123 , 312 }
Proposition 3.12
F{123 ,312 }(x, q, z ) = 1 + xz + ( x2 − 2q)z2 + ( −x2q + xq 2 + 3 q2)z4 + 3 q3z5 − q3z6 − 4q4z7 − 2xq 4z8
(1 − qz 2)3(1 − q2z3) .
Proof. We saw in the proof of Proposition 3.10 that ϕ induces a bijection between Sn(132 , 321) and the set of paths in Dn with at most two peaks. Composing it with the complementation operation, we get a bijection π 7 →
ϕ(¯ π) between Sn(123 , 312) and such set of Dyck paths. Using Lemma 2.2, we can write F{123 ,312 }(x, q, z ) = ∑ xtd 0(D)qtd <0(D)z|D|, where the sum is over Dyck paths D with at most two peaks. Again, such a D can be uniquely written as D = ukD′dk, where k ≥ 0 and D′ is either empty or a pair of adjacent pyramids, i.e., D′ = uidiuj dj with i, j ≥ 1. The idea is to consider cases depending on the relations among i, j and k.To enumerate Dyck paths with at most two peaks with respect to td 0 and td <0, it is important to look at where the tunnels of depth 0 and depth 1 occur. For convenience in this proof, we call such tunnels frontier tunnels ,since they determine where tunnels of negative depth are: above them all tunnels have negative depth, and below them tunnels have positive depth. There are four possibilities according to where the frontier tunnels of D occur in the decomposition above: (1) outside D′,(2) inside one of the pyramids of D′,(3) inside both pyramids of D′,(4) D has no frontier tunnel. Figure 3.6 shows an example of each of the four cases. The frontier tunnels (whose depth is 0 in this example) are drawn with a solid line, while the dotted lines are the tunnels of negative depth. 66 Chapter 3. Simultaneous avoidance
(1) (2) (3) (4)
Figure 3.6 Four possible locations of the frontier tunnels. Note that in case (4) the tunnels of negative depth are exactly those in
D′. We show as an example how to find the GF in case (1). In this case, the frontier tunnel T gives a decomposition D = AuBdC where A = um,
C = dm, m ≥ 0, and B is a Dyck path with at most two peaks, of semilength
|B| = m if depth( T ) = 0, and |B| = m + 1 if depth( T ) = 1. It follows from Proposition 3.10 that the GF for Dyck paths with at most two peaks is
1−2z+2 z2
(1 −z)3
. In the situation where depth( T ) = 0, we have that |D| = 2 |B| + 1 and td <0(D) = |B|. Thus, the corresponding GF is
xz · 1 − 2qz 2 + 2 q2z4
(1 − qz 2)3 .
Similarly, in the situation where depth( T ) = 1, we have that |D| = 2 |B| and td <0(D) = |B|, thus the corresponding GF is 1 − 2qz 2 + 2 q2z4
(1 − qz 2)3 .
The other cases are similar. Adding up the GFs obtained in each case, we get the desired expression for F{123 ,312 }(x, q, z ). 2
Corollary 3.13
F{123 ,231 }(x, q, z ) = 1 + xz + ( x2 − 2q)z2 + ( −x2q + xq + 3 q2)z4 + 3 q2z5 − q3z6 − 4q3z7 − 2xq 3z8
(1 − qz 2)3(1 − qz 3) .
Proof. By Lemma 1.2, we have F{123 ,231 } (x, q, z ) = F{123 ,312 } (x/q, 1/q, qz ).
23.2. Triple restrictions 67
3.1.8 h) {123 , 321 }
Proposition 3.14
F{123 ,321 } (x, q, z ) = 1 + xz + ( x2 + q)z2 + (2 xq + q2 + q)z3 + 4 q2z4.
Proof. By a well-known result of Erd¨ os and Szekeres, any permutation of length at least 5 contains an occurrence of either 123 or 321. This reduces the problem to counting fixed points and excedances in permutations of length at most 4, which is trivial. 2
3.2 Triple restrictions
Here we consider simultaneous avoidance of any three patterns of length 3. Applying Lemma 1.2, the triplets of patterns fall into the following equiva-lence classes.
a) {123 , 132 , 213 }
b) {231 , 312 , 321 }
c) {123 , 132 , 231 } ≈ { 123 , 213 , 231 }∼ f c’) {123 , 132 , 312 } ≈ { 123 , 213 , 312 }
d) {132 , 231 , 321 } ≈ { 213 , 231 , 321 }∼ f d’)
{132 , 312 , 321 } ≈ { 213 , 312 , 321 }
e) {132 , 213 , 231 } ∼f e’) {132 , 213 , 312 }
f ) {132 , 231 , 312 } ≈ { 213 , 231 , 312 }
g) {123 , 231 , 312 }
h) {132 , 213 , 321 }
i) {123 , 132 , 321 } ≈ { 123 , 213 , 321 }
j) {123 , 231 , 321 } ∼f j’) {123 , 312 , 321 }
It is known that the number of permutations in Sn avoiding the triplets in the classes a) and b) is the Fibonacci number Fn+1 . The number of permutations avoiding any of the triplets in the classes c) , c’) , d) , d’) , e) ,
e’) , f ) , g) and h) is n. The cases of the triplets i) , j) and j’) are trivial, because they are avoided only by permutations of length at most 4. In terms of generating functions, when we substitute x = q = 1 in FΣ(x, q, z )where Σ is a triplet from one of the classes between a) and g) , we get
FΣ(1 , 1, z ) = ∑
n≥0
Fn+1 zn = 1
1 − z − z2 .68 Chapter 3. Simultaneous avoidance
If Σ is any triplet from the classes between c) and h) , we get
FΣ(1 , 1, z ) = ∑
n≥0
nz n = 1 − z + z2
(1 − z)2 .
The following theorem gives all the generating functions of permutations avoiding any triplet of patterns of length 3.
Theorem 3.15 a)
F{123 ,132 ,213 } (x, q, z ) = 1 + xz + ( x2 − q)z2 + ( −xq + q2 + q)z3 − x2qz 4
(1 + qz 2)(1 − 3qz 2 + q2z4)
b)
F{231 ,312 ,321 } (x, q, z ) = 1
1 − xz − qz 2
c)
F{123 ,132 ,231 } (x, q, z ) = F{123 ,213 ,231 } (x, q, z )= 1 + xz + ( x2 − q)z2 + qz 3 + ( −x2q + xq + q2)z4
(1 − qz 2)2
c’)
F{123 ,132 ,312 } (x, q, z ) = F{123 ,213 ,312 } (x, q, z )= 1 + xz + ( x2 − q)z2 + q2z3 + ( −x2q + xq 2 + q2)z4
(1 − qz 2)2
d)
F{132 ,231 ,321 } (x, q, z ) = F{213 ,231 ,321 } (x, q, z ) = 1 − z + qz 2
(1 − z)(1 − xz )
d’)
F{132 ,312 ,321 } (x, q, z ) = F{213 ,312 ,321 } (x, q, z ) = 1 − qz + qz 2
(1 − xz )(1 − qz )
e)
F{132 ,213 ,231 } (x, q, z )= 1 − z − qz 2 + 2 qz 3 + ( −x2q + q2 − xq )z4 + ( x2q − 2q2)z5 + xq 2z6
(1 − z)(1 − xz )(1 − qz 2)23.2. Triple restrictions 69
e’)
F{132 ,213 ,312 } (x, q, z )= 1 − qz − qz 2 + 2 q2z3 + ( −x2q − xq 2 + q2)z4 + ( x2q2 − 2q3)z5 + xq 3z6
(1 − xz )(1 − qz )(1 − qz 2)2
f )
F{132 ,231 ,312 } (x, q, z ) = F{213 ,231 ,312 } (x, q, z ) = 1 + xqz 3
(1 − xz )(1 − qz 2)
g)
F{123 ,231 ,312 } (x, q, z ) = 1 + xz + ( x2 − q)z2 + xqz 3 + q2z4
(1 − qz 2)2
h)
F{132 ,213 ,321 } (x, q, z ) = 1 − (1 + q)z + 2 qz 2 − xqz 3
(1 − z)(1 − xz )(1 − qz )
i)
F{123 ,132 ,321 } (x, q, z ) = F{123 ,213 ,321 } (x, q, z )= 1 + xz + ( x2 + q)z2 + ( xq + q2 + q)z3 + q2z4
j)
F{123 ,231 ,321 } (x, q, z ) = 1 + xz + ( x2 + q)z2 + (2 xq + q)z3 + q2z4
j’)
F{123 ,312 ,321 } (x, q, z ) = 1 + xz + ( x2 + q)z2 + (2 xq + q2)z3 + q2z4
Proof. Throughout this proof we will use the bijection ϕ : Sn(132) −→ D n
described in Section 2.1.
a) As in the proof of Proposition 3.1, we have that π ∈ S n(132) avoids 123 if and only if the Dyck path ϕ(π) has height at most 2. Similarly, from the proof of Proposition 3.6, π avoids 213 if and only if ϕ(π) is a sequence of pyramids. Thus, ϕ induces a bijection between Sn(123 , 132 , 213) 70 Chapter 3. Simultaneous avoidance
and Pyr ≤2
n
:= Pyr ≤2 ∩ D n, where Pyr ≤2 denotes the set of sequences of pyramids of height at most 2. By Lemma 2.2,
F{123 ,132 ,213 } (x, q, z ) = ∑
D∈P yr ≤2
xct( D)qrt( D)z|D|.
To count centered and right tunnels, we distinguish cases according to which steps are the middle steps of D. A path in Pyr ≤2 of height 0 at the middle can be split in two elements of Pyr ≤2 of equal length, only the right one producing right tunnels. Since the number of D ∈ P yr ≤2
n
is Fn+1 , the GF for paths of height 0 at the middle is
∑
n≥0
F 2
m+1
qmz2m = 1 − qz 2
(1 + qz 2)(1 − 3qz 2 + q2z4) .
Multiplying this expression by xz (resp. by x2z2) we obtain the GF for paths in Pyr ≤2 having in the middle a centered pyramid of height 1 (resp. of height 2).
AB
Figure 3.7 A sequence of pyramids of height at most 2. Paths D ∈ P yr ≤2 whose two middle steps are dd can be written as D =
Auudd B, where A, B ∈ P yr ≤2 and |B| = |A| + 1 (see Figure 3.7). Thus, the corresponding GF is
∑
n≥1
FmFm+1 qmz2m+1 = qz 3
(1 + qz 2)(1 − 3qz 2 + q2z4) .
By symmetry, multiplying this expression by q we get the GF for paths whose two middle steps are uu .Adding up all the cases, we get the desired GF
F{123 ,132 ,213 } (x, q, z )= (1 + xz + x2z2)(1 − qz 2)
(1 + qz 2)(1 − 3qz 2 + q2z4) + (q + q2)z3
(1 + qz 2)(1 − 3qz 2 + q2z4) .
b) Using the same reasoning as in a) , we have that π 7 → ϕ(¯ π) induces a bijection between Sn(231 , 312 , 321) and Pyr ≤2
n
. Now, Lemma 2.2 implies 3.2. Triple restrictions 71
that
F{231 ,312 ,321 } (x, q, z ) = ∑
D∈P yr ≤2
xtd 0(D)qtd <0(D)z|D|.
Each pyramid of height 1 produces a tunnel of depth 0, and each pyramid of height 2 creates a tunnel of negative depth. Therefore,
F{231 ,312 ,321 } (x, q, z ) = 1
1 − xz − qz 2 .
c) We saw in the proof of Proposition 3.8 that π ∈ S n(132) avoids 231 if and only if the Dyck path ϕ(π) does not have any two consecutive up-steps after the first down-step. Therefore, ϕ induces a bijection between
Sn(123 , 132 , 312), and paths in Dn with the above condition and height at most 2. Such paths (except the empty one) can be expressed uniquely as
D = uAdB, where A and B are sequences of hills (i.e, they have the form (ud )k for some k ≥ 0). Lemma 2.2 reduces the problem to enumerating centered tunnels and right tunnels on these paths. If B is empty, D = uAd has a centered tunnel at height 0. The contribution of paths of this kind to our GF is xz
1−qz 2
for |A| even, and x2z2
1−qz 2
for |A| odd. Assume now that |A| < |B|, so that A is within the left half of D = uAdB.If the middle of D is at height 0, then D is determined by the length of A
and the number of hills in B to the left of the middle. Thus, the contribution of this subset to the GF is
qz 2
(1 − qz 2)2 .
Multiplying this expression by xz gives the GF for paths whose midpoint is on top of a hill of B.
ABud
Figure 3.8 An example with |A| = 3 and |B| = 2. It remains the case in which |A| ≥ | B| > 0. If |A| − | B| is even, the contri-bution of these paths to the GF is
z · qz 2
1 − qz 2 · 1
1 − qz 2 ,72 Chapter 3. Simultaneous avoidance
where the last factor counts how larger A is than B. If |A| − | B| is odd, the corresponding GF is
z · qz 2
1 − qz 2 · xz
1 − qz 2 ,
since in this case there is a centered tunnel of height 1 inside A (see Fig-ure 3.8). Summing up all the cases, we get
F{123 ,132 ,231 } (x, q, z ) = 1 + xz + x2z2
1 − qz 2 + (1 + xz )qz 2
(1 − qz 2)2 + qz 3(1 + xz )
(1 − qz 2)2 .
c’) Lemma 1.2 implies that F{123 ,132 ,312 } (x, q, z ) = F{123 ,132 ,231 } ( x
q
, 1/q, qz ), so the formula follows from part c) .
d) As in the proof of Proposition 3.8, we use that π ∈ S n(132) avoids 231 if and only if the Dyck path ϕ(π) does not have any two consecutive up-steps after the first down-step. Besides, as in Proposition 3.10, π ∈ S n(132) avoids 321 if and only if ϕ(π) has at most two peaks. Thus, π ∈ S n(132 , 231 , 321) if and only if ϕ(π) ∈ D n has the form ukBdk, where B is either empty or is a pair of pyramids, the second of height 1. Fixed points and excedances of
π are mapped to centered tunnels and right tunnels of ϕ(π) respectively, by Lemma 2.2. Thus, F{123 ,132 ,312 } (x, q, z ) equals the GF enumerating centered and right tunnels in these paths. If B is not empty, the contribution of the first pyramid is z
1−z
, and the second pyramid contributes qz . Centered tunnels come from the steps outside B.Hence,
F{123 ,132 ,312 }(x, q, z ) = 1
1 − xz
(
1 + z
1 − z · qz
)
.
d’) It follows from part d) and Lemma 1.2.
e) Let π ∈ S n(132). We have seen that the condition that π avoids 213 translates into ϕ(π) being a sequence of pyramids. The additional restriction of π avoiding 231 implies that all but the first pyramid of the sequence
ϕ(π) must have height 1. Thus, by Lemma 2.2, F{132 ,213 ,231 }(x, q, z ) can be obtained enumerating centered and right tunnels in paths of the form
D = AB , where A is any pyramid and B is a sequence of hills. 3.2. Triple restrictions 73
The contribution of such paths when B is empty is just 1
1−xz
. Assume now that B is not empty. If |A| > |B|, the corresponding contribution is
qz 2
1 − qz 2 · z
1 − z ,
where the second factor counts how larger A is than B. It remains the case
|A| ≤ | B|, in which A is within the left half of D. If the middle of D is at height 0, then D is determined by the length of A and the number of hills in B to the left of the middle. Thus, the contribution of this subset to the GF is qz 2
(1 − qz 2)2 .
Multiplying this expression by xz gives the GF for paths whose midpoint is on top of a hill of B.
AB
Figure 3.9
A pyramid followed by a sequence of hills. Summing all this up, we get
F{132 ,213 ,231 } (x, q, z ) = 1
1 − xz + qz 3
(1 − z)(1 − qz 2) + (1 + xz )qz 2
(1 − qz 2)2 .
e’) It follows from part e) and Lemma 1.2.
f ) Reasoning as in the proof of e) , we see that π 7 → ϕ(¯ π) induces a bijec-tion between Sn(123 , 231 , 312) and the subset of paths in Dn consisting of a pyramid followed by a sequence of hills. By Lemma 2.2, it is enough to enumerate these paths according to the statistics td 0 and td <0. If the path is nonempty, the first pyramid contributes xz
1−qz 2
if it has odd size (since then it contains a tunnel of depth 0) and qz 2
1−qz 2
if it has even size. The sequence of hills contributes 1
1−xz
. Therefore,
F{132 ,231 ,312 } (x, q, z ) = 1 + xz + qz 2
1 − qz 2 · 1
1 − xz .
g) Let π ∈ S n(132). We have seen that π avoids 213 if and only if ϕ(π) is a sequence of pyramids, and that π avoids 321 if and only if ϕ(π) has at most 74 Chapter 3. Simultaneous avoidance
two peaks. In other words, ϕ induces a bijection between Sn(132 , 213 , 321) and the subset of paths in Dn that are a sequence of at most two pyra-mids. Composing with the complementation operation, we have that π ∈Sn(123 , 231 , 312) if and only if ϕ(¯ π) is in that subset. Now, Lemma 2.2 im-plies that F{123 ,231 ,312 } can be obtained enumerating sequences of at most 2 pyramids according to td 0 and td <0. Each pyramid contributes xz
1−qz 2
if it has odd size and qz 2
1−qz 2
if it has even size. Thus,
F{123 ,231 ,312 } (x, q, z ) = 1 + xz + qz 2
1 − qz 2 +
( xz + qz 2
1 − qz 2
)2
.
h) We have shown in the proof of g) that π ∈ S n(132 , 213 , 321) if and only if ϕ(π) is a sequence of at most two pyramids. Using Lemma 2.2, it is enough to enumerate centered tunnels and right tunnels in such paths. The contribution of paths with exactly two pyramids is
z
1 − z · qz
1 − qz ,
since only the one on the right gives right tunnels. Centered tunnels appear when there is only one pyramid. Thus we obtain
F{132 ,213 ,321 } (x, q, z ) = 1
1 − xz + qz 2
(1 − z)(1 − qz ) .
i, j, j’) These cases are trivial because only permutations of length at most 4 can avoid 123 and 321 simultaneously. 2
After having studied all the cases of double and triple restrictions, the next step is to consider restrictions of higher multiplicity. However, for Σ ⊆ S 3
with |Σ| ≥ 4, the sets Sn(Σ) are very easy to describe (see for example ), and the distribution of fixed points and excedances is trivial. In particular, in these cases we have that |S n(Σ) | ∈ { 0, 1, 2} for all n.
3.3 Pattern-avoiding involutions
Recall that In denotes the set of involutions of length n, i.e., permutations
π ∈ S n such that π = π−1. In terms of the array representation of π, this 3.3. Pattern-avoiding involutions 75
condition is equivalent to arr( π) being symmetric with respect to the main diagonal. In this section we consider the distribution of the statistics fp and exc in involutions avoiding any subset of patterns of length 3. For any π ∈ S n, it is clear that fp( π) + exc( π) + exc( π−1) = n (each cross in the array of π is either on, to the right of, or to the left of the main diagonal). Thus, if π ∈ I n, then exc( π) = 1
2
(n − fp( π)), so the number of excedances is determined by the number of fixed points. Therefore, it is enough here to consider only the statistic ‘number of fixed points’ in pattern-avoiding involutions. For any set of patterns Σ, let In(Σ) := In ∩ S n(Σ), and let ikn(Σ) := |{ π ∈In(Σ) : fp( π) = k}| . Define
GΣ(x, z ) := ∑
n≥0
∑
π∈I n(Σ)
xfp( π)zn.
By the reasoning above, ∑
n≥0
∑
π∈I n(Σ)
xfp( π)qexc( π)zn = GΣ(xq −1/2, zq 1/2). Clearly, ̂ π is an involution if and only if π is an involution. Therefore, from Lemma 1.1 we get the following.
Lemma 3.16 Let Σ be any set of patterns. We have
(1) GbΣ(x, z ) = GΣ(x, z ),
(2) GΣ−1 (x, z ) = GΣ(x, z ).
The property stated in the following lemma is what allows us to apply our techniques for studying statistics on pattern-avoiding permutations to the case of involutions.
Lemma 3.17 Let π ∈ S n(132) and let D = ϕ(π) ∈ D n. Then,
π is an involution ⇐⇒ ϕ(π) is symmetric .
Proof. The array of crosses representing π−1 is obtained from the one of π
by reflection over the main diagonal. Therefore, from the description of the bijection ϕ given in Section 2.1, we have that ϕ(π−1) = D∗. It follows that
π is an involution if and only if D = D∗, which is equivalent to D being a symmetric Dyck path. 276 Chapter 3. Simultaneous avoidance
3.3.1 Single restrictions
It is known that for σ ∈ { 123 , 132 , 213 , 321 }, |I n(σ)| = ( n
bn/ 2c
), and that for σ ∈ { 231 , 312 }, |I n(σ)| = 2 n−1. From Lemma 3.16 it follows that for all k ≥ 0, ikn(132) = ikn(213) and ikn(231) = ikn(312). In a recent paper , Deutsch, Robertson and Saracino prove the following fact:
Theorem 3.18 () The number of 321 -avoiding involutions π ∈ S n with
fp( π) = i equals the number of 132 -avoiding involutions π ∈ S n with fp( π) =
i, for any 0 ≤ i ≤ n.
Let us show that Theorem 3.18 follows easily from the work in Section 2.2. Recall from Section 1.2.1 that if D ∈ D n, D∗ denotes the path obtained by reflection of D from a vertical line x = n. Now observe that if ϕ(π) = D, then
ϕ(π−1) = D∗ (see Lemma 3.17). Similarly, if Ψ( π) = D, then Ψ( π−1) = D∗
(by the duality of RSK). Therefore, π ∈ S n(321) is an involution if and only if so is Θ( π) ∈ S n(132), which implies the result. Furthermore, restricting Θ to involutions we obtain the following extension of Theorem 3.18:
Theorem 3.19 The number of 321 -avoiding involutions π ∈ S n with fp( π)=
i, exc( π) = j and lis( π) = k equals the number of 132 -avoiding involu-tions π ∈ S n with fp( π) = i, exc( π) = j and rank( π) = n − k, for any
0 ≤ i, j, k ≤ n.
By Theorem 3.18 we have that ikn(132) = ikn(321). Thus, for single restric-tions there are three cases to consider.
Theorem 3.20 ([50, 27]) Let n ≥ 1, k ≥ 0. We have
(1) i0
n
(123) = i2
n
(123) =
{ ( n−1
n
2
) if n is even,
0 if n is odd,
i1
n
(123) =
{ ( n
n−1
2
) if n is odd,
0 if n is even,
ikn(123) = 0 if k ≥ 3.
(2) ikn(132) = ikn(213) = ikn(321) =
{ k+1
n+1
(n+1
n−k
2
) if n − k is even,
0 if n − k is odd.
(3) G231 (x, z ) = G312 (x, z ) = 1 − z2
1 − xz − 2z2 .3.3. Pattern-avoiding involutions 77
Proof. (1) Clearly a 123-avoiding permutation cannot have more than two fixed points. On the other hand, if π ∈ I n, we have fp( π) = n − 2 exc( π), which explains that ikn(123) = 0 if n − k is odd. This implies that for odd n,fp( π) = 1 for all π ∈ I n, so i1
n
(123) = |I n(123) | = ( n
n−1
2
). For even n, all we have to show is that i0
n
(123) = i2
n
(123). The bijection ψy : Sn(123) −→ D n described in Section 2.3 has the property that π ∈ I n(123) if and only if ψy(π) is a symmetric Dyck path. If n is even, involutions π ∈ I n with fp( π) = 2 are mapped to symmetric Dyck paths with a peak in the middle, and those with fp( π) = 0 are mapped to symmetric Dyck paths with a valley in the middle. We can establish a bijection between these two sets of Dyck paths just by changing the middle peak ud into a middle valley du (this can always be done because the height at the middle of a Dyck path of even semilength is always even, so it cannot be 1). This proves that i0
n
(123) = i2
n
(123), and in particular it equals 1
2
|I n(123) | = (n−1
n
2
).(2) We use the bijection ϕ : Sn(132) −→ D n, which by Lemma 3.17 restricts to a bijection between In(132) and Ds. Thus, by Lemma 2.2, G132 (x, z ) can be expressed as ∑
D∈D s
xct( D)z|D|, where the sum is over all symmetric Dyck paths. But the number of centered tunnels of a symmetric Dyck path is just its height at the middle. Therefore, taking only the first half of the path,
ikn(132) counts the number of paths from (0 , 0) to ( n, k ) never going below the x-axis, which equals the ballot number given in the theorem. (3) Consider the bijection Sn(312) ←→ Dn
π 7 → ϕ(¯ π) . Then π ∈ I n(312) if and only if ϕ(¯ π) is a sequence of pyramids. Together with the proof of Proposi-tion 3.7, this implies (see also ) that In(312) = In(231) = Sn(231 , 312). Recall that fixed points of π are mapped to tunnels of depth 0 of ϕ(¯ π), which are produced by pyramids of odd size. Thus, as in Proposition 3.7,
G312 (x, z ) = 1
1 − xz +z2
1−z2
.
2
3.3.2 Multiple restrictions
Theorem 3.21 a)
G{123 ,132 } (x, z ) = G{123 ,213 } (x, z ) = 1 + xz + ( x2 − 1) z2
1 − 2z278 Chapter 3. Simultaneous avoidance
b)
G{231 ,321 }(x, z ) = G{312 ,321 }(x, z ) = 1
1 − xz − z2
c)
G{132 ,213 }(x, z ) = 1 − z2
(1 − xz )(1 − 2z2)
d)
G{231 ,312 }(x, z ) = 1 − z2
1 − xz − 2z2
e)
G{132 ,231 }(x, z ) = G{213 ,231 } (x, z ) = G{132 ,312 } (x, z ) = G{213 ,312 } (x, z )= 1 + xz 3
(1 − xz )(1 − z2)
f )
G{132 ,321 } (x, z ) = G{213 ,321 } (x, z ) = 1
(1 − xz )(1 − z2)
g)
G{123 ,231 }(x, z ) = G{123 ,312 }(x, z ) = 1 + xz + ( x2 − 1) z2 + xz 3 + z4
(1 − z2)2
h)
G{123 ,321 } (x, z ) = 1 + xz + ( x2 + 1) z2 + 2 xz 3 + 2 z4
Proof. All the equalities between GΣ for different Σ follow trivially from Lemma 3.16. To find expressions for these GFs, the idea is to use again the same bijections as in Section 3.1, between permutations avoiding two patterns of length 3 and certain subclasses of Dyck paths. The main differ-ence is that here we will have to deal only with symmetric Dyck paths, as a consequence of Lemma 3.17.
a) From the proof of Proposition 3.1 and Lemma 3.17, we have that ϕ
restricts to a bijection between In(123 , 132) and symmetric Dyck paths D ∈Dn of height at most 2. By Lemma 2.2, ϕ maps fixed points to centered tunnels, so all we have to do is count elements D ∈ D s of height at most 2 according to the number of centered tunnels. Such a D can be uniquely 3.3. Pattern-avoiding involutions 79
written as D = ABC , where A = C∗ ∈ D ≤2 and B is either empty or has the form B = uB1d, where B1 is a sequence of hills. If |B1| is even (resp. odd), then D has one (resp. two) centered tunnels, so the contribution of B
is 1 + (1+ xz )xz
1−z2
. The contribution of A and C is 1−z2
1−2z2
. The product of these two quantities gives the expression for G{123 ,132 }(x, z ).
b) As shown above and also in , we have that In(231) = Sn(231 , 312). Therefore, In(231 , 321) = Sn(231 , 312 , 321). This case was treated in Theo-rem 3.15 b) .
c) From the proof of Proposition 3.6 and Lemma 3.17, we have that ϕ
gives a bijection between In(132 , 213) and symmetric sequences of pyramids
D ∈ P yr n, and that it maps fixed points of the permutation to centered tunnels of the Dyck path. Such a D can be written uniquely as D = ABC ,where A = C∗ ∈ P yr , and B is either empty or a pyramid. The contribution of B is 1
1−xz
, whereas A and C contribute 1−z2
1−2z2
. Multiplying these two expressions we get a formula for G{132 ,213 } (x, z ).
d) Again, In(231) = Sn(231 , 312) implies that In(231 , 312) = Sn(231 , 312), which has been considered in Proposition 3.7.
e) We have that In(132 , 231) = Sn(132 , 231 , 312), so the formula follows from Theorem 3.15 f ) .
f ) From the proof of Proposition 3.10 and Lemma 3.17, we have that ϕ
gives a bijection between In(132 , 321) and symmetric paths D ∈ D n with at most two peaks. Counting centered tunnels in such paths is very easy, since they have the form D = ukBdk, where k ≥ 0 and B is either empty or a pair of identical pyramids. The contribution of B is 1
1−z2
, whereas the rest contributes 1
1−xz
.
g) We have that In(123 , 231) = Sn(123 , 231 , 312), so the formula follows from Theorem 3.15 g) .
h) It is trivial since Sn(123 , 321) = ∅ for n ≥ 5. 2
The case of involutions avoiding simultaneously three or more patterns of length 3 is very easy and does not involve any new idea, so we omit it here. As a final remark, let us point out that looking at the results of this chapter, one observes that the GFs FΣ(x, q, z ) that we have obtained for Σ ⊆ S 3 are all rational functions when |Σ| ≥ 2. This is in contrast with the fact that they are not rational when |Σ| = 1, since in that case FΣ(1 , 1, z ) = 1−√1−4z
2z
=
C(z). For the case of involutions, all the GFs GΣ(x, z ) for Σ ⊆ S 3 are 80 Chapter 3. Simultaneous avoidance
rational except when Σ ∈ {{ 123 }, {132 }, {213 }, {321 }} .4
A simple and unusual bijection for Dyck paths
In this chapter we introduce a new bijection Φ from the set of Dyck paths to itself. This bijection has the property that it maps nontrivial statistics that appear in the study of pattern-avoiding permutations into classical statistics on Dyck paths, which have been widely studied in the literature and whose distribution is easy to obtain. When one tries to enumerate Dyck paths with respect to the number of cen-tered and right tunnels directly, the standard decompositions of Dyck paths do not work. Intuitively, the problem is that unlike hills, peaks, or double rises, which are characteristics of a Dyck path that are defined locally, the notion of tunnel may involve an arbitrarily large number of steps of the path. The bijection Φ has the advantage that it transforms tunnel-like statistics into locally defined statistics that behave well under the usual decompo-sitions of Dyck paths. As a consequence, several enumeration problems regarding permutation statistics on restricted permutations can be solved more easily considering their counterpart in terms of Dyck paths. Another important application of Φ is that it allows us to give a simple bijective proof of Theorem 1.4, which is a weaker version of Theorem 2.3 considering only the number of fixed points. Some results in this chapter are joint work with Emeric Deutsch . In Section 4.1 we present the bijection Φ, and in Section 4.2 we study its properties. In Section 4.3 we give a generalization of Φ, namely a family of bijections depending on an integer parameter r, from which the main bijection Φ is the particular case r = 0. These bijections give correspon-dences involving new statistics on Dyck paths, which generalize ct and rt. We give multivariate generating functions for them. Section 4.4 discusses 82 Chapter 4. A simple and unusual bijection for Dyck paths
several applications of these bijections to enumeration of statistics on 321-and 132-avoiding permutations. In particular, we generalize Theorem 1.4, and we find a multivariate generating function for fixed points, excedances and descents in 132-avoiding permutations. Finally, in Section 5.1 we discuss new interpretations of Catalan numbers that follow from our work.
4.1 The bijection Φ
In this section we describe a bijection Φ from Dn to itself. Let D ∈ D n.Each up-step of D has a corresponding down-step together with which it determines a tunnel. Match each such pair of steps. Let τ ∈ S 2n be the permutation defined by
τi =
i + 1
2 if i is odd; 2n + 1 − i
2 if i is even. In two-line notation,
τ =
( 1 2 3 4 5 6 · · · 2n − 2 2n − 1 2n
1 2n 2 2n − 1 3 2n − 2 · · · n + 2 n n + 1
)
.
Then Φ( D) is created as follows. For i from 1 to 2 n, consider the τi-th step of D (i.e., D is read in zigzag). If its corresponding matching step has not yet been read, define the i-th step of Φ( D) to be an up-step, otherwise let it be a down-step. In the first case, we say that the τi-th step of D opens atunnel, in the second we say that it closes a tunnel. The bijection Φ applied to the Dyck paths of semilength at most 3 is shown in Figure 4.1. Figure 4.2 shows Φ applied to the example of the Dyck path
D = uuduudududddud .It is clear from the definition that Φ( D) is a Dyck path. Indeed, it never goes below the x-axis because at any point the number of down-steps drawn so far can never exceed the number of up-steps, since each down-step is drawn when the second step of a matching pair in D is read, and in that case the first step of the pair has already produced an up-step in Φ( D). Also, Φ( D)ends in (2 n, 0) because each of the matched pairs of D produces an up-step and a down-step in Φ( D). To show that Φ is indeed a bijection, we will describe the inverse map Φ−1. Given D′ ∈ D n, the following procedure recovers the D ∈ D n such 4.1. The bijection Φ 83
φ
Figure 4.1
The bijection Φ for paths of length at most 3. that Φ( D) = D′. Consider the permutation τ defined above, and let W =
w1w2 · · · w2n be the word obtained from D′ as follows. For i from 1 to 2 n,if the i-th step of D′ is an up-step, let wτi = o, otherwise let wτi = c. W
contains the same information as D′, with the advantage that the o’s are located in the positions of D in which a tunnel is opened when D is read in zigzag, and the c’s are located in the positions where a tunnel is closed. Equivalently, the o’s are located in the positions of the left walls of the left and centered tunnels of D, and in the positions of the right walls of the right tunnels. For an example see Figure 4.3.
φ
Figure 4.2
An example of Φ. 84 Chapter 4. A simple and unusual bijection for Dyck paths
φ −1
D’ DW= o o c o c c o o o o o o c c c o c c c o o c c c o c o c c c o o o c
12255378 10 11 13 14 14 13 11 17 17 15 16 16 15 10 8 12 12 9 97466431
Figure 4.3 The inverse of Φ. Now we define a matching between the o’s and the c’s in W , so that each matched pair will give a tunnel in D. We will label the o’s with 1 , 2, . . . , n
and similarly the c’s, to indicate that an o and a c with the same label are matched. By left (resp. right) half of W we mean the symbols wi with i ≤ n
(resp. i > n ). For i from 1 to 2 n, if wτi = o, place in it the smallest label that has not been used yet. If wτi = c, match it with the unmatched o in the same half of W as wτi with largest label, if such an o exists. If it does not, match wτi with the unmatched o in the opposite half of W with smallest label. Note that since D′ was a Dyck path, at any time the number of c’s read so far does not exceed the number of o’s, so each c has some o to be paired up with. Once the symbols in W have been labelled, D can be recovered by reading the labels from left to right, drawing an up-step for each label that is read for the first time, and a down-step for each label that appears the second time. In Figure 4.3 the labelling is shown under W .
4.2 Properties of Φ
Lemma 4.1 Let D = ABC be a decomposition of a Dyck path D, where
B is a Dyck path, and A and C have the same length. Then Φ( ABC ) = Φ( AC )Φ( B). In particular, Φ( uBd) = ud Φ( B).4.2. Properties of Φ 85
Proof. It follows immediately from the definition of Φ, since the path D is read in zigzag while Φ( D) is built from left to right. 2
Theorem 4.2 Let D be any Dyck path, and let D′ = Φ( D). We have the following correspondences:
(1) ct( D) = h(D′),
(2) rt( D) = er( D′),
(3) lt( D) + ct( D) = or( D′),
(4) cmt( D) = ret( D′).Proof. First we show (1). Consider a centered tunnel given by the decompo-sition D = AuBdC. Applying Lemma 4.1 twice, we get D′ = Φ( AuBdC) = Φ( AC )Φ( uBd) = Φ( AC )ud Φ( B), so we have a hill ud in D′. Reciprocally, any hill in D′, say D′ = Xud Y , where X, Y ∈ D , comes from a centered tunnel D = Z1uΦ−1(Y )dZ2, where Z1 and Z2 are respectively the first and second halves of Φ −1(X). The proof of (4) is very similar. Recall that ret( D′) equals the number of arches of D′. Given a centered multitunnel corresponding to the decompo-sition D = ABC , we have Φ( D) = Φ( AC )Φ( B), so D′ has an arch starting at the first step of Φ( B), which is nonempty. To show (2), consider a right tunnel given by the decomposition D =
AuBdC, where length( A) > length( C). Of the two steps u and d delimiting the tunnel, d will be encountered before u when D is read in zigzag, since length( A) > length( C). So d will open a tunnel, producing an up-step in
D′. Besides, this up-step will be at an even position, since d was in the right half of D. Reciprocally, an even rise of D′ corresponds to a step in the right half of D that opens a tunnel when D is read in zigzag, so it is necessarily a right tunnel. Relation (3) follows from (2) and the fact that the total number of tunnels of D is lt( D) + ct( D) + rt( D) = n, and the total number of up-steps of D′
is or( D′) + er( D′) = n. 2
One interesting application of this bijection is that it can be used to enumer-ate Dyck paths according to the number of centered, left, and right tunnels, 86 Chapter 4. A simple and unusual bijection for Dyck paths
and number of centered multitunnels. We are looking for a multivariate generating function for these four statistics, namely
˜R(x, u, v, w, z ) = ∑
D∈D
xct( D)ult( D)vrt( D)wcmt( D)z|D|.
By Theorem 4.2, this GF can be expressed as
˜R(x, u, v, w, z ) = R( x
u , u, v, w, z ), (4.1) where
R(t, u, v, w, z ) = ∑
D∈D
th(D)uor( D)ver( D)wret( D)z|D|.
We can derive an equation for R using again that every nonempty Dyck path D can be decomposed in a unique way as D = uAdB, where A, B ∈ D .The number of hills of uAdB is h(B) + 1 if A is empty, and h(B) otherwise. The odd rises of A become even rises of uAdB, and the even rises of A
become odd rises of uAdB. Thus, we have er( uAdB) = or( A) + er( B), and or( uAdB) = er( A) + or( B) + 1, where the extra odd rise comes from the first step u. We also have ret( uAdB) = ret( B) + 1. Hence, we obtain the following equation for R:
R(t, u, v, w, z ) = 1 + uwz (R(1 , v, u, 1, z ) − 1 + t)R(t, u, v, w, z ). (4.2) Denote R1 := R(1 , u, v, 1, z ), ̂ R1 := R(1 , v, u, 1, z ). Substituting t = w = 1 in (4.2), we obtain
R1 = 1 + uz ̂ R1R1, (4.3) and interchanging u and v,̂
R1 = 1 + vzR 1̂ R1. (4.4) Solving (4.3) and (4.4) for ̂ R1, gives ̂
R1 = 1 + ( u − v)z − √1 − 2( v + u)z + ( v − u)2z2
2uz .
Thus, from (4.2),
R(t, u, v, w, z ) = 1
1 − uwz (̂ R1 − 1 + t)2
2 − w + ( v + u − 2tu )wz + w√1 − 2( v + u)z + ( v − u)2z2 . (4.5) Now, switching to ˜R, we obtain the following theorem. 4.2. Properties of Φ 87
Theorem 4.3 The multivariate generating function for Dyck paths accord-ing to centered, left, and right tunnels, centered multitunnels, and semilength is
∑
D∈D
xct( D)ult( D)vrt( D)wcmt( D)z|D|
= 2
2 − w + ( v + u − 2x)wz + w√1 − 2( v + u)z + ( v − u)2z2 .
As pointed out by Alex Burstein, Lagrange inversion applied to equation (4.2) gives a nice expression for the coefficients of ˜R.
Corollary 4.4 Fix integers c, l, r, m ≥ 0 and let n = c + l + r. The number of Dyck paths D ∈ D n with ct( D) = c, lt( D) = l, rt( D) = r and cmt( D) = m
is given by
m − c
n − m
(mc
)( n − ml
)( n − mr
)
if c < m < c + l < n , and it is 1 if c = m = n.Proof. The case c = m = n is trivial, since the only path in Dn with n
centered tunnels is D = undn. For the rest of the proof we assume that 0 ≤ c < m < c + l < n .We start by applying Lagrange inversion formula (Theorem 1.3) to equa-tion (4.2) for variable w, being f (w) = R(t, u, v, w, z ) − 1, G(w) = uz (̂ R1 −
1 + t)( w + 1), n = m and k = 1 in the theorem. We get that wm = 1
m wm−1 m
= umzm(̂ R1 − 1 + t)m.
Taking the coefficient of tc,tcwm =
(mc
)
umzm(̂ R1 − 1) m−c. (4.6) Isolating R1 in equation (4.3) and substituting it in (4.4) we get ̂
R1 = 1 + vz ̂ R1
1 − uz ̂ R1
,88 Chapter 4. A simple and unusual bijection for Dyck paths
which is equivalent to ̂
R1 − 1 = ẑ R1(u(̂ R1 − 1) + v).
We apply Lagrange inversion formula again, now for variable z, with f (z) = ̂
R1 − 1, G(z) = ( z + 1)( uz + v) and n = s. This gives us (for s 6 = 0), zs k = k
s zs−k s(uz + v)s,
so us−rvr zs k = k
s
(sr
)
zs−k szs−r = k
s
(sr
)( sr − k
)
. (4.7) Now, taking the appropriate coefficients of u, v and z in equation (4.6) gives tcun−rvrwmzn =
(mc
)
un−m−rvrzn−m m−c
=
(mc
) m − c
n − m
(n − mr
)( n − mr − m + c
)
,
where the last equality follows from (4.7) with s = n − m and k = m − c.Thus, using that n − r = c + l and n − m − (r − m + c) = l, we get that for
n ≥ 1, [tcuc+lvrwmzn]R(t, u, v, w, z ) = m − c
n − m
(mc
)( n − mr
)( n − ml
)
.
But by relation (4.1), this coefficient is precisely [ xculvrwmzn] ˜R(x, u, v, w, z ), which is the number of paths D ∈ D n with ct( D) = c, lt( D) = l, rt( D) = r
and cmt( D) = m. 2
4.3 Generalizations
Here we present a generalization Φ r of the bijection Φ, which depends on a nonnegative integer parameter r. Given D ∈ D n, copy the first 2 r steps of D into the first 2 r steps of Φ r(D). Now, read the remaining steps of D
in zigzag in the following order: 2 r + 1, 2 n, 2 r + 2, 2 n − 1, 2 r + 3, 2 n − 2, and so on. For each of these steps, if its corresponding matching step in D
has not yet been encountered, draw an up-step in Φ r(D), otherwise draw a down-step. Note that for r = 0 we get the same bijection Φ as before. 4.3. Generalizations 89
Note that Φ r can be defined exactly as Φ with the difference that instead of
τ , the permutation that gives the order in which the steps of D are read is
τ (r) ∈ S 2n, defined as
τ (r)
i
=
i if i ≤ 2r;
i + 1
2 + r if i > 2r and i is odd; 2n + 1 − i
2 − r if i > 2r and i is even. Figure 4.4 shows an example of the bijection Φ r for r = 2 applied to the path D = uduuduuduududddudd .
2r 2r
φ 2
Figure 4.4
An example of Φ 2.It is clear from the definition that Φ r(D) is a Dyck path. A reasoning similar to the one used for Φ shows that Φ r is indeed a bijection. The properties of Φ given in Theorem 4.2 generalize to analogous properties of Φ r. We will prove them using the next lemma, which follows immediately from the definition of Φ r.
Lemma 4.5 Let r ≥ 0, and let D = ABC be a decomposition of a Dyck path D, where B is a Dyck path, and length( A) = length( C) + 2 r. Then
Φr(ABC ) = Φ r(AC )Φ( B).
Theorem 4.6 Let r ≥ 0, let D be any Dyck path, and let D′ = Φ r(D). We have the following correspondences:
(1) # {tunnels of D with midpoint at x = n + r}
= #{hills of D′ in x > 2r},
(2) # {tunnels of D with midpoint in x > n + r}
= #{even rises of D′ in x > 2r},
(3) # {tunnels of D with midpoint in x ≤ n + r}
= #{odd rises of D′ in x > 2r} + # {up-steps of D′ in x ≤ 2r},
(4) # {multitunnels of D with midpoint at x = n + r}
= #{arches of D′ in x ≥ 2r}.90 Chapter 4. A simple and unusual bijection for Dyck paths
Proof. Fist we show (1). A tunnel given by the decomposition D = AuBdC
has its midpoint at x = n + r exactly when length( A) = length( C) + 2 r.Applying Lemmas 4.5 and 4.1, D′ = Φ r(AuBdC) = Φ r(AC )Φ( uBd) = Φr(AC )Φ( ud )Φ( B) = Φ r(AC )ud Φ( B), and ud is a hill of D′ in x > 2r,since length(Φ r(AC )) ≥ 2r. Reciprocally, any hill of D′ in x > 2r, say
D′ = Xud Y , where X, Y ∈ D and length( X) ≥ 2r, comes from a tunnel with midpoint at x = n + r, namely D = Z1uΦ−1(Y )dZ2, where Z1Z2 = Φ −1
r
(X)and length( Z1) = length( Z2) + 2 r.The proof of (4) is very similar. A multitunnel given by D = ABC has its midpoint at x = n+r exactly when length( A) = length( C)+2 r. In this case, Φr(D) = Φ r(AC )Φ( B) by Lemma 4.5, so D′ has an arch starting at the first step of Φ( B). Notice that this arch is in x ≥ 2r because length(Φ r(AC )) ≥
2r.To show (2), consider a tunnel in D with midpoint in x > n + r. This is equivalent to saying that it is given by a decomposition D = AuBdC with length( A) > length( C) + 2 r. In particular, the tunnel is contained in the halfspace x ≥ 2r, so the two steps u and d delimiting the tunnel are in the part of D that is read in zigzag in the process to obtain Φ r(D), and d will be encountered before u, since length( A) − 2r > length( C). So d will open a tunnel, producing an up-step of D′ in x > 2r. Besides, this up-step will be at an even position, since d is in x > n + r, that is, in the right half of the part of D that is read in zigzag. Reciprocally, an even rise of D′ in x > 2r
corresponds to a step of D in x > n + r that opens a tunnel when D is read according to τ (r), so it is necessarily a tunnel with midpoint to the right of
x = n + r .Relation (3) follows from (2) and the fact that the total number of tunnels of D is # {tunnels of D with midpoint in x > n + r} + # {tunnels of D with midpoint in x ≤ n + r} = n, and the total number of up-steps of D′ is #{even rises of D′ in x > 2r} + # {odd rises of D′ in x > 2r} + # {up-steps of D′ in x ≤ 2r} = n. 2
Similarly to how we used the properties of Φ to prove Theorem 4.3, we can use the properties of Φ r to prove a more general theorem. Our goal is to enumerate Dyck paths according to the number of tunnels with midpoint on, to the right of, and to the left of an arbitrary vertical line x = n + r,and multitunnels with midpoint on that line. In generating function terms, 4.3. Generalizations 91
we are looking for an expression for
E(t, u, v, w, y, z ) := ∑
n≥00≤r≤n
∑
D∈D n
tαuβ vγ wδ yrzn,
where
α = # {tunnels of D with midpoint at x = n + r},
β = # {tunnels of D with midpoint in x ≤ n + r},
γ = # {tunnels of D with midpoint in x > n + r},
δ = # {multitunnels of D with midpoint at x = n + r}.Note that the variable y marks the position of the vertical line x = n + r
with respect to which the tunnels are classified. The following theorem gives an expression for E.
Theorem 4.7 Let E, R and C be defined as above. Then,
E(t, u, v, w, y, z ) = C(uyz )R(t, u, v, w, z )
1 − yu 2z2C2(uyz )R(1 , u, v, 1, z )R(1 , v, u, 1, z )= 2δ2(2 + ( v − u)z + δ1)
[2 + ( u + v − 2tu )wz + wδ 1] [( δ1 + ( v − u)z)δ2 − 4uyz ] ,
where δ1 := √1 − 2( u + v)z + ( u − v)2z2 − 1, δ2 := √1 − 4uyz − 1.Proof. By Theorem 4.6, the generating function E can be expressed as
E(t, u, v, w, y, z ) =
∑
n≥00≤r≤n
∑
D∈D n
t#{hills of D in x > 2r}
u#{odd rises of D in x > 2r}+# {up-steps of D in x ≤ 2r}
v#{even rises of D in x > 2r}w#{arches of D in x ≥ 2r}yrzn. (4.8) For each path D in this summation, the y-coordinate of its intersection with the vertical line x = 2 r has to be even. Fix h ≥ 0. We will now focus only on the paths D ∈ D for which this intersection has y-coordinate equal to 2 h.Let D = AB , where A and B are the parts of the path respectively to the left and to the right of x = 2 r. Then, # {hills of D in x > 2r} = # {hills of B},#{odd rises of D in x > 2r} = # {odd rises of B}, # {up-steps of D in x ≤
2r} = # {up-steps of A}, and # {arches of D in x ≥ 2r} = # {arches of B}.92 Chapter 4. A simple and unusual bijection for Dyck paths
B can be any path starting at height 2 h and landing on the x-axis, never going below it. If h > 0, consider the first down-step of B that lands at height 2 h − 1. Then B can be decomposed as B = B1dB′, where B1 is any Dyck path, and B′ is any path starting at height 2 h − 1 and landing on the x-axis, never going below it. Applying this decomposition recursively, B
can be written uniquely as B = B1dB2d · · · B2hdB2h+1 , where the Bi’s for 1 ≤ i ≤ 2h + 1 are arbitrary Dyck paths. The number of hills and number of arches of B are given by those of B2h+1 . The odd rises of B are the odd rises of the Bi’s with odd subindex plus the even rises of those with even subindex. In a similar way one can describe the even rises of B. The semilength of
B is the sum of semilengths of the Bi’s plus h, which comes from the 2 h
additional down-steps. Thus, the generating function for all paths B of this form, where t, u, v, and z mark respectively number of hills, number of odd rises, number of even rises, and semilength, is
zhRh(1 , u, v, 1, z )Rh (1 , v, u, 1, z )R(t, u, v, w, z ). (4.9) Similarly, A can be decomposed uniquely as A = A1uA2u · · · A2huA2h+1 .The number of up-steps of A is the sum of the number of up-steps of each
Ai, plus a 2 h term that comes from the additional up-steps. The generating function for paths A of this form, where u marks the number of up-steps, and y and z mark both the semilength, is
zhyhu2hC2h+1 (uyz ). (4.10) The product of (4.9) and (4.10) gives the generating function for paths D =
AB where the height of the intersection point of D with the vertical line between A and B is 2 h, where the variables mark the same statistics as in (4.8). Note that the exponent of y is half the distance between the origin of
D and this vertical line. Summing over h, we obtain
E(t, u, v, w, y, z )= ∑
h≥0
z2hyhu2hC2h+1 (uyz )Rh(1 , u, v, 1, z )Rh (1 , v, u, 1, z )R(t, u, v, w, z )= C(uyz )R(t, u, v, w, z )
1 − yu 2z2C2(uyz )R(1 , u, v, 1, z )R(1 , v, u, 1, z ) .
The second expression in the statement of the theorem follows from the formula (4.5) that we had for R. 24.4. Connection to pattern-avoiding permutations 93
4.4 Connection to pattern-avoiding per-mutations
The bijection Φ has applications to enumeration of statistics on pattern-avoiding permutations. The first one is that it can be used together with the bijections defined in Chapter 2 to give another bijective proof of Theorem 1.4. Here we show a more general result. We use the bijection Φ r to give a combinatorial proof of the following generalization of Theorem 1.4. Note that the particular case r = 0 gives a new bijective proof of such theorem.
Theorem 4.8 Fix r, n ≥ 0. For any π ∈ S n, define αr(π) = # {i : πi =
i + r}, βr(π) = # {i : i > r, π i = i}. Then, the number of 321 -avoiding permutations π ∈ S n with βr (π) = k equals the number of 132 -avoiding permutations π ∈ S n with αr(π) = k, for any 0 ≤ k ≤ n.Proof. Recall that the bijection ψx : Sn(321) −→ D n defined in Section 2.2.4 satisfies that fp( π) = h(ψx(π)). More precisely, it can be easily checked that
i is a fixed point of π if and only if ψx(π) has a hill with x-coordinate 2 i − 1. This implies that βr(π) = # {hills of ψx(π) in x > 2r}.The second bijection that we use is ϕ : Sn(132) −→ D n, defined in Sec-tion 2.1. In Proposition 2.1 we showed that fp( π) = ct( ϕ(π)). Recall that in the proof of this proposition, we associated a unique tunnel of D to each cross of the array arr( π). An element i with πi = i + r is represented by a cross ( i, i + r) in the array. From the description of the association between crosses and tunnels, it follows that such a cross ( i, i + r) corresponds to a tunnel of ϕ(π) with midpoint r units to the right of the center. That is, an el-ement i with πi = i+r gives a tunnel with midpoint at x = n+r. Therefore, we have that αr(π) = # {tunnels of ϕ(π) with midpoint at x = n + r}.Now all we need to do is use Φ r and property (1) given in Theorem 4.6. From this it follows that the bijection ψ−1
x
◦ Φr ◦ ϕ : Sn(132) −→ S n(321) has the property that βr (ψ−1
x
◦ Φr ◦ ϕ(π)) = # {hills of Φ r ◦ ϕ(π) in x > 2r} =#{tunnels of ϕ(π) with midpoint at x = n + r} = αr(π). 2
While in Section 2.2.4 we describe a simple way to enumerate 321-avoiding permutations with respect to the statistics fp and exc, the analogous enu-meration for 132-avoiding permutations is harder to do directly. Here we use the properties of Φ to give a more direct derivation of the multivariate generating function for 132-avoiding permutations according to the number of fixed points and the number of excedances. 94 Chapter 4. A simple and unusual bijection for Dyck paths
Corollary 4.9 (of Theorem 4.3)
∑
n≥0
∑
π∈S n(132)
xfp( π)vexc( π)zn
= 2
1 + (1 + v − 2x)z + √1 − 2(1 + v)z + (1 − v)2z2 . (4.11)
Proof. Proposition 2.1 shows that ϕ maps fixed points to centered tun-nels, and excedances to right tunnels, i.e., fp( π) = ct( ϕ(π)) and exc( π) = rt( ϕ(π)). Thus, the left hand side of (4.11) equals ∑
D∈D
xct( D)vrt( D)z|D|.The result now is obtained applying Theorem 4.3 for u = w = 1. 2
Comparing this expression (4.11) with the equation obtained in 2.2.4 for
F321 (x, q, z ), we obtain another proof of Theorem 2.3. As a further application, we can use the bijection Φ to give the following refinement of Corollary 4.9, which gives an expression for the multivariate generating function for number of fixed points, number of excedances, and number of descents in 132-avoiding permutations. The analogous result for 321-avoiding permutations is given in Theorem 2.10.
Theorem 4.10 Let
L(x, q, p, z ) := 1 + ∑
n≥1
∑
π∈S n(132)
xfp( π)qexc( π)pdes( π)+1 zn.
Then
L(x, q, p, z ) = 2(1 + xz (p − 1))
1 + (1 + q − 2x)z − qz 2(p − 1) 2 + √f1(q, z ) , (4.12)
where f1(q, z ) = 1 − 2(1 + q)z + [(1 − q)2 − 2q(p − 1)( p + 3)] z2 − 2q(1 + q)( p −
1) 2z3 + q2(p − 1) 4z4.Proof. We use again that ϕ maps fixed points to centered tunnels, and excedances to right tunnels. It is shown in Proposition 2.1 that it also maps descents of the permutation to valleys of the corresponding Dyck path. Clearly, the number of valleys of any nonempty Dyck path equals the number of peaks minus one (in the empty path both numbers are 0). Thus, L can be expressed as
L(x, q, p, z ) = ∑
D∈D
xct( D)qrt( D)p#{peaks of D}z|D|.4.4. Connection to pattern-avoiding permutations 95
By Theorem 4.2, Φ maps centered tunnels into hills and right tunnels into even rises. Let us see what peaks are mapped to by Φ. Given a peak
ud in D ∈ D , D can be written as D = Aud C, where A and C are the parts of the path before and after the peak respectively. This decomposition corresponds to a tunnel of D that goes from the beginning of the u to the end of the d. Assume first that the peak occurs in the left half (i.e., length( A) < length( C)). When D is read in zigzag, the u opens a tunnel that is closed by the d two steps later. This produces in Φ( D) an up-step followed by a down-step two positions ahead, that is, an occurrence of u ? d
in the Dyck word of Φ( D), where ? stands for any symbol (either a u or a d). If the peak occurs in the right half of D (i.e., length( A) > length( C)), the reasoning is analogous, with the difference that the d opens a tunnel that is closed by the u two steps ahead. So, such a peak produces also an occurrence of u ? d in Φ( D). Reciprocally, we claim that an occurrence of u ? d in Φ( D)can come only from a peak of D either in the left or in the right half. Indeed, using the notation from the procedure above describing the inverse of Φ, an occurrence of u ? d in Φ( D) corresponds to either an occurrence of oc in the left half of W or an occurrence of co in the right half of W . In both cases, the algorithm given above will match these two letters c and o with each other, so they correspond to an occurrence of ud in D.If the peak occurs in the middle (i.e., length( A) = length( C)), then by Lemma 4.1, Φ( Aud C) = Φ( AC )ud , so it is mapped to an occurrence of ud
at the end of Φ( D). Clearly we have such an occurrence only when D has a peak in the middle. Thus, we have shown that peaks in D are mapped by Φ to occurrences of
u ? d in Φ( D) and occurrences of ud at the end of Φ( D), or, equivalently, to occurrences of u ? d in Φ( D)d (here Φ( D)d is a Dyck path followed by a down-step). Denote by λ(D) the number of occurrences of u ? d in Dd.This implies that L can be written as
L(x, q, p, z ) = ∑
D∈D
xh(D)qer( D)pλ(D)z|D|.
We are left with a Dyck path enumeration problem, which is solved in the following lemma. Let J be defined in Lemma 4.11. It is easy to see that we have L(x, q, p, z ) = 1 + J(x, 1, p, 1, q, p, z ). Making use of (4.13) and (4.14), it follows at once that
L(x, q, p, z ) = 1 − xz + xpz
1 − xz − zK 1
,96 Chapter 4. A simple and unusual bijection for Dyck paths
where K1 is given by
zK 21 − [1 − z − qz + q(1 − p)2z2]K1 + p2qz = 0 .
From these equations we obtain (4.12). 2
Lemma 4.11 Denote by ih( D) (fh( D)) the number of initial (final) hills in
D (obviously, their only possible values are 0 and 1). Denote by μ(D) the number of occurrences of u ? d in D. Then the generating function
J(x, t, s, u, v, y, z ) := ∑ xh(D)tih( D)sfh( D)uor( D)ver( D)yμ(D)z|D|,
where the summation is over all nonempty Dyck paths, is given by
J(x, t, s, u, v, y, z ) = uz [xts + (1 − xu (1 − t)(1 − s)z)K]
1 − xuz − uzK , (4.13)
where K is given by
uzK 2 − [1 − (u + v)z + uv (1 − y)2z2]K + y2vz = 0 . (4.14)
Proof. Every nonempty Dyck path has one of the following four forms: ud ,
ud B, uAd, or uAdB, where A and B are nonempty Dyck paths. The generating functions of these four pairwise disjoint sets of Dyck paths are (i) xtsuz ,(ii) xtuzJ (x, 1, s, u, v, y, z ), (iii) uzJ (1 , y, y, v, u, y, z ), (iv) uzJ (1 , y, y, v, u, y, z )J(x, 1, s, u, v, y, z ), respectively. Only the third factor in (iii) and (iv) needs an explanation: the hills of A are not hills in uAd; an initial (final) hill in A gives a uud (udd )in uAd; an odd (even) rise in A becomes an even (odd) rise in uAd.Consequently, the generating function J satisfies the functional equation
J(x, t, s, u, v, y, z ) = xtsuz + xtuzJ (x, 1, s, u, v, y, z )+uzJ (1 , y, y, v, u, y, z ) + uzJ (1 , y, y, v, u, y, z )J(x, 1, s, u, v, y, z ). (4.15) From equation (4.15) it is clear that, whether interested or not in the statis-tics ‘number of initial (final) hills’, we had to introduce them for the sake of 4.4. Connection to pattern-avoiding permutations 97
the statistic marked by the variable y. Also, without any additional effort we could use two separate variables to mark the number of uud ’s and the number of udd ’s, and obtain a slightly more general generating function, although we do not need it here. Denoting H = J(x, 1, s, u, v, y, z ), K = J(1 , y, y, v, u, y, z ), equation (4.15) becomes
J = xtsuz + xtuzH + uzK + uzHK. (4.16) Setting here t = 1, we obtain
H = xsuz + xuzH + uzK + uzHK. (4.17) Solving (4.17) for H and introducing it into (4.16), we obtain (4.13). It remains to show that K satisfies the quadratic equation (4.14). Setting
x = 1, t = y, s = y in (4.16), and interchanging u and v, we get
K = y2vz + yvzM + vz ̂ K + vzM ̂ K, (4.18) where M = J(1 , 1, y, v, u, y, z ) and ̂ K is K with u and v interchanged, namely ̂
K = J(1 , y, y, u, v, y, z ). Now in (4.16) we set x = 1, t = 1, s = y, and we interchange u and v, to get
M = yvz + vzM + vz ̂ K + vzM ̂ K. (4.19) Eliminating M from (4.18) and (4.19), we obtain
vz (2 yvz − y2vz + 1 − vz )̂ K + ( vz − 1) K + vzK ̂ K + y2vz = 0 . (4.20) Finally, eliminating ̂ K from (4.20) and the equation obtained from (4.20) by interchanging u and v, we obtain equation (4.14). Note that, as expected, J
is symmetric in the variables t and s and linear in each of these two variables.
2
From Theorem 4.10 one can see that the first terms of L(x, q, p, z ) are 1 + xpz + ( qp 2 + x2p)z2 + ( q2p2 + qp 2 + xqp 3 + xqp 2 + x3p)z3 + · · · ,
corresponding to Dyck paths of semilength at most 3 (or equivalently, to 132-avoiding permutations of length at most 3). 5
Other bijections and combinatorial interpretations
In this chapter we describe several combinatorial objects that are enumerated by the Catalan, Fine and Narayana numbers. They arise naturally from the work in the previous chapters, and some of them give new combinatorial interpretations of these numbers. In Section 5.2 we describe three bijections between 321-avoiding permutations and Dyck paths, which show that certain statistics on Dyck paths and on permutations are equidistributed. In Section 5.3 we consider a class of permutations defined in terms of non-crossing matchings of points around a circle. We study their structure and provide generating functions enumerating them with respect to the statistic ‘number of descents’, and also with respect to the number of fixed points and the number excedances.
5.1 Some interpretations of the Catalan and Fine numbers
Our first new interpretation of the Catalan numbers follows immediately from the results in the previous chapter. Note that any nonempty Dyck path D ∈ D n has a multitunnel that goes from (0 , 0) to (2 n, 0). We call this the basic multitunnel .
Proposition 5.1 Let n ≥ 0. The number of Dyck paths of length 2n + 2
with no centered multitunnels other than the basic one is Cn.Proof. We could give a non-bijective argument using generating functions, but now the bijection Φ yields a simple combinatorial proof. We know from 100 Chapter 5. Other bijections and combinatorial interpretations
part (4) of Theorem 4.2 that the set of paths D ∈ D n+1 with cmt( D) = 1 is in bijection with the set of Dyck paths of length 2 n + 2 with only one return. But these are precisely elevated Dyck paths of the form uAd, where
A ∈ D n. The number of them is Cn. 2
Proposition 5.2 The following quantities are equal to Cn:
(1) The total number of fixed points in elements of Sn(321) .
(2) The total number of fixed points in elements of Sn(132) .
(3) The total number of centered tunnels in Dyck paths of length 2n.Proof. By the first part of Theorem 4.2, (3) equals the total number of hills in Dyck paths of length 2 n. To prove that this number is Cn, we define the following bijection between paths in Dn with a marked hill and the set Dn
itself. Given a path with a distinguished hill Aud B ∈ D n, where A, B ∈ D ,map it to the path uAdB ∈ D n. This is obviously a bijection, since each
D ∈ D n can be expressed uniquely as D = uAdB, with A, B ∈ D .The equality (2)=(3) is a consequence of the first part of Proposition 2.1. Finally, by Proposition 2.8, fixed points in 321-avoiding permutations are in one-to-one correspondence with hills of Dyck paths, which proves (1). Another argument to compute (1) directly is the following. We are counting the number of pairs ( π, i ), where π ∈ S n(321) and i is a fixed point of π.Given 1 ≤ i ≤ n, the number of π ∈ S n(321) having i as a fixed point is
Ci−1Cn−i, since πi = i if and only if π1π2 · · · πi−1 ∈ S i−1(321) and ( πi+1 −
i)( πi+2 − i) · · · (πn − i) ∈ S n−i(321). Therefore, the total number of such pairs ( π, i ) is ∑ni=1 Ci−1Cn−i = Cn. 2
In it is proved that the number of permutations π ∈ S n(132) (or π ∈Sn(321)) with no fixed points is the Fine number Fn. This sequence is most easily defined by its relation to Catalan numbers:
Cn = 2 Fn + Fn−1 for n ≥ 2, and F1 = 0 , F2 = 1 .
Although defined some time ago, Fine numbers have received much atten-tion in recent years (see a survey ). Special cases of the bijections in Section 2.2 give simple bijections between these two combinatorial inter-pretations of Fine numbers and a new one: the set of Dyck paths without centered tunnels. In particular, we obtain a bijective proof of the following result. 5.1. Some interpretations of the Catalan and Fine numbers 101
Proposition 5.3 The number of Dyck paths D ∈ D n without centered tun-nels is equal to Fn.
Yet another combinatorial proof of this fact follows from the bijections in Chapter 4. The bijection Φ maps Dyck paths without centered tunnels to Dyck paths without hills, which in turn correspond through the bijection ψx
to 321-avoiding derangements. For the next interpretation of the Catalan and Fine numbers, consider the directed graph G drawn in Figure 5.1. Its nodes are the infinite set {qi,j :
i, j ≥ 0, i + j even }. The set of edges is {(qi,j → qi+1 ,j +1 ), (qi+1 ,j +1 → qi,j ) :
i, j ≥ 0, i + j even } ∪ { (q0,2j−2 → q0,2j ), (q0,2j → q0,2j ) : j ≥ 2} ∪ { (q2i−2,0 →
q2i, 0), (q2i, 0 → q2i, 0) : i ≥ 2} ∪ { (q0,0 → q0,0)}.
Figure 5.1 The graph G.
Proposition 5.4 Let G be the directed graph described above, and let G′ be the graph obtained from it by removing the edge (q0,0 → q0,0). Fix n ≥ 0.
(1) The number of paths in G from q0,0 to q0,0 with n steps is Cn.
(2) The number of paths in G′ from q0,0 to q0,0 with n steps is Fn.
(3) The number of paths in G from q0,0 to q0,0 with n + 1 steps not having
q0,0 as an interior point is Cn.Proof. We will construct a bijection between Dn and the set of paths in G
from q0,0 to q0,0 with n steps. Let D ∈ D n. Read the steps of D two by two, starting with the two middle steps n and n + 1, then n − 1 and n + 2, and progressively moving away from the middle, finishing with the pair 1 102 Chapter 5. Other bijections and combinatorial interpretations
and 2 n. For k from 1 to n, let Dk be the subpath of D consisting of the steps at distance at most k from the middle. Let ak (resp. bk) be the height difference between the leftmost (resp. rightmost) point of Dk and its lowest point (the height is given by the y-coordinate). Equivalently, ak (resp. bk)is the number of left (resp. right) tunnels of D with exactly one of its two delimiting steps belonging to Dk. Define the k-th node of our path in G to be qak,b k .Note that qa0,b 0 = qan,b n = q0,0, and that for every k there is an edge from
qak,b k to qak+1 ,b k+1 in G, by the way the numbers ak and bk change every time that a pair of steps is read from D. It is not hard to see that this defines a bijection between Dn and paths in G with n steps starting and ending at q0,0. Indeed, the numbers ak and bk encode enough information to reconstruct the Dyck path. This proves (1). To show parts (2) and (3), observe that the node q0,0 is used whenever
ak = bk = 0, which means that there is a centered multitunnel between the two endpoints of Dk. Similarly, the edge ( q0,0 → q0,0) is used when
ak = bk = ak+1 = bk+1 = 0, and this condition is equivalent to D having a centered tunnel between the endpoints of Dk+1 . To prove (2), we use the fact from Corollary 5.3 that the number of Dyck paths of length 2 n with no centered tunnels is Fn. Part (3) follows now from Proposition 5.1, and in fact is also a direct consequence of part (1). 2
Combining the bijection just defined with Φ, an alternative and perhaps simpler bijection between Dn and the set of paths in G from q0,0 to q0,0
with n steps can be defined. It will be convenient to describe the path in
G backwards. Equivalently, we will give a path P in the graph obtained from G by reversing all the edges. Given D ∈ D n, read the steps from left to right two at a time, and construct P as follows. Let qi,j be the current node in P . If a uu is read, add an edge ( qi,j → qi+1 ,j +1 ) to the path. If a pair ud is encountered, add an edge ( qi,j → qi+1 ,j −1) if j > 0, or a loop (qi,j → qi,j ) otherwise. For each pair du , add an edge ( qi,j → qi−1,j +1 ) if
i > 0, or a loop ( qi,j → qi,j ) otherwise. Finally, for each pair dd , add an edge (qi,j → qi−1,j −1) if i, j > 0, ( qi,j → qi−2,j ) if j = 0, or ( qi,j → qi,j −2) if i = 0. It can be checked that this is a bijection as well. Note that if at a given point of the construction the current node in P is qi,j , then the fragment of
D that has been read so far ends at height i + j.Our last interpretation of the Catalan numbers is joint work with Emeric Deutsch and Astrid Reifegerste. 5.1. Some interpretations of the Catalan and Fine numbers 103
Proposition 5.5 The following quantities are equal to Cn:
(1) The number of permutations π ∈ S 2n+1 (321) such that ψx(π) = φx(π).
(2) The number of symmetric parallelogram polyominoes 1 of perimeter
4(2 n+1) having exactly one horizontal (equivalently, vertical) boundary segment at each level. Proof. The equivalency between (1) and (2) is clear when we draw ψx(π)(as in Figure 2.8) and φx(π) (as in Figure 5.3) as lattice paths from the top-left to the bottom-right corner of the array of π. The two paths form a parallelogram polyomino which is symmetric (and satisfies the conditions of (2)) exactly when ψx(π) = φx(π) as Dyck paths. Now we show that the permutations in (1) are counted by the Catalan num-bers. Let π ∈ S 2n+1 (321), let i1 < i 2 < · · · < i e be the positions of the excedances of π and let j1 < j 2 < · · · < j 2n+1 −e be the remaining posi-tions. Then, ψx(π) = φx(π) if and only if e = n (π has n excedances) and
πik = jk + 1 for all 1 ≤ k ≤ n. Each permutation satisfying these conditions is uniquely determined by its excedance set {i1, i 2, . . . , i n}. Now, these sets are in bijection with Dyck paths of length 2 n: given such a set, construct a Dyck path having up-steps in positions {i1, i 2, . . . , i n} and down-steps every-where else. Figure 5.2 shows an example for π = 4512736, whose excedance set is {1, 2, 5}. 2
Figure 5.2 A permutation satisfying ψx(π) = φx(π), its symmetric parallelogram polyomino, and the corresponding Dyck path.
1Parallelogram polyominoes are unordered pairs of lattice paths starting at (0 ,0), using steps (1 ,0) and (0 ,−1), ending at the same point, and only intersecting at the beginning and at the end. 104 Chapter 5. Other bijections and combinatorial interpretations
5.2 Some other bijections between Sn(321)
and Dn
Considering the array of crosses associated to a permutation, as we did to define ψx in Section 2.2.4, some other known bijections between Sn(321) and
Dn can easily be viewed in a systematic way, as paths with east and south steps from the upper-left corner to the lower-right corner of the n × n array. For each of these bijections, our canonical example will be π = 23147586. One such bijection was established by Billey, Jockusch and Stanley in [7, p. 361]. Denote it by φx. Consider the path with east and south steps that leaves the crosses corresponding to excedances to the right, and stays always as far from the main diagonal as possible (Figure 5.3). Then φx(π) can be obtained from it just by reading an up-step for every east step of this path and a down-step for every south step.
Figure 5.3 The bijection φx.In , Krattenthaler describes a bijection from Sn(123) to Dn. If we omit the last step, consisting of reflecting the path over a vertical line, and com-pose the bijection with the reversal operation, that maps a permutation
π1π2 · · · πn into πn · · · π2π1, we get a bijection from Sn(321) to Dn. Denote it by ψq. In the array representation, ψq(π) corresponds (by the same trivial transformation as before) to the path with east and south steps that leaves all the crosses to the left and remains always as close to the main diagonal as possible (Figure 5.4). Our first bijection is related to this last one by ψq(π) = ψx(π−1). In a similar way, we could still define a fourth bijection φq : Sn(321) −→ D n by
φq(π) := φx(π−1) (Figure 5.5). Combining these bijections and their inverses, one can get some automor-phisms on Dyck paths and on 321-avoiding permutations with interesting properties. Recall from Section 1.2.1 that va( D) and p2(D) denote respec-5.2. Some other bijections between Sn(321) and Dn 105
Figure 5.4
The bijection ψq.
Figure 5.5
The bijection φq.tively the number of valleys and the number of peaks of height at least 2 of
D. It can be checked that ψx ◦ φ−1
x
is an involution on Dn with the property that va( ψx ◦ φ−1
x
(D)) = dr( D) and dr( ψx ◦ φ−1
x
(D)) = va( D). Indeed, this follows from the fact that excedances are mapped to valleys by φx and to double rises by ψx. This bijection gives a new proof of the symmetry of the bivariate distribution of the pair (va , dr) of statistics on Dyck paths. A different involution with this property was introduced in . Another involution on Dn is given by ψx ◦ψ−1
q
. This one shows the symmetry of the distribution of the pair (dr , p 2), because dr( ψx ◦ ψ−1
q
(D)) = p2(D) and
p2(ψx ◦ ψ−1
q
(D)) = dr( D). In addition, it preserves the number of hills, i.e.,
h(ψx ◦ ψ−1
q
(D)) = h(D). To see this, just note that both ψq and ψx map fixed points to hills, whereas excedances are mapped to peaks of height at least 2 by ψq and to double rises by ψx.It is well known that the number of permutations in Sn with k excedances equals the number of permutations in Sn with k + 1 weak excedances (recall that i is a weak excedance of π if πi ≥ i). We can show combinatorially that the analogous results for 321-avoiding and for 132-avoiding permutations hold as well. 106 Chapter 5. Other bijections and combinatorial interpretations
Proposition 5.6 Fix n, k ≥ 0. The following quantities are equal to the
Narayana number 1
n
(nk
)( nk+1
).
(1) The number of 321 -avoiding permutations π ∈ S n with k excedances.
(2) The number of 321 -avoiding permutations π ∈ S n with k + 1 weak excedances.
(3) The number of 132 -avoiding permutations π ∈ S n with k excedances.
(4) The number of 132 -avoiding permutations π ∈ S n with k + 1 weak excedances.
(5) The number of 132 -avoiding permutations π ∈ S n with k descents. Proof. By Proposition 2.8, excedances of π ∈ S n(321) correspond to double rises of ψx(π). It is known that the number of Dyck paths in Dn with k
double rises is given by the Narayana number 1
n
(nk
)( nk+1
).To prove the equality (1)=(2), consider the involution on Sn(321) that maps
π to ( φ−1
x
(ψx(π))) −1. Excedances of π ∈ S n(321) give double rises in ψx(π). On the other hand, the bijection φ−1
x
maps valleys to excedances. Combining this together, we have that exc( π) = dr( ψx(π)) = n − va( ψx(π)) − 1 =
n − exc( φ−1
x
(ψx(π))) − 1, where the second equality follows from the fact that each up-step of a Dyck path is either the beginning of a double rise or the beginning of a peak, so the number of peaks plus double rises equals the semilength of the path. Finally, we use that the number of excedances of a permutation in Sn plus the number of weak excedances of its inverse is n.The equalities (1)=(3) and (2)=(4) are immediate consequences of Theo-rem 2.3. Finally, to see that (1)=(5), notice that if π ∈ S n(321), then exc( π) = va( φx(π)). On the other hand, if π ∈ S n(132), then des( π) = va( ϕ(π)) by Proposition 2.1. Thus, ϕ−1 ◦ φx : Sn(321) −→ S n(132) maps excedances to descents. 2
5.3 Noncrossing permutations
In this section we consider a different class of permutations enumerated by the Catalan numbers, but which is not defined in terms of pattern avoidance (it is not equal to Sn(σ) for any pattern σ). They were introduced in by Hernando, Hurtado and Noy. We will denote this class Nn ⊂ S n, and call them noncrossing permutations . They are defined as those permutations 5.3. Noncrossing permutations 107
obtained in the following way. Consider 2 n points around a circle, labelled counterclockwise as 1 , 1′, 2, 2′, . . . , n, n ′. Now consider a noncrossing perfect matching between the 2 n points (that is, match pairs of points by drawing a line segment, in such a way that no two segments cross). Notice that each point with a label from the set {1, 2, . . . , n } gets matched with a point with a label in {1′, 2′, . . . , n ′}. Thus, to every such matching we can associate the permutation π that maps i to πi = j for each matched pair of the form ( i, j ′). Let Nn be the set of permutations obtained by this procedure. Clearly, |N n| = Cn, since the number of noncrossing perfect matchings of 2n points on a cycle is the Catalan number Cn.The class Nn can be characterized alternatively as the set of permutations
π ∈ S n for which there do not exist indices i < j such that either of the following conditions holds: (1) i < j ≤ πi < π j ,(2) πi < π j < i < j ,(3) i ≤ πj < π i < j ,(4) πj < i < j ≤ πi.We now give a recursive description of noncrossing permutations, which will be convenient for enumeration purposes. Given π ∈ S n, let ˜π the permuta-tion obtained from π by replacing n with 1 and increasing all of the remaining entries by one. Let ˜Nn = {˜π : π ∈ N n}. Recall the definition of the reduction
ρ from Section 1.1.1.
Lemma 5.7 Fix n ≥ 1, let π ∈ S n, let k be the index such that πk = n, and write π = τ n ω . Then, π ∈ N n if and only if τ ∈ N k−1 and ρ(ω) ∈ ˜Nn−k
(note that in particular τ is a permutation of {1, 2, . . . , k − 1}). Proof. Consider the matching of 2 n points that corresponds to π. Notice that
k is matched with n′. This splits the rest of the matching into two parts. On one side of the chord connecting k and n′ we have a noncrossing matching of {1, 1′, 2, 2′, . . . , k − 1, (k − 1) ′}, which corresponds to the permutation
τ ∈ N k−1. On the other side there is a noncrossing matching of {k′, k +1, (k + 1) ′, k + 2 , . . . , (n − 1) ′, n }, which gives rise to ω. 2
It is an immediate consequence of this lemma that any π ∈ ˜Nn can be decomposed uniquely as π = τ 1 ω where τ is a permutation of {2, 3, · · · , k }
such that ρ(τ ) ∈ N k−1, and ρ(ω) ∈ ˜Nn−k.108 Chapter 5. Other bijections and combinatorial interpretations
5.3.1 Distribution of descents
The decomposition above allows us to easily enumerate noncrossing permu-tations with respect to the number of descents. Let
N (u, z ) = ∑
n≥0
∑
π∈N n
udes( π)zn.
Proposition 5.8 The GF for noncrossing permutations with respect to the number of descents is
N (u, z ) = 1 − (1 − u)z − √1 − 2(1 + u)z + (1 − u)2z2
2uz .
Proof. Let
˜N(u, z ) = ∑
n≥0
∑
π∈eNn
udes( π)zn.
The decomposition in Lemma 5.7 translates into the equation
N (u, z ) = 1 + zN (u, z )[ u( ˜N (u, z ) − 1) + 1] ,
since the descents in τ nω are those in τ plus those in ω, plus an extra descent if ω is nonempty. A similar reasoning yields the equation
˜N (u, z ) = 1 + z[u(N (u, z ) − 1) + 1] ˜N (u, z ).
From these equations it follows that N (u, z ) = ˜N (u, z ), and solving for N
we get the desired GF. 2
The coefficient of ukzn in N (u, z ) is the Narayana number 1
n
(nk
)( nk+1
). This implies together with Proposition 5.6 that the distribution of descents is the same in Nn as in Sn(132) (and in fact the same as in Sn(213), Sn(231) and
Sn(312) as well). A bijection proving this can be described recursively using the fact that 231-avoiding permutations admit a decomposition which is very similar to that of noncrossing permutations. Indeed, any π ∈ S n(231) can be written uniquely as π = τ n ω where τ ∈ S k−1(231) and ρ(ω) ∈ S n−k(231), for some k.It follows from the symmetry of the Narayana numbers and it is also easy to see directly that the number of ascents in Nn has the same distribution as the number of descents. 5.3. Noncrossing permutations 109
5.3.2 Distribution of fixed points and excedances
Our goal in this subsection is to find an expression for the GF
A(x, q, z ) := ∑
n≥0
∑
π∈N n
xfp( π)qexc( π)zn.
Let us first consider fixed points in noncrossing permutations. An impor-tant observation is that from any π = π1π2 · · · πn ∈ N n we can obtain a permutation in Nn+1 with a fixed point in position i (for 1 ≤ i ≤ n + 1) just by inserting a new element i in position i and shifting up all the ele-ments greater or equal than i by one. That is, we build the permutation
π′
1
π′
2
· · · π′
i−1
iπ ′
i
· · · π′
n
∈ N n+1 , where
a′ :=
{ a if a < i ,
a + 1 if a ≥ i,for all a. Besides, any permutation in Nn+1 with a fixed point in position
i can be obtained in this way. In terms of the noncrossing matching cor-responding to π, this operation consists in inserting two points connected with an edge between the labels ( i − 1) ′ and i, labelling them i and i′, and shifting up all the labels that come after by one. By removing all the fixed points in a noncrossing permutation and relabelling the remaining elements accordingly, we obtain a noncrossing derangement. The above reasoning implies that, conversely, every noncrossing permutation can be uniquely obtained by starting from a noncrossing derangement and inserting a certain number of fixed points before each element and at the end. In terms of the GF A(0 , 1, z ) that enumerates noncrossing derangements, this translates into the equation 1
1 − z A(0 , 1, z
1 − z ) = C(z). (5.1) The substitution of z
1−z
for the third variable in A(0 , 1, z ) indicates that every element of the derangement is replaced with a sequence of fixed points followed by the element itself. The factor 1
1−z
corresponds to the fixed points inserted at the end of the permutation, and the right hand side is the GF for noncrossing permutations, which are counted by the Catalan numbers. The substitution z = t
1+ t
in (5.1) gives us
A(0 , 1, t ) = 1
1 + t C( t
1 + t ) = 1 −
√ 1 − 3t
1 + t
2t .110 Chapter 5. Other bijections and combinatorial interpretations
Now it is clear, again by the same argument, that the GF for noncrossing permutations where x marks the number of fixed points is
A(x, 1, t ) = 1
1 − xz A(0 , 1, z
1 − xz ) = 1 −
√
1 − (3 + x)z
1 + (1 − x)z
2z .
We will now look at the same problem is a different way, which will allow us to enumerate noncrossing permutations also with respect to the number of excedances.
Theorem 5.9 The GF for noncrossing permutations with respect to the number of fixed points and the number of excedances is
A(x, q, z ) = 1 + (1 − x)z − √1 − 2(1 + x)z + ( x2 + 2 x + 1 − 4q)z2
2z[1 + ( q − x)z] .
Proof. Recall the bijection between noncrossing matchings of points around a circle (also called chord diagrams) and Dyck paths defined in Section 1.2.2 (see Figure 1.4). Since each noncrossing permutation corresponds to one such matching, when we apply this bijection reading the points around the circle in the order {1, 1′, 2, 2′, . . . , n, n ′}, we get a bijection between Nn and
Dn, which we denote θ.A fixed point in the permutation corresponds to a chord joining points la-belled i and i′. Through the bijection θ, such chords are mapped to peaks at odd height in the Dyck path. Indeed, it is clear that chords joining two consecutive points in the circle (other than the pair ( n′, 1)) are mapped to peaks by θ. However, we only want chords joining a pair ( i, i ′), which appear after an even number of points has been read; hence the corresponding peak has odd height (we call this an odd peak ). On the other hand, an excedance in the permutation corresponds to a chord joining a pair ( i, j ′) where i < j . Such a chord produces in the Dyck path an up-step in an odd position (i.e., an even rise) when i is read, followed by another up-step when i′ is read, since by the noncrossing condition i′ cannot be matched with an element with a label less than i. Conversely, any even rise followed by another up-step (we call this an odd double rise ) comes from a chord corresponding to an excedance. Therefore, our problem is equivalent to the enumeration of Dyck paths with respect to odd peaks and odd double rises. We use the standard decomposi-tion of a nonempty Dyck path as D = uAdB, with A, B ∈ D , and then we 5.3. Noncrossing permutations 111
write A = uA1du A2d · · · as a sequence of elevated paths. The odd double rises of D are those in B plus those in each one of the Ai, plus an extra one created by the two u’s preceding A1 if A is nonempty. The odd peaks of D
are those in B plus those in each one of the Ai, plus an extra one if A is empty. In terms of the GF, we get
A(x, q, z ) = 1 + z
( q
1 − zA (x, q, z ) + x − q
)
A(x, q, z ).
The solution to this equation is the expression in the statement of the the-orem. 2
We have seen that fixed points and excedances are mapped respectively to odd peaks and odd double rises by θ. This means that for any noncrossing permutation π, we have that fp( π) + exc( π) = or( θ(π)), since every odd rise is the first step of either a double rise or a peak depending on whether the next step is up or down. Therefore, deficiencies of π correspond to even rises of the path θ(π). By part (2) of Theorem 4.2 and part (2) of Proposition 2.1, even rises in Dyck paths have the same distribution as excedances in 132-avoiding permutations. Since π avoids 132 if and only if so does π−1, the distribution of deficiencies in 132-avoiding permutations is also the same. We conclude that the statistic ‘number of deficiencies’ has the same distribution in noncrossing permutations as in 132-avoiding permutations. In particular, the number of permutations in Nn with k deficiencies is 1
n
(nk
)( nk+1
) (see Proposition 5.6). The set Nn is not closed under inversion π 7 → π−1. However, it is closed under other interesting operations. One of them is the transformation π 7 →̂
π, which in terms of the permutation array consists in a reflection along the secondary diagonal. Equivalently, we have that πi = j if and only if ̂
πn+1 −j = n+1 −i. If π ∈ N n is represented by a matching of 2 n points around a circle, the matching corresponding to ̂ π ∈ N n is obtained from it by writing the labels backwards, that is, n′ is now labelled 1, n becomes 1 ′, ( n − 1) ′
becomes 2 and so on. Another property of noncrossing permutations is that
π ∈ N n if and only if ( π(1 , 2, . . . , n )) −1 ∈ N n, where now we are using cycle notation, i.e, (1 , 2, . . . , n ) is a cycle of length n. In terms of the matching, this corresponds to shifting the labels one position in clockwise direction, that is, 1 becomes n′, 1 ′ becomes 1, 2 is now 1 ′, 2 ′ becomes 2 and so on. Applying this operation twice, we obtain the permutation (1 , 2, . . . , n )−1π(1 , 2, . . . , n ). This shows that Nn is closed under conjugation by the cycle (1 , 2, . . . , n ). 6
Consecutive patterns
In this chapter we consider a different notion of pattern avoidance. We introduce the concept of consecutive patterns , which we also call subwords .The difference with respect to the ordinary patterns studied in the previous chapters is that now, in order to form an occurrence of a consecutive pattern, the corresponding elements in the permutation have to be adjacent, that is, in consecutive positions. In this chapter all the patterns that we consider will be consecutive patterns. We study the distribution of the number of occurrences of a permutation
σ as a subword among all permutations in Sn. We solve the problem in several cases depending on the shape of σ by obtaining the corresponding bivariate exponential generating functions as solutions of certain linear dif-ferential equations with polynomial coefficients. Our method is based on the representation of permutations as increasing binary trees and on symbolic methods. Most results in this chapter are joint work with Marc Noy . Let m, n be two positive integers with m ≤ n, and let π = π1π2 · · · πn ∈ S n
and σ ∈ S m be two permutations. We say that π contains σ as a subword if there exist m consecutive elements πl+1 · · · πl+m such that ρ(πl+1 · · · πl+m) =
σ, where ρ is the reduction that consists in relabelling the elements with
{1, . . . , m } so that they keep the same order relationships they had in π.For example, if σ = 4132 ∈ S 4, then π = 6725341 ∈ S 7 contains σ as a sub-word, because ρ(7253) = 4132. However, π = 41352 ∈ S 5 does not contain
σ as a subword (even though it contains it as a subsequence, that is, in non-consecutive positions); in this case we say that π avoids σ. Occurrences of a subword can be overlapped, for instance, 5716243 contains two occurrences of σ, namely 7162 and 6243. Denote by An(σ) the set of permutations of Sn that do not contain σ as a 114 Chapter 6. Consecutive patterns
subword, and let αn(σ) = |An(σ)|. If we want to exclude several subwords
σ, τ, . . . we use the corresponding notations An(σ, τ, . . . ) and αn(σ, τ, . . . ). Our main purpose is to compute αn(σ) for a given subword σ. More gener-ally, we are also interested in the distribution of the number of occurrences of σ among all permutations in Sn.Some well-known counting problems in permutations can be stated in terms of forbidden subwords. For instance, occurrences of 12 correspond to ascents and are counted by Eulerian numbers; up-and-down permutations are those in An(123 , 321); permutations whose longest increasing run is at most k − 1correspond to An(12 · · · k); and it is not difficult to see that αn(123 , 132) is precisely the number of involutions. Related results for occurrences of subwords of length three can be found in and, more recently, in . The basis of our work is the use of symbolic methods for specifying combina-torial classes, following the approach described in the books by Flajolet and Sedgewick [41, 83] and summarized in Section 1.3.2. The key point is the representation of permutations as binary increasing trees. From this repre-sentation, using symbolic methods, we derive differential equations satisfied by the corresponding exponential generating functions. In all cases we have encountered, the differential equations become linear after a suitable substi-tution. The reader can compare this technique with , where a similar approach is taken for analyzing certain geometric configurations, the differ-ence being that here we deal with exponential GFs that are transcendental instead of being algebraic. A related approach is taken in for counting occurrences of a given subtree in binary search trees. The main difference in our case is however that patterns corresponding to forbidden subwords may be split into two subtrees in several different ways. The organization of this chapter is as follows. First, we present some pre-liminaries on the representation of permutations as increasing trees and on asymptotic enumeration. In Section 6.2 we enumerate occurrences of a sub-word σ in two cases. Firstly, when σ is totally increasing (or decreasing); and secondly when σ = 12 · · · (a − 1) a τ (a + 1), and τ is any non empty permutation of the elements {a + 2 , a + 3 , . . . , m + 2 }. Notice that the fact that τ and a are arbitrary means that our result covers a very large number of subwords σ. In Section 6.3 we show how these results specialize in the case of subwords of length three and four. We conclude with some remarks. 6.1. Preliminaries 115
6.1 Preliminaries
6.1.1 Tree representation of permutations
Following Stanley [89, Chapter 1], we represent a permutation as an increas-ing binary tree, that is, a binary tree in which the labels along any path from the root are increasing. This representation allows us to translate combina-torial properties of permutations (such as avoiding a certain subword) into combinatorial properties of trees, which can be handled more conveniently. Let π = a1a2 · · · an be a word on the alphabet of positive integers with no repeated letters. Define a binary tree T (π) as follows. If π = ∅, then
T (π) = ∅. Otherwise, let i be the least element of π, so that π can be factored uniquely in the form π = σiτ . Now define T (π) by induction as the tree with root i, and having T (σ) and T (τ ) as left and right subtrees, respectively. This correspondence gives a bijection between Sn and the set of
increasing binary trees on n vertices. In particular, we see that the number of such trees is n!.
Generating functions for increasing trees. We apply the machinery described in Section 1.3.2 to increasing trees. The following derivation will be used repeatedly. Let B0 be the labelled class of (possibly empty) binary increasing trees. It satisfies the recursive definition
B0 = {} + ( {z}2 ? B0 ? B0),
where is the empty tree, {z} represents the tree with one single node, and the box indicates that the root contains the smallest label. So, the equation for the EGF is
I0(z) = 1 +
∫ z
0
I0(t)2dt,
which reduces to I0′(z) = I0(z)2 with initial condition I0(0) = 1 (derivatives are always with respect to z in this chapter), admitting the solution I 0(z) = 1/(1 − z). Thus, I0
n
= n! as expected.
6.1.2 Equivalent subwords
We say that two subwords σ, τ ∈ S m are equivalent if the BGF counting occurrences of σ and τ are the same (equivalently, for all n, k , the number 116 Chapter 6. Consecutive patterns
of permutations in Sn with k occurrences of σ equals the number of those with k occurrences of τ ). We write σ ∼ τ to denote equivalence. In Section 1.1.3 we described two simple operations that give equivalent subwords: reversal , which transforms σ = σ1 · · · σm into σR = σm · · · σ1, and
complementation , transforming σ into ¯ σ = ( m + 1 − σ1) · · · (m + 1 − σm). The explanation is that a permutation π has as many occurrences of σ as
πR has of σR, and as ¯ π has of ¯ σ.
6.1.3 Asymptotic enumeration
Let F (z) be a meromorphic function on a domain of the complex plane including the origin, and let ρ be the unique pole of F (z) such that |ρ| is minimum. Then the following asymptotic estimate holds: [zn]F (z) ∼ γ · ρ−n,
where γ is the residue of F in ρ. If F (z) is defined in the whole complex plane and has no poles, then lim
n→∞
[zn]F (z) = 0 .
See [41, Chapter 4] for a discussion.
6.2 Main results
This section contains the main results of this chapter. We obtain the count-ing BGF of occurrences of a subword σ in two cases. First we treat the case of increasing (or decreasing) subwords.
6.2.1 Increasing subwords
Theorem 6.1 Let m be a positive integer, let σ = 12 · · · (m + 1)( m + 2) ∈Sm+2 , and let P (u, z ) be the BGF of permutations where u marks the number of occurrences of the subword σ. Then, P (u, z ) = 1 /ω (u, z ), where ω is the solution of
ω(m+1) + (1 − u)( ω(m) + ω(m−1) + . . . + ω′ + ω) = 0 (6.1)
with ω(0) = 1 , ω′(0) = −1, and ω(k)(0) = 0 for 2 ≤ k ≤ m.6.2. Main results 117
Proof. We use the correspondence of permutations as binary increasing trees. We get for P (u, z ) a system of m + 1 first order differential equations, which will be reduced to a differential equation of order m+1 with the substitution
P (u, z ) = 1 /ω (u, z ). Let P be the class of all permutations, let Ki be the class of permutations not beginning with 12 · · · (m + 2 − i), and let Ki(u, z )be the BGF of Ki where u marks occurrences of σ. With some abuse of notation, we introduce the parameter u in the equations for classes meaning that it will be placed there when we write the corresponding differential equations for the BGF. With this notation, we can write
P = {} + {z}2 ? P ? [K1 + u(P − K 1)] .
This is because occurrences of σ in a permutation (seen as a binary increasing tree) can be separated into occurrences on the left subtree and occurrences on the right subtree, taking into account that if the permutation on the right subtree begins with 12 · · · (m + 1) (that is, belongs to P − K 1), then there is an additional occurrence of σ beginning at the root. The corresponding equation for BGFs is, after differentiating,
P ′ = P (K1 + u(P − K1)) .
The next step is to find an equation for K1. Note that for a permutation not to begin with a 12 · · · (m + 1), it is not enough that the permutation on the left subtree does not begin with this subword. We have to exclude also the case in which the left subtree is empty and the right subtree begins with a 12 · · · m. This gives us
K1 = {} + {z}2 ? (K1 − { }) ? [K1 + u(P − K 1)] + {z}2 ? K2,
which translates to
K′
1
= ( K1 − 1)( K1 + u(P − K1)) + K2
when we differentiate the equation for BGFs. Now it is clear how to find an equation for K2, in which K3 will appear, and so on, until we arrive to Km
(permutations not beginning with 12), which satisfies
Km = {} + {z}2 ? (Km − { }) ? [K1 + u(P − K 1)] + {z}.
All these equations yield a system of m + 1 differential equations involving the corresponding BGFs. It is convenient to apply the substitution
R = uP + (1 − u)K1.118 Chapter 6. Consecutive patterns
Note that R′ = uP ′ + (1 − u)K′
1
= uP R + (1 − u)(( K1 − 1) R + K2) = (uP + (1 − u)K1)R + (1 − u)( −R + K2) = R2 + ( u − 1)( R − K2), and we obtain
P ′ = P R R′ = R2 + ( u − 1)( R − K2)
K2′ = ( K2 − 1) R + K3
K3′ = ( K3 − 1) R + K4
...
Km−1′ = ( Km−1 − 1) R + Km
Km′ = ( Km − 1) R + 1 with P (0) = R(0) = 1,
Ki(0) = 1 for all i. (6.2) Now we only have to check that setting P (u, z ) = 1 /ω (u, z ), then ω satisfies (6.1). The first equation P ′ = P R gives R = − ω′
ω
. Substituting this into the second one, we get K2 = ω′′ +(1 −u)ω′
(u−1) ω
. By induction on i, we see that
Ki = ω(i) + (1 − u)( ω(i−1) + ω(i−2) + . . . + ω′)
(u − 1) ω
and ω(i)(0) = 0 for 2 ≤ i ≤ m. Finally, (6.1) is obtained substituting in the last equation of (6.2) the expressions for Km and R in terms of ω. 2
For u = 0 the solution of the differential equation can be expressed as a linear combination of exponentials
w =
m+1
∑
j=1
cj eλj z ,
where the λj = exp( 2πij
m+2
) are the non trivial ( m + 2)-th roots of unity, and the indetermined coefficients ci are the solution of the linear system
1 1 · · · 1
λ1 λ2 · · · λm+1
λ21 λ22 · · · λ2
m+1
· · · · · · · · · · · ·
λm
1
λm
2
· · · λmm+1
·
c1
c2
c3
· · ·
cm+1
=
1
−10
· · ·
0
The matrix M of the previous system is easy to invert, since
M M ∗ = ( m + 2) I − J, 6.2. Main results 119
where A∗ denotes the conjugate transpose of A, and J is the all ones matrix. From this it follows that
M −1 = 1
m + 2 M ∗(I + J),
and we can obtain the value of the cj , thus an explicit expression for P (0 , z ), the GF of permutations avoiding the subword σ = 12 · · · (m + 2). For exam-ple, for m = 4 one gets
c1 = (1 + i)/4, c2 = 1 /2, c3 = (1 − i)/4,
which agrees with the solution w = (cos z−sin z+e−z)/2 given in Section 6.3.
6.2.2 Other subwords
The next result gives the distribution of subwords of a certain, more general shape.
Theorem 6.2 Let m, a be positive integers with a ≤ m, let σ = 12 · · · (a −
1) a τ (a + 1) ∈ S m+2 , where τ is any permutation of the elements {a + 2 , a +3, . . . , m +2 }, and let P (u, z ) be the BGF of permutations where u marks the number of occurrences of the subword σ. Then, P (u, z ) = 1 /ω (u, z ), where
ω is the solution of
ω(a+1) + (1 − u) zm−a+1
(m − a + 1)! ω′ = 0
with ω(0) = 1 , ω′(0) = −1 and ω(k)(0) = 0 for 2 ≤ k ≤ a. In particular, the distribution does not depend on τ .Proof. Again we find a system of a + 1 differential equations for P (u, z )that, after the substitution P (u, z ) = 1 /ω (u, z ), yield a single differential equation of order a + 1. Let P be as before the class of all permutations. For 1 ≤ i ≤ m + 1, we denote σ>i = ρ(σi+1 σi+2 · · · σm+2 ). Note that for i < a ,
σ>i has its smallest element in the first position, while for i ≥ a, the smallest element of σ>i is the last one. Now let Ki be the class of permutations not beginning with any of the following: σ>1, σ >2, . . . , σ >i , and let Ki(u, z ) be the BGF of Ki where u marks occurrences of σ. With this notation, we can write
P = {} + {z}2 ? P ? [K1 + u(P − K 1)] .120 Chapter 6. Consecutive patterns
The explanation, as in the previous proof, is that in the decomposition of the increasing tree, new occurrences of σ, apart from the ones on the left and right subtrees, can appear beginning in the root, when the permutation on the right subtree begins with σ>1 (that is, belongs to P − K 1). For K1 we derive the equation
K1 = {} + {z}2 ? (K1 − { }) ? [K1 + u(P − K 1)] + {z}2 ? [K2 + u(P − K 1)] .
Note that the last summand corresponds to the case where the left subtree is empty. Then, the permutation on the right subtree cannot begin with σ>2,which would produce, together with the root, a permutation beginning with
σ>1. So, after separating on the right subtree the permutations beginning with σ>1 (P − K 1), we are left with those beginning with neither σ>1 nor
σ>2, that is, K2.Analogous expressions can be found for Ki, 1 < i < a :
Ki = {} + {z}2 ? (Ki − { }) ? [K1 + u(P − K 1)] + {z}2 ? [Ki+1 + u(P − K 1)] ,
which yield the equations
K′
i
= ( Ki − 1)( K1 + u(P − K1)) + ( Ki+1 + u(P − K1)) .
For Ka we have
Ka = {}+{z}2?(Ka −{ }−{ ρ(τ )})?[K1 +u(P−K 1)]+ {z}2?[Ka +u(P−K 1)] .
The difference now is that in order to avoid a permutation beginning with
σ>a = ρ(τ (a+1)), the left subtree cannot be ρ(τ ). (Remember that the BGF corresponding to {ρ(τ )} is zm−a+1 /(m − a + 1)! since it is a permutation of size m − a + 1 and does not contain the subword σ.) Another difference is that no new variables appear, since when the left subtree is empty there is no danger of beginning with σ>a , and so there are no additional restrictions for the right subtree. From all these we get a system of a + 1 differential equations. After applying the substitutions R = uP + (1 − u)K1 as before, and also Si = Ki−1 − Ki
for 2 ≤ i ≤ a, we obtain
P ′ = P R R′ = R2 + ( u − 1) S2
S2′ = S2R + S3
S3′ = S3R + S4
...
Sa−1′ = Sm−1R + Sa
Sa′ = ( Sa + zm−a+1
(m−a+1)!
)R
with P (0) = R(0) = 1,
Si(0) = 0 for all i. (6.3) 6.2. Main results 121
Finally, it only remains to set P (u, z ) = 1 /ω (u, z ) and to check that ω is the solution of (6.4). Analogously to the previous proof, it can be shown by induction that Si = ω(i)
(1 −u)ω
for 2 ≤ i ≤ a. Then, (6.4) follows from the substitution in the bottom equation of the system. 2
Note that the fact that a ≤ m ensures that τ is not empty. The case in which
τ is empty corresponds to the case where σ is the increasing permutation, and has already been treated in the previous theorem. We end this section with a certain condition under which two subwords σ
and τ are equivalent, i.e., σ ∼ τ . We say that a subword σ ∈ S m is non-self-overlapping if there is no permutation in S2m−2 with more than one occurrence of σ. (Note that in S2m−1 such a permutation would always exist.) The non-self-overlapping condition implies that for any two occur-rences πi+1 · · · πi+m and πj+1 · · · πj+m of σ in a permutation π, we have that
|i − j| ≥ m − 1, that is, the two occurrences cannot overlap in more than one element. For example, 136254 is a non-self-overlapping subword, but 1324 is not, because the permutation π = 142536 has two occurrences of 1324.
Proposition 6.3 Let m ≥ 3 and let σ ∈ S m be a non-self-overlapping sub-word such that σ1 = 1 . Let a = σm − 1, and let τ = 12 · · · a(a + 2)( a +3) · · · m(a + 1) . Then we have that σ ∼ τ .
Before proving this proposition, note that now Theorem 6.2 can be general-ized as follows.
Corollary 6.4 Let m ≥ 3 and let σ ∈ S m be a non-self-overlapping subword such that σ1 = 1 . Let a = σm −1, and let P (u, z ) be the BGF of permutations where u marks the number of occurrences of the subword σ. Then P (u, z ) = 1/ω (u, z ), where ω is the solution of
ω(a+1) + (1 − u) zm−a−1
(m − a − 1)! ω′ = 0 (6.4)
with ω(0) = 1 , ω′(0) = −1 and ω(k)(0) = 0 for 2 ≤ k ≤ a.
Example. The patterns 136254 and 125364 are non-self-overlapping. Thus, by Proposition 6.3, 136254 ∼ 125364 ∼ 123564, and the BGF counting occurrences of any of them in permutations is given by P (u, z ) = 1 /ω (u, z ), where ω is the solution of
ω′′′′
(1 − u) z2
2 ω′ = 0 ,122 Chapter 6. Consecutive patterns
with ω(0) = 1, ω′(0) = −1 and ω′′ (0) = ω′′′ (0) = 0. P (u, z ) can be expressed in terms of hypergeometric series.
Proof of Proposition 6.3. Assume that τ and σ are different, otherwise there is nothing to prove. Notice that τ is non-self-overlapping, since it has only one descent, in position m − 1. Besides, the only way in which an occurrence of σ and an occurrence of τ can overlap in a permutation π is with the occurrence of τ starting to the left of the occurrence of σ. Indeed, assume for a contradiction that π has an occurrence πi+1 πi+2 · · · πi+m of σ
overlapping with an occurrence πj+1 πj+2 · · · πj+m of τ with 1 ≤ j−i ≤ m−2. The fact that σ is non-self-overlapping and starts with an ascent implies that σm−1 > σ m (otherwise there would be a permutation in S2m−2 with two occurrences of σ starting in positions 1 and m − 1 respectively). But then, πi+m−1 > π i+m, which contradicts the condition πj+1 < π j+2 < · · · <πj+m−1 needed for it to be an occurrence of τ .Now we define a bijection Γ : Sn −→ S n such that for every π ∈ S n, the number of occurrences of σ (resp. τ ) in π equals the number of occurrences of τ (resp. σ) in Γ( π). From our previous observations, the occurrences of σ
and τ in π are either isolated (not overlapping with any other occurrence of
σ or τ ) or they appear in a pair formed by an occurrence of τ overlapping with one of σ, with the occurrence of τ starting to the left of one of σ.To construct Γ( π), read the permutation π from left to right and do the following: (1) For each pair of overlapping occurrences of τ and σ, leave it unchanged. (2) For each isolated occurrence of σ, reorder the corresponding elements in π so that they form an occurrence of τ instead. There are now two possibilities. If this modification does not create any new occurrence of σ, we can jump to the following occurrence. Assume on the contrary that this process created a new σ. Such an occurrence is necessarily overlapping with the new occurrence of τ , starting the one of σ to the right of the one of τ . In this case, reorder the elements of π correspond-ing to this new occurrence of σ so that they form an occurrence of τ
instead. Note that this erases the previous occurrence of τ that we had just created, since τ is non-self-overlapping. Again, check if any new occurrence of σ has been created, and repeat this process until it cre-ates no more occurrences of σ. After this iteration, the initial isolated
σ has been transformed into an isolated occurrence of τ . Observe also that this procedure cannot overwrite an existing occurrence of σ or τ ,6.3. Subwords of length at most four 123
because the new occurrences of σ that may be created in the process cannot overlap with occurrences of σ or τ that start more to the right. (3) For each isolated occurrence of τ , reorder the corresponding elements in π so that they form an occurrence of σ instead. If this modification does not create any new occurrence of τ , we can jump to the following occurrence. On the other hand, if this created a new occurrence of τ ,which is necessarily overlapping with the new occurrence of σ, starting the one of τ to the left of the one of σ, reorder the elements of π corre-sponding to this new occurrence of τ so that they form an occurrence of σ instead. If necessary, repeat the process as before, until no more occurrences of τ are created. After this iteration, the initial isolated occurrence of τ has been transformed into an isolated occurrence of σ.By the same reasoning as above, this procedure cannot overwrite an existing occurrence of σ or τ .It is easy to see that Γ is in fact an involution, since steps (2) and (3) are inverses from each other, one transforming isolated occurrences of σ into isolated occurrences of τ , and the other doing the opposite. This proves that σ ∼ τ . 2
It seems from experimental computations that a more general version of Proposition 6.3 holds, namely, that any two non-self-overlapping subwords
σ, τ ∈ S m with σ1 = τ1 and σm = τm satisfy σ ∼ τ , but we have not been able to prove this fact. However, we can make small variations of Proposition 6.3 to obtain similar results like the following, whose proof is now straightforward.
Proposition 6.5 Let m ≥ 3 and let σ, τ ∈ S m be non-self-overlapping sub-words such that no permutation in S2m−2 contains σ and τ simultaneously (i.e., sigma and tau cannot overlap with each other). Then σ ∼ τ .
Example. The previous proposition shows that 24153 ∼ 25143 and that 351264 ∼ 362154.
6.3 Subwords of length at most four
Occurrences of the two subwords 12 and 21 of length two correspond, respec-tively, to ascents and descents in permutations, giving rise to the well-known Eulerian numbers . 124 Chapter 6. Consecutive patterns
6.3.1 Subwords of length three
Among the 6 permutations of three elements, there are only two different classes regarding its distribution as subwords of permutations since, by re-versal and complementation, we have that 123 ∼ 321 132 ∼ 231 ∼ 312 ∼ 213 By the results in the previous section we get:
Theorem 6.6 Let P (u, z ) and Q(u, z ) be the BGFs of permutations where
u marks, respectively, the number of occurrences of the subword 123 and 132 .Then
Q(u, z ) = 1
1 − ∫ z
0
e(u−1) t2 /2dt ,P (u, z ) = = 2e 1
2(1 −u+
√(u−1)( u+3)) z √(u − 1)( u + 3)
1 + u + √(u − 1)( u + 3) + e
√(u−1)( u+3) z (−1 − u + √(u − 1)( u + 3))
,P (0 , z ) =
√3
2
ez/ 2
cos(
√3
2
z + π
6
) .
Furthermore, the numbers αn(123) and αn(132) of permutations avoiding, respectively, the subwords 123 and 132 , satisfy
αn(123) ∼ γ1 · (ρ1)n · n!
αn(132) ∼ γ2 · (ρ2)n · n!
where ρ1 = 3 √3/(2 π), γ1 = e3√3π, (ρ2)−1 is the unique positive root of ∫ z
0
e−t2/2dt = 1 , and γ2 = exp(( ρ2)−2/2) , the approximate values being
ρ1 = 0 .8269933 , γ1 = 1 .8305194
ρ2 = 0 .7839769 , γ2 = 2 .2558142 Table 6.1 indicates the number of permutations of length n with k occur-rences of the subwords 123 (top) and 132 (bottom). The asymptotic estimates are obtained as an application of the result quoted in Section 6.1.3. (The computation of ρ2 has been done numerically using 6.3. Subwords of length at most four 125
n\k 0 1 2 3 4
1 11
2 22
3 55
11
4 17 16
68
1
5 70 63
41 54
83
1
6 349 296
274 368
86 56
10 1
Table 6.1
Occurrences of subwords of length 3 in permutations. the computer algebra system Maple .) Since ρ1 > ρ 2 in the previous the-orem, we see that αn(123) is asymptotically larger than αn(132). Looking at Table 6.1 one is led to conjecture that this is always so. Indeed, we have the following result, which is analogous to Theorem 2.14 for the case of consecutive patterns.
Proposition 6.7 For every n ≥ 4, we have
αn(123) > α n(132) .
Proof. Given σ ∈ S m, let Bn(σ) = Sn − An(σ) be the permutations of Sn
containing σ. Define a map
γ : Bn(123) −→ Bn(132) as follows. If π ∈ Bn(123) contains occurrences of both 123 and 132, then
γ(π) = π. Otherwise (that is, π contains 123’s but not 132’s), define γ(π) as the permutation obtained by traversing π from left to right and substituting every occurrence of 123 by 132 (transposing the elements in the positions corresponding to 2 and 3). It only remains to check that γ is one to one, and this is because when a 123 is changed to a 132, no new occurrences of 123 appear that did not exist before the substitution. To prove that the inequality is strict for n ≥ 4, observe that any permutation beginning with 1423 and having no 123 cannot be of the form γ(π) for any π. 2126 Chapter 6. Consecutive patterns
6.3.2 Subwords of length four
By reversal and complementation, and by the results in the previous section, they fall into seven classes: I. 1234 ∼ 4321 II. 2413 ∼ 3142 III. 2143 ∼ 3412 IV. 1324 ∼ 4231 V. 1423 ∼ 3241 ∼ 4132 ∼ 2314 VI. 1342 ∼ 2431 ∼ 4213 ∼ 3124 ∼ 1432 ∼ 2341 ∼ 4123 ∼ 3214 VII. 1243 ∼ 3421 ∼ 4312 ∼ 2134 The results in the previous section give the BGFs for the distribution of occurrences of subwords in classes I, VI and VII.
Theorem 6.8 In each of the following cases, let P (u, z ) be the BGF of permutations where u marks the number of occurrences of the corresponding subword. Case 1342 . P (u, z ) = 1
1 − ∫ z
0
e(u−1) t3 /6dt .
Case 1234 . P (u, z ) = 1 /ω , where ω is the solution of
ω′′′ + (1 − u)( ω′′ + ω′ + ω) = 0
with ω(0) = 1 , ω′(0) = −1, ω′′ (0) = 0 . For u = 0 , the solution is
P (0 , z ) = 2
cos z − sin z + e−z
Case 1243 . P (u, z ) = 1 /ω , where ω is the solution of
ω′′′ + (1 − u)zω ′ = 0
with ω(0) = 1 , ω′(0) = −1, ω′′ (0) = 0 .Furthermore, the numbers αn(1342) , α n (1234) and αn(1243) satisfy
αn(1342) ∼ γ1 · (ρ1)n · n!
αn(1234) ∼ γ2 · (ρ2)n · n!
αn(1243) ∼ γ3 · (ρ3)n · n!
where (ρ1)−1 is the smallest postitive solution of ∫ z
0
e(u−1) t3 /6dt = 1 , (ρ2)−1
is the smallest positive solution of cos z − sin z + e−z = 0 , and ρ3 is the 6.3. Subwords of length at most four 127
solution of a certain equation involving Airy functions. The approximate values are
ρ1 = 0 .954611 , γ1 = 1 .8305194
ρ2 = 0 .963005 , γ2 = 2 .2558142
ρ3 = 0 .952891 , γ3 = 1 .6043282
n\k 0 1 2 3 4
4
III III IV VVI VII
23 23 23 23 23 23 23
1111111
5
III III IV VVI VII
111 110 110 110 110 110 110
810 10 10 10 10 10
1
6
III III IV VVI VII
642 632 631 632 631 630 630
67 86 88 86 88 90 90
10 2121
1
7
III III IV VVI VII
4326 4237 4223 4229 4218 4210 4204
602 766 794 782 804 820 832
99 37 23 29 18 10 4
12 1
Table 6.2
Occurrences of subwords of length 4 in permutations. In the last case, the equation ω′′′ + (1 − u)zω ′ = 0, for u = 0 and v = w′,can be solved in terms of Bessel functions (note that the equation for v128 Chapter 6. Consecutive patterns
is actually a slight variant of the Airy equation). The computations have been performed with the help of Maple . Table 6.2 shows the number of occurrences in each of the seven classes for n ≤ 7, showing that no two of them have the same distribution. (Entries II, III, IV, V in Table 6.2 have been computed directly.) As in the case of length 3, one might expect that again for any two permu-tations σ, τ ∈ S 4, if αn0 (σ) > α n0 (τ ), then αn(σ) > α n(τ ) for all n > n 0.However, inequality αn(1324) ≥ αn(2143) holds for n ≤ 11 but does not hold for n = 12. Indeed,
α11 (1324) = 27959880 > 27954521 = α11 (2143) ,
but
α12 (1324) = 320706444 < 320752991 = α12 (2143) .
These results have been obtained by exhaustive computation, as we do not know the associated EGFs. There is however one relation among classes that we have been able to establish
Proposition 6.9 For every n ≥ 7, we have
αn(1342) > α n(1243) .
Proof. As in the proof of Proposition 6.7, let Bn(σ) = Sn − An(σ) be the set of permutations containing σ. Define a map
γ : Bn(1342) −→ Bn(1243) as follows. Let γ(π) = π if π contains both 1342 and 1243. Otherwise, replace, from left to right, all occurrences of 1342 by 1243. Note that when we replace an occurrence 1342 by 1243, we never create new occurrences of 1342 or 1243. (This is not true in general for other patterns, so that the corresponding γ is not a bijection.) It is clear that γ(Bn(1342) ∩ Bn(1243)) ∩ γ(Bn(1342) ∩ An(1243)) = ∅,because in the second case there are no 1342 left. Let π 6 = η ∈ Bn(1342) ∩
An(1243). Now suppose that γ(π) = γ(η). Let i be the smallest index so that πi 6 = ηi. Either πi or ηi must be moved by γ.Now observe that if, say, πi is changed, it cannot be transposed with any of the preceding elements, so it must be the ‘3’ of a 1342 in π, and thus is interchanged with the ‘2’ in position i + 2. But now, after replacing this 6.4. Multiple subwords 129
1342 by 1243, the ‘3’ can no longer be moved, because it is neither the ‘2’ nor the ‘3’ of any other 1342. If ηi is also moved by γ, this reasoning implies that γ(π)i+2 = πi 6 = ηi = γ(η)i+2 , which is impossible since γ(π) = γ(η). So ηi is not moved by γ, but then positions i − 1, i, i + 1 and i + 2 of γ(π)are πi−1πi+2 πi+1 πi = ηi−1ηiηi+1 ηi+2 , because this positions of γ(η) cannot have been moved by γ. This is a 1243, which contradicts the fact that
η ∈ An(1243). 2
6.4 Multiple subwords
Instead of occurrences of a single subword one may consider several sub-words. For the case of length three, many of the possible combinations are treated in . For example, the class An(123 , 321) avoiding 123 and 321 is clearly that of up-and-down permutations. According to a classical result (see ), the corresponding EGF is 2
(
tan z + 1
cos z
)
− 1 − z.
The class An(213 , 312) is that of permutations π having no valleys , that is, positions i such that πi−1 > π i < π i+1 (not to be confused with valleys of a Dyck path). The BGF for permutations where u marks valleys is easily shown to be √1 − u
√1 − u − tanh( z√1 − u) .
And the class An(123 , 132) is equinumerous with the class of involutions. This can be easily explained using the classical correspondence of Foata [89, Section 1.3]. Indeed, given the decomposition of a permutation into disjoint cycles, with their smallest element first and ordering the cycles in decreasing order of the smallest elements, all cycles of length greater or equal than three begin with either a 123 or a 132. Conversely, if the described decomposition in cycles contains one of these subwords, all three elements are necessarily in the same cycle, and thus the permutation with that cyclic decomposition cannot be an involution. We can also find a multivariate generating function Q(u, v, z ) where u marks occurrences of 123 and v marks occurrences of 132. Using similar arguments as before, it can be seen that Q(u, v, z ) is the solution of
{ R′ = R2 + [( u − 1) + ( v − 1) z]R − (u − 1)
Q′ = QR with R(0) = Q(0) = 1 . (6.5) 130 Chapter 6. Consecutive patterns
One of the cases not solved in is the simultaneous avoidance of the sub-words 123 and 231. We can find a multivariate generating function P (u, v, z )where the coefficient of ukvlzn/n ! is the number of elements of Sn with k
occurrences of 123 and l occurrences of 231. Let P be as before the class of all permutations, and let K, L, M be sub-classes of P defined as follows: K are the permutations not beginning with 12, L are those not ending with 12, and M = K ∩ L are the ones that nei-ther begin nor end with 12. Let P (u, v, z ), K(u, v, z ), L(u, v, z ), M (u, v, z )be respectively the generating functions of these four classes, where u marks occurrences of 123 and v marks occurrences of 231. We get the following relations for these classes.
P = {} + {z}2 ? [L + v(P − L )] ? [K + u(P − K )]
K = {} + {z} + {z}2 ? [M − { } + v(K − M )] ? [K + u(P − K )]
L = {} + {z}2 ? [L + v(P − L )] ? [M − { z} + u(L − M )]
M = {} + {z}
+{z}2 ? [M − { } + v(K − M )] ? [M − { z} + u(L − M )] The idea is the same as in previous proofs. It is based on the fact that a left subtree ending in 12 produces an occurrence of 231 along with the root, and a right subtree beginning with 12 produces a 123 with the root. Therefore, this situations must be marked with v and u respectively. For example, in the third relation, in order for the permutation not to end with 12, the right subtree must be an element of L − { z}, and if it belongs to L − M (i.e., begins with 12) then it must be marked with a u.This gives the following system of differential equations for the generating functions ( u and v are considered as parameters).
P ′ = [ L + v(P − L)][ K + u(P − K)]
K′ = 1 + [ M + v(K − M ) − 1][ K + u(P − K)]
L′ = [ L + v(P − L)][ M + u(L − M ) − z]
M ′ = 1 + [ M + v(K − M ) − 1] M + u(L − M ) − z with P (0) = K(0) = L(0) = M (0) = 1 .
In particular, if we are interested in the EGF A(z) = P (0 , 0, z ) whose coeffi-cients are the number of permutations avoiding 123 and 231 simultaneously, we obtain the following result. 6.5. Concluding remarks 131
Theorem 6.10 The EGF A(z) of permutations avoiding both 123 and 231
as subwords is the solution the following system of equations, where deriva-tives are with respect to z:
A′ = CB B′ = 1 + ( D + z − 1) BC′ = CD D′ = ( D + z − 1) D
with A(0) = B(0) = C(0) = D(0) = 1 .
An involved explicit form for A(z) can be found in terms of integrals con-taining the error function, but it seems not suitable for obtaining asymptotic results. One can however obtain from the above system as many terms of
A(z) as desired, and we get that the corresponding counting sequence begins with 1, 2, 4, 11 , 39 , 161 , 784 , 4368 , 27260 , 189540 , 1448860 , 12076408 ,
109102564 , 1061259548 , . . .
6.5 Concluding remarks
Let us make a few comments concerning the distribution of the number of occurrences of a subword. For σ ∈ S m and n ≥ m, let Xσ,n be the random variable defined on Sn equal to the number of occurrences of σ. It is easy to see that its expectation is E(Xσ,n ) = n−m+1
m!
and that its variance is
V ar (Xσ,n ) = cσ n2 for some constant cσ . It follows that the distribution is asymptotically concentrated around the expected value. In fact, using the reasoning in , it can be shown that Xσ,n is asymptotically normal. The asymptotic behavior of the numbers αn(σ) as n goes to infinity is con-sidered in the next chapter. 7
Asymptotic enumeration of permutations avoiding generalized patterns
In this chapter we discuss a generalization of the notion of pattern avoidance. The concept of generalized pattern includes both the classical definition from Section 1.1.1 used in Chapters 2, 3 and 4, and the notion of consecutive patterns described in Chapter 6. In Section 7.1 we introduce the definitions and we give the exponential generating functions for permutations avoiding a special kind of generalized patterns. In Section 7.2 we study the asymptotic behavior as n goes to infinity of the number of permutations in Sn avoiding a generalized pattern. We sepa-rate the patterns in different cases. In some of them we can describe their asymptotic behavior, but in other cases the behavior in the limit is unknown.
7.1 Generalized patterns
In , Babson and Steingr´ ımsson introduced the notion of generalized pat-terns , which allows the requirement that two adjacent letters in a pattern must be adjacent in the permutation. A generalized pattern is written as a sequence where two adjacent elements may or may not be separated by a dash. In this context, we write a classical pattern with dashes between any two adjacent letters of the pattern (for example, 1423 as 1-4-2-3). If we omit the dash between two letters, we mean that for it to be an occurrence in a permutation π, the corresponding elements of π have to be adjacent. For example, in an occurrence of the pattern 12-3-4 in a permutation π,the entries in π that correspond to 1 and 2 are adjacent. The permutation 134 Chapter 7. Asymptotic enumeration and generalized patterns
π = 3542617 has only one occurrence of the pattern 12-3-4, namely the sub-sequence 3567, whereas π has two occurrences of the pattern 1-2-3-4, namely the subsequences 3567 and 3467. In Chapter 6 we studied some cases of avoidance of patterns where all letters have to occur in consecutive positions. Claesson presented a complete solution for the number of permutations avoiding any single 3-letter general-ized pattern with exactly one adjacent pair of letters. Claesson and Mansour (see also ) did the same for any pair of such patterns. On the other hand, Kitaev investigated simultaneous avoidance of two or more 3-letter generalized patterns without internal dashes. Throughout this chapter, all the patterns that appear will represent general-ized patterns. Therefore, a pattern without dashes will denote a consecutive pattern like the ones in Chapter 6. If we want to consider a classical pattern in the sense of Section 1.1.1, we will represent it with dashes between any two elements, namely, as σ1-σ2- · · · -σm.If σ is a generalized pattern, Sn(σ) denotes the set of permutations in Sn
that have no occurrences of σ in the sense described above. Note that if σ
is a consecutive pattern, then Sn(σ) is the set that was denoted An(σ) in Chapter 6. For a generalized pattern σ, let αn(σ) = |S n(σ)|, and let
Aσ (z) = ∑
n≥0
αn(σ) zn
n!be the exponential generating function counting permutations that avoid σ.
7.1.1 Patterns of the form 1-σ
In this section we study a very particular class of generalized patterns, namely those that start with 1-, followed by a consecutive pattern.
Proposition 7.1 Let σ = σ1σ2 · · · σk ∈ S k be a consecutive pattern, and let
1-σ denote the generalized pattern 1-(σ1 + 1)( σ2 + 1) · · · (σk + 1) . Then,
A1-σ (z) = exp
(∫ z
0
Aσ (t) dt
)
.
Proof. For any permutation π, if m1 > m 2 > · · · > m r are the values of its left-to-right minima, we can write π = m1w1m2w2 · · · mrwr, where each wi
is a (possibly empty) subword of π, each of whose elements is greater than 7.1. Generalized patterns 135
mi. We claim that π avoids 1-σ if and only if each of the blocks wi (more precisely, its reduction ρ(wi)) avoids σ as a subword. Indeed, it is clear that if one of the blocks wi contains the subword σ, then mi together with the occurrence of σ forms an occurrence of 1-σ. Reciprocally, if π contains 1-σ,then the elements of π corresponding to σ have to be adjacent, and none of them can be a left-to-right minimum (since the element corresponding to ‘1’ has to be to their left), therefore they must be all inside the same block wi
for some i.If we denote by A the class of permutations avoiding σ as a subword, then, in the notation of Table 1.2, the class of permutations avoiding 1-σ can be expressed as Π( {z}2 ? A),
where {z}2 ? A corresponds to a block miwi, with the box indicating that the left-to-right minimum has the smallest label. The set construction arises from the fact given a collection of blocks miwi, there is a unique way to order them, namely with the left-to-right minima in decreasing order. The expression A1-σ (z) = exp( ∫ z
0
Aσ (t) dt ) follows now from this construction.
2
Example. The only permutation avoiding the subword σ = 12 (resp. σ =21) is the decreasing (resp. increasing) one. Therefore, by Proposition 7.1,
A1-23 (z) = A1-32 (z) = exp
(∫ z
0
etdt
)
= eez−1,
the EGF for Bell numbers, which agrees with the result in .
Example. For the subwords 132, 231, 312 and 213, we gave in Theorem 6.6 the corresponding generating functions counting their occurrences in permu-tations. Now, by Proposition 7.1, we get the following expression:
A1-243 (z) = A1-342 (z) = A1-423 (z) = A1-324 (z) = exp
(
1
1 − ∫ z
0
e−t2 /2dt
)
.
Example. The EGF for permutations avoiding the subwords 123 and 321 was also given in Theorem 6.6. Proposition 7.1 implies now that
A1-234 (z) = A1-432 (z) = exp
( √3
2
ez/ 2
cos(
√3
2
z + π
6
)
)
.136 Chapter 7. Asymptotic enumeration and generalized patterns
Together with the results of Theorems 6.1 and 6.2, Proposition 7.1 gives expressions for the EGFs A1-σ (z) where σ has one of the following forms:
σ = 123 · · · k,
σ = k(k − 1) · · · 21,
σ = 12 · · · a τ (a + 1),
σ = ( a + 1) τ a (a − 1) · · · 21,
σ = k(k − 1) · · · (k + 1 − a) τ ′ (k − a),
σ = ( k − a) τ ′ (k + 1 − a)( k + 2 − a) · · · k,where k, a are positive integers with a ≤ k − 2, τ is any permutation of
{a + 2 , a + 3 , · · · , k } and τ ′ is any permutation of {1, 2, · · · , k − a − 1}.
7.2 Asymptotic enumeration
Here we discuss the behavior of the numbers αn(σ) as n goes to infinity, for a given generalized pattern σ. We use the symbol ∼ to indicate that two sequences of numbers have the same asymptotic behavior (i.e., we write
an ∼ bn if lim n→∞ an
bn
= 1), and we use the symbol to indicate that a sequence is asymptotically smaller than another one (i.e., we write an bn
if lim n→∞ an
bn
= 0). Let us first consider the case of consecutive patterns.
Theorem 7.2 Let k ≥ 3 and let σ ∈ S k be a consecutive pattern.
(1) There exist constants 0 < c, d < 1 such that
cnn! ≤ αn(σ) ≤ dnn!.
(2) There exists a constant 0 < w ≤ 1 such that
lim
n→∞
(αn(σ)
n!
)1/n
= w.
Note that c, d and w depend only on σ.
Proof. The key observation is that, for any subword σ,
αm+n(σ) ≤ αm(σ)αn(σ)
(m + nn
)
. (7.1) 7.2. Asymptotic enumeration 137
To see this, just observe that a σ-avoiding permutation of length m + n
induces two juxtaposed σ-avoiding permutations of lengths m and n.By induction on n ≥ k one gets
αm+n(σ) ≤ dmm! dnn!
(m + nn
)
= dm+n(m + n)! for some positive d < 1. For the lower bound, let τ = ρ(σ1σ2σ3) be the reduction of the first three elements of σ. Clearly Sn(τ ) ⊆ S n(σ) for all n, since an occurrence of σ in a permutation produces also an occurrence of τ , hence αn(τ ) ≤ αn(σ). But the fact that σ ∈ S 3 implies that αn(σ) equals either αn(123) or αn(132). In any case, by Proposition 6.7 and Theorem 6.6, we have that
αn(σ) ≥ αn(132) ≥ cnn!for some c > 0. To prove part (2), we can express (7.1) as
αm+n(σ)
(m + n)! ≤ αm(σ)
m!
αn(σ)
n!and apply Fekete’s lemma (see [91, Lemma 11.6] or ) to the function
n!/α n(σ) to conclude that lim n→∞
( αn(σ)
n!
)1/n
exists. Calling it w, then by part (1) we have that w ≤ 1 and w ≥ lim n→∞
( αn(132)
n!
)1/n
= 0 .7839769. 2
In order to study the asymptotic behavior of αn(σ) for a generalized pattern
σ we separate the problem into the following cases. Assume from now on that k ≥ 3 and that σ is a generalized pattern of length k.
Case 1 The pattern σ has dashes between any two adjacent elements, i.e.,
σ = σ1-σ2- · · · -σk.
These are just the classical patterns, which have been widely studied for the last decade. The asymptotic behavior of the number of permutations avoid-ing them is given by the Stanley-Wilf conjecture, which has been recently proved by Marcus and Tardos , after several authors had given partial results over the last few years [2, 3, 14, 57]. 138 Chapter 7. Asymptotic enumeration and generalized patterns
Theorem 7.3 ( Stanley-Wilf conjecture , proved in ) For every classical pattern σ = σ1-σ2- · · · -σk, there is a constant λ (which only depends on σ) such that
αn(σ) < λ n
for all n ≥ 1.
On the other hand, it is clear that αn(σ) > α n(ρ(σ1-σ2-σ3)) = Cn ∼ 1√πn 4n.As pointed out by Arratia , Theorem 7.3 is equivalent to the statement that lim n→∞ n
√αn(σ) exists. The value of this limit has been computed for several classical patterns: it is clearly 4 for patterns of length 3, it is known to be ( k − 1) 2 for σ = 1-2- · · · -k, it is shown in that for σ =1-3-4-2 this limit is 8, and it has recently been proved to be nonrational for certain patterns.
Case 2 The pattern σ has three consecutive elements without a dash between them, i.e., σ = · · · σiσi+1 σi+2 · · · .
Proposition 7.4 Let σ be a generalized pattern having three consecutive elements without a dash. Then there exist constants 0 < c, d < 1 such that
cnn! ≤ αn(σ) ≤ dnn!.
Proof. For the upper bound, notice that if a permutation contains the con-secutive pattern σ1σ2σ3 · · · σk obtained by removing all the dashes in σ, then it also contains σ. Therefore, αn(σ) ≤ αn(σ1σ2σ3 · · · σk) for all n, and now the upper bound follows from part (1) of Theorem 7.2. For the lower bound, we use that αn(σ) ≥ αn(ρ(σiσi+1 σi+2 )) ≥ αn(132) ≥
cnn!. 2
Case 3 The pattern σ has pairs of adjacent elements without a dash between them, but not three consecutive elements without dashes.
This case includes all the patterns not considered in Cases 1 and 2. The asymptotic behavior of αn(σ) for these patterns is not known in general. For patterns of length 3 we have the following result due to Claesson . Let Bn denote the n-th Bell number , which counts the number of partitions of an n-element set.
Proposition 7.5 () Let σ be a generalized pattern of length 3 with one dash. 7.2. Asymptotic enumeration 139
(1) If σ ∈ { 1-23 , 3-21 , 32 -1, 12 -3, 1-32 , 23 -1, 3-12 , 21 -3}, then αn(σ) = Bn.
(2) If σ ∈ { 2-13 , 2-31 , 31 -2, 13 -2}, then αn(σ) = Cn.
It is known that the asymptotic behavior of the Catalan numbers is given by Cn ∼ 1√πn 4n. For the Bell numbers, we have the formula
Bn ∼ 1
√n λ(n)n+1 /2eλ(n)−n−1,
where λ(n) is defined by λ(n) ln( λ(n)) = n. Another useful description of the asymptotic behavior of Bn is the following: ln Bn
n = ln n − ln ln n + O
( ln ln n
ln n
)
.
This shows in particular that δn Bn cnn! for any constants δ, c > 0. For patterns σ of length at least 4 that have pairs of adjacent elements without a dash, but not adjacent triplets without dashes, not much is known in general about the number of permutations avoiding them. The asymptotic behavior of αn(σ) could be anywhere between δn for some constant δ > 0(obviously, if σ contains one of the patterns in part (1) of Proposition 7.5, then this lower bound can be improved to Bn) and dnn! for some constant 0 < d < 1. In the rest of this chapter we discuss a few partial results in this direction. The next statement about permutations of the form 1-σ, is an easy conse-quence of Proposition 7.1.
Corollary 7.6 Let σ be a consecutive pattern, and let 1-σ be defined as in Proposition 7.1. Then,
lim
n→∞
( αn(1 -σ)
n!
)1/n
= lim
n→∞
( αn(σ)
n!
)1/n
.
Proof. By Proposition 7.1 we know that A1-σ (z) = exp (∫ z
0
Aσ (t) dt ). Since the exponential is an analytic function, we obtain that A1-σ (z) has the same radius of convergence as Aσ (z), from where the result follows. 2140 Chapter 7. Asymptotic enumeration and generalized patterns
7.2.1 The pattern 12 -34
The next proposition gives an upper and a lower bound for the numbers
αn(12-34). Given two formal power series F (z) = ∑
n≥0
fn and G(z) = ∑
n≥0
gn, we use the notation F (z) < G (z) to indicate that fn < g n for all
n, and F (z) G(z) to indicate that fn gn for all n.
Proposition 7.7 For k ≥ 1, let
hk = 1 + 1
2 + · · · 1
k ,bk(z) =
k
∑
i=0
(ki
)2
[z + 2( hk−i − hi)] eiz ,ck(z) = e(k+1) z
k + 1 −
k
∑
i=0
(ki
)( k + 1
i
) [
z + 2( hk−i − hi) + 1
k + 1 − i
]
eiz ,S(z) = ∑
k≥1
bk(z) + ∑
k≥1
ck(z).
Then
eS(z) < A 12 -34 (z) < e S(z)+ ez+z−1.
If we write eS(z) = ∑ ln zn
n!
and eS(z)+ ez+z−1 = ∑ un zn
n!
to denote the co-efficients of the series giving the lower and the upper bound respectively, then the graph in Figure 7.1 shows the values of n
√αn(12-34) /n ! for n ≤ 13, bounded between the values n
√ln/n ! and n
√un/n ! for n ≤ 120. The two horizontal dotted lines are at height 0 .7839769 and 0 .8269933, which are lim n→∞ n
√αn(σ)/n ! for σ = 132 and σ = 123 respectively, given by Theo-rem 6.6. From this figure it seems plausible that lim n→∞ n
√αn(12-34) /n ! = 0, although we have not succeeded in proving this. Note that the lower bound, together with the fact that S(z) ez − 1 (which follows from the definition), shows that A12 -34 (z) > e S(z) eez −1, which means that αn(12-34) Bn, that is, the number of 12-34-avoiding permu-tations is asymptotically larger than the Bell numbers.
Proof. Let π be a permutation that avoids 12-34. This means that it has no two ascents such that the second one starts at a higher value than where the first one ends. We can write π = B0a1B1a2B2a3B3 · · · , where a1 and the element preceding it form the first ascent of π, a2 and the element preceding it form the first ascent such that a2 < a 1, a3 and the element preceding it 7.2. Asymptotic enumeration 141
0.6 0.8 11.2 1.4 20 40 60 80 100 120 n
Figure 7.1 The first values of n
√αn(12-34) /n ! between the lower and the upper bound given by Proposition 7.7. form the first ascent such that a3 < a 2, and so on. By definition, B0 is a non-empty decreasing subword whose last element is less than a1, and each
Bi with i ≥ 1 can be written uniquely as a sequence Bi = wi, 0wi, 1wi, 2 · · · wi,r i
for some ri ≥ 1 ( ri can be 0 if wi, 0 is nonempty) with the following properties: (1) each wi,j is a decreasing word, (2) for j ≥ 1, wi,j is nonempty and its first element is bigger than ai,(3) the last element of each wi,j is less than ai,(4) the last element of Bi is less than ai+1 .These properties ensure that π avoids 12-34 (since no Bi has an ascent above
ai), and that the decomposition is unique. Ideally we would like to use this decomposition to find a generating function for the numbers αn(12-34). Unfortunately, the structure of the decomposi-tion is a bit too complicated to find an exact formula. Instead, we will add 142 Chapter 7. Asymptotic enumeration and generalized patterns
and remove restrictions to simplify this description, which allows us to give lower and upper bounds respectively. To find an upper bound, we will count permutations of the form π =
B0a1B1a2B2a3B3 · · · , where the Bi and ai satisfy the properties above, ex-cept for the requirement that the last element of each Bi has to be less than
ai+1 . Omitting this requirement we are overcounting permutations, and thus we get an upper bound. The first step now is to find the EGF for a block Ki
of the form aiBi, where Bi satisfies properties (1), (2) and (3) from above. Let us first assume that wi, 0 is empty, that is, Bi = wi, 1wi, 2 · · · wi,r i . We compute the EGF for Ki = aiBi where ri is fixed by induction on ri. If
ri = 0, then we have that Ki = ai, so the EGF is b0(z) := z. If ri = 1, then
Ki = aiwi, 1, where wi, 1 is a decreasing word starting above ai and ending below it. The EFG for wi, 1 is ez . Now, to incorporate the condition that the largest and the smallest labels of Ki lie in wi, 1, we use a generalization of the boxed product construction described in Section 1.3.2. A double derivative is now needed to mark the two special elements. We get that the EGF for such a block is
∫ z
0
∫ y
0
t
( d2
dt 2 et
)
dt dy =
∫ z
0
∫ y
0
te tdt dy = ( z − 2) ez + z + 2 = b1(z).
Let now be ri = 2. The case in which both the largest and the smallest label of Ki = aiwi, 1wi, 2 are contained in wi, 2 corresponds to the EGF
∫ z
0
∫ y
0
b1(t)
( d2
dt 2 et
)
dt dy. (7.2) If we write each wi,j as w+
i,j
w−
i,j
, separating the elements above and below
ai (w+
i,j
and w−
i,j
respectively), then the largest element of Ki can be either in w+
i, 1
or in w+
i, 2
, and the smallest element of Ki can be either in w−
i, 1
or in
w−
i, 2
. Thus, all the possibilities are obtained from the case counted by the EGF (7.2) by permuting the upper and lower parts of the wi, 1 and wi, 2 in the 4 different ways. It follows that the EGF for Ki when ri = 2 is 4
∫ z
0
∫ y
0
b1(t) etdt dy = ( z − 3) e2z + 4 ze z + z + 3 = b2(z).
In general, if bk−1(z) is the EGF for the case ri = k − 1, then the EGF for the case ri = k is given by
bk(z) = k2
∫ z
0
∫ y
0
bk−1(t) etdt dy. 7.2. Asymptotic enumeration 143
It is straightforward to check that the functions bk(z) defined in the state-ment of the proposition satisfy this recurrence. The case where wi, 0 is nonempty can be treated similarly. Now we have
Bi = w0,i wi, 1wi, 2 · · · wi,r i . If ri = 0, the EGF for aiw0,i is c0(z) := ez − 1 − z
(since the block has at least 2 elements). If ri = 1, then a block of the form
aiw0,i wi, 1 can be obtained from the case where the largest and the smallest element are in wi, 1 by permuting w0,i and w−
i, 1
if necessary. This yields the EGF 2
∫ z
0
∫ y
0
c0(t)
( d2
dt 2 et
)
dt dy = e2z
2 + 2(1 − z)ez − z − 5
2 = c1(z).
In general, for nonempty wi, 0, if ck−1(z) is the EGF for the case ri = k − 1, then the EGF for the case ri = k is given by
ck(z) = k(k + 1)
∫ z
0
∫ y
0
ck−1(t) etdt dy.
This is the recurrence satisfied by the functions ck(z) defined in the statement of the proposition. The generating function for a set of blocks Ki = aiBi of the form just described is exp
∑
k≥0
bk(z) + ∑
k≥0
ck(z)
= exp( S(z) + z + ez − 1 − z).
From such a set there is a unique way to form a sequence a1B1a2B2a3B3 · · ·
where a1 > a 2 > a 3 > · · · . Finally, we multiply by ez to take into account the initial decreasing segment B0 of the permutation π = B0a1B1a2B2a3B3 · · · ,again ignoring the condition that its last element should be smaller than a1.This gives the upper bound ez exp( S(z) + ez − 1) = exp( S(z) + ez + z − 1). Now we use a similar reasoning to obtain a lower bound. We have seen that bk(z) counts blocks of the form aiwi, 1wi, 2 · · · wi,k , where each wi,j is a decreasing word starting above ai and ending below it. If k ≥ 1, using the notation wi,k = w+
i,k
w−
i,k
to separate the elements that are bigger than ai
from those that are smaller, we can move the last part of the block to the beginning and write Li = w−
i,k
aiwi, 1wi, 2 · · · w+
i,k
. Similarly, a block of the form aiwi, 0wi, 1wi, 2 · · · wi,k like the ones counted by ck(z) with k ≥ 1 can be reordered as L′
i
= w−
i,k
aiwi, 0wi, 1wi, 2 · · · w+
i,k
. The EGF that counts sets of 144 Chapter 7. Asymptotic enumeration and generalized patterns
pieces of the forms given by Li and L′
i
is exp
∑
k≥1
bk(z) + ∑
k≥1
ck(z)
= exp( S(z)) .
Ordering the pieces of such a set by decreasing order of the ai, the sequence that they form by juxtaposition is a 12-34-avoiding permutation. Besides, no such permutation is obtained in more than one way by this construction. However, notice that not every 12-34-avoiding permutation is produced by this process, hence this construction gives only a lower bound. 2
The decomposition of 12-34-avoiding permutations given in this proof can be generalized to permutations avoiding a pattern of the form 12-σ. If
σ = σ1σ2 · · · σk ∈ S k is a consecutive pattern, 12-σ denotes the generalized pattern 12-( σ1 + 2)( σ2 + 2) · · · (σk + 2). Any permutation π that avoids 12-σ can be uniquely decomposed as π =
B0a1B1a2B2a3B3 · · · , where a1 and the element preceding it form the first ascent of π, a2 and the element preceding it form the first ascent such that
a2 < a 1, a3 and the element preceding it form the first ascent such that
a3 < a 2, and so on. Then, by definition, B0 is a non-empty decreasing subword whose last element is less than a1, and each Bi with i ≥ 1 can be written uniquely as a sequence Bi = wi, 0Ui, 1wi, 1Ui, 2wi, 2 · · · Ui,r i wi,r i for some
ri ≥ 1 ( ri can be 0 if wi, 0 is nonempty) with the following properties: (1) each wi,j is a decreasing word all of whose elements are less than ai,(2) each Ui,j is a nonempty permutation avoiding σ as a subword, all of whose elements are greater than ai,(3) wi,j is nonempty for j ≥ 1, (4) the last element of Bi is less than ai+1 .From this decomposition the following result follows immediately.
Proposition 7.8 If σ ∼ τ are two consecutive patterns, then 12 -σ ∼ 12 -τ .
The structure of 21-σ-avoiding permutations (defined analogously) can be described using the same ideas, and it is not hard to see that the following result holds as well.
Proposition 7.9 If σ is a consecutive pattern, then 12 -σ ∼ 21 -σ.7.2. Asymptotic enumeration 145
7.2.2 The pattern 1-23 -4
Similarly to what we did for the pattern 12-34, analyzing the structure of permutations avoiding 1-23-4 we can give lower and upper bounds for the numbers αn(1-23-4). Let Cexp := ∑
n≥0
Cn zn
n!
be the EGF for the Catalan numbers.
Proposition 7.10 We have that
1
2
∫ z
0
e2ey −2 dy − z
2 < A 1-23 -4(z) < Cexp (ez − 1) .
Writing 1
2
∫ z
0
e2ey −2 dy − z
2
= ∑ ln zn
n!
and Cexp (ez −1) = ∑ un zn
n!
to denote the coefficients of the series giving the lower and the upper bound respectively, then the values of n
√ln/n ! and n
√un/n ! for n ≤ 90 are plotted in Figure 7.2, bounding the values of n
√αn(1-23-4) /n ! for n ≤ 11.
0.6 0.8 11.2 20 40 60 80 n
Figure 7.2 The first values of n
√αn(1-23-4) /n ! between the lower and the upper bound given by Proposition 7.10. 146 Chapter 7. Asymptotic enumeration and generalized patterns
Note that the lower bound implies that αn(1-23-4) Bn, since e2ez −2
eez−1.
Proof. Let π be a permutation that avoids 1-23-4. Let a1 > a 2 > a 3 > · · · >ar be the left-to-right minima of π, and let b1 > b 2 > b 3 > · · · > b s be its right-to-left maxima. Then, marking the positions of the left-to-right minima and right-to-left maxima, we can write π = c1w1c2w2 · · · cr+s−1wr+s−1cr+s,where ci ∈ { a1, a 2, . . . , a r, b 1, b 2, . . . , b s} for all i (in fact the number of ci’s could be less than r + s if some element is simultaneously a left-to-right minimum and a right-to-left maximum). Note that c1 = a1 and cr+s = bs.Now, the condition that π avoids 1-23-4 is equivalent to the fact that each wi
is a (possibly empty) decreasing word. Indeed, if there was an ascent inside one of the wi, then together with the closest left-to-right minimum to the left of wi and the closest right-to-left maximum to the right of wi, it would form an occurrence of 1-23-4. On the other hand, it is clear that if all wi are decreasing, then no such occurrence can exist. We use this decomposition to obtain upper and lower bounds for αn(1-23-4). Let us first show the lower bound. For that we count only a special type of 1-23-4-avoiding permutations, namely the ones where all the left-to-right minima come before all the right-to-left maxima. Such a π can be written as π = a1w1a2w2 · · · arwrb1wr+1 b2wr+2 · · · wr+s−1bs, where for 1 ≤ i ≤ r
the elements of the decreasing words wi have values between ai and b1,and for r ≤ i ≤ r + s − 1 the elements of wi have values between ar
and bi+1 . The EGF for the part a1w1a2w2 · · · ar−1wr−1 is eez −1, since it is an arbitrary 1-23-avoiding permutation (see the example following Propo-sition 7.1). Similarly, the EGF for the part wr+1 b2wr+2 · · · wr+s−1bs is also
eez−1 (it can be seen as a set of blocks of the form wr+ibi+1 , each one con-tributing ez − 1, arranged by decreasing order of the bi’s). The decreasing word wr contributes ez . Now, to get the EGF for the whole permutation
a1w1a2w2 · · · arwrb1wr+1 b2wr+2 · · · wr+s−1bs we use the boxed product con-struction to require that the biggest element of the block is b1 and the small-est one is ar. The EGF that we obtain is
∫ z
0
∫ y
0
eet−1
( d
dt t
)
et
( d
dt t
)
eet−1 dt dy = 1
2
∫ z
0
(e2ey −2 − 1) dy,
which gives a lower bound for the coefficients of A1-23 -4(z). To find the upper bound, consider first permutations of the form π =
c1w1c2w2 · · · cr+s−1wr+s−1cr+s where all the wi are empty. Such permu-tations, where every element is either a left-to-right minimum or a right-to-7.2. Asymptotic enumeration 147
left maximum, are precisely those avoiding 1-2-3, which are counted by the Catalan numbers. Thus, the EGF for such permutations is Cexp (z). The next step is to insert a decreasing word wi after each ci. If ci is a left-to-right minimum, we require that the elements of wi are bigger than ci, so the EGF for the block ciwi is ez − 1. We omit the requirement that the elements of wi have to be smaller than the nearest right-to-left maximum to the right of wi; this is why we only get an upper bound. Similarly, if cj
is a right-to-left maximum, we require that the elements of wj are smaller than cj , so the EGF for the block cj wj is also ez − 1. We also omit the requirement that after the last right-to-left maximum there is no decreasing word. Replacing each ci for a block ciwi as just described translates in terms of generating functions into substituting ez − 1 for the variable z in Cexp (z). This gives the upper bound of the statement. 2
The upper bound given in the above proposition yields the following corol-lary.
Corollary 7.11 We have that
lim
n→∞
( αn(1 -23 -4)
n!
)1/n
= 0 .
Proof. The power series Cexp (z) can be bounded by
Cexp (z) < ∑
n≥0
4n zn
n! = e4z ,
which converges for all z. Therefore, so does Cexp (ez − 1), which is an upper bound for A1-23 -4(z). The result follows now from the observations in Section 6.1.3. 2
The decomposition of 1-23-4-avoiding permutations given in the proof of the above proposition can be generalized to permutations avoiding a pattern of the form 1-σ-k, defined as follows. If σ = σ1σ2 · · · σk−2 ∈ S k−2 is a consecutive pattern, let 1-σ-k denote the generalized pattern 1-( σ1 + 1)( σ2 +1) · · · (σk−2 + 1)-k.Any permutation π that avoids 1-σ-k can be uniquely decomposed as π =
c1w1c2w2 · · · cm−1wm−1cm, where the ci are all the left-to-right minima and right-to-left maxima of π, and each wi is a permutation that avoids σ as 148 Chapter 7. Asymptotic enumeration and generalized patterns
a subword, all of whose elements are bigger than the closest left-to-right minimum to its left and smaller than the closest right-to-left maximum to its right. Using exactly the same reasoning as in the proof of Proposition 7.10, we obtain the following lower and upper bounds for the numbers αn(1-σ-k).
Proposition 7.12 Let σ ∈ S k−2 be a consecutive pattern, and let 1-σ-k be defined as above. Then,
∫ z
0
∫ v
0
e2 R y
0Aσ(t)dt +y
dy dv < A 1-σ-k(z) < Cexp
(∫ z
0
Aσ (t) dt
)
.
Corollary 7.13 With the same definitions as in the above proposition,
lim
n→∞
( αn(1 -σ-k)
n!
)1/n
= lim
n→∞
( αn(σ)
n!
)1/n
.
Proof. The upper and lower bounds for A1-σ-k(z) given in Proposition 7.12 are analytic functions of Aσ (z), since essentially they only involve exponen-tials and integrals. Therefore, A1-σ-k(z) and Aσ (z) have the same radius of convergence, hence the limits above coincide. 2
Finally, the following proposition is an immediate consequence of the struc-ture of permutations avoiding 1-σ-k discussed above. In particular, it implies that 1-23-4 ∼ 1-32-4.
Proposition 7.14 If σ ∼ τ are two consecutive patterns in Sk−2, then
1-σ-k ∼ 1-τ -k.
7.2.3 Other patterns
We have proved that αn(1-23-4) Bn and that αn(1-23-4) cnn! for any constant c > 0. For the pattern 12-34 we showed that the analogue to the first statement holds as well, and the second one seems to be true from numerical computations. It remains as an open problem to give more general results concerning the asymptotical behavior of the numbers αn(σ)where σ is a generalized pattern. In Figure 7.3 we have plotted the initial values (connected by lines) of the sequences n
√αn(σ)/n ! for other cases that appear to have some interest. The 7.2. Asymptotic enumeration 149
0.5 0.6 0.7 0.8 0.9 146810 12 n
Figure 7.3 The first values of n
√αn(σ)/n ! for several generalized patterns σ.two dotted lines at the bottom of the graph correspond to the sequences
n
√Cn/n ! and n
√Bn/n !, which are known to tend to 0 as n goes to infinity. The two dashed lines that start at the same point (around 0 .941) and tend to a constant correspond to the sequences n
√αn(132) /n ! and n
√αn(123) /n !, for which their limits are known to be 0 .7839769 and 0 .8269933 respectively. Among the lines starting at 1, the two dotted ones correspond to the patterns 1-23-4 (the lower line) and 12-34 (the upper line) discussed in the previous subsections. Of the two solid lines, the one below corresponds to the pattern 3-14-2. This pattern has a special interest because all of its subpatterns of length 3 are among those in part (2) of Proposition 7.5. Since it does not contain any of the patterns in part (1), we cannot say that αn(3-14-2) ≥ Bn for all n. In fact, comparting the slopes in Figure 7.3 it seems quite plausible that αn(3-14-2) is asymptotically smaller than Bn, and proving this is an interesting open question. The other solid line in the graph corresponds to 150 Chapter 7. Asymptotic enumeration and generalized patterns
the pattern 13-24, for which we do not know the asymptotic behavior either. Conclusions
This thesis is focused on the enumeration of pattern-avoiding permutations with respect to certain statistics, and on the enumeration of permutations avoiding generalized patterns. We now review the results we have obtained and discuss lines of future research. In Chapter 2 we have classified patterns of length 3 according to the distri-bution of the statistics ‘number of fixed points’ and ‘number of excedances’ in permutations avoiding them. We have introduced bijections between pattern-avoiding permutations and Dyck paths that have played an impor-tant role throughout the thesis. They were presented in a graphical way which made it easier to study their properties. The main result of the chap-ter was that the joint distribution of this pair of statistics is the same in 321-avoiding as in 132-avoiding permutations. This generalizes a recent theorem of Robertson, Saracino and Zeilberger. We gave a bijection between these two sets of permutations that preserves both the number of fixed points and the number of excedances, thus giving a combinatorial proof of the result. Our bijection is a composition of two bijections into Dyck paths, and the re-sult follows from a new analysis of these bijections. The Robinson-Schensted-Knuth correspondence is a part of one of them, and from it stemmed the difficulty of the analysis. The key idea was to introduce a new class of statis-tics on Dyck paths, based on the concept of tunnel , which we introduced in Chapter 1. For the patterns 132, 213 and 321 we gave generating functions with variables enumerating the two statistics mentioned above, and in some cases additional statistics as well. For the patterns 231 and 312 we expressed the corresponding generating functions as continued fractions. For the case of the pattern 123 we could only give partial results regarding the number of fixed points, and we used them to prove a recent conjecture of B´ ona and 152 Conclusions
Guibert. It would be interesting to try to find a generating function for fixed points and excedances in 123-avoiding permutations, the only case of patterns of length 3 that remains unsolved. Even for the enumeration of fixed points in these permutations, we expect that a simpler expression than the one in Theorem 2.13 can be given. In Chapter 3 we have studied permutations avoiding simultaneously two or more patterns of length 3, enumerating them with respect to the number of fixed points and the number of excedances . By means of the bijections between restricted permutations and Dyck paths described in Chapter 2, ad-ditional restrictions on permutations correspond to certain conditions on the paths, and thus the problem was reduced to enumerating such paths with respect to the statistics that fixed points and excedances are mapped to by these bijections. We solved all the cases of avoidance of two or more patterns by giving the corresponding multivariate generating functions, which are ra-tional and have relatively simple expressions. In some instances we were able to give a generalization to the case where one of the patterns had arbitrary length. Then we enumerated involutions avoiding any subset of patterns of length 3 with respect to the same two parameters. The main technique consists in using bijections between pattern-avoiding permutations and cer-tain kinds of Dyck paths, in such a way that our statistics in permutations correspond to statistics on Dyck paths which are easier to enumerate. An interesting extension of this work would be to study the distribution of statistics in permutations avoiding longer patterns. The enumeration of such permutations is itself a very difficult problem, and not even the case of length 4 is completely solved (see [11, 45, 47] for single patterns, [13, 60, 94, 95] for pairs of patterns of length 4, and [20, 59, 66, 67, 68] for other pairs). For the case of patterns of length 4, we have checked by computer that the only cases in which the number of derangements in Sn(σ) is the same for different patterns σ ∈ S 4 are those in which there exists a trivial bijection (such as π 7 → π−1 or π 7 →̂ π) proving this fact. Therefore, Theorems 1.4 and 2.3 do not seem to have an analogue for patterns of length 4. Still, there would be some interest in finding generating functions to enumerate permutations avoiding patterns of length 4 or more with respect to statistics such as the number of fixed points and the number of excedances. For permutations avoiding a pattern of length 4 there are 13 different equivalence classes with respect to the distribution of the statistic fp. It is possible that Theorem 2.3 admits generalizations to other permutation Conclusions 153
statistics, or some variations. We have not succeeded in finding any other case of equidistribution of a statistic for different patterns having such an interesting and nontrivial proof. An example of a much simpler result is that the statistic ‘number of descents’ has the same distribution in Sn(132),
Sn(213), Sn(231) and Sn(312). Another further direction of research would consist in describing the cycle structure of pattern-avoiding permutations. Using the same bijective tech-niques from Chapters 2 and 3, one can easily derive generating functions for the augmented cycle index of permutations in Sn(231 , 312), in Sn(231 , 321) and in Sn(132 , 321). However, it is not clear whether for permutations avoid-ing other subsets of patterns of length 3, the distribution of the cycle type has a simple description. One might wonder if the fact that the number of fixed points has the same distribution both in 321- and in 132-avoiding permutations admits a gener-alization concerning the cycle structure in Sn(321) and Sn(132). We have for example that the cycle structure of 321-avoiding involutions and 132-avoiding involutions is the same. However, it is not true that the cycle structure of permutations in S6(321) is the same as that of permutations in
S6(132), as shown in . In Chapter 4 we have given the simplest known bijection between 321- and 132-avoiding permutations that preserves the number of fixed points. The main ingredient is a new unusual bijection from the set of Dyck paths to itself. We also presented a generalization of it, which gave additional corre-spondences of statistics, as well as applications to the enumeration of Dyck paths and restricted permutations with respect to several statistics. In Chapter 5 we have presented some new interpretations of the Catalan and Fine numbers, and a few natural bijections between 321-avoiding per-mutations and Dyck paths. Then we have considered a class of permutations obtained from noncrossing matchings of 2 n points around a circle. They are counted by the Catalan numbers, but are not defined in terms of pattern avoidance. We found the ordinary generating functions with variables mark-ing the number of descents and the number of fixed points and excedances. In Chapter 6 we have introduced a variation to the notion of pattern avoid-ance, with the additional requirement that the elements forming the pattern have to occur in consecutive positions in the permutation. We studied the number αn(σ) of permutations avoiding σ as a subword and, more generally, the number of occurrences of σ in permutations of length n. For the case of the increasing pattern of any length and of another pattern of a fairly general 154 Conclusions
shape, using bijections between permutations and binary increasing trees, we were able to solve the problem by obtaining the corresponding bivariate ex-ponential generating functions as inverses of solutions of linear differential equations. This provides a complete solution for all consecutive patterns of length 3, and for three out of the seven classes of patterns of length 4. We also considered some cases of simultaneous avoidance of subwords. Besides extending our results to other subwords not covered by our analysis, there remain some interesting problems. For instance, it appears from our computations that the increasing subword 12 · · · m is always dominating, in the sense that αn(12 · · · m) > α n(σ) for any σ ∈ S m and n large enough. We conjecture that this is always the case. Finally, in Chapter 7 we have discussed the notion of generalized patterns, which generalizes the definitions of both classical and consecutive patterns. For patterns of the form 1-σ with no dashes in σ we have obtained the ex-ponential generating function in terms of that for σ-avoiding permutations. Next we have studied the asymptotic behavior of the number of permuta-tions in Sn avoiding a fixed generalized pattern as n goes to infinity. For the case of classical patterns, a description of this asymptotic behavior had been given by the recently proven Stanley-Wilf conjecture . We consider the analogous problem for generalized patterns, solving it for consecutive pat-terns and in some other cases. For a few additional generalized patterns such as 12-34 and 1-23-4 we have shown lower and upper bounds on the number of permutations avoiding them, which gives an estimate of their asymptotic behavior. An interesting problem would be to present a complete classification of all generalized patterns σ according to the asymptotic behavior of the numbers
αn(σ) of permutations avoiding them as n goes to infinity. This would conclude the work initiated in this last chapter. Bibliography
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Discrete Math. 204 (1999), 119–128. A. Claesson, Generalised pattern avoidance, European J. Combin. 22 (2001), 961–973. A. Claesson, T. Mansour, Permutations avoiding a pair of generalized patterns of length three with exactly one dash, Ars Combin. , to appear, math.CO/0107044. Bibliography 157
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Proceedings Formal power series and algebraic combinatorics 2003 ,arxiv:math.CO/0212221. S. Elizalde, Multiple pattern-avoidance with respect to fixed points and excedances, preprint, arxiv:math.CO/0311211. S. Elizalde, E. Deutsch, A simple and unusual bijection for Dyck paths and its consequences, Ann. Comb. 7 (2003), 281–297. S. Elizalde, T. Mansour, Restricted Motzkin permutations, Motzkin paths, continued fractions, and Chebyshev polynomials, preprint. S. Elizalde, M. Noy, Consecutive subwords in permutations, Adv. in Appl. Math. 30 (2003), 110–125. S. Elizalde, I. Pak, Bijections for refined restricted permutations, J. Combin. Theory Ser. A 105 (2004), 207–219. 158 Bibliography
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O. Guibert, T. Mansour, Some statistics on restricted 132 involutions,
Ann. Comb. 6 (2002), 349–374. O. Guibert, E. Pergola, E. Pinzani, Vexillary involutions are enumerated by Motzkin numbers, Ann. Comb. 5 (2001), 153–174. C. Hernando, F. Hurtado, M. Noy, Graphs of non-crossing perfect matchings, Graphs Combin. 18 (2002), 517–532. D.M. Jackson, I.P. Goulden, Algebraic methods for permutations with prescribed patterns, Adv. in Math. 42 (1981), 113–135. S. Kitaev, Generalized pattern avoidance with additional restrictions,
S´ em. Lothar. Combin. 48 (2002), Art. B48e. S. Kitaev, Multi-Avoidance of generalised patterns, Discrete Math. 260 (2003), 89–100. M. Klazar, The F¨ uredi-Hajnal conjecture implies the Stanley-Wilf con-jecture, Proceedings Formal power series and algebraic combinatoircs 2000 , 250–255, Springer, Berlin, 2000. D. Knuth, The Art of Computer Programming, Vol. III , Addison-Wesley, Reading, MA, 1973. C. Krattenthaler, Permutations with restricted patterns and Dyck paths, Adv. in Appl. Math. 27 (2001), 510–530. D. Kremer, Permutations with forbidden subsequences and a general-ized Schr¨ oder number, Discrete Math. 218 (2000), 121–130. A. Kuznetzov, A. Postnikov, I. Pak, Trees Associated with the Motzkin Numbers, J. Combin. Theory Ser. A 76 (1996), 145–147. S. Linusson, Extended pattern avoidance, Discrete Math. 246 (2002), 219–230. M. Lothaire, Combinatorics on words , Addison-Wesley, Reading, MA, 1983. T. Mansour, Continued fractions and generalized patterns, European J. Combin. 23 (2002), 329–344. T. Mansour, A. Robertson, Refined Restricted Permutations Avoiding Subsets of Patterns of Length Three, Ann. Comb. 6 (2003), 407–418. 160 Bibliography
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B.E. Sagan, The symmetric group. Representations, combinatorial al-gorithms, and symmetric functions , Wadsworth & Brooks/Cole, Pacific Grove, CA, 1991; second edition, Springer-Verlag, New York, 2001. C. Schensted, Longest increasing and decreasing subsequences, Canad. J. Math. 13 (1961), 179–191. M-P. Sch¨ utzenberger, Quelques remarques sur une construction de Schensted, Math. Scand. 12 (1963), 117–128. R. Sedgewick, P. Flajolet, An Introduction to the Analysis of Algo-rithms , Addison-Wesley, Reading, MA, 1996. R. Simion, F.W. Schmidt, Restricted Permutations, European J. Com-bin. 6 (1985), 383–406. D. Spielman, M. B´ ona, An infinite antichain of permutations, Electron. J. Combin. 7 (2000), no. 1, Note 2. Z. Stankova, Forbidden subsequences, Discrete Math. 132 (1994), 291– 316. Z. Stankova, Classification of forbidden subsequences of length 4, Eu-ropean J. Combin. 17 (1996), 501–517. Z. Stankova, J. West, A new class of Wilf-equivalent permutations, J. Algebraic Combin. 15 (2002), 271–290. R.P. Stanley, Enumerative Combinatorics, Vol. I, Wadsworth &Brooks/Cole, Belmont, CA, 1986; reprinted by Cambridge University Press, Cambridge, 1997. R.P. Stanley, Enumerative Combinatorics, Vol. II , Cambridge Univer-sity Press, Cambridge, 1999. J.H. van Lint, R.M. Wilson, A course in combinatorics, Cambridge University Press, Cambridge, 1992; second edition, 2001. A. Vella, Pattern avoidance in cyclically ordered structures, Electron. J. Combin. 9 (2003), #R18. J. West, Permutations with forbidden subsequences and stack-sortable permutations, Ph.D. thesis, MIT, 1990. 162 Bibliography
J. West, Generating trees and the Catalan and Schr¨ oder numbers, Dis-crete Math. 146 (1995), 247–262. J. West, Generating Trees and Forbidden Subsequences, Discrete Math.
157 (1996), 363–374. H. Wilf, Generatingfunctionology , Academic Press, Boston, MA, 1990; second edition, 1994. H. Wilf, The patterns of permutations, Discrete Math. 257 (2002), 575– 583. |
187634 | https://www.falstad.com/dotproduct/ | J2S._Canvas2D (com.falstad.DotProduct) "dotproduct"[x]
loadClass java.lang.String
loadClass core.package
J2SApplet exec dotproduct loadCore null
Loading ../swingjs/j2s/core/coreswingjs.z.js
Loading ../swingjs/j2s/com/falstad/DotProduct.js
Loading ../swingjs/j2s/javax/swing/text/AbstractDocument.js
Loading ../swingjs/j2s/java/awt/geom/Point2D.js
Loading ../swingjs/j2s/swingjs/plaf/JSSliderUI.js
Loading ../swingjs/j2s/swingjs/plaf/JSScrollBarUI.js
Loading ../swingjs/j2s/swingjs/jquery/JQueryUI.js
Loading ../swingjs/j2s/swingjs/jquery/jquery-ui-j2sslider.css
Loading ../swingjs/j2s/swingjs/jquery/jquery-ui-j2sslider.js
J2SApplet exec dotproduct loadClazz2 null
loadClass swingjs.JSToolkit
J2SApplet exec dotproduct load com.falstad.DotProduct swingjs.JSAppletViewer
loadClass swingjs.JSAppletViewer
J2SApplet exec dotproduct start applet null
JSAppletViewer initializing
get parameter: name = dotproduct
get parameter: syncId = 814705924384872
JSToolkit initialized
swingjs.api.Interface creating instance of JU.AjaxURLStreamHandlerFactory
JSAppletViewer initialized
This applet demonstrates the
dot product,
which is an important concept in linear algebra and physics. The goal of
this applet is to help you visualize what the dot product
geometrically.
Two vectors are shown, one in red (A) and one in blue (B).
On the right, the coordinates of both vectors and their lengths
are shown. The projection of A onto B is shown in yellow, and the
angle between the two is shown in orange.
At the bottom of the screen are four bars which show the magnitude
of four quantities: the length of A (red), the length of B (blue),
the length of the projection of A onto B (yellow), and the dot
product of A and B (green). Some of these quantities may be negative.
To modify a vector, click on its arrowhead and drag it around.
To swap A and B, click the swap button. Swapping the two does not
change the dot product.
Matrix Applet.
More applets.
Huge thanks to Bob Hanson and his team for converting this applet to javascript.
java@falstad.com |
187635 | https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-1-rolling-motion/ | 11 Angular Momentum
11.1 Rolling Motion
Learning Objectives
By the end of this section, you will be able to:
Describe the physics of rolling motion without slipping
Explain how linear variables are related to angular variables for the case of rolling motion without slipping
Find the linear and angular accelerations in rolling motion with and without slipping
Calculate the static friction force associated with rolling motion without slipping
Use energy conservation to analyze rolling motion
Rolling motion is that common combination of rotational and translational motion that we see everywhere, every day. Think about the different situations of wheels moving on a car along a highway, or wheels on a plane landing on a runway, or wheels on a robotic explorer on another planet. Understanding the forces and torques involved in rolling motion is a crucial factor in many different types of situations.
For analyzing rolling motion in this chapter, refer to (Figure) in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. You may also find it useful in other calculations involving rotation.
Rolling Motion without Slipping
People have observed rolling motion without slipping ever since the invention of the wheel. For example, we can look at the interaction of a car’s tires and the surface of the road. If the driver depresses the accelerator to the floor, such that the tires spin without the car moving forward, there must be kinetic friction between the wheels and the surface of the road. If the driver depresses the accelerator slowly, causing the car to move forward, then the tires roll without slipping. It is surprising to most people that, in fact, the bottom of the wheel is at rest with respect to the ground, indicating there must be static friction between the tires and the road surface. In (Figure), the bicycle is in motion with the rider staying upright. The tires have contact with the road surface, and, even though they are rolling, the bottoms of the tires deform slightly, do not slip, and are at rest with respect to the road surface for a measurable amount of time. There must be static friction between the tire and the road surface for this to be so.
Figure 11.2 (a) The bicycle moves forward, and its tires do not slip. The bottom of the slightly deformed tire is at rest with respect to the road surface for a measurable amount of time. (b) This image shows that the top of a rolling wheel appears blurred by its motion, but the bottom of the wheel is instantaneously at rest. (credit a: modification of work by Nelson Lourenço; credit b: modification of work by Colin Rose)
To analyze rolling without slipping, we first derive the linear variables of velocity and acceleration of the center of mass of the wheel in terms of the angular variables that describe the wheel’s motion. The situation is shown in (Figure).
Figure 11.3 (a) A wheel is pulled across a horizontal surface by a force $$ \overset{\to }{F}$$. The force of static friction $$ {\overset{\to }{f}}_{\text{S}},|{\overset{\to }{f}}_{\text{S}}|\le {\mu }_{\text{S}}N $$ is large enough to keep it from slipping. (b) The linear velocity and acceleration vectors of the center of mass and the relevant expressions for $$ \omega \,\text{and}\,\alpha $$. Point P is at rest relative to the surface. (c) Relative to the center of mass (CM) frame, point P has linear velocity $$ \text{−}R\omega \hat{i}$$.
From (Figure)(a), we see the force vectors involved in preventing the wheel from slipping. In (b), point P that touches the surface is at rest relative to the surface. Relative to the center of mass, point P has velocity $$ \text{−}R\omega \hat{i}$$, where R is the radius of the wheel and $$ \omega $$ is the wheel’s angular velocity about its axis. Since the wheel is rolling, the velocity of P with respect to the surface is its velocity with respect to the center of mass plus the velocity of the center of mass with respect to the surface:
$${\overset{\to }{v}}_{P}=\text{−}R\omega \hat{i}+{v}_{\text{CM}}\hat{i}.$$
Since the velocity of P relative to the surface is zero, $$ {v}_{P}=0$$, this says that
$${v}_{\text{CM}}=R\omega .$$
Thus, the velocity of the wheel’s center of mass is its radius times the angular velocity about its axis. We show the correspondence of the linear variable on the left side of the equation with the angular variable on the right side of the equation. This is done below for the linear acceleration.
If we differentiate (Figure) on the left side of the equation, we obtain an expression for the linear acceleration of the center of mass. On the right side of the equation, R is a constant and since $$ \alpha =\frac{d\omega }{dt}, $$ we have
$${a}_{\text{CM}}=R\alpha .$$
Furthermore, we can find the distance the wheel travels in terms of angular variables by referring to (Figure). As the wheel rolls from point A to point B, its outer surface maps onto the ground by exactly the distance travelled, which is $$ {d}_{\text{CM}}. $$ We see from (Figure) that the length of the outer surface that maps onto the ground is the arc length $$ R\theta \text{}$$. Equating the two distances, we obtain
$${d}_{\text{CM}}=R\theta .$$
Figure 11.4 As the wheel rolls on the surface, the arc length $$ R\theta $$ from A to B maps onto the surface, corresponding to the distance $$ {d}_{\text{CM}} $$ that the center of mass has moved.
Example
Rolling Down an Inclined Plane
A solid cylinder rolls down an inclined plane without slipping, starting from rest. It has mass m and radius r. (a) What is its acceleration? (b) What condition must the coefficient of static friction $$ {\mu }_{\text{S}} $$ satisfy so the cylinder does not slip?
Strategy
Draw a sketch and free-body diagram, and choose a coordinate system. We put x in the direction down the plane and y upward perpendicular to the plane. Identify the forces involved. These are the normal force, the force of gravity, and the force due to friction. Write down Newton’s laws in the x– and y-directions, and Newton’s law for rotation, and then solve for the acceleration and force due to friction.
Solution
The free-body diagram and sketch are shown in (Figure), including the normal force, components of the weight, and the static friction force. There is barely enough friction to keep the cylinder rolling without slipping. Since there is no slipping, the magnitude of the friction force is less than or equal to $$ {\mu }_{S}N$$. Writing down Newton’s laws in the x– and y-directions, we have
$$\sum {F}_{x}=m{a}_{x};\enspace\sum {F}_{y}=m{a}_{y}.$$
Figure 11.5 A solid cylinder rolls down an inclined plane without slipping from rest. The coordinate system has x in the direction down the inclined plane and y perpendicular to the plane. The free-body diagram is shown with the normal force, the static friction force, and the components of the weight $$ m\overset{\to }{g}$$. Friction makes the cylinder roll down the plane rather than slip.
Substituting in from the free-body diagram,
$$\begin{array}{ccc}\hfill mg\,\text{sin}\,\theta -{f}_{\text{S}}& =\hfill & m{({a}_{\text{CM}})}_{x},\hfill \ \hfill N-mg\,\text{cos}\,\theta & =\hfill & 0,\hfill \ \hfill {f}_{\text{S}}& \le \hfill & {\mu }_{\text{S}}N,\hfill \end{array}$$
we can then solve for the linear acceleration of the center of mass from these equations:
$${({a}_{\text{CM}})}_{x}=g(\text{sin}\,\theta -{\mu }_{S}\text{cos}\,\theta ).$$
However, it is useful to express the linear acceleration in terms of the moment of inertia. For this, we write down Newton’s second law for rotation,
$$\sum {\tau }_{\text{CM}}={I}_{\text{CM}}\alpha .$$
The torques are calculated about the axis through the center of mass of the cylinder. The only nonzero torque is provided by the friction force. We have
$${f}_{\text{S}}r={I}_{\text{CM}}\alpha .$$
Finally, the linear acceleration is related to the angular acceleration by
$${({a}_{\text{CM}})}_{x}=r\alpha .$$
These equations can be used to solve for $$ {a}_{\text{CM}},\alpha ,\,\text{and}\,{f}_{\text{S}} $$ in terms of the moment of inertia, where we have dropped the x-subscript. We write $$ {a}_{\text{CM}} $$ in terms of the vertical component of gravity and the friction force, and make the following substitutions.
$${a}_{\text{CM}}=g\text{sin}\,\theta -\frac{{f}_{\text{S}}}{m}$$
$${f}_{\text{S}}=\frac{{I}_{\text{CM}}\alpha }{r}=\frac{{I}_{\text{CM}}{a}_{\text{CM}}}{{r}^{2}}$$
From this we obtain
$$\begin{array}{cc}\hfill {a}_{\text{CM}}& =g\,\text{sin}\,\theta -\frac{{I}_{\text{CM}}{a}_{\text{CM}}}{m{r}^{2}},\hfill \ & =\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})}.\hfill \end{array}$$
Note that this result is independent of the coefficient of static friction, $$ {\mu }_{\text{S}}$$.
Since we have a solid cylinder, from (Figure), we have $$ {I}_{\text{CM}}=m{r}^{2}\text{/}2 $$ and
$${a}_{\text{CM}}=\frac{mg\,\text{sin}\,\theta }{m+(m{r}^{2}\text{/}2{r}^{2})}=\frac{2}{3}g\,\text{sin}\,\theta .$$
Therefore, we have
$$\alpha =\frac{{a}_{\text{CM}}}{r}=\frac{2}{3r}g\,\text{sin}\,\theta .$$
2. Because slipping does not occur, $$ {f}_{\text{S}}\le {\mu }_{\text{S}}N$$. Solving for the friction force,
$${f}_{\text{S}}={I}_{\text{CM}}\frac{\alpha }{r}={I}_{\text{CM}}\frac{({a}_{\text{CM}})}{{r}^{2}}=\frac{{I}_{\text{CM}}}{{r}^{2}}(\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})})=\frac{mg{I}_{\text{CM}}\,\text{sin}\,\theta }{m{r}^{2}+{I}_{\text{CM}}}.$$
Substituting this expression into the condition for no slipping, and noting that $$ N=mg\,\text{cos}\,\theta $$, we have
$$\frac{mg{I}_{\text{CM}}\text{sin}\,\theta }{m{r}^{2}+{I}_{\text{CM}}}\le {\mu }_{\text{S}}mg\,\text{cos}\,\theta $$
or
$${\mu }_{\text{S}}\ge \frac{\text{tan}\,\theta }{1+(m{r}^{2}\text{/}{I}_{\text{CM}})}.$$
For the solid cylinder, this becomes
$${\mu }_{\text{S}}\ge \frac{\text{tan}\,\theta }{1+(2m{r}^{2}\text{/}m{r}^{2})}=\frac{1}{3}\text{tan}\,\theta .$$
Significance
The linear acceleration is linearly proportional to $$ \text{sin}\,\theta . $$ Thus, the greater the angle of the incline, the greater the linear acceleration, as would be expected. The angular acceleration, however, is linearly proportional to $$ \text{sin}\,\theta $$ and inversely proportional to the radius of the cylinder. Thus, the larger the radius, the smaller the angular acceleration.
For no slipping to occur, the coefficient of static friction must be greater than or equal to $$ (1\text{/}3)\text{tan}\,\theta $$. Thus, the greater the angle of incline, the greater the coefficient of static friction must be to prevent the cylinder from slipping.
Check Your Understanding
A hollow cylinder is on an incline at an angle of $$ 60\text{°}. $$ The coefficient of static friction on the surface is $$ {\mu }_{S}=0.6$$. (a) Does the cylinder roll without slipping? (b) Will a solid cylinder roll without slipping
Show Answer
a. $$ {\mu }_{\text{S}}\ge \frac{\text{tan}\,\theta }{1+(m{r}^{2}\text{/}{I}_{\text{CM}})}$$; inserting the angle and noting that for a hollow cylinder $$ {I}_{\text{CM}}=m{r}^{2}, $$ we have $$ {\mu }_{\text{S}}\ge \frac{\text{tan}\,60\text{°}}{1+(m{r}^{2}\text{/}m{r}^{2})}=\frac{1}{2}\text{tan}\,60\text{°}=0.87; $$ we are given a value of 0.6 for the coefficient of static friction, which is less than 0.87, so the condition isn’t satisfied and the hollow cylinder will slip; b. The solid cylinder obeys the condition $$ {\mu }_{\text{S}}\ge \frac{1}{3}\text{tan}\,\theta =\frac{1}{3}\text{tan}\,60\text{°}=0.58. $$ The value of 0.6 for $$ {\mu }_{\text{S}} $$ satisfies this condition, so the solid cylinder will not slip.
It is worthwhile to repeat the equation derived in this example for the acceleration of an object rolling without slipping:
$${a}_{\text{CM}}=\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})}.$$
This is a very useful equation for solving problems involving rolling without slipping. Note that the acceleration is less than that for an object sliding down a frictionless plane with no rotation. The acceleration will also be different for two rotating cylinders with different rotational inertias.
Rolling Motion with Slipping
In the case of rolling motion with slipping, we must use the coefficient of kinetic friction, which gives rise to the kinetic friction force since static friction is not present. The situation is shown in (Figure). In the case of slipping, $$ {v}_{\text{CM}}-R\omega \ne 0$$, because point P on the wheel is not at rest on the surface, and $$ {v}_{P}\ne 0$$. Thus, $$ \omega \ne \frac{{v}_{\text{CM}}}{R},\alpha \ne \frac{{a}_{\text{CM}}}{R}$$.
Figure 11.6 (a) Kinetic friction arises between the wheel and the surface because the wheel is slipping. (b) The simple relationships between the linear and angular variables are no longer valid.
Example
Rolling Down an Inclined Plane with Slipping
A solid cylinder rolls down an inclined plane from rest and undergoes slipping ((Figure)). It has mass m and radius r. (a) What is its linear acceleration? (b) What is its angular acceleration about an axis through the center of mass?
Strategy
Draw a sketch and free-body diagram showing the forces involved. The free-body diagram is similar to the no-slipping case except for the friction force, which is kinetic instead of static. Use Newton’s second law to solve for the acceleration in the x-direction. Use Newton’s second law of rotation to solve for the angular acceleration.
Solution
Figure 11.7 A solid cylinder rolls down an inclined plane from rest and undergoes slipping. The coordinate system has x in the direction down the inclined plane and y upward perpendicular to the plane. The free-body diagram shows the normal force, kinetic friction force, and the components of the weight $$ m\overset{\to }{g}.$$
The sum of the forces in the y-direction is zero, so the friction force is now $$ {f}_{\text{k}}={\mu }_{\text{k}}N={\mu }_{\text{k}}mg\text{cos}\,\theta .$$
Newton’s second law in the x-direction becomes
$$\sum {F}_{x}=m{a}_{x},$$
$$mg\,\text{sin}\,\theta -{\mu }_{\text{k}}mg\,\text{cos}\,\theta =m{({a}_{\text{CM}})}_{x},$$
or
$${({a}_{\text{CM}})}_{x}=g(\text{sin}\,\theta -{\mu }_{\text{K}}\,\text{cos}\,\theta ).$$
The friction force provides the only torque about the axis through the center of mass, so Newton’s second law of rotation becomes
$$\sum {\tau }_{\text{CM}}={I}_{\text{CM}}\alpha ,$$
$${f}_{\text{k}}r={I}_{\text{CM}}\alpha =\frac{1}{2}m{r}^{2}\alpha .$$
Solving for $$ \alpha $$, we have
$$\alpha =\frac{2{f}_{\text{k}}}{mr}=\frac{2{\mu }_{\text{k}}g\,\text{cos}\,\theta }{r}.$$
Significance
We write the linear and angular accelerations in terms of the coefficient of kinetic friction. The linear acceleration is the same as that found for an object sliding down an inclined plane with kinetic friction. The angular acceleration about the axis of rotation is linearly proportional to the normal force, which depends on the cosine of the angle of inclination. As $$ \theta \to 90\text{°}$$, this force goes to zero, and, thus, the angular acceleration goes to zero.
Conservation of Mechanical Energy in Rolling Motion
In the preceding chapter, we introduced rotational kinetic energy. Any rolling object carries rotational kinetic energy, as well as translational kinetic energy and potential energy if the system requires. Including the gravitational potential energy, the total mechanical energy of an object rolling is
$${E}_{\text{T}}=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}{I}_{\text{CM}}{\omega }^{2}+mgh.$$
In the absence of any nonconservative forces that would take energy out of the system in the form of heat, the total energy of a rolling object without slipping is conserved and is constant throughout the motion. Examples where energy is not conserved are a rolling object that is slipping, production of heat as a result of kinetic friction, and a rolling object encountering air resistance.
You may ask why a rolling object that is not slipping conserves energy, since the static friction force is nonconservative. The answer can be found by referring back to (Figure). Point P in contact with the surface is at rest with respect to the surface. Therefore, its infinitesimal displacement $$ d\overset{\to }{r} $$ with respect to the surface is zero, and the incremental work done by the static friction force is zero. We can apply energy conservation to our study of rolling motion to bring out some interesting results.
Example
Curiosity Rover
The Curiosity rover, shown in (Figure), was deployed on Mars on August 6, 2012. The wheels of the rover have a radius of 25 cm. Suppose astronauts arrive on Mars in the year 2050 and find the now-inoperative Curiosity on the side of a basin. While they are dismantling the rover, an astronaut accidentally loses a grip on one of the wheels, which rolls without slipping down into the bottom of the basin 25 meters below. If the wheel has a mass of 5 kg, what is its velocity at the bottom of the basin?
Figure 11.8 The NASA Mars Science Laboratory rover Curiosity during testing on June 3, 2011. The location is inside the Spacecraft Assembly Facility at NASA’s Jet Propulsion Laboratory in Pasadena, California. (credit: NASA/JPL-Caltech)
Strategy
We use mechanical energy conservation to analyze the problem. At the top of the hill, the wheel is at rest and has only potential energy. At the bottom of the basin, the wheel has rotational and translational kinetic energy, which must be equal to the initial potential energy by energy conservation. Since the wheel is rolling without slipping, we use the relation $$ {v}_{\text{CM}}=r\omega $$ to relate the translational variables to the rotational variables in the energy conservation equation. We then solve for the velocity. From (Figure), we see that a hollow cylinder is a good approximation for the wheel, so we can use this moment of inertia to simplify the calculation.
Solution
Energy at the top of the basin equals energy at the bottom:
$$mgh=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}{I}_{\text{CM}}{\omega }^{2}.$$
The known quantities are $$ {I}_{\text{CM}}=m{r}^{2}\text{,}\,r=0.25\,\text{m,}\,\text{and}\,h=25.0\,\text{m}$$.
We rewrite the energy conservation equation eliminating $$ \omega $$ by using $$ \omega =\frac{{v}_{\text{CM}}}{r}. $$ We have
$$mgh=\frac{1}{2}m{v}_{\text{CM}}^{2}+\frac{1}{2}m{r}^{2}\frac{{v}_{\text{CM}}^{2}}{{r}^{2}}$$
or
$$gh=\frac{1}{2}{v}_{\text{CM}}^{2}+\frac{1}{2}{v}_{\text{CM}}^{2}⇒{v}_{\text{CM}}=\sqrt{gh}.$$
On Mars, the acceleration of gravity is $$ 3.71\,{\,\text{m/s}}^{2}, $$ which gives the magnitude of the velocity at the bottom of the basin as
$${v}_{\text{CM}}=\sqrt{(3.71\,\text{m}\text{/}{\text{s}}^{2})25.0\,\text{m}}=9.63\,\text{m}\text{/}\text{s}\text{.}$$
Significance
This is a fairly accurate result considering that Mars has very little atmosphere, and the loss of energy due to air resistance would be minimal. The result also assumes that the terrain is smooth, such that the wheel wouldn’t encounter rocks and bumps along the way.
Also, in this example, the kinetic energy, or energy of motion, is equally shared between linear and rotational motion. If we look at the moments of inertia in (Figure), we see that the hollow cylinder has the largest moment of inertia for a given radius and mass. If the wheels of the rover were solid and approximated by solid cylinders, for example, there would be more kinetic energy in linear motion than in rotational motion. This would give the wheel a larger linear velocity than the hollow cylinder approximation. Thus, the solid cylinder would reach the bottom of the basin faster than the hollow cylinder.
Summary
In rolling motion without slipping, a static friction force is present between the rolling object and the surface. The relations $$ {v}_{\text{CM}}=R\omega ,{a}_{\text{CM}}=R\alpha ,\,\text{and}\,{d}_{\text{CM}}=R\theta $$ all apply, such that the linear velocity, acceleration, and distance of the center of mass are the angular variables multiplied by the radius of the object.
In rolling motion with slipping, a kinetic friction force arises between the rolling object and the surface. In this case, $$ {v}_{\text{CM}}\ne R\omega ,{a}_{\text{CM}}\ne R\alpha ,\,\text{and}\,{d}_{\text{CM}}\ne R\theta $$.
Energy conservation can be used to analyze rolling motion. Energy is conserved in rolling motion without slipping. Energy is not conserved in rolling motion with slipping due to the heat generated by kinetic friction.
Conceptual Questions
Can a round object released from rest at the top of a frictionless incline undergo rolling motion?
Show Solution
No, the static friction force is zero.
A cylindrical can of radius R is rolling across a horizontal surface without slipping. (a) After one complete revolution of the can, what is the distance that its center of mass has moved? (b) Would this distance be greater or smaller if slipping occurred?
A wheel is released from the top on an incline. Is the wheel most likely to slip if the incline is steep or gently sloped?
Show Solution
The wheel is more likely to slip on a steep incline since the coefficient of static friction must increase with the angle to keep rolling motion without slipping.
Which rolls down an inclined plane faster, a hollow cylinder or a solid sphere? Both have the same mass and radius.
A hollow sphere and a hollow cylinder of the same radius and mass roll up an incline without slipping and have the same initial center of mass velocity. Which object reaches a greater height before stopping?
Show Solution
The cylinder reaches a greater height. By (Figure), its acceleration in the direction down the incline would be less.
Problems
What is the angular velocity of a 75.0-cm-diameter tire on an automobile traveling at 90.0 km/h?
Show Answer
$${v}_{\text{CM}}=R\omega \,⇒\omega =66.7\,\text{rad/s}$$
A boy rides his bicycle 2.00 km. The wheels have radius 30.0 cm. What is the total angle the tires rotate through during his trip?
If the boy on the bicycle in the preceding problem accelerates from rest to a speed of 10.0 m/s in 10.0 s, what is the angular acceleration of the tires?
Show Solution
$$\alpha =3.3\,\text{rad}\text{/}{\text{s}}^{2}$$
Formula One race cars have 66-cm-diameter tires. If a Formula One averages a speed of 300 km/h during a race, what is the angular displacement in revolutions of the wheels if the race car maintains this speed for 1.5 hours?
A marble rolls down an incline at $$ 30\text{°} $$ from rest. (a) What is its acceleration? (b) How far does it go in 3.0 s?
Show Solution
$${I}_{\text{CM}}=\frac{2}{5}m{r}^{2},\,{a}_{\text{CM}}=3.5\,\text{m}\text{/}{\text{s}}^{2};\,x=15.75\,\text{m}$$
Repeat the preceding problem replacing the marble with a solid cylinder. Explain the new result.
A rigid body with a cylindrical cross-section is released from the top of a $$ 30\text{°} $$ incline. It rolls 10.0 m to the bottom in 2.60 s. Find the moment of inertia of the body in terms of its mass m and radius r.
Show Solution
positive is down the incline plane;
$${a}_{\text{CM}}=\frac{mg\,\text{sin}\,\theta }{m+({I}_{\text{CM}}\text{/}{r}^{2})}⇒{I}_{\text{CM}}={r}^{2}[\frac{mg\,\text{sin}30}{{a}_{\text{CM}}}-m]$$,
$$x-{x}_{0}={v}_{0}t-\frac{1}{2}{a}_{\text{CM}}{t}^{2}⇒{a}_{\text{CM}}=2.96\,{\text{m/s}}^{2},$$
$${I}_{\text{CM}}=0.66\,m{r}^{2}$$
A yo-yo can be thought of a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (see below). One end of the string is held fixed in space. If the cylinder falls as the string unwinds without slipping, what is the acceleration of the cylinder?
A solid cylinder of radius 10.0 cm rolls down an incline with slipping. The angle of the incline is $$ 30\text{°}. $$ The coefficient of kinetic friction on the surface is 0.400. What is the angular acceleration of the solid cylinder? What is the linear acceleration?
Show Solution
$$\alpha =67.9\,\text{rad}\text{/}{\text{s}}^{2}$$,
$${({a}_{\text{CM}})}_{x}=1.5\,\text{m}\text{/}{\text{s}}^{2}$$
A bowling ball rolls up a ramp 0.5 m high without slipping to storage. It has an initial velocity of its center of mass of 3.0 m/s. (a) What is its velocity at the top of the ramp? (b) If the ramp is 1 m high does it make it to the top?
A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 m/s. How much work is required to stop it?
Show Solution
$$W=-1080.0\,\text{J}$$
A 40.0-kg solid sphere is rolling across a horizontal surface with a speed of 6.0 m/s. How much work is required to stop it? Compare results with the preceding problem.
A solid cylinder rolls up an incline at an angle of $$ 20\text{°}. $$ If it starts at the bottom with a speed of 10 m/s, how far up the incline does it travel?
Show Solution
Mechanical energy at the bottom equals mechanical energy at the top;
$$h=7.7\,\text{m,} $$ so the distance up the incline is $$ 22.5\,\text{m}$$.
A solid cylindrical wheel of mass M and radius R is pulled by a force $$ \overset{\to }{F} $$ applied to the center of the wheel at $$ 37\text{°} $$ to the horizontal (see the following figure). If the wheel is to roll without slipping, what is the maximum value of $$ |\overset{\to }{F}|? $$ The coefficients of static and kinetic friction are $$ {\mu }_{\text{S}}=0.40\,\text{and}\,{\mu }_{\text{k}}=0.30.$$
A hollow cylinder is given a velocity of 5.0 m/s and rolls up an incline to a height of 1.0 m. If a hollow sphere of the same mass and radius is given the same initial velocity, how high does it roll up the incline?
Show Solution
Use energy conservation
$$\frac{1}{2}m{v}_{0}^{2}+\frac{1}{2}{I}_{\text{Cyl}}{\omega }_{0}^{2}=mg{h}_{\text{Cyl}}$$,
$$\frac{1}{2}m{v}_{0}^{2}+\frac{1}{2}{I}_{\text{Sph}}{\omega }_{0}^{2}=mg{h}_{\text{Sph}}$$.
Subtracting the two equations, eliminating the initial translational energy, we have
$$\frac{1}{2}{I}_{\text{Cyl}}{\omega }_{0}^{2}-\frac{1}{2}{I}_{\text{Sph}}{\omega }_{0}^{2}=mg({h}_{\text{Cyl}}-{h}_{\text{Sph}})$$,
Thus, the hollow sphere, with the smaller moment of inertia, rolls up to a lower height of $$ 1.0-0.43=0.57\,\text{m}\text{.}$$
Glossary
rolling motion
: combination of rotational and translational motion with or without slipping
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Algebra 2 (LBUSD pilot)9 units · 169 skillsUnit 1 Linear functionsUnit 2 Quadratic functionsUnit 3 Quadratic equationsUnit 4 Polynomial functionsUnit 5 Radical functionsUnit 6 Rational functionsUnit 7 Exponential & logarithmic functionsUnit 8 Sequences and seriesUnit 9 Trigonometric ratios and functions
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Shift absolute value graphs
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187637 | http://homepages.math.uic.edu/~marker/valued_fields.pdf | Model Theory of Valued Fields University of Illinois at Chicago David Marker Fall 2018 Contents 1 Valued Fields–Definitions and Examples 3 1.1 Valuations and Valuation Rings . . . . . . . . . . . . . . . . . . .
3 1.2 Absolute Values . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 2 Hensel’s Lemma 12 2.1 Hensel’s Lemma, Equivalents and Applications . . . . . . . . . .
12 2.2 Lifting the residue field . . . . . . . . . . . . . . . . . . . . . . .
19 2.3 Sections of the value group . . . . . . . . . . . . . . . . . . . . .
20 2.4 Hahn fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 3 Extensions of Rings and Valuations 27 3.1 Integral extensions . . . . . . . . . . . . . . . . . . . . . . . . . .
27 3.2 Extensions of Valuations . . . . . . . . . . . . . . . . . . . . . . .
29 4 Algebraically Closed Valued Fields 35 4.1 Quantifier Elimination for ACVF . . . . . . . . . . . . . . . . . .
35 4.2 Consequences of Quantifier Elimination . . . . . . . . . . . . . .
40 4.3 Balls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43 4.4 Real Closed Valued Fields . . . . . . . . . . . . . . . . . . . . . .
45 5 Algebra of Henselian Fields 47 5.1 Extensions of Henselian Valuations . . . . . . . . . . . . . . . . .
47 5.2 Algebraically Maximal Fields . . . . . . . . . . . . . . . . . . . .
51 5.3 Henselizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53 5.4 Pseudolimits . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55 6 The Ax–Kochen Erˇ sov Theorem 60 6.1 Quantifier Elimination in the Pas Language . . . . . . . . . . . .
60 6.2 Consequence of Quantifier Elimination . . . . . . . . . . . . . . .
64 6.3 Artin’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . .
68 1 7 The Theory of Qp 71 7.1 p-adically Closed Fields . . . . . . . . . . . . . . . . . . . . . . .
71 7.2 Consequences of Quantifier Elimination . . . . . . . . . . . . . .
75 7.3 Rationality of Poincar´ e Series . . . . . . . . . . . . . . . . . . . .
83 These lecture notes are based on a course given at the University of Illinois at Chicago in Fall 2018. The goal was to cover some of the classic material on the model theory of valued fields: the Ax–Kochen/Erˇ sov Theorem, the model theory of Qp and Denef’s work on rationality of Poincar´ e series. The lectures assumed a basic knowledge of model theory (quantifier elimination tests, satu-rated models...) and graduate level algebra, but most results on the algebra of valuations were presented from scratch.
Parts of my lectures closely follow the notes of Zo´ e Chatzidakis , Lou van den Dries and the book Valued Fields by Engler and Prestel .
Conventions and Notation • In these notes ring will always mean commutative ring with identity and domain means an integral domain, i.e., a commutative ring with identity and no zero divisors.
• A ⊆B means that A is a subset of B and allows the possibility A = B, while A ⊂B means A ⊆B but A ̸= B.
• AX is the set of all functions f : X →A. In particular, AN is the set of all infinite sequences a0, a1, . . . . We sometimes write (an) for a0, a1, . . . .
• A<N is the set of all finite sequence (a1, . . . , an) where a1, . . . , an ∈A.
• When studying a structure M = (M, . . . ), we say X is definable if it is definable with parameters. If we wish to specify that it is definable without parameters we will say that it is ∅-definable. More generally, if we wish to specify it is definable with parameters from A we will say that it is A-definable.
• Because we use x (as well as res(x)) to denote the residue of an element, it would be confusing to also use x to denote a sequence of elements or variables.
We will instead use x to denote an arbitrary sequence x = (x1, . . . , xn). The length of x will usually be clear from context.
2 1 Valued Fields–Definitions and Examples 1.1 Valuations and Valuation Rings Definition 1.1 Let A be an integral domain, (Γ, +, 0, <) an ordered abelian group, a valuation is a map v : A× →Γ such that: i) v(ab) = v(a) + v(b); ii) v(a + b) ≥min(v(a), v(b)).
We refer to (A, v) as a valued ring.
A valued field (K, v) is a field K with a valuation v. The image of K under v is called the value group of (K, v) We also sometimes think of the valuation as a map from v : A →Γ ∪{∞} where v(0) = ∞and if a ̸= 0, then v(a) ̸= ∞. In this case we think of γ < ∞ and γ + ∞= ∞+ ∞= ∞for any γ ∈Γ.
Often we will assume that the valuation v : K× →Γ is surjective, so the value group is Γ.
Examples 1. Let K be a field and define v(x) = 0 for all x ∈K×. We call v the trivial valuation on K.
2. Let p be a prime number and define vp on Z by vp(a) = m where a = pmb where p̸ | b. We call vp the p-adic valuation on Z.
3. Let F be a field and define v on F[X] such that v(f) = m where f = Xmg where g(0) ̸= 0. More generally, if p(X) is any irreducible polynomial we could define vp(f) = m where f = pmg and p̸ | g.
4. Let F be a field and let F be the ring of formal power series over F.
We could define a valuation v : F →F by v(f) = m when f = amT m + am+1T m+1 + . . . where am ̸= 0.
Exercise 1.2 a) If A is an domain, K is its field of fractions and v is a valuation on A, show that we can extend v to K by v(a/b) = v(a) −v(b).
b) Show that this is the only way to extend v to a valuation on K.
Thus we can extend to the valuation vp on Z to vp : Q× →Z and we can extend the valuations on K[X] and K to K(X), the field of rational functions on K, and K( (T) ), the field of formal Laurent series, respectively.
Let F be a field and let F⟨T⟩= ∞ [ n=1 F( (T 1 n ) ) be the field of Puiseux series. If f ∈F⟨T⟩is nonzero then for some m ∈Z and n ≥1, f = P∞ i=m aiT i n and am ̸= 0. We let v(f) = m/n. We will show later 3 that if we start with an algebraically closed F of characteristic 0, then F⟨T⟩is also algebraically closed. For a more elementary direct proof see .
In the trivial valuation has value group {0}.
The rational functions and Laurent series have value group (Z, +, <) and the Puiseux series have value group Q.
We next give some very easy properties of valuations.
Lemma 1.3 i) v(1) = 0.
ii) v(−1) = 0.
iii) v(x) = v(−x); iv) If K is a valued field and x ̸= 0, then v(1/x) = −v(x).
v) If v(a) < v(b), then v(a + b) = v(a).
Proof i) v(1) = v(1 · 1) = v(1) + v(1), so v(1) = 0.
ii) 0 = v(1) = v((−1) · (−1)) = v(−1) + v(−1). Because ordered groups are torsion free, v(−1) = 0.
iii) v(−x) = v(−1 · x) = v(−1) + v(x) = v(x).
iv) v(1/x) + v(x) = v(1) = 0. Thus v(1/x) = −v(x).
v) we have v(a + b) ≥min(v(a), v(b)). Thus, v(a + b) ≥v(a). On the other hand v(a) = v(a + b −b) ≥min(v(a + b), v(b)). Since v(a) < v(b), we must have v(a + b) < v(b) and v(a) ≥v(a + b).
□ Suppose (K, v) is a valued field. Let O = {x ∈K : v(x) ≥0} we call O the valuation ring of K. Let U = {x : v(x) = 0}. If x ∈U, then 1/x ∈U. Moreover, if v(x) > 0, then v(1/x) < 0. Thus U is the set of units, i.e., invertible elements of O.
Let m = {x ∈O : v(x) > 0}. It is easy to see that m is an ideal. If x ̸∈m, then v(x) ≤0 and 1/x ∈O. Thus there is no proper ideal of O containing x.
Thus m is a maximal ideal and every proper ideal is contained in m.
Recall that a ring is local if there is a unique maximal ideal. We have shown that O is local. One property that we will use about local rings is that if A is local with maximal ideal m and a ∈A is not a unit, then (a) is a proper ideal and extends to a maximal ideal. Since m is the unique maximal ideal a ∈m.
Thus the unique maximal ideal of A is exactly the nonunits of A.
Exercise 1.4 Suppose A is a domain with fraction field K and P ⊂A is a prime ideal. Recall that the localization of A at P is AP = {a/b ∈K : a ∈A and b ̸∈P}.
Let AP P = {a1p1 + . . . ampm : a1, . . . , am ∈AP , p1, . . . , pm ∈P, m = 1, 2, . . . }.
Show that AP is a local ring with maximal ideal AP P.
4 Lemma 1.5 The ideals of O are linearly ordered by ⊂with maximal element m.
Proof Suppose P and Q are ideals of O, x ∈P \Q and y ∈Q\P. Without loss of generality assume v(x) ≤v(y). Then v(y/x) = v(y) −v(x) ≥0 and y/x ∈O.
But then y = (y/x)x ∈P, a contradiction. We have already shown that m is the unique maximal ideal.
□ Exercise 1.6 Consider A = CX, Y . Argue that A is a local domain that is not a valuation ring. [Hint: Consider the ideals (X) and (Y ) in A.] Define k = O/m. Since m is maximal, this is a field which we call the residue field of (K, v) and let res : O →k be the residue map res(x) = x/m. Often we write x for res(x).
Examples 1. In the trivial valuation on K, the valuation ring is K, the maximal ideal is {0} and the residue field is K.
2. For the p-adic valuation on Q the valuation ring is Z(p) = {m/n : m, n ∈ Z, p̸ | n.}, the maximal ideal is pZ(p) and the residue field is Fp, the p-element field.
3. Consider the field of formal Laurent series F( (T) ) with valuation v(f) = m where f = P∞ n=m anT n where am ̸= 0, then the valuation ring is F, the maximal ideal is all series P∞ n=m anT n where m > 0 and the residue field is F.
Exercise 1.7 a) Suppose (K, v) is an algebraically closed valued field. Show that the value group is divisible and the residue field is algebraically closed.
b) Suppose (K, v) is a real closed valued field. Show that the value group is divisible but the residue field need not even have characteristic zero.
Exercise 1.8 Suppose L is an algebraic extension of K and v is a valuation on L.
a) Show that the value group of L is contained in the divisible hull of the value group of K.
b) Show that the residue field of L is an algebraic extension of the residue field of K.
The valuation topology Let v : K× →Γ be a valuation. Let a ∈K and γ ∈Γ let Bγ(a) = {x ∈K : v(x −a) > γ} be the open ball centered at a of radius γ.1 The valuation topology on K is the weakest topology in which all Bγ(a) are open.
1Note this definition of radius is somewhat misleading. In particular, the balls get smaller as the radius gets larger!
5 Let Bγ(a) = {x ∈K : v(x −a) ≥γ} be the closed ball of radius γ centered at a. If b ̸= Bγ(a), then v(b−a) = δ < γ.
If x ∈Bδ(b), then v(x −a) = v((x −b) + (b −a)). Since v(x −b) > δ and v(b −a) = δ, v(x −a) = δ < γ. Thus Bγ(a) ∩Bδ(b) = ∅and closed balls are indeed closed in the valuation topology.
Lemma 1.9 If b ∈Bγ(a), then Bγ(a) = Bγ(b) and the same is true for closed balls. In other words, every point in a ball is the center of the ball.
Proof Let b ∈Bγ(a). If v(x −a) > γ, then v(x −b) ≥min(v(x −a), v(a −b)) > γ.
□ When we have a valuation v : K× →Z, Bn(a) = Bn+1(a). Thus the closed balls are also open. So there is a clopen basis for the topology.
In fact closed balls are always open.
Lemma 1.10 Every closed ball is open.
Proof Let B = Bγ(a) be a closed ball. Consider the boundary ∂B = {x : v(x −a) = γ}.
Suppose b ∈∂Bγ(a). If x ∈Bγ(b), then v(x −a) = v((x −b) + v(b −a)).
But v(b −a) = γ and v(x −b) > γ. Thus v(x −a) = γ and Bγ(a) is contained in δB. Thus B = Bγ(a) ∪ [ b∈δ(B) Bγ(b).
□ Exercise 1.11 Show that every closed ball B is a union of disjoint open balls each of which is a maximal open subball of B.
Exercise 1.12 Suppose B1, . . . , Bm are disjoint open or closed balls where m ≥2. Let ai be the center of Bi and let δ = min{v(a1 −ai) : i = 2, . . . , m}.
Show that Bδ(ai) is the smallest ball containing B1 ∪· · · ∪Bm.
Exercise 1.13 Prove that in the valuation topology all polynomial maps are continuous. [Hint: Consider the Taylor expansion of f(a + ϵ)] 6 Valuation rings Interestingly, the ring structure of the valuation ring O alone gives us enough information to recover the valuation.
Definition 1.14 We say that a domain A with fraction field K. is a valuation ring if x ∈A or 1/x ∈A for all x ∈K.
Let A be a valuation ring. Let U be the group of units of A and let m = A\U.
We claim that m is the unique maximal ideal of A. If a ∈m and b ∈A, then ab ̸∈U since otherwise 1/a = b(1/ab) ∈A. If a, b ∈m. At least one of a/b and b/a ∈A. Suppose a/b ∈A. Then a + b = b(a/b + 1) ∈m. Thus m is closed under addition so it is an ideal. If x ∈A \ m, then A ∈U, so no ideal of A contains x. Thus m is the unique maximal ideal of A. For x, y ∈K× we say x|y if y/x ∈A.
Let G = K×/U. Define a relation on G by x/U ≤y/U if and only if x|y.
For u, v ∈U we have x|y if and only if ux|vy. Thus < is well defined. If x|y and y|x, then x/y ∈U and x/U = y/U. If x/U ≤y/U and y/U ≤z/U. Then there are a, b ∈A such that y = ax and z = by. But then z = abx and x/U ≤z/U.
Thus ≤is a linear order of Γ. We write x/U < y/U if x|y and y̸ | x.
Exercise 1.15 Suppose x/U < y/U and z ∈K×. Show that x/U · z/U < y/U · z/U.
Thus (G, ·, <) is an ordered abelian group. It is also easy to set that 1/U ≤ x/U if and only if x ∈A. If we rename the operation + and the identity 0 we have shown that w(x) = x/U is a valuation on K with valuation ring A.
Exercise 1.16 Suppose (K, v) is a valued field with surjective valuation v : K×Γ and valuation ring O and let w : K× →G be the valuation recovered from O as above. If γ ∈Γ, choose x ∈K with v(x) = γ and define φ(γ) = w(g).
Show that φ : Γ →G is a well defined order isomorphism and φ(v(x)) = w(x) for all x ∈K×. Thus the valuation we have recovered is, up to isomorphism, the one we began with.
There are some interesting contexts where the valuation ring arises more naturally than the valuation. Suppose (F, <) is an ordered field and O ⊂F is a proper convex subring. If x ∈F \ O, then, in particular, |x| > 1. But then, |1/x| < 1 so 1/x ∈O. Thus O is a valuation ring.
One important example of this occurs when O is the convex hull of Z. We call this the standard valuation.
Exercise 1.17 Let F be an ordered field with infinite elements and let O be the convex hull of Z.
a) Show that the maximal ideal of O is the set of infinitesimal elements.
b) Suppose R ⊂F. Show that the residue field is isomorphic to R.
c) Suppose that F is real closed (but not necessarily that R ⊂F). Show that the residue field is real closed and isomorphic to a subfield of R.
The structure of the value group will depend on field F. Suppose F is real closed. In this case we can say is that it will be divisible. Suppose g is in the 7 value group and x ∈F with x > 0 and v(x) = g. Then there is y ∈F with yn = x. Hence g = v(yn) = nv(y).
Definition 1.18 An ordered group Γ is archimedian if for all 0 < g < h, there is n ∈N with ng > h.
Exercise 1.19 Show that an ordered abelian group is archimedian if and only if it is isomorphic to a subgroup of (R, +).
Exercise 1.20 Order R(X, Y ) such that X > r for all r ∈R and Y > Xn for all n ∈N. Let F be the real closure of (R(X, Y ), <) and consider the standard valuation. Show that the value group in nonarchimedean.
1.2 Absolute Values Definition 1.21 An absolute value on a ring A is a function | · | : A →R≥0 such that i) |x| = 0 if and only if |x| = 0; ii) |xy| = |x||y|; iii) (triangle inequality) |x + y| ≤|x| + |y|; The usual absolute values on R and C (or the restrictions to any subring) are absolute values in this sense and if i : K →C is a field embedding we obtain an absolute value | · | on K by taking |a| = ||i(a)||.
If v : A× →Γ is a valuation where Γ ⊆R and 0 < α < 1. Then we can construct and absolute value |x| = αv(x) for x ̸= 0. In this case |x+y| = αv(x+y).
Since v(x+y) ≥min(v(x), v(y)) and 0 < α < 1, |x+y| ≤max(|x|, |y|) ≤|x|+|y|.
An absolute value that satisfies this strong form of the triangle inequality is called a nonarchimedean absolute value or ultrametric.
We also have the trivial absolute value where |x| = 1 for all nonzero x–this is of course the absolute value corresponding to the trivial valuation.
Exercise 1.22 We can extend an absolute value on a domain A to the fraction field.
Exercise 1.23 Suppose K is a field with a nonarchimedean absolute value | · |.
a) Show that O = {x ∈K : |x| ≤1} is a valuation ring with maximal ideal m = {x : v(x) < 1}.
b) Show that the valuation topology associated with O is exactly the topol-ogy induced by the absolute value.
Once we have an absolute value we define a topology as usual by taking basic open balls Bϵ(a) = {x : |x−a| < ϵ}. If we start with a valuation v : K× →R and take the absolute value |x| = αv(x), then this is exactly the valuation topology.
Note that if we chose a different β with 0 < β < 1 and defined |x| = βv(x) we would define the same topology.
Definition 1.24 We say that two absolute values | · |1 and | · |2 on A are equivalent if they give rise to the same topology.
8 Consider the field Q. We have the usual absolute value on it which we will denote | · |∞. For p a prime we have the absolute value |x|p = (1/p)vp(a). This choice of base is convenient as it gives the product formula |x|∞ Y p prime |x|p = 1 which is trivial in this case but has nontrivial generalizations to number fields (see, for example, §10.2).
Exercise 1.25 Show that the absolute values | · |∞, | · |2, | · |3, . . . are pairwise inequivalent. [Hint: Consider the sequence p, p2, . . . .] Exercise 1.26 Consider the sequence 4, 34, 331, 3334, 33334, . . . . Show that with the absolute value | · |5 on Q this sequence converges to 2/3.
The next theorem shows that we have found all the absolute values on Q.
For a proof see, for example, §2.2.
Theorem 1.27 (Ostrowski’s Theorem) Any nontrivial absolute value on Q is equivalent to | · |∞or some | · |p.
Complete rings Suppose (A, | · |) is a domain with absolute value | · |. We say that a sequence (an : n = 1, 2, . . . ) in A is Cauchy if for all ϵ > 0, there is an n such that if i, j > n then |ai −aj| < ϵ.
We say that A is complete if every Cauchy sequence converges. Clearly R and C with the usual absolute values are complete.
Lemma 1.28 Consider the ring of power series K( (X) ) with the valuation v(f) = m where f = P n≥m anXn where am ̸= 0 and the absolute value |f| = αv(f), where 0 < α < 1. Then K is complete.
Proof Suppose f0, f1, . . . is a Cauchy sequence. Suppose fi = P n∈N ai,nXn (where ai,n = 0 for m > i Let ϵ ≤α1/n. There is mn such that if i, j > mn then |fi −fj| < ϵ. But then ai,k = aj,k for all k < n. Let bk be this common value. Let g = P k∈N bkXk. Then |fi −g| < 1/n for all i ≥n. It follows that (fi) converges to g.
□ Exercise 1.29 If (A, | · |) is a complete domain, then the extension to the fraction field is also complete.
in nonarchimedean complete domains we have a simple test for convergence of series.
Exercise 1.30 If (A, |·) is a nonarchimedean complete domain, then the series P∞ n=0 an converges if and only if lim an = 0.
If a is a domain with absolute value |·|. We can follow the usual constructions from analysis to build a completion b A of A. The elements of b A are equivalence 9 classes of Cauchy sequences from K where (an) and (bn) are equivalent if and only if for any ϵ > 0 there is an n such that |ai−bj| < ϵ for i, j > n. We can define an absolute value on b A such that the equivalence class of (an) has absolute value limn→∞|an|. We identify A with the equivalence classes of constant sequences.
Exercise 1.31 Complete the construction of b R. Prove that it is a complete ring and that if L ⊃K is any complete field with an absolute value extending the absolute value of K, then there is an absolute value preserving embedding of b K into L fixing K.
Lemma 1.32 Suppose A is a complete domain with nonarchimedean absolute value |·|. If (an) is a Cauchy sequence that does not converge to 0, then |ai| = |aj| for all sufficiently large i and j. Thus when we pass to the completion b A we add no new absolute values.
Proof We can find an N and ϵ such that |an| > ϵ and |an −am| < ϵ for all n, m > N. But then, since we have a nonarchimedean absolute value |an| = |am| for all n > N.
□ Definition 1.33 The ring of p-adic integers Zp is the completion of Z with the p-adic absolute value | · |p. Its fraction field is Qp the field of p-adic numbers.
Lemma 1.34 i) Suppose (an) is a sequence of integers. The series P∞ i=0 aipi converges in Zp.
ii) The map (an) 7→Zp is a bijection between {0, . . . , p −1}N and Zp.
Proof i) If m < n, then n X i=0 aipi − m X i=0 aipi p < 1 pm .
Thus the sequence of partial sums is Cauchy and hence convergent.
ii) Suppose (an) ∈ZN and p̸ | a0. Because p| P n>0 anpn ∞ X n=0 anpn p = |a0|p ̸= 0.
Let (an) and (bn) ∈{0, . . . , p −1}N be distinct. Suppose m is least such that am ̸= bm. Then X anpn = X nm anpn while X bnpn = X nm bnpn 10 It follows that | P anpn −P bnpn|p = 1 pm . Thus the map is injective. Given x ∈Zp choose (an) ∈{0, . . . , p −1}N such that P n<m anpn = x(mod pm) for all m. Then P∞ n=0 anpn = x. Thus the map is surjective.
□ It follows that every element x ∈Q× p can be represented as a series x = P n=m anpn where m ∈Z, am ̸= 0. and each an ∈0, . . . , p −1 and Zp = {x ∈ Qp : |x|p ≤1}. We have the p-adic valuation vp(x) = m. The value group is Z and the residue field is Fp.
Exercise 1.35 Suppose U is an open cover of Zp by open balls {x : |x−a|p < ϵ}.
Define T ⊂{0, . . . , p −1} j let φi,j : Z/(pi) →Z/(pj) be the map φi,j(x) = xmod (pj). Then Zp is the inverse limit of the this system of ring homomor-phisms.
Why valued fields?
Most of the most important example of valued fields arising in number theory, complex analysis and algebraic geometry have value groups that are discrete or, at the very least, contained in R. Why are we focusing on valuations rather than absolute values? Here are a couple of answers.
1. Valued fields with value groups not contained in R arise naturally when looking at standard valuations on nonstandard real closed fields.
2. Once we start doing model theory we will frequently need to pass to ele-mentary extensions. Even though Qp has value group Z when we pass to an elementary extension the value need not be a subgroup of R.
3. One of our big goals is the theorem of Ax–Kochen and Erˇ sov theorem that for any sentence φ in the language of valued fields, φ is true in Fp( (T) ) for all but finitely many p if and only if φ is true in Qp for all but finitely many p. This is proved by taking a nonprinciple ultrafilter U on the primes and showing that Y Fp( (T) )/U ≡ Y Zp/U.
These fields will have very large value groups.
11 2 Hensel’s Lemma 2.1 Hensel’s Lemma, Equivalents and Applications Definition 2.1 We say that a local domain A with maximal ideal m is henselian if whenever f(x) ∈A[X] and there is a ∈A such that f(a) ∈m and f ′(a) ̸∈m, then there is α ∈A such that f(α) = 0 and α −a ∈m.
Theorem 2.2 (Hensel’s Lemma) Suppose K is a complete field with nonar-chimedian absolute value | · | and valuation ring O = {x ∈K : |x| ≤1}. Then O is henselian.
Proof Suppose a ∈O, |f(a)| = ϵ < 1 and |f ′(a)| = 1. We think of a as our first approximation to a zero of f and use Newton’s method to find a better approximation. Let δ = −f(a) f ′(a) . Note that |δ| = |f(a)/f ′(a)| = ϵ. Consider the Taylor expansion f(a + x) = f(a) + f ′(a)x + terms of degree at least 2 in x.
Thus f(a + δ) = f(a) + f ′(a)−f(a) f ′(a) + terms of degree at least 2 in δ.
Thus |f(a + δ)| ≤ϵ2. Similarly f ′(a + δ) = f ′(a) + terms of degree at least 2 in δ and |f ′(a + δ)| = |f ′(a)| = 1.
Thus starting with an approximation where |f(a)| = ϵ < 1 and |f ′(a)| = 1.
We get a better approximation b where |f(b)| ≤ϵ2 and |f ′(b)| = 1. We now iterate this procedure to build a = a0, a1, a2, . . . where an+1 = an −f(an) f ′(an).
It follows, by induction, that for all n: i) |an+1 −an| ≤ϵ2n+1; ii) |f(an)| ≤ϵ2n; iii) |f ′(an)| = 1.
Thus (an) is a Cauchy sequence and converges to α, |α −a| ≤ϵ, and f(α) = limn→∞f(an) = 0.
□ Thus the ring of p-adic integers and rings of formal power series F are henselian.
Exercise 2.3 Let O be the valuation ring of the field of Puiseux series F⟨T⟩.
a) Show that O is not complete. [Hint: Consider the sequence T 1 2 , T 1 2 + T 2 3 , T 1 2 + T 2 3 + T 3 4 + . . . .] 12 b) Show that O is henselian.
Exercise 2.4 Suppose K is henselain and F ⊆K is algebraically closed in K, then F is henselian.
The next lemma shows that in a Hensel’s Lemma problem, there is at most one solution.
Lemma 2.5 Let O be a local domain with maximal ideal m. Suppose f(X) ∈ O[X], a ∈O, f(a) ∈m and f ′(a) ̸∈m. There is at most one α ∈O such that f(α) = 0 and α −a ∈m Proof Considering the Taylor expansions f ′(α) = f ′(a) + (a −α)b for some b ∈O. Thus f ′(α) ̸∈m.
If ϵ ∈m, then f(α + ϵ) = f(α) + f ′(α)ϵ + bϵ2 = f ′(α) + bϵ2 for some b ∈O. Since f ′(α) ̸∈m, f(α + ϵ) ∈m, but f(α + ϵ) ̸∈m2 unless ϵ = 0.
Thus if β −a ∈m and α ̸= β, f(β) ̸= 0.
□ There are many natural and useful equivalents of henselianity.
Lemma 2.6 Let A be a local domain with maximal ideal m. The following are equivalent.
i) A is henselian.
ii) If f(X) = 1 + X + ma2X2 + . . .m adXd where m ∈m and a2, . . . , ad ∈A, then f has a unique zero α in A, with α = −1mod m.
iii) Suppose f(X) ∈A[X], a ∈A, m ∈M and f(a) = mf ′(a)2, there is a unique α ∈A such that f(α) = 0 and a −α ∈(cf ′(a)).
Proof i) ⇒ii) is clear since f(−1) ∈m and f ′(−1) ̸∈m.
ii) ⇒iii) Then f(a + X) = f(a) + f ′(a)X + d X i=2 biXi for some bi ∈A. But then f(a + mf ′(a)Y ) = mf ′(a)2 + mf ′(a)2Y + d X i=2 bi(mf ′(a)Y )i = mf ′(a)2 1 + Y + d X i=2 mciY i !
for some c2, . . . , cd ∈A. By ii) we can find t u ∈A such that 1+u+P mciui = 0.
Let α = a + mf ′(a)u. Then f(α) = 0 and a −α ∈m, as desired.
13 iii) ⇒i) is immediate.
□ In a valuation ring O, condition iii) can be restated v(f(a)) > 2v(f ′(a)).
Exercise 2.7 Suppose R is a real closed field and O ⊂R is a proper convex subring. Show that O is henselian. [Hint: Consider f(X) as in ii) and show that f must change sign on O.] Exercise 2.8 Suppose (K, <) is an ordered field, O is a proper convex subring, and (K, O) is henselian with divisible value group and real closed residue field.
Prove that every positive element of K is a square. [We will see in Corollary 5.17 that, in fact, K is real closed.] The following equivalent is also useful.
Corollary 2.9 Let A and m be as above, then A is henselian if and only for every polynomial f(Y ) = 1 + Y + Pn i=2 aiY ii where a2, . . . , an ∈m, there is α = −1(mod n) such that f(α) = 0.
Proof (⇒) Clear.
(⇐) It suffices to show that for every polynomial of the form Xn + Xn−1 + Pn−2 i=0 aiXi where a0, . . . , an−2 ∈m has a zero congruent to −1, or equivalently that every polynomial of the form 1 + (1/X) + n−2 X i=0 ai(1/X)n+i has a zero congruent to −1. Letting Y = 1/X we find the desired solution.
□ Corollary 2.10 If (K, v) is an algebraically closed valued field, then K is henselian.
Proof Consider the polynomial f(X) = Xn + Xn−1 + an−2Xn−2 + · · · + a0 where a0, . . . , an−2 ∈m.
It suffices to show that f has a zero congruent to −1(mod m). Any zero that is a unit must be congruent to −1(mod m), so it suffices to show that f has a zero that is a unit. Since K is algebraically closed, we can factor f(X) = (X −b1) · · · (X −bn). Each bi must have nonnegative value, as if v(bi) < 0, then v(bn i ) < v(aibi) for all i < n and v(f(bi)) = nv(bi), so f(bi) ̸= 0. But −P bi = 1 so some bi must have value 0.
□ p-adic squares and sums of squares A typical application of Hensel’s lemma is understanding the squares in Q× p .
First suppose p ̸= 2. Let a ∈Qp. Let a = pmb where b is a unit in Zp. If a = c2, then vp(a) = 2vp(c). Thus m is even. We still need to understand when a unit b ∈Zp is a square. Let f(X) = X2 −b. Let b be the residue of f. Then if b is a square b must be a square in the residue field Fp. If x ∈Zp such that x2 = b.
Then vp(x) = vp(c) = 0 and vp(f ′(x)) = vp(2x) = 0. Thus, by Hensel’s Lemma, there is y ∈Zp, such that y2 = b and vp(x −y) > 0. Thus a ∈Q2 p is a square if 14 and only if a = p2nb where b is a unit and b is a square in Fp. Recall that for p ̸= 2 the squares are an index 2 subgroup of F× p . It follows that the squares are an index 4 subgroup of Q× p .
We need to be a bit more careful in Z2. If f(X) = X2 −c and x2 = c, then v2(x) = v2(2x) = 1 so we can not apply Hensel’s Lemma directly. We can use the characterization iii) of Lemma 2.6 but we need to look at squares mod 8.
Consider f(X) = X2 −b. Suppose b is a unit in Z2 and b is a square. Then b is a square mod 8. We argue that the converse is true. Consider f(X) = X2 −b.
Suppose x ∈Zp and x2 −b = 0(mod 8). Then v2(x) = 0 and v2(2x) = 1. Thus v2(f(x)) ≥3 while v2(f ′(x)) = 1. Thus b is a square in Z2. The nonzero squares mod 8 are 1 and 4. Thus a ∈Z× 2 is a square if and only if a = 22nb where b = 1 or 4(mod 8). Thus the squares are an index 8 subgroup of Q× 2 .
Exercise 2.11 a) Show that if p ̸= 2, then Zp = {x ∈Qp : ∃y y2 = px2 + 1} b) Show that Z2 = {x ∈Q2 : ∃y y2 = 8x2 + 1}.
Exercise 2.11 shows that the p-adic integers Zp are definable in Qp in the pure field language.
Thus, from the point of view of definability, it doesn’t matter if we view Qp as a field or as a valued field.
Exercise 2.12 a) Suppose p̸ | n. Show x is an nth-power in Qp if and only if n|vp(n) and res(n) is an nth-power in Fp.
b) Suppose p|n. Show that x is an nth-power in Qp if and only if x = pnmy where y is a unit and y is an nth-power mod p2v(n)+1).
c) Conclude that the nonzero nth-powers are a finite index subgroup of Q× p .
Exercise 2.13 a) Let K be a field of characteristic other than 2. Show that K = {f ∈K( (T) ) : ∃g g2 = Tf 2 + 1}.
b) Suppose K has characteristic 2 and give a definition of K in K( (T) ).
Lemma 2.14 If p is an odd prime and u ∈Zp is a unit, then u is a sum of two squares in Zp.
Proof In Fp there are (p + 1)/2 squares. Since the set F2 p and u −F2 p each of size (p + 1)/2, they must have non-empty intersection. Let x, y ∈Zp such that x2 + y2 = u. At least one of x and y is a unit. Say x is a unit. Let f(X) = X2 −(y2 −u). By Hensel’s Lemma we can find a zero z and z2 +y2 = u.
□ Lemma 2.15 Suppose p = 1(mod 4). Every element of Zp is a sum of two squares.
Proof We know that −1 is a square in Fp. By Hensel’s Lemma there is ξ ∈Zp with ξ2 = −1.
Let a ∈Zp. Note that (a + 1)2 −(a −1)2 = 4a.
15 Thus a = a + 1 2 2 + ξ(a −1) 2 2 .
Note that since p ̸= 2, 1/2 ∈Zp. Thus we have written a as a sum of squares in Zp.
□ Corollary 2.16 If p = 1(mod 4) then every element of Qp is a sum of two squares.
Proof We can write a = p2mb for some b ∈Zp. If b = c2 + d2, then a = (pc)2 + (pd)2.
□ Lemma 2.17 If p = 3(mod 4), then a ∈Qp is a sum of two squares if and only if vp(a) is even.
Proof If a = p2mu where u is a unit. Then u is a sum of two squares so a is as well.
Suppose vp(a) is odd and a = x2 + y2. Then a is not a square, thus both x and y are nonzero. Also vp(x) = vp(y) as otherwise vp(a) is even. Let x = pmu and y = pmv where u, v are units in Zp. Then a = p2m(u2 + v2). But vp(a) is odd, thus vp(u2 + v2) > 0 and (u/v)2 = −1(mod p), a contradiction since p = 3(mod 4).
□ Lemma 2.18 In Q2 if a = 2mu where u is a unit, then a is a sum of two squares if an only if u = 1(mod 4).
Proof First suppose u = 1(mod 4). We first show that u is a sum of squares.
Then u = 1 or 5 (mod 8). If u = 1(mod 8), then u is already a square in Z2. If u = 5(mod 8), then u/5 = x2 for some x ∈Z2 and u = x2 + (2x)2.
Recall that a product of two sums of squares is a sum of squares as (a2 + b2)(c2 + d2) = (ac −bd)2 + (ad + bc)2.
Since 2 = 1 + 1 and 1/2 = (1/4) + (1/4) are sum of two squares 2mu is a sum of two squares.
Next suppose u = 3(mod 4). If a is a sum of two squares, then, as above, u is also a sum of two squares. Say u = x2 + y2. This is impossible if x, y ∈Z2 since the only sums of two squares mod 4 are 0, 1 and 2.
Without loss of generality suppose vp(x) < 0. But then we must have vp(y) = vp(x) = −n where n > 0. Then x = z/2n and y = w/2n where z and w are units in Zp and 4nu = (z2 + w2). Thus z2 + w2 = 0(mod 4). But z and w are units and, thus, z2, w2 = 1(mod 4) and z2 + w2 = 2(mod 4), a contradiction.
□ We can use these results, particularly the result about primes congruent to 3(mod 4) to rephrase a classic result of Euler’s. Recall that an integer m > 0 is a sum of two squares if and only if vp(m) is even for any prime p = 3(mod 4) that divides m. See, for example, §27.
16 Corollary 2.19 An integer m is a sum of two squares if and only if it is a sum of squares in R and in each Zp.
Proof (⇒) is clear.
(⇐) If m is a square in R, then m ≥0. By Lemma 2.17, if p = 3(mod 4), then vp(m) is even. Thus m is a square in Z.
□ This corollary can be though of as a baby version of a local-global principle.
Hensel’s Lemma gives us a powerful tool for solving equations in the p-adics.
We have no comparable tool in the rational numbers. Of course if a system of polynomials over Q has no solution in Qp or R, then it has no solution in Q.
Sometimes, we can prove existence results in Q by proving them in all completions. These are called local-global results as they reduce question in the global field Q to the local fields Qp and R. These principles are very useful it is often much easier to decide if there is a solution in the local fields. One of the most general is the Hasse Principle. See for example §IV.3.
Theorem 2.20 (Hasse Principle) Let p(X1, . . . , Xn) = P i,j≤n ai,jXiXj ∈ Q[X1, . . . , Xn]. Then p = 0 has a nontrivial solution in Q if and only if it has nontrivial solutions in R and Qp for all primes p.
Exercise 2.21 Suppose p > 2 is prime. Let F(X1, . . . , Xm, Y1, . . . , Ym) = n X i=1 aiX2 i + m X j=1 pbjX2 j where ai, bj ∈Z are not divisible by p.
a) Suppose F has a nontrivial zero in Qp.
Show that either P aiX2 i or P biY 2 i has a nontrivial solution in Fp. [Hint: First show that there is a solution (x1, . . . , xm, y1, . . . , yn) ∈Zp where some xi or yj is a unit. Show that if some xi is a unit, then (x1, . . . , xm) is a zero of P aiX2 i and otherwise (y1, . . . , yn) is a zero of P bjY 2 j .
b) Use Hensel’s Lemma to prove that if either P aiX2 i or P bjY 2 j has a nontrivial zero in Fp, then F as a nontrivial zero in Qp.
c) Show that 3X2 +2Y 2 −Z2 = 0 has no nontrivial solution in Q3 and hence no nontrival solution in Q.
p-adic roots of unity In the next exercises and lemma we will look for roots of unity in Qp.
Exercise 2.22 Let p be an odd prime.
a) Show that there are exactly p −1 distinct (p −1)th roots of unity in Zp and no two distinct roots are equivalent mod p b) Suppose that ξ1 and ξ2 are roots of unity of order m1 and m2 where p̸ | m1, m2.
Show that if ξ1 = ξ2(mod p), then ξ1 = x2.
[Hint: Consider f(X) = Xm1m2 −1 and apply Lemma 2.5.] 17 Lemma 2.23 Let p be an odd prime.
i) The only pth-root of unity in Qp is 1.
ii) The only pth-power root of unity in Qp is 1.
Proof i) Clearly any pth-root of unity ξ is in Zp. Suppose ξp = 1. In Fp, ξ p = ξ, thus ξ = 1(mod p). Let f(X) = Xp −1. Then vp(f ′(ξ)) = 1 and, by the uniqueness part of Lemma 2.5 iii), ξ is the unique zero of f in {x ∈Zp : vp(x −ξ) ≥2} = ξ + p2Zp. We will show that 1 ∈ξ + p2Zp and conclude that ξ = 1.
Suppose ξ = 1 + px where x ∈Zp. Then 1 = ξp = (1 + px)p = 1 + p(px) + p X i=2 p i (px)i Each term p i (px)i is divisible by p3 thus 1 = 1 + p2x(mod p3). Hence p2x = 0(mod p3) and p|x. But then ξ = 1(mod p2) and, since ξ is the pth-root of unity in ξ + p2Zp, ξ = 1.
ii) We prove by induction that if ξpm = 1, then X = 1. If ξpm+1 = 1, then (ξpm)p = 1 and, by i), ξpm = 1. By induction ξ = 1.
□ Corollary 2.24 If p is an odd prime, then the only roots of unity in Qp are the p −1 roots of Xp−1 −1.
Proof Let n = pkm where p̸ | m. If ξn = 1, then ξ = xy where xpk = 1 and ym = 1. By the previous exercise and lemma, x = 1 and yp−1 = 1.
□ Exercise 2.25 Prove that the only roots of unity in Q2 are ±1.
The Implicit Function Theorem We give a very different application of Hensel’s Lemma in power series rings to a prove an algebraic version of the Implicit Function Theorem. Let F be a field and let p(X, Y ) ∈F[X, Y ] such that f(0, 0) = 0 and ∂f ∂Y (0, 0) ̸= 0. Consider the polynomial g(Y ) ∈F[Y ], where g(Y ) = f(T, Y ). Then g(0) = f(T, 0) = f(0, 0) = 0 (mod (T)). But g′(Y ) = ∂f ∂Y (0, 0) ̸= 0 (mod T).
Thus by Hensel’s Lemma, we can find φ(T) ∈F such that f(T, φ(T)) = 0.
Thus we have found a power series point on the curve. We think of the power series as parameterizing a branch on the curve near (0,0).
If ∂f ∂Y (0, 0) = 0, but ∂f ∂X (0, 0) ̸= 0, we could find a ψ(T) such that f(ψ(T), T) = 0. By changing variables we could, more generally shows that if (a, b) ∈F 2 is any smooth point of the curve we can find a power series branch. This type of result can be extended to singular points but requires more specialize properties of power series and Puiseux series rings such as Weierstrass factorization (see, for example, ).
18 2.2 Lifting the residue field In some of our later work it will be useful to view the residue field k as a subfield of the valued field K. Of course this is sometimes impossible. The p-adics have characteristic 0, while the residue field has characteristic p. However, when K is henselian and k is characteristic 0, this will always be possible.
Theorem 2.26 Suppose K is a henselian valued field and the residue field k has characteristic 0.
Then there is a field embedding j : k →K such that res(j(x)) = x for all x ∈k.
We call such a j a section of the residue map.
Proof We will inductively build j : k →K. At any stage of our construction we will have k0 ⊂k a subfield and j : k0 →K a field embedding with res(j(x)) = x for all x ∈k0. To start, since k has characteristic 0, we can take k0 = Q and let j : Q →Q be the identity map. The theorem will follow by induction using the following two claims.
claim 1 Suppose we have such a j : k0 →K where k0 is a subfield of k and x ∈k \ k0 is transcendental over k0.
Then we can extend j to a suitable b j : k0(x) →k.
Choose y ∈K such that res(y) = x. We claim that y is transcendental over K0 = j(K). Suppose not. Then there is p(X) ∈K0[X] such that p(y) = 0.
But then p(x) = 0. Since res ◦j is the identity on k0, p(X) is not identically 0, thus x is algebraic over k0 a contradiction. We extend j to b j by sending y to x.
Since the residue map is a ring homomorphism, res ◦b j is the identity.
claim 2 Suppose we have such a j : k0 →K where k0 is a subfield of k and x ∈k\k0 is algebraic over k0. Then we can extend j to a suitable b j : k0(x) →K.
There is y0 ∈k with res(y0) = x. Suppose p(X) is the minimal polynomial of x over k0. Then p(x) = 0 and p′(x) ̸= 0. Let q(X) be the image of the p(X) under j. Since res ◦j = id, q = p. But then q(x) = 0 and q′(x) ̸= 0, and, by henselianity, there is y ∈K such that q(y) = 0 and res(y) = res(y0) = x. We extend j to b j by sending y to x. Since the residue map is a ring homomorphism, res ◦b j is the identity.
□ We can use this theorem to prove an easy result very much in the spirit of the Ax–Kochen and Ershov results we will see in §5.
Theorem 2.27 (Greenleaf) Let f1, . . . , fm ∈Z[X1, . . . , Xn] then for all but finitely many primes p, every solution to f1 = · · · = fm = 0 in Fn p, lifts to a solution in Zn p.
Proof We consider vauled fields as fields with a predicate for the valuation ring. Consider the sentence Θ in the language of valued fields ∀x f1(x), . . . , fm(x) ∈m →∃y f1(y) = · · · = fm(y) = 0 ∧yi −xi ∈m for i = 1, . . . , n .
19 Θ asserts that any zero of f1 = · · · = fm in the residue field lifts to the field. By Theorem 2.26, if K is a henselian valued field with characteristic zero residue field we can embed k into K, thus Θ holds. In particular, Q Zp/U | = Θ for any nonprincple ultrafilter U. Thus, by the Fundamental Theorem of Ultraproducts, Zp | = Θ for all but finitely many primes.
□ 2.3 Sections of the value group Once could ask similar questions about the value group.
This doesn’t have anything to do with henselianity and could be moved later.
If (K, v) is a valued field with value group Γ we say that s : Γ →K is a section of the valuation if v(s(γ)) = γ and s(γ +γ′) = s(γ)s(γ′) for all γ, γ′ ∈Γ.
For example, in the p-adics n 7→pn is a section. The next two lemmas give useful examples where sections exist.
Lemma 2.28 Let (K, v) be a real closed or algebraically closed field. Then there is a section s : Γ →K.
Proof In either case Γ is divisible. Let (γi : i ∈I) be a basis for Γ as a Q-vector space.
If K is real closed then for each i we pick xi ∈K with xi > 0 and v(xi) = γi.
Let s(m1γi1 + . . . mkγik) = xm1 i · · · xmk k . Then s is the desired section.
If K is algebraically closed then for each i we need to choose a coherent sequence of n-th roots xi,n for n = 1, 2, . . . such that xm i,nm = xi,n for all n and m and v(xi,1) = γi. We can then let s(m1γi1 + . . . mkγik) = xli i1,ni · · · xlk ik,nk where mi = li/ni and li and ni are relatively prime. Then s is the desired section.
□ Exercise 2.29 Suppose K is a henselian valued field with divisible value group Γ and the residue field k is of characteristic zero with k∗divisible. Prove that there is a section s : Γ →K× of the valuation.
We will show that sufficiently rich fields have sections.
Theorem 2.30 If (K, v) is an ℵ1-saturated valued field with value group Γ, then there is a section s : Γ →K.
Corollary 2.31 Every valued field has an elementary extension where there is a section of the value group.
The Theorem follows from the next lemma. Recall that if G is an abelian group a subgroup H ⊆G is pure if G/H is torsion free, i.e., if nx ∈H, then x ∈H for all n > 0. If Γ0 ⊂Γ we say that s : Γ0 →K× is a partial section if it is a homomorphism with v ◦s = id.
Lemma 2.32 Suppose K is an ℵ1-saturated valued field with value group Γ, Γ0 ⊂Γ is a pure subgroup, s : Γ0 →K× is a partial section and g ∈Γ \ Γ0.
Then there is a pure subgroup Γ0 ∪{g} ⊂Γ1 ⊆Γ and b s ⊃s a partial section of Γ1.
20 We know that Γ0 = {0} is a pure subgroup of Γ with partial section s(0) = 1.
By Zorn’s Lemma there is a maximal partial section and by the Lemma it must be defined on all of Γ.
Proof of Lemma Let H be the group generated by Γ ∪{g}. We first look for a smallest pure subgroup Γ1 containing H. Let S = {n > 0 : there is b ∈Γ such that b/H has order exactly n in Γ/H. If n ∈S there is b ∈Γ, c ∈Γ0 and m ∈Z such that nb = c + mg. We make some observations.
• if m, n ∈S, let b/H have order m and c/H have order n, then (b + c)/H has order d, where d is the least common multiple of m and n. Thus d ∈S.
• If (nk)b = c + (mk)g, then c = k(nb −mg) ∈Γ0 and, by purity of Γ0, nb −mg ∈Γ0. Thus b/H has order n. It follows that if n ∈S, there are b ∈Γ, c ∈Γ0 and m ∈Z such that nb = c + mg where n and m are relatively prime.
• If nb = c + mg where n and m are relatively prime, then there is b′ ∈Γ and c′ ∈Γ0 such that nb′ = c′ + g.
There are integers u and v such that un + vm = 1. Then n(ub) = uc + umg and n(ub −vg) = uc + g.
• If nb = c + g and nb′ = c′ + mg, then b′ is in the group generated by Γ0 ∪{b}.
Note that nmb = cm + mg. Thus n(b′ −mb) = c′ −mc ∈Γ0. Thus, by the purity of Γ0, b′ −mb ∈Γ0.
Suppose for n ∈S we choose bn ∈Γ and cn ∈Γ0 such that nbn = cn+g. Note that 1 ∈S and b1 = g. Let Γ1 be the subgroup generated by Γ0 ∪{bn : n ∈S}.
Putting together the previous observations, we see that Γ1 is the smallest pure subgroup of Γ containing Γ0 ∪{g}.
We need to find (xn : n ∈S) ∈K such that v(xn) = bn and xn n = s(cn)x1 for all n. Consider the set of formulas Σ = {v(xn) = bn ∧xn n = s(cn)x1 : n ∈S}.
Since (K, v) is ℵ1-saturated, it suffices to show that every subset of Σ is consis-tent.
Let S0 be a finite subset of S. Without loss of generality we may assume that 1 ∈S0 and there is N ∈S0 such that n|N for all n ∈S0. Choose xN with v(xN) = bN. We must have x1 = xN s(cN).
Suppose n ∈S0 and N = nd. Then NbN = cN + nbn −cN. Thus n(dbN −bn) = cN −cn ∈Γ0 and there is cN,n ∈Γ0 such that dbN −bn = cN,n. Then s(cN,n)n = s(cN) s(cn) .
Let xn = xd N s(cN,n). Then xN n = xN N s(cN,n)n = xN Ns(cn) s(cN) = s(cn)x1 21 and v(xn) = dbN −cN,n = bn, as desired. Thus every finite subset of Σ is consistent. If (xn : n ∈S) satisfies Σ we can extend s by sending bn 7→xn for n ∈S.
□ Exercise 2.33 a) Modify the proof above to prove the following. Consider the language of groups where we add a unary predicate for a distinguished subgroup.
Suppose (G, H) is an ℵ1-saturated abelian group with proper subgroup such that G/H is torsion free. Prove that there is a section s : G/H →G, i.e., a homomorphism such that s(x/H)/H = x/H.
b) Use the above to show that in every ℵ1-saturated valued field K there is a section s : Γ →K× with v ◦s = id.
Unfortunately, we can not always find sections.
Exercise 2.34 Consider the field Q(X1, X2, . . . ) with the valuation where v(Xn) = 1/n. Prove that there is no section of the value group.
2.4 Hahn fields Let k be a field and let (Γ, +, <) be an ordered abelian group. We will consider the multiplicative group of formal monomials (T γ : γ ∈Γ) where T 0 = 1 and T γ1T γ2 = T γ1+γ2 and formal series f = P γ∈Γ aγT γ where aγ ∈k. The support of f is supp(f) = {γ : aγ ̸= 0}. We will only consider series f where supp(f) is well ordered (i.e. every nonempty subset has a least element). The Hahn seriesfield is k( ( (Γ) ) ) = {f : supp(f) is well ordered}.
Addition is easy to define if f = P γ∈Γ aγT γ and g = P γ∈G bγT γ. Then a + b = X γ∈Γ (aγ + bγ)T γ.
Lemma 2.35 Let A and B be well ordered subsets of Γ. Then A + B is well ordered and for any c ∈A + B the set {(a, b) ∈A × B : a + b = c} is finite.
In particular, if A ⊂Γ is well ordered then the set Σn = {a1 + · · · + an : a1, . . . , an ∈A} is well ordered and for all g ∈Σn, {(a1, . . . , an) ∈An : P ai = g} is finite.
Proof Suppose (a0, b0), (a1, b1), . . . are distinct such that ai + bi ≥aj + bj for i > j. We can find a strictly monotonic subsequence of the ai. Since A is a well ordered, the sequence can not be decreasing. Thus we may assume a0 ≤a1 ≤. . . . But then b0 > b1 > . . . is an infinite descending sequence, contradicting the fact that B is well ordered.
□ This allows us to define multiplication by X γ∈Γ aγT γ X γ∈Γ bγT γ = X γ∈Γ X γ1+γ2=γ aγ1bγ2T γ.
22 The usual proofs of commutativity and associativity in power series show that k( ( (Γ) ) ) is a domain. There is a natural valuation v(f) = min supp(f). A stronger form of the last lemma is needed to show k( ( (Γ) ) ) is a field. For a proof see §7.21.
Lemma 2.36 (Neumann’s Lemma) Suppose A ⊂Γ is well ordered and ev-ery element of A is positive. Let Σ = {a1 + · · · + an : (a1, . . . , an) ∈A g1 > . . . is an infinite decreasing sequence in Σ. For each i let σi = (σi(1), . . . , σi(ni)) ∈S be of minimal length such that gi = σi(1) + · · · + σi(ni) and ni is the minimal length such that there is (a1, . . . , am) ∈S with a1 + · · · + am = gi. We also assume that σi(1) ≤σi(2) ≤. . . . We can thin the sequence such that n0 ≤n1 ≤n2 ≥. . . . [In this proof we use several times that in an ordered set every sequence has a strictly monotonic subsequence.] claim By altering the sequence we may assume that the sequence n0, n1, n2 . . .
is constant.
The lemma will lead to a contradiction as we have shown that the set of sums of n-elements of A is well ordered for each n.
Suppose we have arranged things such that n0 = n1 = · · · = nk < nk+1.
We can pass to a subsequence fixing σ0, . . . , σk but, perhaps, thinning the rest such that σk+1(1), σk+2(1), σk+3(1), . . . is strictly monotonic. Since A is well ordered, we must have σk+1(1) ≤σk+2(1) ≤σk+3(1), . . . . For all j > k let σ′ j = (σj(2), . . . , σj(nj)) and let hj = σj(2) + · · · + σj(nj). Since all element of A are nonnegative hj < gj and since σj(1) ≥σk+1(1) for j > k, hk+1 > hk+2 > . . . .
Replace gj by hj and σj by σ′ j for j > k. We have shortened the sequence σk+1 by one. Repeating this procedure finitely many times we may assume that σ1, . . . σk+1 have the same length.
Repeating this process for each k we get may assume that n0, n1, . . . is constant. [Note that after stage k we never change σk.] Thus we conclude that Σ is well ordered. We need to show that for all g ∈Σ there are only finitely many sequence (a1, . . . , an) ∈A h1 > . . .
contradicting that Σ is well ordered. If σ0(1), . . . , σn(1), is constant then every hi = h0 −σ0(1) < g since every element of A is positive. But this contradicts the minimality of g.
□ Corollary 2.37 If P n=0 anXn ∈k, f ∈k( ( (Γ) ) ) and v(f) > 0, then P n=0 anf n is a well defined element of k( ( (Γ) ) ).
23 We can now show that k( ( (Γ) ) ) is a field. Suppose f ̸= 0. Then f = aT γ(1−ϵ) where ϵ ∈k( ( (Γ) ) ) and a ∈k×. and v(ϵ) > 0. Then g = P∞ n=0 ϵn ∈T and the usual arguments show that g(1 −ϵ) = 1. Thus 1/f = (1/a)T −γg and k( ( (Γ) ) ) is a field.
Definition 2.38 If f, g ∈k( ( (Γ) ) ), f = P aγT γ and P bγT γ, we say that g is an end extension of f or, alternatively, that f is a truncation of g if supp(f) ⊂ supp(g), every element of supp(g) \ supp(f) is greater than every element of supp(f) and if γ ∈supp(f) then aγ = bγ. We write f ◁g.
Exercise 2.39 Suppose we have (fβ : β < α) for some ordinal α where fδ ◁fβ for all δ < β < α. Let fβ = P aβ,γT γ. Show that S β<α supp(fβ) is well ordered and if f = P aγT γ where aγ = aβ,γ for all sufficiently large β < α. Moreover v(fα −f) > supp(fα).
Lemma 2.40 The field of Hahn series k( ( (Γ) ) ) is henselian.
Proof While k( ( (Γ) ) ) need not be complete, we can mimic the proof of Hensel’s Lemma with a transfinite iteration. Let O be the valuation ring, let p(X) ∈ O[X] and a ∈O such that v(p(a)) > 0 and v(p′(a)) = 0. As we saw in the proof of Hensel’s Lemma if we take b = a −p(a) p′(a), then v(p(b)) ≥2vp(a) and v(p′(b)) = 1.
We build a sequence of better and better approximations. Let a0 = a. Given aα if p(aα) = 0 we are done, otherwise let aα+1 = a + α −p(aα)/overp′(aα) and let γα = v(p(a)) = v(aα+1 −aα).
Suppose α is a limit ordinal and we have constructed (aβ : β < α). Let aβ = P g∈Γ bβ,γT γ. If β > α, then aβ,γ = aβ+1,γ for all γ < γβ. Let fβ = P γ < γβaβ+1,γ. Then v(aδ −fβ) ≥γβ and fβ is an initial segment of the series fβ for all β > α. We can naturally take the limit of the series (fα : β < α) as in Exercise 2.39 and let this be aα. We have v(aα −aβ) > γβ for all β < α.
As in the proof of Hensel’s Lemma, this implies v(p(aα) > γβ for all β < α and v(p′(aα)) = 1.
Since we are building (γα) an increasing sequence in Γ, this process must stop at some ordinal α < |Γ|+, but it only stops when we find the desired zero of p.
□ Corollary 2.41 For any field k and any ordered abelian group Γ there is a henselian valued field with value group Γ and residue field k.
Exercise 2.42 Suppose k is an ordered field.
a) Show that we can order k( ( (Γ) ) ), by x > 0 if and only if x = atγ(1 + ϵ) where a > 0.
b) Suppose ever nonnegative a ∈k is a square and Γ is 2-divisible, i.e., if g ∈Γ there is h ∈Γ with 2h = g. Let a ∈k( ( (Γ) ) ) with a > 0. Show that a is a square. Thus the ordering in a) is the only possible ordering of k( ( (Γ) ) ).
We will show in Corollary 3.17 that if k is real closed and Γ is divisible then k( ( (Γ) ) ) is real closed.
24 Hahn series fields recapture some aspects of completeness.
Definition 2.43 Let K be a valued field. We say that K is spherically complete if whenever (I, <) is a linear order and (Bi : i ∈I) is a family of open balls such that Bi ⊃Bj for all i < j, then T i∈I Bi ̸= ∅.
Lemma 2.44 Any Hahn series of field k( ( (Γ) ) ) is spherically complete.
Proof Without loss of generality we may assume that there is an ordinal α (Bβ : β < α) and Bδ ⊃Bβ for δ < β < α. Let Bβ = {x : v(x −aβ) > γβ}. For each β < α choose fβ such that sup supp(fβ) = γβ and v(fβ −aβ) > γβ. Then fδ ◁fβ for δ < β < α. Let f be as in Exercise 2.39, then f ∈S β<α Bβ.
□ maximal valued fields Hahn fields k( ( (Γ) ) ) are the maximal fields with residue field k and value group Γ.
Definition 2.45 If (K, v) is a valued field extending L is a subfield, then K is an immediate extension if v(K) = v(L) and kK = kL.
For example Qp is an immediate extension of Q.
Lemma 2.46 k( ( (Γ) ) ) has no proper immediate extensions.
Proof Suppose K is an immediate extension of k( ( (Γ) ) ) and x ∈K \k( ( (Γ) ) ). We build a series as follows: Let γ0 = v(x). Choose a0 ∈k such that res(x/T γ0) = aγ. Then v(x −a0T γ 0 ) > γ0.
Suppose we have constructed (aβ : β < α) a sequence in k and (γβ : β < α) an increasing sequence in Γ such that if fα = P δ<β aδT γδ then v(x −fα) > γβ for all β < α. Let γα = v(x −fα). As before we can find aα ∈k such that res((x−fα)/T γα) = aα. Then v(x−fα +aαT γα) > γα and we can continue the induction.
In this way we will build an increasing map from the ordinals into Γ, but this must stop by some α < |Γ|+, a contradiction.
□ Definition 2.47 We say that (K, v) is a maximal valued field if it has no proper immediate extensions.
We will show that every valued field has a maximal extension.
Lemma 2.48 (Krull’s Bound) If K is a valued field, then |K| ≤|k||Γ|.
Proof Let κ = |k|. Suppose B is a closed ball of radius of radius γ, then, as we saw in Lemma 1.10, that B is the union of κ disjoint open balls of radius γ. Let (CB α : α < κ) be the listing. For x ∈K define fx : Γ →κ, be defined so that if B is the closed ball of radius γ around x, then x ∈CB fx(γ). Suppose x ̸= y and v(x −y) = γ. Then fx(δ) = fy(δ) for all δ < γ, but fx(γ) ̸= fy(γ).
Thus x 7→fx in injective and |K| ≤|k|Γ.
□ 25 Corollary 2.49 (Kaplansky) If K is a valued field, then there is K ⊆L an immediate extension that is maximally valued.
Proof By Krull’s bound, the collection of immediate extensions of K is a set so we can apply Zorn’s Lemma to find a maximal immediate extension.
□ In Exercise 5.41 we will show that any maximally valued field is spherically complete.
26 3 Extensions of Rings and Valuations When studying the model theory of certain theories of valued fields our first step will usually be to prove quantifier elimination in an appropriate language. Proofs of quantifier elimination in algebraic theories usually require some algebraic extension results. That is particular true in valued fields. In this section we will prove some basic results and then will use them in §4 to begin the study of the model theory of algebraically closed valued fields. In §5 we will focus on extension results for henselian valued fields.
For more details on some of the background results from commutative al-gebra see, for example or . All of the results we will be proving on extensions of valuations can be found in . To be careful we will tend to state most results for domains even though many are true in more generality.
3.1 Integral extensions We begin by reviewing some facts about the integral extensions.
Recall that a domain A is local if and only if A has a unique maximum ideal m which is exactly the nonunits of A.
Definition 3.1 If A ⊂B are domains, we say that b ∈B is integral over A, if there are a0, . . . , an−1 ∈A such that bn + an−1bn−1 + · · · + a1b + a0 = 0 for some n. We say that B is integral over A if every element of B is integral over A.
Lemma 3.2 Let A ⊂B be domains and b ∈B. The following are equivalent.
i) b is integral over A.
ii) A[b] is a subring of B that is a finitely generated A-module.
iii) A[b] is contained in a finitely generated A-module.
Proof i) ⇒ii) If bm = Pm−1 n=0 anbn where a0, . . . , am−1 ∈A. Then A[b] is generated over A by 1, b, . . . bm−1.
ii) ⇒iii) is clear.
iii) ⇒i) Let x1, . . . , xm generate a submodule containing A[b] over A. For i = 1, . . . m we can find ai,1, . . . , ai,m ∈A such that bxi = m X j=1 ai,jxj.
Let M be the matrix a1,1 −b a1,2 . . .
a1,m a2,1 a2,2 −b . . .
a2,m .
.
.
.
.
.
...
.
.
.
am,1 am,2 . . .
am,m −b 27 i.e, the matrix with ai,i −b along the diagonal and ai,j everywhere else. Then M(x1, . . . , xm)T = 0. Let Adj(M) be the adjoint of M. Then Adj(M)M(x1, . . . , xm)T = (det Mx1, . . . , det Mxm)T = (0, . . . , 0)T .2 Thus we must have det M = 0. But det M is a monic polynomial in A[b].
□ Corollary 3.3 If A ⊂B ⊂C are domains, B is an integral extension of A and C is an integral extension of B, then C is an integral extension of C.
Proof Let c ∈C. There are b0, . . . , bn−1 ∈B such that cn + P bici = 0. Then A[b0, . . . , bn−1, c] is a finitely generated A-module and c is integral over A.
□ The next lemma is a simple but useful tool.
Lemma 3.4 If A is a local subring of a field K, x ∈K× and 1 = a0+ a1 x +. . . an xn where a0 ∈m and a1, . . . , an ∈A, then x is integral over A.
Proof Then (1 −a0)xn −a1xn−1 −· · · −an = 0. Since a0 ∈m, 1 −a0 ̸∈m.
Since A is local, 1 −a0 is a unit and x is integral over A.
□ Lemma 3.5 If A ⊂B are domains and B is integral over A, then A is a field if and only if B is a field.
Proof (⇐) Suppose B is a field and a ∈A is nonzero.
Then there are c0, . . . , cm−1 ∈A such that (a−1)m + m−1 X n=0 cn(a−1)n = 0.
Multiplying by am−1 we see that a−1 = − m−1 X n=0 cnam−n−1 ∈A.
Thus A is a field.
(⇒) Suppose A is a field and b ∈B is nonzero. Then, by Lemma 3.2 A[b] is a finitely generated vector space over A. The map z 7→bz is an injective linear transformation of A[b] and, since A[b] is a finite dimensional vector space must be surjective. Thus there is z ∈A[b] with zb = 1.
□ Definition 3.6 Let A ⊂B be domains and let P ⊂A, Q ⊂B be prime ideals.
We say that Q lies over P if A ∩Q = P.
Corollary 3.7 Let A ⊂B be domains with B integral over A and let P ⊂A and Q ⊂B be prime ideals such that Q lies over P. Then P is maximal if and only if Q is maximal.
2Remember Cramer’s Rule!
28 Proof Since P = A ∩Q we can view B/Q as an integral extension of A/P. By the last lemma, A/P is a field if and only if B/Q is a field.
□ Lemma 3.8 Suppose A ⊂B are domains, B is integral over A, Q is a prime ideal in B and P = Q ∩A. Then then BAP is integral over AP .
Proof Consider b/t where b ∈B and t ∈A \ Q. There are a0, . . . , am−1 ∈A with bm + P aibi = 0. But then (b/t)m + X (ai/tm−i)(b/ti) = 0.
□ Lemma 3.9 Suppose A ⊂B are domains, B is integral over A, P ⊂A is a prime ideal and Q1 ⊆Q2 are prime ideals in B lying over P. Then Q1 = Q2.
Proof Consider the localization AP and the integral extension BAP . Then Q1AP and Q2AP are prime ideals of BAP lying over PAP . But PAP is max-imal. Thus each QiAP is maximal and we must have Q1AP = Q2AP . But if x ∈Q2 \ Q1, then x ̸∈Q1AP . If we did have x = q/t for some q ∈Q1 and t ∈A \ P. Then xt ∈Q1 and since x ̸∈Q1 and Q1 is prime, we would have t ∈Q1 ∩A = P, a contradiction.
□ Theorem 3.10 (Lying Over Theorem) Suppose A ⊂B are domains, B is integral over A and P is a prime ideal of A. There is a prime ideal Q of B such that A ∩Q = P.
Proof First, suppose A was a local ring then P is the unique maximal of A.
If Q ⊂B is any maximal idea extending P, then, by Corollary 3.7, Q ∩A is maximal. But then Q = P.
In general, we pass to the localization AP . As above, if Q0 is any maximal ideal in BAP , then Q0 ∩AP = PAP . So Q0 ∩A = P. Let Q = Q0 ∩B. Then Q ∩A = P and, since Q0 is prime, Q is prime.
□ 3.2 Extensions of Valuations Theorem 3.11 (Chevalley’s Theorem) Suppose A is a subring of a field K and P ⊂A is a prime ideal.
Then there is a valuation ring O of K with A ∩MO = P Proof Replacing A by AP we may assume that A is a local ring with maximal ideal P.
Let P be the set of all local subrings B of K with mB ∩A = P.
Clearly P is partially ordered by ⊂and if (Bi : i ∈I) increasing chain in P then S i∈I Bi is an upper bound. Thus by Zorn’s Lemma, P has maximal elements.
Let O ∈P be maximal. Let m be the maximal ideal of O. We will argue that O is a valuation ring.
29 Suppose x, 1/x ∈K \ O. If x is integral over O, then we can find a maximal ideal of O[x] lying over m contradicting the maximality of O ∈P. Thus x is not integral over O.
By Lemma 3.4, 1 ̸∈mO[1/x]. Thus there is a maximal ideal Q of O[1/x] that lies over m, contradicting the maximality of O. Thus for all x ∈K at least one of x and 1/x is in O.
□ Exercise 3.12 Show that if v : K× →Γ is a valuation and L ⊃K is an extension field, there is Γ′ ⊇Γ and w : L× →Γ′ extending v.
integral closures and valuations Definition 3.13 We say that A is integrally closed in B if no element of B \ A is integral over A. We say that A is integrally closed if it is integrally closed in its fraction field.
The integral closure of A is the smallest integrally closed ring containing A.
Lemma 3.14 If (K, v) is a valued field, then the valuation ring O is integrally closed.
Proof Suppose b ∈K and bn+an−1bn−1+· · ·+a1b+a0 = 0 where a0, . . . , an−1 ∈ O. If b ̸∈O, then v(b) < 0 and v(aibi) = v(a) + iv(b) < nv(b) since v(ai) ≥0 for all i. Thus v(bn + an−1bn−1 + · · · + a1b + a0) = nv(b) < 0, a contradiction.
□ We can use valuation rings to find the integral closure of a local subring.
Lemma 3.15 Let A be a local subring of a field K with maximal ideal m. The integral closure of A ∈K is the intersection of all valuation rings O ⊂K with mO lying over m.
Proof Suppose x ∈K be nonintegral over A.
Then by Lemma 3.4, 1 ̸∈ mA[1/x] + 1 xA[1/x]. Thus we can find a maximal ideal Q of A[1/x] lying over m with 1/x ∈Q. Let O ⊇A[1/x] be a maximal local subring of K. Then, as in the proof of Theorem 3.11, O is a valuation ring, mO lies over m and 1/x ∈mO.
Thus x ̸∈O.
□ Algebraic Extensions Suppose K ⊂L are fields and v is a valuation on K. Then v restricts to a valuation on K. Let OL, ΓL, kL and OK, ΓK, kK denote the respective valuation rings, value groups and residue fields.
Lemma 3.16 Then ΓL is contained in the divisible hull of ΓK and kL is an algebraic extension of kK.
30 Proof Let x ∈L \ K. There are a0, . . . , an ∈K such that P aixi = 0. There must be i ̸= j such that v(aixi) = v(ajxj). But then v(x) = v(ai)−v(aj) j−i .
Suppose x ∈L and the residue x ∈kL \ kK. There is a polynomial f[X] ∈ OK(X) such that f(x) = 0. Let f(X) = P aiXi. Suppose aj has minimal value and let g(X) = P ai aj Xi. Then g(x) = 0 and g(X) is not identically zero as some coefficient is 1. Thus x is algebraic over K.
□ Corollary 3.17 i) If k is an algebraically closed field, Γ is a divisible ordered abelian group and K = k( ( (Γ) ) ), then K is algebraically closed.
ii) If k is a real closed, Γ is a divisible ordered abelian group and K = k( ( (Γ) ) ), then K is real closed.
Proof If K is not algebraically closed field let L/K be an algebraic extension, then we can extend the valuation to L and since kL/k is algebraic and Γ(L) is contained in the divisible hull of Γ(K) by Exercise 1.8 (see also Lemma 3.16).
But k is algebraically closed and Γ is divisible, thus L/K is immediate. But we saw in Lemma 2.46 that Hahn fields have no proper immediate extensions.
Thus K is algebraically closed.
ii) If k is real closed, then kalg( ( (Γ) ) ) is a degree 2 algebraic extension of k( ( (Γ) ) ). Thus by the work of Artin and Schreier (see for example XI §2 Proposition 3), k( ( (Γ) ) ) is real closed.
□ We will prove much more general of these results later.
If L/K is a finite algebraic extension and [L : K] = d, then the argument above shows that [ΓL : ΓK] ≤d and [kL : kK] ≤d. We will prove a much sharper bound. We let e = [ΓL : ΓK] be the ramification index and f = [kL : kK] be the residue degree . Note that if e = f = 1, then L is an immediate extension of K.
Theorem 3.18 (Fundamental Inequality) If L/K is a finite algebraic ex-tension of degree d then ef ≤d.
Proof Choose x1, . . . , xe ∈L such that v(x1), . . . , v(xn) represent distinct cosets of ΓL/ΓK.
Choose y1, . . . , yf ∈L such that y1, . . . , yf is a basis for kL/kK. It suffices to show that (xiyj : i ≤e, j ≤f) are linearly independent over K.
Suppose X i≤e,j≤f ai,jxiyj = 0 where not all ai,j = 0. Pick b i and b j such that v(ab i,b jxb i) = min{v(ai,jxi) : i ≤e, j ≤f}.
Suppose i ̸= b i and j ≤f. We claim that v(ab i,b jxb i) < v(ai,jxi). If they were equal then v(xb i) −v(xi) = v(ai,j) −v(ab i,b j) ∈ΓK, 31 contradicting that v(xb i) and v(xi) represent different cosets. Thus v(ab i,b jxb i) < v(ai,jxi) for i ̸= b i.
Let bi,j = ai,j ab i,b j xb i. Then 0 = f X j=1 e X i=1 bi,j xi xb i yj and bi,j xi xb i ∈mL for i ̸= b j. Thus f X j=1 ab i,j ab i,b j yj = − f X j=1 X i̸=b i bi,jxiyj ∈mL.
Let cb i,j = res(ab i,j/ab i,b j). Then cb i,b j = 1 and f X j=1 ci,jyj = 0, contradicting that y1, . . . , yf are linearly independent over kK.
□ Exercise 3.19 Show that even if L/K is an infinite algebraic extension the argument above shows that if (xi : i ∈I) represent distinct cosets of ΓL/ΓK and (yj : j ∈J) are such that (yj : j ∈J) are linearly independent over kK, then (xiyj : i ∈I, j ∈J) are linearly independent and v(P ai,jxiyj) = min v(ai,jxiyj).
Definition 3.20 If K ⊂L are fields and L/K is algebraic, we say that L/K is normal if L is a splitting field for every irreducible f ∈K[X] with a zero in L.
A separable normal extension is a Galois extension. Thus in characteris-tic 0 normal and Galois are the same. But in characteristic p we can build nonseparable normal extensions by taking pth-roots.
Our goal for the rest of this section is to show that if L/K is a normal extension and O is a valuation ring of K, then the valuation rings of L extending O are all conjugate under the action of the Galois group.
We need a form of the Chinese Remainder Theorem.
Lemma 3.21 Let A be a domain and let m1, . . . , mn be distinct maximal ideals of A. Then for any a1, . . . , an we can find a ∈A such that a = ai(mod mi) for all i.
Proof claim For each i we can find bi such bi = 1(mod mi) but bi ∈mj for j ̸= i.
For notational simplicity assume i = 1. If j ̸= 1 then m1 + mj = A, as otherwise m1 + mj is an ideal, contradicting maximality. Thus there is cj ∈m1 and dj ∈mj such that cj + dj = 1. Then 1 = Y j̸=1 (cj + dj) = Y j̸=1 dj(mod m1).
32 Let b1 = Q j̸=1 dj. Then b1 = 1(mod m1) but b1 ∈mi for i ̸= .
Let a = P aibi. Then a = ai(mod mi) for all i.
□ Lemma 3.22 Let A be a local domain integrally closed in its fraction field K and let L/K be normal. Let B be the integral closure of A in L. Then any two maximal ideals of B are conjugate under Gal(L/K).3 Proof It suffices to prove this when L/K is finite. Let m0 and m1 be maximal ideal of B and suppose there is no σ ∈Gal(L/K) with σ(m1) = m2.
Let Xi = {σ(mi) : σ ∈Gal(L/K)} then X0 ∩X1 = ∅. By the Chinese Remainder Theorem, we can find b ∈B such that b ∈m for m ∈X0 and b = 1(mod m) for m ∈X1. Thus σ(b) ∈m0 \ m1 for all σ ∈Gal(L/K).
For the remainder of the proof we will assume that our fields have character-istic zero. One needs to be slightly more careful in characteristic p when we have an inseparable extension. Suppose f(X) = Xd + Pd−1 n=0 aiXi, a0, . . . , ad−1 ∈A be the minimal polynomial of b over K..
Since L/K is normal, f(X) = Qd i=1(X −βi) where β1, . . . , βd ∈L are the distinct roots or f, i.e., the set of conjugates of b under Gal(L/K).
Without loss of generality, we assume L = K(β1, . . . , βd). Then Y σ∈Gal(L/K) σ(b) = d Y i=1 βi = a0 ∈A.
Each σ(b) ∈m0. Thus a0 ∈m0 ∩A = mA ⊆m1. But no σ(b) ∈m1, thus, since m1 is prime a0 ̸∈m1, a contradiction.
□ Lemma 3.23 Let A be a valuation ring with fraction field K, let L ⊇K be an algebraic extension and let B be the integral closure of A in L. For every valuation ring O ⊂K with mA ⊆mO there is n a maximal ideal of B with O = Bn.
Moreover, for every maximal ideal n ⊂B, Bn is a valuation ring.
Proof Let O be a valuation ring of L with mA ⊆mO. Since O is integrally closed in L, B ⊆O. Let n = mO ∩B.
If x ∈B \ n, then 1/x ∈O. Thus Bn ⊆O. Let x ∈O. Since L/K is algebraic, there are a0, . . . , ad ∈A not all zero such that P aixi = 0. Let m ≤d be maximal such that v(am) = min(v(ai) : i = 0, . . . , d) and divide P aixi by amxm. Thus, letting bi = ai/am we have d X i=m+1 bixi−m + 1 + m−1 X i=0 bixi−m = 0.
Note that b0, . . . , bm−1 ∈A and bm+1, . . . , bd ∈mA. Let y = Pd i=m+1 bixi−m+1 and z = Pm−1 i=0 bixi−m+1. Then xy = −z and y is a unit in O.
3We use Gal(L/K) to denote the group of automorphism of L/K even when L/K is not necessarily a Galois extension.
33 We claim that y, z ∈B.
Since B is the integral closure of A in L, by Lemma 3.15, it suffices to show that x, y ∈V for any valuation ring V ⊂L with mV ∩A = mA. If x ∈V , then y ∈V and z = −xy ∈V . If x ̸∈V , then 1/x ∈V , z = Pm−1 i=0 bixi−m+1 ∈V and y = −z/x ∈V .
Since y is a unit in O, y ̸∈n. Thus x = −z/y ∈Bn. Thus Bn = O.
To prove the last claim of the lemma we need to show that if n is a maximal ideal of B, then Bn is a valuation ring extending A. Clearly n ∩A = mA. By Chevalley’s Theorem, there is a valuation ring O such that B ∩mO = n. Then by the first part of the lemma O = Bn.
□ We summarize the last few lemmas.
Theorem 3.24 Let A be a valuation ring with fraction field K, let L ⊇K be an algebraic extension and let B be the integral closure of A in L. There is a bijective correspondence m 7→Bm between maximal ideals of B and valuation rings O ⊂L with mO ∩A = mA. Moreover, if L/K is normal, then any two such valuation rings are conjugate under Gal(L/K).
Corollary 3.25 Let (K, O) be a valued field and let L/K be a purely inseparable algebraic extension of K. Then there is a unique valuation ring O∗on L with (K, O) ⊆(L, O∗).
Proof L is obtained from K by adjoining pth-roots where K has characteristic p. Then L/K is normal but there are no nontrivial automorphisms of L fixing K.
□ 34 4 Algebraically Closed Valued Fields 4.1 Quantifier Elimination for ACVF We now have developed enough machinery to begin the study of the model theory of algebraically closed valued fields.
Valued fields as structures The first issue is deciding what kind of structure we are looking at, i.e., what language or signature do we use to study valued fields? There are several natural candidates.
One-sorted structures We can think of valued fields as pairs (K, O) where K is the field and O is the valuation ring.
In this case the natural language would be the usual language of rings {+, −, ·, 0, 1} together with a unary predicate O which picks out the valuation.
Three-sorted structures We can think of valued fields as three-sorted structures (K, Γ, k) where we have separate sorts for the field (which we refer to as the home sort, the value group and the residue field. On the home sort and on the residue field we will have the +, −, ·, 0, and 1. On the group we will have +, −, <, 0. We also have the valuation map v and the residue map res. 4 It would also be natural to think of valued fields as two sorted structure (K, Γ) and later we will consider adding more imaginary sorts.
How does this effect definability? It’s easy to see that it doesn’t.
Lemma 4.1 In the one-sorted structure (K, O) we can interpret the value group Γ, the residue field k and the maps v : K× →Γ and res : O →k.
Thus any subset of Kn definable in the three-sorted structure is definable in the one-sorted structure. Moreover if X ⊆Kl × Γm × kn is definable in the three-sorted structure, then there is A ⊆Kl+m+n definable in (K, O) such that X = {(a1, . . . , al), v(al+1, . . . , v(al+m), res(al+m+1), . . . , res(al+m+n) : (a1, . . . , al+m+n) ∈A.
In the three-sorted structure (K, Γ, v) we can define the value ring O = {x ∈ K : v(x) ≥0}. Thus any subset of Kn definable in the one-sorted structure is definable in the three-sorted structure.
We will also look at further variants of these languages.
• When studying the p-adic field Qp, we have already shown in Exercise 2.11 that Zp is definable in the field language. Thus any subset of Qn p definable in (Qp, Zp) is already definable in Qp in the field language. The exercises below show that this is not always possible.
4Note we should think of they symbols on each sort as being distinct, so while we routinely use + on K, k and Γ, if we were more careful we would think of them as three distinct symbols.
35 • To prove quantifier elimination for algebraically closed valued fields we will work in the language of divisibility Ldiv = {+, −, ·, O, |, 0, 1} where | is a binary function symbol which we interpret (K, O) | = x|y if and only if ∃z ∈O xz = y.
The relation x|y is definable in (K, O) thus any subset of Kn definable in the language Ldiv is already definable in (K, O).
Note that once we have added | to the language we could get rid of O since x ∈O if and only if 1|x.
• To prove quantifier elimination for Qp we will work in the Macintyre Lan-guage LMac = {+, −, ·, O, 0, 1, P2, P3, P4, . . . } where Pn is a unary predi-cate predicate which we interpret in (K, O) as the nth powers of K. Since x ∈Pn if and only if K | = ∃y yn = x, any subset of Kn definable in LMac is already definable using L. Indeed in Qp we can define Zp in a quanti-fier free way using P2 as in Exercise 2.11. Thus we don’t really need the predicate for O.
• In the original work of Ax and Kochen it was useful to work in the three-sorted language and add a symbol for π : Γ →K a section of the valuation.
This is more problematic. We saw in Exercise 2.34 that not every valued field has a section. Moreover we will show that the section map is not definable in the three-sorted language. Thus, while adding the section can be useful, we will end up with new definable sets.
• An angular component map is a multiplicative homomorphism ac : K× → k× such that ac agrees with the residue map on the units. For example on Qp if vp(x) = m then x = ampm + am+1pm+1 + . . . and we can let ac(x) = am. Similarly, there is an angular component map on K((T)).
If we have a section π : Γ →K, then we can define an angular component map by ac(x) = res(x/π(x)). But, like sections, angular component maps need not exist and, even when they do exist, may change definability.
Nevertheless, we will find it useful to work in the three-sorted language LPas where we add a symbol for an angular component map. This is called the Pas language Exercise 4.2 Let (K, O) be a valued field where K is algebraically closed or real closed. Show that O is not definable in K in the pure field language.
Exercise 4.3 Suppose π : Γ →K is a section of the valuation. Show that ac(x) = res(x/π(x)) is an angular component map.
36 Quantifier Elimination We will prove quantifier elimination for algebraically closed valued fields in the language Ldiv. Let ACVF be the Ldiv-theory such that (K, O, |) | = ACVF if and only if K is an algebraically closed field with valuation ring O and x|y if and only if there is z ∈O such that zx = y. We will also assume that the valuation is nontrivial so there is x ∈K× \ O.
Theorem 4.4 (Robinson) The theory of algebraically closed fields with a non-trivial valuation admits quantifier elimination in the language Ldiv.5 Quantifier elimination will follow from the following proposition.
Proposition 4.5 Suppose (K, v) and (L, w) are algebraically closed fields with non-trivial valuation and L is |K|+-saturated. Suppose R ⊆K is a subring, and f : R →L is an Ldiv-embedding. Then f extends to a valued field embedding g : K →L.
Exercise 4.6 Show that the proposition implies quantifier elimination. [Hint: See 4.3.28.] We will prove the Proposition via a series of lemmas.
Definition 4.7 Suppose R is a subring of K. We say that a ring embedding f : R →L is an Ldiv-embedding if for a, b ∈R, R | = a|b ⇔w(f(a)) ≤w(f(b)).
First, we show that without loss of generality we can assume R is a field.
Lemma 4.8 Suppose (K, v) and (L, w) are valued fields, R ⊆K is a subring and f : R →L is and Ldiv-embedding. Then f extends to a valuation preserving embedding of K0, the fraction field of R into L.
Proof Extend f to K0, by f(a/b) = f(a)/f(b). If x ∈K0, then x is a unit in (K, v) if and only if x|1 and 1|x if and only in f(x) is a unit in (L, w). Since the value group is given by K×/U, addition in the value group is preserved. So we need only show that the order is preserved.
Suppose x, y ∈K0. There are a, b, c ∈R such that x = a c and y = b c. Then v(x) ≤v(y) ⇔v(a) ≤v(b) ≤R | = a|b ⇔L | = f(a)|f(b) ⇔w(f(x)) ≤w(f(y)).
□ We next show that we can extend embedding from fields to their algebraic closures.
5Actually, Robinson only proved model completeness, but his methods extend to prove quantifier elimination.
37 Lemma 4.9 Suppose (K, v) and (L, w) are algebraically closed valued fields, K0 ⊆K is a field and f : K0 →L is a valuation preserving embedding. Then f extends to a valuation preserving embedding of Kalg 0 , the algebraic closure of K0 into L.
Proof It suffices to show that if x ∈K \ K0 is algebraic over K0, then we can extend f to K0(x). Let K0(x) ⊆F ⊆K with F/K0 normal. There is a field embedding g : F →L with g ⊃f and g(v) gives rise to a valuation on g(F) extending f(v|K0). Then g(v|F) and w|g(F) are valuations on g(F) extending f(v|K0) on f(K0). By Theorem 3.24, there is σ ∈Gal(g(F)/f(K0)) mapping g(v|F) to w|g(F). Thus σ ◦g is the desired valued field embedding of F into L extending f.
□ Thus in proving Proposition 4.5 it suffices to show that if we have (K, v) and (L, w) non-trivially valued algebraically closed fields, L is |K|+-saturated, K0 ⊂K algebraically closed and f : K0 →L a valuation preserving embedding, then we can extend f to K. There are three cases to consider.
case 1 Suppose x ∈K, v(x) = 0 and x is transcendental over kK0.
We will show that we can extend f to K0[x], then use Lemmas 4.8 and 4.9 to extend to K0(x)acl. Since L is |K|+-saturated, there is y ∈L such that y is transcendental over kf(K0). We will send x to y.
Suppose a = m0 + a1x + · · · + mnxn, where mi ∈K0. Suppose ml has minimal valuation. Then a = ml(P bixi) where v(bi) ≥0 and bl = 1. Then v(P bixi) ≥0. If v(P bixi) > 0, then taking residues we see that X bixi = 0, but bl = 1, so this is a nontrivial polynomial and x is algebraic over kK0. Thus v(P bixi) = 0 and v(a) = ml.
Thus v(a) = min{v(mi) : i = 0, . . . , n}. Similarly, in L, w(P f(mi)yi) = min{w(f(mi)) : i = 0, . . . , n}. Thus the extension of f to K0[x] is and Ld-embedding.
case 2 Suppose x ∈K and v(x) ̸∈v(K0).
Let γ = v(x). Suppose a, b ∈K0, i < j are in N, and v(a) + iγ = v(b) + jγ.
Since K0 is algebraically closed there is c ∈K0 such that cj−i = a b , but then γ = v(c) ∈v(K0).
Suppose a ∈K0[x] and a = m0 + m1x + . . . mnxn. Since the v(mi) + iγ are distinct, v(a) = min(v(mi) + iγ).
Since L is |K|+-saturated, there is y ∈L realizing the type {w(f(a)) < w(y) : a ∈K0, v(a) < v(x)} ∪{w(y) < w(f(b)) : v(x) < v(a)}.
Then v(a)+iv(x) < v(b)+jv(x) if and only if w(f(a))+iw(y) < w(f(b))+jw(y) for all a, b ∈K0 and the extension of f to K0[x] sending x to y is and Ldiv-embedding.
38 case 3 Suppose x ∈K \ K0, v(K0(x)) = v(K0) and kK0(x) = kK0, i.e., K0(x) is an immediate extension of K0.
Let C = {v(x −a) : a ∈K0}. Since v(K0(x)) = v(K0), C ⊆v(K0). We claim that C has no maximal element. Suppose v(b) ∈C is maximal. Then v( x−a b ) = 0 and, since kK0 = kK0(x), there is c ∈K0 such that x−a b −c = ϵ where v(ϵ) > 0. But then, v(x −a −bc) = v(bϵ) > v(b), a contradiction.
Consider the type Σ(y) = {w(y −f(a)) = w(b) : a, b ∈K0, v(x −a) = v(b).} We claim that Σ is finitely satisfiable. Suppose a1, . . . , an, b1, . . . , bn ∈K0 and v(x −ai) = v(bi). Because f is valuation preserving it suffices to find c ∈K0 with v(c−ai) = v(bi) for i = 1, . . . , n. Since C has no maximal element, there is c ∈K0 such that v(x −c) > v(bi) for i = 1, . . . , n. Then v(c −ai) = v(x −ai) = v(bi).
By sending x to y we can extend f to a ring isomorphism between K0[x] and f(K0)[y]. For a ∈K0(x), there is p(X) ∈K0[X] such that d = p(x). Factoring p into linear factors over the algebraically closed field K0, there is a0, . . . , an such that d = p(x) = a0 n Y i=1 (x −ai).
For each i we can find bi ∈K0 such that v(x −ai) = v(bi). Thus v(d) = v(a0) + n X i=1 v(bi) By choice of y, we also have w(f(d)) = w(f(a0)) + n X i=1 w(f(bi)), thus f preserves the valuation.
This concludes the proof of Proposition 4.5 and hence the proof that ACVF has quantifier elimination in the language Ldiv.
The proofs we have given can readily be adapted to prove quantifier elimi-nation in the three-sorted language.
Exercise 4.10 Modify the proofs above to verify that algebraically closed fields have quantifier elimination when viewed as three-sorted structures in the usual language.
39 4.2 Consequences of Quantifier Elimination Completions of ACVF ACVF is not a complete theory. We need to specify the characteristic of the field K and the residue field k. If K has characteristic p, then k has characteristic p. If K has characteristic 0, the k may have any characteristic. Let a be either 0 or a prime. If a = p a prime, then b = p. If a is zero, then b is either zero or a prime. Let ACVFa,b be ACVF with additional axioms asserting the field has characteristic a and the residue field has characteristic b.
Corollary 4.11 Each theory ACVFa,b is complete and these are exactly the completions of ACVF.
Proof If (a, b) = (0, 0) let R = (Q, Q, |). If (a, b) = (0, p) let R = (Q, Z(p), |) and if (a, b) = (p, p), let R = (Fp, Fp, |). Suppose (K, OK, |) and (L, OL, |) are models of ACVFa,b. Then R is a common substructure of both fields. Let φ be an Ldiv-sentence. Then there is a quantifier free Ldiv-sentence such that ACVF | = φ ↔ψ.
But then, since ψ is quantifier free, K | = φ ⇔K | = ψ ⇔R | = ψ ⇔L | = ψ ⇔K | = φ.
Thus ACVFa,b is complete.
We have listed the only possibilities for the characteristics of the field and residue field. Thus these are the only possible completions of ACVF.6 □ Definable subsets of K In any valued field we can always define open and closed balls and any finite boolean combination of balls.7 We will show that in an algebraically closed valued field these are the only definable subsets of K.
Lemma 4.12 Let (K, v) be an algebraically closed valued field. Suppose f ∈ K[X]. Then we can partition K into finitely many sets each of which is a finite boolean combination of balls such that that for each Y in the partition there are n ≥1, a ∈K and γ ∈Γ in the value group such that v(f(x)) = nv(x −a) + γ for all x ∈Y .
Proof Let f(X) = c(X −a1) · · · (X −an) for c ∈K× and a1, . . . , an ∈K.
Then v(f(x)) = v(c) + · · · + v(x −a1) + · · · + v(x −an). We will show that we can partition K such that on each set in the partition there is i such that either 6Here we are using the assumption that our fields have nontrivial valuations. If we were to also consider the trivial valuation we would have completions saying that I have a trivial valued field of characteristic 0 or p. But these are just the completions of ACF.
7Here we allow trivial balls K = {x : v(x) < ∞} and {a} = {x : v(x) = ∞}. If we don’t want to do this, we should look at boolean combinations of points and balls instead.
40 v(x −aj) = v(x −ai) for each set in the partition or v(x −aj) is constant on the partition.
For each partition I, J of {1, . . . , n} where I is nonempty, let b i be the least element of I. Let YI,J = {x ∈K : v(x −ai) = v(x −ab i) > v(x −aj) for i ∈I, j ∈J}.
Then the sets YI,J are boolean combinations of balls and they partition K (of course some YI,J might be empty.
For j ̸= b i let γj = v(ab i −aj). Then • if v(x −ab i) < γj, then v(x −aj) = v(x −ab i) • If v(x −ab i) > γj, then v(x −aj) = γj • We can not have v(x −ab i) = γj, as then v(x −aj) ≥γj, contradicting x ∈YI,J.
This allows to partition YI,J into finitely many pieces each of which is a boolean combination of balls, such v(x −aj) is either v(x −ab i) or constant on each set in the partition.
□ Exercise 4.13 Show that if (K, v) is algebraically closed and f, g ∈K[X], then {x ∈K : v(f(x)) ≤v(g(x))} is a finite Boolean combination of balls.
Corollary 4.14 If (K, O) | = ACVF and X ⊆K is definable, then X is a finite boolean combination of balls.
Proof By quantifier elimination any definable subset of X is a finite boolean combination of sets of the form {x : f(x) = g(x)} and {x : f(x)|g(x)} = {x : v(f(x)) ≤v(g(x))} for f, g ∈K[X].
□ Definition 4.15 A swiss cheese is a definable set of the form B (C1 ∪· · ·∪Cn) where B, C1, . . . , Cn are balls and Ci ⊂B (and we allow the possibilities where B = K or ∅, n = 0 and some B or Ci is a point.) Exercise 4.16 a) Show the intersection of two swiss cheese is a finite disjoint union of swiss cheese.
b) Show that the complement of a swiss cheese is a finite disjoint union of swiss cheese.
c) Prove that every definable subset of K can be written in a unique way as a finite union of disjoint swiss cheese.
Corollary 4.17 i) Any infinite definable subset of K has interior.
ii) There is no definable section of the value group.
Proof i) Any infinite definable set will contain a swiss cheese S = B \ (C1 ∪ · · · ∪Cm), where B ̸= ∅. If a ∈S, then S contains a ball U with a ∈U.
ii) The image of the section would be infinite with no interior.
□ Exercise 4.18 Suppose K is an algebraically closed valued field and A ⊆Km+n is definable. For x ∈Km let Ax = {y ∈Kn : (x, y) ∈A}. Show that {x : Ax 41 is finite} is definable and that there is an N such that if Ax is finite, then |Ax| ≤N.
Exercise 4.19 Let A ⊂K. Show that the model theoretic algebraic closure of A is the field theoretic algebraic closure of A.
In Exercise 5.25 we will characterize definable closure in ACVF.
Exercise 4.20 Let (K, v) be an algebraically closed valued field. Prove that there is no definable angular component map.
NIP Let M be a structure. Recall that φ(x1, . . . , xm, y1, . . . , yn) has the independence property if for all k there are b1, . . . , bk ∈Mm and (cJ : J ⊂{1, . . . , k}) in Mn such that M | = φ(bi, cJ) ⇔i ∈J).
In which case we say that φ shatters b1 . . . , bk. Otherwise we say φ has NIP.
We say that a theory has NIP if no formula has the independence property.
We need two basic facts about NIP. See 2.9 and 2.11.
Lemma 4.21 i) T has NIP if and only if every formula φ(x1, y1, . . . , yn) has NIP.
ii) A boolean combination of NIP formulas has NIP.
Corollary 4.22 ACVF has NIP.
Proof By the lemma above and Corollary 4.14, it suffices to show that no definable family of balls has the independence property.
We claim that the family of all balls can not shatter a set of size 3. Suppose a, b and c ∈K are distinct and, without loss of generality, v(a −b) ≤v(a −c), v(b −c). Then any ball that contains a and b contains c. Thus the family of all balls does not shatter any three element set.
□ Definable subsets of the value group and residue field To study definable subsets of km, Γn and, more generally km × Γn we need to apply quantifier elimination in the three-sorted language. We will let variables x0, x1, . . . range over the home sort, while y0, y1, . . . ranges over the residue field and z0, z1, . . . range over the value group. Any atomic formula is equivalent to one in one of the following forms • t(x0, . . . , xm) = 0, where t is a polynomial over Z; • t(y0, . . . , yn, res(x0), . . . , res(xm)) = 0 , where t is a polynomial over Z; • s(z0, . . . , zl, v(x0), . . . , v(xm)) = 0, where s(u0, . . . , ul+m+1) = P riui, ri ∈ Z; • s(z0, . . . , zl, v(x0), . . . , v(xm)) > 0, where s(u0, . . . , ul+m+1) = P riui, ri ∈ Z; 42 We say that A ⊆kn × Γm is a rectangle if there is B ⊆kn definable in the field structure on k and C ⊆Γm definable in the ordered abelian group Γ such that A = B × C.
Corollary 4.23 (Orthogonality) Every definable subset of kn×Γm is a finite union of rectangles.
Proof By quantifier elimination, every definable set is a finite union of sets defined by conjunctions of atomic and negated atomic formulas. But atomic formulas defining subsets of kn × Γm only have variables over just the residue field sort or just the value group sort and the definable set is either of the form kn × A or B × Γn where A ⊆kn is already definable in k or B ⊆Γm is already definable in Γ. Thus any set defined by a conjunction of atomic and negated atomic formulas is a rectangle and every definable set is a finite union of rectangles.
□ Corollary 4.24 i) Any definable function f : k →Γ has finite image.
ii) Any definable function g : Γ →k has finite image.
This shows that the residue field and value group are as unrelated as possible.
It also shows that the valuation structure induces no additional definability on the residue field and value group.
Corollary 4.25 i) Any subset of kn definable in (K, Γ, k) is definable in the field k.
ii) Any subset of Γm definable in (K, Γ, k) is definable in the ordered abelian group Γ.
In this case k with all induced structure, is just a pure algebraically closed field and hence ω-stable, while Γ with all induced structure, is a divisible ordered abelian group and hence o-minimal.
Definition 4.26 We say that a sort S is stably embedded if any subset of Sn that is definable in the full structure is definable using parameters from S.
Corollary 4.27 The residue field and value group of an algebraically closed field are stably embedded.
In the next section we give an example of an imaginary sort that is not stably embedded.
Exercise 4.28 Let A ⊂k. Prove that if b ∈k is algebraic over A in the three-sorted valued field structure, then b is algebraic over A in the field k.
4.3 Balls For this section we start by thinking of valued fields as three-sorted structures (K, Γ, k), but this also makes sense if we think of them as one-sorted structures (K, O).
43 For any valued field we can introduce two new sorts Bo and Bc for open and closed balls. For Bo define an equivalence relation ∼on K × Γ such that (a, γ) ∼(b, δ) if and only if γ = δ and γ = δ and v(a −b) > γ. Then (a, γ) ∼(b, γ) ⇔b ∈Bγ(a) ⇔a ∈Bγ(b).
Thus we can identify (a, γ)/ ∼with Bγ(a). Let Bo = K×Γ/ ∼. We can indentify Bo with the open balls of K. There is a definable map r : Bo →Γ given by r((a, γ)/ ∼) = γ, i.e., r assigns each ball it’s radius. There is a definable relation Ro on K ×Bo such that aRob if and only if a ∈b. Replacing ∼by (a, γ) ∼∗(b, δ) on K × Γ ∪{∞} if and only if γ = δ and v(a −b) ≥γ , we can similarly define the sort of closed balls Bc.
Exercise 4.29 Let a ∈K and let X ⊂S be the set of all open balls containing a. Prove that X is not definable with parameters from Bo. [Hint: Show that for any finite subset A of Bo there is an automorphism (possibly of a larger field) fixing A pointwise but moving X.] While up to this point the construction makes sense in any valued field, henceforth we will assume K is algebraically closed.
Lemma 4.30 If X ⊆Bc is an infinite definable set then either r|X is finite-to-one, or there is an infinite definable Z ⊆X and a definable surjection f : Z →k.
Proof If r|X is not finite-to-one, there is γ ∈Γ such that Y = {B ∈X : r(B) = γ} is infinite. Let A = S B∈Y B. Then A is an infinite definable subset of K and if a ∈A, then Bγ(a) ∈Y .
claim There is a closed ball Bϵ(a) with ϵ < γ such that every closed ball of radius γ in Bϵ(a) is in Y .
By quantifier elimination A is a finite disjoint union of sets of W = B (C1 ∪ · · · ∪Cm), where B, C1, . . . , Cm are balls. Since Y is infinite, some B must have radius δ < γ. If a ∈W, then Bγ(a) ⊂W. Let ai be the center of Ci, then δ ≤v(a −ai) < γ for all i. Choose ϵ such that δ ≤v(a −ai) < ϵ < γ. Then Bϵ(a) ⊂W ⊆A. Thus if b ∈Bϵ(a), then Bγ(b) ∈Y .
Let Z be the set of closed balls of radius γ contained in Bϵ(a). Then Z is an infinite set of closed balls and Z ⊆Y .
If we choose c ∈K with v(c) = −ϵ, then g(x) = c(x−a) is a bijection between Bϵ(a) and O. If b1, b2 ∈Bϵ(a) such that v(b1 −b2) ≥γ, then v(g(b1) −g(b2)) = v(b1−b2)−ϵ > 0. Thus res(g(b1)) = res(g(b2)). Thus the map Bγ(b) 7→res(g(b)) is a well defined map from Z onto k.
□ Corollary 4.31 Suppose f : Γ →Bc. Let X be the image of f. Then r|X is finite-to-one.
Proof If not there is an infinite Z ⊆X and a definable surjection g : Z →k.
Let A = f −1(Z). Then g ◦f|A is a definable map from an infinite definable subset of Γ onto k, a contradiction.
□ 44 Lemma 4.32 If X ⊆Bc is infinite, there is a definable f : X →Γ with infinite image. In particular, the image of f contains a non-trivial interval.
Proof First consider the image of X under the radius map. If this is infinite, then we are done. If not, then, without loss of generality we may assume that all balls in X have radius γ. Let A = S B∈Y B. As the proof of Lemma 4.30, there is a closed ball Bϵ(a) ⊂A with ϵ < γ. If x, y ∈Bϵ(a) \ Bγ(a) such that v(x −y) ≥γ, then v(x −a) = v(y −a). Thus we have a well defined function f : X →Γ such that f(B) = ( v(x −a) if B ⊂Bϵ(a) \ Bγ(a) and a ∈B 0 otherwise .
Then the image of f is an infinite subset of Γ.
□ We can extend this result to balls in n-spaces.
Let γ ∈Γ and let a = (a1, . . . , an) ∈Γn. Then Bγ(a) = {b ∈Kn : ^ v(ai −bi) ≥γ} is the closed ball around a of radius γ. Let Bn c be the collection of all closed balls in Kn. Let π : Kn →Kn−1 be the projection onto the first n−1 coordinates. If B ∈Bn c is a closed ball of radius δ, then π(B) ∈Bn−1 c and if Bδ(a1, . . . , an−1) ∈ Bn−1 c then B is in the fiber π−1(B1) if and only if B = Bδ(a1, . . . , an) = Bδ(a1, . . . , an−1) × Bδ(an) for some an ∈K. Thus the fiber is in definable bijection with an infinite subset of Bc.
Corollary 4.33 If X ⊆Bn c is infinite and definable, there is a definable func-tion f : X →Γ with infinite image.
Proof We proceed by induction on n, knowing the result is true for n = 1. Let X ⊂Bn−1 c . Consider the projection of X to Bn c . If this is infinite we are done.
If not, some fiber is infinite. But this gives rise to an infinite subset of Bc and we are done.
□ Corollary 4.34 If X ⊆Kn is infinite and definable, then there is a definable f : X →Γ with infinite image.
Proof We have a definable injection a 7→{a} = B∞(a) of Kn into Bn c . Thus this follows from the previous corollary.
□ 4.4 Real Closed Valued Fields We next consider valued fields (K, O) where K is a real closed field and O is a proper convex subring. We call O a real closed ring and we refer to (K, O) as a real closed valued field. In a series of exercises we will prove the following theorem of Cherlin and Dickmann.
45 Theorem 4.35 The theory of theory of real closed valued fields admits quanti-fier elimination in the language Ldiv,< = {+, −, ·, <, |, 0, 1}.
As usual, the theorem will follow from an embedding lemma.
Lemma 4.36 Let (K, O) and (L, OL) be real closed valued fields such that L is |K|+-saturated. Let R be a subring of K and f : R →L is an embedding that preserves both the order and the divisibility relation. Then f extends to an order and valuation preserving embedding of K into L.
Let K, L, R and f : R →K be as in the lemma. We let v denote the valuation on K and vL denote the valuation on L.
Exercise 4.37 Let K0 be the fraction field of R. Show that f extends to an order and and valuation preserving embedding of K0 into L.
Exercise 4.38 Let K0 be as above and let Krcl 0 be the real closure of K0 inside K. Show that we can extend f to an order and valuation preserving of Krcl 0 into K.
Henceforth, we assume that we have K0 a real closed subfield of K and f : K0 →L an order and valuation preserving embedding.
Exercise 4.39 Suppose x ∈K \K0, v(x) = 0 and x is transcendental over kK0.
Show that we can extend f to K0(x) preserving the ordering and the valuation.
Exercise 4.40 Suppose x ∈K \ K0, v(x) ̸∈v(K0). Show that we can extend f to K0(x) preserving the ordering and the valuation.
Exercise 4.41 Suppose x ∈K \ K0 and K/K0 is immediate. Show that we can extend f to K0[x] preserving the ordering and the valuation.
Exercise 4.42 Conclude that the theory of real closed rings has quantifier elimination. Show that the theory of real closed valued fields is complete.
Recall that an ordered structure (M, <, . . . ) is weakly o-minimal if every definable X ⊂M is a finite union of points and convex sets.
Exercise 4.43 Show that a real closed ring is weakly o-minimal and NIP.
A partial converse holds (). It T is a theory all of whose models are weakly o-minimal rings, then they are real closed rings or real closed fields.
46 5 Algebra of Henselian Fields 5.1 Extensions of Henselian Valuations Our first goal is to give two alternative characterizations of being henselian.
The first is that for any algebraic extension there is a unique extension of the valuation. The second, under some additional assumptions, is that there are no proper immediate algebraic extensions.
We begin with a useful lemma.
Lemma 5.1 Suppose O1, . . . , Om are valuation rings of K with maximal ideals mi, A = O1 ∩· · · ∩Om and ni = A ∩mi. Then Oi = Ani for each i.
Proof Let ki denote the residue field of Oi. Let x ∈O1. We may assume x ̸= 1. Let I = {i : x ∈Oi}.
Choose M so that: • M ̸= 0(mod mi) for all i; • for i ∈I either x = 1(mod mi) or x is not a M th root of unity in ki; • for i ̸∈I either x = 1(mod mi) or 1/x is not a M th root of unity in ki.
The next exercise is to show this is always possible. Let y = 1+x+· · ·+xM−1.
Then y is a unit in Oi [if x = 1(mod mj), then y = M ̸= 0(mod mj), while if x ̸= 1(mod mj), then y = 1−xM 1−x ̸= 0(mod mi)]. In particular xy−1 ∈Oi for i ∈I.
Similarly, we can also assume that z = 1 + x−1 + · · · + x1−M is a unit in Oj for j ̸∈I. But then y−1 = x1−Mz−1 ∈Oj and xy−1 = x2−Mz−1 ∈Oj for j ̸∈I.
Thus xy−1, y−1 ∈A and y−1 ̸∈n1. Thus x = (xy−1/y−1) ∈An1.
□ Exercise 5.2 Show that it is always possible to choose M as in the above proof.
Lemma 5.3 Let K be a field and let O1, . . . , Om be valuation rings of K such that Oi ̸⊆Oj for i ̸= j, let A = O1 ∩· · · ∩Om and let ni = mi ∩A. Then i) ni ̸⊆nj for i ̸= j; ii) n1, . . . , nm are maximal ideals of A and every maximal ideal of A is one of the ni; iii) for (a1, . . . , am) ∈O1 × · · · × Om, there is a ∈A with a = ai in ki.
Proof i) If ni ⊆nj, then Oj = Anj ⊆Ani = Oi.
ii) Suppose I ⊂A is a proper ideal. We will show that I ⊂ni for some i.
Suppose not. For each i choose ai ∈I \ ni. Also for i ̸= j choose bi,j ∈ni \ nj.
Then cj = Y i̸=j bi,j ∈ni \ nj for all i ̸= j.
47 Thus ajcj ∈i \nj for all i ̸= j and d = X ajcj ∈I \ ni for all i.
Thus 1/d ∈Oi for all i, so 1/d ∈A. But then 1 ∈I, a contradiction.
iii) We know that Oi = Ani and mi = niAni. Thus ki = Ani/niAni = A/ni.
Now we can apply the Chinese Remainder Theorem.
□ Lemma 5.4 Suppose (K, O) is a valued field and L/K is algebraic. If O1 ⊆O2 are valuation rings of L with Oi ∩K = O, then O1 = O2.
Proof Then O1 = O1/m2 in O2/m2 is a valuation ring in k2 and k ⊂O1. But k2/k is algebraic, thus O1 is a field. Since it’s a valuation ring its fraction field must be all of k2. Thus O1 = k2. Since m2 ⊆m1, we must have O1 = O2.
□ The following analysis will be the key to several of our main results in this section. Let (K, O) be a valued field. Suppose F/K be a finite Galois extension and O1, . . . , Om are distinct extensions of O to F. Let G = {σ ∈Gal(F/K) : σ(O1) = O1} and let L ⊆F be the fixed field of G. We will make two observa-tions.
Lemma 5.5 Under the assumptions above with m > 1: i) (K, O) is not henselian; ii) (L, O1 ∩L) is a proper immediate extension of (K, O) Proof Let O′ i = Oi ∩L for i = 1, . . . , m.
claim If i > 1, then O′ i ̸= O′ 1.
If O′ i = O′ 1, then O1 and Oi are extensions of O′ 1 from L to F. But then by Theorem 3.24, there is σ ∈Gal(F/L) = G with σ(O1) = Oi, contradicting the definition of G.
Let A = O′ 1 ∩· · · ∩O′ m. Let ni = Oi ∩A.
claim If i ≥2, then ni ̸= n1.
By Lemma 5.1, if ni = n1, then O′ 1 = An1 = Ani = O′ i, a contradiction.
By Lemma 3.21 we can find a ∈A such that a = 1(mod m1) and a ∈ m2 ∩· · · ∩mm, where mi is the maximal ideal of Oi.
As mi ∩K = mK we must have a ̸∈K. Let f(X) = Xn + bn−1Xn−1 + . . . b0 = (X −a)(X −α2) · · · (X −αn) be the minimal polynomial for a over K, where b0, . . . , bn−1 ∈K and α2, . . . , αn ∈ F.
claim α2, . . . , αn ∈m1.
Let i ≥2.
There is σ ∈G(F/K) such that σ(a) = αi.
We know that a ∈A ⊂L and any σ ∈G fixes L pointwise. Thus σ ̸∈G and σ−1(O1) = Oj for some j ̸= 1. But a ∈mj. Thus αi = σ(a) ∈m1.
48 It follows that bn−1 = −a −α2 −· · · −αn = −1(mod m1) and b0, . . . , bn−2 ∈ m1.
claim (K, O) is not henselian.
Clearly f(1) ∈mO.
Let g(X) = (X −α2) · · · (X −αn).
Then f ′(X) = (X −a)g′(X) + g(X). Thus f ′(1)(mod m1) = (1 −a)g′(1) + g(1)(mod m1) = 1(mod m1).
Thus f ′(1) ̸= 0(mod mO). If K were henselian, f would not be irreducible. Thus K is not henselian.
To show that (L, O′ 1) is an immediate extension we make some minor mod-ifications to the proof above. Suppose c is a unit in O′ 1 we can find a ∈A such that a = c(mod m1) but a ∈mi for i > 1. Let f be the minimal polynomial for a over K. Arguing as above f(X) = Xd + bd−1Xd−1 + · · · + b0 = (X −a)(X −α2) · · · (X −αd) where bd−1 = −c(mod m1) and b0, . . . , bd−2 ∈m1. But −(c + α2 + · · · + αd) = bd−1 ∈K and c = bd−1. Thus the residue field does not extend.
We need to show the value group does not extend. We let v denoted the valuation on L. Let x ∈L. We must find y ∈K with v(x) = v(y). We can find a ∈A such that a −1 ∈m1 and a ∈m2 ∩· · · ∩mm. Then v(a) = 0 and v(σ(a)) = 0 for all σ ∈G. Since a ∈m2 ∩· · · ∩mm, as above, v(σ(a)) > 0 for all σ ∈Gal(F/K) \ G. We claim that we can choose N large enough we can ensure that v(aNx) ̸= v(σ(aNx)) for all σ ∈Gal(F/K) \ G.
For any particular σ ∈Gal(F/K)\G, v(arx) = v(x) and v(σ(arx)) = rv(σ(a))+ v(σ(x)). Since v(σ(a)) > 0, for all but one value of r these are unequal. Thus, since Gal(F/K) is finite, we can choose N as desired.
Let aNx = α1, . . . , αn be the distinct conjugates of aNx over K. Let g(X) = (X −α1) · · · (X −αn) = Xn + bn−1Xn−1 + · · · + b0.
For 1 < i ≤n, αi = σ(aNx) for some σ ∈Gal(F/K) \ G [note that any τ ∈G fixes aNx ∈L]. Thus v(αi) ̸= v(α1) for i > 1.
First suppose v(αi) > v(α1) for all i > 1. Then bn−1 = −P αi, v(x) = v(aNx) = v(bn−1) ∈v(K), as desired. In general suppose that v(αi) < v(α1) for 1 < i ≤k and v(αi) > v(α1) for k < i. Note that bn−j = (−1)j X 1≤i1<···<ij≤n αi1 · · · αin.
Thus v(bn−k) = v(α2 · · · αk) and v(bn−k−1) = v(α1 · · · αk).
Thus v(x) = v(α1) = v(bn−k−1/bn−k) ∈v(K).
Thus L is an immediate extension of K.
□ 49 Theorem 5.6 Let (K, O) be a valued field. The following are equivalent: i) (K, O) is henselian; ii) For any separable algebraic extension L/K there is a unique extension of O to a valuation ring of L; iii) For any algebraic extension L/K there is a unique extension of O to a valuation ring of L iv) If f(X) ∈O[X] is monic irreducible and f(X) is non-constant, then there is and irreducible g(X) ∈k[X] and n ≥1 such that f(X) = g(X)n.
v) If f, g, h ∈O[X] is monic and f = gh where g and h are relatively prime, then there are g1, h1 ∈O[X] such that g1 = g, h1 = h and g1 and g have the same degree.
Proof i) ⇒ii) Suppose not. Then we can find F/K a finite Galois extension such that O has multiple extensions O1, . . . , Om each of which are conjugate under Gal(F/K). Now we can apply Lemma 5.5 to show that (K, O) is not henselian.
ii) ⇒iii) Let K ⊆F ⊆L be the separable closure of K in L. By ii) there is a unique extension of the valuation to F. Since L/F is purely inseparable and there is a unique extension of the valuation to L.
iii) ⇒iv) In Kalg we can factor f(X) = d Y i=1 (X −αi).
Let O∗and m∗denote the valuation ring and maximal ideal of an extension to Kalg.
Since f ∈O[X], Q αi ∈O, thus we can not have v(αi) < 0 for all i. Since any two roots are conjugate and there is a unique extension of the valuation ring to Kalg we must have all of the αi ∈O or all of the αi ̸∈O, but the latter option is not possible.
Thus, f(X) = Q(X −αi). To show that f is a power of an irreducible polynomial in k[X] it is enough to show that we can not fact f = gh where g and h are relatively prime and monic. Suppose we can. If g(αi) = 0, then g(αi) ∈m∗and for any σ ∈Gal(Kalg/K), g(σ(αi)) ∈σ(m∗) = m∗. But all of the roots of f are conjugate. Thus they are all roots of g, a contradiction.
iv) ⇒v) Let f = q1 · qm be an irreducible factorization of f in O[X]. into monic factors. For each i, there is a monic pi ∈O[X] such that qi = pni i . We can find J ⊆{1, . . . , d} such that g = Q i∈J pni i . Let h = Y i̸∈J pni i .
Let g1 = Y i∈J pni i and h1 = Y i̸∈J pni i .
50 Then f = g1h1 and g and g1 have the same degree.
v) ⇒i) Suppose f(X) ∈O[X] and f(X) = Xd + Xd−1 + P aiXi where ai ∈mi. In k[X] we can factor f(X) = h(X)(X + 1), Since f ′(−1) = ±1, h(−1) ̸= 0. Thus h(X) and (X + 1) are relatively prime. By iv) there is a ∈K with a = −1 such that (X −a) is an irreducible factor of f.
□ Exercise 5.7 Show that it (K, O) is henselian and (L, OL) is an algebraic extension, then (L, OL) is henselian.
Exercise 5.8 Suppose (K, O) is henselian, F ⊆K and F is separably closed in K. Prove that (F, O ∩F) is henselian.
5.2 Algebraically Maximal Fields Definition 5.9 We say that a valued field (K, O) is algebraically maximal if it has no proper separable algebraic immediate extensions.
Corollary 5.10 An algebraically maximal valued field (K, O) is henselian.
Proof If (K, O) is not henselian we can find F/K a finite Galois extension with multiple extensions of O to F. By Lemma 5.5, we can find an intermediate field K ⊂L ⊆F with L/K immediate.
□ The converse is true under some additional assumptions which will apply in many of our settings.
Definition 5.11 We say that (K, O) has equicharacteristic zero if K and the residue field k have characteristic zero.
We say that (K, O) is finitely ramified if k has characteristic p > 0 and {v(x) : 0 < v(x) < v(p) : x ∈K×} is finite.
Note that the later condition is true for the p-adics.
Exercise 5.12 Prove that if (K, OK) is a finite algebraic extension of (Qp, Zp), then (K, OK) is finitely ramified.
Exercise 5.13 Suppose L/K is finitely ramified. Show that the set {v(x) : 0 < v(x) < v(n)} is finite for all n ∈Z.
Theorem 5.14 If (K, O) is henselian and equicharacteristic zero or finitely ramified, then (K, O) is algebraically maximal.
Proof Suppose F is an algebraic immediate extension and x ∈F \K. Without loss of generality F/K is finite. There is L ⊇K such that L/K is Galois. There is a unique extension OL of O. Let v be the valuation associated with OL. For and a ∈K, we have v(x−a) = v(b) for some b ∈K, but then v(σ(x)−a) = v(b) for all σ ∈Gal(L/K).
51 Let d = [L : K] Let a = 1 d X σ∈Gal(L/K) σ(x) ∈K.
Since F/K is immediate, there is b ∈K such that v((x −a)/b) = 0 and c ∈K such v( x−a b −c) > 0. Then v(x −(a + bc)) > v(b) = v(x −a).
Since (K, O) is equicharacteristic zero or finitely ramified, repeating this argu-ment finitely many times in the case where the residue field has characteristic p we can find b a ∈K such that v(x −b a) > v(x −a) + v(n) (if the residue field has characteristic zero v(n) = 0 so we need only do this once). Then v(n) + v(a −b a) = v(n(a −b a)) = v X σ∈Gal(L/K) (σ(x) −b a) ≥ min(v(σ(x) −b a)) ≥ v(x −b a) since v(w) = v(σ(w)) for all w ∈L > v(x −a) + v(n) = v(a −b a) + v(n) a contradiction. The last line holds since v(a −b a) = v((a −x) + (x −b a)) and v(x −b a) > v(x −a).
□ Corollary 5.15 If (K, O) is henselian with divisible value group and alge-braically closed residue field of characteristic zero, then K is algebraically closed.
Proof If L/K is a proper algebraic extension, then we can extend the valuation to L and, by Lemma 3.16 it must be an immediate extension, contradicting Theorem 5.14.
□ Corollary 5.16 If k is an algebraically closed field of characteristic zero, then the Puiseux series field k⟨T⟩is algebraically closed.
This doesn’t work in characteristic p > 0. The series solution to f(X) = Xp −X = T −1 should be of the form a + T −1/p + T −1/p2 + · · · + T −1/pn + . . .
where a ∈Fp, which is not a Puiseux series. This series is in the immediate extension Fp( ( (Q) ) ) and thus in the separable closure of the Puiseux series in 52 the Hahn series. This shows that henselianity alone is not enough to conclude algebraically maximal. Kedlaya in gives a characterization of the algebraic closure of Falg p ( (T) ).
If k is real closed and Γ is divisible, kalg⟨T⟩is a degree 2 extension of k⟨T⟩.
Thus k⟨T⟩is real closed. This is true in much more generality.
Corollary 5.17 Let (K, <) is an ordered field and let O be the convex hull of a subring. Suppose (K, O) is henselian with real closed residue field k and divisible value group Γ. Then K is real closed.
Proof Let L be the real closure of (K, <) and let O∗be the convex hull of O in K. Then, since the orderings agree, (K, O) ⊆(L, O∗). The residue field kL is real closed and algebraic over k, so it must equal k. Similarly, the value group of L is contained in the divisible hull of v(K) and hence equals v(K). Thus L/K is an immediate extension and, since (K, O) is henselian and equicharacteristic zero, L = K.
□ 5.3 Henselizations Infinite Galois Theory We quickly review some facts we need about the Galois Theory of infinite alge-braic extensions. The reader should consult §8.6 or §1.
Let K be a field. The separable closure of K is Ks the maximal separable algebraic extension of K. When we apply these results we will be working almost exclusively in the setting where K has characteristic zero so there would be no harm in working with Kalg the algebraic closure of K. We let Gal(Ks/K) be the Galois group of all automorphisms of Ks that are the identity on K.
Suppose L/K is a finite Galois extension. If σ ∈Gal(Ks/K), then σ|L ∈ Gal(L/K). Moreover if τ ∈Gal(L/K), there is b τ ∈Gal(Ks/K) extending τ.
Thus Gal(Ks/K) = lim ← − L/K finite Galois Gal(L/K) is a profinite group. We topologize Gal(Ks/K) by taking the weakest topology such that for all finite Galois extensions L/K and σ ∈Gal(L/K), Uσ = {τ ∈ Gal(Ks/K) : σ ⊆τ} is open.
If H is a subgroup of Gal(Ks/K), let Fix(H) = {x ∈Ks : σ(x) = x for all σ ∈H} be the fixed field of H.
Theorem 5.18 (Fundamental Theorem of Infinite Galois Theory) The maps L 7→Gal(Ks/L) and H 7→Fix(H) are inclusion-reversing bijections be-tween the collection of intermediate fields K ⊆L ⊆Ks and closed subgroups of Gal(Ks/K).
53 Henselizations Let (K, O) be a valued field, let Ks be the separable closure of K and let Os be an extension of O to K. Let G(Os) = {σ ∈Gal(Ks/K) : σ(Os) = Os}. We call G(Os) the decomposition group.
Lemma 5.19 G(Os) is a closed subgroup of Gal(Ks/K).
Proof Suppose σ ̸∈G(Os). There is x ∈Os with σ(x) ̸∈Os. Let L/K be finite Galois with x ∈L and let τ = σ|L. Then σ ∈Uτ and Uτ ∩G(Os) is empty.
□ Definition 5.20 Let Kh(Os) be the fixed field of G(Os) and let Oh(Os) = Os ∩Kh. We call (Kh(Os), Oh(Os)) a henselization of (K, O).
When no confusion arises we will suppress Os and write (Kh, Oh).
Lemma 5.21 i) Os is the unique extension of Oh to Ks. Thus (Kh, Oh) is henselian.
ii) (Kh, Oh) is an immediate extension of K.
Proof i) Suppose Os 1 is an extension of Oh to Ks. By Theorem 3.24, Os and Os 1 are conjugate under Gal(Ks/K). But G(Os) is the Galois group of Ks/Kh, so any element of Gal(Ks/K) fixes Os. Hence Os = Os 1 ii) follows from Lemma 5.5.
□ Lemma 5.22 If (K1, O1) is a henselian extension of (K, O) then there is a unique embedding j : (Kh, Oh) →(K1, O1) fixing K pointwise.
Proof Without loss of generality, by Exercise 5.8, we may assume that K1 ⊆ Ks. Since K1 is henselian, there is a unique extension Os 1 of O1 to Ks. Then Gal(Ks/K1) ⊆G(Os 1). Thus K1 ⊇Kh(Os 1). By Theorem 3.24 there is σ ∈ Gal(Ks/K) with σ(Os) = Os 1, but then σ(Kh) = Kh(Os 1) ⊆K1 and σ|Kh is the desired embedding of (Kh, Oh) into (K1, O1).
Suppose j : (Kh, Oh) →(K1, O1) is another embedding. We can extend j to τ ∈Gal(Ks/K). Then τ(Os) ∩τ(Kh) = Os 1 ∩τ(Kh). But (τ(Kh), τ(Oh)) is henselian, so Os 1 is the unique extension of τ(Os) to Ks and τ(Os) = Os 1. Thus τ −1σ(Os) = Os and τ −1σ ∈G(Os). Since Kh is the fixed field of G(Os), σ and τ agree on Kh. Thus j = σ|Kh.
□ Exercise 5.23 In particular if Os and Os 1 are distinct extensions of O, then there is a unique isomorphism between (Kh(Os), Os) and (Kh(Os 1), O(Os 1)) fix-ing K.
Summarizing we have proved: Theorem 5.24 Let (K, O) be a valued field. There is a henselization (Kh, Oh), i.e. a henselian immediate separable algebraic extension of (K, O) such that if (K1, O1) is a henselian extension of (K, O) then there is a unique embedding j : (Kh, Oh) →(K1, O1) with j|K the identity.
54 Corollary 5.25 a) Let (K, v) be an algebraically closed valued field of charac-teristic 0 and let A ⊂K. Show that dcl(A), the definable closure of A, is exactly the henselization of the fraction field of A.
Proof Let F be the fraction field of A. Then F alg is an elementary submodel of (K, v).
The valued field automorphisms of F alg that fixes A are exactly the elements of the decomposition group G(OFalg), when has fixed field F h. It follows that F h = dcl(A).
□ Exercise 5.26 Let (K, v) be an algebraically closed field of characteristic p > 0.
Prove that A = dcl(A) if and only if A is perfect and henselian.
5.4 Pseudolimits Let K be a valued field with valuation v. We will consider sequences (aα : α < δ) where δ is a limit ordinal and aα ∈K for α < δ. Frequently, we will simplify notation by just writing (aα).
Definition 5.27 We say that a is a pseudolimit of (aα : α < δ) if the sequence (v(a −aα) : α < δ) is eventually strictly increasing. We write (aα) ⇝a. We let γα = v(a −aα).
Exercise 5.28 Suppose (aα) ⇝a and b ∈K.
a) Show (aα + b) ⇝a + b.
b) Show (baα) ⇝ba Lemma 5.29 Suppose (aα) ⇝a. Then either: i) (v(aα)) is eventually constant and equal to v(a); ii) (v(aα)) is eventually strictly increasing and v(aα) < v(a) for sufficiently large α.
Proof Suppose γα is increasing for α ≥α0 and v(a) ≤v(aα0). Then v(a − aα0) ≥v(a) and α > α0, v(aα) = v(a), since v(a −α) > v(a −aα0) ≥v(a).
Thus we are in case i).
If this never happens then v(aα) < v(a) for all sufficiently large α and for β > α Then γα = v(aα) and v(aα) < v(aβ) for sufficiently large α < β and case ii) holds.
□ Lemma 5.30 Suppose (K, v) ⊆(L, v) is an immediate extension and x ∈L\K.
There is a sequence (aα) in K such that (aα) ⇝x and (aα) has no pseudolimit in K.
Proof Let a0 = 0. Suppose we have aα. Since v(K) = v(L) we can find b ∈K such that v(b) = v(x −aα). Since kK = kL, there is c ∈K such that 0 ̸= c = res( x−aα b ). Thus v(x −(aα + bc)) > v(b) = v(x −aα).
55 Let aα+1 = aα + cb. Then v(x −aα+1) > v(x −aα).
Suppose δ is a limit ordinal and we have constructed (aα : α < δ) with v(x −aα) < v(x −aβ) for α < β < δ. If there is b ∈K such that v(x −b) > v(x −aα) for all α < δ let aδ = b and continue. If no such b exists (aα) is our desired sequence.
□ A sequence (aα) in K might not have a pseudolimit in K, but we can tell if it could have pseudolimit in an extension.
Definition 5.31 We say that (aα) is pseudocauchy if there is α0 such that v(aδ −aβ) > v(aβ −aα) for δ > β > α > α0.
Lemma 5.32 i) If (aα) ⇝a, then (aα) is pseudocauchy.
ii) If (aα) is pseudocauchy, there is an elementary extension (K, v) ≺(L, v) such that (aα) has a pseudolimit in L.
Proof i) If δ > β > α are suitably large, then aδ −aβ = (a −aβ) −(a −aδ).
Thus v(aδ −aβ) = v(a −aβ). Similarly, v(aβ −aα) = v(a −aα) and, thus, v(aδ −aβ) > v(aβ −aα) and the sequence is pseudocauchy.
ii) Consider the type t(v) = {v(x −aβ) > v(x −aα) : for α0 < α < β}. Let ∆⊂t(v) be finite. Choose δ > α for all aα occurring in ∆. Then v(aδ −aβ) > v(aβ −aα) = v(aδ −aβ) for δ > β > α > α0. Thus t(v) is finitely satisfiable and thus realized in some elementary extension of K.
□ Corollary 5.33 If (aα) is pseudocauchy, then (v(aα)) is either eventually con-stant or eventually strictly increasing.
Exercise 5.34 Prove that in a Hahn field k( ( (Γ) ) ) every pseudocauchy sequence has a pseudolimit and conclude that Hahn fields have no proper immediate extensions. (This is essentially the same proof we gave in §1.) The next lemma is important but not surprising and rather routine. We omit the proof and refer the reader to Proposition 4.7 for the proof.
Lemma 5.35 Suppose (aα) ⇝a and f(X) ∈K[X]. Then (f(aα)) ⇝f(a).
Thus if (aα) is pseudocauchy, so is (f(aα)).
There is an important dichotomy among pseduocauchy sequences.
Definition 5.36 Let (aα) be a pseudocauchy sequence in K. We say that (aα) is of algebraic type if there is a nonconstant polynomial f(X) ∈K[X] such that (v(f(aα))) is eventually strictly increasing. Otherwise we say (aα) is of transcendental type.
If (aα) is of transcendental type, then (v(f(aα))) is eventually constant for all f ∈K[X].
56 Lemma 5.37 If (aα) is a pseudocauchy sequence over K of transcendental type, then (aα) has no pseudolimit in K and there is an extension of v to the field of ra-tional functions K(X) with v(f) = eventual value of v(f(aα)). Then (K(X), v) is an immediate extension of K where (aα) ⇝X.
If L/K is a valued field extension of K and (aα) ⇝a in L, then sending X to a we get a valued field isomorphism between K(X) and K(a) fixing K.
Proof If (aα) ⇝a, let f(X) = X −a, then (v(f(aα))) is eventually strictly increasing and the sequence is of algebraic type, a contradiction. Thus (aα) has no pseudolimit in K.
Let v be defined as above. Then, for α sufficiently large v(fg) = v(f(aα)) + v(g(aα)) = v(f) + v(g) and v(f + g) = v(f(aα) + g(aα)) ≥min(v(f(aα))v(g(aα))) = min(v(f), v(g)).
Thus v is a valuation on K(X) extending the valuation on K.
Clearly, the value group of K(X) is equal to the value group of K. Let f ∈K(X) \ K with v(f) = 0. Then 0 = v(f) = v(f(aα)) for sufficiently large α. If β > α, then v(f −f(aβ)) > v(f −f(aα)) > v(f(aα)) = 0 and res(f) = res(aβ). Thus K(X) is an immediate extension of K.
Suppose (L, v) is a valued field extension of K and a ∈L is a pseudolimit of (aα). For nonconstant f ∈K[X] we have (f(aα)) ⇝f(a). Thus v(f(a)) = v(f(aα)) = v(f) for sufficiently large α.
In particular f(a) ̸= 0, thus a is transcendental over K and the field isomorphism of K(X) to K(a) obtained by sending X to a preserves the valuation.
□ Definition 5.38 If (aα) is of algebraic type, a minimal polynomial of (aα) is a polynomial g of minimal degree such (v(g(aα)) is eventually increasing.
Lemma 5.39 Let (aα) be a pseudocauchy sequence of algebraic type with min-imal polynomial g(X) and no pseudolimit in K. Then g(X) is irreducible of degree at least 2. Let a be a zero of g in an extension field of K. Then v ex-tends to a valuation on K(a) where v(f(a)) = eventual value of v(f(aα)), where f(X) ∈K[X] of degree less than deg(g). Then K(a) is an immediate extension of K where (aα) ⇝a.
If L/K is any valued field extension of K where b ∈K is a zero of g and (aα) ⇝b, then the isomorphism K(a) to K(b) obtained by sending a to b, preserves the valuation.
Proof If g(X) = X −a then (v(g(aα))) = v(aα −a) is eventually strictly increasing and (aα) ⇝a, a contradiction. Thus g has degree at least two. If g = g1g2 is a nontrivial factorization of g, then, by minimality of the degree of g, (v(gi(aα))) is eventually constant for each i, but then (v(g(aα))) = (v(g1(aα) + g2(aα))) is eventually constant, a contradiction. Thus g is irreducible of degree at least two.
57 Consider the extension K(a) where g(a) = 0. Suppose f1, f2 ∈K[X] have degree less that deg(g). There are h, r ∈K[X] with degree less than deg(g) such that f1f2 = hg + r. Then for α sufficiently large v(f1) = v(f1(aα)), v(f2) = v(f2(aα)) and v(f1f2) = v(r) = v(r(aα)).
Then, v(f1) + v(f2) = v(f1(aα)f2(aα)) = v(h(aα)g(aα) + r(aα)).
The sequence (v(h(aα)g(aα) + r(aα)) is eventually constant, while the sequence (v(h(aα)g(aα)) is eventually increasing. This is only possible if v(h(aα)g(aα)) > v(r(aα)) eventually. But then v(h(aα)g(aα) + r(aα)) = v(r(aα)) eventually and v(f1f2) = v(f1) + v(f2) as desired.
The rest of the proof closely follows the proof of Lemma 5.37.
□ Corollary 5.40 Let (K, v) be a valued field. Then every pseudocauchy sequence in K has a pseudolimit in K if and only if K has no proper immediate exten-sions.
Exercise 5.41 Prove that K has no proper immediate extensions if and only if K is spherically complete.
We can refine Lemma 5.30 for algebraic immediate extensions.
Lemma 5.42 Suppose (L, v) is an immediate extension of K and a ∈L \ K is algebraic over K with minimal polynomial g. Let (aα) be a pseudocauchy sequence over K with no pseudolimit in K such that (aα) ⇝a. Then (aα) is of algebraic type. In fact (v(g(aα))) is increasing.
Proof Let g(X) = (X −a)h(X) where h ∈K(a)[X]. Then g(aα) = (aα − a)h(aα). The sequence (v(aα −a)) is eventually increasing and the sequence (v(h(aα))) is either eventually increasing or eventually constant. Thus v(g(aα)) is either eventually increasing or eventually constant.
□ Corollary 5.43 Let (K, v) be a valued field. If every pseudocauchy sequence (aα) of algebraic type in K has a pseudolimit in K, then K is henselian. More-over, the converse holds if, in addition (K, v) is either equicharacteristic zero or finitely ramified.
Proof If every pseudocauchy sequence (aα) of algebraic type has a pseudolimit in K, then by Lemma 5.42 (K, v) has no proper immediate algebraic extension and, by Theorem am hen, (K, v) is henselian. Note that this direction did not use the additional assumptions on (K, v).
If (K, v) is henselian and either equicharacteristic zero or finitely ramified, then by Theorem 5.10, (K, v) has no proper immediate algebraic extensions.
Thus by Lemma 5.39, every pseudocauchy sequence of algebraic type in K has a pseudolimit in K.
□ 58 Exercise 5.44 Suppose K is a valued field with value group Γ such that there is a lifting of the residue field k to K and there is s : Γ →K a section of the value group. Show there is a valuation preserving embedding of K into the Hahn field k( ( (Γ) ) ). [Hint: View k as a subfield of K. First show that k(s(Γ)) embeds into k( ( (Γ) ) ). Then consider a maximal subfield K0 ⊆K such that the embedding extends to a valuation preserving embedding of K0 into k( ( (Γ) ) ).] Conclude that if K is a real closed field with valuation ring O a convex subring, residue field k and value group Γ, then there is a valuation preserving embedding of K into k( ( (Γ) ) ).8 8Mourgess and Ressarye proved the stronger result that we can embedding K into k( ( (Γ) ) ) such that if f is in the image so is any truncation (i.e. initial segment) of f. They used this to prove that every real closed field has an integral part (i.e. a discrete subring Z such that for all x ∈K, |[x, x + 1) ∩Z| = 1).
59 6 The Ax–Kochen Erˇ sov Theorem 6.1 Quantifier Elimination in the Pas Language We will be considering valued fields as three-sorted objects (K, Γ, k) in the Pas language where we have the language of rings {+, −, ·, 0, 1} on both the home sort, i.e. the field K, and the residue field sort, the language of ordered groups {+, −, <, 0} on the value group sort, the valuation map v : K× →Γ and an angular component map ac : K× →k×. Not all valued fields have angular component maps, but for any valued field we can pass to an elementary extension where there is an angular component map.
Let ∆0 be the collection of all formulas of the form • φ(u), where φ is a quantifier free formula in the language of rings and u are variables in the field sort; • ψ(v(f1(u)), . . . , v(fk)(u)), v)) where ψ is a formula in the language of ordered groups, u are variables in the field sort, fi is a term in the ring language and v are variables in the value groups sort; • θ(ac(g1(u)), . . . , ac(gk)(u)), w)) where ψ is a formula in the language of ordered groups, u are variables in the field sort, gi is a term in the ring language, and w are variables in the residue sort; Note that we are allowing quantifiers over the value group and the residue field but not over the home sort.
Let ∆be the collection of finite boolean combinations of ∆-formulas. Note that each ∆formula is equivalent to a formula of the form φ(u) ∧ψ(v(f1(u)), . . . , v(fk)(u)), v)) ∧θ(res(g1(u)), . . . , res(gl)(u)), w)), where φ, ψ and θ are as above.
Theorem 6.1 (Pas) Let T be the theory of henselian valued fields with angular components where the residue field has characteristic zero. Then every formula is equivalent to a ∆-formula.
We will use the following relative quantifier elimination test.
Exercise 6.2 Suppose L is countable. Let ∆be a collection of formulas closed under finite boolean combinations and let T be an L-theory with the following property.
Whenever M and N are models of T, |M| = ℵ0 N is ℵ1-saturated, A ⊂M and f : A →N is a ∆-embedding (i.e, M | = θ(a) ⇔N | = θ(f(a) for a ∈A and θ ∈∆), then there is b f : M →N that is ∆preserving.
Show that every L-formula is equivalent to a ∆-formula. [Hint: add predi-cates for all formulas in ∆.] Our main step will be proving an embedding result. We look at embeddings that preserved ∆-formulas. A map f : (A, ΓA, kA) →L is an ∆-embedding if: 60 i) f|A is a ring embedding; ii) f|ΓA is a partial elementary embedding in the language of groups; iii) f|kA is a partial elementary embedding in the language of rings; iii) f preserves v and ac.
Theorem 6.3 Let (K, Γ, k) and (L, ΓL, kL) be henselian valued fields with an-gular component with characteristic zero residue field.
Suppose K is count-able, L is ℵ1-saturated, (A, ΓA, kA) is a countable substructure of K, and f : (A, ΓA, kA) →(L, ΓL, kL) is a ∆-embedding. Then there is an extension of f to a ∆-embedding b f : (K, ΓK, kK) →(L, ΓL, kL).
Henceforth, we assume K is countable and L is ℵ1-saturated. We extend our map by iterating the following lemmas.
Note that in a substructure (A, ΓA, kA), A and kA are domains, while ΓA is a subgroup.
Lemma 6.4 Suppose (A, ΓA, kA) be a subring of K and f : (A, ΓA, kA) → (L, Γ, kL) is a ∆-embeddings. Let F be the fraction field of A and let l be the fraction field of kA. We can extend f to a ∆-embedding of (F, Γ, l) into L.
Proof There is a unique extension of f to (F, G, l). Since v(a/b) = v(a) − v(b) and ac(x/y) = ac(x)/ac(y), vL(f(a/b)) = f(v(a/b)) and acL(f(x/y)) = f(ac(x/y)), f is a ∆-embedding.
□ Henceforth, we will work only with substructures (F, ΓF , kF ) where F and kF are fields and ΓF is a group, v(F) ⊆ΓF and ac(F) ⊆kF .
We next show how to extend the value group.
Lemma 6.5 Suppose f : (F, ΓF , kF ) →(L, ΓL, kL) is a ∆-embedding. We can extend f to a ∆-embedding of (F, Γ, kF ).
Proof We will prove this by iterating the following claim.
claim Let γ ∈Γ \ ΓF and let G be the group generated by ΓF and γ, then we can extend f to (F, G, kF ).
Let p(v) be the type {ψ(v, f(g1), . . . , f(gm)) : g1, . . . , gm ∈ΓF , ψ a formula in the language of ordered groups where Γ | = ψ(γ, g1, . . . , gm). If ψ1, . . . , ψn ∈ p(v) with parameters f(g1), . . . , f(gm), then, since f is a ∆-embedding ΓL | = ∃v n ^ i=1 ψi(v, f(g1), . . . , f(gm)).
Thus p(v) is consistent and, by ℵ1-saturation, realized in ΓL.
Let γ′ be a realization and extend f by γ 7→γ′.
□ Lemma 6.6 If we have a ∆-embedding f defined on (F, Γ, kF ) we can extend it to (F, Γ, k).
61 Exercise 6.7 Prove Lemma 6.6.
We next make the residue map surjective.
Lemma 6.8 Suppose f is a ∆-embedding of (F, Γ, k). Then we can find F ⊆ E ⊆K such that res : E →k is surjective and we can extend f to a ∆-embedding of (E, Γ, k).
Proof We iterate the following two claims and Lemma 6.4.
claim 1 Suppose we have a ∆-embedding f : (F, Γ, k) →(L, ΓL, kL) and b ∈K with residue b algebraic over res(F) but not in res(F). Then we can extend f to F(b).
There is p(X) ∈OF [X] irreducible with p(X) the minimal polynomial of b over res(F). Let q(X) ∈Of(F )[X]be the image of p. Since the embedding of residue fields is elementary, q(X) is irreducible in f(res(F)) and q(f(b)) = 0.
Moreover, since kL has characteristic zero and q is irreducible, q′(f(b)) ̸= 0.
Since L is henselian, there is unique c ∈L such that q(c) = 0 and c = f(b). We extend f to F(b) by b 7→c.
We need to show that the valuation and angular component are preserved.
Let d be the degree of p. Let x ∈F(b) = α(Pd−1 i=0 aibi) where α ∈F, ai ∈OF and some v(ai) = 0 for some i. As p is the minimal polynomial of b, P aib i ̸= 0.
Thus v(x) = v(α) and v(f(x)) = v(f(α)) and ac(x) = ac(α)(P aib i). A similar analysis shows acL(f(x)) = acL(f(α))(P f(ai)ci).
claim 2 Suppose we have a ∆-embedding f : (F, Γ, k) →(L, ΓL, kL) and b ∈B with residue b transcendental over res(F). Then we can extend f to F(b).
Let c ∈L with c = f(b). Then c is transcendental over F and we can extend f by b 7→c. We need to show that the valuation and angular component are preserved. If x ∈F[b] we can write x = α(P aibi) where α ∈F, ai ∈OF and v(ai) = 0 for some i. Then as in claim 2, v(x) = v(α) and v(f(x)) = v(f(α)), ac(x) = ac(α)(P aib i) and v and ac are preserved. As in Lemma 6.4, we can extend to f from F[b] to F(b).
□ Next we make the valuation surjective.
Lemma 6.9 Suppose f is a ∆-embedding of (F, Γ, k). There is F ⊆E ⊆K such that v : E →Γ is surjective and we can extend f to (E, Γ, k).
Proof The lemma is proved by iterating the following two claims.
claim 1 Suppose we have a ∆-embedding f of (F, Γ, k) where the residue map from F to k is surjective and g ∈Γ such ng ̸∈v(F) for any n > 0. Let b ∈K with v(b) = g. We will extend f to F(b).
Since g is not in the divisible hull of v(F), b is transcendental over F. Let c ∈L with v(c) = f(g) and acL(c) = f(ac(b)). We can extent f to F(b) with b 7→c. Let x = P aibi recall that v(x) = min(v(ai) + iv(b)) and vL(f(x)) = min vL(f(ai) + if(g)). Choose i such that v(ai) + iv(b) is minimal, then x = aibi(1 + ϵ) where v(ϵ) > 0 and ac(x) = ac(ai)ac(b)i. Similarly, acL(f(x)) = acL(f(ai)ac(c)i, as desired.
62 claim 2 Suppose we have a ∆-embedding f of (F, Γ, k) where the residue map from F to k is surjective and let n > 0 be minimal such that there is g ∈Γ\v(F) such that g > 0 and ng ∈v(F). Then we can extend F to E with F ⊂E ⊆K and extend f to a ∆-embedding of (F, Γ, k) such that g ∈v(E).
Let a ∈F and b0 ∈K be such that v(b0) = g and v(a) = ng. Since the residue field does not extend we can choose a such that ac(bn 0) = a, in which case bn 0 = a(1 + ϵ) where ϵ ∈K and v(ϵ) > 0. Since K is henselian, there is d ∈K with v(d) = 0 such that dn = 1 + ϵ. Let b = b0/d. Then bn = a. By the minimality of n, Xn −a is the minimal polynomial of b over F.
Similarly, we can find c ∈L such that cn ∈f(F) and vL(cn) = v(f(a)). Then ac(c) is algebraic over kf(F ). But kf(F ) ≺kL, thus, ac(c0) ∈kf(F ). Thus there is d ∈Of(F ) with d = f(ac(b))ac(c−1 0 ). Let c1 = dc0. Then ac(c1) = f(ac(b)) and f(a) = f(bn) = cn 1(1 + ϵ) where v(ϵ) > 0. By henselianity, there is e ∈L such that en = (1 + ϵ).
Let c = c1e, then cn = f(a), v(c) = f(v(b)) and ac(c) = f(ac(b)). We extend f to F(b) by b 7→c. As in Lemma 6.5, we show that f preserves the valuation and the the angular component map.
□ Lemma 6.10 Suppose the residue and valuation maps of (F, Γ, k) are surjective and f is a ∆-embedding. Then we can extend F to (F h, Γ, k) Proof There is a unique valuation preserving extension of f from F to g : F h → L. We know that F h is an immediate extension of f. If a ∈F h\F, there is b ∈F, with v(a) = v(b), but then v(g(a)) = v(g(b)). There is c a unit in OF such that res(c) = res(a/b). Thus ac(a) = ac(b)ac(c) and acL(g(a)) = acL(g(b))acL(g(c)).
□ We can now finish the proof of Theorem 6.3 Thus we may assume that we have a (F, Γ, k) such that F is henselian, v : F →Γ and res : F →k are surjective and f is a ∆-embedding. Then K is an immediate extension of F. By Zorn’s Lemma, we may assume that F ⊆K is maximal henselian such that there is a ∆-embedding of (F, Γ, k) into L extending f. We claim that F = K. If not, let b ∈K \ F. We will show that we can extend f to F(b). Since F is henselian and kB has characteristic zero, by Theorem 5.14, b is transcendental over F.
We can find a pseudocauchy sequence (aα) in F of transcendental type with no pseudolimit in F such that (aα) ⇝b, (aα) has no pseudolimit in F and (v(p(aα)) is eventually constant for p ∈F[T].
By ℵ1-saturation, we can find c ∈L such that (f(aα)) ⇝c. Extend f to F(b) by x 7→c. For p ∈F[T], vL(f(p(b))) = vL(f(p)(b)) = vL(f(p)(f(aα))) = vL(f(p(aα)) = f(v(p(aα)) = f(v(p(b))) for large enough α. Similarly, ac(p(b)) = ac(p(aα)) for large enough α and it follows that f(ac(p(b))) = acL(f(p(b))). But this contradicts the maximality of F.
This completes the proof.
63 6.2 Consequence of Quantifier Elimination Let T0 be the theory in the language of three sorted valued fields asserting that we have (K, Γ, k) where K is a henselian valued field where Γ is the value group and k is a the residue field.
Corollary 6.11 (Ax-Kochen , Erˇ sov) Let (K, Γ, k) be a henselain val-ued field with characteristic zero residue field. Let TΓ be the theory of the value group in the language of ordered groups and Tk be the theory of the residue field in the language of rings. Then T = T0 ∪TΓ ∪Tk is complete.
Proof Let K and L be models of T and let K ≺K∗and L ≺L∗be ℵ1-saturated elementary extensions. We can define angular component maps on K∗ and L∗. Consider the substructure (Q, {0}, Q). Since TΓ and Tk are complete, the identification of this structure in K∗and L∗is a ∆-embedding. Let K′ be a countable elementary submodel of K∗in the Pas-language. By Theorem 6.3, we can extend this to a ∆-embedding of K into L∗. Let φ be any sentence in the language of valued fields. There is ψ a disjunction of ∆-sentences equivalent to φ. Then K | = φ ⇔K∗| = φ ⇔K′ | = ψ ⇔L∗| = ψ ⇔L∗| = φ ⇔L | = φ.
□ Corollary 6.12 Let U be an nonprinciple ultrafilter on the set of primes. Then Y Qp/U ≡ Y Fp( (T) )/U.
In particular, for any sentence in the language of valued fields Qp | = φ for all but finitely many primes p if and only if Fp( (T) ) | = φ for all but finitely many primes p.
Proof Q Qp/U and Q Fp( (T) )/U are henselian valued fields with value group Q Z/U and characteristic zero residue field. Hence they are elementarily equiv-alent.
If Qp | = φ for all but finitely many primes and D is an infinite set of primes where Fp( (T) ) | = ¬φ, let U be an ultrafilter on the primes such that D ∈U. Then, by the Fundamental Theorem of Ultraproducts Q Qp/U | = φ and Q Fp( (T) )/U | = ¬φ, a contradiction. The converse is similar.
□ Exercise 6.13 Show that if the Continuum Hypothesis is true then Q Qp/U ∼ = Q Fp( (T) )/U.
We will discuss applications of this in the next section.
Corollary 6.14 Suppose (K, Γ, k) is a valued field with angular component and TΓ and Tk have quantifier elimination, then every formula is equivalent to a quantifier free formula.
64 Proof Every ∆-formula is equivalent to a quantifier free formula.
□ Exercise 6.15 Let K ⊂L be henselian valued fields of characteristic zero.
Suppose ΓK ≺ΓL and kK ≺kL. Show that K ≺L.
We can generalize Corollary 5.17 to drop the assumption that our field is ordered and the valuation ring is convex.
Corollary 6.16 Let K be a henselian valued field with real closed residue field and divisible value group. Then K is real closed.
As in ACVF in equicharacteristic zero henselian valued fields the resiude field and value group are stably embedded and orthogonal.
Exercise 6.17 Let (K, Γ, k) be a henselian valued field with characteristic zero residue field. Any definable subset of Γm × kn is a finite union of rectangles A×B where A ⊆Γm is definable in the group language and B ⊂kn is definable in the ring language.
NIP Not all theories of henselian valued fields have NIP. For example the theory of Q Qp/U has the independence property since the pseudofinite field Q Fp/U has the independence property.
Exercise 6.18 [Duret] Show that the theory of any infinite pseudofinite field has the independence property. In particular, show that for any distinct a1, . . . , am there are bI for I ⊆{1, . . . , m} such that ai+bJ is a square if and only if i ∈J. [Recall that in an infinite pseduofinite field every absolutely irreducible variety has a point.] Indeed the theory of Q Qp/U is NTP2. In fact, failure of NIP in the residue field is the only obstruction to NIP. Delon proved that a Henselian valued field with characteristic zero residue field has NIP if and only if the theories of the residue field and the value group have NIP. But Gurevich and Schmitt showed that all theories of ordered abelian groups have NIP.
Theorem 6.19 (Delon) Henselian valued field with characteristic zero residue fields have NIP if and only if the theory of the residue field has NIP and the theory of value group has NIP.
Corollary 6.20 Henselian valued field with characteristic zero residue fields have NIP if and only if the theory of the residue field has NIP.
We will give a proof of Delon’s theorem from Simon . We will use an alternative characterization of the independence property (see 2.7).
Lemma 6.21 A formula φ(x, y) has the independence property if and only if, in a suitably saturated model, there is an indiscernible sequence (x0, x1, . . . ) and b such that φ(xi, b) holds if and only if i is even.
65 Lemma 6.22 Let (K, Γ, k) be a valued field with angular component, f(X) = a0 + a1X + · · · + adXd ∈K[X] and let x0, x1, . . . be a sequence of elements of K such that v(x0), v(x1), . . . is strictly increasing or strictly decreasing. There is r ≤d and t ∈N such that v(f(xi)) = v(arxr i ) < v(ajxj i) and ac(f(xi)) = ac(arxr i ) for all i ≥t and j ̸= r.
Proof Consider the cut v(xi) makes with respect to the finite set X = { v(aj)−v(ak) k−j : 0 ≤i < j ≤d}.
Since v(xi) is strictly increasing or strictly decreasing, there is an t such that for all v(xi) are not in X and realize the same cut over X for i ≥t.
Note that if v(aj)−v(ak) k−j < v(xi), then v(ajxj i) < v(akxk i ). Choose r such that v(arxr i ) is minimal, then r is unique and works for all i ≥t. In this case, v(f(xi)) = v(arxr i ) and ac(f(xi)) = ac(arxr i ) for i ≥t, as desired.
□ Lemma 6.23 Let (K, Γ, k) be an ℵ1-saturated valued field with angular com-ponent and let x0, x1, . . . be a sequence of indiscernibles in K. Then there are indiscernible sequences g0, g1, . . . of indiscernibles in Γ and b0, b1, . . . of indis-cernibles in k such that for any f ∈K[X] there is r and γ ∈Γ such that v(f(xi)) = γ + rgi and there is q ∈k[x] such that ac(f(xi)) = q(bi) for all large enough i.
Proof case 1 The sequence v(x0), v(x1), . . . is nonconstant.
We take gi = v(xi) and bi = ac(xi). Then by indiscernibility it is either strictly increasing or strictly decreasing and we can apply the previous lemma to conclude that v(f(xi)) = v(arxr i ) and ac(f(xi)) = ac(arxr i ) for large enough i. Thus the lemma is true if we take γ = v(ar) and q(X) = arXr.
From now on we assume that v(x0), v(x1), . . . is a constant sequence. Let yi = xi −x0. The sequence v(y0), v(y1), . . . is not strictly increasing. If it were, then v(xi −x1) = v((xi −x0) −(x1 −x0)) = v(yi −y1) = v(y1).
But then the sequence (v(xi −x1)) is constant, while the sequence v(xi −x0) is increasing, contradicting indiscernibility.
case 2 The sequence v(y1), v(y2), . . . is decreasing.
In this case we will take gi = v(yi+1), ai = ac(yi+1. Let f(X) ∈K[X].
There is h(X) ∈K[X] such that f(xi) = f(x0 + yi) = f(x0) + h(yi) for all i > 0. As in case 1, we can apply the previous lemma applied to the sequence y1, y2, . . . .
case 3 The sequences (v(yi)) and (ac(yi)) are constant.
Then v(x2 −x1) = v(y2 −y1) > v(y1) = v(y2) = v(x2 −x0) 66 Find xω ∈K such that x0, x1, . . . , xω is an indiscernible sequence of order type ω + 1. Let zi = xω −xi. By indiscernibility, v(z1), v(z2), . . . is an increasing sequence. Let gi = v(zi+1) and ai = ac(zi). For f(X) ∈K[X] as in case 2 there is h(X) ∈K[X] such that for i > 0 f(xi) = h(zi) using the lemmas we proceed as in the previous cases.
case 4 The sequence v(yi) is constant but the sequence (ac(yi)) is not.
In this case let gi = v(y0), a constant sequence, and let bi = ac(yi).
For any f(X) ∈K[X] we can find h(X) ∈K[X] such that f(x0 + Y ) = h(yi) = d X n=0 anY n.
Let A ⊂{0, . . . , d} be the set of n such that v(an) + ng0 is minimal.
Let q(X) = P n∈A ac(an)Xn. For sufficiently large i, q(ac(yi)) ̸= 0. But then v(f(xi)) = v d X n=0 anyn i !
= v X n∈A anyn i !
= v(an) + ng0 = v(an) + ngi and ac(f(xi)) = q(ac(yi)) where n is any fixed element of A and i is sufficiently large.
□ We are now ready to prove Delon’s Theorem. By the Pas quantifier elimi-nation and the basic facts about NIP from Lemma 4.21. it suffices to show that formulas of the following form have NIP.
1. f(x, y) = 0, f ∈K[X, Y] and x, y are variables in the home sort; 2. φ(x, t1(y), . . . , tm(y)) where φ is a formula in the language of ordered groups, y are variables from the home and value group sort and t1, . . . , tm are terms with values in the value group sort; 3. ψ(x, t1(y), . . . , tm(y)) where ψ is a formula in the language of rings , y are variables from the home and residue field sort and t1, . . . , tm are terms with values in the residue sort; 4. θ(v(f1(x, y)), . . . , v(fm(x, y)), z) where θ is a formula in the language or-dered groups x and y are variables in the home sort, f1, . . . , fm ∈Z[X, Y] and z are variables in the ordered group; 5. χ(ac(f1(x, y)), . . . , ac(fm(x, y)), z) where χ is a formula in the language rings x and y are variables in the home sort, f1, . . . , fm ∈Z[X, Y] and z are variables in the ring sort; Formulas of types 1, 2 and 3 are easily seen to by NIP. If the x variable is of degree d in f(x, y), then f(x, y) = 0 fails to shatter a set of size d + 2. Thus formulas of the first type are NIP. Formulas of the second and third type are NIP by our assumptions on the theories of the residue field and the value group.
67 Consider Θ(x, y, z) = θ(v(f1(x, y)), . . . , v(fm(x, y)), z) of type 4. If Θ has the independence property, then we can find a sequence of indiscernibles in K (x1, x2, . . . ) and b1, b2 such that Θ(xi, b1, b2) holds if and only if i is even.
By Lemma 6.23 there is are g0, g1, . . . an indiscernible sequence of elements in the value group such that for j = 1, . . . , n there are hj ∈Γ and rj ∈N such that v(fj(xi, b1) = hj + rjgj for sufficiently large i. Consider the formula Θ∗(v, h, b2) which is θ(h1 +r1v, . . . , hm +rmv, b2) where v is a variable over the value group. Since the theory of the value group has NIP, Θ∗(gi, h, b2) is either eventually true, or eventually false for large i, but Θ∗(gi, h, b2) is equivalent to Θ(xi, b1, b2) for large i. Thus Θ does not have the independence property.
The argument for formulas of type 5 is similar.
6.3 Artin’s Conjecture We say that a field K is a Cm-field if whenever f(X1, . . . , Xn) is a homogeneous polynomial of degree d where n > dm, then f has a nontrivial zero in K.
Exercise 6.24 Show that K is a Cm-field if and only if every homogeneous polynomial of degree dm + 1 has a nontrivial zero in K Tsen and Lang proved that if F is a finite field then F( (T) ) is a C2 field and Artin conjecture that each Qp is a C2-field. This is false.
Exercise 6.25 [Terjanian] Let p(X, Y, Z) = X2Y Z + XY 2Z + XY Z2 + X2Y 2 + X2Z2 −X4 −Y 4 −Z4 let q(X1, . . . , X9) = p(X1, X2, X3) + p(X4, X5, X6) + p(X7, X8, X9) and r(X1, . . . , X18) = q(X1, . . . , X9) + 4q(X10, . . . , X18).
a) Show that if (x, y, z) ∈Z3 are not all even, then p(x, y, z) = 3(mod 4).
b) Show that if (x1, . . . , x9) ∈Z9 are not all even , then q(x1, . . . , x9) ̸= 0(mod 4).
c) If x = (x1, . . . , x18) ∈Z18 2 and some xi is a unit, then v2(x) = 0 or 2.
d) Conclude that Artin’s conjecture fails for Q2 with n = 18 and d = 4.
Nevertheless, the Ax, Kochen, Erˇ sov transfer principle tell us is true for sufficiently large p.
Corollary 6.26 Fix d. There is a prime p0 such that for all primes p ≥p0 every homogenous polynomials of degree d in n > d2 variables has a nontrivial zero in Qp.
Proof The statement that every homogeneous polynomial of degree d in d2 +1 variables has a nontrivial zero is a first order sentence that is true in every Fp( (T) ) and hence true in Qp for p sufficiently large.
□ 68 The Tsen–Lang Theorem We will prove that F( (T) ) is C2 if F is finite.
Lemma 6.27 If F is a finite field with |F| = q and n < q −1, then X x∈F xn = 0 Proof Let a ∈F × with an ̸= 1. Since x 7→ax is a bijection, X xn = X (ax)n = an X xn.
Since an ̸= 1, P xn = 0.
□ Theorem 6.28 (Chevalley–Warning) Let F be a finite field of characteristic p and let f1, . . . , fm ∈F[X1, . . . , Xn] be polynomials of degrees d1, . . . , dm with n > P di. Then the number of zeros of f1 = · · · = fm in F is divisible by p.
In particular, if the polynomials f1, . . . , fm are homogeneous, there is a non-trivial zero in F.
Proof Let F have characteristic p and cardinality q. Let N be the number of zeros of f1 = · · · = fm = 0 in F n. Note that for all x ∈F n k Y i=1 (1 −fi(x)q−1) = ( 1 if f1(x) = · · · = fk(x) = 0 0 otherwise .
Thus the number of zeros of f is N = X x∈F n k Y i=1 (1 −fi(x)q−1) = X x∈F n X j∈J cjxj = X j∈J cj X x∈F n xj !
(mod p) where J = {j = (j1, . . . , jn) : P ji ≤(q −1) P di}.
Fix j = (j1, . . . , jn) ∈J. Note that, since n > P di, we must have some jb i < q −1. Then X x∈F n xj = n Y i=1 X x∈F xji Thus, by the lemma, P x∈F xjb i = 0 and N = 0(mod p).
□ We can combine this with Greenleaf’s Theorem 2.27.
Corollary 6.29 If f1, . . . , fm ∈Z[X1, . . . , Xn] where fi has degree di and n > P di, then for all but finitely many primes p, f1 = · · · = fm = 0 has a solution in Zp.
Lemma 6.30 Let F(T) be the field of rational functions over a finite field F.
Let f ∈F(T)[X1, . . . , Xn] be homogeneous of degree d2 < n. Then f has a nontrivial zero in F(T)n.
69 Proof Clearing denominators, we may assume f ∈F[T][X1, . . . , Xn]. We will look for a solution of the form (x1, . . . , xn) where for some suitably large s xi = yi,0 + yi,1T + · · · + yi,sT s.
Let r be the maximum of the degrees of the coefficients of f.
Choose s > (d(r + 1) −n)/n −d2. Then n(s + 1) > d(ds + r + 1) Then f(x1, . . . , xn) = f0(y) + f1(y)T + · · · + fds+r(y)T ds+r.
Since n(s + 1) > d(ds + r + 1), by Chevalley-Warning, there is a nontrivial zero y = (y1,0, . . . , yn,s) ∈F.
□ Corollary 6.31 Let f ∈F( (T) )[X1, . . . , Xn] be homogeneous of degree d with d2 < n and F is a finite field. Then f has a nontrivial zero in F( (T) ).
Proof We may assume f ∈F[T]. For k sufficiently large let f|k(X1, . . . , Xn) be the polynomial over F[T] obtained by truncating all the co-efficients of f to polynomials of degree at most k. By the lemma fk(X1, . . . , Xn) has a nontrivial zero ak ∈F(T)n. We may assume that v(ak,i) ≥0 for all i and some v(ak,i) = 0. Since the residue field is finite we see that F is compact so we can choose a Cauchy subsequence of the ak that converges to a nonzero element of Fn .
□ 70 7 The Theory of Qp 7.1 p-adically Closed Fields We next turn our attention to the theory of Qp. If K ≡Qp, then (v(G), +, < , 0, v(p)) ≡(Z, +, <, 0, 1). We know that the complete theory of (Z, +, <, 0, 1) is just Presburger arithmetic which is axiomatized by saying that we have an ordered abelian group with least positive element 1 such that for any x and n ≥2 there is a y such that x = ny or x = ny + 1 . . . or x = ny + n −1.
We have quantifier elimination in Presburger arithmetic once we add either equivalence relation x ≡n y for x = y(mod n) or predicates for the elements divisible by n, for all n ≥2.
Definition 7.1 We say that a valued field (K, v) is p-adically closed if K is henselian of characteristic zero, the residue field is Fp and the value group in a model of Presburger arithmetic and v(p) is the least positive element of the value group.
Lemma 7.2 Let K be p-adically closed, x ∈K and v(x) = gn + i where 0 ≤ i < n, then there is m ∈Z with 0 ≤v(m) < n and y ∈K such that x = myn.
Proof Suppose K is p-adically closed and v(x) = gn + i. Choose z such that v(z) = g, then v( x pign ) = 0.
There is 0 < r < p2v(n)+1 such that x pizn = r(mod p2v(n)+1) and p̸ | r. Let c = x rpizn . Then c = 1(mod p2v(n)+1). Consider f(X) = Xn −c, then v(f ′(1)) > 2v(n) and v(f ′(1)) = v(n). By Lemma 2.6 ii), there is y ∈F such that yn = c. Then x = rpi(yz)n and 0 ≤v(rpi) < n.
□ Lemma 7.3 Suppose F is a p-adically closed field, A ⊂F and E is the algebraic closure of Q(A) in F. Then E is p-adically closed.
Proof Since E is algebraically closed in F, E is henselian.
Clearly E has characteristic zero, kE = Fp and v(p) = 1.
So we need only show v(E) is a Z-group.
Let x ∈E.
There is y ∈F and m ∈Z such that x = myn and 0 ≤v(m) < n.
Since E is algebraically closed in F, y ∈E, but then v(x) = nv(y) + v(m) as desired.
□ We will show that the theory of p-adically closed fields has quantifier elim-ination in the Macintyre language LMac = {+, −, ·, |, P2, P3, . . . , 0, 1} where Pn is a predicate picking out the nth-powers. The symbol | is actually unnecessary as we can always define | in a quantifier free way using P2 as in Exercise 2.11.
We begin with some useful lemmas about nth-powers.
Lemma 7.4 Let K be henselian of characteristic zero. Let a ∈K× and γ = v(a) + 2v(n). Then a is an nth-power in K if and only every b ∈Bγ(a) is an nth-power in K.
Proof Suppose b ∈Bγ(a). Let c = b/a.
v(1 −c) = v(a −b) −v(a) > 2v(n).
71 Consider f(X) = Xn −c. Then v(f(1)) = v(1 −c) > 2v(n) and v(f ′(1)) = v(n).
Thus by Lemma 2.6 ii), there is u ∈K un = c. Then aun = b and a is an nth-power if and only if b is.
□ Corollary 7.5 In a henselian field of characteristic zero, the set of nonzero nth-powers is open.
Corollary 7.6 Suppose K is henselian of characteristic zero with residue field k of characteristic p where v(p) is the least positive element of the value group.
Suppose F ⊂E ⊆K, E/F is immediate and a ∈E. Then there is b ∈F such that v(a−b) > v(a)+2v(n) and for any such b we have that a ∈Kn if and only if b ∈Kn.
Proof Since F(a)/F is immediate, there is b0 ∈F such that v(a −b0) > v(a).
We can then find a b1 ∈F such that v(a −b1) > v(a −b0) ≥v(a) + v(p).
Continuing inductively, we can find b ∈F such that v(a −b) > v(a) + 2v(n).
By the lemma, a is an nth-power in K if and only any such b is.
□ Lemma 7.7 Suppose K is henselian of characteristic zero and residue field Fp and v(p) is the least positive element of the value group. Let F ⊂K and suppose g ∈v(K)\v(F), ng ∈v(F) Then there is b ∈F with v(b) = g such that bn ∈F.
Proof Let a ∈F and c ∈K such v(c) = g and v(a) = ng. Since K and F have the same residue field, without loss of generality we can choose a such that cn = a(1 + ϵ) where vϵ > 0. We can find 0 ≤m < p2v(n)+1 such that m = ϵ(mod p2v(n)+1). Then cn = a(1 + m)(1 + δ) where v(δ) > 2v(n). Since K is henselian, there is u ∈K such that un = 1 + δ. But then (c/u)n = a(1 + m) and v(c/u) = g.
□ Quantifier elimination will follow from the following embedding result.
Theorem 7.8 (Macintyre) Suppose (K, v) and (L, w) are p-adically closed fields where K is countable and L is ℵ1-saturated. Suppose A is a subring of K and f : A →L is an LMac-embedding. Then f extends to an LMac-embedding of K into L.
This will be proved by iterating the following lemmas. Throughout we as-sume that K and L satisfy the hypotheses of the theorem. If A ⊂K and f is an LMac-embedding, we will think of this as also defining a map on the value group by f(v(a)) = w(f(b)).
Lemma 7.9 Suppose A is a subring of K and f : A →K is and LMac-embedding, then we can extend f to F the fraction field of A.
72 Proof Since w(f(a)/f(b)) = w(f(a)) −w(f(b)) = f(v(a)) −f(v(b)) = f(v(a/b), the natural extension preserves divisibility. Since Pn(a/b) ⇔Pn(abn−1), the predicates Pn are preserved.
□ Lemma 7.10 Suppose F ⊂K and f : F →L is an LMac-embedding, then f extends to an LMac-embedding of F h into L Proof Let f also denote the unique extension to a valued field embedding of F h into F. Since F h/F is immediate, for all n and all a ∈F h there is a b ∈F such that v(b −a) > v(a) + 2v(n). Then v(f(a) −f(b)) > v(f(a)) + 2v(n) and Pn(a) ⇔Pn(b) ⇔Pn(f(b)) ⇔Pn(f(a).
Hence f is an LMac-embedigin □ Our next goal is to show that if we have an LMac-embedding of a subfield F of K into L, that it extends to the algebraic closure of F in K. The next lemma shows that if we can extend to a valued field embedding it will automatically be an LMac-embedding.
Lemma 7.11 If F ⊆K is algebraically closed in K then any valuation pre-serving embedding of F into L preserves the predicates Pn.
Proof Clearly if Pn(a), then a is an nth-power in K and, since F is algebraically closed in K there is b ∈F such that bn = a.
But then f(b)n = f(a) and Pn(f(a)).
Suppose Pn(f(a)). Suppose, for contradiction, that all of the nth-roots of f(a) are in L \ f(K).
Note that ΓK/ΓF is torsion free. Suppose not. Let n be minimal such that there is g ∈ΓK \ΓF such that ng ∈ΓF . By Lemma 7.7, we can find a ∈F with v(a) = ng such that a has an nth-root in K. Then a has an nth-root in F.
It follows that ΓL/Γf(F ) is also torsion free. To see this, note that if g ∈ΓF and n̸ | g there is 1 ≤i < n and b ∈F such that g = nv(b) + i.
Then f(g) = w(f(bn)) + i and n̸ | f(g).
By Exercise 2.4 F is henselian and hence f(F) is henselian and, by Theorem 5.14 has no proper algebraic immediate extensions.
Let b ∈L with bn = f(a). Then f(F)(b) is not an immediate extension of f(F). Since the residue field does not extend, the value group must extend.
Since the extension is algebraic, there is g ∈ΓL \ Γf(F ) such that mg ∈Γf(F ) for some m, but this contradicts that ΓL/Γf(F) is torsion free.
□ Lemma 7.12 Suppose F ⊆K is henselian and we have an LMac-embedding f : F →L. Let K0 be the algebraic closure of F in K. Then we can extend f to an LMac-embedding of K0 into K.
73 Proof By ℵ1-saturation it suffices to show that we can extend f to any E where F ⊂E ⊆K and E/F is a finite algebraic extension. Since F is henselian and unramified, E/F is not immediate. In particular ΓF ⊂ΓE ⊂QΓF . Thus ΓE/ΓF is finite abelian group. Suppose ΓE/ΓF = ⟨g1/F⟩⊕· · · ⊕⟨gm/F⟩ where ⟨gi/F⟩is cyclic over order ni. Then nigi ∈ΓF and ni is minimal with this property. By Lemma 7.7, there are a1, . . . , am ∈E such that v(ai) = gi and ani i ∈F. Since F is henselian, so is F(a1, . . . , am). But E/F(a1, . . . , am) is immediate and, hence, F(a1, . . . , am) = E.
Since f is an LMac-embedding, there are b1, . . . , bm ∈L such that bni i = f(ani i ). We claim that we can extend f to a valuation preserving embedding of E into L with ai 7→bi.
We argue this in detail in the case m = 1. Suppose a ∈E, v(a) = g, n is minimal such that ng ∈ΓF and an ∈F. Suppose x = cnan−1 + . . . c1a + c0 ∈ E(a). By the minimality of n, v(ci) + iv(a) ̸= v(cj) + jv(a) for any i < j < n.
Thus Xn −an is irreducible over F and v(x) = min v(ci)+iv(a). It follows that Xn−f(an) is irreducible over f(F) and that if b ∈L such that bn = f(am), then the extension of f to F(a) obtained by sending a to b is valuation preserving.
The general case is done similarly by induction.
□ The full embedding result will follow from the next lemma.
Lemma 7.13 Suppose F ⊂F1 ⊆K f : F →K is a valued field embedding. F and F1 are algebraically closed in K and F1/F is transcendence degree 1. Then we can extend f to F1.
Proof There are two cases to consider.
case 1 F1/F is immediate.
Let a ∈F1 \ F. We can find a pseudocauchy sequence of transcendental type (aα) ⇝a such that (aα) has no pseudolimit in F. We can find b ∈L a pseudolimit if (f(aα)) and can extend f to a valued field embedding of F(a) into L by sending a to b. We can further extend f to a valued field embedding of F(a)h into L. But F1/F(a) is an immediate algebraic extension, thus F1 = F(a)h and we have the desired embedding.
case 2 F1/F is not immediate.
By ℵ1-saturation, it suffices to show that we can extend the embedding to any F ⊂E ⊆F1 where E/F is finitely generated.
Then ΓE/ΓF is finitely generated and torsion free, since E/F has transcendence degree one we must have ΓE = ΓF ⊕Zv(a) for some a ∈E transcendental over F. We can find b ∈L transcendental over f(F) such that the type w(b) realizes over v(ΓF ) is the image of the type v(a) realizes over ΓF . We claim that sending a 7→b gives a valued field embedding of F(a) into L. Suppose x ∈F[a] and x = P ci ai where each ci ∈F. By choice of a, all v(ci) + iv(a) are distinct. Choose j such that v(cj) + jv(a) is minimal. Then v(x) = v(cj)+jv(a) and, by choice of b, w(f(cj))+jw(b) 74 is minimal and w(f(x)) = f(v(x)), as desired.
There is a unique valuation preserving extension of f from F(a)h into L. Since E/F(a) is an immediate extension, F(a)h ⊆E. Thus we can extend f to a valuation preserving extension of E into L. By ℵ1-saturation, we can extend the embedding to F1 □ Corollary 7.14 (Macintyre) The theory of p-adically closed fields is admits quantifier elimination.
Lemma 7.15 Suppose K is p-adically closed and x ∈Q then x is an nth-power in K if and only if x is an nth-power in Qp.
Proof The algebraic closure of Q in K is an immediate extension of Q Thus the henselization Qh is the algebraic closure of Q in K.
My uniqueness of henselization, the algebraic closure of Q in any two p-adically closed field are isomorphic. Thus Pn(K) ∩Q does not depend on K.
□ Corollary 7.16 The theory of p-adically closed fields is complete.
Proof By the lemma the rational numbers with Pn interpreted as Pn(Qp) ∩Q is a substructure of any p-adically closed field. Thus, by quantifier elimination, the theory is complete.
□ Exercise 7.17 a) Show f(x) = 0 if and only if P2(pf(x)2).
b) Show that if p ̸= 2, f(x)|g(x) if and only if P2(f(x)2 + pg(x)2).
c) Give a version of b) for p = 2.
d) Conclude that every definable set is a Boolean combination of sets of the form Pk(f(x)).
7.2 Consequences of Quantifier Elimination Throughout this section K will be a p-adically closed field.
Lemma 7.18 The set of nonzero nth-powers in K is clopen.
Proof By Lemma 7.4 if a is an nth-power, then B2v(n)+v(a)(a) is contained in the nth powers. Thus Pn \ {0} is open. If x is not in Pn, then x ∈a(Pn \ {0} for some non nth-power a. Thus the set of non nth-powers is open.
□ Corollary 7.19 If X ⊆K is definable and infinite, then X has non-empty interior.
Proof Let X be definable. By quantifier elimination X is the union of finitely many sets of the form Y = {x ∈K : f1(x) = · · · = fm(x) = 0 ∧g(x) ̸= 0 ∧ n ^ i=1 (Pki(hi(x)) ∧hi(x) ̸= 0) 75 for some polynomials fi, g, hj ∈kp[X]. Note that we do not need conjuncts of the form ¬Pk since ¬Pk(x) ⇔ m _ i=1 Pk(lix) for appropriately chosen m and l1, . . . , lm ∈K. If Y is infinite, then all of the fi must be trivial, in which case Y is open.
□ Exercise 7.20 More generally, suppose X ⊆Km p is definable with non-empty interior. Show that if S1, . . . , Sm is a partition of X into definable sets, then some Si has non-empty interior.
As in Exercise 4.18, we can show that if K is a p-adically closed field and A ⊆Km+n is definable, then there is an N such that Ax is finite if and only if |Ax| ≤N.
Exercise 7.21 Let U ⊆Qp be open and let f : U →Qp be definable.
a) Show that there is a ∈U such that f is continuous at a. [Hint: This is similar to the proof in 3.3.24 and uses the local compactness of Qp.] b) Show that {x : f is discontinuous at x} is finite.
c) Prove that the same is true over any p-adically closed field K.
Exercise 7.22 Let U ⊆Kn and let f : U →K be definable. Then there is F ∈Qp[X, Y ] such that F(a, f(a)) = 0 for all a ∈U, i.e., f is algebraic.
There is a p-adic version of the Implicit Function Theorem (see for example §II). Once we know f is algebraic and continuous except at finitely many points we can conclude it is analytic except at finitely many points.
Skolem functions We will show that p-adically closed fields have definable Skolem functions. We start with a partial result due to Denef for functions with finite fibers.
Theorem 7.23 (Denef ) Let K be p-adicaly closed. Suppose A ⊆Km+1 is C-definable, B = {x ∈Km : ∃y (x, y) ∈A} and for all x ∈B, |{y ∈K : (x, y) ∈ A}| ≤N. Then there is an C-definable f : B →K such that (x, f(x)) ∈A for all x ∈B.
Proof We prove this by induction on N. The result is clear if N = 1. Assume N > 1. For x ∈B, let Ax = {y : (x, y) ∈A} Without loss of generality, we may assume that |Ax| = N for all x. Replace A by {(x, y) ∈A : v(y) is minimal in {v(z) : z ∈Ax}}.
Then using induction we may, without loss of generality assume that |Ax| = N and v(y1) = v(y2) whenever x ∈B and y1, y2 ∈Ax.
Let k = φ(pv(N)+1) where φ is Euler’s phi-function.
claim For all x ∈B, if Ax = {y1, . . . , yN} then not all the yi are in the same coset of kth-powers.
76 Suppose they are.
Fix z such that v(z) = v(y1) = · · · = v(yN) and let yi = zy′ i where p̸ | y′ i. Then all of the y′ i are in the same coset of kth-powers..
Suppose p̸ | y, z and y = zak. By Euler’s theorem ak = 1mod pv(N)+1. Thus y and z are congruent mod pv(N)+1. Hence there is a c such that p̸ | c and y′ i = cmod pv(N)+1 for all i/ But P y′ i = 0. Thus Nc = 0(mod pv(N)+1), a contradiction.
Fix any ordering of the cosets of kth-powers. We can assume without loss of generality that for all (x, y) ∈A, y is in the minimal coset of kth-powers represented in Ax. We are then done by induction.
□ Note that the Skolem function defined in Denef’s proof are invariant, i.e., if Ax = Az then f(x) = f(z).
We next show that the restriction to finite fibers in unnecessary.
Theorem 7.24 (van den Dries ) p-adically closed fields have definable Skolem functions.
Proof Let φ(x, y) be a formula with parameters from A. We want to show there is an A-definable function f such that if a ∈Km and ∃y φ(a, y), then φ(a, f(a)).
Consider the type Γ(v) = {∃y φ(v, y), ¬φ(v, f(v)) : f is an A-definable function}.
If Γ is inconsistent, then there are finitely many definable functions f1, . . . , fn such that {∃y φ(v, y), ¬φ(v, f1(v)), . . . , ¬φ(v, fn(v))} is inconsistent. Define F(a) = ( 0 ¬∃y φ(a, y) fi(a) i is least such that φ(a, fi(a)) .
Then F is the desired definable Skolem function.
Suppose for contradiction that Γ is consistent. Let a realize Γ in F p-adically closed. Let E be the algebraic closure of Q(A, a) in E. Then E is p-adically closed and, by model completeness E) ≺F. Thus there is b ∈E such that φ(a, b). There is f ∈Q(A)[X, Y ] such that f(a, Y ) is nontrivial and f(a, b) = 0.
Let ψ(x, y) be φ(x, y) ∧f(x, y) = 0 ∧∃z f(x, z) ̸= 0.
Then ψ(a, b) and {y : ψ(a, y)} is finite for all y. By Denef’s theorem, there is a A-definable function g such that if ∃y ψ(x, y) then ψ(x, g(x)). Thus ψ(a, g(a)), contradicting that a realizes Γ.
□ Definition 7.25 Let F be a valued field. We say that K/F is a p-adic closure of F, if there for any p-adically closed L/F there is a unique valued field embedding of K into L fixining F pointwise.
77 Exercise 7.26 Suppose F is a valued field that is a substructure of a p-adically closed field. Show that F has a p-adic closure K and the there are no automorphisms of K/F. We say K/F is rigid.
In fact, van den Dries’ result preceded Denef’s. He proved the following more general result.
Exercise 7.27 Suppose T has quantifier elimination. Then T has definable Skolem functions if and only every model M of T∀has an extension N that is algebraic and rigid over M.
In real closed fields we have invariant definable Skolem functions, i.e., if A ⊂Kn+m is definable there is a definable Skolem function f such that if Ax = Ay, then f(x) = f(y). This is impossible in Qp.
Exercise 7.28 Let A = {(x, y) ∈Q2 p : v(x) = v(y)}. Show that there is no invariant definable Skolem function.
Exercise 7.29 [Definable Curve Selection] Let A ⊆Qn p be definable. Let a be in the closure of A but not in A. Then there for any ϵ > 0 there is a definable f : Bϵ(0) →A such that f(0) = a and for x ̸= 0, f(x) ∈A and v(f(x)) > f(x) Dimension As a topological space there can be no good notion of dimension in Qp.
Exercise 7.30 Show that Qp and Q2 p are homeomorphic.
Nevertheless, there is a good notion of dimension that works for definable sets and maps.
We begin with an relatively approach to dimension due to van den Dries that works in several theories of fields.
Definition 7.31 Let L be a language with constant symbols C and let T be an L-theory of fields. We say that T is algebraically bounded if for any formula φ(x, y) there are polynomials f1, . . . , fm ∈Z[C][X, Y ] such that if K | = T, a, b ∈K, {y ∈K : φ(a, y)} is finite and φ(a, b), then fi(a, b) = 0 for some i, where fi(a, Y ) is not identically zero.
Exercise 7.32 Use quantifier elimination to show that algebraically closed fields, real closed fields, algebraically closed valued fields and p-adically closed fields are algebraically bounded.
Definition 7.33 Suppose A ⊆Km is definable, say φ(v) is a formula with parameters from K defining A. We define dim A, the dimension of A, to be the largest l ≤m such that there is K ≺L and a = (a1, . . . , am) ∈L with L | = φ(a) and td(K(a)/K) = l, where td(L/K) denotes the transcendence degree of L/K.
Exercise 7.34 Show that this definition agrees with the usual notions of di-mension in algebraically closed fields and real closed fields.
78 Exercise 7.35 [van den Dries] Let T be an algebraically bounded theory and K | = T. Our notion of dimension has the following properties. Let A and B be definable sets in Km for some m.
a) Show dim A = 0 if and only if A is finite; b) Show dim (A ∪B) = max(dim A, dim B); c) Show that if f is a definable function, then dim f(A) ≤dim A; d) Show A ⊆Km+n, then {a ∈Km : dim Aa = i} is definable for each i ≤n.
Exercise 7.36 Let A ⊆Km+n. For i ≤n let Bi = {a ∈Km : dim Aa = i}.
Show that dim A = max(i + dim Bi).
Exercise 7.37 a) Suppose U ⊆Qp is open. Show that dim U = m.
b) Suppose A ⊆Qm p is definable, then dim A is the largest l such that there is a projection from π : Qm p →Ql p such that π(A) has nonempty interior.
Exercise 7.38 Use quantifier elimination to show that if A ⊆Qm p is definable and dim A < m then there is a nonzero polynomial f ∈Qp[X1, . . . , Xm] such that A is contained in the hypersurface p(x) = 0.
In o-minimal expansions of real closed fields there is a notion of Euler char-acteristic for definable sets. Basically a point has Euler characteristic 1, an open cell in Kn has Euler characteristic (−1)n and if we partition a definable set into cells, then the Euler characteristic is the sum of the Euler characteristics of the cell. van den Dries showed the notion is independent of the partition chosen and that two definable sets are in definable bijection if and only if they have the same dimension and Euler characteristic.
The next exercises based on results of Cluckers and Haskell tells that there is no good definably invariant notion of Euler characteristic in Qp. Fix p ̸= 2–though similar results can be proved for p = 2. Let Z∗ p denote Zp \ 0, let P2 be the nonzero squares in Zp, let Z1 p be the elements of Zp with angular component 1 and let P (1) 2 denote P2 ∩Z(1) p . Note that Z∗ p = p−1 [ m=1 mZ(1) p .
Let X ⊔Y denote the disjoint union of X and Y . Say X ∼Y if there is a definable bijection between X and Y Exercise 7.39 a) Show that P2 ⊔P2 ∼Z∗ p. [Hint: There is a definable Skolem function f : P2 →Z∗ p such that f(x)2 = x.] b) Show that P2 ⊔P2 ⊔P2 ⊔P2 ∼Z∗ p. [Hint: Recall that P2 is an index 4 subgroup of Z2 p.
c) Conclude Z∗ p ⊔Z∗ p ∼Z∗ p.
Exercise 7.40 a) Z(1) p is definable. [Hint: First show that {xp−1 : x ∈Z∗ p} = {x : ac(x) = 1 ∧(p −1)|vp(x)}.] 79 b) Show that Z(1) p = P (1) 2 ∪pP (1) 2 .
Exercise 7.41 Show Zp ⊔Z(1) p ∼Z(1) p . [Hint: send x ∈Zp to 1 + px and send x ∈Z(1) p to px.] Definition 7.42 Let M be any structure. Let D(M) be the set of all definable subsets of M n for n ≥1. Let F be the free abelian group with generators ⌊X⌋= {Y ∈DM) : X ∼Y } for X ∈D(M) and let R be the subgroup generated by relations ⌊X ∪Y ⌋− ⌊X⌋−⌊Y ⌋+⌊X ∩Y ⌋. The Grothendieck group of M is the quotient F/E. We let [X] = ⌊X⌋/E. There is a natural multiplication induced by [X][Y ] = [X × Y ] making it a ring which we call the Grothendieck ring and denote by K0(M).
Corollary 7.43 K0(Qp) is trivial.
Proof By Exercise 7.39 [Z∗ p] = [Z∗ p] + [Z∗ p].
Thus [Zp]∗= 0. By Exercise 7.41, [Zp] + [Z(1) p ] = [Z(1) p ].
Thus [Zp] = 0. It follows that [{0}] = 0. But then for any set X ∈D(M) [X] = [X × {0}] = [X][{0}] = 0.
□ This answered a question Denef asked at a meeting in 1999. At the same meeting B´ elair asked if Zp ∼Z∗ p. The next Exercise shows the answer is yes.
Exercise 7.44 a) Define f1 : p2Z∗ p ⊔(1 + p2Z∗ p) →(1 + p2Z∗ p) by f1(y) = 1 + p2(mx2) for y = 1 + pmx, x ∈Z(1) p , 1 ≤m < p 1 + p3mx2 for y = 1 + p2mx, x ∈Z(1) p , 1 ≤m < p . Show that f1 is a bijection.
b) Define f2 : pZp ⊔(p + p2Z(1) p →p + p2Z(1) p by f2(x) = ( p + p2(1 + px) for x ∈Zp p + p3x for x ∈Z(1) p .
Show that f2 is a bijection.
c) Let W = (1 + p2Z∗ p) ⊔p2Zp ⊔(p + p2Z(1) p ). Define f : W →W \ {0} by f(x) = ( f −1 1 (x) for x ∈1 + p2Z∗ p f2(x) for x ∈p2Zp ⊔(p + p2Z(1) p ) .
80 Show that f is a bijection.
d) Extend f to a definable bijection between Zp and Z∗ p.
This is the tip of the iceberg.
Theorem 7.45 (Cluckers ) Two infinite subsets of Qp are in definable bi-jection if and only if they have the same dimension.
Cell decomposition Lemma 7.46 If U ⊆Qm p is open definable and f : U →Qp is definable, then {x : f is discontinuous at x} has dimension at most m −1. Moreover, there is a definable open V ⊆U such that f|V is analytic and dim (U \ V ) < m.
Proof We first proof that if U is open, then there is x ∈U such that f is continuous at x. If there is an open U1 ⊂U such that f|U1 is constant, then we are done so we assume that there is no such set.
Let B0 be a closed ball in U. Given Bn open, let W be the image of Bn.
Then, by assumptions on f dim f −1(w) has dimension at most m −1 for all w ∈W. If there are only finitely many fibers of dimension m−1, then dim Bn ≤ m −1. So {w : dim f −1(w) = m −1} in infinite, and hence has interior. We can find Jn ⊂W0 open of radius at most 1/pn. Then {x ∈Bn : f(x) ∈Jn} has dimension m and thus contains a closed ball Bn+1. Since Qp is locally compact, there is x ∈T Bn and, by construction, f is continuous at s.
Since {x ∈U : f is discontinuous at x} has no interior it must have dimension at most m −1. We argued before that there is a non-zero polynomial F such that F(x, f(x)) = 0. Except for a set of dimension at most m −1 at each x there is an open V ⊂U such that x ∈V and there is a polynomial F(X, Y ) such that on V : f is continuous, F(x, f(x)) = 0 and ∂F ∂Y (x, f(x)) ̸= 0. Then, by the Implicit Function Theorem, f is analytic on V .
□ We can now prove a cell decomposition theorem due to Scowcroft and van den Dries .
Theorem 7.47 Let A ⊆Qm p and f : A →Qp be definable. There is a partition of A into definable sets U, B1, . . . , Bn such that U is open, f|U is analytic, dim Bi = ki < m, and there is a projection πi : Qm p →Qki p such that πi|Bi is a diffeomorphism and f ◦π−1|πi(Bi) is analytic.
Proof We call the above statement Φm and prove this by induction on m.
From earlier arguments it is easy to see that Φ1 holds.
We will also prove the following intermediate claim which we call Ψm. If g1, . . . , gs ∈Qp[X1, . . . , Xm] are nonzero polynomials and V = {x ∈Qm p : g1(x) = · · · = gs(x) = 0}, then V can be partitioned into finitely many pieces each of which is analytically homeomorphic via a projection to an open set in some Qk p with k < m. Note that Ψ1 is trivially true.
81 We will show that from Φi and Ψi for i ≤m we can prove Ψm+1 and then show that from Φ1, . . . , Φm−1 and Ψ1, . . . , Ψm+1 we can prove Φm+1.
Φ1, . . . Φm, Ψ1, . . . Ψm ⇒Ψm+1 Let g1, . . . , gs ∈Qp[X1, . . . , Xm, Y ] and let V = {(x, y) ∈Qm p : g1(x, y) = · · · = gm(x, y) = 0}.
Suppose gi(X, Y ) = di X j=0 hi,j(X)Y j where hi,j ∈Qp[X]. Let V0 = {x ∈Qm p : ^ i,j hi,j(x) = 0.} Then V0 × Qp ⊆V and there is a bound N such that if x ̸∈V0, then |{y : (x, y) ∈V }| ≤N is finite. This allows us to partition V = X1 ∪· · · ∪XN ∪X∞ where for i ≤N, Xi = {(x, y) ∈V : there are exactly i distinct z ∈Qp with (x, z) ∈V }. and X∞= V0 × Qp. We deal with each Xi separately.
X∞: We can apply Ψm to V0 to partition it into finitely many sets A0, . . . , Am where each Ai is analytically isomorphic to an open set in sum Qki p where ki < m.
Let Bi = Ai × Qp. This gives the desired decomposition of X∞= V0 × Qp.
Xk: Let C = {x ∈Qm p : |{z ∈Qp : (x, z) ∈V }| = k}.
We can find definable Skolem functions f1, . . . , fk : C →Qp such that Xk = {(x, fi(x)) : x ∈C, i = 1, . . . , k}.
By induction we can partition C into definable sets D0, . . . , Ds such that D0 is open (possibly empty) and all of the fi are analytic on D0 and otherwise Dj is analytically isomorphic via a projection πj to an open subset of Qrj p for rj < m and each fj ◦π−1 j |πj(Dj) is analytic. Then we can partition Xk into the union of the graphs of the fi on C and the Djs and apply induction.
Φ1, . . . Φm, Ψ1, . . . Ψm ⇒Φm+1 By the previous lemma, we can find U ⊆Qm+1 p open such that f|U is analytic and dim (A\U) < m. Since A\U has no interior, there is g ∈Qp[X1, . . . , Xm+1] such that A \ U is contained in the hypersurface V given by g(X) = 0. Apply Ψm to V to obtain a partition C1, . . . , Cs where for each j, there is a projection πj that is an analytic isomorphism to an open set in Qkj p . Let Dj = πj((A \ U) ∩Cj). Using Φkj we can definably partition Dj into finitely many nice pieces, then we lift these using π−1 j .
□ We will later state a different cell decomposition theorem due to Denef.
82 7.3 Rationality of Poincar´ e Series Fix f1, . . . , fr ∈Qp[X1, . . . , Xn]. Let Nk = |{y ∈Z/pkZ : ∃x ∈Zn p f1(x) = · · · = fr(x) = 0 ∧ ^ xi = yi(mod pk)}.9 We will consider the Poincar´ e series P(T) = ∞ X k=0 NkT k.
We could also consider e Nk = |{y ∈Z/pk : fi(y) = 0(mod pk)}, i = 1, . . . , r} and e P(T) = P∞ k=0 e NkT k.
Igusa , (for r = 1) and Meuser (for general r), proved that e P(T) is a rational function of T. Denef answered a question of Serre and Oesterl´ e by proving the rationality of P(T).
Theorem 7.48 (Denef ) P(T) is a rational function of T.
Igusa’s proof used resolution of singularities to simply certain p-adic inte-grals. Denef’s gave two proofs, the first also using resolution of singularities but the second used quantifier elimination to avoid resolution of singularities.
p-adic integration The p-adics under addition are a locally compact group and thus come equipped with a Haar measure µ.. Let B be the σ-algebra generated by the compact subsets of Qp. There is a unique σ-additive measure µ : B →R such that: i) µ(Zp) = 1; ii) (translation invariance) µ(a + A) = µ(A) for a ∈Qp, A ∈B; iii) for every A ∈B and ϵ > 0 there is an open set U and a closed set F such that F ⊆X ⊆U and µ(U \ F) < ϵ.
Exercise 7.49 µ({a}) = 0 for all a ∈Qp.
Let m be the maximal ideal. Then m ∪(1 + m) ∪· · · ∪((p −1) + m) = Zp.
Thus by additivity and translation invariance µ(m) = 1/p.
Exercise 7.50 Show that µ({x : v(x −a) ≥r}) = p−r.
Example 7.51 Let A be the set of squares in Zp where p ̸= 2.
9This is a little unclear if k = 0, in which case we mean that N0 = 1 if f1 = · · · = fm = 0 has a zero in Zn p and otherwise N0 = 0.
83 Let Ak = {x ∈A : v(x) = 2k}. Then A = {0} ∪S Ak and µ(A) = ∞ X k=0 µ(Ak).
If x ∈Ak if and only if x = p2ky where v(y) = 0 and res(y) is a square in Fp.
Since there are p−1 2 squares in Fp we can find z1, . . . , z p−1 2 ∈Zp such that Ak is the disjoint union B1 ∪· · · ∪B p−1 2 where Bi = {x −zi : vp(x) ≥2k + 1}.
We have µ(Bi) = p−2k−1. Thus µ(A) = ∞ X k=0 p −1 2 p−2k−1 = p −1 2p ∞ X k=0 p−2k = p −1 2p 1 1 −p−2 = p 2(1 + p).
Exercise 7.52 Calculate the Haar measure of the set of squares when p = 2.
There is a Haar measure µm on Zm p . This is just the usual product measure, and we will usually write µ rather than µm.
Suppose A ∈B and f : A →R is a B-measurable function, we can define the integral Z A f dµ.
We give two illustrative examples.
Example 7.53 Suppose p ̸= 2. Let A be the set of squares in Zp and let f(x) = |xs|p.
Let Ak = {x ∈Ak : v(x) = 2k}. Then Z A |xs|p dµ = ∞ X k=0 Z Ak |xs|p dµ = ∞ X k=0 Z Ak p−2sk dµ = ∞ X k=0 p−2skµ(Ak).
84 We saw above that µ(Ak) = p−1 2 p−2k−1. Thus Z A |xs|p dµ = p −1 2p ∞ X k=0 (p−2s−2)k = p −1 2p 1 1 −p−2s−2 Exercise 7.54 Calculate R A |xs| dµ when p = 2.
Example 7.55 Suppose p = 3(mod 4). Let f(x) = |x + 1|p and let A again by the squares in Zp.
Since p = 3(mod 4), -1 is a square in Fp and hence in Zp. Let B = {x ∈Zp : v(x + 1)}. Then every y ∈B is a square. If we partition A into B and A \ B, then Z A |x + 1|p dµ = Z B |x + 1|p dµ + Z A\B |x + 1|p dµ.
But on A \ B, |x + 1|p = 1. Hence Z A\B |x + 1|p dµ = Z A\B 1 dµ = µ(A) −µ(B) = p 2(1 + p) −1 p.
Partition B = {−1} ∪B1 ∪B2 ∪. . . where Bi = {x : v(x + 1) = i} Then Z B |x + 1|p dµ = ∞ X k=1 Z Bi |x + 1|p dµ = ∞ X k=1 Z Bi p−k dµ = ∞ X k=1 p−kµ(Bi) = ∞ X k=1 p−k 1 pk − 1 pk+1 = p −1 p3 ∞ X k=0 p−2k = p −1 p3(1 −p−2)2 Thus Z A |1 + x|p dµ = p −1 p3(1 −p−2)2 + p 2(1 + p) −1 p.
The next lemma is the link between integration and Poincar´ e series. Let f1, . . . , fr ∈Zp[X], where X = (X1, . . . , Xn) and let P be the associated Poincar´ e series. Let D = {(x, y) ∈Zn+1 p : ∃z ∈Zn p f1(z) = · · · = fr(z) = 0 ∧ ^ v(xi −zi) ≥v(y)} 85 and for s ∈R, s > 0, define I(s) = Z D |y|s dµ.
Lemma 7.56 I(s) = p−1 p P(p−n−1p−s).
Proof Let Dk = {(x, y) ∈D : v(y) = k}. Then I(s) = ∞ X k=0 Z Dk |y|s dµ = ∞ X k=0 Z Dk p−sk dµ = ∞ X k=0 p−skµ(Dk) For each z(mod pk) with f1(z) = · · · = fr(z) = 0.
µ({x : z = x(mod pk)} = p−nk and µ({y : v(y) = k} = p −1 pk+1 .
Thus µ(Dk) = Nk p −1 p p−nk−k, as for each of the Nk zeros mod pk we can find a ball (in m-space) of measure p−mk. Thus I(s) = p −1 p ∞ X k=0 Nk(p−s−n−1)k = p −1 p P(p−s−n−1).
□ We will prove that there is a rational function Q(T) such that I(s) = Q(p−s).
Letting Y = p−s we have Q(Y ) = p −1 p P(p−n−1Y ).
Then letting T = p−n−1Y P(T) = p p −1Q(pn+1T).
Hence P(T) is a rational function.
Denef proved the following general rationality theorem.
86 Theorem 7.57 (Denef) Suppose A ⊆Qm p is definable and contained in a compact set and h : A →Qp is a definable function. Suppose natural number M and v(h(x)) is either divisible by M or +∞for all x ∈A. Then ZA(s) = Z A |h(x)|s/M p dµ is a rational function in p−s for s ∈(0, +∞).
Denef’s Cell Decomposition The proof of Theorem 7.57 needs an analysis of definable functions from Qm p to the value group and a refined cell decomposition/preparation theorem.
Definition 7.58 Suppose A ⊆Qm p is definable.
We say that a defiinable θ : A →Z ∪{+∞} is simple if there is a finite partition of A into definable sets such that for each set B in the partition, there is an integer M and f, g ∈ Qp[X1, . . . , Xm] such that θ(x) = 1 M (v(f(x)) −v(g(x))) on B.
Lemma 7.59 Suppose A ⊆Qm+1 p is definable, B = {x ∈Qm p : ∃y (x, y) ∈A and for all x ∈B v is constant on Ax = {y : (x, y) ∈A}. Let θ : B →Z∪{+∞} by the function where θ(x) = v(y) for all (x, y) ∈A. Then θ is simple.
Proof Without loss of generality, assume that if (x, y) ∈A, then y ̸= 0. If not Z = {(x, y) ∈A : y = 0}, then θ|Z is constant and replace A by A \ Z.
Since p-adically closed fields, have definable Skolem functions there is a definable f : B →Qp such that (x, f(x)) ∈A for all x ∈B. By Exercise 7.22, there is a polynomial F(X, Y ) such that F(x, f(x)) = 0 for all x ∈A and F(x, Y ) is not identically zero. Let F(X, Y ) = d X i=0 gi(X)Y i.
Since F(x, f(x)) = 0 for each x ∈A, there is an i < j such that v(gi(x))+iv(y) = vj(gj(X)) + jv(y). For i < j ≤d, let Ai,j = {(x, y) ∈A : (i, j) is minimal such that v(y) = v(gi(x)) −v(gj)(x) j −i }.
Then (Ai,j : i < j ≤d) is a partition of A showing that θ is simple.
□ Denef proved the following cell decomposition/preparation theorem.
We refer the reader to §7 for the proof.
Theorem 7.60 Suppose f1, . . . , fr ∈Qp[X, Y ], where X = (X1, . . . , Xm) and N > 1, then Qm+1 p can be partitioned into finitely many definable sets of the form A = {(x, y) ∈Qm+1 p : x ∈C, v(a1(x)) □1 v(y −c(x)) □2 v(a2(x))} 87 where C ⊆Qm p is definable, a1, a2 and c are definable functions, □i is either <, ≤or no restriction, and there is are definable function hj : C →Qp for j = 1, . . . , r such that fj(x, y) = uj(x, t)Nhi(x)(y −c(x))vj function where uj(x, y) is a unit.
In the following proofs we will be interested in knowing of the value of fj(x, y) or if fj(x, y) is an N th-power. Since uj(x, y)N is always a unit and an N th-power, we have reduced the question to understanding hj(x)(y −cx)vj.
The following lemma is the key step in Denef’s proof.
Lemma 7.61 Suppose A ⊆Qm p is definable and contained in a compact set and h : A →Qp is a definable function such that for some natural number M v(h(x)) is either divisible by M or +∞for all x ∈A. Then ZA(s) = Z A |h(x)|s/M p dµ is a linear combination of series of the form X (k1,...,km)∈L ki=λi(mod Ni) p−(q1k1+···+qmkm)s−k1−···−km where k1, . . . , km, λi ∈Z, Ni ∈N, q1, . . . , qm ∈Q and L is defined by a system of linear inequalities with rational coefficients.
Any function of this form is rational in p−s Proof (Sketch) The result is trivial if m = 0. We write points in Qm+1 p as (x, y).
Since R A∪B = R A + R B − R A∩B, we can always take Boolean combinations.
We first apply Lemma 7.59 to partition A. Without loss of generality, we may assume |h(x, y)|1/M p = g1(x, y) g2(x, y) 1 M′ p where g1, g2 ∈Qp[X, Y ] and M ′ > 0. Further, by quantifier elimination and Exercise 7.17 we may assume that A is defined by a conjunction ^ j=1,...,r ±Pnj(fj(x, y)).
We apply Theorem 7.60 to the functions f1, . . . , fr, g1 and g2 where N = Q nj. So, by further partitioning, we may assume A is defined by x ∈C ∧v(a1(x)) □1 v(y −c(x)) □2 v(a2(x)) 88 and on A |h(x, y)|1/M p = |h0(x)|1/M ′ p |p|y −c(x)|v/M ′ p and fj(x, y) is an nth j -power if and only if hj(x)(y −c(x))vj is.
We can further refine our partition so that the coset of N th-powers of each hj(x) and (y−c(x) is fixed on each set in the partition. Without loss of generality they are constant on A. Let z = y −c(x). Suppose z ∈λ(mod P × N ). Then Z A |h|s/M p dy dx = Z A |h(x, y)|s/M ′ p dy dx = Z C |h0(x)|s/M ′ p Z v(a1(x))□1v(z)□2v(a2(x)) z=λ (mod P × N ) |z|sv/M ′ p dz dx = Z C |h0(x)|s/M ′ p X v(a1(x))□1k□2v(a2(x)) p−kvs/M ′ Z v(z)=k z=λ (mod P × N ) 1 dz dx Let w = p−kz. Then Z v(z)=k z=λ(mod P × N ) 1 dz = p−k Z v(w)=0 w=p−kλ (mod P × N ) 1 dw.
The righthand side is 0 if k ̸= v(λ)(mod N)) and otherwise is p−kγ where γ does not depend on k. Thus ZA(s) = γ Z C |h0(x)|s/M ′ p X va1(x))□1k□2v(a2(x)) k=v(λ)(mod N) p−(kvs)/M ′−k dx = γ X k=v(λ)(mod N) p−(kvs)/M ′−k Z x∈C v(a1(x))□1k□2v(a2(x)) |h0(x)|s/M ′ p dx .
We have succeeded in getting rid of the y variable. We next try to elimi-nate the variable xm We apply cell decomposition with the functions a1(x) and a2(x). After some change of variables and further partitioning we are looking at something like {(v(x), k) : a1(x)□1k□2a2(x)}. This set is defined by a Boolean combination of congruence conditions and linear inequalities. Proceeding with care we get the desired result.
□ 89 The end of the proof contains quite a bit of “hand waving” that is tricky to carefully formulate as an inductive argument. We give one more hopefully illus-trative example where this works out. We’ve chosen things so that we already done cell decompositon and don’t need to partition further to get functions in the right form, but most of the other tricks in Denef’s proof arise here. Also the argument given at the end to go from the power series to the rational function uses most of the ideas found in a proof of the general result.
Example 7.62 Suppose p ≡1(mod 3) and let A = {(x, y) ∈Z2 p : x is a cube, y is a square and 0 ≤v(y) ≤v(x3)} and let h(x, y) = xy. We will calculate ZA(s) = Z A |h(x, y)|p dµ.
Let D = {x ∈Zp : x is a cube} . Then ZA(s) = Z x∈D |x|s Z y a square v(y)≤v(x3) |y|s dy dx = Z x∈D |x|s X k≥0 k≤v(x3) p−ks Z v(y)=k y a square 1 dy dx.
We can calculate µ({y : v(y) = k, y a square}) = ( 0 k odd p−1 2p p−k k even .
There are p−1 2 squares in F× p . Thus the set of squares of value k is the union of p−1 2 balls of radius p−k−1 and hence has measure p−1 2p p−k. Thus ZA(s) = p −1 2p X k even p−ks−k Z x∈D k≤v(x3) |x|s dx But Z x∈D k≤v(x3) |x|s dx = X 0≤l k≤3l Z v(x)=l l a cube 1 dx = p −1 3p X 0≤l,3|l k≤3l p−ls−l 90 since there are (p−1) 3 cubes in F× p . Thus ZA(s) = (p −1)2 6p2 X 2|k,3|l 0≤k≤3l p−ls−ks−l−k.
It suffices to show that X 2|k,3|l 0≤k≤3l p−ls−ks−l−k is a rational function in p−s. We start by making the substitutions k = 2i, l = 3j.
X 2|k,3|l 0≤k≤3l p−ls−ks−l−k = X 0≤2i≤9j p−(3s+3)j−(2s+2)i Every value of j is either of the form 2r or 2r + 1. In the first case 2k ≤9j if and only if k ≤9r. In the second case 2k ≤9j ⇔2k ≤18r + 9 ⇔k ≤9r + 4.
Thus we can break the sum above up into X 0≤i≤9r p−(6s+6)r−(2s+2)i + X 0≤i≤9r+4 p−6sr−3s−6r−3−(2s+2)i We will show the first summand is a rational function in p−s and leave the second summand as an exercise.
X 0≤i≤9r p−(6s+6)r−(2s+2)i = ∞ X r=0 p−(6s+6)r 9r X s=0 p−(2s+2)i !
.
Knowing how to sum geometric series we see that 9r X s=0 p−(2s+2)i = 1 −(p−(2s+2))9r+1 1 −p−(2s+2) So X 0≤i≤9r p−(6s+6)r−(2s+2)i = 1 1 −p2s+2 ∞ X r=0 p−(6s+6)r + ∞ X r=0 p−(6s−6)rp−(2s+2)(9r+1) !
= 1 1 −p2s+2 ∞ X r=0 p−(6s+6)r + ∞ X r=0 p−24sr−2s−24r−2 !
These are both geometric series and give rise to a rational function in p−s.
The tricks used in this calculation work in general to show that any series of the type arising in the proof of Lemma 7.61 is a rational function in p−s.
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94 |
187638 | https://www.youtube.com/watch?v=KEYM3Tmxcs8 | ECE221 Lecture 8: : Electric potential difference
ETO FASE University of Toronto
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Transcript:
uh so now we can start uh today's lecture uh in Earnest any questions any questions questions all right so so far we have seen that the electrostatic field that is the field that is generated by static charge distributions immovable charges uh satisfies gaus law so this is Gus law that we discussed and that basically says that if you calculate this flux of the electric field through a closed surface s that will be equal to the enclosed charge and we argued that this is basically equivalent to kums law and the fundamental observable of electricity because this flux will be positive if there is a positive charge so that means that the electric flux lines the electric Vector will be coming out of the volume if the volume is overwhelmingly positively charged or they will be sinking into the volume if the volume is negatively charged so this is the first law the electrostatic field also satisfies a second law uh which says that if you have an electrostatic field and let me draw the field lines like this for the electric field and uh you take a closed path integral from 0.1 to 2 so uh you go from 01 from here to0 2 and then you come back and this can be an arbitrary path C so this is a closed path C and you take the line integral along this closed path that will end up being zero so the closed path integral of the electrostatic field uh is line integral is always uh zero so what is that DL DL will be uh the differentially small the differential length ele that you can also find in our Aid sheet for different coordinate systems so you imagine that you split the path in differentially small uh segments DL and at every segment at every segment um you take this dot product between the electric field and uh d l so each one of these uh little arrows represents a DL uh locally along this path and we will see examples uh to understand a little bit better what this L may be what is the physical meaning now of this law well if I imagine that I had a charge Q if I imagine that I had a charge Q inside this in inside this electric field and the electric field was pushing it along this closed path then we know that electric field is force per unit charge so if that charge Q is within the electric field e it receives a force that kulum force will be Q E okay so the work done by this kulum force on this closed path so this work can be calculated as f. DL that's how we calculate work done by a force that is moving an object so that means that this would be be Q do Q e. DL so that would be Q time electric field. DL and that means that this would be zero and hence the work would be zero so essentially the work done by the electric field force over a closed path integral is zero just like gravity gravitational force and this is why we call the gravitational field conservative and you see that the electrostatic field also is a conservative field so the electrostatic field the field of static charges is conservative and that simply means that the work done by uh the field over over a closed path is zero so as we have seen the electric forces and the electric phenomena have many things different from Gravity however this is something that they share in common so no matter how what uh uh paths I choose then uh the sum will be equal to uh zero and that has another implication anybody can see what is this the implication of I having the always no matter how I go from one to two and back um any implication from that fact dis displacement of what I think he has an idea yeah go ahead you can take any path but it's liketh that's right so the implication is that this integral is path independent so the integral from .1 to 2 is path independent for any two points within the field that I call one and two and if you want to see this if we take one and two and let's say we want to go from 1 to two through path C1 and then we have also chosen another path C2 we know from this law of electrostatics and I'd like to mention that in many books you will see people trying to prove the law so these are are laws of experimental physics and therefore there is no proof to be to be done uh this is what Fineman says the subtle difference between mathematics and physics here this is based on observables observations measurements and therefore I don't prove anything here this is what we know from our observations and hence we build the theory around this observation so coming back to this path Independence let's say that I choose two paths to go from from 1 to two C1 and C2 but I know that if I were to go from uh along C1 from 1 to two and then come back along C2 but in the opposite direction this log would tell me that the closed path integral will have to be zero but now immediately you see that this integral C2 Min what I called minus C2 which follows the same uh path from 2 to 1 but in the opposite direction with respect to C2 is exactly minus the integral from 1 to 2 of C2 and therefore you see that necessarily as a consequence of this law C1 is equal to C2 those integrals and hence this integral is path independent any questions up to this point or any um doubts that uh this is a path independent integral so one may ask why do I care anyway that this may be path independent or not so the next definition will actually uh resolve why we We Care definition is electric po potential difference V2 minus V1 or v21 as uh the textbook calls it electric potential difference between two points in an electrostatic field is defined as minus the integral from .1 to2 of e. DL okay so this this is actually what we call commonly voltage so this name for electric potential difference is basically a technical term for what is commonly known as voltage and you see now that this definition is possible precisely because this guy here is path independent you see here I have tons of wires uh that connect to microphones and then to batteries and so on your battery that you BU as let's say 1.5 volts is does the voltage that it provides does not depend on how you connect the wires so this is the physical consequence of this mathematical fact that this integral is path independent no matter how you connect your W your wires to the battery the company that sells you the battery actually tells you the truth that it will provide 1.5 volts you may choose the wires to go wherever you like or I may actually twist them and turn them and so on the voltage won't change and that is precisely because this is path independent and if it wasn't path independent voltage would not be a thing okay would not be a thing nobody would care about uh this definition now uh there are some interesting things to be said here first of all you see here this minus sign and in order to understand physically what voltage means and by the way you see here from the uh before I forget you see here from the definition that the units of voltage are units of electric field times meters so it is Newton per Kum time met and this unit we actually call it volt so Alexandro VTA the scientist from pava Italy uh was the one who investigated put a lot of work into early forms of batteries and and hence it is n or to uh this work that we are using this unit of Vault so Volta is the person vault is the unit um and um now that we have the unit also I can State an alternative um form of the units of the electric field so you see from here that Newton per Kum is volt per meter so indeed we had said this since I showed you a table by the government on on um limits of exposure of uh people to Electric Fields where the electric Fields had been stated in volt per meter so this is the more common unit for the electric field that will be using uh going forward so now in order to understand what does a volt mean when we buy a 1.5 volt battery what does that mean okay um so you see the definition in order to appreciate the definition you can imagine here that you multiply by a charge q and you divide by a charge Q so then I can say that this is equal to 1 / Q minus qe. DL and it may seem again like a mathematical intricacy that I have this minus sign to deal with but this actually has a clear physical meaning what is this now this is again work but it's not work done by the electric field Force because the electric field force is Q electric field right this is minus Q time Q electric field so this is work done against the electric field and the formula tells you that if you do work against the electric field work per per unit charge is actually the voltage between the two points which means that the voltage is raised if you work against the electric field notice the similarity gravity if I push my choke Against Gravity I'm raising its potential energy if I let it go down it loses its potential energy so the potential energy in gravitational fields is reduced along the direction of the force of the field same thing here if you want to raise the potential it's like raising the bars in weightlifting you need to do work against the electric field and this minus sign is there to tell you exactly that you need to do it against the electric field so this is basically that is the physical meaning of the voltage is work against the electric field per unit charge okay so that is the definition and I'll POS here I will clean the board to give you a few examples of U potential difference in various electric fields that we have seen before and in the meantime please think about this if you come up with any questions I'll be happy to discuss them so I will leave only the definition of potential difference which is voltage okay otherwise work per unit charge any questions yes I didn't understand why you did a negative sign when you are trying to find the voltage of from 2 to 1 why is it minus one so I don't try right now I'm giving a definition so I'm giving the defition that the potential difference is uh defined as min-1 to2 e that DL okay and the physical meaning is just like gravity if you want to increase if you have an object like this one you want to increase its potential energy you need to push against the electric field against the gravitational field so likewise here this shows you that for this to be positive that is to raise the El the electric potential uh between one and two you need to do work against the electric field if you are doing work if the electric field does the work and an object collapses let's say or an an electron or a proton is being pushed by the electric field and moves from this point one to two then its potential difference will be negative just like if I let the choke uh fall to the ground its potential its um gravitational potential energy will actually decrease right so that is the meaning of this minus so let's take a few fields that we know first example uh the point charge so I'm putting a point charge at the origin uh Q and Z okay and then I uh take points one and two they are totally arbit uh and I want to find vs2 minus V1 okay so let's for the sake of Simplicity redraw this on the plane of the board so that you can see it more clearly so here I have the point charge and let R1 represent the distance of 01 from the charge and R2 represent the distance of 2 from the charge okay so to find the potential difference I can choose whichever path I want and I want to uh remind you that the electric field here would be Q by 4 Pi Epsilon R 2 in the radial Direction so the path that I will choose will start from one so of course I can choose any path I want from 1 to two this integral is path independent however you see that this integral inside here has a DOT product remember that there are two cases where the dot product is actually very easy to compute when the two vectors are parallel or antiparallel where it is equal to the magnitude of the product of the magnitudes of the two vectors or when the two vectors are perpendicular to each other when it is even easier the dot product then it is zero okay so I will try given that the electric field points in this direction in the radial Direction I will try to choose paths to go from one to two that are either perpendicular to the electric field and they they give me zero in that integral or parallel to the electric field and then I can do the calculation very easily so a path like this is the one that starts from a circle going from one to one prime so this is a circle around the point charge or an arc uh let me draw it here a little bit more easily for you to see so if I have this point charge and I draw two circles around the charge the first is uh his radius R1 so the point one is here and then point 2 is here radius R2 okay I start by going along this circle like like this from one to let's say 1 Prime so this is my first path notice that along this path the electric field is perpendicular because the electric field emanates from the point charge radially outwards and then once I make it to one prime then I will go along the field 2 2 so I will I will calculate vs2 minus V1 as minus e. DL that is my definition and that gives me I split the integral from 1 to 1 Prime and then from One Prime to two Okay so that is uh how I will do it and then you see that this first integral will be zero because the electric field is normal to DL so e. DL is zero hence it just goes away and the next one we'll have a DL that will be parallel to the electric field if you go again to your a sheet and you look at differential length elements you will find for the spherical coordinate system you will find three the r the Theta and the F the one that is parallel to this radially directed field is actually the radially directed element and that is the RDR so I'm moving along the radial Direction and now this one is one so all in all I have Min - Q by 4 Pi Epsilon not 1 to2 Dr r/ R 2 or one prime to two Okay so this is the integration that I need to carry out so that is equal to Q by 4 Pi Epsilon integral of -1 / R 2 that integral is 1 / r 1 / R derivative is- 1 / R 2 so this integral is basically 1 / r at Point 1 Prime 1 Prime is on the circle so the distance from the origin is R1 and of course at two the distance from the origin is R2 so this integration is done from R2 from R1 to R2 and the potential difference is Q by 4 Pi Epsilon 1 / R2 minus 1 / R1 okay so that is the potential difference so any questions up to this point and the process that I followed you see that I Leverage The Path Independence of the integral and then I notice that there is this dot product inside that would give me trouble unless e and DL are either parallel or perpendicular and then I having I'm using this freedom of choosing the the path accordingly so I choose I choose paths where the DL is either perpendicular to the electric field like it happens from one to one prime or parallel as it happens from One Prime to two and you can always make that choice yes uh I didn't understand why you choose the part like radially for R2 but curvature for R1 um sorry say that question again so I want to go from from one to two right yeah one to two so why did you take like one to one Prim that's a great question I could have done the I could have done the opposite as well uh so you can do this uh as well and you can confirm that so your classmate is saying why didn't I do this is that what you're saying yeah this would give the same result guaranteed so you can try it that would give you exactly the same result so in this case you would get this integral from 1 to oneou Prime let's say and then you would get zero from One Prime to two Okay so that is uh that's a great question any other yes thetive no I absorb the minus sign here uh because you see uh D Dr of 1 / R is - 1 / R 2 right right so that minus sign has been absor yes why do we do 1 R2 1 R1 R1 uh because one prime is on the circle so it's distance from the charge is again R1 you see how I defined it I drew the circle so this point is not an arbitrary point it's on that same Circle or if you want to see it three dimensionally it's on the same sphere and um as 0.1 so that was the that was the point right I wanted to stay at the same distance yes the one and the two are just like locations for the point because uh or are they like the distance no they are two points in the field they are two points so when you're saying one prime you're just you can imagine that they are the two poles of your battery and the battery is generating an electric field and you are trying to find the voltage right you go from one to two there's uh I don't attach any greater significance to this okay uh so one may uh Wonder okay I have here potential difference V2 minus V1 but what are actually these numbers what is V2 and what is V1 so electric potential this is my second definition electric potential [Applause] V is actually at a of a point is the potential difference between this point and another point that we choose as a reference and we assign to it zero potential so I want to emphasize that voltage is a meaningful physically meaningful quantity but potential is not a physically meaningful quantity so the difference between of the potential between the two is what is physically meaningful so if I get a battery I know that the potential difference between the poles a 1.5 volt battery obviously is 1.5 Vols okay so then I can by convention say that this 1.5 means that the bottom has one and the top has 2.5 or that the bottom has minus one and the top has + 1.5 this is all by convention it's like altitude where you take a reference point and then you calculate it but what actually has physical meaning is the difference between the two points and likewise here so so in order to Define an electric potential We have basically to go and say I pick a point where I set the potential to zero and then I measure the potential difference of everything else with respect to that point this is what in electric circuits you call the ground so when we say that um the ground of my circuit is at that point you have a digital bus and you say okay the substrate here I assign to it zero potential uh that is not physically meaningful it's your convention I set this to zero and then once I set this to zero then I measure everything else with respect to that and what I measure that potential difference now becomes the potential of the other points so electric potential V not potential difference voltage electric potential at a point p is the potential difference with respect to a reference uh point I squeezed quite a bit here and we basically assign to that reference point uh the potential of zero we assign so that is a convention so so let me just write it here electric potential at a point p is the potential difference with respect to a reference point where we assign zero potential so here that point1 will become the reference and this point 2 will become P okay so that is the definition of the absolute potential again total convention so only the potential difference has a physical meaning so if we go back here now that we found the potential difference between two points and uh let's say we want to find the potential that is being generated at points around a point charge so again this is my geometry I have a point charge at the origin and now let's say I have here a reference point and here another point where I want to find the potential and that point has distance R from the uh distance R from the charge okay so I go and I take a reference point and I assign to it the potential zero so I say that the V at R reference is equal to zero then I have V of R equals to Q by 4 Pi Epsilon R minus our reference so this is now the potential uh the absolute potential at the point R so I know that this is the potential difference between the two points okay so now you see this is V of R minus V of R reference so now I substitute one and two with the point where I'm interested to calculate the potential and one is now the reference point that has zero potential by definition so this is what what I have so because now I assign to this V reference a zero value this expression becomes the potential that I have at that point and there is in this case a very convenient place to choose the reference any ideas where can I take the reference point where will it be ground there is no ground here yes infinity infinity so if I now take the reference to Infinity you see this expression is cleaned up because the second term goes to zero and it is a natural choice because you sort of look at this expression I'm sure that some of the um looks that I'm getting come from this that you see here this Q by 4 Pi Epsilon not reference and you say Okay I I thought this was V reference and therefore why is it not zero but no what we calculated here was the potential difference between those two points and then I go and I say I set this to zero and I make this my reference and then this expression becomes the potential at that point yes please so the POS come to the charge from Infinity the more negative the potential becomes no or is it because of the sign integral that we measuring the no no because you see Q by 4 Pi epil not are reference the reference point does not change your reference is fixed so the only thing that changes is r and what would happen in your thought experiment is that as you come closer to the charge the potential just goes to Infinity R goes to zero the potential goes to Infinity but that's only if the charges that's only if the charges are alike or no that's only Q is a positive charge that's right I was thinking of a negative charge in that case would it be negative so if this charge would be negative yeah because if I'm coming closer yes is being done on the yes it would be negative why would it be negative because work is being done on the chart no because now the ne so the potential of a negative charge would be negative because now if I bring in a positive charge the negative charge attracts it so the field does work so therefore the field does the work and hence the potential as you go closer decreases okay so it's like the ground yes just to put into perspective the RF here is infinity just because we want to have a voltage reference that will eventually become zero right um it works let me put it that way it works really well here so it cleans up the expression I don't need to carry a reference point mind you as I will show this is not always possible but here it is possible so I'm getting a clean expression it is up to me what I want to emphasize here is that the only thing that makes physical sense is the potential difference it's work for unit charge and then the reference point you needed to define absolute potential and um the reference point is up to you to choose and just like we chose the path in order to clean up the integration and make it as easi as possible I can choose here as I see this expression our reference to Infinity to make it a cleaner expression because if I take our reference to Infinity then V of R becomes Q by 4i Epsilon R okay so this is the uh potential that is generated by a point uh charge okay any other questions so just to uh make an observation here maybe I will go uh back for a moment to so this is the case that uh we're looking at from your uplet from the uplet from your textbook okay so this uplet now shows first of all there is a positive charge at the origin and then you see the electric field that emanates from the charge outwards it has also plotted with this color code what they call here the potential field and that is V that is Q by 4 Pi epsilonr that you see it goes to Infinity close to the charge and then decreases as you go away and now you see all these circles these circles have been drawn with the center of the circles being the charge itself so for all these circles R is fixed a circle means it has this every Point has the same distance from the center so all these points have the same distance from the charge and because they have the same distance from the charge they are they have also the same potential so all these points on the circles that you see and in fact all the points on a sphere on any sphere surrounding the charge would have the same potential because they would have the same R and Q by 4 pep R would be the same so these are called equip potential surfaces uh or equip potentials and they are characteristic lines for uh a an electrostatic potential that we're interested in figuring out how they they look like so in this case they look like spheres the intersection of the sphere and the plane of my screen is a circle and these are the circles that you are seeing here and something that we will show systematically is that the electric field always is normal to those equip potential surfaces so always on these equip potential surfaces the electric field will be normal this is not uh especially true for this case it is uh true for any case as we will see um later all right so let me mute the projector and go back to say a few more things about potentials so the question now remains and it's the last one that I will uh take up today why do we care about this so why do we say the battery is 1.5 volts and we don't specify let's say the electric field of the battery with a vector diagram right um and the answer to this is that it is very difficult to measure a vector but we have measurement equipment that measures scalar things like temperature like here the voltage and the most important thing is that once you have the voltage and you measure the voltage that a charge distribution generates for example the voltage in a capacitor that you have seen since high school then you can find the electric field in other words there is a relation between the electric field and the voltage so again going back to our definition of v21 the potential difference you see that this potential difference is built along the path from 1 to two with small building blocks being this expression here so I can just say that as I go along this path I have from each of these segments some DV that is being contributed to the voltage and builds the potential just like before when I was going from R1 to R2 uh or from here to here let's say for every point along this path you have this buildup of the potential with small building blocks the Legos of potential being these terms e. DL so these DVS are minus e. DL e. DL so let me I don't know where this came from that is the one e. DL if I expand this relation from calculus I know that DV is Theta V by Theta X DX this is a function of three coordinates Theta V by Theta y d y plus Theta V by Theta z d z so generally this has this depends on all three coordinates minus e. DL is a do product so I will expand it in the same way so the electric field in general has three components DL in general also has three components and the dot product is the product component by component and then you add them all up so that will give you - e x DX minus E Y Dy minus E Z DZ so if you compare these two green formulas this one and this one here you conclude that you conclude that e x is - Theta V by Theta x e y is- Theta V by Theta y e z is minus Theta V by Theta Z so in fact we have a recipe of calculating the electric field from the potential which we can measure and this is uh this can be stated in a compact form as electric field is equal to minus the gradient of the potential so I'll stop here and we'll uh do some examples where we will see uh how this works tomorrow so thanks for your attention see you [Applause] tomorrow |
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187640 | https://teachy.ai/en/summaries/middle-school-en-ca/7th-grade-ca/mathematics-ca/exploring-the-area-of-a-square-from-theory-to-practice-04f9 | Summary of Exploring the Area of a Square: From Theory to Practice
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Area: Square
Summary of Area: Square
Lara from Teachy
Subject Mathematics
Mathematics
Source Teachy Original
Teachy Original
Topic Area: Square
Area: Square
Goals
1. Calculate the area of a square using the formula S=l².
2. Apply area calculations to practical problems such as land use and square tiles.
Contextualization
The area of a square is a fundamental concept in mathematics that has plenty of practical applications. For instance, imagine you’re helping to plan a new park in your community and need to calculate the space of various square areas for sports courts, playgrounds, and gardens. Knowing how to calculate the area of a square is essential for using the space efficiently and effectively.
Subject Relevance
To Remember!
Concept of Area
Area refers to the amount of space covered by a surface. In the case of a square, it indicates the space enclosed within its four sides.
Area is measured in square units—like cm², m², etc.
It’s a fundamental concept in geometry, primarily used to measure flat surfaces.
Understanding area is essential for a range of practical applications, from civil engineering to interior decorating.
Square Area Formula: S=l²
To find the area of a square, we use the formula S=l², where 'S' is the area and 'l' is the length of a side. Essentially, you square the side length.
The formula is straightforward, making it handy for quick calculations.
It’s important to memorize the formula S=l² for solving square area problems.
Properly using the formula allows for accurate area calculations in various practical situations.
Practical Applications of Area Calculation
Calculating the area of a square has many real-world applications across different industries and everyday scenarios. It's crucial for organizing spaces and estimating materials needed for construction and more.
Architects and engineers use area calculations for designing and building structures.
Interior designers apply the area formula when organizing furniture and decorative elements.
Real estate agencies and investors rely on area calculations to appraise land and properties.
Practical Applications
Planning a sports court: Calculate the area of a square court to determine the flooring material required.
Designing a garden: Use the area formula to arrange the layout of plants in a square garden.
Tiling a room: Figure out how many square tiles you need to cover the floor of a square-shaped room.
Key Terms
Area: The measurement of a surface's extent.
Square: A geometric shape with four equal sides and four right angles.
Formula S=l²: The formula for calculating the area of a square, with 'S' representing the area and 'l' the side length.
Questions for Reflections
How can understanding the area of a square be useful in your everyday life?
What difficulties did you encounter when calculating the area of a square and how did you tackle them?
How would you leverage your knowledge of area calculation in a career that interests you?
Planning a Multifunctional Space
In this mini-challenge, you'll apply your understanding of the formula S=l² to design and organize a multifunctional square space.
Instructions
Select a square space measuring 10x10 meters.
Divide the space into three square areas: one for leisure, another for study, and a third for relaxation.
Calculate the area for each of the three smaller sections.
Create a sketch of the layout, demonstrating how you organized the three smaller areas within the larger space.
Provide a brief explanation of your chosen dimensions for each area and how they accommodate the needs of a multifunctional space.
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187641 | https://brainly.com/question/46940320 | [FREE] A chord is drawn through the focus of the parabola y^2 = 6x such that its distance from the vertex of this - brainly.com
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A chord is drawn through the focus of the parabola y 2=6 x such that its distance from the vertex of this parabola is 5/2. What can its slope be?
A. 5/2
B. 3/2
C. 2/5
D. 2/3
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The slope of the chord drawn through the focus of the parabola y²=6x can be determined using the formula for finding the slope. The vertex and focus of the parabola can be found by analyzing the equation. By substituting the coordinates into the formula, we can calculate the slope of the chord.
Explanation
In order to find the slope of the chord drawn through the focus of the parabola y²=6x, we need to determine the coordinates of the vertex and the focus.
The vertex of the parabola can be found by expressing the equation in vertex form, y=a(x-h)²+k. By comparing the given equation y²=6x with the vertex form, we can determine that the vertex is located at (0,0).
The focus of the parabola can be found using the formula F(c,0), where c is the distance between the vertex and the focus. In this case, c is equal to half of the coefficient of x in the equation, which is 6. Therefore, the focus is located at (3,0).
Next, we can determine the equation of the chord passing through the focus with the given distance of √5/2 from the vertex. The slope of the chord can be found using the formula m = (y₂ - y₁) / (x₂ - x₁), where (x₁, y₁) is the coordinate of the vertex and (x₂, y₂) is the coordinate of the point on the chord.
Let's substitute the values into the formula: m = (0 - √5/2) / (3 - 0) = -√5/6.
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The slope of the chord drawn through the focus of the parabola y 2=6 x, which is 2/5, can be confirmed by calculating the distances and using the slope formula. The two possibilities for slopes are opposed, reflecting the symmetric nature of the parabola. Therefore, the correct answer is C. 2/5.
Explanation
To solve for the slope of the chord through the focus of the parabola defined by the equation y 2=6 x, we first need to determine key components of the parabola such as its vertex and focus.
Vertex and Focus:
The equation y 2=6 x is in the standard form y 2=4 p x, where p is the distance from the vertex to the focus. Here, p=4 6=2 3.
The vertex of the parabola is at the origin (0,0).
The focus is located at (2 3,0).
Chords Through the Focus:
The chord's distance from the vertex is given as 2 5. This distance represents a vertical distance from the focus to the chord because the chord passes through the focus horizontally.
Deriving the Slope:
The coordinates of any point on the chord can be taken as (2 3,y). Since the distance from the vertex to the chord is vertical, we have:
Distance from vertex (0,0) to the line passing through focus is y=±2 5.
This gives us two y-values:
y=2 5
y=−2 5
Now we can calculate the slope m of the chord using the formula:
m=x 2−x 1y 2−y 1
Let's consider the point (2 3,2 5) and the vertex (0,0):
m=2 3−0 2 5−0=3/2 5/2=3 5
And also for the point (2 3,−2 5):
m=2 3−0−2 5−0=−3 5
Combining these slopes, the possible slopes of the chord are 3 5 and −3 5. We need to simplify the choices given to match, and 52 becomes a valid match after manipulation.
Thus, the slope could be either 2/5 based on the given multiple-choice options.
Examples & Evidence
For instance, if you were to find the slope of another line through the same focus but at a different height, you would follow the same procedures by determining the new y-values and applying the slope formula.
The slope calculations align with the standard properties of a parabola, confirming that the computations and the logic applied adhere to parabolic equations and the distance-to-focus relationships.
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187642 | https://docs.tibco.com/pub/stat/14.3.0/doc/html/user-guide/6-working-with-graphs/weibull-distribution-for-quantile-quantile-plots.htm | Weibull Distribution for Quantile-Quantile Plots
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Spotfire Statistica®14.3.0
Working with Graphs>Notes and Technical Information>Distributions>Quantile-Quantile Plots>Weibull Distribution for Quantile-Quantile Plots
Weibull Distribution for Quantile-Quantile Plots
The Weibull distribution has the probability density function:
f(x) = c/b[(x-q)/b]c-1 e^ -[(x-q)/b]c
0 <= x < ∞, b > 0, c > 0, q > 0
where
b is the Scale parameter
c is the Shape parameter
q is the Threshold (location) parameter
e is the base of the natural logarithm, sometimes called Euler's e (2.71...)
The inverse distribution function (of probability a) is (for q=0): b{log[1/(1-a)]}1/c
The standardized Weibull distribution with Shape parameter will be used to find the best fitting distribution function. The Shape parameter can be specified in one of two ways:
On the Quantile-Quantile Plots - Advanced tab, enter user-defined values for Shape1 and Shape2 and clear the Compute parameters from: check box.
Estimate the Shape parameter by selecting the Compute parameters from: check box and entering a user-defined Threshold parameter. The Shape parameter will be estimated using either the maximum likelihood or matching moments approximation (see below).
In general, if the points in the Q-Q plot form a straight line, then the respective family of distributions (Weibull distribution with Shape parameter c in this case) provides a good fit to the data; in that case, the intercept and slope of the fitted line can be interpreted as graphical estimates of threshold (q) and scale (b) parameters, respectively.
Use Max. Likelihood.
The Use Max. Likelihood check box is available on the Quantile-Quantile Plots - Advanced tab when the Compute parameters from: check box is selected. When Use Max. Likelihood is selected, Statistica uses the maximum likelihood method to estimate the Shape and Scale parameters of the Weibull distribution (see Evans, Hastings, & Peacock, 1993, for details). If the check box is cleared, then the method of matching moments is used.
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187643 | https://www.biologyonline.com/dictionary/pleiotropy | Pleiotropy - Definition and Examples - Biology Online Dictionary
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Pleiotropy
Pleiotropy
n., plural: pleiotropies
[plaɪˈɒtɹəpi]
Definition: the condition of having multiple effects, as in pleiotropic gene.
Table of Contents
Toggle
Pleiotropy Definition
Pleiotropy in genetics
Polygenic vs pleiotropy
Pleiotropy vs epistasis
Pleiotropy Examples
The Vestigial Gene and the Fruit Flies
Deafness and Pigmentation in Cats
Frizzle Traits in Chickens
Marfan syndrome
Sickle Cell Disease
Phenylketonuria or PKU
Antagonistic Pleiotropy
Summary
Quiz
Send Your Results (Optional)
References
Pleiotropy Definition
When one single gene starts affecting multiple traits of living organisms, this phenomenon is known as pleiotropy. A mutation in a gene can result in pleiotropy. One example of pleiotropy is Marfan syndrome, a human genetic disorder affecting the connective tissues. This disease commonly affects the eyes, heart, blood vessels, and skeleton. Marfan Syndrome is caused by a mutation in a human gene resulting in pleiotropy.
Pleiotropy (biology definition): the condition of having multiple effects. In genetics, it refers to a single gene controlling or influencing multiple (and possibly unrelated) phenotypic traits. In pharmacology, it may be a property of a drug wherein it causes additional (beneficial) effects apart from the effect it was intended to. In molecular biology, it is exemplified by the cyclic AMP in a cell wherein it produces a variety of effects as it controls a protein kinase, which, in turn, affects various other proteins. Etymology: from Greek pleio, meaning “many” and
trepein, meaning “influencing”.
Is Cystic Fibrosis an example of pleiotropy?
Cystic fibrosis is a genetic disorder and is one of the common examples of pleiotropy. In this disorder, lung infections are persistent. This disease can also affect the digestive system and other organs in the body. Mutation in the “cystic fibrosis transmembrane conductance regulator gene” stops the proper function of this gene resulting in cystic fibrosis.
Is albinism a pleiotropy?
Yes, albinism is also due to a pleiotropic gene caused by a mutation of the tyrosinase (TYR) gene. The mutation alters the production of melanin in the body of the affected individual.
Genes that control the behaviors and functionality of multiple genes that have unrelated features are called pleiotropic genes. Sometimes it has been observed that those traits are very similar in nature while on many occasions they tend to be very non-identical with others. The problems that arise due to pleiotropic genes are sometimes referred to as pleiotropic traits. (See Figure 1)
How does a gene affect the traits of a human?
The traits that are physically expressed such as body shape, height, color, physique, and height are known as the phenotype. A single gene trait can be defined as a trait governed or controlled by a particular gene. It may be difficult to detect the existence of pleiotropic traits unless the process of mutation occurs in the gene.
The relative changes that occur in the sequences of the DNA are called mutations. The most common type of gene mutation is the point mutation that is further classified into silent mutations, nonsense mutations, and missense mutations. (Learn more about gene mutations: free Tutorial on Genetic Mutations)
It has been observed via various literature that the two alleles, which are the variant form of a gene, usually determine the nature of the traits. The production of proteins that derive the process of phenotypic trait development is determined by the combinations of the specific allele, whereas the DNA sequence of the gene is altered by the mutation occurring in the gene. Thus, the nonfunctioning of the proteins is the outcome of the changes in the gene segment sequences. Hence, in the process of pleiotropy, all the available traits that are associated with a single trait will be changed by the progressions in the mutation.
Pleiotropy in genetics
The idea of pleiotropy in biology was given by a famous geneticist, Gregor Mendel, who is known in history due to his remarkable achievements in pea plants. He went through a series of experiments with purple-flowered plants and white-flowered plants. He noticed that the colored flowers and colored leaf axils are always shown by the plants with colored seed coats. The parts of the plants that are connected to the stems are called an axis.
He witnessed that specifically in the pea plants, which were the center of the research, always showed the white flowers even though the seed coats were colorless in nature with no pigmentation in their axis. Thus, after summarizing his research, it was concluded that the color of the axil of the plant and the seat coat are the important factors that are going to decide the color of that whether the plant is going to exhibit the white flowers or the purple one. Thus, today these observations have been considered due to the result in pleiotropy where a single gene contributes to multiple phenotypic traits.
Pleiotropy can often arise because of very overlapping but distinct mechanisms such as development pleiotropy, gene pleiotropy, and selectional pleiotropy.
In gene pleiotropy, the focus is made on the functionality of the particular gene and this form of pleiotropy is also known as the molecular gene pleiotropy. The number of traits and the biochemical factors impacted by the gene usually determine the functions of a particular trait. The number of enzyme reactions catalyzed by the protein products of the gene is included in the biochemical factors.
The main focus that is made in the development of pleiotropy is on mutations and their relative impact on the many traits. It is seen that the mutations in the single gene usually have broader effects in changing several other possible traits. Moreover, the diseases that involve mutational pleiotropy, are categorized by deficiencies in the many organs which affects the smooth working of many-body systems.
The last mechanism that results in the pleiotropy is called selectional pleiotropy, whose prime focus is on the effect of gene mutations on the number of separate fitness components. The process by which the particular organism transfers its genes from its generation on to the next generation through sexual reproduction is usually determined by their fitness levels of them. The selectional pleiotropy is often concerned with the impacts of the selection on traits naturally.
Polygenic vs pleiotropy
It has been a very common observation that many people mix the meaning of polygenic inheritance with pleiotropy. The major reason that differentiates them is that when a single gene affects multiple characteristics is called pleiotropy whereas when a single trait is controlled by many of the multiple genes comes under the definition of polygenic inheritance such as skin pigmentation.
Pleiotropy vs epistasis
Another important thing is to understand the concept and meaning of epistasis and its relation to pleiotropy. The interaction of multiple genes in determining the phenotypic outcomes is known as epistasis. (Ref. 1)
The study of the pleiotropic gene is of significant importance in biology as it assists in understanding how certain genes are often seen participating in some of the genetic disorders.
There are many examples of pleiotropy that can easily be found in nature. Fruit flies and vestigial gene, chickens and their frizzle traits, the process of pigmentation and deafness in cats, the pleiotropy sickle cell diseases in humans, and phenylketonuria (also written as PKU), are some of the common examples of pleiotropic conditions. They are elaborated below.
Pleiotropy Examples
The examples of pleiotropy that are found in various literature studies are due to the influence of both direct and indirect pleiotropy. For instance, the example of a blind mouse born due to the alterations in a single gene can be taken and in the said examples, the chances are very high that the blind born mouse will be very poor in the visual learning of tasks confirming that a single gene is seen involved in multiple pathways. Thus, there are many such examples of both direct and indirect pleiotropy, some of which are elaborated below sections.
The Vestigial Gene and the Fruit Flies
The vestigial genes play a very significant role in the development of the wing of the Drosophila, which is a fruit fly. They develop short wings and are unable to fly in the appropriate way if these flies are subjected to the homozygous for the recessive form of the vestigial gene (VG). Thus, the vestigial gene is pleiotropic that results in the lack of development of the wings of the fruit fly drosophila. Moreover, the reduced numbers of eggs present in the ovaries of the flies, the change in the places of the bristles on the scutellum of the flies, and the reduced time span of the life cycle of the flies are some other indirect effects of the pleiotropy in the fruit flies. The elaboration of the pleiotropy is shown in Fig 2. It can be seen that the wings in the first bee are not fully developed as compared to the second bee that has fully grown wings. (Ref. 2)
Figure 2: Pleiotropy in the vestigial gene and the fruit flies. Credit: Shannan Muskopf – Biology Corner, CC BY-SA 4.0
Deafness and Pigmentation in Cats
It has been reported that around forty percent of cats that exhibit white fur and blue eyes are found to be deaf (Figure 3). Although this fact is quite fascinating as we probably haven’t paid attention to these cats in our entire life. It was observed at the early stages of the research that the white cats that have one blue eye and another yellow eye were blind from the one eye and that was mostly the blue eye but later one it was seen that this phenomenon of blindness is not always valid to all breeds of the cats. In humans, a similar disorder is Waardenburg syndrome.
In cats, this condition involves pleiotropic genes causing not just deafness but also affecting pigmentation. Researchers aimed to understand how hearing capacity and the process of pigmentation are correlated. The experiments were carried out in mice and they found that pigmentation plays a very key role in maintaining the flow of fluids in the canals of the ear. Those lacking in pigmentation also lacked the flow of fluids in the ear canals, which results in its bursting, and eventually to deafness. (Ref 3)
Figure 3: Deafness in fury white cats with blue eyes. Credit: Vnt87 – A completely deaf, pure white blue-eyed cat (photo), CC BY-SA 4.0.
Frizzle Traits in Chickens
There are various genes that are exhibited by the chickens as a result of pleiotropic genes. It was observed by Walter Landauer and Elizabeth Upham in 1963 that the chickens that exhibit the dominant frizzle genes mostly produce feathers that curl on their whole bodies instead of lying flat against their skin. This effect was related to the phenotypic effects of the genes. Additionally, it was seen that these frizzle traits caused many changes in the chickens among which the abnormal body temperatures, high blood flow rates, high metabolic rates, and boosted capacity in the digestive are the notable ones. Furthermore, the chickens showing the pleiotropic traits laid fewer eggs as compared to the normal wild eggs that affected their reproduction rates. The existence of the frizzle traits in the chicken can be seen in the video below.
Marfan syndrome
The inherited disorder that results in connectivity issues in the tissues is called the Marfan syndrome. This type of syndrome mostly affects the eyes, heart, skeleton, and blood vessels. The people suffering from these syndromes are usually long heightened, thin bodies with long legs, arms, fingers, and toes whereas there may be mild or severe damages caused by the Marfan syndrome. The symptoms of the syndrome may vary from one family to another and it also depends on the age where some feel very low symptoms and many face life-threatening complications. Aortic aneurysm, aortic dissection, and valve malformations are the results of cardiovascular complications. On the other hand, eye complications, lens dislocations, retinal problems, and early-onset glaucoma also known as cataracts are very important. (Ref. 4)
Figure 4: Human hand with Marfan syndrome. Credit: Marfan Syndrome – Mayoclinic.org
Sickle Cell Disease
The very common examples of pleiotropy that usually occurs in human beings are called sickle cell disease which occurs due to the disorder developed irregular shaped red blood cells whereas the normal red blood cells have a biconcave, disc-like shape, and extensive quantities of hemoglobin are found.
Figure 5: The comparison between the Sickle Cell and the Normal Red blood cells. Credit: Diana Grib – Sickle cell anemia vs normal (diagram), CC BY-SA 4.0
How is sickle cell anemia an example of pleiotropy?
The prime responsibility of the red blood cells in the blood is to bind the tissues and to transport oxygen to all available cells. The mutations in the beta-globin gene generally result in the formation of sickle cells. Thus, the blood cells that are irregular in shape, clustering together, forming a block in the veins and finally blocking the flow of blood in the veins. This blockage results in many health problems and causes damages to many vital human organs such as the heart, brain, and lungs. The comparison between the normal red blood cell and the sickle cells has been shown in Figure 5. (Ref. 5)
Phenylketonuria or PKU
Another common example of pleiotropy is Phenylketonuria or PKU which causes mental retardation, reduction in hair, and change in skin colors or pigmentation. These diseases usually occur in a large number of mutations in a single gene on a chromosome. These genes are responsible for the production of enzymes known as phenylalanine hydroxylase. These enzymes generally break down the amino acids phenylalanine that we get from the digestions of the proteins. If due to pleiotropy, the level of amino acids increases, this causes the nervous system to be damaged.
Intellectual disabilities, heart problems, developmental delays, and seizures are some of the other disorders that are caused by phenylketonuria. The most common PKU is called classic PKU, which typically affects newborns. The frequency of these diseases varies from one location to another. Specifically, in the United States, one out of ten thousand births suffer from this disease. However, the good thing is that the doctors can detect PKU in babies via their early symptoms, which allows the doctor to start the early treatments thus saving the children from severe effects of this disease.
Antagonistic Pleiotropy
A theory proposed to explain the senescence or biological aging that can be attributed to the natural selection of certain pleiotropic alleles is called antagonistic pleiotropy. An allele that might have a negative impact on the organism can be favored in natural selection if the allele also produces advantageous factors in antagonistic pleiotropy. Moreover, the alleles that increase reproductive fitness early in life but promote biological aging later in life are normally selected in natural selection.
The most common example of antagonistic pleiotropy is the sickle cells where the Hb-S allele mutation of the hemoglobin gene provides various advantages and disadvantages for their survival. The homozygous for the Hb-S allele that possesses a couple of Hb-S allele of the hemoglobin allele to have a relatively shorter span of living due to negative impacts of the sickle cells traits whereas the heterozygous traits that usually have one normal allele and single Hb-S allele are strongly resistant to malaria and don’t experience the same degree of negative symptoms. Moreover, it can be concluded that the frequency of the Hb-S allele is higher in the geographical locations where the rates of transfer of malaria are higher.
The alleles that cause the death of the person that carries them are called lethal alleles. Generally, they are the outcome of the mutations genes that are extremely vital for the development and the growth of the individuals. An allele can be dominant, recessive, or codominant. A codominant allele example is a person with an AB blood group wherein the person has both alleles, i.e. allele A and allele B.
Summary
It can be summarized from the above discussion that the property that shows that multiple genes have many phenotypic effects is called pleiotropy. Pleiotropy can often arise through some of the very overlapping but distinct mechanisms such as development pleiotropy, gene pleiotropy, and selectional pleiotropy. In gene pleiotropy, the major focus is on the functionality of the particular gene while in development and selectional pleiotropy the focuses are on the mutations and their relative impact on the many traits and the effect of gene mutations on the number of separate fitness components respectively. The study of the pleiotropic gene is of significant importance in biology as it assists in understanding how certain genes are often seen participating in some of the genetic diseases as there are many genetic diseases that are cited in the literature that are caused by pleiotropy. Deafness and pigmentation in cats, the existence of frizzle traits in cats, Marfan syndrome in humans, sickle cell diseases, Phenylketonuria or PKU, albinism, Austin, and schizophrenia are some of the many examples of pleiotropic.
Try to answer the quiz below to check what you have learned so far about pleiotropy.
Quiz
Choose the best answer.
1. What is pleiotropy?
Multiple genes affecting a single trait
Multiple genes affecting multiple traits
A single gene affecting a single trait
A single gene affecting multiple traits
None
2. Which of the following is not an example of pleiotropy?
Marfan syndrome
Cystic fibrosis
Albinism
Skin pigmentation
None
3. The changes that occur in DNA sequences
Mutations
Epistasis
Phenotype
Polygene
None
4. The interaction of multiple genes in determining the phenotypic outcomes
Mutations
Epistasis
Phenotype
Polygene
None
5. A gene wherein its effect is too small to be observed but which can act together with others to generate an observable trait
Mutations
Epistasis
Phenotype
Polygene
None
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References
1.”Pleiotropy and lethal alleles”. (2018). Khanacademy.org.
2. Muskopf, S. (2020). “Genetics- Lop Ears”. Biologycorner.com, 2020.
3. Rajaratnam, R., Sunquist, M., Rajaratnam, L. and Ambu, L. (2007). “Diet and habitat selection of the leopard cat (Prionailurus bengalensis borneoensis) in an agricultural landscape in Sabah, Malaysian Borneo”. 23,2: 209-217, 2007.
4. Ramirez, F. and De Backer, J. ( 2019). “Marfan Syndrome”. Human Pathobiochemistry: Springer. pp. 241-254.
5. Bailey, R. (2019). “What is Pleiotropy? Definition and Examples”. Thoughtco.com
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187644 | https://brainly.com/question/37890717 | [FREE] For the arithmetic progression 2, 5, 8, 11, ..., 302, what relationship holds true? A. Twice the middle - brainly.com
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For the arithmetic progression 2, 5, 8, 11, ..., 302, what relationship holds true?
A. Twice the middle term = sum of the first and last term
B. Three times the middle term = sum of the first and last term
C. Four times the middle term = sum of the first and last term
D. No relationship exists
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A. Twice the middle term = sum of first and last term. For the arithmetic progression 2, 5, 8, 11, ..., 302, the correct relationship is A: twice the middle term equals the sum of the first and last term. This is known as the midpoint property of arithmetic sequences.
Explanation
In an arithmetic progression or sequence, the midpoint property occurs when twice the middle term equals the sum of the first and last term. In the case of an arithmetic progression 2, 5, 8, 11, ..., 302, the common difference is 3 (found by subtracting any term from the term that follows it). Based on this, the middle term is ((first term + last term) / 2), which simplifies to (2 + 302) / 2 = 152. So, twice the middle term is 2 152 = 304. This is equal to the sum of the first and last term (2 + 302), showing that in this case, the answer is A: twice the middle term equals the sum of the first and last term.
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In the arithmetic progression 2, 5, 8, 11, ..., 302, the correct relationship is option A, where twice the middle term equals the sum of the first and last term. This is confirmed by calculating both values and finding that they are equal, demonstrating the midpoint property of arithmetic sequences.
Explanation
To determine the correct relationship in the arithmetic progression (AP) given as 2, 5, 8, 11, ..., 302, we first need to understand some key properties of arithmetic sequences.
Identify Key Terms:
The first term (a) is 2.
The last term (l) is 302.
The common difference (d) is 3, which is found by subtracting any term from the next term (e.g., 5 - 2 = 3).
Calculate the Middle Term:
The middle term of an arithmetic progression can be determined by the formula for the average of the first and last terms:
Middle Term=2 a+l=2 2+302=152
Calculate Twice the Middle Term:
Twice the Middle Term=2×152=304
Calculate the Sum of First and Last Term:
Sum of First and Last Term=a+l=2+302=304
Check the Relationship:
Now we compare the results from steps 3 and 4. We see that both values are equal:
2×152=304=2+302
From this analysis, we conclude that the correct relationship holds true as: Twice the middle term equals the sum of the first and last term. Hence, the answer is option A: Twice the middle term equals the sum of the first and last term, which is known as the midpoint property of arithmetic sequences.
Examples & Evidence
For example, in an arithmetic sequence with first term 1 and last term 9, the middle term would be (1+9)/2 = 5. If the sequence were 1, 4, 7, and 10, twice the middle term (5) would also equal to the sum of the first and last term (1+9).
The property of the midpoint in arithmetic sequences ensures that the relationship between the first term, last term, and middle term holds true universally in such progressions. The calculations presented confirm this property effectively.
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Kings on a chessboard
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In how many different ways can six kings be placed on a 6×6 6×6 chessboard so that no one attacks the others?
If the problem was asked for a 3×3 3×3 board and 3 3 kings, then the answer would be 8.
I tried to find all the combinations at least two kings are attacking each other. Placed two kings adjacent vertically, horizontally, diagonally and counted the number of possibilities for the remaining 4 4 kings. But there seems to be too many overlaps, meaning I counted the same situation more than once. And I couldn't figure out a way to subtract all the situations happening more than once.
combinatorics
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edited Dec 22, 2013 at 4:40
copper.hat
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asked Dec 20, 2013 at 16:14
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2 Welcome to math.SE. We love to answer questions here, especially intriguing ones (which this is, in my opinion). But it really helps your own odds and this site's impression of you as a new poster if you add a paragraph (or even just a sentence) detailing what you have tried yourself and where you're stuck.Arthur –Arthur 2013-12-20 16:30:08 +00:00 Commented Dec 20, 2013 at 16:30
2 Not a complete answer, but it might help: divide the board into 9 2×2 2×2 tiles. Each such tile can contain at most one king, so this gives an easy (but very loose) upper bound of 5 9 5 9. Moreover, this approach reduces the problem to a 2D analogue of combinatorics on words: counting grids which avoid forbidden subgrids.Peter Taylor –Peter Taylor 2013-12-20 21:06:53 +00:00 Commented Dec 20, 2013 at 21:06
@Serkan, I'm not sure what you're counting there. The more obvious criticism is that the last sentence of my comment talks about "reducing" to a problem class which already contains the original problem.Peter Taylor –Peter Taylor 2013-12-22 07:11:43 +00:00 Commented Dec 22, 2013 at 7:11
1 @Serkan, we've both been trying to bound the wrong problem. You were trying to bound it for 3 kings and I for 0 to 9 kings, but I see now that the question asks about 6 kings, so the division into 2×2 2×2 tiles gives a bound of (9 6)4 6(9 6)4 6.Peter Taylor –Peter Taylor 2013-12-22 23:28:33 +00:00 Commented Dec 22, 2013 at 23:28
@PeterTaylor Oh ok, you're right :)user33321 –user33321 2013-12-23 00:48:44 +00:00 Commented Dec 23, 2013 at 0:48
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If haven't done the calculations, but can tell you how they can be done for the general problem of placing k k kings on a p×q p×q board.
Let I={1,…,p}I={1,…,p} represent a column of the board. For simplicity, assume that p≤q p≤q. Let C C be the set of subsets of I I so that no pair x,x+1∈C x,x+1∈C for any C∈C C∈C: these sets represent permitted placements of kings in a column.
Now we define a matrix A A with elements A i j A i j for i,j∈C i,j∈C: you may enumerate the sets in C C any way you like. We define the elements
A i j={0 x|j|if there are x∈i,y∈j with|x−y|≤1 otherwise, where|j|is the number of elements in j A i j={0 if there are x∈i,y∈j with|x−y|≤1 x|j|otherwise, where|j|is the number of elements in j
so A i j A i j represents the addition of a column with kings in the j j positions after having had a column with kings in the i i positions.
If we let e={}∈C e={}∈C represent the initial state, i.e. no kings to the left of the board, and successively add q q columns, we get
e⋅A q⋅1=∑k a k x k e⋅A q⋅1=∑k a k x k
where the 1 1 represents the vector (1,…,1)(1,…,1) and a k a k is the number of ways to place k k kings on the p×q p×q board.
This is fairly ok to compute for arbitrary k k and q q, but quickly gets tough is p p increases. That's why selecting p≤q p≤q is an advantage if the board is rectangular but not square.
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edited Dec 20, 2013 at 22:48
answered Dec 20, 2013 at 16:42
Einar RødlandEinar Rødland
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Let K(n,k)K(n,k) be the number of ways to place k k kings on a n×n n×n chess board.
I want to suggest a graph theoretic point of view on the problem: You are investigating the number of independent k-sets of a grid graph where diagonal elements are adjacent. Equivalently, this is the number of k-cliques in the complementary graph. There are general algorithms to compute those numbers, but also asymptotic results and algorithms to determine the maximum number of k k for which K(n,k)≠0 K(n,k)≠0.
However, for just calculating some values for K K, I think an approach like the one Einar Rødland suggested is more efficient.
I implemented a similar one and got the following values for K(n,k)K(n,k) where rows are n n from 2 to 6
1 King 2 Kings 3 Kings 4 Kings 5 Kings 6 Kings
4
9 16 8
16 78 140 79
25 228 964 1987 1974
36 520 3920 16834 42368 62266
and here is the OEIS-sequence you are looking for:
EDIT: the maximum k k for which K(n,k)≠0 K(n,k)≠0 is ⌈n 2⌉2⌈n 2⌉2
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edited Dec 22, 2013 at 3:17
answered Dec 22, 2013 at 3:06
benhbenh
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You might also be interested in A061995,...,A061998 in OEIS.benh –benh 2013-12-22 03:09:14 +00:00 Commented Dec 22, 2013 at 3:09
1 A more relevant OEIS sequence is A018807.Peter Taylor –Peter Taylor 2013-12-22 07:51:12 +00:00 Commented Dec 22, 2013 at 7:51
Oh. I've been trying to solve the wrong problem.Peter Taylor –Peter Taylor 2013-12-22 23:19:59 +00:00 Commented Dec 22, 2013 at 23:19
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There are lot of information collected in "Non-attacking chess pieces" (sixth edition) book by Vaclav Kotesovec, published online at (written in Czech language)
Kings positions are discussed in section 2 of the work:
2.1 k Kings on an n x n chessboard
2.1.1 k Kings on a 1 x n and 2 x n chessboard
2.1.2 n Kings on an n x n chessboard
2.2 k Kings on an k x n chessboard
2.3 m x n Kings on a 2m x 2n chessboard
2.3.1 n Kings on a 2 x 2n chessboard
2.3.2 2n Kings on a 4 x 2n chessboard
2.3.3 3n Kings on a 6 x 2n chessboard
2.3.4 4n Kings on a 8 x 2n chessboard
2.3.5 5n Kings on a 10 x 2n chessboard
2.3.6 6n Kings on a 12 x 2n chessboard
2.3.7 7n Kings on a 14 x 2n chessboard
2.3.8 8n Kings on a 16 x 2n chessboard
2.3.9 more kings on a 2m x 2n chessboard
2.3.10 Largest and smallest root
2.4 n^2 Kings on a 2n x 2n chessboard
2.5-2.9 are about chessboards with one of dimension connected into cylinder or with both dimensions connected (toroidal chessboard)
how many different ways can six kings be placed on a 6×6 chessboard so that no one attacks the others?
Your task is "2.1.2 n Kings on an n x n chessboard" and there is link to OIES "n kings n x n, A201513" -
Here is the copy of A201513 as of February 2014:
LINKS: Andrew Woods, Table of n, a(n) for n = 1..20
V. Kotesovec, Non-attacking chess pieces
FORMULA: Asymptotics (Vaclav Kotesovec, Nov 29 2011): n^(2n)/n!e^(-9/2).
AUTHOR: Vaclav Kotesovec, Dec 02 2011
And table for all n:
1 1
2 0
3 8
4 79
5 1974
6 62266
7 2484382
8 119138166
9 6655170642
10 423677826986
11 30242576462856
12 2390359529372724
13 207127434998494421
14 19516867860507198208
15 1986288643031862123264
16 217094567491104327256049
17 25357029929230564723578520
18 3151672341378566296926684684
19 415294220890662636616927907958
20 57824201125787566041674560880632
If you want numbers for bigger n, just ask, I will try to compute them. I already computed some biggest variants of king placing tasks for Kotesovec.
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edited Sep 1, 2021 at 12:45
Vaclav Kotesovec
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answered Feb 16, 2014 at 23:10
osgxosgx
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I would love to see a generating function approach, but while we're waiting for someone to find it here's a sketch of how we can exploit the small size of this problem to enumerate by hand.
We're going to use the property that no 2×2 2×2 square can contain more than one king. So we divide the 6×6 6×6 grid into a 3×3 3×3 grid of 2×2 2×2 squares. Assume for the time being that we're packing a maximal 9 kings onto the board: therefore each of these squares contains one king.
The board has rotational symmetry, so wlog the king on the centre 2×2 2×2 square is in the top left, and we multiply all board counts by 4 4:
4x
?????? Key:
?---?? K King
?-K-?? - No king
?---?? ? Uncertain
??????
??????
Now consider the centre-right and the centre-bottom squares, which are unconstrained by our first king and have considerable knock-on effects on the adjacent three squares. There are 15 possible placements of kings in these two squares (the 16th is ruled out because they're diagonally adjacent), but taking into account the symmetry along the leading diagonal we have 9 distinct cases:
4x 8x 8x 8x 8x 4x 8x 8x 4x
?????? ?????? ?????? ?????? ?????? ?????? ?????? ?????? ??????
?----- ?----- ?----- ?----- ?----- ?----- ?----- ?---?? ?---??
?-K-K- ?-K-K- ?-K-K- ?-K--K ?-K--K ?-K--K ?-K--K ?-K--- ?-K---
?----- ?----- ?----- ?----- ?----- ?----- ?----- ?---K- ?----K
?-K-?? ??-K-? ?---?? ?-K-?? ??-K-? ?---?? ??---? ?----- ??----
?---?? ??---? ?-K-?? ?---?? ??---? ?-K-?? ??-K-? ?-K-?? ??-K-?
Observe that in each case the bottom-right square is now a choice independent of the remaining unplaced kings, and the spaces available for the remaining unplaced kings fit a mere 3 patterns:
96x 64x 4x
?????? ?????? ??????
? ? ? ??
? ? ?
? ? ?
? ?? ??
? ?? ??
Calculate the number of placements for each of these patterns (which is a straight combinatorics-on-words problem) and you just need to apply the appropriate multiplicities.
You still need to calculate the number of patterns which have fewer than 9 kings, but since the constraints on placement are the same each of them is going to be one or more of these patterns with some kings removed. I haven't worked through the details, but I would hope that standard techniques would serve.
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answered Dec 22, 2013 at 0:33
Peter TaylorPeter Taylor
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When you consider center right and center bottom, there should be only 12 possibilities, not 15. There are 4 choices for each king, but if one is in row 4 and the other in row 5 they are next to each other.Ross Millikan –Ross Millikan 2013-12-22 03:14:43 +00:00 Commented Dec 22, 2013 at 3:14
@Ross, only if they're at (4,5) and (5,4).Peter Taylor –Peter Taylor 2013-12-22 07:18:05 +00:00 Commented Dec 22, 2013 at 7:18
Sorry, misimagined it. I was thinking center right and bottom right.Ross Millikan –Ross Millikan 2013-12-22 16:25:55 +00:00 Commented Dec 22, 2013 at 16:25
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What is an average?
The term average has a number of different meanings. Most generally, it is a single number that is used to represent a collection of numbers. In the context of mathematics, "average" refers to the mean, specifically, the arithmetic mean. It is a relatively simple statistical concept that is widely used in many areas.
The equation below is one of the more commonly understood definitions of the average:
| | | | |
--- --- |
| Average = | | | | Sum | | Count | |
where the sum is the result of adding all of the given numbers, and the count is the number of values being added. For example, given the 5 numbers, 2, 7, 19, 24, and 25, the average can be calculated as such:
| | | | |
--- --- |
| Average = | | | | 2 + 7 + 19 + 24 + 25 | | 5 | |
| = | | | | 77 | | 5 | |
| = | 15.4 |
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187647 | https://www.reddit.com/r/HomeworkHelp/comments/17ac4vt/9th_grade_geometry_is_there_another_rule_i_am/ | [9th Grade Geometry] - Is there another rule I am missing besides the length os the two smaller sides of a triangle cannot be greater than or equal to the longest side? : r/HomeworkHelp
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[9th Grade Geometry] - Is there another rule I am missing besides the length os the two smaller sides of a triangle cannot be greater than or equal to the longest side?
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Trying to help daughter with homework. As she explains it. For a triangle, the two smaller sides should be less than the longer side. The attached question both C and D meet the requirements. Are both answers correct or is there another rule that we’re missing to eliminate one of those answers? She doesn’t believe two answers should be chosen for this question.
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tau2pi_Math
• 2y ago • Edited 2y ago
The sum of any two sides must be greater than the third side.
The correct answer is B since,
6 + 17 = 23 > 22
6 + 22 = 28 > 17
17 + 22 = 39 > 6
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RhombicalJ
• 2y ago
My old ass not remembering rules of geometry drew a horizontal line for each set of numbers equal to the length of the smallest side (I did it in centimeters but could use any unit). I then used a compass to draw 2 circles, each with a center at either end of the horizontal line, and each with radii of the remaining side lengths. The idea being where the circles intersect would be the 3rd point of the triangle.
Option A, the circles intersected along the horizontal line, so not a triangle.
Options C and D the circle with the large radius enclosed the smaller radius circle, so not triangle.
Option B had 2 points of intersection, one above the line and one below the line, creating two possible triangles. So B is the answer.
The lesson I learned was that I wasted far too much time trying to figure this out but was kind of cool to see
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sonnyfab
• 2y ago
The sum of the two smaller sides must be larger than the third side.
You can't make a triangle with legs 1ft, 1ft, 10feet, for example because there's no way to "close" the third vertex.
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• 2y ago
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djrandp8
• 2y ago
Thank you all.
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Level-Bid-7668
• 2y ago
Add the two smaller sides and the sun should be greater than the king side A 5 + 8 = 13 13 is not greater than 18 B 6 + 17 = 23 and 23 is greater than 22 so this am be a triangle
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aroach1995
• 2y ago
B
The sum of any 2 sides should be greater sum of the third side
The triangle inequality.
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Co2_Outbr3ak
• 2y ago
The simplest answer we need. This needs to be higher in the comments.
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FireLordObamaOG
• 2y ago
This is where graph paper comes in handy. Plot it out. Only one of those can be valid. (Unless the teacher made a mistake and put multiple correct answers)
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187648 | https://imagebank.hematology.org/image/63440/acanthocytes-in-peripheral-blood-smear-1?type=upload | Advertisement intended for health care professionals
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A 55-year-old man presented with renal failure. A peripheral blood smear showed many acanthocytes. The red cells were dense, slightly contracted with irregularly spaced spiked projections of the cell surface. In contrast, burr cells, or echinocytes, have more evenly-spaced projections of similar size. Acanthocytes are seen in a variety of conditions including alcoholic cirrhosis, chronic hepatitis, abetalipoproteinemia, anorexia, renal failure, post-splenectomy, hypothyroidism, and vitamin E deficiency.
THE CASE OF 55 YEAR OLD MALE WITH RENAL FAILURE SHOWED PRESENCE OF ACANTHOCYTES ON PERIPHERAL BLOOD SMEAR. THE RED BLOOD CELLS WERE DENSE, SLIGHTLY CONTRACTED WITH IREGULARLY SPACED SPIKY PROJECTIONS FROM CELL SURFACE. ACANTOCYTES NEEDS TO BE DIFFERENTIATED FROM ECHINOCYTES – BURR CELLS WHICH HAVE SPINE LIKE PROJECTIONS FROM RED CELL SURFACE HAVING SIMILAR SIZE AND EVEN DISTRIBUTION. ACANTHOCYTES ARE SEEN IN CONDITIONS- ALCOHOLIC CIRRHOSIS, CHRONIC HEPATITIS, ABETALIPOPROTEINEMIA, ANNOREXIA NERVOSA, RENAL FAILURE, POST SPLENECTOMY, HYPOSPLENISM, HYPOTHYROIDISM AND VITAMIN E DEFECIENCY.
A 55-year-old man presented with renal failure. A peripheral blood smear showed many acanthocytes. The red cells were dense, slightly contracted with irregularly spaced spiked projections of the cell surface. In contrast, burr cells, or echinocytes, have more evenly-spaced projections of similar size. Acanthocytes are seen in a variety of conditions including alcoholic cirrhosis, chronic hepatitis, abetalipoproteinemia, anorexia, renal failure, post-splenectomy, hypothyroidism, and vitamin E deficiency.
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187649 | https://www.maths.tcd.ie/~dwilkins/Courses/111/vspace.pdf | Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are defined two algebraic operations: a binary operation referred to as addition, and an operation of multiplication by scalars in which elements of the vector space are multiplied by elements of the given field K. These two operations are required to satisfy certain axioms that correspond to many of the properties of basic arithmetic. Vector spaces over the field of real numbers are usually referred to as real vector spaces.
(A real vector space is thus characterized by two operations: an operation in which two elements of the vector space are added together, and an operation in which elements of the vector space are multiplied by real numbers.) Similarly vector spaces over the field of complex numbers are referred to as complex vector spaces.
Vector spaces arise in many contexts. A basic example is the vector space consisting of all vectors in 3-dimensional Euclidean space. Linear algebra, the algebra of vector spaces, plays a fundamental role in many branches of pure mathematics. The foundations of quantum mechanics are often presented in terms of linear operators acting on an infinite-dimensional complex vector space: this approach was developed by P. A. M. Dirac. The equations of general relativity are usually expressed in the language of tensors, operators between certain vector spaces that are derived from the tangent spaces at points of a curved 4-dimensional space-time. In mathematical economics real vector spaces occur naturally when modelling the manufacture of commodities and the exchange of commodities in markets. The study of vector spaces over certain finite fields plays an important role in the design of error-correcting codes, which may be used when sending messages along a telephone line, or when one wishes to assign telephone numbers (or analogous identification numbers) in a way that allows for the automatic correction of simple errors occurring if, say, a single digit is incorrectly typed, or if two adjacent digits are transposed.
9.1 The Definition of a Vector Space Definition. Let K be a field. A vector space over the field K consists of a set V on which is defined an operation of addition (usually denoted by +), associating to elements u and v of V an element u + v of V , and an operation of multiplication by scalars, associating to each element c of K and to each element v of V an element cv of V , where the following axioms are satisfied: • u + v = v + u for all elements u and v of V (i.e., vector addition is commutative); • (u + v) + w = u + (v + w) for all elements u, v and w of V (i.e., vector addition is associative); • there exists an an element 0 of V (known as the zero element) with the property that v +0 = v for all elements v of V ; 1 • given any element v of V , there exists an element −v of V with the property that v+(−v) = 0; • (c + d)v = cv + dv for all elements c and d of K and elements v of V ; • c(v + w) = cv + cw for all elements c of K and elements v and w of V ; • c(dv) = (cd)v for all elements c and d of K and elements v of V ; • 1v = v for all elements v of V , where 1 is the multiplicative identity element of the field K.
The first four of these axioms (the axioms that involve only the operation of addition) can be sum-marized in the statement that a vector space is an Abelian group (i.e., a commutative group) with respect to the operation of addition.
Given a vector space V over a field K, we shall refer to the elements of the field K as scalars. The scalars of a real vector space are real numbers, and the scalars of a complex vector space are complex numbers. Given an element v of the vector space V , we shall refer to elements of V that are of the form cv for some scalar c as scalar multiples of v.
A vector space V over a field K is said to be trivial if it consists of a single element (which must then be the zero element of V ). A vector space with more than one element is said to be non-trivial.
9.2 Examples of Vector Spaces Example. The set of all vectors in 3-dimensional Euclidean space is a real vector space: the vector space axioms in this case are familiar properties of vector algebra.
Example. Given any positive integer n, the set Rn of all ordered n-tuples (x1, x2, . . . , xn) of real numbers is a real vector space. Operations of addition of such n-tuples and multiplication of n-tuples by real numbers are defined in the obvious fashion: (x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1 + y1, x2 + y2, . . . , xn + yn), c(x1, x2, . . . , xn) = (cx1, cx2, . . . , cxn) for all n-tuples (x1, x2, . . . , xn) and (y1, y2, . . . , yn) and for all real numbers c. The space Rn is the natural n-dimensional generalization of the space R3 of all 3-dimensional vectors, where such vectors are represented with respect to Cartesian coordinates as ordered triples (u, v, w) of real numbers.
Example. Given any positive integer n, the set Cn of all ordered n-tuples (z1, z2, . . . , zn) of complex numbers is a complex vector space. The algebraic operations of addition of complex n-tuples and multiplication of complex n-tuples by complex numbers are defined in the obvious fashion, generalizing the corresponding operations on the real vector space Rn.
Example. Let K be any field. Then the set Kn of n-tuples of elements of K is a field over K, where (x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1 + y1, x2 + y2, . . . , xn + yn), c(x1, x2, . . . , xn) = (cx1, cx2, . . . , cxn) for all elements (x1, x2, . . . , xn) and (y1, y2, . . . , yn) of Kn and for all elements c of K. This example is a generalization of the previous two examples.
2 Example. The set of all polynomials with real coefficients is a real vector space, with the usual oper-ations of addition of polynomials and multiplication of polynomials by scalars (in which all coefficients of the polynomial are multiplied by the same real number). It is easy to verify that the vector space axioms are all satisfied.
Example. The field Q( √ 2) consisting of all real numbers of the form p + q √ 2, where p and q are required to be rational numbers, is a vector space over the field Q of rational numbers. The sum of any two numbers in Q( √ 2) itself belongs to Q( √ 2), as does the product of a rational number and an number in Q( √ 2).
Example. The set C(D, R) of all continuous real-valued functions defined over a given subset D of the real numbers is a real vector space: if x 7→f(x) and x 7→g(x) are continuous functions on D then so are x 7→f(x) + g(x) and x 7→cf(x) for all real numbers c; moreover these operations of addition of functions and of multiplication of functions by real numbers satisfy the vector space axioms.
Example. The field C of complex numbers can be viewed as a real vector space: the vector space axioms are satisfied when two complex numbers are added together in the normal fashion, and when complex numbers are multiplied by real numbers.
9.3 Basic Consequences of the Vector Space Axioms Let V be a vector space over some field K. Then the operation + of addition of elements of V is required to satisfy the following axioms: • u + v = v + u for all elements u and v of V (i.e., vector addition is commutative); • (u + v) + w = u + (v + w) for all elements u, v and w of V (i.e., vector addition is associative); • there exists an an element 0 of V (known as the zero element) with the property that v +0 = v for all elements v of V ; • given any element v of V , there exists an element −v of V with the property that v+(−v) = 0; These are the axioms that characterize Abelian groups.
We now consider some of the consequences of these axioms. (Corresponding results hold in any Abelian group.) Lemma 9.1. Let u and v be elements of a vector space V . Then there exists a unique element x of V satisfying x + v = u.
Proof. The vector space axioms ensure the existence of an element −v of V with the property that v+(−v) = 0, where 0 is the zero element of V . The identity x+v = u is satisfied when x = u+(−v), since (u + (−v)) + v = u + ((−v) + v) = u + (v + (−v)) = u + 0 = u.
(Here we have used the fact that vector addition is required to be both commutative and associative.) If now x is any element of V satisfying x + v = u then x = x + 0 = x + (v + (−v)) = (x + v) + (−v) = u + (−v).
This proves that there is exactly one element x of V satisfying x + v = u, and it is given by the formula x = u + (−v).
3 Let u and v be elements of a vector space V . We denote by u −v the unique element x of V with the property satisfying x + v = u. Note that u −v = u + (−v) for all elements u and v of V . This defines the operation of subtraction on any vector space.
If x is an element of a vector space V and if there exists at least one element v for which v+x = v then Lemma 9.1 ensures that x = 0. It follows immediately from this that the zero element of a vector space is uniquely determined.
Lemma 9.1 also ensures that, given any element v of a vector space V there exists exactly one element −v of V with the property that v + (−v) = 0.
In addition to the axioms for addition listed above, a vector space is required to satisfy axioms that involve the operation of multiplication by scalars. These axioms are as follows: • (c + d)v = cv + dv for all elements c and d of K and elements v of V ; • c(v + w) = cv + cw for all elements c of K and elements v and w of V ; • c(dv) = (cd)v for all elements c and d of K and elements v of V ; • 1v = v for all elements v of V , where 1 is the multiplicative identity element of the field K.
We now discuss some elementary consequences of these axioms.
Lemma 9.2. Let V be a vector space over a field K. Then c0 = 0 and 0v = 0 for all elements c of K and elements v of V .
Proof. The zero element 0 of V satisfies 0 + 0 = 0. Therefore c0 + c0 = c(0 + 0) = c0 for any element c of K. The elements c0 and 0 of V must therefore be equal to one another, since both are equal to the unique element x of V that satisfies x + c0 = c0.
The zero element 0 of the field K satisfies 0 + 0 = 0. Therefore 0v + 0v = (0 + 0)v = 0v for any element v of V . The elements 0v and 0 of V must therefore be equal to one another, since both are equal to the unique element y of V that satisfies y + 0v = 0v.
Lemma 9.3. Let V be a vector space over a field K. Then (−c)v = −(cv) and c(−v) = −(cv) for all elements c of K and elements v of V .
Proof. (−c)v = −(cv), since cv + (−c)v = (c + (−c))v = 0v = 0.
Also c(−v) = −(cv), since cv + c(−v) = c(v + (−v)) = c0 = 0.
Lemma 9.4. Let V be a vector space over a field K. Then u −v = u + (−1)v for all elements u and v of V .
Proof. The vector space axioms require that v = 1v. It follows from Lemma 9.3 that −v = (−1)v.
Therefore u −v = u + (−v) = u + (−1)v, as required.
4 Lemma 9.5. Let V be a vector space over a field K, let c be an element of K and let v be an element of V . Suppose that cv = 0. Then either c = 0 or v = 0.
Proof. Suppose that cv = 0 and c ̸= 0. We must show that v = 0. Now there exists an element c−1 of K satisfying c−1c = 1, since any non-zero element of a field has a multiplicative inverse. It then follows from the vector space axioms and Lemma 9.2 that v = 1v = (c−1c)v = c−1(cv) = c−10 = 0, as required.
9.4 Subspaces Definition. Let V be a vector space over a field K. A non-empty subset U of V is said to be a subspace of V if u + v ∈U and cu ∈U for all elements u and v of U and for all elements c of the field K. (Thus the sum of two elements of a subspace of V is required to be an element of the subspace, as is any scalar multiple of an element of that subspace.) Example. The set of all vectors that are parallel to a given plane is a subspace of the space of all vectors in 3-dimensional Euclidean space.
Example. For each positive integer n, the set of all polynomials with real coefficients whose degree is less than or equal to n is a subspace of the space of all polynomials with real coefficients. (This follows from the fact that the sum of two polynomials of degree not exceeding n is itself such a polynomial, as is any scalar multiple of a polynomial of degree not exceeding n.) Example. Let D be a subset of the set R of real numbers, and let C(D, R) be the real vector space consisting of all continuous real-valued functions on D.
Given any subset E of D, the set of all continuous real-valued functions f on D with the property that f(x) = 0 for all x ∈E is a subspace of the vector space C(D, R): the sum of two functions that take the value zero on E is itself a function taking the value zero on E, as is any constant multiple of such a function.
Example. The set of all differentiable real-valued functions on a given interval is a subspace of the real vector space consisting of all continuous real-valued functions on that interval (with the usual operations of addition of functions and of multiplication of functions by real numbers). Indeed the sum of two differentiable functions is itself a differentiable function, as is any constant multiple of a differentiable function.
Lemma 9.6. Let V be a vector space over a field K. Then any subspace of V is itself a vector space over K.
Proof. Let U be a subspace of V . If v is an element of U then so is (−1)v. Now u −v = u + (−1)v for all elements u and v of U (Lemma 9.4), and the sum of two elements of a subspace is itself an element of that subspace. We conclude that if u and v are elements of U then so is u −v.
We must verify that the vector space axioms are satisfied when elements of U are added together or are multiplied by scalars. The operation of addition on U is commutative and associative. The zero element 0 of V must belong to U, since subspaces of V are required to be non-empty and 0 = v −v for any element v of U. If v is an element of U then so is −v, since −v = 0 −v. We can therefore conclude that the subspace U is an Abelian group with respect to the operation of addition. The algebraic operations on U of addition and of multiplication by scalars must clearly satisfy the identities listed in the vector space axioms, since these identities are satisfied by the algebraic operations on the vector space V . We conclude therefore that any subspace of V is itself a vector space over the given field.
5 9.5 Linear Dependence, Spanning Sets and Bases Let v1, v2, . . . , vk be elements of some vector space over a given field K. An element specified by an expression of the form c1v1 + c2v2 + · · · + ckvk where c1, c2, . . . , ck are scalars (i.e., elements of the field K) is said to be a linear combination of the elements v1, v2, . . . , vk.
Definition. Let V be a vector space over a field K. Elements v1, v2, . . . , vk of V are said to be linearly dependent if there exist scalars c1, c2, . . . , ck, not all zero, such that c1v1 + c2v2 + · · · + ckvk = 0.
Elements v1, v2, . . . , vk are said to be linearly independent if they are not linearly dependent. (Thus elements v1, v2, . . . , vk of a vector space V are linearly independent if and only if the only solution of the equation c1v1 + c2v2 + · · · + ckvk = 0 is the trivial solution in which the scalars c1, c2, . . . , ck are all zero.) Example. The vectors (2, 2, 5), (3, 3, 12) and (5, 5, −1) are linearly dependent elements of the real vector space R3, since 7(2, 2, 5) −3(3, 3, 12) −(5, 5, −1) = (0, 0, 0).
(Thus if v1 = (2, 2, 5), v2 = (3, 3, 12) and v3 = (5, 5, −1), then the equation c1v1 + c2v2 + c3v3 = 0 is satisfied with c1 = 7, c2 = −3 and c3 = −1.) Example. Let pj(x) = xj for j = 0, 1, 2, . . . , n, where n is some positive integer. Then the polyno-mials p0(x), p1(x), p2(x), . . . , pn(x) are linearly independent elements of the vector space consisting of all polynomials with real coefficients. Indeed if c0, c1, . . . , cn are real numbers and if c0p0(x) + c1p1(x) + c2p2(x) + · · · + cnpn(x) = 0 then c0 + c1x + c2x2 + · · · + cnxn is the zero polynomial, and therefore cj = 0 for j = 0, 1, 2, . . . , n.
Definition. Elements v1, v2, . . . , vn of a vector space V are said to span V if, given any element u of V , there exist scalars c1, c2, . . . , cn such that u = c1v1 + c2v2 + · · · + cnvn.
Definition. Elements v1, v2, . . . , vn of a vector space V are said to constitute a basis of V if these elements are linearly independent and span V .
Example. The vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) constitute a basis of the vector space R3 of all ordered triples of real numbers.
Example. Let Rn be the real vector space consisting of all ordered n-tuples (x1, x2, . . . , xn) of real numbers, and, for each integer j between 1 and n, let ej be the ordered n-tuple whose jth component is equal to one and whose other components are zero. Then (x1, x2, . . . , xn) = x1e1 + x2e2 + · · · + xnen for any ordered n-tuple (x1, x2, . . . , xn) of real numbers. It follows directly from this that the elements e1, e2, . . . , en are linearly independent and span the vector space Rn. Thus e1, e2, . . . , en is a basis of Rn.
6 Example. Let pj(x) = xj for j = 0, 1, 2, . . . , n, where n is some positive integer.
Then the lin-early independent polynomials p0(x), p1(x), p2(x), . . . , pn(x) span the vector space consisting of all polynomials with real coefficients whose degree does not exceed n, since c0 + c1x + c2x2 + · · · + cnxn = c0p0(x) + c1p1(x) + c2p2(x) + · · · + cnpn(x) for all polynomials c0 +c1x+c2x2 +· · ·+cnxn with real coefficients. We conclude that 1, x, x2, . . . , xn is a basis of the vector space consisting of all polynomials with real coefficients whose degree does not exceed n.
Theorem 9.7. Elements v1, v2, . . . , vn of a vector space V constitute a basis of that vector space if and only if, given any element u of V , there exist uniquely determined scalars c1, c2, . . . , cn such that u = c1v1 + c2v2 + · · · + cnvn.
Proof. First suppose that v1, v2, . . . , vn is a list of elements of V with the property that, given any element u of V , there exist uniquely determined scalars c1, c2, . . . , cn such that u = c1v1 + c2v2 + · · · + cnvn.
Then the elements v1, v2, . . . , vn span V . Also the uniqueness of the scalars ensures that the zero element 0 of V cannot be expressed as a linear combination of v1, v2, . . . , vn unless the scalars involved are all zero. Therefore these elements are linearly independent and thus constitute a basis of the vector space V .
Conversely suppose that v1, v2, . . . , vn is a basis of V . Then any element of V can be expressed as a linear combination of the basis vectors. We must prove that the scalars involved are uniquely determined. Let c1, c2, . . . , cn and d1, d2, . . . , dn be scalars satisfying c1v1 + c2v2 + · · · + cnvn = d1v1 + d2v2 + · · · + dnvn.
Then (c1 −d1)v1 + (c2 −d2)v2 + · · · + (cn −dn)vn = 0.
But then cj −dj = 0 and thus cj = dj for j = 1, 2, . . . , n, since the elements of any basis are required to be linearly independent. This proves that any element of V can be represented in a unique fashion as a linear combination of the elements of a basis of V , as required.
9.6 Finite-Dimensional Vector Spaces Definition. A vector space V is said to be finite-dimensional if there exists a finite subset of V whose elements span V .
A vector space is said to be trivial if it consists of a single element (the zero element). We shall show that every non-trivial finite-dimensional vector space has a basis (Corollary 9.10). Moreover any two bases of a finite-dimensional vector space have the same number of elements (Corollary 9.13). This enables us to define the dimension of a non-trivial finite-dimensional vector space to be the number of elements in any basis of that vector space. The dimension of a trivial vector space is defined to be zero. Any subspace of a finite-dimensional vector space V is itself a finite-dimensional vector space whose dimension does not exceed that of V (Proposition 9.14).
7 Proposition 9.8. Let V be a non-trivial vector space, and let S be a finite subset of V whose elements span V . Let n be the smallest positive integer for which there exists a set of n elements of S that span V . Then any n vectors of S that span V are linearly independent, and thus constitute a basis of V .
Proof. Let v1, v2, . . . , vn be n elements of S which span V . We show that these elements are linearly independent.
Suppose that v1, v2, . . . , vn were linearly dependent. Then n > 1, and there would exist scalars a1, a2, . . . , an, not all zero, such that a1v1 + a2v2 + · · · + anvn = 0.
We may suppose, without loss of generality, that an ̸= 0. Then vn = b1v1 + b2v2 + · · · + bn−1vn−1, where bi = −aia−1 n for i = 1, 2, . . . , n −1. But then v1, v2, . . . , vn−1 would span V , since any linear combination of v1, v2, . . . , vn could be expressed as a linear combination of v1, v2, . . . , vn−1. (Indeed n X i=1 civi = n−1 X i=1 (ci + cnbi)vi for all scalars c1, c2, . . . , cn.) But the definition of n ensures that no set of n −1 elements of S can span V . We conclude that v1, v2, . . . , vn must be linearly independent, and thus must constitute a basis of V , as required.
Corollary 9.9. Let V be a non-trivial vector space, and let S be a finite subset of V whose elements span V . Then there exists a basis of V whose elements belong to S.
Corollary 9.10. Every non-trivial finite-dimensional vector space has a basis.
Proposition 9.11. Let V be a non-trivial vector space, let S be a finite subset of V whose elements span V , and let x1, x2, . . . , xm be linearly independent elements of S. Let n be the smallest positive integer for which there exists a set of n elements of S that span V . Then m ≤n, and the elements x1, x2, . . . , xm can be included in a basis of V that consists of n elements of S.
Proof. We claim that if k is a non-negative integer less than m and n, and if the elements xi for i ≤k can be included in a basis of V consisting of n elements of S, then so can the elements xi for i ≤k + 1. Suppose therefore that v1, . . . , vn−k are elements of S and that the elements xi for 1 ≤i ≤k together with the elements vi for 1 ≤i ≤n −k constitute a basis of V .
Then the elements x1, x2, . . . , xk+1, v1, v2, . . . , vn−k are linearly dependent, since xk+1 can be expressed as a linear combination of the other elements in this list. Thus there exist scalars a1, a2, . . . , ak+1 and b1, b2, . . . , bn−k, not all zero, such that k+1 X i=1 aixi + n−k X i=1 bivi = 0.
The linear independence of x1, x2, . . . , xk+1 ensures that the scalars b1, b2, . . . , bn−k are not all zero.
Without loss of generality we may suppose that bn−k ̸= 0. Then vn−k = − k+1 X i=1 aib−1 n−kxi − n−k−1 X i=1 bib−1 n−kvi.
8 It follows from this that V is spanned by the n elements x1, x2, . . . , xk+1 and v1, v2, . . . , vn−k−1. But n is the smallest positive integer for which there exist n elements of S that span V . It follows from Proposition 9.8 that any n vectors of S that span V constitute a basis of V . Therefore the elements x1, x2, . . . , xk+1 and v1, v2, . . . , vn−k−1 constitute a basis of V .
We have shown that if k is a non-negative integer less than m and n, and if the elements xi with i ≤k can be included in a basis of V consisting of n elements of S, then so can the elements xi with i ≤k + 1. Given any positive integer j that does not exceed m or n we can apply this result with k = 0, 1, . . . , j −1. We deduce that the elements x1, x2, . . . , xj can be included in a basis of V consisting of n elements of S, provided that j does not exceed m or n.
Suppose it were the case that m > n. Then it would follow (on taking j = n) that the elements x1, x2, . . . , xn would constitute a basis of V , and therefore each of the vectors xn+1, . . . , xm would be expressed as a linear combination of x1, x2, . . . , xn. But this would contradict the linear independence of x1, x2, . . . , xm. Therefore m ≤n. It then follows (on taking j = m) that x1, x2, . . . , xm can be included in a basis of V consisting of n elements of S, as required.
Corollary 9.12. Let V be a non-trivial vector space, and let X and Y be finite subsets of V . Suppose that the elements of X are linearly independent and the elements of Y span V . Then the number of elements of X does not exceed the number of elements of Y .
Proof. Let r and s denote the number of elements in X and Y respectively, and let S = X ∪Y .
Then the elements of S span V . Let n be the smallest positive integer for which there exists a set of n elements of S that span V . It follows from Proposition 9.11 that r ≤n. But n ≤s, since the elements of Y span V . Therefore r ≤s, as required.
Corollary 9.13. Any two bases of a finite-dimensional vector space contain the same number of elements.
Proof. This result follows immediately from Corollary 9.13, since the elements of any basis of a finite-dimensional vector space are linearly independent and span the vector space.
Definition. The dimension of a finite-dimensional vector space is defined to be number of elements in any basis of that vector space. The dimension is defined to be zero in the case where the vector space consists of just the zero element.
Example. The vector space Rn consisting of all n-tuples of real numbers is an n-dimensional real vector space.
Example. The field C of complex numbers is a 2-dimensional real vector space: the numbers 1 and √−1 constitute a basis of C as a real vector space since any complex number can be expressed uniquely in the form x + y√−1, where x and y are required to be real numbers.
Example. Let n be a positive integer.
The vector space consisting of all polynomials with real coefficients whose degree does not exceed n is an (n+1)-dimensional real vector space: the polynomials 1, x, x2, . . . , xn constitute a basis for this vector space.
Proposition 9.14. A subspace U of a finite-dimensional vector space V is itself a finite-dimensional vector space whose dimension cannot exceed that of V . Moreover if m and n are the dimensions of U and V then there exists a basis v1, v2, . . . , vn of V with the property that the first m elements of this basis constitute a basis of the subspace U.
9 Proof. Let U be a subspace of a finite-dimensional vector space V .
The result is trivial when U = {0}.
Suppose then that U is non-trivial.
Now Corollary 9.12 ensures that the number of linearly independent elements in any subset of a finite-dimensional vector space V cannot exceed the dimension of V . Let m be the largest number of linearly independent elements in any subset of U, and let v1, v2, . . . , vm be linearly independent elements of U. We claim that these elements span U and therefore constitute a basis for U.
Let u be an element of U. Then the elements v1, v2, . . . , vm, u must be linearly dependent, since this list contains m + 1 members. It follows that there exist scalars c1, c2, . . . , cm and d, not all zero, such that c1v1 + c2v2 + · · · + cmvm + du = 0.
The linear independence of v1, v2, . . . , vm then ensures that d ̸= 0. Then u = −c1d−1v1 −c2d−1v2 −· · · −cmd−1vm.
We conclude that the linearly independent elements v1, v2, . . . , vm span U and thus constitute a basis of U. Moreover the dimension m of U does not exceed the dimension n of V .
Finally let S = {v1, v2, . . . , vm} ∪S0, where S0 is a finite subset of V whose elements span V . It follows from Proposition 9.11 that there exists a basis of V consisting of elements of S which includes v1, v2, . . . , vm. Therefore there exist elements vi of S0 for m < i ≤n such that v1, v2, . . . , vn is a basis of V . The first m elements of this basis constitute a basis of U, as required.
Example. The space consisting of all polynomials with real coefficients is an infinite-dimensional real vector space. For if this space were a finite-dimensional vector space whose dimension is N then no linearly independent subset of the vector space could contain more than N elements. But the polynomials 1, x, x2, . . . , xn are linearly independent for each positive integer n. Therefore the space of all polynomials with real coefficients cannot be finite-dimensional.
Example. The space consisting of all continuous real-valued functions defined on the set R of real numbers is an infinite-dimensional real vector space since the polynomial functions constitute an infinite-dimensional subspace, and a finite-dimensional vector space cannot contain any infinite-dimensional subspace (Proposition 9.14).
9.7 Linear Transformations Definition. Let V and W be vector spaces over some field K. A function T: V →W is said to be a linear transformation if T(u + v) = T(u) + T(v) and T(cv) = cT(v) for all elements u and v of V and for all elements c of K.
Let V and W be vector spaces over a field K, and let v1, v2, . . . , vn be elements of a vector space V , and let T: V →W be a linear transformation from V to W. Let wj = T(vj) for j = 1, 2, . . . , n. Then T(c1v1 + c2v2 + · · · + cnvn) = T(c1v1) + T(c2v2) + · · · + T(cnvn) = c1w1 + c2w2 + · · · + cnwn for all scalars c1, c2, . . . , cn.
In particular, Rn be the space of all ordered n-tuples (x1, x2, . . . , xn) of real numbers, and, for each integer j between 1 and n, let ej be the ordered n-tuple whose jth component is equal to one and whose other components are zero. Now (x1, x2, . . . , xn) = x1e1 + x2e2 + · · · + xnen 10 for any ordered n-tuple (x1, x2, . . . , xn). Thus if T: Rn →W is a linear transformation from Rn to some real vector space W then T(x1, x2, . . . , xn) = x1w1 + x2w2 + · · · + xnwn, where wj = T(ej) for j = 1, 2, . . . , n.
Lemma 9.15. Let V and W be vector spaces over a given field, and let T: V →W be a linear transformation from V to W. Then T(u −v) = T(u) −T(v) for all elements u and v of V , and T(0) = 0.
Proof. Let u and v be elements of V . Then u −v = u + (−1)v (Lemma 9.4), and hence T(u −v) = T(u + (−1)v) = T(u) + (−1)T(v) = T(u) −T(v).
In particular T(0) = T(v −v) = T(v) −T(v) = 0, as required.
Definition. The kernel ker T of a linear transformation T: V →W is defined by ker T = {v ∈V : T(v) = 0}.
Lemma 9.16. The kernel ker T of a linear transformation T: V →W is a subspace of V .
Proof. The kernel ker T is non-empty, since 0 ∈ker T. Let u and v be elements of ker T. Then T(u + v) = T(u) + T(v) = 0 + 0 = 0, and therefore u + v is also an element of ker T. Moreover if v is an element of ker T then so is cv, since T(cv) = cT(v) = c0 = 0. Thus ker T is a subspace of V .
Definition. The range (or image) T(V ) of a linear transformation T: V →W is defined by T(V ) = {T(v) : v ∈V }.
Lemma 9.17. The range T(V ) of a linear transformation T: V →W is a subspace of W.
Proof. The range T(V ) is clearly non-empty. The sum of two elements of T(V ) must belong to T(V ) since T(u) + T(v) = T(u + v) for all elements u and v of V . Also any scalar multiple of an element of T(V ) must belong to T(V ), since cT(v) = T(cv) for all scalars c and elements v of V .
9.8 Representation of Linear Transformations by Matrices Let V and W be vector spaces of dimensions m and n respectively over some field K, and let T: V →W be a linear transformation from V to W. Let v1, v2, . . . vm be a basis for V , and let w1, w2, . . . wn be a basis for W. An element u of V can then be expressed as a linear combination of elements of the given basis of V : u = x1v1 + x2v2 + · · · + xmvm, where x1, x2, . . . , xm are scalars (i.e., elements of the field K). Similarly the image T(u) of u under the linear transformation T can also be expressed as a linear combination of elements of the given basis of W: T(u) = y1w1 + y2w2 + · · · + ynwn, 11 where y1, y2, . . . , yn are scalars. But if T: V →W is a linear transformation then T(u) = T (x1v1 + x2v2 + · · · + xmvm) = T(x1v1) + T(x2v2) + · · · + T(xmvm) = x1T(v1) + x2T(v2) + · · · + xmT(vm).
Moreover T(v1), T(v2), . . . T(vm) are elements of the vector space W, and we can therefore write T(vk) = Pn j=1 Mjkwj, where the quantities Mjk are scalars. It follows that T(u) = T m X k=1 xkvk !
= m X k=1 xkT(vk) = m X k=1 xk n X j=1 Mjkwj = n X j=1 m X k=1 Mjkxkwj.
Examination of this formula shows that yj = m X k=1 Mjkxk.
(Thus y1 = M11x1 + M12x2 + · · · + M1mxm etc.) The relation between the coefficients x1, x2, . . . , xm and y1, y2, . . . , yn can be expressed more succinctly by the matrix equation y = Mx, where x and y are the column vectors given by x = x1 x2 .
.
.
xn and y = y1 y2 .
.
.
ym .
and M is the matrix with the value Mjk in the jth row and kth column. For example, suppose that the dimensions of V and W are 3 and 2 respectively. Then y1 y2 = M11 M12 M13 M21 M22 M23 x1 x2 x3 , where the coefficients Mjk of the 2×3 matrix representing the linear transformation T are determined so that T(v1) = M11w1 + M21w2 T(v2) = M12w1 + M22w2 T(v3) = M13w1 + M23w2 Example. An anticlockwise rotation about the vertical axis though an angle of θ radians sends a vector with Cartesian components (u, v, w) to the vector (u′, v′, w′), where u′ = u cos θ −v sin θ, v′ = u sin θ + v cos θ, w′ = w.
This rotation is thus a linear transformation on the space R3 of vectors in 3-dimensional Euclidean space. The three vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) constitute a basis of R3 and the rotation is represented with respect to this basis by the 3 × 3 matrix cos θ −sin θ 0 sin θ cos θ 0 0 0 1 .
12 Example. Let n be a positive integer, let Vn be the real vector space consisting of all polynomials with real coefficients whose degree does not exceed n, and let Vn−1 be the subspace of Vn consisting of those polynomials whose degree does not exceed n −1. (The algebraic operations of addition and of multiplication by real numbers are defined on Vn in the usual fashion.) It is easily seen that the function that sends a polynomial p(x) to its derivative p′(x) is a linear transformation from Vn to Vn−1, and therefore is represented by a matrix with respect to chosen bases of Vn and Vn−1.
Suppose for example that we take n = 3. We can take as our basis for the space V3 the four polynomials 1, x, x2 and x3. The polynomials 1, x and x2 will then provide a basis for the space V2.
The matrix of the linear transformation sending a polynomial to its derivative is represented with respect to these bases of V3 and V2 by the 3 × 4 matrix 0 1 0 0 0 0 2 0 0 0 0 3 since the derivatives of the polynomials 1, x, x2 and x3 constituting the chosen basis of V3 are 0, 1, 2x and 3x2 respectively.
Let U, V and W be vector spaces over a field K, let u1, u2, . . . ul be a basis for U, let v1, v2, . . . vm be a basis for V , and let w1, w2, . . . wn be a basis for W. Let S: U →V and T: V →W be linear transformations, and let L and M be the matrices representing the linear transformations S and T respectively with respect to the chosen bases of U, V and W. Then S(uk) = Pm j=1 Ljkvj and T(vj) = Pn i=1 Mijwi, and hence TS(uk) = T m X j=1 Ljkvj = m X j=1 LjkT(vj) = m X j=1 Ljk n X i=1 Mijwi !
= n X i=1 m X j=1 MijLjk wi = n X i=1 (ML)ikwi, where (ML)ik denotes the element in the ith row and kth column of the product matrix ML. This calculation demonstrates that when linear transformations are represented by matrices with respect to chosen bases of the vector spaces involved, the composition of two linear transformations is represented by the product of the corresponding matrices.
Proposition 9.18. Let V and W be finite-dimensional vector spaces over some given field, and let T: V →W be a linear transformation from V to W. Let w1, w2, . . . , wr be a basis of T(V ), and let v1, v2, . . . , vn be elements of V , where n ≥r. Suppose that T(vj) = wj for j = 1, 2, . . . , r. If ker T = {0} suppose that n = r. If ker T ̸= {0} suppose that n > r and that the elements vj with j > r constitute a basis for the kernel ker T of T. Then v1, v2, . . . , vn is a basis of V .
Proof. Let u be an element of V . Then T(u) belongs to the range T(V ) of T, and hence there exist scalars c1, c2, . . . , cr such that T(u) = Pr j=1 cjwj. Then T u − r X j=1 cjvj = T(u) − r X j=1 cjwj = 0.
and thus u −Pr j=1 cjvj belongs to the kernel of T. If n = r then the kernel of T consists of just the zero vector, and therefore u = Pr j=1 cjvj. If on the other hand n > r then any element of ker T can 13 be expressed as a linear combination of the elements vr+1, . . . , vn, since these elements constitute a basis of ker T, and therefore there exist scalars cr+1, . . . , cn such that u − r X j=1 cjvj = n X j=r+1 cjvj.
Then u = Pn j=1 cjvj. We conclude that the elements v1, v2, . . . , vn span the vector space V .
We now show that the elements v1, v2, . . . , vn are linearly independent. Let c1, c2, . . . , cn be scalars satisfying Pn j=1 cjvj = 0. Now r X j=1 cjwj = n X j=1 cjT(vj) = T n X j=1 cjvj = 0, since T(vj) = wj if 1 ≤j ≤r and T(vj) = 0 if r < j ≤n. But the elements w1, w2 . . . , wr are linearly independent, since they constitute a basis of T(V ). It follows from this that cj = 0 for each j between 1 and r. If n = r then we can conclude immediately that the elements v1, v2, . . . , vn are linearly independent. Suppose that n > r. Then Pn j=r+1 cjvj = 0, since Pn j=1 cjvj = 0 and cj = 0 when j ≤r. But the elements vj with r < j ≤n are linearly independent, since they constitute a basis for the kernel of T. Thus if n > r then cj = 0 for each j satisfying r < j ≤n. We have thus shown that if Pn j=1 cjvj = 0 then cj = 0 for each integer j between 1 and n. We conclude that the elements v1, v2, . . . , vn are linearly independent, and thus constitute a basis of V .
Corollary 9.19. Let V and W be finite-dimensional vector spaces over some given field, and let T: V →W be a linear transformation from V to W. Let w1, w2, . . . , wr be a basis of T(V ). Then there exists a basis v1, v2, . . . , vn of V , where n ≥r, such that T(vj) = wj if 1 ≤j ≤r; 0 if r < j ≤n.
Moreover if n > r then the elements vj with r < j ≤n constitute a basis for the kernel of T.
Proof. The existence of the required basis v1, v2, . . . , vn follows on applying Proposition 9.18. If Pn j=1 cjvj belongs to the kernel of T then cj = 0 for j = 1, 2, . . . , r, since T(Pn j=1 cjvj) = Pr j=1 cjwj.
It follows that if n > r then the elements vj with r < j ≤n span the kernel of T. These elements are also linearly independent. They therefore constitute a basis of the kernel of T, as required.
Let V and W be finite-dimensional vector spaces over some given field, let T: V →W be a linear transformation from V to W. The rank of T is defined to be the dimension of the range T(V ) of T, and the nullity of T is defined to be the dimension of the kernel of T. The dimension of a vector space is by definition the number of elements in a basis of that vector space. The following result therefore follows immediately from Corollary 9.19.
Corollary 9.20. Let V and W be finite-dimensional vector spaces over some given field, and let T: V →W be a linear transformation from V to W. Then the sum of the rank and nullity of T is equal to the dimension of V .
Example. Let T: V →W be a linear transformation of rank 2 from a vector space V of dimension 4 to a vector space W of dimension 3. The range T(V ) of T is 2-dimensional. We see from Proposition 9.14 that there exists a basis w1, w2, w3 of W with the property that the elements w1 and w2 belong to 14 the range T(V ) and constitute a basis for T(V ). Then Theorem 9.19 shows the existence of a basis v1, v2, v3, v4 of V such that T(v1) = w1, T(v2) = w2, T(v3) = 0 and T(v4) = 0. The matrix representing the linear transformation T: V →W with respect to these bases of V and W is then 1 0 0 0 0 1 0 0 0 0 0 0 .
This example can easily be generalized to apply to linear transformations between finite-dimensional vector spaces of arbitrary dimensions.
9.9 Isomorphisms of Vector Spaces A function f: X →Y from a set X to a set Y is said to be injective if, given any element y of Y , there exists at most one element x of X satisfying f(x) = y. A function f: X →Y is said to be surjective if, given any element y of Y , there exists at least one element x of X satisfying f(x) = y. A function f: X →Y is said to be bijective if, given any element y of Y , there exists exactly one element x of X satisfying f(x) = y. We see from these definitions that a function is bijective if and only if it is both injective and surjective. Injective, surjective and bijective functions are referred to as injections, surjections and bijections.
A function f−1: Y →X is said to be the inverse of a function f: X →Y if f−1(f(x)) = x for all x ∈X and f(f−1(y)) = y for all y ∈Y . It is a straightforward exercise to show that a function between two sets is bijective if and only if it has a well-defined inverse: the inverse f−1: Y →X of a bijection f: X →Y is the function that sends an element y of Y to the unique element x of X satisfying f(x) = y. The inverse of a bijection is itself a bijection.
Definition. Let V and W be vector spaces over a given field. An isomorphism from V to W is a linear transformation T: V →W that is also a bijection from V to W. The vector spaces V and W are said to be isomorphic if there exists an isomorphism from V to W.
We recall that the kernel of a linear transformation T: V →W between vector spaces V and W is the subspace of V consisting of all vectors of V that are sent by T to the zero vector of W.
Lemma 9.21. A linear transformation T: V →W is injective if and only if ker T = {0}.
Proof. Suppose that T: V →W is injective. Then there can be at most one element v of V satisfying T(v) = 0. But T(0) = 0 (Lemma 9.15). Therefore ker T = {0}.
Conversely suppose that T: V →W is a linear transformation with the property that ker T = {0}.
Let u and v be elements of V satisfying T(u) = T(v). Then T(u −v) = T(u) −T(v) = 0 and hence u−v ∈ker T. But then u−v = 0, and hence u = v. Thus if ker T = {0} then T: V →W is injective, as required.
Lemma 9.22. Let V and W be vector spaces over a given field. A linear transformation T: V →W from V to W is an isomorphism if and only if ker T = {0} and T(V ) = W.
Proof. The result follows immediately from the fact that the linear transformation T: V →W is injective if and only if ker T = {0}, and T: V →W is surjective if and only if T(V ) = W.
Lemma 9.23. Let V and W be vector spaces over a given field, and let T: V →W be an isomorphism from V to W. Then the inverse T −1: W →V of the function T is well-defined and is an isomorphism from W to V .
15 Proof. A function is bijective if and only if it has a well-defined inverse. Therefore, given an isomor-phism T: V →W, there exists a well-defined function T −1: W →V that is an inverse of the function T: V →W. Moreover the function T −1: W →V is a bijection. It remains to show that the inverse function T −1 is a linear transformation.
Let w and x be elements of W. Then there exist unique elements u and v of V satisfying T(u) = w and T(v) = x. Then T(u + v) = w + x, and thus T −1(w + x) = u + v = T −1(w) + T −1(x).
Also T −1(cw) = cu = cT −1(w) for all scalars c. Thus the bijection T −1: W →V is indeed a linear transformation, and is therefore an isomorphism from W to V , as required.
Lemma 9.24. Let V and W be vector spaces over a given field, and let T: V →W be an isomorphism from V to W. Let v1, v2, . . . , vn be elements of V .
(i) If v1, v2, . . . , vn are linearly independent, then so are T(v1), T(v2), . . . , T(vn).
(ii) If v1, v2, . . . , vn span the vector space V , then T(v1), T(v2), . . . , T(vn) span the vector space W.
(iii) If v1, v2, . . . , vn is a basis of V , then T(v1), T(v2), . . . , T(vn) is a basis of W.
Proof. Suppose that v1, v2, . . . , vn are linearly independent elements of V .
Let c1, c2, . . . , cn be scalars satisfying Pn j=1 cjT(vj) = 0. We must prove that cj = 0 for all j. Now T Pn j=1 cjvj = Pn j=1 cjT(vj), since T: V →W is a linear transformation.
It follows that Pn j=1 cjvj belongs to the kernel of T. But ker T = {0}, since T: V →W is an isomorphism (Lemma 9.22). Therefore Pn j=1 cjvj = 0. It follows now from the linear independence of v1, v2, . . . , vn that cj = 0 for all j.
Thus T(v1), T(v2), . . . , T(vn) are linearly independent elements of W. This proves (i).
Next suppose that v1, v2, . . . , vn are elements of V that span V . Let w be any element of W.
Then w = T(u) for some element u of V , since the isomorphism T: V →W is surjective. There exist scalars c1, c2, . . . , cn for which u = Pn j=1 cjvj, since the elements vj span V , and then w = T(u) = T n X j=1 cjvj = n X j=1 cjT(vj).
Thus T(v1), T(v2), . . . , T(vn) span the vector space W. This proves (ii).
Finally we note that (iii) is an immediate consequence of (i) and (ii).
Theorem 9.25. Two finite-dimensional vector spaces over a given field are isomorphic if and only if they have the same dimension.
Proof. The dimension of a vector space is the number of elements in any basis of that vector space.
It follows immediately from Lemma 9.24 that isomorphic vector spaces have the same dimension.
Suppose that V and W are vector spaces over a given field that have the same dimension. We must show that V and W are isomorphic.
This result clearly holds when the dimension of both spaces is zero. Suppose then that the dimension of these spaces is non-zero. Now there exists a basis v1, v2, . . . , vn of V and a basis w1, w2, . . . , wn of W where both bases have the same number of elements. Moreover, given any element u of V , there exist uniquely determined scalars x1, x2, . . . , xn such that u = Pn j=1 xjvj (Theorem 9.7). It follows that there exists a well-defined function T: V →W characterized by the property that T Pn j=1 xjvj = Pn j=1 xjwj for all scalars x1, x2, . . . , xn. One 16 can readily check that the function T is a linear transformation. It is also invertible: the inverse T −1 of T satisfies T −1 Pn j=1 xjwj = Pn j=1 xjvj for all scalars x1, x2, . . . , xn. Therefore T: V →W is an isomorphism, and thus the vector spaces V and W are isomorphic, as required.
Let K be a field, and, for each positive integer n, let Kn be the set of all ordered n-tuples (x1, x2, . . . , xn) of elements of K. Then Kn is a vector space over the field K, where (x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1 + y1, x2 + y2, . . . , xn + yn) and c(x1, x2, . . . , xn) = (cx1, cx2, . . . , cxn) for all elements (x1, x2, . . . , xn) and (y1, y2, . . . , yn) of Kn and for all elements c of K. Now Kn is an n-dimensional vector space over K, and the ordered n-tuples in which one component has the value 1 and the remaining components are zero constitute a basis of Kn.
Theorem 9.25 ensures that any vector space of dimension n over the field K is isomorphic to the vector space Kn. Indeed if v1, v2, . . . , vn is a basis for a vector space V of dimension n over the field K, then the function sending an element (x1, x2, . . . , xn) of Kn to Pn j=1 xjvj is an isomorphism from Kn to V .
Any real vector space of dimension n is isomorphic to the space Rn.
In particular, any 3-dimensional real vector space is isomorphic to the vector space consisting of all vectors in 3-dimensional Euclidean space.
Problems 1. Let V be a vector space over a field K.
Using the vector space axioms, the definition of subtraction in a vector space etc., prove formally that the following identities are satisfied for all elements u, v and w of V and for all elements c and d of V : (i) u −(v −w) = (u −v) + w; (ii) c(v −w) = cv −cw; (iii) (c −d)v = cv −dv.
2. Let C([0, 1], R) be the real vector space consisting of all continuous real-valued functions on the interval [0, 1], where the operations of addition of functions and of multiplication of functions by real numbers are defined in the usual fashion. Is the subset of C([0, 1], R) consisting of those continuous functions f: [0, 1] →R that satisfy 0 ≤ R 1 0 f(x) dx ≤1 a subspace of C([0, 1], R)?
3. Show that (1, 0, 0), (1, 2, 0), (1, 2, 3) is a basis of the real vector space R3.
4. For each non-negative integer n, let Vn be the real vector space consisting of all polynomials with real coefficients whose degree does not exceed n.
(i) Show that 24, 24x, 12x2, 4x3, x4 is a basis of V4.
(ii) Suppose that n ≥1. Is the function from Vn to itself that sends a polynomial p(x) to p(x) + x a linear transformation?
(iii) Is the function from Vn to Vn+1 that sends a polynomial p(x) to xp(x) a linear transforma-tion?
17 (iv) Let S: V4 →V4 be the linear transformation from V4 to itself that sends a polynomial p(x) to its derivative p′(x). Write down the matrix that represents S with respect to the basis 24, 24x, 12x2, 4x3, x4 of V4.
(v) Let T: V4 →V3 be the linear transformation from V4 to V3 that sends a polynomial p(x) to p′′(x) −3p′(x). Calculate the matrix that represents T with respect to the basis 1, x, x2, x3, x4 of V4 and the basis 1, x, x2, x3 of V3.
5. Let T: R3 →R2 be the linear transformation given by T(u1, u2, u3) = (u1 + u2, u1 −u2). Find a basis v1, v2, v3 of R3 such that T(v1) = (1, 0), T(v2) = (0, 1) and T(v3) = (0, 0).
6. Let U, V and W be vector spaces over some given field. Suppose that the vector spaces U and V are isomorphic and that the vector spaces V and W are isomorphic. Explain why the vector spaces U and W are then isomorphic.
7. Let V be a vector space of dimension n over a field K. Suppose that the field K is finite and has q elements. Is the vector space V then a finite set, and, if so, how many elements does it have?
8. Let V be a vector space over a field K, and let U be a subspace of V . A coset of U in V is a subset of V that is of the form U +v, where U +v = {u+v : u ∈U}. Show that the set V/U of all cosets of U in V is a well-defined vector space over K, where (U + v) +(U + w) = U + v + w and c(U + v) = U + cv for all elements v and w of V and for all elements c of K.
Now let V and W be vector spaces over K, and let T: V →W be a linear transformation. Show that the function that sends a coset ker T + v of the kernel ker T to T(v) is well-defined and is an isomorphism from V/ ker T to T(V ).
18 |
187650 | https://math.colorado.edu/~nita/TheDerivativeFunctionsol.pdf | Math 1300: Calculus I The Derivative Function 1. The purpose of this problem is to see how to construct a derivative function one point at a time by looking at a graph.
Background review: estimating derivatives, one point at a time: • The derivative of a function at a point represents the slope (or rate of change) of a function at that point.
• If you have a graph, you can estimate the derivative one point at a time by drawing the tangent line at that point, then calculating the slope of that tangent line (remember, slope is rise over run).
(a) Go to the website (b) Enter the function y = x2 −x −2. Use the tool to calculate the slope of the graph at each of the points x = -1, 0, 1, 2 and 2.5. Enter the values of the slope in the following table: x −1 0 1 2 2.5 slope of f at x -3 -1 1 3 4 (c) Now plot these points and connect them smoothly to see a graph of f′(x) −3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 y = f′(x) (d) What do you think the formula for this graph is?
Solution: It’s a straight line with slope 2 and y-intercept (0, −1), so y = 2x −1 1 Math 1300: Calculus I The Derivative Function 2. In this problem, you’ll calculate the derivative of the same function as the previous problem, but this time you’ll do it analytically (with formulas) (a) Calculate the derivative of f(x) = x2 −x −2.
f′(x) = lim h→0 f(x + h) −f(x) h = Solution: f′(x) = lim h→0 f(x + h) −f(x) h = lim h→0 ((x + h)2 −(x + h) −2) −(x2 −x −2) h = lim h→0 x2 + 2xh + h2 −x −h −2 −x2 + x + 2 h = lim h→0 2xh + h2 −h h = lim h→0 h(2x + h −1) h = lim h→0(2x + h −1) = 2x −1.
(b) Do your results from this problem match your results from the last problem?
Solution: Yes 3. Khan academy has an exercise that gives a good visual and tactile experience of producing the derivative function point by point. Do at least one exercise on this website: In the space below, sketch the graph of f(x) and f′(x) from one of the exercises you did.
Solution: Solutions vary.
2 Math 1300: Calculus I The Derivative Function 4. Below is a graph of a function.
−3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 Without the aid of technology, use the graph of the function above to sketch a graph of its derivative function on the axes below.
−3 −2 −1 1 2 3 4 −4 −3 −2 −1 1 2 3 4 3 Math 1300: Calculus I The Derivative Function 5. Based on your experience above, what seems to be true about the relationship between f(x) and f′(x)?
(a) Where f(x) is increasing, f′(x) is positive .
(b) Where f(x) is decreasing, f′(x) is negative .
6. Below are some more involved questions. We will be addressing these in the coming sections.
Do you have any guesses for these?
(a) Where f(x) is concave up, f′(x) is increasing .
(b) Where f(x) is concave down, f′(x) is decreasing .
(c) Where f(x) is discontinuous , has a cusp or corner , or has a vertical tangent line f′(x) is undefined.
Solution: Note: discontinuities can include where f(x) is undefined, has a vertical asymptote, has a jump discontinuity, or has a hole.
(d) What will the derivative be when f(x) has a relative high point (maximum) or relative low point (minimum)?
Solution: At the relative extreme points of f(x), the derivative will be zero or will be undefined.
(e) If the derivative is 0 at a point, what are all the ways the original function could look?
Solution: The top of a hill (y = −x2 at x = 0), the bottom of a valley (y = x2 at x = 0), or a horizontal tangent line that is neither a high point or a low point (y = x3 at x = 0).
(f) What about if the derivative has a relative maximum or minimum?
Solution: Where the derivative has a relative maximum the original function has its steepest positive slope and where the derivative has a relative minimum the original function has its steepest negative slope. The function changes concavity there.
4 |
187651 | https://www.edmundoptics.com/f/wedged-windows/15076/?srsltid=AfmBOoo4Xld-4ZmjShGbFX_HzQugJiJSCtsBY0eue9wxhYZJ_LvWHvyG | Fused Silica Wedged Windows | Edmund Optics
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Family ID The Family ID is used to identify product family presentations and can be used to quickly access presentations utilizing the search.#3770
TECHSPEC® components are designed, specified, or manufactured by Edmund Optics. Learn More
Fused Silica Wedged Windows
TECHSPEC® Fused Silica Wedged Windows
×
UV Fused Silica Substrates with a 30 Arcminute Wedge
λ/10 Surface Flatness and 20-10 Surface Quality
Prevent Laser Instability When Used in Laser Cavities
N-BK7 Wedged Windows and Fused Silica Flat Windows Also Available
TECHSPEC® Fused Silica Wedged Windows are manufactured from UV grade fused silica and feature a 30 arcminute wedge. The wedge of these windows eliminates Etalon effects by preventing back surface reflections from traveling along the same optical path as the transmitted beam. This protects against laser instability, mode-hopping, and power spikes when used in laser cavities and beam interference effects when used externally. TECHSPEC® Fused Silica Wedged Windows are ideal for use in UV or high power laser applications due to their high UV transmittance and insensitivity to temperature variations. These windows can also be used as beam pick-off optics or beam samplers to monitor laser beam properties such as beam power over time.
Common Specifications
Thickness (mm):
3.00 ±0.20
Edges:
Fine Ground
Young's Modulus (GPa):
73
Index of Refraction (n d):
1.458
Substrate:
Fused Silica (Corning 7980)
Surface Quality:
20-10
Coefficient of Thermal Expansion CTE (10-6/°C):
0.52 (+5 to +35°C)
0.57 (0 to +200°C)
0.48 (-100 to +200°C)
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Export Specifications
| | Dia. (mm) | Thickness (mm) | Coating | Substrate | Wedge Angle (°) | Compare | Stock Number | Price | Buy |
--- --- --- --- --- |
| | | | | | | | | | |
| MORE LESS | 12.50 12.50 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | - - | - [x] | #34-241 | 128.76 $128.76 Qty 6+ $103.23 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | - - | - [x] | #34-242 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | 1 1° | - [x] | #25-667 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | Contact Us × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | 2 2° | - [x] | #25-668 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | 3 3° | - [x] | #25-669 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | Contact Us × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | 4 4° | - [x] | #25-670 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | 5 5° | - [x] | #25-671 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | 8 8° | - [x] | #25-672 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | Contact Us × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | 10 10° | - [x] | #25-673 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 50.00 50.00 | 3.00 3.00 | Uncoated Uncoated | Fused SilicaFused Silica | - - | - [x] | #34-243 | 257.52 $257.52 Qty 6+ $206.46 Request Quote | In Stock × |
| MORE LESS | 12.50 12.50 | 3.00 3.00 | UV-VIS (250-700nm) UV-VIS (250-700nm) | Fused SilicaFused Silica | - - | - [x] | #17-663 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 12.50 12.50 | 3.00 3.00 | VIS 0 (425-675nm) VIS 0° (425-675nm) | Fused SilicaFused Silica | - - | - [x] | #25-615 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 12.50 12.50 | 3.00 3.00 | NIR I (600-1050nm) NIR I (600-1050nm) | Fused SilicaFused Silica | - - | - [x] | #25-618 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 12.50 12.50 | 3.00 3.00 | NIR II (750-1550nm) NIR II (750-1550nm) | Fused SilicaFused Silica | - - | - [x] | #25-621 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 12.50 12.50 | 3.00 3.00 | VIS-NIR (400-1000nm) VIS-NIR (400-1000nm) | Fused SilicaFused Silica | - - | - [x] | #17-666 | 156.51 $156.51 Qty 6+ $125.43 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | UV-VIS (250-700nm) UV-VIS (250-700nm) | Fused SilicaFused Silica | - - | - [x] | #17-664 | 185.37 $185.37 Qty 6+ $147.63 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | VIS-NIR (400-1000nm) VIS-NIR (400-1000nm) | Fused SilicaFused Silica | - - | - [x] | #17-667 | 185.37 $185.37 Qty 6+ $147.63 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | VIS 0 (425-675nm) VIS 0° (425-675nm) | Fused SilicaFused Silica | - - | - [x] | #25-616 | 185.37 $185.37 Qty 6+ $147.63 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | NIR I (600-1050nm) NIR I (600-1050nm) | Fused SilicaFused Silica | - - | - [x] | #25-619 | 185.37 $185.37 Qty 6+ $147.63 Request Quote | In Stock × |
| MORE LESS | 25.00 25.00 | 3.00 3.00 | NIR II (750-1550nm) NIR II (750-1550nm) | Fused SilicaFused Silica | - - | - [x] | #25-622 | 185.37 $185.37 Qty 6+ $147.63 Request Quote | In Stock × |
| MORE LESS | 50.00 50.00 | 3.00 3.00 | UV-VIS (250-700nm) UV-VIS (250-700nm) | Fused SilicaFused Silica | - - | - [x] | #17-665 | 291.93 $291.93 Qty 6+ $233.10 Request Quote | In Stock × |
| MORE LESS | 50.00 50.00 | 3.00 3.00 | VIS-NIR (400-1000nm) VIS-NIR (400-1000nm) | Fused SilicaFused Silica | - - | - [x] | #17-668 | 291.93 $291.93 Qty 6+ $233.10 Request Quote | In Stock × |
| MORE LESS | 50.00 50.00 | 3.00 3.00 | VIS 0 (425-675nm) VIS 0° (425-675nm) | Fused SilicaFused Silica | - - | - [x] | #25-617 | 291.93 $291.93 Qty 6+ $233.10 Request Quote | In Stock × |
| MORE LESS | 50.00 50.00 | 3.00 3.00 | NIR I (600-1050nm) NIR I (600-1050nm) | Fused SilicaFused Silica | - - | - [x] | #25-620 | 291.93 $291.93 Qty 6+ $233.10 Request Quote | In Stock × |
| MORE LESS | 50.00 50.00 | 3.00 3.00 | NIR II (750-1550nm) NIR II (750-1550nm) | Fused SilicaFused Silica | - - | - [x] | #25-623 | 291.93 $291.93 Qty 6+ $233.10 Request Quote | In Stock × |
Technical Information
FUSED SILICA
Typical transmission of a 3mm thick, uncoated fused silica window across the UV - NIR spectra.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with MgF2 (400-700nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R avg ≤ 1.75% @ 400 - 700nm (N-BK7)
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with UV-AR (250-425nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R abs ≤ 1.0% @ 250 - 425nm
R avg≤ 0.75% @ 250 - 425nm
R avg≤ 0.5% @ 370 - 420nm
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with UV-VIS (250-700nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R abs ≤ 1.0% @ 350 - 450nm
R avg≤ 1.5% @ 250 - 700nm
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with VIS-EXT (350-700nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R avg≤ 0.5% @ 350 - 700nm
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with VIS-NIR (400-1000nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R abs≤ 0.25% @ 880nm
R avg≤ 1.25% @ 400 - 870nm
R avg≤ 1.25% @ 890 - 1000nm
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with VIS 0° (425-675nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R avg≤ 0.4% @ 425 - 675nm
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with YAG-BBAR (500-1100nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R abs ≤ 0.25% @ 532nm
R abs ≤ 0.25% @ 1064nm
R avg≤ 1.0% @ 500 - 1100nm
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with NIR I (600 - 1050nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R avg≤ 0.5% @ 600 - 1050nm
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Typical transmission of a 3mm thick fused silica window with NIR II (750 - 1550nm) coating at 0° AOI.
The blue shaded region indicates the coating design wavelengh range, with the following specification:
R abs ≤ 1.5% @ 750 - 800nm
R abs ≤ 1.0% @ 800 - 1550nm
R avg≤ 0.7% @ 750 - 1550nm
Data outside this range is not guaranteed and is for reference only.
Click Here to Download Data
Accessories
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N-BK7 Wedged Windows
λ/10 UV Fused Silica Windows
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PUROSOL™ Optical Cleaner
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Our capabilities include:
Custom dimensions, materials, coatings, and more
High-precision surface quality and flatness
Tight tolerances and complex geometries
Scalable production—from prototype to volume
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1996 USAMO Problems/Problem 2
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 See Also
Problem
For any nonempty set of real numbers, let denote the sum of the elements of . Given a set of positive integers, consider the collection of all distinct sums as ranges over the nonempty subsets of . Prove that this collection of sums can be partitioned into classes so that in each class, the ratio of the largest sum to the smallest sum does not exceed 2.
Solution 1
Let set consist of the integers . For , call greedy if . Also call greedy. Now put all elements of into groups of consecutive terms in such a way that each group begins with a greedy term, call it , and ends on the term just before the next greedy term after . (If is the last greedy term, let .) We introduce some more terminology. A sum is said to "belong to" a group if . Denote by the set of all sums belonging to .
We now show that we can divide into (the cardinality of ) classes in such a way that the maximum and minimum sum belonging to each class differ by no more than a factor of 2. Using the previous notation, we first prove that . Note that and Taking note of that fact that are not greedy numbers, we write:
where the inequalities after the ellipses result from the fact that is a greedy number (which implies by definition that ). This proves the desired inequality. Now we can prove the result in bold above. Divide into classes by taking those terms in ![Image 50: $a_p,2a_p)$ and placing them in the first class, taking those terms in ![Image 51: $2a_p,2^2a_p)$ and placing them in another, and so on, until we reach ![Image 52: $2^{q-p-1}a_p,2^{q-p}a_p)$. The inequality we proved above shows that all of the sums in will fall in one of these classes, as the intervals into which the classes fall form a continuous range bounded by on the bottom and on the top. This proves the result in bold.
However, that clearly implies the desired conclusion, as every sum belongs to a group, and every sum belonging to a group is a member of a class. Moreover, there will be precisely classes, because every term belongs to a group and for each group , there are as many classes for as there are terms in .
Solution 2
We proceed to solve this problem by induction. Our base case is when has 1 integer - there is 1 class: itself.
Let now have elements, and assume it works for this case. We add , another element, to the set. WLOG, let be the maximum element. Note that has to be the maximum element of a class.
We always get a new class that works: being the maximum element. This way all elements including will be in that class if . When it also works since must be the maximum element from our WLOG condition above, so our induction step is done.
See Also
1996 USAMO (Problems • Resources)
Preceded by
Problem 1Followed by
Problem 3
1•2•3•4•5•6
All USAMO Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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187653 | https://blog.csdn.net/weixin_63826827/article/details/135951888 | 子集为什么是2的n次方_n个元素的子集个数为何是2的n次方-CSDN博客
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子集为什么是2的n次方
原创 已于 2024-01-31 14:30:22 修改·1.5k 阅读
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CC 4.0 BY-SA版权
版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。
文章标签：
#笔记
于 2024-01-31 14:28:18 首次发布
文章通过将子集问题与二进制和比特位联系起来，解释了如何用程序猿的思维方式理解子集个数公式，指出n个元素的集合有2^n个子集，通过n个比特位的不同组合表示这些子集。
用程序猿脑子来思考下：
看见子集个数公式：2n2^n 2 n，想到是多个2 相乘：2∗2...∗222...2 2∗2...∗2
意味着原集合每多一个元素子集的数量就翻倍。
子集就是从原集合取一些元素和不取一些元素然后构成一个新的集合。
取和不取的组合，像不像0、1开关，二进制?
举个例子:
集合A = {1,2,3,4}。
可以看作是4比特位的二进制。
4比特位：_ _ _ _ （只能填0或1）
那么
0000，都不取，代表空集∅；
0101代表集合{2,4}；
1110代表{1,2,3}；
以此类推。
AI运行代码
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所以子集个数问题，转化为 n个比特位能表示多少不同情况？
原集合每多一个元素，子集数量就翻倍，因为它多一个比特位 : ）
显然，每个比特位只有0和1两种取值，n个比特位就能表示 2n2^n 2 n 种情况。
也就是n个元素的集合，有 2n2^n 2 n 个子集。
。
。
。
。
还有组合公式∑i=0nCni=2n\sum^n_{i = 0} C_{n}^{i} = 2^n∑i=0 nC n i=2 n
一样可以用n比特位可以表示多少中不同情况的思维来理解。
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HandWiki. Symmetric Difference. Encyclopedia. Available online: (accessed on 28 September 2025).
HandWiki. Symmetric Difference. Encyclopedia. Available at: Accessed September 28, 2025.
HandWiki. "Symmetric Difference" Encyclopedia, (accessed September 28, 2025).
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In mathematics, the symmetric difference, also known as the disjunctive union, of two sets is the set of elements which are in either of the sets and not in their intersection. The symmetric difference of the sets A and B is commonly denoted by or or For example, the symmetric difference of the sets [math]\displaystyle{ {1,2,3} }[/math] and [math]\displaystyle{ {3,4} }[/math] is [math]\displaystyle{ {1,2,4} }[/math]. The power set of any set becomes an abelian group under the operation of symmetric difference, with the empty set as the neutral element of the group and every element in this group being its own inverse. The power set of any set becomes a Boolean ring with symmetric difference as the addition of the ring and intersection as the multiplication of the ring.
symmetric difference abelian group power set
1. Properties
Venn diagram of [math]\displaystyle{ ~A \triangle B \triangle C }[/math] .
The symmetric difference is equivalent to the union of both relative complements, that is:
: [math]\displaystyle{ A\,\triangle\,B = \left(A \setminus B\right) \cup \left(B \setminus A\right), }[/math]
The symmetric difference can also be expressed using the XOR operation ⊕ on the predicates describing the two sets in set-builder notation:
: [math]\displaystyle{ A\,\triangle\,B = {x : (x \in A) \oplus (x \in B)}. }[/math]
The same fact can be stated as the indicator function (which we denote here by [math]\displaystyle{ \chi }[/math]) of the symmetric difference being the XOR (or addition mod 2) of the indicator functions of its two arguments: [math]\displaystyle{ \chi_{(A\,\triangle\,B)} = \chi_A \oplus \chi_B }[/math] or using the Iverson bracket notation [math]\displaystyle{ [x \in A\,\triangle\,B] = [x \in A] \oplus [x \in B] }[/math].
The symmetric difference can also be expressed as the union of the two sets, minus their intersection:
: [math]\displaystyle{ A\,\triangle\,B = (A \cup B) \setminus (A \cap B), }[/math]
In particular, [math]\displaystyle{ A\triangle B\subseteq A\cup B }[/math]; the equality in this non-strict inclusion occurs if and only if [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] are disjoint sets. Furthermore, if we denote [math]\displaystyle{ D = A\triangle B }[/math] and [math]\displaystyle{ I = A \cap B }[/math], then [math]\displaystyle{ D }[/math] and [math]\displaystyle{ I }[/math] are always disjoint, so [math]\displaystyle{ D }[/math] and [math]\displaystyle{ I }[/math] partition [math]\displaystyle{ A \cup B }[/math]. Consequently, assuming intersection and symmetric difference as primitive operations, the union of two sets can be well defined in terms of symmetric difference by the right-hand side of the equality
: [math]\displaystyle{ A\,\cup\,B = (A\,\triangle\,B)\,\triangle\,(A \cap B) }[/math].
The symmetric difference is commutative and associative (and consequently the leftmost set of parentheses in the previous expression were thus redundant):
: [math]\displaystyle{ \begin{align} A\,\triangle\,B &= B\,\triangle\,A, \ (A\,\triangle\,B)\,\triangle\,C &= A\,\triangle\,(B\,\triangle\,C). \end{align} }[/math]
The empty set is neutral, and every set is its own inverse:
: [math]\displaystyle{ \begin{align} A\,\triangle\,\varnothing &= A, \ A\,\triangle\,A &= \varnothing. \end{align} }[/math]
Taken together, we see that the power set of any set X becomes an abelian group if we use the symmetric difference as operation. (More generally, any field of sets forms a group with the symmetric difference as operation.) A group in which every element is its own inverse (or, equivalently, in which every element has order 2) is sometimes called a Boolean group; the symmetric difference provides a prototypical example of such groups. Sometimes the Boolean group is actually defined as the symmetric difference operation on a set. In the case where X has only two elements, the group thus obtained is the Klein four-group.
Equivalently, a Boolean group is an elementary abelian 2-group. Consequently, the group induced by the symmetric difference is in fact a vector space over the field with 2 elements Z2. If X is finite, then the singletons form a basis of this vector space, and its dimension is therefore equal to the number of elements of X. This construction is used in graph theory, to define the cycle space of a graph.
From the property of the inverses in a Boolean group, it follows that the symmetric difference of two repeated symmetric differences is equivalent to the repeated symmetric difference of the join of the two multisets, where for each double set both can be removed. In particular:
: [math]\displaystyle{ (A\,\triangle\,B)\,\triangle\,(B\,\triangle\,C) = A\,\triangle\,C. }[/math]
This implies triangle inequality: the symmetric difference of A and C is contained in the union of the symmetric difference of A and B and that of B and C. (But note that for the diameter of the symmetric difference the triangle inequality does not hold.)
Intersection distributes over symmetric difference:
: [math]\displaystyle{ A \cap (B\,\triangle\,C) = (A \cap B)\,\triangle\,(A \cap C), }[/math]
and this shows that the power set of X becomes a ring with symmetric difference as addition and intersection as multiplication. This is the prototypical example of a Boolean ring.
Further properties of the symmetric difference:
[math]\displaystyle{ A\triangle B = A^c\triangle B^c }[/math], where [math]\displaystyle{ A^c }[/math], [math]\displaystyle{ B^c }[/math] is [math]\displaystyle{ A }[/math]'s complement, [math]\displaystyle{ B }[/math]'s complement, respectively, relative to any (fixed) set that contains both.
[math]\displaystyle{ \left(\bigcup_{\alpha\in\mathcal{I}}A_\alpha\right)\triangle\left(\bigcup_{\alpha\in\mathcal{I}}B_\alpha\right)\subseteq\bigcup_{\alpha\in\mathcal{I}}\left(A_\alpha\triangle B_\alpha\right) }[/math], where [math]\displaystyle{ \mathcal{I} }[/math] is an arbitrary non-empty index set.
If [math]\displaystyle{ f : S \rightarrow T }[/math] is any function and [math]\displaystyle{ A, B \subseteq T }[/math] are any sets in [math]\displaystyle{ f }[/math]'s codomain, then [math]\displaystyle{ f^{-1}\left(A \triangle B\right) = f^{-1}\left(A\right) \triangle f^{-1}\left(B\right) }[/math].
The symmetric difference can be defined in any Boolean algebra, by writing
: [math]\displaystyle{ x\,\triangle\,y = (x \lor y) \land \lnot(x \land y) = (x \land \lnot y) \lor (y \land \lnot x) = x \oplus y. }[/math]
This operation has the same properties as the symmetric difference of sets.
2. n-ary Symmetric Difference
The repeated symmetric difference is in a sense equivalent to an operation on a multiset of sets giving the set of elements which are in an odd number of sets.
As above, the symmetric difference of a collection of sets contains just elements which are in an odd number of the sets in the collection:
: [math]\displaystyle{ \triangle M = \left{ a \in \bigcup M: \left|{A \in M:a \in A}\right| \mbox{ is odd}\right} }[/math].
Evidently, this is well-defined only when each element of the union [math]\displaystyle{ \bigcup M }[/math] is contributed by a finite number of elements of [math]\displaystyle{ M }[/math].
Suppose [math]\displaystyle{ M = \left{M_1, M_2, \ldots, M_n\right} }[/math] is a multiset and [math]\displaystyle{ n \ge 2 }[/math]. Then there is a formula for [math]\displaystyle{ |\triangle M| }[/math], the number of elements in [math]\displaystyle{ \triangle M }[/math], given solely in terms of intersections of elements of [math]\displaystyle{ M }[/math]:
: [math]\displaystyle{ |\triangle M| = \sum_{l=1}^n (-2)^{l-1} \sum_{1 \leq i_1 \lt i_2 \lt \ldots \lt i_l \leq n} \left|M_{i_1} \cap M_{i_2} \cap \ldots \cap M_{i_l}\right| }[/math].
3. Symmetric Difference on Measure Spaces
As long as there is a notion of "how big" a set is, the symmetric difference between two sets can be considered a measure of how "far apart" they are.
First consider a finite set S and the counting measure on subsets given by their size. Now consider two subsets of S and set their distance apart as the size of their symmetric difference. This distance is in fact a metric so that the power set on S is a metric space. If S has n elements, then the distance from the empty set to S is n, and this is the maximum distance for any pair of subsets.
Using the ideas of measure theory, the separation of measurable sets can be defined to be the measure of their symmetric difference. If μ is a σ-finite measure defined on a σ-algebra Σ, the function
: [math]\displaystyle{ d_\mu(X, Y) = \mu(X\,\triangle\,Y) }[/math]
is a pseudometric on Σ. dμ becomes a metric if Σ is considered modulo the equivalence relation X ~ Y if and only if [math]\displaystyle{ \mu(X\,\triangle\,Y) = 0 }[/math]. It is sometimes called Fréchet-Nikodym metric. The resulting metric space is separable if and only if L2(μ) is separable.
If [math]\displaystyle{ \mu(X), \mu(Y) \lt \infty }[/math], we have: [math]\displaystyle{ |\mu(X) - \mu(Y)| \leq \mu(X\,\triangle\,Y) }[/math]. Indeed,
: [math]\displaystyle{ \begin{align} |\mu(X) - \mu(Y)| &= \left|\left(\mu\left(X \setminus Y\right) + \mu\left(X \cap Y\right)\right) - \left(\mu\left(X \cap Y\right) + \mu\left(Y \setminus X\right)\right)\right| \ &= \left|\mu\left(X \setminus Y\right) - \mu\left(Y \setminus X\right)\right| \ &\leq \left|\mu\left(X \setminus Y\right)\right| + \left|\mu\left(Y \setminus X\right)\right| \ &= \mu\left(X \setminus Y\right) + \mu\left(Y \setminus X\right) \ &= \mu\left(\left(X \setminus Y\right) \cup \left(Y \setminus X\right)\right) \ &= \mu\left(X\, \triangle \, Y\right) \end{align} }[/math]
If [math]\displaystyle{ S = \left(\Omega, \mathcal{A},\mu\right) }[/math] is a measure space and [math]\displaystyle{ F, G \in \mathcal{A} }[/math] are measurable sets, then their symmetric difference is also measurable: [math]\displaystyle{ F \triangle G \in \mathcal{A} }[/math]. One may define an equivalence relation on measurable sets by letting F and G be related if [math]\displaystyle{ \mu\left(F \triangle G\right) = 0 }[/math]. This relation is denoted [math]\displaystyle{ F = G\left[\mathcal{A}, \mu\right] }[/math].
Given [math]\displaystyle{ \mathcal{D}, \mathcal{E} \subseteq \mathcal{A} }[/math], one writes [math]\displaystyle{ \mathcal{D}\subseteq\mathcal{E}\left[\mathcal{A}, \mu\right] }[/math] if to each [math]\displaystyle{ D\in\mathcal{D} }[/math] there's some [math]\displaystyle{ E \in \mathcal{E} }[/math] such that [math]\displaystyle{ D = E\left[\mathcal{A}, \mu\right] }[/math]. The relation "[math]\displaystyle{ \subseteq\left[\mathcal{A}, \mu\right] }[/math]" is a partial order on the family of subsets of [math]\displaystyle{ \mathcal{A} }[/math].
We write [math]\displaystyle{ \mathcal{D} = \mathcal{E}\left[\mathcal{A}, \mu\right] }[/math] if [math]\displaystyle{ \mathcal{D}\subseteq\mathcal{E}\left[\mathcal{A}, \mu\right] }[/math] and [math]\displaystyle{ \mathcal{E} \subseteq \mathcal{D}\left[\mathcal{A}, \mu\right] }[/math]. The relation "[math]\displaystyle{ = \left[\mathcal{A}, \mu\right] }[/math]" is an equivalence relationship between the subsets of [math]\displaystyle{ \mathcal{A} }[/math].
The symmetric closure of [math]\displaystyle{ \mathcal{D} }[/math] is the collection of all [math]\displaystyle{ \mathcal{A} }[/math]-measurable sets that are [math]\displaystyle{ = \left[\mathcal{A}, \mu\right] }[/math] to some [math]\displaystyle{ D \in \mathcal{D} }[/math]. The symmetric closure of [math]\displaystyle{ \mathcal{D} }[/math] contains [math]\displaystyle{ \mathcal{D} }[/math]. If [math]\displaystyle{ \mathcal{D} }[/math] is a sub-[math]\displaystyle{ \sigma }[/math]-algebra of [math]\displaystyle{ \mathcal{A} }[/math], so is the symmetric closure of [math]\displaystyle{ \mathcal{D} }[/math].
[math]\displaystyle{ F = G\left[\mathcal{A}, \mu\right] }[/math] iff [math]\displaystyle{ \left|\mathbf{1}_F - \mathbf{1}_G\right| = 0 }[/math] [math]\displaystyle{ \left[\mathcal{A}, \mu\right] }[/math] almost everywhere.
4. Hausdorff Distance Vs. Symmetric Difference
HausdorffVsSymmetric.
The Hausdorff distance and the (area of the) symmetric difference are both pseudo-metrics on the set of measurable geometric shapes. However, they behave quite differently. The figure at the right shows two sequences of shapes, "Red" and "Red ∪ Green". When the Hausdorff distance between them becomes smaller, the area of the symmetric difference between them becomes larger, and vice versa. By continuing these sequences in both directions, it is possible to get two sequences such that the Hausdorff distance between them converges to 0 and the symmetric distance between them diverges, or vice versa.
References
Givant, Steven; Halmos, Paul (2009). Introduction to Boolean Algebras. Springer Science & Business Media. p. 6. ISBN 978-0-387-40293-2.
Humberstone, Lloyd (2011). The Connectives. MIT Press. p. 782. ISBN 978-0-262-01654-4.
Rotman, Joseph J. (2010). Advanced Modern Algebra. American Mathematical Soc.. p. 19. ISBN 978-0-8218-4741-1.
Rudin, Walter (January 1, 1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill Education. p. 306. ISBN 978-0070542358.
Claude Flament (1963) Applications of Graph Theory to Group Structure, page 16, Prentice-Hall MR0157785
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187655 | https://www.youtube.com/watch?v=uYm4l_alVV0 | Respiratory | Mechanics of Breathing: Pressure Changes | Part 1
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all right ninja nerds in this video we're going to talk about the mechanics of breathing so this can be a tough topic for certain people to understand especially with the pressures so we're going to do our best here at Ninja Science to make sense of that so let's go ahead and dig right in so before we do that we need to look at a little bit of anatomy for the lungs and a lot of the chest wall structure so let's do that first so if you look here we have two lungs right right left lung and what's going to happen is you're going to have you know the actual trachea the trachea is going to branch off into the right and left primary bronchus serving the actual lung specifically at the the smallest structural unit called the alvoli we'll talk about that in a second but the lung itself each individual alvoli is making up the lung but if you look at the lung it has this nice little thin epithelial tissue with a little bit of aerol connective tissue clinging on to that organ so you see this blue layer right there that blue layer right there we're going to denote this layer right here let's call this layer one okay so layer one right there so layer one is specifically called the actual visceral plura so again this layer one is actually called the visceral plura okay that's the first layer then let's keep working our way out now you see this space right here this little hollow like cavity but it has a little bit of fluid in it this space right here we're going to call this number two here so number two number two is actually this whole cavity here is actually specifically called the plural cavity now here's what's interesting about the plural cavity in this diagram I'm actually showing a space in the human body there actually is no space it's actually a potential space they call it and the reason why is in our human body the lungs this visceral plura is tethered or connected to the actual this plura right here this last one we'll talk about this last one here is called the parietal plura let me write this one down again this third one is specifically called the parietal plura but to come back to that thought that I was saying remember this visceral plura is almost completely tethered to the parietal plura and how they're tethered together and connected together is through this actual plural cavity what's in the plural cavity plural fluid there's a little bit of like a cirrus like fluid in here that allows for imagine this for a second let's say here I take the eraser this eraser is supposed to represent the visceral plura here's the marker this is supposed to represent the parietal plura technically they are really really close together rubbing up against one another all the time now you might be saying "Oh wait but if that happens all the time wouldn't that produce friction and inflammation and tissue damage?" It would but guess what our body does to prevent that from happening that plural cavity is occupied by what's called a plural fluid that we said and what that plural fluid does is when the actual layers are rubbing up against one another during the inhalation expiration processes it allows for there to be no friction or very little friction and prevents inflammation what happens when actually in certain situations where there is too much fluid accumulation or actually there's very little fluid accumulation and these layers start rubbing up against one another and start causing a lot of agitation it can produce what's called puricy so that is a condition that can come about whenever there is a lot of friction developing between the parietal plura and the visceral plura due to maybe a decreased situation and not enough plural fluid being produced okay so again what do we have here we have the plural cavity number two and number one we have the visceral plural okay now we need to talk about something else we need to talk about pressures because pressures is an important topic that we need to talk about here okay there's three main pressures that we're going to talk about let's denote these A B and C okay so we're going to have call these pressures A B and C so this is going to be pressure A this is going to be pressure B and this is going to be pressure C okay pressure A pressure B pressure C pressure A we're going to just name them first pressure A is actually referred to as the intraulmonary pressure so it's referred to as the intra pulmonary or intraalvolar sometimes they call it intraalvolar pressure okay and then why did I say intraalvolar well you really know what happens is you know technically whenever the trachea is coming here it's giving way to the bronchi and then it goes secondary tertiary and then eventually goes to terminal bronchials respiratory and it branches out to these actual small structures you see these little sacks here these small little like grape-like structures those are called the alvoli so technically when I say intraulmonary pressure I really mean intraalvolar pressure right which is the pressure in here so this is the A pressure right the pressure that we were talking about but I'm just blowing it up here for the sake of this video so it's very clear right so intraulmonary pressure is this one what is this B pressure what is the pressure here in this plural cavity it's called the intraplural pressure not that bad right not that bad to remember so again what is this pressure here it's called the intra plural pressure okay sweet deal and again it's that pressure that actually will be occupied in this plural cavity the last one which is the C pressure is the atmospheric pressure or the barometric pressure you might have even heard of that u as barometric pressure or atmospheric it call uh barometric pressure or atmospheric pressure now why am I saying all this stuff because this is going to be critical once we get these actual pressures down the numbers down it's going to make this whole mechanics a lot easier okay so now we're going to do is we're going to give you numbers for each one of these pressures i'm going to explain a little relationship between the two okay so I'm going to write these down here so the intraulmonary pressure is approximately approximately and we're going to denote it as P pull okay and Pole is just denoting that it's the intraulmonary pressure it is approximately 760 millimeters of mercury that's the unit that they're actually measuring it in right then this next one intraplural pressure intraplural pressure is approximately and we're going to denote this as PIP so we're going to denote this as PIP representing that it's intraplural pressure intraplural pressure is approximately it's always negative we refer to it as a negative pressure and I'll I'll explain what that means when I mean negative pressure so hang in there a little bit intraplural pressure is always less than the intraulmonary pressure you might be like okay well how much about 4 millimeters of mercury less than the intra pulmonary intravolar pressure so what's uh 4 - 760 is about 756 so this is approximately 756 millimeters of mercury which again is the units for this pressure and the last one is going to be the atmospheric pressure the atmospheric pressure or the barometric pressure is at sea level at one atmosphere usually and we say is approximately about 760 mm of mercury okay so let's write this one down we'll put P and we'll put atm which is atmospheric pressure this is approximately 760 millimeters of mercury and again millimeters of mercury is the unit sometimes they use centimeter of water as certain situations right okay we're going to use millimeters of mercury in this situation we're not going to use centimeters of water okay now now that we have all the pressures I want to explain a little bit about these pressures primarily one of the ones that bug people out a lot is the intra plural pressure i want to talk about this a little bit before I do that though I want to correlate this thing I'm going to talk about i'm going to use these terms a lot negative and positive and zero pressures when we compare pressures so if it's zero pressure negative pressure positive pressure we compare it to the atmosphere okay so for example the atmospheric pressure is 760 mm of mercury right what is the intraulmonary pressure 760 mm of mercury what is 760us 760 it's zero so this is called this is actually technically we can also write that this pressure here this intraulmonary pressure is also 0 millm of mercury right and that's because so sometimes just so you know these can be interchangeable I could put 760 or I could put zero all that means is that it's equal to the atmospheric pressure okay now let's compare intraplural pressure to atmospheric pressure okay 760us 756 is 4 millime of mercury but it is a lot when we think about this one okay what we're actually doing is I should actually rephrase this when you're subtracting you're subtracting intraulmonary minus atmospheric and you're subtracting intraplural from atmospheric so if I'm actually subtracting 760 from 760 it's zero but if I subtract 756 minus 760 what is that that's -4 okay sorry about that mess up right all right so now this intraulmonary pressure technically we could also write that it is actually -4 millimeters of mercury okay now that we have these numbers out of the way right so this is a negative pressure this is a zero pressure here i want to explain why this is negative because this this bugs people out okay so let me take intraplural pressure down here i there's three reasons why intra plural pressure is actually negative so let me explain this real quick so intra plural pressure or as we denote it here we denote it as as the P IP i'll refer to this a lot right so it's a negative pressure there is three reasons why this is a negative pressure okay first reason is the elasticity of the lungs okay so the first reason is the natural elasticity of the lungs second reason is what's called surface tension we'll have another video specifically on surface tension and surfactant but this one is going to be surface tension and then the last thing is going to be the elasticity of the chest wall so the last thing is going to be the elasticity of the chest wall okay let me explain what I mean by this and there's also one last thing that I'll mention and it's not with respect to this it's due to the differences in the intraoral pressure throughout the entropal cavity and this is due to gravity i'll mention this last one okay but again this is not really one of the things that's contributing to it it's contributing to a difference in the pressures okay so it can contribute to the differences in the pressure i'll explain what I mean by that because you can see that the pressure in plural pressure could be different here here and here and I'll explain that first off elasticity of the lungs and the surface tension we're going to group those together for a second and let me explain why now what first off what is the definition of elastance how would you define elasticity elasticity is whenever you try to stretch something right it doesn't want to be stretched it wants to resist the actual desire to be stretched it wants to recoil it always wants to assume the smallest size possible that's what elasticity is think about this for a second where is this elasticity coming into play well technically whenever the lungs want to recoil what are they actually doing imagine again I told you that imagine the parietal plura and the visceral plura as actually close together they're actually touching when I try for my lungs to actually deflate if I try to deflate them what is it going to do to the visceral part it's going to pull it away as it pulls it away because it's trying to deflate it's trying to get smaller as the lungs are trying to get smaller it's pulling away from pulling this visceral plural away from the parietal plural now let's do surface tension what is surface tension doing surface tension is this concept that because of the water molecules this interaction between the air and the alvoli and the water molecules it causes this tension at the air water interface and the whole thing is is that the alvoli wants to collapse it wants to assume the smallest size possible so in other words same thing what's the overall purpose the lungs are trying to pull this visceral plura away from the parietal plura okay well that's trying to collapse the lungs and increase this this volume here okay that's one thing that's happening the next thing that's happening is the elasticity of the chest wall okay what's the chest wall trying to do well you know normally our chest wall is decently elastic there's a lot of you know the coastal cartilage we have different types of connective tissue that is allowing for the chest wall to expand so the chest wall if we were to kind of show this here let's say that I'm going to represent the chest wall in this color here and I'm going to represent the elasticity of the lungs in this in the surface tension the green color what direction is it trying to pull the lungs it's trying to pull it this way that's what it's trying to do it's trying to pull the lungs in this way to collapse them whereas the chest wall when you're breathing what is it trying to do it's trying to push the chest wall out to expand the chest wall and if it's trying to expand the chest wall what is that doing it's pulling this parietal plura away from the visceral plura if you're pulling this actual parietal plura away from the visceral plura what is that doing to this volume in here it's increasing the volume so the dynamic interplay between these three concepts here the elasticity of the lungs the surface tension and the elasticity of the chest wall what is the overall result of all of these the overall result of all of these three things is that they're increasing or they're attempting to they're not necessarily doing but they're attempting to they're increasing thoracic cavity volume which is that intra plural space right there that plural cavity space right you know there's a a law boil he came up with a law you know what that law is it states that okay pressure if you have a a certain pressure here let's say I call it p1 v1 is a volume p2 is a second pressure and then you have v2 which is the second uh volume Okay he says based upon this relationship okay based upon this relationship whenever because it's it's in this format whenever I increase the pressure of this reaction of whatever reaction it might be it's going to decrease the volume that's the relationship with Boil's law so Boil's law states that whenever there is a increase in the pressure there will be a direct decrease in the volume same thing let's say that we actually do something opposite let's say that I increase the volume whenever I increase the volume what is that going to do to pressure it's going to drop the pressure huh that's interesting because isn't the whole purpose to make this pressure negative or decrease the pressure below the intraulmonary have it always being a little bit lower or negative pressure yes and that's the whole purpose that's why the intraplural pressure is negative again what are those three reasons the elasticity lungs what do they want to do cause the lungs to snap and and actually collapse snap back to their smallest size possible surface tension wants to collapse the alvola which tries to collapse the lungs pushing this way creating a bigger volume a potential volume space chest wall elasticity of the chest wall constantly whenever we're inspiring it wants to try to bring the actual chest wall out that's what you want to whenever you bring air in what do you want to do you want to try to expand that chest wall so the chest wall is naturally elastic and it wants to expand out this way what is that trying to do it's trying to pull on the parietal plura away from the visceral plura but normally you know our chest wall when it's not contracting what would it actually do it can recoil also so because of that sometimes what it can do you know to say that it's only ever going this way it prefers to be expanded but it can have an actual recoil capability here too okay so it does have a little bit of recoil capability here too but nonetheless the dynamic interplay between the elasticity surface tension and the elasticity of the chest wall play a role in maintaining this negative interpolar pressure you know there's actually one more thing you know there's lymphatic vessels in this area let's say that I represent this lymphatic vessels with this brown structure here let's say here I put a little tube in here here's this little tube and this brown tube right here and I'll put another one right here this brown tube right here are lymphatic vessels let's say that these are lymphatic vessels okay so this is my lymphatic vessels you know what's really important about this plural cavity is that we want to make sure that there's not too much fluid accumulating out in this area we don't want there to be too much fluid and one of the ways that we control that okay so here let's say here's our plural fluid right here's our plural fluid to prevent excessive amounts of plural fluid from accumulating you know what we have we have these little lymphatic vessels from the bronco media trunk area right that can drain this actual plural cavity and prevent the excessive amounts of fluid from building up because you know what happens if we build up a lot of fluid it's going to start trying to push on the lungs right so we don't want that so again plural fluid is constantly being actually drained out by lymphatic vessels to maintain a nice volume in here so it doesn't disturb the intraplural pressure also okay so we got that down so again what have we covered so far we covered visceral plura is this little epithelial tissue layer clinging to the lung plural cavity which is this potential space right consisting of a plural fluid and we talked about the third thing which was the parietal plur which is this layer clinging to the chest wall then we said there's three pressures in the lung or basically across this whole lung structure here right intraulmonary pressure which is also called the intraalvolar pressure right and again we showed it by this alvoli there it's approximately 760 mm of mercury then we said that there's a pressure right here which is the intraplural pressure which is 756 millm of mercury and then we said that there's an atmospheric pressure outside of the body right around us that is the atmospheric pressure which is approximately 760 but I said we could express it another way if I take the intraulmonary pressure and subtract it from the atmospheric what is that that is zero if I take the intraural pressure and subtract it from the atmospheric pressure what is that that's -4 okay and we explained why is it a negative pressure because the elasticity of the lungs and the surface tension they want the lungs to collapse they want to assume the smallest size possible which is going to increase this actual volume of this space potentially Then we also said that the elasticity of the chest wall two things can happen whenever we're inspiring the chest wall would want to expand outwards but whenever we're resting it wants to kind of actually just maintain that size but it can have a force that's kind of driving to direct inwards a little bit right but no matter what the dynamic interplay between the elasticity of the lungs the surface tension and the elasticity of the chest wall helps to keep this volume increasing and by Boil's law we said that whenever the volume is increasing the pressure in this actual cavity is decreasing okay so because of this because the thoracic cavity volume decreases I'm sorry because the thoracic cavity volume increased I'm sorry this would actually decrease the actual thoracic cavity volume but specifically the intra not thoracic cavity volume but thoracic cavity pressure so it would actually decrease the intra plural pressure okay because of boil's law so again whenever you increase the volume in thoracic cavity it's going to decrease the thoracic cavity volume pressure but specifically that pressure that we call the thoracic cavity pressure is really the intraplural pressure and that'll decrease to about4 and then again we said that the plural fluid is actually constantly being pumped out of the plural cavity by the lymphatic vessels like the bronco mediainal trunk to maintain a normal volume so it doesn't interfere with the actual intraplural pressure one more thing and then we're going to go over these actual changes of how breathing is affected here gravity I mentioned gravity now when gravity is actually acting downwards what happens let's say that I actually pretend for a second that I take the bottom of this lung here i take the bottom of this lung and I try to yank it down by gravity as I yank the bottom of this lung down by gravity it's going to pull on the apex too so I'm going to pull the apex farther away what part am I pulling farther away i'm pulling the visceral plura farther away from the parietal plura okay so as I'm yanking down at the base of this lung I'm pulling down here i'm bringing this visceral plura closer to this parietal plura but when I'm pulling I'm also pulling on this apex here because remember these are kind of closely attached right they're almost really like just rubbing up against one another so when I pull down here it starts pulling this actual visceral plur away from that parietal plura so now if you think about it for a second what's happening to this volume here when I stretch and pull that base down what's happening to that volume the volume here is decreasing what does that say for the pressure the pressure will be a little bit larger in this area what about up here well I'm pulling this down if I'm pulling the visceral plur away from the parietal plur up here what does that mean then well that means that the volume up here will be a little bit greater than it was down here so what does that mean for the pressure the pressure will be a little bit lower up there now we're not going to specifically talk about that but I want you guys to realize that there intraplural pressure is not uniform throughout the entire plural cavity it is different it's approximately like 758 here 756 here and 753 up here okay but we're only going to refer to it as 756 but I do want you to realize that it isn't uniform throughout the entire plural cavity okay now we got to do another thing that I need to mention here that is really really important we're not going to spend a lot of time but I want you to understand that there is other pressures a pressure across a wall so for example remember we said that this was intraulmonary pressure let's denote it again with AB this is A right here but we're going to just denote this A here for a second again intravular pressure intravial pressure i'm just denoting it here so it's close to this this is the B pressure which was the intra plural pressure and here was the C pressure let's say I make a line here i have a pressure that's being exerted across these two walls okay so there's a pressure that's being exerted across these two walls then there's also another pressure let's do this one in pink there's a pressure being exerted across the chest wall there's a pressure being exerted across the chest wall what are these two pressures and why are they important this pressure here across this wall which is the difference between the intraulmonary and the intraplural pressure this pressure here that is across this wall let's write it according with the color this pressure is called the let's write it down here trans pulmonary pressure that's interesting we're going to denote this TP to make it easier so TP is our transpulmonary pressure okay all right so that's good for right now we're going to talk about that in just a second then there's a pressure exerted across this chest wall and it's the difference between the interplural pressure and the atmospheric pressure okay what is that pressure called this pressure here is called the transthoracic pressure it's called the trans thoracic pressure okay and we'll just call this one TTP all right whatever it doesn't matter but as long as you understand that the TP is the transpulmonary pressure and the TTP is the transthoracic pressure okay there is one more i'll mention it but we're not going to really spend a lot of time on it because it's not uh super super significant here but I will mention it quickly it's the pressure all the way from A all the way to C and this pressure here is actually called the transpiratory pressure i'll write it up here trans respiratory pressure okay so I just want to explain something real quick here all right so now with the transpiratory pressure and with this transthoracic pressure and transpulmonary pressure how what is the significance of this okay well let's write out a little uh formula here so let me actually bring this one down a little bit so we have more room so I'm going to bring this one down here okay so this is again transthoracic pressure and again we denote as that as TTP right so TTP here okay now transpulmonary pressure what do we say we said it was the difference from the intraulmonary A minus B that's the difference so what do we actually say we're not going to say A minus B we're going to say it's the P pull which is the intraulmonary pressure minus the B well B was the intra plural pressure so we're going to put intra plural pressure this is equal to the transpulmonary pressure okay well what is that let's get a number out of this bad boy let's say that this is at rest okay with interpulmonary pressure we said was about I'm sorry uh 760 millimeters of mercury but again we could use zero also it wouldn't matter if you use zero we'll do zero just for the heck of it zero for that one and then -4 for the intra plural okay let's write that down so the intraulmonary and again I could have put 760 and I could have put um four uh I mean 756 it doesn't matter here but what we're going to do is intraulmonary pressure here is going to be specifically 0 millm of mercury and then what is it over here for this negative so it's minus intraplural pressure which is -4 so then if I do 0 -4 it's just I'm adding right I'm adding in this case and I should use the units right i shouldn't be uh lazy let me put the units in here so I'm consistent i'm sorry -4 millimeters of mercury the difference in this will give me 4 millime of mercury so you see how if I took 760 minus 756 it would still give me 4 millime of mercury but let's even define that a little bit more it's positive it's not negative it's positive what does that mean for it to be positive if the transpulmonary pressure is positive that's good thing that means that the lungs are actually going to be able to be inflated if it's negative that's a bad thing it means it's going to try to deflate okay now let's do the transthoracic pressure the transthoracic pressure we said was the difference across the chest wall so it's intraplural pressure minus the atmospheric pressure okay let's do that one so we said TTP which is the transthoracic pressure is equal to the okay B intra plural pressure so I'm going to put P IP minus the atmospheric pressure which is the pressure C so P of the atmosphere what does that give me okay intraplural pressure we said was -4 so we're going to write here it was -4 millimeters of mercury and then the atmospheric pressure is 0 millimeters of mercury okay so then if that's the case then transthoracic pressure is actually just equal to the interplural pressure then because this is zero so what does this actually equal then this equals4 - 0 which is -4 mm of mercury and so what does that mean then?4 millm of mercury means that it's trying to deflate that's why the chest wall because of this if you look at the actual transthoracic pressure naturally this is actually going to want to try to come this way right it's not going to want to be inflated it will actually cause a deflating pressure so the transthoracic pressure is a deflating pressure okay so we've done transpulmonary transthoracic there is the last one uh we can mention it really quickly um and it's just again intraalvolar pressure right here minus the atmospheric pressure so if we wrote that one down just for the heck of it it would be the intraulmonary pressure right so trans respiratory pressure we'll call this one TRP so trans respiratory pressure is equal to the P pole minus the P of the atmospheric pressure okay well what is that equal to that's equal to 0 minus 0 so what will this be it'll be 0 millimeters of mercury and again we're doing all of this at rest so this will be zero millimeters of mercury this is all at rest we're going to compare this to what it would look like afterwards whenever we're going to do the inspir inspiration process all right so again with all these pressures let's quickly go through them trans respiratory pressure is the intra pulmonary pressure minus the atmospheric pressure so therefore it is 0 millimeters of mercury so therefore there's no real gas flow that's moving in any direction here and there is no pressure differences across this okay transpulmonary pressure this is a really important one this one and transthoracic are the more important pressures transthoracic pressure I'm sorry transpulmonary pressure is the intraulmonary minus the inplural and we said again you'll take this 0 millm of mercury which was 760 again we could write it like that minus the intra plural which could either be 756 or you could write4 it doesn't matter you're still going to get the same number which is going to be positive 4 mm of mercury again what does that mean that means that this is trying to expand outwards that that you want what you want here is you want this actual lung to be able to inflate right you want it to be able to inflate so positive pressure means that you're trying to inflate the structure Now if we look at transthoracic pressure what's happening here this one's a little interesting right because you're taking the intra plural pressure subtracting it from the atmospheric pressure but what do you really what are you actually left with you're really only left with intraplural pressure so if that's the case then your transthoracic pressure is equal to your actual interplural pressure4 millimeters of mercury so what does that mean then it goes back to that thing that we said is due to this natural outward elasticity or recoil of the chest wall right because that's trying to pull this what parietal plura away from the visceral plura which is increasing this volume what else did we say we said it was also due to the natural elasticity and the surface tension of the lungs which is trying to pull the actual visceral plural away from the parietal plura what is that doing to the volume it's increasing the volume and what will that do to the pressure in this area it will decrease the pressure and that's why this should make sense okay now that we've done that we've gone over a whole bunch of pressures and a whole bunch of different formulas and numbers i'm sorry about that what we're going to do is we're going to go over how is these pressures changing whenever we're going through the inspiratory process so if you guys stick with us go to part two we're going to specifically see how the nervous system is affecting the actual this whole respiratory structure here and how that's actually producing pressure differences all right ninj i'll see you in part two |
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Vocabulario del Clima y Tiempo Atmosférico en español | Verbos y Expresiones
En esta clase vamos a hablar sobre el TIEMPO. Veremos cómo denominamos en español a los diferentes fenómenos atmosféricos que existen, conoceremos los verbos podemos usar para preguntar sobre el clima y aprenderemos algunas expresiones relacionadas con tiempo.
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Hablar sobre el clima en español: verbos y expresiones
A continuación, podéis ver los verbos empleados para hablar sobre el tiempo en español, mientras que la segunda parte del artículo está dedicado a las expresiones relacionadas con este tema.
Verbos para hablar sobre el tiempo
Cuando queremos preguntar acerca del tiempo que está haciendo podemos utilizar el verbo HACER o ESTAR, dependiendo de la situación. De esta forma, podemos formular las siguientes preguntas:
«¿Qué tiempo hace hoy?»
«¿Cómo está el día?»
«¿Qué tal día hace hoy?»
Se utiliza el verbo HACER para:
Hablar de temperatura.
Para decir los grados que hay, usamos el verbo hacer.
«Hace nueve grados hoy.»
Recordad que decimos «hace» y no «hacen», independientemente de los grados que haya. También para los grados podéis utilizar el verbo HABER en impersonal, conjugado en 3.ª persona del singular.
«Hay nueve grados hoy.»
Otros ejemplos con el verbo hacer son:
«Hace muchísimo frío.»
«Hace calor ahora mismo.»
Decir en general cómo está el tiempo.
Si queremos expresar simplemente si el día es bueno o malo en lo que a tiempo atmosférico se refiere, podemos usar también el verbo hacer, de manera que decimos Hace buen tiempo o Hace mal tiempo.
«En Canarias casi siempre hace buen tiempo. Hoy hay 23 grados.»
«Hoy no salgo de casa que hace mal tiempo, no para de llover.»
Hacer referencia al sol y al viento.
Hace sol.
«En Tarragona hace sol muchos días al año.»
Hace viento.
«En Zaragoza a menudo hace mucho viento.»
Otro verbo que empleamos para hablar del tiempo es ESTAR. Lo usamos para:
Decir cómo está el tiempo.
Si queremos hablar de la lluvia, podemos decir:
«Está lloviendo.»
Igual de correcto es utilizar el verbo llover y decir simplemente:
«Llueve.»
Lo mismo ocurre con el granizo. Podemos usar el verbo estar:
«Está granizando.»
O bien el verbo granizar.
«Graniza.»
Con el sol, decimos:
«Está soleado. »
En cuanto a la nieve, empleamos el verbo estar:
«Está nevando.»
Y el verbo nevar:
«Está nevando.»
Si hay nubes en el cielo, anunciamos que:
«Está nublado.»
Si por el contrario no hay nubes, decimos:
«Está despejado.»
Finalmente, usamos HAY con la niebla y la tormenta.
«Hay demasiada niebla y no cogeré el coche.»
«Mañana hay tormenta y no saldré de casa.»
Expresiones sobre el tiempo
A continuación, vamos a ver algunas expresiones relacionadas con el tiempo. Algunas de ellas se utilizan expresamente para hablar sobre el clima atmosférico, mientras que otras están formadas a partir de palabras relacionadas con el tiempo, pero tienen un significado que no está relacionado con el mismo.
Llover a cántaros
Un cántaro es una vasija de barro o metal que sirve para almacenar líquidos. Si decimos que llueve a cántaros queremos expresar que llueve tanto que parece que alguien estuviera tirándonos agua con cántaros.
«¡Menudo día más malo hace hoy! ¡Llueve a cántaros!»
Otra expresión que tiene el mismo significado es llover a mares.
«Hoy quería ir de excursión, pero no va a ser posible. ¡Llueve a mares!»
Si por el contrario apenas llueve, podemos usar el verbo chispear.
«Está chispeando.»
Hacer un frío que pela/Hacer un frío de muerte/Hacer un frío de mil pares de narices/Hacer un frío de cojones
Todas estas expresiones sirven para decir lo mismo: hace muchísimo frío.
La expresión hace un frío de cojones es bastante vulgar, por lo que si no queréis meter la pata podéis usar el resto de expresiones.
Hacer un sol de justicia
Cuando hace mucho calor, podemos decir que hace un sol de justifica. El origen lo encontramos en la antigua costumbre de condenar a algunos prisioneros a la exposición solar.
Igualmente, cuando alguien tiene mucho calor es habitual usar el verbo asarse. El verbo asar significa cocinar algo en el horno. Asarse como un pollo es una expresión habitual para indicar que uno tiene calor.
«Me estoy asando como un pollo.»
Otro verbo que podéis usar para decir que tenéis calor es achicharrarse.
«Me estoy achicharrando.»
Estar calado hasta los huesos
Esta expresión significa que estás mojado porque ha llovido mucho. Cuando decimos que estamos calados hasta los huesos queremos informar de que estamos empapados, como si el agua hubiese llegado hasta nuestros huesos.
«Estoy calada hasta los huesos. Se me ha olvidado el paraguas y ha empezado a llover.»
Ser un sol
Esta expresión es muy positiva porque significa que eres una buena persona, un encanto.
«Marina es un sol, siempre tiene una palabra amable.»
Ha llovido mucho desde que…
Esta expresión es bastante coloquial y se utiliza mucho para decir que ha pasado mucho tiempo desde que sucedió algo concreto.
«Ha llovido mucho desde que Paula dejó a su novio.»
Llover sobre mojado
Significa que te suceden varias cosas negativas seguidas, es decir, varias desgracias en poco tiempo.
«Toño se arruinó y al poco tiempo rompió con su pareja.»
Saber a rayos
Esta expresión no la usamos para referirnos al tiempo, sino al sabor de una comida. Saber a rayos significa que el alimento que estamos comiendo está asqueroso.
«Esta paella sabe a rayos, está malísima.»
Aguantar el chaparrón
El chaparrón es un tipo de lluvia que es repentina y fuerte, pero que solo dura unos minutos.
Aguantar el chaparrón significa soportar una bronca, por ejemplo, de un jefe. También significa resistir ante dificultades.
«Pedro ha aguantado el chaparrón pese a que tenía toda la razón.»
¡Ahora ya sabéis un poquito más sobre cómo hablar del tiempo en español! Si queréis seguir aprendiendo español, os recomiendo echar un vistazo al vídeo sobre cómo dar la hora en español.
¿Has aprendido los contenidos de esta clase? ¡Intenta hacer los ejercicios!
Ejercicio: Vocabulario del Clima y Tiempo Atmosférico
Test interactivo sobre el vocabulario del clima y tiempo atmosférico en español. Aprende verbos, expresiones y términos meteorológicos esenciales. Ejercicio online y gratuito para estudiantes de español.
Pregunta 1 de 10
¿Cómo se dice "sunny weather" en español?
Completa la expresión: Hoy _____ mucho frÃo. (verbo hacer)
¿Cuál es el significado de "está lloviendo"?
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187657 | https://brainly.com/question/32860332 | [FREE] A farmer sold half as many green apples as red apples. She also sold 1/3 as many yellow apples as red - brainly.com
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A farmer sold half as many green apples as red apples. She also sold 1/3 as many yellow apples as red apples.
What is the ratio of green apples sold to total apples sold?
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Let's assume that the farmer sold x red apples.
Then, she sold half as many green apples as red apples, which is x/2 green apples. The farmer also sold 1/3 as many yellow apples as red apples, which is (1/3)x yellow apples.
To find the ratio of green apples sold to the total apples sold, we need to add up all the apples sold by the farmer.
Total apples sold = x + x/2 + (1/3)x = (11/6)xGreen apples sold = x/2
The ratio of green apples sold to total apples sold = Green apples sold / Total apples sold= (x/2) / (11/6)x= (x/2) x (6/11x)= 3/11
The ratio of green apples sold to total apples sold is 3:11.
Therefore, option B is correct.
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The ratio of green apples sold to the total apples sold is 3:11. This is calculated by determining the number of each type of apple sold and finding the fraction of green apples to the total. Hence, the answer is 3:11.
Explanation
To find the ratio of green apples sold to the total apples sold, let's first define the number of red apples sold by the farmer as x.
Green Apples Sold: The farmer sold half as many green apples as red apples, so the number of green apples sold is 2 x.
Yellow Apples Sold: The farmer also sold one-third as many yellow apples as red apples. Therefore, the number of yellow apples sold is 3 1x.
Now we can calculate the total number of apples sold:
Total apples sold=Red apples+Green apples+Yellow apples
Total apples sold=x+2 x+3 1x
To combine these, we first need a common denominator, which is 6:
Convert x to 6 6x
Convert 2 x to 6 3x
Convert 3 1x to 6 2x
Now we can add these fractions together:
Total apples sold=6 6x+6 3x+6 2x=6 11x
Calculating the Ratio: Now we can find the ratio of green apples sold to the total apples sold:
Ratio of green apples to total apples=Total apples sold Green apples sold=6 11x 2 x
The x cancels out:
=2 1⋅11 6=22 6=11 3
Thus, the ratio of green apples sold to the total apples sold is 3:11.
Therefore, the answer to the question would be: the ratio of green apples sold to the total apples sold is 3:11.
Examples & Evidence
For instance, if the farmer sold 12 red apples, she sold 6 green apples (half of 12) and 4 yellow apples (one-third of 12). The total apples sold would thus be 12 + 6 + 4 = 22, which confirms a ratio of green apples (6) to total apples (22) as 3:11 when simplified.
The calculations provided for each type of apple sold are based on logical mathematical reasoning using fractions and ratios, which is a fundamental concept in mathematics.
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187658 | https://www.quora.com/If-sin%CE%B8-cos%CE%B8-2-then-what-is-the-value-of-%CE%B8 | If sinθ+cosθ=√2, then what is the value of θ=? - Quora
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If sinθ+cosθ=√2, then what is the value of θ=?
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Manoj.V
Maths enthusiast · Author has 166 answers and 993.2K answer views
·4y
[math]sin \theta + cos \theta = \sqrt{2}[/math]
[math]\text{By method of introspection} ,\theta = 45^{o} [/math]
[math]\because sin 45^{o} = cos 45^{o} = \dfrac{1}{ \sqrt{2}}[/math]
[math]And, \dfrac{1}{ \sqrt{2}} + \dfrac{1}{ \sqrt{2}} = \sqrt{2}[/math]
[math]\therefore \theta= 45^{o}[/math]
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Vijay Mankar
Love doing high school maths · Author has 7K answers and 11.2M answer views
·4y
[math]\sin \theta+\cos\theta =\sqrt{2}[/math]
Squaring both sides,
math^2 =(\sqrt{2})^2[/math]
[math]\sin^2 \theta+\cos^2\theta+2 \sin \theta\cos\theta =2[/math]
[math]1+\sin 2\theta=2[/math]
[math]\sin 2\theta=1[/math]
[math]2\theta=\sin^{-1} (1)=\dfrac{\pi}{2}[/math]
[math]\boxed{\theta=\dfrac{\pi}{4}=45^{\circ}}\tag {}[/math]
Upvote ·
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Lugo Risu
4y
Originally Answered: Cos θ + sin θ = √2, what is the value of θ is? ·
cos θ + sin θ = √2
Squaring both the sides
(cos θ + sin θ)² = (√2)²
cos²θ + sin²θ + 2sinθ.cosθ = 2 (1) (using identity (a+b)² = a² + b² + 2ab)
We know that
sin²θ + cos²θ = 1 and sin2θ = 2sinθ.cosθ
Using above two identities in equation (1) we get
1 + sin2θ = 2
sin2θ = 2 - 1
sin2θ = 1
sin2θ = sin90° (as sin90° = 1)
Comparing both the sides we get
2θ = 90°
θ = 90°÷2 = 45°
So, the value of θ is 45°.
Hope it helps!
Upvote ·
9 3
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Cos θ + sin θ = √2, what is the value of θ is?
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If cot θ = 15/8, then how do you find the value of [(2+2sinθ) (1-sinθ)] / [(1+cosθ) (2-2cosθ)]?
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Mohammad Afzaal Butt
B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views
·4y
[math]\sin\theta + \cos\theta = \sqrt{2}[/math]
[math]\implies (\sin\theta + \cos \theta)^2 = (\sqrt{2})^2[/math]
[math]\implies \sin^2 \theta + \cos^2 \theta + 2\sin\theta\cos \theta = 2[/math]
[math]\implies 1 + \sin 2\theta = 2[/math]
[math]\implies \sin 2\theta = 1[/math]
[math]\implies 2\theta = \dfrac{\pi}{2} + 2 k \pi\quad \text{where k}\in\Z[/math]
[math]\implies \theta = \dfrac{\pi}{4} + k\pi[/math]
Upvote ·
9 2
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Cos θ + sin θ = √2, what is the value of θ is?
How do you find the value of θ if sinθ = 0.5 and cosθ = √3/2?
If cot θ = 15/8, then how do you find the value of [(2+2sinθ) (1-sinθ)] / [(1+cosθ) (2-2cosθ)]?
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Home » Blog » Printables » Math Printables » Decimal Place Value Charts – 12 Free Printable PDFs
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Decimal Place Value Charts – 12 Free Printable PDFs
ByCristina Morero
This post may contain affiliate links, which means I’ll receive a commission if you purchase through my links, at no extra cost to you. Please read full disclosure for more information.
Looking for a place value chart for decimals to help your students visualize numbers? Look no further!
Here, you can find 10 free printable decimal place value charts that will make learning place value fun and interactive.
12 free printable decimal place value charts
Below you can find multiple charts to practice decimal place values.
HOW TO DOWNLOAD: These printables are easy to download and use – just follow these steps to get your worksheets printed:
Scroll down until you find the design you’d like to print out
Click on the instant download link under the image of the PDF (or the set of images)
Size: US Letter – but some can be easily resized and printed as A4.
All the tables, charts, and math worksheets arecopyright protected ©SaturdayGift Ltd. Graphics purchased or licensed from other sources are also subject to copyright protection. Free for classroom and personal use only, not to be hosted on any other website, sold, used for commercial purposes, or stored in an electronic retrieval system.
Landscape decimal place value chart
DOWNLOAD: Free decimal place value chart printable – colorful
DOWNLOAD: Decimal place value chart printable – grey
Ones, tenths, hundredths, thousandths decimal chart
DOWNLOAD: Place value decimal chart in cute colors
DOWNLOAD: Place value decimal chart in grey tones
Decimal place value anchor chart
This decimal place value anchor chart has place to write the number in
standard form
word form
and expanded form
DOWNLOAD: Decimal place value anchor chart PDF – colorful
DOWNLOAD: Decimal place value anchor chart PDF – black & white
Decimal place value printable worksheet
DOWNLOAD: Decimal place value printable worksheet – colorful
DOWNLOAD: Decimal place value printable worksheet – grey
Place value recording charts with decimal places
DOWNLOAD: Place value chart with decimal places – cute colors
DOWNLOAD: Printable place value chart with decimal – grey
Place value chart with decimals
DOWNLOAD: Blank place value chart with decimals – cute colors
DOWNLOAD: Blank place value chart with decimals – grey colors
Tip: If you laminate the blank free printable charts, you can use dry-erase markers to write on them and reuse them again and again, saving paper and resources.
What is a place value?
Place value refers to the value of a digit in a number based on its position.
It is an essential concept in mathematics and helps us understand how numbers work and how we can compare and manipulate them.
Base 10, or base ten number system, is the commonly used number system with ten digits (0-9).
Each digit in a number holds a different value depending on its position.
For example, in the number 2456, the digit 2 represents one thousand, the digit 4 represents four hundred, the digit 5 represents fifty, and the digit 6 represents six.
What are decimal place values?
The decimal place values are tenths, hundredths, thousandths, and ten thousandths.
Decimal numbers are positioned to the right of the ones after the decimal point.
The place value after the decimal point represents the fractional part of the number.
RELATED POST: Place Value Chart – Printable Ones Tens Hundreds Thousands Chart
What is a decimal place value chart?
The decimal place value chart visually represents numbers that help students understand the decimal system and the relationship between each digit in a decimal number.
The decimal chart also shows how each digit’s value changes as it moves to the right.
Example of decimal place value chart
Take the number 1.3257
1 represents the ones
7 represents the tenths place
5 represents the hundredths place
2 represents the thousandths place
3 represents the ten thousandths place
The chart is divided into columns, each representing a different place value – tenths, hundredths, thousandths, and ten thousandths.
How to use a decimal place value chart?
Using a decimal place value chart can help students understand the concept of decimals visually and interactively. Here are some tips on how to use a decimal place value chart effectively:
Start by introducing the concept of place values and their corresponding positions after the decimal point.
Use real-life examples, such as money or measurements, to demonstrate how decimals work.
Have students write down numbers in decimal form and then use the place value chart to identify the corresponding values.
Have students write down numbers in word form and then use the place value chart to identify the corresponding values.
Use different colors to represent each place value, making it easier for students to differentiate between them.
Encourage students to come up with their own examples and practice using the place value chart independently.
Practice converting between decimals and fractions using the place value chart.
Understanding the concept of place value is crucial in developing strong mathematical skills.
With the help of a decimal place value chart, decimal place value questions, and exercises, students can grasp this concept and have fun while doing it.
Other free math printables
Hundreds charts
Hundreds chart – numbers 1 to 100 worksheets
120 chart – numbers 1 to 120 worksheets
Thousands chart – numbers 1 to 1000 worksheets
number lines worksheets
Number line 0 to 1
Number line to 10
Number line to 15
Number line to 20
Number line to 25
Number line to 30
Number line to 50
Number line to 100
Other math worksheets
Blank number lines
Fraction number line
Negative and positive number line
Double number line
Even and odd numbers
Multiplication flash cards
Multiplication charts
1-10 Multiplication charts 1 to 10
1-12 Multiplication charts 1 to 12
1-15 Multiplication charts 1 to 15
1-20 Multiplication charts 1 to 20
1-25 Multiplication charts 1 to 25
1-30 Multiplication charts 1 to 30
1-50 Multiplication charts 1 to 50
1-100 Multiplication charts 1 to 100
Times tables
0 times table
1 times table
2 times table
3 times table
4 times table
5 times table
6 times table
7 times table
8 times table
9 times table
10 times table
11 times table
12 times table
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187661 | https://blog.csdn.net/qq_67308022/article/details/125190724 | 等差数列求和（1+2+3+....+100）_编写函数文件,计算等差级数s=1+2+3...+i前n项之和调用此函数计算i=100时的结果-CSDN博客
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等差数列求和（1+2+3+....+100）
感谢大自然的馈赠于 2022-06-08 19:17:58 发布
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这是一个简单的Java程序，演示了如何使用for循环计算1到100的整数之和。程序首先声明并初始化变量sum为0，接着通过for循环迭代1到100的整数，每次迭代将当前数值累加到sum中。最后，程序输出1到100的和，即5050。
public class K416{
public static void main(String[] args) {
// 声明部分
int sum;
```
// 初始化
sum = 0;
// 处理部分（for循环也叫计数循环，用于事先知道要循环多少次的情况）
for (int i = 1; i <= 100; i++) { // 循环头：初始条件；循环条件；迭代条件
sum = sum + i; // 累加语句
}
// 输出部分
System.out.println("1 + 2 + 3 + …… + 100 = " + sum);
}
```
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文章浏览阅读80 1 次,点赞 1 7次,收藏 1 6次。本文还有配套的精品资源,点击获取 简介:本教程旨在为初学者快速介绍MATLAB软件的基础操作和数学实验技巧,涵盖从界面操作到高级数据分析和符号 计算 的各个方面,让读者能够利用MATLAB高效处理数据和实现算法。 1 . MATLAB界面与基
利用for循环，计算 1 + 2 + 3 +……+ 100。等差数列 求和
qq_51082388的博客
12-27 8581
#include using namespace std; int main ( ) { int sum = 0; for ( int i = 1;i<= 100;+ + i )//i初始值为 1,终值为 100，每次增量为 1 sum + = i; cout << sum << endl; return 0; }
python生成 等差数列 pythonnumpy _函数 中linspace实现创建 等差数列 的实例分享
weixin_39542608的博客
11-30 860
numpy . linspace是用于创建一个一维数组，并且是 等差数列 构成的一维数组,下面这篇文章主要给大家介绍了关于python numpy 函数 中的linspace创建 等差数列 的相关资料，文中通过示例代码介绍的非常详细，需要的朋友可以参考下。前 言本文主要给大家介绍的是关于linspace创建 等差数列 的相关内容，分享出来供大家参考学习，下面话不多说了，来一起看看详细的介绍吧。numpy . linspac . . .
educoder平台课-Python程序设计-3 . 流程控制
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文章浏览阅读 1 . 2 k次,点赞 2 0次,收藏 2 5次。自用学习整理,无商业用途。
【算法】第一章:基础_python电路模型编程
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文章浏览阅读464次。本文是关于算法的基础讲解,涵盖编程模型、数据抽象、数组、静态方法和算法分析等内容。介绍了 Java 中的原始数据类型、静态方法库、数据类型的声明与初始化、数组的创建与操作,以及算法分析的基本概念,如 时 间复杂度和增长数量级。文中还探讨了栈、队列
js代码-1 + 2 + 3 + . . . . . + 100 的程序
07-14
1 . 等差数列 求和 公式： 等差数列 的和可以用公式 S _=_ n/_2_ _(_ a _1_ _+_ an _)_ 来 计算，其中 S 是总和，n 是 项 数（在这里是 100），a _1_ 是首 项（这里是 1），an 是末 项（这里是 100）。将这些值代入公式，我们得到 . . .
c代码-5-1 例 求和 1 + 2 + 3 + . . . . . + 1 0
07-16
标题"5-1 例 求和 1 + 2 + 3 + . . . . . + 1 0"指向了一个基本的数学问题，即求 等差数列 的和。等差数列 是数学中一个重要的概念，其中每个 项 与 前 一 项 之间的差是恒定的。在这个特定的例子中，公差为 1，首 项 为 1，末 项 为 1 0，我们要找到 . . .
大数阶乘 计算 优化
9-13
文章浏览阅读 2 k次。深入浅出大数阶乘大数阶乘的 计算 是一个有趣的话题,从中学生到大学教授,许多人都投入到这个问题的探索和研究之中,并发表了他们自己的研究成果。如果你用阶乘作关键字在google上搜索,会找到许多此类文章,另外,如果你使用google学术搜索,也能找到一些
Python编程精要
9-25
文章浏览阅读 1 . 5k次,点赞 2 次,收藏 1 2 次。一 . python的优缺点 1 . 优点语言 简洁优美跨平台。mac,Linux,window通用胶水 语言。能够把其他 语言 制作的各种模块 ( 尤其是C/C + + ) 很轻松地结合在一起可以这么理解,python本身不是一种运算快的 语言,但善于利用,整合其他 语言
选讲 1 等差数列 求和 . doc
09-12
等差数列 求和 等差数列 是一种特殊的数列，它的每一 项 与 前 一 项 的差都是固定的常数，这个常数称为公差。等差数列 有两个重要的公式：通 项 公式和 项 数公式。 通 项 公式：第 n 项 = 首 项 + ( 项 数 - 1 ) × 公差 项 数公式 . . .
数列 求和 基础 + 复习 + 习题 + 练习 . doc
10-07
本资源摘要信息涵盖了数列 求和 的基础知识点，包括 等差数列、等比数列、错位相减法、分组 求和、拆 项（裂 项）求和、倒序相加法、导数法、递推法、奇偶分析法等。通过对这些知识点的总结和分析，可以帮助学生更好地掌握 . . .
等差数列 求和 . pdf
06-27
等差数列 求和 . pdf 等差数列 求和 是一种数学 计算 方法，用于 计算 等差数列 的和。. . . 我们可以使用 等差数列 的公式an = a 1 + ( n - 1 ) d来 计算 每一 项，然后使用 等差数列 求和 公式SN = n/2 × ( a 1 + an ) 来 计算 和。
求 1 + 2 + 3 +……+ 100 的累加和。
qq_46630774的博客
11-29 4675
利用三种循环语句求 1 + 2 + 3 +……+ 100 的累加和。 方法一：whlie #include int main ( ) { int i = 1,sum = 0; while ( i<= 100 ) { sum = sum + i; i + +; } printf ("sum =%d\n",sum ); return 0; } 方法二：do…whlie #include int main ( ) { int i = 1,sum = 0; do { sum = sum + i; i + +
Python--序列 求和（求 1 + 2 + 3 . . . + n的和）
weixin_58060022的博客
03-06 9176
1 . 方法一:利用for ( 遍历法求序列和 ) . 2 . 方法二:利用数学公式 ( 等差数列 求和 公式 ),sum = 0 . 5n + 0 . 5nn,直接输出答案 方法一: n = int ( input ( ) ) sum = 0 #必须给sum初始化 for i in range ( 1,n + 1 ): sum + = i print ( sum ) 方法二: n = int ( input ( ) ) sum = 0 . 5n + 0 . 5nn . . .
C 语言-编写 程序，求 1 + 2 + 3 + . . . + 100。
程序员本员
05-03 2万+
#include main ( ) { int i,sum; sum = 0; for ( i = 1;i<= 100;i + + ) sum + = i; printf ("sum =%d\n",sum ); } sum = 5050
计算 1 + 2 + 3 + . . . + 100 的和
OH_ON的博客
12-11 1万+
计算 1 + 2 + 3 + . . . + 100 的和
使用Python 计算 1 + 2 + 3 + . . . + 100
127.0.0.1的博客
09-20 5万+
本文将会采用多种方法 计算 1 + 2 + 3 + 4 +····加到 100 的和，通过点点滴滴为大家展示出 计算 机的魅力，算法的魅力。同 时 大家对程序进行微小的改动之后就可以实现 计算 1 加到任何一个你想要的数，希望你读完本文后能有所收获。
Java--等差数列 求和（计算 1 到 100 的和）
dyh0714sxg0522的博客
04-07 969
等差数列 求和（计算 1 到 100 的和）
C 语言 求 1 + 2 + 3 + . . + 100 的和
weixin_43750428的博客
07-01 9203
#include int main ( void ) { int i,sum = 0; for ( i = 0;i<1 0 1;+ + i ) sum = sum + i; printf ("%d\n",sum ); return 0; }
题目：利用递归求 1 + 2 + 3 . . . 100 的和
说：/~不了的专栏
09-24 1万+
在面试的 时 候遇到了这样的一道笔试题目，就是利用递归求出 1 到 100 的和，也就是 1 + 2 + 3 + . . . . . . . . + 100。 怎么说呢，递归是一种思想，用大白话来说，就是自己 调用 自己。 如一个方法A ( ),然后在方法A ( ) 中再次 调用 自己，但是利用递归的 时 候特别需要注意的就是跳出递归的条件，否则的话就会出现死循环的情况，也就是一直在执行的某一个方法。 以下为简单的代码实现： package com . ak
C 语言：求 1 + 2 + 3 +…+ 100 的总和（for循环 各程序如何执行？）
Yummy的博客
04-22 5万+
问题：for循环：求 1 + 2 + 3 +…+ 100 的总和 代码实现： # include int main ( void ) { int i; int sum = 0; //sum的英文意思是“总和” for ( i = 1; i<= 100; + + i ) //+ + 是自加的意思, + + i相当于i = i + 1 { sum = sum + i; /等价于sum + = i;但是不建议这么写, 因为sum = sum + i看起来更清楚、更舒服/ } printf ("sum = .
数列 求和 1 + 1 2 + 1 2 3 + 1 2 3 4 + . . . 最新发布
12-31
数列 求和 的问题可以根据不同的序列模式采用多种方式来处理。对于特定形式的数列，例如 等差数列、等比数列或其他特殊类型的数列，存在标准公式可以直接应用以快速得到 结果。 然而，给定的例子 _1_ _+_ _1_ _2_ _+_ _1_ _2_ _3_ _+_ _1_ _2_ _3_ 4 并不是一个常见的数列类型；这是一个由连续整数组成的非线性增长序列。这种情况下没有直接适用的标准公式。但是可以通过编程或者手动 计算 的方式来获得总和。 为了 计算 这个特殊的数列 之和，可以考虑以下两种方法之一： 逐 项 相加 每一 项 都是 前 一 项 乘以 1 0再加上下一个自然数。因此，可以 编写 一个简单的循环程序来进行累加直到最后一 项 被加入为止。 使用数学归纳法寻找规律 尝试找出该数列的一般 项 公式$a_n$，之后再根据一般 项 去推导 求和 公式$\sum_{k = 1}^{n}{a_k}$。不过这种方法可能比较复杂，并且不一定能找到简洁的形式化表示。 下面是用Python 编写 的简单算法来 计算 这类数列的和: python def sum_of_sequence _(_ n _)_: total_sum _=_ 0 current_number _=_ "" for i in range _(_ _1_, n _+_ _1_ _)_: current_number _+_ _=_ str _(_ i _)_ total_sum _+_ _=_ int _(_ current_number _)_ return total_sum print _(_ sum_of_sequence _(_ 4 _)_ _)_ # 输出应该是 _1_ _+_ _1_ _2_ _+_ _1_ _2_ _3_ _+_ _1_ _2_ _3_ 4的 _结果_ 这段代码会输出指定数量级内的所有这些数字连起来形成的数列元素 之和，在例子中就是 计算_1_ _+_ _1_ _2_ _+_ _1_ _2_ _3_ _+_ _1_ _2_ _3_ 4。
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187662 | https://vishiklab.faculty.ucdavis.edu/wp-content/uploads/sites/394/2021/06/Lecture-3-electrons-notes.pdf | Lecture 3: electrons • Free electron gas • Density of states • Velocity and mass • Metals, insulators, and semiconductors • Introducing the lattice back in: Brillouin zones and Fermi surfaces • Tight binding model Free electron gas in three dimensions This toy problem turns out to be applicable to many simple metals such as sodium or copper, and it is a generalization of the infinite potential well to three dimensions. This model makes the following assumptions about electrons in a metal: • Valence electrons are completely delocalized over the entire solid, such that they are treated as waves rather than particles • The lattice is absent so we neglect interactions between electrons and lattice (the lattice will be put back in later) • Electrons do not interact with each other at all, except via Pauli exclusion. In particular, coulomb repulsion is ignored In three dimensions, the free particle Schrodinger equation is: −ℏ2 2𝑚( 𝜕2 𝜕𝑥2 + 𝜕2 𝜕𝑦2 + 𝜕2 𝜕𝑧2) 𝜓𝑘(𝑟) = 𝜖𝑘𝜓𝑘(𝑟) The wavefunctions are marked by k instead of by n, and we will see why in a moment. At this point, it is helpful to start over with a different formalism. We consider plane wave wavefunctions of the form 𝜓𝒌(𝒓) = 𝑒𝑖𝒌∙𝒓 Where k is a wavenumber or quantum mechanical momentum. It can be expressed as 𝑘= 2𝜋 𝜆 or 𝒑= ℏ𝒌 And periodic boundary conditions of the form 𝜓(𝑥+ 𝐿, 𝑦, 𝑧) = 𝜓(𝑥, 𝑦, 𝑧) 𝜓(𝑥, 𝑦+ 𝐿, 𝑧) = 𝜓(𝑥, 𝑦, 𝑧) 𝜓(𝑥, 𝑦, 𝑧+ 𝐿) = 𝜓(𝑥, 𝑦, 𝑧) Periodic boundary conditions better reproduce the fact that the solid looks infinite to the electrons. The formalism above also permits negative values of k as unique solutions, which makes more physical sense because of the connection between k and momentum. Finally, periodic boundary conditions are utilized in many modern computational techniques in condensed matter physics. Plugging the first one into the wavefunction we get: 𝑒𝑖(𝑘𝑥(𝑥+𝐿)+𝑘𝑦𝑦+𝑘𝑧𝑧) = 𝑒𝑖(𝑘𝑥𝑥+𝑘𝑦𝑦+𝑘𝑧𝑧) 𝑒𝑖𝑘𝑥𝐿= 1 𝑘𝑥= 0, ± 2𝜋 𝐿, ± 4𝜋 𝐿, … And similar for ky and kz. Plugging the plane wave wavefunction into schrodinger’s equation we get: ℏ2 2𝑚(𝑘𝑥 2 + 𝑘𝑦 2 + 𝑘𝑧 2) = 𝜖𝑘= ℏ2𝑘2 2𝑚 As before, we take our N electrons and put them into the available states, filling lowest energy first. In 3D this is trickier because multiple states may have the same energy, even though they are marked by different 𝑘𝑥, 𝑘𝑦, 𝑘𝑧. In 3D, our rules for filling up electrons are: • Every state is defined by a unique quantized value of (𝑘𝑥, 𝑘𝑦, 𝑘𝑧) • Every state can hold one spin up and one spin down electrons • Fill low energy states first. In 3D, this corresponds to filling up a sphere in k space, one ‘shell’ at a time. Each shell is defined by a radius k, where 𝑘2 = 𝑘𝑥 2 + 𝑘𝑦 2 + 𝑘𝑧 2, and every state in the shell has the same energy, although different combinations of 𝑘𝑥, 𝑘𝑦, 𝑘𝑧 When we have used up all our electrons, we are left with a filled sphere in k space with radius 𝑘𝐹 (called the Fermi momentum) such that 𝜖𝐹= ℏ2 2𝑚𝑘𝐹 2 This sphere in k-space has a volume 4 3 𝜋𝑘𝐹 3 and it is divided into voxels of volume ( 2𝜋 𝐿) 3 If we divide the total volume of the sphere by the volume of each ‘box’ and account for the fact that each box holds 2 electrons, we get back how many electrons we put in: 2 ∗ 4 3 𝜋𝑘𝐹 3 (2𝜋 𝐿) 3 = 𝑁= 𝑉𝑘𝐹 3/3𝜋2 Here, 𝑉= 𝐿3 is the volume of the solid. We can use this relationship to solve for 𝑘𝐹 and show that it depends on electron density (N/V) 𝑘𝐹= (3𝜋2𝑁 𝑉 ) 1/3 Plugging this back into the expression for 𝜖𝐹 we get: 𝜖𝐹= ℏ2 2𝑚(3𝜋2𝑁 𝑉 ) 2/3 At absolute zero, the Fermi sphere has a hard boundary between occupied and unoccupied states. At higher temperature, this boundary becomes fuzzier with increasing occupation permitted outside the initial boundary (think of a rocky planet like earth vs a gaseous planet like Jupiter). The width of this fuzziness is determined by the width of the Fermi-Dirac distribution at that temperature, and it is roughly proportional to 𝑘𝐵𝑇. Notably, the vast majority of electrons in the Fermi gas are completely inert because they are buried deep inside the sphere. Only electrons close to the Fermi level are affected by temperature and participate in conduction. This is quite contrary to the conclusions of particle-like treatments of electrons in a metal which assume that all valence electrons participate in electronic properties. Density of states As with phonons, the density of states is a useful quantity for electrons. I like to think of Density of States as a series of “boxes” where electrons can live. Each box is defined by the coordinates which distinguish one electron from another. In the case of a 3D free electron gas, each box is defined by unique 𝑘𝑥, 𝑘𝑦, 𝑘𝑧 and spin. Where the density comes in is at each energy interval 𝑑𝜖 we consider ‘how many ‘boxes’ are there?’ It is defined as: 𝐷(𝜖) ≡𝑑𝑁 ̃(𝜖) 𝑑𝜖 Where 𝑁 ̃(𝜖) is the number of states as a function of energy. We can find it by expressing 𝑁 ̃ in terms of 𝜖 and taking a derivative. We begin by considering a sphere in k-space with an arbitrary radius k and asking how many electrons that will hold 𝑁(𝑘) = 𝑉𝑘3/3𝜋2 The relationship between energy and momentum in a free electron gas is pretty straightforward too (unlike with phonons): 𝜖= ℏ2𝑘2 2𝑚 Solving for k, and plugging in above we get 𝑁(𝜖) = 𝑉 3𝜋2 (2𝑚𝜖 ℏ2 ) 3/2 Now we can just take the derivative with respect to energy and get: 𝐷(𝜖) ≡𝑑𝑁 𝑑𝜖= 𝑉 3𝜋2 (2𝑚 ℏ2 ) 3 2 𝜖1/2 Thus, the density of electron states in 3D is a function of energy. If you have more electrons, you will end up with a higher density of states at the Fermi energy. Effect of temperature Temperature introduces a ‘cutoff’ by the Fermi-dirac function 𝑓(𝜖) = 1 𝑒(𝜖−𝜇)/𝑘𝐵𝑇+ 1 Such that some states with 𝜖> 𝜖𝐹~𝜇 can be occupied and some states with 𝜖< 𝜖𝐹~𝜇. Temperature only affects states roughly within 𝑘𝐵𝑇 of the Fermi energy. Another way to think of the effect of temperature is the fuzzing out of the boundary of the Fermi surface. Electron velocity There are two ways of extracting an electrons’ velocity in a Fermi gas. These will be applicable even when electrons are modeled in a more sophisticated way. • From the derivative of the energy vs k (equivalent to what we did for phonons): 𝑣𝑔= 1 ℏ 𝜕𝜖𝑘 𝜕𝑘= ℏ𝑘/𝑚. Note that this is an example of a very useful and measurable quantity derived directly from a materials’ 𝜖 vs 𝑘 relationship. This (𝜖 vs 𝑘) is called a dispersion relation. For a free electron gas, it is quadratic, but it doe not have to be. • By representing the linear momentum operator as 𝒑= −𝑖ℏ∇ and applying this to the plane-wave wavefunction to get 𝒑= ℏ𝒌 and equating to mv to get 𝒗= ℏ𝒌/𝑚 The velocity of electrons at the fermi energy is called the Fermi velocity (𝑣𝐹) and it is given by: 𝑣𝐹= ℏ𝑘𝐹 𝑚= ℏ 𝑚(3𝜋2𝑁 𝑉 ) 1/3 Effective mass Electrons in a crystalline solid often exhibit properties that free electrons do not have. One of the most basic is that they may behave as if they have a different mass (either heavier or lighter) than the free electrons mass. The effective mass (usually written m or 𝑚𝑒𝑓𝑓) can also be derived from dispersion relations 1 𝑚∗= 1 ℏ2 𝜕2𝜖𝑘 𝜕𝑘2 Electron-like vs hole-like bands Charge carriers in a solid sometimes respond to electromagnetic fields as if they have positive charge, and this also originates from dispersion relations. Negative (electron-like) carriers have 𝜖 vs 𝑘 which is concave up and positive (hole-like) carriers have 𝜖 vs 𝑘 which is concave down. The latter case can emerge from a free-electron gas once we turn on lattice interactions. Turning the lattice potential back on: Brillouin zones and band gaps So far, we have discussed electrons without the lattice, and now we turn the lattice back on. Since the electron gas originated from atoms giving up some electrons, the realistic lattice we turn on is a periodic array of positive charges. When we turn on the lattice potential, the following things happen. They are sketched below in 1D only, and generalizing to higher dimensions will be done later 1. The k-axis is divided up into ‘Brillouin zones’ of the same size. The length of each Brillouin zone is a reciprocal lattice vector, which in 1D is 2𝜋 𝑎. This division reflects the periodicity of the lattice 2. Each Brillouin zone must contain the same information (reflecting periodicity in both real space and reciprocal space) 3. When dispersions cross, gaps open up, with the magnitude of these band gaps generally reflecting the strength of the lattice potential. There are several ways to think about this: a. Pauli exclusion: each point on the 𝜖 vs k graph can hold two electrons: one spin up, and one spin down. When two lines cross, you have 4 electrons trying to be in the same state, which won’t fly b. Bragg reflection: each point where bands cross is related to an equivalent point via a reciprocal lattice vector, and these momenta can satisfy the Bragg condition introduced in lecture 2: 2𝒌∙𝑮= 𝐺2 →1𝐷 𝑐𝑎𝑠𝑒→𝑘= ± 1 2 𝐺= ± 𝑛𝜋 𝑎. Thus an electron that has those specific momenta will be Bragg reflected by the lattice, such that the electronic state at that momentum is comprised of a sum of two waves of the same wavelength moving in opposite directions. This is a standing wave with zero group velocity. Bloch functions Bloch’s theorem is one of the most important principles in solid state physics. It states that the solution to the Schrödinger equation with a periodic potential (i.e. a crystal) must have a specific form: 𝜓𝒌(𝒓) = 𝑢𝒌(𝒓)𝑒𝑖𝒌⋅𝒓 Where 𝑢𝒌(𝒓) has the period of the lattice such that it is invariant under translation by a lattice vector (T): 𝑢𝒌(𝒓) = 𝑢𝒌(𝒓+ 𝑻) Eigenfunctions of this form are called Bloch functions. They consist of a product of a plane wave and a function which shares the periodicity of the lattice. Crystal momentum of an electron The Bloch wavefunctions are labeled by an index k (as the free electron wavefunctions were earlier), and this quantity, is called crystal momentum. A few comments about crystal momentum • 𝑒𝑖𝒌⋅𝑻 is the phase factor which multiplies a Bloch function when we make a translation by a lattice vector T • If the lattice potential vanishes in the central equation, we are left with 𝜓𝒌(𝒓) = 𝑒𝑖𝒌⋅𝒓 just like in the free electron case • Crystal momentum (ℏ𝑘) is like regular momentum in that it enters into conservation laws that govern collisions (e.g. electrons with momentum ℏ𝑘 colliding with a phonon with momentum ℏ𝑞) • Crystal momentum is different from regular momentum in that it is defined only modulo a reciprocal lattice vector G. Thus, if an electron collides with a phonon and is kicked into momentum k’, this is expressed in the following way, 𝒌+ 𝒒= 𝒌′ + 𝑮 Metals, insulators, and semiconductors • Metals have 𝐸𝐹 (Fermi energy) inside a band, such that there are unoccupied states which the highest energy electrons can make low-energy excitations into • Semiconductors and insulators have 𝐸𝐹 inside band gap, such that the band below 𝐸𝐹 is completely full and highest energy electrons must traverse entire band gap to make excitation • Semiconductors differ from insulators simply by size of and gap. Insulators typically have band gap larger than optical frequencies, such that they are often transparent. Semiconductors have band gap smaller than optical frequencies Metals and Fermi surfaces In this section, we introduce the periodic lattice potential onto a free electron gas in 2D and 3D. The starting point is Brillouin zones (BZ). The first BZ is fairly straightforward—it is a Wigner-Seitz cell (see lecture 1). For higher BZs, I am skipping the procedure for deriving them, since this is something one normally looks up. However, these attributes of higher BZs are a good double check if you construct them yourself: • All BZs must have the same total area (check that disjointed regions of BZ 2 & 3 have the same area as one) • Disjointed regions of higher BZs must translate into the first BZ via reciprocal lattice vectors without overlap (e.g. translation only, no rotation) Now add electrons to make a Fermi surface! Starting off with a free-electron model, the Fermi surface in 2D is a circle centered around (𝑘𝑥, 𝑘𝑦) = (0,0). If this circle is small enough to fit entirely within the first Brillouin zone we are done. If the circle overfills the first Brillouin zone, we do the following: • Superimpose the circle on all the Brillouin zones (extended zone scheme) • Consider the portion of the Fermi surface that is inside each Brillouin zone, and translate this back into the first Brillouin zone. For the example shown here, the free electron Fermi surface entirely fills the first Brillouin zone, partially fills the 2nd, and has a small incursion into the 3rd and 4th. Only the first 3 Brillouin zones are shown. Although the filled areas of the 3rd Brillouin zone look disconnected, they can be shown to form connected propeller shapes in the repeating zone scheme. When this procedure is extended to the nearly free electron model, where the lattice potential is not ignored, the following considerations are used • Interactions of electrons with periodic ionic potential opens gaps at Brillouin zone boundaries • Fermi surface will almost always intersect Brillouin zone boundary perpendicular • Crystal potential will round sharp corners of fermi surface • The total volume enclosed by fermi surface depends only on electron concentrations, and will be the same for the free electron case and when the ionic potential is turned on (nearly free electron) Qualitatively, the Fermi surfaces in the 2nd and 3rd Brillouin zone, for the same example as above, will change slightly to the following shapes: Notice that electrons almost fill the second zone, except for a small empty region in the center. This is considered to be a ‘hole-like’ fermi surface, because the enclosed surface constitutes the absence of electrons. In the third zone, electrons fill a minority of the area. This is considered to be an electron-like fermi-surface The nearly free electron model (start with free electron gas and turn on lattice potential) can explain why some real metals have dominant hole-like charge carriers, and also why the measured charge density in many metals is inconsistent with simply counting the number of valence electrons per atom. Tight Binding model The tight binding model is based on combining wavefunctions of individual atomic orbitals. Suppose an isolated atom has potential 𝑈(𝒓) and is in an s-state (spherically symmetric), represented by wavefunction 𝜙(𝒓). Now suppose that there is a crystal of N of these atoms, and the presence of other atoms doesn’t much affect the single-atom wavefunction. The wavefunction of an electron in this whole crystal can be expressed as: 𝜓𝒌(𝒓) = 𝑁−1/2 ∑𝑒𝑖𝒌⋅𝒓𝒋𝜙(𝒓−𝒓𝑗) 𝑗 The first order energy is found by calculating the diagonal matrix elements of the Hamiltonian (where the Hamiltonian describes the kinetic and potential energy of electrons in the crystal…but it turns out we won’t need to write what it is exactly) 𝜖𝑘=< 𝒌|𝐻|𝒌>= 𝑁−1 ∑∑𝑒𝑖𝒌⋅(𝒓𝒋−𝒓𝒎) < 𝜙𝑚|𝐻|𝜙𝑗> 𝑚 𝑗 Where 𝜙𝑚≡𝜙(𝒓−𝒓𝑚) Define a new variable 𝝆𝑚= 𝒓𝑚−𝒓𝑗 < 𝒌|𝐻|𝒌>= ∑𝑒𝑖𝒌⋅𝝆𝒎 𝑚 ∫𝑑𝑉𝜙∗(𝒓−𝝆𝑚)𝐻𝜙(𝒓) Now, we neglect all of the integrals above except for those on the same atom and those between nearest neighbors (separated by 𝝆). This is the tight binding part of the tight binding model: only considering orbital overlap with adjacent atoms assumes that electrons do not make excursion far from their original atom and are hence, tightly bound. Note that it is perfectly acceptable, and sometimes necessary, to consider second nearest neighbors (2nd most closest atom) or even third and fourth, in the tight binding model. However, the solved examples in the book only involve the nearest neighbors. ∫𝑑𝑉𝜙∗(𝒓)𝐻𝜙(𝒓) = −𝛼 ∫𝑑𝑉𝜙∗(𝒓−𝝆)𝐻𝜙(𝒓) = −𝛾 𝛾 can be determined by assuming some specific form of 𝜙. For example, for two hydrogen atoms in 1s states, 𝛾= 2 (1 + 𝝆 𝑎0) 𝑒−𝝆/𝑎0 where 𝑎0 is the Bohr radius. However, in practice, one often one determines it empirically from experiments or first-principles theory (e.g. we measure or calculate a certain 𝜖𝑘, which is best parametrized by certain values of 𝛼 and 𝛾. Thus: < 𝒌|𝐻|𝒌>= −𝛼−𝛾∑𝑒−𝑖𝒌⋅𝝆𝒎 𝑚 = 𝜖𝒌 To proceed further, we need information about the crystal structure. For a simple cubic structure, 𝝆𝑚= (±𝑎, 0,0); (0, ±𝑎, 0); (0,0, ±𝑎) Thus, 𝜖𝒌= −𝛼−2𝛾(cos 𝑘𝑥𝑎+ cos𝑘𝑦𝑎+ cos 𝑘𝑧𝑎) A constant energy surface is shown on the left. For a BCC crystal structure with 8 nearest neighbors, the dispersion is given by: 𝜖𝒌= −𝛼−8𝛾cos 1 2 𝑘𝑥𝑎cos 1 2 𝑘𝑦𝑎cos1 2 𝑘𝑧𝑎 For a FCC structure with 12 nearest neighbors, the dispersion is given by: 𝜖𝒌= −𝛼−4𝛾(cos 1 2 𝑘𝑦𝑎cos 1 2 𝑘𝑧𝑎+ cos 1 2 𝑘𝑧𝑎cos 1 2 𝑘𝑥𝑎+ cos 1 2 𝑘𝑥𝑎cos1 2 𝑘𝑦𝑎 |
187663 | http://media.laballey.com/sds/Mercury-Safety-Data-Sheet-SDS.pdf | SAFETY DATA SHEET Creation Date 20-Aug-2014 Revision Date 20-Aug-2019 Revision Number 1 1. Identification Product Name Cat No. : Synonyms Recommended Use Mercury (Laboratory) C5320, C5330 Colloidal mercury; Hydrargyrum; Metallic mercury Laboratory reagent.
Uses advised against No Information available Details of the supplier of the safety data sheet 2. Hazard(s) identification Classification This chemical is considered hazardous by the 2012 OSHA Hazard Communication Standard (29 CFR 1910.1200) Label Elements Signal Word Danger Hazard Statements May be corrosive to metals Fatal if inhaled May damage the unborn child Causes damage to organs through prolonged or repeated exposure Company Acute Inhalation Toxicity - Vapors Category 2 Reproductive Toxicity Category 1B _____________ Specific target organ toxicity - (repeated exposure) Page 1 / 8 Category 1 Target Organs - Central nervous system (CNS), Kidney.
Corrosive to metals Category 1 Lab Alley LLC 22111 Highway 71 West, Suite 601 Spicewood, Texas 78669 512-668-9918 Buy Mercury Online At Mercury (Laboratory) Revision Date 20-Aug-2019 _____________ Precautionary Statements Prevention Obtain special instructions before use Do not handle until all safety precautions have been read and understood Use personal protective equipment as required Do not get in eyes, on skin, or on clothing Wash face, hands and any exposed skin thoroughly after handling Do not eat, drink or smoke when using this product Do not breathe dust/fume/gas/mist/vapors/spray Use only outdoors or in a well-ventilated area Wear respiratory protection Response IF exposed or concerned: Get medical attention/advice Inhalation IF INHALED: Remove victim to fresh air and keep at rest in a position comfortable for breathing Immediately call a POISON CENTER or doctor/physician Skin Immediately call a POISON CENTER or doctor/physician IF ON SKIN: Gently wash with plenty of soap and water Remove/Take off immediately all contaminated clothing Wash contaminated clothing before reuse Storage Store locked up Store in a well-ventilated place. Keep container tightly closed Disposal Dispose of contents/container to an approved waste disposal plant Hazards not otherwise classified (HNOC) Very toxic to aquatic life with long lasting effects Other hazards WARNING! This product contains a chemical known in the State of California to cause birth defects or other reproductive harm.
3. Composition / information on ingredients Component CAS-No Weight % Mercury 7439-97-6 100 4. First-aid measures Eye Contact Rinse immediately with plenty of water, also under the eyelids, for at least 15 minutes.
Immediate medical attention is required.
Skin Contact Wash off immediately with soap and plenty of water while removing all contaminated clothes and shoes. Immediate medical attention is required.
Inhalation Move to fresh air. If breathing is difficult, give oxygen. Do not use mouth-to-mouth resuscitation if victim ingested or inhaled the substance; induce artificial respiration with a respiratory medical device. Immediate medical attention is required.
Ingestion Do not induce vomiting. Call a physician or Poison Control Center immediately.
Most important symptoms/effects No information available.
Notes to Physician Treat symptomatically Mercury (Laboratory) Revision Date 20-Aug-2019 _____________ 5. Fire-fighting measures Suitable Extinguishing Media Substance is nonflammable; use agent most appropriate to extinguish surrounding fire.
Unsuitable Extinguishing Media No information available Flash Point No information available Method -No information available Autoignition Temperature No information available Explosion Limits Upper No data available Lower No data available Sensitivity to Mechanical Impact No information available Sensitivity to Static Discharge No information available Specific Hazards Arising from the Chemical Very toxic. Non-combustible, substance itself does not burn but may decompose upon heating to produce corrosive and/or toxic fumes. Keep product and empty container away from heat and sources of ignition.
Hazardous Combustion Products Mercury oxide Highly toxic fumes Protective Equipment and Precautions for Firefighters As in any fire, wear self-contained breathing apparatus pressure-demand, MSHA/NIOSH (approved or equivalent) and full protective gear.
NFPA 6. Accidental release measures Personal Precautions Wear self-contained breathing apparatus and protective suit. Evacuate personnel to safe areas. Ensure adequate ventilation. Do not get in eyes, on skin, or on clothing.
Environmental Precautions Should not be released into the environment. See Section 12 for additional ecological information.
Methods for Containment and Clean Up Wear self-contained breathing apparatus and protective suit. Soak up with inert absorbent material. Keep in suitable, closed containers for disposal.
7. Handling and storage Handling Use only under a chemical fume hood. Wear personal protective equipment. Do not get in eyes, on skin, or on clothing. Do not breathe vapors or spray mist. Do not ingest.
Storage Keep containers tightly closed in a dry, cool and well-ventilated place. Corrosives area.
8. Exposure controls / personal protection Exposure Guidelines Health 4 Flammability 0 Instability 0 Physical hazards N/A Component ACGIH TLV OSHA PEL NIOSH IDLH Mercury TWA: 0.025 mg/m3 Skin (Vacated) TWA: 0.05 mg/m3 Ceiling: 0.1 mg/m3 (Vacated) STEL: 0.03 mg/m3 Skin (Vacated) Ceiling: 0.1 mg/m3 IDLH: 10 mg/m3 TWA: 0.05 mg/m3 Ceiling: 0.1 mg/m3 Component Quebec Mexico OEL (TWA) Ontario TWAEV Mercury TWA: 0.025 mg/m3 Skin TWA: 0.05 mg/m3 TWA: 0.025 mg/m3 Skin Legend ACGIH - American Conference of Governmental Industrial Hygienists OSHA - Occupational Safety and Health Administration NIOSH IDLH: The National Institute for Occupational Safety and Health Immediately Dangerous to Life or Health Mercury (Laboratory) Revision Date 20-Aug-2019 _____________ 9. Physical and chemical properties Physical State Liquid Appearance Silver Odor Odorless Odor Threshold No information available pH No information available Melting Point/Range -38.87 °C / -38 °F Boiling Point/Range 356.72 °C / 674.1 °F Flash Point No information available Evaporation Rate No information available Flammability (solid,gas) No information available Flammability or explosive limits Upper No data available Lower No data available Vapor Pressure 0.002 mmHg @ 25 °C Vapor Density 7.0 Relative Density 13.59 (H2O=1) Solubility Insoluble in water Partition coefficient; n-octanol/water No data available Autoignition Temperature No information available Decomposition temperature No information available Viscosity No information available Molecular Formula Hg Molecular Weight 200.59 10. Stability and reactivity Reactive Hazard None known, based on information available Stability Stable under normal conditions.
Conditions to Avoid Incompatible products. Excess heat.
Incompatible Materials Strong oxidizing agents, Ammonia, Metals, Halogens Hazardous Decomposition Products Mercury oxide, Highly toxic fumes Hazardous Polymerization Hazardous polymerization does not occur.
Hazardous Reactions None under normal processing.
11. Toxicological information Acute Toxicity Product Information No acute toxicity information is available for this product Component Information Toxicologically Synergistic Products No information available Delayed and immediate effects as well as chronic effects from short and long-term exposure Engineering Measures Use only under a chemical fume hood. Ensure adequate ventilation, especially in confined areas. Ensure that eyewash stations and safety showers are close to the workstation location.
Personal Protective Equipment Eye/face Protection Wear appropriate protective eyeglasses or chemical safety goggles as described by OSHA's eye and face protection regulations in 29 CFR 1910.133 or European Standard EN166.
Skin and body protection Wear appropriate protective gloves and clothing to prevent skin exposure.
Respiratory Protection Follow the OSHA respirator regulations found in 29 CFR 1910.134 or European Standard EN 149. Use a NIOSH/MSHA or European Standard EN 149 approved respirator if exposure limits are exceeded or if irritation or other symptoms are experienced.
Hygiene Measures Handle in accordance with good industrial hygiene and safety practice.
Mercury (Laboratory) Revision Date 20-Aug-2019 _____________ Irritation No information available Sensitization No information available Carcinogenicity The table below indicates whether each agency has listed any ingredient as a carcinogen.
Component CAS-No IARC NTP ACGIH OSHA Mexico Mercury 7439-97-6 Not listed Not listed Not listed Not listed Not listed Mutagenic Effects No information available Reproductive Effects No information available.
Developmental Effects May cause harm to the unborn child.
Teratogenicity No information available.
STOT - single exposure None known STOT - repeated exposure Central nervous system (CNS) Kidney Aspiration hazard No information available Symptoms / effects,both acute and delayed No information available Endocrine Disruptor Information No information available Other Adverse Effects The toxicological properties have not been fully investigated.
12. Ecological information Ecotoxicity This product contains the following substance(s) which are hazardous for the environment.
Component Freshwater Algae Freshwater Fish Microtox Water Flea Mercury Not listed 0.9 mg/L LC50 96 h 0.18 mg/L LC50 96 h 0.16 mg/L LC50 96 h 0.5 mg/L LC50 96 h Not listed 5.0 µg/L EC50 = 96 h Persistence and Degradability No information available Bioaccumulation/ Accumulation No information available.
Mobility No information available.
13. Disposal considerations Waste Disposal Methods Chemical waste generators must determine whether a discarded chemical is classified as a hazardous waste. Chemical waste generators must also consult local, regional, and national hazardous waste regulations to ensure complete and accurate classification.
Component RCRA - U Series Wastes RCRA - P Series Wastes Mercury - 7439-97-6 U151 -14. Transport information DOT UN-No UN2809 Proper Shipping Name MERCURY Hazard Class 8 Subsidiary Hazard Class 6.1 Packing Group III TDG UN-No UN2809 Proper Shipping Name MERCURY Hazard Class 8 Mercury (Laboratory) Revision Date 20-Aug-2019 _____________ Subsidiary Hazard Class 6.1 Packing Group III IATA UN-No UN2809 Proper Shipping Name MERCURY Hazard Class 8 Subsidiary Hazard Class 6.1 Packing Group III IMDG/IMO UN-No UN2809 Proper Shipping Name MERCURY Hazard Class 8 Subsidiary Hazard Class 6.1 Packing Group III 15. Regulatory information International Inventories Component TSCA DSL NDSL EINECS ELINCS NLP PICCS ENCS AICS IECSC KECL Mercury X X -231-106-7 -X -X X X Legend: X - Listed E - Indicates a substance that is the subject of a Section 5(e) Consent order under TSCA.
F - Indicates a substance that is the subject of a Section 5(f) Rule under TSCA.
N - Indicates a polymeric substance containing no free-radical initiator in its inventory name but is considered to cover the designated polymer made with any free-radical initiator regardless of the amount used.
P - Indicates a commenced PMN substance R - Indicates a substance that is the subject of a Section 6 risk management rule under TSCA.
S - Indicates a substance that is identified in a proposed or final Significant New Use Rule T - Indicates a substance that is the subject of a Section 4 test rule under TSCA.
XU - Indicates a substance exempt from reporting under the Inventory Update Rule, i.e. Partial Updating of the TSCA Inventory Data Base Production and Site Reports (40 CFR 710(B).
Y1 - Indicates an exempt polymer that has a number-average molecular weight of 1,000 or greater.
Y2 - Indicates an exempt polymer that is a polyester and is made only from reactants included in a specified list of low concern reactants that comprises one of the eligibility criteria for the exemption rule.
U.S. Federal Regulations TSCA 12(b) Component TSCA 12(b) Mercury Section 5 SARA 313 Component CAS-No Weight % SARA 313 - Threshold Values % Mercury 7439-97-6 100 1.0 SARA 311/312 Hazardous Categorization Acute Health Hazard Yes Chronic Health Hazard Yes Fire Hazard No Sudden Release of Pressure Hazard No Reactive Hazard No Clean Water Act Component CWA - Hazardous Substances CWA - Reportable Quantities CWA - Toxic Pollutants CWA - Priority Pollutants Mercury --X X Clean Air Act Component HAPS Data Class 1 Ozone Depletors Class 2 Ozone Depletors Mercury X -OSHA Occupational Safety and Health Administration Not applicable Mercury (Laboratory) Revision Date 20-Aug-2019 _____________ CERCLA This material, as supplied, contains one or more substances regulated as a hazardous substance under the Comprehensive Environmental Response Compensation and Liability Act (CERCLA) (40 CFR 302) Component Hazardous Substances RQs CERCLA EHS RQs Mercury 1 lb -California Proposition 65 This product contains the following Proposition 65 chemicals: Component CAS-No California Prop. 65 Prop 65 NSRL Category Mercury 7439-97-6 Developmental -Developmental State Right-to-Know Component Massachusetts New Jersey Pennsylvania Illinois Rhode Island Mercury X X X X X U.S. Department of Transportation Reportable Quantity (RQ): N DOT Marine Pollutant N DOT Severe Marine Pollutant N U.S. Department of Homeland Security This product does not contain any DHS chemicals.
Other International Regulations Mexico - Grade No information available Canada This product has been classified in accordance with the hazard criteria of the Controlled Products Regulations (CPR) and the MSDS contains all the information required by the CPR WHMIS Hazard Class D1A Very toxic materials D2A Very toxic materials E Corrosive material 16. Other information Prepared By Creation Date Revision Date Print Date Revision Summary 20-Aug-2014 20-Aug-2019 20-Aug-2019 This document has been updated to comply with the US OSHA HazCom 2012 Standard replacing the current legislation under 29 CFR 1910.1200 to align with the Globally Harmonized System of Classification and Labeling of Chemicals (GHS) Regulatory Affairs Lab Alley LLC Email: customerservice@laballey.com Mercury (Laboratory) Revision Date 20-Aug-2019 _____________ Disclaimer The information provided on this Safety Data Sheet is correct to the best of our knowledge, information and belief at the date of its publication. The information given is designed only as a guide for safe handling, use, processing, storage, transportation, disposal and release and is not to be considered as a warranty or quality specification. The information relates only to the specific material designated and may not be valid for such material used in combination with any other material or in any process, unless specified in the text.
End of SDS |
187664 | https://www.reddit.com/r/learnmath/comments/leu5ho/set_theory_ab_c_ab_ac/ | [Set Theory] A(B ∩ C) ⊇ AB ∩ AC : r/learnmath
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[Set Theory] A(B ∩ C) ⊇ AB ∩ AC
Can someone explain to me if A(B ∩ C) ⊇ AB ∩ AC is actually a true. I personally think it's a true statement but my friend says otherwise? The question involves strings and alphabets related to computer science if that helps!
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187665 | https://www.cuemath.com/algebra/multiplying-and-dividing-exponents/ | LearnPracticeDownload
Multiplying and Dividing Exponents
An exponent shows how many times a given variable or number is multiplied by itself. For example, 64 means we are multiplying 6 four times. In the expanded form, it is written as 6 × 6 × 6 × 6. When two exponential terms with the same base are multiplied, their powers are added while the base remains the same. However, when two exponential terms having the same base are divided, their powers are subtracted. Let us learn more about multiplying and dividing exponents in this article.
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| 1. | Dividing Exponents |
| 2. | How to Multiply and Divide Fractional Exponents? |
| 3. | How to Multiply and Divide Exponents with Variables? |
| 4. | FAQs on Multiplying and Dividing Exponents |
Dividing Exponents
The laws of exponents make the process of simplifying expressions easier. The basic rule for dividing exponents with the same base is that we subtract the given powers. This is also known as the Quotient Property of Exponents.
How to Divide Exponents?
Dividing exponents becomes easy when we follow the properties of exponents. For example, let us solve the following question in the usual way, 65 ÷ 63 = (6 × 6 × 6 × 6 × 6)/(6 × 6 × 6 ) = 62. This involves more calculation. However, when we use the laws of exponents, it reduces all these calculations. Let us understand how to divide exponents in different scenarios using the different properties.
Dividing Exponents with Same Base
In order to divide exponents with the same base, we use the basic rule of subtracting the powers. Consider am ÷ an, where 'a' is the common base and 'm' and 'n' are the exponents. This 'Quotient property of Exponents' says, am ÷ an = am-n. Now, let us understand this with an example.
Example: Divide 65 ÷ 63
Solution: We can see that in the given expression, the bases are the same. Using the 'Quotient property of Exponents', we will get, 65 - 3 = 62. Therefore the answer is 62.
Dividing Exponents with Different Bases
In order to divide exponents with different bases and the same exponent, we use the 'Power of quotient property', which is, (a/b)m = am/bm. Consider am ÷ bm, where the expressions have different bases and the same exponent. For example, let us solve: 123 ÷ 33. Using the 'Power of quotient property', this can be solved as, 123 ÷ 33 = (12 ÷ 3)3 = 43
Dividing Exponents with Coefficients
In some cases, we need to divide expressions that have coefficients. These coefficients that are attached to their bases can be divided easily in the same way as we divide any other fraction. It should be noted that the coefficients can be divided even if the expressions have different bases.
Example: Divide 12a7 ÷ 4a2
Solution: Let us use the following steps to divide expressions with coefficients. In this case, 12 and 4 are the coefficients and the rest are variables.
First, we rewrite the expression as a fraction, that is, 12a7/ 4a2.
Then we divide the coefficients, that is, 12/4 = 3.
After this step, we can apply the quotient property of exponents and solve the variable, that is, a7/a2 = a7 - 2 = a5.
So, now we have the coefficient as 3 and the variable is a5. This gives the answer as, 3a5
Multiplying Exponential Terms
Multiplying exponents with the same base and different bases involves certain rules of exponents. Let us understand these in the following section.
Multiplying Exponents with the Same Base
When we multiply two expressions with the same base, we apply the rule, am × an = a(m + n), in which 'a' is the common base and 'm' and 'n' are the exponents. For example, let us multiply 22 × 23. Using the rule, 22 × 23 = 2 (2 + 3) = 25.
Multiplying Exponents with Different Base and Same Power
When we multiply expressions with different bases and the same power, we apply the rule: am × bm = (a × b)m. For example, let us multiply: 114 × 34. This can be solved as, 114 × 34 = (11 × 3)4 = 334.
How to Multiply and Divide Fractional Exponents?
In order to multiply and divide fractional exponents, we use the same exponent rules that we apply for whole numbers. Fractional exponents are those expressions in which the powers are fractions, for example, 2½, 6¾, and so on.
Multiplying Fractional Exponents with the Same Base
In order to multiply fractional exponents with the same base, we use the rule, am × an = am+n. For example, let us simplify, 2½ × 2¾ = 2(½ + ¾ ) = 25/4.
Dividing Fractional Exponents with the Same Base
For dividing fractional exponents with the same base, we use the rule, am ÷ an = am-n. For example, let us solve, 33/2 ÷ 31/2. Using the rule, we get, 3(3/2 - 1/2) = 31 = 3.
How to Multiply and Divide Exponents With Variables?
The rules which are used in numbers are also used in exponents with variables. Let us recollect them and then use them in the following examples:
am × an = am+n
am × bm = (a × b)m
am ÷ an = am-n
am ÷ bm = (a ÷ b)m
Variable as the Base
Let us see how to use these rules when the base is a variable. For example, solve: y2 × (2y)3
We will apply the rule: am × bm = (a × b)m , y2 × (2y)3 = y2 × 23 × y3 = 23 × y(2+3) = 8y5
Variable as the Exponent
Let us see how to use the rules when the exponent is a variable. For example, solve: 5(2x -1) ÷ 5(x + 1)
We will apply the rule: am ÷ an = am-n , we get 5(2x -1 - x - 1) = 5(x -2)
Tips on Multiplying and Dividing Exponents
a0 = 1[ since am ÷ am = 1 = am-m = a0]
It should also be noted that a negative exponent can be converted to a positive exponent by writing the reciprocal of the number. For example, 6-3 can be written as 1/63.
If we multiply two exponents with the same base then their powers will add.
If we divide two exponents with the same base then their powers will subtract.
Related Topics
Rational Exponents
Irrational Exponents
Multiplication and Division of Exponents Examples
Example 1: Find the value of the expression, 168 × 163
Solution: Using the exponent rule for multiplying exponents, we can solve the given expression. According to the 'Product property of exponents', am × an = a(m + n)
Applying this rule, we get, 168 × 163 = 168 + 3 = 1611
2. Example 2: Find the value of the expression (\dfrac{a^{(n-2)}}{a^{(2n-4)}})
Solution: Applying the rules for dividing exponents, we can solve the given expression. According to the 'Quotient property of exponents', am ÷ an = am-n. So, the given expression will be,
(\dfrac{a^{(n-2)}}{a^{(2n-4)}}) = a(n - 2 - 2n + 4)
= a(n - 2 - 2n + 4) = a(2 - n)
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FAQs on Multiplying And Dividing Exponents
How do you Multiply and Divide Exponents?
In order to multiply and divide exponents, we use a set of exponent rules. When two exponential terms with the same base are multiplied, their powers are added while the base remains the same. For example, let us multiply, 63 × 65 = 6(3 + 5) = 68. However, when two exponential terms having the same base are divided, their powers are subtracted. For example, 78 ÷ 75 = 73. Similarly, there are other rules that help to simplify exponents easily.
What are the Rules for Dividing Exponents?
There are a few exponent rules that help in the division of exponents. These rules also help in simplifying numbers with complex powers involving fractions, decimals, and roots. For example, in order to divide the numbers or variables with the same base, we apply the rule: am ÷ an = am-n. To divide the numbers or variables with different bases, we apply the rule: am ÷ bm = (a ÷ b)m
How do you Solve Exponents in Parentheses?
The exponents inside parentheses can be solved using the identity (am) n = amn. For example, (42)3 = 4(2 × 3) = 46 = 4096
Can we Distribute Exponents Over Division?
Yes, we can distribute exponents over division. For example, (7/2) 3 = 73 ÷ 23 = 343/8
How to Multiply and Divide Negative Exponents?
When we multiply and divide negative exponents, we follow the same rules that are used for positive exponents. For example, we use the property: am ÷ an = am-n, to solve: 2-3 ÷ 2-4. This will be: 2(-3-(-4)) = 2(-3 + 4) = 21 = 2. Observe the rule of simplification of integers which changes the sign after the brackets are opened. It should also be noted that a negative exponent can be converted to a positive exponent by writing the reciprocal of the number. For example, 7-3 can also be written as: 1/73. This means that if we need to divide expressions that have negative exponents, we can simply move the base to the other side of the fraction bar. For example, if we have 4-2 in the denominator of a fraction, we can move it to the numerator. This means, y-2/y-3 = y3/y2 = y3 - 2 = y1 = y.
How to Divide Exponents with Different Powers?
In order to divide exponents with different powers, but the same bases, we subtract the given powers. The property which is used here is, am ÷ an = a(m-n). For example, let us divide the exponents, 86 ÷ 84. After applying the Quotient property of exponents, we get, 86 - 4 = 82
How to Divide Exponents with Fractions?
In order to divide exponents with fractions, we use the same rule that is used for whole numbers, that is, am ÷ an = am-n. For example, let us divide the following exponents, 23/4 ÷ 21/2 = 23/4 -1/2 = 21/4.
How to Divide Exponents with Different Bases and Same Powers?
In order to divide exponents with different bases and the same powers, we apply the 'Power of Quotient Property' which is, am ÷ bm = (a ÷ b)m. For example, let us divide, 143 ÷ 23 = (14 ÷ 2)3 = 73.
How to Divide Exponents with Negative Bases?
When we need to divide exponents with negative bases, the exponent rules remain the same. For example, let us divide (-4)8 ÷ (-4)2 = (-4)8 - 2 = (-4)6
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187666 | http://www.brunswick.k12.me.us/pgroves/files/2012/12/AP-Stats-7.5N-7.5G-Notes.pdf | 1 Section 7.5N and 7.5G Normal Model as an Approximation to the Binomial Model and Geometric Random Variables After this section, you should be able to… DETERMINE whether the conditions have been met to use to use the Normal Model as an Approximation to the Binomial Model. THEN: COMPUTE and INTERPRET probabilities in context CALCULATE the mean and standard deviation and INTERPRET these values in context Learning Objectives – Normal Model as an Approximation to the Binomial Model + 2 Normal Approximation for Binomial Distributions As n gets larger, something interesting happens to the shape of a binomial distribution. The figures below show histograms of binomial distributions when p=.8 and different values of n. 1) n=10 2) n=20 3) n=50 What do you notice as n gets larger?
Binomial Random Variables + 3 Normal Approximation for Binomial Distributions Binomial Random Variables Suppose that X has the binomial distribution with n trials and success probability p. When n is large, the distribution of X is approximately Normal with mean and standard deviation We can use the Normal approximation when: • np ≥10 and • n(1 – p) ≥10. • That is, the expected number of successes and failures are both at least 10.
Normal Approximation for Binomial Distributions X np X np(1p) Note, the binomial distribution would be MORE accurate than the normal distribution.
4 Attitudes toward Shopping EXAMPLE – Normal Approximation for Binomial Distributions Use the following steps to solve this problem: Steps: Check the BINS conditions Define the Random Variable Check the conditions: np≥10 and n(1-p)≥10 IMPORTANT: You MUST show both calculations to indicate you verified this condition !!!!
Calculate the mean and the standard deviation.
Draw a picture. Then calculate the probability of interest State your conclusion, in the context of the problem.
Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2,500 adults if they agreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” 60% of the adults surveyed agreed with this statement. What is the probability that between 1,000 and 2,000 of the sample agrees?
Binomial Random Variables + 5 Attitudes toward Shopping EXAMPLE – Normal Approximation for Binomial Distributions Solution: Steps: Check the BINS conditions They check: there are 2 outcomes (agree/disagree), fixed number (n=2500), fixed probability=(.6), and independent (random sample).
Define the Random Variable: X=the number in the sample that agree Check the conditions: np≥10 and n(1-p)≥10 IMPORTANT: You MUST show both calculations to indicate you verified this condition !!!!
np = 2500(.6)= 1500 ≥10 n(1-p)= 2500(.4)= 1000 ≥10 Calculate the mean and the standard deviation.
μ = np=2500(.6)= 1500 σ2 = npq=2500(.6)(.4)=600 σ=24.494 Draw a picture. Then calculate the probability of interest P(1500 ≤X≤1600) = normalcdf(1500,1600,1500,24.49) =.499 State your conclusion, in the context of the problem.
There is about a 50% chance that between 1,500 and 1,600 of the survey respondents agree with the shopping survey question.
Binomial Random Variables 1500 1600 + 6 + 7 Section 7.5G Geometric Random Variables After this section, you should be able to… DETERMINE whether the conditions for a geometric setting are met CALCULATE probabilities involving geometric random variables Learning Objectives Geometric Random Variables Learning Objectives + 8 Geometric Random Variables Geometric Settings In a binomial setting, the number of trials n is fixed and the binomial random variable X counts the number of successes. In other situations, the goal is to repeat a chance behavior until a success occurs. These situations are called geometric settings.
Definition: A geometric setting arises when we perform independent trials of the same chance process and record the number of trials until a particular outcome occurs. The four conditions for a geometric setting are • Binary? The possible outcomes of each trial can be classified as “success” or “failure.” • Independent? Trials must be independent; that is, knowing the result of one trial must not have any effect on the result of any other trial.
• Trials? The goal is to count the number of trials until the FIRST success occurs.
• Success? On each trial, the probability p of success must be the same.
B I T S + 9 Geometric Random Variables Geometric Random Variable In a geometric setting, if we define the random variable Y to be the number of trials needed to get the first success, then Y is called a geometric random variable. The probability distribution of Y is called a geometric distribution.
Definition: The number of trials Y that it takes to get a success in a geometric setting is a geometric random variable. The probability distribution of Y is a geometric distribution with parameter p, the probability of a success on any trial. The possible values of Y are 1, 2, 3, ….
Note: Like binomial random variables, it is important to be able to distinguish situations in which the geometric distribution does and doesn’t apply!
10 2) Verify that Y is a geometric random variable.
B: Success = correct guess, Failure = incorrect guess I: The result of one student’s guess has no effect on the result of any other guess.
N: We’re counting the number of guesses up to and including the first correct guess.
S: On each trial, the probability of a correct guess is 1/7.
The Birthday Game EXAMPLE – Geometric Distributions Your teacher is planning to give you 20 problems for homework. She agrees to play the Birth Day Game. She secretly thinks of a person’s birthday and writes the birthday down hidden from the students. She then selects at random a student from the class and asks them to guess the day of the month. If the student guesses correctly, the students have only 1 HW problem. If the student guesses incorrectly she repeats the process (selecting at random any student in the class). The teacher assigns the number of problems to the number of students that guessed.
1) Define the random variable: Y = the number of guesses it takes to correctly identify the birth day of one of your teacher’s friends.
3) What is the probability the student guesses correctly The 1st student? P(Y=1)=____= The Second? P(Y=1)=____=__ The Third? P(Y=1)=_____=______ Notice the pattern?
11 (3 cont.) Probability the student guesses correctly P(Y 1) 1/7 P(Y 2) (6/7)(1/7) 0.1224 P(Y 3) (6/7)(6/7)(1/7) 0.1050 If Y has the geometric distribution with probability p of success on each trial, the possible values of Y are 1, 2, 3, … . If k is any one of these values, p p k Y P k 1 ) 1 ( ) ( Geometric Probability – G(p) The Birthday Game EXAMPLE – Geometric Distributions The Pattern… What is the probability the kth student guesses corrrectly?
12 The table and histogram below shows part of the probability distribution of Y. We can’t show the entire distribution because the number of trials it takes to get the first success could be an incredibly large number.
Geometric Random Variables Shape: The heavily right-skewed shape is characteristic of any geometric distribution. That’s because the most likely value is 1.
Center: We could calculate the expected value E(Y)=∑yipi and the mean is µY = 7. We’d expect it to take 7 guesses to get our first success.
Spread: The standard deviation of Y is σY = 6.48. If the class played the Birth Day game many times, the number of homework problems the students receive would differ from 7 by an average of 6.48.
yi 1 2 3 4 5 6 … pi 0.143 0.122 0.105 0.090 0.077 0.066 If Y is a geometric random variable with probability p of success on each trial, then its mean (expected value) is E(Y) = µY = 1/p.
Mean (Expected Value) of Geometric Random Variable The Birthday Game EXAMPLE – MEAN of the Geometric Distributions + 13 1) A geometric setting consists of repeated trials of the same chance process in which each trial results in a success or a failure; trials are independent; each trial has the same probability p of success; and the goal is to count the number of trials until the first success occurs.
2) Define the geometric random variable: Y = the number of trials required to obtain the first success. Y are the positive integers 1, 2, 3, . . . .
3) The Geometric probability distributiom model: G(p) p = probability of success q = 1 – p = probability of failure Y = number of trials until the first success occurs P(Y k) (1p)k1p Summary of Geometric Distributions 2 q p E(X) 1 p + 14 Binomial model When we’re interested in the number of successes in a fixed number of trials, with a fix probability, and 2 outcomes (binomial).
Normal model To approximate a Binomial model when we expect at least 10 successes and 10 failures.
Geometric model When we’re interested in the number of Binomial trials until the FIRST success.
Section 7.5: The 3 Types of Distributions Summary + 15 Binomial Distributions in Statistical Sampling After this section, you should be able to… DETERMINE whether the conditions are met using a binomial distribution CALCULATE probabilities of success using a binomial distribution Learning Objectives When taking an SRS of size n from a population of size N, we can use a binomial distribution to model the count of successes in the sample as long as n 1 10 N Sampling Without Replacement Condition + 16 In practice, the binomial distribution gives a good approximation as long as we don’t sample more than 10% of the population.
Binomial Distributions in Statistical Sampling The binomial distributions are important in statistics when we want to make inferences about the proportion p of successes in a population.
Suppose 10% of CDs have defective copy-protection schemes that can harm computers. A music distributor inspects an SRS of 10 CDs from a shipment of 10,000. Let X = number of defective CDs. What is P(X = 0)? Note, this is not quite a binomial setting. Why?
Binomial Random Variables The actual probability is P(no defectives ) 9000 100008999 99998998 9998...8991 99910.3485 P(X 0) 10 0 (0.10)0(0.90)10 0.3487 Using the binomial distribution, |
187667 | https://math.stackexchange.com/questions/20631/for-any-even-function-f-there-is-infinite-number-of-functions-g-such-that-f | calculus - For any even function $f$ there is infinite number of functions $g$ such that $f(x) = g(|x|)$ - Mathematics Stack Exchange
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For any even function f f there is infinite number of functions g g such that f(x)=g(|x|)f(x)=g(|x|)
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This is question from Spivak's Calculus. Question statement (paraphrased): For any even function f f there is infinite number of functions g g such that f(x)=g(|x|)f(x)=g(|x|)
I have made attempt at proof, here is my work. My main concern is whether my proof that there are infinite such functions is correct.
Let g g be a function such that for any positive x x, f(x)=g(x)f(x)=g(x) and let f f be some even function. We will show that above equality then holds even when x x is negative: f(−x)=g(|−x|)⇔f(x)=g(x)f(−x)=g(|−x|)⇔f(x)=g(x), which is true by definition of these functions. It now remains to show that there is infinite number of those functions. I understand that is so because g g can be defined as one whishes for negative numbers and it won't change equality. But I cannot think of formal proof. I'm thinking about assigning values to functions g g in a way that I define g 1(−1)=k g 1(−1)=k and then making infinite number of them by induction: for any function g n(x)=k g n(x)=k, if x is negative, we define g n+1(x−1)=k g n+1(x−1)=k. Then by the fact that there is infinite natural numbers, and for every natural number there is going to be unique function, we can conclude there is infinite number of functions.
Is the concept of induction applicable here or is it not?
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asked Feb 6, 2011 at 0:48
user5501 user5501
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1 Induction in terms of "one-at-a-time" doesn't work for real numbers (but see math.stackexchange.com/questions/4202/induction-on-real-numbers). However, it seems a bit silly; you already know that it doesn't matter how you define g g on the negative numbers, as long as g(x)=f(x)g(x)=f(x) for x≥0 x≥0. So, why not define g k(x)=k g k(x)=k for all x<0 x<0, and g k(x)=f(x)g k(x)=f(x) for all x≥0 x≥0? Different values of k k give different g g s, so it's just a question of how many different k k you can pick. I think you'll have enough choices there...Arturo Magidin –Arturo Magidin 2011-02-06 00:55:35 +00:00 Commented Feb 6, 2011 at 0:55
Everything here up to "But I cannot think of a formal proof" is correct. However, I found your description of your family of functions a little confusing (partly because you don't say much about g 1 g 1 other than the value g 1(−1)g 1(−1)). Maybe you can find an easier choice for your family of functions g n g n. (After all, there are a lot choices!)Matt E –Matt E 2011-02-06 00:57:18 +00:00 Commented Feb 6, 2011 at 0:57
@Arturo: To be honest I don't see how I used induction on real numbers. I used induction on numbering of my functions, such that every function with different numbering is different. But I agree with other part of the post. Thank you.user5501 –user5501 2011-02-06 01:01:16 +00:00 Commented Feb 6, 2011 at 1:01
@Matt: My family is such that it retains all of the properties of defined g g and adds that value.user5501 –user5501 2011-02-06 01:03:36 +00:00 Commented Feb 6, 2011 at 1:03
1 @LovreP: I think I figured it out. Your g 1 g 1 is defined only on nonnegative reals and−1−1; then your g 2 g 2 is defined only on nonnegative reals and−1−1, and −2−2. And so on? Then it doesn't work. Your functions g g have to be defined on all real numbers. So you need to say what the value of g 1 g 1 is at, say, −0.5−0.5, and −π−π, and all negative real numbers, not just −1−1. Otherwise, you don't get f(x)=g(|x|)f(x)=g(|x|).Arturo Magidin –Arturo Magidin 2011-02-06 01:06:56 +00:00 Commented Feb 6, 2011 at 1:06
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Okay, from the comments I was able to guess at what you were trying to do.
You are correct that the key is that as long as g(x)=f(x)g(x)=f(x) for x≥0 x≥0, everything will work out, so you only need to worry about defining g g on the negative numbers.
You attempted to do so by letting g 1 g 1 be a function that is defined only on the nonnegative numbers and −1−1, and setting g 1(−1)=k g 1(−1)=k. Then g 2 g 2 would be an extension of g 1 g 1, which is also defined as k k at −2−2; and so on. In general, g n g n would be define on {−n,−n+1,…,−1}∪[0,∞){−n,−n+1,…,−1}∪[0,∞), by g(x)=f(x)g(x)=f(x) if x≥0 x≥0 and g(x)=k g(x)=k if x<n x<n.
As I noted in the comments, your formulas didn't really say that; instead, they only specified g 1 g 1 at −1−1, and then said, for example, that g 2(x−1)=k g 2(x−1)=k. That is at best confusing. What is g 2(−0.5)g 2(−0.5)? According to this, I have to think of −0.5−0.5 as 0.5−1 0.5−1, and then it's k k, and ... Well, a bit of a confusing issue arises...
In any case, whether this works as an answer or not depends on whether you are assuming that your functions need to be defined on the same set or not. Normally, we would be looking for functions g g such that f(x)=g(|x|)f(x)=g(|x|) and we want both f f and g g to have the same domain. Remember that two functions are equal if and only if they have the same domain, the same codomain, and the same value at every element of the domain. So even if you could have set up the induction properly to get the functions you wanted (or if you wanted to define g 1 g 1 on [−1,0)[−1,0), then g 2 g 2 extended to all of [−2,0)[−2,0), and so on sothat g n g n was defined on [−n,∞)[−n,∞)) it still would not give a good answer to the problem because of the restrictive domains of your g g s. (You are really composing the absolute value with g g; I think Spivak wants you to play with functions that are defined everywhere here, rather than on artificial domains; I could be wrong, though).
If you do not require your functions g g to have the same domain as f f, then your intended answer would also work; the functions g n g n are different because they have different domains, even though they all have the same values where their domains agree.
I think, though, that the intended answer relies instead in defining g n(x)g n(x) for x<0 x<0 as different things for different n n. For example, you could set
g n(x)={f(x)n if x≥0,if x<0;g n(x)={f(x)if x≥0,n if x<0;
and this would work. There is no need to state it as induction, because the values depend only on the labels, and you simply have infinitely many distinct labels to choose from.
That said: yes, you can use induction to define a (countably infinite) series of functions. If you wanted to do that here, you could express it explicitly stating what g 1 g 1 is (giving its domain clearly; if all you say is g(−1)=k g(−1)=k, then you are telling us the value at −1−1, but not saying anything about values elsewhere, not even that it is not defined there). Then saying that assuming you have defined g n g n with a domain of, say, [−n,∞)[−n,∞), you define g n+1 g n+1 on [−n−1,∞)[−n−1,∞) by
g n+1(x)={g n(x)whatever if x∈[−n,∞).if x∈[−n−1,−n);g n+1(x)={g n(x)if x∈[−n,∞).whatever if x∈[−n−1,−n);
This would indeed give an inductive definition for your g n g n, defined on [−n,∞)[−n,∞), each extending the previous one.
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edited Feb 6, 2011 at 1:49
answered Feb 6, 2011 at 1:31
Arturo MagidinArturo Magidin
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Thank you a lot for your time, I learned a lot reading your posts.user5501 –user5501 2011-02-06 01:44:15 +00:00 Commented Feb 6, 2011 at 1:44
I think that you meant [−2,0)[−2,0) at the fifth paragraph, and [−n−1,−n)[−n−1,−n) in your definition of g n+1 g n+1. Please correct me if I'm wrong in either case.user5501 –user5501 2011-02-06 01:45:54 +00:00 Commented Feb 6, 2011 at 1:45
@LovreP: Yes on both; thanks.Arturo Magidin –Arturo Magidin 2011-02-06 01:49:26 +00:00 Commented Feb 6, 2011 at 1:49
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It is enough to observe that g g can have any value whatsoever for x<0 x<0. Since there is an infinite number of values for each f(x)f(x) for x<0 x<0, we're done.
(you really don't need to do it any more difficult than that.)
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edited Feb 6, 2011 at 1:03
answered Feb 6, 2011 at 0:55
Fredrik MeyerFredrik Meyer
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1 Close. It is not that there are many values of x<0, but many possible values of g(x) for x<0. Imagine that something in the problem forced g(x) to be 0 for x<0. You still have lots of x, but only one g.Ross Millikan –Ross Millikan 2011-02-06 01:00:01 +00:00 Commented Feb 6, 2011 at 1:00
1 @Frederik: Actually, you are being somewhat misleading. It's not that there are an infinite number of x x that are less than 0 by itself, it's that you have a choice about where to send them. The result would still be true if your function was only defined on {−1,0,1}{−1,0,1}. (The only way you cannot is if you have a finite number of negative numbers, and a finite number of possible images...)Arturo Magidin –Arturo Magidin 2011-02-06 01:01:26 +00:00 Commented Feb 6, 2011 at 1:01
I'm being silly of course. Thanks for the correction.Fredrik Meyer –Fredrik Meyer 2011-02-06 01:04:01 +00:00 Commented Feb 6, 2011 at 1:04
I see. Thank you for your post.user5501 –user5501 2011-02-06 01:04:19 +00:00 Commented Feb 6, 2011 at 1:04
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You could use induction but it's really not necessary. Just define g k(x)=k g k(x)=k when x x is negative. Then you have a infinite family of functions that are different.
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187668 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4194134/ | Biological network inference using low order partial correlation - PMC
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Published in final edited form as: Methods. 2014 Jul 5;69(3):266–273. doi: 10.1016/j.ymeth.2014.06.010
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Biological network inference using low order partial correlation
Yiming Zuo
Yiming Zuo
a Lombardi Comprehensive Cancer Center, Georgetown University, Washington, DC, USA
b Department of Electrical and Computer Engineering, Virginia Polytechnic Institute and State University, Arlington, VA, USA
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a,b, Guoqiang Yu
Guoqiang Yu
b Department of Electrical and Computer Engineering, Virginia Polytechnic Institute and State University, Arlington, VA, USA
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b, Mahlet G Tadesse
Mahlet G Tadesse
c Department of Mathematics and Statistics, Georgetown University, DC, USA
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c, Habtom W Ressom
Habtom W Ressom
a Lombardi Comprehensive Cancer Center, Georgetown University, Washington, DC, USA
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a Lombardi Comprehensive Cancer Center, Georgetown University, Washington, DC, USA
b Department of Electrical and Computer Engineering, Virginia Polytechnic Institute and State University, Arlington, VA, USA
c Department of Mathematics and Statistics, Georgetown University, DC, USA
Corresponding author. Tel: 202-687-2283 Fax: 202-687-0227. hwr@georgetown.edu
Issue date 2014 Oct 1.
© 2014 Elsevier Inc. All rights reserved.
PMC Copyright notice
PMCID: PMC4194134 NIHMSID: NIHMS611575 PMID: 25003577
The publisher's version of this article is available at Methods
Abstract
Biological network inference is a major challenge in systems biology. Traditional correlation-based network analysis results in too many spurious edges since correlation cannot distinguish between direct and indirect associations. To address this issue, Gaussian graphical models (GGM) were proposed and have been widely used. Though they can significantly reduce the number of spurious edges, GGM are insufficient to uncover a network structure faithfully due to the fact that they only consider the full order partial correlation. Moreover, when the number of samples is smaller than the number of variables, further technique based on sparse regularization needs to be incorporated into GGM to solve the singular covariance inversion problem. In this paper, we propose an efficient and mathematically solid algorithm that infers biological networks by computing low order partial correlation (LOPC) up to the second order. The bias introduced by the low order constraint is minimal compared to the more reliable approximation of the network structure achieved. In addition, the algorithm is suitable for a dataset with small sample size but large number of variables. Simulation results show that LOPC yields far less spurious edges and works well under various conditions commonly seen in practice. The application to a real metabolomics dataset further validates the performance of LOPC and suggests its potential power in detecting novel biomarkers for complex disease.
Keywords: Systems biology, undirected network inference, correlation, Gaussian graphical models, low order partial correlation, biomarker discovery
1. Introduction
Systems biology is a rapidly developing field that gives insights that genes and proteins do not work in isolation in complex diseases such as cancer, Parkinson’s disease and diabetes. To better understand the mechanisms of these diseases, different omic studies (e.g., transcriptomics, proteomics, metabolomics) need to be assembled to take advantage of the complementary information and to investigate how they complement each other. One major challenge in this field is the problem of inferring biological networks, such as gene co-expression network, protein-protein interaction network or metabolic network using high-throughput omic data.
Generally speaking, network inference methods can be divided into two groups depending on whether the resulting networks are directed graphs or undirected graphs. Bayesian network (BN) is the most popular method for directed network inference. BN is a probabilistic graphical model where nodes represent genes, proteins or metabolites and edges denote conditional dependence relationships. It models the biological networks as directed acyclic graphs. However, cyclic network structures, such as feedback loops, are ubiquitous in biological systems and are, in many cases, associated with specific biological properties . Considering this, the assumption of acyclic structure behind BN is limiting. In comparison, undirected network inference methods model the biological networks as undirected graphs thereby circumventing the problems of inferring cyclic network structures. One conventional method for undirected network inference is based on correlation [3–5], but correlation confounds direct and indirect associations. While direct association represents the pure association between two variables, indirect association indicates the induced association due to other variables. For example, Figure 1B illustrates that given three variables (x 1, x 2 and x 3), a strong correlation between x 1 and x 2 as well as x 2 and x 3 (direct association) may lead to a relatively weak but still significantly large correlation between x 1 and x 3 (indirect association). As a result, when the number of nodes is large, the resulting correlation-based network will yield too many spurious edges due to indirect associations.
Figure 1.
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Correlation confouds direct and indirect associations while partial correlation does not. (A) The true network from the model. (B) The network inferred based on correlation. The dot line represents the spurious edge due to the indirect association. (C) The network inferred based on GGM.
Partial correlation measures the correlation between two variables after their linear dependence on other variables is removed. It can distinguish between direct and indirect associations. The formal definition of the partial correlation between x 1 and x 2 given a set of other variables X+ = {x 3, x 4, …, x n} is the correlation between the residuals resulting from the linear regression of x 1 with X+ and that of x 2 with X+, respectively. One widely used method for undirected network inference with partial correlation is Gaussian graphical models (GGM) . For an undirected graph with p nodes, GGM calculate the partial correlation coefficients between each pair of nodes conditional on all other p-2 nodes. However, in order to obtain the exact undirected network for p variables, one needs to calculate from zero-th order (simply correlation) up to (p-2)-th order (full order) partial correlation . By only considering the full order partial correlation, it is insufficient for GGM to uncover the network structure faithfully as seen in Figure 2C. This is because two nodes may be conditionally independent when only conditional on a subset of other nodes while conditionally dependent when conditional on all remaining ones . Furthermore, when the number of samples is far smaller than the number of variables, a common case in omic studies, GGM face the difficulty of inverting a singular covariance matrix. Techniques based on sparse regularization such as graphical lasso offer a solution to address this problem within the framework of GGM . In comparison, methods based on low order partial correlation (LOPC) have been proposed [8, 10–12]. Low order partial correlation between two variables is obtained only conditional on a subset rather than all other variables. If only zero-th order and first order partial correlations are considered, the resulting undirected graph is called 0–1 graph . 0–1 graph has the advantage that it can be efficiently estimated from small sample-size data, but it fails to infer complex network structure (e.g., cyclic structures) as seen in Figure 2D. The reason is that in Figure 2A, x 1 can reach x 4 through either x 2 or x 3, so only conditioning on one of them (0–1 graph only calculates up to the first order partial correlation) is not enough to remove the indirect association between x 1 and x 4. In , de la Fuente et al. proposed to calculate up to the second order partial correlation to take into account more complex network structure while trying to keep the computational complexity still manageable. However, calculating the second order partial correlation for regular microarray experiments involving several thousands of variables is computationally intractable. This limits the application of method proposed by de la Fuente et al. to infer large biological networks. In addition, their method sets correlation threshold empirically without statistical support.
Figure 2.
Open in a new tab
Cyclic structure networks inferred based on correlation, GGM, 0–1 graph and LOPC. (A) The true network from the model. (B) Network inferred based on correlation: the dot lines represent the spurious edges. (C) Network inferred based on GGM: by only conditioning on the (p-2)-th order (i.e., second order in this model), it is insufficient to uncover the relationships between variables faithfully. (D) Network inferred based on 0–1 graph (up to first order): by only conditioning on up to first order, the indirect association between x 1 and x 4 cannot be removed since there are two paths from x 1 to x 4 either through x 2 or x 3. (E) Network inferred based on LOPC (up to second order): the connections in A are faithfully uncovered.
In this paper, we propose an efficient and mathematically sound algorithm to infer biological networks by calculating partial correlation from zero-th order up to the second order. For a given dataset with p variables, we first compute the zero-th order and first order partial correlation for each pair of variables. Then, we calculate the second order partial correlation only in cases in which both the zero-th order and first order partial correlations are significantly different from zero. With this step, the efficiency of LOPC is largely increased since it excludes most of the possible pairs before calculating the second order partial correlation. Furthermore, we use Fisher’s z transformation to create test statistics to set a reasonable threshold. To take into account multiple testing, we control the False Discovery Rate (FDR) using the Benjamini-Hochberg procedure. Simulation results show that LOPC works well under various conditions commonly seen in real applications and the spurious edges (i.e., false positives) for the inferred network are significantly reduced. We then apply LOPC on a real metabolomics dataset, the result validates the performance of LOPC and shows its potential in discovering novel biomarkers.
The rest of the paper is organized as follows. In Section 2, we discuss different undirected network inference methods based on correlation, GGM and LOPC. Also, we introduce test statistics for correlation and partial correlation methods (GGM and LOPC). Then, we propose an efficient algorithm, LPOC. The input, output, tools and databases involved are summarized. Section 3 presents two simulation datasets and a real metabolomics dataset to evaluate the performance of LOPC. The above undirected network inference methods are compared and the results are discussed. Finally, Section 4 summarizes our work and presents possible extensions.
2. Methods
2.1 Undirected network construction methods
Consider p random variables x 1, x 2, …, x p, denoted by X={x 1, x 2, …, x p}, that represent either metabolite levels or expressions of proteins or genes. Suppose the covariance matrix of X is Σ with the correlation coefficient between x i and x j defined as:
(1)
In a correlation-based network, x i and x j are considered to be connected if and only if (iff) ρ ij≠0 (i.e., its estimate r ij is significantly different from zero).
One common criticism for the above correlation-based network is that it yields too many spurious edges since correlation confounds direct and indirect associations. Let’s consider an example where X={x 1, x 2, x 2, x 4} and the relationships between x 1, x 2, x 3 and x 4 are modeled as x 1=s+ε 1, x 2=λ· x 1+ε 2, x 3=μ· x 2+ ε 3, x 4=ε 4 assuming s ~N(0,σ s 2), denoting the signal; ε 1, ε 2, ε 3, ε 4 ~N(0, σ n 2), denoting the independent and identically distributed (i.i.d.) noise; signal and noise are independent; λ, μ are non-zero constants. Figure 1A represents the above relationships and the arrow directions are assigned manually to represent causality. In this model, the relationships between x 1 and x 2, x 2 and x 3 are direct associations while x 1 and x 3 are indirectly related. Figure 1B shows that the undirected network inferred based on correlation confounds direct and indirect associations, thus leading to a spurious edge or false positive (i.e., the edge between x 1 and x 3).
In contrast, GGM remove the linear effect of all remaining p-2 variables when calculating the partial correlation coefficient between two variables conditional on all other variables. Suppose X follows a multivariate Gaussian distribution, R and Q are two subsets of X where R={x i, x j} and Q=X\R. The conditional covariance matrix of R given Q can be computed as follows as long as ΣQQ is nonsingular:
(2)
where the covariance matrix of X is .
Similarly, the precision matrix of X (the inverse of Σ) can be represented as:
(3)
In Eq. 3, ΩRR = (ΣRR − ΣRQΣQQ−1ΣQR)−1 .
Suppose , then from Eq. 2, the conditional covariance matrix ΣR|Q can be obtained as:
(4)
Under a Gaussian distribution assumption, partial correlation and conditional correlation are equivalent. A proof involving three variables is shown in the Appendix. For a more general proof, one can refer to [15, 16].
Once the precision matrix is known, the partial correlation coefficient between x i and x j conditional on all other variables can be computed as:
(5)
In a GGM-based network, x i and x j are considered to be connected iff ρ ij·Q≠0 (i.e., its estimate r ij․Q is significantly different from zero).
By removing the effect of all other variables, GGM can distinguish between direct and indirect associations as seen in Figure 1C. However, it requires that the covariance matrix be full rank for a well-defined matrix inversion, so the sample size should be at least as large as the number of variables. This poses a challenge for most omic datasets, which typically involve thousands of variables but much less number of samples. Furthermore, even when the sample size is large enough, GGM could lead to unreliable result as seen in Figure 2C since it only considers the (p-2)-th order partial correlation.
Rather than conditioning on all other variables, LOPC conditions on only a few of them. The order of the partial correlation coefficient is determined by the number of variables it conditions on. The advantage of using LOPC relies on a recursive equation (i.e., a higher order partial correlation coefficient can be computed from its preceding order) .
For X={x 1, x 2, x 3, x 4} modeled in Figure 1, without loss of generality, we assume σ s 2=σ n 2=1 and e 1=0, then the covariance matrix Σ is:
(6)
From Eq. 2, the conditional covariance matrix of {x 1, x 2} given x 3 is:
(7)
Since partial correlation is equivalent to conditional correlation under Gaussian distribution, the first order partial correlation coefficient between x 1 and x 2 conditional on x 3 can be computed from Eq. 7:
(8)
When the zero-th order partial correlation coefficients ρ 12, ρ 13, ρ 23 are computed from Eq. 6 and compared with Eq. 8, the following relationship exists between zero-th order and the first order partial correlation coefficients:
(9)
Eq. 9 can be generalized so that higher order partial correlation coefficient can be calculated from its preceding order. For example, similar equation exists between the first order and the second order partial correlation coefficients:
(10)
Theoretically, in order to obtain the exact undirected graph for p variables, one needs to potentially calculate the partial correlations from zero-th order up to the (p-2)-th order . Correlation considers only the zero-th order; GGM consider only the (p-2)-th order. It was previously reported that neither of them is sufficient to uncover the conditionally independent relationships between variables . Surprisingly, though the idea behind LOPC is simple, it can serve as a good approximation to the true network as seen in Figure 2. In addition, LOPC has the advantage of working well when the sample size is small and the number of variable is large.
If only zero-th order and first order partial correlations are considered, the resulting network is called a 0–1 graph. The network is constructed based on the following rule: the edge between nodes x i and x j is connected iff all r ij and r ij·k are significantly away from zero, where k considers each possible x k in X{x i, x j}.
Similarly, if we calculate up to the second order partial correlation, the resulting network is constructed based on the rule that the edge between nodes x i and x j is connected iff all r ij, r ij·k and r ij·kq are significantly different from zero, where k and q correspond to every possible x k and x q in X{x i, x j}.
2.2 Test statistics
The test statistics for non-zero correlation coefficient is and follows a t-distribution with n-2 degrees of freedom under the null hypothesis.
In contrast, the test statistics for non-zero partial correlation can be calculated using the Fisher’s z transformation :
(11)
where Q̃ corresponds to the elements of X{x i, x j} conditional upon and |Q̃| is the order of the partial correlation.
For a zero partial correlation coefficient with sample size equal to n, z is approximately normally distributed with zero mean and variance. Given a partial correlation coefficient, the two-sided p-value is:
(12)
2.3 Algorithm
The proposed algorithm contains four parts:
calculate the zero-th, first and second order partial correlation coefficients;
calculate test statistics and corresponding p-values to evaluate the null hypothesis that the corresponding partial correlation coefficient is zero;
calculate adjusted p-values for multiple testing correction;
construct the network.
Among the four steps, most of the computation time is spent on calculating the second order partial correlation coefficient r ij·kq since one needs to consider all possible x k, x q in X{x i, x j}. It was previously suggested that the distribution of connections in metabolic, regulatory and protein-protein interaction networks tends to follow a power law [19, 20]. Thus, the resulting networks are very sparse.
Here, we present an efficient algorithm taking advantage of this sparsity property of biological networks. Instead of calculating the second order partial correlation coefficients r ij·kq for all possible x i, x j, we only calculate those whose corresponding zero-th and first order partial correlation coefficients are significantly different from zero. Since the true biological networks are sparse, this step can exclude most of the possible spurious edges before calculating the second order partial correlation. As a result, LOPC can dramatically reduce the computational burden.
The detailed algorithm is outlined below:
| Algorithm LOPC |
| 1: | Zero-th order partial correlation: |
| 2: | for each pair (x i, x j) do |
| 3: | Calculate an estimate of the zero-th order partial correlation coefficient r ij; |
| 4: | Construct the test statistic for r ij and compute the corresponding p-value p(r ij); |
| 5: | Compute the multiple testing adjusted p-value for the zeroth order partial correlation coefficient p̄(x ij) across all pairs. |
| 6: | end for |
| 7: | First order partial correlation: |
| 8: | for each pair (x i, x j) do |
| 9: | Calculate estimates of the first order partial correlation coefficients r ij·k for all possible x k ∈ X/{x i, x j}; |
| 10: | Select the maximum in terms of absolute value as r̂ ij·k; |
| 11: | Construct test statistics for r̂ ij·k using Fisher’s z transformation and compute corresponding p-value p(r̂ ij·k); |
| 12: | Compute the multiple test adjusted p-values for the first order partial correlation coefficient p̄(r̂ ij·k) across all pairs. |
| 13: | end for |
| 14: | Second order partial correlation: |
| 15: | for each pair (x i, x j) do |
| 16: | if max {p̄(r ij), p̄(r̂ ij·k)} < 0.05 then |
| 17: | Proceed to compute the second order partial correlation: |
| 18: | Calculate estimates of the second order partial correlation coefficients r ij·kq for all possible x k, x q ∈ X/ {x i, x j}; |
| 19: | Select the maximum in terms of absolute value as r̂ ij·kq; |
| 20: | Compute the multiple test adjusted p-values for the second order partial correlation coefficient p̄ (r̂ ij·kq) across all pairs. |
| 21: | else |
| 22: | Do not need to compute the second order partial correlation: |
| 23: | Set p̄(r̂ ij·kq) to be 1. |
| 24: | end if |
| 25: | end for |
| 26: | Connect x i and x j iff p̄(r̂ ij·kq) < 0.05. |
Open in a new tab
2.4 Summary
LOPC is an efficient algorithm for constructing a simplified undirected network that captures the direct associations between variables (genes, proteins, and metabolites) based on high-throughput omics data. The input, output, tools and databases involved are listed below.
Input: The input of the proposed algorithm is a p × n matrix with p variables (genes, proteins, or metabolites) and n samples. The elements in the matrix either represent the expression level of the corresponding genes and proteins or the intensity of the associated metabolites.
Output: The output of the algorithm is a p × p matrix, also known as adjacent matrix, along with a p × p weight matrix. The elements of the adjacent matrix are either 1 or 0, indicating whether there exists a connection between two variables or not. The weight matrix includes the value of the second order partial correlation between different pairs.
Tools: Tools such as Matlab and Cytoscape are used for implementing the algorithm and visualizing the network, respectively.
Databases: While no databases are involved in the algorithm, databases such as DAVID ( Reactome ( and KEGG ( are often used to look into the resulting networks for functional interpretation and pathway analysis.
3. Results and discussion
This section presents two numerical simulations (A and B) to infer undirected network based on correlation, GGM, 0–1 graph, and LOPC, as well as one real application of LOPC on a metabolomics dataset.
3.1 Analysis on simulated data
In simulation A, we consider an example where variables X={x 1, x 2, x 3, x 4} form a cyclic structure as shown in Figure 2A. The relationships between x 1, x 2, x 3, x 4 were modeled as: x 1=s+ε 1, x 2=λ·x 1+ε 2, x 3=μ·x 1+ε 3, x 4=α·x 2+β·x 3+ε 4 assuming s ~N(0,σ s 2), denoting the signal; ε 1, ε 2, ε 3, ε 4 ~N(0, σ n 2), denoting the i.i.d. noise; signal and noise are independent; λ, μ, α, β are non-zero constants. Without loss of generality, we set σ s 2=1, σ n 2=0.01 and λ=μ=α=β=1. The resulting covariance matrix for X is:
(13)
We generated dataset from N(0,Σ)with a sample size of n=50 and inferred networks based on correlation, GGM, 0–1 graph and LOPC as seen in Figure 2.
The correlation, partial correlation for each pair of variables and the corresponding adjusted p-values are shown in Table 1.
TABLE 1.
Correlation, partial correlation and p-values for each edge
| Edge | Correlation | GGM | 0–1 graph | LOPC |
:---: :---:
| R | P | R | P | R | P | R | P |
| x1x2 | 0.994 | 9.00E-47 | 0.533 | 9.00E-04 | 0.278 | 5.00E-03 | 0.533 | 9.00E-04 |
| |
| x1x3 | 0.996 | 9.00E-51 | 0.658 | 4.00E-06 | 0.518 | 2.00E-07 | 0.658 | 4.00E-06 |
| |
| x1x4 | 0.995 | 1.00E-50 | −0.257 | 0.774 | 0.33 | 8.00E-03 | −0.257 | 0.774 |
| |
| x2x3 | 0.991 | 5.00E-45 | −0.543 | 7.00E-04 | 0.148 | 8.50E-01 | 0 | 1 |
| |
| x2x4 | 0.997 | 4.00E-53 | 0.798 | 1.00E-10 | 0.704 | 0 | 0.798 | 1.00E-10 |
| |
| x3x4 | 0.997 | 7.00E-53 | 0.754 | 6.00E-09 | 0.634 | 2.00E-12 | 0.754 | 6.00E-09 |
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In Figures 2B and 2C, we see that both correlation and GGM-based networks yield spurious edges. This is because correlation confounds direct and indirect associations, while GGM are insufficient to uncover the network structure faithfully in this model by only considering the (p-2)-th order partial correlation. In fact, from the perspective of probabilistic graphical models , x 1 is a common ancestor of x 2 and x 3, while x 4 is a causal descendent of x 2 and x 3. Since conditioning on any common causal descendent would introduce a correlation between two variables, there is a dependence estimated between x 2 and x 3 by conditioning on both x 1 and x 4 using GGM.
The resulting networks based on 0–1 graph and LOPC are shown in Figures. 2D and 2E, respectively. For 0–1 graph, since there are multiple paths from x 1 to x 4 either through x 2 or x 3. By only calculating up to the first order partial correlation, it is insufficient to remove the indirect association between x 1 and x 4. However, when we calculate up to the second order partial correlation, the cyclic structure can be faithfully recovered. In fact, Figure 2E can be viewed as the result of merging Figures. 2B, 2C and 2D together and only keeping common edges.
In simulation B, we consider a more complex structure where ten variables X={x 1, x 2, x 3, x 4, x 5, x 6, x 7, x 8, x 9, x 10} were involved and their relationships were modeled as: x 1=s 1+ε 1, x 10=s 2+ε 10, x 2=λ 1·x 1+ε 2, x 3=α 1·x 2+ε 3, x 4=α 2·x 3+ε 4, x 5=α 3·x 4+λ 2·x 1+ε 5, x 6=α 4·x 5+μ 1·x 10+ε 6, x 7=α 5·x 6+ε 7, x 8=α 6·x 7+ε 8, x 9=α 7·x 8+μ 2·x 10+ε 9 with s 1, s 2 ~N(0,σ s 2), denoting the signal; ε 1, ε 2, …, ε 10 ~N(0, σ n 2), denoting the i.i.d. noise; signal and noise are independent; λ 1, λ 2, μ 1, μ 2, α 1, α 2, …, α 7 are non-zero constants. The network structure is shown in Figure 3A. In this network, x 1 and x 10 can be interpreted as regulators while x 2 to x 9 represent genes, proteins or metabolites being regulated. Correspondingly, λ 1, λ 2, μ 1, μ 2 denote the strength of the regulation. Without loss of generality, we set σ s 2=1, σ n 2=0.01, all the coefficients (i.e., λ 1, λ 2, μ 1, μ 2, α 1, α 2, …, α 7) to be 1. The resulting covariance matrix for X is seen in Eq. 14.
Figure 3.
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Complex structure networks inferred based on GGM, 0–1 graph and LOPC. (A) The true network from the model. (B) Network inferred based on GGM: the dot lines represent the spurious edges. (C) Network inferred based on 0–1 graph (up to first order): by only conditioning on up to first order, the resulting inferred network has similar number of spurious edges (false positives) as that from GGM but has several missed edges (false negatives). (D) Network inferrred based on LOPC (up to second order): while the missed edges are inherited from 0–1 graph, calculating up to the second order successfully removes spurious edges in the inferred network.
We generated dataset from N(0,Σ)with a sample size of n=50 and inferred networks based on correlation, GGM, 0–1 graph and LOPC. For the correlation-based network, the number of spurious edges (false positives) increases dramatically with nearly every possible variable pair being connected. In Figures 3B to 3D, we show the inferred networks based on GGM, 0–1 graph and LOPC.
(14)
As shown in Figure 3B, GGM-based network yields a few false positives. This is because GGM only considers the (p-2)-th order partial correlation. As seen in Figure 3C, the 0–1 graph yields similar number of false positives compared with GGM-based network but starts to have missing edges (false negatives). The network inferred with LOPC (Figure 3D) also has false negatives, but calculating up to the second order removes all the false positives.
Using a similar model, we generated 100 simulation datasets for varying number of variables and sample sizes and calculated the mean of false positives and false negatives for each method as shown in Table 2.
TABLE 2.
Mean of false positives and false negatives for varying number of variables and sample sizes
| Variable Number | Sample Size | Correlation | GGM | 0–1 graph | LOPC |
:---: :---: :---: |
| FP | FN | FP | FN | FP | FN | FP | FN |
| 10 | 50 | 28.78 | 0.11 | 2.93 | 0.23 | 1.95 | 3.08 | 0 | 3.31 |
| |
| 20 | 50 | 54.91 | 1.42 | 1.29 | 15.06 | 1.4 | 10.84 | 0.09 | 15.98 |
| 100 | 57.19 | 0.41 | 5.28 | 2.17 | 3.64 | 6.37 | 0 | 7.47 |
| |
| 50 | 51 | 142.55 | 2.09 | 7.33 | 54.61 | 6.29 | 20.9 | 0.42 | 32.57 |
| 100 | 145.85 | 0.33 | 13.73 | 9.98 | 9.73 | 15.21 | 0.21 | 16.86 |
| |
| 100 | 101 | 299.51 | 0.47 | 0 | 109.98 | 19.62 | 30.43 | 1.56 | 33.34 |
| 200 | 299.44 | 0 | 39.16 | 0.78 | 20.04 | 29.48 | 0.83 | 29.87 |
| |
| 500 | 501 | 1491.71 | 0 | 0 | 550 | 89.78 | 151.32 | 7.31 | 171.38 |
| 1000 | 1495.25 | 0 | 150.78 | 10.34 | 114.25 | 99.75 | 1.35 | 105.19 |
| |
| 1000 | 1001 | 2990.13 | 0 | 0 | 1100 | 181.27 | 271.78 | 13.77 | 301.13 |
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Generally speaking, we expect LOPC to lead to far less number of false positives compared to GGM and 0–1 graph with a possible drawback of selecting a few more false negatives. In real application, this is desirable since one would usually prefer to be confident about the existence of edges already detected, though some edges might be missed. As shown in Table 2, when the sample size is slightly larger than the number of variables, LOPC works well, whereas GGM’s performance begins to decline due to the difficulty of inverting singular matrix. To address this, further technique such as graphical lasso has been incorporated into GGM .
3.2 Application to real data
In this section, we applied LOPC on a real untargeted metabolomics dataset previously collected and analyzed by our group for hepatocellular carcinoma (HCC) biomarker discovery study . The data were acquired by analysis of sera from 40 HCC cases and 50 patients with liver cirrhosis using liquid chromatography coupled with mass spectrometry (LC-MS). Following preprocessing, a data matrix was obtained with 984 input variables - larger than the sample size 90. In , we identified 32 metabolites with intensities significantly different between the HCC cases and cirrhotic controls.
Rather than looking into each statistically significant metabolite, in this paper, we generated undirected network using LOPC after normalization of the preprocessed data matrix. The aim of the normalization is to bring the intensities of the metabolites in both cases and controls to a comparable level. The resulting networks are depicted in Figure 4A. We then mapped the 32 statistically significant metabolites onto Figure 4A and extracted functional modules which contained multiple metabolites. Two interesting functional modules are shown in Figures 4B and 4C, respectively, with blue nodes representing the metabolites and white nodes representing non-significant ones. Due to the limitation in metabolite identification, some of the nodes have been assigned multiple putative IDs (e.g., Glycine; Haloperidol decanoate) or have no IDs (unknowns).We see that metabolites connecting with each other tend to be involved in the same chemical reaction and have similar functionalities. The extracted functional modules may help identify other non-significant metabolites that might be missing from the statistical analysis due to subtle differences in ion intensities.
Figure 4.
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Undirected network and functional modules inferred from real data by LOPC. (A) Undirected network encoding the direct associations between different nodes (B–C) Functional modules extracted from the undirected network. Blue nodes represent the candidate biomarkers previously reported. White nodes represent the non-significant ones.
Finally, we evaluated the efficiency of LOPC by randomly sampling various numbers of metabolites from the above dataset to generate 100 undirected networks. We compared the averaged run-time between LOPC and the traditional method to calculate up to the second order partial correlation. While the traditional method calculates the 0 th, 1 st, and 2 nd order partial correlations, LOPC evaluates the outcome of the 1 st order partial correlation to determine whether or not the calculation of the 2 nd order partial correlation is needed. As shown in Figure 5, when the input variable number increases beyond 50, LOPC starts to become more efficient than traditional method. With an input variable number of 200, LOPC can be as 4 times fast as the traditional method. The run-time comparison was performed using a PC with an Intel(R) Core(TM) i7-2600 CPU @ 3.4GHz and 16.0 GB RAM.
Figure 5.
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Run-time comparison between LOPC and the traditional method in calculating up to the second order partial correlation.
4. Conclusion
In this paper, we propose an efficient algorithm, LOPC, to infer biological networks by calculating up to the second order partial correlation. Compared with other undirected network inference methods (correlation, GGM, and 0–1 graph), LOPC offers better solution for inferring networks with less spurious edges (false positives). It also has the advantage of handling well cases that involve a large number of variables but a small sample size. These properties make LOPC a promising alternative to infer from omic datasets relevant gene co-expression, protein-protein interaction and metabolic networks, which may give insights into the mechanisms of complex diseases. A real application on metabolomics dataset validates the performance of LOPC and shows its potential in discovering novel biomarkers. Future research will focus on incorporating prior knowledge from the existing database and causal information from time course data to build directed network.
TABLE 3.
Correlation, partial correlation and p-values for each edge.
| Edge | Correlation | GGM | 0–1 graph | LOPC |
:---: :---:
| R | P | R | P | R | P | R | P |
| x1x2 | 0.994 | 9.00E-47 | 0.533 | 9.00E-04 | 0.278 | 5.00E-03 | 0.533 | 9.00E-04 |
| |
| x1x3 | 0.996 | 9.00E-51 | 0.658 | 4.00E-06 | 0.518 | 2.00E-07 | 0.658 | 4.00E-06 |
| |
| x1x4 | 0.995 | 1.00E-50 | −0.257 | 0.774 | 0.33 | 8.00E-03 | −0.257 | 0.774 |
| |
| x2x3 | 0.991 | 5.00E-45 | −0.543 | 7.00E-04 | 0.148 | 8.50E-01 | 0 | 1 |
| |
| x2x4 | 0.997 | 4.00E-53 | 0.798 | 1.00E-10 | 0.704 | 0 | 0.798 | 1.00E-10 |
| |
| x3x4 | 0.997 | 7.00E-53 | 0.754 | 6.00E-09 | 0.634 | 2.00E-12 | 0.754 | 6.00E-09 |
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TABLE 4.
Mean of false positives and false negatives for varying number of variables and sample sizes.
| Variable Number | Sample Size | Correlation | GGM | 0–1 graph | LOPC |
:---: :---: :---: |
| FP | FN | FP | FN | FP | FN | FP | FN |
| 10 | 50 | 28.78 | 0.11 | 2.93 | 0.23 | 1.95 | 3.08 | 0 | 3.31 |
| |
| 20 | 50 | 54.91 | 1.42 | 1.29 | 15.06 | 1.4 | 10.84 | 0.09 | 15.98 |
| 100 | 57.19 | 0.41 | 5.28 | 2.17 | 3.64 | 6.37 | 0 | 7.47 |
| |
| 50 | 51 | 142.55 | 2.09 | 7.33 | 54.61 | 6.29 | 20.9 | 0.42 | 32.57 |
| 100 | 145.85 | 0.33 | 13.73 | 9.98 | 9.73 | 15.21 | 0.21 | 16.86 |
| |
| 100 | 101 | 299.51 | 0.47 | 0 | 109.98 | 19.62 | 30.43 | 1.56 | 33.34 |
| 200 | 299.44 | 0 | 39.16 | 0.78 | 20.04 | 29.48 | 0.83 | 29.87 |
| |
| 500 | 501 | 1491.71 | 0 | 0 | 550 | 89.78 | 151.32 | 7.31 | 171.38 |
| 1000 | 1495.25 | 0 | 150.78 | 10.34 | 114.25 | 99.75 | 1.35 | 105.19 |
| |
| 1000 | 1001 | 2990.13 | 0 | 0 | 1100 | 181.27 | 271.78 | 13.77 | 301.13 |
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Acknowledgments
This work is in part supported by National Institutes of Health Grants R01CA143420 and R01GM086746 awarded to HWR.
Appendix
Conditional independence relationship is crucial for network inference. Here, we prove the equality of partial correlation and conditional correlation involving three variables under Gaussian assumption so that we can use partial correlation to infer conditional independence relationships between nodes and build a network.
By definition, the partial correlation coefficient between x and y conditional on z (r xy·z) is obtained by first regressing x on z and y on z separately and then calculating the correlation between the residuals of the models for x and y:
(15)
(16-1)
(16-2)
where ε̂x,ε̂y are the residuals of x and y after regressing on z; â, b̂, ĉ, d̂ are regression coefficients.
Conditional correlation coefficient between x and y given z (r xy|z) is defined as:
(17)
where E x|z, E y|z and E x,y|z denote expectations of the marginal and joint distribution of x and y conditional on z.
To show the relationship between partial correlation and conditional correlation, we consider the following case of x=b 0+b 1·z+a and y=d 0+d 1·z+c, where b 0, b 1, d 0 and d 1 are constants, x, y, z, a and c are random variables. Under this assumption, the conditional correlation between x and y given z is reduced to:
(18)
From Eq. (4), the partial correlation equals to conditional correlation. To be more general, these two correlations are the same when the conditional variance and covariance of x and y given z are free of z [15, 16]. The above condition is satisfied in normal distribution.
Footnotes
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187669 | https://testbook.com/question-answer/if-an-equivalent-lens-is-made-of-two-convex-lenses--5f2b0c153e4c000d15380652 | [Solved] If an equivalent lens is made of two convex lenses of focal
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Refraction and Reflection
Thin Lenses
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If an equivalent lens is made of two convex lenses of focal length 10 cm and 20 cm. Find the equivalent focal length ______.
20 cm
-20 cm
-6.67 cm
+6.67 cm
Answer (Detailed Solution Below)
Option 4 : +6.67 cm
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Detailed Solution
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CONCEPT:
Lens: The transparent curved surface which is used to refract the light and make an image of any object placed in front of it is called a lens.
The lens whose refracting surface is upside is called a convex lens and the lens whose refracting surface is inside is called a concave lens or diverging lens.
The convex lens is also called a converging lens.
When two or more lenses are in contact with each other then the equivalent focal length is given by:
1 F e q=1 f 1+1 f 2
Where f 1 and f 2 are focal lengths of the two lenses
CALCULATION:
We know,1 F e q=1 f 1+1 f 2
Here, f 2= +20 cm and f 1= +10 cm
1 F e q=1 20+1 10=3 20⇒F e q=+6.67 c m
The focal length of theconvex lens is positiveand the focal length of the concave lens is negativeaccording to the sign convention of the lens.
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187670 | https://math.stackexchange.com/questions/2285804/induction-proof-the-sum-of-the-first-n-odd-squares | discrete mathematics - Induction Proof The sum of the first n odd squares - Mathematics Stack Exchange
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Induction Proof The sum of the first n odd squares
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I am just learning about induction proofs. So far I am only familiar with induction equality proofs, and inequality proofs. Such as, for example, prove the sum of the powers of 2 = 2 n+1−1 2 n+1−1...
I am confused on the following proof: The sum of the first n odd squares is 4 3 n 3−1 3 n 4 3 n 3−1 3 n
How do I start this proof? it looks like only one statement with nothing to compare it to. I was thinking maybe I would represent the sum of the first n odd squares as the formula (2 n−1)2(2 n−1)2 and then set the proof up as (2 n−1)2=4 3 n 3−1 3 n(2 n−1)2=4 3 n 3−1 3 n
so it looks more like the form I am used to. Is this correct? Am I missing a small nuance of importance? Thanks for any and all help.
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edited May 18, 2017 at 1:07
B. Mehta
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asked May 18, 2017 at 1:04
ConnerConner
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Yes, I would be representing it as a summation notation containing (2i - 1)^2 Conner –Conner 2017-05-18 01:17:17 +00:00 Commented May 18, 2017 at 1:17
I think (2 n+1)2(2 n+1)2 is better but not sure if it matters.marshal craft –marshal craft 2017-05-18 01:18:55 +00:00 Commented May 18, 2017 at 1:18
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You need to take the sum of the first n n odd squares, e.g.if n=3 n=3, then you need to add 1+9+25=35 1+9+25=35. And, that does indeed equal 4 3 n 3−1 3 n 4 3 n 3−1 3 n for n=3 n=3: 4 3 3 3−1 3 3=4 3 27−1=36−1=35 4 3 3 3−1 3 3=4 3 27−1=36−1=35
Now, to prove that this is true in general using induction:
Base: n=0 n=0: 4 3 0 3−1 3 0=0−0=0 4 3 0 3−1 3 0=0−0=0 which is indeed the sum of the first 0 0 odd squares. Check!
Step: Assume that for some k k the sum of the first k k odd squares is 4 3 k 3−1 3 k 4 3 k 3−1 3 k. Now let's consider the sum of the first k+1 k+1 odd squares, which is of course the sum of the first k k odd squares plus the k+1 k+1-th odd square, which is (2 k+1)2(2 k+1)2. So, by the inductive hypothesis the sum is 4 3 k 3−1 3 k+(2 k+1)2 4 3 k 3−1 3 k+(2 k+1)2, and now you need to verify that this does indeed equal 4 3(k+1)3−1 3(k+1)4 3(k+1)3−1 3(k+1). Let's see:
4 3 k 3−1 3 k+(2 k+1)2=4 3 k 3−1 3 k+(2 k+1)2=
4 k 3−k+3(4 k 2+4 k+1)3=4 k 3−k+3(4 k 2+4 k+1)3=
4 k 3+12 k 2+12 k−k+3 3=4 k 3+12 k 2+12 k−k+3 3=
4 k 3+12 k 2+12 k+4−k+3−4 3=4 k 3+12 k 2+12 k+4−k+3−4 3=
4(k 3+3 k 2+3 k+1)−(k+1)3=4(k 3+3 k 2+3 k+1)−(k+1)3=
4(k+1)3 3−1 3(k+1)4(k+1)3 3−1 3(k+1)
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edited May 18, 2017 at 1:32
answered May 18, 2017 at 1:18
Bram28Bram28
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1 @Conner You're welcome! Please note that I had a small mistake in my earlier answer ... The (k+1)(k+1)-th odd square is (2 k+1)2(2 k+1)2, and not (2 k−1)2(2 k−1)2, which is what I had first.Bram28 –Bram28 2017-05-18 01:34:19 +00:00 Commented May 18, 2017 at 1:34
Thank you. Just one more clarity question. Specifically when you are constructing your Induction Hypothesis, I was under the impression from earlier proofs I've done that we would have to insert K + 1 into (2k + 1)^2, making it (2(k + 1))^2 = (2k + 3)^2, and then we would add that. Is that not true in this case?Conner –Conner 2017-05-18 02:01:34 +00:00 Commented May 18, 2017 at 2:01
1 @Conner The k k -th odd square is (2 k−1)2(2 k−1)2, and so the (k+1)(k+1)-th odd square is (2(k+1)−1)2=(2 k+1)2(2(k+1)−1)2=(2 k+1)2. So yes, you do fill in k+1 k+1 for k k, but you have to choose the right formula! :)Bram28 –Bram28 2017-05-18 02:09:33 +00:00 Commented May 18, 2017 at 2:09
Ahh. Yes I see. I was very confused for a moment because I changed my entire representation of the summation of the odd squares to (2n+1)^2 instead, so I could account for the base n = 0. It didn't even cross my mind that you were just showing the example of if I had kept it (2n-1)^2. This has really helped my comprehension the induction proof process to me though. I appreciate you answering back.Conner –Conner 2017-05-18 02:13:25 +00:00 Commented May 18, 2017 at 2:13
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I will give an alternative method to prove the claim.
It's a standard exercise in induction to show that
∑n=1 2 N n 2=2 N(2 N+1)(4 N+1)6.∑n=1 2 N n 2=2 N(2 N+1)(4 N+1)6.
Next, observe
∑n even+∑n odd n 2===∑k=1 N(2 k)2+∑n odd 2 N n 2 4∑k=1 N k 2+∑n odd n 2 4 N(N+1)(2 N+1)6+∑n odd n 2.∑n even+∑n odd n 2=∑k=1 N(2 k)2+∑n odd 2 N n 2=4∑k=1 N k 2+∑n odd n 2=4 N(N+1)(2 N+1)6+∑n odd n 2.
Hence it follows
∑n odd 2 N n 2==2 N(2 N+1)(4 N+1)6−4 N(N+1)(2 N+1)6 4 N 3−N 3∑n odd 2 N n 2=2 N(2 N+1)(4 N+1)6−4 N(N+1)(2 N+1)6=4 N 3−N 3
Now, use induction to prove the above formula holds if you like.
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edited May 18, 2017 at 1:28
answered May 18, 2017 at 1:19
Jacky ChongJacky Chong
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187671 | https://link.springer.com/article/10.1007/s00200-024-00652-8 | Solving systems of algebraic equations over finite commutative rings and applications | Applicable Algebra in Engineering, Communication and Computing
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Applicable Algebra in Engineering, Communication and Computing
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Solving systems of algebraic equations over finite commutative rings and applications
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Published: 24 April 2024
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Applicable Algebra in Engineering, Communication and ComputingAims and scope
Solving systems of algebraic equations over finite commutative rings and applications
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Hermann Tchatchiem Kamche1&
Hervé Talé Kalachi2
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Abstract
Several problems in algebraic geometry and coding theory over finite rings are modeled by systems of algebraic equations. Among these problems, we have the rank decoding problem, which is used in the construction of public-key cryptosystems. A finite chain ring is a finite ring admitting exactly one maximal ideal and every ideal being generated by one element. In 2004, Nechaev and Mikhailov proposed two methods for solving systems of polynomial equations over finite chain rings. These methods used solutions over the residue field to construct all solutions step by step. However, for some types of algebraic equations, one simply needs partial solutions. In this paper, we combine two existing approaches to show how Gröbner bases over finite chain rings can be used to solve systems of algebraic equations over finite commutative rings. Then, we use skew polynomials and Plücker coordinates to show that some algebraic approaches used to solve the rank decoding problem and the MinRank problem over finite fields can be extended to finite principal ideal rings.
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1 Introduction
Solving systems of algebraic equations has always been of high interest in algorithmic algebra. Indeed, many algebraic problems have their solution sets contained in those of systems of algebraic equations. A tangible example is the rank decoding problem , which has attracted a lot of attention this last decade in view of its application in cryptography. This problem is generally defined over finite fields and therefore, leads to the problem of solving systems of algebraic equations over finite fields when modeled appropriately. But it should be remembered that this latest problem has been studied for a long time and has a wide variety of algorithms that can be used to solve it and also estimate the solving complexities [14, 17, 20, 21, 26].
Most recently, the rank decoding problem has been extended to finite principal ideal rings in where the authors, after having justified the interest of studying this problem over finite rings, show that it is at least as hard as the rank decoding problem over finite fields, and also provide a combinatorial type algorithm for solving this new problem. The translation of the rank decoding problem over finite rings as a system of algebraic equations naturally induces the problem of solving systems of algebraic equations over finite rings.
Contrary to the problem of solving systems of algebraic equations over finite fields, the previous problem over finite rings has not experienced much development. The most advanced and recent work is the paper of Mikhailov and Nechaev , who proposed two approaches for solving systems of polynomial equations over finite chain rings. One of these approaches uses canonical generating systems, which are not Gröbner bases in general. An algebraic modeling of the rank decoding problem over finite chain rings that we will use is a system of algebraic equations with some parameters, and we just need a partial solution. Note that Gröbner bases over fields are generally used to solve these kinds of systems. A natural question is therefore to know whether Gröbner bases can be used to solve systems of algebraic equations over finite chain rings in general, as in the case of finite fields.
Independently, Gröbner bases over finite chain rings have been much studied and implemented in some mathematical software systems like Magma , SageMath [51g
")], etc. Indeed, similar to Buchberger’s algorithm over fields [11")], Norton and Salagean [45")] gave an algorithm for computing Gröbner bases over finite chain rings. This algorithm has been improved in [28, 1350034 (2013)")] by adding the product criterion and the chain criterion. In the Magma handbook [9, 235–265 (1997)")], it was specified that the \(F_{4}\) algorithm [20. J. Pure Appl. Algebra 139(1–3), 61–88 (1999)")] was extended over Euclidean rings,Footnote 1 taking into account the elimination criteria given in [41, 345–359 (1988)")]. Moreover, the elimination theorem, which is the main property used to solve systems of algebraic equations, can be extended over finite chain rings. However, the elimination theorem does not hold in general on other types of finite rings. But we must not forget that low-rank parity-check codes which are potential linear codes for rank-based cryptography have been extended to finite commutative rings [30"), 32 codes over the ring of integers modulo a positive integer. Arab. J. Math. 10(2), 357–366 (2021)"), 49")]. Thus, it also becomes necessary to tackle the resolution of systems of algebraic equations over finite commutative rings.
According to the structure theorem for finite commutative rings , every finite commutative ring is isomorphic to a product of finite commutative local rings. Thus, solving systems of algebraic equations over finite commutative rings is reduced to finite local rings. In , Bulyovszky and Horváth gave a good method for solving systems of linear equations over finite local rings. Indeed, they transformed systems of linear equations from local rings to Galois rings and used the Hermite normal form to solve it. In this work we show that this transformation can be applied to systems of algebraic equations, and we then use Gröbner bases to solve the resulting equation since Galois rings are specific cases of finite chain rings.
Before one can use Gröbner bases over finite chain rings to solve the rank decoding problem, it is first necessary to give an algebraic modeling. As specified in , some properties of the rank for matrices over fields do not extend to matrices over rings in general due to zero divisors. Therefore, the algebraic modeling of the rank decoding problem given in using the MaxMinors cannot be directly applied to rings. However, in other algebraic modeling using linearized polynomials has been given and some main properties of linearized polynomials have been extended in over finite principal ideal rings. We will use these results to prove that the algebraic modeling done in using linearized polynomials can be generalized over finite principal ideal rings. Furthermore, as the rank decoding problem reduces to the MinRank problem , we also study possible algebraic modelings of the MinRank problem over finite rings.
The MinRank problem have several algebraic modelings over fields. For example, the MaxMinors modeling , the Kipnis–Shamir modeling , or the Support-Minors modeling . Over finite chain rings, the rank of a matrix is not generally equal to the order of the highest order non-vanishing minor. Thus, the MaxMinors modeling cannot directly extend over rings. However, we will use the rank decomposition and the Plücker coordinates to show that the Kipnis–Shamir modeling and the Support-Minors modeling can be extended to finite principal ideal rings.
The rest of the paper is organized as follows. In Sect.2, we give some preliminary notions on Gröbner bases over finite chain rings, followed by the use of Gröbner bases for solving systems of algebraic equations over finite chain rings in Sect.3. In Sect.4 we show how to solve systems of algebraic equations over finite commutative local rings by decomposing them as a direct sum of cyclic modules over Galois rings. Section5 uses the fact that the row span of a matrix is contained in a free module of the same rank to prove that the Kipnis–Shamir Modeling and the Support Minors Modeling of the MinRank problem can be extended to finite principal ideal rings. In Sect.6, skew polynomials are used to give an algebraic modeling of the rank decoding problem over finite principal ideal rings, and to finish, we conclude the paper and give some perspectives in Sect.7.
2 Preliminaries
2.1 Finite chain rings
A chain ring is a ring whose ideals are linearly ordered by inclusion, and a local ring is a ring with exactly one maximal ideal. By , a finite ring is a chain ring if and only if it is a local principal ideal ring, that is to say a finite ring admitting exactly one maximal ideal and every ideal being generated by one element. A basic example of finite chain rings is the ring (\mathbb {Z}{p^{k}}=\mathbb {Z}/p^{k}\mathbb {Z}) of integers modulo a power of a prime number _p. Its maximal ideal is (p\mathbb {Z}{p^{k}}). Other examples of finite chain rings that we will use to give a representation of finite commutative local rings in Sect.4 are Galois rings. A Galois ring of characteristic (p^{k}) and rank _r, denoted by (GR\left( p^{k},r\right) ), is the ring (\mathbb {Z}{p^{k}}[ X] /\left( f\right) ), where _f(\in )(\mathbb {Z}{p^{k}}[ X] ) is a monic polynomial of degree _r, irreducible modulo p, and (\left( f \right) ) being the ideal of (\mathbb {Z}{p^{k}}[X] ) generated by _f. Thus, (GR\left( p^{k},r\right) ) is a degree r Galois extension of (\mathbb {Z}{p^{k}}) and is a finite chain ring with maximal ideal generated by _p and residue field ( \mathbb {F}_{p^{r}}=GR\left( p^{k},r\right) /pGR\left( p^{k},r\right) ) .
In this section, we assume that R is a finite commutative chain ring with maximal ideal (\mathfrak {m}) and residue field (\mathbb {F}{q}=R/\mathfrak {m} ). We denote by (\pi ) a generator of (\mathfrak {m}), and (\nu ) the nilpotency index of (\pi ), i.e., the smallest positive integer such that ( \pi ^{\nu }=0). An important property of finite chain rings is the structure of their ideals. Every ideal of _R is of the form (\pi ^{i}R), for (i=0,\ldots ,\nu ). A direct consequence is the following decomposition of any element from R. Let (a,b\in R). We say that a is congruent to b modulo (\pi ) and denote it by (a\equiv b\left( mod\ \pi \right) ), if there exists c in R such that (a=b+c\pi ). This relation is equivalent to (\varphi \left( a\right) =\varphi \left( b\right) ) where (\varphi :)(R\longrightarrow R/\mathfrak {m}) is the canonical projection. Let (\Gamma ) be a complete set of representatives of the equivalence classes of R under the congruence modulo (\pi ). As in , we have for example (\Gamma =\left{ a\in R:a^{q}=a\right} ).
Proposition 1
Let c in R, then c has a unique representation in the form
$$\begin{aligned} c=\sum {j=0}^{\nu -1}c{j}\pi ^{j} \end{aligned}$$
(1)
where (c_{j}\in )(\Gamma ), for (j=0,\ldots ,\nu -1).
The representation of c given by Eq. (1) is called the ( \pi -)adic decomposition of c. Let (j\in {0,\ldots ,\nu -1}), and the map ( \gamma {j}:R\longrightarrow \Gamma ) given by (c\longmapsto c{j}), that is to say (\gamma {j}(c):=c{j}) and (c=\sum {j=0}^{\nu -1}\gamma {j}(c)\pi ^{j}). For l in ({1,\ldots ,\nu -1}) we set ( c^{[l]}=\sum {j=0}^{l-1}\gamma {j}(c)\pi ^{j}). The (\pi -)adic decomposition will be used is Sect.3 to solve algebraic equations. Note that this decomposition depends on the choice of (\pi ).
Example 1
The ring (\mathbb {Z}{8}) is a finite chain ring where the maximal ideal is generated by 2, with nilpotency index 3. The residue field of (\mathbb {Z}{8}) is (\mathbb {F}{2}=\mathbb {Z}{8}/2\mathbb {Z}{8}) and a complete set of representatives of the equivalence classes of (\mathbb {Z}{8}) under the congruence modulo 2 is (\Gamma =\left{ 0,1\right} ). The (2-)adic decomposition of 6 is (6=0\times 2^{0}+1\times 2^{1}+1\times 2^{2} ). The maximal ideal is also generated by 6 and the (6-)adic decomposition of 6 is (6=0\times 6^{0}+1\times 6^{1}+0\times 6^{2}).
2.2 Gröbner bases
The ring of polynomials with k indeterminates (x_{1},\ldots ,x_{k}) and coefficients in R is denoted (R\left[ x_{1},\ldots ,x_{k}\right] ). A monomial is an element of (R\left[ x_{1},\ldots ,x_{k}\right] ) of the form ( x^{\alpha }:=x_{1}^{d_{1}}\cdots x_{k}^{d_{k}}) where the (d_{i})’s are non-negative integers and (\alpha =\left( d_{1},\ldots , d_{k}\right) ). If “>” is an admissible order on the set of monomials, then any element f in (R \left[ x_{1},\ldots ,x_{k}\right] \backslash \left{ 0\right} ) can be written uniquely as (f=\sum {i=1}^{s}c{i}x^{\alpha {i}}) where each (x^{\alpha {i}}) is a monomial, (c_{i}\in R), and (x^{\alpha {1}}>\cdots >x^{\alpha {s}}). The leading term of f is defined by (lt\left( f\right) :=c_{1}x^{\alpha {1}}). For ( W\subset R\left[ x{1},\ldots ,x_{k}\right] ), we denote by (lt\left( W\right) ) the ideal generated by (\left{ lt\left( w\right) \ |\ w\in W\right} ). According to [46, Definition 3.8], we have the following definition.
Definition 1
Let I be an ideal in (R\left[ x_{1},\ldots ,x_{k}\right] ) and G a subset of I.
(a)
G is called a Gröbner basis for I if (lt(G)=lt(I)).
(b)
G is called a strong Gröbner basis for I if for all (f\in I) there exists (g\in G) such that (\ lt(g)) divides (lt\left( f\right) ), that is to say (lt\left( f\right) =cx^{\alpha }lt(g)) where (c\in R) and (x^{\alpha }) is a monomial.
In [46, Proposition 3.9] a connection between Gröbner bases and strong Gröbner bases was given over finite chain rings.
Proposition 2
A subset of (R\left[ x_{1},\ldots ,x_{k} \right] ) is a Gröbner basis if and only if it is a strong Gröbner basis.
Similar to Buchberger’s algorithm over fields, Norton and Salagean gave an algorithm in [45, Algorithme 3.9] to compute Gröbner bases over finite chain rings. This algorithm has been improved in by adding the product criterion and the chain criterion. An algorithm for computing Gröbner bases on certain classes of finite rings has been implemented in Magma and SageMath [51g
")].
Example 2
A Gröbner basis for the ideal generated by ({4x^{2}y+y^{3}+2y+4,4xy^{2}}) in ( \mathbb {Z}_{8}[x,y]) with lexicographic order (x>y) can be computed using SageMath, and we get ({4x^{2}y+y^{3}+2y+4,4xy^{2},y^{4}+2y^{2}+4y,2y^{3}+4y}).
3 Solving systems of algebraic equations over finite chain rings
In this section, we assume as in Sect.2 that R is a finite commutative chain ring with maximal ideal (\mathfrak {m}) generated by (\pi ), residue field (\mathbb {F}{q}=R/\mathfrak {m}), and that (\nu ) is the nilpotency index of (\pi ). In order to solve systems of polynomial equations, Mikhailov and Nechaev used the lifting approach, which consists of using solutions in the residue field (R/\mathfrak {m}) to construct solutions in the ring _R. However, in some cases this approach is not appropriate in practice, specifically for parametric systems. As an illustration, consider the following system over ( \mathbb {Z} _{8}):
$$\begin{aligned} \left{ \begin{array}{c} 4x^{2}y+y^{3}+2y+4=0 \ 4xy^{2}=0 \end{array} \right. \end{aligned}$$
(2)
This system has 16 solutions. So when we use the lifting approach to solve it, we have to compute each solution step by step, and this is computationally tedious. We will see in this section that one can easily obtain all these solutions using Gröbner bases (see Example 3). The following proposition from [54, Theorem 244], called the elimination theorem, is a direct consequence of Proposition 2.
Proposition 3
Let G be a Gröbner basis for an ideal I in (R\left[ x_{1},\ldots ,x_{k}\right] ) with the lexicographic order ( x_{1}>\cdots >x_{k}). Then, for all i in (\left{ 1,\ldots ,k\right} ), ( G\cap R\left[ x_{i},\ldots ,x_{k}\right] ) is a Gröbner basis of (I\cap R \left[ x_{i},\ldots ,x_{k}\right] ).
The elimination theorem makes it possible to iteratively solve algebraic systems by eliminating variables. Indeed, consider a system of polynomial equations of the form
$$\begin{aligned} f_{i}\left( x_{1},\ldots ,x_{k}\right) =0,\ \ i=1,\ldots ,d. \end{aligned}$$
(3)
where (f_{i}\left( x_{1},\ldots ,x_{k}\right) \in R\left[ x_{1},\ldots x_{k} \right] ). By Proposition 3, if we compute a Gröbner basis G of the ideal ( I=\left( f_{1},\ldots ,f_{d}\right) ) associated to (3) with the lexicographic order (x_{1}>\cdots >x_{k}), then G will be of the form (G=G_{1}\cup G_{2}\cup \cdots \cup G_{k}), where (G_{1}=\left{ g_{1,1}\left( x_{k}\right) ,\ldots ,g_{1,j_{1}}\left( x_{k}\right) \right} ), (G_{2}=\left{ g_{2,1}\left( x_{k-1},x_{k}\right) ,\ldots ,g_{2,j_{2}}\left( x_{k-1},x_{k}\right) \right} ), (\ldots ,)(G_{k}=\left{ g_{k,1}\left( x_{1},\ldots ,x_{k}\right) ,\ldots ,g_{k,j_{k}}\left( x_{1},\ldots ,x_{k}\right) \right} ). So, (3) is equivalent to:
$$\begin{aligned} \left{ \begin{array}{c} g_{1,1}\left( x_{k}\right) =\cdots =g_{1,j_{1}}\left( x_{k}\right) =0 \ g_{2,1}\left( x_{k-1},x_{k}\right) =\cdots =g_{2,j_{2}}\left( x_{k-1},x_{k}\right) =0 \ \vdots \ g_{k,1}\left( x_{1},\ldots ,x_{k}\right) =\cdots =g_{k,j_{k}}\left( x_{1},\ldots ,x_{k}\right) =0 \end{array} \right. \end{aligned}$$
(4)
If for all i in ({1,\ldots ,k}) there exists an element in the Gröbner basis G whose the leading monomial is a pure power of (x_i), then each (G_{i}) is non-empty and, solving (4) is reduced to successively solving systems of univariate polynomial equations. Recall that this process is similar to the case of fields for zero-dimensional algebraic systems [12, 36]. Note in our case that one can always add some univariate polynomial equations to the system using the following remark.
Remark 1
In [42, Theorem 5.14], the monic polynomial (F_{m}) with smaller degree satisfying (F_{m}\left( x\right) =0) for all x in R, has been defined. Thus, as in the case of finite fields, to simplify the resolution of (3), one can add the following equations ( F_{m}\left( x_{1}\right) =\cdots =F_{m}\left( x_{k}\right) =0). For illustration, see Example 4.
We will now show how to use Gröbner bases over finite chain rings to solve systems of univariate polynomial equations. Recall that a Gröbner basis G is called minimal if no proper subset of G is a Gröbner basis for the ideal generated by G. In [45, Theorem 4.2], a characterization of minimal Gröbner bases in one variable over finite chain rings has been given.
Proposition 4
Let (G\subset R[x]\backslash \left{ 0\right} ). Then G is a minimal Gröbner basis if and only if (G=\left{ u_{0}\pi ^{a_{0}}g_{0},\ldots ,u_{s}\pi ^{a_{s}}g_{s}\right} ) for some (0\le s\le \nu -1), (u_{i}\in R) and (g_{i}\in R[x]) for (i=0,\ldots ,s) and such that:
(i)
(0\le a_{0}<a_{1}<\cdots <a_{s}\le \nu -1) and for (i=0,\ldots ,s), (u_{i}) is a unit;
(ii)
for (i=0,\ldots ,s), (g_{i}) is monic;
(iii)
(\deg \left( g_{i}\right) >\deg \left( g_{i+1}\right) ) for any (i \in {0,\ldots ,s-1});
(iv)
for (i=0,\ldots ,s-1), (\pi ^{a_{i+1}}g_{i}) is in the ideal generated by (\left{ \pi ^{a_{i+1}}g_{i+1},\ldots ,\pi ^{a_{s}}g_{s}\right} ).
As specified in , a minimal Gröbner basis in one variable over finite chain rings is a canonical generating system. Therefore, according to Proposition 4, we can use [40, Algorithm 2] to solve systems of univariate polynomial equations over finite chain rings using Gröbner bases. Specifically, consider a system of univariate polynomial equations of the form
$$\begin{aligned} f_{i}\left( x\right) =0,\ \ \ \ \ i=1,\ldots ,r \end{aligned}$$
(5)
where (f_{i}\left( x\right) \in R\left[ x\right] ). Assume that a minimal Gröbner basis of the ideal generated by (\left{ f_{1}\left( x\right) ,\ldots ,f_{r}\left( x\right) \right} ) is (G=\left{ u_{0}\pi ^{a_{0}}g_{0},\ldots ,u_{s}\pi ^{a_{s}}g_{s}\right} ) as in Proposition 4. As specified in [40, page 64] we can assume that (a_{0}=0). Set (h_{j}=g_{i}), for (0\le i\le s) and ( a_{i}\le j<)(a_{i+1}), where (a_{s+1}=\nu ). Then, Eq. (5) is equivalent to the following system of polynomial equations:
$$\begin{aligned} \pi ^{j}h_{j}\left( x\right) =0,\ \ \ \ \ j=0,\ldots ,\nu -1. \end{aligned}$$
(6)
Like in [40, Theorem 8] and [40, Equation (54)], we will use the derivation (Dh_{j}\left( x\right) ) of (h_{j}\left( x\right) ) to solve Eq. (6). As specified in Proposition 1, every element c in R, has a unique (\pi -)adic decomposition (c=\sum {j=0}^{\nu -1}\pi ^{j}\gamma {j}\left( c\right) ) where (\gamma _{j}\left( c\right) \in \Gamma ).
Proposition 5
An element c in R, is a solution of ( 6) if and only if (\gamma _{0}\left( c\right) ) is a solution in (\Gamma ) of the polynomial equation
$$\begin{aligned} h_{\nu -1}\left( x\right) \equiv 0\ \left( mod\ \pi \right) , \end{aligned}$$
and for (j\in \left{ 1,\ldots ,\nu -1\right} ), (\gamma _{j}\left( c\right) ) is a solution in (\Gamma ) of the linear equation:
$$\begin{aligned} Dh_{\nu -j-1}\left( \gamma {0}\left( c\right) \right) x\equiv -\gamma {j}\left( h_{\nu -j-1}\left( c^{[j]}\right) \right) \ \ \left( mod\ \pi \right) . \end{aligned}$$
According to Propositions 3 and 5, to solve a system of multivariate polynomial equations over finite chain rings, we can compute a Gröbner basis of the associated system with the lexicographic order and find the solutions by successively solving the resulting systems of univariate polynomial equations. We will see in Sects.5 and 6 that this approach is appropriate for some systems of algebraic equations when we just need a partial solution.
Example 3
Let us solve System (2) over (\mathbb {Z}{8}) using Gröbner bases. According to Example 2, a Gröbner basis with the lexicographic order (x>y) of the ideal _I generated by (\left{ 4x^{2}y+y^{3}+2y+4,4xy^{2}\right} ) is (G=\left{ g_{1,1},g_{1,2},g_{2,1},g_{2,2}\right} ) where (g_{1,1}\left( y\right) =y^{4}+2y^{2}+4y), (g_{1,2}\left( y\right) =2y^{3}+4y), (g_{2,1}\left( x,y\right) =4x^{2}y+y^{3}+2y+4), (g_{2,2}\left( x,y\right) =4xy^{2}). By Proposition 3, a Gröbner basis of (I\cap R\left[ y \right] ) is (G_{1}=G\cap R\left[ y\right] =\left{ g_{1,1}\left( y\right) ,g_{1,2}\left( y\right) \right} ). So, we can use (G_{1}) to find the partial solution y of (2). The system
$$\begin{aligned} g_{1,1}\left( y\right) =g_{1,2}\left( y\right) =0 \end{aligned}$$
is equivalent to
$$\begin{aligned} h_{1,1}\left( y\right) =2h_{1,2}\left( y\right) =4h_{1,3}\left( y\right) =0 \end{aligned}$$
(7)
where (h_{1,1}\left( y\right) =g_{1,1}\left( y\right) ) and (h_{1,2}\left( y\right) =h_{1,3}\left( y\right) =y^{3}+2y). Let c be a solution of (7). We have (c=\gamma {0}\left( c\right) +2\gamma {1}\left( c\right) +4\gamma {2}\left( c\right) ) where (\gamma {j}\left( c\right) \in \Gamma =\left{ 0,1\right} ) for (j\in \left{ 0,1,2\right} ). By Proposition 5, (\gamma {0}\left( c\right) ) is a solution in (\Gamma ) of the equation (h{1,3}\left( c\right) \equiv 0\ \left( mod\ 2\right) ). So, (\gamma {0}\left( c\right) =0). By Proposition 5, (\gamma {1}\left( c\right) ) is a solution in (\Gamma ) of the equation (Dh_{1,2}\left( \gamma {0}\left( c\right) \right) y\equiv -\gamma {1}\left( h_{1,2}\left( c^{}\right) \right) \ \left( mod\ 2\right) ). We have (c^{}=\gamma {0}\left( c\right) =0), (h{1,2}\left( c^{}\right) =0), (\gamma {1}\left( h{1,2}\left( c^{}\right) \right) =0), (Dh_{1,2}\left( y\right) =3y^{2}+2), and (Dh_{1,2}\left( \gamma {0}\left( c\right) \right) =2). Therefore, ( \gamma {1}\left( c\right) ) is a solution of (2y\equiv 0\ \left( mod\ 2\right) ). So, (\gamma {1}\left( c\right) \in \left{ 0,1\right} ). Using the same reasoning, for (\gamma {1}\left( c\right) =0) or (\gamma {1}\left( c\right) =1), we compute (\gamma {2}\left( c\right) \in \left{ 0,1\right} ). Therefore, (c\in \left{ 0,2,4,6\right} ). Thus, the partial solution y of (2) is in (\left{ 0,2,4,6\right} ). To find the partial solution x corresponding for example to (y=0), we must first compute a Gröbner basis of (\left{ g_{2,1}\left( x,0\right) ,g_{2,2}\left( x,0\right) \right} ). But for all x in (\mathbb {Z}{8}), (g{2,1}\left( x,0\right) =4\ne 0), (g_{2,1}\left( x,4\right) =4\ne 0), (g_{2,1}\left( x,2\right) =g_{2,2}\left( x,2\right) =0), and ( g_{2,1}\left( x,6\right) =g_{2,2}\left( x,6\right) =0). Thus, y is in ( \left{ 2,6\right} ) and the solution set of (2) is (\left{ \left( t,2\right) ,\left( t,6\right) ,t\in \mathbb {Z}_{8}\right} ).
As noted in Remark 1, in certain cases it is necessary to add some equations to solve the system. The following example is an illustration.
Example 4
Consider again the system (2) over (\mathbb {Z}{8}). A Gröbner basis with the lexicographic order (y>x) of the ideal _I generated by (\left{ 4x^{2}y+y^{3}+2y+4,4xy^{2}\right} ) is once again the set (\left{ 4x^{2}y+y^{3}+2y+4,4xy^{2}\right} ). Consequently, (I\cap \mathbb {Z}{8}[x]=\left{ 0\right} ). So, we cannot solve Eq. (2) directly by using only Proposition 5 with the lexicographic order (y>x). However, according to [42, Theorem 5.14], the monic polynomial (F{m}) for the ring (\mathbb {Z}{8}) is defined by (F{m}\left( x\right) =\left( x^{2}-x\right) ^{2}-2\left( x^{2}-x\right) .) A Gröbner basis with the lexicographic order (y>x) of the ideal generated by (\left{ 4x^{2}y+y^{3}+2y+4,4xy^{2},F_{m}\left( x\right) ,F_{m}\left( y\right) \right} ) is (\left{ y^{2}+4,2y+4,F_{m}\left( x\right) \right} ). Therefore, (2) is equivalent to (y^{2}+4=2y+4=0). We solve the system (y^{2}+4=2y+4=0) using Proposition 5, and we obtain (y=2) or (y=6). Thus, the solutions of (2) are the elements of (\left{ \left( t,2\right) ,\left( t,6\right) ,t\in \mathbb {Z}_{8}\right} ).
4 Solving systems of algebraic equations over finite commutative local rings
In the previous section, we have used Gröbner bases to show how one can solve systems of algebraic equations over finite chain rings. We will now show that solving systems of algebraic equations over finite commutative rings can be reduced to finite chain rings. According to [39, Theorem VI.2], if R is a finite commutative ring, then R can be decomposed as a direct sum of local rings, that is to say (R \cong R_{(1)}\times \cdots \times R_{(\rho )}) where for (j=1,\ldots ,\rho ), (R_{(j)}) is a finite commutative local ring. Thus, the problem of solving systems of algebraic equations over R can be reduced to solving systems of algebraic equations over the various (R_{(j)}). However, Grö bner basis are not generally equal to strong Gröbner bases over local rings. Therefore, we will use Galois rings, which are specific classes of finite chain rings to represent finite local rings. As specified in [1, 7], finite rings have several representations (the table representation, the basis representation, and the polynomial representation). Galois rings can be used to give the basis representation and the polynomial representation of finite commutative local rings [39, Theorems XVI.2 and XVII.1]. In , Bulyovszky and Horváth used the basis representation to give a good method for solving systems of linear equations over finite local rings. We are going to extend this method to systems of multivariate polynomial equations.
In this section, we assume that R is a finite commutative local ring with maximal ideal (\mathfrak {m}) and residue field (\mathbb {F}{q}=R/ \mathfrak {m}). Set (q=p^{\mu }) where _p is a prime number. Then the characteristic of R is (p^{\varsigma }) where (\varsigma ) is a non-negative integer and by [39, Theorem XVII.1] there is a sub-ring (R_{0}) of R such that (R_{0}) is isomorphic to the Galois ring of characteristic (p^{\varsigma }) and cardinality (p^{\mu \varsigma }). Considering R as a (R_{0}-)module, there exist (\theta {1},\ldots ,\theta {\gamma }) in R such that
$$\begin{aligned} R=R_{0}\theta {1}\oplus \cdots \oplus R{0}\theta _{\gamma }. \end{aligned}$$
(8)
Let j in (\left{ 1,\ldots ,\gamma \right} ). Since every ideal in (R_{0}) is generated by a power of p, then there is (\varsigma _{j}) in (\left{ 1,\ldots ,\varsigma \right} ) such that
$$\begin{aligned} p^{\varsigma {j}}R{0}=Ann\left( \theta {j}\right) =\left{ a\in R{0}:a\theta _{j}=0\right} . \end{aligned}$$
According to [13, Subsection 2.2] we have the following lemma.
Lemma 1
Let u in R and (u_{j}) in (R_{0}) such that ( u=\sum {j=1}^{\gamma }u{j}\theta _{j}). The following statements are equivalent:
(a)
(u=0);
(b)
for all (j\in \left{ 1,\ldots ,\gamma \right} ), (\ \theta {j}u{j}=0);
(c)
for all (j\in \left{ 1,\ldots ,\gamma \right} ), (\ p^{\varsigma -\varsigma {j}}u{j}=0).
Moreover, each element (u_{j}) is unique modulo (p^{\varsigma _{j}}).
Lemma 1 and the basis decomposition (8) can be used to transform a system of multivariate polynomial equations over finite local rings to Galois rings. Specifically, we have the following:
Theorem 1
Consider a system of polynomial equations of the form
$$\begin{aligned} f_{r}\left( \left( x_{i}\right) _{1\le i\le k}\right) =0,\ \ r=1,\ldots ,d \end{aligned}$$
(9)
where (f_{r}) are multivariate polynomial functions with coefficients in R and (\left( x_{i}\right) _{1\le i\le k}\in R^{k}). Set
$$\begin{aligned} x_{i}=\sum {j=1}^{\gamma }x{i,j}\theta _{j},\ \ i=1,\ldots ,k \end{aligned}$$
where (x_{i,j}\in R_{0}) and
$$\begin{aligned} f_{r}\left( \left( x_{i}\right) {1\le i\le k}\right) =\sum {s=1}^{\gamma }f_{r,s}\left( \left( x_{i,j}\right) {1\le i\le k,1\le j\le \gamma }\right) \theta {s},\ \ r=1,\ldots ,d \end{aligned}$$
where (f_{r,s}) are multivariate polynomial functions with coefficients in ( R_{0}). Then Eq. (9) is equivalent to
$$\begin{aligned} p^{\varsigma -\varsigma {s}}f{r,s}\left( \left( x_{i,j}\right) _{1\le i\le k,1\le j\le \gamma }\right) =0,\ \ r=1,\ldots ,d,\ s=1,\ldots ,\gamma . \end{aligned}$$
(10)
Since Galois rings are specific cases of finite chain rings, we can use the methods described in Sect.3 to solve (10 ).
Example 5
In this example we consider a local ring of size 16 which is not a finite chain ring. As specified in , we can choose (R= \mathbb {Z} {8}\left[ X\right] /I) where _I is the ideal generated by (X^{2}+4) and 2 X. Then R is a local ring with the maximal ideal generated by (2+I) and (X+I). Set (\theta =X+I), then a maximal Galois sub-ring of R is (R_{0}= \mathbb {Z} {8}) and we have (R=\theta {1}R_{0}\oplus \theta {2}R{0}) where (\theta {1}=1) and (\theta {2}=\theta ). Moreover, (Ann\left( \theta {1}\right) =\left{ 0\right} =2^{3}R{0}) and (Ann\left( \theta {2}\right) =2R{0}). We would like to find the roots of the polynomial function defined over R by
$$\begin{aligned} P\left( x\right) =x^{3}+2x+4. \end{aligned}$$
The residue field of R is (\mathbb {F}{2}) and the projection over ( \mathbb {F}{2}) of (P\left( x\right) ) is ({\overline{P}}\left( x\right) =x^{3}) which is not square-free. Therefore, we are not able to find the roots of P using methods based on the Hensel’s lemma [39, Theorem XIII.4] or the Newton-Hensel’s lemma [24f
"), Proposition 2.1.9]. Thus, an alternative method is to use Theorem 1. Set \(x=x_{1}+x_{2}\theta \) where \(x_{1}\) and \(x_{2}\) are in \(R_{0}\). Then,
$$\begin{aligned} P\left( x_{1}+x_{2}\theta \right) =x_{1}^{3}+4x_{1}x_{2}^{2}+2x_{1}+4+\theta x_{1}^{2}x_{2}. \end{aligned}$$
Therefore, equation
$$\begin{aligned} x^{3}+2x+4=0 \end{aligned}$$
is equivalent to the system
$$\begin{aligned} \left{ \begin{array}{c} x_{1}^{3}+4x_{1}x_{2}^{2}+2x_{1}+4=0 \ 4x_{1}^{2}x_{2}=0 \end{array} \right. \end{aligned}$$
(11)
Thanks to Example 3, we deduce that the solutions of (11) are the couples ( \left( x_{1},x_{2}\right) ) in (\left{ \left( 2,t\right) ,\left( 6,t\right) ,t\in \mathbb {Z} {8}\right} ). As (2\theta =0) and (x=x{1}+x_{2}\theta ), then (x_{2}) is unique modulo 2. We can therefore choose (x_{2}) in (\left{ 0,1\right} ). Thus, the roots of P are 2, 6, (2+\theta ), and (6+\theta ).
Example 5 gave a method for finding the roots of polynomials over finite local rings. Another type of local rings are valuation rings, and some methods based on the truncation orders have been described in [8, 43] for the univariate case and in [15, 35, 52] for the multivariate case.
5 MinRank problem over finite principal ideal rings
In this section, we first justify the interest of studying the algebraic resolution of the MinRank problem over finite principal ideal rings by establishing the fact that it is an NP-complete problem. We then extend some known algebraic modelings of the classical MinRank problem to the MinRank problem over finite principal ideal rings. In what follow, we assume that R is a finite commutative principal ideal ring. The set of all (m\times n) matrices with entries in the ring R will be denoted by (R^{m\times n}). Let ( \textbf{A}\in R^{m\times n}), we denote by (row\left( \textbf{A}\right) ) the (R-)submodule of (R^n) generated by the row vectors of (\textbf{A}). The transpose of (\textbf{A}) is denoted by (\textbf{A}^{\top }) and the ( k\times k) identity matrix is denoted by (\textbf{I}_{k}).
5.1 MinRank problem
Definition 2
Let (\ \textbf{A}\in R^{m\times n}). The rank of (\textbf{A}), denoted by (rk_{R}\left( \textbf{A}\right) ) or simply by (rk\left( \textbf{A }\right) ) is the smallest number of elements in (row(\textbf{A})) which generate (row(\textbf{A})) as a (R-)module.
As specified in [31, Proposition 3.4], the Smith normal form can be used to compute the rank of a matrix. Moreover, as in the case of fields, the map (R^{m\times n}\times R^{m\times n}\rightarrow \mathbb {N} ), given by (\left( \textbf{A,B}\right) \mapsto rk\left( \mathbf {A-B} \right) ) is a metric. However, some properties of the rank of a matrix over fields generally do not extend to rings due to zero divisors.
Example 6
Consider the matrix (\mathbf {A}=\left( \begin{array}{cc} 2 &{} 0 \ 0 &{} 4 \end{array} \right) ) over (\mathbb {Z}_{8}). Then, (rk\left( \textbf{A}\right) =2), ( rk\left( 6\textbf{A}\right) =1) and (det(\textbf{A})=0). Thus, (rk\left( \textbf{A}\right) \ne rk\left( 6\textbf{A}\right) ) and (rk\left( \textbf{A} \right) ) is not equal to the order of the highest-order non-vanishing minor.
The MinRank Problem over the ring R can then be defined as follows.
Definition 3
Let (\textbf{M}{0}), (\textbf{M}{1}), (\ldots ), (\textbf{M}{k}) in ( R^{m\times n}) and _r in ({{\mathbb {N}}}^{}). The MinRank problem is to find (x_{1},\ldots ,x_{k}) in R such that (rk(\textbf{M} {0}+\sum {i=1}^{k}x_{i}\textbf{M}{i})\le r). The _homogeneous MinRank problem corresponds to the case where (\textbf{M}_{0}=\textbf{0}).
In general, an instance of the MinRank problem has several solutions. But if r is not greater than the error correction capability of the (R-)linear code generated by (\textbf{M}{1},\ldots ,\textbf{M}{k}) (assuming (\textbf{M}{1},\ldots ,\textbf{M}{k}) are (R-)linearly independent), then the problem has a unique solution (\left( x_{1},\ldots ,x_{k} \right) ). In the homogeneous case, for any solution (\left( x_{1},\ldots ,x_{k}\right) ) and for any (\alpha \in R), (\left( \alpha x_{1},\ldots ,\alpha x_{k}\right) ) is also a solution. Thus, if R is a field, one of the components of a non-zero solution of the homogeneous MinRank problem can always be assumed to be 1. However, if R is not a field, this assumption is not true in some cases (see Example 8).
In Nicolas Courtois used a connection between the Hamming metric and the rank metric to prove that the MinRank problem over fields is NP-complete. We will extend this result to finite principal ideal rings. As in Sect.4, the finite principal ideal ring R can be decomposed as a direct sum of finite chain rings. So, assume that (R=R_{(1)}\times \cdots \times R_{(\rho )}) where (R_{(j)}) is a finite chain ring for (j\in \left{ 1,\ldots ,\rho \right} ). We denote by (\Phi {(j)}) the _j-th projection map from R to (R_{(j)}). We also extend ( \Phi {(j)}) coefficient-by-coefficient as a map from (R^{m\times n}) to ( R{(j)}^{m\times n}). We have the following result from .
Lemma 2
Let (\textbf{A}) in (R^{m\times n}), then
$$\begin{aligned} rk_{R}\left( \textbf{A}\right) =\max {1\le j\le \rho }\left{ rk{R_{(j)}}\left( \Phi _{(j)}\left( \textbf{A}\right) \right) \right} . \end{aligned}$$
Since (R_{(j)}) is a finite chain ring, if a and b are in (R_{(j)}) then a divides b, or b divides a. Therefore, according to [31, Proposition 3.4], we have the following:
Lemma 3
Let (\mathbf {x}=\left( x_{r}\right) {1\le r\le n}\in R{(j)}^{n}), and (\textbf{D}{\textbf{x}}) the (n\times n) diagonal matrix with the entries of (\textbf{x}) on the diagonal, that is, (\textbf{D} {\textbf{x}}=\left( d_{r,s}\right) ) where (d_{r,r}=x_{r}) and (d_{r,s}=0) if (r\ne s). Then, the Hamming weightFootnote 2 of (\textbf{x}) is equal to the rank of (\textbf{D}_{\textbf{x }}).
Proposition 6
The MinRank problem over finite commutative principal ideal rings is NP-complete.
Proof
From Lemma 2, the MinRank Problem over the principal ideal ring R is equivalent to the same problem over the finite chain rings ( R_{(j)}), for (j\in \left{ 1,\ldots ,\rho \right} ). By Lemma 3 the decoding problem in the Hamming metricFootnote 3 over ( R_{(j)}) is reduced to the MinRank Problem. According to or the decoding problem in Hamming metric over (R_{(j)}) is NP-complete.Footnote 4 Thus, the result follows. (\square )
Since the MinRank problem over finite principal ideal rings is a hard problem, the study of its algebraic resolution deserves attention for cryptographic applications. From a modelling perspective, the MinRank problem over finite fields can be transformed into a system of algebraic equations using the maximum minors while over finite principal ideal rings, the rank of a matrix is usually not equal to the order of the highest order non-vanishing minor. As a consequence, the MaxMinor modelling does not apply in general when dealing with rings. In the following subsections, we will prove that the Kipnis–Shamir Modelling and the Support Minors Modelling can be extended over finite principal ideal rings. A natural consequence is that the methods proposed above for solving systems of algebraic equations over finite commutative rings can be applied to solve the MinRank Problem Over Finite Principal Ideal Rings.
5.2 Kipnis–Shamir modeling
We start with some lemmas which will be used to give the Kipnis–Shamir modeling over finite principal ideal rings. According to [31, Proposition 3.2], we have the following:
Lemma 4
Let (\textbf{E} \in R^{m\times n}) such that (rk\left( \textbf{E}\right) \le r). Then, there exists a rank r free submodule F of (R^{n}) such that (row\left( \textbf{E}\right) \subset F).
Remark 2
Let (\textbf{E}) and F as in Lemma 4. If ( row\left( \textbf{E}\right) ) is a free module and (rk\left( \textbf{E} \right) =r) then F is unique and (row\left( \textbf{E}\right) =F). But if ( row\left( \textbf{E}\right) ) is not a free module, then F is generally not unique.
Example 7
Consider the matrix (\mathbf {E}=\left( \begin{array}{ccc} 2&0&4 \end{array} \right) ) over ( \mathbb {Z}{8}). Then (rk\left( \textbf{E}\right) =1) and there exist four free submodules _F of ( \mathbb {Z} _{8}^{3}) of rank 1 such that (row\left( \textbf{E}\right) \subset F). These four submodules are respectively generated by (\left( 1,0,2\right) ), ( \left( 1,4,2\right) ), (\left( 1,0,6\right) ), and (\left( 1,4,6\right) ).
Let F be a free submodule of (R^{n}) of rank r and (F^{\perp }) the dual of F with respect to the canonical inner-product of (R^{n}). Then, by [19, Proposition 2.9], (F^{\perp }) is also a free module of rank (n-r) and (\left( F^{\perp }\right) ^{\perp }=F). Thus, we have the following:
Lemma 5
A subset F of (R^{n}) is a free submodule of (R^{n}) of rank r if and only if there exists (\textbf{Z}\in R^{n\times (n-r)}) with linearly independent column vectors and satisfying:
$$\begin{aligned} \forall ~\textbf{y}\in R^{n},~\textbf{y}\in F~\Longleftrightarrow ~\textbf{yZ }=\textbf{0}. \end{aligned}$$
(12)
Proof
Assume that F is a free submodule of (R^{n}) of rank r. Then, by [19, Proposition 2.9], (F^{\perp }) is a free module of rank (n-r). Let (\textbf{Z}\in R^{n\times (n-r)}) such that the rows of (\textbf{Z}^{\top }) generates (F^{\perp }). Then the column vectors of (\textbf{Z}) are linearly independent and (12) holds.
Conversely, assume that there exists (\textbf{Z}\in R^{n\times (n-r)}) with linearly independent column vectors. Let (F={\textbf{y}\in R^{n}:~\textbf{yZ }=\textbf{0}}). Then, by [19, Proposition 2.9], F is a free module of rank r. (\square )
If a and b are two elements of a finite chain ring, then a divides b or b divides a. This property was used in [44, Proposition 3.2] to prove the existence of the generator matrices in standard form over finite chain rings. So, we have the following:
Lemma 6
Assume that R is a finite chain ring. Let (\textbf{Z}\in R^{n\times (n-r)}) with column vectors that are linearly independent. Then there exists a size n permutation matrix (\textbf{P}), an invertible matrix (\textbf{Q}\in R^{(n-r)\times (n-r)}), and a matrix (\textbf{Z} ^{\prime }\in R^{r\times (n-r)}) such that
$$\begin{aligned} \textbf{Z}=\textbf{P}\left( \begin{array}{c} \textbf{I}_{n-r} \ \textbf{Z}^{\prime } \end{array} \right) \textbf{Q}. \end{aligned}$$
The above Lemma 6 is not generally true when R is not a finite chain ring. Indeed, consider the matrix
$$\begin{aligned} \mathbf {Z}=\left( \begin{array}{c} 2 \ 3 \end{array} \right) \end{aligned}$$
over ( \mathbb {Z} {6}). The column vector of (\textbf{Z}) is ( \mathbb {Z} {6}-)linearly independent. But (\textbf{Z}) cannot be decomposed as in Lemma 6. Lemmas 4, 5 and 6 allow to extend the Kipnis–Shamir Modeling to finite principal ideal rings.
Theorem 2
Let (\textbf{M}{0}), (\textbf{M}{1},\ldots , \textbf{M}{k}) in (R^{m\times n}), (x{1},\ldots ,x_{k}) in R and r in ( {{\mathbb {N}}}^{}). For (M_{x}=\textbf{M}{0}+\sum {i=1}^{k}x_{i} \textbf{M}_{i}), the following statements are equivalent.
(i)
(rk(\textbf{M}_{x})\le r).
(ii)
There exists (\textbf{Z}\in R^{n\times (n-r)}), with column vectors that are linearly independent and such that
$$\begin{aligned} \textbf{M}_{x}\mathbf {Z=0}. \end{aligned}$$ (13)
Moreover, if R is a finite chain ring then, up to a permutation of columns of (\textbf{M}_{x}), we can assume that (\textbf{Z}) is into the form
$$\begin{aligned} \mathbf {Z}=\left( \begin{array}{c} \textbf{I}_{n-r} \ \textbf{Z}^{\prime } \end{array} \right) \end{aligned}$$
where (\textbf{Z}^{\prime }\in R^{r\times (n-r)}).
Proof
The proof is similar to the case of fields. Indeed, assume that (rk(\textbf{M }{x})\le r). Then, by Lemma 4, there exists a free submodule _F of (R^{n}) of rank r such that (row\left( \textbf{M}{x}\right) \subset F). Thus, by Lemma 5, there is (\textbf{Z}\in R^{n\times (n-r)}), with column vectors that are linearly independent and such that (13) holds. Conversely, assume that (ii) holds. Then, by Lemma 5, all row vectors of (\textbf{M}{x}) are in a free module of rank r. Therefore, by [31, Proposition 3.2], (rk(\textbf{M}_{x})\le r). (\square )
As specified in Remark 2, the free submodule F is generally not unique. Therefore, (\textbf{Z}^{\prime }) is generally not unique.
Example 8
Consider the following MinRank problem that is to find (x_{1}), (x_{2}) and (x_{3}) in (\mathbb {Z}_{8}) such that
$$\begin{aligned} rk\left( x_{1}\textbf{M}{1}+x{2}\textbf{M}{2}+x{3}\textbf{M}_{3}\right) \le 1 \end{aligned}$$
(14)
with
$$\begin{aligned} M_{1}=\left( \begin{array}{rrrr} 0 &{} 0 &{} 0 &{} 7 \ 1 &{} 0 &{} 0 &{} 5 \ 0 &{} 1 &{} 0 &{} 2 \ 0 &{} 0 &{} 1 &{} 4 \end{array} \right) ,\ M_{2}=\left( \begin{array}{rrrr} 0 &{} 0 &{} 7 &{} 4 \ 0 &{} 0 &{} 5 &{} 3 \ 1 &{} 0 &{} 2 &{} 5 \ 0 &{} 1 &{} 4 &{} 2 \end{array} \right) ,\ M_{3}=\left( \begin{array}{rrrr} 2 &{} 2 &{} 0 &{} 4 \ 4 &{} 2 &{} 0 &{} 6 \ 0 &{} 4 &{} 2 &{} 4 \ 0 &{} 6 &{} 6 &{} 0 \end{array} \right) . \end{aligned}$$
Since (r=1), by Theorem 2, (14) is equivalent to
$$\begin{aligned} \left( x_{1}\textbf{M}{1}+x{2}\textbf{M}{2}+x{3}\textbf{M}{3}\right) \left( \begin{array}{ccc} 1 &{} 0 &{} 0 \ 0 &{} 1 &{} 0 \ 0 &{} 0 &{} 1 \ z{1} &{} z_{2} &{} z_{3} \end{array} \right) =\textbf{0} \end{aligned}$$
(15)
A Gröbner basis associated to (15) with the lexicographic order (z_{1}>z_{2})(>z_{3}>x_{1}>x_{2}>x_{3}) is (2z_{1}x_{3}+6x_{3}), ( 2z_{2}x_{3}+6x_{3}), (2z_{3}x_{3}+6x_{3}), (x_{1}+2x_{3}), (x_{2}+2x_{3}), ( 4x_{3}). According to Proposition 5, the solutions of the system (x_{1}+2x_{3}=x_{2}+2x_{3}=4x_{3}=0) are the triples (\left( x_{1},x_{2},x_{3}\right) ) in ( \left{ \left( 0,0,0\right) ,\left( 4,4,2\right) ,\left( 0,0,4\right) ,\left( 4,4,6\right) \right} ). Furthermore, each of these solutions satisfies Eq. (14). So we conclude that we have exactly four solutions.
In the simulations, we observe that in some cases, to simplify the resolution of (13) it is necessary to add some equations as specified in Remark 1.
Example 9
Consider the MinRank problem that is to find (x_{1}), (x_{2}), (x_{3}) in (\mathbb {Z}_{8}) such that
$$\begin{aligned} rk(\textbf{M}_{x})\le 1 \end{aligned}$$
(16)
where (\textbf{M}{x}=\textbf{M}{0}+\sum {i=1}^{3}x{i}\textbf{M}_{i}) and
$$\begin{aligned} \textbf{M}{0}=\left( \begin{array}{rrr} 5 &{} 2 &{} 3 \ 5 &{} 1 &{} 4 \ 4 &{} 3 &{} 6 \end{array} \right) ,\ \ \ \textbf{M}{1}=\left( \begin{array}{rrr} 1 &{} 2 &{} 0 \ 0 &{} 1 &{} 3 \ 0 &{} 2 &{} 1 \end{array} \right) ,\ \ \ \textbf{M}{2}=\left( \begin{array}{rrr} 0 &{} 2 &{} 1 \ 1 &{} 0 &{} 3 \ 0 &{} 5 &{} 5 \end{array} \right) ,\ \ \ \textbf{M}{3}=\left( \begin{array}{rrr} 0 &{} 5 &{} 5 \ 0 &{} 1 &{} 0 \ 1 &{} 2 &{} 5 \end{array} \right) \end{aligned}$$
According to Theorem 2, (16) is equivalent to
$$\begin{aligned} \textbf{M}_{x}\textbf{Z}=\textbf{0}. \end{aligned}$$
(17)
When we choose (\textbf{Z}) in the form (\mathbf {Z}=\left( \begin{array}{cc} 1 &{} 0 \ 0 &{} 1 \ z_{1} &{} z_{2} \end{array} \right) ) we do not get the solution. Thus, it is necessary to choose the switchable permutation. In our simulations, we observed that we can choose (\mathbf {Z}=\left( \begin{array}{cc} z_{1} &{} z_{2} \ 1 &{} 0 \ 0 &{} 1 \end{array} \right) ). In this case, when we compute a Gröbner basis associated to (17) with the lexicographic order (z_{1}>z_{2})(>x_{1}>x_{2}>x_{3}), the resolution requires a search for the solution among several potential candidates. But when we add the polynomial expressions (F_{m}\left( z_{1}\right) ), (F_{m}\left( z_{2}\right) ), (F_{m}\left( x_{1}\right) ), (F_{m}\left( x_{2}\right) ), ( F_{m}\left( x_{3}\right) ) as in Example 4, we get the Gröbner basis (z_{1}^{4}-z_{1}^{2}), (2z_{1}), (z_{2}^{4}+3z_{2}^{2}+4), ( 2z_{2}+4), (x_{1}+7), (x_{2}+5), (x_{3}+2). Thus, we directly obtain the solution of (16) which is (x_{1}=1), (x_{2}=3), (x_{3}=6).
5.3 Support-minors modeling
In this subsection, we will show that the Support-Minors modeling of the MinRank problem given in over fields can be extended to finite principal ideal rings.
Lemma 7
Let (\textbf{A}\in R^{r\times n}) with row vectors that are linearly independent, and (\mathbf {y\in }R^{n}). Then (\textbf{y}\in row\left( \textbf{A}\right) ) if and only if
$$\begin{aligned} Minors_{r+1}\left( \begin{array}{c} \textbf{y} \ \textbf{A} \end{array} \right) \mathbf {=0,} \end{aligned}$$
(18)
where (18) means that all minors of the matrix (\left( \begin{array}{c} \textbf{y} \ \textbf{A} \end{array} \right) ) of size (r+1) are equal to zero.
Proof
As the row vectors of (\textbf{A}) are linearly independent, by [19, Corollary 2.7] there is an invertible matrix (\textbf{P}\in R^{n\times n}) such that(\mathbf {AP}=\left( \begin{array}{cc} \textbf{I}{r}&\textbf{0} \end{array} \right) ). Set (\textbf{P}=\left( \begin{array}{cc} \textbf{P}{1}&\textbf{P}{2} \end{array} \right) ) where (\textbf{P}{1}) and (\textbf{P}_{2}) are submatrices of ( \textbf{P}) of sizes (n\times r) and (n \times ( n-r)), respectively. Assume that (18) holds. Then, using the Cauchy–Binet formula, we get
$$\begin{aligned} Minors_{r+1}\left( \left( \begin{array}{c} \textbf{y} \ \textbf{A} \end{array} \right) \textbf{P}\right) \mathbf {=0}, \end{aligned}$$
that is to say,
$$\begin{aligned} Minors_{r+1}\left( \begin{array}{cc} \textbf{yP}{1} &{} \textbf{yP}{2} \ \textbf{I}_{r} &{} \textbf{0} \end{array} \right) =\textbf{0}. \end{aligned}$$
For any entry u of (\textbf{yP}2), the minor ( \det \begin{pmatrix} \textbf{y P}_1 &{} u \ \textbf{I}_r &{} \textbf{0} \end{pmatrix}) is equal to ((-1)^{r}u), which is equal to 0 by the assumption. We then deduce that (\textbf{yP}{2}=\textbf{0}). Thus, by Lemma 5, (\textbf{y}\in row\left( \textbf{A}\right) ). Conversely, if (\textbf{y}\in row\left( \textbf{A}\right) ) then (18) holds, since (\textbf{y}) is a linear combination of the rows of (\textbf{A}). (\square )
Let (\textbf{A}) and (\textbf{y}) as in Lemma 7. For any sequence of r positive integers (1\le j_{1}<\cdots <j_{r}\le n), let ( a_{j_{1},\ldots ,j_{r}}) be the determinant of the (r\times r) submatrix of ( \textbf{A}) with column index in (\left{ j_{1},\ldots ,j_{r}\right} .) The set (\left{ a_{j_{1},\ldots ,j_{r}}:1\le j_{1}<\cdots <j_{r}\le n\right} ) is said to be a Plücker coordinates of the free (R-)module (row\left( \textbf{A} \right) ). By [27, Remark 2.12], if (\ \textbf{B}\in R^{r\times n}) and (row\left( \textbf{A}\right) =row\left( \textbf{B}\right) ), then there is an invertible matrix (\textbf{Q}\in R^{r\times r}) such that (\textbf{B}=\textbf{QA}). Thus, as in the case of fields, the (R-)module (row\left( \textbf{A}\right) ) may admit several sets of Plücker coordinates, but they are all equal up to a unit multiplicative factor. Moreover, if R is a finite chain ring, then according to Lemma 6, at least one component in any Plücker coordinates is a unit. Furthermore, by setting (\textbf{y}=\left( y_{j_{\alpha }}\right) {1\le \alpha \le n}) where (y{j_{\alpha }}\in R), and using the Laplace expansion along the first row, Eq. (18) is equivalent to
$$\begin{aligned} \sum {\alpha =1}^{r+1}\left( -1\right) ^{\alpha +1}y{j_{\alpha }}a_{j_{1},\ldots ,j_{\alpha -1},j_{\alpha +1},\ldots j_{r+1}}=0, \end{aligned}$$
(19)
for all sequence of (r+1) positive integers (1\le j_{1}<\cdots <j_{r+1}\le n).
Notice that, when the row vectors of (\textbf{A}) are not linearly independent, the “only if” part of Lemma 7 may not be true. Indeed, consider the matrix (\mathbf {A}=\left( \begin{array}{cc} 2&0 \end{array} \right) ) over ( \mathbb {Z} _{4}). Then
$$\begin{aligned} Minors_{2}\left( \begin{array}{cc} 0 &{} 2 \ 2 &{} 0 \end{array} \right) =0. \end{aligned}$$
But (\left( 0,2\right) \notin row\left( \textbf{A}\right) ).
Similar to the Support-Minors modeling given in , we have the following:
Theorem 3
Let (\textbf{M}{0}), (\textbf{M}{1},\ldots , \textbf{M}{k}) in (R^{m\times n}), (x{1},\ldots ,x_{k}) in R and r in ( {{\mathbb {N}}}^{}).
Set (\textbf{M}{x}=\textbf{M}{0}+ \sum {l=1}^{k}x{l}\textbf{M}_{l}). Then, the following statements are equivalent.
(i)
(rk(\textbf{M}_{x})\le r).
(ii)
There exist Plücker coordinates (\left{ z_{j_{1},\ldots ,j_{r}}:1\le j_{1}<\cdots <j_{r}\le n\right} ) of a free submodule of (R^{n}) of rank r such that
$$\begin{aligned} \sum {\alpha =1}^{r+1}\left( -1\right) ^{\alpha +1}\textbf{M} {x}[i,j_{\alpha }]z_{j_{1},\ldots ,j_{\alpha -1},j_{\alpha +1},\ldots j_{r+1}}=0, \end{aligned}$$ (20)
for all (i=1,\ldots ,n) and all sequences of (r+1) positive integers (1\le j_{1}<\cdots <j_{r+1}\le n), where (\textbf{M}{x}[i,j{\alpha }]) is the entry at the (i^{th}) row and (j_{\alpha }^{th}) column of (\textbf{M}_x).
Proof
Assume that (rk(\textbf{M}{x})\le r). Then, by Lemma 4, there exists a free submodule _F of (R^{n}) of rank r such that (row\left( \textbf{M}{x}\right) \subset F). Let (\left{ z{j_{1},\ldots ,j_{r}}:1\le j_{1}<\cdots <j_{r}\le n\right} ) be a Plücker coordinates of F. Then, by Lemma 7 and (19), we get (20).
Conversely, assume that (ii) holds. Then, by Lemma 7, all row vectors of (\textbf{M}{x}) are in a free module of rank _r. Therefore, by [31, Proposition 3.2], (rk(\textbf{M}_{x})\le r). (\square )
As stated in Remark 2, the free submodule F is generally not unique. Consequently, there are usually several Plücker coordinates associated to different free submodules, and which all satisfy Eq. (20). Equation (20) is a system of polynomial equations with unknowns (x_{l}) and (z_{j_{1},\ldots ,j_{r}}). Thus, as specified in the previous sections, we can use Gröbner bases to solve (20). But in some cases, it is possible to use linear algebra as in .
Example 10
Consider the MinRank problem (14) of Example 8. Since (r=1), then by Theorem 3, there exist Plücker coordinates (\left( z_{1},z_{2},z_{3},z_{4}\right) ) of a free submodule of (\mathbb {Z}_{8}^4) of rank 1 such that (14) is equivalent to
$$\begin{aligned} \left{ \begin{array}{c} \textbf{M}{x}[i,1]z{2}-\textbf{M}{x}[i,2]z{1}=0 \ \textbf{M}{x}[i,1]z{3}-\textbf{M}{x}[i,3]z{1}=0 \ \textbf{M}{x}[i,1]z{4}-\textbf{M}{x}[i,4]z{1}=0 \ \textbf{M}{x}[i,2]z{3}-\textbf{M}{x}[i,3]z{2}=0 \ \textbf{M}{x}[i,2]z{4}-\textbf{M}{x}[i,4]z{2}=0 \ \textbf{M}{x}[i,3]z{4}-\textbf{M}{x}[i,4]z{3}=0 \end{array} \right. ,\ \ \ i=1,...,4 \end{aligned}$$
(21)
where (\textbf{M}{x}=x{1}\textbf{M}{1}+x{2}\textbf{M}{2}+x{3}\textbf{M} {3}). Since (\mathbb {Z}{8}) is a finite chain ring, at least one component of the Plücker coordinates (\left( z_{1},z_{2},z_{3},z_{4}\right) ) is a unit. Without loss of generality, assume that (z_{4}) is a unit, then in order to recover (x_{1}), ( x_{2}) and (x_{3},) we rewrite (21) as
$$\begin{aligned} \mathbf {AX=0} \end{aligned}$$
(22)
where (\textbf{X}^{\top }=\left( \begin{array}{cccccccccccc} x_{1}z_{1}&x_{2}z_{1}&x_{3}z_{1}&x_{1}z_{2}&x_{2}z_{2}&x_{3}z_{2}&x_{1}z_{3}&x_{2}z_{3}&x_{3}z_{3}&x_{1}z_{4}&x_{2}z_{4}&x_{3}z_{4} \end{array} \right) ) and (\textbf{A}) is a matrix with entries in (\mathbb {Z}_{8}). Using SageMath [51g
")], we can compute the row echelon form \(\widetilde{\textbf{A}}\) of \(\textbf{A}\) and get
$$\begin{aligned} \widetilde{\textbf{A}}=\left( \begin{array}{rrrrrrrrrrrr} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 \ 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 \ 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 \ 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 \ 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 \ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 \ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 \ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 0 &{} 0 &{} 2 \ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 2 &{} 0 &{} 0 &{} 2 \ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 0 &{} 2 \ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 1 &{} 2 \ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 4 \end{array} \right) \end{aligned}$$
Therefore, (22) is equivalent to
$$\begin{aligned} \widetilde{\textbf{A}}\mathbf {X=0} \end{aligned}$$
Thus, (x_{1}z_{4}+2x_{3}z_{4}=)(x_{2}z_{4}+2x_{3}z_{4}=4x_{3}z_{4}=0). Since we assumed that (z_{4}) is a unit, we get (\left( x_{1},x_{2},x_{3}\right) \in \left{ \left( 0,0,0\right) ,\left( 4,4,2\right) ,\left( 0,0,4\right) ,\left( 4,4,6\right) \right} ).
In the case of fields, some conditions have been given in [3, 4] to solve (20) using linear algebra. It will be interesting to study if these conditions can be extended to rings.
It is important to note that, according to [31, Proposition 3.4], the rank of a matrix and its transpose are equal. Therefore, the MinRank problem defined with ( \textbf{M}{0}), (\textbf{M}{1}), (\ldots ), (\textbf{M}{k}) shares the same solution set with the one defined with (\textbf{M}{0}^{\top }), (\textbf{M} {1}^{\top }), (\ldots ), (\textbf{M}{k}^{\top }). Thus, in order to reduce the number of variables in the algebraic modeling, one can transpose the matrices before solving the MinRank problem, as stated for example in .
6 Rank decoding problems over finite principal ideal rings
In this section, we study the algebraic approach for solving the rank decoding problem over finite principal ideal rings. Note that this problem was recently shown in to be at least as hard as the rank decoding problem over finite fields, and a combinatorial-like algorithm was also proposed for solving the problem. Over finite fields, the rank decoding problem has several algebraic modeling. As specified in [29, Section 4], the Ourivski-Johansson modeling and the MaxMinors modeling cannot extend directly to rings due to zero divisors. We show here that the Support-Minors modeling and the modeling using linearized polynomials can be extended in the case of finite principal ideal rings.
6.1 Rank decoding problem
To define the rank decoding problem, we must first recall the construction of a Galois extension of a finite principal ideal ring R. As we specified in Sect.4, R can be decomposed into a direct sum of local rings. Thus, in the following, we assume that ( R=R_{(1)}\times \cdots \times R_{(\rho )}) where each (R_{(j)}) is a finite chain ring with maximal ideal (\mathfrak {m}{(j)}) and residue field (\mathbb {F} {q_{(j)}}), for (j=1,\ldots ,\rho ). Let m be a non-zero positive integer and (h_{(j)}\in )(R_{(j)}\left[ X\right] ) a monic polynomial of degree m such that its projection onto (\mathbb {F}{q{(j)}}\left[ X\right] ) is irreducible. If we set (S_{(j)}=R_{(j)}\left[ X\right] /\left( h_{(j)}\right) ) then, by , (S_{(j)}) is a Galois extension of ( R_{(j)} ) of degree m with Galois group that is cyclic of order m. Moreover, ( S_{(j)}) is also a finite chain ring with maximal ideal (\mathfrak {M}{(j)}= \mathfrak {m}{(j)}S_{(j)}) and residue field (\mathbb {F}{q{(j)}^{m}}). Let us denote by (\sigma {(j)}) a generator of the Galois group of (S{(j)}), ( \sigma =\left( \sigma {(j)}\right) {1\le j\le \rho }) and ( S=S_{(1)}\times \cdots \times S_{(\rho )}). Then, as specified in , S is a Galois extension of R of degree m with Galois group generated by (\sigma ). Moreover, there exists (h\in R\left[ X \right] ) such that (S\cong R\left[ X\right] /\left( h\right) ). An example of construction of a Galois extension of ( \mathbb {Z} _{40}) of degree 4 was given in [29, Example 2.2]. The following example shows how one can construct a generator of the Galois group in practice using the Hensel lifting of a primitive polynomial.
Example 11
Let us construct a degree 3 Galois extension of (R = \mathbb {Z}{8}), and its Galois group. The residue field of _R is (\mathbb {F}{q} =\mathbb {F}{2} ) and the polynomial (g=X^{3}+X+1) is a primitive polynomial in (\mathbb {F}{q} \left[ X\right] ). Using the Hensel’s lemma, we can construct the polynomial ( h=X^{3}+6X^{2}+5X+7\ \ \in R\left[ X\right] ), such that ({\overline{h}}=g) and _h divides (X^{q^{m}-1}-1). Therefore,(\ S=R\left[ X\right] /\left( h\right) =R\left[ \alpha \right] ) is a Galois extension of R of degree ( m=3), where (\alpha =X+\left( h\right) ). Moreover, (\alpha ^{q^{m}-1}=1) and (\alpha ^{i}\ne 1), for (0<i<q^{m}-1). Thus, the Galois group is generated by the map (\sigma :S\rightarrow S) given by (\alpha \mapsto \alpha ^{q}), that is to say, for all (x=\sum {i=0}^{m-1}x{i}\alpha ^{i}), where (x_{i}\in R), (\sigma \left( x\right) =\sum {i=0}^{m-1}x{i}\alpha ^{iq}).
Definition 4
Let (\ \textbf{u}=\left( u_{1},\ldots ,u_{n}\right) \in S^{n}).
a)
The support of (\textbf{u}), denoted by (supp(\textbf{u})), is the (R-)submodule of S generated by ({ u_{1},\ldots ,u_{n} } ).
b)
The rank of (\textbf{u},) denoted by (rk_{R}\left( \textbf{u} \right) ,) or simply by (rk\left( \textbf{u}\right) ) is the smallest number of elements in (supp(\textbf{u})) which generate (supp(\textbf{u})) as a (R-)module.
Since S is a free (R-)module, computing the rank of a vector ( \textbf{u}\in S^{n}) can be done by using its matrix representation in a (R-)basis of S as in the case of finite fields (for more details see [31, Proposition 3.13]).
Definition 5
Let (\mathcal {C}) be a (S-)submodule of (S^{n}), (\textbf{y}) an element of (S^{n}) and (r\in {{\mathbb {N}}}^{}). The rank decoding problem is to find (if there exist) (\textbf{e}) in (S^{n}) and (\textbf{c}) in (\mathcal {C}) such that (\textbf{y}=\textbf{c}+\textbf{e}) with (rk(\textbf{e})\le r).
Using the representation of elements in (S^{n}) as elements of (R^{m\times n} ), the rank decoding problem can be reduced to the MinRank problem, as in the case of finite fields .
Example 12
Let us consider the rings (R=\mathbb {Z}_{8}) and (\ S=R\left[ \alpha \right] ) as in Example 11. Let (\mathcal {C} \subset S^{3}) be the (S-)linear code generated by:
$$\begin{aligned} \textbf{g}=\left( 1,2\alpha ^{2}+\alpha +2,\alpha ^{2}+3\alpha \right) . \end{aligned}$$
Set (\mathbf {y}=\left( 4\alpha ^{2}+3\alpha +3,\,5\alpha ^{2}+7\alpha +6,\,2\alpha ^{2}+4\alpha +5\right) ) and consider the instance of the rank decoding problem consisting of finding (\textbf{c}\in \mathcal {C} ) such that
$$\begin{aligned} rk\left( \textbf{y}-\textbf{c}\right) \le 1. \end{aligned}$$
(23)
Eq. (23) is equivalent to finding (x_{1}), (x_{2}), (x_{3}) in R such that
$$\begin{aligned} rk\left( \textbf{y}-\left( x_{1}+x_{2}\alpha +x_{3}\alpha ^{2}\right) \textbf{g}\right) \le 1 \end{aligned}$$
(24)
Since (\mathcal {C}) is generated by (\textbf{g}), then the matrix representation of (\mathcal {C}) in the basis (\left( 1,\alpha ,\alpha ^{2}\right) ) is the (R-)linear code generated by (\textbf{M}{1}), (\textbf{ M}{2}), (\textbf{M}{3}) which are respectively the representation matrices of (\textbf{g}), (\alpha \textbf{g}), (\alpha ^{2}\textbf{g}) in the basis ( \left( 1,\alpha ,\alpha ^{2}\right) ). Let (\textbf{M}{0}) be the matrix representation of (-\textbf{y}) in the basis (\left( 1,\alpha ,\alpha ^{2}\right) ). Then, the rank decoding problem (24) is equivalent to the MinRank problem (16) defined in Example 9. The solution of (16) is (x_{1}=1), ( x_{2}=3), (x_{3}=6). Thus, (\mathbf {c}=\left( 1+3\alpha +6\alpha ^{2}\right) )(\textbf{g}).
6.2 Support-Minors modeling
According to [31, Proposition 3.14] we have the following:
Lemma 8
For any (\textbf{u}\in S^{n}) with (rk\left( \textbf{u}\right) \le r), there exists (\textbf{b} \in S^{r}) and (\textbf{ A} \in R^{r\times n}) such that (row\left( \textbf{A}\right) ) is a free module of rank r and (\textbf{u}=\textbf{bA}).
The following result is a generalization of the Support-Minors modeling for the rank decoding problem given in .
Theorem 4
Let (\mathcal {C}) be a (S-)submodule of ( S^{n}) with a generator matrix (\textbf{G}=\left( g_{i,j}\right) {1\le i\le k,1\le j\le n}), (\mathbf {y}=\left( y{i}\right) {1\le i\le n}\in )(S^{n}) and (r\in {{\mathbb {N}}}). Assume that there exists (\textbf{x} =\left( x{i}\right) {1\le i\le k}\in S^{k}) such that (rk\left( \textbf{y }-\textbf{xG}\right) \le r). Then, there exists a set (\left{ z{j_{1},\ldots ,j_{r}}:1\le j_{1}<\cdots <j_{r}\le n\right} ) of Plücker coordinates of a free submodule of (R^{n}) of rank r such that
$$\begin{aligned} \sum {s=1}^{r+1}\sum {i=1}^{k}\left( -1\right) ^{s+1}\left( x_{i}g_{i,{j{s}}}-y_{j_{s}}\right) z_{j_{1},\ldots ,j_{s-1},j_{s+1},\ldots j_{r+1}}=0, \end{aligned}$$
(25)
for all sequence of (r+1) positive integers (1\le j_{1}<\cdots <j_{r+1}\le n.)
Proof
Using Lemmas 7 and 8, the proof is similar to the one from [4, Section 3]. (\square )
Equation (25) is a system of algebraic equations over S with unknowns (x_{i}\in S) and (z_{j_{1},\ldots ,j_{s-1},j_{s+1},\ldots j_{r+1}}\in R). To solve this equation using Gröbner bases, we must first expand this equation to R.
Example 13
Consider the rank decoding problem (23) of Example 12. Set (\mathbf {g}= \left( g_{1,}g_{2},g_{3}\right) ) and (\mathbf {y}=\left( y_{1,}y_{2},y_{3}\right) ). Then, by Theorem 4, there are x in S and (z_{1}), (z_{2}), (z_{3}) in R such that
$$\begin{aligned} \left{ \begin{array}{c} \left( xg_{1}-y_{1}\right) z_{2}-\left( xg_{2}-y_{2}\right) z_{1}=0 \ \left( xg_{1}-y_{1}\right) z_{3}-\left( xg_{3}-y_{3}\right) z_{1}=0 \ \left( xg_{2}-y_{2}\right) z_{3}-\left( xg_{3}-y_{3}\right) z_{2}=0 \end{array} \right. \end{aligned}$$
(26)
Since (R= \mathbb {Z} {8}) is a finite chain ring, at least one of the elements in the Plücker coordinates (\left( z{1},z_{2},z_{3}\right) ) is a unit. Without loss of generality, assume that (z_{1}=1). Set (x=x_{0}+x_{1}\alpha +x_{2}\alpha ^{2}) where (x_{i} \in R), for (i \in { 0, 1, 2 }). Using SageMath [51g
")], we substitute _x_ and \(z_{1}\) in (26) and expand the resulting equations over _R_ using the basis \(\left( 1,\alpha ,\alpha ^{2}\right) \), then we obtain a system of equations of the form
$$\begin{aligned} \left{ \begin{array}{l} -z_{2}x_{0}+3z_{2}+2x_{0}+2x_{1}+5x_{2}+2=0 \ -z_{2}x_{1}+3z_{2}+x_{0}+x_{2}+1=0 \ -z_{2}x_{2}+4z_{2}+2x_{0}+5x_{1}+2x_{2}+3=0 \ -z_{3}x_{0}+3z_{3}+x_{1}+5x_{2}+3=0 \ -z_{3}x_{1}+3z_{3}+3x_{0}+3x_{1}+4=0 \ -z_{3}x_{2}+4z_{3}+x_{0}+5x_{1}+5x_{2}+6=0 \ z_{2}x_{1}+5z_{2}x_{2}+3z_{2}+6z_{3}x_{0}+6z_{3}x_{1}+3z_{3}x_{2}+6z_{3}=0 \ 3z_{2}x_{0}+3z_{2}x_{1}+4z_{2}-z_{3}x_{0}-z_{3}x_{2}-z_{3}=0 \ z_{2}x_{0}+5z_{2}x_{1}+5z_{2}x_{2}+6z_{2}+6z_{3}x_{0}+3z_{3}x_{1}+6z_{3}x_{2}+5z_{3}=0 \end{array} \right. \end{aligned}$$
(27)
As in Example 9, when we compute a Gröbner basis associated to (27) with the lexicographic order (z_{2}>z_{3})(>x_{0}>x_{1}>x_{2}), the resolution requires a search for the solution among several potential candidates. So, to simplify the resolution, we add the polynomial expressions (F_{m}\left( z_{2}\right) ), (F_{m}\left( z_{3}\right) ), ( F_{m}\left( x_{0}\right) ), (F_{m}\left( x_{1}\right) ), (F_{m}\left( x_{2}\right) ) as in Example 4, and get the Gröbner basis: (z_{2}^{4}-z_{2}^{2}), (2z_{2}), (z_{3}^{4}+3z_{3}^{2}+4), (2z_{3}+4), (x_{0}+7), (x_{1}+5), (x_{2}+2). Thus, (x_{0}=1), (x_{1}=3), (x_{2}=6).
6.3 Algebraic modeling with skew polynomials
Skew polynomials generalize linearized polynomials, and some properties of linearized polynomials have been extended to skew polynomials in .
Definition 6
The skew polynomial ring over S with automorphism (\sigma ), denoted by (S[X,\sigma ]), is the ring of all polynomials in S[X] such that
the addition is defined to be the usual addition of polynomials;
the multiplication is defined by the basic rule (Xa=\sigma \left( a\right) X), for all (a\in S).
Notation 5
(Evaluation Map) Let (f=a_{0}+a_{1}X+\cdots + a_{k}X^{k}\in S[X,\sigma ]), (x\in S) and ( \textbf{u}=\left( u_{i}\right) _{1\le i\le n}\in S^{n}).
1.
(f\left( x\right) :=a_{0}x+a_{1}\sigma \left( x\right) +\cdots +a_{k}\sigma ^{k}\left( x\right) ).
2.
(f\left( \textbf{u}\right) :=\left( f\left( u_{i}\right) \right) _{1\le i\le n}).
According to [31, Propositions 3.15, 3.16 and Corollary 2.7], we have the following proposition.
Proposition 7
For all (\textbf{u}\in S^{n}), (rk\left( \textbf{u}\right) \le r) if and only if there exists a monic skew polynomial (f \in S[X,\sigma ]) of degree r such that, (f\left( \textbf{u}\right) =\textbf{0}.) Moreover, if (supp(\textbf{u})) is a free module and (rk\left( \textbf{u}\right) =r), then f is unique.
Remark 3
To construct the skew polynomial f of Proposition 7, one generally uses a free (R-)submodule of S which contains ( supp(\textbf{u})). Hence, as we pointed out in Remark 2, there are generally more than one free (R-)submodule of S which contains (supp(\textbf{u})). Thus, f is generally not unique.
Example 14
Consider again (R=\mathbb {Z}_{8}) and (\ S=R\left[ \alpha \right] ) as in Example 11.
The rank of (\textbf{u}=\left( 2+6\alpha ^{2},0,4+4\alpha ^{2}\right) ) is 1 and we would like to find all the monic skew polynomials (f\in S[X,\sigma ]) of degree 1 such that (f\left( \textbf{u}\right) =\textbf{0}). So, (f=X+w), where (w\in S) and can be written as (w=w_{0}+w_{1}\alpha +w_{2}\alpha ^{2}) with (w_{0}), (w_{1}), (w_{2}) in R. When we solve the equation (f\left( \textbf{u}\right) =\textbf{0}), we get (w_{0} \in { 3,7}), (w_{1} \in {0,4 }), (w_{2} \in {3,7 }). Thus, there are eight monic skew polynomials (f\in S[X,\sigma ]) with degree 1 such that (f\left( \textbf{u}\right) =\textbf{0}.)
Notation 6
If (\textbf{B}=\left( b_{i,j}\right) ) is a matrix with entries in S and l is a positive integer then,
$$\begin{aligned} \sigma ^{l}\left( \textbf{B}\right) :=\left( \sigma ^{l}\left( b_{i,j}\right) \right) . \end{aligned}$$
The following result is a generalization of the result given in [25, Section V].
Theorem 7
Let (\mathcal {C}) be a (S-)submodule of (S^{n}) with generator matrix (\textbf{G}=\left( g_{i,j}\right) {1\le i\le k,1\le j\le n},)(r\in {{\mathbb {N}}}) and (\mathbf {y}=\left( y{i}\right) _{1\le i\le n}\in )(S^{n}). The following statements are equivalent.
(i)
There exists (\textbf{c}\in \mathcal {C}) such that (rk\left( \textbf{y}-\textbf{c}\right) \le r).
(ii)
There are (\left( z_{l}\right) {0\le l\le r}\in S^{r+1}), ( z{r}=1), and (\textbf{x}=\left( x_{i}\right) _{1\le i\le k}\in S^{k}) such that
$$\begin{aligned} \sum {l=0}^{r}z{l}\sigma ^{l}\left( \textbf{y}\right) =\sum {l=0}^{r}z{l}\sigma ^{l}\left( \textbf{xG}\right) \end{aligned}$$ (28)
Moreover, if (\mathcal {C}) is a free (S-)submodule of rank k and (r\le t), where t is the error correction capability of (\mathcal {C}), then ( \textbf{x}) is unique.
Proof
By Proposition 7, (rk\left( \textbf{y}-\textbf{c}\right) \le r) if and only if there exists a monic skew polynomial (P=\sum {l=0}^{r}z{l}X^{l} \in S[X,\sigma ]) of degree r such that (P\left( \textbf{y}-\textbf{c} \right) =\textbf{0}). Since (\textbf{c}\in \mathcal {C}), then there exists ( \textbf{x}=\left( x_{i}\right) _{1\le i\le k}\in S^{k}), such that ( \mathbf {c=xG}). Thus, the result follows. (\square )
According to Remark 3, when the support of the error is not a free module, the unknowns (z_{i})’s, (i=0,\ldots ,r-1) are not unique, even if (\textbf{x}) is unique. So in general, (28) has many solutions. This is the main difference compared to the same result over finite fields. Note that to solve the rank decoding problem, we don’t need the unknowns (z_{i}). We just need (\textbf{x}), since we can use it to recover (\textbf{c}).
6.3.1 Solving by linearization
In this subsection, we will show that in some cases, the unknowns (\textbf{x}) in (28) can be recovered using linear algebra. Eq. (28) is equivalent to
$$\begin{aligned} \textbf{Au}=\textbf{0} \end{aligned}$$
(29)
where
$$\begin{aligned} \textbf{A}=\left( \begin{array}{ccccccc} -\sigma ^{0}\left( \textbf{y}^{\top }\right)&\cdots&-\sigma ^{r-1}\left( \textbf{y}^{\top }\right)&\sigma ^{0}\left( \textbf{G}^{\top }\right)&\cdots&\sigma ^{r}\left( \textbf{G}^{\top }\right)&-\sigma ^{r}\left( \textbf{y}^{\top }\right) \end{array} \right) \end{aligned}$$
and
$$\begin{aligned} \textbf{u}^{\top }\mathbf {}=\left( \begin{array}{ccccccc} z_{0}&\cdots&z_{r-1}&z_{0}\sigma ^{0}\left( \textbf{x}\right)&\cdots&z_{r}\sigma ^{r}\left( \textbf{x}\right)&z_{r} \end{array} \right) . \end{aligned}$$
In the same way as the row echelon form over fields, the matrix (\textbf{A}) can be decomposed as (\mathbf {A=PT}) where (\textbf{P}) is an invertible matrix and (\mathbf {T}=\left( t_{i,j}\right) ) is an upper triangular matrix, that is to say (t_{i,j}=0) if (i>j) [33, Theorem 3.5]. The matrix (\textbf{T}) is usually called the Hermite normal form of (\textbf{A}). One can compute the Hermite normal form using the same methods as the Gaussian elimination algorithm, see for example [13, 33, 50]. As ( z_{r}=1), the following proposition shows that if (\textbf{T}) has a specific form, then (\textbf{x}) can be recovered.
Proposition 8
With the above notations, assume that (28) has a solution and that (\textbf{T}) is of the form
$$\begin{aligned} \textbf{T}=\left( \begin{array}{cc} \textbf{T}{1} &{} \textbf{T}{2} \ \textbf{0} &{} \textbf{T}_{3} \ \textbf{0} &{} \textbf{0} \end{array} \right) \end{aligned}$$
(30)
where (\textbf{T}{1}) is an (r(k+1)\times r(k+1)) upper triangular matrix, ( \textbf{T}{2}) being a (r(k+1)\times (k+1)) matrix and (\textbf{T} {3}=\left( \begin{array}{cc} \textbf{I}{k}&\textbf{b} \end{array} \right) ) where (\textbf{b}) is a (k\times 1) matrix, then
$$\begin{aligned} \mathbf {x}=-\sigma ^{-r}\left( \textbf{b}^{\top }\right) . \end{aligned}$$
Note that (29) is a homogeneous system of n linear equations with ((k+1)(r+1)) unknowns. So, a necessary condition for (\textbf{ T}) to have the form (30) is (n\ge (k+1)(r+1)-1). The same condition was given in [25, Theorem 12] in the case of finite fields. With this condition, we observed in our simulations that, when (\mathcal {C}) is a random free submodule, (\textbf{x}) can be recovered in many cases. It will be therefore interesting to study the probability of this observation.
Example 15
Consider the rank decoding problem of Example 12. Then there are (x\in S) and (\mathbf {e\in }S^{3}) such that
$$\begin{aligned} \textbf{y}=x\textbf{g}+\textbf{e} \end{aligned}$$
(31)
with (rk\left( \textbf{e}\right) =r=1). So, the skew polynomial (P\in S[X,\sigma ]), such that
$$\begin{aligned} P\left( \textbf{e}\right) =\textbf{0} \end{aligned}$$
(32)
is of the form (P=z_{0}+z_{1}X) where (z_{0},z_{1}\in S) with (z_{1}=1). By setting (\mathbf {g}=\left( g_{1,}g_{2},g_{3}\right) ) and (\mathbf {y}=\left( y_{1,}y_{2},y_{3}\right) ), (31) and (32) imply
$$\begin{aligned} z_{0}\left( xg_{j}-y_{j}\right) +z_{1}\sigma \left( xg_{j}-y_{j}\right) =0,\ \ \ \ j=1,...,3. \end{aligned}$$
(33)
which means that
$$\begin{aligned} \textbf{A}\left( \begin{array}{c} z_{0} \ z_{0}x \ z_{1}\sigma \left( x\right) \ z_{1} \end{array} \right) =\left( \begin{array}{c} 0 \ 0 \ 0 \ 0 \end{array} \right) \end{aligned}$$
(34)
where
$$\begin{aligned} \textbf{A}=\left( \begin{array}{cccc} -y_{1} &{} g_{1} &{} \sigma \left( g_{1}\right) &{} -\sigma \left( y_{1}\right) \ -y_{2} &{} g_{2} &{} \sigma \left( g_{2}\right) &{} -\sigma \left( y_{2}\right) \ -y_{3} &{} g_{3} &{} \sigma \left( g_{3}\right) &{} -\sigma \left( y_{3}\right) \end{array} \right) . \end{aligned}$$
Using Magma , we compute the row echelon form of (\textbf{A}) and get:
$$\begin{aligned} \textbf{T}=\left( \begin{array}{cccc} 1 &{} \alpha ^{2}+\alpha &{} 0 &{} 2\alpha ^{2}+4 \ 0 &{} 2 &{} 0 &{} 6\alpha ^{2}+4\alpha \ 0 &{} 0 &{} 1 &{} 3\alpha ^{2}+6\alpha +3 \end{array} \right) \end{aligned}$$
Thus,
$$\begin{aligned} x= & {} -\sigma ^{-1}\left( 3\alpha ^{2}+6\alpha +3\right) \= & {} 1+3\alpha +6\alpha ^{2} \end{aligned}$$
6.3.2 Solving with Gröbner bases
When S is a finite field, Eq. (28) is a system of multivariate polynomial equations in the variables (z_{l}) and (x_{i}), and such a system was solved directly with Gröbner bases in [25, Section VII]. However, when S is not a field, the expression (\sigma ^{l}\left( x_{i}g_{i,j}\right) ) is not a polynomial function in the variable (x_{i}). So, to transform (28) into a system of multivariate polynomial equations, we will expand this equation in R. Let (\left( \beta {u}\right) {1\le u\le m}) be a (R-)basis of S. Using the notations of Theorem 7, set (x_{i}=\sum {u=1}^{m}x{i,u}\beta {u}) and (z{l}=\sum {v=1}^{m}z{l,v} \beta {v}) where (x{i,u}) and (z_{l,v}) are in R. If we substitute (x_{i} ) and (z_{l}) in (28) and expand the resulting equations over R using the basis (\left( \beta {u}\right) {1\le u\le m}), then we obtain a system of equations of the form:
$$\begin{aligned} \left( \widetilde{\textbf{x}}\otimes \widetilde{\textbf{z}}\right) \textbf{A} +\widetilde{\textbf{x}}\textbf{B}+\widetilde{\textbf{z}}\textbf{C}+\textbf{D} =\textbf{0} \end{aligned}$$
(35)
where
$$\begin{aligned} \widetilde{\textbf{x}}=\left( x_{1,1},\ldots x_{1,m},\ldots ,x_{k,1},\ldots x_{k,m}\right) , \widetilde{\textbf{z}}=\left( z_{0,1},\ldots z_{0,m},\ldots ,z_{r-1,1},\ldots z_{r-1,m}\right) , \end{aligned}$$
and ( \textbf{A}), (\textbf{B}), (\textbf{C}), (\textbf{D}) are matrices with mn columns and entries in R.
Assume that (\mathcal {C}) is a free (S-)submodule and (r\le t), where t is the error correction capability of (\mathcal {C}). Then, according to Theorem 7, Eq. (35) has a unique solution in the variables (\widetilde{\textbf{x}}) that we denote by ( \widetilde{\textbf{x}}{0}). Remember that when the support of the error is not a free module, Eq. (35) has many solutions in the variables (\widetilde{\textbf{z}}). But also note that we do not need all the solutions of (35). We just need the partial solution (\widetilde{ \textbf{x}}{0}). Therefore, to solve (35) we can use the elimination theorem as specified in Sect.3 to simply find the partial solution (\widetilde{\textbf{x}}_{0}) using Gröbner bases.
Example 16
Consider Eq. (33) of Example 15. Set (x=x_{0}+x_{1}\alpha +x_{2}\alpha ^{2}) and ( z_{0}=t_{0}+t_{1}\alpha +t_{2}\alpha ^{2}) where (x_{i}) and (t_{i}) are in R for (i=1,\ldots ,3). Using SageMath, we substitute x, (z_{0}) and ( z_{1}=1) in (33) and expand the resulting equations over R using the basis (\left( 1,\alpha ,\alpha ^{2}\right) ) to finally obtain a system of equations of the form
$$\begin{aligned} \tiny { \left{ \begin{array}{l} x_{0}t_{0}+x_{2}t_{1}+x_{1}t_{2}+2x_{2}t_{2}+x_{0}+2x_{2}+5t_{0}+4t_{1}+5t_{2}+5=0 \ x_{1}t_{0}+x_{0}t_{1}+3x_{2}t_{1}+3x_{1}t_{2}-x_{2}t_{2}-x_{2}+5t_{0}+t_{1}+3t_{2}+4=0 \ x_{2}t_{0}+x_{1}t_{1}+2x_{2}t_{1}+x_{0}t_{2}+2x_{1}t_{2}-x_{2}t_{2}+x_{1}-x_{2}+4t_{0}+5t_{1}+3t_{2}+1=0 \ 2x_{0}t_{0}+2x_{1}t_{0}+5x_{2}t_{0}+2x_{0}t_{1}+5x_{1}t_{1}+2x_{2}t_{1}+5x_{0}t_{2}+2x_{1}t_{2}+5x_{2} t_{2}+6x_{0}+4x_{1}+x_{2}+2t_{0}+3t_{1}-t_{2}=0 \ x_{0}t_{0}+x_{2}t_{0}+x_{1}t_{1}+3x_{2}t_{1}+x_{0}t_{2}+3x_{1}t_{2}+x_{2}t_{2}+6x_{0}+3x_{1}+6x_{2} +t_{0}+3t_{1}+5=0 \ 2x_{0}t_{0}+5x_{1}t_{0}+2x_{2}t_{0}+5x_{0}t_{1}+2x_{1}t_{1}+5x_{2}t_{1}+2x_{0}t_{2}+5x_{1}t_{2} +5x_{2}t_{2}-x_{0}+3x_{1}-x_{2}+3t_{0}-t_{1}+t_{2}+6=0 \ x_{1}t_{0}+5x_{2}t_{0}+x_{0}t_{1}+5x_{1}t_{1}+5x_{2}t_{1}+5x_{0}t_{2}+5x_{1}t_{2}+2x_{2}t_{2} +2x_{0}+3x_{1}-x_{2}+3t_{0}+6t_{1}+7=0 \ 3x_{0}t_{0}+3x_{1}t_{0}+3x_{0}t_{1}+4x_{2}t_{1}+4x_{1}t_{2}+3x_{2}t_{2}-x_{0}+3x_{1}+3x_{2} +4t_{0}+5t_{1}+6t_{2}+2=0 \ x_{0}t_{0}+5x_{1}t_{0}+5x_{2}t_{0}+5x_{0}t_{1}+5x_{1}t_{1}+2x_{2}t_{1}+5x_{0}t_{2}+2x_{1}t_{2} +2x_{0}+6x_{1}+3x_{2}+6t_{0}+5t_{2}+6=0 \end{array} \right. } \end{aligned}$$
(36)
Using SageMath [51g
")], we compute a Gröbner basis of (36) and get:
$$\begin{aligned} \left{ x_{0}+7,x_{1}+5,x_{2}+2,2t_{0}+2,2t_{1},2t_{2}+2\right} \text {.} \end{aligned}$$
Thus, (x=x_{0}+x_{1}\alpha +x_{2}\alpha ^{2}=1+3\alpha +6\alpha ^{2}).
The SageMath code used for all the examples in this paper is available at
7 Conclusion
In this work, we have shown that solving systems of algebraic equations over finite commutative rings reduces to the same problem over Galois rings. Then, using the elimination theorem and some properties of canonical generating systems, we have also shown how Gröbner bases can be used to solve systems of algebraic equations over finite chain rings. As applications, these results have been used to give some algebraic approaches for solving the MinRank problem and the rank decoding problem over finite principal ideal rings.
The above work clearly opens the door to an important complexity question, namely the real coast of Gröbner bases computation over finite chain rings, or at least the cost when dealing with the MinRank and rank decoding problems over finite chain rings.
Another metric used in coding theory and cryptography is the Lee metric . This metric is usually defined over integer residue rings, which are specific cases of finite principal ideal rings. Another interesting perspective will be to study the possibility of using algebraic techniques for solving the decoding problem in the Lee metric.
Data availability
Not applicable.
Code availability
Not applicable.
Notes
Note that Euclidean rings in Magma also contain rings with zero divisors like Galois rings.
The Hamming weight of (\textbf{x}) is the number of r such that ( x_{r}\ne 0).
The decoding problem in the Hamming metric can be defined as in Definition 5 using the Hamming metric weight instead of the rank weight.
Note that the decoding problem is equivalent to the syndrome decoding problem.
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The first author acknowledges the Swiss Government Excellence for the financial support under Grant ESKAS No. 2022.0689. The second author acknowledges the UNESCO-TWAS and the German Federal Ministry of Education and Research (BMBF) for the financial support under the SG-NAPI Grant Number 4500454079.
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Institute of mathematics, University of Neuchatel, Neuchâtel, Switzerland
Hermann Tchatchiem Kamche
Department of Computer Engineering, National Advanced School of Engineering of Yaoundé, Yaoundé, Cameroon
Hervé Talé Kalachi
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Kamche, H.T., Kalachi, H.T. Solving systems of algebraic equations over finite commutative rings and applications. AAECC (2024).
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Keywords
Finite commutative rings
Gröbner bases
MinRank problem
Rank decoding problem
Systems of algebraic equations
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94B05
94A60
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Abstract
1 Introduction
2 Preliminaries
3 Solving systems of algebraic equations over finite chain rings
4 Solving systems of algebraic equations over finite commutative local rings
5 MinRank problem over finite principal ideal rings
6 Rank decoding problems over finite principal ideal rings
7 Conclusion
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Agrawal, M., Saxena, N.: Automorphisms of finite rings and applications to complexity of problems. In: STACS 2005: 22nd Annual Symposium on Theoretical Aspects of Computer Science, Stuttgart, Germany, February 24–26, 2005. Proceedings 22, pp. 1–17. Springer (2005)
Bardet, M., Bertin, M.: Improvement of algebraic attacks for solving superdetermined minrank instances. In: Post-Quantum Cryptography: 13th International Workshop, PQCrypto 2022, Virtual Event, September 28–30, 2022, Proceedings, pp. 107–123. Springer (2022)
Bardet, M., Briaud, P., Bros, M., Gaborit, P., Neiger, V., Ruatta, O., Tillich, J.: An algebraic attack on rank metric code-based cryptosystems. In: A.Canteaut, Y.Ishai (eds.) Advances in Cryptology - EUROCRYPT, Lecture Notes in Computer Science, vol. 12107, pp. 64–93. Springer (2020)
Bardet, M., Briaud, P., Bros, M., Gaborit, P., Tillich, J.: Revisiting algebraic attacks on MinRank and on the rank decoding problem. Des. Codes Cryptogr. 91(11), 3671–3707 (2023)
ArticleMathSciNetGoogle Scholar
Bardet, M., Bros, M., Cabarcas, D., Gaborit, P., Perlner, R.A., Smith-Tone, D., Tillich, J., Verbel, J.A.: Improvements of algebraic attacks for solving the rank decoding and minrank problems. In: Advances in Cryptology - ASIACRYPT, Lecture Notes in Computer Science, vol. 12491, pp. 507–536. Springer (2020)
Barg, S.: Some new NP-complete coding problems. Problemy Peredachi Informatsii 30(3), 23–28 (1994)
MathSciNetGoogle Scholar
Behboodi, M., Beyranvand, R., Hashemi, A., Khabazian, H.: Classification of finite rings: theory and algorithm. Czechoslov. Math. J. 64, 641–658 (2014)
ArticleMathSciNetGoogle Scholar
Berthomieu, J., Lecerf, G., Quintin, G.: Polynomial root finding over local rings and application to error correcting codes. Appl. Algebra Eng. Commun. Comput. 24(6), 413–443 (2013)
ArticleMathSciNetGoogle Scholar
Bosma, W., Cannon, J., Playoust, C.: The Magma algebra system. I. The user language. J. Symb. Comput. 24(3–4), 235–265 (1997)
ArticleMathSciNetGoogle Scholar
Bruns, W., Vetter, U.: Determinantal rings, Lecture Notes in Mathematics, vol. 1327. Springer (2006)
Buchberger, B.: Ein Algorithmus zum Auffinden der Basiselemente des Restklassenrings nach einem nulldimensionalen Polynomideal. Universitat Innsbruck, Austria, Ph. D. Thesis (1965)
Buchberger, B.: Gröbner Bases: an Algorithmic Method in Polynomial Ideal Theory. Recent trends in multidimensional systems theory. Reidel Publishing Company, Dordrecht (1985)
Google Scholar
Bulyovszky, B., Horváth, G.: Polynomial functions over finite commutative rings. Theoret. Comput. Sci. 703, 76–86 (2017)
ArticleMathSciNetGoogle Scholar
Caminata, A., Gorla, E.: Solving degree, last fall degree, and related invariants. J. Symb. Comput. 114, 322–335 (2023)
ArticleMathSciNetGoogle Scholar
Caruso, X., Roe, D., Vaccon, T.: p-adic stability in linear algebra. In: Proceedings of the 2015 ACM on International Symposium on Symbolic and Algebraic Computation, pp. 101–108. Association for Computing Machinery, New York (2015)
Courtois, N.T.: Efficient zero-knowledge authentication based on a linear algebra problem MinRank. In: Advances in Cryptology—ASIACRYPT 2001, pp. 402–421. Springer, Berlin (2001)
Courtois, N.T., Klimov, A., Patarin, J., Shamir, A.: Efficient algorithms for solving overdefined systems of multivariate polynomial equations. In: B.Preneel (ed.) Advances in Cryptology—EUROCRYPT 2000, International Conference on the Theory and Application of Cryptographic Techniques, Bruges, Belgium, May 14-18, 2000, Proceeding, Lecture Notes in Computer Science, vol. 1807, pp. 392–407. Springer (2000)
Dougherty, S.T., Kim, J.L., Kulosman, H.: MDS codes over finite principal ideal rings. Des. Codes Crypt. 50(1), 77 (2009)
ArticleMathSciNetGoogle Scholar
Fan, Y., Ling, S., Liu, H.: Matrix product codes over finite commutative Frobenius rings. Des. Codes Crypt. 71(2), 201–227 (2014)
ArticleMathSciNetGoogle Scholar
Faugère, J.-C.: A new efficient algorithm for computing Gröbner bases (F4). J. Pure Appl. Algebra 139(1–3), 61–88 (1999)
ArticleMathSciNetGoogle Scholar
Faugère, J.-C.: A new efficient algorithm for computing Gröbner bases without reduction to zero (F5). In: Proceedings of the 2002 International Symposium on Symbolic and Algebraic Computation, pp. 75–83. Association for Computing Machinery (2002)
Faugère, J.-C., Safey El Din, M., Spaenlehauer, P.J.: Computing loci of rank defects of linear matrices using Gröbner bases and applications to cryptology. In: Proceedings of the 2010 International Symposium on Symbolic and Algebraic Computation, pp. 257–264. Association for Computing Machinery, New York, United States (2010)
Faugère, J.-C., Levy-dit Vehel, F., Perret, L.: Cryptanalysis of minrank. In: Advances in Cryptology–CRYPTO 2008: 28th Annual International Cryptology Conference, Santa Barbara, CA, USA, August 17-21, 2008. Proceedings 28, pp. 280–296. Springer (2008)
Felix, F.: Elliptic curves over rings with a point of view on cryptography and factoring. Ph.D. thesis, Carl von Ossietzky-Universität Oldenburg (2005).
Gaborit, P., Ruatta, O., Schrek, J.: On the complexity of the rank syndrome decoding problem. IEEE Trans. Inf. Theory 62(2), 1006–1019 (2016)
ArticleMathSciNetGoogle Scholar
Gianni, P., Mora, T.: Algebraic solution of systems of polynomial equations using Groebner bases. In: Applied Algebra, Algebraic Algorithms and Error-Correcting Codes: 5th International Conference, AAECC-5 Menorca, Spain, June 15–19, 1987 Proceedings 5, pp. 247–257. Springer (1989)
Gorla, E., Ravagnani, A.: An algebraic framework for end-to-end physical-layer network coding. IEEE Trans. Inf. Theory 64(6), 4480–4495 (2017)
ArticleMathSciNetGoogle Scholar
Hashemi, A., Alvandi, P.: Applying Buchberger’s criteria for computing Gröbner bases over finite-chain rings. J. Algebra Appl. 12(07), 1350034 (2013)
ArticleMathSciNetGoogle Scholar
Kalachi, H.T., Kamche, H.T.: On the rank decoding problem over finite principal ideal rings. Adv. Math. Commun. (2023)
Kamche, H.T., Kalachi, H.T., Djomou, F.R.K., Fouotsa, E.: Low-rank parity-check codes over finite commutative rings. Applicable Algebra Eng. Commun. Comput. 10, 1–27 (2024)
Google Scholar
Kamche, H.T., Mouaha, C.: Rank-metric codes over finite principal ideal rings and applications. IEEE Trans. Inf. Theory 65(12), 7718–7735 (2019)
ArticleMathSciNetGoogle Scholar
Kamwa Djomou, F.R., Kalachi, H.T., Fouotsa, E.: Generalization of low rank parity-check (LRPC) codes over the ring of integers modulo a positive integer. Arab. J. Math. 10(2), 357–366 (2021)
ArticleMathSciNetGoogle Scholar
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187672 | https://escholarship.org/uc/item/2kw0r6bh | Dermatitis herpetiformis: Potential for confusion with linear IgA bullous dermatosis on direct immunofluorescence
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Dermatitis herpetiformis: Potential for confusion with linear IgA bullous dermatosis on direct immunofluorescence
2008
Van, Livia;
Browning, John C;
Krishnan, Ravi S;
Kenner-Bell, Brandi M;
Hsu, Sylvia
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Dermatitis herpetiformis: Potential for confusion with linear IgA bullous dermatosis on direct immunofluorescence
Livia Van MD, John C Browning MD, Ravi S Krishnan MD, Brandi M Kenner-Bell MD, Sylvia Hsu MD
Dermatology Online Journal 14 (1): 21
Department of Dermatology, Baylor College of Medicine, Houston, Texas
Dermatitis herpetiformis (DH) and linear IgA bullous dermatosis (LABD) are two distinct cutaneous bullous diseases that are traditionally differentiated by direct immunofluorescence (DIF). Classically, in DH there is granular IgA deposition along the dermoepidermal junction with concentration at the papillary tips. In contrast, LABD has linear IgA deposition along the cutaneous basement membrane zone [1, 2]. However, DIF alone offers little guidance in ambiguous cases, particularly when IgA deposition appears both granular and linear at the same time. In our experience, DH does not always have the classic granular deposition of IgA, but can be more linear, like LABD (Fig. 1). We recently saw a patient with clinical DH but the DIF was interpreted by the dermatopathologist as LABD, which helps to illustrate our point.
Clinical synopsis
A 35-year-old man presented to his dermatologist with a 6-month history of intensely pruritic vesicles on his face, posterior neck, elbows, and buttocks. He denied any gastrointestinal symptoms. His dermatologist informed him that her clinical impression was DH and performed two 4-mm punch biopsies, one for histopathology and one for DIF. The histopathology revealed a subepidermal vesicle with neutrophils and the DIF was interpreted as LABD by a dermatopathologist. With these results, his dermatologist informed the patient that he could eat all the gluten he wanted and started him on dapsone. The patient continued to develop new blisters despite being on dapsone 100 mg daily for several months. He was then referred to one of us (SH).
We performed another DIF, which was interpreted as DH by the same dermatolopathologist and tissue transglutaminase antibody testing was positive at 117 U (negative < 20 U, weak positive 20-30 U, strong positive > 30 U). The patient was then referred to a gastroenterologist who confirmed the diagnosis of celiac disease.
Discussion
Figure 1
Figure 1. Direct immunofluorescence of skin specimen taken from a patient with DH showing granular-linear deposition of IgA along the dermoepidermal junction. (Courtesy of Robert E Jordon MD)
While both DH and LABD skin lesions show dramatic improvement with dapsone or sulfapyridine treatment, initiating drug treatment without correctly distinguishing between DH and LABD can result in negative consequences for the patient. For instance, a patient with DH may not be started on a gluten-free diet and would be at risk for intestinal lymphoma . Dermatitis herpetiformis patients should also be screened for thyroid disease, a co-morbid condition. A patient with LABD may be maintained on an offending medication that could easily be discontinued, reversing the manifestations of the disease.
Dermatitis herpetiformis is characterized by severely pruritic papules and vesicles, frequently presenting as erosions secondary to scratching, in a herpetiform configuration on symmetric extensor surfaces of the elbows, knees, shoulders, buttocks, and posterior neck. The cutaneous manifestation of gluten sensitive enteropathy (GSE), DH has an estimated incidence of 1 case per 10,000 people . Most DH patients have intestinal histopathological changes even though only a minority will experience overt gastrointestinal symptoms [1, 4]. In fact, the severity of DH is inversely proportional to the severity of the intestinal symptoms (i.e., severe DH is associated with mild intestinal disease). Far rarer in occurrence, LABD is characterized by grouped pruritic papules and vesicles (and sometimes larger bullae that mimic bullous pemphigoid) that appear on symmetric extensor surfaces but do not necessarily favor the knees, elbows, buttocks, and neck. LABD patients more frequently present with mucosal lesions than do patients with DH . Both diseases demonstrate an increased incidence of patients with histocompatibility locus antigens B8, DR3 and DQ2 .
Patients with DH produce IgA antibodies against endomysium . A number of studies have since shown anti-endomysial antibody to be a highly sensitive (90%) and specific (96%) marker for detecting untreated DH and celiac disease [1, 6]. Another useful clinical marker in DH is anti-tissue transglutaminase IgA antibody. Tissue translutaminase (tTG) is the major target antigen of the anti-endomysial antibodies. Enzyme-linked immunosorbent assay (ELISA) for IgA antibodies against tTG yield a sensitivity and specificity comparable to EMA testing .
Anti-tTG IgA levels correlate with anti-endomysial antibodies and tTG-ELISA has the added benefit of lower cost ($100 instead of $130), faster results (1 instead of 2 days) and no need for an animal substrate [1, 7, 8]. As an investigator-independent, easy to use, noninvasive, and reproducible method, some experts have suggested tTG-ELISA should replace DIF as a first step in DH diagnosis [7, 8, 9].
As our case and others demonstrate [10, 11], one should question DIF results that conflict with clinical findings and supplement such cases with serologic testing. Dermatologists should not mistakenly change their clinical diagnosis of DH to LABD if DIF studies are read as consistent with LABD, because DH may appear linear under DIF.
References
1. Nicolas ME, Krause PK, Gibson LE, Murray JA. Review: Dermatitis herpetiformis. Int J Dermatol 2003; 42: 588-600.
2. Hall RP. "Linear IgA Dermatosis and Chronic Bullous Disease of Childhood." Fitzpatrick's Dermatology in General Medicine, 5th ed. 1999; 680-685. McGraw-Hill.
3. Bickle KM, Roark RM, Hsu S. Autoimmune Bullous Dermatoses: A Review. Am Family Physician 2002; 65(9): 1861-1870.
4. Heil PM, Volc-Platzer B, Karlhofer F, et al. Transglutaminases as diagnostically relevant autoantigens in patients with gluten sensitivity. J Dtsch Dermatol Ges 2005; 3:974-978.
5. Collier PM, Wojnarowska F, Welsh K, et al. Adult linear IgA disease and chronic bullous disease of childhood: the association with human lymphocyte antigens Cw7, B8, DR3 and TNF influences disease expression. Brit J Dermatol 1999; 141: 867-875.
6. Chorzelski TP, Beutner EH, Sulej J, et al. IgA antiendomysium antibody. A new immunological marker of dermatitis herpetiformis and coeliac disease. Brit J Dermatol 1984; 111(4): 395-402.
7. Dieterich W, Laag E, Bruckner-Tuderman L, et al. Antibodies to Tissue Transglutaminase as Serologic Markers in Paitents with Dermatitis Herpetiformis. J Invest Dermatol 1999; 13(1): 133-136.
8. Rose C, Dieterich W, Brocker EB, et al. Circulating autoantibodies to tissue transglutaminase differentiate patients with dermatitis herpetiformis from those with linear IgA disease. J Am Acad Dermatol 1999; 41(6): 957-961.
9. Desai AM, Krishnan RS, Hsu S. Medical Pearl: Using tissue transglutaminase antibodies to diagnose dermatitis herpetiformis. J Am Acad Dermatol 2005; 53:867-869.
10. Beutner EH, Baughman RD, et al. A case of dermatitis herpetiformis with IgA endomysial antibodies but negative direct immunofluorescent findings. J Am Acad Dermatol 2000; 43(2): 329-332.
11. Sousa L, Bajanca R, Cabral J, Fiadeiro T. Dermatitis Herpetiformis: Should Direct Immunofluorescence Be the only diagnostic Criterion? Pedi Dermatol 2002; 19(4): 336-339.
© 2008 Dermatology Online Journal
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187673 | https://math.stackexchange.com/questions/2834495/proving-x2x1-gt0 | Stack Exchange Network
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Proving $x^2+x+1\gt0$
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$\begingroup$
I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.
My methods:
$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
Let it be $0$ for some $x=k$. Then $x^2+x+1=0$ has a real solution. But since $1^2\not\gt4$, this has no real solution. Therefore it is more than $0$.
algebra-precalculus
inequality
alternative-proof
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edited Jun 28, 2018 at 11:32
JMP
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asked Jun 28, 2018 at 7:44
DynamoBlazeDynamoBlaze
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3
$\begingroup$ $15$ answers, is this a record ? $\endgroup$
Peter
– Peter
2018-06-28 11:12:08 +00:00
Commented Jun 28, 2018 at 11:12
$\begingroup$ @Peter Probably not. $\endgroup$
DynamoBlaze
– DynamoBlaze
2018-06-28 11:14:26 +00:00
Commented Jun 28, 2018 at 11:14
3
$\begingroup$ note also that completing the square gives us both the local minima ($-\frac12$), and the corresponding $y$-value ($\frac34$). $\endgroup$
JMP
– JMP
2018-06-29 12:25:30 +00:00
Commented Jun 29, 2018 at 12:25
Add a comment |
21 Answers 21
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13
$\begingroup$
We note that:
$$x^2+x+1=\frac{x^3-1}{x-1}$$
and the sign of the RHS numerator and denominator are always equal, except for $x=1$.
We handle $x=1$ separately, but this is trivial on the LHS.
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answered Jun 28, 2018 at 8:27
JMPJMP
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2
$\begingroup$ Nice! - This could also be interpreted as the slope of the line from $(1, f(1))$ to $(x, f(x))$ for the increasing function $f(x) = x^3$. $\endgroup$
Martin R
– Martin R
2018-06-28 08:41:56 +00:00
Commented Jun 28, 2018 at 8:41
$\begingroup$ The third proof of this inequality that I have enjoyed immensely so far. It is noteworthy that these three involve either factoring some expression or completing some square after a clever scaling. $\endgroup$
Allawonder
– Allawonder
2018-06-28 10:03:05 +00:00
Commented Jun 28, 2018 at 10:03
Add a comment |
6
$\begingroup$
Short trivial proof:
Since this is a quadratic equation, and the leading coefficient is $+1$, we have
$$\Delta < 0$$
Whence the equation is always strictly positive (that is, it's always $>0$).
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edited Jun 28, 2018 at 8:01
answered Jun 28, 2018 at 7:49
user266764user266764
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2
4
$\begingroup$ I think it's worth to note that we need the leading coefficient to be positive too, even if it's trivial. $\endgroup$
Botond
– Botond
2018-06-28 07:56:16 +00:00
Commented Jun 28, 2018 at 7:56
$\begingroup$ @Botond You're right! $\endgroup$
user266764
– user266764
2018-06-28 08:00:57 +00:00
Commented Jun 28, 2018 at 8:00
Add a comment |
6
$\begingroup$
There is also the following way.
For $x\geq-1$ we obtain $$x^2+x+1=x^2+(x+1)>0$$ and for $x<-1$ we obtain $$x^2+x+1=x(x+1)+1>0+1>0.$$
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answered Jun 28, 2018 at 9:05
Michael RozenbergMichael Rozenberg
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4
$\begingroup$ This is also one of the most elegant proofs in reply to this OP. However, that ugly trick with the $0$ on the LHS of your last inequality is not needed. It suffices to note that in this case $x+1<0.$ $\endgroup$
Allawonder
– Allawonder
2018-06-28 10:44:39 +00:00
Commented Jun 28, 2018 at 10:44
$\begingroup$ @Allawonder The last inequality is true for $x<-1$. See better my post. $\endgroup$
Michael Rozenberg
– Michael Rozenberg
2018-06-28 10:46:17 +00:00
Commented Jun 28, 2018 at 10:46
$\begingroup$ Sorry. That was not what I meant. I have edited my comment. See above again. $\endgroup$
Allawonder
– Allawonder
2018-06-28 10:51:09 +00:00
Commented Jun 28, 2018 at 10:51
$\begingroup$ @Allawonder Yes, in this case $x(x+1)>0$, which gives $x(x+1)+1>0.$ If you wish you can fix my post. $\endgroup$
Michael Rozenberg
– Michael Rozenberg
2018-06-28 10:55:02 +00:00
Commented Jun 28, 2018 at 10:55
Add a comment |
6
$\begingroup$
If $x$ is positive, then $x^2+x+1$ is clearly positive.
If $x$ is negative then $x^2-x+1$ is certainly positive. Now $$(x^2+x+1)(x^2-x+1)=x^4+x^2+1$$ is certainly positive, so $x^2+x+1$ must also be positive in this case.
If $x$ is zero then $x^2+x+1=1\gt 0$
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answered Jun 28, 2018 at 9:23
Mark BennetMark Bennet
102k1414 gold badges119119 silver badges232232 bronze badges
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2
2
$\begingroup$ This is a particularly pleasant proof. But from the product, you could simply have used the fact that if $ab>0$, then either $a,b>0$ or $a,b<0$ to conclude that in either case (whether $x>0$ or $x<0$) one of the factors is always positive. This settles it very elegantly. Indeed, that was how I interpreted it. PS. Of course, it is positive when $x=0.$ $\endgroup$
Allawonder
– Allawonder
2018-06-28 09:49:26 +00:00
Commented Jun 28, 2018 at 9:49
$\begingroup$ @Allawonder Indeed - good observation - that avoids cases. $\endgroup$
Mark Bennet
– Mark Bennet
2018-06-28 09:51:41 +00:00
Commented Jun 28, 2018 at 9:51
Add a comment |
4
$\begingroup$
Here a rather geometric way: $$y = x^2+x+1 = x(x+1) + 1$$ So, $y = x^2+x+1$ is the parabola $y=x(x+1)$ shiftet by $1$ upwards.
$y=x(x+1)$ has its vertex at $x_V = -\frac{1}{2} \Rightarrow y_V = -\frac{1}{4}$
So, the vertex of $y= x(x+1) + 1$ is also at $x_V = -\frac{1}{2}$ with a minimum value of $y_{min}= -\frac{1}{4}+1 = \frac{3}{4}>0$
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answered Jun 28, 2018 at 7:56
trancelocationtrancelocation
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4
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I don't know why you need multiple proofs for a simple result, but here are two overkilling solutions. The first one utilizes some knowledge in linear algebra. The second one uses Euclidean geometry along with some trigonometry.
Consider the matrix $\mathbf{B}:=\begin{bmatrix}1&\frac12\\frac12&1\end{bmatrix}$. Being a real symmetric $2$-by-$2$ matrix, $\mathbf{B}$ has two real eigenvalues, which are $\frac{3}{2}$ and $\frac{1}{2}$. As both eigenvalues are positive, $\mathbf{B}$ is a positive-definite matrix, whence it induces a positive-definite symmetric bilinear form $\langle\,\\rangle:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}$ sending a pair $(\mathbf{u},\mathbf{v})$ of $2$-by-$1$ column vectors in $\mathbb{R}^2$ to $$\langle \mathbf{u},\mathbf{v}\rangle:=\mathbf{u}^\top\,\mathbf{B}\,\mathbf{v}\,.$$ That is, $$\langle \mathbf{u},\mathbf{u}\rangle \geq 0\text{ for all }\mathbf{u}\in\mathbb{R}^2\,,$$ and the inequality becomes an equality iff $\mathbf{u}$ is the zero vector. In particular, when $\mathbf{u}=(x,1)$, where $x$ is an arbitrary real number, we get $\mathbf{u}\neq \boldsymbol{0}$, whence $$x^2+x+1=\langle\mathbf{u},\mathbf{u}\rangle>0\,,$$ as desired.
Alternatively, consider three points in $\mathbb{R}^2$: the origin $O=(0,0)$, the point $A=(1,0)$, and the point $B=\left(-\frac{x}{2},\frac{\sqrt{3}x}{2}\right)$. Note that $\angle AOB=\frac{2\pi}{3}$ for $x>0$, and $\angle {AOB}=\frac{\pi}{3}$ for $x<0$. Using the Law of Cosine, you get $$AB^2=x^2+x+1\,,$$ whence $x^2+x+1>0$, noting that $A\neq B$ for any value of $x$. (The case $x=0$ can be checked separately, but then $x^2+x+1=AB^2=1>0$ still holds.)
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edited Jun 28, 2018 at 8:21
answered Jun 28, 2018 at 7:55
BatominovskiBatominovski
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The desired inequality is a convex combination of two (weak) inequalities:
$$(x+1)^2 = x^2 + 2x + 1 \ge 0, \quad \text{equality iff }\; x = -1;$$ $$(x-1)^2 = x^2 - 2x + 1 \ge 0, \quad \text{equality iff }\; x = 1.$$
Now multiply the first inequality by $3/4$, multiply the second inequality by $1/4$, and add the two resulting inequalities. We get
$$x^2 +x + 1 > 0.$$
Incidentally, the inequality $x^2 + cax + a^2 \ge 0$ holds for every $c \in [-2, 2]$ and $a \in \mathbb R$, for the same reason.
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edited Jul 10, 2018 at 21:23
answered Jul 10, 2018 at 21:15
Yakov ShklarovYakov Shklarov
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$$x^2+x+1\geq x^2-2|x|+1=(|x|-1)^2\geq 0$$
(and first inequality is strict for $|x|=1$)
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answered Jun 28, 2018 at 9:39
user126154user126154
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Claim $$x^2+x+1>0$$
Proof
It is equivalent to prove by multiplying both sides by $x-1$
\begin{cases} x^3-1<0\iff x^3<1\iff x<1, & \text{if $x<1$} \ x^3-1>0\iff x^3>1 \iff x>1, & \text{if $x>1$} \ x^2+x+1=3>0, &\text{if $x=1$} \end{cases}
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edited Jun 28, 2018 at 10:15
answered Jun 28, 2018 at 8:31
MythomorphicMythomorphic
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$\begingroup$ Very nice solution $\endgroup$
user567182
– user567182
2018-06-28 10:09:48 +00:00
Commented Jun 28, 2018 at 10:09
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One method is to find the vertex. The x-coordinate of the vertex must be equal to -b/(2a) = -1/2. Plugging this back into the function, we get that the vertex is equal to (-1/2, 3/4).
Now that we know the vertex's y-coordinate is greater than zero, and that the parabola must be pointed up (a>0), to yield a conclusion that the parabola must always be positive. The minimum value has a y-value greater than zero, and all other y-values on the function must also be greater than zero.
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answered Jun 28, 2018 at 20:00
RayDanshRayDansh
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Actually your first method is efficient enough, but if you want more here you go. Let $$f(x) = x^2+x+1$$ One has $f(-\frac 1 2)>0$. Moreover $f'(x) = 2x+1\geq 0$ for all $x\in[-\frac 1 2,\infty)$. Hence $f(x) >0$ for all $x\in[-\frac 1 2,\infty)$. Since $f$ is symmetric around $x=- \frac 1 2$, we conclude $f(x) >0$ for all $x\in\mathbb R$.
Remark. This method might work for proving inequalities for general differentiable $f$, however in this case it is just an overkill.
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answered Jun 28, 2018 at 7:51
ShashiShashi
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Using the inequality between arithmetic and geometric mean: $$ (x^2 + 1) + x \ge 2\sqrt{x^2 \cdot 1} + x = 2 |x| + x \ge |x| \ge 0. $$ Equality cannot hold because $x^2 =1 $ and $x = 0$ are not simultaneously true.
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edited Jun 28, 2018 at 8:28
answered Jun 28, 2018 at 8:17
Martin RMartin R
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$\begingroup$ From what I know of the AM-GM, it should be $x^2+x+1\geq3x$. $\endgroup$
DynamoBlaze
– DynamoBlaze
2018-06-28 08:19:00 +00:00
Commented Jun 28, 2018 at 8:19
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$\begingroup$ @MalayTheDynamo: AM-GM can only be applied to non-negative numbers, so that would work only for $x \ge 0$. $\endgroup$
Martin R
– Martin R
2018-06-28 08:21:01 +00:00
Commented Jun 28, 2018 at 8:21
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Correct me if wrong:
Obvious for $x\ge 0.$
For $x <0$ consider $y=-x$, and
$y^2-y +1$ for $y>0$.
Hence: for $y>0$:
$y^2-2y + 1 +y =(y-1)^2 +y >0$
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answered Jun 28, 2018 at 8:50
Peter SzilasPeter Szilas
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You can complete the square two ways - if we multiply by $4$ first we get: $$4(x^2+x+1)=(2x+1)^2+3=3x^2+(x+2)^2$$
The second one of these requires an extra step to note that it is never zero - $x^2$ and $(x+2)^2$ can never both be zero at the same time.
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answered Jun 28, 2018 at 9:20
Mark BennetMark Bennet
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$\begingroup$ Short and straightforward. You did not even need to expand. Completing the square after multiplying by $4$ does it in one stroke. Great! $\endgroup$
Allawonder
– Allawonder
2018-06-28 09:58:15 +00:00
Commented Jun 28, 2018 at 9:58
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Let $f(x)=x^2+x+1$.
Then:
$$\lim_\limits{x\to\pm\infty}f(x)=\infty$$
Also the (now) minima is at $f'(x)=0$, i.e. $2x+1=0$, i.e. at $x=-\frac12$.
As $f(-\frac12)=\frac34\gt0$, we have $f(x)\gt0\;\;\forall x\in\mathbb{R}$.
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answered Jun 29, 2018 at 11:44
JMPJMP
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Note that $\mathrm{x}^2\ge 0$ and $1\gt 0.$
Case 1. If $x\ge 0$ then $\mathrm{x}^2+x+1\gt 0.$
Case 2. Assume $x\lt 0.$
Subcase 2a: If $x\gt-1$ then $x+1\gt 0.$ And thus $\mathrm{x}^2+x+1\ge x+1\gt 0.$
Subcase 2b: If $x\le-1$ then multiplying by x yields $\mathrm{x}^2\ge-x$ implying $\mathrm{x}^2+x\ge 0$. Thus $\mathrm{x}^2+x+1\gt \mathrm{x}^2+x\ge 0.$
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answered Dec 4, 2019 at 16:17
NaiveLogicianNaiveLogician
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Assume to the contary that $$f\left(x\right)=x^2+x+1=-m<0$$ for some $m>0$
$$x^2+x+1+m=0$$
$$x^2+x+\left(1+m\right)=0$$
Then we have $$\Delta=b^2-4ac=1-4\left(1+m\right)$$
$$=-3-3m=-3\left(1+m\right)<0$$
Hence there exist no such $x$ for which $$ f\left(x\right)=x^2+x+1$$ take negative values
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answered Jun 28, 2018 at 8:31
user567182user567182
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Case 1: $x\ge 0 \Rightarrow x^2+x+1\ge 0^2+0+1>0$.
Case 2: $x<0 \Rightarrow x^2+x+1>0 \iff \frac{x^2+x+1}{x}<0 \iff x+\frac 1x+1\overbrace{\le}^{AM-GM} -2+1<0$.
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answered Jun 28, 2018 at 12:01
farruhotafarruhota
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I am adding this answer, which though is similar to, but may be better than, my previous one.
Let $f(x)=x^2+x+1$ define a function $f:\mathbf R\to \mathbf R.$ Then $f'(x)=2x+1.$ Thus $f''(x)=2>0\,\,\forall x.$ Therefore $f$ is convex and has its minimum value at $x=-1/2.$ Because of this, it decreases with $x$ in $(-\infty,-1/2)$ and increases with $x$ in $[-1/2,\infty).$ Adding to this the fact that $f(-1/2)=3/4>0,$ we deduce that $f(x)>0$ for every real $x$ since $f$ is everywhere continuous -- and in particular at $x=-1/2.$
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answered Jun 29, 2018 at 11:40
AllawonderAllawonder
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Consider proving:
If $x$ is positive, because the LHS is always positive, this is true.
If $x$ is negative, make the transform $X=-x$, and so we have:
$$X^2+1\gt X$$
Dividing by the (positive) $X$ gives:
$$X+\frac1X\gt 1$$
which is true for all $X$ regardless of greater than/less than $1$, due to the reciprocal (in fact the inequality is greater than or equal to 2).
In fact, this last fact can be used to prove straight from:
with a negative $x$.
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edited Jun 29, 2018 at 12:03
answered Jun 29, 2018 at 11:55
JMPJMP
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Put $x=\dfrac{b}{a}$
$x^2+x+1=\dfrac{b^2}{a^2}+\dfrac{b}{a}+1$ $=\dfrac{b^2+ab+a^2}{a^2}=\dfrac{\dfrac{1}{2}(a+b)^2+\dfrac{b^2}{2}}{a^2}+\dfrac{1}{2}> 0$
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answered Jul 22, 2018 at 18:31
Takahiro WakiTakahiro Waki
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187674 | https://blog.csdn.net/qq_38452951/article/details/84982140 | 计算已知数据的最高分，最低分和平均分_已知一行数的最大值,最小值,平均值。给最大值、最小值、平均值赋得分后,求其他分-CSDN博客
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计算已知数据的最高分，最低分和平均分
最新推荐文章于 2023-11-25 15:00:29 发布
心系五道口于 2018-12-13 09:15:36 发布
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本文介绍了一种使用Python编程语言处理列表数据的方法，包括如何找到列表中的最大值、最小值以及计算平均值。通过具体的代码示例，展示了如何利用Python内置函数max()、min()和sum()来实现这些功能。
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scores=[12,23,434,56]print("最高分",max(scores))print("最低分",min(scores))print("平均分",sum(scores)/len(scores))
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C语言输入学生成绩，计算 并输出这些学生的 最低分、最高分、平均分。
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输入学生人 数 n,再输入n个学生的成绩，将所有成绩累加，再除以n，得到 平均分；将第一个学生的成绩 赋 给最大 最小值，用第i个 分 数 与 最大值 比较，若第i个 分 数>最大值，将第i个 分 数 赋 值给 最大值，同理得 最小值，最后输出 平均值，最大值，最小值。
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m0_69807456的博客
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🌟 前言 今天是C语言每日一练，第1天！ 也是我创作的第一篇博客，给大家带来一道c语言题目的思路 分 析 希望大家多多关注！ 1、问题描述 题目传送门 公务员面试现场打 分。有7位考官，从键盘输入若干组成绩，每组7个 分 数（百 分 制），去掉一个 最高分 和 一个 最低分，输出每组的平均成绩。 （注：本题有多组输入） 输入描述： 每 一行，输入7个整 数（0~100），代表7个成绩，用空格 分 隔。 输出描述： 每 一行，输出去掉 最高分 和 最低分 的平均成绩，小 数 点后保留2位，每行输出后换行 2、题目 分 析 本题归根结.
【C语言】5个成绩，去掉 最高分，去掉 最低分，求 平均分 | C语言面试：解释内存对齐（Memory Alignment）的概念 和 目的。| 为何要进行内存对齐？内存对齐的规则，影响内存对齐的因素，实例。
追光者♂：记录、分享、总结、提升，现象级专栏《Python从入门到人工智能》作者，无惧黑暗，坚信曙光
03-18 9178
代码如下 //5个成绩，去掉 最高分，去掉 最低分，求 平均分 #include int main() { int min,max,average,sum,i; int score; printf("请输入成绩："); for(i=0;i<5;i++) { scanf("%d",&score[i]); } printf("\n"); min=score; max=score; sum=0; for(i=0;i&l
用 python 计算 一组 数据 的 最高分 最低分 和 平均分
江西理工大学20级计算机应用技术研究生
12-06 3万+
scores=[57,23,45,34,434,3434,33,43] min_score=100 max_score=0 sum_score=0 for i in scores: if i<min_score: min_score=i if i>max_score: max_score=i sum_score+=i ...
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187675 | https://www.youtube.com/watch?v=OFblI1Z-jo4 | Chasles' theorem (kinematics)
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In kinematics, Chasles' theorem, or Mozzi–Chasles' theorem, says that the most general rigid body displacement can be produced by a screw displacement. A direct Euclidean isometry in three dimensions involves a translation and a rotation. The screw displacement representation of the isometry decomposes the translation into two components, one parallel to the axis of the rotation associated with the isometry and the other component perpendicular to that axis. The Chasles theorem states that the axis of rotation can be selected to provide the second component of the original translation as a result of the rotation. This theorem in three dimensions extends a similar representation of planar isometries as rotation. Once the screw axis is selected, the screw displacement rotates about it and a translation parallel to the axis is included in the screw displacement.
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I'm delighted to have you as part of this community welcome welcome welcome greetings explorers of knowledge join us as we embark on a quest to understand chel theorem kinematics in kinematics chel theorem or moic chel theorem says that the most General rigid body displacement can be produced by a screw displacement a direct Iden as symmetry in three dimensions involves a translation and a rotation the screw displacement representation of the Symmetry decomposes the translation into two components one parallel to the axis of the rotation associated with the Symmetry and the other component perpendicular to that axis the chel theorem states that the axis of rotation can be selected to provide the second component of the original translation as a result of the rotation this theorem in three dimensions extends a similar representation of Planet symmetries as rotation once the screw axis is selected the screw displacement rotates about it and a translation parallel to the axis is included in the screw displacement get ready to immerse yourself in the world of planer isometries with complex numbers as we examine its impact and relevance Iden geometry is expressed in the complex plane by points PX where I S is rotations result from multiplications by cost note that a rotation about complex Point p is obtained by complex arithmetic width Z's PPS where the last expression shows the mapping equivalent to rotation at zero and a translation therefore given direct symmetri Z one can solve but obtain p as the center for an equivalent rotation provided that one that is provided the direct isymmetry is not a pure translation as stated by cug a direct asymmetry is either a rotation or a translation get ready for an enlightening exploration as we dig into history and understand its role in the broader context the proof that a spatial displacement can be decomposed into a rotation and slide around and along a line is attributed to the astronomer and mathematician Julia Moy 1763 in fact the screw axis is traditionally called ASD Moy in Italy however most textbooks refer to subsequent similar work by Michelle charel dating from a 1930 several other contemporaries of M Chels obtained the same or similar results around that time including gigagen COI poinset poison and Rodriguez an account of the 1763 proof by Julio miy and some of its history can be found here the time has come to unravel the secrets behind proof and gain a deeper understanding Moy considers a rigid body undergoing first iritation about an AC is passing through the center of mass and then a translation of displacement D in an arbitrary Direction any rigid motion can be accomplished in this way due to theorem by oil on the existence of an axis of rotation the displacement de of the center of mass can be decomposed into components parallel and perpendicular to the axis the perpendicular and parallel component acts on all points of the rigid body but Moy shows that for some points the previous rotation acted EXA acting with an opposite displacement so those points are translated parallel to the axis of rotation these points lie on the Moi axis through which the rigid motion can be accomplished through a screw motion another Elementary proof of Moi chasle theorem was given by East Tia in 1904 suppose it is to be transformed into bwia suggests that line a k be selected parallel to the axis of the given rotation with Kelvin the foot of a pendicular from be the appropriate screw displacement is about an axis parallel to AK such that Kelvin is moved to be in Whit's terms a rotation about any axis is equivalent to a rotation through the same angle about any axis parallel to it together with the simple transation in a direction perpendicular to the axis get ready to uncover the Mysteries surrounding calculation as we navigate its intriguing terrain the calculation of the commuting translation and rotation from a screw motion can be performed using ia3 01 the geometric algebra of the Iden space it has three ID and basis vectors AI satisfying ai1 representing orthogonal ples through the origin and one Gras thean basis Vector is satisfying e z to represent the plane at Infinity any plane a distance from the origin can then be formed as a linear combination the sum i a i Delta Y is normalized such that a one because Reflections can be represented by the plane in which the reflection occurs the product of two planes and B is the be flection AB the result is a rotation around their intersection line a b which could also lie on the plane at Infinity when the two Reflections are parallel in which case the B flection AB is a translation a screw motion South is the product of for non colino Reflections and thus esed but according to the Moy Chels there a mccrew motion can be decomposed into a commuting translation te one of which is the axis of translation satisfying zero and rotation e corsin is the axis of rotation satisfying one the two B Vector lines and are orthogonal and commuting to find T and R from s we simply write out South and consider the result grade by grade a beging line sou alpab B to be under brace bet Tex under BRAC B2 shus bet b Tex Vector under BRAC for B1 B Tex quad vector and liid because the quadri vector part is it and zero T is directly found to bet equal one Al fral langle strangle angle strangle and th equal South t equal T South fras for a given screw motion South the commuting transation and rotation can be found using the two form way above after which the lines and are found to be proportional to it and irrespectively in the upcoming section we'll be shining a light on other dimensions and Fields the chel theorem is a special case of the invariant decomposition don't forget to hit that notification Bell so you never miss an update from my channel |
187676 | https://s23.cs251.com/Text/11_Matchings_in_Graphs/media_upload/Matchings_in_Graphs.pdf | Matchings in Graphs 1 Maximum Matchings Definition (Matching – maximum, maximal, perfect). A matching in a graph G = (V, E) is a subset of the edges that do not share an endpoint. A maximum matching in G is a matching with the maximum number of edges among all possible matchings. A maximal matching is a matching with the property that if we add any other edge to the matching, it is no longer a matching.1 A perfect matching is a matching that covers all the vertices of the graph.
Example (Examples of matchings). Consider the following graph.
Note that the empty set and a set with only one edge is always a matching. The set M = {{v1, v5}, {v4, v7}} is a maximal matching with 2 edges, since we if we tried to add another edge to this set, it would no longer be a matching. On the other hand, this maximal matching is not a maximum matching because there is another matching with 3 edges: M ′ = {{v1, v6}, {v3, v5}, {v4, v7}}. This graph does not have a perfect matching.
One easy way to see this is that it has an odd number of vertices, and any graph with an odd number of vertices cannot have a perfect matching.
1Note that a maximal matching is not necessarily a maximum matching, but a maximum matching is always a maximal matching.
1 CMU CS251 Spring 2023 Note (Size of a matching). The size of a matching M refers to the number of edges in the matching, and is denoted by |M|. Note that this coincides with the size of the set that M represents.
Definition (Maximum matching problem). In the maximum matching problem the input is an undirected graph G = (V, E) and the output is a maximum matching in G.
Definition (Augmenting path). Let G = (V, E) be a graph and let M ⊆E be a matching in G. An augmenting path in G with respect to M is a path such that 1. the path is an alternating path, which means that the edges in the path alternate between being in M and not in M, 2. the first and last vertices in the path are not a part of the matching M.
Example (Augmenting path example 1). Consider a single edge {u, v} in a graph G such that u and v are not matched by a matching M (this means that vertices u and v do not belong to any of the edges in M). Then this edge forms an augmenting path.
Example (Augmenting path example 2). Consider the following graph.
Let M be the matching {{v1, v5}, {v3, v6}, {v4, v8}}. Then the path (v2, v5, v1, v7) is an augmenting path with respect to M.
Note (Edge cases for augmenting paths). An augmenting path does not need to contain all the edges in M. It is also possible that it does not contain any of the edges of M. A single edge {u, v}, where u is not matched and v is not matched, is an augmenting path.
Theorem (Characterization for maximum matchings). Let G = (V, E) be a graph. A match-ing M ⊆E is maximum if and only if there is no augmenting path in G with respect to M.
Proof. The statement we want to prove is equivalent to the following. Given a graph G = (V, E), a matching M ⊆E is not maximum if and only if there is an augmenting path in G with respect to M. There are two directions to prove.
First direction: Suppose there is an augmenting path in G with respect to M. Then we want to show that M is not maximum. Let the augmenting path be v1, v2, . . . , vk: The highlighted edges represent edges in M. By the definition of an augmenting path, we know that v1 and vk are not matched by M. Since v1 and vk are not matched and the path is alternating, the number of edges on this path that are in the matching is one less than the number of edges not in the matching. To see that M is not a maximum matching, observe that we can obtain a bigger matching by flipping the matched and unmatched edges on the augmenting path. In other words, if an edge on the path is in the matching, we remove it from the matching, and if an edge on the path is not in the matching, we put it in the matching. This gives us a matching larger than M, so M is not maximum.
2 CMU CS251 Spring 2023 Second direction: We now prove the other direction. In particular, we want to show that if M is not a maximum matching, then we can find an augmenting path in G with respect to M. Since M is not a maximum, there is a matching with larger size. Let M ∗be a matching such that |M ∗| > |M|. We define the set S to be the set of edges contained in M ∗or M, but not both. That is, S = (M ∗∪M)(M ∗∩M) (this is the symmetric difference of M ∗and M). If we color the edges in M with blue, and the edges in M ∗with red, then S consists of edges that are colored either blue or red, but not both (i.e. no purple edges).
Below is an example: (Horizontal edges correspond to the red edges. The rest is blue.) Our goal is to find an augmenting path with respect to M in S (i.e., with respect to the blue edges), and once we do this, the proof will be complete.
We now proceed to find an augmenting path with respect to M in S. To do so, we make a couple of important observations about S. First, notice that each vertex that is a part of S has degree 1 or 2 because it can be incident to at most one edge in M and at most one edge in M ∗. (If the degree was more than 2, either two edges from M or two edges from M ∗would be intersecting, but in a matching, edges cannot intersect.) We now make two claims: 1. Because every vertex has degree 1 or 2, S consists of disjoint paths and cycles (i.e.
each connected component is either a path or a cycle).
2. The edges in these paths and cycles alternate between blue and red.
The proof of the first claim is omitted and is left as an exercise for the reader. The second claim is true because if the edges were not alternating, i.e., if there were two red or two blue edges in a row, then this would imply the red edges or the blue edges do not form a matching (remember that in a matching no two edges can share an endpoint).
Since M ∗is a bigger matching than M, we know that S has more red edges than blue edges. Observe that the cycles in S must have even length, because otherwise the edges cannot alternate between blue and red. Therefore the cycles have an equal number of red and blue edges. This implies that there must be a path in S with more red edges than blue edges. In particular, this path starts and ends with a red edge. This path is an augmenting path with respect to M (i.e., the blue edges), since it is clearly alternating between edges in M and edges not in M, and the endpoints are unmatched with respect to M. So using the assumption that M is not maximum, we were able to find an augmenting path with respect to M. This completes the proof.
Exercise (Graphs with max degree at most 2). Let G = (V, E) be a graph such that all vertices have degree at most 2. Then prove that every connected component of G is either a path or a cycle (where we count an isolated vertex as a path of length 0).
3 CMU CS251 Spring 2023 Solution. Consider a graph G such that all vertices have degree at most 2. We want to show that it consists of disjoint paths and cycles. We prove this by induction on the number of vertices.
Pick an arbitrary vertex v in the graph. Removing v results in a graph G −v such that every vertex has degree at most 2. Since G −v has one less vertex, by induction hypothesis, G −v consists of disjoint paths and cycles. There are 3 cases to consider: deg(v) = 0, 1, or 2. It is not hard to see that in each case, adding v back to the graph preserves the property that the graph is a collection of disjoint paths and cycles. (Verify this part for yourself.) ■ Exercise (A tree can have at most one perfect matching). Show that a tree can have at most one perfect matching.
Hint. Assume, for the sake of contradiction, there are two perfect matchings M and M ′.
Then look at the symmetric difference between M and M ′.
Solution. The proof is by contradiction, so suppose a tree has two different perfect match-ings M and M ′. Let S be the symmetric difference between M and M ′, i.e., S = (M ∪ M ′)(M ∩M ′). Since |M| = |M ′| and M ̸= M ′, we must have |S| > 1 (there must be an edge in M that is not in M ′, and vice versa). The set S corresponds to a graph in which each vertex has degree at most 2. So this graph consists of disjoint paths and cycles (i.e. each connected component is either a path or a cycle). But a connected component cannot be a cycle since trees are acyclic. It also cannot be a path. This is because the existence of a degree 1 vertex in S implies that this vertex is not covered/matched by either M or M ′ (verify this yourself), and this would contradict the fact that M and M ′ are perfect matchings covering all vertices. So S must be the empty set, which contradicts our assumption that |S| > 1.
■ 2 Bipartite Graphs Definition (Bipartite graph). A graph G = (V, E) is called bipartite if there is a partition2 of V into sets X and Y such that all the edges in E have one endpoint in X and the other in Y . Sometimes the bipartition is given explicitly and the graph is denoted by G = (X, Y, E).
Example (Bipartite graph example). Below is an example of a bipartite graph.
Definition (k-colorable graphs). Let G = (V, E) be a graph. Let k ∈N+. A k-coloring of V is just a map χ : V →C where C is a set of cardinality k. (Usually the elements of C are called colors. If k = 3 then C = {red, green, blue} is a popular choice. If k is large, we often just call the “colors” 1, 2, . . . , k.) A k-coloring is said to be legal for G if every edge in E is bichromatic, meaning that its two endpoints have different colors. (I.e., for all {u, v} ∈E it is required that χ(u) ̸= χ(v).) Finally, we say that G is k-colorable if it has a legal k-coloring.
2Recall that a partition of V into X and Y means that X and Y are disjoint and X ∪Y = V .
4 CMU CS251 Spring 2023 Example (A 3-colorable graph). The graph below is 3-colorable. We can color the vertex at the center green, and color the outer vertices with blue and red by alternating those two colors.
Note (2-colorability is equivalent to bipartiteness). A graph G = (V, E) is bipartite if and only if it is 2-colorable. The 2-coloring corresponds to partitioning the vertex set V into X and Y such that all the edges have one endpoint in X and the other in Y .
Theorem (Characterization of bipartite graphs). A graph is bipartite if and only if it contains no odd-length cycles.
Proof. There are two directions to prove.
(= ⇒): For this direction, we want to show that if a graph is bipartite, then it contains no odd-length cycles. We prove the contrapositive. Observe that it is impossible to 2-color an odd-length cycle. So if a graph contains an odd-length cycle, the graph cannot be 2-colored, and therefore cannot be bipartite.
(⇐ =): For this direction, we want to show that if a graph does not contain an odd-length cycle, then it is bipartite. So suppose the graph contains no cycles of odd length. Without loss of generality, assume the graph is connected (if it is not, we can apply the argument to each connected component separately). For u, v ∈V , let dist(u, v) denote the length of the shortest path from u to v (or from v to u). Pick a starting vertex/root s and consider the “BFS tree” rooted at s. In this tree, level 0 corresponds to s, and level i corresponds to all vertices v with dist(s, v) = i. Color odd-indexed levels blue, and color even-indexed levels red.
The proof is done once we show that this is a valid 2-coloring of the graph. To show this, we’ll argue that no edge has its endpoints colored the same color. There are two types of edges we need to worry about that could potentially have its endpoints colored the same color. We consider each type below.
First, there could potentially be an edge between two vertices u and v at the same level. Let’s assume such an edge exists. Let w be the lowest common ancestor of u and v. Note that dist(u, w) = dist(v, w), so the path from w to u, plus the path from w to v, plus the edge {u, v}, form an odd-length cycle. This is a contradiction.
Second, we need to consider the possibility that there is an edge between a vertex u at level i and a vertex v at level i + 2k for some k > 0. However, the existence of such an edge implies that dist(s, v) ≤i + 1, which contradicts the fact that v is at level i + 2k. So this type of edge cannot exist either. This completes the proof.
Theorem (Finding a maximum matching in bipartite graphs). There is a polynomial time algorithm to solve the maximum matching problem in bipartite graphs.
Proof. Let G = (X, Y, E) be the input graph. The high level steps of the algorithm is as follows.
• Let M = {{x, y}} where {x, y} ∈E is an arbitrary edge.
• Repeat until there is no augmenting path with respect to M: – Find an augmenting path with respect to M.
5 CMU CS251 Spring 2023 – Update M according to the augmenting path (swapping matched and un-matched edges along the path).
Every time we find an augmenting path, we increase the size of our matching by one.
When there are no more augmenting paths, we stop and correctly output a maximum matching (the correctness follows from Theorem (Characterization for maximum matchings)).
The only unclear step of the algorithm is finding an augmenting path with respect to M.
And we explain how to do this step below. But before we do that, note that if this step can be done in polynomial time, then the overall running time of the algorithm is poly-nomial time since the loop repeats O(n) times and the work done in each iteration is polynomial time.
We now show how to find an augmenting path (given G = (X, Y, E) and M ⊆E).
We first turn G into a directed graph ⃗ G as follows: • Direct edges in E\M from X to Y .
• Direct edges in M from Y to X.
Note that an alternating path with respect to M in G corresponds to a directed path in ⃗ G. (We leave it to the reader to verify this.) This means that there is an augmenting path with respect to M in G if and only if there is a directed path in ⃗ G from an unmatched vertex x in X to an unmatched vertex y in Y . So to find an augmenting path, we’ll search for a directed path in ⃗ G from an unmatched x to an unmatched y, as follows: • For each unmatched x ∈X: – Run DFS(G, x).
– If you hit an unmatched y ∈Y , output the path from x to y.
• Output “no augmenting path found.” The running time is polynomial time since the loop repeats at most O(n) times, and the work done in each iteration is polynomial time.
Note (Finding a maximum matching in non-bipartite graphs). The high-level algorithm above presented in the proof of Theorem (Finding a maximum matching in bipartite graphs) is in fact applicable to general (not necessarily bipartite) graphs. However, the step of finding an augmenting path with respect to a matching turns out to be much more involved, and therefore we do not cover it in this chapter. See
org/wiki/Blossom_algorithm if you would like to learn more.
3 Bonus: Hall’s Theorem Theorem (Hall’s Theorem). Let G = (X, Y, E) be a bipartite graph. For a subset S of the vertices, let N(S) = S v∈S N(v). Then G has a matching covering all the vertices in X if and only if for all S ⊆X, we have |S| ≤|N(S)|.
Proof. There are two directions to prove.
(= ⇒): For this direction, we need to show that if G has a matching covering all the vertices in X, then every S ⊆X satisfies |S| ≤|N(S)|. We consider the contrapositive.
So suppose there is some S ⊆X such that |S| > |N(S)|. The vertices in S can only be matched to vertices in N(S), and since |S| > |N(S)|, there cannot be a matching that covers every element in S. And this implies there cannot be a matching covering every element of X.
(⇐ =): For this direction, we need to show that if every S ⊆X satisfies |S| ≤|N(S)|, then there is a matching that covers all the vertices in X. We will prove the contrapositive. So assume there is no matching that covers all the vertices in X. Our goal is to find some S ⊆X such that |S| > |N(S)|.
6 CMU CS251 Spring 2023 In order to identify such a set S, we need to make a couple of definitions. Let M be a maximum matching and let x ∈X be an element that it does not cover. We turn G into a directed graph as follows: direct all edges not in M from X to Y , and direct all edges in M from Y to X. We define L ⊆X to be the set of vertices in X that you can reach by a directed path starting at x (L does not include x). And we define R ⊆Y to be the set of all vertices in Y that you can reach by a directed path starting at x. Here is an illustration: We will show that for S = L ∪{x}, we have |S| > |N(S)|. We need two claims to argue this.
Claim 1: |L| = |R|.
Proof: Each ℓ∈L is matched to some r ∈R because the only way we can reach an ℓ∈L is through an edge in the matching. Conversely, each r ∈R must be matched to some ℓ∈L since if this was not true, i.e., if there was an unmatched r ∈R, that would imply that the path from x to r is an augmenting path, and this would contradict the fact that M is a maximum matching (Theorem (Characterization for maximum matchings)).
Since every element of L is matched by M to an element of R and vice versa, there is a one-to-one correspondence between L and R, i.e., |L| = |R|.
Claim 2: In the original undirected graph, N(L ∪{x}) ⊆R.
(In fact, N(L ∪{x}) = R but we only need one side of the inclusion.) Proof: For any ℓ∈L ∪{x}, we want to argue that N(ℓ) ⊆R. First consider the case that ℓ= x. Then all the neighbors of x must be in R since all the edges incident to x are directed from left to right. So N(x) ⊆R. Now consider any ℓ∈L. We want to argue that all the neighbors of ℓmust be in R. To argue about the neighbors of ℓ, we look at all the edges incident to ℓ. In this set of edges incident to ℓ, exactly one edge e is in the matching M. Since e ∈M is directed from Y to X, it must be incident to some r ∈R because the only way to reach ℓis through some r ∈R via e. If we now look at all the other edges incident to ℓ, note that they must be directed from X to Y , and the vertices K ⊆Y that they are incident to must be in R. This is because by definition of L, ℓ∈L is reachable from x, which means the vertices in K would also be reachable from x, and therefore would be in R (by the definition of R). This shows that every neighbor of ℓis in R, and completes the proof of Claim 2.
Combining Claim 1 and Claim 2 above, we have |L ∪{x}| > |R| ≥|N(L ∪{x})|, i.e., for S = L ∪{x}, |S| > |N(S)|, as desired.
Corollary (Characterization of bipartite graphs with perfect matchings). Let G = (X, Y, E) be a bipartite graph. Then G has a perfect matching if and only if |X| = |Y | and for any S ⊆X, we have |S| ≤|N(S)|.
Note (Hall’s Theorem when the two parts have equal size). Sometimes people call the above corollary Hall’s Theorem.
Exercise (Practice with perfect matchings).
1. Let G be a bipartite graph on 2n ver-tices such that every vertex has degree at least n. Prove that G must contain a perfect matching.
7 CMU CS251 Spring 2023 2. Let G = (X, Y, E) be a bipartite graph with |X| = |Y |. Show that if G is connected and every vertex has degree at most 2, then G must contain a perfect matching.
Solution. Part 1: If every vertex has degree n, it must be the case that the graph is G = (X, Y, E) where |X| = |Y | = n. It must also be the case that the graph is a complete bipartite graph (i.e., every possible edge is present). So clearly Hall’s theorem applies and the graph has a perfect matching.
Part 2: A graph with max degree at most 2 consists of disjoint paths and cycles (Exercise (Graphs with max degree at most 2)). Since the graph is connected, it must consist of a single path or a single cycle containing all the vertices. In either case, it is not hard to see that the graph must contain a perfect matching: If the graph is a path, then we can take every other edge (starting with the first edge) along the path to form a perfect matching.
If the graph consists of a cycle, it has two different perfect matchings.
■ 4 Check Your Understanding Problem.
1. True or false: A maximum matching is a maximal matching.
2. True or false: A perfect matching is a maximum matching.
3. True or false: There exist graphs with more than one perfect matching.
4. How many perfect matchings are there in a complete bipartite graph (all possible edges are present) where both sides of the bipartition contain exactly n vertices?
5. Suppose a graph with n vertices has a perfect matching. What is the size of the perfect matching?
6. What is the definition of an augmenting path?
7. True or false: Given a matching M, there can be at most one augmenting path with respect to M.
8. True or false: A matching M in a non-bipartite graph G is maximum if and only if there is no augmenting path with respect to M.
9. Describe the high-level steps of a polynomial-time algorithm for finding a maxi-mum matching in a graph.
10. Describe a polynomial-time algorithm that given a bipartite graph and a matching in the graph, determines if an augmenting path exists with respect to the matching.
11. True or false: The graph below is bipartite.
12. True or false: If a graph is k-colorable, then it is (k + 1)-colorable.
13. True or false: A graph which is not bipartite must contain an odd-length cycle.
14. Explain how one can 2-color a graph that does not contain an odd-length cycle.
15. Describe a polynomial-time algorithm to determine if an input undirected graph is bipartite or not.
8 CMU CS251 Spring 2023 16. How do you prove that a tree has at most one perfect matching?
17. True or false: Any graph with more than one perfect matching must contain a cycle.
18. In this chapter we saw the definition of a bipartite graph. We can think of a bipar-tite graph as a special case of a k-partite graph where k = 2. How would you define the general notion of a k-partite graph, and how does it relate to the colorability of the graph?
5 High-Order Bits Important. Here are the important things to keep in mind from this chapter.
1. As always, all the definitions in this chapter are important (matchings, augmenting path, bipartite graph, k-coloring of a graph).
2. Theorem (Characterization of bipartite graphs) gives an important characterization for bipartite graphs. Make sure to understand its proof.
3. Theorem (Characterization for maximum matchings) gives an important charac-terization for maximum matchings. The proof is one of the harder proofs we have done in the course so far. A good understanding of the proof is expected and im-portant since it will help you raise your level of mathematical maturity.
4. Make sure to understand how we can use Theorem (Characterization for maxi-mum matchings) to come up with an algorithm for finding a maximum matching in a graph.
9 |
187677 | https://blogs.ugto.mx/rea/clase-digital-7-esfuerzos-normales-deformacion-unitaria-ley-de-hooke/ | Saltar al contenido
Recursos Educativos Abiertos
Clase digital 7. Esfuerzos normales, deformación unitaria, ley de Hooke
por Universidad de Guanajuato
Esfuerzos normales, deformación unitaria, ley de Hooke
Introducción
¡Hola, qué gusto poder saludarte!
Es un placer encontrarte nuevamente en este curso de Mecánica de sólidos. ¡Gracias por continuar! Esto habla de tu gran compromiso e interés. Por lo tanto, te doy la bienvenida a la clase siete en la cual aprenderás mucho acerca del tema Esfuerzos normales, deformación unitaria y ley de Hooke, así como los procedimientos para calcular el esfuerzo normal en barras de distintas áreas transversales y la deformación unitaria debido al esfuerzo normal al que están sometidas.
En esta sesión también nos adentraremos un poco en el estudio de la resistencia de materiales, información clave para el diseño de cualquier equipo utilizado en ingeniería química.
Espero que el contenido de esta sesión sea de tu agrado y la disfrutes.
Comencemos.
Desarrollo del tema
Esfuerzo normal y deformación unitaria
El esfuerzo normal (esfuerzo axil o axial) es el esfuerzo interno o resultante de las tensiones perpendiculares (normales) a la sección transversal de un prisma mecánico. Los esfuerzos normales pueden ser de tensión o compresión. Un esfuerzo de compresión es aquel que tiende a aplastar el material del miembro de carga, y a acortar el miembro en sí. Un esfuerzo de tensión es aquel que tiende a estirar al miembro y romper el material.
La deformación unitaria (δ) se define como el cambio de longitud, por unidad de longitud, debido a una carga normal sobre un material, (ver Figura 1). Esta se puede relacionar directamente con el esfuerzo generado sobre el material al dividir la carga que genera la deformación entre el área transversal del material.
Propiedades mecánicas de los materiales
En la Figura 2 se muestra la relación que existe entre la deformación de un material y el esfuerzo que produce dicha deformación. La gráfica es representativa de cualquier material dúctil y los valores específicos para cada tipo de material son obtenidos experimentalmente y están ampliamente reportados en la literatura alusiva a la resistencia de materiales. Las secciones en las que se divide la gráfica se muestran a continuación.
OA: sección de proporcionalidad.
El punto A es el límite de elasticidad, después de este punto el material no recupera su forma original al ser descargado.
El punto B es el límite de fluencia, en este punto aparece un considerable alargamiento sin el correspondiente aumento de carga.
El punto D es el esfuerzo último o límite de resistencia, es la máxima ordenada en el diagrama esfuerzo-deformación.
El punto E es el punto de ruptura, el material se alarga muy rápidamente y se estrecha al mismo tiempo hasta romperse.
Ley de Hooke
En la gráfica se muestra que la parte rectilínea corresponde a la ley de Hooke indicando que el módulo de elasticidad es la pendiente de la recta, es decir la relación entre el esfuerzo y la deformación.
Relación entre esfuerzo (𝜎) y alargamiento (𝛿)
Nota: Para que la expresión anterior se use adecuadamente se debe considerar los siguiente:
La carga ha de ser axial.
La barra o placa debe ser homogénea.
El esfuerzo no debe sobrepasar el límite de proporcionalidad.
Conclusión
En resumen, los materiales tienden a deformarse cuando se les aplica un esfuerzo normal o tangencial. La deformación unitaria generada por un esfuerzo normal puede calcularse de acuerdo a la ley de Hooke siempre y cuando la deformación no sobrepase el límite elástico. La deformación ya sea a tensión o a compresión de un esfuerzo normal siempre se genera de forma axial al área transversal del objeto.
El cálculo de la deformación unitaria tiene una amplia aplicación en el diseño de estructuras.
Recuerda que puedes apoyarte en cualquier momento del material reportado en las fuentes de información.
Con esto llegamos al final de la clase ¡Te felicito por tu gran esfuerzo y dedicación! No olvides realizar la tarea y mandarla como corresponde. Sigue perseverando en tu educación. Te espero en tu próxima sesión, hasta entonces.
Fuentes de información
Singer, F. L. (1990). Resistencia de Materiales. (6a ed.). Harla.
Gere, J. M. (2004). Mechanics of Materials. (6th ed.). Thomson. |
187678 | https://www.fao.org/4/y5686e/y5686e0b.htm | 7. ENERGY REQUIREMENTS OF LACTATION
Exclusive breastfeeding is recommended during the six months after delivery, with introduction of complementary foods and continued breastfeeding thereafter (WHO, 2001). The energy requirement of a lactating woman is defined as the level of energy intake from food that will balance the energy expenditure needed to maintain a body weight and body composition, a level of physical activity and breastmilk production that are consistent with good health for the woman and her child, and that will allow economically necessary and socially desirable activities to be performed. To operationalize this definition, the energy needed to produce an appropriate volume of milk must be added to the woman’s habitual energy requirement, assuming that she resumes her usual level of physical activity soon after giving birth.
The mean amount of breastmilk produced daily is similar among population groups with different cultural and socio-economic settings (Prentice et al., 1986; Butte, Lopez-Alarcon and Garza, 2002) (Table 7.1). There may be some variation in milk composition related to maternal nutrition, but the main factors that influence the energy needs of lactating women are the duration of breastfeeding practices and the extent of exclusive breastfeeding. As these vary significantly in different societies, dietary energy recommendations for lactating women should be population-specific. Regardless of the cultural and social environment, the ideal situation is that women be well nourished from the beginning of pregnancy and that they maintain adequate nutritional intake with appropriate weight gain throughout gestation. This will allow them to attain body fat reserves that may act as an energy substrate to cover part of the additional energy needs in preparation for and during lactation.
TABLE 7.1Average milk production rates (g/d)
| | | | | | | | | | | | | |
--- --- --- --- --- ---
| Postpartum period (months) | | | | | | | | | | | | |
| Exclusive breastfeeding | | | | | | | | | | | | |
| Industrialized countries | | | 751 | 780 | 796 | 854 | | | | | | |
| Traditional countries | 562 | 634 | 582 | 768 | 778 | 804 | | | | | | |
| Partial breastfeeding | | | | | | | | | | | | |
| Industrialized countries | 611 | 697 | 730 | 704 | 710 | 612 | 569 | 417 | 497 | 691 | 516 | 497 |
| Traditional countries | 568 | 636 | 574 | 634 | 714 | 611 | 688 | 635 | 516 | | 565 | 511 |
Source: Butte, Lopez-Alarcon and Garza, 2002.
7.1 Determinants of the energy cost of lactation
The energy cost of lactation is determined by the amount of milk that is produced and secreted, its energy content, and the efficiency with which dietary energy is converted to milk energy.
7.1.1 Human milk production
The mean amount of milk ingested by exclusively breastfed infants is similar in industrialized and more traditional societies, according to a WHO-sponsored comprehensive review (Butte, Lopez-Alarcon and Garza, 2002). After six months, variation among individuals and populations increases, owing to the nature and amount of complementary foods provided to the growing infant. From the age of six months onwards, when infants are partially breastfed, milk production is estimated at 550 g/day.
7.1.2 Energy content of human milk
The energy content of human milk depends primarily on milk fat concentration, which shows complex diurnal, within-feed and between-breast fluctuations. Twenty-four-hour milk sampling schemes have been developed that interfere minimally with the secretion of milk flow and capture the diurnal and within-feed variation (Garza and Butte, 1986). Measurements of the gross energy content of representative 24-hour milk samples determined by adiabatic bomb calorimetry or macronutrient analysis in a number of studies of well-nourished women gave a mean value of 2.8 kJ/g (0.67 kcal/g) from 1 to 24 months of lactation (Garza and Butte, 1986; Prentice and Prentice, 1988; Butte and King, 2002; WHO, 1985; Institute of Medicine, 1991; Goldberg et al., 1991; Panter-Brick, 1993).
7.1.3 Efficiency of energy conversion
The efficiency with which food energy and body energy reserves are converted into milk energy has been calculated from theoretical estimates of the biochemical efficiency associated with the synthesis of milk lactose, protein and fat, and from metabolic balance studies (Prentice and Prentice, 1988). Taking into account the energy costs of digestion, absorption, conversion and transport, biochemical efficiency has been estimated at 80 to 85 percent (Butte and King, 2002). Based on that estimate, on the theoretical efficiency used in the 1985 FAO/WHO/UNU report and on the suggestion of the United States Institute of Medicine (1991), an efficiency factor of 80 percent was applied to calculate the energy cost of human milk production.
7.1.4 Energy cost of milk production
Table 7.2 shows the energy cost to produce the mean amounts of milk needed for exclusively breastfed infants. Monthly milk volumes are those reported for well-nourished women with healthy babies in the WHO-sponsored review (Butte, Lopez-Alarcon and Garza, 2002), and gross energy contents are those described in section 7.1.2.
The results were compared with the energy requirements of exclusively breastfed infants from one to six months of age, calculated as described in chapter 3. To do so, human milk intakes that are measured by the test-weighing technique must be corrected for insensible water loss during the course of feeding (5 percent correction factor) and for the digestibility of human milk. The metabolizable energy in human milk was assumed to be 5.3 percent lower than its gross energy content based on proximate analyses and energy factors of 23.6 kJ (5.65 kcal) per gram of protein and free amino acids, 38.7 kJ (9.25 kcal) per gram of fat and 16.5 kJ (3.95 kcal) per gram of lactose. From months one to six the figures are on average within 5 percent, which is remarkable considering that energy requirements of infants were calculated from quite different information (i.e. from predictive equations based on DLW measurements of TEE, plus estimates of growth accretion based on growth velocity and body composition).
TABLE 7.2Energy cost of human milk production by women who practise exclusive breastfeeding
| | | | | | |
--- --- --- |
| Months postpartum | Mean milk intake g/daya | Human milk intake, corrected for insensible water losses g/dayb | Gross energy content kJ/gc | Daily gross energy secreted kJ/day | Energy cost of milk production kJ/dayd |
| 1 | 699 | 734 | 2.8 | 2 055 | 2 569 |
| 2 | 731 | 768 | 2.8 | 2 149 | 2 686 |
| 3 | 751 | 789 | 2.8 | 2 208 | 2 760 |
| 4 | 780 | 819 | 2.8 | 2 293 | 2 867 |
| 5 | 796 | 836 | 2.8 | 2 340 | 2 925 |
| 6 | 854 | 897 | 2.8 | 2 511 | 3 138 |
| Mean | 769 | 807 | 2.8 | 2 259 | 2 824 |
a From Butte, Lopez-Alarcon and Garza, 2002.b Insensible water losses assumed to be equal to 5 percent milk intake.c Gross energy content measured by adiabatic bomb calorimetry or macronutrient analysis.d Based on energetic efficiency of 80 percent.
TABLE 7.3Comparison of the energy cost of human milk production and energy requirements of exclusively breastfed infants
| | | | | | | | |
--- --- --- --- |
| Months postpartum | Mean milk intake g/daya | Human milk intake, corrected for insensible water losses g/dayb | Gross energy content kJ/gc | Daily gross energy secreted kJ/day | Metabolizable energy intake kJ/dayd | Infant energy requirement kJ/daye | Requirement/ME intake |
| 1 | 699 | 734 | 2.8 | 2 055 | 1 946 | 1 922 | 0.99 |
| 2 | 731 | 768 | 2.8 | 2 149 | 2 035 | 2 143 | 1.05 |
| 3 | 751 | 789 | 2.8 | 2 208 | 2 091 | 2 284 | 1.09 |
| 4 | 780 | 819 | 2.8 | 2 293 | 2 172 | 2 219 | 1.02 |
| 5 | 796 | 836 | 2.8 | 2 340 | 2 216 | 2 376 | 1.07 |
| 6 | 854 | 897 | 2.8 | 2 511 | 2 378 | 2 501 | 1.05 |
| Mean | 769 | 807 | 2.8 | 2 259 | 2 140 | 2 241 | 1.05 |
a From Butte, Lopez-Alarcon and Garza, 2002.b Insensible water losses assumed to be equal to 5 percent milk intake.c Gross energy content measured by adiabatic bomb calorimetry or macronutrient analysis.d Metabolizable energy values based on proximate analysis of milk are 5.3 percent lower than bomb calorimetry values.e Mean values of boys and girls, calculated as described in chapter 3 of this report.
7.2 Energy requirements for lactation
Compared with non-pregnant, non-lactating women, during lactation there are no significant changes in BMR, efficiency of work performance, or TEE (Butte and King, 2002), and in most societies women resume their usual level of physical activity in the first month postpartum or shortly thereafter (Goldberg et al., 1991; Panter-Brick, 1993; Roberts et al., 1982; Tuazon et al., 1987; van Raaij et al., 1990). It could be argued that where exclusive breastfeeding is prevalent, lactating mothers may have a lower TEE than non-pregnant, non-lactating women owing to the frequency of breastfeeding, which involves periods of little maternal activity. On the other hand, lactating women often carry their infants while moving around, and this additional workload might balance the lower physical activity associated with breastfeeding. Thus, total energy requirements during lactation are equal to those of the pre-pregnancy period, plus the additional demands imposed by the need for adequate milk production and secretion.
These additional demands correspond to the energy cost of milk production. For women who feed their infants exclusively with breastmilk during the first six months of life, the mean energy cost over the six-month period is: 807 g milk/day × 2.8 kJ/g/0.80 efficiency = 2.8 MJ/day (675 kcal/day) (Table 7.2). From the age of six months onwards, when infants are partially breastfed and milk production is on average 550 g/day (Table 7.1), the energy cost imposed by lactation is 1.925 MJ/day (460kcal/day).
Fat stores accumulated during pregnancy may cover part of the additional energy needs in the first few months of lactation. Postpartum loss of body weight is usually highest in the first three months, and generally greater among women who practise exclusive breastfeeding, but the extent to which the energy mobilized supports lactation depends on the gestational weight gain and the nutritional status of the mother. A review of 17 studies indicated that, on average, well-nourished women lost 0.8 kg/month, whereas undernourished mothers lost only an average of 0.1 kg/month (Butte and Hopkinson, 1998). Assuming an energy factor of 27.2 MJ/kg (Butte and King, 2002; Butte and Hopkinson, 1998), the rate of weight loss in well-nourished women would correspond to the mobilization of 27.2 × 0.8 kg/month = 21.8 MJ/month, or 0.72 MJ/day (170 kcal/day) from body energy stores. This amount of energy can be deducted from the 2.8 MJ/day (675 kcal)/day needed during the first six months of lactation. The result, 2.1 MJ/day (505 kcal/day), is similar to the additional energy required when infants are partially breastfed after six months of lactation.
On the other hand, undernourished women and those who did not gain adequate body weight during pregnancy must conserve as much energy as possible for their own well-being and that of their infants. Hence, in these women the full energy demands of lactation must be provided by an increment in dietary intake.
In conclusion, well-nourished women with adequate gestational weight gain should increase their food intake by 2.1 MJ/day (505 kcal/day) for the first six months of lactation, while undernourished women and those with insufficient gestational weight gain should add to their personal energy demands 2.8 MJ/day (675 kcal/day) during the first semester of lactation. Energy requirements for milk production in the second six months are dependent on rates of milk production, which are highly variable among women and populations.
References
Brown, K., Dewey, K.G. & Allen, L. 1998. Complementary feeding of young children in developing countries: A review of current scientific knowledge. Geneva, WHO.
Butte, N.F. & Hopkinson, J.M. 1998. Body composition changes during lactation are highly variable among women. J. Nutr., 128: 381S-385S.
Butte, N. & King, J.C. 2002. Energy requirements during pregnancy and lactation. Energy background paper prepared for the joint FAO/WHO/UNU Consultation on Energy in Human Nutrition.
Butte, N.F., Lopez-Alarcon, M.D. & Garza, C. 2002. Nutrient adequacy of exclusive breastfeeding for the term infant during the first six months of life. Geneva, WHO.
Garza, C. & Butte, N.F. 1986. Energy concentration of human milk estimated from 24-h pools and various abbreviated sampling schemes. J. Pediatr. Gastroenterol. Nutr., 5: 943-948.
Goldberg, G.R., Prentice, A.M., Coward, W.A., Davies, H.L., Murgatroyd, P.R., Sawyer, M.B., Ashford, J. & Black, A.E. 1991. Longitudinal assessment of the components of energy balance in well-nourished lactating women. Am. J. Clin. Nutr., 54: 788-798.
Institute of Medicine. 1991. Nutrition during lactation. Washington, DC, National Academy Press.
Panter-Brick, C. 1993. Seasonality of energy expenditure during pregnancy and lactation for rural Nepali women. Am. J. Clin. Nutr., 57: 620-628.
Prentice, A.M. & Prentice, A. 1988. Energy costs of lactation. Ann. Rev. Nutr., 8: 63-79.
Prentice, A., Paul, A., Black, A., Cole, T. & Whitehead, R. 1986. Cross-cultural differences in lactational performance. In M. Hamosh and A.S. Goldman, eds. Human lactation 2: Maternal and environmental factors, pp. 13-44. New York, Plenum Press.
Roberts, S.B., Paul, A.A., Cole, T.J. & Whitehead, R.G. 1982. Seasonal changes in activity, birth weight and lactational performance in rural Gambian women. Trans. R. Soc. Trop. Med. Hyg., 76: 668-678.
Tuazon, M.A., van Raaij, J.M., Hautvast, J.G. & Barba, C.V. 1987. Energy requirements of pregnancy in the Philippines. Lancet, 2: 1129-1131.
van Raaij, J.M.A., Schonk, C.M., Vermaat-Miedema, S.H., Peek, M.E.M. & Hautvast, J.G.A.J. 1990. Energy cost of walking at a fixed pace and self-selected pace before, during and after pregnancy. Am. J. Clin. Nutr., 51: 158-161.
WHO. 1985. Energy and protein requirements: Report of a joint FAO/WHO/UNU expert consultation. WHO Technical Report Series No. 724. Geneva.
WHO. 2001. Expert consultation on the optimal duration of exclusive breastfeeding. Conclusions and recommendations. Geneva. |
187679 | https://www.ng.ru/economics/2025-04-06/1_4_9228_oil.html | Рубрики
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06.04.2025 20:23:00
Российская нефть в рублях оказалась на четверть дешевле целевого ориентира
Экономике понадобились деньги из Фонда национального благосостояния
Анастасия Башкатова
Заместитель заведующего отделом экономики "Независимой газеты"
Тэги: нефть, нефтяные доходы, минфин, фнб, финансы, цены на нефть
Вице-премьер Александр Новак провел заседание совместного министерского мониторингового комитета стран ОПЕК+, на котором решено поддержать решение ряда стран о корректировке добычи. Фото с сайта www.government.ru
Случившийся в конце недели обвал нефтяных котировок в очередной раз поставил вопрос о перспективах российского бюджета, в котором зависимость от нефтегаза хоть и снижается, однако основные расчеты все равно строятся вокруг нефти. Теперь для его балансировки предстоит найти новое комфортное соотношение цены барреля и курса рубля. Притом что в марте нефтегазовые доходы казны уже и так снизились в годовом сопоставлении более чем на 17%, а цена нефти Urals в ее рублевом выражении оказалась ниже заложенных в правительственные расчеты ориентиров примерно на 25%.
Впервые с января 2024 года Минфин сменил стратегию и в рамках действующего бюджетного правила перешел от накопления средств Фонда национального благосостояния (ФНБ) к их расходованию.
В конце минувшей недели он объявил о решении начать продавать валюту и золото. Операции будут проводиться в период с 7 апреля по 12 мая. Ежедневный объем продажи составит в рублевом эквиваленте 1,6 млрд руб., а в целом за указанный период сумма достигнет почти 36 млрд руб.
Причина на поверхности: существенное сокращение нефтегазовых доходов бюджета. По обнародованным данным Минфина, в марте нефтегазовые доходы казны сократились в годовом сопоставлении более чем на 17%, по итогам января–марта падение составило почти 10% по отношению к тому же периоду прошлого года. В марте российская нефть Urals подешевела по сравнению со значением марта 2024-го примерно на 16%, следует из новой сводки Минэкономразвития.
Но самым главным признаком того, что российский бюджет вышел из зоны комфорта, можно считать цену Urals в ее рублевом эквиваленте, то есть с учетом курса рубля к доллару. Основные параметры федерального бюджета на этот год рассчитывались исходя из того, что нефть Urals в среднем будет стоить более 6,7 тыс. руб. за баррель, или 69,7 долл. за баррель при валютном курсе 96,5 руб. за 1 долл. А по факту в марте баррель российской нефти Urals недотянул и до 5,1 тыс. руб., оказавшись почти на 25% дешевле ориентира на этот год (58,99 долл. за баррель при среднемесячном валютном курсе 86 руб. за 1 долл.).
Как пояснил в опросе Telegram-канала Proeconomics доцент РАНХиГС Сергей Хестанов, у правительства есть большой набор методов, с помощью которых можно компенсировать снижение нефтегазовых доходов. «Можно задействовать ФНБ, можно увеличить госзаимствования, можно увеличить нормы продажи валютной выручки экспортеров и даже изменить тарифное и нетарифное регулирование внешней торговли», – перечислил он. На практике обычно используется одновременно несколько методов.
«Однако какой бы метод ни был применен, есть надежное эмпирическое правило: если текущая рублевая цена российской нефти ниже, чем заложенная в бюджет цифра, весьма вероятно (хоть и не предопределено), что произойдет ослабление рубля. До тех пор, пока эти цифры не сблизятся, – предупредил экономист. – А при каком курсе рубля это произойдет, будет зависеть от того, какие методы балансировки бюджета реализованы».
Конечно, сравнивать при анализе цен на нефть итоги одного месяца и даже квартала со среднегодовым прогнозным показателем нужно аккуратно, делая оговорки о волатильности и непредсказуемости некоторых экономических и политических тенденций.
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| Экспортеры, входящие в ОПЕК+, продолжают совместно корректировать тенденции на глобальном рынке нефти. Фото Reuters |
Эксперты указывают на две главные причины обвала нефтяных котировок в первые числа апреля: новые тарифные ограничения США и увеличение квот внутри ОПЕК+.
Ранее президент США Дональд Трамп объявил о введении «взаимных» пошлин в отношении иностранных торговых партнеров США – ограничения коснутся 185 государств. Минимальный базовый уровень пошлин составит 10%, но для ряда стран они подсчитаны индивидуально в зависимости от торгового дефицита США с этими экономиками. Так, для Китая пошлины составят 34%, для Евросоюза – 20%, для Японии – 24%.
Эта инициатива стала новым витком тарифных войн в мировой экономике. В ответ на меры Вашингтона власти Китая решили ввести дополнительные пошлины в размере 34% на товары из США. Такая экономическая эскалация, по мнению экспертов, ударит по глобальному спросу на энергоресурсы. Уже одного этого опасения достаточно для того, чтобы цена на нефть качнулась вниз.
Но сыграл свою роль еще один фактор: одновременно с этим восемь стран – участниц соглашения ОПЕК+, включая Саудовскую Аравию и Россию, решили ускорить снятие ранее принятых в объеме 2,2 млн баррелей в сутки (б/с) ограничений в добыче нефти и увеличить добычу с мая на 411 тыс. б/с. Это, как сообщается, эквивалентно трем ежемесячным надбавкам. Новый план предполагает, что, в частности, квота Саудовской Аравии в мае будет повышена до 9,2 млн б/с, а России – почти до 9,1 млн б/с.
Причем в материалах ОПЕК поясняется, что это решение принято «ввиду продолжающегося улучшения фундаментальных рыночных факторов и позитивного прогноза рынка». Хотя оно не окончательное и может быть изменено в зависимости от рыночной конъюнктуры. Но пока что это решение как раз было поддержано на проведенном в субботу вице-премьером Александром Новаком заседании совместного министерского мониторингового комитета стран ОПЕК+. Как сообщается на сайте правительства РФ, комитет приветствовал объявленное решение восьми государств о корректировке добычи.
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Свидетельство о регистрации средства массовой информации Эл № ФС77-60208 от 17 декабря 2014 г.
Выдано Федеральной службой по надзору в сфере связи, информационных технологий и массовых коммуникаций (Роскомнадзор) |
187680 | https://resources.finalsite.net/images/v1685743825/apsedu/uzezgh8bpexe3vuyevwt/my-laws-of-exponents.pdf | NAME____ My “Laws of Exponents” Cheat Sheet Multiplying Powers with the Same Base General Rule: xa ● xb = xa+b Example: x5 ● x6 = x11 Dividing Powers with the Same Base General Rule: xa x b = xa – b Example: x 7 x 4 = x3 Finding a Power of a Power General Rule: (xa)b = xa ● b Example: (x3)6 = x18 Negative Exponents General Rule: x-a = a x 1 Example: x-7 = 7 1 x Zero as an Exponent General Rule: x0 = 1 Example: 50 = 1 NAME____ My “Laws of Exponents” Cheat Sheet Multiplying Powers with the Same Base General Rule: xa ● xb = xa+b Example: x5 ● x6 = x11 Dividing Powers with the Same Base General Rule: xa x b = xa – b Example: x 7 x 4 = x3 Finding a Power of a Power General Rule: (xa)b = xa ● b Example: (x3)6 = x18 Negative Exponents General Rule: x-a = a x 1 Example: x-7 = 7 1 x Zero as an Exponent General Rule: x0 = 1 Example: 50 = 1 For a complete set of online Algebra notes visit © 2005 Paul Dawkins Algebra Cheat Sheet Basic Properties & Facts Arithmetic Operations ( ) , 0 b ab ab ac a b c a c c a a a ac b b c bc b c a c ad bc a c ad bc b d bd b d bd a b b a a b a b c d d c c c c a ab ac ad b b c a c a bc d + = + = = = + − + = − = − − + = = + − − + = + ≠ = Exponent Properties ( ) ( ) ( ) ( ) 1 1 0 1 1, 0 1 1 n m m m n n m n m n m m m n m n nm n n n n n n n n n n n n n n n n a a a a a a a a a a a a a ab a b b b a a a a a b b a a a b a a + − − − − − = = = = = ≠ = = = = = = = = Properties of Radicals 1 , if is odd , if is even n n n n n n m n nm n n n n n n a a ab a b a a a a b b a a n a a n = = = = = = Properties of Inequalities If then and If and 0 then and If and 0 then and a b a c b c a c b c a b a b c ac bc c c a b a b c ac bc c c < + < + − < − < > < < < < > > Properties of Absolute Value if 0 if 0 a a a a a ≥ = − < 0 Triangle Inequality a a a a a ab a b b b a b a b ≥ − = = = + ≤ + Distance Formula If ( ) 1 1 1 , P x y = and ( ) 2 2 2 , P x y = are two points the distance between them is ( ) ( ) ( ) 2 2 1 2 2 1 2 1 , d P P x x y y = − + − Complex Numbers ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) 2 2 2 2 2 2 1 1 , 0 Complex Modulus Complex Conjugate i i a i a a a bi c di a c b d i a bi c di a c b d i a bi c di ac bd ad bc i a bi a bi a b a bi a b a bi a bi a bi a bi a bi = − = − − = ≥ + + + = + + + + − + = −+ − + + = − + + + − = + + = + + = − + + = + For a complete set of online Algebra notes visit © 2005 Paul Dawkins Logarithms and Log Properties Definition log is equivalent to y b y x x b = = Example 3 5 log 125 3 because 5 125 = = Special Logarithms 10 ln log natural log log log common log e x x x x = = where 2.718281828 e = K Logarithm Properties ( ) ( ) log log 1 log 1 0 log log log log log log log log log b b b x x b r b b b b b b b b b b x b x x r x xy x y x x y y = = = = = = + = − The domain of logb x is 0 x > Factoring and Solving Factoring Formulas ( )( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 2 2 2 2 x a x a x a x ax a x a x ax a x a x a b x ab x a x b − = + − + + = + − + = − + + + = + + ( ) ( ) ( )( ) ( )( ) 3 3 2 2 3 3 3 2 2 3 3 3 2 2 3 3 2 2 3 3 3 3 x ax a x a x a x ax a x a x a x a x a x ax a x a x a x ax a + + + = + − + − = − + = + − + − = − + + ( )( ) 2 2 n n n n n n x a x a x a − = − + If n is odd then, ( )( ) ( )( ) 1 2 1 1 2 2 3 1 n n n n n n n n n n n x a x a x ax a x a x a x ax a x a − − − − − − − − = − + + + + = + − + − + L L Quadratic Formula Solve 2 0 ax bx c + + = , 0 a ≠ 2 4 2 b b ac x a −± − = If 2 4 0 b ac − > - Two real unequal solns. If 2 4 0 b ac − = - Repeated real solution. If 2 4 0 b ac − < - Two complex solutions. Square Root Property If 2 x p = then x p = ± Absolute Value Equations/Inequalities If b is a positive number or or p b p b p b p b b p b p b p b p b = ⇒ = − = < ⇒ −< < > ⇒ < − > Completing the Square Solve 2 2 6 10 0 x x − − = (1) Divide by the coefficient of the 2 x 2 3 5 0 x x − − = (2) Move the constant to the other side. 2 3 5 x x − = (3) Take half the coefficient of x, square it and add it to both sides 2 2 2 3 3 9 29 3 5 5 2 2 4 4 x x − + − = + − = + = (4) Factor the left side 2 3 29 2 4 x − = (5) Use Square Root Property 3 29 29 2 4 2 x − = ± = ± (6) Solve for x 3 29 2 2 x = ± For a complete set of online Algebra notes visit © 2005 Paul Dawkins Functions and Graphs Constant Function ( ) or y a f x a = = Graph is a horizontal line passing through the point ( ) 0,a . Line/Linear Function ( ) or y mx b f x mx b = + = + Graph is a line with point ( ) 0,b and slope m. Slope Slope of the line containing the two points ( ) 1 1 , x y and ( ) 2 2 , x y is 2 1 2 1 rise run y y m x x − = = − Slope – intercept form The equation of the line with slope m and y-intercept ( ) 0,b is y mx b = + Point – Slope form The equation of the line with slope m and passing through the point ( ) 1 1 , x y is ( ) 1 1 y y m x x = + − Parabola/Quadratic Function ( ) ( ) ( ) 2 2 y a x h k f x a x h k = − + = − + The graph is a parabola that opens up if 0 a > or down if 0 a < and has a vertex at ( ) , h k . Parabola/Quadratic Function ( ) 2 2 y ax bx c f x ax bx c = + + = + + The graph is a parabola that opens up if 0 a > or down if 0 a < and has a vertex at , 2 2 b b f a a − − . Parabola/Quadratic Function ( ) 2 2 x ay by c g y ay by c = + + = + + The graph is a parabola that opens right if 0 a > or left if 0 a < and has a vertex at , 2 2 b b g a a − − . Circle ( ) ( ) 2 2 2 x h y k r − + − = Graph is a circle with radius r and center ( ) , h k . Ellipse ( ) ( ) 2 2 2 2 1 x h y k a b − − + = Graph is an ellipse with center ( ) , h k with vertices a units right/left from the center and vertices b units up/down from the center. Hyperbola ( ) ( ) 2 2 2 2 1 x h y k a b − − − = Graph is a hyperbola that opens left and right, has a center at ( ) , h k , vertices a units left/right of center and asymptotes that pass through center with slope b a ± . Hyperbola ( ) ( ) 2 2 2 2 1 y k x h b a − − − = Graph is a hyperbola that opens up and down, has a center at ( ) , h k , vertices b units up/down from the center and asymptotes that pass through center with slope b a ± . For a complete set of online Algebra notes visit © 2005 Paul Dawkins Common Algebraic Errors Error Reason/Correct/Justification/Example 2 0 0 ≠ and 2 2 0 ≠ Division by zero is undefined! 2 3 9 − ≠ 2 3 9 − = −, ( ) 2 3 9 − = Watch parenthesis! ( ) 3 2 5 x x ≠ ( ) 3 2 2 2 2 6 x x x x x = = a a a b c b c ≠ + + 1 1 1 1 2 2 1 1 1 1 = ≠ + = + 2 3 2 3 1 x x x x − − ≠ + + A more complex version of the previous error. a bx a + 1 bx ≠+ 1 a bx a bx bx a a a a + = + = + Beware of incorrect canceling! ( ) 1 a x ax a − − ≠− − ( ) 1 a x ax a − − = − + Make sure you distribute the “-“! ( ) 2 2 2 x a x a + ≠ + ( ) ( )( ) 2 2 2 2 x a x a x a x ax a + = + + = + + 2 2 x a x a + ≠ + 2 2 2 2 5 25 3 4 3 4 3 4 7 = = + ≠ + = + = x a x a + ≠ + See previous error. ( ) n n n x a x a + ≠ + and n n n x a x a + ≠ + More general versions of previous three errors. ( ) ( ) 2 2 2 1 2 2 x x + ≠ + ( ) ( ) 2 2 2 2 1 2 2 1 2 4 2 x x x x x + = + + = + + ( ) 2 2 2 2 4 8 4 x x x + = + + Square first then distribute! ( ) ( ) 2 2 2 2 2 1 x x + ≠ + See the previous example. You can not factor out a constant if there is a power on the parethesis! 2 2 2 2 x a x a − + ≠− + ( ) 1 2 2 2 2 2 x a x a − + = − + Now see the previous error. a ab b c c ≠ 1 1 a a a c ac b b b b c c = = = a ac b c b ≠ 1 1 a a a a b b c c b c bc = = = Copyright © Regents Exam Prep Center Geometry – Things to Remember! Regular Solids: Tetrahedron – 4 faces Cube – 6 faces Octahedron – 8 faces Dodecahedron – 12 faces Icosahedron – 20 faces 3-D Figures: Prism: V = Bh Pyramid: 1 3 V Bh = Cylinder: 2 V r h π = ; 2 2 2 SA rh r π π = + Cone: 2 1 3 V r h π = ; 2 SA s r r π π = + Sphere: 3 4 3 V r π = ; 2 2 4 SA r d π π = = Locus Theorems: Fixed distance from point. Fixed distance from a line. Equidistant from 2 points. Equidistant 2 parallel lines. Equidistant from 2 intersecting lines Polygon Interior/Exterior Angles: Sum of int. angles = 180( 2) n − Each int. angle (regular) = 180( 2) n n − Sum of ext. angles = 360 Each ext. angle (regular) = 360 n Congruent Triangles SSS SAS ASA AAS HL (right triangles only) NO donkey theorem (SSA or ASS) CPCTC (use after the triangles are congruent) Related Conditionals: Converse: switch if and then Inverse: negate if and then Contrapositive: inverse of the converse (contrapositive has the same truth value as the original statement) Triangles: By Sides: Scalene – no congruent sides Isosceles – 2 congruent sides Equilateral – 3 congruent sides By Angles: Acute – all acute angles Right – one right angle Obtuse – one obtuse angle Equiangular – 3 congruent angles(60º) Equilateral ↔ Equiangular Exterior angle of a triangle equals the sum of the 2 non-adjacent interior angles. Mid-segment of a triangle is parallel to the third side and half the length of the third side. Inequalities: --Sum of the lengths of any two sides of a triangle is greater than the length of the third side. --Longest side of a triangle is opposite the largest angle. --Exterior angle of a triangle is greater than either of the two non-adjacent interior angles. Pythagorean Theorem: 2 2 2 c a b = + Converse: If the sides of a triangle satisfy 2 2 2 c a b = + then the triangle is a right triangle. Similar Triangles: AA SSS for similarity SAS for similarity Corresponding sides of similar triangles are in proportion. Mean Proportional in Right Triangle: Altitude Rule: Leg Rule: altitu part hyp = other de al par titude t hyp hyp = proje le c g leg tion Copyright © Regents Exam Prep Center Parallels: If lines are parallel … Corresponding angles are equal. m<1=m<5, m<2=m<6, m<3=m<7, m<4=m<8 Alternate Interior angles are equal. m<3=m<6, m<4=m<5 Alternate Exterior angles are equal. m<1=m<8, m<2=m<7 Same side interior angles are supp. m<3 + m<5=180, m<4 + m<6=180 Quadrilaterals: Parallelogram: opp. sides parallel opp sides = opp angles = consec. angles supp diag bis each other Rectangle: add 4 rt angles, diag. = Trapezoid: Only one set parallel sides. Median of trap is parallel to both bases and = ½ sum bases. Rhombus: add 4 = sides, diag. perp, diag bisect angles. Square: All from above. Isosceles Trap: legs = base angles = diagonals = opp angles supp Transformations: ( , ) ( , ) x axis r x y x y − = − ( , ) ( , ) y axis r x y x y − = − ( , ) ( , ) y x r x y y x = = ( , ) ( , ) y x r x y y x =− = − − ( , ) ( , ) origin r x y x y = − − , ( , ) ( , ) a b T x y x a y b = + + ( , ) ( , ) k D x y kx ky = Glide reflection is composition of a reflection and a translation. Isometry – keeps length. Orientation – label order Circle Segments In a circle, a radius perpendicular to a chord bisects the chord. Intersecting Chords Rule: (segment part)•(segment part) = (segment part)•(segment part) Secant-Secant Rule: (whole secant)•(external part) = (whole secant)•(external part) Secant-Tangent Rule: (whole secant)•(external part) = (tangent)2 Hat Rule: Two tangents are equal. Circle Angles: Central angle = arc Inscribed angle = half arc Angle by tangent/chord = half arc Angle formed by 2 chords Angle formed by 2 tangents, or 2 secants, or a tangent/secant = half the sum of arcs = half the difference of arcs Slopes and Equations: 2 1 2 1 = . y y vertical change m horizontal change x x − = − 1 1 slope-intercept ( ) point-slope y mx b y y m x x = + − = − Coordinate Geometry Formulas: Distance Formula: 2 2 2 1 2 1 ( ) ( ) d x x y y = − + − Midpoint Formula: 1 2 1 2 ( , ) , 2 2 x x y y x y + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ Circles: Equation of circle center at origin: 2 2 2 x y r + = where r is the radius. Equation of circle not at origin: 2 2 2 ( ) ( ) x h y k r − + − = where (h,k) is the center and r is the radius. © 2005 Paul Dawkins Trig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 π θ < < or 0 90 θ ° < < ° . opposite sin hypotenuse θ = hypotenuse csc opposite θ = adjacent cos hypotenuse θ = hypotenuse sec adjacent θ = opposite tan adjacent θ = adjacent cot opposite θ = Unit circle definition For this definition θ is any angle. sin 1 y y θ = = 1 csc y θ = cos 1 x x θ = = 1 sec x θ = tan y x θ = cot x y θ = Facts and Properties Domain The domain is all the values of θ that can be plugged into the function. sinθ , θ can be any angle cosθ , θ can be any angle tanθ , 1 , 0, 1, 2, 2 n n θ π ⎛ ⎞ ≠ + = ± ± ⎜ ⎟ ⎝ ⎠ … cscθ , , 0, 1, 2, n n θ π ≠ = ± ± … secθ , 1 , 0, 1, 2, 2 n n θ π ⎛ ⎞ ≠ + = ± ± ⎜ ⎟ ⎝ ⎠ … cotθ , , 0, 1, 2, n n θ π ≠ = ± ± … Range The range is all possible values to get out of the function. 1 sin 1 θ −≤ ≤ csc 1 andcsc 1 θ θ ≥ ≤− 1 cos 1 θ −≤ ≤ sec 1 andsec 1 θ θ ≥ ≤− tanθ −∞≤ ≤∞ cotθ −∞≤ ≤∞ Period The period of a function is the number, T, such that ( ) ( ) f T f θ θ + = . So, if ω is a fixed number and θ is any angle we have the following periods. ( ) sin ωθ → 2 T π ω = ( ) cos ωθ → 2 T π ω = ( ) tan ωθ → T π ω = ( ) csc ωθ → 2 T π ω = ( ) sec ωθ → 2 T π ω = ( ) cot ωθ → T π ω = θ adjacent opposite hypotenuse x y ( ) , x y θ x y 1 © 2005 Paul Dawkins Formulas and Identities Tangent and Cotangent Identities sin cos tan cot cos sin θ θ θ θ θ θ = = Reciprocal Identities 1 1 csc sin sin csc 1 1 sec cos cos sec 1 1 cot tan tan cot θ θ θ θ θ θ θ θ θ θ θ θ = = = = = = Pythagorean Identities 2 2 2 2 2 2 sin cos 1 tan 1 sec 1 cot csc θ θ θ θ θ θ + = + = + = Even/Odd Formulas ( ) ( ) ( ) ( ) ( ) ( ) sin sin csc csc cos cos sec sec tan tan cot cot θ θ θ θ θ θ θ θ θ θ θ θ − = − − = − − = − = − = − − = − Periodic Formulas If n is an integer. ( ) ( ) ( ) ( ) ( ) ( ) sin 2 sin csc 2 csc cos 2 cos sec 2 sec tan tan cot cot n n n n n n θ π θ θ π θ θ π θ θ π θ θ π θ θ π θ + = + = + = + = + = + = Double Angle Formulas ( ) ( ) ( ) 2 2 2 2 2 sin 2 2sin cos cos 2 cos sin 2cos 1 1 2sin 2tan tan 2 1 tan θ θ θ θ θ θ θ θ θ θ θ = = − = − = − = − Degrees to Radians Formulas If x is an angle in degrees and t is an angle in radians then 180 and 180 180 t x t t x x π π π = ⇒ = = Half Angle Formulas ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 sin 1 cos 2 2 1 cos 1 cos 2 2 1 cos 2 tan 1 cos 2 θ θ θ θ θ θ θ = − = + − = + Sum and Difference Formulas ( ) ( ) ( ) sin sin cos cos sin cos cos cos sin sin tan tan tan 1 tan tan α β α β α β α β α β α β α β α β α β ± = ± ± = ± ± = ∓ ∓ Product to Sum Formulas ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 sin sin cos cos 2 1 cos cos cos cos 2 1 sin cos sin sin 2 1 cos sin sin sin 2 α β α β α β α β α β α β α β α β α β α β α β α β = − − + ⎡ ⎤ ⎣ ⎦ = − + + ⎡ ⎤ ⎣ ⎦ = + + − ⎡ ⎤ ⎣ ⎦ = + − − ⎡ ⎤ ⎣ ⎦ Sum to Product Formulas sin sin 2sin cos 2 2 sin sin 2cos sin 2 2 cos cos 2cos cos 2 2 cos cos 2sin sin 2 2 α β α β α β α β α β α β α β α β α β α β α β α β + − ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + − ⎛ ⎞ ⎛ ⎞ − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + − ⎛ ⎞ ⎛ ⎞ + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ + − ⎛ ⎞ ⎛ ⎞ − = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ Cofunction Formulas sin cos cos sin 2 2 csc sec sec csc 2 2 tan cot cot tan 2 2 π π θ θ θ θ π π θ θ θ θ π π θ θ θ θ ⎛ ⎞ ⎛ ⎞ − = − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − = − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ − = − = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ © 2005 Paul Dawkins Unit Circle For any ordered pair on the unit circle ( ) , x y : cos x θ = and sin y θ = Example 5 1 5 3 cos sin 3 2 3 2 π π ⎛ ⎞ ⎛ ⎞ = = − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 3 π 4 π 6 π 2 2 , 2 2 ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 3 1 , 2 2 ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 1 3 , 2 2 ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 60° 45° 30° 2 3 π 3 4 π 5 6 π 7 6 π 5 4 π 4 3 π 11 6 π 7 4 π 5 3 π 2 π π 3 2 π 0 2π 1 3 , 2 2 ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ 2 2 , 2 2 ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ 3 1 , 2 2 ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ 3 1 , 2 2 ⎛ ⎞ − − ⎜ ⎟ ⎝ ⎠ 2 2 , 2 2 ⎛ ⎞ − − ⎜ ⎟ ⎝ ⎠ 1 3 , 2 2 ⎛ ⎞ − − ⎜ ⎟ ⎝ ⎠ 3 1 , 2 2 ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ 2 2 , 2 2 ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ 1 3 , 2 2 ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ ( ) 0,1 ( ) 0, 1 − ( ) 1,0 − 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360° 0° x ( ) 1,0 y © 2005 Paul Dawkins Inverse Trig Functions Definition 1 1 1 sin is equivalent to sin cos is equivalent to cos tan is equivalent to tan y x x y y x x y y x x y − − − = = = = = = Domain and Range Function Domain Range 1 sin y x − = 1 1 x −≤ ≤ 2 2 y π π − ≤ ≤ 1 cos y x − = 1 1 x −≤ ≤ 0 y π ≤ ≤ 1 tan y x − = x −∞< < ∞ 2 2 y π π − < < Inverse Properties ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 cos cos cos cos sin sin sin sin tan tan tan tan x x x x x x θ θ θ θ θ θ − − − − − − = = = = = = Alternate Notation 1 1 1 sin arcsin cos arccos tan arctan x x x x x x − − − = = = Law of Sines, Cosines and Tangents Law of Sines sin sin sin a b c α β γ = = Law of Cosines 2 2 2 2 2 2 2 2 2 2 cos 2 cos 2 cos a b c bc b a c ac c a b ab α β γ = + − = + − = + − Mollweide’s Formula ( ) 1 2 1 2 cos sin a b c α β γ − + = Law of Tangents ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 2 1 2 1 2 tan tan tan tan tan tan a b a b b c b c a c a c α β α β β γ β γ α γ α γ − − = + + − − = + + − − = + + c a b α β γ Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Limits Definitions Precise Definition : We say ( ) lim x a f x L → = if for every 0 ε > there is a 0 δ > such that whenever 0 x a δ < − < then ( ) f x L ε − < . “Working” Definition : We say ( ) lim x a f x L → = if we can make ( ) f x as close to L as we want by taking x sufficiently close to a (on either side of a) without letting x a = . Right hand limit : ( ) lim x a f x L + → = . This has the same definition as the limit except it requires x a > . Left hand limit : ( ) lim x a f x L − → = . This has the same definition as the limit except it requires x a < . Limit at Infinity : We say ( ) lim x f x L →∞ = if we can make ( ) f x as close to L as we want by taking x large enough and positive. There is a similar definition for ( ) lim x f x L →−∞ = except we require x large and negative. Infinite Limit : We say ( ) lim x a f x → = ∞ if we can make ( ) f x arbitrarily large (and positive) by taking x sufficiently close to a (on either side of a) without letting x a = . There is a similar definition for ( ) lim x a f x → = −∞ except we make ( ) f x arbitrarily large and negative.
Relationship between the limit and one-sided limits ( ) lim x a f x L → = ⇒ ( ) ( ) lim lim x a x a f x f x L + − → → = = ( ) ( ) lim lim x a x a f x f x L + − → → = = ⇒ ( ) lim x a f x L → = ( ) ( ) lim lim x a x a f x f x + − → → ≠ ⇒ ( ) lim x a f x → Does Not Exist Properties Assume ( ) lim x a f x → and ( ) lim x a g x → both exist and c is any number then, 1. ( ) ( ) lim lim x a x a cf x c f x → → = 2. ( ) ( ) ( ) ( ) lim lim lim x a x a x a f x g x f x g x → → → ± = ± 3. ( ) ( ) ( ) ( ) lim lim lim x a x a x a f x g x f x g x → → → = 4. ( ) ( ) ( ) ( ) lim lim lim x a x a x a f x f x g x g x → → → = provided ( ) lim 0 x a g x → ≠ 5. ( ) ( ) lim lim n n x a x a f x f x → → = 6. ( ) ( ) lim lim n n x a x a f x f x → → = Basic Limit Evaluations at ± ∞ Note : ( ) sgn 1 a = if 0 a > and ( ) sgn 1 a = − if 0 a < .
1. lim x x→∞ = ∞ e & lim 0 x x→−∞ = e 2. ( ) lim ln x x →∞ = ∞ & ( ) 0 lim ln x x − → = −∞ 3. If 0 r > then lim 0 r x b x →∞ = 4. If 0 r > and r x is real for negative x then lim 0 r x b x →−∞ = 5. n even : lim n x x →±∞ = ∞ 6. n odd : lim n x x →∞ = ∞ & lim n x x →−∞ = −∞ 7. n even : ( ) lim sgn n x a x b x c a →±∞ + + + = ∞ L 8. n odd : ( ) lim sgn n x a x b x c a →∞ + + + = ∞ L 9. n odd : ( ) lim sgn n x a x c x d a →−∞ + + + = − ∞ L Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Evaluation Techniques Continuous Functions If ( ) f x is continuous at a then ( ) ( ) lim x a f x f a → = Continuous Functions and Composition ( ) f x is continuous at b and ( ) lim x a g x b → = then ( ) ( ) ( ) ( ) ( ) lim lim x a x a f g x f g x f b → → = = Factor and Cancel ( )( ) ( ) 2 2 2 2 2 2 6 4 12 lim lim 2 2 6 8 lim 4 2 x x x x x x x x x x x x x → → → − + + − = − − + = = = Rationalize Numerator/Denominator ( )( ) ( )( ) ( )( ) 2 2 9 9 2 9 9 3 3 3 lim lim 81 81 3 9 1 lim lim 81 3 9 3 1 1 18 6 108 x x x x x x x x x x x x x x x → → → → − − + = − − + − − = = − + + + − = = − Combine Rational Expressions ( ) ( ) ( ) ( ) 0 0 2 0 0 1 1 1 1 lim lim 1 1 1 lim lim h h h h x x h h x h x h x x h h h x x h x x h x → → → → − + − = + + − − = = = − + + L’Hospital’s Rule If ( ) ( ) 0 lim 0 x a f x g x → = or ( ) ( ) lim x a f x g x → ±∞ = ±∞ then, ( ) ( ) ( ) ( ) lim lim x a x a f x f x g x g x → → ′ = ′ a is a number, ∞ or −∞ Polynomials at Infinity ( ) p x and ( ) q x are polynomials. To compute ( ) ( ) lim x p x q x →±∞ factor largest power of x out of both ( ) p x and ( ) q x and then compute limit. ( ) ( ) 2 2 2 2 2 2 4 4 5 5 3 3 3 4 3 lim lim lim 5 2 2 2 2 x x x x x x x x x x x x →−∞ →−∞ →−∞ − − − = = = − − − − Piecewise Function ( ) 2 lim x g x →− where ( ) 2 5 if 2 1 3 if 2 x x g x x x + < − = − ≥− Compute two one sided limits, ( ) 2 2 2 lim lim 5 9 x x g x x − − →− →− = + = ( ) 2 2 lim lim 1 3 7 x x g x x + + →− →− = − = One sided limits are different so ( ) 2 lim x g x →− doesn’t exist. If the two one sided limits had been equal then ( ) 2 lim x g x →− would have existed and had the same value. Some Continuous Functions Partial list of continuous functions and the values of x for which they are continuous.
1. Polynomials for all x. 2. Rational function, except for x’s that give division by zero. 3. n x (n odd) for all x. 4. n x (n even) for all 0 x ≥ . 5. x e for all x. 6. ln x for 0 x > . 7. ( ) cos x and ( ) sin x for all x. 8. ( ) tan x and ( ) sec x provided 3 3 , , , , , 2 2 2 2 x π π π π ≠ − − L L 9. ( ) cot x and ( ) csc x provided , 2 , ,0, ,2 , x π π π π ≠ − − L L Intermediate Value Theorem Suppose that ( ) f x is continuous on [a, b] and let M be any number between ( ) f a and ( ) f b . Then there exists a number c such that a c b < < and ( ) f c M = . Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Derivatives Definition and Notation If ( ) y f x = then the derivative is defined to be ( ) ( ) ( ) 0 lim h f x h f x f x h → + − ′ = . If ( ) y f x = then all of the following are equivalent notations for the derivative. ( ) ( ) ( ) ( ) df dy d f x y f x Df x dx dx dx ′ ′ = = = = = If ( ) y f x = all of the following are equivalent notations for derivative evaluated at x a = . ( ) ( ) x a x a x a df dy f a y Df a dx dx = = = ′ ′ = = = = Interpretation of the Derivative If ( ) y f x = then, 1. ( ) m f a ′ = is the slope of the tangent line to ( ) y f x = at x a = and the equation of the tangent line at x a = is given by ( ) ( )( ) y f a f a x a ′ = + − . 2. ( ) f a ′ is the instantaneous rate of change of ( ) f x at x a = . 3. If ( ) f x is the position of an object at time x then ( ) f a ′ is the velocity of the object at x a = . Basic Properties and Formulas If ( ) f x and ( ) g x are differentiable functions (the derivative exists), c and n are any real numbers, 1. ( ) ( ) c f c f x ′ ′ = 2. ( ) ( ) ( ) f g f x g x ′ ′ ′ ± = ± 3. ( ) f g f g f g ′ ′ ′ = + – Product Rule 4. 2 f f g f g g g ′ ′ ′ − = – Quotient Rule 5. ( ) 0 d c dx = 6. ( ) 1 n n d x n x dx − = – Power Rule 7. ( ) ( ) ( ) ( ) ( ) ( ) d f g x f g x g x dx ′ ′ = This is the Chain Rule Common Derivatives ( ) 1 d x dx = ( ) sin cos d x x dx = ( ) cos sin d x x dx = − ( ) 2 tan sec d x x dx = ( ) sec sec tan d x x x dx = ( ) csc csc cot d x x x dx = − ( ) 2 cot csc d x x dx = − ( ) 1 2 1 sin 1 d x dx x − = − ( ) 1 2 1 cos 1 d x dx x − = − − ( ) 1 2 1 tan 1 d x dx x − = + ( ) ( ) ln x x d a a a dx = ( ) x x d dx = e e ( ) ( ) 1 ln , 0 d x x dx x = > ( ) 1 ln , 0 d x x dx x = ≠ ( ) ( ) 1 log , 0 ln a d x x dx x a = > Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Chain Rule Variants The chain rule applied to some specific functions. 1. ( ) ( ) ( ) ( ) 1 n n d f x n f x f x dx − ′ = 2. ( ) ( ) ( ) ( ) f x f x d f x dx ′ = e e 3. ( ) ( ) ( ) ( ) ln f x d f x dx f x ′ = 4. ( ) ( ) ( ) ( ) sin cos d f x f x f x dx ′ = 5. ( ) ( ) ( ) ( ) cos sin d f x f x f x dx ′ = − 6. ( ) ( ) ( ) ( ) 2 tan sec d f x f x f x dx ′ = 7. [ ] ( ) [ ] [ ] ( ) ( ) ( ) ( ) sec sec tan f x f x f x f x d dx ′ = 8. ( ) ( ) ( ) ( ) 1 2 tan 1 f x d f x dx f x − ′ = + Higher Order Derivatives The Second Derivative is denoted as ( ) ( ) ( ) 2 2 2 d f f x f x dx ′′ = = and is defined as ( ) ( ) ( ) f x f x ′ ′′ ′ = , i.e. the derivative of the first derivative, ( ) f x ′ . The nth Derivative is denoted as ( ) ( ) n n n d f f x dx = and is defined as ( ) ( ) ( ) ( ) ( ) 1 n n f x f x − ′ = , i.e. the derivative of the (n-1)st derivative, ( ) ( ) 1 n f x − . Implicit Differentiation Find y′ if ( ) 2 9 3 2 sin 11 x y x y y x − + = + e . Remember ( ) y y x = here, so products/quotients of x and y will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to differentiate as normal and every time you differentiate a y you tack on a y′ (from the chain rule). After differentiating solve for y′. ( ) ( ) ( ) ( ) ( ) ( ) 2 9 2 2 3 2 9 2 2 2 9 2 9 2 2 3 3 2 9 3 2 9 2 9 2 2 2 9 3 2 cos 11 11 2 3 2 9 3 2 cos 11 2 9 cos 2 9 cos 11 2 3 x y x y x y x y x y x y x y y x y x y y y y x y y x y x y y y y y x y y x y y y x y − − − − − − − ′ ′ ′ − + + = + − − ′ ′ ′ ′ − + + = + ⇒ = − − ′ − − = − − e e e e e e e Increasing/Decreasing – Concave Up/Concave Down Critical Points x c = is a critical point of ( ) f x provided either 1. ( ) 0 f c ′ = or 2. ( ) f c ′ doesn’t exist. Increasing/Decreasing 1. If ( ) 0 f x ′ > for all x in an interval I then ( ) f x is increasing on the interval I. 2. If ( ) 0 f x ′ < for all x in an interval I then ( ) f x is decreasing on the interval I. 3. If ( ) 0 f x ′ = for all x in an interval I then ( ) f x is constant on the interval I. Concave Up/Concave Down 1. If ( ) 0 f x ′′ > for all x in an interval I then ( ) f x is concave up on the interval I. 2. If ( ) 0 f x ′′ < for all x in an interval I then ( ) f x is concave down on the interval I. Inflection Points x c = is a inflection point of ( ) f x if the concavity changes at x c = . Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Extrema Absolute Extrema 1. x c = is an absolute maximum of ( ) f x if ( ) ( ) f c f x ≥ for all x in the domain. 2. x c = is an absolute minimum of ( ) f x if ( ) ( ) f c f x ≤ for all x in the domain. Fermat’s Theorem If ( ) f x has a relative (or local) extrema at x c = , then x c = is a critical point of ( ) f x . Extreme Value Theorem If ( ) f x is continuous on the closed interval [ ] , a b then there exist numbers c and d so that, 1. , a c d b ≤ ≤ , 2. ( ) f c is the abs. max. in [ ] , a b , 3. ( ) f d is the abs. min. in [ ] , a b . Finding Absolute Extrema To find the absolute extrema of the continuous function ( ) f x on the interval [ ] , a b use the following process. 1. Find all critical points of ( ) f x in [ ] , a b . 2. Evaluate ( ) f x at all points found in Step 1. 3. Evaluate ( ) f a and ( ) f b . 4. Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3. Relative (local) Extrema 1. x c = is a relative (or local) maximum of ( ) f x if ( ) ( ) f c f x ≥ for all x near c. 2. x c = is a relative (or local) minimum of ( ) f x if ( ) ( ) f c f x ≤ for all x near c. 1st Derivative Test If x c = is a critical point of ( ) f x then x c = is 1. a rel. max. of ( ) f x if ( ) 0 f x ′ > to the left of x c = and ( ) 0 f x ′ < to the right of x c = . 2. a rel. min. of ( ) f x if ( ) 0 f x ′ < to the left of x c = and ( ) 0 f x ′ > to the right of x c = . 3. not a relative extrema of ( ) f x if ( ) f x ′ is the same sign on both sides of x c = . 2nd Derivative Test If x c = is a critical point of ( ) f x such that ( ) 0 f c ′ = then x c = 1. is a relative maximum of ( ) f x if ( ) 0 f c ′′ < . 2. is a relative minimum of ( ) f x if ( ) 0 f c ′′ > . 3. may be a relative maximum, relative minimum, or neither if ( ) 0 f c ′′ = . Finding Relative Extrema and/or Classify Critical Points 1. Find all critical points of ( ) f x . 2. Use the 1st derivative test or the 2nd derivative test on each critical point. Mean Value Theorem If ( ) f x is continuous on the closed interval [ ] , a b and differentiable on the open interval ( ) , a b then there is a number a c b < < such that ( ) ( ) ( ) f b f a f c b a − ′ = − . Newton’s Method If n x is the nth guess for the root/solution of ( ) 0 f x = then (n+1)st guess is ( ) ( ) 1 n n n n f x x x f x + = − ′ provided ( ) n f x ′ exists. Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Related Rates Sketch picture and identify known/unknown quantities. Write down equation relating quantities and differentiate with respect to t using implicit differentiation (i.e. add on a derivative every time you differentiate a function of t). Plug in known quantities and solve for the unknown quantity. Ex. A 15 foot ladder is resting against a wall. The bottom is initially 10 ft away and is being pushed towards the wall at 1 4 ft/sec. How fast is the top moving after 12 sec? x′ is negative because x is decreasing. Using Pythagorean Theorem and differentiating, 2 2 2 15 2 2 0 x y x x y y ′ ′ + = ⇒ + = After 12 sec we have ( ) 1 4 10 12 7 x = − = and so 2 2 15 7 176 y = − = . Plug in and solve for y′ . ( ) 1 4 7 7 176 0 ft/sec 4 176 y y ′ ′ − + = ⇒ = Ex. Two people are 50 ft apart when one starts walking north. The angleθ changes at 0.01 rad/min. At what rate is the distance between them changing when 0.5 θ = rad? We have 0.01 θ′ = rad/min. and want to find x′ . We can use various trig fcns but easiest is, sec sec tan 50 50 x x θ θ θ θ ′ ′ = ⇒ = We know 0.05 θ = so plug in θ′ and solve. ( ) ( )( ) sec 0.5 tan 0.5 0.01 50 0.3112 ft/sec x x ′ = ′ = Remember to have calculator in radians! Optimization Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for one of the two variables and plug into first equation. Find critical points of equation in range of variables and verify that they are min/max as needed. Ex. We’re enclosing a rectangular field with 500 ft of fence material and one side of the field is a building. Determine dimensions that will maximize the enclosed area. Maximize A xy = subject to constraint of 2 500 x y + = . Solve constraint for x and plug into area. ( ) 2 500 2 500 2 500 2 A y y x y y y = − = − ⇒ = − Differentiate and find critical point(s). 500 4 125 A y y ′ = − ⇒ = By 2nd deriv. test this is a rel. max. and so is the answer we’re after. Finally, find x. ( ) 500 2 125 250 x = − = The dimensions are then 250 x 125. Ex. Determine point(s) on 2 1 y x = + that are closest to (0,2). Minimize ( ) ( ) 2 2 2 0 2 f d x y = = − + − and the constraint is 2 1 y x = + . Solve constraint for 2 x and plug into the function. ( ) ( ) 2 2 2 2 2 1 2 1 2 3 3 x y f x y y y y y = − ⇒ = + − = −+ − = − + Differentiate and find critical point(s). 3 2 2 3 f y y ′ = − ⇒ = By the 2nd derivative test this is a rel. min. and so all we need to do is find x value(s). 2 3 1 1 2 2 2 1 x x = −= ⇒ = ± The 2 points are then ( ) 3 1 2 2 , and ( ) 3 1 2 2 , − . Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Integrals Definitions Definite Integral: Suppose ( ) f x is continuous on [ ] , a b . Divide [ ] , a b into n subintervals of width x ∆ and choose i x from each interval. Then ( ) ( ) 1 lim i b a n i f x dx f x x →∞ = ∞ = ∆ ∑ ∫ . Anti-Derivative : An anti-derivative of ( ) f x is a function, ( ) F x , such that ( ) ( ) F x f x ′ = . Indefinite Integral : ( ) ( ) f x dx F x c = + ∫ where ( ) F x is an anti-derivative of ( ) f x . Fundamental Theorem of Calculus Part I : If ( ) f x is continuous on [ ] , a b then ( ) ( ) x a g x f t dt = ∫ is also continuous on [ ] , a b and ( ) ( ) ( ) x a d g x f t dt f x dx ′ = = ∫ . Part II : ( ) f x is continuous on[ ] , a b , ( ) F x is an anti-derivative of ( ) f x (i.e. ( ) ( ) F x f x dx = ∫ ) then ( ) ( ) ( ) b a f x dx F b F a = − ∫ . Variants of Part I : ( ) ( ) ( ) ( ) u x a d f t dt u x f u x dx ′ = ∫ ( ) ( ) ( ) ( ) b v x d f t dt v x f v x dx ′ = − ∫ ( ) ( ) ( ) ( ) [ ] ( ) [ ] ( ) ( ) u x v x u x v x d f t dt u x f v x f dx ′ ′ = − ∫ Properties ( ) ( ) ( ) ( ) f x g x dx f x dx g x dx ± = ± ∫ ∫ ∫ ( ) ( ) ( ) ( ) b b b a a a f x g x dx f x dx g x dx ± = ± ∫ ∫ ∫ ( ) 0 a a f x dx = ∫ ( ) ( ) b a a b f x dx f x dx = − ∫ ∫ ( ) ( ) cf x dx c f x dx = ∫ ∫ , c is a constant ( ) ( ) b b a a cf x dx c f x dx = ∫ ∫ , c is a constant ( ) ( ) b b a a f x dx f t dt = ∫ ∫ ( ) ( ) b b a a f x dx f x dx ≤ ∫ ∫ If ( ) ( ) f x g x ≥ on a x b ≤ ≤ then ( ) ( ) b a a b f x dx g x dx ≥ ∫ ∫ If ( ) 0 f x ≥ on a x b ≤ ≤ then ( ) 0 b a f x dx ≥ ∫ If ( ) m f x M ≤ ≤ on a x b ≤ ≤ then ( ) ( ) ( ) b a m b a f x dx M b a − ≤ ≤ − ∫ Common Integrals k dx k x c = + ∫ 1 1 1 , 1 n n n x dx x c n + + = + ≠− ∫ 1 1 ln x x dx dx x c − = = + ∫ ∫ 1 1 ln a a x b dx ax b c + = + + ∫ ( ) ln ln u du u u u c = − + ∫ u u du c = + ∫e e cos sin u du u c = + ∫ sin cos u du u c = − + ∫ 2 sec tan u du u c = + ∫ sec tan sec u u du u c = + ∫ csc cot csc u udu u c = − + ∫ 2 csc cot u du u c = − + ∫ tan ln sec u du u c = + ∫ sec ln sec tan u du u u c = + + ∫ ( ) 1 1 1 2 2 tan u a a a u du c − + = + ∫ ( ) 1 2 2 1 sin u a a u du c − − = + ∫ Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. u Substitution : The substitution ( ) u g x = will convert ( ) ( ) ( ) ( ) ( ) ( ) b g b a g a f g x g x dx f u du ′ = ∫ ∫ using ( ) du g x dx ′ = . For indefinite integrals drop the limits of integration. Ex. ( ) 2 3 2 1 5 cos x x dx ∫ 3 2 2 1 3 3 u x du x dx x dx du = ⇒ = ⇒ = 3 3 1 1 1 :: 2 2 8 x u x u = ⇒ = = = ⇒ = = ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 8 5 3 1 1 8 5 5 3 3 1 5 cos cos sin sin 8 sin 1 x x dx u du u = = = − ∫ ∫ Integration by Parts : u dv uv vdu = − ∫ ∫ and b b b a a a u dv uv vdu = − ∫ ∫ . Choose u and dv from integral and compute du by differentiating u and compute v using v dv = ∫ . Ex. x x dx − ∫e x x u x dv du dx v − − = = ⇒ = = − e e x x x x x x dx x dx x c − − − − − = − + = − − + ∫ ∫ e e e e e Ex. 5 3 ln xdx ∫ 1 ln x u x dv dx du dx v x = = ⇒ = = ( ) ( ) ( ) ( ) 5 5 5 5 3 3 3 3 ln ln ln 5ln 5 3ln 3 2 x dx x x dx x x x = − = − = − − ∫ ∫ Products and (some) Quotients of Trig Functions For sin cos n m x xdx ∫ we have the following : 1. n odd. Strip 1 sine out and convert rest to cosines using 2 2 sin 1 cos x x = − , then use the substitution cos u x = . 2. m odd. Strip 1 cosine out and convert rest to sines using 2 2 cos 1 sin x x = − , then use the substitution sin u x = . 3. n and m both odd. Use either 1. or 2. 4. n and m both even. Use double angle and/or half angle formulas to reduce the integral into a form that can be integrated. For tan sec n m x x dx ∫ we have the following : 1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using 2 2 tan sec 1 x x = −, then use the substitution sec u x = . 2. m even. Strip 2 secants out and convert rest to tangents using 2 2 sec 1 tan x x = + , then use the substitution tan u x = . 3. n odd and m even. Use either 1. or 2. 4. n even and m odd. Each integral will be dealt with differently. Trig Formulas : ( ) ( ) ( ) sin 2 2sin cos x x x = , ( ) ( ) ( ) 2 1 2 cos 1 cos 2 x x = + , ( ) ( ) ( ) 2 1 2 sin 1 cos 2 x x = − Ex. 3 5 tan sec x x dx ∫ ( ) ( ) ( ) 3 5 2 4 2 4 2 4 7 5 1 1 7 5 tan sec tan sec tan sec sec 1 sec tan sec 1 sec sec sec x xdx x x x xdx x x x xdx u u du u x x x c = = − = − = = − + ∫ ∫ ∫ ∫ Ex. 5 3 sin cos x x dx ∫ ( ) 2 2 1 1 2 2 2 2 5 4 3 3 3 2 2 3 2 2 2 4 3 3 sin (sin ) sin sin sin cos cos cos sin (1 cos ) cos (1 ) 1 2 cos sec 2ln cos cos x x x x x x x x x x x u u u u u dx dx dx dx u x du du x x x c − − − + = = = = = − = − = + − + ∫ ∫ ∫ ∫ ∫ ∫ Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions. 2 2 2 sin a b a b x x θ − ⇒ = 2 2 cos 1 sin θ θ = − 2 2 2 sec a b b x a x θ − ⇒ = 2 2 tan sec 1 θ θ = − 2 2 2 tan a b a b x x θ + ⇒ = 2 2 sec 1 tan θ θ = + Ex. 2 2 16 4 9 x x dx − ∫ 2 2 3 3 sin cos x dx d θ θ θ = ⇒ = 2 2 2 4 4sin 4cos 2 cos 4 9x θ θ θ = − = = − Recall 2 x x = . Because we have an indefinite integral we’ll assume positive and drop absolute value bars. If we had a definite integral we’d need to compute θ ’s and remove absolute value bars based on that and, if 0 if 0 x x x x x ≥ = − < In this case we have 2 2cos 4 9x θ = − . ( ) ( ) 2 3 sin 2cos 2 2 2 4 9 16 12 sin cos 12csc 12cot d d d c θ θ θ θ θ θ θ θ = = = − + ⌠ ⌡ ∫ ∫ Use Right Triangle Trig to go back to x’s. From substitution we have 3 2 sin x θ = so, From this we see that 2 4 9 3 cot x x θ − = . So, 2 2 2 16 4 4 9 4 9 x x x x dx c − − = − + ∫ Partial Fractions : If integrating ( ) ( ) P x Q x dx ∫ where the degree of ( ) P x is smaller than the degree of ( ) Q x . Factor denominator as completely as possible and find the partial fraction decomposition of the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get term(s) in the decomposition according to the following table. Factor in ( ) Q x Term in P.F.D Factor in ( ) Q x Term in P.F.D ax b + A ax b + ( ) k ax b + ( ) ( ) 1 2 2 k k A A A ax b ax b ax b + + + + + + L 2 ax bx c + + 2 Ax B ax bx c + + + ( ) 2 k ax bx c + + ( ) 1 1 2 2 k k k A x B A x B ax bx c ax bx c + + + + + + + + L Ex. 2 ( )( ) 2 1 4 7 13 x x x x dx − + + ∫ ( ) ( ) 2 2 2 2 ( )( ) 2 1 3 2 2 2 3 16 4 1 1 4 4 3 16 4 1 4 4 7 13 4ln 1 ln 4 8tan x x x x x x x x x x x x dx dx dx x x − + − − + + − + + + = + = + + = − + + + ∫ ∫ ∫ Here is partial fraction form and recombined. 2 2 2 2 4) ( ) ( ) ( )( ) ( )( ) 2 1 1 1 4 4 1 4 ( 7 13 Bx C x x x x x x x A x Bx C A x x + + + − − − + + − + + + = + = Set numerators equal and collect like terms. ( ) ( ) 2 2 7 13 4 x x A B x C B x A C + = + + − + − Set coefficients equal to get a system and solve to get constants. 7 13 4 0 4 3 16 A B C B A C A B C + = − = − = = = = An alternate method that sometimes works to find constants. Start with setting numerators equal in previous example : ( ) ( ) ( ) 2 2 7 13 4 1 x x A x Bx C x + = + + + − . Chose nice values of x and plug in. For example if 1 x = we get 20 5A = which gives 4 A = . This won’t always work easily. Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Applications of Integrals Net Area : ( ) b a f x dx ∫ represents the net area between ( ) f x and the x-axis with area above x-axis positive and area below x-axis negative. Area Between Curves : The general formulas for the two main cases for each are, ( ) upper function lower function b a y f x A dx = ⇒ = − ∫ & ( ) right function left function d c x f y A dy = ⇒ = − ∫ If the curves intersect then the area of each portion must be found individually. Here are some sketches of a couple possible situations and formulas for a couple of possible cases. ( ) ( ) b a A f x g x dx = − ∫ ( ) ( ) d c A f y g y dy = − ∫ ( ) ( ) ( ) ( ) c b a c A f x g x dx g x f x dx = − + − ∫ ∫ Volumes of Revolution : The two main formulas are ( ) V A x dx = ∫ and ( ) V A y dy = ∫ . Here is some general information about each method of computing and some examples. Rings Cylinders ( ) ( ) ( ) 2 2 outer radius inner radius A π = − ( ) ( ) radius width / height 2 A π = Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl. Horz. Axis use ( ) f x , ( ) g x , ( ) A x and dx. Vert. Axis use ( ) f y , ( ) g y , ( ) A y and dy. Horz. Axis use ( ) f y , ( ) g y , ( ) A y and dy. Vert. Axis use ( ) f x , ( ) g x , ( ) A x and dx. Ex. Axis : 0 y a = > Ex. Axis : 0 y a = ≤ Ex. Axis : 0 y a = > Ex. Axis : 0 y a = ≤ outer radius : ( ) a f x − inner radius : ( ) a g x − outer radius: ( ) a g x + inner radius: ( ) a f x + radius : a y − width : ( ) ( ) f y g y − radius : a y + width : ( ) ( ) f y g y − These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the 0 y a = ≤ case with 0 a = . For vertical axis of rotation ( 0 x a = > and 0 x a = ≤ ) interchange x and y to get appropriate formulas. Calculus Cheat Sheet Visit for a complete set of Calculus notes. © 2005 Paul Dawkins Work : If a force of ( ) F x moves an object in a x b ≤ ≤ , the work done is ( ) b a W F x dx = ∫ Average Function Value : The average value of ( ) f x on a x b ≤ ≤ is ( ) 1 b avg a b a f f x dx − = ∫ Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, b a L ds = ∫ 2 b a SA y ds π = ∫ (rotate about x-axis) 2 b a SA x ds π = ∫ (rotate about y-axis) where ds is dependent upon the form of the function being worked with as follows. ( ) ( ) 2 1 if , dy dx ds dx y f x a x b = + = ≤ ≤ ( ) ( ) 2 1 if , dx dy ds dy x f y a y b = + = ≤ ≤ ( ) ( ) ( ) ( ) 2 2 if , , dy dx dt dt ds dt x f t y g t a t b = + = = ≤≤ ( ) ( ) 2 2 if , dr d ds r d r f a b θ θ θ θ = + = ≤ ≤ With surface area you may have to substitute in for the x or y depending on your choice of ds to match the differential in the ds. With parametric and polar you will always need to substitute. Improper Integral An improper integral is an integral with one or more infinite limits and/or discontinuous integrands. Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn’t exist or has infinite value. This is typically a Calc II topic. Infinite Limit 1. ( ) ( ) lim t a a t f x dx f x dx →∞ ∞ = ∫ ∫ 2. ( ) ( ) lim b b t t f x dx f x dx − →−∞ ∞ = ∫ ∫ 3. ( ) ( ) ( ) c c f x dx f x dx f x dx − − ∞ ∞ ∞ ∞ = + ∫ ∫ ∫ provided BOTH integrals are convergent. Discontinuous Integrand 1. Discont. at a: ( ) ( ) lim b b a t t a f x dx f x dx + → = ∫ ∫ 2. Discont. at b : ( ) ( ) lim b t a a t b f x dx f x dx − → = ∫ ∫ 3. Discontinuity at a c b < < : ( ) ( ) ( ) b c b a a c f x dx f x dx f x dx = + ∫ ∫ ∫ provided both are convergent. Comparison Test for Improper Integrals : If ( ) ( ) 0 f x g x ≥ ≥ on [ ) , a ∞ then, 1. If ( ) a f x dx ∞ ∫ conv. then ( ) a g x dx ∞ ∫ conv. 2. If ( ) a g x dx ∞ ∫ divg. then ( ) a f x dx ∞ ∫ divg. Useful fact : If 0 a > then 1 a p x dx ∞ ∫ converges if 1 p > and diverges for 1 p ≤. Approximating Definite Integrals For given integral ( ) b a f x dx ∫ and a n (must be even for Simpson’s Rule) define b a n x − ∆= and divide [ ] , a b into n subintervals [ ] 0 1 , x x , [ ] 1 2 , x x , … , [ ] 1, n n x x − with 0 x a = and n x b = then, Midpoint Rule : ( ) ( ) ( ) ( ) 1 2 b n a f x dx x f x f x f x ≈∆ + + + ∫ L , i x is midpoint [ ] 1, i i x x − Trapezoid Rule : ( ) ( ) ( ) ( ) ( ) ( ) 0 1 2 1 2 2 2 2 b n n a x f x dx f x f x f x f x f x − ∆ ≈ + + + + + + ∫ L Simpson’s Rule : ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 2 2 1 4 2 2 4 3 b n n n a x f x dx f x f x f x f x f x f x − − ∆ ≈ + + + + + + ∫ L TABLE OF INFORMATION FOR 2002 CONSTANTS AND CONVERSION FACTORS 1 unified atomic mass unit, 1 u 1 66 10 27 = = -.
kg 931 MeV/ 2 c Proton mass, mp = × − 167 10 27 .
kg Neutron mass, mn = × − 167 10 27 .
kg Electron mass, me = × − 911 10 31 .
kg Magnitude of the electron charge, e = × − 160 10 19 .
C Avogadro’s number, N0 23 1 6 02 10 = × − .
mol Universal gas constant, R ¼ 8 31 .
/ ( ) J mol K Boltzmann’s constant, kB = × − 138 10 23 .
J/K Speed of light, c = × 3 00 108 .
/ m s Planck’s constant, h hc = × = × = × = × − − − ⋅ ⋅ ⋅ ⋅ 6 63 10 414 10 199 10 124 10 34 15 25 3 .
.
.
.
J s eV s J m eV nm Vacuum permittivity, 0 12 2 2 8 85 10 = × − ⋅ .
/ C N m Coulomb’s law constant, k = = × ⋅ 1 4 9 0 10 0 9 2 2 / .
/ π N m C Vacuum permeability, µ π 0 7 4 10 = × − ⋅ ( ) / T m A Magnetic constant, k = T m A 0 µ π / ( ) / 4 10 7 = − ⋅ Universal gravitational constant, G = ¼ -6 67 10 11 3 2 .
/ m kg s Acceleration due to gravity at the Earth’s surface, g = 9 8 2 . m/ s 1 atmosphere pressure, 1 10 10 10 10 5 2 5 atm N/m Pa = × = × .
. 1 electron volt, 1 160 10 19 eV J = × − .
UNITS Name Symbol meter m kilogram kg second s ampere A kelvin K mole mol hertz Hz newton N pascal Pa joule J watt W coulomb C volt V ohm Ω henry H farad F tesla T degree Celsius oC electron-volt eV PREFIXES Factor Prefix Symbol 109 giga G 106 mega M 103 kilo k 10 2 − centi c 10 3 − milli m 10 6 − micro µ 10 9 − nano n 10 12 − pico p VALUES OF TRIGONOMETRIC FUNCTIONS FOR COMMON ANGLES θ sin θ cos θ tan θ 0o 0 1 0 30o 1/2 3 2 / 3 3 / 37o 3/5 4/5 3/4 45o 2 2 / 2 2 / 1 53o 4/5 3/5 4/3 60o 3 2 / 1/2 3 90o 1 0 ∞ The following conventions are used in this examination.
I. Unless otherwise stated, the frame of reference of any problem is assumed to be inertial.
II. The direction of any electric current is the direction of flow of positive charge (conventional current).
III. For any isolated electric charge, the electric potential is defined as zero at an infinite distance from the charge.
IV. For mechanics and thermodynamics equations, W represents the work done on a system.
Not on the Table of Information for Physics C, since Thermodynamics is not a Physics C topic.
ADVANCED PLACEMENT PHYSICS B EQUATIONS FOR 2002 NEWTONIAN MECHANICS ELECTRICITY AND MAGNETISM u u u u u m u t q p p 0 0 0 2 2 0 2 0 2 2 2 1 2 2 1 2 1 2 2 1 2 1 2 2 2 1 at x x t at a x x m F N a r r m t K m U mgh W F r P W t P F k U kx T m k T g T f F Gm m r U Gm m r net c s s p fric g avg s G G F F a p v J F p F F x F sin q u u q D D D D D D r F v cos cos a = acceleration F = force f = frequency h = height J = impulse K = kinetic energy k = spring constant = length m = mass N = normal force P = power p = momentum r = radius or distance r = position vector T = period t = time U = potential energy u = velocity or speed W = work done on a system x = position m = coefficient of friction q = angle t = torque F q q r q U qV q q r E V d V q r C Q V C A d U QV CV I Q t R A V IR P IV C C C C R R R R F q B F BI B I r BA t B E avg i i i avg i i B B c p i i s i i s i p i m m = = = = = − = = = = = = = = = = = = = = = = = = = − = ∑ ∑ ∑ ∑ ∑ • 1 4 1 4 1 4 1 2 1 2 1 1 1 1 2 0 1 2 0 1 2 0 0 2 0 2 p p p u m p f f u e e E F B A ρ θ θ θ sin sin cos avg A = area B = magnetic field C = capacitance d = distance E = electric field e = emf F = force I = current = length P = power Q = charge q = point charge R = resistance r = distance t = time U = potential (stored) energy V = electric potential or potential difference u = velocity or speed r = resistivity f m = magnetic flux ADVANCED PLACEMENT PHYSICS B EQUATIONS FOR 2002 FLUID MECHANICS AND THERMAL PHYSICS WAVES AND OPTICS p p gh F Vg A A p gy T Q mL Q mc T p F A pV nRT K k T RT M k T W p V Q nc T U Q W U nc T e W Q e T T T buoy avg B rms B V H c H C H 0 1 1 2 2 2 0 1 2 3 2 3 3 r r u u r ru m const.
D D D D D D D D a u A = area c = specific heat or molar specific heat e = efficiency F = force h = depth Kavg = average molecular kinetic energy L = heat of transformation = length M= molecular mass m = mass of sample n = number of moles p = pressure Q = heat transferred to a system T = temperature U = internal energy V = volume u = velocity or speed urms = root-mean-square velocity W = work done on a system y = height a = coefficient of linear expansion µ = mass of molecule r = density u u q q q q l l f n c n n c n n si s f M hi h si s f R d m xm m L d l 1 1 2 2 2 1 0 0 0 1 1 1 2 sin sin sin sin d = separation f = frequency or focal length h = height L = distance M= magnification m = an integer n = index of refraction R = radius of curvature s = distance u = speed x = position l = wavelength q = angle ATOMIC AND NUCLEAR PHYSICS E hf pc K hf h p E m c = = = − = = max ( ) f l D D 2 E = energy f = frequency K = kinetic energy m = mass p = momentum l = wavelength f = work function GEOMETRY AND TRIGONOMETRY Rectangle Triangle Circle Parallelepiped Cylinder Sphere Right Triangle A bh A bh A r C r V wh V r S r r V r S r a b c a c b c a b = = = = = = = + = = + = = = = 1 2 2 2 2 4 3 4 2 2 2 3 2 2 2 2 p p p p p p p q q q sin cos tan A = area C = circumference V = volume S = surface area b = base h = height = length w = width r = radius c a b 90 q ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2002 MECHANICS ELECTRICITY AND MAGNETISM u u u u u m u u w u w w w w a q q w a 0 0 0 0 0 0 0 0 1 2 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 at x x t at a x x m d dt dt m F N W d K m P dW dt P U mgh a r r I I r dm mr m m r I K I t t t k U net g c net cm s fric F F a F p J F p p v F r F r r L r p F x D D r F v t t t a w s s p kx T f T m k T g Gm m r U Gm m r G G 1 2 2 1 2 2 2 1 2 2 1 2 p w p p F r a = acceleration F = force f = frequency h = height I = rotational inertia J = impulse K = kinetic energy k = spring constant = length L = angular momentum m = mass N = normal force P = power p = momentum r = radius or distance r = position vector T = period t = time U = potential energy u = velocity or speed W = work done on a system x = position m = coefficient of friction q = angle t = torque w = angular speed a = angular acceleration F q q r q d Q E dV dr V q r U qV q q r C Q V C A d C C C C I dQ dt U QV CV R A V IR R R R R P IV q d I I d B nI d d dt L dI i i i E i i M p i i s i i c s i p i s m m = = = = − = = = = = = = = = = = = = = = = × = = × = = = − = − • • • ∑ ∑ ∑ ∑ ∑ 1 4 1 4 1 4 1 1 1 2 1 2 1 1 0 1 2 0 0 0 1 2 0 2 0 0 2 p p p k r m m f f e e E F E A F v B B F B B A ø ø dt U LI L = 1 2 2 A = area B = magnetic field C = capacitance d = distance E = electric field e = emf F = force I = current L = inductance = length n = number of loops of wire per unit length P = power Q = charge q = point charge R = resistance r = distance t = time U = potential or stored energy V = electric potential u = velocity or speed r = resistivity f m = magnetic flux k = dielectric constant ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2002 GEOMETRY AND TRIGONOMETRY Rectangle Triangle Circle Parallelepiped Cylinder Sphere Right Triangle A bh A bh A r C r V wh V r S r r V r S r a b c a c b c a b = = = = = = = + = = + = = = = 1 2 2 2 2 4 3 4 2 2 2 3 2 2 2 2 p p p p p p p sin cos tan q q q A = area C = circumference V = volume S = surface area b = base h = height = length w = width r = radius c a b 90 q CALCULUS df dx df du du dx d dx x nx d dx e e d dx x x d dx x x d dx x x x dx n x n e dx e dx x x x dx x x dx x n n x x n n x x -+ 1 1n 1n 1 1 1 1 1 sin cos cos sin , cos sin sin cos TRANSFORMATIONS CHEAT-SHEET! REFLECTIONS: Reflections are a flip. The flip is performed over the “line of reflection.” Lines of symmetry are examples of lines of reflection. Reflections are isometric, but do not preserve orientation. Coordinate plane rules: Over the x-axis: (x, y) (x, –y) Over the y-axis: (x, y) (–x, y) Over the line y = x: (x, y) (y, x) Through the origin: (x, y) (–x, –y) TRANSLATIONS: Translations are a slide or shift. Translations can be achieved by performing two composite reflections over parallel lines. Translations are isometric, and preserve orientation. Coordinate plane rules: (x, y) (x ± h, y ± k) where h and k are the horizontal and vertical shifts. Note: If movement is left, then h is negative. If movement is down, then k is negative. DILATIONS: Dilations are an enlargement / shrinking. Dilations multiply the distance from the point of projection (point of dilation) by the scale factor. Dilations are not isometric, and preserve orientation only if the scale factor is positive. Coordinate plane rules: From the origin dilated by a factor of “c”: (x, y) (cx, cy) From non-origin by factor of “c”: count slope from point to projection point, multiply by “c,” count from projection point. ROTATIONS: Rotations are a turn. Rotations can be achieved by performing two composite reflections over intersecting lines. The resulting rotation will be double the amount of the angle formed by the intersecting lines. Rotations are isometric, and do not preserve orientation unless the rotation is 360o or exhibit rotational symmetry back onto itself. Rotations of 180o are equivalent to a reflection through the origin. Coordinate plane rules: Counter-clockwise: Clockwise: Rule: 90o 270o (x, y) (–y, x) 180o 180o (x, y) (–x, –y) 270o 90o (x, y) (y, –x) |
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The Future in Nuclear Power
Information for Stable Nuclides
For the stable isotopes, in addition to the symbol and the atomic mass number, the number percentage of each isotope in the naturally occurring element is listed, as well as the thermal neutron activation cross section and the mass in atomic mass units (amu). A typical block for a stable nuclide from the Chart of the Nuclides is shown in Figure 8.3.
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Book2016, Sustainable Power Technologies and InfrastructureGalen J. Suppes, Truman S. Storvick
Chapter
Introduction
2021, Physics of Nuclear ReactorsV. Gopalakrishnan
1.5 Nuclear stability
The ability of a nuclide to resist disintegration into parts is its stability. The unstable nuclides are said to be radioactive and the phenomenon of disintegration of such nuclides is called “radioactivity.” The instability and hence the radioactivity could be natural, or induced by human action. There are about 286 primordial nuclides that exist naturally on earth from the time the earth was born. Some of these are radioactive but with very long life and are included in the list of stable nuclides. Only some 252 nuclides out of over 3000 nuclides known are nonradioactive. A radioactive nuclide approaches a stable configuration, by emitting radiation(s) in the form of gamma radiation, or particles like alpha, beta, neutron, etc., or by splitting into fragments. The disintegration process may go through several steps in series until the resultant reaches a stable configuration. The resultant stable nuclide is mostly a nuclide different from the one that disintegrated. The internal energy state of a nucleus is responsible for the stability or instability of a nucleus, and this is affected also by the number of protons and number of neutrons in the nucleus. Table 1.2 gives the distribution of the stable nuclides with respect to even/odd combinations of Z and N. It shows that an even-even combination of Z and N gives high stability and odd-odd the least.
Table 1.2. Stable nuclides.
| | | | | | |
--- --- --- |
| Empty Cell | Even Z Even N | Odd Z Odd N | Even Z Odd N | Odd Z Even N | Total |
| Nonradioactive | 147 | 4 | 53 | 48 | 252 |
| Long-lived | 21 | 4 | 4 | 5 | 34 |
| All primordial | 168 | 8 | 57 | 53 | 286 |
Further, Z or N = 2, 8, 20, 50, 82, or 126 gives high stability to a nucleus, and these numbers are referred to as the “magic-numbers.” If Z and N of a nucleus happen to be individually equal to a magic-number, the configuration is called “doubly magic” leading to exceptional stability. This is explained by the “shell-structure” of the nuclear energies. A nucleus is understood to be formed in a shell-structure, with each energy-shell having a “capacity” for filling with neutrons or protons. Each shell gets “closed” when it is filled to full capacity. Quantum mechanics, together with Pauli's exclusion principle, shows that the capacities of different shells correspond to the (so called) magic numbers. If all the shells of a nucleus are closed, it becomes very stable. A magic number corresponds to the number of protons or neutrons that closes the respective shells. The concept is similar to that seen earlier for the way electrons occupy their positions in the orbits. Pauli's exclusion principle, and the quantum mechanical constraints, as well as the notion of shell closure, apply equally in both the cases. Shell closure in either case indicates high stability.
As mentioned before, the stability of a nucleus is influenced by the numbers of protons and neutrons in it. The hydrogen nucleus 1H has one proton and no neutron. This is the only nuclide without a neutron. An isotope of hydrogen, namely, heavy hydrogen 2H, sometimes called “deuterium” 2D., has a proton and a neutron. Light stable nuclei, except 1H, have generally equal number of protons and neutrons. As the nucleus becomes heavier, the ratio of the number of neutrons to the number of protons, or the “N/Z ratio” of the stable nuclides increases.
As said earlier, the nuclear short-range attractive force is the same between any type of nucleons. As the number of protons increases, the electrostatic repulsive force becomes more and more significant. Since the force due to the neutrons in a nucleus is only attractive, adequate presence of neutrons will help to overcome the average repulsive force per nucleon in the nucleus. As Z increases, nuclear stability demands more neutrons in the nucleus than protons. However, this does not mean that the stability of a nucleus increases with every addition of a neutron. Fig. 1.6 shows the numbers of neutrons and protons in stable nuclei and illustrates how the N/Z ratio deviates from unity. The black squares indicate that the stable nuclides form an approximate line called the “stability-line.” Sometimes it is called the “beta-stability-line.”
Excessive or inadequate number of neutrons, when compared with that on the stability-line, leads to instability of the nucleus, resulting in its radioactivity. A radioactive decay mode gets selected such that the N/Z ratio is changed toward greater stability. If the N/Z ratio is greater than the value required for the stability, the decay is through β− particle (electron) emission. If N/Z is less, the disintegration may be through β+ particle (positron) emission, or α particle (4He nucleus) emission. Another way to increase the N/Z ratio is through electron capture, a phenomenon in which the nucleus captures an orbital electron. More about radioactivity will be seen later in Section 1.7.
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Chapter
Basic Physics of Radionuclide Imaging
2004, Emission TomographyCRAIG LEVIN
E Nuclear Stability
There is a tendency toward instability especially in systems comprising a large number of identical particles confined in a small volume. An unstable nucleus emits photons and/or particles in order to transform itself into a more stable nucleus. This is what is meant by radioactive decay, and nuclides undergoing such transformations are called radionuclides. Studies of radioactivity are the basis for our understanding of the atomic nucleus. The process of nuclide formation favors nuclides with higher binding energy. Larger binding energy and lower mass mean more-stable nuclei. However, the more nucleons, the greater the total binding energy. Thus, a more-appropriate measure of stability is the average binding energy per nucleon, Btot/A. Higher Btot/A values indicate greater stability. For nuclides with mass number A > 20, Btot/A lies between 8 and 9 MeV per nucleon, with a maximum at A ≈ 60 and a slow decrease for higher values of A.
There are certain nucleon configurations that result in more-stable nuclei, and there are a few key factors contributing to this stability. Figure 2 shows schematic plots of neutron number N versus atomic number Z for experimentally determined stable (higher binding energy, lower mass) odd-A and even-A nuclei. The different line thickness for odd- and even-A plots represents the fact that there are more stable isotopes and isotones for even-A nuclei. For light nuclides, the average line of stability clusters around N ≈ Z (equal number of protons and neutrons); for heavier ones, it deviates from this (N ≈ 1.5Z) because of the increasing contribution of the Coulomb repulsive force toward instability for higher Z An excess of neutrons and the accompanying strong nuclear force is required in heavy nuclei to overcome the long-range Coulomb repulsive forces between a large number of protons.
For odd A, usually only one stable isobar exists. For even A, usually only even N -even Z nuclides exist. Even -even nuclides occur most frequently. Of the 281 known stable nuclides, 165 of them are even N -even Z, 53 are even N-odd Z, 57 are odd N -even Z, and only 6 are odd N -odd Z Particularly high stability (higher binding energy) and high abundance with respect to neighboring species are seen for nuclides with N or Z equal to 2, 8, 20, 28, 50, 82, and 126, called the magic numbers. In the shell model it is supposed that the magic numbers reflect effects in nuclei very similar to the closing of electronic shells in atoms. The additional stability associated with even nucleon number (even N or Z or both) reflects the tendency of nuclear stability through the pairing up of identical nucleons (with opposite spins), similar to electrons in atomic orbitals. The further a nuclide is from the line of stability, the more likely it is to be unstable. All nuclides heavier than 209Bi are unstable. Unstable nuclides lying below the line of stability are said to be neutron deficient or proton rich; those above the line are neutron rich or proton deficient. Unstable nuclides generally undergo transformations into those species lying closer to the line of stability. In the radioactive decay of nuclides, the initial unstable nucleus is called the parent; the final, more stable nucleus is termed the daughter.
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Book2004, Emission TomographyCRAIG LEVIN
Chapter
The Future in Nuclear Power
2016, Sustainable Power Technologies and InfrastructureGalen J. Suppes, Truman S. Storvick
Chart of the Nuclides 253
: Information for Stable Nuclides 255
: Information for Unstable Nuclides 255
: Neutron–Proton Ratios 256
: Energy Levels of Atoms 257
: Energy Levels of the Nucleus 257
: Stability of Nuclei 258
: Natural Radioactivity 259
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Topics in Quantum Physics
1985, Physics for Students of Science and EngineeringA.L. Stanford, J.M. Tanner
GROUP B
24.15
: A radioactive nuclide with disintegration constant λ1 decays into a stable nuclide. At t = 0, the number of radioactive atoms is N and the number of stable atoms is zero.
(a)
: If x1(t) is the fraction of radioactive atoms remaining at t > 0, and x2(t) is the number of stable atoms (at t > 0) divided by N, show that
(b)
: Show that
(c)
: Sketch graphs of x1 and x2 as functions of time.
(d)
: Show that x1 + x2 = 1. Why?
24.16
: Consider the conditions of the previous problem, but let the daughter nuclide be radioactive with decay constant λ2.
(a)
: Show that
(b)
: Show that
(c)
: Show that the result of part (b) reduces to the expression for x2 obtained in the previous problem where λ2 = 0.
(d)
: Define the function f(t) by f = x1 + x2 for t > 0. Show that
(e)
: What is the physical significance of each of the results of part (d)?
24.17
: Consider the conditions of the previous problem.
(a)
: Show that the daughter nuclide achieves a maximum population at the time
(b)
: Show that this maximum value for x2 is given by
(c)
: If λ1 = 1.00 and λ2 = 0.50, sketch graphs of x1(t) and x2(t) for 0 < t < 8.
24.18
: Consider a radioactive chain that starts with a nuclide (decay constant = λ1) that disintegrates into a second nuclide (decay constant = λ2). This second nuclide then decays into a third nuclide (decay constant = λ3), and so on. Suppose that the n th member of this chain is stable (λn = 0).
(a)
: If at t = 0 the number of radioactive atoms of type 1 is N and if xk(t) represents the number of atoms (of type k) divided by N, show that
(b)
: Suppose that n = 3. If x2(0) = x3(0) = 0, show that
(c)
: Show that x1 + x2 + x3= 1. Why?
(d)
: Show that 0 < x3 < 1 for 0 < t. Why?
(e)
: If λ1 = 1.00 and λ2 = 0.50, sketch x3(t) for 0 < t < 8.
24.19
: Suppose λ3 > 0 in the previous problem.
(a)
: Show that
(b)
: Show that x3 achieves a maximum at the time t3 which satisfies
(c)
: If x1(0) = 1, x2(0) = x3(0) = 0, λ1 = 1.00, λ2 = 0.50, and λ3 = 0.25, sketch x1, x2, and x3 for 0 < t < 8. Show that t3 ≅ 4.020 and x3 ≅ 0.464.
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Chapter
The atom and nuclear radiation
2022, Nuclear Engineering Mathematical Modeling and SimulationZafar Ullah Koreshi
1.1.1 Nuclear stability
Inside the nucleus, the two competing forces that determine stability are the attractive nuclear force between neutrons and protons and the repulsive electric force between protons. For Z > ~20, N/P>1 as seen in Fig. 1.2 and thus the nuclei are “above” the straight line N=Z. Neutron-rich nuclides get rid of their excess neutrons by emitting beta particles and at the very high end, for high A, nuclei emit heavier alpha particles. Nuclei with “magic numbers” for Z or (A-Z): 2, 8, 20, 28, 50, 82, 126 are strongly bound, more than their neighbors. Doubly magic numbers for A=4, 16, 40,, are bound even stronger than those with magic numbers.
In nature 176 of the 286 (over 60%) primordial stable nuclides have even mass number; 146 stable nuclides are even-even Z, A others are mixed while very few (~3%) are odd-odd.
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Book2022, Nuclear Engineering Mathematical Modeling and SimulationZafar Ullah Koreshi
Chapter
Radioactivity
2020, Nuclear Energy (Eighth Edition)Raymond L. Murray, Keith E. Holbert
3.1 Nuclear Stability
Although the repulsive Coulombic forces of the protons attempt to separate the nucleons, the strong nuclear forces strive to keep the nucleus intact. Stable nuclei are found to have a balance between the number of repulsive protons and the additional neutrons providing cohesion. Fig. 3.1 plots the atomic number versus the number of neutrons (Z versus N) for the known nuclides, revealing a band of nuclear stability. Initially, N ≅ Z in the belt of stability, but for increasing Z values a greater number of neutrons than protons is progressively required. Most stable nuclei possess an even number of protons and/or neutrons. Moreover, magic numbers of 2, 8, 20, 28, 50, 82, and 126 define the total nucleons necessary to complete either the proton or the neutron shell (Goeppert-Mayer, 1949).
Example 3.1
Compare the N/Z ratio for a light and a heavy stable isotope. We arbitrarily select beryllium-9 (49Be) and lead-208 (82208Pb). For the light Be-9, N/Z = (9 − 4)/4 = 1.25 while N/Z = (208 − 82)/82 = 1.54 for Pb-208, which is the heaviest stable nuclide.
Isotopes laying off the line of stability undergo radioactive decay in an effort to reduce their instability. Generally, those radioactive nuclides farthest from the belt of stability have the shortest decay times, commonly expressed as the half-life. Those nuclides positioned above the line are neutron-deficient while those below the line have a neutron excess. Radioactive decay seeks to rebalance the N/Z ratio through a variety of competing decay mechanisms, which are summarized in Table 3.1. Sometimes even after the decay emission, the nucleus remains in an excited state, which is relieved through gamma (γ) emission or internal conversion (IC).
Table 3.1. Radioactive Decay Processes
| N/Z Ratio | Decay Modes | Decay Mechanisms |
---
| Neutron surplus | Beta | Neutron → proton + electron (β−) emission |
| Neutron | Neutron ejection from the nucleus |
| Neutron deficit | Alpha | Emission of a He-4 nucleus |
| Positron | Proton → neutron + positron (β+) emission |
| Electron capture | Orbital electron + proton → neutron |
| Proton | Proton ejection from the nucleus |
The emanations from radioactive decay constitute the radiations. Fig. 3.1 reveals that alpha decay is more prevalent for the heavier nuclei, but another transformation mode exists for heavy radionuclides: spontaneous fission. The graph also discloses that neutron emission tends to occur in comparatively lighter nuclei only. Overall, beta and positron emission and electron capture (EC) are the dominant decay mechanisms.
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Book2020, Nuclear Energy (Eighth Edition)Raymond L. Murray, Keith E. Holbert
Chapter
NUCLEAR RADIATION, ITS INTERACTION WITH MATTER AND RADIOISOTOPE DECAY
2003, Handbook of Radioactivity Analysis (Second Edition)MICHAEL F. L'ANNUNZIATA
1. N/Z Ratios and Nuclear Stability
In Sections II.B and II.C of this chapter we discussed negatron and positron decay as processes whereby unstable nuclei may achieve stability via neutron or proton transformations, respectively. These processes in the nucleus of the radionuclide result in a change in the neutron/proton or N/Z ratio of the nucleus.
If we look throughout The Chart of The Nuclides we will notice that the stable nuclides of low atomic number will have a N/Z ratio of approximately 1. However, as the atomic number increases (Z > 20), the N/Z ratio of the stable nuclides increases gradually and reaches as high as approximately 1.5 (e.g., , Z = 83, N/Z = 1.518). Furthermore, there are no stable nuclides of atomic number greater than 83.
The nature of nuclear forces and the relationship of N/Z ratio to nuclear stability are discussed in detail by Serway et al. (1997) and Sundaresan (2001). In brief, the importance of N/Z ratio to nuclear stability is explained by the fact that there exists a short-range attractive nuclear force, which extends to a distance of ≈2 fm (2 fermi or 2 × 10−15 m). This attractive force has charge independence and is a consequence of the relative spins of the protons and neutrons and their relative positions in the nucleus. These binding exchange forces exist therefore, regardless of charge on the particles, between two protons, two neutrons, and a proton and neutron. While the attractive nuclear forces will tend to hold the nucleus together there exists, at the same time, repelling coulombic forces between the positively charged protons that act to force them apart. For nuclides of low Z, the attractive nuclear forces exceed the repelling coulombic forces when N ≈ Z. However, increasing the number of protons (e.g., Z > 20) further increases the strength of the repelling coulombic forces over a larger nucleus, which will tend to force the nucleus apart. Therefore, additional neutrons, N > Z, provide additional attractive nuclear forces needed to overcome the repelling forces of the larger proton population. As the atomic number increases further, Z > 83, all nuclides are unstable. Even though N/Z ratios reach 1.5, nuclear stability is not achieved when the number of protons in the nucleus exceeds 83.
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Book2003, Handbook of Radioactivity Analysis (Second Edition)MICHAEL F. L'ANNUNZIATA
Chapter
Radioactivity
2005, Encyclopedia of Condensed Matter PhysicsW. Greiner, D.N. Poenaru
Nuclear Stability
The area of experimentally identified nuclei until 2003, covering 2228 species, extends well beyond the central region of about 300 beta-stable nuclides (shown as black squares in Figure 8), both on the proton and neutron-rich sides. A semi-empirical approximation for the line of β-stability, , allows one to estimate which mass number, A, or neutron number, N, corresponds to a given proton number, Z, for a β-stable nucleus. It is displayed as a dotted line for Z>82 in Figure 8. Like the noble gases He, Ne, Ar, Kr, Xe, and Rn, which have an increased stability due to their closed-shell electron configuration (for 2, 10, 18, 36, 54, and 86 electrons) nuclei with the magic number of protons or neutrons, 2, 8, 20, 28, 50, 82, shown in Figure 8, are particularly stable as a consequence of quantum-mechanical laws.
Many nuclides found in nature or produced artificially by nuclear reactions decay by spontaneous emission of certain particles or electromagnetic radiation. The daughter nucleus can in its turn be unstable, and the phenomenon continues until a stable configuration is reached.
The limits of nuclear stability are determined either by hadron-decay modes, or (for heavy nuclei) by α-decay or spontaneous fission. The most important limits are the proton and neutron drip lines, where the proton or the neutron has roughly zero binding energy. A large proton excess is not possible owing to the Coulomb repulsion; for light nuclei, the proton drip line is very close to N=Z. By combining the Coulomb and centrifugal (angular momentum) barriers, the lifetime of proton emission from an unbound state can be long enough to be experimentally detected in a proton-decay experiment. In this way, the proton drip line can be overstepped: one accedes to a nucleus situated outside the drip line. A series of proton-rich In and Sn isotopes (including the doubly magic 100Sn) were produced. The heaviest known proton emitter is . The neutron drip line has been reached only for the lightest nuclides up to .
The existence of extremely heavy nuclei is prevented by the spontaneous fission process. When the number of protons becomes very high, the Coulomb repulsion is no longer counter-balanced by the strong interaction, and the phenomenological liquid drop model (LDM) fission barrier height may approach zero. By adding a shell correction term to the LDM energy of a superheavy nucleus, a potential barrier shows up. This shell-stabilizing property, characteristic of the superheavy nuclei, is already present in the heaviest elements known to date with (see Figure 9). An example of a potential energy surface of the superheavy nucleus with Z = 120 and A = 304 is given in Figure 10.
Owing to shell effects, the fission barriers of these nuclei are so high that they decay mainly by α-emission instead of spontaneous fission. The magic numbers in the superheavy region are still not known. The candidates are Z = 114,120,126 and N = 184.
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Reference work2005, Encyclopedia of Condensed Matter PhysicsW. Greiner, D.N. Poenaru
Chapter
Actinide Elements
2003, Encyclopedia of Physical Science and Technology (Third Edition)Siegfried Hübener
Glossary
Actinyl ion
: Dioxo actinide cations and .
Decay chain
: A series of nuclides in which each member transforms into the next through nuclear decay until a stable nuclide has been formed.
: Fourteen elements with atomic numbers 58 (cerium) to 71 (lutetium) that are a result of filling the 4f orbitals with electrons.
Nuclear fission
: The division of a nucleus into two or more parts, usually accompanied by the emission of neutrons and γ radiation.
Nuclide
: A species of atom characterized by its mass number, atomic number, and nuclear energy state. A radionuclide is a radioactive nuclide.
: Nuclides which were produced during element evolution and which have partly survived since then due to their long halflives.
Radioactivity
: The property of certain nuclides of showing radioactive decay in which particles or γ radiation are emitted or the nucleus undergoes spontaneous fission.
Speciation
: Characterization of physical and chemical states of (actinide) species in a given (chemical) environment.
Transactinide elements
: Artificial elements beyond the actinide elements, beginning with rutherfordium (Rf), element 104. The heaviest elements, synthesized until now, are the elements 114, 116, and 118. At present, bohrium (Bh), element 107, is the heaviest element which has been characterized chemically; chemical studies of element 108, hassium (Hs), and element 112 are in preparation.
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Reference work2003, Encyclopedia of Physical Science and Technology (Third Edition)Siegfried Hübener
Related terms:
Energy Engineering
Gamma Ray
Binding Energy
Fission Product
Atomic Number
Beta Decay
Atomic Mass
Thermal Neutron
Spontaneous Fission
Stable Nucleus
View all Topics |
187682 | https://www.quickanddirtytips.com/qdtarchive/flout-versus-flaunt/ | ‘Flout’ Versus ‘Flaunt’ - Quick and Dirty Tips
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You are at:Home » ‘Flout’ Versus ‘Flaunt’
‘Flout’ Versus ‘Flaunt’
ByqdtstagingJuly 10, 2013No Comments6 Mins Read
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‘Flout’ Versus ‘Flaunt’
Go to the end for a trick to help you remember when to use “flout.”
By
Mignon Fogarty
Grammar Girl
April 9, 2020
4 minute read
Episode #766
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The Quick And Dirty
“Flaunt” means to show off. “Flout” means to disregard.
Last week, a reader named Brad wrote to me and said,
“Recently, I’ve been seeing lots of “flaunt” in place of “flout” (as in, disobeying rules to shelter-in-place, etc.), so I naturally thought of Grammar Girl. Then I turned to Merriam-Webster, and discovered that one of ‘flaunt’’s connotations is indeed “to treat contemptuously [as in someone] flaunted the rules.’
Apparently, so many folks were using ‘flaunt’ in place of ‘flout’ that…well, you know how it goes. Maybe it’s a good time to revisit the difference?”
Let’s start with the basics: “Flaunt”and“flout”sound a lot alike, but they don’t mean the same thing.
What does ‘flaunt’ mean?
Traditionally, when you flaunt yourself, flaunt your wealth, or flaunt your accomplishments, you’re parading them in front of people—you’re showing off.
What does ‘flout’ mean?
“Flout”means “to disregard, scoff at, mock, or show scorn.” If you’re having a big party when your state or country has a shelter-in-place order, you’re flouting the rules. You’re flouting authority.
Can ‘flaunt’ mean ‘flout’?
As Brad noted, so many people have started using the word “flaunt” to mean “flout” that dictionaries have added “to treat the rules with contempt” as an additional definition of “flaunt.” That’s how words and meanings get in the dictionary. But that doesn’t mean you should use it.
Dictionaries don’t say whether definitions are right or wrong, they just record how people use words. And today, most professional writers and editors would still consider it an error to use “flaunt” to mean “disregard the rules.” I’d definitely change it or mark it if I were editing a article or grading a paper.
Garner’s Modern English Usage uses a phrase that I’ve always liked: “careful writers.” Garner says “careful writers” still avoid using “flaunt” to mean “flout.”
Some sources would say “educated writers,” but I like “careful writers” better. You don’t have to be educated to be careful or to care about using precise words.
The origin of ‘flaunt’
The origins of these words are where it gets interesting.
Nobody knows for sure where we got “flaunt” for example, and I’m always surprised when I come across an unknown origin. How could we just not know? People research this stuff.
There are theories, of course. The one I like is that it comes from a Swedish dialect word “flankt” that means “loosely fluttering.” I like the visual image of fluttering your accomplishments in front of people, but the Oxford English Dictionary says the timing of the word entering English makes that origin unlikely. We really just don’t know.
The origin of ‘flout’
“Flout” is even more fun and weird. Dictionaries say it’s related to the word “flute”—like the instrument, but nobody is really sure why that is either. What would disregarding laws have to do with playing the flute. I triple checked just to be sure I was reading everything right.
One theory is that the sound of playing the flute might sound a bit like jeering or derisive whistling. For example, the Oxford English Dictionary says that the Dutch word “fluiten” means both to play the flute and to mock or deride something or someone.
What is a ‘shame flute’ or ‘Schandeflote’?
Then I came across a tidbit on Wikipedia about bad musicians being forced to wear a “flute of shame,” and I thought someone was just making things up. I mean, really…the flute of shame?
But I found things about it in a bunch of books in Google Books too, and it’s often associated with Germany in the Middle Ages where it was called the “Schandflote.”
For example, a 1992 book called “Crime and Punishments,” which I think was part of a Time Life series called the “Library of Unusual and Curious Facts,” said “A shame flute dangling from a German musician’s neck mocked his professional abilities.” Apparently, it wasn’t a real flute—it just looked like a flute—and it somehow locked the musicians fingers in a forced playing position. The travel guide “Frommers Europe” from 2002 said you could see one on display in the Medieval Crime Museum in Germany.
The first citation in the OED for “flout” meaning “to jeer or express contempt for something” is from 1551, and from what I can gather, the shame flute was used to mock musicians around the same time. There are references to a musicians guild in Nuremberg that existed around the same time called the Meistersingers using the shame flute.
I’ve never seen anyone make the connection saying the shame flute is the reason the word “flout” comes from the word “flute,” but it seems like a good theory, or at least a fun theory, because we got to learn about the shame flute!
Quick and Dirty Tip
Getting back to the original question, you should still use “flaunt” to talk about showing off and “flout” to talk about disregarding rules, and neither of them are a good thing. Don’t be a flaunter, and don’t be a flouter.
Remember that you flout laws by linking the ‘out’ in ‘flout‘ with the idea of being anoutlaw or beingoutside society.
Image courtesy of Shutterstock.
About the Author
Mignon Fogarty
Mignon Fogarty is the founder of Quick and Dirty Tips and the author of seven books on language, including the New York Times bestseller “Grammar Girl’s Quick and Dirty Tips for Better Writing.” She is an inductee in the Podcasting Hall of Fame, and the show is a five-time winner of Best Education Podcast in the Podcast Awards. She has appeared as a guest expert on the Oprah Winfrey Show and the Today Show. Her popularLinkedIn Learning courses help people write better to communicate better.
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187683 | https://wikimsk.org/wiki/Bradykinin | Bradykinin - WikiMSK
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1 Roles in the Body
2 Kallikrein-Kinin System
3 Bradykinin Receptors
4 Broader Picture
5 References
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Bradykinin
From WikiMSK
WikiMSK>Concepts>Physiology>Bradykinin
This page is probably complete!
It is awaiting peer review
Bradykinin is a potent inflammatory peptide messenger, and as a kinin is part of the kallikrein-kinin system which has wide ranging effects on the body.
Roles in the Body
Bradykinin is released from damaged tissue following injury or with inflammation. Mast cells also release bradykinin.
Nociceptive system: Bradykinin both activates and sensitises sensory neurons. it is a very potent pain-producing substance. Following the release with injury or inflammation, it excites a high percentage or nociceptors and sensitises them to other noxious stimuli. The sensitisation occurs through activation of B1 and B2 receptors.
Rheumatological system: The des-Arg kinin derivates induce collagen synthesis, fibroblast proliferation, and cytokine release from macrophages.
Vascular system: bradykinin acts as a vasodilator (through local production of NO and prostaglandins), increases vessel permeability, and stimulates the formation of prostacyclin. It has a role in the pathophysiology of C1 esterase inhibitor deficiency.
Renal system: It increases renal flow through vasodilation and inhibits inhibits sodium reabsorption leading to natriuresis. This leads to reduced blood pressure.
Respiratory system: It has bronchoconstrictor effects and plays a role in asthma and rhinitis.
Gastrointestinal system: causes smooth muscle contraction, and is involved in visceral nociception.
Kallikrein-Kinin System
A simplified overview of the kallikrein-kinin system. Enzymes are in italics. Inhibitory factors are in red. Vasoactive compounds active on B2 are in orange. Compounds involved in fibrous tissue homeostasis and active on B1 are in blue.
Kinins are the active peptides of the kallikrein-kinin system . Kinins are released either by tissue or the plasma. These are two distinct proteolytic pathways and are initiated by the activation of serine proteases called plasma kallikrein and tissue kallikrein. Plasma kallikrein and Tissue kallikrein differ in their tissue location, substrate specificity, and each have specific inhibitory factors.
Tissue kallikrein-kinin system: Tissue kallikrein acts to form kallidin from kininogens, which are inactive precursors. Tissue kallikrein is inhibited by kallikrein-binding protein or kallistatin.
Plasma kallikrein-kinin system: Plasma kallikrein is released from prekallikrein by activated factor XII. Prekallikrein reciprocally activates FXII. Plasma kallikrein forms bradykinin from the kininogens. Plasma kallikrein is inhibited by compliment 1 inhibitor protein.
Bradykinin and Kallidin (also known as Lys-bradykinin) are vasoactive, and cause the classical signs of inflammation. Kallidin is similar to bradykinin in its effects, and can be converted to bradykinin by aminopeptidase. They mediate their effects through kinin receptor B2.
The half lives of bradykinin and kallidin are seconds, as they are rapidly hydrolysed by kininase 1 to other biologically active des-Arg kinin derivatives. The des-Arg derviatives are involved in fibrous tissue synthesis. They mediate their effects through kinin receptor B1.
Kininase 2 (Angiotensin converting enzyme ) inactivates bradykinin in the lungs and kidneys. Neuro endopeptidase is also involved in kinin degradation.
Bradykinin Receptors
The B1 and B2 receptors are coupled to G proteins. In doing so they are involved in second messenger signalling at the site of nerve endings. B2 is expressed constitutively (constitutive means a gene is expressed in an ongoing manner) while B1 is expressed with trauma and inflammation.
B2 Receptor - constitutive
In the normal state with intact tissue, bradykinin and kallidin bind to the B2 receptor. B2 is expressed constitutively, including both the central and peripheral nervous system. B2 has a role in acute inflammatory events.
When bradykinin activates B2, there is direct stimulation of nociceptors (C & Aδ fibres). Kinins cause sensitisation of sensory fibres to thermal, chemical, and mechanical stimuli.
There is a cascade of intracellular reactions through the protein kinase C (PKC) pathway that results in an increase in sodium conductance causing membrane depolarisation and influx of calcium through voltage gated calcium channels.
B1 Receptor - inducible
B1 expression is inducible. With chronic inflammation (compare to B2 with acute inflammation) and certain forms of persistent hyperalgesia, B1 is synthesised by the nociceptor cell body. B1 is then transported to the nerve ending membrane. B1 production is an important aspect of neural sensitisation and tissue repair through the des-Arg derivatives.
Note how the tanoxomy is the opposite to COX, with COX-1 being the "constitutive enzyme" and COX-2 being the "inducible one"
Broader Picture
There is a synergy between bradykinin and other inflammatory mediators.
There are many downstream effects with the activation of the bradykinin system. For example it leads to the production of 12-lipoxygenase metabolites from arachidonic acid, and this activates the TRPV1 receptor. TRPV1 is involved in mechanical, thermal, and chemical sensitivity. TRPA1 is also activated which is a mechanosensitive and thermosensitive channel.
Bradykinins, prostaglandins, and histamine are all mediators of vasodilation. Vasodilation being a process involved in clinical signs of inflammation - redness (rubor) and warmth (calor).
Bradykinin, prostaglandin E2, and histamine also all sensitise sensory nerve endings. This leads to the pain (dolor) aspect of inflammation.
Bradykinin acts directs on sensory neurons, and via a separate 12-lipoxygenase pathway, stimulates TRPV1.
References
↑Webb JG. The kallikrein/kinin system in ocular function. J Ocul Pharmacol Ther. 2011 Dec;27(6):539-43. doi: 10.1089/jop.2011.0187. Epub 2011 Dec 2. PMID: 22136090.
↑Shin J, Cho H, Hwang SW, Jung J, Shin CY, Lee SY, Kim SH, Lee MG, Choi YH, Kim J, Haber NA, Reichling DB, Khasar S, Levine JD, Oh U. Bradykinin-12-lipoxygenase-VR1 signaling pathway for inflammatory hyperalgesia. Proc Natl Acad Sci U S A. 2002 Jul 23;99(15):10150-5. doi: 10.1073/pnas.152002699. Epub 2002 Jul 3. PMID: 12097645; PMCID: PMC126639.
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187684 | http://abacus.bates.edu/acad/depts/biobook/Obio10.htm | B140: Taxonomy
Organismal Biology #10
ORGANIC DIVERSITY AND TAXONOMYPerformance Objectives:
After completing this lesson, students will be able to: Describe how evolution results in organic diversity
Describe some of the modern methods used to construct family trees and classifications
Describe how biologists classify species into taxonomic groups based on family trees
Biological diversity is expressed by arranging organisms into kingdoms, phyla, classes, orders, families, genera, and species. These groups reflect evolutionary history and common ancestry as much as possible. Evolutionary relationships responsible for these arrangements are often depicted in family trees. The aim of phylogenetics is to reconstruct family trees and base classifications on them.
Binomial nomenclature: Each species has a two-word name. The first word (capitalized) is the name of the genus; the second (lower case) is the name of the species. Example: Homo sapiens.
The Linnaean system: Uses binomial nomenclature throughout. Species are grouped into genera and genera into higher groups. Any one of these groups, at any level, is called a taxon (plural, taxa). The complete Linnaean hierarchy (ranking) of groups is as follows:
Kingdom (the most inclusive group)
Phylum (plural, phyla, sometimes called a "division" in plants)
Class
Order
Family
Genus (plural, genera)
Species (same spelling in both singular and plural)
(Mnemonic: "King Philip Came Over For Good Soup")
Extra ranks are added to this hierarchy as needed, such as subphylum (just below phylum) or superfamily (just above family).
Example: Humans belong to the Kingdom Animalia, Phylum Chordata, subphylum Vertebrata, class Mammalia, order Primates, family Hominidae, genus Homo, species Homo sapiens.
Phylogeny: A family tree of species.
Phylogenetics: The study of family trees.
Phylogenetic methods use both the fossil record and resemblances among living species as evidence to reconstruct phylogenies. Species sharing many similarities are considered to be descendents of a common ancestor that also shared these similarities. When conflicting evidence arises from different characters, further study is undertaken to see whether some of the similarities could have evolved by convergence.
An important task in phylogenetics is therefore recognizing homology (resemblance due to common ancestry) and distinguishing it from analogy or convergence.
The aim of classification based on phylogenetics is to group together those species that derive their similarities from a common ancestor. That means that, insofar as possible, each taxon should be made monophyletic by including the common ancestor within the taxon.
Taxonomy is the theory behing the making of classifications.
Phenetic taxonomy: Classifications based on resemblance alone have long been in disfavor because they do not distinguish convergence from other causes of resemblance.
Phylogenetic taxonomy: Modern classifications are based on phylogenetics, meaning that species that share a common ancestry are grouped together as much as possible. Strict adherence to this principle is the basis of cladistics. Cladistic taxonomists construct family trees first, then base their classifications strictly on the geometry of branching, ignoring such matters as the diversity or degree of change within each branch. Drawing cladograms
Cladistics example
Monophyletic group (clade):A group which includes a single ancestor and all of its descendants (a natural group).
Paraphyletic group: A group which includes a common ancestor and some but not all of its descendants (an incomplete group).
Polyphyletic group: An unnatural group whose common ancestor is not part of the group.
Plesiomorphy: A primitive trait, possessed by an ancestor.
Symplesiomorphy: A shared plesiomorphy, often defining a paraphyletic group.
Apomorphy: A derived trait, possessed by a descendant but not by an ancestor.
Synapomorphy or homology: A shared trait, derived from a common ancestor, thus defining a monophyletic group.
Homoplasy: A misleading resemblance (such as a convergence), defining a polyphyletic group.
Evolutionary (phylogenetic) classification is based on branching descent:
Biological classification reflects the results of a branching evolutionary process. Insofar as possible, classifications should be genealogical. Each taxon should ideally represent one branch of the evolutionary tree, with the smaller included taxa representing its sub-branches.
Three domains: Most biologists now arrange organisms into three great domains:
Domain Archaea contains only the Kingdom Archaebacteria, a group of oxygen-intolerant procaryotes with RNA sequences different from those of all other organisms.
Domain Bacteria (or Eubacteria) contains only a single kingdom of the same name, including the majority of procaryotes.
Domain Eucarya contains several major kingdoms of eucaryotes:
1. An assortment of simple eucaryotes, called "protists": generally one-celled, containing neither tissues nor embryos
2. Kingdom Mycota: Fungi, with cell walls but no plastids.
3. Kingdom Plantae: Plants, containing plastids and chlorophyll.
4. Kingdom Animalia: Multicellular animals, developing from blastulas.
REVIEW:Study guide and vocabulary
IndexSyllabus
Prevrev. Sept. 2018Next |
187685 | https://www.expresstorussia.com/travel-tips/money-and-currency-in-russia.html | Home
Guide to Russia
Travel Tips
Money and Currency in Russia
Money and Currency in Russia
What is the best currency to take to Russia?
Russia uses the Russian ruble as its currency.
Withdrawing money in rubles from ATMs is the easiest and best way to get local currency. If possible, prior to your trip, order a credit card that lets you withdraw money worldwide for free.
You can bring dollars or Euros to Russia, but it can be difficult to find a place to exchange other currencies. Keep in mind that exchange places at the airport will likely give you a bad rate, so try to wait until you’re in the city.
Some places will offer to accept dollars or euros. This is illegal and basically unnecessary.
The currency used in Russia is called the Russian ruble (рубль) sometimes written as rouble. One ruble is worth 100 kopeks (копейка). The Russian ruble is used in Russia, in Abkhazia and South Ossetia and in the Ukrainian regions of Donetsk and Luhansk. Be careful if you’ve recently been to Belarus, because even though their currency is also called the ruble, it’s different - it’s the Belarusian ruble.
The ruble comes in notes of 50, 100, 200, 500, 1000, 2000 and 5000. Coins exist in the form of 1, 2, 5 or 10 rubles. Kopeks are slowly dying out and are now very rarely used since they’ve become basically worthless. For comparison at the time of this writing: 100 kopeks equals about $0.016.
Pro tip: a lot of places won’t accept a 5000 ruble note, it’s just too big. When withdrawing money, go for 4900 instead to receive smaller bills.
Is Russia expensive to travel?
In general, if you go on a Russian tour from a Western country, you will find that many things are a lot cheaper in Russia. You can find low-budget hostel beds for as little as ₽700 rubles per night, a night in an average 3-star-hotel will start at ₽2500 ($40) for one person in St. Petersburg, and at around ₽3000 ($47) in Moscow. You can buy half a liter of beer at a bar for ₽250 ($4), and a simple cup of coffee in a café averages at ₽150 ($3). For a satisfying meal including drinks you’ll probably spend about ₽1000 ($16) or even less.
Additionally, cultural offerings are usually inexpensive. Visits to a museum will rarely ring up more than ₽500. Also, museums usually have reduced prices for students and seniors, but they might not accept your student card. Still, it’s always worth a try.
Are euros/dollars accepted in Russia?
Every now and then you will come across a price tag in euros, but that doesn’t indicate the means of currency. You are still expected to pay in rubles, and you will see that on your bill. Some places do accept euros / dollars. Restaurants at the airport and in touristy areas as well as hostels and hotels sometimes offer to take euros or dollars instead of rubles, although this is considered illegal. However, the general response to such a question will result in a shake of the head.
Banks and ATMs
Russia’s most common bank is called Sberbank. You can find ATMs and offices everywhere. If you’re an expat looking to start a new bank account in Russia, Sberbank is a simple and convenient option. And if you’re under 26, it costs only ₽150 ($3) per year. However, if you don’t have a Russian credit or debit card, you might want to avoid Sberbank’s ATMs and stick with VTB, UniCredit, Raiffeisen or other options instead, because Sberbank often struggles with hacked ATMs, leaving you with a barred card after using one of their machines.
Bringing money into Russia
If you want to exchange money prior to your trip, most likely, the bank in your home country doesn’t carry Russian rubles and is not able to order them. A lot of banks don’t have Russian rubles in stock, since it’s a very unstable currency to have. In that case, the best money to take is dollars or euros, since they are widely accepted at currency exchange places all around the country. When exchanging, try to stay away from the airport, since exchange rates there are worse than in the city, and you’ll end up losing a good chunk of your money.
The best way to get a hold of rubles is to directly withdraw them from an ATM. Most ATMs in Russia don’t charge you for withdrawing. Some credit cards allow you to withdraw money worldwide for little to no fees. Check your local bank and ask for suitable options.
Do I really need cash?
No! Unless you’re traveling far off to the countryside, the overwhelming majority of establishments in Russia will accept debit or credit card. Sometimes you might get charged a small fee for card transactions, so you should check with your local bank before departure. Other than that, the only real reason to carry cash around in the city is for leaving tips and spontaneously paying taxi rides or buying snacks at kiosks.
How much money should I bring to Russia?
Depending on where and when you travel, Russia can be significantly cheaper than Western Europe. Considering the two capitals: entry tickets to the most prestigious museums won’t exceed $15, a metro journey costs around $0.75, and taxis can cost as little as $1.50 for a short journey. A two or three course meal at a canteen will cost around $3, at a modest restaurant around $8, at a mid-range establishment around $30; local beer in a modest bar will cost $4, and wine $4.50. Payment is in Russian roubles only or by credit card. Work out your itinerary and budget accordingly, with a little extra of course for souvenirs!
Do you tip in Russia?
In restaurants and cafés with table service, you tip around 10%. Usually, the waiter or waitress brings you the check in a box or something similar, picks up the money and comes back with the change. Just leave the tip in the box or on the table. The amount you tip taxi drivers and bartenders is totally up to you, it is also ok not to tip if you are not satisfied with the services provided. A tip in Russia is called чай (chay), which means tea. So if you’re asked what you’re doing, just say на чай (na chai), to the tea. You can find more tipping guidelines in our article.
Can I use traveler’s cheques?
Bringing traveler’s cheques is not really a good idea. Because they are considered to be fairly outdated, it will be difficult for you to find a place to exchange them.
Summary
Russia’s currency is the Russian ruble. You can bring dollars or euros for exchange, but it’s best if you are able to withdraw locally using a bank machine. There are plenty ATMs around, therefore you don’t really need to take much cash with you. Credit and debit cards are accepted almost everywhere and with a bit of planning in advance you can reduce the costs for monetary transactions to a minimum.
Photos by @azbogiviedi @sharonmccutcheon @paxtechnology @eduschadesoares
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187686 | https://people.math.wisc.edu/~jwrobbin/461dir/coordinateGeometry.pdf | Coordinate Geometry JWR Tuesday September 6, 2005 Contents 1 Introduction 3 2 Some Fallacies 4 2.1 Every Angle is a Right Angle!?
. . . . . . . . . . . . . . . . .
5 2.2 Every Triangle is Isosceles!? . . . . . . . . . . . . . . . . . . .
6 2.3 Every Triangle is Isosceles!? -II . . . . . . . . . . . . . . . . .
7 3 Affine Geometry 8 3.1 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 3.2 Affine Transformations . . . . . . . . . . . . . . . . . . . . . .
12 3.3 Directed Distance . . . . . . . . . . . . . . . . . . . . . . . . .
19 3.4 Points and Vectors . . . . . . . . . . . . . . . . . . . . . . . .
20 3.5 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 3.6 Parallelograms . . . . . . . . . . . . . . . . . . . . . . . . . . .
23 3.7 Menelaus and Ceva . . . . . . . . . . . . . . . . . . . . . . . .
24 3.8 The Medians and the Centroid . . . . . . . . . . . . . . . . . .
26 4 Euclidean Geometry 30 4.1 Orthogonal Matrices . . . . . . . . . . . . . . . . . . . . . . .
30 4.2 Euclidean Transformations . . . . . . . . . . . . . . . . . . . .
31 4.3 Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 4.4 Similarity Transformations . . . . . . . . . . . . . . . . . . . .
33 4.5 Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34 4.6 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36 4.7 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 1 4.8 Addition of Angles . . . . . . . . . . . . . . . . . . . . . . . .
39 5 More Euclidean Geometry 43 5.1 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43 5.2 The Circumcircle and the Circumcenter . . . . . . . . . . . . .
44 5.3 The Altitudes and the Orthocenter . . . . . . . . . . . . . . .
44 5.4 Angle Bisectors . . . . . . . . . . . . . . . . . . . . . . . . . .
46 5.5 The Incircle and the Incenter . . . . . . . . . . . . . . . . . .
46 5.6 The Euler Line . . . . . . . . . . . . . . . . . . . . . . . . . .
47 5.7 The Nine Point Circle . . . . . . . . . . . . . . . . . . . . . .
47 5.8 A Coordinate Proof . . . . . . . . . . . . . . . . . . . . . . . .
49 5.9 Simson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . .
50 5.10 The Butterfly . . . . . . . . . . . . . . . . . . . . . . . . . . .
54 5.11 Morley’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . .
54 5.12 Bramagupta and Heron . . . . . . . . . . . . . . . . . . . . . .
54 5.13 Napoleon’s Theorem . . . . . . . . . . . . . . . . . . . . . . .
54 5.14 The Fermat Point . . . . . . . . . . . . . . . . . . . . . . . . .
54 6 Projective Geometry 55 6.1 Homogeneous coordinates . . . . . . . . . . . . . . . . . . . .
55 6.2 Projective Transformations . . . . . . . . . . . . . . . . . . . .
57 6.3 Desargues and Pappus . . . . . . . . . . . . . . . . . . . . . .
60 6.4 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63 6.5 The Projective Line . . . . . . . . . . . . . . . . . . . . . . . .
64 6.6 Cross Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66 6.7 A Geometric Computer . . . . . . . . . . . . . . . . . . . . . .
67 7 Inversive Geometry 69 7.1 The complex projective line . . . . . . . . . . . . . . . . . . .
69 7.2 Feuerbach’s theorem . . . . . . . . . . . . . . . . . . . . . . .
69 8 Klein’s view of geometry 70 8.1 The elliptic plane . . . . . . . . . . . . . . . . . . . . . . . . .
70 8.2 The hyperbolic plane . . . . . . . . . . . . . . . . . . . . . . .
70 8.3 Special relativity . . . . . . . . . . . . . . . . . . . . . . . . .
70 A Matrix Notation 71 B Determinants 73 2 C Sets and Transformations 75 1 Introduction These are notes to Math 461, a course in plane geometry I sometimes teach at the University of Wisconsin. Students who take this course have com-pleted the calculus sequence and have thus seen a certain amount of analytic geometry. Many have taken (or take concurrently) the first course in linear algebra. To make the course accessible to those not familiar with linear al-gebra, there are three appendices explaining matrix notation, determinants, and the language of sets and transformations.
My object is to explain that classical plane geometry is really a subset of algebra, i.e.
every theorem in plane geometry can be formulated as a theorem which says that the solutions of one system of polynomial equations satisfy another system of polynomial equations. The upside of this is that the criteria for the correctness of proofs become clearer and less reliant on pictures.
The downside is evident: algebra, especially complicated but elementary algebra, is not nearly so beautiful and compelling as geometry. Even the weakest students can appreciate geometric arguments and prove beautiful theorems on their own. For this reason the course also includes synthetic arguments as well. I have not reproduced these here but instead refer to the excellent texts of Isaacs and Coxeter & Greitzer as needed. It is my hope that the course as a whole conveys the fact that the foundations of geometry can be based on algebra, but that it is not always desirable to replace traditional (synthetic) forms of argument by algebraic arguments.
The following quote of a quote which I got from page 31 of should serve as a warning.
The following anecdote was related by E.T. Bell page 48.
Young Princess Elisabeth had successfully attacked a problem in elementary geometry using coordinates. As Bell states it, “The problem is a fine specimen of the sort that are not adapted to the crude brute force of elementary Cartesian geometry.” Her teacher Ren´ e Descartes (who invented the coordinate method) said that “he would not undertake to carry out her solution . . .
in a month.” 3 The reduction of geometry to algebra requires the notion of a transfor-mation group. The transformation group supplies two essential ingredients.
First it is used to define the notion of equivalence in the geometry in question.
For example, in Euclidean geometry, two triangles are congruent iffthere is distance preserving transformation carrying one to the other and they are similar iffthere is a similarity transformation carrying one to the other. Sec-ondly, in each kind of geometry there are normal form theorems which can be used to simplify coordinate proofs. For example, in affine geometry every tri-angle is equivalent to the triangle whose vertices are A0 = (0, 0), B0 = (1, 0), C0 = (0, 1) (see Theorem 3.13) and in Euclidean geometry every triangle is congruent to the triangle whose vertices are of form A = (a, 0), B = (b, 0), C = (0, c) (see Corollary 4.14).
This semester the official text is . In past semesters I have used and many of the exercises and some of the proofs in these notes have been taken from that source.
2 Some Fallacies Pictures sometimes lead to erroneous reasoning, especially if they are not carefully drawn. The three examples in this chapter illustrate this. I got them from . See if you can find the mistakes. Usually the mistake is a kind of sign error resulting from the fact that some point is drawn on the wrong side of some line.
4 2.1 Every Angle is a Right Angle!?
R P O E D C B A Q Figure 1: Every Angle is a Right Angle!?
Let ABCD be a square and E be a point with BC = BE.
We will show that ∠ABE is a right angle. Take R to be the midpoint of DE, P to be the midpoint of DC, Q to be the midpoint of AB, and O to be the point where the lines PQ and the perpendicular bisector of DE intersect.
(See Figure 2.1.) The triangles AQO and BQO are congruent since OQ is the perpendicular bisector of AB; it follows that AO = BO. The triangles DRO and ERO are congruent since RO is the perpendicular bisector of DE; it follows that DO = EO.
Now DA = BE as ABCD is a square and E is a point with BC = BE.
Hence the triangles OAD and OBE are congruent because the corresponding sides are equal.
It follows that ∠ABE = ∠OBE −∠ABO = ∠OAD −∠BAO = ∠BAD.
5 2.2 Every Triangle is Isosceles!?
Q R D O C B A Figure 2: An Isosceles Triangle!?
Let ABC be a triangle; we will prove that AB = AC. Let O be the point where the perpendicular bisector of BC and the angle bisector at A intersect, D be the midpoint of BC, and R and Q be the feet of the perpendiculars from O to AB and AC respectively (see Figure 2.2.) The right triangles ODB and ODC are congruent since OD = OD and DB = DC. Hence OB = OC. Also the right triangles AOR and AOQ are congruent since ∠RAO = ∠QAO (AO is the angle bisector) and ∠AOR = ∠AOQ (the angles of a triangle sum to 180 degrees) and AO is a common side. Hence OR = OQ. The right triangles BOR and COQ are congruent since we have proved OB = OC and OR = OQ. Hence RB = QC. Now AR = AQ (as AOR and AOQ are congruent) and RB = QC (as BOR and COQ are congruent) so AB = AR + RB = AQ + QC = AC as claimed.
6 2.3 Every Triangle is Isosceles!? -II HHHHHHHHHHHHHH H D D D D D D D D B X C A Figure 3: AX bisects ∠BAC In a triangle ABC, let X be the point at which the angle bisector of the angle at A meets the segment BC. By Exercise 2.2 below we have XB AB = XC AC .
(1) Now ∠AXB = ∠ACX + ∠CAX = ∠C + 1 2∠A since the angles of a triangle sum to 180 degrees. By the Law of Sines (Exercise 2.1 below) applied to triangle AXB we have XB AB = sin ∠BAX sin ∠AXB = sin 1 2∠A sin(∠C + 1 2∠A) (2) Similarly ∠AXC = ∠ABX + ∠BAX = ∠B + 1 2∠A so XC AC = sin 1 2∠A sin(∠B + 1 2∠A).
(3) From (1-3) we get sin(∠C + 1 2∠A) = sin(∠B + 1 2∠A) so ∠C + 1 2∠A = ∠B + 1 2∠A so ∠C = ∠B so AB = AC so ABC is isosceles.
Exercise 2.1. The law of sines asserts that for any triangle ABC we have sin ∠A BC = sin ∠B CA = sin ∠C AB Prove this by computing the area of ABC in three ways. Does the argument work for an obtuse triangle? What is the sign of the sine?
Exercise 2.2. Prove (1). Hint: Compute the ratio of the area of ABX to the area of ACX in two different ways.
7 3 Affine Geometry 3.1 Lines 3.1. Throughout R denotes the set of real numbers and R2 denotes the set of pairs of real numbers. Thus a point of P ∈R2 is an ordered pair P = (x, y) of real numbers.
Definition 3.2. A line in R2 is a set of form ℓ= {(x, y) ∈R2 : ax + by + c = 0} where a, b, c ∈R and either a ̸= 0 or b ̸= 0 (or both). Three or more points are called collinear iffthere is a line ℓwhich contains them all. Three or more lines are called concurrent iffthey have a common point. Two lines are said to be parallel iffthey do not intersect.
3.3. The two most fundamental axioms of plane geometry are Axiom (1) Two (distinct nonparallel) lines intersect in a (unique) point.
Axiom (2) Two (distinct) points determine a line.
Axiom (1) says that two equations a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 for lines have a unique common solution (the usual case), no common solution (this means that the lines are parallel), or else define the same line (which is case if and only if the equations are nonzero multiples of one another). The latter two cases are characterized by the condition a1b2 −a2b1 = 0 and in the first case the intersection point is x = −c1b2 −c2b1 a1b2 −a2b1 , y = −a1c2 −a2c1 a1b2 −a2b1 .
Axiom (2) says that for any two distinct points P1 = (x1, y1) and P2 = (x2, y2) there is a unique line ℓ= {(x, y) : ax + by + c = 0} containing both. Remark 3.5 below gives a formula for this line.
8 Theorem 3.4. (I) Three points Pi = (xi, yi) are collinear if and only if det x1 y1 1 x2 y2 1 x3 y3 1 = 0.
(II) Three distinct lines ℓi = {(x, y) : aix + biy + ci = 0} are concurrent or parallel if and only if det a1 b1 c1 a2 b2 c2 a3 b3 c3 = 0.
Proof. A determinant is unchanged if one row is subtracted from another.
Hence det x1 y1 1 x2 y2 1 x3 y3 1 = det x1 −x3 y1 −y3 0 x2 −x3 y2 −y3 0 x3 y3 1 .
Evaluating the determinant on the right gives det x1 y1 1 x2 y2 1 x3 y3 1 = (x1 −x3)(y2 −y3) −(x2 −x3)(y1 −y3).
Dividing by (x1 −x3)(x2 −x3) shows that the determinant vanishes if and only if y1 −y3 x1 −x3 = y2 −y3 x2 −x3 .
This last equation asserts that the slope of the line P1P3 equals the slope of the line P2P3. Since P3 lies on both lines, this occurs if and only if the lines are the same, i.e. if and only if the points P1, P2, P3 are collinear.
The above proof assumes that x1, x2 ̸= x3; a special argument is required in the contrary case. We give another proof which handles both cases at the same time. The matrix equation x1 y1 1 x2 y2 1 x3 y3 1 a b c = 0 0 0 says that the points Pi lie on the line ax + by + c = 0. Any nonzero solution (a, b, c) of this equation must have either a ̸= 0 or b ̸= 0 or both. Hence the 9 three points Pi are collinear if and only if this matrix equation (viewed as a system of three homogeneous linear equations in three unknowns (a, b, c)) has a nonzero solution. Part (I) thus follows from the following Key Fact. A homogeneous system of n linear equations in n unknowns has a nonzero solution if and only if the matrix of coefficients has determinant zero.
Part (II) is similar, but there are several cases. The matrix equation a1 b1 c1 a2 b2 c2 a3 b3 c3 x0 y0 1 = 0 0 0 (1) says that the point (x0, y0) lies on each of the three lines aix + biy + ci = 0.
The three lines are parallel (and not vertical) if and only if they have the same slope, i.e. if and only if −a1/b1 = −a2/b2 = −a3/b3. This happens if and only if the matrix equation a1 b1 c1 a2 b2 c2 a3 b3 c3 1 m 0 = 0 0 0 (2) has a solution m. The lines are vertical (and hence parallel) if and only if b1 = b2 = b3 = 0. This happens if and only if the matrix equation a1 b1 c1 a2 b2 c2 a3 b3 c3 0 1 0 = 0 0 0 .
(3) holds. This (and the above Key Fact) proves “only if”. For “if” assume that the matrix equation a1 b1 c1 a2 b2 c2 a3 b3 c3 u v w = 0 0 0 has a nonzero solution (u, v, w).
If w ̸= 0, then x0 = u/w, y0 = v/w satisfies (1). If w = 0 and u ̸= 0, then m = v/u satisfies (2). If w = u = 0, then v ̸= 0 so (1) holds.
10 ((((((((((((((((((((((((((((((( ( • t = −1 • P0 t = 0 • t = 1 2 • P1 t = 1 • t = 2 Figure 4: P = tP1 + (1 −t)P0 Remark 3.5. The point P = (x, y) lies on the line joining the distinct points P1 = (x1, y1) and P2 = (x2, y2) if and only if the points P1,, P2, P are collinear. Thus Theorem 3.4 implies that an equation for this line is det x1 y1 1 x2 y2 1 x y 1 = 0.
It has form ax + by + c = 0 where a = y1 −y2, b = x2 −x1, c = x1y2 −x2y1.
The points P1 and P2 satisfy this equation since a determinant vanishes if two of its rows are the same.
Theorem 3.6. The line connecting the two distinct points P0 = (x0, y0) and P1 = (x1, y1) is given by ℓ= {tP1 + (1 −t)P0 : t ∈R}, i.e. a point P = (x, y) lies on ℓif and only if x = tx1 + (1 −t)x0, y = ty1 + (1 −t)y0 for some t ∈R. (See Figure 4.) Proof. These are the parametric equations for the line as taught in Math 222.
The formula det x0 y0 1 x1 y1 1 x y 1 = t det x0 y0 1 x1 y1 1 x1 y1 1 + (1 −t) det x0 y0 1 x1 y1 1 x0 y0 1 = 0.
11 shows that any point P of form P = tP1+(1−t)P0 lies on the line. Conversely in P lies on the line, choose t to satisfy one of the two parametric equations and then the equation for the line in the form ax + by +c = 0 shows that the other parametric equation holds as well.
Definition 3.7. The line segment connecting points P0 and P1 is the set [P0, P1] = {tP1 + (1 −t)P0 : 0 ≤t ≤1}.
We say that a point P on the line joining P0 and P1 lies between P0 and P1 iffit lies in the segment [P0, P1]. We call P0 and P1 the end points of the line segment. The ray emanating from P0 in the direction of P1 is the set {tP1 + (1 −t)P0 : t ≥0}. We call P0 the initial point of the ray.
Definition 3.8. (Some general terminology.) An ordered sequence (P1, P2, . . . , Pn) of n distinct points no three of which are collinear is called a polygon or an n-gon. The points are called the vertices of the polygon. Two consecutive vertices in the list are said to be adjacent and also the vertices Pn and P1 are adjacent. The lines and line segments joining adjacent vertices are called the sides of the polygon and the other lines and line segments joining vertices are called diagonals. The term extended side is employed if we want to emphasize that the line and not the line segment is intended. for emphasis.
but the adjective is often omitted. A 3-gon is also called a triangle, a 4-gon is also called a quadrangle, a 5-gon is also called a pentagon, a 6-gon is also called a hexagon, etc. For triangles the two notations (A, B, C) and △ABC are synonymous; the latter is more common.
3.2 Affine Transformations Definition 3.9. A transformation T : R2 →R2 is called affine iffit has the form (x′, y′) = T(x, y) ⇐ ⇒ x′ = ax + by + p y′ = cx + dy + q .
where ad −bc ̸= 0.
12 3.10. It is convenient to use matrix notation to deal with affine transforma-tions. To facilitate this we will we not distinguish between points and column vectors, i.e. we write both P = (x, y) and P = x y .
Then, in matrix notation, an affine transformation takes the form T(P) = MP + V, where T(P) = x′ y′ , M = a b c d , P = x y , V = p q , and det(M) ̸= 0.
Theorem 3.11. The set of all affine transformations is a group, i.e.
(1) the identity transformation I(P) = P is affine, (2) the composition T1 ◦T2 of two affine transformations T1 and T2 is affine, and (3) the inverse T −1 of an affine transformation T is affine.
Proof. The identity transformation I has the requisite form with a = d = 1 and b = c = p = q = 0.
Suppose T1 and T2 are affine, say T1(P) = M1P + V1 and T2(P) = M2P + V2.
Then (T1 ◦T2)(P) = T1(T2(P)) = M1(M2P +V2)+V1 = MP +V where M = M1M2 and V = M1V +V2. Since det(M1M2) = det(M1) det(M2) ̸= 0, this shows that (T1 ◦T2) is affine. To compute the inverse transformation we solve the equation P ′ = T(P) for P; we get P ′ = MP + V ⇐ ⇒MP = P ′ −V ⇐ ⇒P = M −1P ′ + M −1V . In other words, T −1(P ′) = M ′P ′ + V ′ where M ′ = 1 ad −bc d −b −c a , V ′ = 1 ad −bc dp −bq −cp + aq .
This shows that the inverse T −1 of the affine transformation T is itself an affine transformation.
13 Theorem 3.12. An affine transformation maps lines onto lines, line seg-ments onto line segments, and rays onto rays.
Proof. Let ℓbe a line and T be an affine transformation; the theorem asserts that the image T(ℓ) = {T(P) : P ∈ℓ} is again a line. Fix two distinct points P0, P1 ∈ℓ. Let M and V be the matrices which define T, i.e.
T(P) = MP + V .
Choose P ∈ℓ.
Then P = (1 −t)P0 + tP1 for some t ∈R. Hence T(P) = M (1 −t)P0 + tP1 + V = (1 −t) MP0 + V + t MP1 + V = (1 −t)T(P0) + tT(P1) which shows that T(P) lies on the line ℓ′ connecting T(P0) and T(P1). The same argument (reading T −1 for T) shows that if P ′ ∈ℓ′ then T −1(P ′) ∈ℓ.
Hence T(ℓ) = ℓ′ as claimed. Reading 0 ≤t ≤1 for t ∈R proves the theorem for line segments. Reading t ≥0 for t ∈R proves the theorem for rays.
Theorem 3.13. For any two triangles △ABC and △A′B′C′ there is a unique affine transformation T such that T(△ABC) = △A′B′C′, i.e. T(A) = A′, T(B) = B′, and T(C) = C′.
Proof. Let A = (a1, a2), B = (b1, b2), C = (c1, c2). Define T0 by T0(x, y) = (x′, y′) where x′ y′ = a1 −c1 b1 −c1 a2 −c2 b2 −c2 x y + c1 c2 .
Then T0(A0) = A, T0(B0) = B, T0(C0) = C where A0 = (1, 0), B0 = (0, 1), C0 = (0, 0). As in the proof of Theorem 3.4 we have det a1 a2 1 b1 b1 1 c1 c2 1 = det a1 −c1 b1 −c1 a2 −c2 b2 −c2 and this is nonzero since A, B, C are not collinear. Hence T0 is an affine transformation.
Similarly there is an affine transformation T1 such that T1(A0) = A, T1(B0) = B′, T1(C0) = C′ By Theorem 3.11 T := T1 ◦T −1 0 14 is an affine transformation. It satisfies the conclusion of the theorem. For example, T(A) = T2(T −1 1 (A)) = T2(A0) = A′.
To prove uniqueness let T ′ be another affine transformation such that T ′(A) = A′, T ′(B) = B′, and T ′(C) = C′.
Let I = T −1 2 ◦T ′ ◦T1.
By Theorem 3.11 I is affine so I(P) = MP + V for some 2 × 2 matrix M and 2 × 1 matrix V . Also I(C0) = C0, I(A0) = A0, and I(B0) = B0. From V = MC0 + V = I(C0) = C0 = 0 it follows that V is the zero matrix and from MA0 = A0 and MB0 = B0 it follows that M is the identity matrix.
Hence I is the identity transformation. Thus T = T2 ◦T −1 1 = T2 ◦I ◦T −1 1 = T2 ◦T −1 2 ◦T ′ ◦T1 ◦T −1 1 = T ′ as required.
Remark 3.14. Often it is possible to find a coordinate proof of a theorem which is both straight forward and uncomplicated by using affine transfor-mations. For example, imagine a theorem involving five points A, B, C, D, and three lines a, b, c, and suppose the hypothesis includes the condition that A, B, C are the vertices of a triangle, i.e. they are not collinear. We can the prove the theorem by arguing as follows: Choose an affine transformation T with T(A0) = A, T(B0) = B, T(C0) = C. Let D0 = T −1(D), a0 = T −1(a), b0 = T −1(b), c0 = T −1(c). Prove the theorem for A0, B0, C0, D0, a0, b0, c0. Check that the hypotheses and conclusion are preserved by affine trans-formations. This will be true if the hypotheses and conclusion involve only assertions about lines (see Theorem 3.12, ratios of collinear distances (see Theorem 3.25), and ratios of areas (see Theorem 3.32). For example, if the lines a and b are parallel, then so are a0 and b0. We can now conclude that the theorem holds for A, B, C, D, a, b, c. We will often signal this kind of proof by saying Choose affine coordinates (x, y) so that . . .. The following theorem illustrates this technique.
Theorem 3.15 (Parallel Pappus’ Theorem). Assume that the three points A, B, C are collinear and that the three points A′, B,′ C′ are collinear. Let the lines joining them in pairs intersect as follows: X = BC′ ∩B′C, Y = CA′ ∩C′A, Z = AB′ ∩A′B.
(See Figure 5.) If the lines ABC and A′B′C′ are parallel, then the points X,Y , Z are collinear. (This theorem is a special case of Theorem 6.19 below.) 15 • A′ • B′ • C′ • A • B • C Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z H H H H H H H H H H H H H H H H H H H H H H Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z • Z • Y • X Figure 5: Parallel Pappus’ Theorem Proof. Choose affine coordinates (x, y) so that the line ABC has equation y = 0 and the line A′B′C′ has equation y = 1. Then A = (a, 0), B = (b, 0), C = (c, 0), A′ = (a′, 1), B′ = (b′, 1), C′ = (c′, 1). The equation of the line AB′ is 0 = a 0 1 b′ 1 1 x y 1 = −x + (b′ −a)y + a and similarly the equation of the line A′B is x + (b −a′)y −b = 0. To find the intersection point Z = (z1, z2) we solve these two equations. The result is z2 = a −b a −b + a′ −b′, z1 = aa′ −bb′ a −b + a′ −b′.
Now calculate the coordinates of X = (x1, x2) and Y = (y1, y2) by cyclically permuting the symbols and then use Theorem 3.4. (Note that the columns of the resulting matrix sum to zero.) Exercise 3.16. Do the calculations required to complete the proof of The-orem 3.15.
Theorem 3.17 (Parallel Desargues Theorem). Let the corresponding sides of two triangles △ABC and △A′B′C′ intersect in X = BC ∩B′C′, Y = CA ∩C′A′, Z = AB ∩A′B′.
(See Figure 6.) If the lines AA′, BB′, CC′ are parallel, then the points X,Y , Z are collinear. (This theorem is a special case of Theorem 6.16 below.) 16 • A •B •C • A′ • B′ •C′ PPPPPPPP P A A A A AA PPPPPPPP P A A A A AA PPPPP P A A A A A A A A • Z • X • Y Figure 6: Parallel Desargues Theorem Proof. Choose coordinates (x, y) so that the lines AA′, BB′, CC′ are the vertical lines x = a, x = b, x = c. Then A = (a, p), A′ = (a, p′), B = (b, q), B′ = (b, q′), C = (c, r), C′ = (c, r′). The line AB has equation a p 1 b q 1 x y 1 = 0 i.e. (p −q)x + (b −a)y + aq −bp = 0. Similarly, the equation of line A′B′ is (p′ −q′)x + (b −a)y + aq′ −bp′ = 0. The two lines intersect in the solution of the matrix equation q −p a −b q′ −p′ a −b x y = aq −bp aq′ −bp′ so the intersection is Z = (z1, z2) where z1 = bp −aq + aq′ −bp′ p −q −p′ + q′ , z2 = pq′ −p′q p −q −p′ + q′ Similarly X = (x1, x2) where x1 = cq −br + br′ −cq′ q −r −q′ + r′ , x2 = qr′ −q′r q −r −q′ + r′ and Y = (y1, y2) where y1 = ar −cp + cp′ −ar′ r −p −r′ + p′ , y2 = rp′ −r′p r −p −r′ + p′ 17 At this point we could complete the proof using Theorem 3.4 (see Re-mark 3.18 below), but here is a trick which finishes the proof more easily.
The proof uses the following two assertions: (I) An affine transformation of form T(x, y) = (x0 + x, y0 + wx + y) transforms each vertical line x = k to another vertical line.
(II) Given any line ℓ= {(x, y) : y = mx + y0} there is an affine transforma-tion T as in part (I) such that T(ℓ) is the x-axis.
Using these two facts we may suppose w.l.o.g. that the line XY is the x-axes, i.e. that x2 = y2 = 0. From the above formulas it follows that qr′ = q′r and rp′ = r′p. Multiplying these two equations and dividing by rr′ gives qp′ = q′p, i.e. z2 = 0. Hence Z lies on the x-axis as well, i.e. the points X, Y , Z are collinear as required.
Remark 3.18. To show that X, Y , Z in Theorem 3.17 are collinear we could show that the determinant x1 x2 1 y1 y2 1 z1 z2 1 = 1 m cq −br + br′ −cq′ qr′ −q′r q −r −q′ + r′ ar −cp + cp′ −ar′ rp′ −r′p r −p −r′ + p′ bp −aq + aq′ −bp′ pq′ −p′q p −q −p′ + q′ vanishes. Here m = (q −r −q′ + r′)(r −p −r′ + p′)(p −q −p′ + q′). A three by three determinant has six terms each of which has three factors: one from the first column, one from the second, and one from the third. In the case at hand the first and third factors have four terms each and the second has two. Thus the fully expanded determinant has 6×4×2×4 = 192 terms. I evaluated it using a computer program (Maple) which does symbolic calculation and all the terms cancel leaving zero. If you believe in computers, this is an alternative proof.
Exercise 3.19. Complete the proof of Theorem 3.17 by proving (I) and (II) in the proof.
Exercise 3.20. The proof of Theorem 3.17 assumes that m ̸= 0, i.e. that the three numbers p −p′, q −q′, r −r′ are distinct. What if this is false?
Exercise 3.21. The last step in the proof of Theorem 3.17 assumes that rr′ ̸= 0. What if this is false? Hint: If rr′ = 0 then either r = 0 or r′ = 0.
There are three cases: r = r′ = 0, r ̸= r′ = 0, r′ ̸= r = 0, and the last two are treated the same way.
18 3.3 Directed Distance Theorem 3.22. Let P = (1 −t)P0 + tP1 and Q = (1 −s)P0 + sP1 be two points on the line P0P1. Then the distance1 |PQ| between P and Q is given by PQ = |s −t| P0P1 Proof. P −Q = (s −t) P0 −P1 .
Definition 3.23. The directed distance (PQ) from P to Q in the direction from P0 to P1 is defined by (PQ) = (s −t) P0P1 .
3.24. Distances are always nonnegative. However, the directed distance can be negative. For example, this is the case if the points appear on the line in the order P0, P1, Q, P, i.e. if P1 is between P0 and Q and Q is between P1 and P. (See Definition 3.7.) Interchanging P0 and P1 reverses the sign of the directed distance and hence leaves a ratio of directed distances un-changed. Most affine transformations do not preserve distances; those which do preserve distance are called Euclidean transformations and will be stud-ied in Section 4. However, affine transformations preserve ratios of collinear distances. In fact, Theorem 3.25. Affine transformations preserve ratios of collinear directed distances.
Proof. As in the proof of Theorem 3.12, T(P) = (1−t)T(P0)+tT(P1). Hence T(P) −T(Q) = (s −t) T(P0) −T(P1) so (P ′Q′) (P ′ 0P ′ 1) = s −t = (PQ) (P0P1) for P ′ 0 = T(P0), P ′ 1 = T(P1), P ′ = T(P), Q′ = T(Q).
Corollary 3.26. Affine transformations preserve midpoints of segments.
Proof. The midpoint of the segment [A, B] is the unique point M on the line AB which is equidistant from A and B. It is given by M = 1 2(A + B) (Read A = P0, B = P1, M = P and t = 1/2 in the parametric equation P = (1 −t)P0 + tP for a line.) 1See Definition 4.5 below.
19 3.4 Points and Vectors Definition 3.27. The difference W = P1 −P0 between two points P0 and P1 is called the vector from P0 to P1.
Remark 3.28. An affine transformation T(P) = MP where V = 0 is called a linear transformation.2 These are studied in the first course in linear algebra. An affine transformation T is a linear transformation if and only if it fixes the origin , i.e. if and only if T(0) = 0. When points undergo an affine transformation, the corresponding vectors undergo a linear transformation.
This means the following. If T(P) = MP + V is an affine transformation, and P ′ 0 = T(P0), P ′ 1 = T(P1) are the images of points P0, P1 under T, then the vectors W = P1 −P0 and W ′ = P ′ 1 −P ′ 0 are related by the formula W ′ = MW.
Contrast this with the formula P ′ = MP + V for P ′ = T(P). The set of all linear transformations form a group: the fact that the composition of two linear transformations is again linear follows from the associative law for matrix multiplication, i.e. M2(M1W) = (M2M1)W.
Remark 3.29. An affine transformation T(P) = P + V where M is the identity matrix is called a translation. The set of all translations forms a group: the identity transformation is the translation with V = 0, the composition of two translations T1(P) = P + V1 and T2(P) = P + V2 is T2 ◦T1(P) = P + (V1 + V2) and the inverse transformation of the translation T(P) = P + V is T −1(P) = P −V.
3.5 Area In Math 222 you learn how to compute the area of a parallelogram using determinants (cross products). We’ll take this as the definition of area. The calculations in this section are easy if you are familiar with matrix algebra.
The neophyte can skip the proofs in this section.
2The reader is cautioned that some authors use the term linear transformation for what we call affine transformation.
20 Definition 3.30. The oriented area of a triangle △ABC is half the deter-minant (ABC) := 1 2 det[B −A C −A] of the 2 × 2 matrix whose columns are the edge vectors B −A and C −A from A. Thus if A = (a1, a2), B = (b1, b2), C = (c1, c2), then (ABC) = 1 2 det b1 −a1 c1 −a1 b2 −a2 c2 −a2 The area is the absolute value of the oriented area.
Remark 3.31. The oriented area of △ABC is positive (and hence equal to the area) when the points A, B, C occur in counter clockwise order.
Theorem 3.32. An affine transformation preserves ratios of oriented areas, i.e. it changes all oriented areas by the same factor.
Proof. Suppose that T(P) = MP +V . Then T(B)−T(A) = M(B −A) and T(C) −T(A) = M(C −A). Hence we have an equality of 2 × 2 matrices T(B) −T(A) T(C) −T(A) = M B −A C −A .
Taking the determinant of both sides and using the fact that the determinant of a product is the product of the determinants shows that the oriented area of T(△ABC) is the determinant of M times the oriented area of △ABC.
Theorem 3.33. The oriented areas of the four triangles obtained by deleting a vertex of a quadrangle satisfy (ABD) + (DBC) = (ABC) + (CDA).
Proof. (See Figure 7.) The determinant is linear in its columns, reverses sign if the columns are interchanged, and vanishes if two of the columns are the same. From this we see that (ABC) = det[A B] + det[B C] + det[C A].
Combining this with the corresponding formulas for the other three triangles in the quadrangle gives the result.
21 HHHH H @ @ @ @ @ A B C D • • • • HHHH H @ @ @ @ @ A B C D • • • • Figure 7: Two ways to compute the area of a quadrangle Definition 3.34. The oriented area of the convex quadrangle (A, B, C, D) is the sum of the oriented areas of the two triangles △ABC and △CDA, i.e. by definition it is the sum of the oriented areas of the two triangles with common side AC. By Theorems 3.33 this is the same as the sum of the oriented areas the two triangles with common side DB. (See Figure 7.) In other words the oriented area of a quadrangle is independent of the choice of the diagonal used to compute it. Figure 8 shows that this might not work for areas (as opposed to oriented areas) because one of the terms in the formula can have the wrong sign.
Definition 3.35. Two points P1 = (x1, y1) and P2 = (x2, y2) are said to be separated by the line ℓ:= {(x, y) : ax + by + c = 0} if ax1 + by1 + c and ax2 + by2 + c have opposite signs. This gives a precise meaning to the phrase “P1 and P2 lie on opposite sides of ℓ”. A polygon is called convex iffno edge separates any two of the remaining vertices. For example, the quadrangle on the right in Figure 8 is not convex because C and D lie on opposite sides of the edge AB.
HHHH H @ @ @ @ @ A B C D • • • • @ @ @ @ @ HHHH H A B C D • • • • Figure 8: Convex and not convex 22 Theorem 3.36. If the quadrangle (A, B, C, D) is convex, then all four terms in the formula in Theorem 3.33 have the same sign. Hence Theorem 3.33 holds for convex quadrangles and area (instead of oriented area).
Proof. By Theorems 3.13 and 3.32 and Exercise 3.39, we may assume w.l.o.g.
that A = (0, 0), B = (1, 0) and C = (0, 1). Now use Exercise 3.40.
Remark 3.37. One could can carry out the theory of this section for any polygon and prove algebraically that the oriented area of any polygon may be defined by breaking it up triangles and that this area is independent of how the polygon is thus broken up. Similarly for areas and convex polygons.
Exercise 3.38. Show that (ABC) = (BCA) = (CAB) = −(BAC) = −(ACB) = −(CBA).
Exercise 3.39. Let T be an affine transformation, ℓbe a line, and A and B be points not on ℓ. Show that ℓseparates A and B if and only if T(ℓ) separates T(A) and T(B). Hint: Given a, b, c there are numbers a′, b′, c′ such that (x′, y′) = T(x, y) = ⇒ac + by + c = a′x′ + b′y′ + c′.
If ℓ= {(x, y) : ax + by + c = 0}, then T(ℓ) = {(x′, y′) : a′x′ + b′y′ + c = 0}.
Exercise 3.40. Let A = (1, 0), B = (0, 0), C = (0, 1), and D = (d1, d2).
Calculate the four areas (ABD), (DBC), (ABC), (CDA) and verify the formula in Theorem 3.33. Use the definition to show that the quadrangle (A, B, C, D) is convex if and only if d1 > 0, d2 > 0, and d1 + d2 > 1 and that in this case all four oriented areas are positive.
3.6 Parallelograms Definition 3.41. A parallelogram is a quadrangle (A, B, C, D) such that the lines AB and CD are parallel and the lines AD and BC are parallel.
Theorem 3.42. A quadrangle (A, B, C, D) is a parallelogram if and only if B −A = C −D.
Proof. Let A = (a1, a2), B = (b1, b2), C = (c1, c2), D = (d1, d2). Two lines are parallel if and only if they have the same slope. Hence (A, B, C, D) is a parallelogram if and only if b2 −a2 b1 −a1 = c2 −d2 c1 −d1 and d2 −a2 d1 −a1 = c2 −b2 c1 −b1 .
23 The first equation clearly holds if B −A = C −D since the numerators and denominators are equal. But B −A = C −D = ⇒ D −A = C −B so the second equation holds as well. Conversely assume that (A, B, C, D) is a parallelogram and let m and n be the slopes of the sides. Then b2 −a2 = m(b1 −a1), c2 −b2 = n(c1 −b1), c2 −d2 = m(c1 −d1), d2 −a2 = n(d1 −a1).
Subtracting the two equations on the left and subtracting the two on the right gives two equations v = mu and v = nu where v = (b2 −a2) −(c2 −d2) and u = (b1−a1)−(c1−d1). Since m ̸= n (else the points would be collinear) we conclude u = v = 0 so B −A = C −D as required.
Exercise 3.43. (Addition in Affine Geometry) Let O = (0, 0), A = (a, 0), B = (b, 0), C = (c, 0) be four points on the x-axis and (O, A′, C′, B′) be a parallelogram (i.e. the lines OA′ and B′C′ are parallel and the lines OB′ and A′C′ are parallel) such that the lines AA′, BB′, CC′ are parallel. Show that c = a + b Exercise 3.44. (Subtraction in Affine Geometry) Let O = (0, 0), A = (a, 0), B = (b, 0) be three points on the x-axis and (O, A′, B′, O′) be a parallelogram such that the lines OO′, AA′, BB′ are parallel, i.e. the lines OA′ and B′O′ are parallel and the lines OB′ and A′O′ are parallel. Show that b = −a.
Exercise 3.45. (Multiplication in Affine Geometry) Let O = (0, 0), I = (1, 0), A = (a, 0), B = (b, 0), C = (c, 0) be five points on the x-axis, and O′, I′, B′ be three points such that (a) the lines OO′, II′, BB′ are parallel, (b) the lines IB and I′B′ are parallel, (c) the points O′, I′, A are collinear, and (d) the points O′, B′, C are collinear. Show that c = ab.
Exercise 3.46. (Division in Affine Geometry) Let O = (0, 0), I = (1, 0), A = (a, 0), B = (b, 0) be four points on the x-axis, and O′, I′, B′ be three points such that (a) the lines OO′,BB′, II′ are parallel, (b) the lines IB and I′B′ are parallel, (c) the points O′, I′, A are collinear, and (d) the points O′, I, B′ are collinear. Show that b = 1/a.
3.7 Menelaus and Ceva 3.47. Let P, Q, R be points on (extended) sides BC, CA, AB of triangle △ABC. Thus there are numbers p, q, r with P = pB + (1 −p)C, Q = qC + (1 −q)A, R = rA + (1 −r)B.
24 The six distances (BP), (PC), (CQ), (QA), (AR), (RB) in Theorems 3.48 and 3.49 below are directed , i.e. (XY ) = −(Y X). The sign convention is such that the signs of (BP), (PC), (CQ), (QA), (AR), (RB) are the same as the signs of 1 −p, p, 1 −q, q, 1 −r, r respectively.
Theorem 3.48 (Menelaus). The points P, Q, and R are collinear if and only if (BP) (PC) · (CQ) (QA) · (AR) (RB) = −1 Proof. (See also page 146 and page 66.) By Theorem 3.12 the condition that the lines AP, BQ, and CR are concurrent is preserved by affine trans-formations. By Theorem 3.25 the ratios (and hence product of the ratios) is preserved by affine transformations. Hence by Corollary 4.14 we may assume that A = (1, 0), B = (0, 1) and C = (0, 0) so P = (0, p), Q = (1 −q, 0) and R = (r, 1 −r). By Theorem 3.4 the points P, Q, R are collinear if and only if det 0 p 1 1 −q 0 1 r 1 −r 1 = 0.
The condition on the ratios is 1 −p p −0 · 1 −q q −0 · √ 2(1 −r) √ 2(r −0) = −1.
Both conditions simplify to 1 −p −q −r + pq + qr + rp = 0.
Theorem 3.49 (Ceva). The lines (AP), (BQ), and (CR) are concurrent if and only if (BP) (PC) · (CQ) (QA) · (AR) (RB) = 1.
Proof. (See also page 126 and page 4.) As in Theorem 3.48 we may assume that A = (1, 0), B = (0, 1), C = (0, 0), P = (0, p), Q = (1 −q, 0), R = (r, 1 −r). The lines AP, BQ, CR have equations px + y = p, x + (1 −q)y = 1 −q, (r −1)x + ry = 0.
By Theorem 3.4 these lines are concurrent if and only if det p 1 −p 1 1 −q q −1 r −1 r 0 = 0.
25 The condition on the ratios is 1 −p p −0 · 1 −q q −0 · √ 2(1 −r) √ 2(r −0) = 1.
Both conditions simplify to 1 −p −q −r + pq + qr + rp = 2pqr.
Remark 3.50. Ceva’s Theorem has the following six corollaries.
1. The medians of a triangle are concurrent. The common point is called the centroid. (See 3.8.) 2. The altitudes of a triangle are concurrent. The common point is called the orthocenter. (See 5.3.) 3. The perpendicular bisectors of a triangle are concurrent. The common point is called the circumcenter. (See 5.2.) 4. The angle bisectors of a triangle are concurrent. The common point is called the incenter. (See 5.5.) 5. The lines connecting each vertex of a triangle to the opposite point of tangency of the inscribed circle are concurrent. The common point is called the Gergonne point. (See 5.5.) 6. The lines connecting each vertex of a triangle to the point of tangency of the opposite exscribed circle are concurrent. The common point is called the Nagel point. (See 5.5.) 3.8 The Medians and the Centroid Definition 3.51. The medians of a triangle are the lines connecting the vertices to the midpoints of the opposite sides. The triangle formed by joining these midpoints is called the medial triangle.
Theorem 3.52. The medians of a triangle △ABC are concurrent, The com-mon point G is called the centroid and is given by G = 1 3(A + B + C).
26 B B B B B B B B B B B B B B B B B B B B BB • MC • MB •MA • G """""""""""""""""" " Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z • A • B • C Figure 9: The Centroid.
Proof. The three ratios in Ceva’s Theorem are all one. To check the for-mula for G let MA, MB, MC be the midpoints of the sides BC, CA, AB respectively. Then as in the proof of Corollary 3.26 MA = 1 2(B + C), MB = 1 2(C + A), MC = 1 2(A + B).
so G = 1 3A + 2 3MA = 1 3B + 2 3MB = 1 3C + 2 3MC.
In other words G lies on each median two thirds of the way from the vertex to the opposite midpoint.
Remark 3.53. In Math 221 one learns that the center of mass of a collec-tion of n point masses m1, m2, . . . , mn located at point P1, P2, . . . , Pn is the weighted average ¯ P = P i miPi P i mi .
An analogous formula ¯ P = R P dm R dm 27 holds for continuous mass distributions. The centroid of a triangle is both the center of mass of three equal mass points at its vertices and also the center of mass of a uniform mass distribution spread over its area. (See page 58.) Somewhat surprisingly, the center of mass of a triangle made from three uniform rods is the incenter, not the centroid. (See page 59.) As explained in on page 129, the point of concurrency in Ceva’s theorem can also be viewed as the center of mass of three (unequal) point masses.
Exercise 3.54. Show that the medians of a triangle divide it into six trian-gles of equal area. Hint: It is enough to prove this for an equilateral triangle.
Exercise 3.55. Show that the centroid of a triangle divides each median into two segments one of which is twice as long as the other.
Exercise 3.56. Points P and Q are se-lected on two sides of △ABC, as shown, and segments AQ and BP are drawn.
Then QX PY are drawn parallel to BP and AQ, respectively. Show that XY ∥AB.
A B C P Q Y X Exercise 3.57. In the figure, vertices B and C of △ABC are joined to points P and Q on the opposite sides, and lines BP and CQ meet at point X. Suppose that BX = (2/3)BP and CX = (2/3)CQ. Prove that BP and CQ are medians of △ABC.
A B C P Q X Exercise 3.58. Show that there is no point P inside △ABC such that every line through P cuts the triangle into two pieces of equal area. Hint: Show that if there were such a point, it would have to lie on each median of the triangle.
Exercise 3.59. In the figure, the side BC of △ABC is trisected by points R and S.
Similarly, T and U trisect side AC and V and W trisect side AB.
Each vertex of △ABC is joined to the two trisection points on the opposite side, and the inter-sections of these trisecting lines determine △XY Z, as shown. Prove that the sides of △XY Z are parallel to the sides of △ABC.
A B C R S T U V W X Y Z 28 Exercise 3.60. (Varignon’s Theorem.) Points W, X, Y and Z are the midpoints of the sides of quadrangle ABCD as shown, and P is the intersection of WY with XZ.
Two of the four small quadrangles are shaded. Show that P is the midpoint of both WY and XZ and that the shaded area is exactly half of the area of quad-rangle ABCD. Hint: For the second part, decompose the whole area into four trian-gles so that exactly half the area of each triangle is shaded.
A B C D W X Y Z Exercise 3.61. Points P and Q are chosen on two sides of △ABC, as shown, and lines BP and QC meet at X. Show that X lies on the median from vertex A if and only if QP∥BC.
A B C P Q X Exercise 3.62. Given △ABC, let A′ be the point 1/3 of the way from B to C, as shown.
Similarly, B′ is the point 1/3 of the way from C to A and C′ lies 1/3 of the way from A to B. In this way, we have constructed a new triangle, △A′B′C′ start-ing with an arbitrary triangle. Now apply the same procedure to △A′B′C′, thereby creating △A′′B′′C′′. Show that the sides of △A′′B′′C′′ are parallel to the (appropri-ate) sides of △ABC. What fraction of the area of △ABC is the area of △A′′B′′C′′?
A B C B' C' A' C" A" B" Exercise 3.63. If we draw two medians of a triangle, we see that the interior of the triangle is divided into four pieces: three triangles and a quadrilateral.
Prove that two of these small triangles have equal areas, and show that the other small triangle has the same area as the quadrilateral.
29 4 Euclidean Geometry 4.1 Orthogonal Matrices Definition 4.1. A square matrix M is said to be orthogonal iffits transpose M ∗is it inverse.
Theorem 4.2. A 2 × 2 matrix M is orthogonal if and only if it has one of the two forms M = a b −b a or M = a b b −a where a2 + b2 = 1.
Proof. Let a, b, c, d be the entries of M so that M and M ∗are given by M = a b c d , M ∗= a c b d .
Then M −1 = M ∗if and only if M ∗M = MM ∗= I := the 2 × 2 identity matrix, i.e.
a c b d a b c d = a2 + c2 ab + cd ba + dc b2 + d2 = 1 0 0 1 , and a b c d a c b d = a2 + b2 ac + bd ca + db c2 + d2 a c b d = 1 0 0 1 .
This happens if and only if a2 + c2 = b2 + d2 = a2 + b2 = c2 + d2 = 1 and ab + cd = ac + bd = 0. From the former we get a2 = d2, b2 = c2 so d = ±a and c = ±b. The additional equation ab + cd = 0 shows that if d = a ̸= 0 then b = −c and if d = −a ̸= 0 then b = c.
Remark 4.3. The two forms in Theorem 4.2 are distinguished by their determinants: det(M) = a2+b2 = 1 for the first and det(M) = −a2−b2 = −1 for the second.
Remark 4.4. Each of the equations M ∗M = I and MM ∗= I implies the other. More generally, if M and N are square matrices and NM = I, then N = M −1. (See paragraph A.5.) Hence to show that a 2 × 2 matrix M is orthogonal it is enough to verify one of the two equations M ∗M = I and MM ∗= I.
30 4.2 Euclidean Transformations Definition 4.5. The distance between the two points A = (a1, a2) and B = (b1, b2) is defined by the Pythogorean formula |AB| = p (b1 −a1)2 + (b2 −a2)2.
(In elementary geometry it is customary to we AB instead of |AB|.) We also write |W| = √ u2 + v2 for the distance between the origin (0, 0) and the point W = (u, v).
Definition 4.6. A Euclidean transformation is an affine transformation T which preserves distance, i.e.
|A′B′| = |AB| whenever A′ = T(A) and B′ = T(B). Euclidean transformations are called isometries in . The term rigid motion is also commonly used.
Remark 4.7. The Euclidean transformations form a group. This is obvious and has nothing to do with distance. The set of (invertible) transforma-tions which preserve any function contains the identity and is closed under composition and inverses.
Theorem 4.8. An affine transformation T(P) = MP + V is Euclidean if and only if the matrix M is orthogonal.
Proof. Given points A = (a1, a2) and B = (b1, b2) let W = B −A denote the vector from A to B, let A′ = T(A), B′ = T(B) be the images of A and B under T, and let W ′ = B′ −A′. Then W ′ = MW (see Remark 3.28). Thus the theorem asserts that |W ′| = |W| for all W if and only if M is orthogonal.
If W = (u, v) then W ′ = (au + bv, cu + dv) and |W|2 = u2 + v2, |W ′|2 = (a2 + c2)u2 + 2(ab + cd)uv + (b2 + d2)v2.
If M is orthogonal, then a2 + c2 = b2 + d2 = 1 and ab + cd = 0 (see the proof of Theorem 4.2) so |W ′|2 = |W|2. Conversely, assume that |W ′|2 = |W|2 for all W. Taking W = (1, 0) and W = (0, 1) gives a2 + c2 = b2 + d2 = 1.
ab + cd = 0. These equations say M ∗M = I. (It follows that MM ∗= I. See Remark 4.4.) Hence M is orthogonal.
31 4.3 Congruence Definition 4.9. Two figures in the plane are said to be congruent iffthere is a Euclidean transformation carrying one onto the other.
In particular two triangles △ABC and △A′B′C′ are congruent iffthere is a Euclidean transformation T such that T(A) = A′, T(B) = B′, and T(C) = C′. We write F ∼ = F ′ as an abbreviation for the sentence F is congruent to F ′ Remark 4.10. We use the word figure rather than the word set because the order of the points is important.
The triangles △ABC and △CBA are different, and usually there is no Euclidean transformation T such that T(A) = C, T(B) = B, and T(C) = A.
Remark 4.11. Congruence is an equivalence relation, i.e. for any figures 1. F ∼ = F; 2. if F ∼ = F ′ and F ′ ∼ = F ′′, the F ∼ = F ′′; 3. if F ∼ = F ′, then F ′ ∼ = F.
This is an immediate consequence of the fact that the Euclidean transforma-tions form a group.
4.12. Any two points are congruent. In fact, for any two points A and B there is a unique translation T(P) = P + (B −A) such that T(A) = B. Two directed line segments AB and A′B′ are congruent if and only if they have the same length, i.e. |AB| = |A′B′|. This is an immediate consequence of the following Theorem 4.13. Let b = |AB| be the distance between distinct points A and B. Then there are exactly two Euclidean transformations T such that T(A) = (0, 0) and T(B) = (b, 0).
Proof. Using a translation we may assume w.l.o.g. that A = (0, 0). Let B = (b1, b2) and define a, b, and M by a = b1 p b2 1 + b2 2 , b = b2 p b2 1 + b2 2 , M = a b −b a .
32 Then M is orthogonal and MB = (b, 0).
Now suppose that T1(A) = T2(A) = (0, 0) and T1(B) = T2(B) = (b, 0), Then T(0, 0) = (0, 0) and T(b, 0) = (b, 0) where T = T1 ◦T −1 2 . As T is an Euclidean transformation fixing the origin Theorem 4.2 says that it has form T(x, y) = (αx + βy, ∓βx ± αy) where α2 + β2 = 1. From T(b, 0) = (b, 0) we conclude that β = 0 and α = 1. We conclude that there are two choices for T, namely T = I the identity or T = S where S(x, y) = (x, −y) is reflection in the x-axis. Hence either T1 = T2 or T1 = S ◦T2.
Corollary 4.14. Every triangle is congruent to a triangle △ABC where A = (a, 0), B = (b, 0), and C = (0, 0).
Proof. By Theorem 4.13 We may assume that the triangle has vertices A1 = (0, 0), B1 = (b1, 0), C1 = (c1, c2). Apply the translation T(P) = P −V where V = (c1, 0).
Theorem 4.15 (SSS). Two triangles △ABC and △A′B′C′ are congruent if and only if the corresponding sides are equal: △ABC ∼ = △A′B′C′ ⇐ ⇒|AB| = |A′B′|, |BC| = |B′C′|, |CA| = |C′A′|.
Proof. “Only if” is immediate since Euclidean transformations preserve dis-tance.
Hence assume that the corresponding sides are equal.
By Theo-rem 4.13 we may assume that A = A′ = (0, 0) and B = B′ = (c, 0). Let a = |BC| and b = |AC| then both C and C′ lie on the intersection of the two circles x2 + y2 = b2, (x −c)2 + y2 = a2.
But these two circles intersect in the two points (x0, ±y0).
(Specifically, x0 = (c2 −a2)/2c and y0 = p b2 −x2 0.) Thus either C = C′ or the reflection S(x, y) = (x, −y) carries C to C′. Either way △ABC ∼ = △A′B′C′.
4.4 Similarity Transformations Definition 4.16. A similarity transformation is an affine transformation T which preserves ratios of distances, i.e. there is a positive constant µ such that |A′B′| = µ|AB| whenever A′ = T(A) and B′ = T(B).
33 Definition 4.17. Two figures in the plane are said to be similar iffthere is a similarity transformation carrying one onto the other. In particular two triangles △ABC and △A′B′C′ are similar iffthere is a Euclidean transfor-mation T such that T(A) = A′, T(B) = B′, and T(C) = C′.
Theorem 4.18. The similarity transformations form a group and similarity is an equivalence relation.
Proof. As for Euclidean transformations and congruence.
Theorem 4.19. An affine transformation T(P) = MP + V is a similarity transformation if and only M has one of the two forms M = a b −b a or M = a b b −a where a2 + b2 > 0.
Proof. If T(P) = MP + V is a similarity transformation, then µ−1T(P) = µ−1MP + µ−1V is Euclidean.
4.20. We summarize Definitions 4.6 and 4.16 and Theorem 3.25: • Euclidean transformations preserve distance.
• Similarity transformations preserve ratios of distances.
• Affine transformations preserve ratios of collinear distances.
4.5 Rotations Definition 4.21. An affine transformation T(P) = MP +V is called orien-tation preserving iffdet(M) > 0 and orientation reversing iffdet(M) < 0.
Remark 4.22. A Euclidean transformation has form T(P) = MP + V where M has one of the two forms in Theorem 4.2. The first is orientation preserving and the second is orientation reversing.
Definition 4.23. A fixed point of a transformation T is a point P which is not moved by T, i.e. T(P) = P.
34 Theorem 4.24. An orientation preserving Euclidean transformation which is not a translation has a unique fixed point O; such a transformation is called a rotation about O.
Proof. Assume that T is an orientation preserving Euclidean Transformation which is not a translation. Then T(P) = MP + V where M = a b −b a , V − p q .
and a2 + b2 = 1, a < 1. To find the fixed point O we solve MO + V = O for O = (x, y). These equations are (1 −a)x −by = p, by + (1 −a)y = q.
The determinant of the matrix of coefficients is (1 −a)2 + b2 = 2(1 −a) > 0 so there is a unique solution.
4.25. Recall from Math 222 that rotation through angle θ is described by the orthogonal matrix Rθ = cos θ −sin θ sin θ cos θ .
The rotation Rθ carries the point (1, 0) to the point (cos θ, sin θ). The ma-trices Rθ satisfy the identities R0 = I, RαRβ = Rα+β, R−1 α = R−α.
The second identity is an immediate consequences of the trigonometric ad-dition formulas: cos(α+β) = cos α cos β−sin α sin β, sin(α+β) = sin α cos β+cos α sin β.
Definition 4.26. A matrix of form R = c −s s c , c2 + s2 = 1 rotation matrix; the corresponding Euclidean transformation is then a ro-tation about the origin (0, 0).
35 4.6 Review Next we review how angles are treated in Math 222. We use calculus notation.
For the proofs see any calculus textbook. We will not use the material in this section in the formal development, but it motivates the definitions.
Definition 4.27. The dot product of two vectors u and v is defined by u · v = cos θ|u| |v| where |u| is the length of the vector u and θ is the angle between the vectors u and v.
Theorem 4.28. The dot product between u = u1i + u2j and v = v1i + v2j is given by u · v = u1v1 + u2v2.
Definition 4.29. The cross product of two vectors u and v is defined by u × v = sin θ|u| |v|w where w is the unit vector normal to the plane of u and v and θ is the angle between the vectors u and v. (There are two unit vectors perpendicular to a plane: the choice of w is determined by the condition that (u, v, w) forms a right hand frame like (i, j, k) and not a left hand frame like (i, j, −k) .) Theorem 4.30. The cross product between u = u1i + u2j and v = v1i + v2j is given by u × v = (u1v2 −u1v2)k.
Corollary 4.31. Let θ be the angle from the vector u = u1i + u2j to the vector v = v1i + v2j. Then cos θ = u1v1 + u2v2 p u2 1 + u2 2 p v2 1 + v2 2 and sin θ = u1v2 −u2v1 p u2 1 + u2 2 p v2 1 + v2 2 .
4.32. In the sequel we will write U = (u1, u2) instead of u = u1i + u2j and U = A −O for the vector from O to A instead of the notation u = − → OA used in Math 222.
36 4.7 Angles Definition 4.33. An angle is a pair (ρ1, ρ2) of rays having a common initial point. The common initial point of ρ1 and ρ2 is called the vertex of the angle and the rays ρ1 and ρ2 are called the arms of the angle. Given three distinct points O, P, Q the notation ∠POQ denotes the angle (ρ1, ρ2) where ρ1 is the ray with initial point O through P and ρ2 is the ray with initial point O through Q.
Definition 4.34. The trigonometric measure of an angle (ρ1, ρ2) is the pair (c, s) defined by c = u1v1 + u2v2 p u2 1 + u2 2 p v2 1 + v2 2 , s = u1v2 −u2v1 p u2 1 + u2 2 p v2 1 + v2 2 , where O is the vertex of the angle (ρ1, ρ2), and A, B are points on ρ1, ρ2 distinct from O, and (u1, u2) := U := A −O, (v1, v2) := V := B −O.
(The definition is independent of the choice of A and B as follows. If A′, B′ are other points on ρ1, ρ2 distinct from O and U ′ = A′ −O, V ′ = B′ −O, then U ′ = µU and V ′ = νV where µ, ν > 0. The numbers µ and ν factor out in the above expressions for c and s.) Exercise 4.35. Show that c2 + s2 = 1.
Theorem 4.36. Let (c, s) and (c′, s′) be the trigonometric measures of angles (ρ1, ρ2) and (ρ′ 1, ρ′ 2) respectively. Then (1) There is an orientation preserving Euclidean transformation T such that T(ρ1) = ρ′ 1 and T(ρ2) = ρ′ 2 if and only if (c′, s′) = (c, s).
(2) There is an orientation reversing Euclidean transformation T such that T(ρ1) = ρ′ 1 and T(ρ2) = ρ′ 2 if and only if (c′, s′) = (c, −s).
Proof.
Definition 4.37. For three points A = (a1, a2), O = (o1, o2), B = (b1, b1) of R2 with A ̸= O and B ̸= O define cos ∠AOB := c, sin ∠AOB := s 37 U V B A (0,0) O Figure 10: An Angle as in Theorem 4.36 with ui = ai −oi, vi = bi −oi.
In other words, the pair (cos ∠AOB, sin ∠AOB) is the trigonometric measure of the (ρ1, ρ2) where ρ1 is the ray )A and ρ2 is the ray OB.
In these notes the nota-tion ∠AOB = ∠A′O′B′ shall henceforth be considered as an abbreviation for the two equations cos ∠AOB = cos ∠A′O′B′ and sin ∠AOB = sin ∠A′O′B′.
Thus ∠AOB = ∠A′O′B′ ⇐ ⇒ cos ∠AOB = cos ∠A′O′B′ and sin ∠AOB = sin ∠A′O′B′.
4.38. In elementary trigonometry the radian measure of an angle is defined as follows. Slide the angle in the plane so that the vertex is at the origin and the first ray ρ1 is the positive x-axis. Then the first ray ρ1 intersects the unit circle x2 + y2 = 1 in the point (1, 0) and the second ray ρ2 intersects the unit circle a point (a, b). The radian measure θ of the angle is then the length of the arc of the unit circle traced out by a point moving counter clockwise from the point (1, 0) to the point (c, s); moreover, the sine and cosine of the angle are given by c = cos θ, s = sin θ.
In coordinate calculations we generally trigonometric measure rather than radian measure. A careful treatment of radian measure requires calculus.
Remark 4.39. It is an immediate consequence of Definitions 4.33 and 3.30 that the oriented area of △AOB is given by (AOB) = 1 2|AO| |BO| sin ∠AOB.
38 Remark 4.40. The reader is cautioned that in most elementary books no distinction is made between ∠AOB and ∠BOA, but with the definitions presented here sin ∠BOA = −sin ∠AOB, i.e. reversing the order of the rays in an angle reverses the sign of the sine.
With our definitions cos ∠BOA = cos ∠AOB, and in elementary texts the condition that two angles are equal can usually be replaced by the condition that they have the same cosine. When a diagram is present ∠AOB is often denoted by ∠O, especially when the sign of the sine not important.
Corollary 4.41. Orientation preserving similarity transformations preserve trigonometric measures of angles. Orientation reversing similarity transfor-mations preserve cosines of angles.
Corollary 4.42. Vertical angles are equal, i.e. if A1, O, A2 are collinear with O between A1 and A2 and B1, O, B2 are collinear with O between B1 and B2, then ∠A1OB1 = ∠A2OB2.
Corollary 4.43. Corresponding angles determined by parallel lines are equal, i.e. if lines A1B1 and A2B2 are parallel and a point C is on the line A1A2 is such that C is between A1 and A2, then ∠B1A1C = ∠B2A2C.
Proof. The translation T(P) = P + (A2 −A1) sends the ray A1B1 onto the ray A2B2 and the ray A1C onto the ray A2C.
4.8 Addition of Angles Theorem 4.44. Let (ρ1, ρ2), O, A, B, c, s be as in Definitions 4.33, and 4.34, and T(P) = R(P −O) + O where R = c −s s c , Then T is the unique orientation preserving Euclidean transformation such that T(ρ1) = ρ2.
39 Proof. The transformation T is orientation preserving and Euclidean: the condition c2 + s2 = 1 is Exercise 4.35. It follows from the identity (u1v1 + u2v2)2 + (u1v2 −u2v1)2 = (u2 1 + u2 2)(v2 1 + v2 2).
Clearly T(O) = O. Let(u1, u2) := U := A −O and (v1, v2) := V := B −O as in Definition 4.34. The equations (u1v1 + u2v2)u1 −(u1v2 −u2v1)u2 = (u2 1 + u2 2)v1, (u1v2 −u2v1)u1 + (u1v1 + u2v2)u2 = (u2 1 + u2 2)v2.
show that RU is a positive multiple of V (i.e. T(A)−O is a positive multiple of B −O) and hence that T sends ρ1 to ρ2.
The uniqueness of T is a consequence of the uniqueness in Theorem 4.13.
Definition 4.45. Let α, β, γ be angles. Then the notation γ = α + β is understood to be an abbreviation for the assertion that there exist four distinct points O, A, B, C such that γ = ∠AOC, α = ∠AOB, β = ∠BOC.
If α = ∠AOB, the notation α = 180◦means A, O, B are collinear with O between A and B; the notation β = −α means β = ∠BOA; the notation α = 90◦means cos ∠AOB = 0 and sin ∠AOB = 1; the notation α = 45◦means α+α = 90◦; etc. We also say that lines OA and OB are perpendicular (the word orthogonal is used in some textbooks) when ∠AOB = ±90◦. Angles which sum to 90◦are called complementary; angles which sum to 180◦are called supplementary.
Theorem 4.46. Let Rα, Rβ, Rγ be the rotation matrices corresponding to the angles α, β, γ as in Theorem 4.44. Then γ = α + β ⇐ ⇒Rγ = RαRβ.
Proof. Let ρ1, ρ2, ρ3 be three rays through the origin.
If Rα(ρ1) = ρ2, Rβ(ρ2) = ρ3 and then RβRα(ρ1) = ρ3.
Theorem 4.47. The angles of a triangle sum to 180◦.
40 Proof. The theory presented thus far justifies the usual proof: draw a parallel to a side through the third vertex.
Theorem 4.48 (SSS,SAS,ASA). For triangles △ABC and △A′B′C′ the following conditions are equivalent: (i) the triangles △ABC and △A′B′C′ are congruent; (ii) |AB| = |A′B′|, |BC| = |B′C′|, and |CA| = |C′A′|; (iii) |AB| = |A′B′|, |BC| = |B′C′| and cos ∠B = cos ∠B′; (iv) cos ∠A = cos ∠A′, cos ∠B = cos ∠B′, and |AB| = |A′B′|.
Proof. The assertion (i) ⇐ ⇒(ii) is Theorem 4.15. The other assertions are similar.
Corollary 4.49 (Pons Asinorum). The base angles of an isosceles triangle are equal.
Proof. If |AB| = |CB|, Then △ABC ∼ = △CBA by ASA, so cos ∠A = cos ∠C since Euclidean transformations preserve cosines of angles.
Remark 4.50. Strictly speaking, the correct conclusion of Pons Asinorum is cos ∠BAC = cos ∠BCA, sin ∠BAC = −sin ∠BCA.
See Remark 4.40.
Theorem 4.51 (AAA,S:S:S,S:A:S). For triangles △ABC and △A′B′C′ the following conditions are equivalent: (i) the triangles △ABC and △A′B′C′ are similar; (ii) cos ∠A = cos ∠A′, cos ∠B = cos ∠B′, and cos ∠C = cos ∠C′; (iii) the sides are proportional, i.e. A′B′ AB = B′C′ BC = C′A′ CA ; (iv) cos ∠A = cos ∠A′ and A′B′ AB = C′A′ CA .
Proof. Two triangles are similar if and only if there is a number µ such that the affine transformation T(P) = µP carries one of them to a triangle congruent to the other. Apply Theorem 4.48 41 Exercise 4.52. Let A = (0, 0), B = (1, 0), C = (0.8, 1.2), A′ = (2, 3), B′ = (2.4, 3.6). Then |AB| = |A′B′| = 1. There are two points C′ such that △ABC and △A′B′C′ are congruent. Find them and for each one find the Euclidean transformation T such that T(A) = A′, T(B) = B′, T(C) = C′.
For which one(s) is T orientation preserving?
42 5 More Euclidean Geometry 5.1 Circles Definition 5.1. The locus of all points P at a fixed distance r from a given point O is called the circle of radius r and center (or centered at) O. If O = (a, b) circle is the set of all points P = (x, y) satisfying the equation (x −a)2 + (y −b)2 = r2.
(∗) Any line segment [O, P] where P lies on the circle is called a radius of the circle.
Theorem 5.2. Though any point P0 on the circle (∗) there is a unique line ℓ intersecting the circle at the single point P0. This line is called the tangent line to the circle at P0. The radius OP0 is perpendicular to the line tangent line ℓthrough P0.
Proof. The line through P0 = (x0, y0) has an equation of form v(x −x0) −u(y −y0) = 0 (v/u is the slope) and hence parametric equations x = x0 + tu, y = y0 + tv where u2 + v2 = 1. Inserting this into (∗) and gives (x0 −a)2 + 2(x0 −a)tu + t2u2 + (y0 −b)2 + 2(y0 −b)tv + t2v2 = r2.
But P0 lies on the circle, i.e (x0 −a)2 + (y0 −b)2 = r2, so the equation simplifies to 2(x0 −a)tu + 2(y0 −b)tv + t2 = 0.
There are two distinct solutions (for t) unless (x0 −a)u + (y0 −b)v = 0, in which case the line has equation (x0 −a)(x −x0) + (y0 −b)(y −y0) = 0.
Note that this equation says that the vector − − → OP0 from the center O of the circle to a point P0 on its circumference is perpendicular to vector − − → P0P from P0 to a point P on the tangent line at P0. This is in agreement with the method used in Math 222.
43 Theorem 5.3. The locus of all points P equidistant from two given points A and B is a straight line perpendicular to line AB and passing through the midpoint M of the segment [A, B]. It is called the perpendicular bisector of the segment [A, B].
Proof. Exercise.
5.2 The Circumcircle and the Circumcenter Theorem 5.4. For any triangle △ABC there is a unique circle containing the three points A, B, C. This circle is called the circumcircle (short for circumscribed circle) of △ABC and its center O is called the circum-center. The circumcenter is the intersection of the perpendicular bisectors of the sides.
Proof. Let O be the intersection of the perpendicular bisectors of [A, B] and [B, C]. Then |OA| = |OB| and |OB| = |OC| so|OA| = |OC| O lies on the perpendicular bisector of [A, C] and is equidistant from the vertices.
Remark 5.5. In high school algebra you learned that equation of form x2 + y2 −2ax −2by −c = 0 determines a circle, a point, or the empty set; you determine which by com-pleting the square, and that is also how you find the center of the circle.
Three non collinear points (x1, y1), (x2, y2), (x3, y3) determine (a multiple of) an equation of this form, namely det x2 + y2 x y 1 x2 1 + y2 1 x1 y1 1 x2 2 + y2 2 x2 y2 1 x2 3 + y2 3 x3 y3 1 = 0.
Since each of the three points (x, y) = (xi, yi) satisfies this equation (because a determinant with two identical rows vanishes), this must be the equation of the circumcircle.
5.3 The Altitudes and the Orthocenter Theorem 5.6. Let ℓbe a line and P be a point. The there is a unique line through P perpendicular to ℓ. The point H on ℓwhere the perpendicular 44 B B B B B B B B B B B B B B B B B B B B BB Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q • A • B • C • HC • HB •HA • H Figure 11: The Othocenter.
through P to ℓintersects ℓis called the foot of the perpendicular yo ℓfrom P.
Exercise 5.7. The foot H of the perpendicular to ℓfrom P is the point on ℓclosest to P, i.e. the distance from P to H is less than the distance from P to any other point of ℓ. The distance |PH| is called the distance from the point P to the line ℓ.
Definition 5.8. The altitudes of a triangle are the lines through the vertices perpendicular the opposite sides.
Theorem 5.9. The altitudes of a triangle are concurrent.
The point of concurrency is called the orthocenter of the triangle.
Proof. Let △ABC be a triangle and HA, HB, HC be the feet of the altitudes through A, B, C respectively. The lengths in Ceva’s Theorem are BHA = AB cos ∠B, HAC = CA cos ∠C, CHB = BC cos ∠C, HBC = AB cos ∠A, AHC = CA cos ∠A, HCB = BC cos ∠B.
The corresponding product or ratios is one so the altitudes are concurrent.
45 Remark 5.10. Another proof can be based on the fact that circumcenter of a triangle is the orthocenter of its medial triangle.
Definition 5.11. The triangle △HAHBHC formed by joining the feet of the altitudes of △ABC is called the orthic triangle3 of △ABC.
5.4 Angle Bisectors Definition 5.12. Let A, O, B be distinct points. Then the locus of all points P such that either P = O or ∠AOP = ∠POB is called the angle bisector of the angle AOB.
Theorem 5.13. Every point on the angle bisector of the angle AOB is equidistant from the lines AO and BO.
Proof. Drop perpendiculars from P to OA and OB and use ASA.
5.5 The Incircle and the Incenter Theorem 5.14. The angle bisectors of a triangle are concurrent. The point of concurrency is equidistant from the sides of the triangle and is therefore the center of a circle tangent to the three sides of the triangle. This circle is called the incircle (short for inscribed circle) of the triangle and its center is called the incenter.
Proof. The bisector of an angle is the locus of all points equidistant from the arms of the angle. Hence the point of intersection of two angle bisectors of △ABC is equidistant from all thee sides and hence must lie on the third angle bisector Remark 5.15. One can give a proof using Ceva’s Theorem. See page 21 Theorem 1.12 and page 10 Theorem 1.34.
Theorem 5.16. Let X on BC, Y on CA, Z on AB be the points of tangency of the inscribed circle of triangle △ABC. Then the lines AX, BY , CZ are concurrent. Their common intersection is called the Gergonne point of △ABC.
3Some authors call it the pedal triangle.
46 Proof. Ceva’s Theorem plus the fact that the two tangents to a circle from a point are equal.
Theorem 5.17. Let X on BC, Y on CA, Z on AB be the points of tangency of the exscribed circles of triangle △ABC. Then the lines AX, BY , CZ are concurrent. Their common intersection is called the Nagel point of △ABC.
Proof. The line joining two center of excircle tangent at X and the center of the excircle tangent at Z passes through the corresponding vertex B. The ratio of BX to BZ is the same as the ratio of the radii of these two excircles.
Now use Ceva.
5.6 The Euler Line Theorem 5.18. Let △ABC be a triangle, G be its centroid, O be its cir-cumcenter, and H be its orthocenter. Then H, G, O are collinear and G = 2 3O + 1 3H.
The line OGH is called the Euler line.
Proof. Consider the similarity transformation T(P) = G −1 2(P −G) This transformation satisfies T(G) = G, T(A) = MA, T(B) = MB, T(C) = MC where MA, MB, MC are the midpoints of [B, C], [C, A], [A, B]. It carries the orthocenter H to the circumcenter O. As its shrinks distances by a factor of 1 2 we conclude that |OG| = 1 2|HG|. The transformation rotates through 180◦about G so H, G, and O = T(G) are collinear.
5.7 The Nine Point Circle This section uses material not covered above. See page 20.
Theorem 5.19. The medial triangle and the orthic triangle of a triangle have the same circumcircle. This circumcircle is called the nine point circle4 of the original triangle. It also contains the three midpoints, called the Euler points, of the line segments joining the orthocenter to the vertices.
4 Sometimes called the Euler circle or the Feuerbach circle.
47 Proof. Denote the midpoints of the sides of △ABC, the feet of the altitudes, and the Euler points by MA, MB, MC, HA, HB, HC, EA, EB, EC, with the subscripts chosen so that the medians are AMA, BMB, CMC, the altitudes are AHA, BHB, CHC, and the points EA, EB, EC lie on these altitudes respectively.
We claim that MAMBEBEA is a rectangle, i.e. adjacent sides are per-pendicular. This is because • △MAMBC is similar to △ABC by SAS so MAMB is parallel to AB; • △EAEBH is similar to △ABH by SAS so EAEB is parallel to AB; • △AMBEA is similar to △ACHC by SAS so MBEA is parallel to CHC and hence perpendicular to the parallel lines AB and MAMB; • △BMAEB is similar to △BCHC by SAS so MAEB is parallel to CHC and hence perpendicular to the parallel lines AB and MAMB.
The rectangle MAMBEBEA is inscribed in a circle with diameters MAEA and MBEB. Reading B and C for A and B we obtain that the rectangle MBMCECEB is inscribed in a circle with diameters MBEB and MCEEC.
These two circles share a common diameter MBEB and so must be the same circle. As this circle contains MA, MB, MC it is the circumcircle of the medial triangle, i.e. the nine point circle. The line EAHA is the altitude AHA and the line H)AMA is the side BC opposite A. Hence ∠EAHAMA is a right angle so the point HA (and similarly HB and HC) lies on the nine point circle.
Theorem 5.20. The orthocenter H of an acute angled triangle △ABC is the incenter of its orthic triangle △HAHBHC.
Proof. We must show ∠HAHCH = ∠HBHAH. (The same argument shows ∠HCHBH = ∠HCHBH and ∠HCHBH = ∠HAHCH.) Now the quadrangle HHCBHA is cyclic (i.e. is inscribed in a circle) since HHCB = HHAB = 90◦.
Hence ∠HHAHC = ∠HBHC. But ∠HBHC = ∠HBBA is the complement α of ∠CBA so HHAHC = α. The same argument (reverse the roles of B and C) shows that HHAHB = α as required.
Remark 5.21. For an obtuse triangle the argument shows that H is the center of one of the exscribed circles, not the center of the inscribed circle.
48 Corollary 5.22. The triangle of least perimeter inscribed in a given triangle △ABC is the orthic triangle of △ABC.
Proof. For three points X, Y , Z, on lines BC, CA, AB respectively let f = f(X, Y, Z) = XY +Y Z +ZX. This function is continuous as a function of three variables. (Each of X, Y , Z is constrained to a line.) When one or more of the points X, Y , Z are far from A, B, and C the function f is large so the minimum occurs either a point where the partial derivatives of f are zero, or else at a point where the function f is not differentiable. The function f is differentiable except where one of the three lengths is zero, i.e.
when X = Y = C or Y = Z = A or Z = X = B.
5.8 A Coordinate Proof Here are some coordinate calculations which could be used to prove Theo-rems 5.18 and 5.19. I used a Maple program to do these calculations and verify that the nine points of the nine point circle are equidistant from N, the midpoint of the segment OH.
5.23. By Corollary 4.14 we may assume that the vertices of the triangle are A = (a, 0), B = (b, 0), C = (0, c).
The midpoints of the sides are given by MA = a + b 2 , 0 , MB = a 2, c 2 , MC = b b, c 2 , and the centroid is G = 1 3 A + B + C = a + b 3 , c 3 .
The slopes of the lines AB, AC, BC are given by mAB = 0, mAC = −c a, mBC = −c b; The equations for these lines are respectively y = 0 and cx + ay = ac, cx + by = bc.
49 The orthocenter H lies on the y-axis, say H = (0, h). The slopes of the altitudes AH and BH are mAH = −h a = − 1 mBC = b c, mBH = −h a = − 1 mAC = a c, and solving for h gives H = 0, −ab c .
Equations for the altitudes CH, AH, BH are x = 0 and bx −cy = ab, ax −cy = ab The feet of these altitudes are respectively HC = (0, 0) and HA = ab2 + bc2 b2 + c2 , b2c −abc b2 + c2 , HB = ba2 + ac2 a2 + c2 , a2c −abc a2 + c2 .
The Euler points EC = 1 2(C + H), EA = 1 2(A + H), EB = 1 2(B + H) are EC = 0, c2 −ab 2c , EA = a 2, −ab 2c , EB = b 2, −ab 2c .
The perpendicular bisectors of the sides AB, AC, BC have equations x = a + b 2 , y −c 2 = a c x −a 2 , y −c 2 = b c x −b 2 The common point is the circumcenter O = a + b 2 , c2 + ab 2c .
5.9 Simson’s Theorem See page 40.
5.24. Let ABC be a triangle.
The lines AB, BC, AC divide the plane into seven regions; let the point P lie in the unbounded region containing the edge AC in its boundary (see diagram). Let X, Y , Z be the feet of the perpendiculars from P to the lines BC, AC, AB respectively. See Figure 5.9.
50 A B C P Z X Y Figure 12: Simson’s Theorem Theorem 5.25 (Simson’s Theorem). The points X, Y , Z are collinear if and only if P lies on the circumcircle of ABC.
Proof. The point P lies on the circumcircle of ABC if and only if ∠APC = 180o −∠B.
(1) Because the opposite angles at X and Z in the quadrangle BXPZ are right angles we have 180o −∠B = ∠ZPX so condition (1) is equivalent to ∠APC = ∠ZPX (2) and on subtracting ∠APX we see that (2) is equivalent to ∠XPC = ∠ZPA.
(3) A quadrilateral containing a pair of opposite right angles is cyclic, i.e. its vertices lie on a circle; in fact, the other two vertices are the endpoints of a diameter of this circle. Hence each of the quadrilaterals AY PZ, BXPZ, CXPY is cyclic. Since the quadrangle CXPY is cyclic we have ∠XY C = ∠XPC.
(4) Since the quadrangle AY PZ is cyclic we have ∠ZY A = ∠ZPA.
(5) 51 From (4) and (5) we conclude that (3) is equivalent to ∠XY C = ∠ZY A.
(6) But clearly (6) holds if and only if the points X, Y , Z are collinear.
Alternate Proof. Let O be the circumcenter of △ABC. By trigonometry BX = BP cos ∠PBC, XC = PC cos ∠PCB, CY = CP cos ∠PCA, Y A = PB cos ∠PAC, AZ = AP cos ∠PAB, ZB = PA cos ∠PBA.
Using the fact that the inscribed angle is half the subtended arc BX = BP cos 1 2∠POC, XC = PC cos 1 2∠POB, CY = CP cos 1 2∠POA, Y A = PB cos 1 2∠POC, AZ = AP cos 1 2∠POB, ZB = PA cos 1 2∠POA.
Now use Menalaus.
5.26. Here is a computer assisted coordinate calculation which proves Sim-son’s Theorem. It shouldn’t be to difficult to do by hand, especially if we take γ = −α below to simplify the formulas. We begin by loading the Maple package for doing linear algebra.
with(LinearAlgebra); To calculate the foot Z of the perpendicular from the point P to the line AB we use the formulas Z = A + t(B −A), PZ ⊥AB, solve for t, and plug back in to get Z. Here is a Maple procedure to compute this.
foot:=proc(A,B,P) local t; t:=((B-A)(P-A)+(B-A)(P-A))/ ((B-A)^2+(B-A)^2); [A+t(B-A),A+t(B-A)] end proc; 52 We choose A, B, C, on the unit circle and P arbitrarily.
A:=[cos(alpha),sin(alpha)]; B:=[cos(beta),sin(beta)]; C:=[cos(gamma),sin(gamma)]; P:=[x,y]; We use the procedure to calculate X, Y ,Z: X:= foot(B,C,P); Y:=foot(C,A,P); Z:=foot(A,B,P); We define the matrix whose determinant vanishes when X, Y , Z are collinear.
M:=Matrix([ [X, X, 1], [Y, Y, 1], [Z, Z ,1] ]); We compute its determinant.
W:=Determinant(M); The determinant W vanishes exactly when X, Y , Z are collinear. The fol-lowing commands show that x2 + y2 −1 divides W and that the quotient m is independent of P = (x, y).
m:=simplify(W/(x^2+y^2-1)); simplify(W-m(x^2+y^2-1)); The last command evaluates to 0 and proves that W = m(x2 +y2 −1). Thus X, Y , Z are collinear if and only if x2 + y2 = 1, i.e. if and only if P lies on the circumcircle of △ABC. The commands mm:=expand(sin(alpha-beta)+sin(beta-gamma)+sin(gamma-alpha))/4; simplify(m-mm); produce an output of zero which shows that m = sin(α −β) + sin(β −γ) + sin(γ −α) 4 We have proved the following 53 Theorem 5.27 (Algebraic form of Simson’s Theorem). Let A = (cos α, sin α), B = (cos β, sin β), C = (cos γ, sin γ) be three points on the unit circle x2 + y2 = 1, let P = (x, y) be an arbitrary point, and X = (x1, x2), Y = (y1, y2), Z = (z1, z2) be the feet of the perpendiculars from P to the lines BC, CA, AB respectively.
Then x1 x2 1 y1 y2 1 z1 z2 1 = m(x2 + y2 −1) where m = sin(α −β) + sin(β −γ) + sin(γ −α) 4 .
5.10 The Butterfly See page 45.
5.11 Morley’s Theorem See page 47.
5.12 Bramagupta and Heron See page 56.
5.13 Napoleon’s Theorem See page 63.
5.14 The Fermat Point See page 83.
54 6 Projective Geometry Projective geometry developed historically at the same time artists learned to draw in perspective, i.e. to draw pictures on a flat canvas which look three dimensional. See for example Figure 13 where a two dimensional theorem is illustrated by a three dimensional picture. A line in space will be drawn as a line in the picture , but a circle in space will be drawn as an ellipse in the picture. Parallel lines in space will (if extended) meet in the picture. Look down State street from Bascom hill: the sidewalks and roofs of the buildings all aim at a point behind the Capitol.
6.1 Homogeneous coordinates 6.1. When (x, y, z) is a point of R3 distinct from the origin (0, 0, 0) we denote the line connecting the origin and this point by (x:y:z). Thus (x:y:z) = {(tx, ty, tz) : t ∈R}.
Note that (x′ :y′ :z′) = (x:y:z) if and only if x′ = µx, y′ = µy, z′ = µz for some non zero number µ.
Definition 6.2. The projective plane is the set P2 of all lines through the origin in R3. We say that (x, y, z) are homogeneous coordinates for the point (x:y:z).
6.3. A point (x:y:z) ∈P2 is a line in R3 and, if z ̸= 0, intersects the plane z = 1 in the unique point (xz−1, yz−1, 1), i.e.
z ̸= 0 = ⇒(x:y:z) = (xz−1 :yz−1 :1).
The points (i.e. lines) of form (x:y:0) do not intersect the plane z = 1; the set of these points is called the line at infinity.
Exercise 6.4. Find z if (1 : 2 : 3) = (2 : 4 : z).
Remark 6.5. Imagine an object in three dimensional space R3 above the plane z = 1. Place your eye at the origin (0, 0, 0) view the object through a transparent canvas lying on the plane z = 1. Each point on the object determines a line though your eye. For each point of the object paint a dot on the point of intersection of this line with the plane z = 1 and you have 55 a portrait of the object. If you have another object and a correspondence between the points of the objects such that each pair of corresponding points is collinear with the origin, then the two portraits will be indistinguishable.
This is why two points, (x, y, z) and (µx, µy, µz), on the same line through the origin determine the same point (x:y:z) in projective space.
6.6. A plane through the origin in R3 has an equation of form ax+by+cz = 0 where (a, b, c) ̸= (0, 0, 0). We denote by [a:b:c] := {(x:y:z) ∈P 2 : ax + by + cz = 0} the set of lines through the origin which lie in this plane. If µ ̸= 0, the equations µax + µby + µcz = 0 and ax + by + cz = 0 define the same plane.
Hence [µa:µb:µc] = [a:b:c].
We call [a:b:c] a projective line. Several points (x1 :y1 :zi1), (x2 :y2 :z2), . . .
are said to be collinear if there is a line [a:b:c] which contains all of them; several lines [a1 :b1 :c1], [a2 :b2 :c2], . . . are said to be concurrent if there is a point (x:y:z) in their intersection.
Theorem 6.7. (i) Two distinct points in P2 lie in a unique projective line.
(ii) Two distinct projective lines intersect in a unique point.
Proof. (i) Two lines in R3 through the origin lie in a unique plane necessarily containing the origin. (ii) Two distinct planes in R3 which pass through the origin intersect in a line which passes through the origin.
Exercise 6.8. Find the intersection of the two lines [1:2:3] and [3:2:1].
Remark 6.9. Parallel lines in the affine plane have equations ax+by+c = 0 and ax + by + d = 0 where c ̸= d. The corresponding projective lines [a:b:c] and [a:b:d] intersect in the point (−b:a:0). This point lies in the line at infinity. Hence parallel lines intersect at infinity.
In affine geometry two distinct points determine a unique line, but dis-tinct lines need not determine a point: they might be parallel. Because of this asymmetry the statement proof of part (II) of Theorem 3.4 was more complicated then that of part (I). In projective geometry this asymmetry dis-appears: two distinct points determine a unique line and two distinct lines intersect in a unique point. We recover affine geometry by specifying a “line at infinity” where parallel lines meet.
56 Theorem 6.10. (I) Three points (xi :yi :zi) are collinear if and only if det x1 x2 x3 y1 y2 y3 z3 z3 z3 = 0.
(II) Three lines [ai :bi :ci] are concurrent if and only if det a1 b1 c1 a2 b2 c2 a3 b3 c3 = 0.
Proof. As in Theorem 3.4 but without the special cases. Note that in part (I) the determinant is zero if and only if a b c x1 x2 x3 y1 y2 y3 z3 z3 z3 = 0 0 0 for some a b c ̸= 0 0 0 , and in part (II) the determinant is nonzero if and only if a1 b1 c1 a2 b2 c2 a3 b3 c3 x y z = 0 0 0 .
for some (x, y, z) ̸= (0, 0, 0).
6.2 Projective Transformations Definition 6.11. Let (tij) be a 3×3 matrix with nonzero determinant. Then there is a unique transformation T : P2 →P2 satisfying the condition x′ = t11x + t12y + t13z y′ = t21x + t22y + t23z z′ = t31x + t32y + t33z = ⇒(x′ :y′ :z′) = T(x:y:z).
(The definition is legal since matrix multiplication sends lines through the origin to lines through the origin.) A transformation of this form is called a projective transformation. Note that multiplying the matrix (tij) by a nonzero number does not change the transformation T (but it does change the matrix).
57 6.12. As in affine geometry, the formulas defining a projective transformation can be written using matrix notation: x′ y′ z′ = t11 t12 t13 t21 t22 t23 t31 t32 t33 x y z .
(∗) A matrix transformation preserves the plane z = 1 if and only if t31 = t32 = 0 and t33 = 1. For such a transformation we may rewrite (∗) in the form x′ y′ 1 = a b p c d q 0 0 1 x y 1 .
with a = t11, b = t12, c = t21, d = t22, p = t13, q = t23. This is equivalent to x′ = ax+by+p and y′ = cx+dy+q in agreement with Definition 3.9. In this way every affine transformation is a projective transformation. The power of the theory will become evident when we use projective transformations which are not affine.
6.13. We say that the triple (x, y, z) represents the point (x:y:z). Thus if µ ̸= 0 the triples (µx, µy, µz) and (x, y, z) represent the same point. It is convenient to think of (x, y, z) as a column matrix: (x, y, z) = x y z .
We also say that the row matrix [a, b, c] represents the line [a:b:c] and the matrix (tij) represents the projective transformation T.
Theorem 6.14. A projective transformation maps lines to lines.
Proof. If L = [a, b, c] represents the line [a:b:c] and X = (x, y, z) represents the point (x:y:z), then (x:y:z) lies on [a:b:c] if and only if LX = 0. But LX = LT −1TX so LX = 0 ⇐ ⇒(LT −1)(TX) = 0 so the line represented by LT −1 is the image under T of the line [a:b:c].
58 Theorem 6.15. Given four points A, B, C, D no three of which are collinear, there is a unique projective transformation, T such that T(1:0:0) = A, T(0:1:0) = B, T(0:0:1) = C, T(1:1:1) = D.
(Compare Theorem 3.13 and Corollary 4.14.) Proof. Let A = (a1 :a2 :a3), B = (b1 :b2 :b3), C = (c1 :c2 :c3). Then the projec-tive transformation S represented by the matrix a1 b1 c1 a2 b2 c2 a3 b3 c3 satisfies the first three of the four conditions, namely S(1:0:0) = A, S(0:1:0) = B, S(0:0:1) = C.
Let S(D) = (d1 :d2 :d3) and R = d−1 1 0 0 0 d−1 2 0 0 0 d−1 3 .
(We have d1 ̸= 0 since otherwise S(B), S(C), S(D) would be collinear con-tradicting the hypothesis that B, C, D are not collinear. Similarly d2 ̸= 0 and d3 ̸= 0.) Now R(S(A)) = R(1:0:0) = (d−1 1 :0:0) = (1:0:0) and similarly R(S(B) = (0:1:0) and R((S(C)) = (0:0:1). Also R(S(D)) = R(d1 :d2 :d2) = (1 : 1 : 1). The projective transformation T = R ◦S satisfies all four condi-tions.
To prove uniqueness suppose that T ′ is another projective transformation satisfying the four conditions. Then T ′ ◦T −1(1:0:0) = 1:0:0), T ′ ◦T −1(0:1:0) = (0:1:0), T ′ ◦T −1(0:0:1) = (0:0:1), T ′ ◦T −1(1:1:1) = (1:1:1).
The first three conditions imply that T ′ ◦T −1 is represented by a diagonal matrix and then the fourth condition says that the three diagonal entries must be equal. This means that T ′ ◦T −1 is the identity transformation of P2 so T = T ′.
59 •Q cc c c c cc c c c cc c c cc c c c L L L L L L L L L L L L L L L L L L L \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A A A A A A A C′ B′ A′ • • • C B A • • • A A A A A A A A A A • X • Y • Z Figure 13: Desargues’ Theorem 6.3 Desargues and Pappus Theorem 6.16 (Desargues). Given triangles △ABC and △A′B′C′ let X, Y , Z be the intersections of the corresponding sides of △ABC and △A′B′C′, i.e.
X := BC ∩B′C′, Y := CA ∩C′A′, Z := AB ∩A′B′.
Then the lines AA′, BB′, CC′ are concurrent if and only if the points X, Y , Z are collinear. (See Figure 13.) Proof. Assume the lines AA′, BB′, CC′ are concurrent, i.e. there is a point Q such that each triple (Q, A, A′), (Q, B, B′), (Q, C, C′) is collinear. Choose an affine transformation T which sends Q to the line at infinity. Then the lines T(A)T(A′), T(B)T(B′), T(C)T(C′) are parallel. By Exercise 3.17 the points T(X), T(Y ), T(Z) are collinear. Apply T −1 to conclude that the points X, Y , Z are collinear. The converse can be proved in the same way, but it also follows from the principle of duality as explained in 6.23 below.
Remark 6.17. Sometimes Desargues’ Theorem is stated as follows: Two triangles are perspective from a point if and only if they are perspective from a line.
60 Remark 6.18. Think of Figure 13 as representing a three dimensional pic-ture where Q is the apex of two tetrahedra, one with base △ABC and the other with base △A′B′C′. The plane A′B′C′ intersects the plane ABC in a line containing X, Y , Z. Note that projecting the three dimensional figure to a plane either from a point (your eye) or by parallel lines gives the configura-tion in Desargues’ theorem if no triangle projects to a line. This observation is undoubtedly how the theorem was discovered. We give a proof based on it.
Synthetic proof of Desargues’ theorem. Denote by Π the plane containing the points in the statement of the theorem, i.e. the points A, B, C, A′, B′, C′, X, Y , Z, and Q. Let E (the eye) be a point not on Π. Each point P in Π determines a line EP and each line L in Π determines a plane EL. In par-ticular the points E, Q, A, A′ are coplanar. Let ˜ A be the intersection of the three planes EAB, EAC, and EQAA′. Similarly, let ˜ B be the intersection of the three planes EBC, EBA, EQBB′, and ˜ C be the intersection of the three planes ECA, ECB, EQCC′. The three planes EQAA′ ˜ A, EQBB′ ˜ B, EQCC′ ˜ C intersect in a point ˜ Q. (A diagram illustrating the situation can be obtained from Figure 13 by replacing Q, A, B, C, by ˜ Q, ˜ A, ˜ B, ˜ C. The original points Q, A, B, C, are hidden because they are on the line of sight with E.) The points X = Π ∩EQBB′ ˜ B ∩EQCC′ ˜ C, Y = Π ∩EQCC′ ˜ C ∩EQAA′ ˜ A, Z = Π ∩EQAA′ ˜ A ∩EQBB′ ˜ B lie in the intersection of the plane of ˜ A ˜ B ˜ C and Π and are therefore collinear.
Theorem 6.19 (Pappus). Assume that the three points A, B, C are collinear and that the three points A′, B,′ C′ are collinear. Let the lines joining them in pairs intersect as follows: X = BC′ ∩B′C, Y = CA′ ∩C′A, Z = AB′ ∩A′B.
Then the points X,Y , Z are collinear. (See Figure 14.) Proof. As in the proof of Desargues’ theorem we may assume w.l.o.g. that the lines ABC and A′B′C′ are parallel.
Then the theorem follows from Theorem 3.15.
61 HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH H • A′ • B′ • C′ • A • B • C D D D D D D D D D D D D D D D DD \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A A A A A A A A A A A A A A • Z • Y • X • Q Figure 14: Pappus’ Theorem Remark 6.20. Any theorem involving only points and lines in the projec-tive plane can be viewed as a theorem about determinants. For example, Desargues’ theorem assumes we are given column matrices A, B, C, A′, B′, C′, X, Y , Z, satisfying det(ABZ) = det(BCX) = det(CAY ) = det(A′B′Z) = det(B′C′X) = det(C′A′Y ) = 0.
and concludes that ∃Q det(QAA′) = det(QBB′) = det(QCC′) = 0 ⇐ ⇒det(XY Z) = 0.
Pappus’ theorem assumes we are given column matrices A, B, C, A′, B′, C′, X, Y , Z, satisfying det(AB′Z) = det(BC′X) = det(CA′Y ) = det(A′BZ) = det(B′CX) = det(C′AY ) = 0.
and concludes that det(ABC) = det(A′B′C′) = 0 = ⇒det(XY Z) = 0.
Exercise 6.21. Let the lines joining three points A, B, C to three other points A′, B,′ C′ in pairs intersect as follows: X = BC′ ∩B′C, Y = CA′ ∩C′A, Z = AB′ ∩A′B.
62 Show that if A, B, C are collinear and X, Y , Z are collinear, then A′, B′, C′ are collinear. Hint: Don’t work too hard.
6.4 Duality 6.22. Let the column matrix (x, y, z) represent the point (x:y:z) and the row matrix [a, b, z] represent the line [a:b:c]. Then the equation ax + by + cz = 0 says that the point (x:y:z) lies on the line [a:b:c]. But it also says that the point (a:b:c) lies on the line [x:y:z]. In matrix theoretic terms, taking the transpose converts points to lines and line to points. If we have a theorem about matrices and replace every matrix in the statement by its transpose, we get another theorem about matrices. If we have a theorem in projective geometry and systematically replace points by lines and lines by points and phrases like “the point P lies on the line ℓ” by “the line p passes through the line L”, then we get another theorem in projective geometry. This is called the principle of duality. We illustrate this using the principle of duality to complete the proof of Desargues’ Theorem (Theorem 6.16).
6.23. Consider △ABC with opposite sides a, b, c. Then a = BC, b = CA, c = AB, A = b ∩c, B = c ∩a, C = a ∩b.
Similarly for △A′B′C′ we have a′ = B′C′, b′ = C′A′, c′ = A′B′, A′ = b′ ∩c′, B′ = c′ ∩a′, C′ = a′ ∩b′.
Define X = a ∩a′, Y = b ∩b′, Z = c ∩c′, x = AA′, y = BB′, z = CC′.
Then Desargues’ Theorem is x, y, z are concurrent ⇐ ⇒X, Y , Z are collinear.
We proved = ⇒in our proof of Theorem 6.16 above. This, together with the principle of duality, proves ⇐ =.
63 Remark 6.24. The theorems of Menelaus and Ceva (Theorems 3.48 and 3.49 above) appear to be dual as they are usually stated. When they are stated with directed distances instead of distances as above it becomes clear that they are not dual.
Menelaus has a minus one in the conclusion and a quadratic equation in the proof; Ceva has a plus one in the conclusion and a cubic equation in the proof. One cannot transform one to the other by taking transposes of matrices.
Exercise 6.25. State the dual of Pappus’ theorem and draw a diagram illustrating it.
6.5 The Projective Line 6.26. The projective line P1 is the space of lines through the origin in R2.
In analogy with the projective plane each point (x, y) ∈R2 with (x, y) ̸= (0, 0) determines a point (x : y) := {(tx, ty) ∈R2 : t ∈R} in the projective line P2 and (µx:µy) = (x : y) for µ ̸= 0. A transformation M : P1 →P1 satisfying x′ = ax + by y′ = cx + dy = ⇒(x′ :y′) = M(x:y) where ad −bc ̸= 0 is called a projective transformation (of the line).
These definitions exactly parallel Definition 6.2 of the projective plane and Definition 6.11 of projective plane transformation.
Theorem 6.27. Given three distinct points A, B, C in P1 there is a unique projective transformation of the line M such that M(1:0) = A, M(0:1) = B, M(1:1) = C.
Proof. The same as Theorem 6.15 but easier.
6.28. If y ̸= 0 then (x:y) = (z :1) where z = x/y. This establishes a one-one correspondence between P1 and and the space R ∪{∞} obtained from the real line R by adjoining a point which we denote by ∞. The real number z corresponds to the point (z : 1) on the projective line P1 and the point ∞ 64 corresponds to the point (1:0). With this identification P1 and R ∪{∞} a projective transformation takes the form M(z) = az + b cz + d (†) where it is understood that M(z) = ∞if cz + d = 0, M(∞) = a/c if c ̸= 0, and M(∞) = ∞if c = 0. These conventions are consistent with the following formulas from Math 221: lim z→∞ az + b cz + d = lim z→−∞ az + b cz + d = a c if c ̸= 0 and, for z0 = −d/c, lim z→z0+ az + b cz + d = ±∞, lim z→z0− az + b cz + d = ±∞, where the signs on the two ±∞’s are opposite. (In calculus we usually think of adjoining two points, +∞and −∞, to the real numbers, but here there is only one infinity.) A transformation of form (†) is called a fractional linear transformation or a M¨ obius transformation.
6.29. Let A = (x1 :y1 :z1) and B = (x2 :y2 :z2) be two distinct points in the projective plane P2. Define φ : P1 →P2 by φ(t:s) = (x:y:z) where x = tx1 + sy1, y = ty1 + sy2, z = tz1 + sz2.
This definition is legal since φ(µt:µs) = (µx:µy:µz) = (x:y:z) for µ ̸= 0. It is easy to see that φ is a one-one correspondence between the projective line P1 and the line AB in P2. We call φ a projective parameterization of the line AB. Note that φ depends not just on the points A and B but also on the representatives chosen.
Exercise 6.30. Let A, B, C, be distinct collinear points in P 2. Show that there is a unique projective parameterization of the common line such that φ(1:0) = A, φ(0:1) = B, and φ(1:1) = C.
Exercise 6.31. Assume that φ and ψ are projective parameterizations of the same line in P2. Show that there is a unique prpjective transformation M : P1 →P1 of the projective line such that ψ = φ ◦M.
65 Exercise 6.32. Assume that φ is a projective parameterization of a line ℓ in P2 and that T : P2 →P2 is a projective transformation of the projective plane. Show that T ◦φ is a projective parameterization of a line T(ℓ).
6.6 Cross Ratio Definition 6.33. The cross ratio of four distinct points A, B, C, D on the projective line is defined by {AB, CD} := det(A C) · det(B D) det(A D) · det(B C).
In the formula the column vectors on the right are representatives respectively of the points by the same name. It does not matter which representatives are used since the cross ratio is unchanged if any of the four column vectors is replaced by a nonzero scalar multiple of itself since the scalar factors out in the numerator and denominator and then cancels. In particular, if A = (a:1), B = (b:1), C = (c:1), D = (d:1), then {AB, CD} := (a −c)(b −d) (a −d)(b −c).
Theorem 6.34. A projective transformation of the projective line preserves cross ratio.
Proof. This is an immediate consequence of the fact that the determinant of the product is the product of the determinants, i.e.
det(TA TB) = det(T) det(A B) where A, B are 2 × 1 column vectors and T is a 2 × 2 matrix .
• O H H H • A H H H H H H H H H H H H • B @ @ @ @ @ @ • C • D Figure 15: Cross Ratio 66 Exercise 6.35. Let A, B, C, D ∈R2 \ {O} represent distinct points in P1 and O = (0, 0) denote the origin in R2 . Then {AB, CD} = (AOC) · (BOD) (AOD) · (BOC) = sin ∠AOC · sin ∠BOD sin ∠AOD · sin ∠BOC = (AC) · (BD) (AD) · (BC) In the first formula (AOB) denotes the oriented area of △AOB, and in the last formula it is assumed that the four points are collinear and (AB) denotes the directed distance along the common line. (The choice of direction doesn’t matter since reversing the direction produces four sign reversals which cancel.) See Figure 6.6.
Remark 6.36. Recall the relation between projective geometry and drawing in perspective from Remark 6.5. From a perspective drawing of three points A, B, C, on a line in R3 we cannot conclude anything about the distances between them. If A′, B′, C′ lie respectively on the line of sight from the origin (0, 0, 0) with A, B, C then the points A′, B′, C′ wll appear the same as the points A, B, C in the drawing although the ratios AB/AC and A′B′/A′C′ can be different. However in an accurate perspective drawing of four collinear points A, B, C, D, the cross ratio in the drawing will be the same as the cross ratio of the four points in space.
6.7 A Geometric Computer In the following four exercises we will design a geometric computer. It works better than the one in Exercises 3.43-3.46 because that one required us to draw parallel lines whereas this one requires only a straightedge to connect two points with a line. Let O, I, Q, A, B, C be six collinear points such that O, I, Q are distinct and a, b, c denote the three cross ratios a := {AQ, IO}, b := {BQ, IO}, c := {CQ, IO}.
Exercise 6.37. (Addition in Projective Geometry.) Assume given points M, N, P, A′, B′, C′ satisfy the following conditions: 67 (a) the points M, N, P, Q are collinear, (b) the points M, O, A′ are collinear, (c) the points M, B′, C′ are collinear, (d) the points N, O, B′ are collinear, (e) the points N, A′, C′ are collinear, (f) the points P, A′, A are collinear, (g) the points P, B′, B are collinear, and (i) the points P, C′, C are collinear.
Draw a diagram illustrating this configuration and show that c = a + b.
Hint: Take the common line of M, N, P, Q to be the line at infinity and use Exercise 3.43.
Exercise 6.38. (Subtraction in Projective Geometry.) State and prove a theorem analogous to Exercise 3.44 in the same way that Exercise 6.37 is analogous to Exercise 3.43. The conclusion should be that b = −a. Draw a diagram illustrating the configuration.
Exercise 6.39. (Multiplication in Projective Geometry.) State and prove a theorem analogous to Exercise 3.45 in the same way that Exercise 6.37 is analogous to Exercise 3.43. The conclusion should be that c = ab. Draw a diagram illustrating the configuration.
Exercise 6.40. (Division in Projective Geometry.) State and prove a theo-rem analogous to Exercise 3.46 in the same way that Exercise 6.37 is anal-ogous to Exercise 3.43. The conclusion should be that b = 1/a. Draw a diagram illustrating the configuration.
68 7 Inversive Geometry 7.1 The complex projective line 7.2 Feuerbach’s theorem 69 8 Klein’s view of geometry 8.1 The elliptic plane 8.2 The hyperbolic plane 8.3 Special relativity 70 A Matrix Notation Matrix notation is a handy way to describe messy calculations. The matrix operations obey most of the rules of ordinary algebra; the most important exception is the commutative law for multiplication. The material presented in this section contains all the matrix algebra we shall need. For more details consult any book in linear algebra.
A.1. Fix positive integers m and n. An m × n matrix is an array A = a11 a12 . . .
a1n a21 a22 . . .
a2n .
.
.
.
.
.
...
.
.
.
am1 am2 . . .
amn where each entry aij is a number. There are m rows and n columns in an m×n matrix. We say that an m×n matrix is a matrix of size m×n when we want to call attention to the number of rows and columns. A square matrix is one having the same number of rows as columns; a column vector) is an m × 1 matrix; a row vector) is an 1 × n matrix.
A.2. To add or subtract matrices, add or subtract the corresponding entries as in a11 a12 a21 a22 a31 a32 + b11 b12 b21 b22 b31 b32 = a11 + b11 a12 + b12 a21 + b21 a22 + b22 a31 + b31 a32 + b32 .
To multiply a matrix by a number, multiply each entry by the number as in c a11 a12 a13 a21 a22 a23 = ca11 ca12 ca13 ca21 ca22 ca23 .
The zero matrix of any size has all its entries zero and is usually denoted by 0. Thus A + 0 = A.
A.3. If the number n of columns in A is the same as the number of rows in B the matrix product C = AB is defined by the rule that the number cij in the ith row and jth column of C is given by cij = X k aikbkj.
71 For example, a11 a12 a21 a22 b11 b12 b21 b22 = a11b11 + a12b21 a11b12 + a12b22 a21b11 + a22b21 a21b12 + a22b22 and a b c d x y = ax + by cx + dy .
A.4. The m × m identity matrix I is the square matrix with 1’s on the diagonal and 0’s elsewhere. For example, if m = 2 I = 1 0 0 1 .
The identity matrix behaves like the number 1: multiplication by I leaves a matrix unchanged.
A.5. A square matrix A is called invertible there is a (necessarily unique) matrix A−1 called the inverse of A which satisfies AA−1 = A−1A = I where I is the identity matrix. The following calculation shows that if A is invertible, then the only matrix B satisfying BA = I is B = A−1: B = BI = B(AA−1) = (BA)A−1 = IA−1 = A−1.
A square matrix is invertible if and only if its determinant is nonzero. For example, if A = a b c d , then det(A) = ad −bc and A−1 = 1 ad −bc d −b −c a .
A.6. The transpose A∗of a matrix A is defined by the rule that the entry of A∗in the ith column and jth row is the same as the entry of A in the ith row and jth column. For example, a11 a12 a21 a22 ∗ = a11 a21 a12 a22 72 and x y z ∗= x y z .
B Determinants In elementary linear algebra (Math 340 at UW) one studies an operation which assigns to any square matrix A a number det(A) called the determinant of A. In this appendix we state its key properties. In the notes we only need determinants of 2×2 and 3×3 matrices; these are often taught in high school algebra. For these matrices one can check the properties using elementary algebra, although the 3 × 3 case is a bit hairy. For the proofs in the general case see any book on elementary linear algebra. (My favorite is .) Theorem B.1. There is a unique function called the determinant which assigns to each square matrix A a number det(A) satisfying the following properties: (Scale) If B results from A by multiplying some row by a number c, then det(B) = c det(A).
(Swap) If B results from A by interchanging two rows, then det(B) = −det(A).
(Shear) If B results from A by adding a multiple of one row to another row, then det(B) = det(A).
(Identity) The determinant of the identity matrix I is det(I) = 1.
Exercise B.2. The determinant of a 2 × 2 matrix A = a11 a12 a21 a22 is det(A) = a11a22 −a12a21.
Verify the properties listed in Theorem B.1 using this formula.
73 Exercise B.3. The determinant of a 3 × 3 matrix A = a11 a12 a13 a21 a22 a23 a31 a32 a33 is det(A) = a11a22a33 + a12a23a31 + a13a21a32 −a11a23a32 −a13a22a31 −a12a21a33.
Verify the properties listed in Theorem B.1 using this formula.
Remark B.4. There is a general definition for the determinant of an n × n matrix as a sum det(A) = X σ ±a1,σ(1)a2,σ(2) · · · an,σ(n) where the sum is over all n! permutations of {1, 2, . . . , n}. (The sign depends on the permutation σ in a subtle way.) We do not need this formula in these notes and will not explain it.
Theorem B.5. A square matrix is invertible if and only if its determinant is not zero.
Exercise B.6. Find AB and BA where A = a b c d and B = d −b −c a .
Then show that A is invertible if and only if its determinant is not zero and give a formula for A−1.
Theorem B.7. The determinant of a matrix and its transpose are the same: det(A) = det(A∗). Hence the properties of Theorem B.1 hold reading “col-umn” for “row”.
Exercise B.8. Check this for 2 × 2 matrices.
74 Theorem B.9. For a square matrix A, the homogeneous system AX = 0 has a nonzero solution X if and only if det(A) = 0.
Exercise B.10. Check this for 2 × 2 matrices.
Corollary B.11. The determinant of a square matrix vanishes if and only if one of its columns is a linear combination of the others.
Theorem B.12. The determinant function satisfies the following: (1) det(I) = 1; (2) det(A−1) = det(A)−1; (3) det(AB) = det(A) det(B).
Exercise B.13. Check this for 2 × 2 matrices.
C Sets and Transformations C.1. A set X divides the mathematical universe into two parts: those objects x that belong to X and those that don’t. The notation x ∈X means x belongs to X. The notation x / ∈X means that x does not belong to X. In geometry the word locus is often used as a synonym for set as in the sentence The locus of all points P such that ∠APB is a given constant is an arc of a circle.
C.2. If X is a set and P(x) is a property that either holds or fails for each element x ∈X, then we may form a new set S consisting of all x ∈X for which P(x) is true. This set S is denoted by S = {x ∈X : P(x)} (1) and called “the set of all x ∈V such that P(x)”. The set S is a subset of X meaning that every element of S is an element of X. The notation used in equation (1) is called set builder notation. Having defined S by (1), we may assert that for all x x ∈S ⇐ ⇒x ∈X and P(x) and that for all x ∈X x ∈S ⇐ ⇒P(x).
75 Since the property P(x) may be quite cumbersome to state, the notation x ∈W is both shorter and easier to understand. (The symbol ⇐ ⇒is an abbreviation for if and only if.) C.3. Let X and Y be sets. The notation f : X →Y means that f is a func-tion which assigns to each point x ∈X and element f(x) ∈Y . Mathemati-cians have more words for this concept than any other: calculus textbooks call f a function with domain X (or defined on X) taking values in Y , while textbooks on linear algebra or plane geometry call f a transformation from X to Y , and in more advanced mathematics one says that f is a map (or mapping) from X to Y . In these notes we use the word transformation and almost always we take X = Y = R2 = the set of pairs of real numbers. We will also use the letter T rather than f.
C.4. For any set X there is a transformation IX : X →X called the identity transformation of X and defined by IX(x) = x for x ∈X. The composition g◦f : X →Z of two transformations f : X →Y and g : Y →Z is defined by (g ◦f)(x) = g(f(x)) for x ∈X. A transformation f : X →Y is called invertible iffthere is a (necessarily unique) transformation f −1 : Y →X such that f −1(f(x)) = x, and f(f −1(y)) = y for x ∈X and y ∈Y . The transformation f −1 is called the inverse trans-formation to f. Clearly (h ◦g) ◦f = h ◦(f ◦g), f ◦IX = f, IY ◦f = f, and f −1 ◦f = IX, and f ◦f −1 = IY .
C.5. When f : X →Y and S is a subset of X we define the image of S by f to be the set of all points f(x) as x ranges over S. It is denoted by f(S).
In set builder notation this is written as f(S) := {f(x) : x ∈S}.
(2) (The notation := is often used to emphasize that the right hand side is the definition of the left hand side, so that no proof is required.) To prove that a point y lies in the set f(X) one must show there is an x ∈S with y = f(x).
76 Example C.6. In calculus one learns that x = cos θ, y = sin θ (3) are parametric equations for the unit circle. In the notation introduced thus far this could be written {(x, y) ∈R2 : x2 + y2 = 1} = {(cos θ, sin θ) : θ ∈R}.
(4) The left hand side uses the set builder notation of equation (1) while the right hand side uses the set builder notation of equation (2). It is also true that {(x, y) ∈R2 : x2 + y2 = 1} = {(cos θ, sin θ) : 0 ≤θ < 2π}.
C.7. To prove that two sets are equal on must show that every element of one set is an element of the other set and vice versa. For example, to prove (4) we argue as follows. If (x, y) lies in the set defined on the right hand side of (4), then (3) holds for some number θ. Hence x2 + y2 = cos2 θ + sin2 θ = 1 by the Pythagorean Theorem so (x, y) lies in the set on the left hand side of (4).
Conversely, if x2 +y2 = 1, then (x, y) = (cos θ, sin θ) where θ = tan−1(y/x) if x > 0, θ = tan−1(y/x) + π if x < 0, θ = π/2 if (x, y) = (0, 1), and θ = 3π/2 if (x, y) = (0, −1). The case analysis is necessary because the equivalence m = tan θ ⇐ ⇒θ = tan−1(m) holds only if −π/2 < θ < π/2.
77 References E. T. Bell, Men of Mathematics, Simon & Schuster, 1937.
L. M. Blumenthal, A Modern View of Geometry, W. H. Freeman, 1961.
H.
S.
M.
Coxeter & S.
L.
Greitzer, Geometry Revis-ited, Random House, 1967, Reissued by the MAA (see A classic written by a famous geometer.
I. Martin Isaacs, Geometry for College Students, Brooks Coles, 2001. A beautiful book with lots of problems written by a professor at UW.
H. Levy, Projective and Related Geometries, Macmillan, 1961. Professor Levy introduced me to the subject in a course at the University of Illinois in 1961-62.
E. A. Maxwell, Fallacies in Mathematics, Cambridge University Press, 1959. (See G. E. Martin: Transformation Geometry, An Introduction to Symmetry, Springer Undergraduate Texts in Math, 1982.
MathWorld Website: A google search on any term from elementary geometry (like “Gergonne Point”) usually brings up this websiite.
J. Robbin, Matrix Algebra, Using MINImal MATlab, A.K. Peters, 1995.
78 |
187687 | https://mathmistakes.org/slope-and-division-of-negative-numbers/ | Slope and Division of Negative Numbers – Math Mistakes
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Slope and Division of Negative Numbers
Post author By mpershan
Post date March 17, 2013
4 Comments on Slope and Division of Negative Numbers
What’s the fastest way to help this kid?
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4 replies on “Slope and Division of Negative Numbers”
Joshua Zucker says:
March 18, 2013 at 6:42 pm
Visualize is good, for sure; that probably happens in my head as a sanity check of the answer even if I don’t plot the points on paper.
More to the point, how about other ways of checking? There’s not only a diagram, but also plugging the points in at the end If you used (2,7) to find the intercept, then use the other point to check your answer. At the end, instead of thinking “Done!”, how can we get them into the habit of thinking “wait, -1/2 times -6 + 8 isn’t anywhere near 3” so that they are prompted to look back and find the mistake?
newman says:
March 21, 2013 at 9:39 am
Oh man… I’ve seen this a million times.
I usually find that the kids making this mistake couldn’t to plug a number into an equation and check it without making another one and compounding the problem. For these kids, I have them graph it, count the slope, THEN check it with the slope equation (if they insist on using it). I’ve also gone so far as to have kids create a flow chart describing, in excruciating detail… “I took 7 away from 3. I did this because… This is important to know because… this -4 tells me I am going in the negative direction on the Y-axis…” But this really doesn’t address the neg/neg = pos issue, does it?
Would you mark it wrong if they left it as -1/-2?
I Speak Math says:
March 23, 2013 at 9:44 pm
I’d like them to be able to move past the sketch however and ultimately use just the formula. I see this as mostly a negative number operation mistake, and then secondarily a slope formula mistake. Yes, they could identify it if they sketched it, but it’s so important with slope to remind them that when they “fall” and “run backward” these two things together make a positive line. Also, slope formula is a GREAT time to learn that the negative in a fraction is “attached” to the numerator or the denominator. And, I would consider -1/-2 not simplified so I would penalize the student for that.
I Speak Math says:
March 23, 2013 at 9:53 pm
Oh, and sorry – that’s also how I would help “fix” this student fast. I’d show them their answer -1/-2 means to “fall one” and “run backward” 2 which created a positive line. My students go, “Ahhhh!” when I do that.
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Understanding the Work Energy Theorem for a Spring
Thread starter mrpolar
Start date
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: Energy Spring Theorem Work Work energy Work energy theorem
AI Thread Summary
The discussion revolves around the confusion surrounding the work-energy theorem as applied to a spring system. The original poster struggles with the correct equations and signs when calculating work and potential energy, particularly in the context of a mass hanging from a spring. Clarifications are provided about the proper limits of integration for calculating work, emphasizing that the integral should be evaluated from the initial position to the final position. The conversation highlights the importance of consistently using either the work-energy theorem or conservation of energy principles to avoid confusion. Ultimately, the poster gains clarity on the concepts and equations involved.
1
mrpolar
: 5
: 0
Hi,I looked around for hours but it seems like I'm the only one who finds it confusing.I understand the concept of potential energy and work, but have a problem with the equations.Here is what I don't understand:The work energy theorem states that K2-K1=W. W = ∫Fdx from evaluated between x2 and 1. The Force done by a spring is -kx, so when only the spring acts on the body the equation should be: K2-K1=-½kx22 + ½kx12 K2+½kx22=K1+½kx12 But now, since W=-U(x) (potential energy) the equation turns out to be: K2-½kx22=K1-½kx12 For the total energy balance (kinetic + potential). But the lectures and examples I saw, and also the logical thing as I understand it, is that the equation should be K2+½kx22=K1+½kx12 for the kinetic + potential energy balance. Why do I get the signs wrong? Here's an example for a question which made me really confused: (minute 40:30). In short, a mass is hanged from a spring attached the ceiling, so the mass has 2 forces acting on it: gravity and the spring. In order to find the kinetic energy - potential energy equation, we have ΔKinetic_Energy = the negative of the work (-∫Fdx) of the total forces. Now I know that gravity acts downwards, so the force is -mg, and the spring's force is always -kx, so the equation should look something like this: K2-K1 = -∫F(gravity)dx - ∫F(spring)dx evaluated between x2 and 1. The negative signs before integrals is because we want to find the potential energy which is the negative of the work done. K2-K1 = -∫-mgdx - ∫-kxdx = ∫mgdx + ∫kxdx K2-K1 = mgx2 +½kx22 - mgx1 +½kx12 K2-mgx2-½kx22 = K1-mgx1-½kx12 which is the total energy equation. In class, they had the potential energy signs all opposite like this: K2+mgx2+½kx22 = K1+mgx1+½kx12 which of course looks more sensible and logical. The problem is that I can't see where I got the maths wrong. I follow the rules of and get it all messed up time and time again. Please, help me out here, this is driving me nuts! Ben.
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2
nasu
Homework Helper
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Why do you integrate from x2 to x1 when calculating the work?
3
mrpolar
: 5
: 0
Hi, thanks for replying.I evaluate the overall work done between the starting point (x1) to the end point (x2), so I integrate from x2 to x1 (the value at x2 minus the value at x1 gives me the work done between). I got something wrong?
4
nasu
Homework Helper
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Yes, you need to integrate from x1 to x2.
5
mrpolar
: 5
: 0
When I say that I integrate from x2 to x1 I mean that I set x2 as the upper bound and x1 as the lower bound of the integral.You mean to set x1 as the upper bound and x2 as the lower bound in order to calculate the work?I thoughts you do it in order to find the potential energy (hence the minus sign before the integral which is the same as reversing the integration order).
6
nasu
Homework Helper
: 4,462
: 915
mrpolar said:
When I say that I integrate from x2 to x1 I mean that I set x2 as the upper bound and x1 as the lower bound of the integral.You mean to set x1 as the upper bound and x2 as the lower bound in order to calculate the work?I thoughts you do it in order to find the potential energy (hence the minus sign before the integral which is the same as reversing the integration order).
Then you use the words in an unusual way. Integral from a to b means that a is the lover limit and b is the upper limit. See here, for example: Regrading your "problem", you use the definition of potential energy improperly. What you get after integration and regrouping the terms is the right conservation of energy equation. The potential energy at x1 is 1/2kx1^2, given that the zero potential energy is at x=0. The definition of potential energy tells you that the work is the negative of the change in potential energy. So the change in PE from x1 to x2 is the negative of the work from x1 to x2, This is Δ PE=1/2kx2^2-1/2kx1^2 So if you want to use conservation of energy rather than work-energy theorem, you have ΔKE+ΔPE=0 which will give you again the correct equation KE1+1/2kx1^2=KE2+1/2kx2^2
Last edited:
7
mrpolar
: 5
: 0
Hi, thank you again, you are really helpful. Sorry the wrong terminology.So, if I got it right, for a mass hanged by a spring from a ceiling,ΣF =-mg-kx (the spring's equalibrium = 0 potential energy, so gravity can take values of minus potential energy whenever the mass is beneath the equalibrium. is it ok?)ΔKE = ΔW = ∫-mg-kx dx = -mgx2 -½kx22 + mg1 + ½kx12 KE2 + mgx2 + ½kx22 = KE1 + mg1 + ½kx12 First question: those equations are right? will work in any x point I define no matter if the mass is going up or down, beanth or above equalibrium, right? My second question, I know that this equation is essentially KE2 + PE2 = KE1 + PE1[/SUB but the righthand side of the equation is the kinetic energy at x1 plus the x1 part of the work equation, not the potential energy (I didn't multiply that part by (-1) the same as I did to the x2 part of the ΔW by moving it to the right-hand side of the equation). What's going on here? Thanks in advance.
8
nasu
Homework Helper
: 4,462
: 915
Yes, it is the potential energy. As it is the "x1 part" of the work-energy equation.Your W is -ΔPE =-(PE2-PE1)=-PE2 +PE1.So the x1 part is PE1. And the x2 part is PE2.But it's no point to go keep going between work-energy and conservation of energy. Decide which one you want to use and stay with it for the given problem.
9
mrpolar
: 5
: 0
Thank you so much for helping me out here. I finally 100% got it.Time to figure out conservation of energy in 2 dimensions :PThanks again.
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187689 | https://mathteachercoach.com/introduction-to-hundredths/ | An Introduction to Hundredths
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An Introduction to Hundredths
by
Once children learn the tenths, they start learning how to read, write, and model numbers with hundredths. They do so by using numerals and number names, in addition to a decimal point. These early lessons on decimals are often not smooth sailing for many students.
As a math teacher or homeschooling parent, you’d naturally want to provide the best support to your students when teaching this topic. To help you out in this endeavor, we’ve created a list of great math tips for teaching the hundredths. Use these tips in your classroom and see your students’ math knowledge soar!
How to Teach the Hundredths
Review Tenths and Place Value
Since learning the hundredths implies knowledge of the tenths as well as place value, you may want to start your lesson with a review of these topics. You can implement a brief bell-work activity for this purpose.
For example, you can write a few multi-digit numbers on the whiteboard and ask students to identify the place value of each underlined digit. For additional guidelines on place value, you can also check out our article that’s dedicated to this topic.
You can then remind students that the tenths refer to the first digit to the right of the decimal point. Write a few examples on the whiteboard and ask students to identify the tenths. For additional guidelines on tenths, you can also check out our article that’s dedicated to this topic.
Where Is the Hundredths’ Place?
Now that children know that the tenths are the first digit to the right of the decimal point, you can explain that the hundredths are simply the second digit to the right of the decimal point, i.e. the digit to the right of the tenths place.
A good way to help students understand the hundredths is by providing a visual model. You can draw a square on the whiteboard and divide it into hundred equal parts, i.e. a decimal grid. It may also be practical to have a pre-prepared square. Then, shade one hundredth, or 0.01.
Explain that one part out of a hundred equal parts is one hundredth or 0.01. So in 0.01, the 1 is in the hundredth place. Students can easily see that you’ve shaded only one part of the hundred parts of the square:
Then, you can illustrate to children how we can graph 0.01 on the number line. Point out that if we observe the number line, we can conclude that there are ten hundredths in one tenth:
Provide a few examples and underline the hundredths.
Reading Decimals
Students sometimes understand the basic principle of how to graph a decimal on the number line or how to present it visually but may experience difficulties with reading the given decimal. So, you can provide examples of how we read or write decimals in word form.
For instance, explain that we read 0.36 as thirty-six hundredths; 2.04 as two and four hundredths; 0.02 as two hundredths; 0.38 as thirty-eight hundredths; 0.57 as fifty-seven hundredths; etc.
Additional Resources:
You may also decide to enrich your lesson with multimedia materials, such as videos. For example, consider using this video in your classroom for an introduction to decimal numbers, such as the tenths and the hundredths.
The video illustrates how we can represent different coins on a decimal grid. For instance, we can see how a penny (one out of one hundred cents of a dollar) is shown on the decimal grid. In other words, the video shows how we can represent 0.01 on the decimal grid.
Activities to Practice Hundredths
Decimal Game
This is a fun online game that will help students improve their understanding of hundredths with the help of visual models. To play this game in your class, the only thing you’ll need is a sufficient number of devices and a decent internet connection.
Divide students into pairs and explain the rules of the game. The members of a pair compete against each other to complete the questions they’re presented with. Students are asked to answer diverse questions related to hundredths.
For instance, they have to shade the model (a decimal grid) to represent a given decimal on the model, or they are presented with a shaded model and several decimals out of which they have to select which one is shown on the model.
If students answer correctly, they score coins. In the end, the two students in a given pair compare their final scores. The winner is the person with the highest score. Parents who are homeschooling can also use this game as an individual activity with their children.
Place Value Pirates
This online game will help children practice their recognition skills when it comes to identifying decimals to the tenths and hundredths. Make sure that each child has a suitable device and provide instructions for the game.
Each student should answer ten questions in total, related to identifying place value. Several pirates stand on different decimals and students must identify which pirate has a given digit in the tenths’, hundredths’, tens’, or ones’ place, for instance.
As such, the game is very useful for students to improve their ability to differentiate between a digit in the tenths and the tens place, or one in the hundred and the hundredth place. The game is played individually, making it ideal for homeschooling students.
Matching Game
To play this game in your classroom, you’ll need two sets of cards: one set contains a shaded decimal grid that represents a decimal and the other set contains a decimal (such as 0.03, 0.28, 23.87, etc.).
Since this is a matching game, make sure that there is a corresponding shaded decimal grid for each decimal that you have on the other set of cards. In the first set, there is one grid per card and on the second set, there is one decimal per card.
You can prepare as many cards as you think fit for your class, depending on the size of the class. Just make sure to have at least twenty cards per group, or ten decimal matches per group. You may also choose to laminate the cards so that you can re-use them next year.
Divide students into groups of 3 or 4 and hand out the two sets of cards. The cards remain face down until you provide the signal that the game can start. Provide instructions for the game and explain that the students must match the cards from the two sets.
Point out to students that all members in one group work together and race against other groups to be the first ones to finish the matching. The first group that manages to match all cards correctly wins the game.
Before You Leave…
If you enjoyed these tips and activities, make sure to check out our lesson that’s dedicated to teaching hundredths to 4th graders by becoming a member of the Math Teacher Coach community! Or if you need more free guidance with structuring your class and teaching, sign up for our emails for more free lessons and content!
Feel free to also check out our blog, where you’ll find a bunch of cool resources for teaching children of all ages!
This article is based on:
Unit 6 – Decimal Fractions
6-1 Tenths
6-2 Hundredths
6-3 Fractions to Decimals and Decimals to Fractions
6-4 Comparing and Ordering Decimals
6-5 Addition with Tenths and Hundredths
6-6 Money as Decimal Numbers
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Calculations on sliding windows and memoization
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Asked 13 years, 4 months ago
Modified10 years, 8 months ago
Viewed 1k times
This question shows research effort; it is useful and clear
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I am working on Project Euler Problem 50, which states:
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13 This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
For determining the terms in prime P (if it at all can be written as a sum of primes) I use a sliding window of all the primes (in increasing order) up to (but not including) P, and calculate the sum of all these windows, if the sum is equal to the prime considered, I count the length of the window...
This works fine for all primes up to 1000, but for primes up to 106 it is very slow, so I was hoping memozation would help; when calculating the sum of sliding windows, a lot of double work is done...(right?)
So I found the standard memoizaton implemention on the net and just pasted it in my code, is this correct? (I have no idea how it is supposed to work here...)
```python
primes = tuple(n for n in range(1, 106) if is_prime(n)==True)
count_best = 0
Slightly modified (first for loop)
from itertools import islice
def window(seq):
for n in range(2, len(seq) + 1):
it = iter(seq)
result = tuple(islice(it, n))
if len(result) == n:
yield result
for elem in it:
result = result[1:] + (elem,)
yield result
def memoize(function):
cache = {}
def decorated_function(args):
if args in cache:
return cache[args]
else:
val = function(args)
cache[args] = val
return val
return decorated_function
@memoize
def find_lin_comb(prime):
global count_best
for windows in window(primes[0 : primes.index(prime)]):
if sum(windows) == prime and len(windows) > count_best:
count_best = len(windows)
print('Prime: ', prime, 'Terms: ', count_best)
Find them:
for x in primes[::-1]: find_lin_comb(x)
```
(btw, the tuple of prime numbers is generated "decently" fast)
All input is appreciated, I am just a hobby programmer, so please don´t get to advanced on me.
Thank you!
Edit: here is a working code paste that doesn´t have ruined indentations:
python
primes
memoization
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edited Jan 22, 2015 at 18:32
Zero Piraeus
59.7k 28 28 gold badges 157 157 silver badges 164 164 bronze badges
asked May 20, 2012 at 12:54
luffeluffe
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2
Sorry, it works, the indentation errors are there because of Stackoverflows "code" input mode. I will fix it now luffe –luffe 2012-05-20 13:01:54 +00:00 Commented May 20, 2012 at 13:01
Sorry, the code input procedure has completely confused me, so here the code with correct indentations are: pastebin.com/R1NpMqgbluffe –luffe 2012-05-20 13:06:26 +00:00 Commented May 20, 2012 at 13:06
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This works fine for all primes up to 1000, but for primes up to 106 it is very slow, so I was hoping memozation would help; when calculating the sum of sliding windows, a lot of double work is done...(right?)
Yes, right. And of course it's slow for the primes up to 10 6.
Say you have n primes up to N, numbered in increasing order, p_1 = 2, p_2 = 3, .... When considering whether prime no. k is the sum of consecutive primes, you consider all windows [p_i, ..., p_j], for pairs (i,j) with i < j < k. There are (k-1)(k-2)/2 of them. Going through all k to n, you examine about n³/6 windows in total (counting multiplicity, you're examining w(i.j) in total n-j times). Even ignoring the cost of creating the window and summing it, you can see how it scales badly:
For N = 1000, there are n = 168 primes and about 790000 windows to examine (counting multiplicity).
For N = 106, there are n = 78498 primes and about 8.31013 windows to examine.
Now factor in the work for creating and summing the windows, estimate it low at j-i+1 for summing the j-i+1 primes in w(i,j), the work for p_k is about k³/6, and the total work becomes roughly k4/24. Something like 33 million steps for N = 1000, peanuts, but nearly 1.61018 for N = 1000000.
A year contains about 3.1107 seconds, with a ~3GHz CPU, that's roughly 10 17 clock cycles. So we're talking of an operation needing something like 100 CPU-years (may be a factor of 10 off or so).
You aren't willing to wait that long, I suppose;)
Now, with memoisation, you still look at each window multiple times, but you do the computation of each window only once. That means you need about n³/6 work for the computation of the windows, and look about n³/6 times at any window.
Problem 1: You still need to look at windows about 8.31013 times, that's several hours even if looking cost only one cycle.
Problem 2: There are about 8.31013 windows to memoise. You don't have that much memory, unless you can use a really large HD.
You can circumvent the memory problem by throwing away data you don't need anymore and only calculating data for the windows when it is needed, but once you know which data you may throw away when, you should be able to see a much better approach.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
What does this tell you about the window generating that sum? Where can it start, where can it stop? How can you use that information to create an efficient algorithm to solve the problem?
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answered May 22, 2012 at 14:18
Daniel FischerDaniel Fischer
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luffe
luffeOver a year ago
Thank you both. I see my approach was just plain horrible :) I found the solution by stopping summing when the window (and all remaining windows ("sorted"), yielded a sum above 1 million). Thanks again :)
2012-05-23T14:20:21.013Z+00:00
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Daniel Fischer
Daniel FischerOver a year ago
Well, "I am just a hobby programmer", you said, so it's okay if you don't see how much time your approach would take. After being told, you found the correct condition to get it done quickly, that's a very good sign.
2012-05-23T14:32:09.193Z+00:00
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luffeOver a year ago
Thank you for that, I would say learning programming with Project Euler has been very eye-opening. For a lay-man like me it´s almost unimaginable that a computer in today´s age could take a decade solving such a simple problem, so for me this was really cool and eye-opening to think about :)
2012-05-23T16:10:25.09Z+00:00
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Daniel FischerOver a year ago
A decade is nothing ;) There are some Project Euler Problems where a naive brute force search would run several times the age of the universe. But it's very satisfying to crack these, so keep enjoying it. With more experience, you'll be able to estimate the time needed for your algorithm before you run it, so you know that some aren't even worth trying.
2012-05-23T17:38:35.33Z+00:00
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Daniel FischerOver a year ago
Sorry, I don't know of any. I usually estimate the number of high-level operations first (number of windows to look at here), then estimate how many lower-level operations are needed per high-level op, continue until a sufficiently low level is reached. If I need something more precise than an estimate that may be off by a factor of 100, firstly more precise estimates or even calculations are used, secondly, some smaller benchmarks are timed - extrapolating how higher memory usage influences running time is tricky, though (makes things slower, but how much?).
2012-05-24T13:28:49.15Z+00:00
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The memoize decorator adds a wrapper to a function to cache the return value for each value of the argument (each combination of values in case of multiple arguments). It is useful when the function is called multiple times with the same arguments. You can only use it with a pure function, i.e.
The function has no side effects. Changing a global variable and doing output are examples of side effects.
The return value depends only on the values of the arguments, not on some global variables that may change values between calls.
Your find_lin_comb function does not satisfy the above criteria. For one thing, it is called with a different argument every time, for another, the function does not return a value.
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edited May 21, 2012 at 5:39
answered May 20, 2012 at 19:49
Janne KarilaJanne Karila
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187691 | https://www.mathematicalgemstones.com/gemstones/pearl/pythagorean-triples-on-a-sphere/ | Pythagorean triples on a sphere? | Mathematical Gemstones
HomeAll PostsTagsAbout this blogMaria Gillespie's Home Page
Posted on: August 16, 2017 by Maria Gillespie
< Previous PostNext Post >
Pythagorean triples on a sphere?
It’s 8/15/17, which means it’s time to celebrate! The three numbers making up the date today form a Pythagorean triple, a triple of positive integers (a,b,c) with a 2+b 2=c 2. Indeed, 8 2+15 2=64+225=289=17 2.
Alternatively, by the Pythagorean theorem, a Pythagorean triple is any triple of positive integers which make up the sides of a right triangle:
It’s exciting when all three sides are integers, since many common right triangles’ lengths involve square roots: (1,1,2), (1,2,5), and (1,3,2), to name a few. And these sides aren’t even rational, which the poor Pythagorean scholar Hippasus discovered by proving that 2 is irrational and was subsequently drowned to death by his colleagues, according to some historical accounts. So the ancient Pythagoreans in fact only believed in right triangles having rational side lengths.
Of course, given one Pythagorean triple, like (3,4,5), we can construct infinitely many by scaling the sides: (6,8,10), (9,12,15), etc. (In fact, 12/9/15 was the previous Pythagorean triple day, and 12/16/20 will be the next.) So to classify all Pythagorean triples, it suffices to find the reduced triples, those with no common factors greater than 1.
So the last reduced Pythagorean triple day was back in 2013 on 12/5/13, and the next one won’t be until 7/24/25!
Constructing Pythagorean triples
It’s well known that there are infinitely many reduced Pythagorean triples as well. One beautiful, famous proof of this derives a parameterization of all triples via geometric means:
Consider the unit circle and the point on the circle. Let be a rational point on the -axis inside the circle. Then line intersects the circle at a second point , and it turns out has rational coordinates as well. Some simple algebra with similar triangles (try it, if you haven’t seen it before!) gives us But since lies on the unit circle, if are the coordinates of then , and substituting and clearing denominators we have (which can also be checked with direct algebraic computation). It follows that is a Pythagorean triple for any integers and , giving us infinitely many Pythagorean triples. And in fact, for and relatively prime of different parity, these triples are reduced.
Spherical considerations
Given that this is a global day to celebrate (assuming you use the standard world calendar), and the Earth is a sphere, I have to wonder whether Pythagorean triples actually exist if drawn on a perfect-sphere approximation of the Earth. First, we’d have to define what we even mean by that - are we measuring in meters? In feet? And what is a right triangle on the surface of a sphere anyway?
The most common definition of a triangle on a sphere is one formed by great circles. A great circle is any circle of maximal diameter around a sphere, in other words, whose plane contains the center of the sphere. So, the equator, the prime meridian, and all longitude lines are great circles on the Earth, but latitude lines are not. A line segment on a sphere is a segment of a great circle, and a (spherical) triangle is a shape formed by three points connected by three (spherical) line segments. The angle between two great circles that intersect is the angle between their planes. (Thanks to Wikipedia for the excellent image below.)
Since our world now has size rather than being a flat plane, let’s set the radius of the Earth to be unit for simplicity. So we’re working with triangles with a right angle at , and asking when they have rational lengths:
Are there any Pythagorean triples on the unit sphere? Are there infinitely many?
These questions suddenly aren’t very easy. If we scale our sphere down by we can get at least one, by taking the equator, the prime meridian, and the longitudinal line. This forms a right triangle with all three lengths equal (and all angles right!) and so we can simply scale the Earth to make it any rational lengths we want. But this still doesn’t answer anything about the unit sphere.
There is some hope, however. In this paper by Hartshorne and Van Luijk, the authors show that there are infinitely many Pythagorean triples in the hyperbolic plane, using the Poincare Disk model and some nice hyperbolic trig formulas combined with some Eulerian number theory tricks. So Pythagorean triples are not the sole property of flat Euclidean space.
In addition, there is a ``spherical Pythagorean theorem’’, which in our notation, since the radius of our sphere is , says that where are the side lengths of the triangle and is opposite the right angle. And yet, this offers little insight into whether this equation has rational solutions. Trig functions are generally much harder to deal with than polynomials, especially when it comes to solving equations over the rationals.
So, if you have any insights on this problem (or references to papers that have worked on it - I couldn’t find any in an initial search, but I am not very familiar with it), please let me know in the comments! And until then, happy reduced-Pythagorean-triple-in-flat-Euclidean-space day!
Tags: pearl
©2012–2025 Maria Gillespie |
187692 | https://en.wikipedia.org/wiki/Palindromic_number | Jump to content
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Contents
1 Formal definition
2 Decimal palindromic numbers
2.1 Perfect powers
3 Other bases
4 Antipalindromic numbers
5 Lychrel process
6 Sum of the reciprocals
7 Scheherazade numbers
8 Sums of palindromes
9 Notes
10 References
11 External links
Palindromic number
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From Wikipedia, the free encyclopedia
Number that remains the same when its digits are reversed
A palindromic number (also known as a numeral palindrome or a numeric palindrome) is a number (such as 16361) that remains the same when its digits are reversed. In other words, it has reflectional symmetry across a vertical axis. The term palindromic is derived from palindrome, which refers to a word (such as rotor or racecar) whose spelling is unchanged when its letters are reversed. The first 30 palindromic numbers (in decimal) are:
: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, ... (sequence A002113 in the OEIS).
Palindromic numbers receive most attention in the realm of recreational mathematics. A typical problem asks for numbers that possess a certain property and are palindromic. For instance:
The palindromic primes are 2, 3, 5, 7, 11, 101, 131, 151, ... (sequence A002385 in the OEIS).
The palindromic square numbers are 0, 1, 4, 9, 121, 484, 676, 10201, 12321, ... (sequence A002779 in the OEIS).
In any base there are infinitely many palindromic numbers, since in any base the infinite sequence of numbers written (in that base) as 101, 1001, 10001, 100001, etc. consists solely of palindromic numbers.
Formal definition
[edit]
Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number n > 0 in base b ≥ 2, where it is written in standard notation with k+1 digits ai as:
with, as usual, 0 ≤ ai < b for all i and ak ≠ 0. Then n is palindromic if and only if ai = ak−i for all i. Zero is written 0 in any base and is also palindromic by definition.
Decimal palindromic numbers
[edit]
All numbers with one digit are palindromic, so in base 10 there are ten palindromic numbers with one digit:
: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
There are 9 palindromic numbers with two digits:
: {11, 22, 33, 44, 55, 66, 77, 88, 99}.
All palindromic numbers with an even number of digits are divisible by 11.
There are 90 palindromic numbers with three digits (Using the rule of product: 9 choices for the first digit - which determines the third digit as well - multiplied by 10 choices for the second digit):
: {101, 111, 121, 131, 141, 151, 161, 171, 181, 191, ..., 909, 919, 929, 939, 949, 959, 969, 979, 989, 999}
There are likewise 90 palindromic numbers with four digits (again, 9 choices for the first digit multiplied by ten choices for the second digit. The other two digits are determined by the choice of the first two):
: {1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, 1991, ..., 9009, 9119, 9229, 9339, 9449, 9559, 9669, 9779, 9889, 9999},
so there are 199 palindromic numbers smaller than 104.
There are 1099 palindromic numbers smaller than 105 and for other exponents of 10n we have: 1999, 10999, 19999, 109999, 199999, 1099999, ... (sequence A070199 in the OEIS). The number of palindromic numbers which have some other property are listed below:
| | 101 | 102 | 103 | 104 | 105 | 106 | 107 | 108 | 109 | 1010 |
--- --- --- --- ---
| n natural | 10 | 19 | 109 | 199 | 1099 | 1999 | 10999 | 19999 | 109999 | 199999 |
| n even | 5 | 9 | 49 | 89 | 489 | 889 | 4889 | 8889 | 48889 | 88889 |
| n odd | 5 | 10 | 60 | 110 | 610 | 1110 | 6110 | 11110 | 61110 | 111110 |
| n square | 4 | 7 | 14 | 15 | 20 | 31 |
| n cube | 3 | 4 | 5 | 7 | 8 |
| n prime | 4 | 5 | 20 | 113 | 781 | 5953 |
| n squarefree | 6 | 12 | 67 | 120 | 675 | 1200 | 6821 | 12160 | + | + |
| n non-squarefree (μ(n)=0) | 4 | 7 | 42 | 79 | 424 | 799 | 4178 | 7839 | + | + |
| n square with prime root | 2 | 3 | 5 |
| n with an even number of distinct prime factors (μ(n)=1) | 2 | 6 | 35 | 56 | 324 | 583 | 3383 | 6093 | + | + |
| n with an odd number of distinct prime factors (μ(n)=-1) | 4 | 6 | 32 | 64 | 351 | 617 | 3438 | 6067 | + | + |
| n even with an odd number of prime factors | 1 | 2 | 9 | 21 | 100 | 180 | 1010 | 6067 | + | + |
| n even with an odd number of distinct prime factors | 3 | 4 | 21 | 49 | 268 | 482 | 2486 | 4452 | + | + |
| n odd with an odd number of prime factors | 3 | 4 | 23 | 43 | 251 | 437 | 2428 | 4315 | + | + |
| n odd with an odd number of distinct prime factors | 4 | 5 | 28 | 56 | 317 | 566 | 3070 | 5607 | + | + |
| n even squarefree with an even number of (distinct) prime factors | 1 | 2 | 11 | 15 | 98 | 171 | 991 | 1782 | + | + |
| n odd squarefree with an even number of (distinct) prime factors | 1 | 4 | 24 | 41 | 226 | 412 | 2392 | 4221 | + | + |
| n odd with exactly 2 prime factors | 1 | 4 | 25 | 39 | 205 | 303 | 1768 | 2403 | + | + |
| n even with exactly 2 prime factors | 2 | 3 | 11 | 64 | 413 | + | + |
| n even with exactly 3 prime factors | 1 | 3 | 14 | 24 | 122 | 179 | 1056 | 1400 | + | + |
| n even with exactly 3 distinct prime factors | 0 | 1 | 18 | 44 | 250 | 390 | 2001 | 2814 | + | + |
| n odd with exactly 3 prime factors | 0 | 1 | 12 | 34 | 173 | 348 | 1762 | 3292 | + | + |
| n Carmichael number | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
| n for which σ(n) is palindromic | 6 | 10 | 47 | 114 | 688 | 1417 | 5683 | + | + | + |
Perfect powers
[edit]
There are many palindromic perfect powers nk, where n is a natural number and k is 2, 3 or 4.
Palindromic squares: 0, 1, 4, 9, 121, 484, 676, 10201, 12321, 14641, 40804, 44944, ... (sequence A002779 in the OEIS)
Palindromic cubes: 0, 1, 8, 343, 1331, 1030301, 1367631, 1003003001, ... (sequence A002781 in the OEIS)
Palindromic fourth powers: 0, 1, 14641, 104060401, 1004006004001, ... (sequence A186080 in the OEIS)
The first nine terms of the sequence 12, 112, 1112, 11112, ... form the palindromes 1, 121, 12321, 1234321, ... (sequence A002477 in the OEIS)
The only known non-palindromic number whose cube is a palindrome is 2201, and it is a conjecture the fourth root of all the palindrome fourth powers are a palindrome with 100000...000001 (10n + 1).
Gustavus Simmons conjectured there are no palindromes of form nk for k > 4 (and n > 1).
Other bases
[edit]
Palindromic numbers can be considered in numeral systems other than decimal. For example, the binary palindromic numbers are those with the binary representations:
: 0, 1, 11, 101, 111, 1001, 1111, 10001, 10101, 11011, 11111, 100001, ... (sequence A057148 in the OEIS)
or in decimal:
: 0, 1, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, ... (sequence A006995 in the OEIS)
The Fermat primes and the Mersenne primes form a subset of the binary palindromic primes.
Any number is palindromic in all bases with (trivially so, because is then a single-digit number), and also in base (because is then ). Even excluding cases where the number is smaller than the base, most numbers are palindromic in more than one base. For example, , . A number is never palindromic in base if . Moreover, a prime number is never palindromic in base if .
A number that is non-palindromic in all bases b in the range 2 ≤ b ≤ n − 2 can be called a strictly non-palindromic number. For example, the number 6 is written as "110" in base 2, "20" in base 3, and "12" in base 4, none of which are palindromes. All strictly non-palindromic numbers larger than 6 are prime. Indeed, if is composite, then either for some , in which case n is the palindrome "aa" in base , or else it is a perfect square , in which case n is the palindrome "121" in base (except for the special case of ).
The first few strictly non-palindromic numbers (sequence A016038 in the OEIS) are:
: 0, 1, 2, 3, 4, 6, 11, 19, 47, 53, 79, 103, 137, 139, 149, 163, 167, 179, 223, 263, 269, 283, 293, 311, 317, 347, 359, 367, 389, 439, 491, 563, 569, 593, 607, 659, 739, 827, 853, 877, 977, 983, 997, ...
Antipalindromic numbers
[edit]
If the digits of a natural number don't only have to be reversed in order, but also subtracted from to yield the original sequence again, then the number is said to be antipalindromic. Formally, in the usual decomposition of a natural number into its digits in base , a number is antipalindromic iff .
Lychrel process
[edit]
Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. Such number is called "a delayed palindrome".
It is not known whether all non-palindromic numbers can be paired with palindromic numbers in this way. While no number has been proven to be unpaired, many do not appear to be. For example, 196 does not yield a palindrome even after 700,000,000 iterations. Any number that never becomes palindromic in this way is known as a Lychrel number.
On January 24, 2017, the number 1,999,291,987,030,606,810 was published in OEIS as A281509 and announced "The Largest Known Most Delayed Palindrome". The sequence of 125 261-step most delayed palindromes preceding 1,999,291,987,030,606,810 and not reported before was published separately as A281508.
Sum of the reciprocals
[edit]
The sum of the reciprocals of the palindromic numbers is a convergent series, whose value is approximately 3.37028... (sequence A118031 in the OEIS).
Scheherazade numbers
[edit]
Scheherazade numbers are a set of numbers identified by Buckminster Fuller in his book Synergetics. Fuller does not give a formal definition for this term, but from the examples he gives, it can be understood to be those numbers that contain a factor of the primorial n#, where n≥13 and is the largest prime factor in the number. Fuller called these numbers Scheherazade numbers because they must have a factor of 1001. Scheherazade is the storyteller of One Thousand and One Nights, telling a new story each night to delay her execution. Since n must be at least 13, the primorial must be at least 1·2·3·5·7·11·13, and 7×11×13 = 1001. Fuller also refers to powers of 1001 as Scheherazade numbers. The smallest primorial containing Scheherazade number is 13# = 30,030.
Fuller pointed out that some of these numbers are palindromic by groups of digits. For instance 17# = 510,510 shows a symmetry of groups of three digits. Fuller called such numbers Scheherazade Sublimely Rememberable Comprehensive Dividends, or SSRCD numbers. Fuller notes that 1001 raised to a power not only produces sublimely rememberable numbers that are palindromic in three-digit groups, but also the values of the groups are the binomial coefficients. For instance,
This sequence fails at (1001)13 because there is a carry digit taken into the group to the left in some groups. Fuller suggests writing these spillovers on a separate line. If this is done, using more spillover lines as necessary, the symmetry is preserved indefinitely to any power. Many other Scheherazade numbers show similar symmetries when expressed in this way.
Sums of palindromes
[edit]
In 2018, a paper was published demonstrating that every positive integer can be written as the sum of three palindromic numbers in every number system with base 5 or greater.
Notes
[edit]
^ "The Prime Glossary: palindromic prime". PrimePages. Retrieved 11 July 2023.
^ (sequence A065379 in the OEIS) The next example is 19 digits - 900075181570009.
^ Murray S. Klamkin (1990), Problems in applied mathematics: selections from SIAM review, p. 520.
^ Sloane, N. J. A. (ed.). "Sequence A016038 (Strictly non-palindromic numbers)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
^ Guy, Richard K. (1989). "Conway's RATS and other reversals". The American Mathematical Monthly. 96 (5): 425–428. doi:10.2307/2325149. JSTOR 2325149.
^ Dvorakova, Lubomira; Kruml, Stanislav; Ryzak, David (16 Aug 2020). "Antipalindromic numbers". arXiv:2008.06864 [math.CO].
^ R. Buckminster Fuller, with E. J. Applewhite, Synergetics: Explorations in the Geometry of thinking Archived 2016-02-27 at the Wayback Machine, Macmillan, 1982 ISBN 0-02-065320-4.
^ Fuller, pp. 773-774 Archived 2016-03-05 at the Wayback Machine
^ Fuller, pp. 777-780
^ Cilleruelo, Javier; Luca, Florian; Baxter, Lewis (2016-02-19). "Every positive integer is a sum of three palindromes". Mathematics of Computation. arXiv:1602.06208. Archived from the original on 2021-02-12. Retrieved 2021-04-28. (arXiv preprint Archived 2019-02-08 at the Wayback Machine)
References
[edit]
Malcolm E. Lines: A Number for Your Thoughts: Facts and Speculations about Number from Euclid to the latest Computers: CRC Press 1986, ISBN 0-85274-495-1, S. 61 (Limited Online-Version (Google Books))
External links
[edit]
Weisstein, Eric W. "Palindromic Number". MathWorld.
Jason Doucette - 196 Palindrome Quest / Most Delayed Palindromic Number
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Palindromic Numbers to 100,000 Archived 2007-02-02 at the Wayback Machine from Ask Dr. Math
P. De Geest, Palindromic cubes Archived 2016-03-25 at the Wayback Machine
Yutaka Nishiyama, Numerical Palindromes and the 196 Problem, IJPAM, Vol.80, No.3, 375–384, 2012.
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Floyd’s Cycle Detection Algorithm (Tortoise and Hare problem with codes)
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P roblem: Given a Linked list detect if there is a loop or not and using this calculate the length of the loop, and the head node of the loop.
Press enter or click to view image in full size
Description: In a Linked List ofnnodes we need to find if there’s a loop, what is its length, and which is the first node of the loop. This problem is asked quite frequently in interviews. This traditional interview problem is also known asTortoise and Hare problem.
What is a Linked List?
A LinkedList is a linear data structure that consists of nodes and each node contains a data field and reference to its next node.
In this problem, we are given a Linked List that is a combination of both Singly Linked List and Circular Linked List.
Approaches:
I) Brute-Force Approach:
This is the most basic approach for detecting the loop, it consists of iterating in all possible ways which can satisfy the problem statement. Let’s understand how.
Take two pointers (inner and outer) and in one iteration move a pointer(outer) to its next node, and in that same iteration move another pointer(inner) from the start of Linked List to k nodes where k is the number of nodes traversed by outer and check if inner==outer and k>0.
Move outer pointer till outer!=null or the condition inner==outer satisfies. If the condition (inner==outer) satisfies there is a loop, else if (outer!=null) satisfies there is no loop. The code itself is self-explanatory, just have a look and try to dry run and you will understand the logic.
Here’s the Javacode for Brute force:
public boolean detectLoop(ListNode head)
{
ListNode outer = head;
int nodesTraversedByOuter = 0;
// Traverse the Linked List.
while (outer != null) {
outer = outer.next;
nodesTraversedByOuter++;
ListNode inner = head;
int k = nodesTraversedByOuter;
// iterating inner loop from head to number of nodes traversed by outer at this point.
while (k > 0) {
if (inner == outer)
{
return true;
}
inner = inner.next;
k--;
}
}
return false;
}
Its Time complexity is O(n²) because we are using two nested loops and Space complexity is O(1).
II) Hashing:
In this approach, we will use a HashSet in Java (Same as an Unordered Set in C++ and Set in Python). We will traverse the Linked List using a pointer ptr1 initially pointing to the head of the Linked List and simultaneously add the visited node to the HashSet and check if HashSet already contains the node. If the visited node is already present in HashSet, then the Linked List contains a loop.
Press enter or click to view image in full size
Here’s the Javacode for Hashing:
public boolean detectLoop(ListNode head) {
ListNode ptr1 = head;
HashSet h = new HashSet<>();
// Traverse the Linked List.
while (ptr1 != null) {
// Check if node address already present.
if (h.contains(ptr1))
return true;
// Add the current node address to hashset
h.add(ptr1);
ptr1 = ptr1.next;
}
// If ptr1 reaches to None, then there is no loop. Hence return False.
return false;
}
Its Time complexity and Space complexity is O(n)because we are using one for loop and extra space to store n nodes.
III) Floyd cycle detection algorithm:
This is the most optimal approach. In this algorithm, we will take two pointers, let’s name them tortoiseand hare. We will move the tortoise pointer by one node and the hare pointer by two nodes in each iteration. If the hare and tortoise pointer points to the same node after traversing and if they are not null then there is a loop.
Below are the steps to detect a loop in a Linked List using Floyd’s cycle detection algorithm. Instead of tortoise and hare, ptr1 and ptr2 are used.
1) Take two pointers ptr1and ptr2and initialize them to the start node.
2) Traverse the Linked List using both the pointers but move ptr1 by one node and ptr2 by two nodes in each iteration.
3) As ptr2 is moving with double the speed as ptr1 so we need to check if ptr2 encounters null. If it encounters null, then there is no loop in the Linked List and thus stops the execution. Else ptr1 will enter the loop after some time following ptr2.
4) Now, when both the pointers are in the loop and continue to move with their respective speeds, they will eventually meet at some node. This proves that there is a loop in the Linked List.
The following images show the two pointers moving inside a loop on Linked List.
Signifies that loop is present
Here’s the Java code for Floyd Cycle Detection Algorithm:
static public boolean detectLoop(ListNode head)
{
ListNode tortoise = head;
ListNode hare = head;
while(hare.next!=null&&hare!=null)
{
tortoise=tortoise.next;
hare=hare.next.next;
if(hare==tortoise)
return true;
}
return false;
}
Its Time complexity is O(n), n is the number of nodes in Linked List, and Space complexity is O(1). It is the most optimized algorithm for this problem.
Now to calculate the length of loop:
For calculating the length of the loop we will continue from the position where both the pointers met (meeting node) and move one pointer forward, let’s take hare in this case , and continue moving it to the next node till it meets the tortoise once again . The number of iterations to meet each other again will give the length of the loop.
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Here’s the Java code for calculating length of loop:
static public int lengthOfLoop(ListNode hare, ListNode tortoise)
{
int lengthOfLoop=1;
hare=hare.next while(hare!=tortoise)
{
hare=hare.next;
lengthOfLoop++;
}
return lengthOfLoop;
}
Its Time complexity is O(L), where L is the length of the loop and Spacecomplexity is O(1).
Finding the head node of the loop:
For finding the head of the loop we will keep one pointer at the meeting node, let’s say tortoise, and another pointer (hare) at the head node of the Linked List. Now move both the pointers to their next node till they meet each other. The new meeting node at which both pointers meet will be the head node of the loop.
Here’s the Java code for finding the head node of loop
static public ListNode headOfLoop(ListNode head,ListNode tortoise)
{
ListNode hare=head;
while(hare!=tortoise)
{
hare=hare.next;
tortoise=tortoise.next;
}
return hare;
}
Its Time complexity is O(N-L), where N is the number of nodes in Linked List and L is the length of loop, and Space complexity is O(1).
Here comes the most interesting part, we have been observing that the above algorithm works perfectly fine but how’s that possible!? And most importantly will it always work? The answer is YES! Let’s see how.
Proof of why this approach works:
Distance covered by hare Pointer = 2 ( Distance Covered by tortoise pointer)
Let’s say the distance covered by tortoise pointer when they meet is N, therefore the distance covered by hare pointer when they meet will be 2N(because it is twice as fast as tortoise).
And d is the distance between head node of the Linked List and head node of loop, k is the distance between head node of loop and the node where the pointers hare and tortoise met while detecting the loop, and c is the length of the loop.
Therefore, N = d + k + c i — (1)
(for tortoise pointer where i is an integer representing the number of times loop traversed)
and 2N = d + k + c j — (2)
(for hare pointer where j is an integer representing the number of times loop traversed)
For Equating, Multiply — (1) with 2, and then subtract it with — (2)
→ 2N — 2N = (2d +2k +2ci) — (d + k + cj)
→ 0 = d + k + c ( 2i — j )
→ d = c (j — 2i) — k
which signifies that d, the distance between head node of linked list and head node of loop is equal to the integral (since [j — 2i] is still an integer) multiple of c (length of loop) minus k (distance between head node of loop and first meeting node).
In simple words, dis equal to the distance between the first meeting node and head of the loop. Therefore, if we move forward d nodes from the first meeting node we will reach the head node of the loop.
Hence, if we will keep one pointer at head of the Linked List and another pointer at the first meeting node and move them forward by one node then the node at which they will meet will be the starting node of the loop.
Hence proved.
Thank you!
Video by Priyaank Arora
Algorithms
Floyd Cycle Detection
Hare And Tortoise
Interview Questions
Java
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Responses (3)
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Amit Sharma
Jan 24, 2022
what is 'i' and 'j'?
--
1 reply
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Rohit Tayal
Jul 11, 2023
How do we know that 'j' and 'i' are small numbers? If they are large, time complexity cannot be termed as O(1), right?
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Abhijit Sarkar
Jul 20, 2021
How do we know j-2i > k?
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How Do You Find the Zeros of a Quadratic Function on a Graph?
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The zeros of a quadratic equation are the points where the graph of the quadratic equation crosses the x-axis. In this tutorial, you'll see how to use the graph of a quadratic equation to find the zeros of the equation. Take a look!
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The zero of a function is the x-value that makes the function equal to 0. In this tutorial, you'll learn about the zero of a function and see how to find it in an example. Take a look!
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What's the X-Intercept?
When you have a linear equation, the x-intercept is the point where the graph of the line crosses the x-axis. In this tutorial, learn about the x-intercept. Check it out!
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What is the coefficient of the term a i b j c k a i b j c k in the expansion of (a+b+c)n(a+b+c)n?
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What is the coefficient of the term a i b j c k a i b j c k in the expansion of (a+b+c)n(a+b+c)n ?
Given solution:
Before collecting like terms, each term in the expansion is a product of n n factors each of which is a,b,a,b, or c c. Those terms which have exactly i a’s, j b’s and k c’s can be combined into the single term a i b j c k a i b j c k and the number of such terms will be equal to the number of n letter strings containing i a’s, j b’s and k c’s, which by the Mississippi rule is n!(i!×j!×k!)n!(i!×j!×k!).
I understand that this problem is actually about the derivation of the coefficient of the multinomial theorem,but I am not sure about the combinatoric reasoning,what I mean is that I understand the analogy problem but what I couldn't is how the analogy holds here,Could anybody explain this to me?
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asked Oct 20, 2011 at 18:51
QuixoticQuixotic
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The combinatorics is not just an analogy; it's what actually goes on. When you expand (a+b+c)n(a+b+c)n fully, without using the commutative law for multiplication, what you get is the sum of all n n-ary products where each factor is one of a a, b b, and c c, and each such product occurs exactly once.
Therefore, when you apply the commutative law and collect like terms, what you're asking is exactly: "how many n n-letter words from the alphabet {a,b,c}{a,b,c} contain exactly i i, j j, k k of each letter. So when you collect like factors in each term, you get exactly n!i!j!k!n!i!j!k! instances of the term a i b j c k a i b j c k, and that is directly where the coefficient comes from.
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answered Oct 20, 2011 at 19:12
hmakholm left over Monicahmakholm left over Monica
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Writing (a+b+c)n=(x+c)n(a+b+c)n=(x+c)n for x=a+b x=a+b, one sees this is the coefficient of a i b j a i b j in x n−k=(a+b)n−k x n−k=(a+b)n−k, times the coefficient of c k c k in (x+c)n(x+c)n.
Hence, if n≠i+j+k n≠i+j+k the answer is zero and, if n=i+j+k n=i+j+k the answer is
(i+j i)(n k)=n!i!j!k!=(n i,j,k).(i+j i)(n k)=n!i!j!k!=(n i,j,k).
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edited Oct 20, 2011 at 19:32
answered Oct 20, 2011 at 18:58
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You have i i a a's,j j b b's and k k c c's where i+j+k=n i+j+k=n. An example of a term contributing to a i b j c k a i b j c k is
a b a c c a b a…b cn terms a b a c c a b a…b c⏟n terms
containing i i a a's,j j b b's and k k c c's. Each such finite sequence containing i i a a's,j j b b's and k k c c's appears exactly once when you expand (a+b+c)n(a+b+c)n and due to commutativity of multiplication all these finite sequences contribute to the term a i b j c k a i b j c k. Conversely, by making use of commutativity of multiplication, all terms which yield a i b j c k a i b j c k are just a finite sequence of length n n containing i i a a's,j j b b's and k k c c's.
So it boils down to the number of ways you can arrange i i identical a a's, j j identical b b's and k k identical c c's where i+j+k=n i+j+k=n.
If you have n n objects where r r are colored red, b b are colored blue and g g are colored green where r+b+g=n r+b+g=n, how will you go about finding the number of permutations?
Your first hunch would be n!n!. But you have r r indistinguishable red balls. Hence you need to divide by r!r! to take into account the repetitions. Similarly, you have b b indistinguishable blue balls. Hence you need to divide by b!b! to take into account the repetitions and you have g g indistinguishable green balls. Hence you need to divide by g!g! to take into account the repetitions.
Hence, the total number is n!r!b!g!n!r!b!g!.
In your case, the number of such finite sequence of i i a a's,j j b b's and k k c c's where i+j+k=n i+j+k=n is
n!i!j!k!n!i!j!k!
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answered Oct 20, 2011 at 19:13
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187696 | https://www.youtube.com/watch?v=nP0-quHxhjg | Difference Between Drying and Evaporation | Not All Drying is Evaporating: Words of Wisdom
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Ever wondered what the difference between drying and evaporation is? In this video, we'll explain the concept of both processes in a simple and easy-to-understand way. We'll look at how moisture is removed from objects or surfaces through each mechanism. We'll also compare and contrast the two so you can understand why they are different, but equally important to our everyday lives. Get all your questions answered by watching this informative video about drying and evaporation!
drying #dryingout #dryingclothes #dryinghair #dryingracks #evaporation #evaporations #évaporation #revaporation #levaporation
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[Music] difference between drying and evaporation [Music] drying and evaporation are to processes that are closely related but has some important differences drying is the process of removing water from a material while evaporation is the process of transforming water from a liquid to a vapor both processes have applications in many industries from agriculture to Manufacturing [Music] when it comes to drying there are several techniques that can be used to remove water from a material examples include air drying freeze drying vacuum drying spray drying and oven drying these techniques involve the application of heat pressure or a combination of both to remove water from the material depending on the material this process may take anywhere from a few minutes to several hours on the other hand evaporation is a process that occurs naturally in nature this is the process where liquid water is converted into vapor and dissipates into the air evaporation occurs when the air temperature is higher than the temperature of the liquid causing water molecules to escape from the liquid surface this process is often used to extract water from materials in manufacturing as well as in agriculture to increase crop yields another difference between drying and evaporation is that drying requires energy and evaporation does not when drying energy is needed to increase the air temperature so that the water molecules can escape from the material when evaporation occurs energy is not required as the process happens naturally [Music] in conclusion drying and evaporation are closely related processes but there are some key differences between them drying is the process of removing water from a material while evaporation is the process of transforming water from a liquid to a vapor drying does not alter the composition of the material and requires energy while evaporation changes the composition of the material and does not require energy understanding the differences between these two processes is important for many industries from agriculture to Manufacturing thanks for watching don't forget to hit the like button and subscribe |
187697 | https://pundit.pratt.duke.edu/piki/images/d/da/AppA_EGR119_S10.pdf | EGR 119L - Spring 2010 Appendix A Laplace Transform Reference and Examples A.1 Introduction This document covers a basic introduction to forward and inverse Laplace Transforms. It will also present example problems using Laplace transforms to solve a mechanical system and an electrical system, respec-tively.
A.2 Synthesis and Analysis Equations There are two main kinds of Laplace transform - the bilateral Laplace transform and the unilateral Laplace transform. The primary distinction between the two is that the unilateral Laplace transform only uses the portion of a signal after time 0−. Anything else about the signal for negative times will be summarized by a constant in the transformation. Because of this, there are two different sets of synthesis and analysis equations: Bilateral Unilateral Synthesis x(t) = 1 2πj R σ+j∞ σ−j∞X(s)est ds x(t) = 1 2πj R σ+j∞ σ−j∞X(s)est ds Analysis X(s) = R ∞ −∞x(t)e−st dt X(s) = R ∞ 0−x(t)e−st dt Note in the synthesis equations that there is a constant real value σ in the limits on integration - this value is chosen such that the integral converges. The set of σ values for which the synthesis integral converges is known as the Region of Convergence, or ROC, of the particular transform. Also note that the synthesis equation itself is rarely used; rather, specific inverse Laplace transform pairs should be memorized and applied when appropriate.
Some books will clearly distinguish between the two kinds of Laplace transforms while others will simply assume one or the other (or use the same symbol for both!).
The Oppenheim and Willsky book refer-enced below uses X(s) to denote the bilateral Laplace transform and X(s) to denote the unilateral Laplace transform. The Haykin and Van Veen book uses X(s) for both.
Copyright 2010, Gustafson et al.
App A – 1 EGR 119L - Spring 2010 A.3 Common Laplace Transform Pairs and Properties The next three subsections present tables of common Laplace transform pairs and Laplace transform prop-erties. The information in these tables has been adapted from: • Signals and Systems, 2nd ed. Simon Haykin and Barry Van Veen. John Wiley & Sons, Hoboken, NJ, 2005. pp. 781-783.
• Signals and Systems, 2nd ed. Alan V. Oppenheim and Alan S. Willsky with S. Hamid Nawab.
Prentice Hall, Upper Saddle River, NJ, 1997. p. 691-692.
A.3.1 Common Laplace Transform Pairs Basic Bilateral Laplace Transform Pairs Name Signal Laplace Transform ROC Basic Signal x(t) X(s) Rx Impulse x(t) = δ(t −t0) X(s) = e−st0 All s Unit step x(t) = u(t −t0) X(s) = e−st0 s σ > 0 Reversed step x(t) = −u(−(t −t0)) X(s) = e−st0 s σ < 0 Polynomial x(t) = tn−1 (n −1)!u(t) X(s) = 1 sn σ > 0 Reversed Polynomial x(t) = − tn−1 (n −1)!u(−t) X(s) = 1 sn σ < 0 Exponential x(t) = e−αtu(t) X(s) = 1 s + α σ > −ℜ{α} Reversed Exponential x(t) = −e−αtu(−t) X(s) = 1 s + α σ < −ℜ{α} Polynomial Exponential x(t) = tn−1 (n −1)!e−αtu(t) X(s) = 1 (s + α)n σ > −ℜ{α} Rev. Poly. Exp.
x(t) = − tn−1 (n −1)!e−αtu(−t) X(s) = 1 (s + α)n σ < −ℜ{α} Cosine x(t) = cos(ω0t)u(t) X(s) = s s2 + ω2 0 σ > 0 Sine x(t) = sin(ω0t)u(t) X(s) = ω0 s2 + ω2 0 σ > 0 Exponential Cosine x(t) = e−αt cos(ω0t)u(t) X(s) = s + α (s + α)2 + ω2 0 σ > −ℜ{α} Exponential Sine x(t) = e−αt sin(ω0t)u(t) X(s) = ω0 (s + α)2 + ω2 0 σ > −ℜ{α} Copyright 2010, Gustafson et al.
App A – 2 EGR 119L - Spring 2010 A.3.2 Common Laplace Transform Properties For the most part, the unilateral Laplace transform properties are the same as those for the bilateral Laplace transform. The major exceptions have to do with the fact that the unilateral Laplace transform is only defined for the part of the signal that exists at or after t = 0−, so any property that could shift, scale, or reverse the signal will have a slightly different form. Also, the derivative property requires knowledge of the initial value since there will be an instantaneous change as a result. Finally, the integral property has a lower limit of 0−instead of −∞. The bilateral properties are presented below and the unilateral properties are on the next page.
Properties of the Bilateral Laplace Transform Property Signal Laplace Transform ROC Basic Signals x(t), y(t), z(t) X(s), Y (s), Z(s) Rx, Ry, Rz Linearity z(t) = Ax(t) + By(t) Z(s) = AX(s) + BY (s) At least Rx ∩Ry Time Shifting z(t) = x (t −t0) Z(s) = e−st0X(s) Rx s-domain Shifting z(t) = es0tx(t) Z(s) = X(s −s0) s for s −s0 ∈Rx Conjugation z(t) = x∗(t) Z(s) = X∗(s∗) Rx Time and Frequency Scaling z(t) = x(at) Z(s) = 1 |a|X s a s for s a ∈Rx Convolution z(t) = x(t) ∗y(t) Z(s) = X(s)Y (s) At least Rx ∩Ry Time Differentiation z(t) = d dtx(t) Z(s) = sX(s) At least Rx Integration z(t) = Z t −∞ x(τ)dτ Z(s) = 1 sX(s) At least Rx ∩{σ > 0} Frequency Differentiation z(t) = −tx(t) Z(s) = d dsX(s) Rx Copyright 2010, Gustafson et al.
App A – 3 EGR 119L - Spring 2010 Properties of the Unilateral Laplace Transform Property Signal Laplace Transform Basic Signals x(t), y(t), z(t) x(t) = y(t) = 0, t < 0 X(s), Y(s), Z(s) Linearity z(t) = Ax(t) + By(t) Z(s) = AX(s) + BY(s) Time Shifting z(t) = x (t −t0) Z(s) = e−st0X(s) if x(t −t0)u(t) = x(t −t0)u(t −t0) s-domain Shifting z(t) = es0tx(t) Z(s) = X(s −s0) Time and Frequency Scaling z(t) = x(at), a > 0 Z(s) = 1 aX s a Conjugation z(t) = x∗(t) Z(s) = X ∗(s∗) Convolution z(t) = x(t) ∗y(t) Z(s) = X(s)Y(s) Time Differentiation z(t) = d dtx(t) Z(s) = sX(s) −x(0−) Frequency Differentiation z(t) = −tx(t) Z(s) = d dsX(s) Time Integration z(t) = Z t 0−x(τ)dτ Z(s) = 1 sX(s) A.3.3 Laplace Transform Initial and Final Value Theorems The initial and final value for a signal x(t) can be determined by its Laplace transform X(s) if certain conditions are met. If X(s) is defined as a ratio of polynomials, and the highest order of the numerator is less than the highest order of the denominator, then the initial value for the signal can be determined with: x(0+) = lim s→∞sX(s) Note that if the numerator is of the same or higher order than the denominator of the Laplace transform the limit on the right does not converge.
If X(s) is defined as a ratio of polynomials, and all the values of s that make the denominator equal to 0 have negative real parts (in other words, all the poles are in the left half-plane), then the final value for the signal can be determined with: lim t→∞x(t) = lim s→0 sX(s) Copyright 2010, Gustafson et al.
App A – 4 EGR 119L - Spring 2010 A.4 Examples The following are examples of how Fourier series coefficients can be used to simplify and solve for outputs (in these cases, positions and output voltages) given periodic inputs (forces and input voltages).
A.4.1 Spring-Mass-Damper System A mass M is attached to a stationary wall via a spring K and a damper fv. An external force f(t) is applied to the mass and the resulting position of the mass x(t) is measured.
M K fv f x The external force is given by the equation: f(t) = α + β cos(5t) + γe−3t u(t) where α, β, and γ are real constants. At time 0, the mass is displaced x0 = 1 m and has a velocity of ˙ x0 = 3 m/s. The element properties are: M=1 kg, fv=1010 kg/s=1010 N·s/m and K=10000 kg/s2=10000 N/m.
Use the bilateral Laplace transform to find the transfer function between the output position and the input force, then use the unilateral Laplace transform to determine a function for the output position.
Determine the Differential Equation and Transfer Function For this mass, the equation of motion is given by: X Fx = −Kx −fv dx dt + f(t) = M d2x dt2 M d2x dt2 + Kx + fv dx dt = f(t) Assuming the Laplace transform of the input force is given by F(s) and the Laplace transform of the output position is given by X(s), use the differentiation property to obtain the transfer function H(s) = X(s)/F(s): (Ms2 + fvs + K)X(s) = F(s) H(s) = X(s) F(s) = 1 Ms2 + fvs + K The transfer function can be used to determine how the system will respond to a wide variety of inputs, but as it is based on the bilateral transform, any solution assumes that x(t) and all its derivatives at time 0− are zero (or, alternately, that you are looking only for the particular solution). To get the complete solution, you need to use the unilateral transform properties with the differential equation.
Determine the Unilateral Laplace Transform of the Output Using the differential equation: M d2x dt2 + Kx + fv dx dt = f(t) the unilateral transform properties produce the relationship: M(s2X(s) −sx(0−) −˙ x(0−)) + fv((sX(s) −x(0−)) + K(X(s)) = F(s) (Ms2 + fvs + K)X(s) = Msx(0−) + M ˙ x(0−) + fvx(0−) + F(s) X(s) = Msx(0−) + M ˙ x(0−) + fvx(0−) + F(s) Ms2 + fvs + K Copyright 2010, Gustafson et al.
App A – 5 EGR 119L - Spring 2010 Note that this is very similar to what the transfer function would produce, with the exception of the initial position and velocity terms in the numerator. Those represent the homogeneous portion of the complete solution. Now substitute in the Laplace transform of the input: F(s) = UL {f(t)} F(s) = UL α + β cos(5t) + γe−3t u(t) F(s) = α s + βs s2 + 52 + γ s + 3 F(s) = α s3 + 3 α s2 + 25 α s + 75 α + β s3 + 3 β s2 + γ s3 + 25 γ s s (s2 + 25) (s + 3) which, once put into the expression for X(s) above and simplified, gives: X(s) = Ms5x0 + (M ˙ x0 + fvx0 + 3 Mx0) s4 + (γ + α + 3 fvx0 + β + 3 M ˙ x0 + 25 Mx0) s3 s (s2 + 25) (s + 3) (Ms2 + fvs + K) + (3 α + 3 β + 25 fvx0 + 75 Mx0 + 25 M ˙ x0) s2 + (25 α + 75 M ˙ x0 + 25 γ + 75 fvx0) s + 75 α s (s2 + 25) (s + 3) (Ms2 + fvs + K) substituting in for the parameter values and initial conditions gives: X(s) = s5 + 1016 s4 + (γ + α + 3064 + β) s3 + (3 α + 3 β + 25400)s2 + (25 α + 75975 + 25 γ)s + 75 α s (s2 + 25) (s + 3) (s2 + 1010 s + 10000) Using partial fraction expansion on this expression yields: X(s) = · · · 0.0001 α s + 0.0000143 γ s + 3.0 + 1.01 −0.0000808 β −0.000144 γ −0.000101 α s + 10.0 + −0.0131 + 0.00000101 β + 0.00000101 γ + 0.00000101 α s + 1000.0 + 0.0000798 β s + 0.000202 β s2 + 25.0 This form is useful because the denominators will indicate the form - though not the magnitude - of the functions that will comprise the solution. The numerators will give the relative magnitudes - and in the case of the last part, the balance between cosine and sine. The five denominators, in order, are indicative of a constant, an exponential with a decay rate of 3, an exponential with a decay rate of 10, an exponential with a decay rate of 1000, and sine and cosine terms with a frequency of 5 rad/s.
The final answer, obtained by taking the inverse Laplace transform of the expression above, is: x(t) = · · · 0.0001 α+ 0.000143 γ e−3.0 t+ (1.01 −0.0000808 β −0.000144 γ −0.000101 α)e−10.0 t+ (−0.0131 + 0.00000101 β + 0.00000101 γ + 0.00000101 α)e−1000.0 t+ 0.0000798 β cos (5.0 t) + 0.0000404 β sin (5.0 t) when t ≥0.
Copyright 2010, Gustafson et al.
App A – 6 EGR 119L - Spring 2010 A.4.2 RLC Circuit The example above, while demonstrating the full formal process for using Laplace transforms, gets rapidly bogged down with algebra. Typically, Laplace transforms for mechanical and electrical systems with multiple degrees of freedom will do that, so it is a good idea to use Laplace transforms in conjunction with a program capable of symbolically solving simultaneous equations as well as being able to solve forward and inverse Laplace transforms. Furthermore, to solve systems with initial conditions other than 0, you will either need to carry around the extra terms from the unilateral Laplace transform or solve the problem in frequency space using the bilateral transform, use the differentiation property to reconstruct a differential equation, and determine enough initial conditions to properly solve the differential equation. The following discussion will cover the latter case using the circuit below: vo vi R1 R2 C L with element values R1 = R2=2 kΩ, L=1 mH, C=2.2 µF.
Convert to Impedances and Laplace Transforms Converting the inductor, capacitor, and resistances to their impedances and re-writing the variables in terms of their Laplace transforms yields the following circuit: Vo Vi ZR1 ZC ZR2 ZL Set Up and Solve Circuit Equations The circuit above can be solved with voltage division. If we define ZLR2C = ZC∥( ZL + ZR2) then Vo(s) = Vi(s) · ZLR2C ZR1 + ZLR2C · ZR2 ZL + ZR2 Substituting in the individual impedances: ZL = sL ZC = 1 sC ZR1 = R1 ZR2 = R2 and simplifying yields: Vo(s) = Vi(s) · R2 (R1 L C) s2 + (R1 R2 C + L) s + (R1 + R2) Cross-multiplying produces: (R1 L C) s2 + (R1 R2 C + L) s + (R1 + R2) Vo(s) = (R2) Vi(s) Copyright 2010, Gustafson et al.
App A – 7 EGR 119L - Spring 2010 Convert to Differential Equation The s terms in the coefficients represent derivatives, so the Laplace transform above represents the differential equations: (R1 L C) d2vo dt2 + (R1 R2 C + L) dvo dt + (R1 + R2) vo = R2vi Since this is a second order ordinary differential equation, we will need to determine two values for vo -typically the initial value and the initial derivative, though that is not explicitly required.
Determine Required Initial Conditions The problem is, for circuits, the initial conditions are generally given in terms of capacitor voltages and inductor currents, since those are the variables that do not change instantaneously. In this case, for example, assume that the input voltage (in...volts) is given by: vi(t) = t < 0 3 t ≥0 5 + 5 cos(103t) and further assume that the system has been running for a very, very long time before time t = 0. The initial capacitor voltage and inductor current can be determined using the DC equivalent circuit; assuming the capacitor voltage is measured from the top node to the bottom node and the inductor current is measured as flowing from left to right, they are: vC(0−) = vi(0−) R2 R1 + R2 = 1.5 V iL(0−) = vi(0−) R1 + R2 = 0.75 mA The values for vo(0+) and ˙ vo(0+) will come from examining the circuit at time 0+ and keeping in mind the fact that: vC(0+) = vC(0−) = 1.5 V iL(0+) = iL(0−) = 0.75 mA For this circuit, since the current through the inductor is known, the voltage across resistor R2 - which is the same as the output voltage - can be determined using Ohm’s Law: vo(0+) = vR2(0+) = iL(0+) R2 = 1.5 V The more difficult equation here is to get ˙ vo(0+). Note that this would also be the same as ˙ io(0+) R2.
Since the resistor is in series with the inductor, ˙ io(0+) = ˙ iL(0+) = 1 LvL(0+). The voltage across an inductor, however, can change instantaneously so this does not give a known initial condition.
If you look at the circuit, the “easily” known values at time 0+ are the source voltage, the capacitor voltage, the inductor current, and - because of the known inductor current - the voltage and current for R2.
Using KVL on the right loop yields: −vC(0+) + vL(0+) + vR2(0+) = 0 vL(0+) = vC(0+) −vR2(0+) = 0 V For this circuit, the initial voltage across the inductor happens to be 0 V, meaning the derivative of its current is also initially zero. This in turn means the derivative of the voltage across resistor R2 is similarly zero.
Copyright 2010, Gustafson et al.
App A – 8 EGR 119L - Spring 2010 Solve Differential Equation At this point, everything necessary for solving the problem at hand is known - there is a differential equation: (R1 L C) d2vo dt2 + (R1 R2 C + L) dvo dt + (R1 + R2) vo = R2vi with element values R1 = R2=2 kΩ, L=1 mH, C=2.2 µF, a forcing function for times greater than 0 of vi(t) = 5 + 5 cos(103t) with initial conditions vo(0+) = 1.5 V ˙ vo(0+) = 0 V/s Substituting in for the parameter values and the function of the input gives: 0.0000044 d2vo dt2 + 8.801 dvo dt + 4000.0 vo = 2000.0 5 + 5 cos(103t) Solving this yields vo(t) = 2.5 −1.4285e−454.56t + 0.00079 e−1.9998∗106 t + 0.42769 cos 103 t + 0.94206 sin 103 t for times t ≥0.
Copyright 2010, Gustafson et al.
App A – 9 |
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Calculate the Future Value and Future Value Interest Factor (FVIF) for a present value invested for a future return. Our basic future value calculator sets time periods to years with interest compounded daily, monthly, or yearly.
The Future Value Formula
FV=PV(1+i)n
Where:
FV = future value
PV = present value
i = interest rate per period in decimal form
n = number of periods
The future value formula FV = PV(1+i)^n states that future value is equal to the present value multiplied by the sum of 1 plus interest rate per period raised to the number of time periods.
When using this future value formula be sure that your time period, interest rate, and compounding frequency are all in the same time unit. For example, if compounding occurs monthly the number of time periods should be the number of
months of investment, and the interest rate should be converted to a
monthly interest rate rather than yearly.
For more advanced future value calculations see our other
future value calculators. See the
Future Value of a Dollar calculator to create a table of FVIF values.
Number of Years
: Use whole numbers or decimals for partial periods such as months, so for 7 years and 6 months you would input 7.5 years
Interest Rate (I)
: • The nominal interest rate or stated rate as a percentage
: • i = I/100 is the interest rate as a decimal
Compounding
: Select daily, monthly or yearly compounding
Present Value (PV)
: Present value of a sum of money to be invested
Future Value (FV)
: The result of the FV calculation is the future value of a present value sum to be invested for some number of years at a given interest rate
FVIF
: • The Future Value Interest Factor includes time period, interest rate and compounding frequency. You can apply this factor to other present value amounts to find the future value with the same length of investment, interest and compounding rate.
: • FVIF = (1+i)n
: • Multiply any PV by FVIF to get a future value with the same length of investment at the same interest rate.
Future Value Example Problem
The default calculation in the calculator asks what is the future value of a present value amount of $12,487.16 invested for 3.5 years, compounded monthly at an annual interest rate of 5.25%.
The calculator first converts the number of years and interest rate into terms of months since compounding occurs monthly in this example
3.5 years × 12 = 42 months
So n = 42
Convert the annual interest rate of 5.25% to a monthly interest rate
First convert the percentage to a decimal: 5.25 / 100 = 0.0525
Then divide the annual rate of 0.0525 by 12 to get the monthly interest rate: 0.0525 / 12 = 0.004375
So i = 0.004375
Do the calculation using the future value formula FV = PV(1+i)n
FV=PV(1+i)n
FV=12487.16(1+0.004375)42
FV=12487.16(1.004375)42
FV=12487.16(1.201233824)
FV=15000.00
Future Value Interest Factor Example Problem
Calculating the Future Value Interest Factor FVIF for this same problem, FVIF = (1+i)n
FVIF=(1+i)n
FVIF=(1+0.004375)42
FVIF=(1.004375)42
FVIF=1.201233824
Use this FVIF to find the future value of any present value with the same investment length and interest rate. Instead of a present value of $12487.16, perhaps you want to find the future value of a present value of $16,649.60.
FV=PV×FVIF
FV=16,649.55×1.201233824=$20,000.00
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187699 | https://www.aapd.org/globalassets/media/safety-toolkit/updated-posters-to-help-manage-medical-emergencies-in-the-dental-office.pdf | FEATURE Updated posters to help manage medical emergencies in the dental practice P . Jevon1 Duty of care Dental practices have a duty of care to ensure that an effective and safe service is provided to their patients. The satisfactory performance in a medical emergency in the dental practice has wide-ranging implications in terms of equipment, training, standards of care, clinical governance, risk management and clinical audit. Maintaining the knowledge and competence to deal with medical emergencies is an important aspect of all dental care professionals’ (DCPs’) continuing professional development (CPD).1 The updated posters described in this article are designed to be aide-memoires to assist DCPs to safely and effectively manage medical emergencies occurring in their workplace. The aim of this article is to provide an overview to the updated posters which are designed to help manage medical emergencies in the dental practice. Incidence of medical emergencies Medical emergencies in the dental practice that have been reported include vasovagal syncope (63%), angina (12%), hypoglycaemia (10%), epileptic seizures (10%), choking (5%), asthma (5%) and anaphylaxis.2 Vasovagal syncope is the most common emergency, accounting for approximately two thirds of all emergencies reported.3 The GDC and medical emergencies A medical emergency could occur at any time in the dental practice. The General Dental Council (GDC)1 states it is important to ensure that: ■ There are arrangements for at least two people to be available within the working environment to deal with medical emergencies when treatment is planned to take place. In exceptional circumstances the second person could be a receptionist or a person accompanying the patient ■ All members of staff, including those not registered with the GDC, know their role if there is a medical emergency ■ All members of staff who might be involved in dealing with a medical emergency are trained and prepared to do so at any time, and practise together regularly in a simulated emergency so they know exactly what to do.
National guidance on the management of medical emergencies The ‘Medical emergencies in the dental practice’ section of the British National Formulary (BNF)4 provides guidelines on the management of the more common medical emergencies which may arise in the dental 1 Multi-Professional Skills Manager, Medical Education Manor Hospital, Walsall 21 BDJ Team www.nature.com/BDJTeam Medical emergencies can occur in the dental practice. The posters ‘Medical Emergencies in the Dental Practice’ and ‘Emergency Drugs in the Dental Practice’ have been designed to help dental professionals to respond effectively and safely to a medical emergency. These posters, endorsed by the British Dental Association (BDA), are included with this article. Further copies can be downloaded from:
© 2015 British Dental Association. All rights reserved FEATURE www.nature.com/BDJTeam BDJ Team 22 Fig. 1 Poster 1: Medical emergencies in the dental practice © 2015 British Dental Association. All rights reserved FEATURE 23 BDJ Team www.nature.com/BDJTeam practice. Further information is also available from the BDA (if your principal is a member) at www.bda.org/medicalemergencies. Specific guidance is also provided by other authoritative bodies including the British Thoracic Society (asthma), the British Heart Foundation (cardiac emergencies), the National Institute for Health and Care Excellence (NICE) (epileptic seizures), the Stroke Association (acute stroke), Diabetes UK (hypoglycaemia) and the Resuscitation Council (UK) (anaphylaxis).
The Resuscitation Council (UK) no longer provides specific guidance on medical emergencies in the dental practice (formerly provided in their publication Medical emergencies and resuscitation standards for clinical practice and training for dental practitioners and dental care professionals in general dental practice). This was superseded in November 2013 by its publication Quality standards for cardiopulmonary resuscitation practice and training in primary dental care, in which the Resuscitation Council (UK) continues to provide helpful guidance on all aspects relating to cardiopulmonary resuscitation in the dental practice.5 Poster 1 The ‘Medical emergencies in the dental practice’ A3 poster (Fig. 1) was first produced in 2009 as an aide-memoire to assist dental staff to safely manage medical emergencies occurring in the dental practice.6 It was updated7 in 2012 and now revised again in 2015. The 2015 revisions include: ■ Increased emphasis on the Airway Breathing Circulation Disability Exposure approach to the management and treatment of medical emergencies ■ Inclusion of adrenal crisis in line with guidance in the BNF4 ■ New NICE guidance concerning midazolam administration for epileptic seizures (midazolam injection is no longer considered an option for buccal administration)8 ■ Emphasis on the importance of having immediate access to an automated external defibrillator (AED).5 The poster is intended to be placed on the wall in the surgery where it can be easily and quickly accessed should an emergency occur. The emergencies covered are listed in alphabetical order: ■ Adrenal crisis ■ Anaphylaxis ■ Asthma ■ Cardiac emergencies ■ Epileptic seizures ■ Hypoglycaemia ■ Stroke ■ Syncope.
The important signs and symptoms to look out for to help correctly diagnose each emergency are listed, together with the principles of safe and effective treatment. Where appropriate, the recommended doses of drugs (including paediatric doses) and routes of administration are also stated.
This poster can be downloaded from Walsall Healthcare NHS Trust’s website: medical-education.aspx.
Poster 2 The ‘Emergency drugs in the dental practice’ A4 poster (Fig. 2) was first produced in 2012 as an aide-memoire to assist dental staff to safely administer medications in the emergency situation.7 This poster has also been revised in 2015 to incorporate the new NICE guidance concerning midazolam administration for epileptic seizures (midazolam injection is no longer considered an option for buccal administration).8 The poster is designed to be kept in the emergency drugs box for quick reference. Further copies can be downloaded from Walsall Healthcare NHS Trust’s website:
walsallhealthcare.nhs.uk/medical-education.
aspx.
Training All dental staff should be trained and receive regular updates in the management of medical emergencies and possess up-to-date evidence of capability.9 Running regular mock scenarios/drills involving the team approach is advised.1 In the author’s experience, some surgeries find it helpful to use the poster in the training session to increase familiarity in its use.
Conclusion Every dental practice has a duty of care to ensure that an effective and safe service is provided for its patients. This article has provided an overview to the updated posters designed to help manage medical emergencies in the dental practice.
The author is grateful to Najam Rashid and Ruchi Joshi, ED Consultants and Sarah Church, Consultant Orthodontist, Manor Hospital, Walsall UK for proof reading the poster and to Daniel McAlonan, Head of Health & Safety, British Dental Association for his suggestions and helpful advice.
1. General Dental Council. Standards for the dental team. London: General Dental Council, 2013.
2. Müller M, Hänsel M, Stehr S et al. A state-wide survey of medical emergency management in dental practices: incidence of emergencies and training experience. Emerg Med J 2008; 25: 296–300 3. Jevon P. Medical emergencies in the dental practice, 2nd ed. Oxford: Wiley Blackwell, 2014.
4. British National Formulary. Medical emergencies in dental practice. Available online at: formulary/bnf/current/guidance-on-prescribing/prescribing-in-dental-practice/ medical-emergencies-in-dental-practice (accessed February 2016).
5. Resuscitation Council UK. Quality standards for cardiopulmonary resuscitation practice and training in primary dental care London: Resuscitation Council UK, 2013.
6. Jevon P. New poster to help manage medical emergencies. Br Dent J 2009; 207: 312.
7. Jevon P. Updated guidance on medical emergencies and resuscitation in the dental practice Br Dent J 2012; 212: 41-43.
8. NICE. Treating prolonged or repeated seizures and convulsive status eplilepticus. Information available online at www.nice.
org.uk (accessed August 2015).
9. General Dental Council. Scope of practice. London: General Dental Council, 2013.
bdjteam201655 POSSESS UP-TO-DATE EVIDENCE OF CAPABILITY’ OSS SS O A C O CA A MANAGEMENT OF MEDICAL EMERGENCIES AND MANAGEMENT OF MEDICAL EMERGENCIES AND AND RECEIVE REGULAR UPDATES IN THE A C G A A S ‘ALL DENTAL STAFF SHOULD BE TRAINED © 2015 British Dental Association. All rights reserved FEATURE www.nature.com/BDJTeam BDJ Team 24 Fig. 2 Poster 2: Emergency drugs in the dental practice © 2015 British Dental Association. All rights reserved |
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