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7900 | https://math.ou.edu/~cremling/teaching/lecturenotes/alg/ln1.pdf | ALGEBRA CHRISTIAN REMLING 1. Some elementary number theory 1.1. Primes and divisibility. We denote the collection of integers by Z = {. . . , −2, −1, 0, 1, . . .}. Given a, b ∈Z, we write a|b if b = ac for some c ∈Z. We then say that a divides b or that a is a divisor of b. For example, 3|21 or −5|100, but 2 ∤−3.
Exercise 1.1. Show that divisibility has the following general properties: (1) if a|b and b|c, then a|c; (2) if a|b and b|a, then a = ±b; (3) if a|b and a|c, then also a|(bx + cy) for arbitrary x, y ∈Z; (4) if 1 ≤b < a, then a ∤b; (5) a|x for all x ∈Z precisely if a = ±1; (6) a and −a have the same divisors; (7) a|0 for all a ∈Z.
An integer p ≥2 is called a prime if it has no divisors other than ±1, ±p. The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, . . .. The key tool for studying divisibility in Z is division with remainder: given a, b ∈Z, b ≥1, there are unique integers m, r, with 0 ≤r < b, such that a = mb + r. To see that this is true, just consider the integers a −xb, for fixed a, b ∈Z and with x varying over Z. These are equally spaced, with distance b between consecutive numbers, so exactly one choice of x produces an integer in {0, 1, . . . , b −1}.
For example, if we divide 23 by 6, with remainder, then we obtain 23 = 3 · 6 + 5. Or let’s divide −6 by 23: this gives −6 = (−1) · 23 + 17.
Let a, b ∈Z, not both equal to zero. The greatest common divisor of a and b is defined as the greatest integer d that divides both a and b. We denote it by gcd(a, b) or just by (a, b). For example, (6, 10) = 2.
If (a, b) = 1, then we call a, b relatively prime. If a and b are distinct primes, they will be relatively prime: there aren’t any individual di-visors, let alone common divisors. However, this sufficient condition for two integers to be relatively prime is certainly not necessary: for example, a = 6 and b = 25 are also relatively prime, but neither a nor b is a prime.
The Euclidean algorithm finds (a, b) by repeated division with re-mainder. By Exercise 1.1, sign changes do not affect divisors; in par-ticular, they leave the gcd unchanged, and thus it’s enough to be able 1 2 CHRISTIAN REMLING to handle (a, b) in the case a, b ≥1. It will be convenient to assume this when running the Euclidean algorithm (by the way, what is (a, 0)?).
We put r−1 = a, r0 = b and then define a sequence r1, r2, . . . , rN recursively by rn−2 = mnrn−1 + rn; in other words, in the nth step, we take the two previous members of the sequence rn−2, rn−1 and divide the former by the latter, to produce the new remainder rn. Since the remainders decrease strictly (why is that true?), we must eventually reach rN = 0, after N steps. When this happens, the algorithm stops and outputs rN−1 = (a, b) as the desired gcd.
I’ll give a general proof that this works in a moment, but let’s first work an example. Let a = 4620, b = 126. We start out by dividing a by b: 4620 = 36 · 126 + 84, so r1 = 84. Next, we divide r0 = b by r1. Since 126 = 84 + 42, this gives the new remainder r2 = 42. We continue in this way and divide 84 by 42. Since this doesn’t leave a remainder, we have that r3 = 0 and thus N = 3 and (4620, 126) = r2 = 42.
Exercise 1.2. Find (4680, 756) and (81, 210) by running the Euclidean algorithm by hand.
Exercise 1.3. Show that if a < b, then r1 = a (in other words, the first step just swaps a and b if they originally appeared in the “wrong” order).
Let’s now discuss the general theory.
Theorem 1.1. The Euclidean algorithm, as described above, computes rN−1 = (a, b).
Proof. Since rN = 0, we have that rN−2 = mN−1rN−1, so rN−1|rN−2, and then rN−3 = mN−2rN−2+rN−1, so also rN−1|rN−3. We can continue in this way to see that rN−1|rn for all n ≤N−1, and since the algorithm started with r−1 = a and r0 = b in the first two steps, we arrive at the conclusion that rN−1|a, rN−1|b.
It remains to be shown that rN−1 is the largest number that divides both a and b. To see this, let c be any common divisor of a and b.
I claim that then c|rn for all n ≥−1.
This again follows by just keeping track of what the algorithm does, step by step. Clearly, since a = m1b + r1, we have that c|r1. Next, from b = m2r1 + r2, we then see that c|r2, and we can continue until the algorithm stops to establish my claim. In particular, d = (a, b) must satisfy d|rN−1, so d ≤rN−1 and it follows that d = rN−1, as claimed.
□ ALGEBRA 3 From the Euclidean algorithm, we obtain a very useful decription of the gcd: Theorem 1.2. The gcd of a, b can be characterized as the smallest positive integer of the form ax + by, x, y ∈Z.
Corollary 1.3. The integers a, b are relatively prime precisely if there are x, y ∈Z so that ax + by = 1.
Proof of Theorem 1.2. Write d = (a, b). We first show that there are x, y ∈Z with d = ax + by. In fact, x, y can also be extracted from the Euclidean algorithm. Starting at the end again, we first observe that (1.1) d = rN−1 = rN−3 −mN−1rN−2.
In fact, it is true in general that rn = rn−2 −mnrn−1, that is, each remainder is a linear combination of the previous two remainders. We can now successively plug these linear combinations into (1.1) to (even-tually) express d = rN−1 as a linear combination of a = r−1 and b = r0.
To conclude the proof, we must now show that the set of numbers ax + by, x, y ∈Z, does not contain positive integers that are smaller than d. That is clear, however, because d divides any such linear com-bination ax + by, so ax + by ≥d if ax + by is positive.
□ Exercise 1.4. We defined the Euclidean algorithm only for a, b ≥1.
Convince yourself that Theorem 1.2 and its Corollary hold for arbitrary a, b ∈Z, not both zero.
Example 1.1. We saw above that (4620, 126) = 42, so by the Theorem, there are x, y ∈Z so that 4620x + 126y = 42. Moreover, by the proof, we can systematically find x, y from the Euclidean algorithm. Let us do this here; I’ll make use of the divisions with remainder we did above, when we ran the Euclidean algorithm. We must start at the end. The algorithm terminated in N = 3 steps, and we obtained d = r2 = 42 from 126 = 84 + 42 or, equivalently, 42 = 126 −84. Working our way from the bottom up towards the top, we recall that we obtained r1 = 84 from the division 4620 = 36 · 126 + 84, so 84 = 4620 −36 · 126. Plug this into the previous equation to obtain that 42 = 126 −(4260 −36 · 126) = −4260 + 37 · 126, which is the desired representation of the gcd, with x = −1 and y = 37.
Lemma 1.4. If (a, b) = (a, c) = 1, then also (a, bc) = 1.
Proof. By the Corollary, ax+by = au+cv = 1 for suitable u, v, x, y ∈Z.
It follows that bcvy = (1 −ax)(1 −au) = 1 −az for some z ∈Z, by multiplying out the RHS. This says that (a, bc) = 1, as desired.
□ 4 CHRISTIAN REMLING Proposition 1.5. (a) If (a, b) = 1 and a|c, b|c, then also ab|c.
(b) If p is a prime and p|ab, then p|a or p|b.
Exercise 1.5. Give (simple) examples that show that both parts fail if the assumptions ((a, b) = 1 and p prime, respectively) are dropped.
Proof. (a) By assumption, c = ad = be and also ax+by = 1 for suitable x, y ∈Z, so c = c · ax + c · by = beax + adby, and it is now clear that ab|c, as desired.
(b) A prime has no divisors ≥2 other than itself, so (p, a) = 1 unless p|a.
So if what we are trying to show were false, then p would be relatively prime to both a and b, but then also to ab, by Lemma 1.4.
This contradicts our assumption that p|ab, that is, (p, ab) = p.
□ With these preparations out of the way, we can now give a proof, from scratch, of the fundamental theorem of arithmetic: Theorem 1.6. Every integer a ≥1 can be written as a product of primes a = p1p2 · · · pn. Moreover, this factorization is essentially unique in the sense that if also a = p′ 1p′ 2 · · · p′ m, then m = n and, after relabel-ing suitably, pj = p′ j.
Proof. We first establish the existence of such a factorization into primes.
If a is itself a prime, then there’s nothing to show: take n = 1, p1 = a.
(Also, in the trivial case a = 1, we take n = 0, which works formally because the empty product is, by definition, equal to 1.) If a ≥2 is not a prime, then a = bc with 1 < b, c < a, and now either b, c are both primes and we’re done, with n = 2, p1 = b, p2 = c, or at least one of them has non-trivial divisors, which again must be smaller than the original number. We then factorize further, and since the numbers keep getting smaller as long as we still have divisors, this must stop at some point, and we obtain the desired factorization.
Exercise 1.6. Give a more formal version of this very simple argument; more precisely, prove the existence of a factorization into primes by an induction on a.
We now prove the uniqueness claim, by induction on n. If n = 0, then a = 1, and we must have that m = 0 also, since p′ j ≥2. Now suppose that n ≥1 and that the claim holds for n−1. Clearly, p1|a, and by repeatedly applying Proposition 1.5(b), we see that p1|p′ j for some j; for convenience, let’s assume that j = 1. Since p1, p′ 1 are both primes, this can only happen if p1 = p′ 1. It follows that p2 · · · pn = p′ 2 · · · p′ m, and since the product on the LHS has n−1 prime factors, the induction ALGEBRA 5 hypothesis now gives that n−1 = m−1 and the primes in both products agree, after relabeling.
□ In the factorization from Theorem 1.6, repeated primes are of course possible, for example 12 = 2 · 2 · 3. We usually reorganize slightly and write prime factorizations in the form a = pe1 1 pe2 2 · · · pen n , with exponents ej ≥0.
Theorem 1.7 (Euclid). There are infinitely many primes.
Proof. Given primes p1, . . . pN, consider a = p1p2 · · · pN + 1.
Then a > 1, so some prime at least must occur in its factorization (a itself could be prime), but clearly none of p1, . . . , pN divides a, so there must be an additional prime.
□ 1.2. Congruences.
Definition 1.8. An equivalence relation on a set A is a binary relation ∼that is: (1) reflexive: a ∼a for all a ∈A; (2) symmetric: if a ∼b, then b ∼a; (3) transitive: if a ∼b and b ∼c, then a ∼c.
For example, equality = is an equivalence relation on any set A, and so is the relation defined by letting a ∼b hold for any two a, b ∈A.
We will see a more interesting example in a moment.
In general, equivalence relations are essentially the same thing as partitions. Here, we call a collection of non-empty subsets Aα ⊆A (finitely or infinitely many, and the sets themselves may be finite or infinite) a partition of A if Aα ∩Aβ = ∅for α ̸= β and S α Aα = A.
Exercise 1.7. Elaborate on the claim made above. More specifically, show the following: (a) If ∼is an equivalence relation on A and we define Aa = {b ∈A : b ∼a}, for a ∈A, then, for any two a, b ∈A, either Aa ∩Ab = ∅or Aa = Ab. Moreover, S a∈A Aa = A, so if we view the {Aa : a ∈A} as a collection of sets, then we have a partition (which is not indexed by a ∈A).
(b) Conversely, given a partition {Aα}, define a relation by declaring a ∼b precisely if a, b ∈Aα for some α (note that a, b are required to lie in the same set). Show that ∼is an equivalence relation on A.
(c) Parts (a) and (b) provide operations that produce a partition from a given equivalence relation and vice versa.
Show that these operations are inverses of each other. Maybe you want to set up some notation to formulate this clearly: for an equivalence relation R, let F(R) be the partition defined in part (a). Similarly, given a partition 6 CHRISTIAN REMLING P, let G(P) be the equivalence relation defined in part (b). Then you want to show that G(F(R)) = R and F(G(P)) = P.
General comment: Everything in this problem is completely straight-forward, you should be able to do it in your head. Just make sure you don’t get intimidated by the notation.
In the sequel, we will pass freely between equivalence relations and partitions, as spelled out in this Exercise. Given an equivalence rela-tion, we call Aa = {b ∈A : b ∼a} the equivalence class of a; also, it is common to denote this by Aa = (a) or by a. A member b ∈(a) is called a representative of the equivalence class of a; in other words, a representative is just a b with b ∼a. Finally, we will often want to consider the set {(a) : a ∈A} of equivalence classes; it is common to denote this by A/ ∼.
Now fix an integer k ≥1.
A particularly important equivalence relation on Z is obtained by defining m ≡n mod k if k|m −n.
If this holds, we say that m, n are congruent modulo k.
Exercise 1.8. Show that congruence modulo k indeed is an equivalence relation.
Exercise 1.9. What is the equivalence class of an n ∈Z with respect to congruence modulo k?
Congruence is an important equivalence relation because it respects the algebraic structure of Z. Let’s make this more precise. I claim that if m ≡m′ mod k and n ≡n′ mod k, then also m + n ≡m′ + n′ mod k and mn ≡m′n′ mod k.
Let’s verify the second claim: we know that m′ = m + kx and n′ = n + ky for some x, y ∈Z. Thus m′n′ = mn + k(nx + my + kxy), and this shows that m′n′ ≡mn, as desired.
Exercise 1.10. Prove the claim about addition in the same way.
These properties allow us to define addition and multiplication on the equivalence classes, as follows: (m) + (n) := (m + n), (m)(n) := (mn).
Here, m + n and mn on the right-hand sides will depend on the choice of representatives; recall in this context that m is not determined by its equivalence class. Rather, we have that (m′) = (m) if (and only if) m′ ∈(m). However, the argument from the preceding paragraph and the Exercise show that this will not cause problems here: while m, n are not uniquely determined by their equivalence classes, and thus m + n isn’t either, the equivalence class (m + n) of m + n is independent of ALGEBRA 7 these choices, and of course all the same remarks apply to the product (mn).
This collection of equivalence classes mod k is denoted by Zk or (for reasons that will become clear later) by Z/(k). The equivalence classes themselves are also called residue classes in this context. We will later develop a more abstract view of these matters. For now, let me just present one result (Fermat’s little theorem) as a teaser.
Exercise 1.11. Show that +, · in Zk obey (most of) the usual rules.
More precisely, let x, y, z ∈Zk. Show that: (1) x + y = y + x and xy = yx; (2) (x + y) + z = x + (y + z) and (xy)z = x(yz); (3) x + 0 = x and x · 1 = x; (4) there is an additive inverse −x ∈Zk such that x + (−x) = 0; in fact, we can obtain −x as −x = (−1)x; (5) x(y + z) = xy + xz.
We have followed the usual practice of blurring the distinction be-tween equivalence classes (= members of Zk) and their representatives (which are integers, from Z): 0 in part (3) really stands for the equiv-alence class (0) ∈Zk of 0 ∈Z, and similar remarks apply to parts (4), (5).
By and large this says that we can just manipulate algebraic ex-pressions in Zk the way we are used to. Some care must definitely be exercised, however.
For example, 2 · 3 ≡0 mod 6 even though 2 ̸≡0, 3 ̸≡0 mod 6, so Z6 (unlike Z) has the property that zero may be written as a product of non-zero factors: 0 = 2 · 3 Proposition 1.9. Let k ≥1 and a ∈Z be given. Then the congruence ax ≡1 mod k has a solution x precisely if a and k are relatively prime. In particular, this holds if k = p is a prime and a ̸≡0 mod p.
Proof. When written out, the congruence requires us to find x, y ∈Z so that ax + ky = 1, and such x, y exist precisely if a, k are relatively prime, by Corollary 1.3.
□ Theorem 1.10 (Fermat). If p is a prime and a ̸≡0 mod p, then ap−1 ≡1 mod p.
Proof. In this proof, I am again not going to carefully distinguish be-tween residue classes and their representatives in the notation.
Observe that there are exactly p residue classes modulo p: more specifically, by division by p with remainder, we see that each n ∈Z is congruent to exactly one of 0, 1, . . . , p −1. Now multiply the non-zero residue classes by a to obtain a, 2a, . . . , (p −1)a. I claim that these 8 CHRISTIAN REMLING are still the residue classes 1, 2, . . . , p −1, possibly in a different order.
Indeed, Proposition 1.9 guarantees that any 1 ≤n ≤p −1 can be written as n ≡ax mod p, for some 1 ≤x ≤p −1; note also that x = 0 can not be the solution that the Proposition says exists because a0 ≡0. So the list a, 2a, . . . , (p −1)a contains all non-zero residue classes, and this implies that there can’t be repetitions because there are p −1 entries in the list and also p −1 residue classes that we know are listed.
In particular, this implies that 1 · 2 · · · (p −1) ≡a · (2a) · · · ((p −1)a) mod p because on both sides, we are multiplying the same residue classes, only in a different order (perhaps). Manipulating the RHS further, we see that A ≡Aap−1 mod p, A ≡1 · 2 · · · (p −1); in these steps, I’ve made repeated use of the properties from Exercise 1.11. With the help of Proposition 1.9 again, we can now successively “divide through” by 1, 2, . . . , p−1 to conclude that 1 ≡ap−1, as claimed (more precisely, for each n = 1, 2 . . . , p −1, I multiply both sides by an x with nx ≡1).
□ Exercise 1.12. Compute 25 mod 6. So Fermat’s little theorem may fail if p is not a prime. Where in the proof did I use this assumption?
(I really used it twice, find both instances please.) Exercise 1.13. As a by-product of this proof we saw that if ax ≡ay mod p, a ̸≡0 mod p, then x ≡y mod p (for combinatorial reasons).
Show this directly. |
7901 | https://www.ncbi.nlm.nih.gov/books/NBK537118/ | Pleural Friction Rub - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Pleural Friction Rub
Nicola Adderley; Jennifer Goldin; Sandeep Sharma.
Author Information and Affiliations
Authors
Nicola Adderley 1; Jennifer Goldin 2; Sandeep Sharma 3.
Affiliations
1 University of Calgary, Calgary, AB
2 Wright State University School of Medicine
3 Mery Fitzgerald Hospital
Last Update: October 6, 2024.
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Continuing Education Activity
Pleural friction rubs are distinctive sounds akin to "the sound made by walking on fresh snow"when the pleural surfaces become rough or inflamed due to conditions like pleural effusion, pleurisy, or serositis. Common underlying causes include autoimmune diseases, pneumonia, malignancy,and pulmonary emboli. Patients typically present with dyspnea and a sudden, sharp pain that worsens with breathing. Affected patients may pinpoint the rub based on the pain's location. Treatment involves addressing the underlying cause, such as antibiotics for infections, oncological treatments for malignancies,and anticoagulation for emboli. Analgesics can provide symptomatic relief, and interventional procedures like thoracentesis may be necessary for effusion-related rubs.
This activity for healthcare professionals is designed to enhance learners' proficiency in evaluating and managing pathological conditions giving rise to pleural friction rubs. Participants gain a deeper understanding of the diverse etiologies and underlying mechanisms of this clinical sign, providing essential knowledge for its recognition and management. The potential complications of the conditions associated with pleural friction rubs are also discussed. Greater competence promotes earlier and more comprehensive involvement of healthcare teams, ultimately reducing the morbidity and mortality associated with conditions linked to pleural friction rubs.
Objectives:
Screen patients presenting with respiratory complaints for the presence of pleural friction rubs, which are potential indicators of pleural inflammation.
Implement appropriate diagnostic measures, such as imaging studies and laboratory tests, to confirm the etiology of pleural friction rubs.
Apply evidence-based interventions for managing pleural friction rubs, including pharmacological therapies and procedures like thoracentesis if indicated.
Develop interprofessional team strategies for improving care coordination and communication to advance the detection and management of pleural friction rubs and improve patient outcomes.
Access free multiple choice questions on this topic.
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Introduction
A pleural friction rub is an audible respiratory sound, often likened to the creaking of leather or the squeaking of a shoe on wet surfaces, that serves as a distinctive clinical sign indicative of pleural inflammation and is commonly associated with conditions such as pleurisy, pneumonia, a pulmonary embolism,and malignancy. For clinicians, encountering this sound during auscultation can provide a valuable clue in diagnosing various underlying respiratory conditions.
Characterized by its localized, discontinuous, and grating nature, a pleural friction rub typically occurs during inspiration and expiration, resulting from the inflamed and roughened pleural surfaces rubbing against each other. Understanding the significance of this auscultatory finding and its potential implications enables clinicians to diagnose and treat patients presenting with related respiratory complaints.
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Etiology
The pleurae comprise visceral and parietal tissues that overlay the lungs and line the thoracic cavity. A small amount of serous fluid typically coats the surfaces of these layers, allowing them to slide easily over one another and creating surface tension that pulls the parietal and visceral pleura together. This mechanism ensures that the lungs expand with the thorax during inspiration (seeImage.Lung Anatomy). A pleural friction rub occurs when the usually smooth surfaces of the visceral and parietal layers roughen due to inflammation.
Any condition that causes pleurisy or a pleural effusion can produce a pleural friction rub. The following list contains possible underlying conditions associated with a pleural friction rub:
Bacterial pneumonia
Pulmonary embolism
Viral infections
Autoimmune conditions like systemic lupus erythematosus, systemic sclerosis, and rheumatoid arthritis
Pulmonary infarction
Tuberculosis
Empyema
Drugs that cause pleural disease
End-stage renal disease
Pancreatitis
Malignancy
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Epidemiology
A friction rub is a common finding in patients with pleurisy.This clinical sign may also be detected in approximately 4% of patients with pneumonia and pulmonary embolism.Human enteroviruses, such as poliovirus, group A and B coxsackieviruses, and echoviruses, cause a pleural friction rub in up to 25% of patients. Approximately 20% to 40% of patients who die of chronic renal insufficiency have fibrinous pleuritis, and 10% or less of patients with pancreatic pseudocysts develop pancreatic pulmonary effusions, which can also lead to a pleural friction rub.
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Pathophysiology
The pleural surfaces become thick and rough in conditions marked by inflammation or neoplastic growth, impeding easy motion and giving rise to a pleural friction rub.Prolonged inflammation causes pleural macrophage accumulation, triggering fibrocyte proliferation and producing the characteristic grating sound of a pleural rub as inflamed pleural surfaces slide past each other.
Pleural effusions, whether transudative or exudative, can lead to pleural friction rubs with various underlying causes. Transudative effusions, often associated with conditions like heart failure and liver cirrhosis, contrast with exudative effusions, which stem from inflammatory or infectious conditions such as pneumonia, malignancy, connective tissue disorders,and pulmonary emboli. Complicated effusions, a subtype of exudative effusions, result from microorganisms directly invading the pleura, which activate the coagulation cascade, triggering fibrin deposition and loculation formation within the effusion. Without drainage, complicated effusions may progress to empyema, and marked fibroblast activity leads to pleural thickening and fibrosis.
Malignancy can disrupt lymphatic drainage due to tumor invasion or lymph node involvement, leading to pleural effusion and subsequent friction rubs. The accumulation of tumor cells within the pleurae increases pleural fluid production, worsening the effusion.Patients with chronic pancreatitis may have a disruption in the pancreatic duct,forming a fistula in the chest, followed by a pleural effusion.Uremic pleuritis results from chronic fibrinous inflammation caused by unknown pathogenetic mechanisms. The typical patient with a uremic pleural effusion undergoes hemodialysis or peritoneal dialysis for over 1 to 2 years.
While the visceral pleura lacks somatic innervation and nociceptors, the sensation of pleuritic pain,characterized by sudden, intense, and sharp discomfort exacerbated by breathing, is relayed by somatic nerves innervating the parietal pleura. This pain may be referred to the neck or shoulder if the inflammation is near the diaphragm. Inflammatory mediators released into the pleural space trigger local pain receptors.
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History and Physical
On auscultation, a pleural friction rub is a nonmusical, short sound described as creaking or grating and likened to walking on fresh snow. These sounds may vary in intensity, and clinicians may accentuate them by applying additional pressure to their stethoscope. While typically heard during inspiration and expiration, pleural friction rubs may be transient and occur only during a specific part of the respiratory cycle. Pleural friction rubs usually localize to a small area on the chest, do not change after coughing, and may be palpable, feeling like sandpaper or cracking eggshells.
Interestingly, pleural friction rubs may still be audible even in the presence of a large pleural effusion, which prevents direct contact between the pleural surfaces.Pleural effusion often presents with dullness to percussion, decreased tactile fremitus, and reduced or delayed chest expansion on the affected side. Clinicians typically find diminished or inaudible breath sounds and often appreciate egophony at the superior aspect of the effusion.
Patients may have chest pain, described as sudden and intense, sharp, stabbing, or burning pain when inhaling and exhaling.The accompanying presenting features depend on the underlying cause of the pleural friction rub. Patients with hemoptysis, weight loss, fatigue, night sweats, a history of smoking, change in bowel habits, breast lumps, or skin abnormalities may have an underlying malignancy.Individuals with a respiratory tract infection usually present with cough, sputum production, hemoptysis, fever, history of aspiration, or, possibly, immunosuppression.
Edema, orthopnea, paroxysmal nocturnal dyspnea, and jugular venous distension suggest heart failure. Jaundice and ascites suggest liver failure, while tachycardia, tachypnea, hypoxia, and hemoptysis suggest a pulmonary embolism. Historical features like diffuse skin thickening and hardening, arthritis, rash, photosensitivity, and oral ulcers may suggest an autoimmune disorder.
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Evaluation
Because a pleural friction rub indicates an underlying process, all patients warrant an in-depth history and complete physical examination to guide further evaluation. Healthcare professionals should inquire about heart, liver, or kidney disease and conditions like systemic lupus erythematosus, hypothyroidism, amyloidosis, pancreatitis, lymphangioleiomyomatosis, and rheumatoid arthritis.
Imaging begins with chest radiography.If a pleural effusion is present, an ultrasound can measure the size and evaluate its characteristics. Computed tomography with contrast is helpful when loculations are likely present, the effusion size is small, or more detailed information about the surrounding anatomy is necessary.
Diagnostic thoracentesis under ultrasound guidance is typically the next step in evaluating patients with a pleural effusion.Pleural fluid is tested for pH, total protein, lactate dehydrogenase (LDH), glucose, cholesterol, cell counts, and differential, along with serum protein and LDH levels, to determine whether the fluid is a transudate or exudate. The pH for transudates generally ranges from 7.40 to 7.55, while most exudates have pH values between 7.30 and 7.45. The Light criteria define an exudative effusion as:
Pleural fluid-to-serum protein ratio greater than 0.5
Pleural fluid-to-serum LDH ratio greater than 0.6
Pleural fluid LDH greater than 0.67 or 2/3 of the upper limits of the laboratory's normal serum LDH
The appearance of the fluid can help determine whether additional testing is necessary. For instance, green fluid should undergo testing for bilirubin levels, cytology, and rheumatoid factor. Black fluid should undergo fungal and bacterial cultures and testing for amylase, cytology, and hematocrit or pleurocrit.Milky fluid should be evaluated for lipid levels, while bloody fluid requires hematocrit and cytological analysis.Turbid fluid requires pH determination and microbiological assessment.
Malignancy
Clinicians perform a cell count and cytology on pleural fluid collected from patients with suspected malignancy. Leukemic involvement of the pleura is likely when basophils comprise more than 10% of nucleated cells in the pleural fluid. Eosinophils greater than 15% may also signify the presence of malignancy.Flow cytometry and immunohistochemical staining are essential for patients with suspected lymphoma or multiple myeloma. High pleural fluid protein concentrations of 7.0 to 8.0 g/dL may suggest Waldenström macroglobulinemia or multiple myeloma. See StatPearls' companion references, "Multiple Myeloma"and "Lymphoma," for a detailed discussion regarding evaluating patients with suspected multiple myeloma and lymphoma.Patients with suspected mesothelioma or solid pleural lesions or thickening may be evaluated with thoracoscopy or image-guided biopsy.
Infection
Empyema generally results from pneumonia. No specific laboratory tests establish the diagnosis of empyema. However, the expected findings are leukocytosis with a left shift on a complete blood count with differential and an elevated C-reactive protein. Blood cultures may be helpful in the case of concurrent bacteremia. Pleural fluid should undergo pH testing, Gram staining, cell counts, and aerobic and anaerobic cultures.
Bacterial pneumonia is most often associated with a predominance of neutrophils, while lymphocyte predominance may indicate tuberculous or fungal etiologies.Polymorphonuclear cell counts above 50,000/µL from an exudative pleural effusion may suggest a complicated parapneumonic infection like empyema. Patients with counts above 10,000/µL from an exudative pleural effusion likely have bacterial pneumonia, acute pancreatitis, or lupus pleuritis.Individuals with counts below 5000/µL are likelier to have chronic exudates due to tuberculous pleurisy or malignancy.
Lymphocyte predominance may indicate a reactive process, a benign condition, or a neoplasm. A lymphocyte count of 85% to 90% of the nucleated cell count suggests tuberculous pleurisy or lymphoma. However, these findings may also indicate sarcoidosis, chronic rheumatoid pleurisy, yellow nail syndrome, chylothorax, or drug-induced pleural effusion. Malignant pleural effusions are generally associated with 50% to 70% lymphocytes.
Patients with pleural effusion and risk factors for tuberculous infection should undergo tuberculosis evaluation beginning with a chest radiograph and 3 sputum specimens obtained 8 hours apart for acid-fast bacillus (AFB), mycobacterial culture, and nucleic acid amplification testing. One sample should be an early morning sample.
Patients with poor immune status, including people with human immunodeficiency virus and CD4 counts below 100 cells/mm³, should also undergo blood and urine mycobacterial cultures. Clinicians obtain a pleural fluid specimen for AFB smears, mycobacterial cultures, and measurement of adenosine deaminase (ADA), an enzyme produced from lymphocytes and involved in purine metabolism.AFB smears and mycobacterial cultures of pleural fluid are only positive in approximately 50% of cases, and the results can take several weeks. Pleural fluid ADA levels can establish a presumptive diagnosis in patients with lymphocyte-predominant effusions and who live in areas with a high tuberculosis incidence. The pleural fluid in patients with tuberculosis almost always has a total protein content above 4 g/dL.
Autoimmunity
Rheumatoid arthritis, idiopathic juvenile arthritis, systemic lupus erythematosus (SLE), inflammatory bowel disease, and Sjögren syndrome are autoimmune conditions that can produce pleural friction rubs due to serositis or pleural effusion. Basic laboratory tests for suspected Sjögren syndrome are a complete blood count with differential, erythrocyte sedimentation rate (ESR), urinalysis, basic metabolic profile, and a spot urine-to-creatinine ratio. An ophthalmologist should evaluate the patient for dry eyes using Schirmer testing and tear break-up time.Additional assessments include anti-Sjögren-syndrome-related antigens A and B antibodies, antinuclear antibodies (ANA), and rheumatoid factor.
Besides meeting diagnostic criteria, clinicians evaluate patients with suspected SLE with ANA, anti-double-stranded deoxyribonucleic acid, antiphospholipid antibodies, immunoglobulins G (IgG) and M (IgM), anticardiolipin antibodies, IgG and IgM anti-β2-glycoprotein,C3 and C4 or CH50 complement levels,ESR or CRP levels, and a urine protein-to-creatinine ratio. Anti-Smith antibodies are highly specific for SLE. Pleural fluid ANA titers of 1:160 or higher are a good indicator of pleuritis in a patient with known SLE.
Individuals with symptoms consistent with rheumatoid arthritis, such as symmetric pain, swelling, and morning stiffness of multiple joints, should undergo testing for rheumatoid factor and anticyclic citrullinated peptide antibodies. Pleural fluid may have elongated macrophages, unique multinucleated giant cells called "tadpole cells," and rheumatoid factor. Patients with suspected irritable bowel disease should undergo stool Clostridioides difficile toxin polymerase chain reaction, routine stool cultures, microscopy for ova and parasites, and Giardia stool antigen.Fecal calprotectin and lactoferrin levels are elevated in irritable bowel disease and help distinguish this condition from irritable bowel syndrome. Endoscopy and biopsy follow.
Heart Failure
A pleural N-terminal pro-B-type natriuretic peptide (NT-proBNP) level exceeding 1500 pg/mL supports the presence of a transudative pleural effusion consistent with heart failure. However, this finding has limited utility since a serum NT-proBNP is equally reliable.
Renal Disease
Uremic pleuritis is a diagnosis of exclusion. Typical findings of uremia include elevated blood urea nitrogen and creatinine levels, metabolic acidosis, and altered mental status. However, kidney disease often coexists with other lung pathologies, including infectious and autoimmune conditions.Clinicians should carefully rule out other causes of pleuritis and consider uremia when no other potential cause is identified.
Pulmonary Embolism
Clinicians determine the pretest probability of a pulmonary embolism in hemodynamically stable patients using the Wells criteria.Individuals with a low probability or a Wells score of less than 2 should have the pulmonary embolism rule-out criteria applied.Patients who meet all 8 criteria do not require any further testing, while the rest undergo a sensitive D-dimer. No further testing is necessary if the D-dimer is negative or less than 500 ng/mL.Patients with a positive D-dimer should undergo a computed tomography pulmonary angiogram (CTPA). Patients with an intermediate score of 2 to 6 undergo D-dimer testing with CTPA if positive. Some authors recommend proceeding directly with imaging if the Wells score is in the upper range of 4 to 6. Patients with a high probability or a Wells score greater than 6 proceed directly to CTPA.
Additional testing includes electrocardiography,chest radiography, and B-type natriuretic peptide and troponin levels to exclude other possible diagnoses. Hemodynamically unstable patients can undergo immediate CTPA if successfully resuscitated.Otherwise, a bedside echocardiogram and lower extremity Doppler are alternatives.
Medications
Pleural reactions from medications manifest as pleural effusions, pleural thickening, or pleuritic chest pain.Nitrofurantoin, dantrolene, methysergide, dasatinib, amiodarone, interleukin-2, procarbazine, methotrexate, clozapine, phenytoin, β-blockers, and ergot drugs can all cause pleuropulmonary reactions. When an apparent cause for pleural findings is not evident, in the setting of a pleural effusion with pleural fluid eosinophilia, clinicians should take a careful drug history, paying close attention to the temporal relationship between the start of therapy and the onset of symptoms.A therapeutic trial of drug withdrawal before additional evaluation may be in order.
Gastrointestinal Causes
Patients with a history of chronic pancreatitis and a left-sided pleural effusion should undergo a pleural fluid pancreatic amylase level. A pancreaticopleural fistula should be ruled out in cases of large, rapidly recurring pleural effusion in the setting of chronic pancreatic disease.
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Treatment / Management
Addressing pleural friction rubs begins with identifying the underlying cause. Since these rubs often arise from various conditions, managing the underlying pathology should be the primary focus.
In pneumonia cases, appropriate antibiotic therapy is essential to resolve the infection and alleviate associated inflammation. Similarly, in malignancy-related pleural friction rubs, oncological treatment modalities such as chemotherapy, radiation therapy,and surgical excision may be necessary to target and reduce tumor burden, thereby alleviating pleural fluid accumulation and friction rubs.
Patients with pulmonary emboli require appropriate anticoagulation.Individuals with autoimmune disorders warrant a suitable disease-modifying therapy. Heart failure management requires treatment of the underlying cause and associated comorbid conditions. See StatPearls' companion references, "Acute Pulmonary Embolism," "Typical Bacterial Pneumonia," "Rheumatoid Arthritis," "Systemic Lupus Erythematosus,"and "Active Tuberculosis," for additional information regarding further management of some of the underlying causes of pleural friction rubs.
Symptomatic relief may be provided by administering analgesics to manage pleuritic chest pain commonly associated with pleural friction rubs. Nonsteroidal anti-inflammatory drugs or opioids may be used, depending on the severity of pain and individual patient factors. Some patients may require corticosteroids due to underlying autoimmune conditions or inflammation, causing pleurisy. However, clinicians must monitor patients closely for adverse events associated with these medications, particularly in people with comorbidities such as renal impairment or gastrointestinal bleeding. Clinicians may perform interventional procedures like thoracentesis or pleurodesis to drain pleural fluid or prevent reaccumulation, especially in patients with large or recurrent effusions causing the friction rub.
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Differential Diagnosis
Distinguishing a pleural friction rub from a pericardial rub is critical. A pericardial rub is highly specific for pericarditis, characterized by a grating sound as inflamed pericardial layers slide against each other. Unlike pleural friction rubs, pericardial rubs persist even while a patient is holding their breath. A pleural friction rub typically has 2 sounds, heard during inspiration and expiration. In comparison, pericardial rubs consist of 3 sounds heard during atrial systole, ventricular systole, and early ventricular diastole.
Coarse crackles and rhonchi are 2 additional lung sounds that may resemble a pleural friction rub. Crackles are explosive, nonmusical sounds lasting for 25 ms or less and are usually heard during inspiration and, sometimes, during expiration. A proposed etiology of crackles is that small airways that collapse during expiration snap open during inspiration. Other authors suggest that crackles originate from the vibration in the walls of small airways. Coarse crackles are loud, low-pitched, and occur less frequently per breath, whereas fine crackles are soft, higher-pitched, and more numerous per breath.
Coarse crackles often resolve after coughing, but fine crackles may persist in some patients. Body position changes can alter the sound of fine crackles but not coarse crackles. Crackles are associated with pneumonia, chronic obstructive pulmonary disease, pulmonary edema, interstitial lung disease, and heart failure.Rhonchi are harsh, low-pitched rattling sounds heard during inspiration or expiration due to airway obstruction from secretions, edema, or inflammation.
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Prognosis
The prognosis of a pleural friction rub depends on the underlying cause. Patients with a friction rub due to malignancy-related pleural effusion generally have a guarded prognosis and are likely terminal, with a 1-year mortality rate of 77%. Study results have revealed a 20% to 90% incidence of pleuropulmonary disease in patients with systemic lupus erythematous, ranging from subclinical pleural effusions to diffuse alveolar hemorrhage, with a mortality rate of 68% to 75%.
Patients with rheumatoid arthritis who develop interstitial lung disease have a 5-year mortality rate of 36% compared to 18% in patients without lung disease. The prognosis associated with empyema is good if treated early and aggressively.One study revealed that patients with uremic pleuritis have a high mortality of 60% at 3 years. The prognosis for a pulmonary embolism varies based on timing, embolus size, and treatment. The overall mortality is 30%if left untreated. In patients treated with anticoagulation, the mortality is between 2% and 11%.
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Complications
The complications associated with pleural friction rubs are linked to their underlying causes and the treatment regimens used, listed below.
Corticosteroid use:Adrenal insufficiency, diabetes, cataracts, peptic ulcers, glaucoma, hypertension, mood changes, and decreased bone density
Thoracentesis: Infection, intercostal nerve or artery injury, injury to the lung, diaphragm, heart, liver, or spleen, re-expansion pulmonary edema, pneumothorax, and death
Pleurodesis: Systemic inflammation with hypotension and acute respiratory distress syndrome,empyema, decreased lung volume, and death
Anticoagulant use:Intracranial hemorrhage, gastrointestinal bleeding, hemarthrosis, and hemothorax
Immunosuppressant medications:Increased risk of skin cancer and infection, neutropenia, and lymphopenia
Therapy with disease-modifying antirheumatic medications: Cirrhosis, hepatic fibrosis, neutropenia, megaloblastic anemia, thrombocytopenia, and pancytopenia
Chemotherapy:Neuropathy, pancytopenia, infection, myocardial dysfunction, pulmonary fibrosis, secondary malignancy, and neurocognitive impairment
Antibiotic use:Allergic reactions, cardiac conduction abnormalities with macrolides and fluoroquinolones
Tendon rupture can occur with fluoroquinolones.
Autoimmune diseases:Episcleritis, scleritis, osteopenia, osteoporosis, interstitial lung disease, pericarditis, myocarditis, heart failure, coronary artery disease, vasculitis, stroke, glomerulonephritis, Felty syndrome, anemia of chronic disease, lymphoma, monoclonal gammopathy,cryoglobulinemia, and colorectal cancer
Pulmonary emboli:Death, shock, pulmonary hypertension, and stroke
Empyema:Sepsis, lung abscess, fibrothorax, pneumothorax, and extension into the chest wall and soft tissue
Understanding these complications and their pathophysiological mechanisms enables clinicians to formulate proactive measures to ensure optimal care.
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Deterrence and Patient Education
Clinicians are pivotal in educating patients about the significance of pleural friction rubs to enhance awareness and promote proactive management. Healthcare professionals should inform patients that pleural friction rubs are distinctive respiratory sounds indicating inflammation of the pleural surfaces. A pleural friction rub indicates an underlying condition, such as pneumonia, autoimmune disease, or pulmonary embolism, that requires further evaluation and management. Patients must understand that the root causes of this clinical sign may require long-term therapies and careful monitoring to prevent complications.
Understanding the significance of these sounds empowers patients to recognize potential respiratory issues early and seek prompt medical attention. Additionally, clinicians should discuss the importance of adhering to prescribed treatment regimens, maintaining regular follow-up appointments, and reporting new or worsening symptoms like dyspnea, hemoptysis, joint pain,and chest pain to allow clinicians to address and manage the underlying cause effectively. Moreover, lifestyle modifications, including smoking cessation and maintaining a healthy weight, can contribute to overall respiratory health and reduce the risk of complications associated with pleural friction rubs. By fostering patient engagement and providing comprehensive education, clinicians can empower individuals to actively manage their respiratory health and mitigate the impact of pleural friction rubs.
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Enhancing Healthcare Team Outcomes
A pleural friction rub, characterized by its distinct grating sound reminiscent of creaking leather or squeaking shoes on wet surfaces, is a valuable diagnostic clue for pleural inflammation and associated underlying conditions. Roughened pleural surfaces moving against each other during breathing cause pleural friction rubs, which may arise from conditions like pneumonia, pulmonary embolism, malignancy, autoimmune disorders, or uremic pleuritis in patients undergoing long-term dialysis. Evaluation begins with a chest radiograph, followed by an ultrasound if a pleural effusion occurs. A thoracentesis helps differentiate the possible etiologies of pleural effusion. Further testing is guided based on concurrent symptoms.
Treatment involves addressing the underlying pathology, such as administering antibiotics for infections or initiating anticancer therapies for tumors, while providing symptomatic relief with analgesics to manage pleuritic chest pain. In some cases, interventional procedures like thoracentesis and pleurodesis may be necessary to drain pleural fluid and prevent reaccumulation, underscoring the need for comprehensive care and vigilant monitoring for potential complications.
An interprofessional approach involving clinicians, advanced clinicians, nurses, pharmacists, and other healthcare professionals is necessary to optimize patient-centered care, outcomes, patient safety, and team performance. Clinicians and advanced clinicians are crucial in accurately diagnosing and initiating treatment for underlying conditions contributing to pleural friction rubs. The clinical expertise of these healthcare professionals is essential in determining the appropriate diagnostic tests, interpreting results, and devising individualized treatment plans.
Nurses are instrumental in monitoring patients' respiratory status, administering medications, providing patient education on symptom management, and promptly recognizing and reporting any patient status changes. Pharmacists contribute by ensuring medication safety, reviewing drug interactions, and optimizing medication regimens to minimize adverse effects and enhance treatment efficacy.
Effective interprofessional communication is paramount, facilitated through precise documentation and open dialogue to discuss patient progress, address concerns, and adjust treatment plans collaboratively. By working cooperatively, healthcare team members can pool their diverse skills and knowledge to provide holistic care, address patient needs comprehensively, and improve outcomes for individuals with pleural friction rubs. This interprofessional approach fosters synergy within the healthcare team and enhances overall outcomes, reducing morbidity and mortality.
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Figure
Lung Anatomy.This illustration includes the visceral and parietal pleurae, the right and left lungs,the right and left pleural cavities, and the mediastinum. The anatomic relationships of these structures (more...)
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Disclosure:Nicola Adderley declares no relevant financial relationships with ineligible companies.
Disclosure:Jennifer Goldin declares no relevant financial relationships with ineligible companies.
Disclosure:Sandeep Sharma declares no relevant financial relationships with ineligible companies.
Continuing Education Activity
Introduction
Etiology
Epidemiology
Pathophysiology
History and Physical
Evaluation
Treatment / Management
Differential Diagnosis
Prognosis
Complications
Deterrence and Patient Education
Enhancing Healthcare Team Outcomes
Review Questions
References
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Continuing Education Activity
Introduction
Etiology
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Treatment / Management
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Deterrence and Patient Education
Enhancing Healthcare Team Outcomes
Review Questions
References
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7902 | https://www.vfmgua.ru/uploads/files/kaf/kazusi/%D0%9A%D0%B0%D0%B7%D1%83%D1%81%D1%8B%20%D0%BF%D0%BE%20%D1%8D%D0%BA%D0%BE%D0%BD%D0%BE%D0%BC%D0%B8%D0%BA%D0%B5.pdf | 1 Примеры решения задач по дисциплине «Экономика для юристов» Умение решать экономические задачи – важный навык не только на экзамене, но и в жизни. Решение экономических задач в рамках существенно сокращенного курса «Экономика для юристов», читаемого обучающимся юридических направлений обучения, дают возможность лучше понять сущность теоретического материала, а также показывают обучающимся, что экономика – наука весьма полезная и практическая. Обучение в вузе закончится, а необходимость рассчитывать проценты по вкладам и кредитам, искать оптимальные решения в различных хозяйственных (в т.ч. финансовых) ситуациях остается с человеком практически на всю сознательную жизнь. Тема «Издержки производства» Задача 1. Фирма производит 125 ед. товара, при этом общие постоянные издержки составляют 500 руб., общие переменные издержки – 1125 руб. Определите: - средние переменные издержки; - минимальную цену реализации товара, обеспечивающую безубыточность производства Решение задачи базируется на четком понимании сущности таких видов издержек, как постоянные, переменные, общие и средние. В частности, Общие издержки, это совокупность постоянных и переменных издержек, рассчитанных на весь объем производимой продукции, а средние издержки – это издержки в расчете на единицу продукции. Тогда, 1) средние переменные издержки находим путем деления общих переменных издержек на объем производства: 1125 руб. : 125 ед. = 9 руб. (средние переменные издержки в расчете на 1 ед. продукции) 2) минимальная цена реализации товара устанавливается таким образом, чтобы были возмещены (покрыты) все затраты на производство продукции. Другими словами, она должна быть не ниже, чем величина средних общих издержек. Следовательно, задачу можно решить двумя способами: А) Минимальная цена реализации товара = Средние общие издержки = средние постоянные издержки (500 руб. : 125 ед) + средние переменные издержки (9 руб.) = 4 + 9 = 13 руб. 2 Б) Минимальная цена реализации товара = Общие издержки : кол-во продукции = (общие постоянные + общие переменные) : количество продукции = (500 руб. + 1125 руб.) : 125 ед. = 13 руб. Тема «Сущность и виды прибыли» Задача 1. Совокупный доход предприятия – 1200 тыс. руб., заработная плата работников составила 400 тыс. руб., затраты на сырье и материалы – 450 тыс. руб., неявные издержки – 80 тыс. руб. Рассчитайте бухгалтерскую и экономическую прибыль Решение задачи базируется на понимании различия между бухгалтерской и экономической прибылью, а также явными и неявными издержками Прибыль – это конечный положительный финансовый результат (разница между доходами и расходами) хозяйствующего субъекта Бухгалтерская прибыль = совокупный доход – явные издержки = 1200 тыс. руб. – 400 тыс. руб. – 450 тыс. руб. = 350 тыс. руб. Экономическая прибыль = совокупный доход – все издержки (явные и неявные) = 1200 тыс. руб. - 400 тыс. руб. – 450 тыс. руб. – 80 тыс. руб. = 270 тыс. руб. Экономическая прибыль всегда будет меньше, чем бухгалтерская, т.к. учитывает большее количество издержек Задача 2. Рассчитайте годовую прибыль предприятия, а также рентабельность продаж, если доход за год составил 2,5 млн. рублей, годовые переменные издержки составили 0,5 млн. рублей, постоянные издержки составили 1,2 млн. рублей. Решение задачи А) расчет прибыли: Прибыль рассчитывается по формуле: Прибыль=Доход – Общие издержки Следовательно, прибыль = 2,5- (0,5+1,2)=0,8 млн. руб. или 800 тыс. руб. Б) расчет рентабельности продаж: Рентабельность продаж рассчитывается по формуле = Прибыль / Общий доход 100% Следовательно, 0,8 млн. руб. / 2,5 млн. руб. 100% = 32%. Вывод по расчету рентабельности: Работа предприятия является эффективной, т.к. показатель рентабельности более 0%. Рентабельность в 32% означает, что каждый рубль, вложенный в производство, не только вернулся 3 предприятию вместе с выручкой от реализации продукции, но и принес прибыль в размере 32 копеек. Задача 3. Определите себестоимость продукции и прибыль от реализации продукции, если переменные издержки на единицу продукции в 2016 году составили 100 рублей на единицу. Всего изготовлено 5000 изделий. Постоянные издержки за год составили 20000 рублей. Предполагается установить на изделия после изготовления 20% наценку. Рассчитайте себестоимость единицы изделия. Продажную цену изделия. Выручку предприятия при продаже данных 5000 изделий. Рассчитать валовую прибыль предприятия, прибыль после налогообложения и валовую маржу (маржинальную прибыль). Решение задачи Переменные издержки на единицу продукции составили 100 рублей. Постоянные издержки на единицу продукции составили 20000/5000=4 рубле Себестоимость единицы продукции рассчитывается по формуле: Себестоимость единицы продукции = переменные издержки на единицу + Постоянные издержки на единицу продукции = 100+4=104 рублей. Общая себестоимость продукции = 1045000=520000 рублей. При 20% наценке цена изделия будет рассчитываться следующим образом: Цена = %наценки + себестоимость единицы продукции = 0,2104+104=124,8 рублей. Выручка рассчитывается по формуле: Выручка = ЦенаКоличество проданной продукции = 124,85000=624000 рублей (за год). Валовая прибыль рассчитывается по формуле: Выручка – Переменные затраты – Постоянные затраты. Валовая прибыль=624000-520000=104000 рублей. Налог на прибыль = 20%. Значит прибыль после налогообложения = 104000-1040000,2=83200 рублей. Маржинальная прибыль рассчитывается по формуле: Маржинальная прибыль = Выручка – Переменные затраты = 624000-1005000=124000 рублей. 4 Тема «Конкуренция и монополия» Задача 1. Определить степень концентрации экономической власти (коэффициент Херфиндаля-Хиршмана), если доля первой и второй фирм – по 25%, третьей - 15%, остальных двух - по 10%. Как изменится ситуация на рынке, если доля первой фирмы снизится до 20%, а доля четвертой увеличится на 5%? Решение задачи базируется на знании формулы расчета коэффициента и его значений. В частности, максимальное значение коэффициента = 10000; пороговое (выше которого ситуация на рынке считается уже неблагополучной) = 1800. Расчет коэффициента в динамике показывает улучшение или ухудшение ситуации. HHI = S12 + S22 + … + Sn2 где S – доля фирмы на отраслевом рынке 1) HHI = 25 2 + 252 + 152 + 102 + 102 = 625 + 625 + 225 + 100 + 100 = 1675 2) HHI = 202 + 252 + 152 + 152 + 102 = 400 + 625 + 225 + 225 + 100 = 1575 Вывод: так как на данном сегменте рынка функционируют всего 5 фирм, т.е. существует олигополия, что уже свидетельствует об определенных проблемах при вступлении новых хозяйствующих субъектов в данную сферу деятельности. Вместе с тем, значение коэффициента ниже порогового значения (1800), что свидетельствует о том, что ни одна из фирм не имеет главенствующего положения. Снижение коэффициента во втором периоде оценивается положительно, так как свидетельствует о снижении уровня концентрации (монополизации) экономической власти. Тема «Земля как фактор производства. Земельная рента» Задача 1. На своем участке фермер ежегодно выращивает и продает картофель в среднем на 150 тыс. руб. Затраты на выращивание, сбор и реализацию картофеля составляют 100 тыс. руб., а банковский процент равен 8%. Определите возможную цену данного земельного участка. Цена земли определяется по формуле: Цена земли = Рента (доход) / банковский процент В нашем случае доход (рента) от земли представлен в виде прибыли, которая рассчитывается по формуле: выручка – Затраты, следовательно: Цена земли = 150 тыс. руб. – 100 тыс. руб. / 0,08 (8%:100%) 5 Цена земли = 625 тыс. руб. Вывод: фермер готов продать свой земельный участок за 625 тыс. руб., так как положив эти деньги (625 тыс. руб.) в банк под 8% годовых, он получит тот же самый доход (50 тыс. руб.), что раньше получал в виде ренты (прибыли). Тема «Сбережение и инвестиции» Задача 1. Инвестор вложил в бизнес 1,5 млн. руб. В течение годa по результатам деятельности была получена чистая прибыль в размере 300 тыс. руб. За это время цены выросли на 4%, а банки привлекали вклады под 6% годовых. Оцените: 1. Эффективность инвестиций; 2. Целесообразность инвестиций; 3. Срок окупаемости инвестиций. Решение задачи 1. Доходность инвестиций отражает показатель рентабельности, который рассчитывается по формуле: Рентабельность инвестиций = чистая прибыль / затраты х 100% Соответственно, рентаб. = 300 т. р. / 1500 т. р. х 100% = 20% Вывод: каждый рубль, вложенный в бизнес, не только полностью окупился (т.е. вернулся в виде выручки), но и принес прибыль в размере 20 копеек. 2. Целесообразность осуществления инвестиций можно оценить в сравнении с инфляцией и возможным доходом по депозитным операциям. А) в сравнении с инфляцией: 20% > 4% Б) в сравнении с доходом по депозиту 20% > 6% Вывод: При заданных условиях инвестиции в бизнес являются целесообразными 3. Срок окупаемости инвестиций рассчитывается по формуле: СО = КВ / ЧП Соответственно, Со = 1500 тыс. руб. / 300 тыс. руб. = 5 лет Вывод: средства, вложенные в бизнес, полностью окупятся (вернутся инвестору в виде прибыли) через 5 лет 6 Тема «Капитал» Задача 1. ООО «Маяк» взяло кредит в размере 2,5 млн. руб. на выплату заработной платы под 16% годовых на 21 день. Какую сумму должен вернуть заемщик в конце указанного срока? Сколько составит сумма уплачиваемых процентов? При решении задачи применяем схему начисления простых процентов. Схема простых процентов предполагает неизменность базы, с которой происходит начисление. В финансовых вычислениях базовым периодом является год, поэтому обычно говорят о годовой ставке. Вместе с тем достаточно широко распространены краткосрочные операции продолжительностью до года. FV = P x (1 + t/T x i); I = P x t/T x i I = FV – P где: FV - сумма платежа, или наращенная сумма, руб.; P - начальная сумма кредита (вклада), полученная заемщиком, руб.; t - срок кредита (депозита), дней T - число дней в году, дней i - процентная ставка, доли единиц; I - сумма процентных денег, руб. Соответственно: FV = 2.5 x (1 + 21/365 x 0,16) = 2,523 млн. руб.; I = FV – P = 2,523… - 2.5 = 23013,5 руб. Вывод: через 21 день заемщик должен вернуть банку 2 миллиона 523 тысячи 13 рублей и 50 копеек. При этом сумма основного долга составила 2,5 млн. рублей; сумма уплаченных процентов – 23013,5 руб. Задача 2. На счет в банк положена сумма в размере 300 тыс. руб. Определить сумму процентных денег, которую можно получить через год, если процентная ставка равна 8% годовых, а начисление процента производится ежеквартально (проценты капитализируются). В данном случае используется схема сложных процентов, которая предполагает начисление процента на процент: с каждым разом расчетная база, на которую он начисляется, становится все больше. FV = P x (1 + i)n ; I = P x [(1 + i)n - 1] I = FV – P где: n - число единичных периодов времени от момента получения кредита 7 до его погашения. Единичным является период времени между двумя начислениями процентов; i - ставка процентов в расчете на один единичный период времени. Соответственно: Ежеквартальная % ставка = 8% : 4 кв = 2% FV = 300 х (1 + 0,02)4 = 300 х 1,08243 = 324,729 тыс.руб. I = FV – P = 324,729 - 300 = 24729 руб. Вывод: через год на депозитном счете будет находиться вкладе будет находится 324729 рублей, из них сумма начисленных процентов за период нахождения на счете 300 тыс. рублей (процентных денег) составит 24729 рублей. |
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From Wikipedia, the free encyclopedia
Grammar of the Standard Chinese language
This article concerns Modern Standard Chinese. For the grammars of other forms or varieties of Chinese, see their respective articles via links on Chinese language and varieties of Chinese.
The grammar of Standard Chinese shares many features with other varieties of Chinese. The language almost entirely lacks inflection; words typically have only one grammatical form. Categories such as number (singular or plural) and verb tense are often not expressed by grammatical means, but there are several particles that serve to express verbal aspect and, to some extent, mood.
The basic word order is subject–verb–object (SVO), as in English. Otherwise, Chinese is chiefly a head-final language, meaning that modifiers precede the words that they modify. In a noun phrase, for example, the head noun comes last, and all modifiers, including relative clauses, come in front of it. This phenomenon, however, is more typically found in subject–object–verb languages, such as Turkish and Japanese.
Chinese frequently uses serial verb constructions, which involve two or more verbs or verb phrases in sequence. Chinese prepositions behave similarly to serialized verbs in some respects,[a] and they are often referred to as coverbs. There are also location markers, which are placed after nouns and are thus often called postpositions; they are often used in combination with coverbs. Predicate adjectives are normally used without a copular verb ("to be") and so can be regarded as a type of verb.
As in many other East Asian languages, classifiers (or measure words) are required when numerals (and sometimes other words, such as demonstratives) are used with nouns. There are many different classifiers in the language, and each countable noun generally has a particular classifier associated with it. Informally, however, it is often acceptable to use the general classifier gè (个; 個) in place of other specific classifiers.
Word formation
[edit]
In Chinese, the difference between words and Chinese characters is often not clear,[b] this is one of the reasons the Chinese script does not use spaces to separate words. A string of characters can be translated as a single English word, but these characters have some kind of independence. For example, tiàowǔ (跳舞; 'jump-dance'), meaning 'to dance', can be used as a single intransitive verb, or may be regarded as comprising two single lexical words. However, it does in fact function as a compound of the verb tiào (跳; 'to jump') and the object wǔ (舞; 'a dance'). Additionally, the present progressive aspect marker zhe (着) can be inserted between these two parts to form tiàozhewǔ (跳着舞; 'to be dancing').
Chinese morphemes (the smallest units of meaning) are mostly monosyllabic. In most cases, morphemes are represented by single characters. However, two or more monosyllabic morphemes can be translated as a single English word. These monosyllabic morphemes can be either free or bound – that is, in particular usage, they may or may not be able to stand independently. Most two-syllable compound nouns often have the head on the right (e.g. 蛋糕; dàngāo; 'egg-cake' means "cake"), while compound verbs often have the head on the left (e.g. 辩论; biànlùn; 'debate-discuss' means "debate").
Some Chinese morphemes are polysyllabic; for example, the loanwords shāfā (沙发; 沙發; 'sofa') is the compound of shā (沙; 'sand') and fā (发; 發; 'to send', 'to issue'), but this compound is actually simply a transliteration of "sofa". Many native disyllabic morphemes, such as zhīzhū (蜘蛛; 'spider'), have consonant alliteration.[citation needed]
Many monosyllabic words have alternative disyllabic forms with virtually the same meaning, such as dàsuàn (大蒜; 'big-garlic') for suàn (蒜; 'garlic'). Many disyllabic nouns are produced by adding the suffix zi (子; 'child') to a monosyllabic word or morpheme. There is a strong tendency for monosyllables to be avoided in certain positions; for example, a disyllabic verb will not normally be followed by a monosyllabic object. This may be connected with the preferred metrical structure of the language.
Reduplication
[edit]
Reduplication (the repetition of a syllable or word) is a common feature in Chinese:
Family members: māma (妈妈; 媽媽, "mother"); dìdi (弟弟, "younger brother")
Adjectives or adverb: to emphasize the state described by the adjective/adverb, or as a childish expression.
: hóng-hóng-de (红红的; 紅紅的 "red"), from hóng (红; 紅, "red");
: ex. 手心看起来红红的; 手心看起來紅紅的(Shǒuxīn kàn qǐlái hóng-hóng-de, "palm looks red")
: gāo-gāo-xìng-xìng-de (高高兴兴地; 高高興興地 "very happily"), from gāo-xìng (高兴; 高興, "happy, happiness");
: ex. 高高兴兴地吃; 高高興興地吃 (Gāo-gāo-xìng-xìng-de chī, "eat happily")
: bīng-bīng-liáng-liáng-de (冰冰凉凉的, "ice-cool" ), from bīng-liáng (冰凉, "ice-cool");
: ex. 冰冰凉凉的饮料; 冰冰涼涼的飲料 (Bīng-bīng-liáng-liáng de yǐnliào, "ice-cold drink")
Other adjectives have ABB reduplication structure:
: xiāng-pēn-pēn (香喷喷; 香噴噴, literally" good smell spray out", means "smell very good"), from xiāng (香, "to smell good, smell") and pēn (喷; 噴, "spray");
: liàng-jīng-jīng (亮晶晶, "shining, bright and clear"), from liàng (亮, "bright") and jīng (晶, "shiny like a star");
Verbs: to mark the delimitative aspect ("to do something for a little bit") or for general emphasis – see the § Aspects section:
: xiě-xiě-zuòyè (写写作业; 寫寫作業 "write homework / write homework for a while"), from the verb xiě (写; 寫 "write") and the noun zuò-yè (作业; 作業 "homework")
Single morphemes:
: xīngxīng (星星, "star"), from xīng (星, "star");
: chángcháng (常常, "often"), from cháng (常, "constant");
: gǒugǒu (狗狗, "puppy/doggy"), from gǒu (狗, "dog")
Chinese classifiers, to produce a phrase meaning "every" or "many":
: Nǐmen yī gè gè dōu zhǎng dé yī fù cōng-míng xiāng (你們一個個都長得一副聰明相, "You all look smart", from Crystal Boys), where ordinarily gè (个; 個) is the general classifier. Literally, the phrase 一個個; yī gè gè means "every", and the character 都; dōu means "all".
: Yī-zuò-zuò qīng-shān (一座座青山, "many green hills"), where ordinarily zuò (座) is a proper classifier for shān (山, "hill").
Prefixes
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可 — "-able"
可 靠 — "reliable"
可 敬 — "respectable"
反 — "anti-"
反 恐 [反恐] — "anti-terror"
反 教权的 [反教權的] — "anti-clerical"
反 法西斯 [反法西斯] — "anti-fascist"
Suffixes
[edit]
化 — used to form verbs from nouns or adjectives
国际 化 [國際化] — "internationalize", from 国际 ("internationality")
恶 化 [惡化] — "worsen", from 恶 ("bad")
性 — "attribute"
安全 性 — "safety"
有效 性 — "effectiveness"
Intrafixes
[edit]
得 — "can" and 不 — "cannot"
听得懂 — "can understand"
听不懂 — "cannot understand"
Sentence structure
[edit]
Chinese, like Spanish or English, is classified as an SVO (subject–verb–object) language. Transitive verbs precede their objects in typical simple clauses, while the subject precedes the verb. For example:
他 打 人。
tā dǎ rén
He hit person
He hits someone.
Chinese can also be considered a topic-prominent language: there is a strong preference for sentences that begin with the topic, usually "given" or "old" information; and end with the comment, or "new" information. Certain modifications of the basic subject–verb–object order are permissible and may serve to achieve topic-prominence. In particular, a direct or indirect object may be moved to the start of the clause to create topicalization. It is also possible for an object to be moved to a position in front of the verb for emphasis.
Another type of sentence is what has been called an ergative structure, where the apparent subject of the verb can move to object position; the empty subject position is then often occupied by an expression of location. Compare locative inversion in English. This structure is typical of the verb yǒu (有, "there is/are"; in other contexts the same verb means "have"), but it can also be used with many other verbs, generally denoting position, appearance or disappearance. An example:
[院子裡停著車。/ 院子裏停着車。]
院子 里 停着 车。
yuànzi lǐ tíngzhe chē
Courtyard in park vehicle
In the courtyard is parked a vehicle.
Chinese is also to some degree a pro-drop or null-subject language, meaning that the subject can be omitted from a clause if it can be inferred from the context. In the following example, the subject of the verbs for "hike" and "camp" is left to be inferred—it may be "we", "I", "you", "she", etc.
Today
爬
pá
climb
山
shān
mountain,
明天
míngtiān
tomorrow
露
lù
outdoors
营。
yíng
camp
[今天爬山，明天露營。]
今天 爬 山 明天 露 营。
jīntiān pá shān míngtiān lù yíng
Today climb mountain, tomorrow outdoors camp
Today hike up mountains, tomorrow camp outdoors.
In the next example the subject is omitted and the object is topicalized by being moved into subject position, to form a passive-type sentence. For passive sentences with a marker such as 被; bèi, see the passive section.
饭
fàn
Food
做
zuò
make
好
hǎo
complete
了。
le
PFV
[飯做好了。]
饭 做 好 了。
fàn zuò hǎo le
Food make complete PFV
The food has been made or the food is ready.
Adverbs and adverbial phrases that modify the verb typically come after the subject but before the verb, although other positions are sometimes possible; see Adverbs and adverbials. For constructions that involve more than one verb or verb phrase in sequence, see Serial verb constructions. For sentences consisting of more than one clause, see Conjunctions.
Objects
[edit]
Some verbs can take both an indirect object and a direct object. Indirect normally precedes direct, as in English:
我
wǒ
I
给
gěi
give
了
le
PFV
她
tā
her
六
liù
six
本
běn
book-CL
书。
shū
books
[我給了她六本書。]
我 给 了 她 六 本 书。
wǒ gěi le tā liù běn shū
I give PFV her six book-CL books
I gave her six books.
With many verbs, however, the indirect object may alternatively be preceded by prepositional gěi (给; 給); in that case it may either precede or follow the direct object. (Compare the similar use of to or for in English.)
To emphasize the direct object, it can be combined with the accusative marker bǎ (把, literally "hold") to form a "bǎ + direct object" phrase. This phrase is placed before the verb. For example:
我
wǒ
I
打
dǎ
hit
破
pò
broken
了
le
PFV
盘子。
pánzi
plate
[我打破了盤子。]
我 打 破 了 盘子。
wǒ dǎ pò le pánzi
I hit broken PFV plate
I broke a plate.
我
wǒ
I
把
bǎ
ba
盘子
pánzi
plate
打
dǎ
hit
破
pò
broken
了。
le
PFV
[我把盤子打破了。]
我 把 盘子 打 破 了。
wǒ bǎ pánzi dǎ pò le
I ba plate hit broken PFV
I made the plate break.
Other markers can be used in a similar way as bǎ, such as the formal jiāng (将; 將, literally "lead") :
将
Jiāng
Jiāng
办理
bàn-lǐ
handle
情形
qíng-xíng
status
签
qiān
sign
报
bào
report
长官。
zhǎng-guān
superior
[將辦理情形簽報長官。]
将 办理 情形 签 报 长官。
Jiāng bàn-lǐ qíng-xíng qiān bào zhǎng-guān
Jiāng handle status sign report superior
Submit the implementation status report to the superior, and ask for approval.
and colloquial ná (拿, literally "get")
他
Tā
he
能
néng
can
拿
ná
ná
我
wǒ
me
怎样?
zěn-yàng
what
[他能拿我怎樣?]
他 能 拿 我 怎样?
Tā néng ná wǒ zěn-yàng
he can ná me what
What can he do to me? (He can't do anything to me.)
To explain this kind of usage, some linguists assume that some verbs can take two direct objects, called the called "inner" and "outer" object. Typically, the outer object will be placed at the start of the sentence (which is the topic) or introduced via the bǎ phrase. For example:
我
wǒ
I
把
bǎ
ba
橘子
júzi
tangerine
剥
bō
peel
了
le
PFV
皮。
pí
skin
[我把橘子剝了皮。]
我 把 橘子 剥 了 皮。
wǒ bǎ júzi bō le pí
I ba tangerine peel PFV skin
I make the tangerine peeled.[c]
Noun phrases
[edit]
The head noun of a noun phrase comes at the end of the phrase; this means that everything that modifies the noun comes before it. This includes attributive adjectives, determiners, quantifiers, possessives, and relative clauses.
Chinese does not have articles as such; a noun may stand alone to represent what in English would be expressed as "the ..." or "a[n] ...". However the word yī (一, "one"), followed by the appropriate classifier, may be used in some cases where English would have "a" or "an". It is also possible, with many classifiers, to omit the yī and leave the classifier on its own at the start of the noun phrase.
The demonstratives are zhè (这; 這, "this"), and nà (那, "that"). When used before a noun, these are often followed by an appropriate classifier (for discussion of classifiers, see Classifiers below and the article Chinese classifiers). However this use of classifiers is optional. When a noun is preceded by a numeral (or a demonstrative followed by a numeral), the use of a classifier or measure word is in most cases considered mandatory. (This does not apply to nouns that function as measure words themselves; this includes many units of measurement and currency.)
The plural marker xiē (些, "some, several"; also used to pluralize demonstratives) is used without a classifier. However jǐ (几; 幾, "some, several, how many") takes a classifier.
For adjectives in noun phrases, see the Adjectives section. For noun phrases with pronouns rather than nouns as the head, see the Pronouns section.
Possessives are formed by adding de (的)—the same particle that is used after relative clauses and sometimes after adjectives—after the noun, noun phrase or pronoun that denotes the possessor.
Relative clauses
[edit]
Chinese relative clauses, like other noun modifiers, precede the noun they modify. Like possessives and some adjectives, they are marked with the final particle de (的). A free relative clause is produced if the modified noun following the de is omitted. A relative clause usually comes after any determiner phrase, such as a numeral and classifier. For emphasis, it may come before the determiner phrase.
There is usually no relative pronoun in the relative clause. Instead, a gap is left in subject or object position as appropriate. If there are two gaps—the additional gap being created by pro-dropping—ambiguity may arise. For example, chī de (吃的) may mean "[those] who eat" or "[that] which is eaten". When used alone, it usually means "things to eat".
If the relative item is governed by a preposition in the relative clause, then it is denoted by a pronoun, e.g. tì tā (替他, "for him"), to explain "for whom". Otherwise the whole prepositional phrase is omitted, the preposition then being implicitly understood.
For example sentences, see Relative clause → Mandarin.
Classifiers
[edit]
Main article: Chinese classifier
See also: List of Chinese classifiers
Some English words are paired with specific nouns to indicate their counting units. For example, Bottle in "two bottles of wine" and sheet in "three sheets of paper". However, most English nouns can be counted directly without specifying units, while counting of most Chinese nouns must be associated with a specific classifier, namely liàng-cí (量词; 量詞, "measure words"), to represent their counting units. Every Chinese noun can only be associated with a limited number of classifiers. For example
一
yī
one
瓶
píng
bottle
酒
jiǔ
wine
[一瓶酒]
一 瓶 酒
yī píng jiǔ
one bottle wine
a bottle of wine
两
liǎng
two
杯
bēi
cup
酒
jiǔ
wine
[兩杯酒]
两 杯 酒
liǎng bēi jiǔ
two cup wine
two glasses of wine
píng (瓶, "bottle") and bēi (杯, "cup") are both proper classifiers of the countable noun jiǔ (酒), while liǎng zuò jiǔ (两座酒) and liǎng-jiǔ (两酒) are unacceptable.
While there are dozens of classifiers, the general classifier gè (个; 個) is colloquially (i.e. in informal conversations) acceptable for most nouns. However, there are still some exceptions. For example, liǎng gè jiǔ (两个酒) is weird and unacceptable.
Most classifiers originated as independent words in Classical Chinese, so they are generally associated with certain groups of nouns with common properties related to their own classical meaning, for example:
| Classifier (Original meaning) | Common Properties | Examples |
| tiáo (条; 條, "twig") | long or thin (twigs are long and thin) | yī-tiáo-shéngzi (一条绳子; 一條繩子, "a rope") liǎng-tiáo-shé (两条蛇; 兩條蛇, "two snakes") |
| bǎ (把, "hold") | with a handle (a handle to hold) | yī-bǎ-dāo (一把刀, "a knife") liǎng-bǎ-sǎn (两把伞; 兩把傘, "two umbrellas") |
| zhāng (张; 張, "draw a bow") | flat or sheet-like ("extended" like a bow) | yī zhāng zhào-piàn (一张照片; 一張照片, "a photograph") liǎng zhāng máo-pí (两张毛皮; 兩張毛皮, "two furs") |
Therefore, collocation of classifiers and noun sometimes depends on how native speakers realize them. For example, the noun zhuōzi (桌子, "table") is associated with the classifier zhāng (张; 張), due to the sheet-like table-top. Additionally, yǐ-zi (椅子, "chair") is associated with bǎ (把, "hold"), because a chair can be moved by holding its top like a handle. Furthermore, due to the invention of the folding chair, yǐ-zi (椅子, "chair") is also associated with the classifier zhāng (张; 張) to express a folding chair can be "extended" (unfolded).
Classifiers are also used optionally after demonstratives, and in certain other situations. See the Noun phrases section, and the article Chinese classifier.
Numerals
[edit]
Main article: Chinese numerals
Pronouns
[edit]
Main article: Chinese pronouns
The Chinese personal pronouns are wǒ (我, "I, me"), nǐ (你; 你/妳,[d] "you"), and tā (他/她/牠/它, "he, him / she, her / it (animals) / it (inanimate objects)". Plurals are formed by adding men (们; 們): wǒmen (我们; 我們, "we, us"), nǐmen (你们; 你們, "you"), tāmen (他们/她们/它们/它们; 他們/她們/牠們/它們, "they/them"). There is also nín (您), a formal, polite word for singular "you", as well as a less common plural form, nínmen (您们). Some northern dialects have a third-person formal, polite word 怹 (他+心, he/him + heart) similar to 您 (你+心, you + heart). The alternative inclusive word for "we/us"—zán (咱) or zá[n]men (咱们; 咱們), specifically including the listener —is used colloquially. The third-person pronouns are not often used for inanimates, with demonstratives used instead.
Possessives are formed with de (的), such as wǒde (我的, "my, mine"), wǒmende (我们的; 我們的, "our[s]"), etc. The de may be omitted in phrases denoting inalienable possession, such as wǒ māma (我妈妈; 我媽媽, "my mom").
The demonstrative pronouns are zhè (这; 這, "this", colloquially pronounced zhèi as a shorthand for 这一; 這一) and nà (那, "that", colloquially pronounced nèi as a shorthand for 那一). They are optionally pluralized by the addition of plural quantifiers xiē (些) or qún (群). There is a reflexive pronoun zìjǐ (自己) meaning "oneself, myself, etc.", which can stand alone as an object or a possessive, or may follow a personal pronoun for emphasis. The reciprocal pronoun "each other" can be translated from bǐcǐ (彼此), usually in adverb position. An alternative is hùxiāng (互相, "mutually").
Adjectives
[edit]
Main article: Chinese adjectives
Adjectives can be used attributively, before a noun. The relative marker de (的)[e] may be added after the adjective, but this is not always required; "black horse" may be either hēi mǎ (黑马; 黑馬) or hēi de mǎ (黑的马; 黑的馬). When multiple adjectives are used, the order "quality/size – shape – color" is followed, although this is not necessary when each adjective is made into a separate phrase with the addition of de.
Gradable adjectives can be modified by words meaning "very", etc.; such modifying adverbs normally precede the adjective, although some, such as jíle (极了; 極了, "extremely"), come after it.
When adjectives co-occur with classifiers, they normally follow the classifier. However, with most common classifiers, when the number is "one", it is also possible to place adjectives like "big" and "small" before the classifier for emphasis.
ex:
一
一
yí
one
大
大
dà
big
个
個
ge
CL
西瓜
西瓜
xīguā
watermelon
一 大 个 西瓜
一 大 個 西瓜
yí dà ge xīguā
one big CL watermelon
Adjectives can also be used predicatively. In this case they behave more like verbs; there is no need for a copular verb in sentences like "he is happy" in Chinese; one may say simply tā gāoxìng (他高兴; 他高興, "he happy"), where the adjective may be interpreted as a verb meaning "is happy". In such sentences it is common for the adjective to be modified by a word meaning "very" or the like; in fact the word hěn (很, "very") is often used in such cases with gradable adjectives, even without carrying the meaning of "very".
It is nonetheless possible for a copula to be used in such sentences, to emphasize the adjective. In the phrase tā shì gāoxìng le, (他是高兴了; 他是高興了, "he is now truly happy"), shì is the copula meaning "is", and le is the inceptive marker discussed later. This is similar to the cleft sentence construction. Sentences can also be formed in which an adjective followed by de (的) stands as the complement of the copula.
Adverbs and adverbials
[edit]
Adverbs and adverbial phrases normally come in a position before the verb, but after the subject of the verb. In sentences with auxiliary verbs, the adverb usually precedes the auxiliary verb as well as the main verb. Some adverbs of time and attitude ("every day", "perhaps", etc.) may be moved to the start of the clause, to modify the clause as a whole. However, some adverbs cannot be moved in this way. These include three words for "often", cháng (常), chángcháng (常常) and jīngcháng (经常; 經常); dōu (都, "all"); jiù (就, "then"); and yòu (又, "again").
Adverbs of manner can be formed from adjectives using the clitic de (地).[f] It is generally possible to move these adverbs to the start of the clause, although in some cases this may sound awkward, unless there is a qualifier such as hěn (很, "very") and a pause after the adverb.
Some verbs take a prepositional phrase following the verb and its direct object. These are generally obligatory constituents, such that the sentence would not make sense if they were omitted. For example:
放
fàng
put
本
běn
book-CL
书
shū
book
在
zài
in
桌子
zhuōzi
table
上
shàng
on
[放本書在桌子上]
放 本 书 在 桌子 上
fàng běn shū zài zhuōzi shàng
put book-CL book in table on
Put a book on the table
There are also certain adverbial "stative complements" which follow the verb. The character de (得)[g] followed by an adjective functions the same as the phrase "-ly" in English, turning the adjective into an adverb. The second is hǎo le (好了, "complete"). It is not generally possible for a single verb to be followed by both an object and an adverbial complement of this type, although there are exceptions in cases where the complement expresses duration, frequency or goal. To express both, the verb may be repeated in a special kind of serial verb construction; the first instance taking an object, the second taking the complement. Aspect markers can then appear only on the second instance of the verb.
The typical Chinese word order "XVO", where an oblique complement such as a locative prepositional phrase precedes the verb, while a direct object comes after the verb, is very rare cross-linguistically; in fact, it is only in varieties of Chinese that this is attested as the typical ordering.
Locative phrases
[edit]
Expressions of location in Chinese may include a preposition, placed before the noun; a postposition, placed after the noun; both preposition and postposition; or neither. Chinese prepositions are commonly known as coverbs – see the Coverbs section. The postpositions—which include shàng (上, "up, on"), xià (下, "down, under"), lǐ (里; 裡, "in, within"), nèi (内, "inside") and wài (外, "outside")—may also be called locative particles.
In the following examples locative phrases are formed from a noun plus a locative particle:
桌子
zhuōzi
table
上
shàng
on
桌子 上
zhuōzi shàng
table on
on the table
房子
fángzi
house
里
lǐ
in
[房子裡]
房子 里
fángzi lǐ
house in
in the house
The most common preposition of location is zài (在, "at, on, in"). With certain nouns that inherently denote a specific location, including nearly all place names, a locative phrase can be formed with zài together with the noun:
在
zài
in
美国
měiguó
America
[在美國]
在 美国
zài měiguó
in America
in America
However other types of nouns still require a locative particle as a postposition in addition to zài:
在
zài
in
报纸
bàozhǐ
newspaper
上
shàng
on
[在報紙上]
在 报纸 上
zài bàozhǐ shàng
in newspaper on
in the newspaper
If a noun is modified so as to denote a specific location, as in "this [object]...", then it may form locative phrases without any locative particle. Some nouns which can be understood to refer to a specific place, like jiā (家, home) and xuéxiào (学校; 學校, "school"), may optionally omit the locative particle. Words like shàngmiàn (上面, "top") can function as specific-location nouns, like in zài shàngmiàn (在上面, "on top"), but can also take the role of locative particle, not necessarily with analogous meaning. The phrase zài bàozhǐ shàngmiàn (在报纸上面; 在報紙上面; 'in newspaper-top'), can mean either "in the newspaper" or "on the newspaper".
In certain circumstances zài can be omitted from the locative expression. Grammatically, a noun or noun phrase followed by a locative particle is still a noun phrase. For instance, zhuōzi shàng can be regarded as short for zhuōzi shàngmiàn, meaning something like "the table's top". Consequently, the locative expression without zài can be used in places where a noun phrase would be expected – for instance, as a modifier of another noun using de (的), or as the object of a different preposition, such as cóng (从; 從, "from"). The version with zài, on the other hand, plays an adverbial role. However, zài is usually omitted when the locative expression begins a sentence with the ergative structure, where the expression, though having an adverbial function, can be seen as filling the subject or noun role in the sentence. For examples, see sentence structure section.
The word zài (在), like certain other prepositions or coverbs, can also be used as a verb. A locative expression can therefore appear as a predicate without the need for any additional copula. For example, "he is at school" (他在学校; 他在學校; tā zài xuéxiào, literally "he at school").
Comparatives and superlatives
[edit]
Comparative sentences are commonly expressed simply by inserting the standard of comparison, preceded by bǐ (比, "than"). The adjective itself is not modified. The bǐ (比, "than") phrase is an adverbial, and has a fixed position before the verb. See also the section on negation.
If there is no standard of comparison—i.e., a than phrase—then the adjective can be marked as comparative by a preceding adverb bǐjiào (比较; 比較), jiào (较; 較) or gèng (更), all meaning "more". Similarly, superlatives can be expressed using the adverb zuì (最, "most"), which precedes a predicate verb or adjective.
Adverbial phrases meaning "like [someone/something]" or "as [someone/something]" can be formed using gēn (跟), tóng (同) or xiàng (像) before the noun phrase, and yīyàng (一样; 一樣) or nàyàng (那样; 那樣) after it.
The construction yuè ... yuè ... 越...越... can be translated into statements of the type "the more ..., the more ...".
Copula
[edit]
Further information: Chinese copula
The Chinese copular verb is shì (是). This is the equivalent of English "to be" and all its forms—"am", "is", "are", "was", "were", etc. However, shì is normally only used when its complement is a noun or noun phrase. As noted above, predicate adjectives function as verbs themselves, as does the locative preposition zài (在), so in sentences where the predicate is an adjectival or locative phrase, shì is not required.
For another use of shì, see shì ... [de] construction in the section on cleft sentences. The English existential phrase "there is" ["there are", etc.] is translated using the verb yǒu (有), which is otherwise used to denote possession.
Aspects
[edit]
Chinese does not have grammatical markers of tense. The time at which action is conceived as taking place—past, present, future—can be indicated by expressions of time—"yesterday", "now", etc.—or may simply be inferred from the context. However, Chinese does have markers of aspect, which is a feature of grammar that gives information about the temporal flow of events. There are two aspect markers that are especially commonly used with past events: the perfective-aspect le (了) and the experiential guo (过; 過). Some authors, however, do not regard guo (or zhe; see below) as markers of aspect. Both le and guo immediately follow the verb.
There is also a sentence-final inchoative le (了), which is an aspect-marking particle that indicates a change in state. Following a convention used by some textbooks, it is listed with the modal particles below, even though it does not indicate a grammatical mood.
The perfective le presents the viewpoint of "an event in its entirety". It is sometimes considered to be a past tense marker, although it can also be used with future events, given appropriate context. Some examples of its use:
我
wǒ
I
当
dāng
serve as
了
le
le
兵。
bīng
soldier.
[我當了兵。]
我 当 了 兵。
wǒ dāng le bīng
I {serve as} le soldier.
I became a soldier.
Using le (了) shows this event that has taken place or took place at a particular time.
他
tā
He
看
kàn
watch
了
le
le
三
sān
three
场
chǎng
sports-CL
球赛。
qiúsài
ballgames.
[他看了三場球賽。]
他 看 了 三 场 球赛。
tā kàn le sān chǎng qiúsài
He watch le three sports-CL ballgames.
He watched three ballgames.
This format of le (了) is usually used in a time-delimited context such as "today" or "last week".
The above may be compared with the following examples with guo, and with the examples with sentence-final le given under Particles.
The experiential guo "ascribes to a subject the property of having experienced the event".
我
wǒ
I
当
dāng
serve-as
过
guo
guo
兵。
bīng
soldier.
[我當過兵。]
我 当 过 兵。
wǒ dāng guo bīng
I serve-as guo soldier.
I have been a soldier before.
This also implies that the speaker no longer is a soldier.
他
tā
He
看
kàn
watch
过
guo
guo
三
sān
three
场
chǎng
sports-CL
球赛。
qiúsài
ballgames.
[他看過三場球賽。]
他 看 过 三 场 球赛。
tā kàn guo sān chǎng qiúsài
He watch guo three sports-CL ballgames.
He has watched three ballgames up to now.
There are also two imperfective aspect markers: zhèngzài (正在) or zài (在), and zhe (着; 著), which denote ongoing actions or states. Zhèngzài and zài precede the verb, and are usually used for ongoing actions or dynamic events – they may be translated as "[be] in the process of [-ing]" or "[be] in the middle of [-ing]". Zhe follows the verb, and is used mostly for static situations.
我
wǒ
I
[正] 在
zhèng zài
in-middle-of
挂
guà
hang
画。
huà
pictures
[我[正]在掛畫。]
我 {[正] 在} 挂 画。
wǒ {zhèng zài} guà huà
I in-middle-of hang pictures
I'm hanging pictures up.
墙
qiáng
Wall
上
shàng
on
挂
guà
hang
着
zhe
ongoing
一
yì
one
幅
fú
picture-CL
画。
huà
picture
[牆上掛著一幅畫。]
墙 上 挂 着 一 幅 画。
qiáng shàng guà zhe yì fú huà
Wall on hang ongoing one picture-CL picture
A picture is hanging on the wall.
Both markers may occur in the same clause, however. For example, tā zhèngzai dǎ [zhe] diànhuà, "he is in the middle of telephoning someone" (他正在打[着]电话; 他正在打[著]電話; 'he [', 'in-middle-of]', '[', 'verb form]', '[', 'ongoing]', 'telephone').
The delimitative aspect denotes an action that goes on only for some time, "doing something 'a little bit'". This can be expressed by reduplication of a monosyllabic verb, like the verb zǒu (走 "walk") in the following sentence:
我
wǒ
I
到
dào
to
公园
gōngyuán
park
走
zǒu
walk
走。
zǒu
walk
[我到公園走走。]
我 到 公园 走 走。
wǒ dào gōngyuán zǒu zǒu
I to park walk walk
I'm going for a walk in the park.
An alternative construction is reduplication with insertion of "one" (一 yī). For example, zǒu yi zǒu (走一走), which might be translated as "walk a little walk". A further possibility is reduplication followed by kàn (看 "to see"); this emphasizes the "testing" nature of the action. If the verb has an object, kàn follows the object.
Some compound verbs, such as restrictive-resultative and coordinate compounds, can also be reduplicated on the pattern tǎolùn-tǎolùn (讨论讨论; 討論討論), from the verb tǎolùn (讨论; 討論), meaning "discuss". Other compounds may be reduplicated, but for general emphasis rather than delimitative aspect. In compounds that are verb–object combinations, like tiào wǔ (跳舞; 'to jump a dance', "dance"), a delimitative aspect can be marked by reduplicating the first syllable, creating tiào-tiào wǔ (跳跳舞), which may be followed with kàn (看).
Passive
[edit]
As mentioned above, the fact that a verb is intended to be understood in the passive voice is not always marked in Chinese. However, it may be marked using the passive marker 被 bèi, followed by the agent, though bèi may appear alone, if the agent is not to be specified.[h] Certain causative markers can replace bèi, such as those mentioned in the Other cases section, gěi, jiào and ràng. Of these causative markers, only gěi can appear alone without a specified agent. The construction with a passive marker is normally used only when there is a sense of misfortune or adversity. The passive marker and agent occupy the typical adverbial position before the verb. See the Negation section for more. Some examples:
我们
wǒmen
We
被
bèi
by
他
tā
him
骂
mà
scolded
了。
le
PFV
[我們被他罵了。]
我们 被 他 骂 了。
wǒmen bèi tā mà le
We by him scolded PFV
We were scolded by him.
他
tā
He
被
bèi
by
我
wǒ
me
打
dǎ
beaten
了
le
PFV
一
yí
one
顿。
dùn
event-CL
[他被我打了一頓。]
他 被 我 打 了 一 顿。
tā bèi wǒ dǎ le yí dùn
He by me beaten PFV one event-CL
He was beaten up by me once.
Negation
[edit]
The most commonly used negating element is bù (不), pronounced with second tone when followed by a fourth tone. This can be placed before a verb, preposition or adverb to negate it. For example: "I don't eat chicken" (我不吃鸡; 我不吃雞; wǒ bù chī jī; 'I not eat chicken'). For the double-verb negative construction with bù, see Complement of result, below. However, the verb yǒu (有)—which can mean either possession, or "there is/are" in existential clauses—is negated using méi (没; 沒) to produce méiyǒu (没有; 沒有; 'not have').
For negation of a verb intended to denote a completed event, méi or méiyǒu is used instead of bù (不), and the aspect marker le (了) is then omitted. Also, méi[yǒu] is used to negate verbs that take the aspect marker guo (过; 過); in this case the aspect marker is not omitted.
In coverb constructions, the negator may come before the coverb (preposition) or before the full verb, the latter being more emphatic. In constructions with a passive marker, the negator precedes that marker; similarly, in comparative constructions, the negator precedes the bǐ phraseNot clear (unless the verb is further qualified by gèng (更, "even more"), in which case the negator may follow the gèng to produce the meaning "even less").
The negator bié (别) precedes the verb in negative commands and negative requests, such as in phrases meaning "don't ...", "please don't ...".
The negator wèi (未) means "not yet". Other items used as negating elements in certain compound words include wú (无; 無),wù (勿), miǎn (免) and fēi (非).
A double negative makes a positive, as in sentences like wǒ bú shì bù xǐhuān tā (我不是不喜欢她; 我不是不喜歡她, "It's not that I don't like her" ). For this use of shì (是), see the Cleft sentences section.
Questions
[edit]
In wh-questions in Chinese, the question word is not fronted. Instead, it stays in the position in the sentence that would be occupied by the item being asked about. For example, "What did you say?" is phrased as nǐ shuō shé[n]me (你说什么?; 你說什麼？, literally "you say what"). The word shénme (什么; 什麼, "what" or "which"), remains in the object position after the verb.
Other interrogative words include:
"Who": shuí/shéi (谁; 誰)
"What": shénme (什么; 什麼); shá (啥, used informally)
"Where": nǎr (哪儿; 哪兒); nǎlǐ (哪里; 哪裡); héchù (何处; 何處)
"When": shénme shíhòu (什么时候; 什麼時候); héshí (何时; 何時)
"Which": nǎ (哪)
When used to mean "which ones", nǎ is used with a classifier and noun, or with xiē (些) and noun. The noun may be omitted if understood through context.
"Why": wèishé[n]me (为什么; 為什麼); gànmá (干吗; 幹嘛)
"How many": duōshǎo (多少)
When the number is quite small, jǐ (几; 幾) is used, followed by a classifier.
"How": zěnme[yang] (怎么[样]; 怎麼[樣]); rúhé (如何).
Disjunctive questions can be made using the word háishì (还是; 還是) between the options, like English "or". This differs from the word for "or" in statements, which is huòzhě (或者).
Yes–no questions can be formed using the sentence-final particle ma (吗; 嗎), with word order otherwise the same as in a statement. For example, nǐ chī jī ma? (你吃鸡吗?; 你吃雞嗎？; 'you eat chicken MA', "Do you eat chicken?").
An alternative is the A-not-A construction, using phrases like chī bu chī (吃不吃, "eat or not eat").[i] With two-syllable verbs, sometimes only the first syllable is repeated: xǐ-bu-xǐhuān ( 喜不喜欢; 喜不喜歡, "like or not like"), from xǐhuān (喜欢; 喜歡, "like"). It is also possible to use the A-not-A construction with prepositions (coverbs) and phrases headed by them, as with full verbs.
The negator méi (没; 沒) can be used rather than bù in the A-not-A construction when referring to a completed event, but if it occurs at the end of the sentence—i.e. the repetition is omitted—the full form méiyǒu (没有; 沒有) must appear.
For answering yes–no questions, Chinese has words that may be used like the English "yes" and "no" – duì (对; 對) or shì de (是的) for "yes"; bù (不) for "no" – but these are not often used for this purpose; it is more common to repeat the verb or verb phrase (or entire sentence), negating it if applicable.
Imperatives
[edit]
Second-person imperative sentences are formed in the same way as statements, and like in English, the subject "you" is often omitted.
Orders may be softened by preceding them with an element such as qǐng (请, "to ask"), in this use equivalent to English "please". See Particles for more. The sentence-final particle ba (吧) can be used to form first-person imperatives, equivalent to "let's...".
Serial verb constructions
[edit]
Chinese makes frequent use of serial verb constructions, or verb stacking, where two or more verbs or verb phrases are concatenated together. This frequently involves either verbal complements appearing after the main verb, or coverb phrases appearing before the main verb, but other variations of the construction occur as well.
Auxiliaries
[edit]
A main verb may be preceded by an auxiliary verb, as in English. Chinese auxiliaries include néng and nénggòu (能 and 能够; 能夠, "can"); huì (会; 會, "know how to"); kéyǐ (可以, "may"); gǎn (敢, "dare"); kěn (肯, "be willing to"); yīnggāi (应该; 應該, "should"); bìxū (必须; 必須, "must"); etc. The auxiliary normally follows an adverb, if present. In shortened sentences an auxiliary may be used without a main verb, analogously to English sentences such as "I can."
Verbal complements
[edit]
The active verb of a sentence may be suffixed with a second verb, which usually indicates either the result of the first action, or the direction in which it took the subject. When such information is applicable, it is generally considered mandatory. The phenomenon is sometimes called double verbs.
Complement of result
[edit]
A complement of result, or resultative complement (结果补语; 結果補語; jiéguǒ bǔyǔ) is a verbal suffix which indicates the outcome, or possible outcome, of the action indicated by the main verb. In the following examples, the main verb is tīng (听; 聽 "to listen"), and the complement of result is dǒng (懂, "to understand/to know").
听
tīng
hear
懂
dǒng
understand
[聽懂]
听 懂
tīng dǒng
hear understand
to understand something you hear
Since they indicate an absolute result, such double verbs necessarily represent a completed action, and are thus negated using méi (没; 沒):
没
méi
not
听
tīng
hear
懂
dǒng
understand
[沒聽懂]
没 听 懂
méi tīng dǒng
not hear understand
to have not understood something you hear
The morpheme de (得) is placed between the double verbs to indicate possibility or ability. This is not possible with "restrictive" resultative compounds such as jiéshěng (节省, literally "reduce-save", meaning "to save, economize").
听
tīng
hear
得
de
possible/able
懂
dǒng
understand
[聽得懂]
听 得 懂
tīng de dǒng
hear possible/able understand
to be able to understand something you hear
This is equivalent in meaning to néng tīng dǒng (能听懂; 能聽懂), using the auxiliary néng (能), equivalent to "may" or "can".[j]
To negate the above construction, de (得) is replaced by bù (不):
听
tīng
hear
不
bù
impossible/unable
懂
dǒng
understand
[聽不懂]
听 不 懂
tīng bù dǒng
hear impossible/unable understand
to be unable to understand something you hear
With some verbs, the addition of bù and a particular complement of result is the standard method of negation. In many cases the complement is liǎo, represented by the same character as the perfective or modal particle le (了). This verb means "to finish", but when used as a complement for negation purposes it may merely indicate inability. For example: shòu bù liǎo (受不了, "to be unable to tolerate").
The complement of result is a highly productive and frequently used construction. Sometimes it develops into idiomatic phrases, as in è sǐ le (饿死了; 餓死了, literally "hungry-until-die already", meaning "to be starving") and qì sǐ le (气死了; 氣死了, literally "mad-until-die already", meaning "to be extremely angry"). The phrases for "hatred" (看不起; kànbùqǐ), "excuse me" (对不起; 對不起; duìbùqǐ), and "too expensive to buy" (买不起; 買不起; mǎi bùqǐ) all use the character qǐ (起, "to rise up") as a complement of result, but their meanings are not obviously related to that meaning. This is partially the result of metaphorical construction, where kànbùqǐ (看不起) literally means "to be unable to look up to"; and duìbùqǐ (对不起; 對不起) means "to be unable to face someone".
Some more examples of resultative complements, used in complete sentences:
他
tā
he
把
bǎ
object-CL
盘子
pánzi
plate
打
dǎ
hit
破
pò
break
了。
le
PRF
[他把盤子打破了。]
他 把 盘子 打 破 了。
tā bǎ pánzi dǎ pò le
he object-CL plate hit break PRF
He hit/dropped the plate, and it broke.
Double-verb construction where the second verb, "break", is a suffix to the first, and indicates what happens to the object as a result of the action.
这
zhè(i)
this
部
bù
电影
diànyǐng
movie
我
wǒ
I
看
kàn
watch
不
bù
impossible/unable
懂。
dǒng
understand
[這部電影我看不懂。]
这 部 电影 我 看 不 懂。
zhè(i) bù diànyǐng wǒ kàn bù dǒng
this {} movie I watch impossible/unable understand
I can't understand this movie even though I watched it.
Another double-verb where the second verb, "understand", suffixes the first and clarifies the possibility and success of the relevant action.
Complement of direction
[edit]
A complement of direction, or directional complement (趋向补语; 趨向補語; qūxiàng bǔyǔ) indicates the direction of an action involving movement. The simplest directional complements are qù (去, "to go") and lái (来; 來, "to come"), which may be added after a verb to indicate movement away from or towards the speaker, respectively. These may form compounds with other verbs that further specify the direction, such as shàng qù (上去, "to go up"), gùo lái (过来; 過來, "to come over"), which may then be added to another verb, such as zǒu (走, "to walk"), as in zǒu gùo qù (走过去; 走過去, "to walk over"). Another example, in a whole sentence:
他
tā
he
走
zǒu
walk
上
shàng
up
来
lái
come
了。
le
PRF
[他走上來了。]
他 走 上 来 了。
tā zǒu shàng lái le
he walk up come PRF
He walked up towards me.
: The directional suffixes indicate "up" and "towards".
If the preceding verb has an object, the object may be placed either before or after the directional complement(s), or even between two directional complements, provided the second of these is not qù (去).
The structure with inserted de or bù is not normally used with this type of double verb. There are exceptions, such as "to be unable to get out of bed" (起不来床; 起不來床; qǐ bù lái chuáng or 起床不来; 起床不來; qǐ chuáng bù lái).
Coverbs
[edit]
Chinese has a class of words, called coverbs, which in some respects resemble both verbs and prepositions. They appear with a following object (or complement), and generally denote relationships that would be expressed by prepositions (or postpositions) in other languages. However, they are often considered to be lexically verbs, and some of them can also function as full verbs. When a coverb phrase appears in a sentence together with a main verb phrase, the result is essentially a type of serial verb construction. The coverb phrase, being an adverbial, precedes the main verb in most cases. For instance:
我
wǒ
I
帮
bāng
help
你
nǐ
you
找
zhǎo
find
他。
tā.
him
[我幫你找他。]
我 帮 你 找 他。
wǒ bāng nǐ zhǎo tā.
I help you find him
I will find him for you.
Here the main verb is zhǎo (找, "find"), and bāng (帮; 幫) is a coverb. Here bāng corresponds to the English preposition "for", even though in other contexts it might be used as a full verb meaning "help".
我
wǒ
I
坐
zuò
sit
飞机
fēijī
airplane
从
cóng
from
上海
Shànghǎi
Shanghai
到
dào
arrive(to)
北京
Běijīng
Beijing
去。
qù.
go
[我坐飛機從上海到北京去。]
我 坐 飞机 从 上海 到 北京 去。
wǒ zuò fēijī cóng Shànghǎi dào Běijīng qù.
I sit airplane from Shanghai arrive(to) Beijing go
I'll go from Shanghai to Beijing by plane.
Here there are three coverbs: zuò (坐 "by"), cóng (从; 從, "from"), and dào (到, "to"). The words zuò and dào can also be verbs, meaning "sit" and "arrive [at]" respectively. However, cóng is not normally used as a full verb.
A very common coverb that can also be used as a main verb is zài (在), as described in the Locative phrases section. Another example is gěi (给), which as a verb means "give". As a preposition, gěi may mean "for", or "to" when marking an indirect object or in certain other expressions.
我
wǒ
I
给
gěi
to
你
nǐ
you
打
dǎ
strike
电话。
diànhuà
telephone
[我給你打電話。]
我 给 你 打 电话。
wǒ gěi nǐ dǎ diànhuà
I to you strike telephone
I'll give you a telephone call
Because coverbs essentially function as prepositions, they can also be referred to simply as prepositions. In Chinese they are called jiè cí (介词; 介詞), a term which generally corresponds to "preposition", or more generally, "adposition". The situation is complicated somewhat by the fact that location markers—which also have meanings similar to those of certain English prepositions—are often called "postpositions".
Coverbs normally cannot take aspect markers, although some of them form fixed compounds together with such markers, such as gēnzhe (跟着; 跟著; 'with +[aspect marker]'), ànzhe (按着; 按著, "according to"), yánzhe (沿着; 沿著, "along"), and wèile (为了; 為了 "for").
Other cases
[edit]
Serial verb constructions can also consist of two consecutive verb phrases with parallel meaning, such as hē kāfēi kàn bào, "drink coffee and read the paper" (喝咖啡看报; 喝咖啡看報; 'drink coffee read paper'). Each verb may independently be negated or given the le aspect marker. If both verbs would have the same object, it is omitted the second time.
Consecutive verb phrases may also be used to indicate consecutive events. Use of the le aspect marker with the first verb may imply that this is the main verb of the sentence, the second verb phrase merely indicating the purpose. Use of this le with the second verb changes this emphasis, and may require a sentence-final le particle in addition. On the other hand, the progressive aspect marker zài (在) may be applied to the first verb, but not normally the second alone. The word qù (去, "go") or lái (来; 來, "come") may be inserted between the two verb phrases, meaning "in order to".
For constructions with consecutive verb phrases containing the same verb, see under Adverbs. For immediate repetition of a verb, see Reduplication and Aspects.
Another case is the causative or pivotal construction. Here the object of one verb also serves as the subject of the following verb. The first verb may be something like gěi (给, "allow", or "give" in other contexts), ràng (让; 讓, "let"), jiào (叫, "order" or "call") or shǐ (使, "make, compel"), qǐng (请; 請, "invite"), or lìng (令, "command"). Some of these cannot take an aspect marker such as le when used in this construction, like lìng, ràng, shǐ. Sentences of this type often parallel the equivalent English pattern, except that English may insert the infinitive marker "to". In the following example the construction is used twice:
他
tā
he
要
yào
want
我
wǒ
me
请
qǐng
invite
他
tā
him
喝
hē
drink
啤酒。
píjiǔ
beer
[他要我請他喝啤酒。]
他 要 我 请 他 喝 啤酒。
tā yào wǒ qǐng tā hē píjiǔ
he want me invite him drink beer
He wants me to treat him [to] beer.
Particles
[edit]
See also: Chinese particles and Chinese exclamative particles
Chinese has a number of sentence-final particles – these are weak syllables, spoken with neutral tone, and placed at the end of the sentence to which they refer. They are often called modal particles or yǔqì zhùcí (语气助词; 語氣助詞), as they serve chiefly to express grammatical mood, or how the sentence relates to reality and/or intent. They include:
ma (吗; 嗎), which changes a statement into a yes–no question
ne (呢), which expresses surprise, produces a question "with expectation", or expresses a currently ongoing event when answering a question
ba (吧), which serves as a tag question, e.g. "don't you think so?"; produces a suggestion e.g. "let's..."; or lessens certainty of a decision.
a (啊),[k] which reduces forcefulness, particularly of an order or question. It can also be used to add positive connotation to certain phrases or inject uncertainty when responding to a question.
ou (呕; 噢), which signals a friendly warning
zhe (着; 著), which marks the inchoative aspect, or need for change of state, in imperative sentences. Compare the imperfective aspect marker zhe in the section above)
le (了), which marks a "currently relevant state". This precedes any other sentence-final particles, and can combine with a (啊) to produce la (啦); and with ou (呕; 噢) to produce lou (喽; 囉).
This sentence-final le (了) should be distinguished from the verb suffix le (了) discussed in the Aspects section. Whereas the sentence-final particle is sometimes described as an inceptive or as a marker of perfect aspect, the verb suffix is described as a marker of perfective aspect. Some examples of its use:
我
wǒ
I
没
méi
no
钱
qián
money
了。
le
PRF
[我沒錢了。]
我 没 钱 了。
wǒ méi qián le
I no money PRF
I have no money now or I've gone broke.
我
wǒ
I
当
dāng
work
兵
bīng
soldier
了。
le
PRF
[我當兵了。]
我 当 兵 了。
wǒ dāng bīng le
I work soldier PRF
I have become a soldier.
The position of le in this example emphasizes his present status as a soldier, rather than the event of becoming. Compare with the post-verbal le example given in the Aspects section, wǒ dāng le bīng. However, when answering a question, the ending should be 呢 instead of 了. For example, to answer a question like "你现在做什么工作？" (What's your job now?), instead of using le, a more appropriate answer should be
我
wǒ
I
当
dāng
work
兵
bīng
soldier
呢。
ne
ongoing
[我當兵呢。]
我 当 兵 呢。
wǒ dāng bīng ne
I work soldier ongoing
I am being a soldier.
他
tā
He
看
kàn
watch
三
sān
three
场
chǎng
sports-CL
球赛
qiúsài
ballgames
了。
le
PRF
[他看三場球賽了。]
他 看 三 场 球赛 了。
tā kàn sān chǎng qiúsài le
He watch three sports-CL ballgames PRF
He [has] watched three ballgames.
Compared with the post-verbal le and guo examples, this places the focus on the number three, and does not specify whether he is going to continue watching more games.
The two uses of le may in fact be traced back to two entirely different words. The fact that they are now written the same way in Mandarin can cause ambiguity, particularly when the verb is not followed by an object. Consider the following sentence:
妈妈
māma
来
lái
了！
le
[媽媽來了！]
妈妈 来 了！
māma lái le
Mom come le
This le might be interpreted as either the suffixal perfective marker or the sentence-final perfect marker. In the former case it might mean "mother has come", as in she has just arrived at the door, while in the latter it might mean "mother is coming!", and the speaker wants to inform others of this fact. It is even possible for the two kinds of le to co-occur:
他
tā
He
吃
chī
eat
了
le
PFV
饭
fàn
food
了。
le
PRF
[他吃了飯了]。
他 吃 了 饭 了。
tā chī le fàn le
He eat PFV food PRF
He has eaten.
Without the first le, the sentence could again mean "he has eaten", or it could mean "he wants to eat now". Without the final le the sentence would be ungrammatical without appropriate context, as perfective le cannot appear in a semantically unbounded sentence.
Plural
[edit]
Chinese nouns and other parts of speech are not generally marked for number, meaning that plural forms are mostly the same as the singular. However, there is a plural marker men (们; 們), which has limited usage. It is used with personal pronouns, as in wǒmen (我们; 我們, "we" or "us"), derived from wǒ (我, "I, me"). It can be used with nouns representing humans, most commonly those with two syllables, like in péngyoumen (朋友们; 朋友們, "friends"), from péngyou (朋友, "friend"). Its use in such cases is optional. It is never used when the noun has indefinite reference, or when it is qualified by a numeral.
The demonstrative pronouns zhè (这; 這, "this"), and nà (那, "that") may be optionally pluralized by the addition of xiē (些，"few"), making zhèxiē (这些; 這些, "these") and nàxiē (那些, "those").
Cleft sentences
[edit]
There is a construction in Chinese known as the shì ... [de] construction, which produces what may be called cleft sentences. The copula shì (是) is placed before the element of the sentence which is to be emphasized, and the optional possessive particle de (的) is placed at the end of the sentence if the sentence ends in a verb, or after the last verb of the sentence if the sentence ends with a complement of the verb. For example:
他
tā
He
是
shì
shi
昨天
zuótiān
yesterday
来
lái
come
[的]。
[de]
[de].
[他是昨天來[的]。]
他 是 昨天 来 [的]。
tā shì zuótiān lái [de]
He shi yesterday come [de].
It was yesterday that he came.
Example with a sentence that ends with a complement:
他
tā
He
是
shì
shi
昨天
zuótiān
yesterday
[的]
[de]
[de]
food
[他是昨天買[的]菜。]
他 是 昨天 买 [的] 菜。
tā shì zuótiān mǎi [de] cài
He shi yesterday buy [de] food
It was yesterday that he bought food.
If an object following the verb is to be emphasized in this construction, the shì precedes the object, and the de comes after the verb and before the shì.
他
tā
He
昨天
zuótiān
yesterday
买
mǎi
buy
de
de
是
shì
shi
菜。
cài
[他昨天買的是菜。]
他 昨天 买 的 是 菜。
tā zuótiān mǎi de shì cài
He yesterday buy de shi vegetable.
What he bought yesterday was vegetable.
Sentences with similar meaning can be produced using relative clauses. These may be called pseudo-cleft sentences.
昨天
zuótiān
yesterday
是
shì
他
tā
he
买
mǎi
buy
cài
food
的
de
de
[昨天是他買菜的時間。]
昨天 是 他 买 菜 的 时间。
zuótiān shì tā mǎi cài de shíjiān
yesterday is he buy food de time
Yesterday was the time he bought food.
Conjunctions
[edit]
Chinese has various conjunctions (连词; 連詞; liáncí) such as hé (和, "and"), dànshì (但是, "but"), huòzhě (或者, "or"), etc. However Chinese quite often uses no conjunction where English would have "and".
Two or more nouns may be joined by the conjunctions hé (和, "and") or huò (或 "or"); for example dāo hé chā (刀和叉, "knife and fork"), gǒu huò māo (狗或貓, "dog or cat").
Certain adverbs are often used as correlative conjunctions, where correlating words appear in each of the linked clauses, such as búdàn ... érqiě (不但 ... 而且; 'not only ... (but) also'), suīrán ... háishì (虽然 ... 还是; 雖然...還是; 'although ... still'), yīnwèi ... suǒyǐ (因为 ... 所以; 因為...所以; 'because ... therefore'). Such connectors may appear at the start of a clause or before the verb phrase.
Similarly, words like jìrán (既然, "since/in response to"), rúguǒ (如果) or jiǎrú (假如) "if", zhǐyào (只要 "provided that") correlate with an adverb jiù (就, "then") or yě (也, "also") in the main clause, to form conditional sentences.
In some cases, the same word may be repeated when connecting items; these include yòu ... yòu ... (又...又..., "both ... and ..."), yībiān ... yībiān ... (一边...一边..., "... while ..."), and yuè ... yuè ... (越...越..., "the more ..., the more ...").
Conjunctions of time such as "when" may be translated with a construction that corresponds to something like "at the time (+relative clause)", where as usual, the Chinese relative clause comes before the noun ("time" in this case). For example:
当
dāng
我
wǒ
I
的
de
de
time
[當我回家的時候...]
当 我 回 家 的 时候...
dāng wǒ huí jiā de shíhòu...
At I return home de time
When I return[ed] home...
Variants include dāng ... yǐqián (当...以前; 當...以前 "before ...") and dāng ... yǐhòu (当...以后; 當...以後, "after ..."), which do not use the relative marker de. In all of these cases, the initial dāng may be replaced by zài (在), or may be omitted. There are also similar constructions for conditionals: rúguǒ /jiǎrú/zhǐyào ... dehuà (如果/假如/只要...的话, "if ... then"), where huà (话; 話) literally means "narrative, story".
See also
[edit]
Classical Chinese grammar
Cantonese grammar
China portal
Notes
[edit]
^ Several of the common prepositions can also be used as full verbs.
^ The first Chinese scholar to consider the concept of a word (词; 詞; cí) as opposed to the character (字; zì) is claimed to have been Shizhao Zhang in 1907. However, defining the word has proved difficult, and some linguists consider that the concept is not applicable to Chinese at all. See San, Duanmu (2000). The Phonology of Standard Chinese. Oxford University Press. ISBN 9780198299875.
^ A more common way to express this would be wǒ bǎ júzi pí bō le (我把橘子皮剥了; 我把橘子皮剝了, "I BA tangerine's skin peeled"), or wǒ bō le júzi pí (我剥了橘子皮; 我剝了橘子皮, "I peeled tangerine's skin").
^ 妳 is an alternative character for nǐ (你, "you") when referring to a female; it is used mainly in script written in traditional characters.
^ Also used after possessives and relative clauses
^ Not the same character as the de used to mark possessives and relative clauses.
^ This is a different character again from the two types of de previously mentioned.
^ This is similar to the English "by", though it is always followed by an agent.
^ Either the verb or the whole verb phrase may be repeated after the negator bù; it is also possible to place bù after the verb phrase and omit the repetition entirely.
^ Néng (能) does not mean "may" or "can" in the sense of "know how to" or "have the skill to".
^ alternately ya (呀), wa (哇), etc. depending on the preceding sound
References
[edit]
^ However, like 'dance', 舞 can also be used as a verb: for example, 「項莊舞劍」; "Xiang Zhuang danced with a sword"
^ Sun (2006), p. 50.
^ Melloni, Chiara; Basciano, Bianca (2018). "Reduplication across boundaries: The case of Mandarin". The Lexeme in Theoretical and Descriptive Morphology. 4: 331 – via OAPEN.
^ Sun (2006), p. 147.
^ Sun (2006), p. 184.
^ Sun (2006), p. 185.
^ Li (1990), p. 234 ff..
^ Sun (2006), p. 161.
^ Li & Thompson (1981), pp. 463–491.
^ Li (1990), p. 195.
^ Sun (2006), p. 159.
^ Jump up to: a b Sun (2006), p. 165.
^ Sun (2006), p. 188.
^ However, classifiers are not commonly used in Classical Chinese, for example 三人行 (sān-rén-xíng, literally "three-person-walk", means "three persons walk together", from Analects).
^ The following original meaning in Classical Chinese are referenced from Shuowen Jiezi, an old dictionary written during the Eastern Han dynasty.
^ ""怹"字的解释 | 汉典". www.zdic.net (in Chinese (China)). Retrieved 14 May 2023.
^ "汉语我们和咱们有区别吗？". Retrieved 2022-01-08.
^ ""这"字的解释 | 汉典". www.zdic.net (in Chinese (China)).
^ ""那"字的解释 | 汉典". www.zdic.net (in Chinese (China)).
^ Sun (2006), pp. 152, 160.
^ Sun (2006), p. 151.
^ Sun (2006), p. 154.
^ Sun (2006), p. 163.
^ Sun (2006), p. 203.
^ "Chapter 84: Order of Object, Oblique, and Verb". World Atlas of Language Structures. 2011.
^ Sun (2006), p. 81 ff.
^ Sun (2006), p. 85.
^ Sun (2006), p. 199.
^ Yip & Rimmington (2004), p. 107.
^ Li & Thompson (1981), p. 185.
^ Sun (2006), p. 70.
^ Yip & Rimmington (2004), p. 109.
^ Li & Thompson (1981), pp. 29, 234.
^ Sun (2006), p. 211.
^ Yip & Rimmington (2004), p. 110.
^ Sun (2006), pp. 209–211.
^ Sun (2006), p. 181.
^ Sun (2006), p. 52.
^ Sun (2006), p. 53.
^ Sun (2006), p. 208.
^ Sun (2006), p. 200.
^ Sun (2006), p. 205.
^ Sun (2006), p. 76 ff.
^ Li & Thompson (1981), quoted in Sun (2006), p. 80.
^ Li & Thompson (1981), pp. 296–300.
^ Chao (1968), p. 246.
^ Sun (2006), p. 80.
^ Sun (2006), p. 64.
^ Yip & Rimmington (2004), p. 8.
^ Sun (2006), p. 190.
^ Sun (2006), p. 191.
^ Yip & Rimmington (2004), p. 12.
^ Sun (2006), p. 197.
^ Sun (2006), p. 198.
Bibliography
[edit]
Chao, Yuen Ren (1968). A Grammar of Spoken Chinese. Berkeley: University of California Press. ISBN 978-0-520-00219-7.
Li, Charles N.; Thompson, Sandra A. (1981). Mandarin Chinese: A functional reference grammar. Berkeley: University of California Press. ISBN 978-0-520-06610-6.
Li, Yen-hui Audrey (1990). Order and Constituency in Mandarin Chinese. Springer. ISBN 978-0-792-30500-2.
Lin, Helen T. (1981). Essential Grammar for Modern Chinese. Cheng & Tsui. ISBN 978-0-917056-10-9.
Ross, Claudia; Ma, Jing-Heng Sheng (2006). Modern Mandarin Chinese Grammar: A Practical Guide. Routledge. ISBN 978-0-415-70009-2.
Sun, Chaofen (2006). Chinese: A Linguistic Introduction. Cambridge University Press. ISBN 978-0-521-82380-7.
Yip, Po-Ching; Rimmington, Don (2004). Chinese: A Comprehensive Grammar. Routledge. ISBN 0-415-15031-0.
Yip, Po-Ching; Rimmington, Don (2006). Chinese: An Essential Grammar (2nd ed.). Routledge. ISBN 978-0-203-96979-3.
Lü Shuxiang (吕叔湘) (1957). Zhongguo wenfa yaolüe 中国文法要略 [Summary of Chinese grammar]. Shangwu yinshuguan. OCLC 466418461.
Wang Li (1955). Zhongguo xiandai yufa 中国现代语法 [Modern Chinese grammar]. Zhonghua shuju.
Further reading
[edit]
W. Lobscheid (1864). Grammar of the Chinese language: in two parts, Volume 2. Office of Daily Press. p. 178. Retrieved 2011-07-06.
Joshua Marshman, Confucius (1814). Elements of Chinese grammar: with a preliminary dissertation on the characters, and the colloquial medium of the Chinese, and an appendix containing the Tahyoh of Confucius with a translation. Printed at the Mission press. p. 622. Retrieved 2011-07-06.
External links
[edit]
A Summary of Chinese Grammar
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7905 | https://pubmed.ncbi.nlm.nih.gov/40373315/ | Pharmacologic Interventions for Endometriosis-Related Pain: A Systematic Review and Meta-analysis - PubMed
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Meta-Analysis
Obstet Gynecol
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. 2025 May 15;146(2):e23-e35.
doi: 10.1097/AOG.0000000000005923.
Pharmacologic Interventions for Endometriosis-Related Pain: A Systematic Review and Meta-analysis
Liqiu Kou1,Changyou Huang,Di Xiao,Songjie Liao,Yaling Li,Qing Wang
Affiliations Expand
Affiliation
1 Department of Pharmacy, Zigong Maternal and Child Health Care Hospital, Zigong, and the Department of Pharmacy, The Affiliated Hospital of Southwest Medical University, and the School of Pharmacy, Southwest Medical University, Luzhou, Sichuan, China.
PMID: 40373315
DOI: 10.1097/AOG.0000000000005923
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Meta-Analysis
Pharmacologic Interventions for Endometriosis-Related Pain: A Systematic Review and Meta-analysis
Liqiu Kou et al. Obstet Gynecol.2025.
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. 2025 May 15;146(2):e23-e35.
doi: 10.1097/AOG.0000000000005923.
Authors
Liqiu Kou1,Changyou Huang,Di Xiao,Songjie Liao,Yaling Li,Qing Wang
Affiliation
1 Department of Pharmacy, Zigong Maternal and Child Health Care Hospital, Zigong, and the Department of Pharmacy, The Affiliated Hospital of Southwest Medical University, and the School of Pharmacy, Southwest Medical University, Luzhou, Sichuan, China.
PMID: 40373315
DOI: 10.1097/AOG.0000000000005923
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Abstract
Objective: To evaluate the effectiveness of various medications for the treatment of endometriosis-related pain through a network meta-analysis.
Data sources: A comprehensive search was conducted in PubMed, Embase, Web of Science, and the Cochrane Controlled Trials Register until July 22, 2024. We also searched ClinicalTrials.gov for additional data on recently completed trials or potentially eligible randomized controlled trials (RCTs) but found nothing.
Methods of study selection: The analysis included randomized RCTs that used pharmacologic interventions for managing endometriosis-related pain. The primary efficacy outcome was endometriosis-associated pelvic pain, which included dysmenorrhea, dyspareunia, and nonmenstrual pelvic pain. The analysis adhered to the PRISMA (Preferred Reporting Items for Systematic Reviews and Meta-analyses) guidelines.
Tabulation, integration, and results: A total of 31 RCTs involving 8,665 patients were included in the analysis. In terms of endometriosis-associated pelvic pain, four interventions demonstrated significantly greater efficacy compared with placebo: leuprolide combined with combined oral contraceptive pills (OCPs) (standardized mean difference [SMD] -1.40, 95% CI, -2.41 to -0.38), dienogest (SMD -1.20, 95% CI, -1.78 to -0.61), leuprolide alone (SMD -1.05, 95% CI, -1.64 to -0.45), and combined OCP (SMD -0.67, 95% CI, -1.25 to -0.09). Leuprolide combined with combined OCP emerged as the most effective treatment modality. In addition, elagolix and the combination of vitamin C and vitamin E were identified as the most effective interventions for dysmenorrhea and dyspareunia. For nonmenstrual pelvic pain, gestrinone demonstrated superior efficacy compared with placebo and all other interventions.
Conclusion: This network meta-analysis indicates that leuprolide in combination with combined OCP, elagolix, vitamins C and E, and gestrinone may represent the most effective therapeutic options for alleviating endometriosis-associated pelvic pain, dysmenorrhea, dyspareunia, and nonmenstrual pelvic pain. These findings enhance our understanding of the relative efficacy of treatment strategies for pain associated with endometriosis. Future research should focus on conducting larger-scale and rigorously designed clinical trials within the target patient populations to further validate these results.
Copyright © 2025 by the American College of Obstetricians and Gynecologists. Published by Wolters Kluwer Health, Inc. All rights reserved.
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Conflict of interest statement
Financial Disclosure The authors did not report any potential conflicts of interest.
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7906 | https://math.stackexchange.com/questions/1686728/can-i-write-sum-n-1n-xn-as-sum-n-1n-x-cdot-xn-1-and-then-apply-t | summation - Can I write $\sum_{n=1}^N x^n$ as $\sum_{n=1}^N x\cdot x^{n-1}$ and then apply the geometric series formula for sum of first $n$ terms with $a=x$? - Mathematics Stack Exchange
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Can I write ∑N n=1 x n∑n=1 N x n as ∑N n=1 x⋅x n−1∑n=1 N x⋅x n−1 and then apply the geometric series formula for sum of first n n terms with a=x a=x?
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If the answer is yes, then what about the fact that
∑n=1 N x n∑n=1 N x n
has a=1,r=x a=1,r=x, which implies a sum of 1−x N 1−x 1−x N 1−x whereas ∑N n=1 x⋅x n−1∑n=1 N x⋅x n−1 has a sum of
x(1−x N−1)1−x=x−x N 1−x x(1−x N−1)1−x=x−x N 1−x
The numerators are different but the denominators are same, so the two formulas have different sums?
I think I'm missing something dumb here, like a division or multiplication by zeros somewhere, or something.
Also, assume |x|<1|x|<1
Thanks.
Edit: My formulas in this post are incorrect. See comments below for correct ones, and explanation of why mine are incorrect.
summation
geometric-series
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edited Mar 7, 2016 at 17:08
majmunmajmun
asked Mar 7, 2016 at 7:48
majmunmajmun
1,554 1 1 gold badge 15 15 silver badges 33 33 bronze badges
3
2 The first term of the first series is x x, not 1 1.pancini –pancini 2016-03-07 07:51:36 +00:00 Commented Mar 7, 2016 at 7:51
1 As @Elliot G noted, the first one should be S N=x(1−x N)1−x S N=x(1−x N)1−x. Also, in the second one you have first term which is 1 1 and still there are n n terms, thus S N=x⋅1(1−x N)1−x=x(1−x N)1−x S N=x⋅1(1−x N)1−x=x(1−x N)1−x.Galc127 –Galc127 2016-03-07 07:59:07 +00:00 Commented Mar 7, 2016 at 7:59
All these sums are finite, so that the condtion x≠1 x≠1 suffices.user65203 –user65203 2016-03-07 08:08:56 +00:00 Commented Mar 7, 2016 at 8:08
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The basic formula is
∑n=0 N x n=1−x N+1 1−x,∑n=0 N x n=1−x N+1 1−x,
so that
∑n=1 N x n=1−x N+1 1−x−1∑n=1 N x n=1−x N+1 1−x−1
(first term missing) and
∑n=1 N x⋅x n−1=x∑n=0 N−1 x n=x 1−x N 1−x∑n=1 N x⋅x n−1=x∑n=0 N−1 x n=x 1−x N 1−x
(by shifting of the index).
You can verify that both expressions equal
x−x N+1 1−x.x−x N+1 1−x.
For instance,
3+9+27+81=3(1+3+9+27)=2−243 1−3.3+9+27+81=3(1+3+9+27)=2−243 1−3.
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edited Mar 7, 2016 at 8:11
answered Mar 7, 2016 at 8:05
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Let a n a n be geometric series and S n S n the sum of first n n terms, thus S n=a 1(1−q n)1−q S n=a 1(1−q n)1−q.
In your case we have
∑n=1 N x n=x+x 2+⋯+x N=x(1−x N)1−x∑n=1 N x n=x+x 2+⋯+x N=x(1−x N)1−x
where a 1=x a 1=x and there are n n terms.
∑n=1 N x n=x∑n=1 N x n−1=x(1+x+⋯+x N−1)=x⋅1(1−x N)1−x∑n=1 N x n=x∑n=1 N x n−1=x(1+x+⋯+x N−1)=x⋅1(1−x N)1−x
where a 1=1 a 1=1 and there are n n terms.
Easy to understand that both sums are equal
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answered Mar 7, 2016 at 8:13
Galc127Galc127
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WLOG, N=4 N=4.
S:=x+x 2+x 3+x 4=x(1+x+x 2+x 3)=x(1+S−x 4)=x S+x−x 5,S:=x+x 2+x 3+x 4=x(1+x+x 2+x 3)=x(1+S−x 4)=x S+x−x 5,
then
S=x−x 5 1−x.S=x−x 5 1−x.
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answered Mar 7, 2016 at 8:45
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7907 | https://arxiv.org/pdf/2406.01296 | arXiv:2406.01296v1 [math.NT] 3 Jun 2024
ALGEBRAIC RELATIONS BETWEEN SINE AND COSINE VALUES
by
B. Adamczewski & ´E. Delaygue
Abstract . — The aim of this note is to show that any algebraic relation over Q between the values of the trigonometric functions sine and cosine at algebraic points can be derived from the Pythagorean identity and the angle addition formulas. This result is obtained as a consequence of the Lindemann– Weierstrass theorem.
Introduction.
In their trigonometry course, all high school students learn the Pythagorean identity (1.1) cos 2 α + sin 2 α = 1 ,
as well as the angle addition formulas (1.2)
{
cos( α + β) = cos α cos β − sin α sin β ,
sin( α + β) = sin α cos β + cos α sin β .
These fundamental identities hold for all complex numbers α and β. Of course, they can be used recursively to produce various polynomial relations between the values of the trigonometric functions sine and cosine, such as cos(2 α) cos 2(α) + 4 sin(2 α) sin
( α
2
)
cos
( α
2
)
cos( α) − sin 2(α) cos(2 α) = 1 ,
2020 Mathematics Subject Classification . — 11J81, 11J85.
Key words and phrases . — Transcendence, algebraic independence, Lindemann– Weierstrass theorem, trigonometric functions, Siegel E-functions. This project has received funding from the ANR project De Rerum Natura (ANR-19-CE40-0018). 2 B. ADAMCZEWSKI & ´E. DELAYGUE
which is valid for all complex numbers α. The aim of this note is to study the following question:
Is it true that any polynomial relation with algebraic coefficients between sine and cosine values can be derived from the fundamental geometric identities
(1.1) and (1.2) ?
In general, the answer is no. Indeed, there are some sporadic relations such as sin( π) = 0 and cos 2 ( π
5
)
− 1
2 cos
( π
5
)
− 1
4 = 0 ,
that cannot be obtained in this way. The reason is that, if they could, they would remain valid by replacing the transcendental numbers π and π/ 5 respec-tively by any complex number α, which is obviously not the case. However, as the following result shows, the answer to our question turns out to be positive if we restrict ourselves to the values of sine and cosine at algebraic points . In the sequel, we fix an embedding of the field of algebraic numbers Q into C.
Vague Theorem . — Any algebraic relation over Q between the values of sine and cosine at algebraic points can be derived from the Pythagorean identity and the angle addition formulas.
Since the expression “can be derived from the Pythagorean identity and the angle addition formulas” is somewhat imprecise, our first task is to formalize our vague theorem into a precise one which expresses the very same idea (cf. Theorem 1). In order to do that, we introduce the Q-algebra T formed by all polynomial expressions in algebraic numbers and the values of sine and cosine at algebraic points, that is
T := Q[cos( α), sin( α) : α ∈ Q] ⊆ C .
Then we introduce the ring of polynomials with algebraic coefficients in count-ably many variables Xα, Yα, α ∈ Q:
A := Q[( Xα, Y α)α∈Q] .
We recall that each element of A is a polynomial with algebraic coefficients in only finitely many of the variables Xα and Yα. The ring A is also a Q-algebra. Let ev denote the evaluation map from A to T defined by ev( Xα) = cos( α) and ev( Yα) = sin( α) , α ∈ Q .
This defines a surjective (but non-injective) homomorphism of Q-algebras from
A to T . Let I denote the ideal of A spanned by the polynomials (1.3) X2
α
Y 2
α
− 1, Xα+β − XαXβ + YαYβ , and Yα+β − YαXβ − XαYβ ,
where α and β run along Q. The identities (1.1) and (1.2) imply that ev( I) =
{0}. Hence the map ev allows us to define a homomorphism of Q-algebras ev ALGEBRAIC RELATIONS BETWEEN SINE AND COSINE VALUES 3
from the quotient algebra A/I to T . Now, our vague theorem can be properly formalized as follows.
Theorem 1 . — The map ev : A/I → T is an isomorphism.
Although it is deduced quite directly from the famous Lindemann– Weierstrass theorem using some basic notions of commutative algebra, we were unable to find this result in the literature. We felt this statement was interesting enough for a broad audience of mathematicians to write this note. We will also draw a parallel with some of the deeper conjectures and results that permeate transcendental number theory at the end of this note.
Proof of Theorem 1. 2.1. Reflection formulas and angle difference identities.— Beyond the Pythagorean identity and the angle addition formulas, there are other classical algebraic relations between the values of sine and cosine that come to mind. For example, one can think of the initial values sin(0) = 0 and cos(0) = 1, the reflection formulas cos( −α) = cos( α) and sin( −α) = − sin( α), and the angle difference identities
{
cos( α − β) = cos( α) cos( β) + sin( α) sin( β) ,
sin( α − β) = sin( α) cos( β) − cos( α) sin( β) .
The following lemma shows that all of them belong to I.
Lemma 2 . — For every α and β in Q, the polynomials
Xα − X−α ,Yα + Y−α ,Xα−β − XαXβ − YαYβ ,Yα−β − YαXβ + XαYβ ,
as well as the polynomials Y0 and X0 − 1, all belong to I.Proof . — Throughout this proof, the congruences hold modulo I. We first infer from (1.3) with α = β = 0 that (2.4) X20 + Y 20 ≡ 1, X0 ≡ X20 − Y 20 and Y0 ≡ 2X0Y0 .
We obtain the system (2.5)
( X0 Y0
X0 −Y0
) ( X0
Y0
)
≡
( 1
X0
)
,4 B. ADAMCZEWSKI & ´E. DELAYGUE
with determinant −2X0Y0 ≡ − Y0. Left multiplying System (2.5) by its adju-gate matrix, we obtain that
−Y0
( X0
Y0
)
≡
( −Y0 −Y0
−X0 X0
) ( 1
X0
)
≡
( −Y0 − Y0X0
−X0 + X20
)
.
As a consequence, we obtain that Y0 ≡ 0. By (2.4), it follows that X20 ≡ 1and X0 ≡ X20 ≡ 1. Hence both Y0 and X0 − 1 belong to I.Let α ∈ Q. Using the angle addition formulas with β = −α, we obtain that 1 ≡ XαX−α − YαY−α and 0 ≡ YαX−α + XαY−α ,
that is ( X−α −Y−α
Y−α X−α
) ( Xα
Yα
)
≡
( 10
)
.
By the Pythagorean identity, this linear system is invertible and we have
( Xα
Yα
)
≡
( X−α Y−α
−Y−α X−α
) ( 10
)
≡
( X−α
−Y−α
)
,
which gives that Xα − X−α and Yα + Y−α belong to I. Then, using the angle addition formulas, we deduce that the polynomials
Xα−β − XαXβ − YαYβ and Yα−β − YαXβ + XαYβ , α, β ∈ Q ,
also belong to I, as expected.
2.2. The Lindemann–Weierstrass theorem and basic commutative algebra.—
One of the gems of transcendental number theory is the Lindemann-Weierstrass theorem.
Theorem 3 (Lindemann-Weierstrass) . — Let α1, . . . , α n be algebraic num-bers that are linearly independent over Q. Then the complex numbers
eα1 , . . . , e αn are algebraically independent over Q.
We will first deduce from the Lindemann–Weierstrass theorem the following result. Given a tuple of complex numbers E := ( ζ1, . . . , ζ m), we let (2.6) Alg Q(E) := {P (X1, . . . , X m) ∈ Q[X1, . . . , X m] : P (ζ1, . . . , ζ m) = 0 }
denote the ideal of algebraic relations over Q between the coordinates of E.
Corollary 4 . — Let α1, . . . , α n be Q-linearly independent algebraic numbers and set
E := (cos( α1), sin( α1), . . . , cos( αn), sin( αn)) .
Then {X2
i
Y 2
i
− 1 : 1 ≤ i ≤ n} is a set of generators of Alg Q(E), viewed as an ideal of Q[X1, Y 1, . . . , X n, Y n].ALGEBRAIC RELATIONS BETWEEN SINE AND COSINE VALUES 5
This corollary will probably come as no surprise to specialists and may seem obvious to them, but as this note is intended for a wide audience we have chosen to give a detailed proof. To this end, we begin by recalling some basic notions of commutative algebra, for which we refer the reader to [ 5,Chap. 8]. Let R be a unitary commutative ring. We say that a chain of prime ideals of R of the form p0 ( p1 ( . . . ( pn has length n. The Krull dimension of
R, denoted by dim R, is defined as the supremum of the lengths of all chains of prime ideals. For example, if K is a field, then dim K[X1, . . . , X n] = n.Given a prime ideal p of R, the height of p, denoted by ht( p), is defined as the supremum of the lengths of all chains of prime ideals contained in p.Let K be a field and L be a field extension of K. A subset S of L is said to be algebraically independent over K, if the elements of S do not satisfy any non-trivial polynomial equation with coefficients in K. The transcendence degree of L over K, denoted by tr .deg KL, is then defined as the maximal cardinality among all algebraically independent subsets S ⊆ L.Let A be a finitely generated integral K-algebra, say A := K[a1, . . . , a r]. Then, the Noether normalization lemma implies that (2.7) dim A = tr .deg KK(a1, . . . , a r ) ≤ r .
Furthermore, if p is a prime ideal of A then (2.8) dim A/ p = dim A − ht( p) .
Proof of Corollary 4 . — By Euler’s formula, we have
eiα = cos( α) + i sin( α) ,
for all complex numbers α. It follows that the complex numbers eiα 1 , . . . , e iα n
all belong to
Q[E] = Q[cos( α1), sin( α1), . . . , cos( αn), sin( αn)] .
Since the numbers iα 1, . . . , iα n are linearly independent over Q, the Lindemann–Weierstrass theorem implies that the numbers eiα 1 , . . . , e iα n are algebraically independent over Q. We deduce that (2.9) tr .deg QQ(E) ≥ n .
The Q-algebra Q[E] is isomorphic to Q[X1, Y 1, . . . , X n, Y n]/Alg Q(E). Since
Q[E] is a finitely generated integral Q-algebra, Equalities (2.7) and (2.8) yield degtr QQ(E) = dim Q[E]= dim Q[X1, Y 1, . . . , X n, Y n]/Alg Q(E)= 2 n − ht(Alg Q(E)) ,6 B. ADAMCZEWSKI & ´E. DELAYGUE
where the last equality holds because Alg Q(E) is a prime ideal. It follows from (2.9) that (2.10) ht(Alg Q(E)) ≤ n .
For every k, 1 ≤ k ≤ n, we let pk denote the ideal of Q[X1, Y 1, . . . , X n, Y n]spanned by the polynomials X2
i
Y 2
i
− 1, 1 ≤ i ≤ k. Thus, we want to prove that pn = Alg Q(E). By the Pythagorean identity (1.1), we have the ascending chain
{0} p1 · · · pn ⊆ Alg Q(E) .
By (2.10), if all these ideals are prime, we obtain that pn = Alg Q(E), as wanted. It thus remains to prove that pk is a prime ideal for all k ∈ { 1, . . . , n }.To that purpose, we first prove the following claim:
if R is an integral domain of characteristic 6 = 2 , then R[X, Y ]/(X2 + Y 2 − 1)
is also an integral domain.
Indeed, let P, Q ∈ R[X, Y ] be such that P Q ∈ (X2 + Y 2 − 1). Substituting
Y 2 by 1 − X2 in P and Q, we see that there exist polynomials P0, P 1, Q 0, Q 1 ∈
R[X] such that we have P ≡ P0+P1Y and Q ≡ Q0+Q1Y modulo ( X2+Y 2−1). It follows that 0 ≡ P Q ≡ P0Q0 + P1Q1(1 − X2) + ( P0Q1 + Q0P1)Y mod ( X2 + Y 2 − 1) ,
which yields P0Q0 + P1Q1(1 − X2) = P0Q1 + Q0P1 = 0. We thus obtain the linear system ( Q0 Q1(1 − X2)
Q1 Q0
) ( P0
P1
)
= 0 .
Hence either P0 = P1 = 0 and then P ∈ (X2 +Y 2 −1), or Q20 −Q21(1 −X2) = 0. In the latter case, we thus have
Q20 = Q21(1 − X2) ,
but, since the characteristic of R is not equal to 2, the valuation at (1 − X) is even on the left-hand side and odd on the right-hand side, unless Q0 = Q1 = 0. We deduce that Q0 = Q1 = 0 and hence Q ∈ (X2 + Y 2 − 1). This proves our claim. Let R be an integral domain and p be an ideal of R. Then the extended ideal pe in R[X1, . . . , X r] is defined as the ideal of R[X1, . . . , X r ] generated by the elements of p, that is the set of polynomials in the variables X1, . . . , X r
whose coefficients belong to p. This definition implies that (2.11) (R/ p) [ X1, . . . , X r ] ∼= R[X1, . . . , X r]/pe .
If p is prime, we have that R/ p is integral and hence ( R/ p)[ X1, . . . , X r] is integral too, and then by (2.11), pe is also prime. Furthermore, if p and q are ALGEBRAIC RELATIONS BETWEEN SINE AND COSINE VALUES 7
two ideals of R, then (2.12) (R/ p)/q ∼= R/ (p + q) ,
where q is the image of the ideal p + q by the natural projection from R to
R/ p(∗).Now, let qk denote the ideal of Rk := Q[X1, Y 1, . . . , X k , Y k] spanned by the polynomials X2
i
Y 2
i
− 1, 1 ≤ i ≤ k. Let us prove that these ideals are all prime by induction on k. Using the claim, we obtain that R1/q1 is an integral domain of characteristic zero and hence q1 is prime. Now, let k < n and let us assume that qk is prime. We infer from (2.11) and (2.12) that (Rk/qk )[ Xk+1 , Y k+1 ]/(X2
k+1
Y 2
k+1
− 1) ∼= ( Rk+1 /qek) /(X2
k+1
Y 2
k+1
− 1)
∼= Rk+1 /qk+1 .
Since qk is prime, we get that Rk/qk is an integral domain of characteristic zero, and the claim ensures that the left-hand side is an integral domain. We deduce that Rk+1 /qk+1 is an integral domain and hence qk+1 is prime. Thus, all the
qk are prime. Since pk is the extended ideal of qk in K[X1, Y 1, . . . , X n, Y n], it follows that all the pk are prime, as desired.
2.3. Proof of Theorem 1.— We are now ready to deduce our main result.
Proof of Theorem 1 . — The map ev is clearly surjective, so we only have to prove that ker(ev) = I. We have already observed that I ⊆ ker(ev), so now we just have to prove the reverse inclusion. Let P ∈ A be such that P ∈ ker(ev) and let us prove that P ∈ I.We define the support of a polynomial Q ∈ A = Q[Xα, Y α : α ∈ Q] as the smallest set S1 × S 2 ⊆ Q × Q such that Q can be written as a polynomial in the variables Xα, α ∈ S 1, and Yβ , β ∈ S 2, meaning that the support is the intersection of all such finite sets. We let S := {α1, . . . , α n} ⊆ Q be a finite set such that the support of P is included in S × S .Let k denote the dimension of the Q-vector space V generated by the el-ements of S. Reordering if necessary, we can assume without any loss of generality that α1, . . . , α k is a basis of V . Then there exists a positive integer
d such that all the elements of S belong to the Z-module spanned by the num-bers αi/d , 1 ≤ i ≤ k. By Lemma 2, we have the following relations modulo
I:
Xα+β ≡ XαXβ − YαYβ ,Yα+β ≡ YαXβ + XαYβ ,Xα−β ≡ XαXβ + YαYβ ,Yα−β ≡ YαXβ − XαYβ .
(∗)We recall that there is a one-to-one correspondence between the ideals of R/ pand the ideals of Rthat contain p.8B. ADAMCZEWSKI & ´E. DELAYGUE
These congruences show that, for every γ in S, there exist polynomials Pγ and
Qγ whose support is included in {α1/d, . . . , α k /d }2 and such that Xγ ≡ Pγ
and Yγ ≡ Qγ modulo I. It follows that there exists a polynomial U with support in {α1/d, . . . , α k/d }2 such that P ≡ U mod I. Since P ∈ ker(ev) and I ⊆ ker(ev), it follows that U ∈ ker(ev) too. Since the numbers αi/d ,1 ≤ i ≤ k, are Q-linearly independent, Corollary 4 implies that U belongs to the ideal spanned by the polynomials X2
αi/d
Y 2
αi/d
− 1, 1 ≤ i ≤ k. We deduce that U ∈ I and hence P ∈ I, as desired.
Finding explicit relations.
Theorem 1 provides the raison d’ˆ etre of the algebraic relations over Q be-tween the values of sine and cosine at algebraic points: they all come from the fundamental identities (1.1) and (1.2). We now show that, given algebraic numbers α1, . . . , α n, they can be used to explicitly find generators of the ideal of the algebraic relations Alg Q(E) defined in (2.6), where
E := (cos( α1), sin( α1), . . . , cos( αn), sin( αn)) .
More precisely, we describe an algorithm to compute a set of generators of the ideal Alg Q(E). Observe that Theorem 1 does not imply that the ideal spanned by the polynomials (1.3) with support in {α1, . . . , α n} is equal to Alg Q(E). However, as described below, only a finite number of polynomials of the form (1.3) are needed to span this ideal. Our input is a finite set of algebraic numbers α1, . . . , α n. We assume that algebraic numbers are given by their minimal integer polynomial and a ratio-nal approximation precise enough to separate it from the other roots of this polynomial.
First step.— The data concerning the αi’s provide an explicit bound on their naive height ( i.e. , the maximum of the absolute values of the coefficients of their minimal integer polynomial). We can thus use Siegel lemma [ 11 ] to com-pute an explicit integer M such that if there is a Q-linear relation between some of the αi’s, then one exists whose coefficients have height at most M . Then we can use the algorithm given in [ 8] to find a basis ( β1, . . . , β k ) ⊆ { α1, . . . , α n}
of the Q-vector space span Q(α1, . . . , α n), and, for each i ∈ { 1, . . . , n }, an ex-pression of αi as a Q-linear combination of the βj , 1 ≤ j ≤ k. Let d be the lcm of the denominator of the coefficients of these linear combinations, then we can explicitly find rational integers ai,j such that
αi = ai, 1
β1
d + · · · + ai,k
βk
d , 1 ≤ i ≤ n . ALGEBRAIC RELATIONS BETWEEN SINE AND COSINE VALUES 9
Second step.— Let us fix an integer i in {1, . . . , n }. Assume that ai, 1 > 0. Then writing αi = ( αi−β1/d )+ β1/d and using the angle addition formulas, one can write both cos( αi) and sin( αi) as polynomials in cos( αi − β1/d ), sin( αi −
β1/d ), cos( β1/d ), and sin( β1/d ). Using this procedure ai, 1 − 1 times, we end up with expressions of cos( αi) and sin( αi) as polynomials in cos( ai, 2 β1
d
· · · +
ai,k βk
d
), sin( ai, 2 β1
d
· · · + ai,k βk
d
), cos( β1/d ), and sin( β1/d ). If ai, 1 is negative, one just replaces the angle addition formulas by the angle difference formulas. Then one can use the same procedure with ai, 2 and so on. In the end, we can explicitly find polynomials Pi and Qi in Z[W1, Z 1, . . . , W k, Z k], 1 ≤ i ≤ n,such that, for every i ∈ { 1, . . . , n }, one has cos( αi) = Pi
(
cos
( β1
d
)
, sin
( β1
d
)
, . . . , cos
( βk
d
)
, sin
(βk
d
))
and sin( αi) = Qi
(
cos
( β1
d
)
, sin
( β1
d
)
, . . . , cos
( βk
d
)
, sin
( βk
d
))
.
Third step.— At this point, the polynomials
Pi, Q i ∈ Z[W1, Z 1, . . . , W k, Z k] , 1 ≤ i ≤ n ,
have been computed. Now consider the tuple F of complex numbers formed by the concatenation of E and (cos( β1/d ), sin( β1/d ), . . . , cos( βk/d ), sin( βk/d )) .
We claim that the ideal Alg Q(F) of Q[X1, Y 1, . . . , X n, Y n, W 1, Z 1, . . . , W k, Z k]is generated by the polynomials Xi − Pi, Yi − Qi, 1 ≤ i ≤ n, and W 2
j
Z2
j
− 1, 1 ≤ j ≤ k. Let p denote the ideal generated by these polynomials. Clearly,
p ⊆ Alg Q(F). Let P ∈ Alg Q(F). Since Xi ≡ Pi and Yi ≡ Qi modulo p, there exists a polynomial R ∈ Q[W1, Z 1, . . . , W k, Z k] such that P ≡ R mod p. Since
β1/d, . . . , β k /d are Q-linearly independent, Corollary 4 implies that R belongs to the ideal spanned by the polynomials W 2
j
Z2
j
− 1, 1 ≤ j ≤ k. It follows that R ∈ p and hence P ∈ p. We deduce that Alg Q(F) = p, as claimed. The polynomials Xi − Pi, Yi − Qi, 1 ≤ i ≤ n, and W 2
j
Z2
j
− 1, 1 ≤
j ≤ k form a set of generators of Alg Q(F), so we can consider a monomial order eliminating the variables Wj and Zj , 1 ≤ j ≤ k, and use Buchberger’s algorithm to find an associated Gr¨ obner basis G of Alg Q(F), and hence G ∩
Q[X1, Y 1, . . . , X n, Y n] is a Gr¨ obner basis of Alg Q(E) (see, for example, [ 3]). Then the output of our algorithm is simply G ∩ Q[X1, Y 1, . . . , X n, Y n], which provides an explicit set of generators of the ideal Alg Q(E), as wanted. 10 B. ADAMCZEWSKI & ´E. DELAYGUE
Concluding remarks.
In this final section, we place our results in a broader context, with a few references to more advanced work that motivated the writing of this note. Transcendental number theory has two main driving forces. Given a set of complex numbers Ξ, one wishes, on the one hand, to be able to determine, for all ξ1, . . . , ξ r ∈ Ξ, the algebraic relations over Q between these numbers, and, on the other, to find the raison d’ˆ etre for these putative relations. These two problems are naturally linked, and the second, more ambiguous, depends of course on how the elements of Ξ are defined. To make our point a little clearer, we begin by briefly recalling three famous conjectures. Solving the first two would answer this second problem in the case of the ring of periods, (†) while solving the third would answer it in the case of (normalized) values of the Euler Gamma function at rational points. In the case where Ξ denotes the ring of periods, a conjecture due to Kont-sevich and Zagier [ 9] predicts that any algebraic relation between periods can be derived from the fundamental rules of integral calculus–additivity, change of variables and the Stokes formula. We point out to the interested reader that this conjecture is essentially equivalent to the famous Grothendieck pe-riod conjecture, although the latter is expressed in terms too elaborate to be defined here. (‡) An in-depth discussion of these two conjectures and their links can be found in [ 7]. In the second case, we choose Ξ :=
{ Γ( r)
√2π : r ∈ Q
}
,
where Γ is the Euler gamma function. Then the Rohrlich–Lang conjecture predicts that all algebraic relations come from specializations of standard func-tional relations associated with Γ (see, for example, [ 12 , Conjecture 22]). In addition to sharing a common ambition, these three famous conjectures have in common the fact that they are considered to be totally beyond the reach of current methods. It is therefore remarkable that the two main objec-tives we just described have recently been achieved in a few cases. Namely, when Ξ is: (a) the set of values at algebraic points of Siegel E-functions (see [ 1, 6 ]); (b) the set of values at algebraic points of Mahler M -functions (see [ 1, 2 ]).
(†)We recall that a period is a complex number whose real and imaginary parts are values of absolutely convergent integrals of rational functions with rational coefficients, over domains in
Rngiven by polynomial inequalities with rational coefficients. Many classical mathematical constants such as the algebraic numbers, π, log 2, and ζ(3) are periods.
(‡)It involves a more geometric definition of periods associated with algebraic varieties, their (Betti) homology and de Rham cohomology, and their motivic Galois group. ALGEBRAIC RELATIONS BETWEEN SINE AND COSINE VALUES 11
A Siegel E-function is defined as a power series of the form
f (z) =
∞
∑
n=0
an
n! zn ,
where ( an)n≥0 is a sequence of algebraic numbers satisfying some arithmetic growth condition and such that f (z) satisfies a linear differential equation with coefficients in Q[z] (see [ 11 ] for a precise definition). These functions can be seen as generalizations of the exponential function, which corresponds to the case where ( an)n≥0 is the constant function equal to 1. The Siegel-Shidlovskii theorem (see [ 10 ]) is a fundamental result concerning the algebraic relations between values of E-functions at nonzero algebraic points that provides a broad generalization of the Lindemann-Weierstrass theorem. An important refinement of the Siegel-Shidlovskii theorem was obtained by Beukers in [ 4] and it was recently observed in [ 1] that it can be used to prove that all algebraic relations over Q between values of E-functions at nonzero algebraic points have a functional origin. More concretely, if f1(z), . . . , f r(z) are E-functions whose values at a nonzero algebraic point α satisfy a polynomial relation with algebraic coefficients, then this relation can be obtained as the specialization at α of a polynomial relation with coefficients in Q[z] between the functions
f1(z), . . . , f r(z) and a finite number of their successive derivatives. In this context, this result provides a satisfactory answer to the second problem, while it can also be used to solve the first one as also explained in [ 1]. Indeed, it allows us to prove that given a finite number of values (at nonzero algebraic points) of E-functions, there exists an algorithm to find an explicit basis of the ideal of the algebraic relations over Q between these numbers. The existence of such an algorithm was first proved in [ 6] in a slightly different way. A Mahler M -function is defined as a power series
f (z) = ∑
n≥0
anzn ,
where an ∈ Q and such that f (z) satisfies a linear difference equation with coefficients in Q[z] of the form
a0(z)f (z) + a1(z)f (zq ) + · · · + am(z)f (zqm
) = 0 ,
where q ≥ 2 is an integer. The results obtained in [ 1, 2 ] in the case of Mahler
M -functions are similar to those obtained in the case of E-functions, the role played by the derivative d
dz
being replaced by the Mahler operators σq : z 7 → zq .This note was inspired by these recent results. Indeed, Theorem 1 and the result in Section 3 solve the two problems in the case where Ξ ={cos( α), sin( α) : α ∈ Q}. Since the functions sine and cosine are E-functions, 12 B. ADAMCZEWSKI & ´E. DELAYGUE
this is a special case of (a), but one for which we can provide an elemen-tary proof and also a concrete and beautiful description of the algebraic re-lations. In the same vein, a similar result could be obtained for the set Ξ = {Jλ(α) : α ∈ Q, λ ∈ Q \ Z<0} using the theorem proved by Siegel in his famous memoir [ 11 ] concerning the algebraic independence of values at algebraic points of the Bessel functions
Jλ(z) =
∞
∑
n=0
(−1) n
n!Γ( n + λ + 1)
( z
2
)2n+λ
.
References
Adamczewski, B., Faverjon, C. (2023) Relations alg´ ebriques entre valeurs de E-fonctions ou de M -fonctions, preprint, arXiv:2303.05997 [math.NT], to appear in C. R. Math. Acad. Sci. Paris . Adamczewski, B., Faverjon, C. (2020). Mahler’s method in several variables and finite automata, preprint, arXiv :2012.08283 [math.NT], to appear in Ann. of Math. . Becker, T., Weispfenning, V. (1993). Gr¨ obner bases. A computational approach to commutative algebra, Graduate Texts in Mathematics 141, Springer-Verlag, New York. Beukers, F. (2006). A refined version of the Siegel–Shidlovskii theorem, Annals of Math. 163, 369–379. Bourbaki, N. (2006). ´El´ ements de math´ ematique. Alg` ebre commutative. Chapitres 8 et 9 , Springer, Berlin. Fischler, S., Rivoal, T. (2023). Effective algebraic independence of values of E-functions, Math. Z. 305, Paper No. 48, 17 pp. Fres´ an J. (2022). Une introduction aux p´ eriodes. P´ eriodes et transcendance. Journ´ ees math´ ematiques X-UPS 2019 , ´Editions de l’ ´Ecole polytechnique, Palaiseau. Just, B. (1990). Integer relations among algebraic numbers, Math. Comp. 54, 467– 477. Kontsevicth, M., Zagier, D. (2001). Periods, Mathematics unlimited–2001 and beyond , Springer-Verlag, Berlin, p. 771–808. Shidlovskii, A.B. (1989), Transcendental numbers, De Gruyter Studies in Math-ematics 12, Walter de Gruyter & Co., Berlin. Siegel, C. L. (1929). ¨ Uber einige Anwendungen diophantischer Approximationen,
Abh. Preuß. Akad. Wiss., Phys.-Math. Kl. 1, 1–70. Waldschmidt, M. (2006). Transcendence of periods: the state of the art, Pure Appl. Math. Q. 2, 435–463. ALGEBRAIC RELATIONS BETWEEN SINE AND COSINE VALUES 13
B. Adamczewski , Universit´ e Claude Bernard Lyon 1, CNRS, ´Ecole Centrale de Lyon, INSA Lyon, Universit´ e Jean Monnet, ICJ UMR5208, 69622 Villeurbanne, France.
E-mail : Boris.Adamczewski@math.cnrs.fr
´E. Delaygue , Universit´ e Claude Bernard Lyon 1, CNRS, ´Ecole Centrale de Lyon, INSA Lyon, Universit´ e Jean Monnet, ICJ UMR5208, 69622 Villeurbanne, France.
E-mail : delaygue@math.univ-lyon1.fr |
7908 | https://www.sciencedirect.com/science/article/pii/S0022391300831923 | Immediate lingual flange extension - ScienceDirect
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References (8)
The Journal of Prosthetic Dentistry
Volume 84, Issue 5, November 2000, Pages 583-584
Immediate lingual flange extension☆,☆☆,★,★★
Author links open overlay panel Yuuji Sato DDS, PhD a, Ryuji Hosokawa DDS, PhD b, Takayasu Kubo DDS, PhD c
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Abstract
J Prosthet Dent 2000;84:583-4.
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Section snippets
Procedure
Add soft wax to the underextended lingual flange, and shape it as desired (Fig. 1). Try-in the denture. Evaluate for proper extension and modify as necessary. If more precise extension is necessary, use stick modeling plastic impression compound instead of soft wax, and perform border molding.
Mix a silicon putty (Exafine, GC Co, Tokyo, Japan), and press it firmly around the extended portion, allowing it to set for 3 to 4 minutes (Fig. 2).
Remove the silicone putty core and soft wax from the
Discussion
The procedure described achieves immediate lingual flange extension for improved stability without sending the prosthesis to a laboratory. Placing the resin with a silicone putty core can produce a smooth surface for the resin. However, a smooth junction of the intaglio surface of the extended resin is difficult to achieve because of excess resin invasion. Therefore, this procedure is designed for provisional extension for tissue conditioning and relining.
To eliminate the irritation of mucosa
Summary
A procedure for lingual flange extension has been described. This simple chairside procedure achieves immediate recovery of stability for an underextended mandibular denture before tissue conditioning.
Recommended articles
References (8)
BJ Roberts
Mylohyoid ridge reductions as an aid to success in complete lower dentures
J Prosthet Dent (1977)
RA Fitzloff
Functional impressions with thermoplastic materials for reline procedures
J Prosthet Dent (1984)
JE Ward et al.
Effect of repair surface design, repair material, and process method on the transverse strength of repaired acrylic denture resin
J Prosthet Dent (1992)
Y Sato et al.
Immediate maxillary denture base extension for posterior palatal seal
J Prosthet Dent (2000)
There are more references available in the full text version of this article.
Cited by (0)
☆
a Associate Professor, Department of Removable Prosthodontics.
☆☆
b Assistant Professor, Department of Removable Prosthodontics.
★
c Assistant Professor, Department of Removable Prosthodontics.
★★
Reprint requests to: Dr Yuuji Sato, Department of Removable Prosthodontics, Hiroshima University Faculty of Dentistry, Kasumi 1-2-3, Minami-Ku, Hiroshima 734-8553, Japan, Fax: +(81)82-257-5679, E-mail: sato@hiroshima-u.ac.jp
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Copyright © 2000 Editorial Council of The Journal of Prosthetic Dentistry. Published by Mosby, Inc. All rights reserved.
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7910 | https://www.britannica.com/science/natural-selection | SUBSCRIBE
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Key People:
: Charles Darwin
: Alfred Russel Wallace
: Francisco J. Ayala
: Patrick Matthew
: H.W. Bates
Related Topics:
: survival of the fittest
: adaptation
: sexual selection
: group selection
: Darwinian fitness
On the Web:
: PNAS - Natural selection and phylogenetic analysis (July 18, 2025)
See all related content
natural selection, process that results in the adaptation of an organism to its environment by means of selectively reproducing changes in its genotype, or genetic constitution.
A brief treatment of natural selection follows. For full treatment, see evolution: The concept of natural selection.
In natural selection, those variations in the genotype (the entire complex of genes inherited from both parents) that increase an organism’s chances of survival and procreation are preserved and multiplied from generation to generation at the expense of less advantageous variations. Evolution often occurs as a consequence of this process. Natural selection may arise from differences in survival, in fertility, in rate of development, in mating success, or in any other aspect of the life cycle. All such differences result in natural selection to the extent that they affect the number of progeny an organism leaves.
More From Britannica
evolution: The concept of natural selection
Gene frequencies tend to remain constant from generation to generation when disturbing factors are not present. Factors that disturb the natural equilibrium of gene frequencies include mutation, migration (or gene flow), random genetic drift, and natural selection. A mutation is a spontaneous change in the gene frequency that takes place in a population and occurs at a low rate. Migration is a local change in gene frequency when an individual moves from one population to another and then interbreeds. Random genetic drift is a change that takes place from one generation to another by a process of pure chance. Mutation, migration, and genetic drift alter gene frequencies without regard to whether such changes increase or decrease the likelihood of an organism surviving and reproducing in its environment. They are all random processes.
Natural selection moderates the disorganizing effects of these processes because it multiplies the incidence of beneficial mutations over the generations and eliminates harmful ones, since their carriers leave few or no descendants. Natural selection enhances the preservation of a group of organisms that are best adjusted to the physical and biological conditions of their environment and may also result in their improvement in some cases. Some characteristics, such as the male peacock’s tail, actually decrease the individual organism’s chance of survival. To explain such anomalies, Darwin posed a theory of “sexual selection.” In contrast to features that result from natural selection, a structure produced by sexual selection results in an advantage in the competition for mates.
The Editors of Encyclopaedia BritannicaThis article was most recently revised and updated by John P. Rafferty.
evolution
Table of Contents
Introduction
General overview
The evidence for evolution
The fossil record
Structural similarities
Embryonic development and vestiges
Biogeography
Molecular biology
History of evolutionary theory
Early ideas
Charles Darwin
Modern conceptions
The Darwinian aftermath
The synthetic theory
Molecular biology and Earth sciences
The cultural impact of evolutionary theory
Scientific acceptance and extension to other disciplines
Religious criticism and acceptance
Intelligent design and its critics
The science of evolution
The process of evolution
Evolution as a genetic function
The concept of natural selection
Genetic variation in populations
The gene pool
Genetic variation and rate of evolution
Measuring gene variability
The origin of genetic variation: mutations
Gene mutations
Chromosomal mutations
Dynamics of genetic change
Genetic equilibrium: the Hardy-Weinberg law
Processes of gene-frequency change
Mutation
Gene flow
Genetic drift
The operation of natural selection in populations
Natural selection as a process of genetic change
Selection against one of the homozygotes
Overdominance
Frequency-dependent selection
Types of selection
Stabilizing selection
Directional selection
Diversifying selection
Sexual selection
Kin selection and reciprocal altruism
Species and speciation
The concept of species
The origin of species
Reproductive isolation
Ecological isolation
Temporal isolation
Ethological (behavioral) isolation
Mechanical isolation
Gametic isolation
Hybrid inviability
Hybrid sterility
Hybrid breakdown
A model of speciation
Geographic speciation
Adaptive radiation
Quantum speciation
Polyploidy
Genetic differentiation during speciation
Patterns and rates of species evolution
Evolution within a lineage and by lineage splitting
Convergent and parallel evolution
Gradual and punctuational evolution
Diversity and extinction
Evolution and development
Reconstruction of evolutionary history
DNA and protein as informational macromolecules
Evolutionary trees
Distance methods
Maximum parsimony methods
Maximum likelihood methods
Evaluation of evolutionary trees
Molecular evolution
Molecular phylogeny of genes
Multiplicity and rate heterogeneity
The molecular clock of evolution
The neutrality theory of molecular evolution
References & Edit History
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evolution
scientific theory
Also known as: descent
Written by
Francisco Jose Ayala
Francisco J. Ayala is a University Professor and Donald Bren Professor of Biological Sciences, School of Biological Sciences; Professor of Philosophy, School of Humanities; and Professor of Logic and the...
Francisco Jose Ayala
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Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree. They write new content and verify and edit content received from contributors.
The Editors of Encyclopaedia Britannica
Last Updated:
•
Article History
Key People:
: Hugo de Vries
: Robert Broom
: Charles Darwin
: Thomas Henry Huxley
: Charles Lyell
Related Topics:
: evolutionary tree
: catastrophe theory
: punctuated equilibrium model
: divergence
: molecular clock
On the Web:
: Khan Academy - Evolution and classification link (July 31, 2025)
See all related content
Top Questions
What is evolution in scientific terms?
Who proposed the theory of evolution by natural selection?
What is natural selection, and how does it work?
Why is the study of fossils important for understanding evolution?
How do variations within a species affect evolution?
What role do genetics play in the process of evolution?
News •
400-million-year-old fish exposes big mistake in how we understood evolution
• July 29, 2025, 5:23 AM ET (ScienceDaily)
Tired of put-downs, Tennessee town corrects the record with play about the Scopes trial it hosted
• July 18, 2025, 9:22 AM ET (AP)
evolution, theory in biology postulating that the various types of plants, animals, and other living things on Earth have their origin in other preexisting types and that the distinguishable differences are due to modifications in successive generations. The theory of evolution is one of the fundamental keystones of modern biological theory.
The diversity of the living world is staggering. More than 2 million existing species of organisms have been named and described; many more remain to be discovered—from 10 million to 30 million, according to some estimates. What is impressive is not just the numbers but also the incredible heterogeneity in size, shape, and way of life—from lowly bacteria, measuring less than a thousandth of a millimetre in diameter, to stately sequoias, rising 100 metres (300 feet) above the ground and weighing several thousand tons; from bacteria living in hot springs at temperatures near the boiling point of water to fungi and algae thriving on the ice masses of Antarctica and in saline pools at −23 °C (−9 °F); and from giant tube worms discovered living near hydrothermal vents on the dark ocean floor to spiders and larkspur plants existing on the slopes of Mount Everest more than 6,000 metres (19,700 feet) above sea level.
The virtually infinite variations on life are the fruit of the evolutionary process. All living creatures are related by descent from common ancestors. Humans and other mammals descend from shrewlike creatures that lived more than 150 million years ago; mammals, birds, reptiles, amphibians, and fishes share as ancestors aquatic worms that lived 600 million years ago; and all plants and animals derive from bacteria-like microorganisms that originated more than 3 billion years ago. Biological evolution is a process of descent with modification. Lineages of organisms change through generations; diversity arises because the lineages that descend from common ancestors diverge through time.
The 19th-century English naturalist Charles Darwin argued that organisms come about by evolution, and he provided a scientific explanation, essentially correct but incomplete, of how evolution occurs and why it is that organisms have features—such as wings, eyes, and kidneys—clearly structured to serve specific functions. Natural selection was the fundamental concept in his explanation. Natural selection occurs because individuals having more-useful traits, such as more-acute vision or swifter legs, survive better and produce more progeny than individuals with less-favourable traits. Genetics, a science born in the 20th century, reveals in detail how natural selection works and led to the development of the modern theory of evolution. Beginning in the 1960s, a related scientific discipline, molecular biology, enormously advanced knowledge of biological evolution and made it possible to investigate detailed problems that had seemed completely out of reach only a short time previously—for example, how similar the genes of humans and chimpanzees might be (they differ in about 1–2 percent of the units that make up the genes).
This article discusses evolution as it applies generally to living things. For a discussion of human evolution, see the article human evolution. For a more complete treatment of a discipline that has proved essential to the study of evolution, see the articles genetics, human and heredity. Specific aspects of evolution are discussed in the articles coloration and mimicry. Applications of evolutionary theory to plant and animal breeding are discussed in the articles plant breeding and animal breeding. An overview of the evolution of life as a major characteristic of Earth’s history is given in community ecology: Evolution of the biosphere. A detailed discussion of the life and thought of Charles Darwin is found in the article Darwin, Charles.
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General overview
The evidence for evolution
Darwin and other 19th-century biologists found compelling evidence for biological evolution in the comparative study of living organisms, in their geographic distribution, and in the fossil remains of extinct organisms. Since Darwin’s time, the evidence from these sources has become considerably stronger and more comprehensive, while biological disciplines that emerged more recently—genetics, biochemistry, physiology, ecology, animal behaviour (ethology), and especially molecular biology—have supplied powerful additional evidence and detailed confirmation. The amount of information about evolutionary history stored in the DNA and proteins of living things is virtually unlimited; scientists can reconstruct any detail of the evolutionary history of life by investing sufficient time and laboratory resources.
Evolutionists no longer are concerned with obtaining evidence to support the fact of evolution but rather are concerned with what sorts of knowledge can be obtained from different sources of evidence. The following sections identify the most productive of these sources and illustrate the types of information they have provided.
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PNAS - Natural selection and phylogenetic analysis
University of Utah - Learn.Genetics - How Natural Selection Works
Palomar College - Natural Selection
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7911 | https://askfilo.com/user-question-answers-maths/find-all-the-quadratic-residues-of-13-35383937393038 | Question asked by Filo student
Find all the quadratic residues of 13 .
Views: 5,035 students
Updated on: Oct 27, 2023
Text SolutionText solutionverified iconVerified
To find all the quadratic residues of 13, we can follow the following steps:
Step 1. Define the base set of numbers to find quadratic residues for
Since we are finding the quadratic residues of 13, we only need to consider numbers from 1 to (13-1)/2 = 6. This is because the results will start repeating after that point for bigger numbers. So, the base set of numbers is: {1, 2, 3, 4, 5, 6}.
Step 2. Calculate the squares of the base set numbers modulo 13
We will square each number from the base set and then take the result modulo 13. Using modular arithmetic, we get:
12≡1 (mod 13)22≡4 (mod 13)32≡9 (mod 13)42≡3 (mod 13)52≡12 (mod 13)62≡10 (mod 13)
Step 3. Write down the unique quadratic residues
After calculating the squares of the base set numbers modulo 13, list down these values and make sure not to count any duplicates. The unique quadratic residues are {1, 3, 4, 9, 10, 12}. Therefore, the quadratic residues of 13 are {1, 3, 4, 9, 10, 12}.
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| Question Text | Find all the quadratic residues of 13 . |
| Updated On | Oct 27, 2023 |
| Topic | All Topics |
| Subject | Maths |
| Class | Grade 12 |
| Answer Type | Text solution:1 |
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7912 | https://etd.auburn.edu/bitstream/handle/10415/2687/Guven_Yucturk_PhD_Dissertation.pdf;sequence=2 | Gregarious Path Decompositions of Some Graphs
by Guven Yuceturk A dissertation submitted to the Graduate Faculty of Auburn University in partial fulfillment of the requirements for the Degree of Doctor of Philosophy Auburn, Alabama August 6, 2011 Keywords: gregarious, path, decomposition Copyright 2011 by Guven Yuceturk Approved by Dean G. Hoffman, Chair, Professor of Mathematics & Statistics Curt Lindner, Professor of Mathematics & Statistics Chris Rodger, Professor of Mathematics & Statistics Peter Johnson, Professor of Mathematics & Statistics Abstract Let G be a simple graph and f (v) a positive integer for each vertex v of G. Form Gf
by replacing each v by a set F (v) of f (v) vertices, and each edge uv by complete bipartite graph on bipartition ( F (u), F (v)). Can we partition Gf into paths of length 2 which are gregarious, that is, meet three different F (u)’s? ii Acknowledgments I would first like to thank Dr. Dean G. Hoffman for his invaluable knowledge, guidance, and patience. I am indebted to him. I would also like to thank Dr. Chris Rodger and the rest of the Auburn faculty for guiding me throughout my graduate career. I would like to thank Dan Roberts for his insightful comments. Finally, I would like to thank my family and my friends for their love and support throughout my life. iii Table of Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii List of Abbreviations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 History of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Tutte’s f-Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1 Tutte’s f −Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Applying Tutte’s f-Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . 73 Parity Balanced Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.2 Bipartite Graphs with Four Degrees . . . . . . . . . . . . . . . . . . . . . . . 13 3.3 Parity balanced Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . 18 4 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.1 Complete Multipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.1.1 Complete Tripartite Graph . . . . . . . . . . . . . . . . . . . . . . . . 31 4.2 Star Multipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4.3 Cycle Multipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 4.3.1 Even Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 4.3.2 Odd Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 4.4 Path Multipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 4.5 Some Tree Multipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . 42 iv 4.5.1 Necessary And Sufficient Conditions for T (A1, A 2, A 3; B1, . . . , B n) . . 46 4.5.2 Necessary And Sufficient Conditions for T (C1, . . . , C m; A1, A 2; B1, . . . , B n) 48 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 vList of Figures 1.1 Example of Gregarious Path Decomposition . . . . . . . . . . . . . . . . . . . . 11.2 Example of Gf and G[h] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Example of an f-factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Example of an ¯f -factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Definition of λ(S, T ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.4 Example of a G-triple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.5 Component of B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.6 Tutte’s f-factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.7 Gf and G[h] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.8 G[h] and Lh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.9 LM (G) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.10 Example of how to apply Tutte’s f-Factor Theorem . . . . . . . . . . . . . . . . 10 2.11 Counter Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.1 Bipartite Graph with Four Degrees . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.2 Adding balanced distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 vi 3.3 Example of a parity balanced bipartite graph . . . . . . . . . . . . . . . . . . . 18 3.4 Example of a bipartite complement . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.5 Translating PBBG to a BGwFDs . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.6 Bipartite graphs with 2 edges . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.7 Exception for b = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.1 K(A, B, C ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.2 Multipartite Star S(A; B1, B 2, ..., B n) . . . . . . . . . . . . . . . . . . . . . . . . 33 4.3 Multipartite Even Cycle C2n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 4.4 Multipartite Odd Cycle C2n+1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 4.5 Multipartite Path Pn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 4.6 T (C1, . . . C m; A1, A 2; B1, . . . , B n) . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.7 T (A1, A 2, A 3; B1, . . . , B n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.8 Orientation of G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.9 An Example for Theorem 4.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 4.10 How to get conditions 2 - 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.11 How to get conditions 2 - 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 4.12 Types of paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 vii List of Tables 3.1 Distribution of the edges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.2 Exceptions for Theorem 3.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 4.1 Exceptions for Theorem 4.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 viii List of Abbreviations PBBG Parity Balanced Bipartite Graph BGwFD Bipartite Graph with Four Degrees L(G) Line Graph of G GDDs Group Divisible Designs NASCs Necessary and Sufficient Conditions TFAE The following are equivalent ix Chapter 1 Introduction Let G = ( V, E ) be a simple graph and f : V → N, where f (v) is a positive integer for each vertex v of G. Form the graph Gf by replacing each v by a set F (v) of f (v) vertices, and each edge uv by a complete bipartite graph on bipartition ( F (u), F (v)). Our question is: “Can we partition Gf into paths of length 2, P3, which are gregarious , that is, each vertex of P3 is in a different F (u)?”
Example 1.1. Let G = ( V, E ) be a graph where |V | = 5 and f : V (G) → P such that
f : ( v1, v 2, v 3, v 4) → (2 , 3, 2, 2). 2
v1
3
v2
2
v3
2
v4
G
GfGf
Figure 1.1: Example of Gregarious Path Decomposition Each color represents a different class of P3’s. 11.1 History of the problem Non-Gregarious Case:
In a recent paper , the complete solution is given for a decomposition of any complete multipartite graph into paths of lengths 3 and 4. Dr. Hoffman & Dr. Billington introduce the problem “what if we try to solve the same problem with gregarious paths?”
Previous work in Gregarious Decompositions:
In , Dean G. Hoffman & Elizabeth Billington give necessary and sufficient conditions to decompose a complete tripartite graph into gregarious 4-cycles. They use the notion of gregarious decompositions as “a cycle is said to be gregarious if its vertices occur in as many different parts of the multipartite graph as possible”. 2. In , Dean G. Hoffman & Elizabeth Billington give the necessary and sufficient con-ditions for gregarious 4-cycle decompositions of the complete equipartite graph Kn(m)
(with n > 4 parts of size m) whenever a 4-cycle decomposition (gregarious or not) is possible, and also of a complete multipartite graph in which all parts but one have the same size. 3. In , Benjamin R. Smith give necessary and sufficient conditions for the existence of a gregarious 5-cycle decomposition of the complete equipartite graph Km(n).4. In , Elizabeth J. Billington, Benjamin R. Smith & D.G. Hoffman give necessary and sufficient conditions for gregarious cycle decomposition of the complete equipartite graph Kn(m) (with n parts, n > 6 or n > 8, of size m) into both 6-cycles and 8-cycles. 5. In Jung R. Cho & Ronald J. Gould give necessary and sufficient conditions for the existence of decompositions of the complete multipartite graph Kn(2 t) into gregarious 6-cycles if n ≡ 0, 1, 3 or 4 (mod 6) . They used the method of a complete set of differences in Zn.26. In , Jung R. Cho gives another proof of the problem of decomposing the complete multipartite graph Kn(2 t) into gregarious 6-cycles for the case of n ≡ 0 or 3 (mod 6). 7. In , Benjamin R. Smith gives necessary and sufficient conditions for the existence of gregarious k− cycle decomposition of a complete equipartite graph, having n parts of size m, and either n ≡ 0, 1 (mod k), or k is odd and m ≡ 0 (mod k). 8. In , Elizabeth J. Billington , Dean G. Hoffman & Chris A. Rodger give necessary and sufficient conditions for decomposing a complete equipartite graph Kn(m) with n parts of size m into n-cycles in such a way that each cycle meets each part of Kn(m); that is, each cycle is said to be gregarious. Furthermore, they give gregarious decompositions which are also resolvable. 9. In , Saad I. El-Zanati, Narong Punnim & Chris A. Rodger give necessary and sufficent conditions for the existence of Gregarious GDDs with Two Associate Classes having block size 3.
Definition 1.2. Let G = ( V, E ) be simple graph and h : E → P. Define G[h] on vertex set
V as follows: if u, v ∈ V and uv = e ∈ E then put h(e) edges in between u and v in G[h].
Example 1.3. Let’s use the example 1.1 and define h(vivj ) := f (vi)f (vj ). 2
v1
3
v2
2
v3
2
v4
G
GfG[h]Gf
Figure 1.2: Example of Gf and G[h]
In G[h], any P3 will be a gregarious path of length two. 3Chapter 2 Tutte’s f-Factor Theorem
2.1 Tutte’s f −Factor Theorem Definition 2.1. Let G = ( E, V ) be a graph. G is called a k-regular graph if for every v ∈ V ,
deg (v) = k for some k ∈ N.
Definition 2.2. Let G = ( V, E ) be a graph, then 1. A factor of a graph G is a spanning subgraph of G.2. A k−factor is a spanning k−regular subgraph. 3. Given a function f : V (G) → Z, an f −factor of a graph G is a spanning subgraph H
such that dH (v) = f (v) for all v ∈ V .
Example 2.3. Let G = ( V, E ) be a graph with f : ( v1, v 2, v 3, v 4, v 5, v 6) → (3 , 1, 1, 2, 3, 2), then we can get the f − f actor :G
v1v2
v3v4
v5v6f−f actor
G
31
12
32
Figure 2.1: Example of an f-factor 4Definition 2.4. If f : V (G) → Z, define ¯f : V (G) → Z by ¯f (v) = deg G(v) − f (v).
Example 2.5. From the previous example ¯f : ( v1, v 2, v 3, v 4, v 5, v 6) → (0 , 2, 2, 1, 1, 2) f − f actor G
31
12
32
¯f − f actor
02
21
12
1
Figure 2.2: Example of an ¯f -factor
Definition 2.6. Let S, T ⊆ V (G), with S∩T = ∅, then λ(S, T ) = the set of edges with one end in S
and the other end in T .G
ST
λ(S, T )
Figure 2.3: Definition of λ(S, T )
Definition 2.7. B = ( S, T, U ) is a G − triple if 1. S, T, U ⊆ V (G), 2. S ∪ T ∪ U = V (G)53. S ∩ T = S ∩ U = T ∩ U = ∅S T
U
1
Figure 2.4: Example of a G-triple
Definition 2.8. Let G = ( V, E ) be a graph and f : V → P be a function. Then define f (S) = ∑
s∈S
f (s) for any S ⊆ V .
Definition 2.9. By a component of B, we mean a component of G \ (S ∪ T ). S T
U
c
Figure 2.5: Component of B
If c is a component of B, let J(B, f, c ) = f (c) + λ(c, T ). 2. c is called odd or even according to if J(B, f, c ) is odd or even. 3. k(B, f ) = # of odd components of B.6Theorem 2.10. G has an f − f actor , iff for each G − triple B = ( S, T, U )
k(B, f ) + λ(S, T ) 6 f (S) + ¯f (T )S T
U
1
Figure 2.6: Tutte’s f-factor Theorem
2.2 Applying Tutte’s f-Factor Theorem Definition 2.11. The line graph of a graph G, written L(G), is the graph whose vertices are the edges of G, with ef ∈ E(L(G)) whenever e and f are different edges of G having at least one vertex of G in common. Let G = ( V, E ) be a simple graph and f : V → N, where f (v) is a positive integer for each vertex v of G. Then, let h : E → P where h(vivj ) := f (vi)f (vj ). In this way we can get
Gf and G[h] with given G and f .7Gf
B1B2
B3
B4
⇒
G[h]
b1b2
b2b3
b3b4
b2b4Figure 2.7: Gf and G[h]
Now, if we have a gregarious decomposition of Gf into P3’s (denoted by Gf g
↪→ P3), then getting decomposition of G[h] into P3’s (denoted by G[h] ↪→ P3) is trivial. So Gf g
↪→ P3 ⇒ G[h] ↪→ P3.The opposite direction is not true, but it will still give us some of the necessary conditions for Gf g
↪→ P3. We can apply Tutte’s f-factor theorem to solve G[h] ↪→ P3.G[h]
B1B2
B3
B4
e12
e23
e34
e24
⇒Lh
e12 e23
e34 e24
x1
x2
x3
x4
x5
Figure 2.8: G[h] and LhIn G[h], assume that there are eij = bibj edges in between any pair of sets of vertices
Bi and Bj where |Bi| = bi for any i. Firstly, get the line graph of G, L(G), then blow the edges of L(G) in such a way that for each vertex eij ∈ V (L(G)), deg (eij ) is bibj in the new graph, Lh. In Lh, each edge will represent a P3 in G[h]. For example, in figure 2.8,
x1 represents the number of P3’s passing through sets B1 → B2 → B3 in G[h].8Therefore, if we find each xi, we can find the P3 decomposition of G[h]. To find the xi’s, we will use Tutte’s f-factor theorem. Start with G, get L(G), let M be a sufficiently large number, then get LM (G) by replacing every edge of L(G) with M edges (fig. 2.9). So if we find an h−factor of LM (G) such that h(eij ) = bibj , then we can get G[h] ↪→ P3.e12 e23
e34 e24
M
M
M
M
M
Figure 2.9: LM (G)
Theorem 2.12. LM (G) has an h − f actor for all sufficiently large M , iff for each
L(G) − triple B = ( S, T, U ) where T is independent and λ(T, U ) = 0 ,
k(B, h ) + h(T ) 6 h(S).
Proof. Let M be a sufficiently large number. From Tutte’s f-factor theorem, for all
L(G)-triples B = ( S, T, U ):
k(B, h ) + λLM (S, T ) 6 h(S) + ¯h(T )
k(B, h ) + M λ L(S, T ) 6 h(S) + M deg L(T ) − h(T )
k(B, h ) + h(T ) − h(S) 6 M (deg L(T ) − λL(S, T )) Here, deg L(T ) > λL(S, T ) for any L(G)−triple since deg L(T ) = ∑
t∈T
deg (t) and
λL(S, T ) = number of edges in between S & T . In addition, when deg L(T ) > λ L(S, T )the inequality holds since we can choose M sufficiently large and the left hand side of the 9inequality doesn’t depend on M . So the only problem is when deg L(T ) = λL(S, T ). Which means T is independent and there is no edge in between T & U (λL(T, U ) = 0). So the condition we need to check for each L(G)−triple reduces to:
k(B, h ) + h(T ) 6 h(S)when T is independent and λL(T, U ) = 0.
Example 2.13. Let G be the underlying graph in Figure 2.8. Let’s find the necessary conditions for G[h] to have a gregarious P3 decomposition. We need to check
k(B, h ) + h(T ) 6 h(S)for all L(G) triples B = ( S, T, U ) where T is independent and λL(T, U ) = 0. S TU
e23
e24
e12
e34 STU
e23
e24
e12
e34 STU
e23
e24
e34
e12
Figure 2.10: Example of how to apply Tutte’s f-Factor Theorem where deg (eij ) = bibj . So we can get conditions: 1. b1b2 + b3b4 6 b2b3 + b2b4
b1b2 6 b2b3 + b2b4
b3b4 6 b2b3 + b2b4
10 Therefore the necessary condition to decompose G[h] into gregarious P3’s is:
b1b2 + b3b4 6 b2b3 + b2b4
since 1 is stronger than 2 and 3. In summary, if we have a gregarious decomposition of Gf into P3’s ( Gf g
↪→ P3), then getting a decomposition of G[h] into P3’s is trivial ( G[h] ↪→ P3). So Gf g
↪→ P3 ⇒ G[h] ↪→ P3.The opposite direction is not true ( G[h] ↪→ P3 ; Gf g
↪→ P3, see the following counter example), but it will still give us some of the necessary conditions for Gf g
↪→ P3.
Example 2.14. Let G = S3 be a tristar and define f : V (G) → P by
f : ( a, v 1, v 2, v 3) → (2 , 1, 1, 1). S3GfG[h]
Figure 2.11: Counter Example
Gf doesn’t have a gregarious P3 decomposition since both vertices in the root have odd degree. On the other side, G[h] has a gregarious P3 decomposition, each color gives a different gregarious P3.11 Chapter 3 Parity Balanced Bipartite Graphs Let a, b ∈ P and e ∈ N , and let a, b ∈ { 0, 1}. We say the simple bipartite graph
G on bipartition ( A, B ), where |A| = a and |B| = b, with e edges, is parity balanced with parameters ( a, b, e, a, b) if
∀u ∈ A, deg (u) ≡ a (mod 2), and further ∀v ∈ A, |deg (u) − deg (v)| 6 2,
∀u ∈ B, deg (u) ≡ b (mod 2), and further ∀v ∈ B, |deg (u) − deg (v)| 6 2. We will give necessary and sufficient conditions on the parameters ( a, b, e, a, b) for the existence of such graphs.
3.1 Introduction
All the graphs are simple, i.e., they have no loops or multiple edges. Let P be the set of positive integers and N be the set of non-negative integers.
Definition 3.1. The integer vector ( x1, x 2, ... , x t) is said to be balanced if |xi − xj | 6 1for all 1 6 i, j 6 t. Two vectors are equivalent if one can be obtained from the other by permuting the entries.
Definition 3.2. Let G be a bipartite graph on bipartition ( A, B ). If for all v ∈ A, deg G(v) =
d1 and for all w ∈ B, deg G(w) = e1, then we will call G a ( d1, e 1) − regular bipartite graph. The following lemmas are proved in , p. 399.
Lemma 3.3. Let v and w be balanced vectors with the same number of coordinates. Then, for some vector w′ equivalent to w, v + w′ is balanced.
Lemma 3.4. Let a, b ∈ P, and let e 6 ab be a non-negative integer. Then there is a bipartite graph G on bipartition ( A, B ) with both ( deg G(v) | v ∈ A) and ( deg G(y) | y ∈ B) balanced. 12 3.2 Bipartite Graphs with Four Degrees
The theorems we will be proving here can be proven by using the Ryser-Gale theorem (, p. 185), but the proof is much harder.
Theorem 3.5. Let a1, a 2, b 1, b 2, d 1, d 2, e 1, e 2 be non-negative integers. Then: There is a simple bipartite graph on bipartition (A, B ), where A consists of a1 vertices of degree d1 and a2 vertices of degree d2, and B consists of b1 vertices of degree e1 and b2 vertices of degree e2, if and only if
(∗) a1d1 + a2d2 = b1e1 + b2e2
a1d1 6 a1b1 + b2e2, or, equivalently, b1e1 6 a1b1 + a2d2
a1d1 6 a1b2 + b1e1, or, equivalently, b2e2 6 a1b2 + a2d2
b1e1 6 a2b1 + a1d1, or, equivalently, a2d2 6 a2b1 + b2e2
b2e2 6 a2b2 + a1d1, or, equivalently, a2d2 6 a2b2 + b1e1
either a1 = 0, or d1 6 b1 + b2
either a2 = 0, or d2 6 b1 + b2
either b1 = 0, or e1 6 a1 + a2
either b2 = 0, or e2 6 a1 + a2
Necessity:
Proof. Each side of ( ∗) counts the total number of edges, hence they must be equal. Condi-tions 5 - 8 come from the fact that maximum degree of any vertex is less than the number of vertices in the other part. Now for conditions 1 - 4: 13 A
d1
A1
d2
A2
x
B
e1
B1
e2
B2Figure 3.1: Bipartite Graph with Four Degrees For i = 1 , 2, let Ai (resp. Bi) be the vertices of degree di (resp. ei) in Ai (resp. Bi). In addition, let’s assume that there are x edges in between vertices of A1 and B1. If we look at the other pairs, we will get:
B1 B2
A1 x a1d1 − xa2d2 − b1e1 + xA2 b1e1 − x =
b2e2 − a1d1 + x
Table 3.1: Distribution of the edges Using table 3.1, we can get the following inequalities: 0 6 x 6 a1b1
0 6 a1d1 − x 6 a1b2
0 6 b1e1 − x 6 a2b1
0 6 a2d2 − b1e1 + x 6 a2b2
14 So, we can get: 0 6 x 6 a1b1
a1d1 − a1b2 6 x 6 a1d1
b1e1 − a2b1 6 x 6 b1e1
b1e1 − a2d2 6 x 6 a2b2 − a2d2 + b1e1
We can get sixteen inequalities on the variables ( a1, a 2, b 1, b 2, d 1, d 2, e 1, e 2) from above since we have x in the middle of all of the four inequalities. If we use the left side of the first inequality and right sides of the all them, then we can get: 0 6 a1b1
0 6 a1d1
0 6 b1e1
0 6 a2b2 − a2d2 + b1e1 ⇒ a2d2 6 a2b2 + b1e1 , cond. 4 X
From the second one:
a1d1 − a1b2 6 a1b1 ⇒ d1 6 b1 + b2
a1d1 − a1b2 6 a1d1 ⇒ 0 6 a1b1
a1d1 − a1b2 6 b1e1 ⇒ a1d1 6 a1b2 + b1e1 , cond. 2 X
a1d1 − a1b2 6 a2b2 − a2d2 + b1e1 ⇒ a1d1 + a2d2 6 a2b2 + b1e1 + a1b2
If we use ( ∗), we will get a1d1 + a2d2 = b1e1 + b2e2 6 a2b2 + b1e1 + a1b2, so this reduces to
e1 6 a1 + a2.15 From the third one:
b1e1 − a2b1 6 a1b1 ⇒ e1 6 a1 + a2
b1e1 − a2b1 6 a1d1 ⇒ b1e1 6 a2b1 + a1d1 , cond. 3 X
b1e1 − a2b1 6 b1e1 ⇒ 0 6 a2b1
b1e1 − a2b1 6 a2b2 − a2d2 + b1e1 ⇒ d2 6 b1 + b2
From the fourth one:
b1e1 − a2d2 6 a1b1 ⇒ b1e1 6 a1b1 + a2d2 , cond. 1 X
b1e1 − a2d2 6 a1d1 ⇒ b1e1 6 a1d1 + a2d2
b1e1 − a2d2 6 b1e1 ⇒ 0 6 a2d2
b1e1 − a2d2 6 a2b2 − a2d2 + b1e1 ⇒ 0 6 a2b2
In the second equation, if we use a1d1 + a2d2 = b1e1 + b2e2, then we will get
b1e1 6 b1e1 + b2e2 ⇒ 0 6 b2e2.16 Sufficiency :
Proof. Assume there is bipartite graph ( A, B ) satisfying the necessary conditions and there are x edges in between the vertices of A1 and B1 like in figure 3.1. Therefore, we can find the number of edges in between other vertices using the remaining edges like in table 3.1. Now using the construction in on pg. 399, 1. distribute x edges on a1 vertices with balanced degrees. 2. distribute a1d1 − x on a1 vertices with balanced degrees. So we will get two balanced vectors with the same number of entries. Balanced Distrubution of x edges on a1 vertices
Balanced Distrubution of a1d1−xedges on a1vertices
1 less 1 more Equal
Balanced Balanced Balanced
Figure 3.2: Adding balanced distributions At the end, we will have one of these three cases from figure 3.2 and all of them will still have balanced distributions since both distributions were balanced to begin with. However, the first two cases are impossible, since we have a balanced distribution of a1d1 edges on a1
vertices, this means that each vertex will be incident with d1 edges. In the same manner we can prove that we can distribute the remaining edges with the desired degrees. 17 3.3 Parity balanced Bipartite Graphs Definition 3.6. Let a, b ∈ P and e ∈ N , and let a, b ∈ { 0, 1}. We say the bipartite graph
G with e edges on bipartition ( A, B ), with |A| = a and |B| = b, is parity balanced with parameters ( a, b, e, a, b) if
∀ u ∈ A, deg (u) ≡ a (mod 2) and further ∀ v ∈ A, |deg (u) − deg (v)| 6 2. 2. ∀ u ∈ B, deg (u) ≡ a (mod 2) and further ∀ v ∈ B, |deg (u) − deg (v)| 6 2.
Example 3.7. Let |A| = a = 6, |B| = b = 5, e = 14, a = 1 and b = 0. e = 14
A
deg = 1
a= 1
deg = 3
|A|=a= 6
B
deg = 2
b= 0
deg = 4
|B|=b= 5
Figure 3.3: Example of a parity balanced bipartite graph
Definition 3.8. If A and B are disjoint sets, we denote KA,B to be the complete bipartite graph on bipartition ( A, B ).
Definition 3.9. Let KA,B be a complete bipartite graph on bipartition ( A, B ). A bipartite complement of a bipartite graph G on bipartition ( A, B ) with edge set E is the bipartite graph G′ on bipartition ( A, B ) with the edge set E′ where E′ = E(KA,B )\E.
Fact 3.10. If G is a parity balanced bipartite graph with parameters ( a, b, e, a , b), then
G′ is a parity balanced bipartite graph with parameters ( a, b, e ′ = ab − e, ′
a
, ′
b
) where
a + ′
a
≡ b (mod 2) and b + ′
b
≡ a (mod 2) 18 Example 3.11. The bipartite complement of G is G′ with parameters ( a = 6 , b = 5 , e ′ =
ab − 14 = 30 − 14 = 16 , ′
a
= 0 , ′
b
= 0): e′ = 16
A
deg = 4
′
a= 0
deg = 2
|A|=a= 6
B
deg = 4
′
b= 0
deg = 2
|B|=b= 5
Figure 3.4: Example of a bipartite complement where a + ′
a
= 1 + 0 = 1 ≡ 5 (mod 2) and b + ′
b
= 0 + 0 ≡ 6 (mod 2).
Theorem 3.12. Let a, b ∈ P, e ∈ N, a, b, ′
a
, ′
b
∈ { 0, 1},
a + ′
a
≡ b (mod 2), b + ′
b
≡ a (mod 2). Then, there is a parity balanced bipartite graph G on bipartition ( A, B ) with parameters (a, b, e, a, b) if and only if
aa 6 e 6 ab − ′
a
a , bb 6 e 6 ab − ′
b
b, and all of these are congruent (mod 2), with the following exceptions: 19 e a b a ′
a
b ′
b
22 2 0 0 0 02 3 0 1 0 03 2 0 0 0 1odd > 3 odd > 3 0 1 0 1odd > 3 even > 3 0 0 0 10 0 1 0even > 3 odd > 3 0 1 0 01 0 0 0even > 3 even > 3 1 1 1 10 0 0 0
ab −22 2 0 0 0 02 3 1 0 0 03 2 0 0 1 0odd > 3 odd > 3 1 0 1 0odd > 3 even > 3 0 0 1 00 0 0 1even > 3 odd > 3 1 0 0 00 1 0 0even > 3 even > 3 1 1 1 10 0 0 0Table 3.2: Exceptions for Theorem 3.12 20 Necessity:
Proof. For any u ∈ A we have deg G(u) + deg G′ (u) = b so deg G(u) + deg G′ (u) ≡ b (mod 2) where deg G(u) ≡ a (mod 2) and deg G′ (u) ≡ ′
a
(mod 2) by definition, and a + ′
a
≡ b
(mod 2) follows. In the same way, we can get b + ′
b
≡ a (mod 2). To get aa 6 e 6 ab − ′
a
a and bb 6 e 6 ab − ′
b
b, if one of a, ′
a
, b, ′
b
is 1, then we have to have enough edges in either G or G′.Finally, to get aa, e, ab − ′
a
a, bb, ab − ′
b
b all congruent (mod 2);
e = ∑
u∈A
deg G(u) ≡ a a (mod 2),and
a + ′
a
≡ b (mod 2) ⇒ a a + a ′
a
≡ ab (mod 2) ⇒ a a ≡ ab − a ′
a
(mod 2). In the same way we can get the other conditions. For the exceptions, it is easy to prove that there is no parity balanced bipartite graph with parameters given in table 4.1. Figure 3.6 shows all possible parity balanced bipartite graphs with 2 edges and and the ones with
ab − 2 edges will be bipartite complement of these graphs.
Sufficiency:
Proof. We can use theorem 3.5 for this proof. Define n, m, q a, r a, q b, r b ∈ N by,
e = 2 n + aa = 2 m + bbn = aq a + ra, m = bq b + rb
0 6 ra 6 a − 1, 0 6 rb 6 b − 1. So e = 2 aq a + 2 ra + aa = 2 bq b + 2 rb + bb.
qa = e − aa − 2ra
2a = e − aa
2a − ra
a =
⌊e − aa
2a
⌋
. In the same way, qb =
⌊e − bb
2b
⌋
.Now let’s translate this problem into“bipartite graphs with four degrees” since we already know NASCs for those graphs (figure 3.5). 21 A
a1 = a − ra
d1 = 2 qa + a
a2 = ra
d2 = 2 qa + a + 2
B
b1 = b − rb
e1 = 2 qb + b
b2 = rb
e2 = 2 qb + b + 2 Figure 3.5: Translating PBBG to a BGwFDs Note that ( ∗) holds since: (a − ra)(2 qa + a) + ra(2 qa + a) = ( b − rb)(2 qb + b) + rb(2 qb + b)
Case 1 : Assume a2 = 0 = b2, then we will get a ( d1, e 1)−regular bipartite graph . Since
a2 = 0, and b2 = 0, we only need to prove 1, 5 and 7 in theorem 3.5. Let’s start proving conditions 5 and 7 which say:
d1 6 b1 + b2
2qa + a 6 b − rb + rb = b
and
e1 6 a1 + a2
2qb + b 6 a − ra + ra = a
So for 5 if we prove 2 qa + a 6 b, we are done. Using 2 qa = e − aa − 2ra
a = ea − a − 2ra
a ;we can get 2 qa + a = ea − a − 2ra
a + a = ea − 2ra
a 6 b − 2ra
a < b . We can prove 7 in the same way. 22 Now, let’s prove 1:
a1d1 6 a1b1 + b2e2
a1d1 6 a1b1 since b2 = 0
d1 6 b1
2qa + a 6 b and we just proved this in 5
Case 2: Assume a2 = 0 and b2 6 = 0 ( a2 6 = 0 and b2 = 0 is just the symmetric case). Since
a2 = 0, we only need to prove 1, 2, 5, 7 and 8 in theorem 3.5. 5 and 7 are the same as in Case 1. 1 and 2 will reduce to 7 and 8, respectively since a2 = 0. So just proving 8 is enough which says:
e2 6 a1 + a2
2qb + b + 2 6 a1 = a since a2 = 0 So 2 qb = eb − b − 2rb
b , using this: 2qb + b + 2 = eb − b − 2rb
b + b + 2 = eb − 2rb
b + 2 6 a − ′
b
− 2rb
b + 2 < a + 2. The only problem is when a = 2 qb + b + 1, then b + ′
b
≡ a = 2 qb + b + 1 (mod 2). So ′
b
= 1. In this case: 2qb + b + 2 = eb − b − 2rb
b + b + 2 = eb − 2rb
b + 2 6 a − ′
b
− 2rb
b + 2 < a + 1.
Case 3: We can assume a2 6 = 0 6 = b2. Let’s start proving conditions 5 through 8 in theorem 3.5. So 5 and 6 say:
d1 6 b1 + b2
2qa + a 6 b − rb + rb = b
23 and
d2 6 b1 + b2
2qa + a + 2 6 b − rb + rb = b
If we prove 2 qa + a + 2 6 b, this will cover both cases. However, this is the same as (8) in Case 2, just switch a and b. We can prove (7) and (8) in the same way. Now let’s prove conditions 1 through 4 of theorem 3.5. For 1, we need to prove:
a1d1 6 a1b1 + b2e2
(a − ra)(2 qa + a) 6 (a − ra)( b − rb) + rb(2 qb + b + 2) First, suppose rb 6 b − 2qa − a, then we get: (a − ra)(2 qa + a) 6 (a − ra)( b − rb) + rb(2 qb + b + 2) (a − ra)(2 qa + a − b + rb) 6 rb(2 qb + b + 2) (a − ra)( rb − (b − 2qa − a)) 6 rb(2 qb + b + 2) So rb − (b − 2qa − a) 6 0 and the inequality is automatically satisfied since a − ra > 0, r b >
0, 2qb + b + 2 > 0. So we can assume rb > b−2qa −a +1. In addition, recall that e = 2 n+aa = 2 aq a +2 ra +aa.24 Need to prove:
a1d1 6 a1b1 + b2e2
(a − ra)(2 qa + a) 6 (a − ra)( b − rb) + rb(2 qb + b + 2) 2aq a + a a − ra(2 qa + a) 6 ab − ar b − br a + rarb + rb(2 qb + b + 2) 2aq a + a a − ra(2 qa + a) + 2 ra − 2ra 6 ab − ar b − br a + rarb + rb(2 qb + b + 2) (2 aq a + a a + 2 ra) − ra(2 qa + a + 2) 6 ab − ar b − br a + rarb + rb(2 qb + b + 2)
e + ra(b − (2 qa + a + 2)) + rb(a − (2 qb + b + 2)) − rarb 6 ab
So if we show e + ra(b − (2 qa + a + 2)) + rb(a − (2 qb + b + 2)) − rarb 6 ab , we are done. Using rb > b − 2qa − a + 1, we can get ra(b − (2 qa + a + 2)) < r arb. So
e + ra(b − (2 qa + a + 2)) + rb(a − (2 qb + b + 2)) − rarb < e + rarb + rb(a − (2 qb + b + 2)) − rarb
= e + rb(a − (2 qb + b + 2)) where we can use, e = ( b − rb)(2 qb + b) + rb(2 qb + b + 2). = e + rb(a − (2 qb + b + 2)) = (b − rb)(2 qb + b) + rb(2 qb + b + 2) + rb(a − (2 qb + b + 2)) = (b − rb)(2 qb + b) + ar b
here using the fact that 2 qb + b 6 a (which we just proved in 7), we will get: = ( b − rb)(2 qb + b) + ar b
< (b − rb)a + ar b = ab.
25 Now let’s prove 2:
a1d1 6 a1b2 + b1e1
(a − ra)(2 qa + a) 6 (a − ra)rb + ( b − rb)(2 qb + b)First suppose 2 qa + a 6 rb, then we get: (a − ra)(2 qa + a) 6 (a − ra)rb + ( b − rb)(2 qb + b)(a − ra)(2 qa + a − rb) 6 (b − rb)(2 qb + b).
So 2 qa + a − rb 6 0 and the inequality is automatically satisfied since a − ra > 0, (b − rb) >
0, 2qb + b ≥ 0. We can assume 2 qa + a + 1 > rb. We need to prove:
a1d1 6 a1b2 + b1e1
(a − ra)(2 qa + a) 6 (a − ra)rb + ( b − rb)(2 qb + b)2aq a + a a − ra(2 qa + a) 6 ar b − rarb + 2 bq b + b b − rb(2 qb + b)2aq a + a a − ra(2 qa + a) + 2 ra − 2ra 6 ar b − rarb + 2 bq b + b b − rb(2 qb + b) + 2 rb − 2rb
(2 aq a + a a + 2 ra) − ra(2 qa + a + 2) 6 ar b − rarb + (2 bq b + b b + 2 rb) − rb(2 qb + b + 2)
e − ra(2 qa + a + 2) 6 ar b − rarb + e − rb(2 qb + b + 2)
rarb + rb(2 qb + b + 2) 6 ar b + ra(2 qa + a + 2) If we show rarb + rb(2 qb + b + 2) 6 ar b + ra(2 qa + a + 2), then we have shown (2). Using 2qa + a + 1 > rb we can get rarb < r a(2 qa + a + 2). In addition, we can use the previously 26 proved fact in 8 that 2 qb + b + 2 6 a. So
rarb + rb(2 qb + b + 2) 6 ra(2 qa + a + 2) + rb(2 qb + b + 2)
6 ra(2 qa + a + 2) + rba
= ar b + ra(2 qa + a + 2) The proof of 3 is exactly the same as the proof of 2, if we switch parts A ←→ B.Now, let’s prove the last condition, 4. We need to prove b2e2 6 a2b2 + a1d1, or, equivalently, a2d2 ≤ a2b2 + b1e1.
b2e2 6 a2b2 + a1d1
rb(2 qb + b + 2) 6 rarb + ( a − ra)(2 qa + a)
rb(2 qb + b + 2 − ra) 6 (a − ra)(2 qa + a)which is equivalent to
a2d2 6 a2b2 + b1e1
ra(2 qa + a + 2) 6 rarb + ( b − rb)(2 qb + b)
ra(2 qa + a + 2 − rb) 6 (b − rb)(2 qb + b).
First, assume 2 qb + b + 2 6 ra or 2 qa + a + 2 6 rb.Then b2e2 6 a2b2 + a1d1 or a2d2 ≤ a2b2 + b1e1 will be automatically satisfied. So we can assume 2 qb + b + 2 > r a and 2 qa + a + 2 > r b ⇒ 2qa + a + 1 > rb.If we turn back to the problem and use the fact, which follows from 8, that 2 qb + b + 2 6 a,27 then:
rb(2 qb + b + 2) 6 rba
= rarb + ( a − ra)rb
6 rarb + ( a − ra)(2 qa + a + 1) So the only problem is when rb = 2 qa + a + 1. Similarly, we can assume ra = 2 qb + b + 1. Need to show:
b2e2 6 a2b2 + a1d1
rb(2 qb + b + 2) 6 rarb + ( a − ra)(2 qa + a)(2 qa + a + 1)(2 qb + b + 2) 6 (2 qa + a + 1)(2 qb + b + 1) + ( a − ra)(2 qa + a)2qa + a + 1 6 (a − ra)(2 qa + a)1 6 (a − ra − 1)(2 qa + a)1 6 (a − ra − 1)( rb − 1) Here rb > 1 since ra 6 = 0 6 = rb. In this case both are positive and we can assume a − ra > 1since 0 6 ra < a . Therefore, we only need to prove rb 6 = 1 or a − ra 6 = 1. First, suppose
rb = 1, then rb = 2 qa + a + 1 = 1 so qa = 0 and a = 0.
e = 2 aq a + 2 ra + a a = 2bq b + 2 rb + b b
2ra = 2bq b + 2 + b b
2(2 qb + b + 1) = b(2 qb + b) + 2 2(2 qb + b) = b(2 qb + b)0 = (b − 2)(2 qb + b)0 = (b − 2)( ra − 1) 28 Therefore, in this case, either b = 2 or ra = 1. If ra = 1, then e = 2 ra = 2 and this is not possible since when e = 2 there are only two bipartite graphs with 2 edges (see figure 3.6) and both of them have either b2 = 0 or a2 = 0 = b2, which contradicts the assumption
a2 6 = 0 6 = b2.We can get the exceptions in table 4.1 with parameters ( a, b, e = 2 , a, ′
a
, b, ′
b
) easily since no other bipartite graphs exist with 2 edges but the ones in figure 3.6. A BA B
Figure 3.6: Bipartite graphs with 2 edges Now, assume b = 2 where rb = 1 and ra > 2. BA
d2= 2
d1= 0
Figure 3.7: Exception for b = 2 There are only two vertices in B and d2 = 2 which means every vertex in a2 will be adjacent to the vertices in B. This implies b1 = 2, and b2 = 0, and contradicts the assumption b2 6 = 0. So we finished proving the case where rb 6 = 1. Therefore we can assume rb ≥ 2. 29 Now assume a = ra + 1: If ra = 1, then a = 2 and it will be the same case as b = 2. Assume ra > 2, so a > 3 where
ra = 2 qb + b + 1 = a − 1.
e = 2bq b + 2 rb + b b
= b(2 qb + b) + 2 rb
= b(a − 2) + 2 rb
= ab − 2( b − rb)On the other side,
e = 2 bq b + 2 rb + b b = 2aq a + 2 ra + a a
(b − rb)(2 qb + b) + rb(2 qb + b + 2) = 2aq a + 2( a − 1) + a a
(b − rb)( a − 2) + ar b = a(2 qa + a + 2) − 2(b − rb)( a − 2) + ar b = a(rb + 1) − 2(b − rb)( a − 2) + ar b = ar b + a − 2(b − rb)( a − 2) = a − 2(b − rb − 1)( a − 2) = 0We know a > 3, b = rb + 1, which means e = ab − 2( b − rb) = ab − 2, which is the bipartite complement of the exception e = 2. So we proved rb 6 = 1 or a − ra 6 = 1. This completes the proof. 30 Chapter 4 Results
4.1 Complete Multipartite Graphs 4.1.1 Complete Tripartite Graph Theorem 4.1. For a complete tripartite graph K(A, B, C ) with |A| = a, |B| = b and |C| = c,assume a > b > c, then the NASCs are: 1. 2 | (ab + ac + bc )
ab 6 ac + bc A
BC
Figure 4.1: K(A, B, C )
Necessity:
Proof. 2 | (ab + ac + bc ) comes from the fact that the total number of edges must be divisible by 2 since there are two edges in P3. For ab 6 ac + bc :We have three kinds of paths, let x, y, z be the number of the paths C → A → B,
A → B → C and A → C → B respectively, then 31 x + y = ac x + y = ab y + z = bc
So
x = 12(ac + ab − bc ) ⇒ bc 6 ac + ab y = 12(ab + bc − ac ) ⇒ ac 6 ab + bc z = 12(ac + bc − ab ) ⇒ ab 6 ac + cb
So if we have ab 6 ac + cb , the other two follow easily since a > b > c.
Sufficiency:
Proof. Let A, B, C be sets of size a, b, c respectively. Assume the necessary conditions are satisfied, then we can find proper x, y, z . Find subgraphs S1 of K(C, A ) and S2 of K(A, B )with x edges, as in Lemma 3.4, so that their degrees agree on A (thus S1 ∪ S2 is a union of x gregarious paths). Do the same for y paths in K(A, B ) ∪ K(B, C ) and z paths in
K(A, C ) ∪ K(C, B ). Now we take the union of these three collections of paths, taking care to rename vertices as in Lemma 3.3. Thus the resulting graph will be the required complete tripartite graph.
4.2 Star Multipartite Graphs Definition 4.2. A star is a tree consisting of one vertex (called the root) adjacent to the all others. So a star multipartite graph S = ( A; B1, B 2, . . . , B n) has |A| = a non-adjacent 32 vertices in the root which are adjacent to all the other sets of vertices ( B1, B 2, . . . , B n) where
|Bi| = bi for any i.
Theorem 4.3. Let S = ( A; B1, B 2, . . . , B n) be a star multipartite graph and assume
b1 > b2 > · · · > bn. The NASCs are:
2 | (b1 + b2 + · · · + bn)2. b1 6 b2 + b3 + · · · + bnA
B1B2· · ·
· · ·
Bn
Figure 4.2: Multipartite Star S(A; B1, B 2, ..., B n)
Necessity:
Proof. Let v be a vertex in A. So all the gregarious paths passing through v have both ends in B1 ∪ B2 ∪ · · · ∪ Bn. So 2 | (b1 + b2 + · · · + bn). For the second condition, the number of the vertices in any bi should be less than the number of remaining vertices, because if you fix a vertex, say v in a, then the gregarious paths passing through v gives a one-to-one matching in between vertices. So the number of vertices in any part, bi, should be less than the sum of the number of vertices in the remaining parts. So for any 1 6 i 6 n,
bi 6 b2 + · · · + bi−1 + bi+1 + · · · + bn. Therefore, if b1 6 b2 + b3 + · · · + bn is true, then for any 1 6 i 6 n, bi 6 b2 + · · · + bi−1 + bi+1 + · · · + bn is also true since b1 > b2 > · · · > bn.
Sufficiency:
33 Proof. First, take any vertex v in a, then find the gregarious decomposition of ( v; b1, b 2 , ..., b n). Afterwards, we put the copies of this decomposition on the remaining vertices in A (every vertex in a has the same degree). To find the gregarious decomposition of ( v; b1, b 2 , ..., b n): 1. take a P3 between the first (biggest) two parts. 2. reorder ( b1 − 1, b 2 − 1, ..., b n) so it is non-increasing. 3. repeat steps 1 and 2 until there are no edges left. Now we need to prove that in each step the graph we get still satisfies the necessary conditions. The proof of the first condition is easy since we start with an even number of vertices and in each step we just remove two vertices, so in the next step we should still have an even number of vertices. Now we need prove that in each step we preserve the second condition. We will use induction. Assume that in the kth step we have ( b(k)1 , b (k)2 , . . . , b (k)
n
). For k = 1 the second condition holds since ( b(1) 1 , b (1) 2 , . . . , b (1)
n
) = ( b1, b 2, . . . , b n) and for any 1 6 i 6 n, bi 6 b2 + · · · + bi−1 + bi+1 + · · · + bn .To use induction, assume the condition holds for k:for any 1 6 i 6 n, b(k)
i
6 b(k)2 + · · · + b(k)
i−1
b(k)
i+1
· · · + b(k)
n
.So, we need to prove it holds for k + 1: for any 1 6 i 6 n, b(k+1)
i
6 b(k+1) 2 + · · · + b(k+1)
i−1
b(k+1)
i+1
· · · + b(k+1)
n
Fix i, 1 6 i 6 n.
Case 1: b(k+1)
i
= b(k)
i
− 1: If we remove one vertex from bki , there exists an m with 1 6 m 6 n such that b(k+1)
m
= bkm − 1. In addition, b(k+1)
j
= bkj for any j except j = m, i . So,
b(k+1)
i
= b(k)
i
− 1 6 b(k)1 + · · · + b(k)
m
− 1 + · · · + b(k)
n
6 b(k+1) 1 + · · · + b(k+1)
m
· · · + b(k+1)
n
34 Case 2: b(k+1)
i
= b(k)
i
:Since we removed two vertices in each step, there exist b(k)
p
and b(k)
q
such that b(k)
p
b(k)
q
b(k)
i
.If b(k)
i
2, then b(k+1)
i
= 2 6 b(k+1)
p
b(k+1)
q
· · · .If b(k)
i
= 1 and b(k)
p
= b(k)
q
= 1, then we should have at least one more b(k)
w
= 1 since we have an even number of vertices in each step. So
b(k+1)
i
= 1 6 b(k+1)
p
b(k+1)
q
b(k+1)
w
· · ·
6 b(k)
p
− 1 + b(k)
q
− 1 + b(k)
w
· · ·
6 1 − 1 + 1 − 1 + 1 + · · ·
6 1 + · · ·
Note that the following are equivalent: 1. There is a gregarious P3 decomposition of S(A; B1, B 2, . . . , B n). 2. There is a loopless multigraph with degree sequence ( b1, b 2, . . . , b n). 3. The complete multigraph K(B1, B 2, . . . , B n) has a perfect matching.
4.3 Cycle Multipartite Graphs 4.3.1 Even Cycles Theorem 4.4. For an even cycle multipartite graph C(B1, . . . , B 2n), the NASCs are: 1. b1b2 + b3b4 + · · · + b2n−1b2n = b2b3 + b4b5 + · · · + b2nb1
for any 1 6 i 6 2n, bibi+1 6 bi−1bi + bi+1 bi+2
35 C2n
B1B2
B3
B4B2n−1
B2n
· · · Figure 4.3: Multipartite Even Cycle C2n
Necessity:
Proof. For any 1 6 i 6 2n let xi be the number of gregarious paths that have their middle vertex in Bi. Then,
x1 + x2 = b1b2
x2 + x3 = b2b3
...
x2n−1 + x2n = b2n−1b2n
x2n + x1 = b2nb1
If we add the first, third, fifth, ..., and (2 n − 1) th equations, we get,
x1 + x2 + · · · + x2n = b1b2 + b3b4 + · · · + b2n−1b2n
36 and if we add the second, fourth, sixth, ..., and (2 n)th equations and rearrange the xi’s, we get:
x1 + x2 + · · · + x2n = b2b3 + b4b5 + · · · + b2nb1
So these two equations give the first condition. For the second condition, let x1 = x and
x > 0, then:
x2 = b1b2 − xx3 = b2b3 − x2
x4 = b3b4 − x3
...
x2n = b2n−1b2n − x2n−1
and if we get all equations in terms of x,
x2 = b1b2 − xx3 = b2b3 − b1b2 + xx4 = b3b4 − b2b3 + b1b2 − x
...
x2n = b2n−1b2n − b2n−2b2n−1 + · · · + x
If we use x > 0 and the equations we have above, then we get:
bibi+1 6 bi−1bi + bi+1 bi+2 for any 1 6 i 6 2n.
Sufficiency:
37 Proof. If the necessary conditions are satisfied, we can find all the xi’s for 1 6 i 6 2n, then we use the same technique that we used in the proof of Theorem 4.1 to find gregarious a P3
decomposition.
4.3.2 Odd Cycles Theorem 4.5. For an odd cycle multipartite graph C(B1, . . . , B 2n+1 ), the NASCs are: 1. 2 |
i=2 n
∑
i=1
bibi+1 (# of the edges) 2. for any 1 6 i 6 2n + 1 , bibi+1 6 bi−1bi + bi+1 bi+2
for any 1 6 i 6 2n + 1 ,
bi+1 bi+2 + bi+3 bi+4 + · · · + bi+2 n−2b(i+1)+2 n−2 6 bibi+1 + bi+2 bi+3 + · · · + bi+2 nb(i+1)+2 n
where the subscripts of the b’s are taken (mod 2 n + 1) .C2n+1
B0
B1
Bi−1
BiBi+1
Bi+2
B2n
· · ·
· · ·
Figure 4.4: Multipartite Odd Cycle C2n+1
Necessity:
Proof. The first condition comes from the fact that the total number of edges is divisible by 2. To get the second condition, for any 1 6 i 6 2n + 1 let xi be the number of gregarious 38 paths that have their middle vertex in Bi . Then
x1 + x2 = b1b2
x2 + x3 = b2b3
...
x2n + x2n+1 = b2nb2n+1
x2n+1 + x1 = b2n+1 b1
To find x1;(+) x1 + x2 = b1b2
(−)x2 + x3 = b2b3
...(−)x2n + x2n+1 = b2nb2n+1
(+) x2n+1 + x1 = b2n+1 b1
then we get
x1 = b1b2 − b2b3 + b3b4 − · · · − b2nb2n+1 + b2n+1 b1
2= b1b2 + b3b4 + · · · + b2n+1 b1 − (b2b3 + · · · + b2nb2n+1 )2Using the same technique we can get all the xi’s along with condition 3 since each xi > 0. Condition 2 is the same as the even cycle case.
Sufficiency:
Proof. After finding xi, constructing the gregarious P3 decomposition is the same as for the even cycle case. 39 4.4 Path Multipartite Graphs Theorem 4.6. For a path multipartite graph P (B1, B 2, . . . , B n), the NASCs are: 1. b3 > b1 and bn−2 > bn
b1b2 + b3b4 · · · + bk−1bk = b2b3 + b4b5 + · · · + bl−1bl
for any 2 6 i 6 n − 2, bibi+1 6 bi−1bi + bi+1 bi+2
where k is the largest even number such that k 6 n and l is the largest odd number such that l 6 n.Pn
· · · B1 B2 B3 Bn−2 Bn−1 Bn
Figure 4.5: Multipartite Path Pn
40 Necessity:
Proof. For any 2 6 i 6 n − 1 let xi be the number of gregarious paths that have the middle vertex in Bi .
x2 = b1b2
x2 + x3 = b2b3
x3 + x4 = b3b4
...
xn−2 + xn−1 = bn−2bn−1
xn−1 = bn−1bn
The first condition comes from the fact that x3 = b2b3 − x2 = b2b3 − b1b2 = b2(b3 − b1). So we get b3 > b1 since x2 > 0. We can get bn > bn−2 in the same way. We can find the remaining
xi’s easily. If we add the first, third, fifth,... , and ( k − 1) th equations, we get,
x2 + x3 · · · + xn−1 = b1b2 + b3b4 + · · · + bk−1bk
and if we add the second, fourth, sixth,... , and ( l − 1) th equations, we get:
x2 + x3 + · · · + xn−1 = b2b3 + b4b5 + · · · + bl−1bl
So these two equations give the second condition. The third condition is the same as the condition in the cycle case. 41 Sufficiency:
Proof. If the necessary conditions are satisfied, we can find all the xi’s for 2 6 i 6 n − 1, then we use the same technique that we used in the proof of Theorem 4.1 to find gregarious a P3 decomposition.
4.5 Some Tree Multipartite Graphs Definition 4.7. Let T (C1, . . . C m; A1, A 2; B1, . . . , B n) be a multipartite graph such that two multipartite stars S(A1; C1, . . . , C m) and S(A2; B1, . . . , B n) are attached to each other via putting a complete bipartite graph on bipartition ( A1, A 2). See figure 4.6. C1
C2
Cm
A1A2
B1
B2
Bn
Figure 4.6: T (C1, . . . C m; A1, A 2; B1, . . . , B n)
Definition 4.8. Define T (A1, A 2, A 3; B1, . . . , B n) by using definition 4.7 as T (A1; A2, A 3; B1, . . . , B n). See figure 4.7. 42 ...
A1 A2 A3
B1
B2
BnFigure 4.7: T (A1, A 2, A 3; B1, . . . , B n)
Lemma 4.9. Let G = ( E, V ) be a graph. There is an orientation of G such that for all
v ∈ V , |out (v) − in (v)| 6 1.
Proof. We can assume that G is connected.
Case 1: If all vertices have even degree, then there exists an Euler trail, we can orient the graph this way.
Case 2: If G has some vertices with odd degree, make an extra vertex u and connect all those vertices to u, then find an Euler trail on G ∪ { u} and remove the edges at the end. For all v ∈ V , we still have |out (v) − in (v)| 6 1 since we remove one edge from each vertex with odd degree . G
odd
u
Figure 4.8: Orientation of G
43 Theorem 4.10. Let A, B, I be finite non-empty sets, let f : B × I → N be such that for all t ∈ B, ∑
i∈I
f (t, i ) = |A|. Then the edges of K(A, B ) can be partitioned into spanning subgraphs Gi, i ∈ I, such that for each i ∈ I, Gi is balanced on A, and for each t ∈ B, the degree of t in Gi is f (t, i ).Proof. If |A| = 1, then the proof is trivial. Now suppose |A| = 2. Let A = {s1, s 2}. Form a graph H on vertex set I as follows: For each t ∈ B, H has an edge et: If f (t, i ) = 2, (and so f (t, j ) = 0 for all other j ∈ I) then
et is a loop at vertex i of H. If f (t, i ) = 1 = f (t, j ), i 6 = j, ( f (t, k ) = 0 for the other k ∈ I), then et joins the vertices i and j in H.Orient H so that at each vertex of H the indegree and outdegree differ by at most 1 using Lemma 4.9. For each t ∈ B, if et is directed from i to j in the oriented H, place the edge between t and s1 in Gi, and the edge between t and s2 in Gj (see example 4.11). If |A| > 3, then partition the edges of K(A, B ) into spanning subgraphs Gi whose degrees on B are given by f (this is certainly possible by the sum condition on f ). If everything is balanced on |A|, then we are done. Otherwise degrees in some Gi differ by 2 or more. Fix i, and let s1, s 2 be two vertices in A, whose degrees differ by 2 or more in Gi. So use the previous case where |A| = 2 on this graph to find the balanced distribution. Using this method repeatedly for each unbalanced pair of vertices of Gi in A, finally we can get the balanced distribution on A. Afterwards, we can repeat the same process for the other
Gj for each j ∈ I.Now we need to prove that this process will stop after finitely many steps. Let V1 =(a1, . . . , a i, . . . , a j , . . . , a n) be a integer vector with fixed sum ∑ ai = a. So the shortest integer vector with respect to the Euclidean metric with the fixed sum of the entries is the balanced one. To see this assume aj > ai + 2, then if we balance ai and aj , we get 44 V2 = ( a1, . . . , a i + 1 , . . . , a j − 1, . . . , a n) and |V2|2 6 |V1|2 − 2 since,
|V2|2 = a21 + · · · + ( ai + 1) 2 + · · · + ( aj − 1) 2 + · · · + a2
n
= a21 + · · · + a2
i
2 ai + 1 + · · · + a2
j
− 2aj + 1 + · · · + a2
n
= (a21 + · · · + a2
i
· · · + a2
j
· · · + a2
n
) + 2( ai − aj ) + 2 = |V1|2 + 2( ai − aj ) + 2
6 |V1|2 + 2( −2) + 2
6 |V1|2 − 2This means that when we balance a pair of entries in the vector at a time, the vector gets shorter, and after finitely many steps we will find the shortest one. This completes the proof.
Example 4.11. Let G be a bipartite graph on bipartition ( A, B ) where A = {s1, s 2} and
B = {v1, v 2, v 3, v 4}. Let I ={green, blue, red }. We want to get green and blue balanced on
A without changing the color census on B (see the first picture in Figure 4.9). Then using the method defined in Theorem 4.10 build a graph H (see the second picture in Figure 4.9) and orient H so that |in (w) − out (w)| 6 1 for every vertex w in H. Finally, we can swap edges of G with respect to the orientation on H to get a balanced coloring on A.45 s1
s2
v1
v2
v3
v4
Gv2
v4
v1
v3
Hv2
v4
v1
v3
Oriented Hs1
s2
v1
v2
v3
v4
Colors are balanced on AFigure 4.9: An Example for Theorem 4.10
4.5.1 Necessary And Sufficient Conditions for T (A1, A 2, A 3; B1, . . . , B n)
Theorem 4.12. For a graph T (A1, A 2, A 3; B1, . . . , B n) assume bn 6 · · · 6 b2 6 b1, the NASCs are: 1. 2 | [a2(a1 + a3) + a3(b1 + b2 + · · · + bn)]
a1 6 a3
a1a2 + b1a3 6 a3(a2 + b2 + b3 + · · · + bn)
a2a3 6 a1a2 + a3(b1 + b2 + · · · + bn)
If
• a2 + ( b1 + · · · + bn) is even, then a1a2 is even.
• a2 + ( b1 + · · · + bn) is odd, then a1a2 − a3 is even and non-negative.
46 Necessity:
Proof. Let G = T (A1, A 2, A 3; B1, . . . , B n). Condition 1 comes from the fact that the number of edges is even. We can get conditions 2 - 4 using Tutte’s f-factor Theorem on L(G).
L(G) is union of a complete graph on n + 1 vertices and an edge attached at the vertex
A2A3. If we check all the possible L(G) triples B = ( S, T, U ) where T is independent and
λ(T, U ) = 0 from theorem 2.12, all will reduce to the the following three cases in figure 4.10. We need to check k(B, h ) + h(T ) 6 h(S) for each case. From the first picture in the figure 4.10, we will get a2a3 6 a1a2 which gives condition 2. From the second picture, we get a1a2 + b1a3 6 a2a3 + b2a3 + · · · + bna3 which gives condition 3. From the last picture, we get a2a3 6 a1a2 + a3b1 + a3b2 + · · · + a3bn which gives condition 4. S T
U
a1a2
a2a3S
T
U
a1a2
b1a3
a2a3
b2a3
bna3S
T
U
a1a2
b1a3
a2a3b2a3
bna3
Figure 4.10: How to get conditions 2 - 4 For condition 5, we need to consider all the types of paths we have and the degree of any vertex in A3. Firstly, the degree of any vertex v in A3 is deg (v) = a2 + b1 + · · · + bn. Let x1
be the number of paths passing through the sets of vertices A1 → A2 → A3 so x1 = a1a2.In the same way, yi : A2 → A3 → Bi for any 1 6 i 6 n, and wij : Bi → A3 → Bj
for any 1 6 i 6 j 6 n. Here, both yi and wij have their middle vertices in A3, so if
deg (v) is even, then x1 must be even. If deg (v) is odd, then we should have enough x1 type paths which means a1a2 > a3. That gives a1a2 − a3 non-negative. To see that a1a2 − a3 is even, consider the vertices in A3 and distribution of a1a2 edges on A3. There are αi’s for 47 1 6 i 6 a3 such that
a3
∑
i=1
αi = a1a2 where each αi = 2 βi + 1, an odd number. βi = αi − 12 .
a3
∑
i=1
βi =
a3
∑
i=1
αi − 12 = 12(a1a2 − a3). Therefore a1a2 − a3 is an even number.
Sufficiency:
Proof. If the necessary conditions are satisfied we can find proper x1, yi’s and wij ’s. In between pairs of sets ( A1, A 2) and ( A3, B i) for any i, we can find balanced edge distributions with the required numbers as we did for the sufficiency case of the Theorem 4.1. The only problem is finding a construction for ( A2, A 3) since we need to find a parity balanced distribution of x1 edges on A3 with respect to the parity of a2 + ( b1 + · · · + bn) (see condition 5 in Theorem 4.12). We also need to find a balanced distribution for the remaining yi’s. To be able to find this special distribution we can use theorem 4.10 and choose the degrees on
A3 to get balanced degrees on A2.
4.5.2 Necessary And Sufficient Conditions for T (C1, . . . , C m; A1, A 2; B1, . . . , B n)
Theorem 4.13. For a graph T (C1, . . . , C m; A1, A 2; B1, . . . , B n) assume cm 6 · · · 6 c2 6 c1
and bn 6 · · · 6 b2 6 b1, and let C = C1 ∪ C2 ∪ · · · ∪ Cm and |C| = c, B = B1 ∪ B2 ∪ · · · ∪ Bn,
|B| = b and d1 = c + a2, d2 = b + a1. The NASCs are: 1. 2 | [a1(c1 + c2 + · · · + cm) + a1a2 + a2(b1 + b2 + · · · + bn)]
c1 6 a2 + ( c2 + · · · + cm) and b1 6 a1 + ( b2 + · · · + bn)
a1c1 + a2b1 6 a1(c2 + · · · + cm) + a1a2 + a2(b2 + · · · + bn)
a1a2 6 a1(c1 + c2 + · · · + cm) + a2(b1 + b2 + · · · + bn)
If
• d1 and d2 are even, then a1a2 is even.
48 • d1 is even and d2 is odd, then either both a1 and a2 are odd, both even or a1 is odd and a2 even. In addition, ca 1 − a2 > 0.
• d1 is odd and d2 is even, then either both a1 and a2 are odd, both even or a1 is even and a2 odd. In addition, ba 2 − a1 > 0.
• d1 and d2 are odd, then both a1 and a2 are even. In addition, ca 1 − a2 > 0 and
ba 2 − a1 > 0.with the following exceptions:
d1 d2 a1 a2 c1 & b1
even even 2 2 any c1 with 1 + ( b2 + · · · + bn) 6 b1 6 2 + ( b2 + · · · + bn)even 1 any c1 with 1 + ( b2 + · · · + bn) 6 b1 6 a1 + ( b2 + · · · + bn)1 even any b1 with 1 + ( c2 + · · · + cm) 6 c1 6 a2 + ( c2 + · · · + cm)even odd odd 1 any c1 with 1 + ( b2 + · · · + bn) 6 b1 6 a1 + ( b2 + · · · + bn)1 even any c1 with b1 = 1 + ( b2 + · · · + bn)odd even 1 odd any b1 with 1 + ( c2 + · · · + cm) 6 c1 6 a2 + ( c2 + · · · + cm)even 1 any b1 with c1 = 1 + ( c2 + · · · + cm)Table 4.1: Exceptions for Theorem 4.13
Necessity:
Proof. Let G = T (C1, . . . , C m; A1, A 2; B1, . . . , B n). Condition 1 comes from the fact that the number of edges is even. We can get conditions 2 - 4 using Tutte’s f-factor Theorem on
L(G). L(G) is union of two complete graphs on m and n vertices attached at the vertex
A1A2. If we check all the possible L(G) triples B = ( S, T, U ) where T is independent and
λ(T, U ) = 0 from theorem 2.12, all will reduce to the the following three cases in figure 4.11. We need to check k(B, h ) + h(T ) 6 h(S) for each case. From the first picture in the figure 4.11, we will get c1 6 a2 + ( c2 + · · · + cm) and in the same picture if we replace B’s with C’s 49 and C’s with B’s then we get b1 6 a1 + ( b2 + · · · + bn) which gives condition 2. From the second picture, we get a1c1 + a2b1 6 a1(c2 + · · · + cm) + a1a2 + a2(b2 + · · · + bn) which gives condition 3. From the last picture, we get a1a2 6 a1(c1 + c2 + · · · + cm) + a2(b1 + b2 + · · · + bn)which gives condition 4. S
T
A1C2
A1A2
A1C1
A1C3
U
A2B1
A2BnS
T
U
A1C2
A1C1
A2B1
A2Bn
A1A2S
T
U
A1C1
A1Cm
A2B1
A2Bn
A1A2
Figure 4.11: How to get conditions 2 - 4 For condition 5, we need to consider all the types of paths we have and the degree of any vertex in A1 and A2 . Firstly, the degree of any vertex v1 in A1 is:
d1 = deg (v1) = a2 + c1 + · · · + cm = a2 + c
and the degree of any vertex v2 in A2 is:
d2 = deg (v2) = a1 + b1 + · · · + bn = a1 + b.Let xi be the number of paths passing through the sets of vertices Ci → A1 → A2 for any 1 6 i 6 m and let x = ∑mi=1 xi. In the same way, yj : Bj → A2 → A1 for any 1 6 j 6 n
and y = ∑nj=1 yj . So x + y = a1a2. In addition, we have wij : Ci → A1 → Cj for any 1 6 i 6 j 6 m and zkl : Bk → A2 → Bl for any 1 6 k 6 l 6 n. Here wij ’s have their middle vertex in A1 and zkl ’s have their middle vertex in A2, so we have four cases with respect to the parity of d1 and d2. So the parity of x and d2, and y and x1 must be consistent (see figure 4.12). 50 A1 A2
wij
xi
yj
zkl
CB
Figure 4.12: Types of paths
Case 1: If d1 and d2 are even, then y and x are even so a1a2 is even since x + y = a1a2.
Case 2: If d1 is even and d2 is odd, then y is even and x ≡ a2 (mod 2) and x > a2. So we can get ca 1 − a2 > 0 since ca 1 > x. To get either both a1 and a2 odd, both even or or a1 is odd and a2 even, see:
x + y = a1a2
x + y ≡ a1a2 (mod 2)
x ≡ a1a2 (mod 2) since y is even
Case 3: If d1 is odd and d2 is even, then this is the same as case 2, just replace a2 with a1.
Case 4: If d1 is odd and d2 is odd, then y ≡ a1 (mod 2) and y > a1, and x ≡ a2 (mod 2) and x > a2. We can get ca 1 − a2 > 0 and ba 2 − a1 > 0 in the same way as in case 2. To get both a1 and a2 even, see:
x + y = a1a2
x + y ≡ a1a2 (mod 2)
a1 + a2 ≡ a1a2 (mod 2) since x ≡ a2 (mod 2) and y ≡ a1 (mod 2) 51 a1 + a2 ≡ a1a2 (mod 2) is only satisfied when both a1 and a2 are even. Note that theorem 4.12 is a special case of theorem 4.13. In theorem 4.13, if we get
c2 = c3 = · · · = cm = 0 and replace c1 with a1, a1 with a2 and a2 with a3 we will get exactly the same conditions as in theorem 4.12.
Sufficiency:
Proof. If the necessary conditions are satisfied we can find proper xi’s, yj ’s, wij ’s and vkl ’s. First we find a proper x and y then we will find wij ’s and vkl ’s since we have more restriction on x and y. In between pairs of sets ( Ci, A 1) for any 1 6 i 6 m, and ( A2, B j ) for any 1 6 j 6 n, we can find balanced edge distributions with the required numbers as we did for the sufficiency case of theorem 4.1. The only problem is finding a construction for ( A1, A 2)since we need to find a parity balanced distribution of x + y edges on A1 and A2 with respect to the parity of d1 and d2 (see condition 5 in theorem 4.13). To find this parity balanced distribution, we will use theorem 3.12.
Case 1: Assume d1 and d2 are even, then y and x are even so a1a2 is even since x+y = a1a2.So there are three cases for ( a1, a 2): (even, even), (even, odd) and (odd, even). If a1 and a2 are both even, then we need to find a parity balanced bipartite graph (PBBG) with pararameters ( a = a1, b = a2, e = x, a = 0 , b = 0) with bipartite complement (a = a1, b = a2, e = ab − x = y, ′
a
= 0 , ′
b
= 0) so that the distribution of x on A2 has even parity ( b = 0) and the distribution of y on A1 has even parity ( ′
a
= 0). We can we can find such a PBBG since the necessary conditions of theorem 3.12 are satisfied.
a + ′
a
= 0 + 0 = 0 ≡ b (mod 2) and b + ′
b
= 0 + 0 = 0 ≡ a (mod 2)
aa 6 e 6 ab − ′
a
a ⇒ 0 6 x 6 a1a2
bb 6 e 6 ab − ′
b
b ⇒ 0 6 x 6 a1a2
aa ≡ ab − ′
a
a ≡ e = x ≡ bb ≡ ab − ′
b
b ≡ 0 (mod 2) 52 For the exceptions in table 4.1 the only problem concerning this case is when a = a1 =even,
b = a2 = even , e = x = 2, a = 0, ′
a
= 0, b = 0, ′
b
= 0. We can solve this problem by choosing x > 4 since a1 > 2 and a2 > 2 in the exceptions. In the case of a1 = 2 = a2,we will not have any y’s which means we need to put a gregarious P3 decomposition of star multipartite graph on S = ( A2; B2, B 2, . . . , B n). The first condition of theorem 4.3 is satisfied since 2 | b = b1 + b2 + . . . + bn (d1 =even= a1 + b and a1 is even so b is even). For the second condition of theorem 4.3, we can get b1 6 2 + b2 + . . . + bn from condtion 2 of theorem 4.13. From here we get two exceptions: b1 = 1 + b2 + · · · + bn and b1 = 2 + b2 + · · · + bn.So T (C1, . . . C m; 2 , 2; B1, . . . , B n) doesn’t exist when either b1 = 1 + b2 + · · · + bn or b1 =2 + b2 + · · · + bn.If a1 is even and a2 is odd, then we need to find a PBBG with pararameters ( a =
a1, b = a2, e = x, a = 1 , b = 0) with bipartite complement ( a = a1, b = a2, e = ab − x =
y, ′
a
= 0 , ′
b
= 0) so that the distribution of x edges on a2 has even parity ( b = 0) and the distribution of y edges on a1 has even parity( ′
a
= 0). We can we can find such a PBBG since the necessary conditions of theorem 3.12 are satisfied.
a + ′
a
= 1 + 0 = 1 ≡ b (mod 2) and b + ′
b
= 0 + 0 = 0 ≡ a (mod 2)
aa 6 e 6 ab − ′
a
a ⇒ a1 6 x 6 a1a2
bb 6 e 6 ab − ′
b
b ⇒ 0 6 x 6 a1a2
aa ≡ ab − ′
a
a ≡ e = x ≡ bb ≡ ab − ′
b
b ≡ 0 (mod 2) If we check the exceptions in table 4.1, then we see ( a = even > 4, b = odd > 3, e = 2 , a =1, ′
a
= 0 , b = 0 , ′
b
= 0). However, in this case e = x > a1 and a1 > 4 so we don’t have any exception for e = 2. There are other exceptions coming from x > a1. If a2 = 1, then we don’t have any y’s which means we need to put a gregarious P3 decomposition of a star multipartite graph on S = ( A2 = 1; B2, B 2, . . . , B n). The first condition of theorem 4.3 is satisfied since 2 | b = b1 + b2 + . . . + bn (d2 =even= a1 + b and a1 is even so b is even). For the second 53 condition of theorem 4.3, we can get b1 6 a1 + ( b2 + . . . + bn) from condition 2 of theorem 4.13. From here we get exceptions when 1 + ( b2 + · · · + bn) 6 b1 6 a1 + ( b2 + · · · + bn). So
T (C1, . . . C m; A1, 1; B1, . . . , B n) doesn’t exist when 1+( b2 +· · · +bn) 6 b1 6 a1 +( b2 +· · · +bn). If a1 is odd and a2 is even, then this case is the same as the previous case, just switch
a2 with a1.
Case 2: If d1 is even and d2 is odd, then y is even and x ≡ a2 (mod 2) and x > a2. So there are three cases for ( a1, a 2): (even, even), (odd, odd) and (odd, even). If a1 and a2 are both even, then x and y are both even. We need to find a PBBG with parameters ( a = a1, b = a2, e = x, a = 0 , b = 1) with bipartite complement ( a = a1, b =
a2, e = a1a2 − x = y, ′
a
= 0 , ′
b
= 1) so that the distribution of x on A2 has odd parity (b = 1) and the distribution of y on A1 has even parity ( ′
a
= 0). We can we can find such a PBBG since the necessary conditions of theorem 3.12 are satisfied.
a + ′
a
= 0 + 0 = 0 ≡ b (mod 2) and b + ′
b
= 1 + 1 = 0 ≡ a (mod 2)
aa 6 e 6 ab − ′
a
a ⇒ 0 6 x 6 a1a2
bb 6 e 6 ab − ′
b
b ⇒ a2 6 x 6 a1a2 − a2
aa ≡ ab − ′
a
a ≡ e = x ≡ bb ≡ ab − ′
b
b ≡ 0 (mod 2) If we check the exceptions in table 4.1, we see that we don’t have any exception for this case. If a1 and a2 are both odd, then x is odd and y is even. We need to find a PBBG with parameters ( a = a1, b = a2, e = x, a = 1 , b = 1) with bipartite complement ( a = a1, b =
a2, e = a1a2 − x = y, ′
a
= 0 , ′
b
= 0) so that the distribution of x on A2 has odd parity (b = 1) and the distribution of y on A1 has even parity ( ′
a
= 0). We can we can find such 54 a PBBG since the necessary conditions of theorem 3.12 are satisfied.
a + ′
a
= 0 + 1 = 1 ≡ b (mod 2) and b + ′
b
= 1 + 0 = 1 ≡ a (mod 2)
aa 6 e 6 ab − ′
a
a ⇒ a1 6 x 6 a1a2
bb 6 e 6 ab − ′
b
b ⇒ a2 6 x 6 a1a2
aa ≡ ab − ′
a
a ≡ e = x ≡ bb ≡ ab − ′
b
b ≡ 1 (mod 2) If we check the exceptions in table 4.1, then we see ( a = odd > 3, b = odd > 3, e = 2 , a =1, ′
a
= 0 , b = 1 , ′
b
= 0). However, in this case e = x > max {a1, a 2} and a1, a 2 > 3 so we don’t have any exception for e = 2. There are other exceptions coming from x > max {a1, a 2}.If a2 = 1, then x = a1 = a1a2 so we don’t have any y’s which means we need to put a gregarious P3 decomposition of a star multipartite graph on S = ( A2 = 1; B2, B 2, . . . , B n). The first condition of theorem 4.3 is satisfied since 2 | b = b1 + b2 + . . . + bn (d2 =odd=
a1 + b and a1 is odd so b is even). For the second condition of theorem 4.3, we can get
b1 6 a1 + ( b2 + . . . + bn) from condition 2 of theorem 4.13. From here we get exceptions when 1 + ( b2 + · · · + bn) 6 b1 6 a1 + ( b2 + · · · + bn). So T (C1, . . . C m; A1, 1; B1, . . . , B n) doesn’t exist when 1 + ( b2 + · · · + bn) 6 b1 6 a1 + ( b2 + · · · + bn). If a1 is odd and a2 is even, then x and y are both even. We need to find a PBBG with parameters ( a = a1, b = a2, e = x, a = 0 , b = 1) with bipartite complement ( a = a1, b =
a2, e = a1a2 − x = y, ′
a
= 0 , ′
b
= 0) so that the distribution of x on A2 has odd parity (b = 1) and the distribution of y on A1 has even parity ( ′
a
= 0). We can we can find such a PBBG since the necessary conditions of theorem 3.12 are satisfied.
a + ′
a
= 0 + 0 = 0 ≡ b (mod 2) and b + ′
b
= 1 + 0 = 1 ≡ a (mod 2)
aa 6 e 6 ab − ′
a
a ⇒ 0 6 x 6 a1a2
bb 6 e 6 ab − ′
b
b ⇒ a2 6 x 6 a1a2
aa ≡ ab − ′
a
a ≡ e = x ≡ bb ≡ ab − ′
b
b ≡ 0 (mod 2) 55 If we check the exceptions in table 4.1, then we see ( a = odd > 3, b = even > 3, e =2, a = 0 , ′
a
= 0 , b = 1 , ′
b
= 0). However, in this case e = x > a2 and a2 > 4 so we don’t have any exception for e = 2. There is an exception coming from x > a2. If a1 = 1, then x = a2 = a1a2 so we don’t have any y’s which means we need to put a gregarious P3
decomposition of a star multipartite graph on S = ( A2; B2, B 2, . . . , B n). The first condition of theorem 4.3 is satisfied since 2 | b = b1 + b2 + . . . + bn (d2 =odd= a1 + b and a1 is odd so b
is even). For the second condition of theorem 4.3, we can get b1 6 a1 + ( b2 + . . . + bn) from condition 2 of theorem 4.13. From here we get exceptions when b1 = 1 + ( b2 + · · · + bn). So
T (C1, . . . C m; 1 , A 2; B1, . . . , B n) doesn’t exist when b1 = 1 + ( b2 + · · · + bn).
Case 3: If d1 is odd and d2 is even, then this case is the same as case 2, just switch a1 and
a2.
Case 4: If d1 and d2 are both odd, then x, y, a 1, a 2 are even and y > a1, x > a2. We need to find a PBBG with parameters ( a = a1, b = a2, e = x, a = 1 , b = 1) with bipartite complement ( a = a1, b = a2, e = a1a2 − x = y, ′
a
= 1 , ′
b
= 1) so that the distribution of x on
A2 has odd parity ( b = 1) and the distribution of y on A1 has odd parity too ( ′
a
= 1). We can we can find such a PBBG since the necessary conditions of theorem 3.12 are satisfied.
a + ′
a
= 1 + 1 = 0 ≡ b (mod 2) and b + ′
b
= 1 + 1 = 0 ≡ a (mod 2)
aa 6 e 6 ab − ′
a
a ⇒ a1 6 x 6 a1a2 − a1
bb 6 e 6 ab − ′
b
b ⇒ a2 6 x 6 a1a2 − a2
aa ≡ ab − ′
a
a ≡ e = x ≡ bb ≡ ab − ′
b
b ≡ 0 (mod 2) If we check the exceptions in table 4.1, we see that we don’t have any exception for this case. 56 Bibliography D.G Hoffman , Cores of Class II Graphs, Journal of Graph Theory, Vol. 20, No. 3, 397-402 (1995) Douglas B. West, Introduction to Graph Theory, Prentice Hall 2001. Elizabeth J. Billington, Dean G. Hoffman, Short path decompositions of arbitrary com-plete multipartite graphs, Proc. of the 38th Southeastern Int. Conf. on Comb., Graph Theo. and Comp., Congr. Numer. 187 (2007), 161 - 173. W.T. Tutte, Graph Factors, Combinatorica 1 (1981) 79-97 Billington, Elizabeth J. ; Hoffman, D. G. , Decomposition of complete tripartite graphs into gregarious 4-cycles. Papers on the occasion of the 65th birthday of Alex Rosa. Discrete Math. 261 (2003), no. 1-3, 87–111. Billington, Elizabeth J. ; Hoffman, D. G. , Equipartite and almost-equipartite gregarious 4-cycle systems. Discrete Math. 308 (2008), no. 5-6, 696–714. Smith, Benjamin R. , Equipartite gregarious 5-cycle systems and other results. Graphs Combin. 23 (2007), no. 6, 691–711. Billington, Elizabeth J. ; Smith, Benjamin R. ; Hoffman, D. G. Equipartite gregarious 6- and 8-cycle systems. Discrete Math. 307 (2007), no. 13, 1659–1667. Cho, Jung R. ; Gould, Ronald J. Decompositions of complete multipartite graphs into gregarious 6-cycles using complete differences. J. Korean Math. Soc. 45 (2008), no. 6, 1623–1634. Cho, Jung Rae . A note on decomposition of complete equipartite graphs into gregarious 6-cycles. Bull. Korean Math. Soc. 44 (2007), no. 4, 709–719. Smith, Benjamin R. Some gregarious cycle decompositions of complete equipartite graphs. Electron. J. Combin. 16 (2009), no. 1, Research Paper 135, 17 pp. Billington, Elizabeth J. ; Hoffman, D. G. ; Rodger, C. A., Resolvable gregarious cycle decompositions of complete equipartite graphs. Discrete Math. 308 (2008), no. 13, 2844– 2853. El-Zanati, Saad I. ; Punnim, Narong ; Rodger, Chris A. Gregarious GDDs with two associate classes. Graphs Combin. 26 (2010), no. 6, 775–780. Bryant, Darryn; Private Communication 57 |
7913 | https://www2.isye.gatech.edu/~nemirovs/Vi29Nov2013-ai.pdf | Solving Variational Inequalities with Monotone Operators on Domains Given by Linear Minimization Oracles Anatoli Juditsky∗ Arkadi Nemirovski† December 1, 2013 Abstract The standard algorithms for solving large-scale convex-concave saddle point problems, or, more generally, variational inequalities with monotone operators, are proximal type algorithms which at every iteration need to compute a prox-mapping, that is, to minimize over problem’s domain X the sum of a linear form and the specific convex distance-generating function underly-ing the algorithms in question. (Relative) computational simplicity of prox-mappings, which is the standard requirement when implementing proximal algorithms, clearly implies the possibil-ity to equip X with a relatively computationally cheap Linear Minimization Oracle (LMO) able to minimize over X linear forms. There are, however, important situations where a cheap LMO indeed is available, but where no proximal setup with easy-to-compute prox-mappings is known.
This fact motivates our goal in this paper, which is to develop techniques for solving variational inequalities with monotone operators on domains given by Linear Minimization Oracles. The techniques we are developing can be viewed as a substantial extension of the proposed in method of nonsmooth convex minimization over an LMO-represented domain.
1 Introduction This paper is motivated by the desire to develop First Order algorithms for solving convex-concave saddle point problems on convex domains with “difficult geometry,” meaning that the domain does not allow for a proximal setup resulting in easy-to-compute prox-mappings. Whenever this is the case, proximal type algorithms (which form the vast majority of First Order methods) become impractical.
In what follows, we relax the assumption on problem’s domain to admit “computation-friendly” proximal setup to the weaker assumption that the domain allows for computationally cheap Linear Minimization Oracle (LMO) – a routine able to minimize a linear function over the domain.
It is easily seen that an easy-to-compute proximal setup automatically implies computationally cheap LMO, but not vice versa. For example, when the domain is the ball Bn of nuclear norm in Rn×n, computing prox-mapping, for all known proximal setups, requires full singular value decomposition ∗LJK, Universit´ e J. Fourier, B.P. 53, 38041 Grenoble Cedex 9, France, Anatoli.Juditsky@imag.fr †Georgia Institute of Technology, Atlanta, Georgia 30332, USA, nemirovs@isye.gatech.edu Research of the second author was supported by the NSF grant CMMI 1232623.
1 of an n × n matrix, which can be prohibitively time consuming when n is large. In contrast to this, minimizing a linear form over Bn only requires finding the leading singular vectors and the singular value of an n × n matrix, which is much easier than full-fledged singular value decomposition.
Recently, there was significant interest in solving convex minimization problems on domains given by LMO’s. The emphasis in this line of research is on smooth/smooth norm-regularized convex min-imization [8, 10, 11, 12, 13, 23], where the main “working horse”is the classical Conditional Gradient (a.k.a. Frank-Wolfe) algorithm originating from and intensively studied in 1970’s (see [6, 7, 22] and references therein). Essentially, Conditional Gradient is the only traditional convex optimization technique capable to handle convex minimization problems on LMO-represented domains. To the best of our knowledge, Conditional Gradient algorithm is not applicable beyond the smooth mini-mization setting; we are not aware on any attempt to apply this algorithm even to the simplest – bilinear – saddle point problems. The approach proposed in this paper is different and is inspired by our recent paper , where a method for nonsmooth convex minimization over an LMO-represented convex domain was developed. The latter method method utilizes Fenchel-type representations of the objective in order to pass from the problem of interest to its special dual. In many important cases the domain of the dual problem admits proximal setup with easy-to-compute prox-mapping, so that the dual problem can be solved by proximal-type algorithms. We then use the machinery of accuracy certificates originating from allowing to recover a good solution to the problem of interest from the information accumulated when solving the dual problem. In this paper we follow the same strategy in the context of variational inequalities (v.i.’s) with monotone operators (this covers, in particular, convex-concave saddle point problems). Specifically, we introduce the notion of a Fenchel-type representation of a monotone operator, allowing to associate with the v.i. of interest its dual, which is again a v.i. with monotone operator with the values readily given by the represen-tation and the LMO representing the domain of the original v.i.. Then we solve the dual v.i. (e.g., by a proximal-type algorithm) and use the machinery of accuracy certificates to recover a good solution to the v.i. of interest from the information gathered when solving the dual v.i.
The main body of the paper is organized as follows. Section 2 outlines the background of convex-concave saddle point problems, variational inequalities with monotone operators and accuracy cer-tificates. In section 3, we introduce the notion of a representation of a monotone operator and the induced by this notion concept of v.i. dual to a given v.i. This section also contains a simple fully algorithmic “calculus” of Fenchel-type representations of monotone operators: it turns out that basic monotonicity-preserving operations with these operators (summation, affine substitution of argu-ment, etc.) as applied to operands given by Fenchel-type representations yield similar representation for the result of the operation. As a consequence, our abilities to numerically evaluate these operator representations are comparable with our abilities to evaluate the operators themselves. Section 4 contains our main result – Theorem 4.1. It shows how certain information information, collected when solving the dual v.i. to a certain accuracy, can be used to point out an approximate solution, of the same accuracy, to the primal v.i. In Section 4 we present a self-contained description of two well known proximal type algorithms for v.i.’s with monotone operators – Mirror Descent (MD) and Mirror Prox (MP) – which indeed are capable to collect the required information. In the concluding Section 5, we illustrate the proposed approach by applying it to the “matrix completion problem 2 with spectral norm fit” – to the problem min u∈Rn×n, ∥u∥≤1 ∥Au −b∥∗, where ∥x∥= P i σi(x) is the nuclear norm, σ(x) being the singular spectrum of x, u 7→Au is a linear mapping from Rn×n to Rm×m, and ∥· ∥∗is the spectral norm (the largest singular value) of a matrix.
2 Preliminaries Variational inequalities and related accuracy measures.
Let Y be a convex compact set in Euclidean space Ey and H(y) : Y →Ey be a monotone operator: ⟨H(y) −H(y′), y −y′⟩≥0 ∀y, y′ ∈Y.
The variational inequality (v.i.) associated with (H, Y ) is find y∗∈Y : ⟨H(z), z −y∗⟩≥0 ∀z ∈Y ; VI(H, Y ) (every) y∗∈Y satisfying the target relation in VI(H, Y ) is called a weak solution to the v.i., and under our assumption (Y is convex and compact, H is monotone on Y ) weak solutions always exist.
A strong solution to v.i. is a point y∗∈Y such that ⟨H(y∗), y −y∗⟩≥0 for all y ∈Y ; from the monotonicity of H is follows that a strong solution is a weak one as well. Note that when H is monotone and continuous on Y (this is the only case we will be interested in), weak solutions are exactly the strong solutions.
The accuracy measure naturally quantifying the inaccuracy of a candidate solution y ∈Y to VI(H, Y ) is the gap function ϵvi(y|H, Y ) = sup z∈Y ⟨H(z), y −z⟩; this (clearly nonnegative for y ∈Y ) quantity is zero if and only if y is a weak solution to the v.i.
We will be interested also in the Special case where Y = V × W is the direct product of convex compact subsets V ⊂Ev and W ⊂Ew of Euclidean spaces Ev, Ew, and H is associated with Lipschitz continuous function f(v, w) : Y = V × W →R convex in v ∈V and concave in w ∈W: H(y = [v; w]) = [Hv(v, w); Hw(v, w)] with Hv(v, w) ∈∂vf(v, w), Hw(v, w) ∈∂w[−f(v, w)].
We can associate with the Special case two optimization problems Opt(P) = minv∈V f(v) = maxw∈W f(v, w) (P) Opt(D) = maxw∈W f(w) = minv∈V f(v, w) (D) with equal optimal values, and with a pair (v, w) ∈V × W – saddle point inaccuracy (duality gap) ϵsad(v, w|f, V, W) = [f(v) −Opt(P)] + [Opt(D) −f(w)] = f(v) −f(w).
3 Accuracy certificates.
Given Y, H, let us call a collection CN = {yt ∈Y, λt ≥0, H(yt)}N t=1 with P t λt = 1, an N-step accuracy certificate, and the quantity Res CN = max y∈Y N X t=1 λt⟨H(yt), yt −y⟩ the resolution of the certificate CN.
Let us make two observations coming back to : Lemma 2.1 Let Y be a convex compact set in Euclidean space Ey, H be a monotone operator on Y , and CN = {yt ∈Y, λi ≥0, H(yt)}N t=1 be an accuracy certificate. Setting b y = N X t=1 λtyt, we have b y ∈Y and ϵvi(b y|H, Y ) ≤Res CN (1) In the Special case we have also ϵsad(b y|f, V, W) ≤Res CN (2) Proof. For z ∈Y we have ⟨H(z), z −P t λtyt⟩= P t λt⟨H(z), z −yt⟩ [since P t λt = 1] ≥ P t λt⟨H(yt), z −yt⟩ [since H is monotone and λt ≥0] ≥ −Res CN [by definition of resolution] Thus, ⟨H(z), b y−z⟩≤Res CN for all z ∈Y , and (1) follows. In the Special case, setting yt = [vt; wt], for every y = [v; w] ∈Y = V × W we have Res(CN) ≥ P t λt⟨H(yt), yt −y⟩ [by definition of resolution] = P t λt [⟨Hv(vt, wt), vt −v⟩+ ⟨Hw(vt, wt), wt −w⟩] ≥ P t λt [by origin of H and since f(v, w) is convex in v and concave in w] = P t λt[f(vt, w) −f(v, wt)] ≥ f(b v, w) −f(v, b w) [since f(v, w) is convex in v and concave in w] Since the resulting inequality holds true for all v ∈V , w ∈W, we get f(b v) −f( b w) ≤Res(CN), and (2) follows.
Lemma 2.1 can be inverted in the case of skew-symmetric operator H, that is, 4 H(y) = a + Sy (3) with skew-symmetric (S = −S∗) linear operator S. A skew-symmetric H clearly satisfies the identity ⟨H(y), y −y′⟩= ⟨H(y′), y −y′⟩, y, y′ ∈Ey.
Lemma 2.2 Let Y be a convex compact set in Euclidean space Ey, H(y) = a+Sy be skew-symmetric, and let CN = {yt ∈Y, λt ≥0, H(yt)}N t=1 be an accuracy certificate. Then for b y = P t λtyt it holds ϵvi(b y|H, Y ) = Res CN .
(4) Proof. We already know that ϵvi(b y|H, Y ) ≤Res CN . To prove the inverse inequality, note that for every y ∈Y we have ϵvi(b y|H, Y ) ≥⟨H(y), b y −y⟩= ⟨H(b y), b y −y⟩ [since H is skew-symmetric] = ⟨a, b y −y⟩−⟨Sb y, y⟩+ ⟨Sb y, b y⟩= ⟨a, b y −y⟩−⟨Sb y, y⟩= P t λt[⟨a, yt −y⟩−⟨Syt, y⟩] [due to S∗= −S] [due to b y = P t λtyt and P t λt = 1] = P t λt[⟨a, yt −y⟩+ ⟨Syt, yt −y⟩] = P t λt⟨H(yt), yt −y⟩.
[due to S∗= −S] Thus, P t λt⟨H(yt), yt −y⟩≤ϵvi(b y|H, Y ) for all y ∈Y , so that Res CN ≤ϵvi(b y|H, Y ).
Corollary 2.1 Assume we are in the Special case, so that Y = V × W is a direct product of two convex compact sets, and the monotone operator H is associated with a convex-concave function f(v, w). Assume also that f is bilinear: f(v, w) = ⟨a, v⟩+ ⟨b, w⟩+ ⟨w, Av⟩, so that H is affine and skew-symmetric. Then for every y ∈Y it holds ϵsad(y|f, V, W) ≤ϵvi(y|H, Y ).
(5) Proof. Consider accuracy certificate C1 = {y1 = y, λ1 = 1, H(y1)}; for this certificate, b y as defined in Lemma 2.2 is just y. Therefore, by Lemma 2.2, Res{C1) = ϵvi(y|H, Y ). This equality combines with Lemma 2.1 to imply (5).
3 Representations of Monotone Operators 3.1 Outline To explain the origine of the developments to follow, let us summarize the approach to solving convex minimization problems on domains given by Linear Minimization Oracles (LMOs), developed in . The principal ingredient of this approach is a Fenchel-type representation of a convex function 5 f : X →R defined on a convex subset X of Euclidean space E; by definition, such a representation is f(x) = min y∈Y [⟨x, Ay + a⟩−ψ(y)] , (6) where Y is a convex subset of Euclidean space F and ψ : Y →R is convex. Assuming for the sake of simplicity that X, Y are compact and ψ is continuously differentiable on Y , representation (6) allows to associate with the primal problem Opt(P) = min x∈X f(x) (P) its dual Opt(D) = max y∈Y f∗(y) = min x∈X⟨x, Ay + a⟩−ψ(y) (D) with the same optimal value. Observe that the first order information on the (concave) objective of (D) is readily given by the first order information on ψ and the information provided by an LMO for X. As a result, we can solve (D) by, say, a proximal type First Order Method, provided that Y is “proximal-friendly,” i.e., allows for “implementable” proximal algorithms. The crucial in this approach question of how to recover a good approximate solution to the problem of interest (P) from the information collected when solving (D) is addressed via the machinery of accuracy certificates [20, 5].
In the sequel, we intend to apply a similar scheme to the situation where the role of (P) is played by a variational inequality with monotone operator on a convex compact domain X given by an LMO. Our immediate task is to outline informally what a Fenchel-type representation of a monotone operator is and how we intend to use such a representation. To this end note that (P) and (D) can be reduced to variational inequalities with monotone operators, specifically • the “primal” v.i.
stemming from (P).
The domain of this v.i.
is X, and the operator is f ′(x) = Ay(x) + a, where y(x) is a maximizer of the function ⟨x, Ay⟩−ψ(y) over y ∈Y , or, which is the same, a (strong) solution to the v.i. given by the domain Y and the monotone operator y 7→G(y) −A∗x, where G(y) = ψ′(y); • the “dual” v.i. stemming from (D). The domain of this v.i. is Y , and the operator is y 7→ G(y) −A∗x(y), where x(y) is a minimizer of ⟨x, Ay + a⟩over x ∈X.
Observe that both operators in question are described in terms of a monotone operator G on Y and affine mapping y 7→Ay + a : F →E; in the above construction G was the gradient field of ψ, but the construction of the primal and the dual v.i.’s makes sense whenever G is a monotone operator on Y satisfying minimal regularity assumptions. The idea of the approach we are about to develop is as follows: in order to solve a v.i. with a monotone operator Φ and domain X given by an LMO, A. We represent Φ in the form of Φ(x) = Ay(x) + a, where y(x) is a strong solution to the v.i. on Y given by the operator G(y) −A∗x, G being an appropriate monotone operator on Y .
It can be shown that a desired representation always exists, but by itself existence does not help much – we need the representation to be suitable for numerical treatment, to be available in a “closed computation-friendly form.” We show that “computation-friendly” representations of 6 monotone operators admit a kind of fully algorithmic calculus which, for all basic monotonicity-preserving operations, allows to get straightforwardly a desired representation of the result of an operation from the representations of the operands. In view of this calculus, “closed analytic form” representations, allowing to compute efficiently the values of monotonous operators, automatically lead to required computation-friendly representations.
B. We use the representation from A to build the “dual” v.i. with domain Y and the operator Ψ(y) = G(y)−A∗x(y), with exactly the same x(y) as above, that is, x(y) ∈Argminx∈X⟨x, Ay+ a⟩. We sill see that Ψ is monotone, and that usually there is a significant freedom in choosing Y ; in particular, we typically can choose Y to be “proximal-friendly.” C. We solve the dual v.i. by an algorithm, like Mirror Descent or Mirror Prox, which produce necessary accuracy certificates. We will see – and this is our main result – that such a certificate CN can be converted straightforwardly into a feasible solution xN to the v.i. of interest such that ϵvi(xN|Φ, X) ≤Res(CN). As a result, if the certificates in question are good, meaning that the resolution of CN as a function of N obeys the standard efficiency estimates of the algorithm used to solve the dual v.i., we solve the v.i. of interest with the same efficiency estimate as the one for the dual v.i. It remains to note that most of the existing first order algorithms for solving v.i.’s with monotone operators (various versions of polynomial time cutting plane algorithms, like the Ellipsoid method, Subgradient/Mirror Descent, and different bundle-level versions of Mirror Descent) indeed produce good accuracy certificates, see [20, 5].
3.2 The construction 3.2.1 Situation Consider the situation where we are given • a continuous monotone operator G(y) : F →F on a Euclidean space F, • a nonempty convex compact set Y ⊂F, • a nonempty convex compact set X in Euclidean space E, • an affine mapping y 7→Ay + a : F →E.
These data give rise to two operators: “primal” Φ : E →E which is monotone, and “dual” Ψ : F →F which is antimonotone (that is, −Ψ is monotone).
7 3.2.2 Primal monotone operator The primal operator Φ : E →E is defined by Φ(x) = Ay(x) + a : y(x) ∈Y, ⟨A∗x −G(y(x)), y(x) −y⟩≥0, ∀y ∈Y.
(7) Observe that required y(x) do exist: these are just strong solutions to the variational inequality given by the monotone operator G(y) −A∗x and the domain Y .
Now, with x′, x′′ ∈E, setting y(x′) = y′, y(x′′) = y′′, so that y′, y′′ ∈Y , we have ⟨Φ(x′) −Φ(x′′), x′ −x′′⟩= ⟨Ay′ −Ay′′, x′ −x′′⟩= ⟨y′ −y′′, A∗x′ −A∗x′′⟩ = ⟨y′ −y′′, A∗x′⟩+ ⟨y′′ −y′, A∗x′′⟩ = ⟨y′ −y′′, A∗x′ −G(y′)⟩+ ⟨y′′ −y′, A∗x′′ −G(y′′)⟩+ ⟨G(y′), y′ −y′′⟩+ ⟨G(y′′), y′′ −y′⟩ = ⟨y′ −y′′, A∗x′ −G(y′)⟩ | {z } ≥0 due to y′ = y(x′) + ⟨y′′ −y′, A∗x′′ −G(y′′)⟩ | {z } ≥0 due to y′′ = y(x′′) + ⟨G(y′) −G(y′′), y′ −y′′⟩ | {z } ≥0 since G is monotone ≥0.
Thus, Φ(x) is monotone. We call (7) a representation of the monotone operator Φ, and the data F, A, a, y(·), G(·), Y – the data of the representation. We also say that these data represent Φ.
Given a domain X ⊂E and a monotone operator ¯ Φ on this domain, we say that data F, A, a, y(·), G(·), Y of the above type represent ¯ Φ on X, if the monotone operator Φ represented by these data coincides with ¯ Φ on X.
3.2.3 Dual operator Operator Ψ : F →F is given by Ψ(y) = A∗x(y) −G(y) : x(y) ∈X, ⟨Ay + a, x(y) −x⟩≤0, ∀x ∈X (8) (in words: Ψ(y) = A∗x(y) −G(y), where x(y) minimizes ⟨Ay + a, x⟩over x ∈X). For y′, y′′ ∈F, setting x′ = x(y′), x′′ = x(y′′), so that x′, x′′ ∈X, we have ⟨Ψ(y′) −Ψ(y′′), y′ −y′′⟩= ⟨A∗x′ −A∗x′′, y′ −y′′⟩−⟨G(y′) −G(y′′), y′ −y′′⟩ = ⟨x′ −x′′, Ay′ + a⟩ | {z } ≤0 due to x′ = x(y′) + ⟨x′′ −x′, Ay′′ + a⟩ | {z } ≤0 due to x′′ = x(y′′) −⟨G(y′) −G(y′′), y′ −y′′⟩ | {z } ≥0 since G is monotone ≤0.
Thus, Ψ(·) is antimonotone.
Remark 3.1 Note that computing the value of Ψ at a point y reduces to computing G(y), Ay + a, a single call to the Linear Minimization Oracle for X to get x(y), and computing A∗x(y).
3.3 Calculus of representations 3.3.1 Multiplication by nonnegative constants Let F, A, a, y(·), G(·), Y represent a monotone operator Φ : E →E: Φ(x) = Ay(x) + a : y(x) ∈Y and ⟨A∗x −G(y(x)), y(x) −y⟩≥0 ∀y ∈Y.
For λ ≥0, we clearly have λΦ(x) = [λA]y(x) + [λa] : ⟨[λA]∗x −[λG(y(x))], y(x) −y⟩≥0∀y ∈Y, that is, a representation of λΦ is given by F, λA, λa, y(·), λG(·), Y .
8 3.3.2 Summation Let Fi, Ai, ai, yi(·), Gi(·), Yi, 1 ≤i ≤m, represent monotone operators Φi(x) : E →E: Φi(x) = Aiyi(x) + ai : yi(x) ∈Yi and ⟨A∗ i x −Gi(yi(x)), yi(x) −yi⟩≥0 ∀yi ∈Yi.
Then P i Φi(x) = [A1, ..., Am][y1(x); ...; ym(x)] + [a1; ...; am], y(x) := [y1(x); ...; ym(x)] ∈Y := Y1 × ... × Ym, ⟨[A1, ..., Am]∗x −[G1(y1(x)); ...; Gm(ym(x))], [y1(x); ...; ym(x)] −[y1; ...; ym]⟩ = P i⟨A∗ i x −Gi(yi(x)), yi(x) −yi⟩≥0 ∀y = [y1; ...; ym] ∈Y, so that the data F = F1 × ... × Fm, A = [A1, ..., Am], [a1; ...; am], y(x) = [y1(x); ...; ym(x)], G(y) = [G1(y1); ...; Gm(ym)], Y = Y1 × ... × Ym represent P i Φi(x).
3.3.3 Affine substitution of argument Let F, A, a, y(·), G(·), Y represent Φ : E →E, let H be a Euclidean space and h 7→Qh + q be an affine mapping from H to E. We have b Φ(h) := Q∗Φ(Qh + q) = Q∗(Ay(Qh + q) + a) : y(Qh + q) ∈Y and ⟨A∗[Qh + q] −G(y(Ah + q)), y(Qh + q) −y⟩≥0 ∀y ∈Y ⇒with b A = Q∗A, b a = Q∗a, b G(y) = G(y) −A∗q, b y(h) = y(Qh + q) we have b Φ(h) = b Ab y(h) + b a : b y(h) ∈Y and ⟨b A∗h −b G(b y(h)), b y(h) −y⟩≥0 ∀y ∈Y, that is, F, b A,b a, b y(·), b G(·), Y represent b Φ.
3.3.4 Direct sum Let Fi, Ai, ai, yi(·), Gi(·), Yi, 1 ≤i ≤m, represent monotone operators Φi(xi) : Ei →Ei. Then Φ(x := [x1; ...; xm]) = [Φ1(x1); ...; Φm(xm)] = Diag{A1, ..., Am}y(x) + [a1; ...; am] : y(x) := [y1(x1); ...; ym(xm)] ∈Y := Y1 × ... × Ym and ⟨Diag{A∗ 1, ..., A∗ m}[x1; ...; xm] −[G1(y1(x1)); ...; Gm(ym(xm))], y(x) −[y1; ...; ym]⟩ = P i⟨A∗ i x −Gi(yi(xi)), yi(xi) −yi⟩≥0 ∀y = [y1; ...; ym] ∈Y, so that F = F1 × ... × Fm, A = Diag{A1, ..., Am}, a = [a1; ...; am], y(x) = [y1(x1); ...; ym(xm)], G(y = [y1; ...; ym]) = [G1(y1); ...; Gm(ym)], Y = Y1 × ... × Ym represent Φ : E1 × ... × Em →E1 × ... × Em.
9 3.3.5 Representing affine monotone operators Representing a skew-symmetric operator.
Let X ⊂E be a compact convex set in Euclidean space E, and x 7→Sx : E →E be a skew-symmetric (S∗= −S) linear map. Let us set F = E × E, A = [ 1 2I, S], G = I −I , G(y) = Gy : F →F, and let Y ∈E be a compact convex set which contains the image of X under the linear mapping x 7→G−1A∗x = S 1 2I x.
For x ∈X, setting ¯ y(x) = G−1A∗x we ensure that ¯ y(x) ∈Y and G(¯ y(x)) −A∗x = 0, meaning that ¯ y(x) is a strong solution to the v.i. to which y(x) should be a strong solution. Thus, setting y(x) = G−1A∗x when x ∈X, the data F, A, a, y(·), G(·), Y represent a monotone operator Φ such that x ∈X ⇒Φ(x) = A[G−1A∗]x + a = Sx + a.
Thus, the restriction of a skew-symmetric operator onto a given convex compact set X admits an explicit representation.
Representing affine operator with positive semidefinite symmetric matrix.
Let X ⊂E be a compact convex set in Euclidean space E, let x 7→Qx : E →F be a linear mapping from E to Euclidean space F, and let a convex compact set Y ⊂F be such that Qx ∈Y whenever x ∈X.
Setting G(y) = y : F →F, for x ∈X, the solution y(x) to the variational inequality given by the monotone operator G(y)−Qx : F →F and Y is just y(x) = Qx, so that the data F, A = Q∗, a, y(·), G(·), Y represent the operator which, for x ∈X, is given by Φ(x) = Ay(x) + a = Q∗Qx + a.
Thus, the restriction of an affine monotone operator with symmetric positive semidefinite matrix onto a convex compact set admits an explicit representation.
Combined with the calculus rules, the last two constructions imply that the restriction of an affine monotone operator on a convex compact set admits an explicit representation.
3.3.6 Representing gradient fields Let f(x) be a convex function given by Fenchel-type representation f(x) = max y∈Y {⟨x, Ay + a⟩−ψ(y)} , (9) where Y is a convex compact set in Euclidean space F, and ψ(·) : F →R is continuously differentiable convex function. Denoting by y(x) a maximizer of ⟨x, Ay⟩−ψ(y) over y, observe that Φ(x) := Ay(x) + a is a subgradient field of f, and that this monotone operator is given by a representation with the data F, A, a, y(·), G(·) := ∇ψ(·), Y .
10 4 Main result Consider the situation described in section 3.2. Thus, we are given Euclidean space E, a convex compact set X ⊂E and a monotone operator Φ : E →E represented according to (7), the data being F, A, a, y(·), G(·), Y . We denote by Ψ : F →F the dual (antimonotone) operator induced by the data X, A, a, y(·), G(·), see (8). Our goal is to solve variational inequality given by Φ, X, and our main observation is that a good accuracy certificate for the variational inequality given by −Ψ, Y induces an equally good solution to the variational inequality given by (Φ, X). The exact statement is as follows.
Theorem 4.1 Let X ⊂E be a convex compact set and Φ : E →E be a monotone operator repre-sented, in the above sense, by data F, A, a, y(·), G(·), Y . Let also Ψ : F →F be the antimonotone operator as defined above by the data X, A, a, y(·), G(·). Let, finally, CN = {yt, λt, −Ψ(yt)}N t=1 be an accuracy certificate associated with the monotone operator [−Ψ] and Y . Setting xt = x(yt) (these points are byproducts of computing Ψ(yt), 1 ≤t ≤N) and b x = N X t=1 λtxt ∈X, we ensure that ϵvi(b x|Φ, X) ≤Res CN .
(10) The proof of the theorem is given in Section A.1 of the appendix.
In view of Theorem 4.1, given a representation of the monotone operator Φ participating in the v.i. of interest VI(Φ, X), we can reduce solving the v.i. to solving the dual v.i. VI(−Ψ, Y ) by an algorithm producing good accuracy certificates. Below we discuss in details the situation when the latter algorithm is either Mirror Descent (MD) [14, Chapter 5], or Mirror Prox (MP) (, see also [14, Chapter 6] and ).
4.1 Mirror Descent and Mirror Prox algorithms Preliminaries.
Saddle Point MD and MP are algorithms for solving convex-concave saddle point problems and variational inequalities with monotone operators1. The algorithms are of proximal type, meaning that in order to apply the algorithm to a v.i. VI(H, Y ), where Y is a convex compact set in Euclidean space Ey and H is a monotone operator on Y , one needs to equip Ey with a norm ∥· ∥, and Y - with a continuously differentiable distance generating function (d.-g.f.) ω(·) : Y →R compatible with ∥· ∥, meaning that ω is strongly convex, modulus 1, w.r.t. ∥· ∥. We call ∥· ∥, ω(·) proximal setup for Y . This setup gives rise to 1MD algorithm originates from [17, 18]; its modern proximal form was developed in . MP was proposed in .
For the most present exposition of the algorithms, see [14, Chapters 5,6] and .
11 • ω-center yω = argminy∈Y ω(y) of Y , • Bregman distance Vy(z) = ω(z) −ω(y) −⟨ω′(y), z −y⟩≥1 2∥z −y∥2 where the concluding inequality is due to strong convexity of ω, • ω-size of Y Ω= 2 q max Y ω −min Y ω.
Due to the origin of yω, we have Vyω(y) ≤1 2Ω2 for all y ∈Y , implying that ∥y −yω∥≤Ωfor all y ∈Y .
Given y ∈Y , the prox-mapping with center u is defined as Proxy(ξ) = argmin z∈Y [Vy(z) + ⟨ξ, z⟩] = argmin z∈Y [ω(z) + ⟨ξ −ω′(y), z⟩] : Ey →Y.
The algorithms.
Let Y be a convex compact set in Euclidean space Ey, H : Y →Ey be a vector field, and ∥· ∥, ω(·) be a proximal setup for Y . As applied to (H, Y ), MD is the recurrence y1 = yω; yt 7→ yt+1 = Proxyt(γtH(yt)), t = 1, 2, ...
(11) MP is the recurrence y1 = yω; yt 7→ zt = Proxyt(γtH(yt)) 7→yt+1 = Proxyt(γtH(zt)), t = 1, 2, ...
(12) In both MD and MP, γt > 0 are stepsizes. The most important to us properties of these recurrences are as follows.
Proposition 4.1 For N = 1, 2, ..., consider the accuracy certificate CN = yt ∈Y, λN t := γt " N X τ=1 γτ #−1 , H(yt) N t=1 , associated with (11). Then Res(CN) ≤Ω2 + PN t=1 γ2 t ∥H(yt)∥2 ∗ 2 PN t=1 γt , (13) where ∥· ∥∗is the norm conjugate to ∥· ∥: ∥ξ∥∗= max∥x∥≤1⟨ξ, x⟩.
In particular, if ∀y ∈Y : ∥H(y)∥∗≤M (14) with some finite M ≥0, then setting, N being fixed, (a) : γt = Ω M √ N , 1 ≤t ≤N, or (b) : γt = Ω ∥H(yt)∥∗ √ N , 1 ≤t ≤N, (15) one has Res(CN) ≤ΩM √ N .
(16) 12 Proposition 4.2 For N = 1, 2, ..., consider the accuracy certificate CN = zt ∈Y, λN t := γt " N X τ=1 γτ #−1 , H(zt) N t=1 , associated with (12). Then, setting dt = γt⟨H(zt), zt −yt+1⟩−Vyt(yt+1), (17) we have Res(CN) ≤ 1 2Ω2 + PN t=1 dt PN t=1 γt (18) dt ≤ 1 2 γ2 t ∥H(zt) −H(yt)∥2 ∗−∥yt −zt∥2 , (19) where ∥· ∥∗is the norm conjugate to ∥· ∥. In particular, if ∀y, y′ ∈Y : ∥H(y) −H(y′)∥∗≤L∥y −y′∥+ M (20) with some finite L ≥0, M ≥0, then setting, N being fixed, γt = 1 √ 2 min 1 L, Ω M √ N , 1 ≤t ≤N, (21) one has Res(CN) ≤1 √ 2 max Ω2L N , ΩM √ N .
(22) To make the text self-contained, we present the proofs of these known results in Appendix.
5 Illustration The problem.
We illustrate our construction by considering the following problem (“matrix com-pletion with spectral norm fit”):2 Opt(P) = min u∈Rpu×qu:∥u∥≤1 f(u) := ∥Au −b∥∗ (23) where Rp×q is the space of p × q real matrices, ∥x∥= P i σi(x) is the nuclear norm on this space (sum of the singular values σi(x)), ∥· ∥∗is the spectral norm (the largest singular value), and A is a linear mapping from Rpu×qu to Rpb×qb. We are interested in the “large-scale” case, where the sizes of pu, qu of u are large enough to make the full singular value decomposition of a pu × qu matrix prohibitively time consuming, but, at the same time, computing the leading singular vectors and the leading singular value of a pu × qu or a pb × qb matrix (which, computationally, is by far easier task than finding full singular value decomposition) still can be carried out in reasonable time. The first assumption seemingly rules out the possibility to solve (23) by proximal type First Order algorithms, not speaking about interior point ones.
2A more interesting for applications problem (cf. [2, 3, 16]) would be Opt = min u∈Rpu×qu {∥u∥: ∥Au −b∥∗≤δ} ; applying the approach from , this problem can be reduced to a “small series” of problems (23).
13 Processing the problem.
We rewrite (23) as a bilinear saddle point problem Opt(P) = min u∈U max v∈V ⟨v, [Au −b]⟩ | {z } f(u,v) U = {u ∈Rpu×qu : ∥u∥≤1}, V = {v ∈Rpb×qb : ∥v∥≤1} (24) where ⟨·, ·⟩is the Frobenius inner product. The domain X of the problem is the direct product of two unit nuclear norm balls; minimizing a linear form over this domain reduces to minimizing, given ξ and η, the linear forms Tr(uξT), Tr(vηT) over {u ∈Rpu×qu : ∥u∥≤1}, resp., {v ∈Rpv×qv : ∥v∥≤1}, which, in turn, reduces to computing the leading singular vectors and singular values of ξ and η.
The monotone operator associated with (24) is affine and skew-symmetric: Φ(u, v) = [∇uf(u, v); −∇vf(u, v)] = [A∗v; −Au] + [0; b] : Rpu×qu × Rpb×qb | {z } E →E, where A∗v : Rpb×qb →Rpu×qu is the mapping conjugate to A: ∀(u ∈Rpu×qu, v ∈Rpb×qb) : ⟨A∗v, u⟩F = ⟨v, Au⟩F.
Assuming from now on that A is of spectral norm at most 1 (i.e., ∥Au∥F ≤∥u∥F for all u; the latter can always be achieved by scaling), we can represent the restriction of Φ on X by the data (cf.
section 3.3.5) F = Rpu×qu × Rpu×qu Ay + a = ξ Aη + b , y = [ξ; η] ∈F (ξ ∈Rpu×qu, η ∈Rpu×qu), G([ξ; η] |{z} y ) = [−η; ξ] : F →F Y = {y = [ξ; η] ∈F : ∥ξ∥F ≤1, ∥η∥F ≤1} (25) Indeed, in the notation from section 3.2, for x = [u; v] ∈X = {u ∈Rpu×qu × Rpb×qb : ∥u∥≤ 1, ∥v∥≤1}, the solution y(x) = [ξ(x); η(x)] to the linear system A∗x = G(y) is given by η(x) = −u, ξ(x) = A∗v, so that both components of y(x) are of Frobenius norm ≤1 (recall that spectral norm of A is ≤1, and therefore y(x) ∈Y ). Besides this, Ay(x) + a = [ξ(x); Aη(x) + b] = [A∗v; b −Au] = Φ(u, v).
We conclude that when x = [u; v] ∈X, just defined y(x) meets all requirements from (7), and thus the data F, A, a, y(·), G(·), Y given by (25) indeed represent the monotone operator Φ on X.
The dual (antimonotone) operator Ψ given by the data F, A, a, y(·), G(·), Y is Ψ( y z}|{ [ξ; η]) = A∗x(y) −G(y) = [u(y) + η; A∗v(y) −ξ], u(y) ∈Argmin ∥u∥≤1 ⟨u, ξ⟩, v(y) ∈Argmin ∥v∥≤1 ⟨v, Aη + b⟩.
(26) We use the Euclidean proximal setup for Y , i.e., we equip the space F embedding Y with the Frobenius norm and take, as the d.-g.f. for Y , the function ω(ξ, η) = 1 2 ∥ξ∥2 F + ∥η∥2 F : F := Rpu×qu × Rpu×qu →R, 14 resulting in Ω= √ 2. Furthermore, from (26) and the fact that the spectral norm of A is bounded by 1 it follows that the monotone operator Θ(y) = −Ψ(y) : Y →F satisfies (14) with M = 2 √ 2 and (20) with L = 0 and M = 4 √ 2.
Remark.
Theorem 4.1 combines with Corollary 2.1 to imply that when converting an accuracy certificate CN for the dual v.i. VI(−Ψ, Y ) into a feasible solution b xN to the primal v.i. VI(Φ, X), we ensure that ϵsad(b xN|f, U, V ) ≤Res(CN), (27) with f, U, V given by (24). In other words, in the representation b xN = [b uN; b vN], b uN is a feasible solution to problem (23) (which is the primal problem associated with (24)), and b vN is a feasible solution to the problem Opt(D) = max v∈V min u∈U ⟨v, Au −b⟩= max v∈V f(v) := −∥A∗v∥∗−⟨b, v⟩ , (which is the dual problem associated with (24)) with the sum of non-optimalities, in terms of respective objectives, ≤Res(CN). Computing f(b v) (which, together with computing f(b u), takes a single call to LMO for X), we get a lower bound on Opt(P) = Opt(D) which certifies that f(b u) −Opt(P) ≤Res(CN).
Numerical illustration.
Here we report on some numerical experiments with problem (23). In these experiments, we used pb = qb =: m, pu = qu =: n, with n = 2m, and the mapping A given by Au = k X i=1 ℓiurT i , (28) with generated at random m×n factors ℓi, ri scaled to get ∥A∥∗≈1. In all our experiments, we used k = 2. Matrix b in (23) was built as follows: we generated at random n × n matrix ¯ u with ∥¯ u∥≈1 and Rank(¯ u) ≈√n, and took b = A¯ u+δ, with randomly generated m×m matrix δ of spectral norm about 0.01.
In our experiments, the dual v.i. VI(−Ψ, Y ) was solved by MD algorithm with N = 512 steps for all but the largest instance, where N = 257 was used. The stepsizes were proportional, with coefficient of order of 1, to those given by (15.b) with ∥· ∥≡∥· ∥∗= ∥· ∥F and Ω= √ 2 3; the coefficient was tuned empirically in pilot runs on small instances and was never changed afterwards.
We also used two straightforward “tricks”: • Instead of considering one accuracy certificate, CN = {yt, λN t = 1/N, −Ψ(yt)}N t=1, we built a “bunch” of certificates Cν µ = yt, λt = 1 ν −µ + 1, −Ψ(yt) ν t=µ , where µ runs through a grid in {1, ..., N} (in this implementation, a 16-element equidistant grid), and ν ∈{µ, µ + 1, ..., N} runs through another equidistant grid (e.g., for the largest 3As we have already mentioned, with our proximal setup, the ω-size of Y is ≤ √ 2, and (14) is satisfied with M = 2 √ 2.
15 problem instance, the grid {1, 9, 17, ..., 256}).
For every of these certificates, we computed its resolution and identified the best (with the smallest resolution) certificate obtained so far.
Every 8 steps, the best certificate was used to compute the current approximate solution to (24) along with the saddle point inaccuracy of this solution.
• When applying MD to problem (24), the “dual iterates” yt = [ξt; ηt] and the “primal iterates” xt := x(yt) = [ut; vt] are pairs of matrices, with n × n matrices ξt, ηt, ut and m × m matrix vt (recall that we are in the case of pu = qu = n, pb = qb = m). It is easily seen that with A given by (28), the matrices ξt, ηt, ut are linear combinations of rank 1 matrices αiβT i , 1 ≤i ≤(k +1)t, and vt are linear combinations of rank 1 matrices δiϵT i , 1 ≤i ≤t, with on-line computable vectors αi, βi, δi, ϵi; every step of MD adds k + 1 new α- and k + 1 new β-vectors, and a pair of new δ- and ϵ-vectors. Then the matrix iterates are represented by the vectors of coefficients in the above rank 1 decompositions (let us call this representation incremental), so that the computations performed at a step of MD, including computing the leading singular vectors by straightforward power iterations, are as if the standard representations of matrices were used, but all these matrices were of the size (at most) n×[(k+1)N], and not n×n and m×m, as they actually are. In our experiments, for k = 2 and N ≤512, this incremental representation of iterates yields meaningful computational savings (e.g., by factor of 6 for n = 8192) as compared to the plain representation of iterates by 2D arrays.
Typical results of our preliminary experiments are presented in Table 1. There Ct stands for the best certificate found in course of t steps, and Gap(Ct) denotes the saddle point inaccuracy of the solution to (24) induced by this certificate (so that Gap(Ct) is a valid upper bound on the inaccuracy, in terms of the objective, to which the problem of interest (23) was solved in course of t steps). The comments are as follows: 1. The results clearly demonstrate “nearly linear”, and not quadratic, growth of running time with m, n; this is due to the incremental representation of iterates.
2. When evaluating the “convergence patterns” presented in the table, one should keep in mind that we are dealing with a method with slow O(1/ √ N) convergence rate, and from this per-spective, 50-fold reduction in resolution in 512 steps is not that bad.
3. A natural alternative to the proposed approach would be to solve the saddle point problem (24) “as it is,” by applying to the associated primal v.i. (where the domain is the product of two nuclear norm balls and the operator is Lipschitz continuous and even skew symmetric) a proximal type saddle point algorithm and computing the required prox-mappings via full singular value decompositions. The state-of-the-art MP algorithm when applied to this problem exhibits O(1/N) convergence rate;4 yet, every step of this method would require 2 SVD’s of n × n, and 2 SVD’s of m × m matrices. As applied to the primal v.i., MD exhibits O(1/ √ N) convergence rate, but the steps are cheaper – we need one SVD of n × n, and one SVD of an m × m matrix, and we are unaware of a proximal type algorithm for the primal v.i. with cheaper iterations. For the sizes m, n, k we are interested in, the computational effort required 4For the primal v.i., (20) holds true for some L > 0 and M = 0. Moreover, with properly selected proximal setup for (23) the complexity bound (22) becomes Res(CN) ≤O(1) p ln(n) ln(m)/N.
16 Iteration count t 1 65 129 193 257 321 385 449 512 Res(Ct) 1.5402 0.1535 0.0886 0.0621 0.0487 0.0389 0.03288 0.0293 0.0278 n = 1024 Res(C1)/Res(Ct) 1.00 10.04 17.38 24.79 31.61 39.61 46.84 52.64 55.41 m = 512 Gap(Ct) 0.1269 0.0239 0.0145 0.0103 0.0075 0.0063 0.0042 0.0040 0.0040 k = 2 Gap(Ct)/Gap(Ct) 1.00 5.31 8.78 12.38 17.03 20.20 29.98 31.41 31.66 cpu, sec 0.2 9.5 27.6 69.1 112.6 218.1 326.2 432.6 536.4 Res(Ct) 1.4809 0.1559 0.0842 0.0607 0.0471 0.0391 0.0337 0.0306 0.0285 n = 2048 Res(C1)/Res(Ct) 1.00 9.50 17.59 24.38 31.43 37.88 43.89 48.36 51.96 m = 1024 Gap(Ct) 0.1329 0.0196 0.0119 0.0075 0.0053 0.0041 0.0036 0.0034 0.0027 k = 2 Gap(Ct)/Gap(Ct) 1.00 6.79 11.21 17.81 25.09 32.29 37.23 38.70 50.06 cpu, sec 0.7 38.0 101.1 206.3 314.1 508.9 699.0 884.9 1070.0 Res(Ct) 1.4845 0.1476 0.0891 0.0605 0.0491 0.0395 0.0329 0.0292 0.0275 n = 4096 Res(C1)/Res(Ct) 1.00 10.06 16.66 24.53 30.25 37.60 45.17 50.85 53.95 m = 2048 Gap(Ct) 0.1239 0.0222 0.0139 0.0108 0.0086 0.0041 0.0037 0.0035 0.0035 k = 2 Gap(Ct)/Gap(Ct) 1.00 5.57 8.93 11.48 14.40 30.48 33.14 35.76 35.77 cpu, sec 2.2 103.5 257.6 496.9 742.5 1147.8 1564.4 1981.4 2401.0 Res(Ct) 1.4778 0.1391 0.0888 0.0590 0.0469 0.0386 0.0324 0.0289 0.0270 n = 8192 Res(C1)/Res(Ct) 1.00 10.63 16.64 25.06 31.53 38.29 45.68 51.10 54.76 m = 4096 Gap(Ct) 0.1193 0.0232 0.0134 0.0108 0.0054 0.0040 0.0035 0.0034 0.0034 k = 2 Gap(Ct)/Gap(Ct) 1.00 5.14 8.90 11.08 22.00 29.83 33.93 34.85 35.14 cpu, sec 6.5 289.9 683.8 1238.1 1816.0 2724.5 3648.3 4572.2 5490.8 Res(Ct) 1.4566 0.1154 0.0767 0.0556 0.0447 n = 16384 Res(C1)/Res(Ct) 1.00 12.62 19.00 26.22 32.60 m = 8192 Gap(Ct) 0.1196 0.0214 0.0146 0.0101 0.0085 k = 2 Gap(Ct)/Gap(Ct) 1.00 5.60 8.19 11.82 14.01 cpu, sec 21.7 920.4 2050.2 3492.4 4902.2 Table 1: MD on problem (23). Platform: 3.40 GHz i7-3770 desktop with 16 GB RAM, 64 bit Windows 7 OS.
by the outlined SVD’s is, for all practical purposes, the same as the overall effort per step.
Taking into account the actual SVD cpu times on the platform used in our experiments, the overall running times presented in Table 1, i.e., times required by 512 steps of MD as applied to the dual v.i., allow for the following iteration counts M for MP as applied to the primal v.i.: n 1024 2048 4096 8192 M 406 72 17 4 and for twice larger iteration counts for MD. From our experience, for n = 1024 (and perhaps for n = 2048 as well), MP algorithm as applied to the primal v.i. would yield solutions of better quality than those obtained with our approach. It, however, would hardly be the case, for both MP and MD, when n = 4096, and definitely would not be the case for n = 8192. Finally, with n = 16384, CPU time used by the 257-step MD as applied to the dual v.i. is hardly enough to complete just one iteration of MD as applied to the primal v.i. We believe these data demostrate that the approach developed in this paper has certain practical potential.
References Beck, A., Teboulle, M. “Mirror descent and nonlinear projected subgradient methods for convex optimization” – Operations Research Letters 31:3 (2003), 167–175.
17 Candes, E. J., Plan, Y. Matrix completion with noise. Proceedings of the IEEE, 98(6) (2010), 925-936.
Candes, E. J., Plan, Y. Tight oracle inequalities for low-rank matrix recovery from a minimal number of noisy random measurements. Information Theory, IEEE Transactions on, 57(4) (2011), 2342-2359.
Chen, G., Teboulle, M., “Convergence analysis of a proximal-like mini- mization algorithm using Bregman functions” –SIAM J. Optim. 3(3) (1993), 538–543.
Cox, B., Juditsky, A., Nemirovski, A. “Dual subgradient algorithms for large-scale nons-mooth learning problems” – Mathematical Programming Series B (2013), Online First, DOI 10.1007/s10107-013-0725-1. E-print: arXiv:1302.2349.
Demyanov, V., Rubinov, A. Approximate Methods in Optimization Problems – Elsevier, Ams-terdam 1970.
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Dudik, M., Harchaoui, Z., Malick, J. “Lifted coordinate descent for learning with trace-norm regularization” – In AISTATS (2012).
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Freund, R., Grigasy. P. “New Analysis and Results for the Conditional Gradient Method” (2013) – submitted to Mathematical Programming, E-print: Harchaoui, Z., Douze, M., Paulin, M., Dudik, M., Malick, J. “Large-scale image classification with trace-norm regularization” – In CVPR (2012).
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Juditsky, A., Kilin¸ c Karzan, F., Nemirovski, A., “Randomized first order algorithms with appli-cations to ℓ1-minimization” – Mathematical Programming Online First (2012), DOI: 10.1007/s10107-012-0575-2 18 Juditsky, A., Kilin¸ c Karzan, F., Nemirovski, A. On unified view of nullspace-type conditions for recoveries associated with general sparsity structures. Linear Algebra and its Applications, see also arXiv preprint arXiv:1207.1119 (2012) .
Nemirovski, A., Yudin, D., Problem Complexity and Method Efficiency in Optimization – Nauka Publishers, Moscow, 1978 (in Russian); John Wiley, New York (1983) (in English) Nemirovskii, A. “Efficient iterative algorithms for variational inequalities with monotone op-erators” – Ekonomika i Matematicheskie Metody 17:2 (1981), 344–359 (in Russian; Engllish translation: Matekon) Nemirovski, A., “Prox-method with rate of convergence O(1/t) for variational inequalities with Lipschitz continuous monotone operators and smooth convex-concave saddle point problems” – SIAM Journal on Optimization 15 (2004), 229–251.
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A Proofs A.1 Proof of Theorem 4.1 Proof. Let us set ϵ = Res(CN). Observe that (a) : Φ(x) = Ay(x) + a : (a.1) : y(x) ∈Y, (a.2) : ⟨y(x) −y, A∗x −G(y(x))⟩≥0 ∀y ∈Y ; (b) : Ψ(y) = A∗x(y) −G(y) : (b.1) : x(y) ∈X, (b.2) : ⟨Ay + a, x −x(y)⟩≥0 ∀x ∈X; (c) : P t λt⟨A∗xt −G(yt), z −yt⟩≤ϵ, ∀z ∈Y ; where xt = x(yt).
Let for ¯ x ∈X, ¯ y = y(¯ x), and let b y = P t λtyt, with b y, ¯ y ∈Y by (a.1). Since G is monotone, for all t ∈{1, ..., N} we have ⟨¯ y −yt, G(¯ y) −G(yt)⟩≥0 ⇒ ⟨¯ y, G(¯ y)⟩≥⟨yt, G(¯ y)⟩+ ⟨¯ y, G(yt)⟩−⟨yt, G(yt)⟩∀t ⇒ ⟨¯ y, G(¯ y)⟩≥P t λt [⟨yt, G(¯ y)⟩+ ⟨¯ y, G(yt)⟩−⟨yt, G(yt)⟩] [since λt ≥0 and P t λt = 1], 19 and we conclude that ⟨¯ y, G(¯ y)⟩−⟨b y, G(¯ y)⟩≥ X t λt [⟨¯ y, G(yt)⟩−⟨yt, G(yt)⟩] .
(29) We now have ⟨Φ(¯ x), ¯ x −P t λtxt⟩ = ⟨A¯ y + a, ¯ x −P t λtxt⟩ [by (a)] = ⟨¯ y, A∗¯ x −P t λtA∗xt⟩+ ⟨a, ¯ x −P t λtxt⟩ = ⟨¯ y, A∗¯ x −G(¯ y)⟩+ ⟨¯ y, G(¯ y) −P t λtA∗xt⟩+ ⟨a, ¯ x −P t λtxt⟩ ≥ ⟨b y, A∗¯ x −G(¯ y)⟩+ ⟨¯ y, G(¯ y) −P t λtA∗xt⟩+ ⟨a, ¯ x −P t λtxt⟩ [by (a.2) with y = b y and due to ¯ y = y(¯ x)] = ⟨b y, A∗¯ x⟩+ [⟨G(¯ y), ¯ y⟩−⟨G(¯ y), b y⟩] −⟨¯ y, P t λtA∗xt⟩+ ⟨a, ¯ x −P t λtxt⟩ ≥ ⟨b y, A∗¯ x⟩+ P t λt [⟨¯ y, G(yt)⟩−⟨yt, G(yt)⟩] −⟨¯ y, P t λtA∗xt⟩+ ⟨a, ¯ x −P t λtxt⟩ [by (29)] = P t λt⟨yt, A∗¯ x⟩+ P t λt [⟨¯ y, G(yt)⟩−⟨yt, G(yt)⟩−⟨¯ y, A∗xt⟩+ ⟨a, ¯ x −xt⟩] [since b y = P t λtyt and P t λt = 1] = P t λt [⟨Ayt, ¯ x −xt⟩+ ⟨Ayt, xt⟩+ ⟨¯ y, G(yt)⟩−⟨yt, G(yt)⟩−⟨¯ y, A∗xt⟩+ ⟨a, ¯ x −xt⟩] = P t λt [⟨yt, A∗xt⟩+ ⟨¯ y, G(yt)⟩−⟨yt, G(yt)⟩−⟨¯ y, A∗xt⟩] + P t λt⟨Ayt + a, ¯ x −xt⟩ = P tλt⟨A∗xt −G(yt), yt −¯ y⟩ | {z } ≥−ϵ by (c) + P tλt⟨Ayt + a, ¯ x −xt⟩ | {z } ≥0 [by (b.2) and due to xt = x(yt)] ≥ −ϵ.
The bottom line is that ⟨Φ(¯ x), b x −¯ x⟩≤ϵ = Res CN ∀¯ x ∈X, and (10) follows.
A.2 Proof of Proposition 4.1 Observe, first, that the optimality conditions in the optimization problem specifying v = Proxy(ξ) imply that ⟨ξ −ω′(y) + ω′(v), z −v⟩≥0, ∀z ∈Y, or ⟨ξ, v −z⟩≤⟨ω′(v) −ω′(y), z −v⟩= ⟨V ′ y(v), z −v⟩, ∀z ∈Y, which, using a remarkable identity ⟨V ′ y(v), z −v⟩= Vy(z) −Vv(z) −Vy(v), can be rewritten equivalently as v = Proxy(ξ) ⇒⟨ξ, v −z⟩≤Vy(z) −Vv(z) −Vy(v) ∀z ∈Y.
(30) 20 Setting y = yt, ξ = γtH(yt), which results in v = yt+1, we get ∀z ∈Y : γt⟨H(yt), yt+1 −z⟩≤Vyt(z) −Vyt+1(z) −Vyt(yt+1), whence, ∀z ∈Y : γt⟨H(yt), yt −z⟩ ≤ Vyt(z) −Vyt+1(z) + [γt⟨H(yt), yt −yt+1⟩−Vyt(yt+1)] | {z } ≤γt∥H(yt)∥∗∥yt−yt+1∥−1 2 ∥yt−yt+1∥2 ≤ Vyt(z) −Vyt+1(z) + 1 2γ2 t ∥H(yt)∥2 ∗.
Summing up these inequalities over t = 1, ..., N and taking into account that for z ∈Y , we have Vy1(z) ≤1 2Ω2 and that VyN+1(z) ≥0, we get (13).
A.3 Proof of Proposition 4.2 Applying (30) to y = yt, ξ = γtH(zt), which results in v = yt+1, we get ∀z ∈Y : γt⟨H(zt), yt+1 −z⟩≤Vyt(z) −Vyt+1(z) −Vyt(yt+1), whence, by the definition (17) of dt, ∀z ∈Y : γt⟨H(zt), zt −z⟩≤Vyt(z) −Vyt+1(z) + dt.
Summing up the resulting inequalities over t = 1, ..., T and taking into account that Vy1(z) ≤1 2Ω2 for all z ∈Y and VyN+1(z) ≥0, we get n X t=1 λN t ⟨H(zt), zt −z⟩≤ 1 2Ω2 + PN t=1 dt PN t=1 γt .
The right hand side in the latter inequality is independent of z. Taking supremum of the left hand side over z ∈Y , we arrive at (18).
Moreover, invoking (30) with y = yt, ξ = γtH(yt) and specifying z as yt+1, we get γt⟨H(yt), zt −yt+1⟩≤Vyt(yt+1) −Vzt(yt+1) −Vyt(zt), whence dt = γt⟨H(zt), zt −yt+1⟩−Vyt(yt+1) ≤γt⟨H(yt), zt −yt+1⟩+ γt⟨H(zt) −H(yt), zt −yt+1⟩−Vyt(yt+1) ≤ −Vzt(yt+1) −Vyt(zt) + γt⟨H(zt) −H(yt), zt −yt+1⟩ ≤ γt∥H(zt) −H(yt)∥∗∥zt −yt+1∥−1 2∥zt −yt+1∥2 −1 2∥yt −zt∥2 ≤ 1 2 γ2 t ∥H(zt) −H(yt)∥2 ∗−∥yt −zt∥2 , as required in (19).
21 |
7914 | https://www.iso.org/standard/62277.html | ISO 20795-1:2013 - Dentistry — Base polymers — Part 1: Denture base polymers
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ISO 20795-1:2013
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ISO 20795-1:2013
Dentistry — Base polymers — Part 1: Denture base polymers
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ISO 20795-1:2013 Dentistry — Base polymers Part 1: Denture base polymers
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Abstract
ISO 20795-1:2013 classifies denture base polymers and copolymers and specifies their requirements. It also specifies the test methods to be used in determining compliance with these requirements. It further specifies requirements with respect to packaging and marking the products and to the instructions to be supplied for use of these materials. Furthermore, it applies to denture base polymers for which the manufacturer claims that the material has improved impact resistance. It also specifies the respective requirement and the test method to be used.
ISO 20795-1:2013 applies to denture base polymers such as those listed below:
poly(acrylic acid esters);
poly(substituted acrylic acid esters);
poly(vinyl esters);
polystyrene;
rubber modified poly(methacrylic acid esters);
polycarbonates;
polysulfones;
poly(dimethacrylic acid esters);
polyacetals (polyoxymethylene);
copolymers or mixtures of the polymers listed in 1 to 9.
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7915 | https://teachy.ai/en/project/high-school-en-US/us-12th-grade/math/exploring-patterns-and-properties-of-pascals-triangle | Activities of Exploring Patterns and Properties of Pascal's Triangle
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Pascal’s Triangle
Project: Exploring Patterns and Properties of Pascal's Triangle
Lara from Teachy
Subject Math
Math
Source Teachy Original
Teachy Original
Topic Pascal’s Triangle
Pascal’s Triangle
Contextualization
Introduction to Pascal’s Triangle
Pascal’s Triangle is a unique numerical construction named after the French mathematician Blaise Pascal. It is an array of binomial coefficients in a triangular pattern that has captivated mathematicians for centuries due to its fascinating and intriguing properties. The triangle starts with a 1 at the top, and each subsequent number is the sum of the two numbers directly above it.
The numbers in Pascal's Triangle have an array of applications in mathematics, number theory, probability theory, algebra, and calculus. They represent the coefficients of the binomial expansion of a binomial raised to a power, which is a fundamental concept in algebra. Moreover, the triangle exhibits various patterns and symmetries, which make it an interesting subject for mathematical exploration.
Real-World Applications of Pascal’s Triangle
The seemingly abstract concept of Pascal’s Triangle has surprising practical applications in our everyday life. It is used in various fields, including computer science, physics, and engineering. For instance, in computer science, it is used in the design of algorithms and data structures. In physics, it is used to study waveforms and calculate probabilities. In engineering, it is used in signal processing and image compression.
One of the most famous applications of Pascal's Triangle is in the field of probability theory, where it is used to calculate the probabilities of various outcomes in a given experiment. It is also used in the study of fractals, a fascinating branch of mathematics that deals with infinitely complex patterns.
Resources for Further Understanding
To delve deeper into the topic and to gain a more thorough understanding of Pascal’s Triangle, students can refer to the following resources:
Pascal's Triangle - Math is Fun
Pascal's Triangle - Brilliant
Pascal's Triangle - Khan Academy
Book: "A Beginner's Guide to Pascal's Triangle" by PatrickJMT
Video: Pascal's Triangle: Introduction and Patterns by Math Antics on YouTube
These resources provide a comprehensive overview of the concept, its history, and its applications, making them ideal for aiding in the understanding of Pascal’s Triangle.
Practical Activity
Activity Title: Exploring the Patterns and Properties of Pascal's Triangle
Objective of the Project:
The project aims to enhance students' understanding of Pascal’s Triangle and its properties through a hands-on and interactive exploration. It will involve constructing the triangle, identifying patterns, and applying the knowledge to solve real-world problems.
Detailed Description of the Project:
Students will be divided into groups of 3 to 5 members. Each group will be given a large sheet of paper, a marker, and a table of numbers. The table will contain the first ten rows of Pascal's Triangle, with the numbers clearly labeled. The students' task is to recreate the triangle on the sheet of paper, filling in the numbers correctly.
After recreating the triangle, students will then be guided through a series of activities to explore the patterns and properties of Pascal's Triangle. These activities will involve identifying symmetries, finding the sums of specific rows or diagonals, and calculating the probabilities of certain outcomes using the triangle.
Necessary Materials:
A large sheet of paper for each group
A marker
A table of numbers representing the first ten rows of Pascal's Triangle
Calculators
Notebook for each student for documenting the process and findings
Detailed Step-by-step for Carrying Out the Activity:
Formation of Groups and Distribution of Materials: Divide the class into groups of 3 to 5 members. Provide each group with the necessary materials.
Recreating Pascal's Triangle: Each group should recreate the triangle on their large sheet of paper using the provided table of numbers. Encourage students to work together and discuss their strategies.
Discussing the Triangle's Properties: After recreating the triangle, guide students through a discussion on its properties. Encourage them to identify patterns, such as the symmetry along the middle column, the sum of each row, and the sums of the diagonals.
Activities on Pascal's Triangle: Assign the following activities to the groups:
Activity 1: Find the sum of each row and each diagonal of the triangle.
Activity 2: Calculate the probabilities of getting a specific number of heads when flipping a coin a certain number of times.
Activity 3: Identify any other unique patterns or properties in the triangle and explain their significance.
Documentation: Throughout the activities, students should document their process, findings, and insights in their notebooks. This will be used to write the final report.
Discussion: At the end of the activities, groups will discuss their findings and insights with the whole class. This will be an opportunity for students to learn from each other and deepen their understanding of the topic.
Report Writing: After the discussion, each group will write a report detailing their process, findings, and conclusions based on the activities.
Project Deliveries:
The final delivery of the project will be a written report containing the following sections:
Introduction: Contextualize the theme of the project (Pascal's Triangle), its real-world applications, and the objective of the project.
Development: Detail the theory behind Pascal's Triangle, explain the activities in detail, present the methodology used, and discuss the results obtained.
Conclusion: Revisit the main points of the project, explicitly state the learnings obtained, and draw conclusions about the project.
Bibliography: Indicate the sources relied on to work on the project, such as books, web pages, videos, etc.
Each group will submit one report, and the report should reflect the collective effort of all members of the group. This project will be graded based on the completeness of the report, the accuracy of the recreated triangle, the understanding demonstrated in the activities, and the quality of the discussion and collaboration in the group. The project report is due one week after the practical activity.
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7916 | https://en.wikipedia.org/wiki/3D_rotation_group | Jump to content
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Contents
1 Length and angle
2 Orthogonal and rotation matrices
3 Group structure
3.1 Complete classification of finite subgroups
4 Axis of rotation
5 Topology
6 Connection between SO(3) and SU(2)
6.1 Using quaternions of unit norm
6.2 Using Möbius transformations
7 Lie algebra
7.1 A note on Lie algebras
7.2 Isomorphism with 𝖘𝖚(2)
8 Exponential map
9 Logarithm map
10 Uniform random sampling
11 Baker–Campbell–Hausdorff formula
12 Infinitesimal rotations
13 Realizations of rotations
14 Spherical harmonics
15 Generalizations
16 See also
17 Footnotes
18 References
19 Bibliography
3D rotation group
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From Wikipedia, the free encyclopedia
Group of rotations in 3 dimensions
In mechanics and geometry, the 3D rotation group, often denoted SO(3), is the group of all rotations about the origin of three-dimensional Euclidean space under the operation of composition.
By definition, a rotation about the origin is a transformation that preserves the origin, Euclidean distance (so it is an isometry), and orientation (i.e., handedness of space). Composing two rotations results in another rotation, every rotation has a unique inverse rotation, and the identity map satisfies the definition of a rotation. Owing to the above properties (along composite rotations' associative property), the set of all rotations is a group under composition.
Every non-trivial rotation is determined by its axis of rotation (a line through the origin) and its angle of rotation. Rotations are not commutative (for example, rotating R 90° in the x-y plane followed by S 90° in the y-z plane is not the same as S followed by R), making the 3D rotation group a nonabelian group. Moreover, the rotation group has a natural structure as a manifold for which the group operations are smoothly differentiable, so it is in fact a Lie group. It is compact and has dimension 3.
Rotations are linear transformations of and can therefore be represented by matrices once a basis of has been chosen. Specifically, if we choose an orthonormal basis of , every rotation is described by an orthogonal 3 × 3 matrix (i.e., a 3 × 3 matrix with real entries which, when multiplied by its transpose, results in the identity matrix) with determinant 1. The group SO(3) can therefore be identified with the group of these matrices under matrix multiplication. These matrices are known as "special orthogonal matrices", explaining the notation SO(3).
The group SO(3) is used to describe the possible rotational symmetries of an object, as well as the possible orientations of an object in space. Its representations are important in physics, where they give rise to the elementary particles of integer spin.
Length and angle
[edit]
Besides just preserving length, rotations also preserve the angles between vectors. This follows from the fact that the standard dot product between two vectors u and v can be written purely in terms of length (see the law of cosines):
It follows that every length-preserving linear transformation in preserves the dot product, and thus the angle between vectors. Rotations are often defined as linear transformations that preserve the inner product on , which is equivalent to requiring them to preserve length. See classical group for a treatment of this more general approach, where SO(3) appears as a special case.
Orthogonal and rotation matrices
[edit]
Main articles: Orthogonal matrix and Rotation matrix
Every rotation maps an orthonormal basis of to another orthonormal basis. Like any linear transformation of finite-dimensional vector spaces, a rotation can always be represented by a matrix. Let R be a given rotation. With respect to the standard basis e1, e2, e3 of the columns of R are given by (Re1, Re2, Re3). Since the standard basis is orthonormal, and since R preserves angles and length, the columns of R form another orthonormal basis. This orthonormality condition can be expressed in the form
where RT denotes the transpose of R and I is the 3 × 3 identity matrix. Matrices for which this property holds are called orthogonal matrices. The group of all 3 × 3 orthogonal matrices is denoted O(3), and consists of all proper and improper rotations.
In addition to preserving length, proper rotations must also preserve orientation. A matrix will preserve or reverse orientation according to whether the determinant of the matrix is positive or negative. For an orthogonal matrix R, note that det RT = det R implies (det R)2 = 1, so that det R = ±1. The subgroup of orthogonal matrices with determinant +1 is called the special orthogonal group, denoted SO(3).
Thus every rotation can be represented uniquely by an orthogonal matrix with unit determinant. Moreover, since composition of rotations corresponds to matrix multiplication, the rotation group is isomorphic to the special orthogonal group SO(3).
Improper rotations correspond to orthogonal matrices with determinant −1, and they do not form a group because the product of two improper rotations is a proper rotation.
Group structure
[edit]
The rotation group is a group under function composition (or equivalently the product of linear transformations). It is a subgroup of the general linear group consisting of all invertible linear transformations of the real 3-space .
Furthermore, the rotation group is nonabelian. That is, the order in which rotations are composed makes a difference. For example, a quarter turn around the positive x-axis followed by a quarter turn around the positive y-axis is a different rotation than the one obtained by first rotating around y and then x.
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Complete classification of finite subgroups
[edit]
The finite subgroups of are completely classified.
Every finite subgroup is isomorphic to either an element of one of two countably infinite families of planar isometries: the cyclic groups or the dihedral groups , or to one of three other groups: the tetrahedral group , the octahedral group , or the icosahedral group .
Axis of rotation
[edit]
Main article: Axis–angle representation
Every nontrivial proper rotation in 3 dimensions fixes a unique 1-dimensional linear subspace of which is called the axis of rotation (this is Euler's rotation theorem). Each such rotation acts as an ordinary 2-dimensional rotation in the plane orthogonal to this axis. Since every 2-dimensional rotation can be represented by an angle φ, an arbitrary 3-dimensional rotation can be specified by an axis of rotation together with an angle of rotation about this axis. (Technically, one needs to specify an orientation for the axis and whether the rotation is taken to be clockwise or counterclockwise with respect to this orientation).
For example, counterclockwise rotation about the positive z-axis by angle φ is given by
Given a unit vector n in and an angle φ, let R(φ, n) represent a counterclockwise rotation about the axis through n (with orientation determined by n). Then
R(0, n) is the identity transformation for any n
R(φ, n) = R(−φ, −n)
R(π + φ, n) = R(π − φ, −n).
Using these properties one can show that any rotation can be represented by a unique angle φ in the range 0 ≤ φ ≤ π and a unit vector n such that
n is arbitrary if φ = 0
n is unique if 0 < φ < π
n is unique up to a sign if φ = π (that is, the rotations R(π, ±n) are identical).
In the next section, this representation of rotations is used to identify SO(3) topologically with three-dimensional real projective space.
Topology
[edit]
Main article: Hypersphere of rotations
The Lie group SO(3) is diffeomorphic to the real projective space
Consider the solid ball in of radius π (that is, all points of of distance π or less from the origin). Given the above, for every point in this ball there is a rotation, with axis through the point and the origin, and rotation angle equal to the distance of the point from the origin. The identity rotation corresponds to the point at the center of the ball. Rotations through an angle 𝜃 between 0 and π (not including either) are on the same axis at the same distance. Rotation through angles between 0 and −π correspond to the point on the same axis and distance from the origin but on the opposite side of the origin. The one remaining issue is that the two rotations through π and through −π are the same. So we identify (or "glue together") antipodal points on the surface of the ball. After this identification, we arrive at a topological space homeomorphic to the rotation group.
Indeed, the ball with antipodal surface points identified is a smooth manifold, and this manifold is diffeomorphic to the rotation group. It is also diffeomorphic to the real 3-dimensional projective space so the latter can also serve as a topological model for the rotation group.
These identifications illustrate that SO(3) is connected but not simply connected. As to the latter, in the ball with antipodal surface points identified, consider the path running from the "north pole" straight through the interior down to the south pole. This is a closed loop, since the north pole and the south pole are identified. This loop cannot be shrunk to a point, since no matter how it is deformed, the start and end point have to remain antipodal, or else the loop will "break open". In terms of rotations, this loop represents a continuous sequence of rotations about the z-axis starting (by example) at the identity (center of the ball), through the south pole, jumping to the north pole and ending again at the identity rotation (i.e., a series of rotation through an angle φ where φ runs from 0 to 2π).
Surprisingly, running through the path twice, i.e., running from the north pole down to the south pole, jumping back to the north pole (using the fact that north and south poles are identified), and then again running from the north pole down to the south pole, so that φ runs from 0 to 4π, gives a closed loop which can be shrunk to a single point: first move the paths continuously to the ball's surface, still connecting north pole to south pole twice. The second path can then be mirrored over to the antipodal side without changing the path at all. Now we have an ordinary closed loop on the surface of the ball, connecting the north pole to itself along a great circle. This circle can be shrunk to the north pole without problems. The plate trick and similar tricks demonstrate this practically.
The same argument can be performed in general, and it shows that the fundamental group of SO(3) is the cyclic group of order 2 (a fundamental group with two elements). In physics applications, the non-triviality (more than one element) of the fundamental group allows for the existence of objects known as spinors, and is an important tool in the development of the spin–statistics theorem.
The universal cover of SO(3) is a Lie group called Spin(3). The group Spin(3) is isomorphic to the special unitary group SU(2); it is also diffeomorphic to the unit 3-sphere S3 and can be understood as the group of versors (quaternions with absolute value 1). The connection between quaternions and rotations, commonly exploited in computer graphics, is explained in quaternions and spatial rotations. The map from S3 onto SO(3) that identifies antipodal points of S3 is a surjective homomorphism of Lie groups, with kernel {±1}. Topologically, this map is a two-to-one covering map. (See the plate trick.)
Connection between SO(3) and SU(2)
[edit]
In this section, we give two different constructions of a two-to-one and surjective homomorphism of SU(2) onto SO(3).
Using quaternions of unit norm
[edit]
Main article: Quaternions and spatial rotation
The group SU(2) is isomorphic to the quaternions of unit norm via a map given by restricted to where , , , and , .
Let us now identify with the span of . One can then verify that if is in and is a unit quaternion, then
Furthermore, the map is a rotation of Moreover, is the same as . This means that there is a 2:1 homomorphism from quaternions of unit norm to the 3D rotation group SO(3).
One can work this homomorphism out explicitly: the unit quaternion, q, with is mapped to the rotation matrix
This is a rotation around the vector (x, y, z) by an angle 2θ, where cos θ = w and |sin θ| = ‖(x, y, z)‖. The proper sign for sin θ is implied, once the signs of the axis components are fixed. The 2:1-nature is apparent since both q and −q map to the same Q.
Using Möbius transformations
[edit]
The general reference for this section is Gelfand, Minlos & Shapiro (1963). The points P on the sphere
can, barring the north pole N, be put into one-to-one bijection with points S(P) = P' on the plane M defined by z = −1/2, see figure. The map S is called stereographic projection.
Let the coordinates on M be (ξ, η). The line L passing through N and P can be parametrized as
Demanding that the z-coordinate of equals −1/2, one finds
We have Hence the map
where, for later convenience, the plane M is identified with the complex plane
For the inverse, write L as
and demand x2 + y2 + z2 = 1/4 to find s = 1/1 + ξ2 + η2 and thus
If g ∈ SO(3) is a rotation, then it will take points on S to points on S by its standard action Πs(g) on the embedding space By composing this action with S one obtains a transformation S ∘ Πs(g) ∘ S−1 of M,
Thus Πu(g) is a transformation of associated to the transformation Πs(g) of .
It turns out that g ∈ SO(3) represented in this way by Πu(g) can be expressed as a matrix Πu(g) ∈ SU(2) (where the notation is recycled to use the same name for the matrix as for the transformation of it represents). To identify this matrix, consider first a rotation gφ about the z-axis through an angle φ,
Hence
which, unsurprisingly, is a rotation in the complex plane. In an analogous way, if gθ is a rotation about the x-axis through an angle θ, then
which, after a little algebra, becomes
These two rotations, thus correspond to bilinear transforms of R2 ≃ C ≃ M, namely, they are examples of Möbius transformations.
A general Möbius transformation is given by
The rotations, generate all of SO(3) and the composition rules of the Möbius transformations show that any composition of translates to the corresponding composition of Möbius transformations. The Möbius transformations can be represented by matrices
since a common factor of α, β, γ, δ cancels.
For the same reason, the matrix is not uniquely defined since multiplication by −I has no effect on either the determinant or the Möbius transformation. The composition law of Möbius transformations follow that of the corresponding matrices. The conclusion is that each Möbius transformation corresponds to two matrices g, −g ∈ SL(2, C).
Using this correspondence one may write
These matrices are unitary and thus Πu(SO(3)) ⊂ SU(2) ⊂ SL(2, C). In terms of Euler angles[nb 1] one finds for a general rotation
| | |
--- |
| | 1 |
one has
| | |
--- |
| | 2 |
For the converse, consider a general matrix
Make the substitutions
With the substitutions, Π(gα, β) assumes the form of the right hand side (RHS) of (2), which corresponds under Πu to a matrix on the form of the RHS of (1) with the same φ, θ, ψ. In terms of the complex parameters α, β,
To verify this, substitute for α. β the elements of the matrix on the RHS of (2). After some manipulation, the matrix assumes the form of the RHS of (1).
It is clear from the explicit form in terms of Euler angles that the map
just described is a smooth, 2:1 and surjective group homomorphism. It is hence an explicit description of the universal covering space of SO(3) from the universal covering group SU(2).
Lie algebra
[edit]
Associated with every Lie group is its Lie algebra, a linear space of the same dimension as the Lie group, closed under a bilinear alternating product called the Lie bracket. The Lie algebra of SO(3) is denoted by and consists of all skew-symmetric 3 × 3 matrices. This may be seen by differentiating the orthogonality condition, ATA = I, A ∈ SO(3).[nb 2] The Lie bracket of two elements of is, as for the Lie algebra of every matrix group, given by the matrix commutator, [A1, A2] = A1A2 − A2A1, which is again a skew-symmetric matrix. The Lie algebra bracket captures the essence of the Lie group product in a sense made precise by the Baker–Campbell–Hausdorff formula.
The elements of are the "infinitesimal generators" of rotations, i.e., they are the elements of the tangent space of the manifold SO(3) at the identity element. If denotes a counterclockwise rotation with angle φ about the axis specified by the unit vector then
This can be used to show that the Lie algebra (with commutator) is isomorphic to the Lie algebra (with cross product). Under this isomorphism, an Euler vector corresponds to the linear map defined by
In more detail, most often a suitable basis for as a 3-dimensional vector space is
The commutation relations of these basis elements are,
which agree with the relations of the three standard unit vectors of under the cross product.
As announced above, one can identify any matrix in this Lie algebra with an Euler vector
This identification is sometimes called the hat-map. Under this identification, the bracket corresponds in to the cross product,
The matrix identified with a vector has the property that
where the left-hand side we have ordinary matrix multiplication. This implies is in the null space of the skew-symmetric matrix with which it is identified, because
A note on Lie algebras
[edit]
Main article: Angular momentum operator
See also: Representation theory of SU(2) and Jordan map
In Lie algebra representations, the group SO(3) is compact and simple of rank 1, and so it has a single independent Casimir element, a quadratic invariant function of the three generators which commutes with all of them. The Killing form for the rotation group is just the Kronecker delta, and so this Casimir invariant is simply the sum of the squares of the generators, of the algebra
That is, the Casimir invariant is given by
For unitary irreducible representations Dj, the eigenvalues of this invariant are real and discrete, and characterize each representation, which is finite dimensional, of dimensionality . That is, the eigenvalues of this Casimir operator are
where j is integer or half-integer, and referred to as the spin or angular momentum.
So, the 3 × 3 generators L displayed above act on the triplet (spin 1) representation, while the 2 × 2 generators below, t, act on the doublet (spin-1/2) representation. By taking Kronecker products of D1/2 with itself repeatedly, one may construct all higher irreducible representations Dj. That is, the resulting generators for higher spin systems in three spatial dimensions, for arbitrarily large j, can be calculated using these spin operators and ladder operators.
For every unitary irreducible representations Dj there is an equivalent one, D−j−1. All infinite-dimensional irreducible representations must be non-unitary, since the group is compact.
In quantum mechanics, the Casimir invariant is the "angular-momentum-squared" operator; integer values of spin j characterize bosonic representations, while half-integer values fermionic representations. The antihermitian matrices used above are utilized as spin operators, after they are multiplied by i, so they are now hermitian (like the Pauli matrices). Thus, in this language,
and hence
Explicit expressions for these Dj are,
where j is arbitrary and .
For example, the resulting spin matrices for spin 1 () are
Note, however, how these are in an equivalent, but different basis, the spherical basis, than the above iL in the Cartesian basis.[nb 3]
For higher spins, such as spin 3/2 ():
For spin 5/2 (),
Main article: Spin (physics) § Higher spins
Isomorphism with 𝖘𝖚(2)
[edit]
The Lie algebras and are isomorphic. One basis for is given by
These are related to the Pauli matrices by
The Pauli matrices abide by the physicists' convention for Lie algebras. In that convention, Lie algebra elements are multiplied by i, the exponential map (below) is defined with an extra factor of i in the exponent and the structure constants remain the same, but the definition of them acquires a factor of i. Likewise, commutation relations acquire a factor of i. The commutation relations for the are
where εijk is the totally anti-symmetric symbol with ε123 = 1. The isomorphism between and can be set up in several ways. For later convenience, and are identified by mapping
and extending by linearity.
Exponential map
[edit]
Since SO(3) is a matrix Lie group, its exponential map is defined using the standard matrix exponential series,
For any skew-symmetric matrix A ∈ 𝖘𝖔(3), eA is always in SO(3). The proof uses the elementary properties of the matrix exponential
since the matrices A and AT commute, this can be easily proven with the skew-symmetric matrix condition. This is not enough to show that 𝖘𝖔(3) is the corresponding Lie algebra for SO(3), and shall be proven separately.
The level of difficulty of proof depends on how a matrix group Lie algebra is defined. Hall (2003) defines the Lie algebra as the set of matrices
in which case it is trivial. Rossmann (2002) uses for a definition derivatives of smooth curve segments in SO(3) through the identity taken at the identity, in which case it is harder.
For a fixed A ≠ 0, etA, −∞ < t < ∞ is a one-parameter subgroup along a geodesic in SO(3). That this gives a one-parameter subgroup follows directly from properties of the exponential map.
The exponential map provides a diffeomorphism between a neighborhood of the origin in the 𝖘𝖔(3) and a neighborhood of the identity in the SO(3). For a proof, see Closed subgroup theorem.
The exponential map is surjective. This follows from the fact that every R ∈ SO(3), since every rotation leaves an axis fixed (Euler's rotation theorem), and is conjugate to a block diagonal matrix of the form
such that A = BDB−1, and that
together with the fact that 𝖘𝖔(3) is closed under the adjoint action of SO(3), meaning that BθLzB−1 ∈ 𝖘𝖔(3).
Thus, e.g., it is easy to check the popular identity
As shown above, every element A ∈ 𝖘𝖔(3) is associated with a vector ω = θ u, where u = (x,y,z) is a unit magnitude vector. Since u is in the null space of A, if one now rotates to a new basis, through some other orthogonal matrix O, with u as the z axis, the final column and row of the rotation matrix in the new basis will be zero.
Thus, we know in advance from the formula for the exponential that exp(OAOT) must leave u fixed. It is mathematically impossible to supply a straightforward formula for such a basis as a function of u, because its existence would violate the hairy ball theorem; but direct exponentiation is possible, and yields
where and . This is recognized as a matrix for a rotation around axis u by the angle θ: cf. Rodrigues' rotation formula.
Logarithm map
[edit]
Given R ∈ SO(3), let denote the antisymmetric part and let Then, the logarithm of R is given by
This is manifest by inspection of the mixed symmetry form of Rodrigues' formula,
where the first and last term on the right-hand side are symmetric.
Uniform random sampling
[edit]
is doubly covered by the group of unit quaternions, which is isomorphic to the 3-sphere. Since the Haar measure on the unit quaternions is just the 3-area measure in 4 dimensions, the Haar measure on is just the pushforward of the 3-area measure.
Consequently, generating a uniformly random rotation in is equivalent to generating a uniformly random point on the 3-sphere. This can be accomplished by the following
where are uniformly random samples of .
Baker–Campbell–Hausdorff formula
[edit]
Main article: Baker–Campbell–Hausdorff formula
Suppose X and Y in the Lie algebra are given. Their exponentials, exp(X) and exp(Y), are rotation matrices, which can be multiplied. Since the exponential map is a surjection, for some Z in the Lie algebra, exp(Z) = exp(X) exp(Y), and one may tentatively write
for C some expression in X and Y. When exp(X) and exp(Y) commute, then Z = X + Y, mimicking the behavior of complex exponentiation.
The general case is given by the more elaborate BCH formula, a series expansion of nested Lie brackets. For matrices, the Lie bracket is the same operation as the commutator, which monitors lack of commutativity in multiplication. This general expansion unfolds as follows,[nb 4]
The infinite expansion in the BCH formula for SO(3) reduces to a compact form,
for suitable trigonometric function coefficients (α, β, γ).
The trigonometric coefficients
The (α, β, γ) are given by
where
for
The inner product is the Hilbert–Schmidt inner product and the norm is the associated norm. Under the hat-isomorphism,
which explains the factors for θ and φ. This drops out in the expression for the angle.
See also: Rotation formalisms in three dimensions § Rodrigues parameters and Gibbs representation
It is worthwhile to write this composite rotation generator as
to emphasize that this is a Lie algebra identity.
The above identity holds for all faithful representations of 𝖘𝖔(3). The kernel of a Lie algebra homomorphism is an ideal, but 𝖘𝖔(3), being simple, has no nontrivial ideals and all nontrivial representations are hence faithful. It holds in particular in the doublet or spinor representation. The same explicit formula thus follows in a simpler way through Pauli matrices, cf. the 2×2 derivation for SU(2).
The SU(2) case
The Pauli vector version of the same BCH formula is the somewhat simpler group composition law of SU(2),
where
the spherical law of cosines. (Note a', b', c' are angles, not the a, b, c above.)
This is manifestly of the same format as above,
with
so that
For uniform normalization of the generators in the Lie algebra involved, express the Pauli matrices in terms of t-matrices, σ → 2i t, so that
To verify then these are the same coefficients as above, compute the ratios of the coefficients,
Finally, γ = γ' given the identity d = sin 2c'.
For the general n × n case, one might use Ref.
The quaternion case
The quaternion formulation of the composition of two rotations RB and RA also yields directly the rotation axis and angle of the composite rotation RC = RBRA.
Let the quaternion associated with a spatial rotation R is constructed from its rotation axis S and the rotation angle φ this axis. The associated quaternion is given by,
Then the composition of the rotation RR with RA is the rotation RC = RBRA with rotation axis and angle defined by the product of the quaternions
that is
Expand this product to obtain
Divide both sides of this equation by the identity, which is the law of cosines on a sphere,
and compute
This is Rodrigues' formula for the axis of a composite rotation defined in terms of the axes of the two rotations. He derived this formula in 1840 (see page 408).
The three rotation axes A, B, and C form a spherical triangle and the dihedral angles between the planes formed by the sides of this triangle are defined by the rotation angles.
Infinitesimal rotations
[edit]
This section is an excerpt from Infinitesimal rotation matrix.[edit]
An infinitesimal rotation matrix or differential rotation matrix is a matrix representing an infinitely small rotation.
While a rotation matrix is an orthogonal matrix representing an element of (the special orthogonal group), the differential of a rotation is a skew-symmetric matrix in the tangent space (the special orthogonal Lie algebra), which is not itself a rotation matrix.
An infinitesimal rotation matrix has the form
where is the identity matrix, is vanishingly small, and .
For example, if , representing an infinitesimal three-dimensional rotation about the x-axis, a basis element of , then
The computation rules for infinitesimal rotation matrices are the usual ones except that infinitesimals of second order are dropped. With these rules, these matrices do not satisfy all the same properties as ordinary finite rotation matrices under the usual treatment of infinitesimals. It turns out that the order in which infinitesimal rotations are applied is irrelevant.
Realizations of rotations
[edit]
Main article: Rotation formalisms in three dimensions
See also: Charts on SO(3)
We have seen that there are a variety of ways to represent rotations:
as orthogonal matrices with determinant 1,
by axis and rotation angle
in quaternion algebra with versors and the map 3-sphere S3 → SO(3) (see quaternions and spatial rotations)
in geometric algebra as a rotor
as a sequence of three rotations about three fixed axes; see Euler angles.
Spherical harmonics
[edit]
Main article: Spherical harmonics
See also: Representations of SO(3)
The group SO(3) of three-dimensional Euclidean rotations has an infinite-dimensional representation on the Hilbert space
where are spherical harmonics. Its elements are square integrable complex-valued functions[nb 5] on the sphere. The inner product on this space is given by
| | |
--- |
| | H1 |
If f is an arbitrary square integrable function defined on the unit sphere S2, then it can be expressed as
| | |
--- |
| | H2 |
where the expansion coefficients are given by
| | |
--- |
| | H3 |
The Lorentz group action restricts to that of SO(3) and is expressed as
| | |
--- |
| | H4 |
This action is unitary, meaning that
| | |
--- |
| | H5 |
The D(ℓ) can be obtained from the D(m, n) of above using Clebsch–Gordan decomposition, but they are more easily directly expressed as an exponential of an odd-dimensional su(2)-representation (the 3-dimensional one is exactly 𝖘𝖔(3)). In this case the space L2(S2) decomposes neatly into an infinite direct sum of irreducible odd finite-dimensional representations V2i + 1, i = 0, 1, ... according to
| | |
--- |
| | H6 |
This is characteristic of infinite-dimensional unitary representations of SO(3). If Π is an infinite-dimensional unitary representation on a separable[nb 6] Hilbert space, then it decomposes as a direct sum of finite-dimensional unitary representations. Such a representation is thus never irreducible. All irreducible finite-dimensional representations (Π, V) can be made unitary by an appropriate choice of inner product,
where the integral is the unique invariant integral over SO(3) normalized to 1, here expressed using the Euler angles parametrization. The inner product inside the integral is any inner product on V.
Generalizations
[edit]
The rotation group generalizes quite naturally to n-dimensional Euclidean space, with its standard Euclidean structure. The group of all proper and improper rotations in n dimensions is called the orthogonal group O(n), and the subgroup of proper rotations is called the special orthogonal group SO(n), which is a Lie group of dimension n(n − 1)/2.
In special relativity, one works in a 4-dimensional vector space, known as Minkowski space rather than 3-dimensional Euclidean space. Unlike Euclidean space, Minkowski space has an inner product with an indefinite signature. However, one can still define generalized rotations which preserve this inner product. Such generalized rotations are known as Lorentz transformations and the group of all such transformations is called the Lorentz group.
The rotation group SO(3) can be described as a subgroup of E+(3), the Euclidean group of direct isometries of Euclidean This larger group is the group of all motions of a rigid body: each of these is a combination of a rotation about an arbitrary axis and a translation, or put differently, a combination of an element of SO(3) and an arbitrary translation.
In general, the rotation group of an object is the symmetry group within the group of direct isometries; in other words, the intersection of the full symmetry group and the group of direct isometries. For chiral objects it is the same as the full symmetry group.
See also
[edit]
Angular momentum
Charts on SO(3)
Coordinate rotations
Euler angles
Infinitesimal rotation
Lie group
Orthogonal group
Pauli matrix
Pin group
Plane of rotation
Plate trick
Quaternions and spatial rotations
Representations of SO(3)
Rigid body
Rodrigues' rotation formula
Spherical harmonics
Spherical symmetry
Three-dimensional rotation operator
Footnotes
[edit]
'^ This is effected by first applying a rotation through φ about the z-axis to take the x-axis to the line L, the intersection between the planes xy and x'y, the latter being the rotated xy-plane. Then rotate with through θ about L to obtain the new z-axis from the old one, and finally rotate by through an angle ψ about the new z-axis, where ψ is the angle between L and the new x-axis. In the equation, and are expressed in a temporary rotated basis at each step, which is seen from their simple form. To transform these back to the original basis, observe that Here boldface means that the rotation is expressed in the original basis. Likewise,Thus
^ For an alternative derivation of , see Classical group.
^ Specifically, for
^ For a full proof, see Derivative of the exponential map. Issues of convergence of this series to the correct element of the Lie algebra are here swept under the carpet. Convergence is guaranteed when and The series may still converge even if these conditions are not fulfilled. A solution always exists since exp is onto in the cases under consideration.
^ The elements of L2(S2) are actually equivalence classes of functions. two functions are declared equivalent if they differ merely on a set of measure zero. The integral is the Lebesgue integral in order to obtain a complete inner product space.
^ A Hilbert space is separable if and only if it has a countable basis. All separable Hilbert spaces are isomorphic.
References
[edit]
^ Jacobson (2009), p. 34, Ex. 14.
^ n × n real matrices are identical to linear transformations of expressed in its standard basis.
^ Coxeter, H. S. M. (1973). Regular polytopes (Third ed.). New York. p. 53. ISBN 0-486-61480-8.{{cite book}}: CS1 maint: location missing publisher (link)
^ Hall 2015 Proposition 1.17
^ Rossmann 2002 p. 95.
^ These expressions were, in fact, seminal in the development of quantum mechanics in the 1930s, cf. Ch III, § 16, B.L. van der Waerden, 1932/1932
^ Hall 2015 Proposition 3.24
^ Rossmann 2002
^ a b Engø 2001
^ Hall 2015 Example 3.27
^ See Rossmann 2002, theorem 3, section 2.2.
^ Rossmann 2002 Section 1.1.
^ Hall 2003 Theorem 2.27.
^ Shoemake, Ken (1992-01-01), Kirk, DAVID (ed.), "III.6 - Uniform Random Rotations", Graphics Gems III (IBM Version), San Francisco: Morgan Kaufmann, pp. 124–132, ISBN 978-0-12-409673-8, retrieved 2022-07-29
^ Hall 2003, Ch. 3; Varadarajan 1984, §2.15
^ Curtright, Fairlie & Zachos 2014 Group elements of SU(2) are expressed in closed form as finite polynomials of the Lie algebra generators, for all definite spin representations of the rotation group.
^ Rodrigues, O. (1840), Des lois géométriques qui régissent les déplacements d'un système solide dans l'espace, et la variation des coordonnées provenant de ses déplacements con- sidérés indépendamment des causes qui peuvent les produire, Journal de Mathématiques Pures et Appliquées de Liouville 5, 380–440.
^ (Goldstein, Poole & Safko 2002, §4.8)
^ a b c Gelfand, Minlos & Shapiro 1963
^ In Quantum Mechanics – non-relativistic theory by Landau and Lifshitz the lowest order D are calculated analytically.
^ Curtright, Fairlie & Zachos 2014 A formula for D(ℓ) valid for all ℓ is given.
^ Hall 2003 Section 4.3.5.
Bibliography
[edit]
Boas, Mary L. (2006), Mathematical Methods in the Physical Sciences (3rd ed.), John Wiley & sons, pp. 120, 127, 129, 155ff and 535, ISBN 978-0471198260
Curtright, T. L.; Fairlie, D. B.; Zachos, C. K. (2014), "A compact formula for rotations as spin matrix polynomials", SIGMA, 10: 084, arXiv:1402.3541, Bibcode:2014SIGMA..10..084C, doi:10.3842/SIGMA.2014.084, S2CID 18776942
Engø, Kenth (2001), "On the BCH-formula in 𝖘𝖔(3)", BIT Numerical Mathematics, 41 (3): 629–632, doi:10.1023/A:1021979515229, ISSN 0006-3835, S2CID 126053191
Gelfand, I.M.; Minlos, R.A.; Shapiro, Z.Ya. (1963), Representations of the Rotation and Lorentz Groups and their Applications, New York: Pergamon Press
Goldstein, Herbert; Poole, Charles P.; Safko, John L. (2002), Classical Mechanics (third ed.), Addison Wesley, ISBN 978-0-201-65702-9
Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Graduate Texts in Mathematics, vol. 222 (2nd ed.), Springer, ISBN 978-3319134666
Hall, Brian C. (2003). Lie groups, Lie algebras, and representations : an elementary introduction. Graduate Texts in Mathematics. Vol. 222. New York: Springer. ISBN 0-387-40122-9.
Jacobson, Nathan (2009), Basic algebra, vol. 1 (2nd ed.), Dover Publications, ISBN 978-0-486-47189-1
Joshi, A. W. (2007), Elements of Group Theory for Physicists, New Age International, pp. 111ff, ISBN 978-81-224-0975-8
Rossmann, Wulf (2002), Lie Groups – An Introduction Through Linear Groups, Oxford Graduate Texts in Mathematics, Oxford Science Publications, ISBN 0-19-859683-9
van der Waerden, B. L. (1952), Group Theory and Quantum Mechanics, Springer Publishing, ISBN 978-3642658624{{citation}}: CS1 maint: ignored ISBN errors (link) (translation of the original 1932 edition, Die Gruppentheoretische Methode in Der Quantenmechanik).
Varadarajan, V. S. (1984). Lie groups, Lie algebras, and their representations. New York: Springer-Verlag. ISBN 978-0-387-90969-1.
Veltman, M.; 't Hooft, G.; de Wit, B. (2007). "Lie Groups in Physics (online lecture)" (PDF). Retrieved 2016-10-24..
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8.4: Combinations
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8.3: Permutations
8.5: The Binomial Theorem
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: 8431
Harris Kwong
State University of New York at Fredonia via OpenSUNY
( \newcommand{\kernel}{\mathrm{null}\,})
In many counting problems, the order of arrangement or selection does not matter. In essence, we are selecting or forming subsets.
Example
Determine the number of ways to choose 4 values from 1, 2, 3, …, 20, in which the order of selection does not matter.
Solution
: Let be the number of ways to choose the 4 numbers. Since the order in which the numbers are selected does not matter, these are not sequences (in which order of appearance matters). We can change a selection of 4 numbers into a sequence. The 4 numbers can be arranged in ways. Therefore, all these 4-number selections together produce sequences. The number of 4-number sequences is . Thus, , or equivalently, .
Definition: combinations
The number of -element subsets in an -element set is denoted by
where is read as “ choose .” It determines the number of combinations of objects, taken at a time (without replacement). Alternate notations such as and can be found in other textbooks. Do not write it as ; this notation has a completely different meaning.
Recall that counts the number of ways to choose or select objects from a pool of objects in which the order of selection does not matter. Hence, -combinations are subsets of size .
Example
The 2-combinations of are
Therefore . What are the 1-combinations and 3-combinations of ? What can you say about the values of and ?
Solution
: The 1-combinations are the singleton sets , , , and . Hence, . The 3-combinations are Thus, .
Theorem
For all integers and satisfying , we have
Proof
: The idea is similar to the one we used in the alternate proof of Theorem 8.3.2. Let be the set of all -permutations, and let be the set of all -combinations. Define to be the function that converts a permutation into a combination by “unscrambling” its order. Then is an -to-one function because there are ways to arrange (or shuffle) objects. Therefore Since , and , it follows that .
Example
There are ways to choose 5 numbers, without repetitions, from the integers . To compute its numeric value by hand, it is easier if we first cancel the common factors in the numerator and the denominator. We find
which gives .
hands-on Exercise
Compute by hand.
hands-on Exercise
A three-member executive committee is to be selected from a group of seven candidates. In how many ways can the committee be formed?
hands-on Exercise
How many subsets of have five elements?
Corollary
For , we have .
Proof
: According to Theorem 8.4.1, we have which is precisely .
Example
To compute the numeric value of , instead of computing the product of 47 factors as indicated in the definition, it is much faster if we use from which we obtain .
hands-on Exercise
Compute, by hand, the numeric value of .
Now we are ready to look at some mixed examples. In all of these examples, sometimes we have to use permutation, other times we have to use combination. Very often we need to use both, together with the addition and multiplication principles. You may ask, how can I figure out what to do? We suggest asking yourself these questions:
Use the construction approach. If you want to list all the configurations that meet the requirement, how are you going to do it systematically?
Are there several cases involved in the problem? If yes, we need to list them first, before we go through each of them one at a time. Finally, add the results to come up with the final answer.
Do we allow repetitions or replacements? This question can also take the form of whether the objects are distinguishable or indistinguishable.
Does order matter? If yes, we have to use permutation. Otherwise, use combination.
Sometimes, it may be easier to use the multiplication principle instead of permutation, because repetitions may be allowed (in which case, we cannot use permutation, although we can still use the multiplication principle). Try drawing a schematic diagram and decide what we need from it. If the analysis suggests a pattern that follows the one found in a permutation, you can then use the formula for permutation.
Do not forget: it may be easier to work with the complement.
It is often not clear how to get started because there seem to be several ways to start the construction. For example, how would you distribute soda cans among a group of students? There are two possible approaches:
From the perspective of the students. Imagine you are one of the students, which soda would you receive?
From the perspective of the soda cans. Imagine you are holding a can of soda, to whom would you give this soda?
Depending on the actual problem, usually only one of these two approaches would work.
Example
Suppose we have to distribute 10 different soda cans to 20 students. It is clear that some students may not get any soda. In fact, some lucky students could receive more than one soda (the problem does not say this cannot happen). Hence, it is easier to start from the perspective of the soda cans.
Solution
: We can give the first soda to any one of the 20 students, and we can also give the second soda to any one of the 20 students. In fact, we always have 20 choices for each soda. Since we have 10 sodas, there are ways to distribute the sodas.
Example
In how many ways can a team of three representatives be selected from a class of 885 students? In how many ways can a team of three representatives consisting of a chairperson, a vice-chairperson, and a secretary be selected?
Solution
: If we are only interested in selecting three representatives, order does not matter. Hence, the answer would be . If we are concerned about which offices these three representative will hold, then the answer should be .
hands-on Exercise
Mike needs some new shirts, but he has only enough money to purchase five of the eight that he likes. In how many ways can he purchase the five shirts by choosing them at random?
hands-on Exercise
Mary wants to purchase four shirts for her four brothers, and she would like each of them to receive a different shirt. She finds ten shirts that she thinks they will like. In many ways can she select them?
Playing cards provide excellent examples for counting problems. Just in case you are not familiar with them, let us briefly review what a deck of playing cards contains.
There are 52 playing cards, each of them is marked with a suit and a rank.
There are four suits: spades (), hearts (), diamonds () and clubs ().
Each suit has 13 ranks, labeled A, 2, 3, …, 9, 10, J, Q, and K, where A means ace, J means jack, Q means queen, and K means king.
Each rank has 4 suits (see above).
hands-on Exercise
Determine the number of five-card poker hands that can be dealt from a deck of 52 cards.
Solution
: All we care is which five cards can be found in a hand. This is a selection problem. The answer is .
hands-on exercise
In how many ways can a 13-card bridge hand be dealt from a standard deck of 52 cards?
Example
In how many ways can a deck of 52 cards be dealt in a game of bridge? (In a bridge game, there are four players designated as North, East, South and West, each of them is dealt a hand of 13 cards.)
Solution
: The difference between this problem and the last example is that the order of distributing the four bridge hands makes a difference. This is a problem that combines permutations and combinations. As we had suggested earlier, the best approach is to start from scratch, using the addition and/or multiplication principles, along with permutation and/or combination whenever it seems appropriate.
There are ways to give 13 cards to the first player. Now we are left with 39 cards, from which we select 13 to be given to the second player. Now, out of the remaining 26 cards, we have to give 13 to the third player. Finally, the last 13 cards will be given to the last player (there is only one way to do it). The number of ways to deal the cards in a bridge game is .
We could have said the answer is The last factor is the number of ways to give the last 13 cards to the fourth player. Numerically, , so the two answers are the same. Do not dismiss this extra factor as redundant. Take note of the nice pattern in this answer. The bottom numbers are 13, because we are selecting 13 cards to be given to each player. The top numbers indicate how many cards are still available for distribution at each stage of the distribution. The reasoning behind the solution is self-explanatory!
Example
Determine the number of five-card poker hands that contain three queens. How many of them contain, in addition to the three queens, another pair of cards?
Solution
: 1. The first step is to choose the three queens in ways, after which the remaining two cards can be selected in ways. Therefore, there are altogether hands that meet the requirements.
2. As in part (a), the three queens can be selected in ways. Next, we need to select the pair. We can select any card from the remaining 48 cards (therefore, there are 48 choices), after which we have to select one from the remaining 3 cards of the same rank. This gives choices for the pair, right? The answer is NO!
The first card we picked could be , and the second could be . However, the first card could have been , and the second . These two selections are counted as different selections, but they are actually the same pair! The trouble is, we are considering “first,” and “second” cards, which in effect imposes an ordering among the two cards, thereby turning it into a sequence or an ordered selection. We have to divide the answer by 2 to overcome the double-counting. The answer is therefore .
Here is a better way to count the number of pairs. An important question to ask is
Which one should we pick first: the suit or the rank?
Here, we want to pick the rank first. There are 12 choices (the pair cannot be queens) for the rank, and among the four cards of that rank, we can pick the two cards in ways. Therefore, the answer is . Numerically, the two answers are identical, because . In summary: the final answer is .
hands-on Exercise
How many bridge hands contain exactly four spades?
hands-on Exercise
How many bridge hands contain exactly four spades and four hearts?
hands-on Exercise
How many bridge hands are there containing exactly four spades, three hearts, three diamonds, and three clubs?
Example
How many positive integers not exceeding 99999 contain exactly three 7s?
Solution
: Regard each legitimate integer as a sequence of five digits, each of them selected from 0, 1, 2, …, 9. For example, the integer 358 can be considered as 00358. Three out of the five positions must be occupied by 7. There are ways to select these three slots. The remaining two positions can be filled with any of the other nine digits. Hence, there are such integers.
Example
How many five-digit positive integers contain exactly three 7s?
Solution
: Unlike the last example, the first of the five digits cannot be 0. Yet, the answer is not . Yes, there are choices for the placement of the three 7s, but some of these selections may have put the 7s in the last four positions. This leaves the first digit unfilled. The nine choices counted by 9 allows a zero to be placed in the first position. The result is, at best, a four-digit number. The correct approach is to consider two cases:
Case 1. If the first digit is not 7, then there are eight ways to fill this slot. Among the remaining four positions, three of them must be 7, and the last one can be any digit other than 7. So there are integers in this category.
Case 2. If the first digit is 7, we still have to put the other two 7s in the other four positions. There are such integers.
Together, the two cases give a total of integers.
hands-on Exercise
Five balls are chosen from a bag of eight blue balls, six red balls, and five green balls. How many of these five-ball selections contain exactly two blue balls?
Example
Find the number of ways to select five balls from a bag of six red balls, eight blue balls and four yellow balls such that the five-ball selections contain exactly two red balls or two blue balls.
Solution
: The keyword “or” suggests this is a problem that involves the union of two sets, hence, we have to use PIE to solve the problem.
How many selections contain two red balls? Following the same argument used in the last example, the answer is .
How many selections contain two blue balls? The answer is .
According to PIE, the final answer is In each term, the upper numbers always add up to 18, and the sum of the lower numbers is always 5. Can you explain why?
How many selections contain two red balls and 2 blue balls? The answer is .
Example
We have 11 balls, five of which are blue, three of which are red, and the remaining three are green. How many collection of four balls can be selected such that at least two blue balls are selected? Assume that balls of the same color are indistinguishable.
Solution
: The keywords “at least” mean we could have two, three, or four blue balls. There are ways to select four balls, with at least two of them being blue.
hands-on Exercise
Jerry bought eight cans of Pepsi, seven cans of Sprite, three cans of Dr. Pepper, and six cans of Mountain Dew. He want to bring 10 cans to his pal’s house when they watch the basketball game tonight. Assuming the cans are distinguishable, say, with different expiration dates, how many selections can he make if he wants to bring
Exactly four cans of Pepsi?
At least four cans of Pepsi?
At most four cans of Pepsi?
Exactly three cans of Pepsi, and at most three cans of Sprite?
The proof of the next result uses what we call a combinatorial or counting argument. In general, a combinatorial argument does not rely on algebraic manipulation. Rather, it uses the combinatorial significance of the situations to solve the problem.
Theorem
Prove that for all nonnegative integers .
Proof
: Since counts the number of -element subsets selected from an -element set , the summation on the left is the sum of the number of subsets of of all possible cardinalities. In other words, this is the total number of subsets in . We learned earlier that has subsets, which establishes the identity immediately.
Summary and Review
Use permutation if order matters, otherwise use combination.
The keywords arrangement, sequence, and order suggest using permutation.
The keywords selection, subset, and group suggest using combination.
It is best to start with a construction. Imagine you want to list all the possibilities, how would you get started?
We may need to use both permutation and combination, and very likely we may also need to use the addition and multiplication principles.
Exercise
If the Buffalo Bills and the Cleveland Browns have eight and six players, respectively, available for trading, in how many ways can they swap three players for three players?
Exercise
In the game of Mastermind, one player, the codemaker, selects a sequence of four colors (the “code”) selected from red, blue, green, white, black, and yellow.
How many different codes can be formed?
How many codes use four different colors?
How many codes use only one color?
How many codes use exactly two colors?
How many codes use exactly three colors?
Exercise
Becky likes to watch DVDs each evening. How many DVDs must she have if she is able to watch every evening for 24 consecutive evenings during her winter break?
A different subset of DVDs?
A different subset of three DVDs?
Exercise
Bridget has friends from her bridge club. Every Thursday evening, she invites three friends to her home for a bridge game. She always sits in the north position, and she decides which friends are to sit in the east, south, and west positions. She is able to do this for 200 weeks without repeating a seating arrangement. What is the minimum value of ?
Exercise
Bridget has friends from her bridge club. She is able to invite a different subset of three of them to her home every Thursday evening for 100 weeks. What is the minimum value of ?
Exercise
How many five-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 7? How many of them do not have repeated digits?
Exercise
The Mathematics Department of a small college has three full professors, seven associate professors, and four assistant professors. In how many ways can a four-member committee be formed under these restrictions:
There are no restrictions.
At least one full professor is selected.
The committee must contain a professor from each rank.
Exercise
A department store manager receives from the company headquarters 12 football tickets to the same game (hence they can be regarded as “identical”). In how many ways can she distribute them to 20 employees if no one gets more than one ticket? What if the tickets are for 12 different games?
Exercise
A checkerboard has 64 distinct squares arranged into eight rows and eight columns.
In how many ways can eight identical checkers be placed on the board so that no two checkers can occupy the same row or the same column?
In how many ways can two identical red checkers and two identical black checkers be placed on the board so that no two checkers of the same color can occupy the same row or the same column?
Exercise
Determine the number of permutations of that satisfy the following conditions:
occupies the first position.
occupies the first position, and the second.
appears before .
Exercise
A binary string is a sequence of digits chosen from 0 and 1. How many binary strings of length 16 contain exactly seven 1s?
Exercise
In how many ways can a nonempty subset of people be chosen from eight men and eight women so that every subset contains an equal number of men and women?
Exercise
A poker hand is a five-card selection chosen from a standard deck of 52 cards. How many poker hands satisfy the following conditions?
There are no restrictions.
The hand contains at least one card from each suit.
The hand contains exactly one pair (the other three cards all of different ranks).
The hand contains three of a rank (the other two cards all of different ranks).
The hand is a full house (three of one rank and a pair of another).
The hand is a straight (consecutive ranks, as in 5, 6, 7, 8, 9, but not all from the same suit).
The hand is a flush (all the same suit, but not a straight).
The hand is a straight flush (both straight and flush).
Exercise
A local pizza restaurant offers the following toppings on their cheese pizzas: extra cheese, pepperoni, mushrooms, green peppers, onions, sausage, ham, and anchovies.
How many kinds of pizzas can one order?
How many kinds of pizzas can one order with exactly three toppings?
How many kinds of vegetarian pizza (without pepperoni, sausage, or ham) can one order?
8.3: Permutations
8.5: The Binomial Theorem |
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How can we find the point of continuity of a logarithmic function, e.g., log(1+x)?
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Biraj Chhetri
B.Sc from Sunshine School, Birpara (Graduated 2021)
·4y
First you have learn finding limit of log function after that check questions for continuity in which interval we have to find continuity. If you are familiar with the question you will get knowledge about in which interval you have to check continuity without question being asked.
Checkout this website for help: mathexpertsolutions.blogspot.com
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Is x/(x-1) (x-2) [1,4] a continuous function?
Why does the value of log (3 X 10^-3) and log (0.003) come different?
What can you say about the function continuity of (x^2 - 1)/( x - 1) at x = 1?
What is Log (x-1) +log(x+8) =2log (x+2)?
Is the following function even or odd: f(x) = (log (1-x/ 1+x)) ^3?
Peter Shea
B. Sc in Mathematics&Computer Science, Monash University (Graduated 1972) · Author has 5.2K answers and 1.2M answer views
·4y
There is no point of continuity. A function with a single point of continuity is discontinuous.
Your question only has meaning if you actually mean is point of discontinuity. Huge difference.
Functions with 1 or more points of discontinuity. have an infinite factor. Log (1+x) is discontinuous when 1+x = 0, because log 0 is undefined. The closest you could come to defining a value would be negative infinity, but that's not really a number. Other common points of discontiuity are when you divide by a value which can be zero
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Rhea Saikia
Lady Shri Ram College for Women, Delhi University · Upvoted by
Rahul Trivedi
, studied at Tezpur University · Author has 166 answers and 669.4K answer views
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Is log x always increasing?
A2A
Let me explain to you in detail, so that you can understand and apply these concepts on your own for other problems too.
When you say log x, whether it is an always increasing function or an always decreasing function depends upon the base of the logarithm which is a in the function given below-
log a x here by definition x > 0 (always!)log ax here by definition x > 0 (always!)
log a x log ax is always increasing when a>1
log a x log ax is always decreasing when 0<a<1.(x>0 according to the definition of the log x function)
Remember log function is not defined for negative or zero values of x. x always
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A2A
Let me explain to you in detail, so that you can understand and apply these concepts on your own for other problems too.
When you say log x, whether it is an always increasing function or an always decreasing function depends upon the base of the logarithm which is a in the function given below-
log a x here by definition x > 0 (always!)log ax here by definition x > 0 (always!)
log a x log ax is always increasing when a>1
log a x log ax is always decreasing when 0<a<1.(x>0 according to the definition of the log x function)
Remember log function is not defined for negative or zero values of x. x always will be positive according to definition of log x.
Why? Well, look at the graphs below.
Graph-1 is always decreasing.
Graph-2 is always increasing.
But, if you do not know the graph then don’t worry. Read the fun fact. (although I highly recommend to learn thess handful of such basic graphs such as sin x, cos x, tan x and |x|, exponential function etc)
Fun fact:
So, Suppose you do not know the graph of log. But, you know the graph of exponential function, right? Its easy. Log function is nothing but the inverse of exponential function. Confused? Don’t be..
Consider this.. y=l o g a x y=l o g a x
To obtain inverse of y, we interchange x and y.
So, we have x=l o g a y⟹a x=y x=l o g a y⟹a x=y
This means l o g a x l o g a x and a x a x are inverses of each other.
We know that inverse functions are mirror images of each other about the line y=x.
So, if you know how to plot the graph of one then you can easily plot the graph of its inverse.
Alternatively,
We can also confirm the increasing or decreasing nature of the function log a x log ax by differentiating it. (This is the same approach we use to find maxima or minima of a function.. we differentiate and then equate the differentiated function to zero and obtain critical points. Then we differentiate the function again and check value of this double differentiated equation by putting x= Values of critical points.)
Let, y=l o g a x l o g a x
⟹a y=x⟹a y=x
Taking l o g e l o g e both sides we have-
⟹y log e a=log e x⟹y log ea=log ex
Now differentiating w.r.t x we have-
⟹y′log e a=1 x⟹y′log ea=1 x
⟹y′=1 x(l o g e a)⟹y′=1 x(l o g e a)
Here, x is always greater than zero, therefore sign of y’ depends on the sign of log e a log ea
When we say log x, then a=e,
Notice how y′=1 x(l o g e e)=1 x>0 y′=1 x(l o g e e)=1 x>0 whenever a=e for all x belonging to the domain of the function (i.e x>0)
(By always increasing/decreasing, I mean monotonically increasing or decreasing..)
Most commonly, when base isn’t mentioned then the base is either said to be 10 (log x) or it is said to be e e (ln x) both of which are greater than 1 and hence log x is an increasing function whenever the base is greater than 1.
Let me know in the comment section below if you find this answer useful or if you have any related queries. :)
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Abhik Mukherjee
Used to teach physics & mathematics · Author has 297 answers and 2M answer views
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Is f(x)=e 1/x−1 e 1/x+1 f(x)=e 1/x−1 e 1/x+1 continuous at x=0 x=0 ?
f(x) is not continuous at x=0.
Explanation:
By the Defn. of Continuity,
A function f(x) is continuous at x=0, if and only if, (limx→0-) f(x) = f(0) = (limx→0+) f(x).
Observe that, f(x) is not defined at x=0; as (1/x) is undefined at x=0.
Therefore, f(x) can not be continuous at x=0.
(e^(1/x)-1)/(e^(1/x)+1) Graph Plot:
(e^(1/x)-1) Graph Plot:
(e^(1/x)+1) Graph Plot:
Hope it helped!
Thank You.
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f(x) is not continuous at x=0.
Explanation:
By the Defn. of Continuity,
A function f(x) is continuous at x=0, if and only if, (limx→0-) f(x) = f(0) = (limx→0+) f(x).
Observe that, f(x) is not defined at x=0; as (1/x) is undefined at x=0.
Therefore, f(x) can not be continuous at x=0.
(e^(1/x)-1)/(e^(1/x)+1) Graph Plot:
(e^(1/x)-1) Graph Plot:
(e^(1/x)+1) Graph Plot:
Hope it helped!
Thank You.
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More answers below
Why is log (base 1) (x) not a function?
Why does log x+log y=log(x×y)logx+logy=log(x×y)?
If 2^x = 3, how can I find the value of x without using a logarithm?
What is a logarithm?
How would one write this logarithmic expression as a single logarithm? (log z - log 3)/4 - 5(log x/2) ?
Mohammad Afzaal Butt
B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views
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How do I find f(0) if the function f=log(1+x) /sin-1x, x not equal to 0 is continuous at x=0?
f(x)=log(1+x)sin−1 x f(x)=log(1+x)sin−1x
f(0)=lim x→0 log(1+x)sin−1 x f(0)=lim x→0 log(1+x)sin−1x
=lim x→0 d d x log(1+x)d d x(sin−1 x)We have indeterminate form 0 0 L’Hospital s’ rule applies=lim x→0 d d x log(1+x)d d x(sin−1x)We have indeterminate form 0 0 L’Hospital s’ rule applies
=lim x→0 1 1+x 1√1−x 2=lim x→0 1 1+x 1 1−x 2
=1=1
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Michal Forišek
Ph. D. in theoretical Computer Science · Upvoted by
Jay Wacker
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Justin Rising
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How do we solve x log x=1 x logx=1?
We don't. Saying "the number x x for which x log x=1 x logx=1" is the simplest form we know.
The Lambert W function has been defined to express numbers related to our x x. In terms of the W function, our x x can be written as e W(1)e W(1).
The approximate value of our x x is 1.76322 1.76322. A very simple way to determine this approximate value is to realize that the solution must lie in [1,∞)[1,∞) and that x log x x logx is increasing on this interval so the solution is unique. We can then check that x=2 x=2 is already too much and thus we can find our x x to any required precision by Bisection of the interval [1,2][1,2].
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Logarithm Functions: Why is log b(x y)=log b(x)+log b(y)log b(x y)=log b(x)+log b(y)?
I'll provide an answer only with the natural logarithm, hereafter denoted by ln(x).ln(x). The definition I'll use is "The natural logarithm of a number x x greater than 1 is the area underneath the curve 1/t 1/t between the values t=1 t=1 and t=x t=x on the t t-axis." The representing equation is
ln(x)=∫x 1 d t t.ln(x)=∫1 x d t t.
Page on Fooplot
The link is a low tech depiction of ln(e)ln(e) , which defines the number e e as the unique number which makes that area equal to 1. If you imagine the second vertical line being moved around to any value x x greater than 1, then you can imagine the natural l
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I'll provide an answer only with the natural logarithm, hereafter denoted by ln(x).ln(x). The definition I'll use is "The natural logarithm of a number x x greater than 1 is the area underneath the curve 1/t 1/t between the values t=1 t=1 and t=x t=x on the t t-axis." The representing equation is
ln(x)=∫x 1 d t t.ln(x)=∫1 x d t t.
Page on Fooplot
The link is a low tech depiction of ln(e)ln(e) , which defines the number e e as the unique number which makes that area equal to 1. If you imagine the second vertical line being moved around to any value x x greater than 1, then you can imagine the natural logarithm. The variable t t on the right hand side is a "dummy variable" meaning it can be replaced by any letter and still have the same meaning.
Let y y be another number greater than 1 and fix the previous x x value to be constant. The dummy variable I'll choose below will be s=t/x s=t/x and t t will be the main dummy variable. Since the logarithm of y y is the area underneath 1/s 1/s between s=1 s=1 and s=y s=y, it is also equal to the area underneath 1/(t/x)=x/t 1/(t/x)=x/t between t=x t=x and t=x y t=x y corresponding to the values of s=1,y s=1,y respectively. If you think of d s d s as the change in s s, then it changes just as much as t t changes, except it carries a 1/x 1/x factor along with it. So d s=d t/x d s=d t/x because x x is constant. This is a form of substitution.
ln(y)=∫y 1 d s s=∫x y x d t x⋅x t=∫x y x d t t.ln(y)=∫1 y d s s=∫x x y d t x⋅x t=∫x x y d t t.
I've labored with the definition because adding two logarithms together become simple now.
ln(x)+ln(y)=∫x 1 d t t+∫y 1 d s s ln(x)+ln(y)=∫1 x d t t+∫1 y d s s
=∫x 1 d t t+∫x y x d t t.=∫1 x d t t+∫x x y d t t.
The last part is if we add the areas between t=1 t=1 and t=x t=x with the area between t=x t=x and t=x y t=x y, then this is equal to the sum from t=1 t=1 and t=x y t=x y without stopping
ln(x)+ln(y)=∫x y 1 d t t=ln(x y).ln(x)+ln(y)=∫1 x y d t t=ln(x y).
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Aman Singh
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How do I solve x/(x+1) < log(1+x) < x by Cauchy's theorem?
Hope its help.
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Hope its help.
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Gaurav Kumar
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How do we solve log (x + y) = 1?
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Alon Amit
PhD in Mathematics; Mathcircler. · Upvoted by
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Is f(x)=e 1 x f(x)=e 1 x at x=0 x=0 a continuous or discontinuous function?
A function cannot be said to be continuous at a point where it's not defined.
Let me say this again: for a function to be continuous at a point, it must first of all have a value at that point. Otherwise there's no meaning to the question.
If someone hands you a function f f, defined wherever but not defined at x=0 x=0, and then asks you whether the function is continuous at x=0 x=0, immediately turn around and walk away, or if you're not in a position to do that, calmly quote the phrase above and tell them they'll need to try harder.
The expression e 1/x e 1/x defines a function whose domain is all real numbe
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A function cannot be said to be continuous at a point where it's not defined.
Let me say this again: for a function to be continuous at a point, it must first of all have a value at that point. Otherwise there's no meaning to the question.
If someone hands you a function f f, defined wherever but not defined at x=0 x=0, and then asks you whether the function is continuous at x=0 x=0, immediately turn around and walk away, or if you're not in a position to do that, calmly quote the phrase above and tell them they'll need to try harder.
The expression e 1/x e 1/x defines a function whose domain is all real numbers except 0 0. It does not define a function at x=0 x=0, so the question is moot.
If you wish, you can extend this function to a function such as this:
f(x)={e 1 x if x≠0 0 if x=0 f(x)={e 1 x if x≠0 0 if x=0
This is a different function, and it is defined at x=0 x=0, and it is discontinuous there because e 1/x e 1/x is huge when x x is positive and small.
You should prove this rigorously, but most of all you must remember the Fundamental Mantra of Asking Questions about Continuity: the function must first and foremost exist at the point you're wondering about.
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Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
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What is the differential of logarithm in (1_e^x) _in (1+e^X)?
It’s not easy to cope with the way you confuse mathematical symbols. First, there is no “in” function; there’s only ln, the natural logarithm function. Second, the underscore _ is not an operator for subtraction: that symbol is the dash - . Third, you’re confusing two variables, x and X. These are not the same variable! They denote two quantities, not one.
In the future, take better care to write mathematical expressions properly, using the proper notation!
This derivative is not difficult. When u represents a function of x, use the rule to differentiate ln(u), shown below; apply the Chain Rule
Continue Reading
It’s not easy to cope with the way you confuse mathematical symbols. First, there is no “in” function; there’s only ln, the natural logarithm function. Second, the underscore _ is not an operator for subtraction: that symbol is the dash - . Third, you’re confusing two variables, x and X. These are not the same variable! They denote two quantities, not one.
In the future, take better care to write mathematical expressions properly, using the proper notation!
This derivative is not difficult. When u represents a function of x, use the rule to differentiate ln(u), shown below; apply the Chain Rule where needed, then simplify. (Click on the image to expand it as necessary.)
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Awnon Bhowmik
Studied at University of Dhaka · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) · Author has 3.7K answers and 11.2M answer views
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Can we break log(x+y)?
A2A
Okay, think like this
Can you factor x+y x+y? Don’t think so.
I did see the other answers. It goes like this….
No you cannot factor x+y x+y at least in the usual sense, but if you want you can force out some x y x y out of it. But I don’t really think that’s how we would usually factor something.
x+y=x y(1 x+1 y)x+y=x y(1 x+1 y)
Now, what happens when you take logs?
log(x+y)=log x y(1 x+1 y)log(x+y)=logx y(1 x+1 y)
⟹log(x+y)=log x+log y+log(1 x+1 y)⟹log(x+y)=logx+logy+log(1 x+1 y)
This is what some of the other answers have shown.
But it is simply done by doing this…
log(x+y)=log(x+y log(x+y)=log(x+y
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A2A
Okay, think like this
Can you factor x+y x+y? Don’t think so.
I did see the other answers. It goes like this….
No you cannot factor x+y x+y at least in the usual sense, but if you want you can force out some x y x y out of it. But I don’t really think that’s how we would usually factor something.
x+y=x y(1 x+1 y)x+y=x y(1 x+1 y)
Now, what happens when you take logs?
log(x+y)=log x y(1 x+1 y)log(x+y)=logx y(1 x+1 y)
⟹log(x+y)=log x+log y+log(1 x+1 y)⟹log(x+y)=logx+logy+log(1 x+1 y)
This is what some of the other answers have shown.
But it is simply done by doing this…
log(x+y)=log(x+y)+log x y−log x y log(x+y)=log(x+y)+logx y−logx y, for x>0,y>0 x>0,y>0
Play with this using the properties of sum and difference of logarithms, you’ll see.
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Ajesh K C
Studied at Mar Athanasius College of Engineering, Kothamangalam · Upvoted by
David Joyce
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How do you prove log(x+y) =logx+log (1+y/x)?
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7919 | https://www.formulo.org/wp-content/uploads/2016/02/FdITMmini.pdf | УСЛОВИЯ ЗАДАЧ 2012–2013 год Формула Единства Отборочный этап 7–8 классы 13.1.
Можно ли нарисовать четыре треугольника так, что внутри каждого из них содержится ровно по одной вершине каждого из остальных треугольников? Вершины не могут ле-жать на сторонах других треугольников.
13.2.
Петя бегает по круговой дорожке. Каждые 5 минут он пробегает мимо Маши, качающейся на качелях, а каждые 15 минут обгоняет пенсионера Михаила Ивановича, который тоже бегает по кругу. В некоторый момент Петя развернулся и побежал с той же скоростью в противоположном направлении.
Как часто он теперь встречается с Михаилом Ивановичем?
13.3.
Дети загадали натуральное число и произнесли сле-дующие девять фраз: «Число делится на 2», «Число делится на 3, но не делится на 2», «Число делится на 4, но не делится на 3», . . . , «Число делится на 10, но не делится на 9». Какое наи-большее количество фраз могут быть верными одновременно?
Сборник задач олимпиады, включающий также подробные решения, исторические и статистические сведения, вы можете заказать у организа-торов. Все подробности можно узнать, кликнув по этой ссылке.
2 УСЛОВИЯ ЗАДАЧ 13.4.
Можно ли разбить числа от 1 до 2012 на пары так, чтобы сумма чисел в каждой паре содержала в десятичной за-писи только нули и четверки?
13.5.
В некоторых клетках доски 8 × 8 стоит по фишке, причем в каждой строке и в каждом столбце фишек не менее четырех. Всегда ли можно снять часть фишек так, чтобы в каж-дой строке и в каждом столбце осталось ровно по 4 фишки?
13.6.
В ряд стоят 15 слонов, каждый из которых весит целое число килограммов. Если к весу любого слона, кроме по-следнего, прибавить удвоенный вес стоящего за ним, то полу-чится 15 тонн. Найдите вес каждого из слонов (и докажите, что он не может быть другим).
9–10 классы 13.7.
Толик умножил пятизначное число на сумму его цифр. Потом Толик умножил результат на сумму его (результа-та) цифр. Удивительно, но получилось опять пятизначное чис-ло. Какое число Толик умножал в первый раз? (Найдите все возможные варианты ответа.) 13.8.
В квадрате 7×7 каждый квадратик 1×1 покрасили в красный, желтый или зеленый цвет. Докажите, что существуют строка, столбец и цвет такие, что и в строке, и в столбце есть по крайней мере 3 квадратика этого цвета.
13.9.
Сколько есть способов выписать в строку 𝑛букв A и 𝑛букв B так, чтобы в строке не встречался фрагмент ABB?
13.10.
В остроугольном треугольнике 𝐴𝐵𝐶угол 𝐶равен 45 градусам, 𝐴𝐴1 и 𝐵𝐵1 — высоты. Докажите, что 𝐴1𝐵1 = √︂ 𝐴1𝐵2 + 𝐴1𝐶2 2 .
13.11.
Вдоль окружности расставлено 100 чисел, каждое из которых равно либо 2, либо 5, либо 9, причем никакие два 2012–2013 ГОД, ФОРМУЛА ЕДИНСТВА 3 равных числа не стоят рядом. Числа разбили на 50 пар рядом стоящих. Числа в парах перемножили и полученные 50 произ-ведений записали на первую доску. Затем эти же 100 чисел раз-били на 50 пар рядом стоящих другим способом, числа в парах снова перемножили и произведения записали на вторую доску.
Докажите, что суммы чисел, написанных на первой и второй досках, равны.
13.12.
Натуральные числа 𝑛и 𝑘таковы, что 𝑘2 + 𝑛2 −𝑘 делится на 𝑘𝑛. Докажите, что 𝑘— точный квадрат.
Заключительный этап 7 класс 13.13.
На рисунке изображен квадрат, разби-тый на прямоугольники. Докажите, что если пло-щади четырех «угловых» прямоугольников равны, то «центральный» — квадрат.
13.14.
Ежедневно Андрей спускается в метро по эскалато-ру. Если он сбегает на одну ступень в секунду, то спускается за 96 секунд, а если на две ступени в секунду, то за 60 секунд. За какое время спустится Андрей, если будет стоять неподвижно?
13.15.
У марсиан есть голова, спина, рука, нога и хвост.
Каждая из этих частей тела может быть красной, желтой, зе-леной или синей. Однажды компания марсиан собралась и об-наружила, что у каждого из них хотя бы одна из перечисленных частей тела уникальна (то есть окрашена в цвет, в который эта часть тела не окрашена больше ни у кого из компании). Какова максимально возможная численность компании?
13.16.
Сколько из чисел от 1 до 2013 делятся на 5 и при этом имеют сумму цифр, кратную 5?
4 УСЛОВИЯ ЗАДАЧ 13.17.
На столе стоят десять коробок, в одной из них 550 конфет, в другой 450, остальные пусты. За один ход разреша-ется выбрать любые две коробки и переложить часть конфет из одной в другую так, чтобы в этих коробках стало поровну конфет (если число конфет получается дробным, то такая опе-рация запрещена). Можно ли с помощью таких ходов добиться того, чтобы в двух коробках было по 200 конфет, а в остальных по 75?
8 класс 13.18.
В шестиугольнике 𝐴𝐵𝐶𝐷𝐸𝐹противоположные сто-роны параллельны, и к тому же 𝐴𝐵= 𝐷𝐸. Докажите, что 𝐵𝐶= 𝐸𝐹, а 𝐶𝐷= 𝐹𝐴.
13.19.
Смотри задачу 13.14.
13.20.
Смотри задачу 13.15.
13.21.
Смотри задачу 13.16.
13.22.
Коле подарили набор из шести палочек, причем все палочки имеют разную длину. Коля сложил из них два подоб-ных треугольника (при этом он использовал каждую палочку в качестве стороны одного из треугольников). Тут к нему подо-шла Даша и сложила из тех же палочек два других треугольни-ка, которые тоже оказались подобны между собой. Приведите пример такого набора палочек (то есть укажите длину каждой палочки).
9–10 классы 13.23.
В числе 123456789 можно взять любые 2 цифры оди-наковой четности и заменить каждую из них на их среднее арифметическое. Можно ли такими операциями получить чис-ло, большее 800000000?
2012–2013 ГОД, ТРЕТЬЕ ТЫСЯЧЕЛЕТИЕ 5 13.24.
Сумма трех различных натуральных чисел равна 2013. Какое наибольшее значение может принимать их НОД (наибольший общий делитель)?
13.25.
На сторонах 𝐴𝐵и 𝐵𝐶треугольника 𝐴𝐵𝐶выбраны точки 𝑋и 𝑌соответственно. Пусть 𝑍— точка пересечения 𝐴𝑌 и 𝐶𝑋. Оказалось, что 𝐴𝑌= 𝑌𝐶и 𝐴𝐵= 𝐶𝑍. Докажите, что точки 𝐵, 𝑌, 𝑍, 𝑋лежат на одной окружности.
13.26.
Для положительных чисел 𝑥, 𝑦и 𝑧докажите нера-венство (1 + 𝑥+ 2𝑥2)(2 + 3𝑦+ 𝑦2)(4 −11𝑧+ 8𝑧2) ⩾7𝑥𝑦𝑧.
13.27.
В конструкторе есть 𝑛плоских фигурок из 9 квад-ратиков, каждая из которых выглядит так: Докажите, что при нечетном 𝑛из них нельзя сложить прямо-угольник 9 × 𝑛.
Третье тысячелетие 5 класс 13.28.
Маша хочет разложить 9 карандашей в 5 разных ко-робок так, чтобы количество карандашей в коробках было по-парно различным. Как это сделать? (Если это невозможно, то объясните, почему.) 13.29.
В любую клетку квадрата 5 × 5 разрешается поста-вить желтую, красную или синюю фишку, но так, чтобы ника-кие две фишки разных цветов не оказались на одной вертикали или горизонтали. Выставьте наименьшее возможное количество фишек, к которым (с учетом этого запрета) нельзя было бы до-бавить ни одной еще.
6 УСЛОВИЯ ЗАДАЧ 13.30.
Даны квадраты 3 × 3 и 4 × 4. На какое наименьшее общее число частей нужно их разрезать, чтобы из них можно было сложить квадрат 5 × 5?
13.31.
Ян коллекционирует геометрические модели. Любые две из его моделей отличаются либо по размеру, либо по форме, либо по цвету, либо сразу по нескольким признакам. Есть моде-ли трех размеров (мелкие, средние и крупные), причем их ко-личество попарно различно. Есть модели четырех форм (шары, кубы, пирамиды и цилиндры), причем их количество попарно различно. Есть модели пяти цветов (желтые, синие, красные, белые, зеленые), причем их количество попарно различно. Че-му равно наименьшее возможное число моделей в коллекции, удовлетворяющей этим условиям?
13.32.
Найдите наибольшее пятизначное число, нацело де-лящееся на 2013, все цифры которого различны.
13.33.
На турнир приезжают 9 шахматистов, каждые два из которых должны будут сыграть одну партию между собой.
Организаторы хотят провести турнир в 3 городах в течение 4 дней. Важно, чтобы ежедневно все игроки играли одинаковое число партий, и никому из них не пришлось бы переезжать в другой город в течение игрового дня. Составьте расписание тур-нира, удовлетворяющее этим требованиям. (Если это невозмож-но сделать, то объясните, почему.) 6 класс 13.34.
Маша хочет разложить 13 карандашей в 6 разных коробок так, чтобы количество карандашей в коробках было попарно различным. Как это сделать? (Если это невозможно, то объясните, почему.) 13.35.
В любую клетку квадрата 6 × 6 разрешается поста-вить желтую, красную или синюю фишку, но так, чтобы ника-кие две фишки разных цветов не оказались на одной вертикали 2012–2013 ГОД, ТРЕТЬЕ ТЫСЯЧЕЛЕТИЕ 7 или горизонтали. Выставьте наименьшее возможное количество фишек, к которым (с учетом этого запрета) нельзя было бы до-бавить ни одной еще.
13.36.
Даны квадраты 6 × 6 и 8 × 8. На какое наименьшее общее число частей нужно их разрезать, чтобы из них можно было сложить квадрат 10 × 10?
13.37.
Вова записал несколько многочленов, возвел каж-дый в квадрат и сложил результаты. В итоге он получил выра-жение 𝑥2 + 𝑦2 + 𝑧2 + 2013(𝑥𝑦+ 𝑥𝑧+ 𝑦𝑧) + 1.
Коля не знает, какие именно многочлены использовал Вова, но уверен, что тот ошибся. Кто из них прав и почему?
13.38.
Найдите наибольшее шестизначное число, нацело делящееся на 2013, все цифры которого различны.
13.39.
Смотри задачу 13.33.
7 класс 13.40.
Маша хочет разложить 20 карандашей в 7 разных коробок так, чтобы количество карандашей в коробках было попарно различным. Как это сделать? (Если это невозможно, то объясните, почему.) 13.41.
В любую клетку квадрата 7 × 7 разрешается поста-вить желтую, красную или синюю фишку, но так, чтобы ника-кие две фишки разных цветов не оказались на одной вертикали или горизонтали. Выставьте наименьшее возможное количество фишек, к которым (с учетом этого запрета) нельзя было бы до-бавить ни одной еще.
13.42.
Даны квадраты 1 × 1 и 7 × 7. На какое наименьшее общее число частей нужно их разрезать, чтобы из них можно было сложить два квадрата 5 × 5?
8 УСЛОВИЯ ЗАДАЧ 13.43.
Ян коллекционирует геометрические модели. Любые две из его моделей отличаются либо по размеру, либо по форме, либо по цвету, либо сразу по нескольким признакам. Есть моде-ли трех размеров (мелкие, средние и крупные), причем их коли-чества попарно различны. Есть модели четырех форм (шары, кубы, пирамиды и цилиндры), причем их количества попарно различны. Есть модели пяти цветов (желтые, синие, красные, белые, зеленые), причем их количества попарно различны. Че-му равно наибольшее возможное число моделей в коллекции, удовлетворяющей этим условиям?
13.44.
Найдите наибольшее семизначное число, нацело де-лящееся на 2013, все цифры которого различны.
13.45.
Смотри задачу 13.33.
8 класс 13.46.
Маша хочет разложить 25 карандашей в 8 разных коробок так, чтобы количество карандашей в коробках было попарно различным. Как это сделать? (Если это невозможно, то объясните, почему.) 13.47.
В любую клетку квадрата 8 × 8 разрешается поста-вить желтую, красную или синюю фишку, но так, чтобы ника-кие две фишки разных цветов не оказались на одной вертикали или горизонтали. Выставьте наименьшее возможное количество фишек, к которым (с учетом этого запрета) нельзя было бы до-бавить ни одной еще.
13.48.
Смотри задачу 13.37.
13.49.
Смотри задачу 13.43.
13.50.
Марк последовательно выписывает числа 122, 122122, 122122122 и так далее. На каком шаге он запишет число, нацело делящееся на 2013?
13.51.
Смотри задачу 13.33.
2012–2013 ГОД, ТРЕТЬЕ ТЫСЯЧЕЛЕТИЕ 9 9 класс 13.52.
Пусть 𝑇(𝑥) — сумма всех простых чисел, меньших 𝑥. Найдите все корни уравнения 𝑇(𝑥) = 𝑥2 2 .
13.53.
В квадрате 9×9 разрешается делать разрезы длины 1 по общей границе любых двух соседних единичных квадрати-ков, но так, чтобы он не распался на части. Найдите наибольшее возможное число таких разрезов. Приведите пример.
13.54.
При каких значениях 𝑝оба корня уравнения 𝑥2 − −𝑝𝑥+ 2013 = 0 — целые?
13.55.
Смотри задачу 13.43.
13.56.
Марк последовательно выписывает числа 61, 6161, 616161 и так далее. На каком шаге он запишет число, нацело делящееся на 2013?
13.57.
На турнир приезжают 16 шахматистов, каждые два из которых должны будут сыграть одну партию между собой.
Организаторы хотят провести турнир в 4 городах в течение 5 дней. Важно, чтобы ежедневно все игроки играли одинаковое число партий, и никому из них не пришлось бы переезжать в другой город в течение игрового дня. Составьте расписание тур-нира, удовлетворяющее этим требованиям. (Если это невозмож-но сделать, то объясните, почему.) 10 класс 13.58.
Смотри задачу 13.52.
13.59.
В квадрате 10×10 разрешается делать разрезы дли-ны 1 по общей границе любых двух соседних единичных квад-ратиков, но так, чтобы он не распался на части. Найдите наи-большее возможное число таких разрезов. Приведите пример.
13.60.
Смотри задачу 13.54.
10 УСЛОВИЯ ЗАДАЧ 13.61.
Пусть точки 𝑄и 𝑅делят отрезок 𝑃𝑆на три рав-ные части, а точки 𝐵, 𝑋, 𝑌, 𝑍, 𝑇служат серединами отрезков 𝐴𝐶, 𝐴𝑆, 𝐵𝑅, 𝐵𝑄и 𝐶𝑃соответственно. Какие значения может принимать отношение длин отрезков 𝑋𝑇и 𝑌𝑍?
13.62.
На какое наименьшее число частей нужно разрезать куб с ребром 6, чтобы из них можно было сложить кубы с реб-рами 3, 4 и 5?
13.63.
Смотри задачу 13.57.
11 класс 13.64.
Пусть 𝑇(𝑥) — сумма всех простых чисел, меньших 𝑥. Найдите все корни уравнения 𝑇(𝑥) = 2𝑥2 5 .
13.65.
Смотри задачу 13.62.
13.66.
Чему равно количество таких пар чисел (𝐴, 𝐵), по-сле подстановки которых уравнение 𝑥3 + 𝐴𝑥2 + 𝐵𝑥+ 2013 = 0 имеет три различных целых корня?
13.67.
Окружности радиусов 1, 2 и 3 попарно касаются друг друга. Какие значения может принимать площадь обла-сти, граница которой состоит из дуг этих трех окружностей?
13.68.
Решите уравнение ((𝑥+ 1)𝑥−𝑥) ((𝑥−1)𝑥+ 𝑥) = 2013 𝑥 .
13.69.
Смотри задачу 13.57.
12 класс В некоторых странах дети учатся в школе на год дольше, чем в России, поэтому в олимпиаде «Третье Тысячелетие» вводилось отдельное условие для этого «дополнительного» класса.
2013–2014 ГОД, ОТБОРОЧНЫЙ ЭТАП 11 13.70.
Пусть 𝑇(𝑥) — сумма всех простых чисел, меньших 𝑥. Найдите все корни уравнения 𝑇(𝑥) = 𝑥2 4 .
13.71.
Смотри задачу 13.62.
13.72.
При каких 𝐴и 𝐵уравнение 𝑥3 +𝐴𝑥2 +𝐵𝑥+2013 = 0 имеет три различных целых корня?
13.73.
Смотри задачу 13.67.
13.74.
Смотри задачу 13.68.
13.75.
Смотри задачу 13.57.
2013–2014 год Отборочный этап 5 класс 14.1.
Назовем год лихим, если в записи его номера есть одинаковые цифры. Например, все годы с 1988 по 2012 были лихими. Каково максимальное количество лихих лет, идущих подряд, среди уже прошедших лет нашей эры?
14.2.
На круглом торте стоит 6 свечей. Тремя разрезами торт разрезали на части, причем в каждой части оказалась ров-но одна свеча. Сколько свечей могло стоять в каждой из частей, которые образовались после первого разреза? Объясните, поче-му никакие другие варианты невозможны.
14.3.
Даны три нечетных положительных числа 𝑝, 𝑞, 𝑟.
Про них известно, что 𝑝> 2𝑞, 𝑞> 2𝑟, 𝑟> 𝑝−2𝑞. Докажите, что 𝑝+ 𝑞+ 𝑟⩾25.
14.4.
У Кости есть шесть кубиков, грани которых раскра-шены в шесть разных цветов (каждая грань полностью в один 12 УСЛОВИЯ ЗАДАЧ цвет). Все кубики раскрашены одинаково. Костя составил из кубиков столбик и смотрит на него с четырех сторон. Может ли он сделать это таким образом, чтобы с каждой стороны все шесть граней были разного цвета?
14.5.
В одном доме провели перепись населения. Выясни-лось, что в каждой квартире живет супружеская пара (мать и отец) и в каждой семье есть хотя бы один ребенок. У каждого мальчика в доме есть сестра, но всего мальчиков больше, чем девочек. Детей же в доме меньше, чем взрослых. Докажите, что в результаты переписи вкралась ошибка.
14.6.
Фокусник хочет сложить колоду из 36 карт так, что-бы у любых двух подряд идущих карт совпадало либо достоин-ство, либо масть. При этом начать он хочет с пиковой дамы, а закончить бубновым тузом. Как это сделать?
6 класс 14.7.
Смотри задачу 14.1.
14.8.
На круглом торте стоит 7 свечей. Тремя разреза-ми торт разрезали на части, причем в каждой части оказалась ровно одна свеча. Сколько частей было после второго разреза и сколько свечей стояло в каждой из них?
14.9.
Смотри задачу 14.3.
14.10.
Смотри задачу 14.4.
14.11.
Смотри задачу 14.5.
14.12.
На продажу выставлены 20 книг по цене от 7 до 10 евро и 20 обложек по цене от 10 центов до 1 евро, причем все цены разные. Смогут ли Том и Леопольд купить по книге с обложкой, заплатив одну и ту же сумму денег?
2013–2014 ГОД, ОТБОРОЧНЫЙ ЭТАП 13 7 класс 14.13.
В стопке лежат одинаковые карточки, на каждой из которых записаны числа от 1 до 9. Билл взял одну карточку и тайно отметил на ней 4 числа. Марк может сделать то же самое с несколькими карточками. Затем карточки открывают.
Если на одной из карточек Марка хотя бы два из четырех отме-ченных чисел совпадут с числами Билла, то Марк выигрывает.
Какое наименьшее число карточек должен взять Марк и как их заполнить, чтобы наверняка выиграть?
14.14.
На круглом торте стоит 10 свечей. Четырьмя разре-зами торт разрезали на части, причем в каждой части оказалась ровно одна свеча. Сколько свечей могло стоять в каждой из ча-стей, которые образовались после первого разреза? Объясните, почему никакие другие варианты невозможны.
14.15.
У фокусника есть два комплекта по 7 карточек. На розовых карточках записаны целые числа от 0 до 6. На первой голубой карточке написано 1, а число на каждой следующей голубой карточке в 7 раз больше предыдущего. Фокусник рас-кладывает карточки попарно (розовую с голубой). Затем зри-тели перемножают числа в каждой паре и находят сумму всех 7 произведений. Фокус состоит в том, что в сумме должно полу-читься простое число. Подскажите фокуснику, какие карточки можно для этого объединить в пары (или докажите, что у него ничего не получится).
14.16.
Смотри задачу 14.4.
14.17.
По кругу в каком-то порядке выписаны числа от 1 до 77. Какова минимально возможная сумма модулей разностей между соседними числами?
14.18.
Смотри задачу 14.12.
14 УСЛОВИЯ ЗАДАЧ 8 класс 14.19.
В стопке лежат одинаковые карточки, на которых записаны числа от 1 до 12. Билл взял одну карточку и тайно отметил на ней 4 числа. Марк может сделать то же самое с несколькими карточками. Затем карточки открывают. Если на одной из карточек Марка хотя бы два из четырех отмеченных чисел совпадут с числами Билла, то Марк выигрывает. Какое наименьшее число карточек должен взять Марк и как их за-полнить, чтобы наверняка выиграть?
14.20.
Дан прямоугольник 𝐴𝐵𝐶𝐷. На луче 𝐷𝐶отложен отрезок 𝐷𝐾, равный 𝐵𝐷. Точка 𝑀— середина отрезка 𝐵𝐾.
Докажите, что 𝐴𝑀— биссектриса угла 𝐵𝐴𝐶.
14.21.
У фокусника есть два комплекта по 8 карточек. На розовых карточках записаны целые числа от 0 до 7. На первой голубой карточке написано 1, а число на каждой следующей голубой карточке в 8 раз больше предыдущего. Фокусник рас-кладывает карточки попарно (розовую с голубой). Затем зри-тели перемножают числа в каждой паре и находят сумму всех 8 произведений. Фокус состоит в том, что в сумме должно полу-читься простое число. Подскажите фокуснику, какие карточки можно для этого объединить в пары (или докажите, что у него ничего не получится).
14.22.
На плоскости нарисовали 5 красных точек. Все сере-дины отрезков между ними отметили синим цветом. Располо-жите красные точки так, чтобы синих точек было минимально возможное количество. (Точка может оказаться красной и си-ней одновременно.) 14.23.
По кругу в каком-то порядке выписаны числа от 1 до 88. Какова минимально возможная сумма модулей разностей между соседними числами?
14.24.
Смотри задачу 14.12.
2013–2014 ГОД, ОТБОРОЧНЫЙ ЭТАП 15 9 класс 14.25.
В стопке лежат одинаковые карточки, на которых записаны числа от 1 до 33. Билл взял одну карточку и тайно отметил на ней 10 чисел. Марк может сделать то же самое с несколькими карточками. Затем карточки открывают. Если на одной из карточек Марка хотя бы три из десяти отмеченных чисел совпадут с числами Билла, то Марк выигрывает. Какое наименьшее число карточек должен взять Марк и как их за-полнить, чтобы наверняка выиграть?
14.26.
Смотри задачу 14.20.
14.27.
Назовем основание системы счисления комфорт-ным, если существует простое число, запись которого в этой системе счисления ровно по одному разу содержит каждую из ее цифр. Например, 3 — комфортное основание, так как троич-ное число 102 — простое. Найдите все комфортные основания, не превосходящие 10.
14.28.
На плоскости нарисовали 5 красных точек, никакие три из которых не лежат на одной прямой. Все середины отрез-ков между ними отметили синим цветом. Расположите красные точки так, чтобы синих точек было минимально возможное ко-личество.
14.29.
По кругу в каком-то порядке выписаны числа от 1 до 99. Какова минимально возможная сумма модулей разностей между соседними числами?
14.30.
Решите систему уравнений: {︃ 𝑥+ 𝑦+ 𝑥𝑦= 11 𝑥2𝑦+ 𝑥𝑦2 = 30 10 класс 14.31.
Назовем год лихим, если в записи его номера есть одинаковые цифры. Например, все годы с 1988 по 2012 были 16 УСЛОВИЯ ЗАДАЧ лихими. Докажите, что в каждом столетии, начиная с двадцать первого, есть хотя бы 44 лихих года.
14.32.
Азимутом называется угол от 0 до 360∘, отсчитан-ный по часовой стрелке от направления на север до направле-ния на нужный ориентир. Алекс видит телебашню под азиму-том 60∘, водонапорную башню под азимутом 90∘, а колокольню под азимутом 120∘. Для Бориса те же азимуты соответственно равны 270∘, 240∘и 𝑋. Какие значения может принимать 𝑋?
14.33.
Назовем основание системы счисления комфорт-ным, если существует простое число, запись которого в этой системе счисления ровно по одному разу содержит каждую из ее цифр. Например, 3 — комфортное основание, так как троич-ное число 102 — простое. Найдите все комфортные основания, не превосходящие 12.
14.34.
У Кости есть 𝑛одинаковых кубиков. У каждого ку-бика на двух противоположных гранях написаны числа 5 и 6, а на остальных — 1, 2, 3 и 4 (именно в этом порядке по кру-гу). Костя склеил из этих кубиков столбик — параллелепипед 1 × 1 × 𝑛— и покрыл лаком все шесть граней этого столбика.
После этого он расклеил кубики и обнаружил, что сумма чи-сел на покрытых лаком гранях меньше, чем на остальных. При каком наименьшем 𝑛такое могло произойти?
14.35.
𝐶𝐻— высота в треугольнике 𝐴𝐵𝐶, а 𝑂— центр его описанной окружности. Из точки 𝐶опустили перпендикуляр на 𝐴𝑂, а его основание обозначили через 𝑇. Наконец, через 𝑀 обозначили точку пересечения 𝐻𝑇и 𝐵𝐶. Найдите отношение длин отрезков 𝐵𝑀и 𝐶𝑀.
14.36.
Смотри задачу 14.30.
11 класс 14.37.
Смотри задачу 14.31.
2013–2014 ГОД, ЗАКЛЮЧИТЕЛЬНЫЙ ЭТАП 17 14.38.
Для исследования подводного мира соорудили пря-молинейную штангу, уходящую под углом 45∘к поверхности во-ды на глубину 100 метров. Водолаз связан со штангой гибким тросом, позволяющим ему удаляться от любой точки штанги на расстояние не более 10 метров. Считая размеры водолаза нуле-выми (точечными), найдите объем доступной ему части под-водного пространства. Дайте точный ответ и округлите его до ближайшего целого значения в кубических метрах.
14.39.
Назовем основание системы счисления комфорт-ным, если существует простое число, запись которого в этой системе счисления ровно по одному разу содержит каждую из ее цифр. Например, 3 — комфортное основание, так как троич-ное число 102 — простое. Найдите все комфортные основания.
14.40.
Смотри задачу 14.34.
14.41.
Смотри задачу 14.35.
14.42.
Пусть 𝑝1, . . . , 𝑝𝑛— различные простые числа. Пусть 𝑆— сумма всевозможных произведений четного (ненулевого) количества различных простых из этого набора. Докажите, что 𝑆+ 1 делится на 2𝑛−2.
Заключительный этап 5 класс 14.43.
Разрежьте шахматную доску по клеточкам на две фигуры так, что в первой фигуре на 4 клетки больше, чем во второй, но во второй фигуре на 4 черных клетки больше, чем в первой. Обе фигуры должны быть связными, то есть не должны распадаться на части.
14.44.
Известно, что в понедельник маляр красил вдвое медленнее, чем во вторник, среду и четверг, а в пятницу — вдвое быстрее, чем в эти три дня, но работал 6 часов вместо 18 УСЛОВИЯ ЗАДАЧ 8. В пятницу он покрасил на 300 метров забора больше, чем в понедельник. Сколько метров забора маляр покрасил с поне-дельника по пятницу?
14.45.
Найдите количество четырехзначных чисел, у кото-рых все цифры различны, первая цифра делится на 2, а сумма первой и последней цифры делится на 3.
14.46.
В семье Олимпионовых принято особо отмечать день, когда человеку исполняется столько лет, какова сумма цифр его года рождения. У Коли Олимпионова такой праздник настал в 2013 году, а у Толи Олимпионова — в 2014. Кто из них старше и на сколько лет?
14.47.
Карлсон купил в буфете несколько блинов (по 25 рублей за штуку) и несколько банок меда (по 340 рублей за штуку). Когда он сообщил Малышу, какую сумму потратил в буфете, тот сумел только на основании этой информации опре-делить, сколько банок меда и сколько блинов купил Карлсон.
Могла ли эта сумма превысить 2000 рублей?
14.48.
Братья нашли клад из золота и серебра. Они раз-делили его так, что каждому досталось по 100 кг. Старшему досталось больше всего золота — 30 кг — и пятая часть всего серебра. Сколько золота было в кладе?
6 класс 14.49.
Разрежьте шахматную доску по клеточкам на две фигуры так, чтобы в первой фигуре было на 6 клеток больше, чем во второй, но во второй фигуре было на 6 черных клеток больше, чем в первой. Обе фигуры должны быть связными, то есть не должны распадаться на части.
14.50.
Смотри задачу 14.46.
2013–2014 ГОД, ЗАКЛЮЧИТЕЛЬНЫЙ ЭТАП 19 14.51.
Найдите количество таких пятизначных чисел, у ко-торых все цифры различны, первая цифра делится на 2, а сумма первой и последней цифр делится на 3.
14.52.
В начале года американский доллар стоил 80 евро-пейских центов. Эксперт дал прогноз, что в течение года курс евро по отношению к рублю вырастет на 8% (то есть за 1 евро можно будет купить на 8% рублей больше, чем в начале года), а курс доллара по отношению к рублю упадет на 10%. Если прогноз сбудется, то сколько американских центов будет сто-ить евро в конце года?
14.53.
Смотри задачу 14.47.
14.54.
Смотри задачу 14.48.
7 класс 14.55.
Мила и Женя придумали по числу и выписали на доску все натуральные делители своих чисел. Мила написала 10 чисел, Женя выписала 9 чисел, а максимальное число, на-писанное на доске дважды, оказалось равно 50. Сколько всего различных чисел выписано на доске?
14.56.
На клетчатой бумаге нарисован многоугольник с пе-риметром 36, стороны которого проходят по линиям сетки. Ка-кую наибольшую площадь он может иметь?
14.57.
В окружности проведены три равные хорды, прохо-дящие через одну точку. Докажите, что эти хорды являются диаметрами.
14.58.
Братья нашли клад из золота и серебра. Они раз-делили его так, что каждому досталось по 100 кг. Старшему досталось больше всего золота — 25 кг — и восьмая часть всего серебра. Сколько золота было в кладе?
14.59.
Три человека хотят приехать из города 𝐴в город 𝐵, расположенный в 45 километрах от 𝐴. У них есть два велосипе-20 УСЛОВИЯ ЗАДАЧ да. Скорость велосипедиста равна 15 км/ч, пешехода — 5 км/ч.
За какое минимальное время они смогут добраться до 𝐵, если велосипед нельзя оставлять на дороге без присмотра?
14.60.
Лев взял два натуральных числа, прибавил их сум-му к их произведению и в результате получил 1000. Какие числа мог взять Лев? Найдите все варианты.
8 класс 14.61.
Мила и Женя придумали по числу и выписали на доску все натуральные делители своих чисел. Мила написала 10 чисел, Женя — 9, а число 6 оказалось написано дважды.
Сколько всего различных чисел на доске?
14.62.
Смотри задачу 14.57.
14.63.
Несколько братьев нашли клад из золота и серебра.
Они разделили его так, что каждому досталось по 100 кг. Стар-шему досталась 1/5 всего золота и 1/7 всего серебра, а младше-му — 1/7 всего золота. А какая доля общего серебра досталась младшему брату?
14.64.
Три человека хотят приехать из города 𝐴в город 𝐵, расположенный в 45 километрах от 𝐴. У них есть два вело-сипеда. Скорость велосипедиста 15 км/ч, пешехода — 5 км/ч.
За какое минимальное время они смогут добраться до 𝐵, если велосипед можно оставлять на дороге без присмотра?
14.65.
Смотри задачу 14.47.
14.66.
На клетчатой бумаге нарисован многоугольник с пе-риметром 2014, стороны которого проходят по линиям сетки.
Какую наибольшую площадь он может иметь?
9 класс 14.67.
В выпуклом пятиугольнике провели все диагонали.
Для каждой пары диагоналей, пересекающихся внутри пяти-2013–2014 ГОД, ЗАКЛЮЧИТЕЛЬНЫЙ ЭТАП 21 угольника, нашли меньший из углов между ними. Какие значе-ния может принимать сумма этих пяти углов?
14.68.
Смотри задачу 14.61.
14.69.
Смотри задачу 14.63.
14.70.
Докажите, что из круга радиуса 1 можно вырезать три части, из которых можно составить прямоугольник 1 × 2,4.
Части можно поворачивать и переворачивать.
14.71.
Пусть 𝑎и 𝑛— натуральные числа, причем известно, что 𝑎𝑛— 2014-значное число. Найдите наименьшее натуральное 𝑘такое, что 𝑎не может быть 𝑘-значным числом.
14.72.
Павел придумал новый способ сложения чисел — он называет «павлосуммой» чисел 𝑥и 𝑦значение выражения 𝑥⊕𝑦= 𝑥+ 𝑦 1 −𝑥𝑦, если оно определено. Однажды он «сложил» своим способом числа 𝑎и 𝑏и «прибавил» к ним 𝑐, а друга попросил «сложить» числа 𝑏и 𝑐и «прибавить» к ним 𝑎. Могли ли у них получиться разные результаты?
10 класс 14.73.
Смотри задачу 14.67.
14.74.
Пусть 𝑓(𝑥) = 𝑥3 + 9𝑥2 + 27𝑥+ 24. Решите уравнение 𝑓(𝑓(𝑓(𝑓(𝑥)))) = 0.
14.75.
Докажите, что из круга радиуса 1 можно выре-зать четыре части, из которых можно составить прямоугольник 1 × 2,5. Части можно поворачивать и переворачивать.
14.76.
Внутри квадрата со стороной 100 нарисовали 100 000 квадратов. Диагонали разных квадратов не имеют общих точек.
Докажите, что сторона хотя бы одного квадрата меньше 1.
14.77.
Смотри задачу 14.71.
22 УСЛОВИЯ ЗАДАЧ 14.78.
Павел придумал новый способ сложения чисел — он называет «павлосуммой» чисел 𝑥и 𝑦значение выражения 𝑥⊕𝑦= 𝑥+ 𝑦 1 −𝑥𝑦, если оно определено. Однажды он «сложил» своим способом числа 𝑎и 𝑏, «прибавил» к ним 𝑐, а к результату — 𝑑. В то же время его друг «сложил» числа 𝑐и 𝑑, «прибавил» к ним 𝑏, а к результату — 𝑎. Могли ли у них получиться разные результаты?
11 класс 14.79.
Смотри задачу 14.67.
14.80.
Смотри задачу 14.74.
14.81.
Докажите, что из круга радиуса 1 можно вырезать пять частей, из которых можно составить прямоугольник 1×2,7.
Части можно поворачивать и переворачивать.
14.82.
Существует ли треугольная пирамида, у которой вы-сота равна 60, высота каждой боковой грани, проведенная к стороне основания, равна 61, а периметр основания равен 62?
14.83.
Смотри задачу 14.71.
14.84.
Павел придумал новый способ сложения чисел — он называет «павлосуммой» чисел 𝑎и 𝑏значение выражения 𝑎⊕𝑏= 𝑎+ 𝑏 1 −𝑎𝑏, если оно определено. Как и в обычной арифметике, умножение на натуральное число Павел понимает как сложение соответ-ствующего числа одинаковых слагаемых: 𝑎⊗𝑏= ((𝑎⊕𝑎) ⊕𝑎) ⊕. . . ) ⊕𝑎(здесь 𝑏«слагаемых»).
Существуют ли в арифметике Павла такие неравные натураль-ные числа 𝑥и 𝑦, для которых равны «произведения» 𝑥⊗𝑦и 𝑦⊗𝑥?
2014–2015 ГОД, ОТБОРОЧНЫЙ ЭТАП 23 2014–2015 год Отборочный этап 5 класс 15.1.
Назовем тяжелым месяц, в котором пять понедель-ников. Сколько тяжелых месяцев может быть в течение года?
15.2.
Андрей перемножил две последовательные цифры и получил в итоге двузначное число, записываемое двумя после-довательными цифрами. Найдите все такие примеры.
15.3.
Саша зачеркнул на 25-й странице учебника все сло-ва, в которых нет буквы А, потом он зачеркнул все слова, в которых нет буквы Б, а потом он нашел все слова, где есть и буква О, и буква А, и тоже зачеркнул их. Костя на той же стра-нице своего учебника зачеркнул слова, где нет Б, но есть А или О (возможно, обе сразу), и после этого он зачеркнул все слова, где нет ни буквы А, ни буквы О. Могло ли у Саши остаться незачеркнутыми больше слов, чем у Кости?
15.4.
В каждом из двух классов по 30 учеников. Мальчи-ков в первом классе вдвое больше, чем во втором, а девочек — втрое меньше, чем во втором. Сколько мальчиков и девочек в каждом классе?
15.5.
Три ручки, четыре карандаша и линейка вместе сто-ят 26 рублей, а пять ручек, шесть карандашей и три линейки — 44 рубля. Сколько стоят вместе две ручки и три карандаша?
15.6.
Первоначально на доске написано число 1. Разре-шается любое написанное на доске число умножить на 3 или переставить в нем цифры. Можно ли таким образом получить 999?
24 УСЛОВИЯ ЗАДАЧ 6 класс 15.7.
Смотри задачу 15.1.
15.8.
Смотри задачу 15.2.
15.9.
Смотри задачу 15.3.
15.10.
Смотри задачу 15.4.
15.11.
Смотри задачу 15.5.
15.12.
Первоначально на доске написано число 1. Разре-шается любое написанное на доске число умножить на 2 или переставить в нем цифры. Можно ли таким образом получить 209?
7 класс 15.13.
Смотри задачу 15.1.
15.14.
Смотри задачу 15.2.
15.15.
Сумма трех натуральных чисел равна 100. Какое наименьшее возможное значение может принимать НОК этих чисел?
15.16.
Докажите, что при любой расстановке чисел 1, 2, . . ., 10 по кругу найдутся три соседних числа с суммой не менее 18.
15.17.
Смотри задачу 15.5.
15.18.
Найдите наименьшее натуральное число, которое начинается на 11, заканчивается на 11 и делится на 7. Объ-ясните, почему это число является наименьшим из удовлетво-ряющих условию.
8 класс 15.19.
Докажите, что для любого 𝑛> 3 существует 𝑛-угольник, у которого никакие две диагонали не параллельны.
2014–2015 ГОД, ОТБОРОЧНЫЙ ЭТАП 25 15.20.
𝐵𝐾— биссектриса треугольника 𝐴𝐵𝐶. Известно, что 𝐴𝐵= 𝐴𝐶, а 𝐵𝐶= 𝐴𝐾+ 𝐵𝐾. Найдите углы треугольника 𝐴𝐵𝐶.
15.21.
Каждый из трех землекопов, работая в одиночку, может вырыть траншею за целое число дней. А если ту же траншею они будут рыть все втроем, на это у них уйдет соот-ветственно на 2, 5 и 10 дней меньше, чем при рытье вдвоем (то есть без первого, второго и третьего соответственно). За сколь-ко дней может выкопать траншею самый медленный из них?
15.22.
Даны 15 составных чисел, не превосходящих 2014.
Докажите, что какие-то два из них имеют общий делитель, больший 1.
15.23.
Дан квадрат 100×100 без угловой клетки. Можно ли разрезать его по клеткам на 33 фигуры, у которых одинаковые площади и одинаковые периметры?
15.24.
В шестизначном числе поставили знак умножения после первых трех цифр, и оказалось, что произведение двух полученных трехзначных чисел в 7 раз меньше исходного числа.
Какое число было написано?
15.25.
Есть набор из 𝑁2 карточек, на каждой карточке с одной стороны написано число, с другой стороны пусто. На-писанные числа попарно различны. Эти карточки выложены в виде квадрата 𝑁× 𝑁пустой стороной (рубашкой) вверх. Раз-решается перевернуть любую карточку и тем самым узнать на-писанное на ней число. Докажите, что не более чем за 8𝑁пере-ворачиваний можно найти карточку, число на которой меньше, чем число на каждой из соседних с ней (по стороне) карточек.
15.26.
Назовем натуральное число возрастающим, если цифры в его записи идут в порядке строгого возрастания (на-пример, числа 7 и 1589 — возрастающие, а 2447 — нет). Ка-26 УСЛОВИЯ ЗАДАЧ кое наименьшее количество возрастающих чисел надо сложить, чтобы получить 2014?
15.27.
Найдите все натуральные 𝑎, 𝑏и 𝑐, для которых 2𝑎−2𝑏−2𝑏+𝑐= 2014.
15.28.
В треугольнике 𝐴𝐵𝐶углы 𝐵и 𝐶равны 30∘и 105∘, а 𝑃— середина стороны 𝐵𝐶. Найдите угол 𝐵𝐴𝑃.
9 класс 15.29.
Смотри задачу 15.19.
15.30.
Смотри задачу 15.15.
15.31.
Смотри задачу 15.21.
15.32.
Андрей перемножил два последовательных нату-ральных числа и получил в некоторой системе счисления дву-значное число, записываемое двумя последовательными цифра-ми, не превосходящими 9. Найдите эти цифры.
15.33.
Дан квадрат 100 × 100 без угловой клетки. Можно ли разрезать его на 33 фигуры, у которых одинаковые площади и одинаковые периметры?
15.34.
Смотри задачу 15.27.
15.35.
В таблице 30×30 клеток поставлено 162 плюса и 144 минуса (в каждой клетке не более одного знака) так, что в каж-дой строке и каждом столбце таблицы стоит не более 17 знаков.
Для каждого плюса подсчитали, сколько минусов находится в той же строке. Для каждого минуса подсчитали, сколько плю-сов находится в том же столбце. Какое наибольшее значение может иметь сумма найденных чисел?
15.36.
В треугольнике 𝐴𝐵𝐶выбрана точка 𝐷на стороне 𝐴𝐵так, что углы 𝐴𝐶𝐷и 𝐴𝐵𝐶равны. Пусть 𝑆— центр опи-санной окружности треугольника 𝐵𝐶𝐷. Докажите, что точки 𝐴, 𝐶, 𝑆и середина 𝐵𝐷лежат на одной окружности.
2014–2015 ГОД, ОТБОРОЧНЫЙ ЭТАП 27 15.37.
Треугольники 𝐴𝐵𝐶и 𝐴1𝐵1𝐶1 таковы, что sin 𝐴= = cos 𝐴1, sin 𝐵= cos 𝐵1, sin 𝐶= cos 𝐶1. Какие значения может принимать наибольший из шести углов?
15.38.
Пусть 𝐻— такая точка внутри треугольника 𝐴𝐵𝐶, что ∠𝐻𝐴𝐵= ∠𝐻𝐶𝐵и ∠𝐻𝐵𝐶= ∠𝐻𝐴𝐶. Докажите, что 𝐻— точка пересечения высот треугольника 𝐴𝐵𝐶.
10 класс 15.39.
Выберите на каждой стороне квадрата по одной точ-ке так, чтобы образованный ими четырехугольник имел наи-меньший периметр.
15.40.
Смотри задачу 15.21.
15.41.
Смотри задачу 15.32.
15.42.
Костя выписал на доску 30 последовательных чле-нов арифметической прогрессии с разностью 2061. Докажите, что в ней содержится не более 20 точных квадратов.
15.43.
Вещественные числа 𝑥и 𝑦таковы, что 𝑥4𝑦2 + 𝑥2 + 2𝑥3𝑦+ 6𝑥2𝑦+ 8 ⩽0.
Докажите, что 𝑥⩾−1/6.
15.44.
Решите систему уравнений в целых числах: {︃ 2𝑎+ 3𝑏= 5𝑏 3𝑎+ 6𝑏= 9𝑏 15.45.
Маша красит клетки белой доски 10×10. Она может покрасить любой вертикальный ряд клеток синей краской или любой горизонтальный ряд красной краской (каждый ряд кра-сят не более одного раза). Если синяя краска ложится поверх красной, получается синяя клетка, а если красная поверх синей, то краски вступают в реакцию и обесцвечиваются, получается белая клетка. Может ли на доске оказаться 33 красных клетки?
15.46.
Смотри задачу 15.36.
28 УСЛОВИЯ ЗАДАЧ 15.47.
Смотри задачу 15.37.
15.48.
Решите уравнение в простых числах: 100𝑞+ 80 = 𝑝3 + 𝑝𝑞2.
11 класс 15.49.
Смотри задачу 15.21.
15.50.
Смотри задачу 15.32.
15.51.
Смотри задачу 15.42.
15.52.
Смотри задачу 15.43.
15.53.
Смотри задачу 15.45.
15.54.
Можно ли утверждать, что log√𝑎(𝑎+ 1) + log𝑎+1 √𝑎⩾ √ 6 при 𝑎> 1?
15.55.
Докажите, что количество способов разрезать пря-моугольник 200 × 3 на домино (прямоугольники 1 × 2) делится на 3.
15.56.
Случайным образом выбираются три числа от 1 до 𝑁(возможно, совпадающие) и располагаются в порядке возрас-тания. С какой вероятностью они образуют арифметическую прогрессию?
15.57.
Смотри задачу 15.37.
15.58.
Пусть 𝑑(𝑘) — число делителей натурального числа 𝑘, а квадратные скобки означают целую часть вещественного числа. Докажите, что числа 𝑑(1)+𝑑(2)+. . .+𝑑(𝑛) и [√𝑛] имеют одинаковую четность.
2014–2015 ГОД, ЗАКЛЮЧИТЕЛЬНЫЙ ЭТАП 29 Заключительный этап 5 класс 15.59.
В некотором языке есть 3 гласных и 5 согласных букв. Слог может состоять из любой гласной буквы и любой согласной в любом порядке, а слово — из любых двух слогов.
Сколько слов в этом языке?
15.60.
Приведите пример таких целых чисел 𝑎и 𝑏, что 𝑎𝑏(2𝑎+ 𝑏) = 2015.
15.61.
Маша с Леной вышли из дома и пошли в магазин за мороженым. Маша шла быстрее и дошла до магазина за 12 минут. Потратив 2 минуты на покупку мороженого, она пошла назад и встретила Лену еще через 2 минуты. Сколько времени потребовалось Лене, чтобы дойти до магазина? Скорости дево-чек постоянны.
15.62.
Шестеро школьников решили посадить в школьном дворе 5 деревьев. Известно, что каждое дерево сажало разное число школьников и каждый школьник участвовал в посадке одинакового количества деревьев. Могло ли так случиться?
15.63.
В плоском мире есть два прямоугольных острова.
Прибрежными водами каждого острова считается часть моря, удаленная от берега не более чем на 50 км. Может ли случиться, что площадь первого острова больше, чем второго, а площадь прибрежных вод у второго острова больше, чем у первого? Счи-тайте, что ближайшая к каждому острову суша находится на расстоянии больше 50 км.
15.64.
Аня, Галя, Даша, Соня и Лена приехали в лагерь «Формула Единства» из разных городов: Курска, Вологды, Но-вороссийска, Петрозаводска и Чебоксар. Знакомясь с другими членами отряда, они рассказали о себе следующее: Соня и Даша никогда не были в Курске; 30 УСЛОВИЯ ЗАДАЧ Галя и Соня были вместе в прошлом лагере с девочкой из Новороссийска; Аня и Соня подарили девочке из Чебоксар по сувенир-чику; Галя и Соня помогли девочке из Вологды занести вещи в комнату; Галя и Лена общаются по скайпу с девочкой из Чебоксар, а девочка из Новороссийска переписывается в контакте с Аней.
Кто где живет?
6 класс 15.65.
В некотором языке есть 5 гласных и 7 согласных букв. Слог может состоять из любой гласной буквы и любой согласной в любом порядке, а слово — из любых двух слогов.
Сколько слов в этом языке?
15.66.
Смотри задачу 15.60.
15.67.
На столе лежат три конфеты. У Ани и Лены есть мешок с неограниченным количеством конфет, и они играют в игру. Каждая из них своим ходом добавляет некоторое коли-чество конфет из мешка на стол, но при этом не может поло-жить больше конфет, чем уже лежит на столе. Девочки ходят по очереди, начинает Аня. Выигрывает та, после хода которой на столе окажется ровно 2015 конфет. Кто из девочек может обеспечить себе победу, как бы ни играла соперница?
15.68.
Смотри задачу 15.62.
15.69.
В плоском мире есть два прямоугольных острова.
Прибрежными водами каждого острова считается часть моря, удаленная от берега не более чем на 50 км. Может ли случиться, что периметр первого острова больше, чем второго, а площадь прибрежных вод у второго острова больше, чем у первого? Счи-тайте, что ближайшая к каждому острову суша находится на расстоянии больше 50 км.
2014–2015 ГОД, ЗАКЛЮЧИТЕЛЬНЫЙ ЭТАП 31 15.70.
Дима варит кашу. Чтобы каша получилась вкусной, ему нужно варить крупу ровно 24 минуты. Обычных часов у Димы нет, но есть двое песочных часов: одни — на 20 минут, другие — на 7 минут. Как Диме точно отмерить требуемое вре-мя?
7 класс 15.71.
В некотором языке есть 3 гласных и 7 согласных букв. Слог может состоять из любой гласной буквы и любой согласной в любом порядке, а слово — из любых трех слогов.
Слово называется забавным, если в нем встречаются две оди-наковые буквы подряд. Сколько забавных слов в этом языке?
15.72.
Приведите пример таких целых чисел 𝑎и 𝑏, что (10𝑎+ 𝑏)(𝑎+ 10𝑏)(𝑎+ 𝑏+ 1) = 2015.
15.73.
В равнобедренном треугольнике 𝐴𝐵𝐶(какие две из сторон треугольника равны, неизвестно) проведены медианы 𝐴𝐴1 и 𝐵𝐵1, которые пересекаются в точке 𝑂. Известно, что ∠𝐴𝑂𝐵= 120∘. Найдите углы треугольника 𝐴𝐵𝐶.
15.74.
По вновь придуманным правилам в каждом матема-тическом бою участвуют одновременно 3 команды. Организа-торы хотят провести турнир из нескольких (более одного) боев так, чтобы каждые две команды встречались между собой ров-но один раз. Какое наименьшее число команд нужно для этого пригласить?
15.75.
В плоском мире есть два треугольных острова. При-брежными водами каждого острова считается часть моря, уда-ленная от берега не более чем на 50 км. Может ли случиться, что периметры этих островов одинаковы, а площадь прибреж-ных вод у них различается? Считайте, что ближайшая к каж-дому острову суша находится на расстоянии больше 50 км.
15.76.
На плоскости нарисован 2015-угольник со всеми диа-гоналями. Дима с Сашей играют в следующую игру. Они по-32 УСЛОВИЯ ЗАДАЧ очередно стирают либо от 1 до 10 соседних сторон нарисован-ного многоугольника, либо от 1 до 9 его диагоналей. Тот, кто не может сделать ход, проигрывает. Первым ходит Дима. Кто из играющих может обеспечить себе победу при любой игре со-перника? Как он сможет это сделать?
8 класс 15.77.
В некотором языке есть 3 гласных и 8 согласных букв. Слог может состоять из любой гласной буквы и любой согласной в любом порядке, а слово — из любых трех слогов.
Слово называется забавным, если в нем встречаются две оди-наковые буквы подряд. Сколько забавных слов в этом языке?
15.78.
Один из концов отрезка закрасили в синий цвет, а другой — в красный. Внутри отрезка выбрали 2015 точек и каж-дую из них произвольным образом закрасили в какой-то из этих же цветов. В результате отрезок разбился на 2016 частей. Мо-жет ли количество таких частей, у которых оба конца красные, равняться количеству частей, у которых оба конца синие?
15.79.
Смотри задачу 15.73.
15.80.
Натуральные числа 𝑎, 𝑏, 𝑐и 𝑑таковы, что 2015𝑎+ 2015𝑏= 2015𝑐+ 2015𝑑.
Могут ли быть различными числа 𝑎2015 + 𝑏2015 и 𝑐2015 + 𝑑2015?
15.81.
В плоском мире есть два треугольных острова. При-брежными водами каждого острова считается часть моря, уда-ленная от берега не более чем на 50 км. Может ли случиться, что периметр первого острова больше, чем второго, а площадь прибрежных вод у второго острова больше, чем у первого? Счи-тайте, что ближайшая к каждому острову суша находится на расстоянии больше 50 км.
15.82.
Марк задумал число 𝑚и нашел число 𝑘диагона-лей у выпуклого 𝑚-угольника. Затем Марк сообщил Кириллу 2014–2015 ГОД, ЗАКЛЮЧИТЕЛЬНЫЙ ЭТАП 33 число 𝑘и предложил ему найти 𝑚. Перепутав вопрос, Кирилл пересчитал диагонали у выпуклого 𝑘-угольника. Их оказалось 2015. Найдите 𝑚.
9 класс 15.83.
Смотри задачу 15.80.
15.84.
Смотри задачу 15.78.
15.85.
На сторонах 𝐴𝐵и 𝐵𝐶квадрата 𝐴𝐵𝐶𝐷выбраны соответственно такие точки 𝑀и 𝑃, что 𝐴𝑀= 𝐶𝑃. Окружность на диаметре 𝐷𝑃пересекает отрезок 𝐶𝑀в точке 𝐾. Докажите, что 𝑀𝐾и 𝐵𝐾перпендикулярны.
15.86.
Даны 10 последовательных целых чисел, превосхо-дящих 1. Каждое из них разложили на простые множители, а через 𝑝обозначили наибольший из всех множителей. Какое наименьшее значение может принимать 𝑝?
15.87.
В плоском мире есть два острова, которые име-ют форму выпуклых многоугольников. Прибрежными водами каждого острова считается часть моря, удаленная от берега не более чем на 50 км. Может ли случиться, что периметр перво-го острова больше, чем второго, а площадь прибрежных вод у второго острова больше, чем у первого? Считайте, что ближай-шая к каждому острову суша находится на расстоянии больше 50 км.
15.88.
Смотри задачу 15.82.
10 класс 15.89.
Смотри задачу 15.80.
15.90.
Сколько пятизначных чисел делятся на свою послед-нюю цифру?
15.91.
Точки 𝐻, 𝐾и 𝑀лежат соответственно на сторонах 𝐵𝐶, 𝐴𝐶и 𝐴𝐵треугольника 𝐴𝐵𝐶, в котором 𝐴𝐻является вы-34 УСЛОВИЯ ЗАДАЧ сотой. Докажите, что 𝐴𝐻служит биссектрисой угла 𝐾𝐻𝑀то-гда и только тогда, когда 𝐴𝐻, 𝐵𝐾и 𝐶𝑀пересекаются в одной точке.
15.92.
Смотри задачу 15.86.
15.93.
Смотри задачу 15.87.
15.94.
Смотри задачу 15.82.
11 класс 15.95.
Смотри задачу 15.80.
15.96.
Смотри задачу 15.90.
15.97.
Смотри задачу 15.91.
15.98.
Смотри задачу 15.86.
15.99.
Ребро правильного тетраэдра 𝐴𝐵𝐶𝐷равно 1. Через точку 𝑀, лежащую на грани 𝐴𝐵𝐶(но не на ребре), проведены плоскости, параллельные трем другим граням. Эти плоскости делят тетраэдр на части. Найдите сумму длин ребер той части, которая содержит точку 𝐷.
15.100. Смотри задачу 15.82.
2015–2016 год Отборочный этап 5 класс 16.1.
Петя, Вася и Толик в складчину купили футбольный мяч. Известно, что каждый из них заплатил не больше полови-ны суммы, заплаченной двумя другими. Мяч стоил 9 рублей.
Сколько денег заплатил Петя?
2015–2016 ГОД, ОТБОРОЧНЫЙ ЭТАП 35 16.2.
Полина написала на доске два числа 𝐴и 𝐵. Вика их стерла, записав числа 𝐶и 𝐷, равные сумме и произведению чисел, записанных Полиной. Затем Полина стерла числа, запи-санные Викой, также записав их сумму и произведение — 𝐸и 𝐹. Из последних двух чисел одно оказалось нечетным. Какое именно и почему?
16.3.
Считается, что ученик 𝐴учится лучше ученика 𝐵, если в большинстве контрольных работ оценка у ученика 𝐴вы-ше, чем у ученика 𝐵. В классе провели три контрольные рабо-ты. Могло ли оказаться, что ученик 𝐴учится лучше, чем ученик 𝐵, ученик 𝐵— лучше, чем ученик 𝐶, 𝐴ученик 𝐶— лучше, чем ученик 𝐴?
16.4.
По вечерам Лева врет маме, если в этот день полу-чил двойку, а в остальных случаях говорит правду. А еще у Левы есть сестра, которой мама дает конфеты в те дни, когда она не получает двоек. Однажды вечером Лева сказал маме: «Сегодня я получил больше двоек, чем моя сестра». Достанут-ся ли сестре конфеты?
16.5.
Волшебный календарь показывает правильную дату по четным числам месяца и неправильную по нечетным. Какое максимальное количество дней подряд он может показывать од-ну и ту же дату? Укажите все возможные числа месяца, кото-рые он при этом может показывать.
16.6.
Сколько существует десятизначных чисел, в кото-рых нет повторяющихся цифр, а цифры 0, 1, 2, 3 стоят подряд в порядке возрастания?
16.7.
Алексей разрезал квадрат 8 × 8 по границам кле-ток на 7 частей с равными периметрами. Покажите, как он это сделал (достаточно привести один пример).
36 УСЛОВИЯ ЗАДАЧ 6 класс 16.8.
По кругу сидят 14 человек. Петя, Вика, Толик и Чингиз сидят подряд, у каждого из них есть по монете: у Пети 1 рубль, у Вики 2 рубля, у Толика 5 рублей, а у Чингиза 10 рублей. У других ребят денег нет. Любой человек из сидящих в кругу может передать свою монету другому, если между ними сидят ровно три человека. Оказалось, что через некоторое вре-мя монеты опять оказались у Пети, Вики, Толика и Чингиза.
У кого теперь какая монета?
16.9.
Смотри задачу 16.2.
16.10.
Смотри задачу 16.3.
16.11.
Смотри задачу 16.4.
16.12.
Смотри задачу 16.5.
16.13.
Сколько существует десятизначных чисел, в кото-рых нет повторяющихся цифр, а цифры 0, 1, 2, 3 стоят подряд в порядке возрастания или в порядке убывания?
16.14.
Смотри задачу 16.7.
7 класс 16.15.
Смотри задачу 16.5.
16.16.
Расставьте в клетках квадрата 5×5 различные нату-ральные числа так, чтобы их суммы в каждой строке и каждом столбце были равны между собой и (при этом условии) как мож-но меньшими. На одной из диагоналей уже стоят числа 1, 2, 3, 4 и 2015 (повторно их использовать нельзя).
16.17.
Смотри задачу 16.7.
16.18.
В тараканьих бегах участвуют 27 тараканов. В каж-дом забеге бегут три таракана. Скорости всех тараканов различ-ны и постоянны в течение всех забегов. После каждого забега мы узнаем, в каком порядке его участники пришли к финишу.
2015–2016 ГОД, ОТБОРОЧНЫЙ ЭТАП 37 Мы хотели бы узнать двух самых быстрых тараканов (в пра-вильном порядке). Хватит ли для этого 14 забегов?
16.19.
Считается, что ученик 𝐴учится лучше ученика 𝐵, если в большинстве контрольных работ оценка у ученика 𝐴вы-ше, чем у ученика 𝐵. В классе провели несколько работ (боль-ше трех). Может ли по их результатам оказаться, что ученик 𝐴 учится лучше, чем ученик 𝐵, ученик 𝐵— лучше, чем ученик 𝐶, а ученик 𝐶— лучше, чем ученик 𝐴?
16.20.
Натуральное число называется красивым, если оно равно произведению факториалов простых чисел (не обязатель-но различных). Положительное рациональное число называет-ся практичным, если оно равно отношению двух красивых на-туральных чисел. Докажите, что любое положительное рацио-нальное число — практичное.
16.21.
Назовем натуральное число возрастающим, если его цифры идут в порядке строгого возрастания (например, 1589 — возрастающее, а 447 — нет). Какое наименьшее количество воз-растающих чисел надо сложить, чтобы получить 2015?
8 класс 16.22.
Смотри задачу 16.16.
16.23.
Смотри задачу 16.18.
16.24.
Найдите хотя бы одно натуральное число, произве-дение натуральных делителей которого равно 1090.
16.25.
У Васи есть 12 палочек, длина каждой из которых — натуральное число, не превосходящее 56. Докажите, что из каких-то трех палочек можно сложить треугольник.
16.26.
Смотри задачу 16.20.
16.27.
В треугольнике 𝐴𝐵𝐶угол 𝐵равен 30∘, а угол 𝐶 равен 105∘. Точка 𝐷— середина стороны 𝐵𝐶. Найдите угол 𝐵𝐴𝐷.
38 УСЛОВИЯ ЗАДАЧ 16.28.
Смотри задачу 16.19.
9 класс 16.29.
Вершины правильного 12-угольника покрашены в красный и синий цвета. Известно, что если выбрать любые три вершины, образующие равносторонний треугольник, то как ми-нимум две из них будут окрашены в красный цвет. Докажи-те, что найдется квадрат, как минимум три вершины которого красные.
16.30.
Смотри задачу 16.20.
16.31.
Смотри задачу 16.18.
16.32.
Смотри задачу 16.27.
16.33.
Смотри задачу 16.25.
16.34.
Смотри задачу 16.24.
16.35.
Хорошо известно, что 32 + 42 = 52. Менее известно, что 102 + 112 + 122 = 132 + 142. А существует ли 2015 после-довательных натуральных чисел, таких что сумма квадратов первых 1008 из них равна сумме квадратов последних 1007?
10 класс 16.36.
Багз Банни и Кролик Роджер поспорили, кто из них быстрее прыгает. Чтобы выяснить это, они решили провести соревнование: каждый должен прыжками преодолеть 50-метро-вую дистанцию, затем развернуться и вернуться к месту старта.
Известно, что Багз Банни прыгает на 50 см, а Роджер на 60 см, но за то время, за которое Багз делает 6 прыжков, Роджер де-лает 5. Кто же из кроликов финиширует первым?
16.37.
При каких 𝑛можно разрезать квадрат на 𝑛подоб-ных прямоугольников, не все из которых равны?
16.38.
Существуют ли такие натуральные числа 𝑎и 𝑏, что НОК (𝑎, 𝑏) = НОК (𝑎+ 2015, 𝑏+ 2016)?
2015–2016 ГОД, ОТБОРОЧНЫЙ ЭТАП 39 16.39.
Смотри задачу 16.27.
16.40.
Расставьте в клетках квадрата 10 × 10 различные натуральные числа так, чтобы их суммы в каждой строке и каждом столбце были равны между собой и (при этом условии) как можно меньшими. На одной из диагоналей уже стоят числа 1, 2, 3, 4, 5, 6, 7, 8, 9 и 2015 (повторно их использовать нельзя).
16.41.
Вписанная окружность касается сторон 𝐴𝐵, 𝐵𝐶и 𝐴𝐶треугольника 𝐴𝐵𝐶в точках 𝐶1, 𝐴1 и 𝐵1 соответственно.
Докажите неравенство: 𝐴𝐶 𝐴𝐵1 + 𝐶𝐵 𝐶𝐴1 + 𝐵𝐴 𝐵𝐶1 > 4.
16.42.
Хорошо известно, что 32 + 42 = 52. Менее известно, что 102 + 112 + 122 = 132 + 142. А для любого ли натурального 𝑘 существуют 2𝑘+1 последовательных натуральных чисел таких, что сумма квадратов первых 𝑘+1 из них равна сумме квадратов последних 𝑘?
11 класс 16.43.
Смотри задачу 16.36.
16.44.
Смотри задачу 16.37.
16.45.
Смотри задачу 16.38.
16.46.
Смотри задачу 16.27.
16.47.
В каждой целочисленной точке плоскости растет де-рево диаметром 10−6. Дровосек срубил дерево, стоящее в точке (0, 0), и встал в центр пенька. Ограничена ли часть плоскости, которую он сумеет увидеть? Считайте каждое дерево бесконеч-ной цилиндрической колонной, ось симметрии которой прохо-дит через целочисленную точку плоскости.
16.48.
Приведите пример четырех положительных чисел, которые не могут служить радиусами четырех попарно касаю-щихся сфер.
40 УСЛОВИЯ ЗАДАЧ 16.49.
Смотри задачу 16.42.
Заключительный этап 5 класс 16.50.
Придумайте пять различных натуральных чисел, произведение которых равно 1000.
16.51.
Каждая клетка доски 10×10 покрашена в синий или белый цвет. Назовем клетку радостной, если ровно две соседних с ней клетки синие. Закрасьте доску так, чтобы все клетки были радостными. (Клетки считаются соседними, если имеют общую сторону.) 16.52.
Вот задача из задачника С. А. Рачинского (конец XIX века): «Сколько досок длиною в 6 аршин, шириною в 6 вершков нужно, чтобы замостить пол в квадратной комнате, коей сторона — 12 аршин?» Ответ к задаче — 64 доски. Устано-вите по этим данным, сколько вершков в аршине.
16.53.
На поляне на расстоянии 20 метров одна от другой растут две ели высотой по 30 метров. Ветки елей растут очень густо и среди них есть направленные точно навстречу друг дру-гу, а длина каждой ветки вдвое меньше расстояния от нее до вершины. Паук может ползти по стволу (вверх или вниз строго по вертикали), по веткам (строго по горизонтали), либо спус-каться вертикально вниз по паутине с одной ветки на другую.
Какое наименьшее расстояние ему придется проползти, чтобы добраться с вершины одной ели на вершину другой?
16.54.
У Никиты есть волшебная банка. Если в банку по-ложить 𝑛конфет и закрыть на час, то количество лежащих в ней конфет увеличится на сумму цифр числа 𝑛. Например, если было 137 конфет, то станет 137 + 1 + 3 + 7 = 148. Какое макси-мальное количество конфет Никита может получить за 20 часов 16 минут, если вначале у него одна конфета?
2015–2016 ГОД, ЗАКЛЮЧИТЕЛЬНЫЙ ЭТАП 41 6 класс 16.55.
Смотри задачу 16.51.
16.56.
Смотри задачу 16.52.
16.57.
Смотри задачу 16.54.
16.58.
Назовем типичным любой прямо-угольный параллелепипед, все размеры которо-го (длина, ширина и высота, на рисунке обозна-ченные 𝑎, 𝑏и 𝑐) различны. На какое наимень-шее число типичных параллелепипедов можно разрезать куб? Не забудьте доказать, что это действительно наименьшее количество.
16.59.
Пятизначное число нравится Лидии, если ни одна из цифр в его записи не делится на 3. Найдите общую сумму цифр всех пятизначных чисел, которые нравятся Лидии.
7 класс 16.60.
Смотри задачу 16.50.
16.61.
На тетрадном листе обведены два прямоугольника.
У первого прямоугольника вертикальная сторона короче гори-зонтальной, а у второго — наоборот. Найдите максимально воз-можную площадь их общей части, если первый прямоугольник содержит 2015 клеток, а второй — 2016.
16.62.
Смотри задачу 16.58.
16.63.
Смотри задачу 16.59.
16.64.
Каждая клетка доски 100 × 100 покрашена в синий или белый цвет. Назовем клетку равновесной, если среди ее со-седей поровну синих и белых. Для каких 𝑛можно раскрасить доску так, чтобы на ней было ровно 𝑛равновесных клеток?
(Клетки считаются соседними, если имеют общую сторону.) 42 УСЛОВИЯ ЗАДАЧ 8 класс 16.65.
Существуют ли три таких различных цифры 𝐴, 𝐵и 𝐶, что 𝐴𝐵𝐶, 𝐶𝐵𝐴, 𝐶𝐴𝐵— квадраты натуральных чисел?
16.66.
Каждая клетка доски 100 × 100 покрашена в синий или белый цвет. Назовем клетку равновесной, если среди ее со-седей поровну синих и белых. Какое максимальное количество равновесных клеток может оказаться на доске? (Клетки счита-ются соседними, если имеют общую сторону.) 16.67.
На сторонах 𝐴𝐵и 𝐴𝐶треугольника 𝐴𝐵𝐶отмечены точки 𝑀и 𝑁соответственно, причем 𝐴𝑀= 𝐴𝑁. Отрезки 𝐶𝑀 и 𝐵𝑁пересекаются в точке 𝑂, причем 𝐵𝑂= 𝐶𝑂. Докажите, что 𝐴𝐵𝐶равнобедренный.
16.68.
На тетрадном листе обведены два прямоугольника.
У первого прямоугольника вертикальная сторона короче гори-зонтальной, а у второго — наоборот. Найдите максимально воз-можную площадь их общей части, если каждый прямоугольник содержит больше 2010, но меньше 2020 клеток.
16.69.
В игре «сет» участвуют всевозможные четырехзнач-ные числа, состоящие из цифр 1, 2, 3 (каждое число по одному разу). Говорят, что тройка чисел образует сет, если в каждом разряде либо все три числа содержат одну и ту же цифру, либо все три числа содержат разные цифры.
Например, числа 1232, 2213, 3221 образуют сет (в первом раз-ряде встречаются все три цифры, во втором — только двойка, в третьем — все три цифры, в четвертом — все три цифры). А числа 1123, 2231, 3311 не образуют сета (в последнем разряде встречаются две единицы и тройка).
Сколько всего сетов существует в игре?
Примечание: перестановка чисел не приводит к образованию нового сета, например, 1232, 2213, 3221 и 2213, 1232, 3221 — один и тот же сет.
2015–2016 ГОД, ЗАКЛЮЧИТЕЛЬНЫЙ ЭТАП 43 9 класс 16.70.
Найдите все такие числа 𝑘, для которых (︂𝑘 2 )︂ ! · 𝑘 4 = 2016 + 𝑘2.
16.71.
Смотри задачу 16.67.
16.72.
Смотри задачу 16.59.
16.73.
На координатной плоскости нарисовали равнобед-ренный треугольник 𝐴𝐵𝐶, у которого 𝐴𝐵= 2016, 𝐵𝐶= 𝐴𝐶= 1533, а вершины 𝐴и 𝐵лежат в узлах на одной горизонтали.
Определите, сколько узлов лежит в треугольнике 𝐴𝐵𝐶(вклю-чая узлы, лежащие на сторонах).
Примечание: Узлом называется точка координатной плоско-сти, у которой обе координаты целые.
16.74.
На плоскости расположено 100 прямоугольников, стороны которых параллельны координатным осям. Каждый пересекается хотя бы с 90 другими. Докажите, что найдется прямоугольник, пересекающийся со всеми.
10 класс 16.75.
В некотором треугольнике сумма тангенсов углов оказалась равна 2016. Оцените (хотя бы с точностью до 1 гра-дуса) величину наибольшего из его углов.
16.76.
Смотри задачу 16.58.
16.77.
Найдите все натуральные числа 𝑛, для которых 2𝑛+ + 𝑛2016 — простое число.
16.78.
В выпуклом четырехугольнике 𝐴𝐵𝐶𝐷внутри тре-угольника 𝐴𝐷𝐶выбрана точка 𝐸, причем ∠𝐵𝐴𝐸= ∠𝐵𝐸𝐴= = 80∘, ∠𝐶𝐴𝐷= ∠𝐶𝐷𝐴= 80∘, ∠𝐸𝐴𝐷= ∠𝐸𝐷𝐴= 50∘. Докажи-те, что △𝐵𝐸𝐶равносторонний.
16.79.
В игре «сет» участвуют всевозможные четырехзнач-ные числа, состоящие из цифр 1, 2, 3 (каждое число по одному 44 УСЛОВИЯ ЗАДАЧ разу). Говорят, что тройка чисел образует сет, если в каждом разряде либо все три числа содержат одну и ту же цифру, либо все три числа содержат разные цифры.
Сложностью сета будем называть количество таких разрядов, где все три цифры различны.
Например, числа 1232, 2213, 3221 образуют сет сложности 3 (в первом разряде встречаются все три цифры, во втором — только двойка, в третьем — все три цифры, в четвертом — все три циф-ры); числа 1231, 1232, 1233 — сет сложности 1 (в первых трех разрядах цифры совпадают, и только в четвертом все цифры различны). А числа 1123, 2231, 3311 вообще не образуют сета (в последнем разряде встречаются две единицы и тройка).
Сетов какой сложности в игре больше всего и почему?
11 класс 16.80.
Каждая клетка доски 1000×1000 покрашена в синий или белый цвет. Назовем клетку равновесной, если среди ее со-седей поровну синих и белых. Можно ли раскрасить доску так, чтобы на ней было более 600 000 синих равновесных клеток?
(Клетки считаются соседними, если имеют общую сторону.) 16.81.
Смотри задачу 16.77.
16.82.
В трехмерном пространстве задана стандартная си-стема координат. Найдите площадь множества точек, удовле-творяющих следующим условиям: ⎧ ⎪ ⎨ ⎪ ⎩ 𝑥2 + 𝑦2 = 5 |𝑥−𝑦| < 1 |𝑦−𝑧| < 1 16.83.
Смотри задачу 16.78.
16.84.
Смотри задачу 16.79. |
7920 | https://www.pearson.com/channels/organic-chemistry/learn/johnny/34-nucleic-acids/base-pairing | Base Pairing Explained: Definition, Examples, Practice & Video Lessons
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A Review of General Chemistry 5h 5m Worksheet
Summary 23m
Intro to Organic Chemistry 5m
Atomic Structure 16m
Wave Function 9m
Molecular Orbitals 17m
Sigma and Pi Bonds 9m
Octet Rule 12m
Bonding Preferences 12m
Formal Charges 6m
Skeletal Structure 14m
Lewis Structure 20m
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Degrees of Unsaturation 15m
Constitutional Isomers 14m
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Molecular Geometry 16m
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Molecular Representations 1h 14m Worksheet
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How To Determine Solubility 11m
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Acids and Bases 2h 46m Worksheet
Organic Chemistry Reactions 7m
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Equilibrium Constant 9m
pKa 25m
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Ranking Acidity 54m
Alkanes and Cycloalkanes 4h 17m Worksheet
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Epoxidation 8m
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Radical Stability 7m
Free Radical Halogenation 19m
Radical Selectivity 23m
Calculating Radical Yields 19m
Anti Markovnikov Addition of Br 11m
Free Radical Polymerization 8m
Allylic Bromination 12m
Radical Synthesis 8m
Alcohols, Ethers, Epoxides and Thiols 2h 42m Worksheet
Alcohol Nomenclature 4m
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Alcohol Synthesis 7m
Leaving Group Conversions - Using HX 11m
Leaving Group Conversions - SOCl2 and PBr3 13m
Leaving Group Conversions - Sulfonyl Chlorides 7m
Leaving Group Conversions Summary 4m
Williamson Ether Synthesis 3m
Making Ethers - Alkoxymercuration 4m
Making Ethers - Alcohol Condensation 4m
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Making Ethers - Cumulative Practice 10m
Ether Cleavage 8m
Alcohol Protecting Groups 3m
t-Butyl Ether Protecting Groups 5m
Silyl Ether Protecting Groups 10m
Sharpless Epoxidation 9m
Thiol Reactions 6m
Sulfide Oxidation 4m
Alcohols and Carbonyl Compounds 2h 17m Worksheet
Oxidizing and Reducing Agents 9m
Oxidizing Agent 26m
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Nucleophilic Addition 8m
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Grignard Reaction 13m
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Synthetic Techniques 1h 26m Worksheet
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Retrosynthesis 16m
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H NMR Table 24m
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NMR Integration 18m
NMR Practice 14m
Carbon NMR 4m
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Conjugated Systems 6h 13m Worksheet
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Reactions at the Allylic Position 39m
Conjugated Hydrohalogenation (1,2 vs 1,4 addition) 26m
Diels-Alder Reaction 9m
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Molecular Orbital Theory 9m
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Pericyclic Reaction 10m
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Ultraviolet Spectroscopy 51m Worksheet
The UV-Vis Spectroscopy 17m
The Beer-Lambert Law 9m
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Woodward-Fieser Rules 16m
Aromaticity 2h 34m Worksheet
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Huckel's Rule 10m
Pi Electrons 7m
Aromatic Hydrocarbons 18m
Annulene 17m
Aromatic Heterocycles 20m
Frost Circle 17m
Naming Benzene Rings 13m
Acidity of Aromatic Hydrocarbons 10m
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Ionization of Aromatics 18m
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Electrophilic Aromatic Substitution 9m
Benzene Reactions 11m
EAS:Halogenation Mechanism 6m
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EAS:Any Carbocation Mechanism 7m
Electron Withdrawing Groups 22m
EAS:Ortho vs. Para Positions 4m
Acylation of Aniline 9m
Limitations of Friedel-Crafts Alkyation 19m
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Blocking Groups - Sulfonic Acid 12m
EAS:Synergistic and Competitive Groups 13m
Side-Chain Halogenation 6m
Side-Chain Oxidation 4m
Reactions at Benzylic Positions 31m
Birch Reduction 10m
EAS:Sequence Groups 4m
EAS:Retrosynthesis 29m
Diazo Replacement Reactions 6m
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Diazo Retrosynthesis 13m
Nucleophilic Aromatic Substitution 28m
Benzyne 16m
Phenols 55m Worksheet
Phenol Acidity 15m
Oxidation of Phenols to Quinones 14m
Cleavage of Phenyl Ethers 10m
Kolbe-Schmidt Reaction 15m
Aldehydes and Ketones: Nucleophilic Addition 4h 56m Worksheet
Naming Aldehydes 8m
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Oxidizing and Reducing Agents 9m
Oxidation of Alcohols 28m
Ozonolysis 7m
DIBAL 5m
Alkyne Hydration 9m
Nucleophilic Addition 8m
Cyanohydrin 11m
Organometallics on Ketones 19m
Overview of Nucleophilic Addition of Solvents 13m
Hydrates 6m
Hemiacetal 9m
Acetal 12m
Acetal Protecting Group 16m
Thioacetal 6m
Imine vs Enamine 15m
Addition of Amine Derivatives 5m
Wolff Kishner Reduction 7m
Baeyer-Villiger Oxidation 39m
Acid Chloride to Ketone 7m
Nitrile to Ketone 9m
Wittig Reaction 18m
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Carboxylic Acid Derivatives: NAS 2h 51m Worksheet
Carboxylic Acid Derivatives 7m
Naming Carboxylic Acids 9m
Diacid Nomenclature 6m
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Acid Chloride Nomenclature 5m
Naming Anhydrides 7m
Naming Amides 5m
Nucleophilic Acyl Substitution 18m
Carboxylic Acid to Acid Chloride 6m
Fischer Esterification 5m
Acid-Catalyzed Ester Hydrolysis 4m
Saponification 3m
Transesterification 5m
Lactones, Lactams and Cyclization Reactions 10m
Carboxylation 5m
Decarboxylation Mechanism 14m
Review of Nitriles 46m
The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides 1h 10m Worksheet
Intro to Thioesters 10m
Hydrolysis of Thioesters 13m
Hydrolysis of Phosphate Esters 12m
Intro to Phosphate Anhydrides 13m
Chemical Reactions of Phosphate Anhydrides 20m
Enolate Chemistry: Reactions at the Alpha-Carbon 1h 53m Worksheet
Tautomerization 9m
Tautomers of Dicarbonyl Compounds 6m
Enolate 4m
Acid-Catalyzed Alpha-Halogentation 4m
Base-Catalyzed Alpha-Halogentation 3m
Haloform Reaction 8m
Hell-Volhard-Zelinski Reaction 3m
Overview of Alpha-Alkylations and Acylations 5m
Enolate Alkylation and Acylation 12m
Enamine Alkylation and Acylation 16m
Beta-Dicarbonyl Synthesis Pathway 7m
Acetoacetic Ester Synthesis 13m
Malonic Ester Synthesis 15m
Condensation Chemistry 2h 9m Worksheet
Condensation Reactions 5m
Aldol Condensation 18m
Directed Condensations 8m
Crossed Aldol Condensation 16m
Claisen-Schmidt Condensation 4m
Claisen Condensation 17m
Intramolecular Aldol Condensation 15m
Conjugate Addition 9m
Michael Addition 16m
Robinson Annulation 17m
Amines 1h 43m Worksheet
Amine Alkylation 11m
Gabriel Synthesis 10m
Amines by Reduction 12m
Nitrogenous Nucleophiles 8m
Reductive Amination 10m
Curtius Rearrangement 15m
Hofmann Rearrangement 10m
Hofmann Elimination 11m
Cope Elimination 12m
Heterocycles 2h 0m Worksheet
Nomenclature of Heterocycles 15m
Acid-Base Properties of Nitrogen Heterocycles 10m
Reactions of Pyrrole, Furan, and Thiophene 13m
Directing Effects in Substituted Pyrroles, Furans, and Thiophenes 16m
Addition Reactions of Furan 8m
EAS Reactions of Pyridine 17m
SNAr Reactions of Pyridine 18m
Side-Chain Reactions of Substituted Pyridines 20m
Carbohydrates 5h 53m Worksheet
Monosaccharide 20m
Monosaccharides - D and L Isomerism 9m
Monosaccharides - Drawing Fischer Projections 18m
Monosaccharides - Common Structures 6m
Monosaccharides - Forming Cyclic Hemiacetals 12m
Monosaccharides - Cyclization 18m
Monosaccharides - Haworth Projections 13m
Mutarotation 11m
Epimerization 9m
Monosaccharides - Aldose-Ketose Rearrangement 8m
Monosaccharides - Alkylation 10m
Monosaccharides - Acylation 7m
Glycoside 6m
Monosaccharides - N-Glycosides 18m
Monosaccharides - Reduction (Alditols) 12m
Monosaccharides - Weak Oxidation (Aldonic Acid) 7m
Reducing Sugars 23m
Monosaccharides - Strong Oxidation (Aldaric Acid) 11m
Monosaccharides - Oxidative Cleavage 27m
Monosaccharides - Osazones 10m
Monosaccharides - Kiliani-Fischer 23m
Monosaccharides - Wohl Degradation 12m
Monosaccharides - Ruff Degradation 12m
Disaccharide 30m
Polysaccharide 11m
Amino Acids 4h 20m Worksheet
Proteins and Amino Acids 19m
L and D Amino Acids 14m
Polar Amino Acids 14m
Amino Acid Chart 1h 18m
Acid-Base Properties of Amino Acids 33m
Isoelectric Point 14m
Amino Acid Synthesis: HVZ Method 12m
Synthesis of Amino Acids: Acetamidomalonic Ester Synthesis 16m
Synthesis of Amino Acids: N-Phthalimidomalonic Ester Synthesis 13m
Synthesis of Amino Acids: Strecker Synthesis 13m
Reactions of Amino Acids: Esterification 7m
Reactions of Amino Acids: Acylation 3m
Reactions of Amino Acids: Hydrogenolysis 6m
Reactions of Amino Acids: Ninhydrin Test 11m
Peptides and Proteins 2h 42m Worksheet
Peptides 12m
Primary Protein Structure 4m
Secondary Protein Structure 17m
Tertiary Protein Structure 11m
Disulfide Bonds 17m
Quaternary Protein Structure 10m
Summary of Protein Structure 7m
Intro to Peptide Sequencing 2m
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Peptide Sequencing: Partial Hydrolysis with Cyanogen Bromide 7m
Peptide Sequencing: Edman Degradation 28m
Merrifield Solid-Phase Peptide Synthesis 18m
Catalysis in Organic Reactions 1h 30m Worksheet
Introduction to Catalysis 3m
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Nucleophilic Catalysis 11m
Metal Ion Catalysis 7m
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Rates of Intramolecular Reactions 14m
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Intramolecular Nucleophilic Catalysis 9m
Lipids 2h 50m Worksheet
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Fatty Acids 21m
Physical Properties of Fatty Acids 19m
Waxes 6m
Triacylglycerols 15m
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Sphingomyelins 13m
Steroids 28m
The Organic Chemistry of Metabolic Pathways 2h 52m Worksheet
Intro to Metabolism 6m
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Coenzymes in Metabolism 16m
Energy Production in Biochemical Pathways 5m
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Catabolism of Carbohydrates: Glycolysis 27m
Glycolysis Summary 15m
Pyruvate Oxidation (Simplified) 4m
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Intro to Citric Acid Cycle 7m
Structures of the Citric Acid Cycle 19m
The Citric Acid Cycle 35m
Nucleic Acids 1h 32m Worksheet
Intro to Nucleic Acids 4m
Nitrogenous Bases 22m
Naming Nucleosides and Nucleotides 14m
Hydrolysis of Nucleosides 11m
Primary Structure of Nucleic Acids 11m
Base Pairing 10m
DNA Double Helix 17m
Transition Metals 6h 14m Worksheet
Electron Configuration of Elements 45m
Coordination Complexes 20m
Ligands 24m
Electron Counting 10m
The 18 and 16 Electron Rule 13m
Cross-Coupling General Reactions 40m
Heck Reaction 40m
Stille Reaction 13m
Suzuki Reaction 25m
Sonogashira Coupling Reaction 17m
Fukuyama Coupling Reaction 15m
Kumada Coupling Reaction 13m
Negishi Coupling Reaction 16m
Buchwald-Hartwig Amination Reaction 19m
Eglinton Reaction 17m
Catalytic Allylic Alkylation 18m
Alkene Metathesis 23m
Synthetic Polymers 1h 49m Worksheet
Introduction to Polymers 6m
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Radical Polymerization 15m
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Polymer Stereochemistry 3m
Ziegler-Natta Polymerization 4m
Copolymers 6m
Step-Growth Polymers 11m
Step-Growth Polymers: Urethane 6m
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Polymers Structure and Properties 8m
Nucleic Acids
Base Pairing
Nucleic Acids
Base Pairing: Videos & Practice Problems
Video LessonsPracticeWorksheet
Topic summary
DNA and RNA base pairing relies on hydrogen bonding, where adenine (A) pairs with thymine (T) or uracil (U) through 2 hydrogen bonds, while guanine (G) pairs with cytosine (C) through 3 hydrogen bonds, enhancing structural stability. Chargaff's rule states that in any species, the percentage of A equals T, and G equals C, ensuring proper base pairing. This balance is crucial for the integrity of nucleic acids, with A and T, G and C contributing to the overall 100% of nitrogenous bases in DNA and RNA.
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Base Pairing Concept 1
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Base Pairing Concept 1 Video Summary
Understanding the structure of DNA and RNA involves recognizing the crucial role of hydrogen bonding in base pairing. Hydrogen bonds, while individually weak, collectively provide significant stability to the DNA structure. This stabilizing effect is essential for maintaining the integrity of the DNA molecule, as the numerous hydrogen bonds present contribute to its overall strength.
Complementary base pairing is a key concept in this context, referring to the specific bonding preferences between nitrogenous bases. In DNA, adenine (A) pairs with thymine (T) through two hydrogen bonds, while guanine (G) pairs with cytosine (C) through three hydrogen bonds. This difference in the number of hydrogen bonds is important; the stronger G-C pairing, with its three hydrogen bonds, contributes to regions of increased stability within the DNA structure.
In RNA, the pairing changes slightly due to the presence of uracil (U) instead of thymine. Here, adenine pairs with uracil, while cytosine still pairs with guanine. Thus, the pairing rules can be summarized as follows: in DNA, A pairs with T, and G pairs with C; in RNA, A pairs with U, and C pairs with G. This distinction is vital for understanding the differences between DNA and RNA structures and their respective functions in biological systems.
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Base Pairing Example 1
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Base Pairing Example 1 Video Summary
In the context of RNA structure, understanding the base pairing and hydrogen bonding is crucial. In RNA, the nitrogenous bases include adenine (A), uracil (U), cytosine (C), and guanine (G). The base pairing rules dictate that adenine pairs with uracil, while cytosine pairs with guanine.
When examining the base pairs, adenine (A) forms two hydrogen bonds with uracil (U). Therefore, in a given RNA sequence, wherever there is an A, it will bond with a U. Conversely, cytosine (C) pairs with guanine (G) through three hydrogen bonds, which is a stronger interaction compared to the A-U pairing.
To summarize the hydrogen bonding: A and U form two hydrogen bonds, while C and G form three hydrogen bonds. This distinction is essential for the stability and structure of RNA molecules, influencing their function in biological processes.
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Four species shown below give the percentages of A–T pairings vs G–C pairings. Based on only the information given, which species would have the most significant strength in their base interactions?
A
Drosophila melanogaster (fruit fly) (55% : 45%)
B
Zea mays (corn) (51% : 49%)
C
Neurospora crassa (fungus) (46% : 54%)
D
Escherichia coli (bacteria) (49% : 51%)
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Base Pairing Concept 2
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Base Pairing Concept 2 Video Summary
Chargaff's rule, established by Erwin Chargaff in the early 1950s, is a fundamental principle in molecular biology that describes the composition of DNA. This rule states that in any given species, the percentage of adenine (A) is approximately equal to the percentage of thymine (T), and the percentage of guanine (G) is approximately equal to the percentage of cytosine (C). This equality is crucial because A and T, as well as G and C, form hydrogen bonds with each other, ensuring the stability of the double-stranded DNA structure.
To visualize this, consider a scale where the amount of A nitrogenous bases must balance with the amount of T bases, reflecting their equal percentages. Similarly, G and C must also balance each other out. This means that while the total percentages of A and T combined with G and C will equal 100%, the individual amounts of A and T do not need to be equal to the amounts of G and C. Instead, the relationships are:
In practical terms, when analyzing the DNA of a species, one can apply Chargaff's rule to determine the unknown percentages of these nitrogenous bases. For example, if the percentage of A is known to be 30%, then the percentage of T must also be 30%. If the total percentage of A and T is 60%, then G and C together must account for the remaining 40%, with G and C being equal to each other. This foundational concept is essential for understanding the structure and function of DNA in various biological contexts.
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Base Pairing Example 2
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Base Pairing Example 2 Video Summary
In human DNA, the composition of nitrogenous bases follows specific pairing rules, known as Chargaff's rules. These rules state that adenine (A) pairs with thymine (T), and guanine (G) pairs with cytosine (C). Given that adenine constitutes approximately 20% of the DNA, thymine must also account for 20%, as their percentages are equal.To determine the percentage of guanine in the DNA, we first calculate the total percentage contributed by adenine and thymine:This leaves us with 60% of the total nitrogenous bases to be accounted for by guanine and cytosine:Since guanine and cytosine must also be present in equal amounts, we divide the remaining percentage by 2:Thus, the percentage of guanine in a human DNA sample is 30%. This understanding of base pairing and the application of Chargaff's rules are crucial for analyzing DNA composition and structure.
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Cytosine (C) makes up 42% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be thymine (T)?
A
8%
B
16%
C
21%
D
60%
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Base Pairing quiz #1 34. Nucleic Acids 40 Term s
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What is the significance of hydrogen bonding in DNA and RNA base pairing?
Hydrogen bonding is crucial in DNA and RNA base pairing because it provides stability to the nucleic acid structures. In DNA, adenine (A) pairs with thymine (T) through 2 hydrogen bonds, and guanine (G) pairs with cytosine (C) through 3 hydrogen bonds. In RNA, adenine pairs with uracil (U) instead of thymine. Although individual hydrogen bonds are weak, collectively, they create a strong stabilizing effect, holding the DNA or RNA strands together. This stability is essential for the integrity and proper functioning of genetic material.
How does Chargaff's rule apply to DNA base pairing?
Chargaff's rule states that in any double-stranded DNA, the percentage of adenine (A) is equal to thymine (T), and the percentage of guanine (G) is equal to cytosine (C). This rule is based on the fact that A pairs with T and G pairs with C through hydrogen bonding. Therefore, if a DNA sample has 30% adenine, it must also have 30% thymine. Similarly, if it has 20% guanine, it must have 20% cytosine. This balance ensures proper base pairing and the overall stability of the DNA structure.
Why does adenine pair with thymine in DNA and uracil in RNA?
In DNA, adenine (A) pairs with thymine (T) through 2 hydrogen bonds, while in RNA, adenine pairs with uracil (U) instead of thymine. This difference is due to the presence of uracil in RNA, which replaces thymine. The hydrogen bonding preferences remain the same, with adenine forming 2 hydrogen bonds with either thymine or uracil. This substitution is a key distinction between DNA and RNA, reflecting their different roles and structures in the cell.
How many hydrogen bonds are formed between guanine and cytosine?
Guanine (G) and cytosine (C) form 3 hydrogen bonds when they pair together. This triple hydrogen bonding provides greater stability compared to the 2 hydrogen bonds formed between adenine (A) and thymine (T) or uracil (U). The increased number of hydrogen bonds between G and C contributes to regions of DNA with higher stability and strength.
What is complementary base pairing and why is it important?
Complementary base pairing refers to the specific hydrogen bonding between nitrogenous bases in DNA and RNA. In DNA, adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C). In RNA, adenine pairs with uracil (U) instead of thymine. This specificity ensures accurate replication and transcription of genetic information. Complementary base pairing is essential for maintaining the integrity of genetic material and for the proper functioning of biological processes such as DNA replication and RNA transcription.
Previous Topic: Primary Structure of Nucleic AcidsNext Topic: DNA Double Helix
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7921 | https://artofproblemsolving.com/wiki/index.php/Harmonic_sequence?srsltid=AfmBOorqOV8qWx41cJbwqY69jav_GPP0MqIIPZAIe1aTs3Yd87mqKlLz | Art of Problem Solving
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Harmonic sequence
In algebra, a harmonic sequence, sometimes called a harmonic progression, is a sequence of numbers such that the difference between the reciprocals of any two consecutive terms is constant. In other words, a harmonic sequence is formed by taking the reciprocals of every term in an arithmetic sequence.
For example, and are harmonic sequences; however, and are not.
More formally, a harmonic progression biconditionally satisfies A similar definition holds for infinite harmonic sequences. It appears most frequently in its three-term form: namely, that constants , , and are in harmonic progression if and only if .
Contents
[hide]
1 Properties
2 Sum
3 Examples
3.1 Example 1
3.2 Example 2
3.3 Example 3
4 More Problems
4.1 Introductory
5 See Also
Properties
Because the reciprocals of the terms in a harmonic sequence are in arithmetic progression, one can apply properties of arithmetic sequences to derive a general form for harmonic sequences. Namely, for some constants and , the terms of any finite harmonic sequence can be written as
A common lemma is that a sequence is in harmonic progression if and only if is the harmonic mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of harmonic sequences.
Sum
A harmonic series is the sum of all the terms in a harmonic series. All infinite harmonic series diverges, which follows by the limit comparison test with the series . This series is referred to as theharmonic series. As for finite harmonic series, there is no known general expression for their sum; one must find a strategy to evaluate one on a case-by-case basis.
Examples
Here are some example problems that utilize harmonic sequences and series.
Example 1
Find all real numbers such that is a harmonic sequence.
Solution: Using the harmonic mean properties of harmonic sequences, Note that would create a term of —something that breaks the definition of harmonic sequences—which eliminates them as possible solutions. We can thus multiply both sides by to get . Expanding these factors yields , which simplifies to . Thus, is the only solution to the equation, as desired.
Example 2
Let , , and be positive real numbers. Show that if , , and are in harmonic progression, then , , and are as well.
Solution: Using the harmonic mean property of harmonic sequences, we are given that , and we wish to show that . We work backwards from the latter equation.
One approach might be to add to both sides of the equation, which when combined with the fractions returns Because , , and are all positive, . Thus, we can divide both sides of the equation by to get , which was given as true.
From here, it is easy to write the proof forwards. Doing so proves that , which implies that , , is a harmonic sequence, as required.
Example 3
2019 AMC 10A Problem 15: A sequence of numbers is defined recursively by , , and for all Then can be written as , where and are relatively prime positive integers. What is ?
Solution: We simplify the series' recursive formula. Taking the reciprocals of both sides, we get the equality Thus, . This is the harmonic mean, which implies that is a harmonic progression. Thus, the entire sequence is in harmonic progression.
Using the tools of harmonic sequences, we will now find a closed expression for the sequence. Let and . Simplifying the first equation yields and substituting this into the second equation yields . Thus, and so . The answer is then .
More Problems
Here are some more problems that utilize harmonic sequences and series. Note that harmonic sequences are rather uncommon compared to their arithmetic and geometric counterparts.
Introductory
1959 ASHME Problem 33
See Also
Arithmetic sequence
Geometric sequence
Harmonic series
Sequence
Series
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7922 | https://pradeepresearch.org/pdf-files/teaching-and-presentations/teaching/undergraduate%20courses/cy102-equilibrium-and-dynamics-of-chemical-systems/lectures-2004/L5.pdf | Lecture 5 Entropy and Disorder Third law S(T) = S(0) + CP ln T Background information: ∫2 1 dS = n Cp ln T2/T1 – nR ln P2/P1 The change in entropy can be calculated using graphical method. In all these cases, the only parameter which is impossible to measure is S(0). This parameter has to be evaluated for determining the value of S(T) [note: absolute entropy, not ∆S]. Here comes third law of thermodynamics.
“Value of S(0) is zero for every pure, perfectly crystalline substance” M. Plank, 1913 Then, S (T) = CP ln T S(T) is called the third law entropy, ST or simply entropy at temperature T and pressure P. If pressure is 1 atm, it is called standard entropy, So T To measure S(T), we need to measure CP as a function of T. Measuring CP down to absolute zero is not done as CP obeys Debye law, CP = aT3, a is a constant for a material. So CP is measured up to a temperature, T’and S (T) is calculated from, ∫T’ T CP/T dT = ∫T’ T CP d(lnT) = 2.303 ∫T’ T CPd(lnT) Is evaluated graphically. These numbers are tabulated.
1 2 3 log T Cp ∫Cp d(logT) Absolute entropy: evaluation Nernst heat theorem The Nernst heat theorem states that, “the entropy change accompanying a physical or chemical change approaches zero as temperature tends to zero. ∆S Æ 0 as T Æ 0”. This form of third law is less restrictive than the Plank’s statement.
Another form: “If entropy of every element at absolute zero in their standard state is taken to be zero, all substances have positive entropy which may become zero at T = 0 and it will indeed become zero for all perfectly crystalline solids.” For a non-crystalline state, the entropy could become nonzero as the first part of the statement reads.
Entropy change For a reaction, standard entropy is defined just as we define standard enthalpy.
∆reactionS = ΣSproducts - ΣSreactants (∂∆S0/ ∂T)P = (∂Sproducts/ ∂T)P - (∂Sreactants/ ∂T)P =CP0Products/T - CP0Products/T = ∆CP0/T This equation can be written in differential form can be integrated.
∫ToT d(∆S0) = ∫ToT ∆CP0/T dT ∆S0T = ∆S0T0 + ∫ToT ∆CP0/T dT How do you reach low temperature?
Adiabatic demagnetization Adiabatic nuclear demagnetization Laser cooling The meaning of entropy: Entropy of mixing Consider that a number of ideal gases are separated which are present in a vessel. Let ni be the number of moles of each gas and Vi is the volume it occupies. The total entropy, S1 = ∑ni (Cv ln T + R ln Vi + Si) 1 Note the term Si. It is because, dS = Cv(dT/T) + R(dV/V) S = Cv ln T + R ln V + So Let the partitions between the gases are removed and they are allowed to mix. Let the volume the gases occupy is V. The entropy, S2 = ∑Ni (Cv ln T + R ln V + Si) 2 Assume that the pressure has not changed and there is no change in temperature. Ratio of volume, Vi/V = ni/n = Ni Where n is the total number of moles.
Substituting Vi = VNi in Eqn.1 S1 = ∑ni (Cv ln T + R ln V + R ln Ni + Si) The increase in entropy, the entropy of mixing, S2 – S1 = –∑ni R ln Ni Entropy of mixing of 1 mole of the ideal gas, ∆Sm = –R ∑ni/n ln Ni = –R ∑Ni ln Ni The fraction Ni is less than unity in all cases, the logarithm is negative and thus ∆Sm is always positive. Thus mixing of gases (eg. by diffusion), always results in increase in entropy. Mixing is spontaneous!
This is in general true of any material, liquid or solid.
Entropy and disorder Spontaneous processes Æ net increase in entropy Æ increase in randomness of distribution. The diffusion of initially separated gases result in an increase in entropy. The process has increased the random distribution of molecules. Spontaneous conduction of heat results in the random distribution of kinetic energy of the atoms. Thus spontaneous processes increase randomness, at the same time increases entropy. Therefore, it is appropriate to suggest a relationship between entropy and randomness.
This definition of entropy, that it is a measure of randomness, is one of great value. Apart form the quantitative relationship, the concept is of great value in understanding chemical processes quantitatively. A measure of entropy changes gives an indication of structural changes. The process of fusion involves increase in disorder and therefore, the entropy increase. Greater the disorder, greater the entropy increase. Therefore, the heat of fusion of ice and benzene are 5.26 and 8.27 cal/deg/mol. The difference is because in water there is considerable amount of order due to hydrogen bonding while there is no such interaction in benzene.
Entropy and probability A correlation is possible between entropy and probability. Suppose there are two bulbs, connected through a valve, and assume that one bulb is evacuated and the other is full of gas. As the stopcock is opened, according to second law, the gas will distribute uniformly. We can explain this using the theory of probability. Assume that there is just one molecule in the system. Here, the probability that it can be found in one bulb is ½ ie. one chance in two. If we have 2 molecules, the probability that two will be found in one bulb is ¼ ie. (½)2. If there are N molecules, the probability that all the molecules will be found in one bulb is (½)N. For a bulb having one litre volume and the pressure is one atmosphere, the number of molecules is 1022. Even for 10-6 atm, the number of molecules is 1016. Thus the probability for the molecules to reside in one bulb is very small. This is also the probability for the molecules to return to one bulb after being distributed uniformly.
Thus it can be considered that the spontaneous process in which the gas is distributed uniformly in two bulbs is associated with the large probability. All spontaneous processes represent changes from a less portable state to a more portable state. Since processes are accompanied by increase in entropy it is possible that there is a relation between entropy and probability.
If S is the entropy and W is the probability, S = f(W) If there are two systems of probability, WA and WB, with entropies SA and SB, the combined system will have entropy SA + SB and probability WA x WB.
SAB = SA + SB = f(WA x WB) f(WA) + f(WB) = f(WA x WB) Thus the function has to be logarithmic.
S = k ln W + constant Where k is the Botzmann constant, R/N. It has been shown later that the constant is zero and S = k ln W.
There is no complete proof for this expression and therefore has to be regarded as a postulate.
Now the task to calculate entropy is to determine probability.
Use of statistical mechanics.
Tombstone of Ludwig Boltzmann Boltzmann paradox and a discussion on entropy and probability Let us assume that we have a small box in a room containing some gas. Assume that the walls of the box are made open and the gas was free to mix. Sometime later the gas spread uniformly in the box. For each of the molecule originally in the box, the probability of one motion is just the same as the motion in the opposite direction. So why is that the reverse of the process, namely the opposite of random diffusion, is not observed? The fact that some motions of a system are never observed is called Boltzmann paradox.
The puzzle is solved by looking at the following fact. The probability of motion for one molecule is just the same as the reverse motion. However, the probability of a group of molecules for uniform filling of available space is enormously greater than the probability for occupation in one part of the box. Assume that we have two balls and two cells. There is only one way to fill the cells uniformly, so probability of uniform filling is one. Let us say we have four cells and two balls. After two cells, there is a partition. Consider various kinds of arrangements. We see that there are two kinds of non uniform filling possibilities and four kinds of uniform filling possibilities. So we see that probability of uniform filling is, 4/6. The probability of finding two balls in one side of the box is 2/6 or 1/3. Two balls and two cells Four cells Uniform filling Now let us say we have eight cells. There are 28 arrangements and 16 of them correspond to uniform filling. Thus probability of uniform filling is 16/28 or 4/7. Now one can see that as the number of cells increase, the probability of uniform filling approaches ½. The entropy of the system is related to the number of arrangements of particles of the system, called the complexion, Ωof the system. The Boltzmann definition of entropy, S = k ln Ω For two balls in one box, S1= k ln Ω= k ln (1) = 0 For the six arrangements, S2 = k ln 6 The entropy difference upon expansion from 2 cells to six cells, ∆S = S2-S1 = k ln 6 for 2 balls = ½ k ln 6 for 1 ball How about N cells?
For the first ball, there are N possibilities. For the second, there are N-1 possibilities. Thus there are N (N-1) possibilities. Since we cannot distinguish between the two balls, we will have to take the total number of distinct arrangements as, N(N-1)/2.
So, entropy for this arrangement, S1 = k ln [N(N-1)/2] For N’ cells, S2 = k ln [N’(N’-1)/2] ∆S = S2-S1 = k ln {N’(N’-1)/N(N-1)} Let us say N is very large. Then N and N-1 are the same. ∆S = S2-S1 = k ln (N’/N)2 = 2k ln (N’/N) For ideal gases, the number of cells correspond to volume, ∆S = 2k ln (V’/V) for two molecules, ∆S = k ln (V’/V) for one molecule.
∆S = R ln (V’/V) for one mole.
This is what we got for the expansion of one mole of ideal gas. |
7923 | https://rodensapchemistry.weebly.com/uploads/5/1/3/6/5136176/thermo_frq.pdf | The Advanced Placement Examination in Chemistry Part II - Free Response Questions & An-swers 1970 to 2005 Thermodynamics Teachers may reproduce this publication, in whole or in part, in limited print quantities for non-commercial, face-to-face teaching purposes. This permission does not apply to any third-party copyrights contained with-in this publication. Advanced Placement Examination in Chemistry. Questions copyright© 1970-2005 by the College Entrance Ex-amination Board, Princeton, NJ 08541. Reprinted with permission. All rights reserved. apcen-tral.collegeboard.com. This material may not be mass distributed, electronically or otherwise. This publica-tion and any copies made from it may not be resold. Portions copyright © 1993-2005 by Unlimited Potential, Framingham, MA 01701-2619. Compiled for the Macintosh and PC by: Harvey Gendreau Framingham High School 115 “A” Street Framingham, MA 01701-4195 508-620-4963 419-735-4782 (fax) 508-877-8723 (home office) www.apchemistry.com hgendrea@framingham.k12.ma.us hgendreau@rcn.com Requests for copies of these questions and answers as e-mail attachments for either the Macin-tosh or the PC (MS-Office files) should be sent to: apchemfiles@apchemistry.com. Please include your name, school, school phone, name of principal/headmaster and school website address. Don’t forget to include the file format you want, Mac or PC. Thermodynamics, ΔH, ΔS, ΔG page 2 1970 Consider the first ionization of sulfurous acid: H2SO3(aq) H+(aq) + HSO3 -(aq) Certain related thermodynamic data are provided be-low: H2SO3(aq) H+(aq) HSO3 -(aq) ----------------- ---------- ------------------ Hf kcal/mole -145.5 0 -151.9 S cal/mole K 56 0 26 (a) Calculate the value of G at 25C for the ioniza-tion reaction. (b) Calculate the value of K at 25C for the ionization reaction. (c) Account for the signs of S and H for the ioni-zation reaction in terms of the molecules and ions present. Answer: (a) H H f _ ( products ) H f _ ( reactants ) = [-159.9] - [-145.5] kcal = -14.4 kcal S S (products )S (reactants) = (26 - 56) cal = -30 cal/K G = H - TS = -14400 - (298)(-30) cal = -5.46 kcal (b) K = e-G/RT = e-(-5460/(1.9872)(298)) = 10100 (c) 1971 Given the following data for graphite and diamond at 298K. S(diamond) = 0.58 cal/mole deg S(graphite) = 1.37 cal/mole deg Hf CO2(from graphite) = -94.48 kilocalories/mole Hf CO2(from diamond) = -94.03 kilocalories/mole Consider the change: C(graphite) = C(diamond) at 298K and 1 atmosphere. (a) What are the values of S and H for the con-version of graphite to diamond. (b) Perform a calculation to show whether it is ther-modynamically feasible to produce diamond from graphite at 298K and 1 atmosphere. (c) For the reaction, calculate the equilibrium constant Keq at 298K Answer: (a) S = S(dia.) - S(graph.) = (0.58 - 1.37) cal/K = -0.79 cal/K CO2 C(dia.) + O2 H = + 94.03 kcal/mol C(graph.) + O2 CO2 H = - 94.48 kcal/mol C(graph.) C(dia.) H = -0.45 kcal/mol (b) G = H - TS = -450 - (298)(-0.79) cal = -223.52 cal/mol; a G < 0 indicates feasible conditions (c) Keq = e-G/RT = e-(-223.52/(1.9872)(298)) = -0.686 1972 Br2 + 2 Fe2+(aq) 2 Br-(aq) + 2 Fe3+(aq) For the reaction above, the following data are available: 2 Br-(aq) Br2(l) + 2e- E = -1.07 volts Fe2+(aq) Fe3+(aq) + e- E = -0.77 volts S, cal/mole C Br2(l) 58.6 Fe2+(aq) -27.1 Br-(aq) 19.6 Fe3+(aq) -70.1 (a) Determine S (b) Determine G (c) Determine H Answer: (a) S S (products )S (reactants) = [(19.6)(2)+(-70.1)(2)]-[58.6+(-27.1)(2)] cal = -105.4 cal = -441 J/K (b) Ecell = [+1.07 + (-0.77)] v = 0.30 v G=-nE=-(2)(96500)(0.30v)=-57900 J/mol (c) H = G + TS = 57900 + 298(-441) J = -73.5 kJ/mol 1974 WO3(s) + 3 H2(g) W(s) + 3 H2O(g) Tungsten is obtained commercially by the reduction of WO3 with hydrogen according to the equation above. The following data related to this reaction are available: WO3(s) H2O(g) Hf (kilocalories/mole) -200.84 -57.8 Gf (kilocalories/mole) -182.47 -54.6 (a) What is the value of the equilibrium constant for the system represented above? Thermodynamics, ΔH, ΔS, ΔG page 3 (b) Calculate S at 25C for the reaction indicated by the equation above. (c) Find the temperature at which the reaction mixture is in equilibrium at 1 atmosphere. Answer: (a) G = [3(-54.6) + 0] - [-182.47 + 0] = 18.7 kcal Keq = e-G/RT = e-(18700/(1.9872)(298)) = 1.9310-8 (b) H = [3(-57.8) + 0] - [-200.84 + 0] = 27.44 kcal S H G T 27440 18670 298 29. 2 cal mol K (c) G 0 ; assume K P H 2 O P H 2 3 1 at equilibrium T = H / S = 27440 / 29.2 = 938K 1975 B 2 NO(g) + O2 2 NO2(g) A rate expression for the reaction above is: d [ O 2 ] dt k [ NO ] 2 [ O 2 ] Hf S Gf kcal/mole cal/(mole)(K) kcal/mole NO(g) 21.60 50.34 20.72 O2(g) 0 49.00 0 NO2(g) 8.09 57.47 12.39 (a) For the reaction above, find the rate constant at 25C if the initial rate, as defined by the equation above, is 28 moles per liter-second when the con-centration of nitric oxide is 0.20 mole per liter and the concentration of oxygen is 0.10 mole per liter. (b) Calculate the equilibrium constant for the reaction at 25C. Answer: (a) k rate [ NO ] 2 [ O 2 ] 28 ( 0 . 20) 2 ( 0 . 10) = 7000 mol-2L2sec-1 (b) G = [2(12.39)] - [2(20.72) + 0] = -16.66 kcal Keq = e-G/RT = e-(-16660/(1.9872)(298)) = 1.651012 1975 D 2 Cu + S Cu2S For the reaction above, H, G, and S are all neg-ative. Which of the substances would predominate in an equilibrium mixture of copper, sulfur, and copper(I) sul-fide at 298K? Explain how you drew your conclusion about the predominant substance present at equilibri-um. Why must a mixture of copper and sulfur be heat-ed in order to produce copper(I) sulfide? Answer: Copper(I) sulfide. The forward reaction involves bond formation and is, therefore, exothermic (H<0). The forward reaction produces 1 molecule from 3 atoms and, therefore, decreases in entropy (S<0). But since G is <0 and G = H -TS, this reaction is spontane-ous at low temperatures. This mixture must be heated because both reactants are solids and they react only when the copper atoms and sulfur atoms collide, an infrequent occurrence in the solid state. 1977 B CH3OH(l) + 3/2 O2(g) 2 H2O(l) + CO2(g) The value of S for the reaction is -19.3 cal/mol-degree at 25C. Hf S kcal/mole at 25C cal/mole-degree at 25C ------------------------- ---------------------------------- CH3OH(l) -57.0 30.3 H2O(l) -68.3 16.7 CO2(g) -94.0 51.1 (a) Calculate G for the complete combustion of methanol shown above at 25C. (b) Calculate the value for the equilibrium constant for this reaction at 25C. (c) Calculate the standard absolute entropy, S, per mole of O2(g). Answer: (a) H _ H f ( products ) H f ( reactants ) = [2(-68.3) + (-94.0)] - [-57.0] = -173.6 kcal G = H - TS = -173.6 + (298)(0.0193) kcal = -167.8 kcal (b) Keq = e-G/RT = e-(-167800/(1.9872)(298)) = 1.1510123 (c) S S (products )S (reactants) -19.3 = [2(16.7) + 51.1] - [30.3 + 3/2 X] X = 49.0 cal/mol K 1978 B Standard Entropy Substance cal/deg mole Thermodynamics, ΔH, ΔS, ΔG page 4 N2(g) 45.8 H2(g) 31.2 NH3(g) 46.0 Ammonia can be produced by the following reaction: N2(g) + 3 H2(g) 2 NH3(g) The Gibbs free energy of formation Gf of NH3(g) is -3.94 kilocalories per mole. (a) Calculate the value for H for the reaction above 298K. (b) Can the yield of ammonia be increased by raising the temperature? Explain. (c) What is the equilibrium constant for the reaction above at 298K? (d) If 235 milliliters of H2 gas measured at 25C and 570 millimeters Hg were completely converted to ammonia and the ammonia were dissolved in suffi-cient water to make 0.5000 liter of solution, what would be the molarity of the resulting solution? Answer: (a) S S (products )S (reactants) = [2(46.0)] - [45.8 + 3(31.2)] = -47.4 cal/K H = G + TS = -7.88 + (298)(-0.0474) kcal = -22.0 kcal (b) No, since H > 0, an increase in T shifts equilib-rium to left and decreases equilibrium yield of NH3. (c) Keq = e-G/RT = e-(-7880/(1.9872)(298)) = 6.01105 (d) n P V R T ( 570 )( 235 ) ( 62400 )( 298 ) 0 . 00720 mol H 2 0 . 00720 mol H 2 2 mol N H 3 3 mol H 2 1 0.500 L = 0.00960 M NH3 1979 B Hf S Compound (kilocalories/mole) (calories/mole K) H2O(l) -68.3 16.7 CO2(g) -94.1 51.1 O2(g) 0.0 49.0 C3H8 ? 64.5 When 1.000 gram of propane gas, C3H8, is burned at 25C and 1.00 atmosphere, H2O(l) and CO2(g) are formed with the evolution of 12.03 kilocalories. (a) Write a balanced equation for the combustion re-action. (b) Calculate the molar enthalpy of combustion, Hcomb, of propane. (c) Calculate the standard molar enthalpy of for-mation, Hf, of propane gas. (d) Calculate the entropy change, Scomb, for the reac-tion and account for the sign Scomb. Answer: (a) C3H8 + 5 O2(g) 3 CO2(g) + 4 H2O(l) (b) H _ comb . 12. 03kcal 1.000 g 44.10 g 1 mol 530 . 5 kcal mol (c) H comb. 3 H f CO 2 4 H f H 2 O H f C 3 H 8 -530.8 kcal = [3(-94.1) + 4(-68.3) - X] kcal Hcomb. = -25.0 kcal/mol (d) S comb. [ 3 S _ CO 2 4 S _ H 2 O ] [ S _ C 3 H 8 5 S _ O 2 ] = [3(51.1) + 4(16.7)] - [64.5 + 5(49.0)] = -89.4 cal/mol.K Entropy decreases due to loss of highly disordered gaseous species upon combustion. 1980 D (a) State the physical significance of entropy. (b) From each of the following pairs of substances, choose the one expected to have the greater abso-lute entropy. Explain your choice in each case. As-sume 1 mole of each substance. (1) Pb(s) or C(graphite) at the same temperature and pressure. (2) He(g) at 1 atmosphere or He(g) at 0.05 atmos-phere, both at the same temperature. (3) H2O(l) or CH3CH2OH(l) at the same tempera-ture and pressure. (4) Mg(s) at 0C or Mg(s) at 150C both at the same pressure. Answer: (a) Entropy is a measure of randomness, disorder, etc. in a system. (b) (1) Pb has greater molar entropy, Pb, with metal-lic bonding, forms soft crystals with high am-plitudes of vibration; graphite has stronger Thermodynamics, ΔH, ΔS, ΔG page 5 (covalent) bonds, is more rigid, and thus is more ordered. (2) He(g) at 0.05 atmosphere has greater molar entropy. At lower pressure (greater volume) He atoms have more space in which to move are so are more random. (3) CH3CH2OH has greater molar entropy. Etha-nol molecules have more atoms and thus more vibrations; water exhibits stronger hydrogen bonding. (4) Mg(s) at 150C has greater molar entropy. At the higher temperature the atoms have more kinetic energy and vibrate faster and, thus, show greater randomness. 1981 D PCl5(g) PCl3(g) + Cl2(g) For the reaction above, H = +22.1 kilocalories per mole at 25C (a) Does the tendency of reactions to proceed to a state of minimum energy favor the formation of the products of this reaction? Explain (b) Does the tendency of reactions to proceed to a state of maximum entropy favor the formation of the products of this reaction? Explain. (c) State whether an increase in temperature drives this reaction to the right, to the left, or has no ef-fect. Explain. (d) State whether a decrease in the volume of the sys-tem at constant temperature drives this reaction to the right, to the left or has no effect. Explain? Answer: (a) No, since reaction is endothermic, the products must be at higher energy than the reactants. OR ln KP = -H/RT + constant; if H>0, ln KP is less than if H<0. OR G = H - TS. Low free energy (G0) is not favored by H>0. (b) Yes, S>0 since 1 mol gas yields 2 mol gas, which means increased disorder. OR At equilibrium H = TS and a positive H means a positive S. (c) Application of heat favors more products. Predict-able from LeChatelier’s principle. OR TS term here increases as T is increased resulting in a more negative G. (d) Reduction of volume favors more reactants. Pre-dictable from LeChatelier’s principle. Increased pressure is reduced by 2 gas molecules combining to give 1 molecule. 1983 B CO(g) + 2 H2(g) CH3OH(l) H = -128.1 kJ Hf Gf S (kJ mol-1) (kJ mol-1) (J mol-1 K-1) CO(g) -110.5 -137.3 +197.9 CH3OH(l) -238.6 -166.2 +126.8 The data in the table above were determined at 25C. (a) Calculate G for the reaction above at 25C. (b) Calculate Keq for the reaction above at 25C. (c) Calculate S for the reaction above at 25C. (d) In the table above, there are no data for H2. What are the values of Hf, Gf, and of the absolute entropy, S, for H2 at 25C? Answer: (a) G _ G f ( products ) G f ( reactants ) = -166.2 - [-137.3 + 2(0)] = -28.9 kJ/mol (b) Keq = e-G/RT = e-(-28900/(8.3143)(298)) = 1.16105 (c) S _ H _ G _ T 128100 ( 28900 ) J 298 K = -333 J/K (d) Both the standard enthalpy of formation and the standard free energy of formation of elements = 0. S S (products )S (reactants) -333 J/K = 126.8 J/K - 197.9 J/K - 2 SH2 SH2 = 131 J/mol.K 1984 B Standard Heat of Absolute Formation, Hf, Entropy, S, Substance in kJ mol-1 in J mol-1 K-1 ------------------ ----------------------------- ------------------------ C(s) 0.00 5.69 CO2(g) -393.5 213.6 H2(g) 0.00 130.6 H2O(l) -285.85 69.91 O2(g) 0.00 205.0 Thermodynamics, ΔH, ΔS, ΔG page 6 C3H7COOH(l) ? 226.3 The enthalpy change for the combustion of butyric acid at 25C, Hcomb, is -2,183.5 kilojoules per mole. The combustion reaction is C3H7COOH(l) + 5 O2(g) 4 CO2(g) + 4 H2O(l) (a) From the above data, calculate the standard heat of formation, Hf, for butyric acid. (b) Write a correctly balanced equation for the for-mation of butyric acid from its elements. (c) Calculate the standard entropy change, Sf, for the formation of butyric acid at 25C. The entropy change, S, for the combustion reaction above is -117.1 J K-1 at 25C. (d) Calculate the standard free energy of formation, Gf, for butyric acid at 25C. Answer: (a) H H f _ ( products ) H f _ ( reactants ) = [4(393.5) + 4(205.85) - 2183.5] kJ = -533.8 kJ (b) 4 C(s) + 4 H2(g) + O2(g) C3H7COOH(l) (c) Sf (butyric acid) = S(butyric acid) - [4 S(C) + 4 S(H2) + S(O2)] = 226.3 -[4(5.69) + 4(130.6) + 205] = -523.9 J/K (d) Gf = Hf - TSf = 533.8 - (298)(-0.5239) kJ = -377.7 kJ 1985 D (a) When liquid water is introduced into an evacuated vessel at 25C, some of the water vaporizes. Pre-dict how the enthalpy, entropy, free energy, and temperature change in the system during this pro-cess. Explain the basis for each of your predic-tions. (b) When a large amount of ammonium chloride is added to water at 25C, some of it dissolves and the temperature of the system decreases. Predict how the enthalpy, entropy, and free energy change in the system during this process. Explain the basis for each of your predictions. (c) If the temperature of the aqueous ammonium chlo-ride system in part (b) were to be increased to 30C, predict how the solubility of the ammonium chloride would be affected. Explain the basis for each of your predictions. Answer: (a) H>0 since heat is required to change liquid water to vapor S>0 since randomness increases when a liquid changes to vapor. G<0 since the evaporation takes place in this sit-uation. T<0 since the more rapidly moving molecules leave the liquid first. The liquid remaining is cool-er. (b) H>0. The system after dissolving has a lower temperature and so the change is endothermic. S>0, since the solution is less ordered than the separate substances are. G<0. The solution occurred and so is spontane-ous. (c) Solubility increases. The added heat available pushes the endothermic process toward more dis-solving. 1986 D The first ionization energy of sodium is +496 kilojoules per mole, yet the standard heat of formation of sodium chloride from its elements in their standard state is -411 kilojoules per mole. (a) Name the factors that determine the magnitude of the standard heat of formation of solid sodium chloride. Indicate whether each factor makes the reaction for the formation of sodium chloride from its elements more or less exothermic. (b) Name the factors that determine whether the reac-tion that occurs when such a salt dissolves in water is exothermic or endothermic and discuss the effect of each factor on the solubility. Answer: (a) heat of sublimation of sodium : endothermic first ionization energy of sodium: endothermic heat of dissociation of Cl2: endothermic electron affinity of chlorine: exothermic lattice energy of NaCl: exothermic (b) lattice energy of NaCl: endothermic to solution hydration energy of the ions: exothermic solvent expansion is endothermic. OR Thermodynamics, ΔH, ΔS, ΔG page 7 increased exothermicity is associated with in-creased solubility. 1987 D When crystals of barium hydroxide, Ba(OH)2 .8H2O, are mixed with crystals of ammonium thiocyanate, NH4SCN, at room temperature in an open beaker, the mixture liquefies, the temperature drops dramatically, and the odor of ammonia is detected. The reaction that occurs is the following: Ba(OH)2 .8H2O(s) + 2 NH4SCN(s) Ba2+ + 2 SCN- + 2 NH3(g) + 10 H2O(l) (a) Indicate how the enthalpy, the entropy, and the free energy of this system change as the reaction occurs. Explain your predictions. (b) If the beaker in which the reaction is taking place is put on a block of wet wood, the water on the wood immediately freezes and the beaker adheres to the wood. Yet the water inside the beaker, formed as the reaction proceeds, does not freeze even though the temperature of the reaction mix-ture drops to -15C. Explain these observations. Answer (a) The enthalpy increases (H>0) since the reaction absorbs heat as in shown by the decrease in tem-perature. The entropy increases (S>0) since solid reactants are converted to gases and liquids, which have a much higher degree of disorder. The free energy decreases (G<0) as is shown by the fact that the reaction is spontaneous. (b) The water on the wood froze because the endo-thermic reaction lowered the temperature below the freezing point of water. The solution in the beaker did not freeze because the presence of ions and dissolved gases lowered the freezing point of the solution below -15C. The freezing point depression is given by the equa-tion T = Kfm where m = the molality of the solu-tion and Kf = the molal freezing point constant for water. 1988 B Enthalpy of Absolute Combustion, H Entropy, S Substance (kiloJoules/mol) (Joules/mol-K) C(s) -393.5 5.740 H2(g) -285.8 130.6 C2H5OH(l) -1366.7 160.7 H2O(l) - - 69.91 (a) Write a separate, balanced chemical equation for the combustion of each of the following: C(s), H2(g), and C2H5OH(l). Consider the only products to be CO2 and/or H2O(l). (b) In principle, ethanol can be prepared by the follow-ing reaction: 2 C(s) + 2 H2(g) + H2O(l) C2H5OH(l) Calculate the standard enthalpy change, H, for the preparation of ethanol, as shown in the reac-tion above. (c) Calculate the standard entropy change, S, for the reaction given in part (b). (d) Calculate the value of the equilibrium constant at 25C for the reaction represented by the equation in part (b). Answer: (a) C + O2 CO2 2 H2 + O2 2 H2O C2H5OH + 3 O2 2 CO2 + 3 H2O (b) 2 C + 2 O2 2 CO2 H = 2(-393.5) = -787.0 kJ 2 H2 + O2 2 H2O H = 2(-285.8) = -571.6 kJ 2 CO2 + 3 H2O C2H5OH + 3 O2 H = -(-1366.7) = +1366.7 kJ 2 C + 2 H2 + H2O C2H5OH H = +8.1 kJ OR Hcomb. C(s) = Hf CO2(g) Hcomb. H2(g) = Hf H2O(l) C2H5OH + 3 O2 2 CO2 + 3 H2O H = -1366.7 kJ H H f _ ( products ) H f _ ( reactants ) = [2(-393.5) + 3(-258.8)] - [Hf C2H5OH + 0] kJ = -277.7 kJ/mol 2 C + 2 H2 + H2O C2H5OH H H f _ ( products ) H f _ ( reactants ) = [-277.7] - [0 + 0 + (-285.8)] kJ = +8.1 kJ (c) S S (products )S (reactants) Thermodynamics, ΔH, ΔS, ΔG page 8 = [160.7] - [11.5 + 261.2 + 69.9] J/mol.K = -181.9 J/mol.K (d) G = H - TS = 8100 -(298)(-181.9) J = 62300 J Keq = e-G/RT = e-(62300/(8.3143)(298)) = 1.210-11 1988 D An experiment is to be performed to determine the standard molar enthalpy of neutralization of a strong acid by a strong base. Standard school laboratory equipment and a supply of standardized 1.00 molar HCl and standardized 1.00 molar NaOH are available. (a) What equipment would be needed? (b) What measurements should be taken? (c) Without performing calculations, describe how the resulting data should be used to obtain the stand-ard molar enthalpy of neutralization. (d) When a class of students performed this experi-ment, the average of the results was -55.0 ki-lojoules per mole. The accepted value for the standard molar enthalpy of neutralization of a strong acid by a strong base is -57.7 kilojoules per mole. Propose two likely sources of experimental error that could account for the result obtained by the class. Answer: (a) Equipment needed includes a thermometer, and a container for the reaction, preferably a container that serves as a calorimeter, and volumetric glass-ware (graduated cylinder, pipet, etc.). (b) Measurements include the difference in tempera-tures between just before the start of the reaction and the completion of the reaction, and amounts (volume, moles) of the acid and the base. (c) Determination of head (evolved or absorbed): The sum of the volumes (or masses) of the two solu-tions, and change in temperature and the specific heat of water are multiplied together to determine the heat of solution for the sample used. q = (m)(cp)(T). Division of the calculated heat of neutralization by moles of water produced, or moles of H+, or moles of OH-, or moles of limiting reagent. (d) Experimental errors: heat loss to the calorimeter wall, to air, to the thermometer; incomplete trans-fer of acid or base from graduated cylinder; spat-tering of some of the acid or base so that incom-plete mixing occurred, … Experimenter errors: dirty glassware, spilled solution, misread volume or temperature, … 1989 B Br2(l) Br2(g) At 25C the equilibrium constant, Kp, for the reaction above is 0.281 atmosphere. (a) What is the G298 for this reaction? (b) It takes 193 joules to vaporize 1.00 gram of Br2(l) at 25C and 1.00 atmosphere pressure. What are the values of H298 and S298 for this reaction? (c) Calculate the normal boiling point of bromine. As-sume that H and S remain constant as the temperature is changed. (d) What is the equilibrium vapor pressure of bromine at 25C? Answer: (a) G = -RTlnK = -(8.314 J.mol-1K-1)(298 K)(ln 0.281) = 3.14103 J.mol-1 (b) H = (193 J/g)(159.8 g/mol) = 3.084104 J/mol S H G T ( 30840 3140 ) J 298 K = 92.9 J/mol.K (c) At boiling point, G = 0 and thus, T H S 3 . 08 104 92. 9 332 K (d) vapor pressure = 0.281 atm. 1990 B Standard Free Energies of Formation at 298 K Substance Gf 298 K, kJ mol-1 C2H4Cl2(g) -80.3 C2H5Cl(g) -60.5 HCl(g) -95.3 Cl2(g) 0 Average Bond Dissociation Energies at 298 K Thermodynamics, ΔH, ΔS, ΔG page 9 Bond Energy, kJ mol-1 C-H 414 C-C 347 C-Cl 377 Cl-Cl 243 H-Cl 431 The tables above contain information for determining thermodynamic properties of the reaction below. C2H5Cl(g) + Cl2(g) C2H4Cl2(g) + HCl(g) (a) Calculate the H for the reaction above, using the table of average bond dissociation energies. (b) Calculate the S for the reaction at 298 K, using data from either table as needed. (c) Calculate the value of Keq for the reaction at 298 K. (d) What is the effect of an increase in temperature on the value of the equilibrium constant? Explain your answer. Answer: (a) H = energy of bonds broken - energy of bonds formed C2H5Cl + Cl2 C2H4Cl2 + HCl H = (2794 + 243) - (2757 + 431) kJ = -151 kJ OR CH + Cl-Cl C-Cl + HCl (representing the changes) H = (414) + 243) - (377 + 431) = -151 kJ (b) G G ( products ) G ( reactants ) = [-80.3 + (-95.3)] - [-60.5 + 0] = -115 kJ S H G T 151 ( 115 ) kJ 298 K 0 . 120 kJ K (c) Keq=e-G/RT = e-(-115100/(8.3143)(298)) = 1.501020 (d) Keq will decrease with an increase in T because the reverse (endothermic) reaction will be favored with the addition of heat. OR G will be less negative with an increase in tem-perature (from G = H - TS), which will cause Keq to decrease. 1991 D (Required) BCl3(g) + NH3(g) Cl3BNH3(s) The reaction represented above is a reversible reaction. (a) Predict the sign of the entropy change, S, as the reaction proceeds to the right. Explain your pre-diction. (b) If the reaction spontaneously proceeds to the right, predict the sign of the enthalpy change, H. Ex-plain your prediction. (c) The direction in which the reaction spontaneously proceeds changes as the temperature is increased above a specific temperature. Explain. (d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific temperature at which the direction of the sponta-neous reaction changes? Explain. Answer: (a) Because a mixture of 2 gases produces a single pure solid, there is an extremely large decrease in entropy, S < 0, i.e. the sign of S is negative. (b) In order for a spontaneous change to occur in the right direction, the enthalpy change must over-come the entropy change which favors the reac-tants (left), since nature favors a lower enthalpy, then the reaction must be exothermic to the right, H < 0. (c) G = H - TS, the reaction will change direction when the sign of G changes, since H < 0 and S < 0, then at low temperatures the sign of G is negative and spontaneous to the right. At some higher T, H = TS and G = 0, thereafter, any higher temperature will see G as positive and spontaneous in the left direction. (d) At equilibrium, K = e-G/RT, where G = 0, K = eo = 1 1992 B Cl2(g) + 3 F2(g) 2 ClF3(g) ClF3 can be prepared by the reaction represented by the equation above. For ClF3 the standard enthalpy of formation, Hf, is -163.2 kilojoules/mole and the standard free energy of formation, Gf, is -123.0 ki-lojoules/mole. (a) Calculate the value of the equilibrium constant for the reaction at 298K. (b) Calculate the standard entropy change, S, for the reaction at 298K. Thermodynamics, ΔH, ΔS, ΔG page 10 (c) If ClF3 were produced as a liquid rather than as a gas, how would the sign and the magnitude of S for the reaction be affected? Explain. (d) At 298K the absolute entropies of Cl2(g) and ClF3(g) are 222.96 joules per mole-Kelvin and 281.50 joules per mole-Kelvin, respectively. (i) Account for the larger entropy of ClF3(g) rela-tive to that of Cl2(g). (ii) Calculate the value of the absolute entropy of F2(g) at 298K. Answer: (a) Keq = e-G//RT = e-(-246000/(8.314)(298)) = 1.321043 (b) S H G T [ 326400 ( 246000 )]J 298 K = -270 J/K (c) S is a larger negative number. ClF3(l) is more or-dered (less disordered) than ClF3(g). (d) Entropy of ClF3 > entropy of Cl2 because (i) 1) larger number of atoms OR 2) more complex praticle OR 3) more degrees of freedom (ii) S S (products )S (reactants) -270 = [2(281.5)] - [222.96 + 3(SF2)] SF2 = 203 J/mol.K 1993 D 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l) The reaction represented above is spontaneous at 25C. Assume that all reactants and products are in their standard state. (a) Predict the sign of S for the reaction and justify your prediction. (b) What is the sign of G for the reaction? How would the sign and magnitude of G be affected by an increase in temperature to 50C? Explain your answer. (c) What must be the sign of H for the reaction at 25C? How does the total bond energy of the reac-tants compare to that of the products? (d) When the reactants are place together in a contain-er, no change is observed even though the reaction is known to be spontaneous. Explain this observa-tion. Answer: (a) S<0. The number of moles of gaseous products is less than the number of moles of gaseous reac-tants. OR A liquid is formed from gaseous reac-tants. (b) G<0. G becomes less negative as the tempera-ture is increased since S < 0 and G = H -TS. The term “-TS” adds a positive number to H. (c) H<0. The bond energy of the reactants is less than the bond energy of the products. (d) The reaction has a high activation energy; OR is kinetically slow; OR a specific mention of the need for a catalyst or spark. 1994 D 2 H2S(g) + SO2(g) 3 S(s) + 2 H2O(g) At 298 K, the standard enthalpy change, H for the reaction represented above is -145 kilojoules. (a) Predict the sign of the standard entropy change, S, for the reaction. Explain the basis for your prediction. (b) At 298 K, the forward reaction (i.e., toward the right) is spontaneous. What change, if any, would occur in the value of G for this reaction as the temperature is increased? Explain your reasoning using thermodynamic principles. (c) What change, if any, would occur in the value of the equilibrium constant, Keq, for the situation de-scribed in (b)? Explain your reasoning. (d) The absolute temperature at which the forward reaction becomes nonspontaneous can be predict-ed. Write the equation that is used to make the prediction. Why does this equation predict only an approximate value for the temperature? Answer: (a) S is negative (-). A high entropy mixture of two kinds of gases forms into a low entropy solid and a pure gas; 3 molecules of gas makes 2 molecules of gas, fewer gas molecules is at a lower entropy. (b) G < 0 if spontaneous. G = H - TS Since S is neg. (-), as T gets larger, -TS will become larger than +145 kJ and the sign of G becomes pos. (+) and the reaction is non-spontanseous. (c) When -TS < +145 kJ, Keq > 1, when -TS = +145 kJ, Keq = 1, Thermodynamics, ΔH, ΔS, ΔG page 11 when -TS > +145 kJ, Keq < 1, but > 0 (d) G = 0 at this point, the equation is T = H/S; this assumes that H and/or S do not change with temperature; not a perfect assumption leading to errors in the calculation. 1995 B Propane, C3H8, is a hydrocarbon that is commonly used as fuel for cooking. (a) Write a balanced equation for the complete com-bustion of propane gas, which yields CO2(g) and H2O(l). (b) Calculate the volume of air at 30C and 1.00 at-mosphere that is needed to burn completely 10.0 grams of propane. Assume that air is 21.0 percent O2 by volume. (c) The heat of combustion of propane is -2,220.1 kJ/mol. Calculate the heat of formation, Hf, of propane given that Hf of H2O(l) = -285.3 kJ/mol and Hf of CO2(g) = -393.5 kJ/mol. (d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00 kilo-grams of water (specific heat = 4.18 J/g.K), calcu-late the increase in temperature of water. Answer: (a) C3H8 + 5 O2 3 CO2 + 4 H2O (b) 10.0 g C3H8 1 mol C3H8/44.0 g 5 mol O2/1 mol C3H8) = 1.14 mol O2 V O 2 nRT P 1.14 mol 0 . 0821 L atm mol K 303 K 1 . 00atm = 28.3 L O2 ; f(28.3 L,21.0%) = 135 L of air (c) H comb o H f (CO 2 ) o H f (H 2 O ) o H f (C3 H 8 ) o H f (O 2 ) o -2220.1 = [3(-393.5) + 4(-285.3)] - [X+ 0] X = Hcomb) = -101.6 kJ/mol (d) q = 30.0 g C3H8 1 mol/44.0 g 2220.1 kJ/1 mol = 1514 kJ q = (m)(Cp)(T) 1514 kJ = (8.00 kg)(4.184 J/g.K)(T) T = 45.2 1995 D (repeated in the solid, liquid, solutions section) Lead iodide is a dense, golden yellow, slightly soluble solid. At 25C, lead iodide dissolves in water forming a system represented by the following equation. PbI2(s) Pb2+ + 2 I- H = +46.5 kilojoules (a) How does the entropy of the system PbI2(s) + H2O(l) change as PbI2(s) dissolves in water at 25C? Explain (b) If the temperature of the system were lowered from 25C to 15C, what would be the effect on the value of Ksp? Explain. (c) If additional solid PbI2 were added to the system at equilibrium, what would be the effect on the con-centration of I- in the solution? Explain. (d) At equilibrium, G = 0. What is the initial effect on the value of G of adding a small amount of Pb(NO3)2 to the system at equilibrium? Explain. Answer: (a) Entropy increases. At the same temperature, liq-uids and solids have a much lower entropy than do aqueous ions. Ions in solutions have much greater “degrees of freedom and randomness”. (b) Ksp value decreases. Ksp = [Pb2+][I-]2. As the tem-perature is decreased, the rate of the forward (endo-thermic) reaction decreases resulting in a net de-crease in ion concentration which produces a smaller Ksp value. (c) No effect. The addition of more solid PbI2 does not change the concentration of the PbI2 which is a constant (at constant temperature), therefore, nei-ther the rate of the forward nor reverse reaction is affected and the concentration of iodide ions re-mains the same. (d) G increases. Increasing the concentration of Pb2+ ions causes a spontaneous increase in the reverse reaction rate (a “shift left” according to LeChate-lier’s Principle). A reverse reaction is spontaneous when the G>0. 1996 B C2H2(g) + 2 H2(g) C2H6(g) Information about the substances involved in the reac-tion represented above is summarized in the following tables. Substance S (J/molK) Hf (kJ/mol) Thermodynamics, ΔH, ΔS, ΔG page 12 C2H2(g) 200.9 226.7 H2(g) 130.7 0 C2H6(g) - - - - -84.7 Bond Bond Energy (kJ/mol) C-C 347 C=C 611 C-H 414 H-H 436 (a) If the value of the standard entropy change, S, for the reaction is -232.7 joules per moleKelvin, calculate the standard molar entropy, S, of C2H6 gas. (b) Calculate the value of the standard free-energy change, G, for the reaction. What does the sign of G indicate about the reaction above? (c) Calculate the value of the equilibrium constant, K, for the reaction at 298 K. (d) Calculate the value of the CC bond energy in C2H2 in kilojoules per mole. Answer: (a) -232.7 J/K = S(C2H6) - [2(130.7) + 200.9] J/K S(C2H6) = 229.6 J/K (b) H = Hâ(products) - Hâ(reactants) = -84.7 kJ - [226.7 + 2(0)] kJ = -311.4 kJ G = H - TS = -311.4 kJ - (298K)(-0.2327 kJ/K) = -242.1 kJ A G < 0 (a negative G) indicates a spontane-ous forward reaction. (c) Keq = e-G/RT = e-(-242100/(8.314)(298)) = 2.741042 (d) H = bond energy of products - bond energy of reactants -311.4 kJ = [(2)(436) + Ecc + (2)(414)] - [347 + (6)(414)] kJ Ecc = 820 kJ 1997 D For the gaseous equilibrium represented below, it is observed that greater amounts of PCl3 and Cl2 are pro-duced as the temperature is increased. PCl5(g) PCl3(g) + Cl2(g) (a) What is the sign of S for the reaction? Explain. (b) What change, if any, will occur in G for the re-action as the temperature is increased? Explain your reasoning in terms of thermodynamic princi-ples. (c) If He gas is added to the original reaction mixture at constant volume and temperature, what will happen to the partial pressure of Cl2? Explain. (d) If the volume of the reaction mixture is decreased at constant temperature to half the original vol-ume, what will happen to the number of moles of Cl2 in the reaction vessel? Explain. Answer: (a) The sign of S is (+). There is an increase in the number of gas molecules as well as a change from a pure gas to a mixture of gases. (b) G = H - TS. Both S and H are (+). As temperature increases, at some point the sign of G will change from (+) to (-), when the system will become spontaneous. (c) There will be no change in the partial pressure of the chlorine. Without a volume or temperature change, the pressure is independent of the other gases that are present. (d) The number of moles of Cl2 will decrease. The de-crease in volume will result in an increase in pres-sure and, according to LeChatelier’s Principle, the equilibrium system will shift to the left (the side with fewer gas molecules) to reduce this increase in pressure. This will cause a decrease in the num-ber of moles of products and an increase in the number of moles of reactant. 1998 B C6H5OH(s) + 7 O2(g) 6 CO2(g) + 3 H2O(l) When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow. Standard Heat of Absolute Entropy, Thermodynamics, ΔH, ΔS, ΔG page 13 Substance Formation, Hâ; at 25C (kJ/mol) S, at 25C (J/molòK) C(graphite) 0.00 5.69 CO2(g) -393.5 213.6 H2(g) 0.00 130.6 H2O(l) -285.85 69.91 O2(g) 0.00 205.0 C6H5OH(s) ? 144.0 (a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25C. (b) Calculate the standard heat of formation, Hâ, of phenol in kilojoules per mole at 25C. (c) Calculate the value of the standard free-energy change, G, for the combustion of phenol at 25C. (d) If the volume of the combustion container is 10.0 liters, calculate the final pressure in the container when the temperature is changed to 110.C. (As-sume no oxygen remains unreacted and that all products are gaseous.) Answer (a) Hcomb = kJ mass molar mass 64.98kJ 2.000g 94.113 g mol = -3058 kJ (b) Hcomb = Hâ(products) - Hâ(reactants) -3058 kJ/mol = [(6)(-393.5)+(3)(-285.85)]-[X+0] = -161 kJ (c) S = S(products) - S(reactants) = [(6)(213.6)+(3)(69.91)]-[144.0+(7)(205.0)] = -87.67 J G = H - TS = -3058 - (298)(-0.08767) kJ = -3032 kJ (d) 2.000 g phenol 1 mol phenol 94.113 g 7 mol O 2 1 mol phenol = 0.1488 mol O2 mol of gaseous product = 6 7 ( 0 . 1489 ) 3 7 ( 0 . 1489 ) = 0.1913 mol of gas P = nRT V (0.1913 mol) 0 . 08205 L • atm mol • K (383K) 10.0 L = 0.601 atm (or 457 mm Hg, or 60.9 kPa) 2001 B 2 NO(g) + O2(g) 2 NO2(g) H°= -114.1 kJ, S°= -146.5 J K-1 The reaction represented above is one that contributes significantly to the formation of photochemical smog. (a) Calculate the quantity of heat released when 73.1 g of NO(g) is converted to NO2(g). (b) For the reaction at 25C, the value of the standard free-energy change, G, is -70.4 kJ. (i) Calculate the value of the equilibrium con-stant, Keq, for the reaction at 25C. (ii) Indicate whether the value of G would be-come more negative, less negative, or remain unchanged as the temperature is increased. Justify your answer. (c) Use the data in the table below to calculate the value of the standard molar entropy, S, for O2(g) at 25C. Standard Molar Entropy, S (J K-1 mol-1) NO(g) 210.8 NO2(g) 240.1 (d) Use the data in the table below to calculate the bond energy, in kJ mol-1, of the nitrogen-oxygen bond in NO2 . Assume that the bonds in the NO2 molecule are equivalent (i.e., they have the same energy). Bond Energy (kJ mol-1) Nitrogen-oxygen bond in NO 607 Oxygen-oxygen bond in O2 495 Nitrogen-oxygen bond in NO2 ? Answer: (a) 73.1 g 1 mol NO 30.007 g 114.1 kJ 2 mol NO = 139 kJ (b) (i) Keq = e–G/RT = e–(–70400/(8.31)(298)) = 2.221012 (ii) less negative; G = H – TS; as tempera-ture increases, –TS becomes a larger positive value causing an increase in G (less negative). (c) S = S(products) – S(reactants) Thermodynamics, ΔH, ΔS, ΔG page 14 -146.5 = [(2)(240.1)] – [(210.8)(2)+ Soxygen] J/K Soxygen = +205.1 J/K (d) 2 NO(g) + O2(g) 2 NO2(g) + 114.1 kJ H = enthalpy of bonds broken – enthalpy of bonds formed -114.1 = [(607)(2) + 495] - 2X X = 912 kJ / 2 N=O bonds 456 kJ = bond energy for N=O bond 2002 D Required (repeated in lab procedures) A student is asked to determine the molar enthalpy of neutralization, ∆Hneut, for the reaction represented above. The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup calo-rimeter. The heat released by the reaction is determined by using the equation q = mc∆T. Assume the following. • Both solutions are at the same temperature before they are combined. • The densities of all the solutions are the same as that of water. • Any heat lost to the calorimeter or to the air is neg-ligible. • The specific heat capacity of the combined solu-tions is the same as that of water. (a) Give appropriate units for each of the terms in the equation q = mc∆T. (b) List the measurements that must be made in order to obtain the value of q. (c) Explain how to calculate each of the following. (i) The number of moles of water formed during the experiment (ii) The value of the molar enthalpy of neutraliza-tion, ∆Hneut, for the reaction between HCl(aq) and NaOH(aq) (d) The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl and 2.0 M NaOH. (i) Indicate whether the value of q increases, de-creases, or stays the same when compared to the first experiment. Justify your prediction. (ii) Indicate whether the value of the molar en-thalpy of neutralization, ∆Hneut, increases, de-creases, or stays the same when compared to the first experiment. Justify your prediction. (e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect would this have on the calculated value of the mo-lar enthalpy of neutralization, ∆Hneut? Justify your answer. Answer: (a) q in J, m in grams, C in J/g˚C, T in ˚C (b) mass or volume of each solution starting temperature of each reagent ending temperature of mixture (c) (i) both are 1 M acid and base and react on a 1:1 basis volume 1 mol HCl 1000 mL 1 mol H+ 1 mol HCl = mol of H+ H+ + OH– H2O (ii) joules released mol H2O produced (d) (i) increases. Twice as much water is produced so it is twice the energy released in the same volume of solution (ii) same. twice energy twice mol water = same result (e) smaller. heat lost to the air gives a smaller amount of temperature change in the solution, which leads to a smaller measured heat release 2003 D Answer the following questions that relate to the chem-istry of nitrogen. (a) Two nitrogen atoms combine to form a nitrogen molecule, as represented by the following equa-tion. 2 N(g) N2(g) Using the table of average bond energies below, determine the enthalpy change, ∆H, for the reac-tion. Bond Average Bond Energy (kJ mol–1) N–N 160 Thermodynamics, ΔH, ΔS, ΔG page 15 N=N 420 NN 950 (b) The reaction between nitrogen and hydrogen to form ammonia is represented below. N2(g) + 3 H2(g) 2 NH3(g) ∆H˚ = –92.2 kJ Predict the sign of the standard entropy change, ∆S˚, for the reaction. Justify your answer. (c) The value of ∆G˚ for the reaction represented in part (b) is negative at low temperatures but posi-tive at high temperatures. Explain. (d) When N2(g) and H2(g) are placed in a sealed con-tainer at a low temperature, no measurable amount of NH3(g) is produced. Explain. Answer: (a) a triple bond is formed, an exothermic process ∆H = –950 kJ mol–1 (b) (–); the mixture of gases (high entropy) is convert-ed into a pure gas (low entropy) and the 4 mole-cules of gas is reduced to 2, a smaller number of possible microstates is available (c) ∆G˚ = ∆H˚ – T∆S˚; enthalpy favors spontaneity (∆H < 0), negative entropy change does not favor spontaneity. Entropy factor becomes more signifi-cant as temperature increases. At high tempera-tures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G > 0). (d) at low temperatures, the kinetic energy of the mol-ecules is low and very few molecules have enough activation energy 2004 B (repeated in stoichiometry) 2 Fe(s) + 3 2 O2(g) Fe2O3(s) ∆Hf˚ = -824 kJ mol–1 Iron reacts with oxygen to produce iron(III) oxide as represented above. A 75.0 g sample of Fe(s) is mixed with 11.5 L of O2(g) at 2.66 atm and 298 K. (a) Calculate the number of moles of each of the fol-lowing before the reaction occurs. (i) Fe(s) (ii) O2(g) (b) Identify the limiting reactant when the mixture is heated to produce Fe2O3. Support your answer with calculations. (c) Calculate the number of moles of Fe2O3 pro-duced when the reaction proceeds to comple-tion. (d) The standard free energy of formation, ∆Gf˚ of Fe2O3 is –740. kJ mol–1 at 298 K. (i) Calculate the standard entropy of formation ∆Sf˚ of Fe2O3 at 298 K. Include units with your answer. (ii) Which is more responsible for the spontaneity of the formation reaction at 298K, the stan-dard enthalpy or the standard entropy? The reaction represented below also produces iron(III) oxide. The value of ∆H˚ for the reaction is –280 kJ per mol. 2 FeO(s) + 1 2 O2(g) Fe2O3(s) (e) Calculate the standard enthalpy of formation, ∆Hf˚ of FeO(s). Answer: (a) (i) 75.0 g Fe 1 mol 55.85 g = 1.34 mol Fe (ii) PV = nRT, n = PV RT (2.66 atm)(11.5 L) (0.0821 L atm mol K) (298 K) = 1.25 mol O2 (b) Fe; 1.34 mol Fe 3 2 mol O2 2 mol Fe = 1.01 mol O2 excess O2, limiting reagent is Fe (c) 1.34 mol Fe 1 mol Fe2O3 2 mol Fe = 0.671 mol Fe2O3 (d) (i) ∆Gf˚ = ∆Hf˚ – T∆Sf˚ –740 kJ mol–1 = –824 kJ mol–1 – (298 K)(∆Sf˚) ∆Sf˚ = 0.282 kJ mol–1 K–1 (ii) standard enthalpy; entropy decreases (a non-spontaneous process) so a large change in enthalpy (exothermic) is need to make this reaction sponta-neous (e) ∆H = ∆Hf(products) – ∆Hf(reactants) –280 kJ mol–1 = –824 kJ mol–1 – [2(∆Hf˚ FeO) – 1/2(0)] = -272 kJ mol–1 Thermodynamics, ΔH, ΔS, ΔG page 16 2005 D [repeated in electrochem] AgNO3(s) Ag+(aq) + NO3–(aq) The dissolving of AgNO3(s) in pure water is rep-resented by the equation above.. (a) Is ∆G for the dissolving of AgNO3(s) positive, neg-ative, or zero? Justify your answer. (b) Is ∆S for the dissolving of AgNO3(s) positive, neg-ative, or zero? Justify your answer. (c) The solubility of AgNO3(s) increases with increas-ing temperature. (i) What is the sign of ∆H for the dissolving pro-cess? Justify your answer. (ii) Is the answer you gave in part (a) consistent with your answers to parts (b) and (c) (i)? Ex-plain. The compound NaI dissolves in pure water according to the equation NaI(s) Na+(aq) + I–(aq). Some of the information in the table of standard reduction potentials given below may be useful in answering the questions that follow. Half-reaction E˚ (V) O2(g) + 4 H+ + 4 e- 2 H2O(l) 1.23 I2(s) + 2 e- 2 I– 0.53 2 H2O(l) + 2 e- H2(g) + 2 OH– -0.83 Na+ + e- Na(s) -2.71 (d) An electric current is applied to a 1.0 M NaI solu-tion. (i) Write the balanced oxidation half reaction for the reaction that takes place. (ii) Write the balanced reduction half-reaction for the reaction that takes place. (iii) Which reaction takes place at the anode, the oxida-tion reaction or the reduction reaction? (iv) All electrolysis reactions have the same sign for ∆G˚. Is the sign positive or negative? Justify your answer. Answer: (a) sign of ∆G = (–); since the dissolving of silver ni-trate is spontaneous, then ∆G < 0 (b) sign of ∆S = (+); an increase in entropy occurs when a solid becomes aqueous and the products contain more particles than the reactants. (c) (i) sign of ∆H = (+); an endothermic process will be favored when the temperature is increased. (ii) yes; ∆G = ∆H – T∆S, as the temperature in-creases the –T∆S term will increase, keeping ∆G negative. (d) (i) 2 I– I2(s) + 2 e- (ii) 2 H2O(l) + 2 e- H2(g) + 2 OH– (iii) anode = oxidation (iv) sign ∆G˚ = (+); by definition, an electrolysis is a non-spontaneous process and requires the input of energy to get it to proceed. |
7924 | https://www.ck12.org/flexi/chemistry/bronsted-lowry-acid-base-reactions/what-are-the-functions-of-the-proton-motive-force/ | Flexi answers - What are the functions of the proton motive force? | CK-12 Foundation
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What are the functions of the proton motive force?
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The proton motive force (PMF) serves two main functions in cells: 1. ATP Synthesis: The PMF is used by ATP synthase to generate ATP, the main energy currency of the cell. As protons flow down their concentration gradient through ATP synthase, the energy released is used to add a phosphate group to ADP, forming ATP. 2. Active Transport: The PMF is also used to drive the active transport of nutrients and ions across the cell membrane. This process involves the movement of substances against their concentration gradient, which requires energy. The energy is provided by the PMF as protons move down their concentration gradient, coupled with the transport of the nutrient or ion in the opposite direction.
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7925 | https://byjus.com/biology/nitrogenous-bases/ | Nitrogenous bases, also called nucleobases, are nitrogenous compounds that form an important part of the nucleotides. Nucleotides are building blocks of DNA and RNA that are composed of a sugar, nitrogenous base and a phosphate group. There are a total of five bases found in the DNA and RNA world, namely – Adenine (A), Cytosine (C), Guanine (G), Thymine (T) and Uracil (U).
Let us look at the five nucleotides found in DNA and RNA.
Adenine
Adenine is a two ringed purine derived nucleobase that has an amino group attached to the C6 position. In the nucleotide structure it forms a covalent bond with the ribose/deoxyribose sugar and hydrogen bond with the adjacent nucleobase, that is either a thymine or uracil.
Other compounds formed by adenine include vitamin B12, adenosine triphosphate (ATP), nicotinamide adenine dinucleotide (NAD) and flavin adenine dinucleotide (FAD).
Guanine
Guanine is another two ringed purine derived nucleobase composed of a fused pyrimidine-imidazole ring system that is conjugated with double bonds. It forms hydrogen bonds with cytosine in the nucleotide sequence.
Guanine combines with ribose to form guanosine and with deoxyribose to form deoxyguanosine.
Thymine
Thymine is an organic compound that belongs to the pyrimidine family. It forms double hydrogen bonds with adenine in the DNA helix. It is also known as 5-methyluracil because it is methylated at the C5 position in the molecule. It is not found in RNA strands.
Cytosine
Cytosine is a pyrimidine derived nitrogenous base that has an amino group at the C4 position. It forms triple hydrogen bonds with guanine in the DNA helix.
Uracil
Uracil is another pyrimidine derived nitrogenous base that is only found in RNA molecules in place of thymine. It is a demythlated form of thymine that is substituted with oxo groups at C2 and C4.
Visit BYJU’S Biology for more information.
Also Read:
Difference between Nucleotide and Nucleoside
MCQs on Nucleotide for NEET 2022
Nucleic Acid and Genetic Code – Structure and the Functions
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7926 | https://www.youtube.com/watch?v=I6TBBzIvgB8 | Factoring algebraic expressions using the distributive property | Algebra I | Khan Academy
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Transcript:
In earlier mathematics
that you may have done, you probably got familiar
with the idea of a factor. So for example, let me just pick an arbitrary number, the number 12. We could say that the
number 12 is the product of say two and six; two
times six is equal to 12. So because if you take the product of two and six, you get 12, we could say that two is a factor of
12, we could also say that six is a factor of 12. You take the product of
these things and you get 12! You could even say that
this is 12 in factored form. People don't really talk that way but you could think of it that way. We broke 12 into the things
that we could use to multiply. And you probably remember
from earlier mathematics the notion of prime factorization, where you break it up into
all of the prime factors. So in that case you could break the six into a two and a three, and you have two times two times three is equal to 12. And you'd say, "Well, this would be 12 "in prime factored form or the
prime factorization of 12," so these are the prime factors. And so the general idea,
this notion of a factor is things that you can multiply together to get your original thing. Or if you're talking about factored form, you're essentially taking the
number and you're breaking it up into the things
that when you multiply them together, you get
your original number. What we're going to do now is extend this idea into the algebraic domain. So if we start with an expression, let's say the expression is two plus four X, can we break this up into the product of two either numbers or two expressions or the product of a
number and an expression? Well, one thing that might jump out at you is we can write this as two times one plus two X. And you can verify if you like that this does indeed
equal two plus four X. We're just going to distribute the two. Two times one is two, two times two X is equal to four X, so plus four X. So in our algebra brains, this will often be reviewed as or referred to as this expression factored
or in a factored form. Sometimes people would say that we have factored out the two. You could just as easily say that you have factored out a one plus two X. You have broken this thing
up into two of its factors. So let's do a couple of examples of this and then we'll
think about, you know, I just told you that we
could write it this way but how do you actually figure that out? So let's do another one. Let's say that you had, I don't know, let's say you had, six, let
me just in a different color, let's say you had six X six X plus three, no, let's write it six X plus 30, that's interesting. So one way to think about it is can we break up each of these terms so that they have a common factor? Well, this one over here,
six X literally represents six times X, and then 30,
if I want to break out a six, 30 is divisible
by six, so I could write this as six times five, 30 is the same thing as six times five. And when you write it this way, you see, "Hey, I can factor out a six!" Essentially, this is the reverse of the distributive property! So I'm essentially undoing
the distributive property, taking out the six, and you
are going to end up with, so if you take out the six,
you end up with six times, so if you take out the six here, you have an X, and you take out the six here, you have plus five. So six X plus 30, if you factor it, we could write it as
six times X plus five. And you can verify with
the distributive property. If you distribute this six, you get six X + five times six or six X + 30. Let's do something that's a little bit more interesting where we might want to factor out a fraction. So let's say we had the situation ... Let me get a new color here. So let's say we had 1/2 minus 3/2, minus 3/2 X. How could we write this in a, I guess you could say, in a factored form, or if we wanted to factor out something? I encourage you to pause the video and try to figure it out,
and I'll give you a hint. See if you can factor out 1/2. Let's write it that way. If we're trying to factor out 1/2, we can write this first term as 1/2 times one and this second one we could write as minus 1/2 times three X. That's what this is,
3/2 X is the same thing as three X divided by
two or 1/2 times three X. And then here we can see
that we can just factor out the 1/2 and you're going to get 1/2 times one minus three X. Another way you could
have thought about it is, "Hey, look, both of these are products "involving 1/2," and
that's a little bit more confusing when you're
dealing with a fraction here. But one way to think about it is, I can divide out a 1/2
from each of these terms. So if I divide out a 1/2 from this, 1/2 divided by 1/2 is one. And if I take 3/2 and divide it by 1/2, that's going to be
three, and so I took out a 1/2, that's another
way to think about it. I don't know if that confuses you more or it confuses you less, but hopefully this gives you the sense of
what factoring an expression is. I'll do another example,
where we're even using more abstract things, so
I could say, "AX plus AY." How could we write this in factored form? Well, both of these terms have products of A in it, so I could write
this as A times X plus Y. And sometimes you'll hear people say, "You have factored out the A," and you can verify it if
you multiply this out again. If you distribute the A,
you'd be left with AX plus AY. |
7927 | https://www.ck12.org/flexi/cbse-math/polynomials-in-standard-form/ | Polynomials in Standard Form | Flexi Homework help & answers | CK-12 Foundation
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CBSE Math
Polynomials in Standard Form
Polynomials in Standard Form
Concept Summary:
A function f(x)of the form f(x)=a 0+a 1 x+a 2 x 2+…⋯+a n x n,where a 0,a 1,a 2,…a n are real numbers,a n≠0 and n is a non negative integer is called a polynomial.
If the coefficients of a polynomial are integers, then it is called a polynomial over integers.
Ask your own question
For questions 28-30, classify each polynomial as either a monomial, binomial, or trinomial. Then, give the degree of the polynomial. 28. Unexpected text node: '-5a^6' This is a monomial with a degree of 6. 29. Unexpected text node: '-x^3 + 9x^2 - 4x' This is a trinomial with a degree of 3. 30. Unexpected text node: '4x - 7' This is a binomial with a degree of 1.
Find the degree of the polynomial and indicate whether the polynomial is a monomial, binomial, trinomial, or none of these: Unexpected text node: '4x^2 + 0.2'.
How to create a polynomial written in standard form?
Write the polynomial in standard form. Then name the polynomial based on its degree and number of terms. 3x + 2x^2 – 6
Form a polynomial whose zeros and degree are given: Zeros: -1, 1, 9; degree: 3.
Is x^4+5x^2+6 a strictly linear polynomial?
Is x^4−16 a strictly linear polynomial?
How can one determine if a function is a polynomial?
How to factorize a quadratic expression?
How to solve polynomial inequalities?
How do I solve polynomial equations?
How to solve a polynomial equation?
How to find the degree of a polynomial graph?
How do I solve a polynomial equation?
How do you determine the end behavior of a polynomial?
Which graph is an example of a cubic function?
What is a polynomial of degree 3?
Define a polynomial equation.
What is a complex polynomial?
What is a real polynomial?
Define a polynomial.
How to calculate the number of one-one functions?
What is meant by the codomain of a function?
What is the standard form of a polynomial?
What is the degree of the zero polynomial?
How do you solve polynomial equations?
How do you solve a cubic polynomial?
What is a cubic polynomial?
What is a zero polynomial?
What is the Polynomial?
What is a polynomial in one variable?
How do I find a polynomial?
What is the meaning of polynomial?
What is a linear polynomial?
How do I find the zeros of a polynomial?
How can I find a polynomial equation?
How can a polynomial equation be created?
Calculate the least common multiple (LCM) of 18 and 30.
How to change a polynomial into standard form?
How to rewrite polynomials in standard form?
Define the leading coefficient of a polynomial.
How to find the leading coefficient?
How to find the leading coefficient of a polynomial?
What types of algebraic expressions fall under the category of binomials?
How to identify which algebraic expressions are polynomials?
How to identify a polynomial function?
Define terms, coefficients, and constants.
What is a polynomial with a degree of 3 called?
Explain what a polynomial function is.
Define monomials in mathematics.
Can a constant be classified as a polynomial?
Is a binomial considered a type of polynomial?
How to write polynomials in standard form?
How to write any polynomial in standard form?
How many constants are required in a cubic polynomial?
What criteria are used to identify a polynomial?
How is the degree of a polynomial determined?
What is the process of factoring a polynomial function?
How can polynomials with four terms be factored?
Which algebraic expressions can be polynomials?
What is the degree of a monomial?
How do you find the degree of a monomial?
Are monomials also polynomials?
A polynomial with a degree of 3 is ___.
Is a constant a polynomial?
How do you change a polynomial into standard form?
How can one rewrite polynomials in standard form?
What is the degree and leading coefficient of the polynomial?
What effect does the leading coefficient have on the graph?
Is the leading coefficient positive or negative?
How do you write polynomials in standard form?
How does one determine the degree of a polynomial?
What would be the quadratic polynomial if its roots are 3 and 4?
What are the categories of polynomials (linear, quadratic, and cubic)?
How is a polynomial written in standard form?
How is a polynomial written in its standard form?
How do you identify polynomials?
How do you identify a polynomial?
How do you find the leading coefficient of a polynomial on a graph?
How do you find the constant coefficient in a polynomial?
How do we identify which algebraic expressions are polynomials?
Define a constant polynomial.
What is the constant in a polynomial?
How can a polynomial be constructed given its zeros and degree?
Can the leading coefficient be a negative value?
What are monomials in mathematics?
Is a binomial a type of polynomial?
How do you determine the leading coefficient in a polynomial?
What are the types of polynomials (linear, quadratic, and cubic)?
Can the leading coefficient be negative?
How do you write a polynomial function in standard form?
What are linear, quadratic and cubic polynomials?
What is a polynomial?
How can a polynomial with given zeros and degree be formed?
What is a constant polynomial?
What are the characteristics of polynomial equations?
What is a polynomial equation of degree 2?
What is a polynomial function?
What is a polynomial with a degree of 3?
What is a square of a binomial?
What is the L.C.M. of 18 and 30?
What is the lowest common multiple of 6 and 10?
How do you find the degree of a polynomial?
How do you know if something is a polynomial?
How do you write a polynomial in standard form?
How to find the degree of a polynomial?
What are all the different types of polygons?
What are Polynomials?
What is the procedure for finding a polynomial in standard form?
What is the meaning of degree of a polynomial?
What is the degree of a polynomial in one variable?
What is the definition of the degree of a polynomial?
What are polynomials in algebra?
How to find greatest common factor?
What is the meaning of the degree of a polynomial?
What are the like terms in polynomials?
What are Polynomials in math?
How to divide polynomials using long division?
How do we convert a polynomial into standard form?
Define the degree of a polynomial.
Who invented Algebra?
Who is known as the Father of Algebra?
Which algebraic expression is a polynomial?
Which polynomial can be factored completely?
What is zero polynomial?
What is the LCM of 8 and 10?
What is the LCM of 3 and 8?
What is the LCM for 6 and 8?
What is the LCM for 4 and 6?
What is the general form of cubic polynomial?
What is the degree of a polynomial?
What is the exponential form in algebra?
What is the degree of zero polynomial?
What is the degree of a quadratic polynomial?
What is the definition of a polynomial?
What is exponential form in algebra?
What is an identity in algebra?
What is a constant in algebra?
How to write a polynomial in standard form?
How to simplify monomials?
How to know if an expression is a polynomial?
How to calculate the degree of a polynomial?
How do you find the leading coefficient of a polynomial?
How do you compare two whole numbers?
How can you tell if something is a polynomial?
What is the method to find the roots of a polynomial?
Define function in algebra.
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7928 | https://math-angel.io/lessons/nth-root-fractional-indices/ | Numbers and Decimals
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The nth Root and Fractional Indices
🎬 Video: nth Roots and Simplifying Indices
What is the nth Root? (0:01)
🔮 Definition of nth Root:
The nth root of a number is the value that, when raised to the power of $ n $, gives the original number. It is written as:
$$ \Large \sqrt[n]{x} $$
🔮 Examples of nth Root:
Square root ($ 2^{nd} $ root):$$ \sqrt{9} = 3 \quad \text{(Since } 3^2 = 9 \text{)} $$
Cube root ($ 3^{rd} $ root):$$ \sqrt{125} = 5 \quad \text{(Since } 5^3 = 125 \text{)} $$
Fourth root ($ 4^{th} $ root):$$ \sqrt{16} = 2 \quad \text{(Since } 2^4 = 16 \text{)} $$
Understanding the nth Root Rule (0:40)
🔮 What is the nth Root Rule?
The nth root rules tells us, if you multiply the nth root of a number exactly n times, you get back the original number.
$$\Large \underbrace{\sqrt[n]{a} \times \sqrt[n]{a} \times \cdots \times \sqrt[n]{a}}_{\text{n times}} = a$$
Example 1: If you multiply $ \sqrt{8}$ three times, you get 8.
$$ \sqrt{8} \times \sqrt{8} \times \sqrt{8} = 8 $$
Example 2: If you multiply $ \sqrt{16}$ four times, you get 16.
$$ \sqrt{16} \times \sqrt{16} \times \sqrt{16} \times \sqrt{16} = 16 $$
Understanding the Fractional Exponent Rule (1:20)
🔮 What Is the Fractional Exponent Rule?
The Fractional Exponent Rule connects exponents and roots:
$$ \Large a^{\frac{1}{n}} = \sqrt[n]{a} $$
It tells us that taking the nth root of a number is the same as raising it to the power of $ \frac{1}{n}$.
🔍 Example of Using This Rule:
$$ 125^{\frac{1}{3}} = \sqrt{125} = 5 $$
How to Simplify Fractional Indices? (2:15)
The fractional index rule states that an exponent in the form of a fraction can be rewritten using roots:
$$ \Large a^{\frac{m}{n}} = \left( \sqrt[n]{a} \right)^m $$
This means:
The denominator ($ n $) represents the nth root.
The numerator ($ m $) represents the power applied after taking the root.
🔍 Example of Simplifying Fractional Index:
Rewrite the fractional exponent: $$ 125^{\frac{2}{3}} = 125^{\frac{1}{3} \times 2} = \left(125^{\frac{1}{3}}\right)^2 $$
Find the cube root of 125:$$ 125^{\frac{1}{3}} = \sqrt{125} = 5 $$
Square the result:$$ 5^2 = 25 $$
Thus:$$ 125^{\frac{2}{3}} = 25 $$
📂 Flashcards: nth Root and Fractional Index Rules
🍪 Quiz: Practice Roots and Fractional Exponents
0%
🎩 Stuck on Roots or Indices? Try AI Math Solver
Need math help? Chat with our AI Math Solver at the bottom right — available 24/7 for instant answers.
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7929 | https://www.sciencedirect.com/science/article/pii/S0022316623279559 | Protein Requirements of Normal Infants at the Age of about 1 Year: Maintenance Nitrogen Requirements and Obligatory Nitrogen Losses - ScienceDirect
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Abstract
References (24)
Cited by (21)
The Journal of Nutrition
Volume 110, Issue 9, September 1980, Pages 1727-1735
Protein Requirements of Normal Infants at the Age of about 1 Year: Maintenance Nitrogen Requirements and Obligatory Nitrogen Losses1,2
Author links open overlay panel Huang P.C.1, Lin C.P.1, Hsu J.Y.1
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With 34 normal, healthy male infants aged 9–17 months, a total of 61 nitrogen (N) balance studies was conducted with N intake between 12 and 180 mg/kg/day. By regression analysis, the crude N maintenance requirements, either with whole egg or cow's milk protein, were estimated to be about the same, 106 and 103 mg/kg/day, respectively. The 97.5% confidence limits for the requirements were 128 (egg) and 142 (milk) mg/kg, respectively. Sums of the obligatory urinary and fecal N for the egg and milk protein series were 75 and 71 mg/kg as compared with 76 mg/kg of actually measured figure. Ratio of the maintenance N requirement to the obligatory N loss was 1.4. In another 15 N balance study, for which N intake from milk formulae ranged between 220 and 320 mg/kg/day, the mean apparent N retention was 25% of the intake. Total integumental N losses (skin + hair + nail) of infants fed 217–522 mg N/kg/day amounted to 7.8 ± 2.9 mg/kg daily. Egg protein had somewhat higher digestibility than cow's milk protein, 92 versus 87%, but has lower biological value, 76 versus 82. Net protein utilization (NPU) estimated from the regression line was about the same for both proteins, 71 and 69, respectively.
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LITERATURE CITED (24)
Ziegler E.E. et al.
Nitrogen balance studies with normal children
Am. J. Clin. Nutr (1977)
Rand W.M. et al.
Determination of protein allowances in human adults from nitrogen balance data
Am. J. Clin. Nutr (1977)
Fomon S.J. et al.
Urinary and fecal excretion of endogenous nitrogen by infants and children
J. Nutr (1965)
Inoue G. et al.
Studies on protein requirements of young men fed egg protein and rice protein with excess and maintenance energy intakes
J. Nutr (1973)
Young V.R. et al.
Protein requirements of man: comparative nitrogen balance response within the submaintenanceto-maintenance range of intakes of wheat and beef proteins
J. Nutr (1975)
Calloway D.H. et al.
Variation in endogenous nitrogen excretion and dietary nitrogen utilization as determinants of human protein requirement
J. Nutr (1971)
Calloway D.H. et al.
Nitrogen balance as related to calorie and protein intake in active young men
Am. J. Clin. Nutr (1954)
Joint FAO/WHO Expert Committee on Energy and Protein Requirements (1973) Energy and protein requirements. World Health...
Irwin M.I. et al.
A conspectus of research on protein requirements of man
J. Nutr (1971)
Chan H. et al.
The protein requirement of infants at the age of about 1 year
Br. J. Nutr (1966)
Fomon S.J. et al.
Determination of nitrogen balance of infants less than 6 months of age
Pediatrics (1958)
Hawk, P. B., Oser, B. L. & Summerson, W. H. (1965) Practical physiological chemistry, 13th ed., pp. 874–875,...
View more references
Cited by (21)
Protein and amino acid requirements and the composition of complementary foods
2003, Journal of Nutrition Citation Excerpt :
3) Unaccounted losses: In the analysis of maintenance nitrogen intake (requirement), an important problem is the estimation of unaccounted nitrogen losses, usually assumed to be due to nitrogen loss in desquamation, hair loss and sweat. The literature on this point is limited (31–34) but indicates that the average value for unaccounted losses is about 6.5 mg at intakes close to maintenance. 4) Efficiency of dietary protein utilization: The estimates of the efficiency of protein utilization derived from the analysis of published values (Fig. 2) include the effect of protein quality (i.e., chemical score), the relationship between the source of dietary protein and the bioavailability as well as variations related to the genotype and physiological state of the subjects. Show abstract In this paper, factorial models of the dietary requirements for protein, nitrogen and individual indispensable amino acids are developed from published information on the relationship between age and protein deposition and between protein (amino acid) intake and nitrogen balance. The results are used to develop recommendations on the protein–energy ratio and the amino acid pattern of the diet. As part of the development of the models, factors affecting dietary protein digestibility, bioavailability and efficiency of utilization are discussed. Over the age range of 6–24 mo the models predict a fall in the weight-specific protein and amino acid requirement that results almost entirely from the changes in the growth rate of the children. It is also concluded that the requirement for the maintenance of body protein equilibrium (so-called maintenance) changes little with age. This contrasts markedly with the relationship between age and energy requirements. The amino acid modeling implies that the optimum pattern of individual essential amino acids also changes only marginally across the age range considered in the report. The calculations of the dietary requirement for whole protein imply that achieving a minimum protein–energy ratio of 6.3% is desirable. The amount of protein needed from complementary foods for breast-fed children is discussed.
### Meta-analysis of nitrogen balance studies for estimating protein requirements in healthy adults
2003, American Journal of Clinical Nutrition Show abstract The most recent international dietary protein recommendations for healthy adults are those developed and proposed by the 1985 FAO/WHO/UNU Joint Expert Consultation. The objective was to analyze available nitrogen balance data to establish new recommendations for the protein required by healthy adults. Data were gathered from published nitrogen balance studies that had as their primary objective either the estimation of basal or maintenance requirements or the testing of the adequacy of specific nitrogen intakes in healthy adults. These data were synthesized to characterize the distribution of individual protein requirements; the effects of climate of the study site, adult age, sex, and dietary protein source on individual requirements; and the midpoint of and the variability between the protein requirements of healthy persons. Data for 235 individual subjects, each studied at ≥ 3 test protein intakes, were gathered from 19 studies. The median estimated average requirement (EAR) of nitrogen from these data was 105 mg N · kg−1 · d−1. Individual requirements were found to fit a log-normal distribution. The median EAR was estimated as the median of this distribution, 105 mg N · kg−1 · d−1, whereas the 97.5th percentile (the recommended dietary allowance; RDA) was estimated from the distribution of the log of the requirement (after correction of the total observed variability to remove within-individual variability) as 132 mg N · kg−1 · d−1. No significant differences between the climate of the study site, adult age class, sex, or source of dietary protein were observed, although there was an indication that women might have a lower requirement than do men. This meta-analysis provides new recommendations for dietary reference values, ie, an EAR (median) and RDA (97.5th percentile) for healthy adults of 105 and 132 mg N · kg−1 · d−1 (0.65 and 0.83 g good-quality protein · kg−1 · d−1), respectively.
### The effect of phytic acid on the levels of blood glucose and some enzymes of carbohydrate and lipid metabolism
2005, West Indian Medical Journal
### Recommended dietary allowances (RDAs), recommended dietary intakes (RDIs), recommended nutrient intakes (RNIs), and population reference intakes (PRIs) are not 'recommended intakes'
1997, Journal of Pediatric Gastroenterology and Nutrition
### Protein requirements of infants and children
1996, European Journal of Clinical Nutrition
### Requirements and recommended dietary intakes of protein during infancy
1991, Pediatric Research
View all citing articles on Scopus
1
Supported in part by a WHO Nutrition Research Grant (1970-1971).
2
Apart of this report was presented at the IX International Congress of Nutrition held in Mexico in 1972.
View full text
Copyright © 1980 American Society for Nutrition. Published by Elsevier Inc. All rights reserved.
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7930 | https://forums.autodesk.com/t5/revit-architecture-forum/distance-between-ceiling-and-floor-directly-below/td-p/7697246 | Solved: Distance between ceiling and floor directly below - Autodesk Community
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Distance between ceiling and floor directly below
Anonymous
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01-17-2018 06:07 AM
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Message 1 of 11
Distance between ceiling and floor directly below
Anonymous
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01-17-2018 06:07 AM
Hi,
Is there a way to display height of ceiling which would be measured to floor which is exactly below? Keep in mind that floor doesn't (and isn't in my case) corresponding to the levels set in model. I'm working on a project where floors everywhere are at different levels and so do ceilings follow - we can't find a simple solution to display ceiling height measured above floor level other than writing it by hand.
Solved! Go to Solution.
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Message 2 of 11
Sahay_R
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01-17-2018 06:13 AM
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Use a ceiling tag.A Ceiling Tag w Height.rfa can be found in the OOTB Library under Annotations.
Rina Sahay
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Anonymous
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01-17-2018 06:26 AM
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Hi Rina,
Thanks for the input but I'm afraid that maybe I didn't make myself clear. Ceiling tag only shows ceiling height above defined floor level. If the floor though is placed with height offset from level then the ceiling height to that specific floor (object) is incorrect.
I think the image below shows better what I mean - I need a ceiling tag showing height 2746 (distance to the modeled floor below) while it shows distance 1277 which is distance to defined floor LEVEL. So as there are modeled floors at different heights and ceilings at different offsets we want to make it as automatic as possible. Currently only option we see is to display either absolute value or write it each by hand (which obviously isn't the best way).
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Sahay_R
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01-17-2018 06:30 AM
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Are the secondary floors associated with a secondary level?
Rina Sahay
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Anonymous
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01-17-2018 06:32 AM
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No, both elements (ie. floor and ceiling) are associated with the same level. Floor is at a negative offset while ceiling at positive.
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Sahay_R
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01-17-2018 08:03 AM
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I would associate the floors with a secondary level then.
Rina Sahay
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01-17-2018 11:45 PM
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What do you mean by secondary? For display purposes we need these elements referenced to the level on which they should be displayed. Could you show an example of how would you do it? (ie. show ceiling tag distance to floor which isn't at level's height?)
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ToanDN
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01-17-2018 11:59 PM
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Create a ceiling based generic family and add a reference line a distance from the ceiling. Dimension the reference line to the ceiling and give it an instance length shared parameter. Load it in and host on a ceiling, go yo a 3d view or section to stretch the bottom of the family to meet the floor. Tag the family with the shared length parameter above to report the distance between the ceiling and the floor.
P/s: use a reference line instead of a reference plane do that you can have the stretching grip in 3d.
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01-18-2018 12:04 AM
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But wouldn't I need to stretch the reference every time I place the family? I could even handle that since the project isn't that big if I could lock the reference to the floor... Good suggestion - I'll try to check this.
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ToanDN
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01-18-2018 12:36 AM
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Accepted solution
Scratch that. I have a better solution using a two point adaptive family. Place it by picking one point on the floor and one point on the ceiling on a 3d view. See screencast and sample 2017 file.
2017 ceiling ht aff.rvt
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Anonymous
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01-18-2018 12:47 AM
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Thanks a million! Too bad though that there's no such solution OOTB.
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7931 | https://www.youtube.com/watch?v=f09PDzgLZVU | The distance between two numbers on a number line is 19. One number is 8, what’s the other number?
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Transcript:
okay so you might be surprised to find out that your answer to what appears to be a pretty simple math question here is not going to be correct so be careful with this question but uh let's take a look at the problem it is the following the distance between two numbers on a number line is 19 one number is eight what is the other number all right now feel free to use a calc calcor but if you can figure this out go ahead and put your answer into the comment section I'll show you the correct answer in just one second then of course I'm going to walk through exactly how to solve this problem step by step and the key word here is obviously distance and we're talking about distance uh from a mathematics standpoint but before we get started let me quickly introduce myself my name is John and I have been teaching middle and high school math for decades and if you need help learning math check out my math he help program at TCM academy.com you can find a link to that in the description below and if this video helps you out or if you just enjoy this content make sure to like And subscribe as that definitely helps me out all right so let's take another uh look at the problem before I show you the answer so the distance between two numbers on a number line is 19 one number is eight what is the other number now some of you out there might be saying he Mr YouTube Math Man uh this is is a trick question isn't it well yes indeed you're kind of on to something because the answer is not another number there's actually two numbers right so the question is what is the other number where we're actually have um we actually have two numbers as the answer so if you picked up on that that is fantastic and let's take a look at the correct answer the correct answer is ne1 and 27 now I think most people probably uh guessed 27 as their answer and that is one of the numbers but this is only 50% of the total answer you also need1 all right now if you got both of these numbers that is fantastic and you definitely get a happy face and an A+ so congratulations because it seems to me that you know a thing or two about distance between numbers on a number line all right now this is not that difficult and let's go ahead and get into it right now and obviously we are talking about distance but uh we all U we're also talking about a number line so I think a good starting point here is to re review a basic number line and then we'll talk about distance in just one second and of course we'll solve the problem all right so what is a number line well here is a basic number line and we're talking about the set of real numbers now I don't want to get all Technical and whatnot but this is important stuff especially for those of you that are studying oh I would say like um pre algebra and Beyond but let's just do a quick review of the type of numbers that are on this number line and then we'll talk about distance all right so uh there is a subset of numbers within the total span of the real number set and that's what this is right here and the first numbers is what we call the counting or natural numbers so that would be like one 2 3 4 this is what we see naturally right so we you know we might say well we see like this is a terrible little picture uh let me see here oh there is one cat right we don't see 2/3 or 75 we see things naturally there's two trees there's three cars Etc so these are Etc these are naturally occurring numbers or the counting numbers all right so that's what these numbers are so we'll put this as natural numbers and county numbers now uh someone figured out that hey you know zero is a good uh symbol to represent hey we don't have anything here you know we're empty so zero uh with the uh counting numbers or natural numbers is what we call the whole numbers now these are the positive whole numbers right here and if we take the positive and the negative whole numbers well then we have what we call the set of integers SO3 -2 negative 1 0 1 2 3 Etc these type of numbers are called integers and it is important that you kind of understand these basic um uh subsets of the real numbers now any fraction that you can create from integers like let's say 2/3 or negative one2 any number that you can express as a fraction of integers is called the what well that's called the rational numbers and uh the symbol for that is a q okay so you might be confused about that but that's what it is and then we have the irrational numbers which are numbers that you cannot express as a fraction on the uh fraction of uh integers okay so that's just a quick review of the real number line and that's important for those of you to understand but really that's not going to be totally relevant to our problem so now let's go ahead and talk about distance so here is a two and four now if I asked you what is the distance between two and four what you want to do is think of this distance in terms of like a ruler or maybe like a tape measure right so when you measure you're going to uh be measuring in what in positive units right your your room for example is not going to be F negative 15 ft long right so when we measure things are in positive units now you might be saying hey Mr YouTu Math Man you can have negative units well negative U or except negative distance excuse me negative um distance uh typically um is kind of basically uh indicating Direction let me show you a simple example because some of you may be confused so let's suppose let me do this like this let's say this is the ocean and we're at sea level so that's zero feet now here is our little ship up here okay okay and let's say there's an airplane flying up in the sky and that's going to be above the zero feet so maybe that's flying at 1,000 ft so this would be a positive number but maybe we have a Little Submarine down here and it's under zero so maybe it's at -200 FT so yes indeed you can have negative distance but that generally indicates like Direction okay but in general okay distance is going to be a positive number all right so uh kind of keeping that in mind let's go back to this problem so what is the distance between two and four well most of you out there saying hey Mr you mathat uh this is a very simple problem uh that is two and you would be correct but how we could um how could we figure this out okay so uh you might be saying well this is super easy all we have to do is subtract you know two and four so maybe we're going to go 2 - 4 right this number from this number we'll get the distance well 2 - 4 is Nega 2 but the answer is pos2 so when we're calculating the distance between two numbers on a number line it's always going to be the absolute value of the difference of those numbers so 2 - 4 is -2 but when we take the absolute value of Nega -2 we get a positive2 or we can go 4 minus 2 which of course is a positive2 now um two uh so from from two okay uh um and we have two as a distance away from two I should probably use the different example here because this is a bit confusing but what number is two units away from two well four is right but there is another number because we can measure two units in this direction right so two units in this way well what other number is two units away from two well it's zero right because this distance of two to zero is also two so of course we can measure that as 0 - 2 that's -2 but we're talking about the absolute value so that's positive2 or 2 - 0 all right so this is basically how we're going to solve our problem and if you're a bit confused uh you know about distance well hopefully this kind of clears things up but let's go ahead and take the next step which of course is to read our problem again now that we understand uh something about the distance between numbers on a number line and uh we also understand what a number line is well we can simply kind of model this situation so let's go ahead and take a look at a simple model because the question is saying uh the distance between two numbers on a number line is 19 so that's the distance right and one number is eight what is the other number so if you kind of visualize this one number is eight and uh we're looking for another number that's 19 units away from or 19 U units in terms of distance away from eight well we can go 19 units this way or 19 units uh 19 units this way so here is one number and the other number is way over here so how can we calculate this well that is what we're going to be doing next but before we do those calculations we're going to do this which is to take a quick math commercial break now I definitely need your help to continue to grow my channel I've been on YouTube I think it's like 14 years but that's when I started my YouTube channel so you know the path you know that I've taken on YouTube it kind of looks like this because when I started I really didn't do much I posted video here or there and uh you know I was distracted with other things you know quite busy but it wasn't maybe until like 5 6 seven years ago that really started putting a lot of effort in but when I started doing that guess what happened so did my you know YouTube channel started to grow and I put more effort in and that is just the same with anything in life especially mathematics so if you are trying to learn math a little bit here a little bit there you're not going to get great results the best way to learn math is is to just really get yourself in complete immersion and it takes time it takes effort right so if anyone's trying to tell you oh you know I can uh teach you one two three little shortcuts yeah they might be able to give you a quick tutorial on something but if you truly want deep comprehension in mathematics it takes a long time and sustain effort so don't give up and uh you know don't feel bad you know if you don't understand something now if you need additional math help now my YouTube videos are kind of like tutorial videos but if you're really need serious uh math instruction in you know algebra geometry basic math and whatnot check out my full main math courses you can find links to those in the description of this video but I have some momentum going here and it's exciting because I'm trying to reach as many people as possible on YouTube I'm trying to always teach math in a way that's clear and understandable but I need your support so go ahead and hit that subscribe button and that notification Bell so you can get my latest videos all right so let's go ahead and finish up this problem so we know that we have two numbers that are going to be 19 units away in terms of distance from eight because we can go from uh we could start at eight this way we can go 19 units to the right or 19 units to the left so what are these two numbers well let's uh discuss the harder one to figure out and that is 19 units away to the left from eight so how can we kind of figure this out well you can see I have the math here it's 8 plus or 8 minus uh 19 but what we would have to do is we have to start walking backwards right so I'm like all right well here's eight units this way and now I need to go another 11 units right to have 19 total units so um I'm I'm going to be going less than zero so these are going to be negative numbers I'm going to end up at a negative 11 so another way you can kind of think of this is 8 minus 19 or 8 plus ative 19 so 19 in this direction is uh going in the negative Direction so 8 + -19 is -11 so that is one number and the other number is pretty easy to figure out because if we're at eight and we add 19 we're going to get to the other number so 8 plus 19 of course is 27 all right so don't feel bad if you uh you know got this wrong my videos are intended to really you know encourage people to uh you know stick with math okay making mistakes and getting math problems wrong you know that's all part of learning math you know I know for myself uh I probably and I'm being quite uh you know honest here you know when you're studying very difficult levels of mathematics you struggle it's all relative I know uh like my professors you know over theories were rown topnotch math uh you know uh phds and whatnot and I kind of felt like this person all the time because I was learning such a sophisticated mathematics so just because you say Oh Mr YouTu Math Man you you know you might be you know Smart in math well listen I've been doing this stuff for a long time it takes time it takes effort and it takes practice so stick with it all right so with all that being said I definitely wish you all the best in your math Adventures thank you for your time and have a great day |
7932 | https://www.changjiangcai.com/mystudynotes/matrix-rank/ | Changjiang Cai's Blog
Hi, this is Changjiang. I'm actively posting my study notes. Those blogs are powered by Jekyll and GitHub.
Matrix Rank
Linear independence
In the theory of vector spaces, a set of vectors is said to be linearly independent if there exists no nontrivial linear combination of the vectors that equals the zero vector. If such a linear combination exists, then the vectors are said to be linearly dependent. These concepts are central to the definition of dimension.
A vector space can be of finite dimension or infinite dimension depending on the maximum number of linearly independent vectors. The definition of linear dependence and the ability to determine whether a subset of vectors in a vector space is linearly dependent are central to determining the dimension of a vector space.
Definition
Linearly dependent
A sequence of vectors ${ v_i }_{i=1}^{k}$ from a vector space $V$ is said to be linearly dependent, if there exist scalars $a_1, a_2, \dots, a_k$ not all zero, such that:
[a_{1}\mathbf {v} _{1}+a_{2}\mathbf {v} _{2}+\cdots +a_{k}\mathbf {v} _{k}=\mathbf {0}]
where $\mathbf {0}$ denotes the zero vector.
This implies that at least one of the scalars is nonzero, say $a_{1}\neq 0$, and the above equation is able to be written as
[{ \mathbf {v} _{1}={\frac {-a_{2}}{a_{1}}}\mathbf {v} _{2}+\cdots +{\frac {-a_{k}}{a_{1}}}\mathbf {v} _{k},} \text{ if } k>1.]
Or $\mathbf {v} _{1}=\mathbf {0}$ if $k=1$.
Thus, a set of vectors is linearly dependent if and only if one of them is zero or a linear combination of the others.
Linearly Independent
A sequence of vectors $\mathbf {v}{1},\mathbf {v}{2},\dots,\mathbf {v}_{n}$ is said to be linearly independent if it is not linearly dependent, that is, if the equation
[a_{1}\mathbf {v} _{1}+a_{2}\mathbf {v} _{2}+\cdots +a_{n}\mathbf {v} _{n}=\mathbf {0}]
can only be satisfied by ${\displaystyle a_{i}=0}$ for $i=1,\dots, n$.
This implies that no vector in the sequence can be represented as a linear combination of the remaining vectors in the sequence. In other words, a sequence of vectors is linearly independent if the only representation of $\mathbf {0}$ as a linear combination of its vectors is the trivial representation in which all the scalars $a_{i}$ are zero. Even more concisely, a sequence of vectors is linearly independent if and only if $\mathbf{0}$ can be represented as a linear combination of its vectors in a unique way.
If a sequence of vectors contains the same vector twice, it is necessarily dependent. The linear dependency of a sequence of vectors does not depend of the order of the terms in the sequence. This allows defining linear independence for a finite set of vectors: A finite set of vectors is linearly independent if the sequence obtained by ordering them is linearly independent. In other words, one has the following result that is often useful.
A sequence of vectors is linearly independent if and only if it does not contain the same vector twice and the set of its vectors is linearly independent.
Rank of matrix
In linear algebra, the rank of a matrix $A$ is the dimension of the vector space generated (or spanned) by its columns. This corresponds to the maximal number of linearly independent columns of $A$. This, in turn, is identical to the dimension of the vector space spanned by its rows. Rank is thus a measure of the "nondegenerateness" of the system of linear equations and linear transformation encoded by $A$. There are multiple equivalent definitions of rank. A matrixs rank is one of its most fundamental characteristics.
The rank is commonly denoted by rank(A) or rk(A); sometimes the parentheses are not written, as in rank A.
Matrix Rank Definition
In this section, we give some definitions of the rank of a matrix.
The column rank of A is the dimension of the column space of A, while the row rank of A is the dimension of the row space of A.
A fundamental result in linear algebra is that the column rank and the row rank are always equal. This number (i.e., the number of linearly independent rows or columns) is simply called the rank of A.
A matrix is said to have full rank if its rank equals the largest possible for a matrix of the same dimensions, which is the lesser of the number of rows and columns. I.e., given a matrix $A_{m \times n}$ (\text {rank}A \leq min(m, n))
If $\text {rank}A \leq min(m, n)$, then $A$ is called full rank matrix.
A matrix is said to be rank-deficient if it does not have full rank. The rank deficiency of a matrix is the difference between the lesser of the number of rows and columns, and the rank.
The rank of a linear map or operator $\Phi$ is defined as the dimension of its image:
(\operatorname {rank} (\Phi ):=\dim(\operatorname {img} (\Phi ))) where $\dim$ is the dimension of a vector space, and $\operatorname {img}$ is the image of a map.
Recall: In mathematics, for a function $f:X\to Y$, the image of an input value $x$ is the single output value produced by $f$ when passed $x$. The preimage of an output value $y$ is the set of input values that produce $y$.More generally, evaluating $f$ at each element of a given subset $A$ of its domain $X$ produces a set, called the image of $A$ under (or through) $f$. Similarly, the inverse image (or preimage) of a given subset $B$ of the codomain $Y$ is the set of all elements of $X$ that map to a member of $B$.
Examples
The matrix
[A={\begin{bmatrix}1&0&1\0&1&1\0&1&1\end{bmatrix}}]
has rank 2: the first two columns are linearly independent, so the rank is at least 2, but since the third is a linear combination of the first two (the first column plus the second), the three columns are linearly dependent so the rank must be less than 3.
The matrix
[A={\begin{bmatrix}1&1&0&2\-1&-1&0&-2\end{bmatrix}}]
has rank 1: there are nonzero columns, so the rank is positive, but any pair of columns is linearly dependent.
Similarly, the transpose
[A^{\mathrm {T} }={\begin{bmatrix}1&-1\1&-1\0&0\2&-2\end{bmatrix}}]
of $A$ has rank 1. Indeed, since the column vectors of $A$ are the row vectors of the transpose of $A$, the statement that the column rank of a matrix equals its row rank is equivalent to the statement that the rank of a matrix is equal to the rank of its transpose, i.e., $\operatorname {rank} (A) = \operatorname {rank} (A^T)$.
Written on December 26, 2024 |
7933 | https://www.khanacademy.org/math/cc-sixth-grade-math/x0267d782:cc-6th-rates-and-percentages/cc-6th-percentages/v/describing-the-meaning-of-percent | The meaning of percent (video) | Khan Academy
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6th grade math
Course: 6th grade math>Unit 3
Lesson 2: Intro to percents
The meaning of percent
Meaning of 109%
Intro to percents
Percents from fraction models
Percents from fraction models
Math>
6th grade math>
Rates and percentages>
Intro to percents
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The meaning of percent
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About About this video Transcript
In this math lesson, we learn about percentages by shading 20% of a square. The square is divided into 100 smaller squares in a 10x10 grid. To represent 20%, we shade 20 of these smaller squares, which can be done by coloring two rows of 10 squares each. This visual approach helps us understand the concept of percentages.Created by Sal Khan and Monterey Institute for Technology and Education.
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Partha Sarker 10 years ago Posted 10 years ago. Direct link to Partha Sarker's post “How will you know what pe...” more How will you know what percent a number will be if the number is something like 40 out of a number that isn't 100?
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Kimi 5 years ago Posted 5 years ago. Direct link to Kimi's post “what is 12 percent of 123” more what is 12 percent of 123
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woodhamlauralee 3 months ago Posted 3 months ago. Direct link to woodhamlauralee's post “We can represent 12% as 1...” more We can represent 12% as 12/100. I think, 12 parts of 100. So the question is asking, how many parts (n) of 123 is equal to 12 parts out of 100? This can be set up as 12/100 = n/123.
Solve for n.
n =12/100 × 123
n = 14.76
My answer 14.76 seems reasonable because 10% of 123 is 12.3 and 1% of 123 is 1.23. (12.3+1.23+1.23 =14.76)
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Jo21Mixels 8 years ago Posted 8 years ago. Direct link to Jo21Mixels's post “Is a percent in anyway re...” more Is a percent in anyway related to a ratio?
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louisaandgreta 3 years ago Posted 3 years ago. Direct link to louisaandgreta's post “Is 0.4% the same as sayin...” more Is 0.4% the same as saying 4 tenths of 1 percent?
And if so, doesn’t that algebraically translate to: (4/10) times (1/100)?
But that gives us 0.004
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201060008 2 years ago Posted 2 years ago. Direct link to 201060008's post “One square is 1% so 20 sq...” more One square is 1% so 20 square is 20%
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Zoey 2 years ago Posted 2 years ago. Direct link to Zoey's post “ok so he a larger square ...” more ok so he a larger square divide into 100 equal pieces but i dont get the divide part im confused on 2:45 and 2:48
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corgeek25 2 years ago Posted 2 years ago. Direct link to corgeek25's post “In 20%, 20 100ths are sha...” more In 20%, 20 100ths are shaded on a pictograph, right? So that would mean, ratios, rates, percentages, and fractions are interconnected?
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Video transcript
We're asked to shade 20% of the square below. Before doing that, let's just even think about what percent means. Let me just rewrite it. 20% is equal to-- I'm just writing it out as a word-- 20 percent, which literally means 20 per cent. And if you're familiar with the word century, you might already know that cent comes from the Latin for the word hundred. This literally means you can take cent, and that literally means 100. So this is the same thing as 20 per 100. If you want to shade 20%, that means, if you break up the square into 100 pieces, we want to shade 20 of them. 20 per 100. So how many squares have they drawn here? So if we go horizontally right here, we have one, two, three, four, five, six, seven, eight, nine, ten squares. If we go vertically, we have one, two, three, four, five, six, seven, eight, nine, ten. So this is a 10 by 10 square. So it has 100 squares here. Another way to say it is that this larger square-- I guess that's the square that they're talking about. This larger square is a broken up into 100 smaller squares, so it's already broken up into the 100. So if we want to shade 20% of that, we need to shade 20 of every 100 squares that it is broken into. So with this, we'll just literally shade in 20 squares. So let me just do one. So if I just do one square, just like that, I have just shaded 1 per 100 of the squares. 100 out of 100 would be the whole. I've shaded one of them. That one square by itself would be 1% of the entire square. If I were to shade another one, if I were to shade that and that, then those two combined, that's 2% of the entire square. It's literally 2 per 100, where 100 would be the entire square. If we wanted to do 20, we do one, two, three, four-- if we shade this entire row, that will be 10%, right? One, two, three, four, five, six, seven, eight, nine, ten. And we want to do 20, so that'll be one more row. So I can shade in this whole other row right here. And then I would have shaded in 20 of the 100 squares. Or another way of thinking about it, if you take this larger square, divide it into 100 equal pieces, I've shaded in 20 per 100, or 20%, of the entire larger square. Hopefully, that makes sense.
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Introduction to Test of Independence
2,464 Views01:21 min May 22, 2025
Overview
In statistics, the term independence means that one can directly obtain the probability of any event involving both variables by multiplying their individual probabilities. Tests of independence are chi-square tests involving the use of a contingency table of observed (data) values.
The test statistic for a test of independence is similar to that of a goodness-of-fit test:
where:
O = observed values
E = expected values (which should be at least 5)
A test of independence determines whether two factors are independent or not. The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit.
The number of degrees of freedom for the test of independence is:
The following formula calculates the expected number (E):
This text is adapted fromOpenstax, Introductory Statistics, Section 11.3 Test of Independence
Transcript
A test of independence determines whether a contingency table's two variables are independent.
In this case, independence means that the probability of any event involving both variables can be directly obtained by multiplying their individual probabilities.
For example, to understand the relationship between alcohol consumption and road accident fatality, arrange the data in a two-by-two contingency table. The rows represent the subjects'sobriety or intoxication, while the columns represent the fatality or nonfatality of road accidents.
Data from randomly selected samples represent the observed frequencies arranged in the two-way table.
Here,E represents the expected frequency,r indicates the number of rows, and c indicates the number of columns. The expected frequency for each cell must be atleast 5.
The chi-square test statistic is calculated using these expected and observed frequencies. The critical value and P-values are calculated using suitable degrees of freedom from the chi-square table or software.
Finally, a hypothesis test is performed to determine whether alcohol consumption and road accident fatality are independent events.
Explore More Videos
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Can we apply p V γ p V γ=const only for quasistatic adiabatic process?
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But If this is so, I have often seen people applying this formula for quick processes where no heat exchange is possible.
Here is an example from where I am quoting the following:
we will assume this happens quickly enough that no heat can enter or leave the gas [...]
It may be to make Δ Q Δ Q=0, But does not that "quick" word limit the applicability of the formula
p V γ=const?p V γ=const?
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edited Jun 27, 2014 at 17:33
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"Quick" and "slow" always has to be compared to something. It is perhaps a shortcoming of most books on thermodynamics that they do not explicitly state what that something is, though perhaps it is also because it's a bit tricky to explain.
The scenario that one always imagines is a gas in a compartment in a piston. A realistic system of this sort will have an obvious time scale, on which heat will leave or enter due to imperfect insulation. For the adiabatic equation to apply, the change in volume will have to be done "quickly" relative to that speed.
There is however another time scale which occurs due to the gas having constituents and therefore having internal dynamics, such as eddies, or local density variations. Imagine pulling the piston out so fast that the wall moves significantly faster than the speed of sound in the gas --- you will create a vacuum which then causes a shockwave to be set up and some time will elapse before the system settles back into equilibrium. This will occur roughly on the order of size of box divided by speed of sound. For the "quasi-static" limit to apply, we have to move the piston slower than this speed.
For this system, hopefully you can see that if it is close to equilibrium, then the state of the system can be expressed as pressure and volume, and you get to ignore all the internal dynamics of the gas.
All applications of statistical mechanics and thermodynamics contain either explicitly or implicitly such a coarse graining of microscopic information, and therefore something of a floor on what the fastest macroscopic transition can be. A lot of confusion, e.g. over entropy, tends to occur when people forget this and start believing that coarse grained descriptions are somehow complete and exact.
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answered Jun 28, 2014 at 8:22
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You are correct. In a real process, one would modify this formula to include the so-called polytropic exponent n n such that P V n=c o n s t P V n=c o n s t. This reflect that the process is not perfectly isentropic. For a fixed final volume this means the final temperature and pressure will be higher than in the ideal case, and more work need to be expended.
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answered Jul 30, 2014 at 8:18
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Intro
Ideal types of thermodynamic processes are quasistatic and reversible and they are imprinted as constant graphic functions.
!(
Quasi static adiabatic processes are the ideal types of adiabatic processes. An ideal or fictive quasi-static adiabatic transfer of energy as work that occurs without friction or viscous dissipation within the system is said to be is-entropic, with ΔS = 0 ,as well as reversible. Nevertheless a natural adiabatic process is irreversible and is not is-entropic.
Bottom line Poisson's Law works either the process is slow or quick as long as there is no heat transfer.
Fun-Fact
Poisson's law is actually utilized in order to understand the changes in temperature as air rises up through the atmosphere or subsides downward under the conditions where there is no heat gain, e.g from solar radiation, or heat loss, e.g. from terrestrial radiation.
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answered Jun 27, 2014 at 20:28
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7936 | https://brainly.com/question/48597349 | [FREE] The measure of one angle of a rhombus is 60 degrees, and the perimeter is 24 inches. 1. Sketch the - brainly.com
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The measure of one angle of a rhombus is 60 degrees, and the perimeter is 24 inches.
Sketch the rhombus.
Determine the side lengths.
Find the area.
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The rhombus with an angle of 60 degrees has four equal side lengths of 6 inches and an area of 36∙√3÷ 2 square inches. The perimeter of the rhombus is 24 inches.
Explanation
The measure of one angle of a rhombus is 60 degrees, which implies that the other acute angle is also 60 degrees because the angles occur in pairs. The obtuse angles will be 120 degrees. Since a rhombus has all sides equal, and the perimeter is given as 24 inches, each side of the rhombus will be 24 inches ÷ 4, which is 6 inches long.
Next, we can find the area of the rhombus. We can draw the rhombus with one of the 60-degree angles at the bottom. To find the area, we can divide the rhombus into two congruent equilateral triangles by drawing one of its diagonals.
The formula for the area of an equilateral triangle with side length 'a' is (a²∙√3÷ 4). Since the side length 'a' is 6 inches; the area of one triangle is (6²∙√3÷ 4). To find the entire area of the rhombus, we multiply the area of one triangle by 2: Area = 2× (6²∙√3÷ 4) = 36∙√3÷ 2 square inches.
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The rhombus has four sides of 6 inches each and an area of 18√3 square inches. The perimeter, being 24 inches, confirms the equal side lengths. The area was calculated using the diagonals derived from the interior angles of the rhombus.
Explanation
Sketching the Rhombus:
To sketch a rhombus with one angle measuring 60 degrees, start by drawing one angle of 60 degrees at a corner. Each rhombus has two pairs of equal angles. Thus, the opposite angle will also be 60 degrees, while the other two angles will be 120 degrees each.
Determining the Side Lengths:
A rhombus has four equal sides. Given the perimeter is 24 inches, we can find the length of one side by dividing the perimeter by 4:
Side length=4 Perimeter=4 24=6 inches
Finding the Area:
To find the area of the rhombus, we can use the formula for the area of a rhombus:
Area=2 d 1×d 2
where d 1 and d 2 are the lengths of the diagonals. In our rhombus, we can find the diagonals using some trigonometry.
Create two right triangles from the rhombus by drawing a diagonal. One triangle will have legs formed by half of one diagonal and a side of the rhombus. The angles of the rhombus provide that each triangle has angles of 60 degrees, 30 degrees, and 90 degrees.
Using the sin function for the triangle,
Side length=6=2 sin(60)d 1
so the length of half of diagonal 1 d 1 can be calculated as:
d 1=6×2×sin(60)=6×2×2 3=6 3
Now use cosine to find diagonal 2, which is perpendicular to diagonal 1:
d 2=6×2×cos(60)=6×2×2 1=6
The area will then be:
Area=2 d 1×d 2=2(6 3×6)=18 3square inches
Examples & Evidence
For example, if you have a rhombus with equal sides measuring 4 inches, the perimeter would be 16 inches, and you could also use the diagonal method to determine the area just like we did with 6 inches per side. Understanding how to apply the properties of angles and side lengths of a rhombus helps in visualizing and calculating other geometric shapes as well.
The calculation of perimeter by multiplying the side length by 4 and the area calculation from the diagonals follows standard geometric formulas and properties relating to rhombuses.
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Community Answer 4.8 7 Given rhombus TURN , with angle T measuring 60 degrees and the side length equal to 10 cm, calculate the area of the rhombus. Leave your answer in simplest radical form.
Community Answer In rhombus PQRS.If m angle QRS=60 and QS=15, find the perimeter of the rhombus
Community Answer A rhombus has a diagonals of 6 inches and 8 inches. what is the perimeter and the area of the rhombus?
Community Answer 4.8 65 A quilt piece is designed with four congruent triangles to form a rhombus so that one of the diagonals is equal to the side length of the rhombus. A rhombus with diagonals is shown. The diagonals form 4 triangles. All sides are 4 inches long. The distance from the top point to the middle point is 2 inches. The distance from the right point to the middle point is x. One interior angle is 30 degrees and another is a degrees. Which measures are true for the quilt piece? Select three options. a = 60° x = 3 in. The perimeter of the rhombus is 16 inches. The measure of the greater interior angle of the rhombus is 90°. The length of the longer diagonal is approximately 7 inches.
Community Answer 4.5 12 if a rhombus has diagonals of length 20 inches and 48 inches, what is the length of its perimeter? show the work you used to find your answer.
Community Answer "ABCD is a rhombus with perimeter 52 m. The length of diagonal AC is 24 m. What is the area of rhombus ABCD? m2"
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Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
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I am looking for a faster way to solve this problem:
Let's suppose we have n boxes and n marbles (each of them has a different kind). Every box can contain only some kinds of marbles (It is shown it the example below), and only one marble fits inside one box. Please read the edits. The whole algorithm has been described in the post linked below but it was not precisely described, so I am asking for a reexplenation.
The question is: In how many ways can I put marbles inside the boxes in polynomial time?
Example:
```
n=3
Marbles: 2,5,3
Restrictions of the i-th box (i-th box can only contain those marbles): {5,2},{3,5,2},{3,2}
The answer: 3, because the possible positions of the marbles are: {5,2,3},{5,3,2},{2,5,3}
```
I have a solution which works in O(2^n), but it is too slow. There are also one limitation about the boxes tolerance, which I don't find very important, but I will write them also. Each box has it's own kind-restrictions, but there is one list of kinds which is accepted by all of them (in the example above this widely accepted kind is 2).
Edit: I have just found this question but I am not sure if it works in my case, and the dynamic solution is not well described. Could somebody clarify this? This question was answered 4 years ago, so I won't ask it there.
Edit#2: I also have to mention that excluding widely-accepted list the maximum size of the acceptance list of a box has 0 1 or 2 elements.
Edit#3: This question refers to my previos question(Allowed permutations of numbers 1 to N), which I found too general. I am attaching this link because there is also one more important information - the distance between boxes in which a marble can be put isn't higher than 2.
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edited Oct 31, 2021 at 18:19
Chate
asked Oct 30, 2021 at 18:02
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Some things are not clear to me. The condition "only one marble fits inside one box" how is honoured if in your example marbel "5" can fit in boxes 1 and 2? If box 3 only allows marbels 3 and 2 does its order matters?
– Ripi2
Commented
Oct 30, 2021 at 18:28
@Ripi2 this condition just says that you can't put 2 marbles in one box. These box restricions just say which of the marbles you can put on this place. I don't understand the second part of your question. What order do you mean?
– Chate
Commented
Oct 30, 2021 at 18:32
Putting each of n different marbels in each of n boxes has n! solutions. The restrictions will low this number, say to (n-k)! for some k. But its complexity is still factorial and not polynomial.
– Ripi2
Commented
Oct 30, 2021 at 19:06
@Ripi2 Yes so the problem is to find this k, because I just need the number, not all the combinations. I have also found out an expression which let's me solve this, but it sometimes skips some cases
– Chate
Commented
Oct 30, 2021 at 19:12
1
The linked thread is hard to piece together. However, given the information in your 2nd edit, I believe this can be solved in polynomial time with the FKT algorithm. If you choose the 'all accepted' marble and some box B, and remove them from the graph, the remaining graph doesn't contain K_3,3 as a subgraph, so the FKT algorithm should work on it to count perfect matchings, repeating this process for each choice of initial box B.
– kcsquared
Commented
Oct 30, 2021 at 23:15
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As noted in the comments, is this problem, with links to papers on how to tackle it, and counting the number of matchings is #P-complete. I would recommend finding those papers.
As for dynamic programming, simply write a recursive solution and then memoize it. (That's top down, and is almost always the easier approach.) For the stack exchange problem with a fixed (and fairly small) number of boxes, that approach is manageable. Unfortunately in your variation with a large number of boxes, the naive recursive version looks something like this (untested, probably buggy):
```
def solve (balls, box_rules):
ball_can_go_in = {}
for ball in balls:
ball_can_go_in[ball] = set()
for i in range(len(box_rules)):
for ball in box_rules[i]:
ball_can_go_in[ball].add(i)
def recursive_attempt (n, used_boxes):
if n = len(balls):
return 1
else:
answer = 0
for box in ball_can_go_in[balls[n]]:
if box not in used_boxes:
used_boxes.add(box)
answer += recursive_attempt(n+1, used_boxes)
used_boxes.remove(box)
return answer
return recursive_attempt(0, set())
```
In order to memoize it you have to construct new sets, maybe use bit strings, BUT you're going to find that you're calling it with subsets of n things. There are an exponential number of them. Unfortunately this will take exponential time AND use exponential memory.
If you replace the memoizing layer with an LRU cache, you can control how much memory it uses and probably still get some win from the memoizing. But ultimately you will still use exponential or worse time.
If you go that route, one practical tip is sort the balls by how many boxes they go in. You want to start with the fewest possible choices. Since this is trying to reduce exponential complexity, it is worth quite a bit of work on this sorting step. So I'd first pick the ball that goes in the fewest boxes. Then I'd next pick the ball that goes in the fewest new boxes, and break ties by fewest overall. The third ball will be fewest new boxes, break ties by fewest boxes not used by the first, break ties by fewest boxes. And so on.
The idea is to generate and discover forced choices and conflicts as early as possible. In fact this is so important that it is worth a search at every step to try to discover and record forced choices and conflicts that are already visible. It feels counterintuitive, but it really does make a difference.
But if you do all of this, the dynamic programming approach that was just fine for 5 boxes will become faster, but you'll still only be able to handle slightly larger problems than a naive solution. So go look at the research for better ideas than this dynamic programming approach.
(Incidentally the inclusion-exclusion approach has a term for every subset, so it also will blow up exponentially.)
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answered Nov 1, 2021 at 18:29
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Real algebraic sets bounded away from integer points
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A subset S S of R n R n is "bounded away from integer points" if for some positive ϵ ϵ every point in S S lies at a distance of at least ϵ ϵ from Z n Z n. For example the line x+y=1/2 x+y=1/2 in R 2 R 2 is bounded away from integer points, but the curve x 2+y=1/2 x 2+y=1/2 is not, because the points (n+1 4 n,−n 2−1 16 n 2)(n+1 4 n,−n 2−1 16 n 2) for n=1,2,…n=1,2,… lie on this curve.
Question: Can anyone give an algorithm to determine whether a system of polynomial equations with real algebraic coefficients cuts out a subset S S of R n R n that is bounded away from integer points? Is there a simple description of all such subsets S S?
Remark: I have a vague notion that if S S is bounded away from integer points then this must be ``trivially'' verifiable, perhaps because S S projects on a linear affine subset of R n R n that is obviously bounded away from integer points, but this is little more than guesswork. I don't actually know that the problem is decidable, but I would be surprised if it were not.
UPDATE: (I'll use this section to collect my latest thoughts on the problem.)
The situation is fairly transparent in R 2 R 2, and the real problem is how things generalize to higher dimensions. Let S S be as above. Let ⌊⋅⌋⌊⋅⌋ be the floor function, which will be applied to points coordinatewise. Then I propose the following conjecture:
There exists some translate of S S bounded away from integer points if and only if the set of all points ⌊p⌋⌊p⌋ for p∈S p∈S is contained in a finite union of linear-affine subspaces of R n R n (which will be defined over the rationals).
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edited Feb 1, 2011 at 9:26
Sidney RafferSidney Raffer
asked Jan 28, 2011 at 12:21
Sidney RafferSidney Raffer
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2 There are examples other than hyperplanes: consider the hyperbola in R 2 R 2 whose asymptotes are lines x=1/2 x=1/2 and y=1/2 y=1/2.Boris Bukh –Boris Bukh 2011-01-28 13:43:43 +00:00 Commented Jan 28, 2011 at 13:43
Interesting question, but I'm not sure I share your optimism. Is there any reason why this problem should be considered over the reals (versus, say, bounded away from the Gaussian integers over the complexes)?Thierry Zell –Thierry Zell 2011-01-28 16:09:24 +00:00 Commented Jan 28, 2011 at 16:09
@Thierry: I suspect that the problems is decidable only because I can't find any hard-to-verify examples of sets bounded away from integer points. For example in the plane we have certain lines with rational slope, and examples like Boris mentioned in his comment, such as (x−1/2)(y−1/2)=1(x−1/2)(y−1/2)=1, where the branches at infinity are asymptotic to certain lines with rational slope. I don't know any more subtle examples. Does the problem get harder in higher dimensions? I don't know. As for the Gaussians and the complexes, maybe this boils down to the same problem... I'll have to think about that.Sidney Raffer –Sidney Raffer 2011-01-28 16:42:39 +00:00 Commented Jan 28, 2011 at 16:42
A trivial but rather large set of examples that I think has gone unmentioned would be bounded sets, e.g., that determined by x 2+y 2=3 x 2+y 2=3. Gerry Myerson –Gerry Myerson 2011-01-29 05:28:05 +00:00 Commented Jan 29, 2011 at 5:28
@Gerry: Yes, and by quantifier elimination over real closed fields, we can recognize all such examples algorithmically. So only unbounded sets pose a problem Sidney Raffer –Sidney Raffer 2011-01-29 06:08:49 +00:00 Commented Jan 29, 2011 at 6:08
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This is not a really an answer, but you might want to look at the papers
MR1852803 (2002e:11085) Vâjâitu, Marian(R-AOS); Zaharescu, Alexandru(R-AOS) Integer points near hyperelliptic curves. (French summary) C. R. Math. Acad. Sci. Soc. R. Can. 23 (2001), no. 3, 84–90. 11J25 (11D75 11G30)
and
MR1689552 (2001a:11117) Huxley, M. N.(4-WALC-SM) The integer points close to a curve. III. Number theory in progress, Vol. 2 (Zakopane-Kościelisko, 1997), 911–940, de Gruyter, Berlin, 1999. 11J54 (11P21)
The latter seems particularly relevant to your specific question.
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answered Jan 28, 2011 at 17:14
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Thanks Igor. If you have a link to Huxley's paper I could sure use it.... I have no way around the paywalls at present.Sidney Raffer –Sidney Raffer 2011-01-28 17:38:57 +00:00 Commented Jan 28, 2011 at 17:38
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Find the critical points of $f(x) = |x^2 - 2x|$
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Find the critical points of $f(x) = |x^2 - 2x|$
The only way I was able to evaluate this was to draw the graph of $y = x^2 - 2x$ and when the parabola was going under the $x$ axis I mirrored it above the $x$ axis. I observed that the critical points were $x$ = 0, 1, 2.
How do I evaluate this function properly using maybe left and right hand side limits or some kind of calculus/differentiation? If I was going to justify my critical points beyond "because they are on my graph". Perhaps I can justify $0$ and $2$ by factorizing $f(x)$ to $|x(x - 2)|$, but how about $x = 1$?
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Generally speaking, you want to write absolute values of functions piecewise to get your hands dirty and solve for critical points. This means we have to find the sign of the function between zeros and apply the definition of absolute value.
As you noted, there are zeros at 0 and 2; these are the only two zeros of $x^2-2x$. We can check signs between and outside these points to find that
$x^2-2x>0$ if $x>2$ or $x<0$
$x^2-2x<0$ if $0
Therefore our function $f$ may be written
$$f(x)=\begin{cases}x^2-2x & \text{if } x\leq 0 \ 2x-x^2 & \text{if } 0\leq x\leq 2 \ x^2-2x & \text{if } 2\leq x \end{cases}$$
To find critical points, we must find points where the derivative does not exist or equals zero. In this case, it suffices to differentiate each piece; if the derivatives do not agree at the endpoints, then the derivative does not exist at the endpoint. $$f'(x)=\begin{cases}2x-2 & \text{if } x< 0 \ \text{undefined} & \text{if } x= 0 \ 2-2x & \text{if } 0< x\leq 2 \ \text{undefined} & \text{if } x= 2 \ 2x-2 & \text{if } 2\leq x \end{cases}$$
So we see that the derivative is undefined at $0$ and $2$. It is defined everywhere else, but it is easy to check the only point where the derivative is zero is when $x=1$.
(By the way, the comparison of the derivatives to the left and right of $0$, $2$ is the same as evaluating the left-hand derivative and right-hand derivative).
(EDIT: The definition of critical point may vary in what cases we include, but many standard calculus texts include non-existence in the definition for optimization.)
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answered Apr 30, 2012 at 13:24
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Writing $x^2-2\,x=x(x-2)$, we see that $x^2-2\,x\ge0$ if $x\ge2$ or $x\le0$, and $x^2-2\,x<0$ if $00$ if $x\ne0,2$.
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answered Apr 30, 2012 at 13:22
Julián AguirreJulián Aguirre
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$\begingroup$ From all the books I have read, critical points are those points where the original function is defined and the derivative is 0 OR the derivative is undefined. $\endgroup$
GeoffDS
– GeoffDS
2012-04-30 13:44:39 +00:00
Commented Apr 30, 2012 at 13:44
$\begingroup$ Perhaps the distinction is in "calculus" and "real analysis". For example, the calculus books by Larson, Hostetler and Edwards and also Edwards and Penney, and also Varberg, Purcell, and Rigdon all include where the derivative is undefined (as long as the original function is defined). $\endgroup$
GeoffDS
– GeoffDS
2012-04-30 13:58:49 +00:00
Commented Apr 30, 2012 at 13:58
$\begingroup$ @Graphth I understand that from the point of view of teaching calculus it might be desirable to define a critical point as one where the derivative vanishes or exits and equals 0. But I am more at ease with the "real analysis" definition. $\endgroup$
Julián Aguirre
– Julián Aguirre
2012-04-30 15:01:32 +00:00
Commented Apr 30, 2012 at 15:01
$\begingroup$ Sure, that's fine, but telling someone about the "usual" definition of critical point when the one they usually see is different, at best won't confuse and will do no good, and at worst will confuse. If they take real analysis, they'll learn the "usual" definition. For now, their definition is not the "usual". $\endgroup$
GeoffDS
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2012-05-01 15:39:03 +00:00
Commented May 1, 2012 at 15:39
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This solution is different than the others shown. First, consider the simple function $f(x) = |x|$. We want to know $f'(x)$. That's actually pretty simple to do, since $f'(x)$ gives the slope of the tangent line. That is -1 for $x < 0$ and 1 for $x > 0$. At $x = 0$, there is a sharp point in $f(x)$ so $f'(x)$ does not exist. Or, another way to see it, the left hand limit of the difference quotient is -1 and the right hand limit is +1, so the limit itself (the derivative) does not exist. Therefore, a very nice shorthand way of writing the derivative is
$$f'(x) = \frac{x}{|x|}.$$
This is -1 when $x$ is negative, undefined when $x$ is 0, and 1 when $x$ is positive, just as I described above. When you have a more complicated function inside the absolute values, just use the chain rule.
If $f(x) = |x^2 - 2x|$, then $$f'(x) = \frac{x^2 - 2x}{|x^2 - 2x|} \cdot \frac{d}{dx} (x^2 - 2x) = \frac{x^2 - 2x}{|x^2 - 2x|} (2x - 2) = \frac{2x(x - 1)(x - 2)}{|x^2 - 2x|}.$$
Now, critical points are those where the derivative is 0 or undefined, and the original function is defined. Since the function $f(x)$ is defined for all $x$, find where the derivative is 0 or undefined. Just find the places where the numerator is 0 or the denominator is 0. That's pretty simple here, $x = 0, 1, 2$.
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answered Apr 30, 2012 at 13:51
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The complexes [C o(N H 3)5 B r]S O 4 and [C o(N H 3)5 S O 4]Br are the examples of ionization isomers.
Ionization isomerism: This trait, known as ionisation isomerism, refers to the compound's ability to produce distinct ions in the solution despite having the same chemical makeup.
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this it's just typing over here a function but with parameters for example you might be teaching turning-point form of quadratic equations or parabola and you type y equals a bracket notice it starts to graph what you've got X minus B bracket the caret symbol for squared error cross to get down plus C notice that's suggesting it adds sliders for those three parameters you can click individually I just click all and there we go let's set them all at one so pretty easily honest change a we increase it you see it affects the width of the parabola but it goes negative and so forth so that's very powerful they the same horizontal shift and when it's four that's where the turning point is the x-coordinate see the up or down you can say say is two in the area you can also plot points just by typing a bracket bracket excuse my shift bracket and the turning point should be B comma C press ENTER and then there's a turning point over here you can see the turning point if I drag that around you can see that the slide is then chose if I drag it to minus 2 comma minus 4 little pretty close you can see that's where the sliders are as well you can also adjust the sliders busted increments by clicking on one of the extremities and saying look I only want it to go from negative 5 to 5 whatever you click away in by dragging it it will go for nearly 5 so I didn't alter the step so I'll click on my extremity again the step might want to go in one and click away and they are it'll just go in ones you can do the same for the other sliders another function that's a little bit counterintuitive but it sort of makes sense when you see it is how to restrict the domain basically it if to multiply the whole four let's put brackets around the entire function notice put in the right-hand bracket for me I don't this multiply it by curly bracket standard with the model flight symbol in say 1 is less than X which is less than 3 in the curly all curly brackets put in at the right you can see there it is let's restrict the domain from 1 to 3 if you want to restrict the while I don't mind it's the same sort of thing and go peps negative 1 it's nicer than Y just less than that shall make it a small one negative 2 to negative 1 and scroll down here clicking you can see restricted the Y domain I'll have to take a break so sorry about that you'll notice a few things have changed in the meantime I'll widen this window here just by clicking and dragging so I can see more of the equations and over adjusted the slide is simply changing increments of 1 and I also remove the brackets here I didn't need brackets around the entire equation when I put in these domain and range constraints so that's something else I just learned also I can put a label in with this point and it labels the coordinates there now if I'm going to hide that graph by clicking the color off and this point and show you a tree graph the same sliders second alter the amplitude I can alter the period and it will to see where it starts from I can move over here and use the mouse wheel to scroll to zoom in or click and drag to change the coordinate system origin placement there's quite a lot you can do with desmos I think I'm writing this one here and let's make some others for example how to generate tables of values in another video in the meantime thanks for watching |
7943 | https://flexbooks.ck12.org/cbook/ck-12-interactive-middle-school-math-6-for-ccss/section/6.2/primary/lesson/using-exponents-msm6-ccss/ | Skip to content
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6.2
Using Exponents
Written by:Katie Sinclear
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Lesson
How many bubbles can you make?
Use the interactive below to make patterns with bubbles. How can you figure out the numbers of bubbles without counting them all individually?
In this lesson, you will learn how to use exponents and make exponential patterns.
INTERACTIVE
How Many Bubbles Can You Make?
How are the numbers related to the number of bubbles on the screen?
Drag the sliders to change the number of bubbles.
Your device seems to be offline.
Please check your internet connection and try again.
+
Do you want to reset the PLIX?
Yes
No
Discussion Question
How can you figure out the numbers of bubbles without counting them all individually?
How many branches are there on the fractal tree?
Fractals are mathematical patterns that can make beautiful artwork. Try building your own fractal tree in the interactive below.
You can learn more about Fractals from this CK-12 lesson: Self-Similarity and Fractals.
INTERACTIVE
How Many Branches Are on the Fractal Tree?
Drag the slider to see the number of branches added to the tree.
Fill in the table with the correct number of branches added on each step.
Your device seems to be offline.
Please check your internet connection and try again.
+
Do you want to reset the PLIX?
Yes
No
INTERACTIVE
Exponent Table
Fill in the blanks to complete the exponent table.
Make sure to follow the correct format.
Your device seems to be offline.
Please check your internet connection and try again.
+
Do you want to reset the PLIX?
Yes
No
How can you shorten repeated multiplication?
As the number of steps gets bigger, the number of times you have to multiply a number gets to be too much. You can show repeated multiplication in a shorter way with exponents.
Try making expressions with exponents in the interactive below.
Can you fit the teddy bear in the gift box?
You can find the volume of a box by multiplying the length times the width times the height of the box. When a box is a cube, each of the dimensions of the box (length, width, and height) are the same.
In the interactive below, try to find a box size that would snuggly fit a teddy bear. Check out all of the ways that volume of the box can be expressed.
INTERACTIVE
Evaluate Numerical Expressions with Exponents
Resize the box so that it does not squish the bear.
How can you express the volume of the box?
Your device seems to be offline.
Please check your internet connection and try again.
+
Do you want to reset the PLIX?
Yes
No
Remember that an inch is a measure of length and in2 is a measure of area. There are other units to measure volume including cm3 and ft3.
CK-12 PLIX Interactive: Teddy Bear Box
In this PLIX, you again resize a box for a teddy bear and you will answer 5 questions about exponential questions. The different forms of the exponential expression do not show up in the interactive. The last question asks you to think up an example of a real life application for exponential expressions.
| |
| Summary |
| Exponents represent repeated multiplication. For example, in + 7 is the base and 3 is the exponent. + This can be read as “seven to the third” |
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Overview
| |
| + Exponents represent repeated division. For example, in @$7^3=7\cdot7\cdot7@$ - 7 is the base and 3 is the exponent. - This can be read as “seven to the third” |
Test Your Knowledge
Question 1
What is the result of @$\begin{align}p^{13} \div p^{8}\end{align}@$ for a non-zero rational number @$\begin{align}p\end{align}@$?
a@$\begin{align}p^{5}\end{align}@$
b@$\begin{align}p^{21}\end{align}@$
c@$\begin{align}p^{-5}\end{align}@$
d@$\begin{align}p^{3}\end{align}@$
When you divide two powers with the same base, you subtract the exponents. So, @$\begin{align}p^{13} \div p^{8} = p^{13-8} = p^{5}.\end{align}@$ Therefore, the correct answer is @$\begin{align}p^{5}.\end{align}@$
Question 2
What is the value of @$\begin{align}x^{4} \div y^{4}\end{align}@$ for any two non-zero rational numbers @$\begin{align}x\end{align}@$ and @$\begin{align}y\end{align}@$?
a@$\begin{align}(x \div y)^{2}\end{align}@$
b@$\begin{align}(x \div y)^{1}\end{align}@$
c@$\begin{align}(x \div y)^{4}\end{align}@$
d@$\begin{align}(x \div y)^{8}\end{align}@$
When you divide two numbers with the same exponent, you divide the bases and keep the exponent the same. Here, both exponents are 4, so the result is @$\begin{align}(x \div y)^{4}.\end{align}@$
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Related Content
Write Repeated Multiplication in Exponential Form
Memory Matters
Teddy Bear Box
Whole Number Exponents: Building Blocks
Values Written as Powers: Arranging Circles
Whole Number Exponents
| Image | Reference | Attributions |
---
| | | Credit: CK-12 Foundation Source: CK-12 Foundation License: CC BY-NC 3.0 |
| | | Credit: CK-12 Foundation Source: CK-12 Foundation License: CC BY-NC 3.0 |
| | | Credit: CK-12 Foundation Source: CK-12 Foundation License: CC BY-NC 3.0 |
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7944 | https://www.robhammerphotography.com/blog/tag/basketball+court | basketball court — Photographer based in Denver, Colorado
American Backcourts currently exhibiting at Western Spirit - Scottsdale’s Museum of the West
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Basketball in Vietnam
Rob HammerJune 16, 2023
Photographs of basketball courts in Vietnam
Vietnamese Basketball Culture
When you think about the humble beginning of basketball by Dr. Naismith back in 1891, it’s mind blowing to see how far the game has come. While there have been monumental economic changes with shoe contracts and television coverage, the basics of the game are still unchanged. A hoop is still a hoop. But the sport, an American original, has influenced people all over the world. After ten years of working on this project, there’s one line that seems to reoccur - “basketball is everywhere”. Truly everywhere. Even Vietnam. Which is particularly fascinating when you stop to consider the tumultuous relationship we once shared with them.
Click here to see more of my basketball hoop photography
Contact me directly about prints of my basketball photography - rob@robhammerphotography.com
Vietnamese basketball hoop
Basketball hoop in Ha Noi, Vietnam
Hoi An, Vietnam basketball hoop
Basketball court in Vietnam
Colorful basketball court at a school in Vietnam
Black and white photograph of a Vietnamese basketball hoop
Photograph of an old basketball hoop at a school in a small Vietnamese town
InBasketball, Documentary PhotographyTagsVietnam, Basketball, basketball hoop, basketball court, travel, Hanoi, Hoi An, basketball never stops, basketball is everywhere
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American Basketball Culture
Rob HammerMay 28, 2022
Basketball Hoop Photography - American Sports Culture
10 years into this series and it’s still just as much fun documenting the sport of basketball as it was initially. It’s always interesting to think about the games played on hoops in different parts of the country. It’s also enjoyable to see the images and realize that each one was an experience in itself to make. The first photo here in Primm was taken on a day so windy that I had to brace myself with one leg five feet in front of the other. You can see how the net is being pushed backwards. The second shot is from a high school gym in the middle of a remodel. Door was wide open and not a sole in sight. The hoop in Santa Rosa is actually one I photographed 10 or so years ago under completely different conditions. That image from all those years ago is in the book. Crazy how a location so random can be unintentionally revisited. And shocking to see that there is still a chain net hanging from the rim. The last image was made on a road I’ve driven a hundred times and never noticed before.
Click here to grab a copy of the book
Primm, Nevada
Minden, Nevada
Santa Rosa, New Mexico
Olancha, California
InDocumentary Photography, BasketballTagsbasketball, sport, hoop dreams, ball is life, basketball never stops, basketball court, basketball hoop, American Backcourts, photo book, documentary photography, Sports Culture, Basketball culture
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European Basketball Hoop Photos
Rob HammerApril 30, 2022
Basketball Hoop Photography - Europe
Over the past few years, I’ve been documenting basketball hoops throughout Europe—not the polished ones inside arenas, but the forgotten, improvised, and well-worn hoops you find in alleyways, apartment courtyards, village parks, and industrial edges of cities. These photos show the global reach of the game in its most local forms.
Each image carries the texture of its surroundings—cobblestone streets, Soviet-era apartment blocks, medieval stone walls, graffiti-tagged fences. The contrast between architecture and asphalt makes these hoops feel different from their American counterparts but no less soulful. They reflect how basketball adapts to the space it’s given, whether in a quiet plaza or behind an old church.
This series is available for editorial and commercial licensing, and it’s a great fit for brands or publications looking to tap into global basketball culture in an authentic, visually arresting way. Whether you're working on a story about international basketball, designing a campaign around street culture, or sourcing imagery for a global brand, these photographs offer a fresh take on the game—and its reach far beyond the NBA.
Interested in licensing these photos?
Please reach out for pricing and usage details. High-resolution files are available, and custom image sets can be curated based on your project’s needs - rob@robhammerphotography.com
Click here to see more basketball imagery from my American Backcourts series.
Photograph of a basketball hoop with a plastic net in Denmark
Photograph of a European style metal grate basketball hoop
Gritty basketball court in Europe
Black and white photograph of a European basketball hoop
Basketball hoop in a rural part of the Netherlands
Moss growing on a basketball court in Europe
Basketball court in Copenhagen
City basketball court in Denmark
Black and white photo of a basketball hoop in a unique European location
Copenhagen basketball court
Basketball court near a European castle
Basketball hoop in Sweden
Basketball court at a church in Europe
InDocumentary Photography, Basketball, Street PhotographyTagsbasketball, european basketball, sport, euro ball, basketball hoop, travel, basketball court, Denmark, Germany, Sweden, Netherlands, fine art photography, documentary photography
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Culture Brewing Company - Encinitas
Rob HammerMay 25, 2021
Feels so good to see the world opening back up. Over the last week especially there has been so much life out on the street and in the local shops, restaurants, and bars. Grateful to have the opportunity to be social again while sharing some work from American Backcourts. So if you’re in San Diego during the month of June, stop on by Culture Brewing Company in Encinitas to check out some fine art prints while enjoying some delicious craft beer in the sunshine.
InFine art photography, BasketballTagsAmerican Backcourts, Basketball, Culture Brewing, Encinitas, San Diego, California, hoop dreams, wall art, photo book, ball is life, Basketball never stops, basketball court, America, sport, fine art prints
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Basketball Never Stops
Rob HammerApril 12, 2021
Basketball during Covid-19
It’s hard getting people to stop playing basketball. They will find a way. These image are from a court in south east San Diego during Covid that were shut down for obvious reasons. Still though, people wanted to ball and wouldn’t let anything stop them. Hard to get angry at that. They climbed the fences, went through the holes, whatever it took. So the city had to take further measures. No doubt they were only temporary solutions. Basketball never stops.
Click here to pick up a copy of American Backcourts
Colina Del Sol - San Diego
Basketball hoop with a chained padlock on it - San Diego, CA
Tagsbasketball, basketball never stops, American Backcourts, Project Backboard, Hoop Dreams, Ball is Life, Sport, NBA, basketball court, basketball hoop
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AMERICAN BACKCOURTS - FINE ART PRINTS
Rob HammerFebruary 28, 2021
There is no greater compliment to a project than when a person decides to hang an image on their walls. Buying art for your wall is a commitment and an expense, so it makes me extremely grateful when my image(s) is chosen out of all the other options in the world. It’s also satisfying to think about the personal connection to people across the world. Even though I may never meet most collectors in person, it’s fun to think about my images in their homes and the joy they receive from looking at them on a daily basis.
Click here to grab a fine art print from American Backcourts for yourself
-
Tagsbasketball, hoop dreams, wall art, fine art prints, basketball art, nba art, National Basketball Association, American Backcourts, Rob Hammer, photo book, basketball photo book, basketball hoop, basketball court
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Troy, New York Photography
Rob HammerApril 6, 2020
Basketball - Troy, New York
At some point I have to wonder if I’m at all capable of working on short term projects? That’s not a complaint. I really love long term projects. Everything about them really, but with the recent amount of time that’s fallen in my lap (the whole COVID-19 thing), it’s given way for a lot of thought. Also something I spend quite a bit of time doing, which has me thinking that maybe I draw things out a bit too long. Started reading Rick Rubin’s book a few days ago, and one thing he talks about is that his work gets done when it gets done. He’s not concerned about deadlines or any other outside influences because he doesn’t want them to affect the final product. If he were to rush a record, it wouldn’t allow the project enough time to breath. He feels like the space and time are necessary to properly pull things together in the way they are naturally supposed to. Reading all that I felt myself understanding and agreeing with everything he was saying. Still though, my natural tendency is to string things along a bit too far. Or maybe it’s just because I don’t devote enough time to certain aspects of each project. There are hard drives of images from 5 years ago that still haven’t been touched much because I’m not sure how they fit in. The process of understanding a group of images is very complex if you really want it to work. And sometimes that means letting go of your favorites because they just don’t work well with the series. Creating a cohesive body of work is quite hard to do when you’re so attached to the images. And it’s not been till recently that I’m starting to get even the slightest bit of handle on it. There are so many factors that dictate why an image works on it’s own, let alone with a group of 30-100 other images for say a gallery show or a book. Anyway, the down time that’s been created by the “Stay home” order has allowed me to focus more on certain projects and helped me to feel like I’m pulling them together in a way that finally make sense. And trimming the fat is starting to become easier too. The Hoops Project was started 8+ years ago, which in itself blows my mind. Hoops have been a major focus on every road trip since 2012. Some of those trips have been shockingly productive. And others don’t yield the most satisfying results. As time goes on I continue to raise the bar, which makes it harder and harder to find a hoop that works. One that fits. One that’s unique. The web gallery for this series hasn’t been updates in quite some time. That’s not out of negligence, but rather from purpose. My efforts over the past couple months have been focused specifically on a few “products” (for the lack of a better word) pertaining to this series, and I want to keep some fresh content for the time when that is finally released. The ones you see below are from an 8,000+/- mile road trip in December/January. Most of that trip did not present me with hoops that turned me on, and it wasn’t until a day of shooting around home that much happened. All 3 of these were made in Troy, NY, which is a few miles from my mothers house. Funny to think that sometimes you drive all the way to the other side of the country before finding something that works.
Click here to purchase a fine art print from this series.
InDocumentary Photography, BasketballTagsbasketball, hoop dreams, basketball never stops, nba, ncaa, sport, athlete, troy, new york, upstate, basketball court, basketball hoop, The Hoops Project, Nikon, street photography
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Project Backboard
Rob HammerMay 13, 2019
Community basketball court renovations
Been saying this for a while now, but personal projects are the best, especially when they connect you with other like minded people. Which is certainly the case with Dan Peterson of Project Backboard. He’s been doing amazing things with outdoor basketball courts all over the country. Taking broken down courts and turning them into beautiful works of art that locals are excited to play on. Recently we visited a few of his courts in Los Angeles together, and I was able to talk with him first hand about the process and how things have developed over the years. I really applaud this project and hope that it continues to grow. If you want to check out more of what PB has done, go to their WEBSITE or follow along on their INSTAGRAM PAGE.
If you recognize the bridge in the Watts Oasis images, that’s because it is the very bridge from those famous scenes in White Men Can’t Jump. I personally love that movie and was ecstatic when Dan told me what it was.
Click here to see more of my basketball photography from the American Backcourts series
1) Where are you from and what place has basketball taken in your life (prior to Project Backboard) ?
I grew up in suburban NY during the heyday of the great 1990s Knicks teams and ultimately played a year of basketball at Iona College before leaving my official playing days behind.
2) When did you come up with the idea for Project Backboard(PB)?
Project Backboard wasn't really my idea! I started the work just by painting lines on public courts in Memphis that did not have any just because I loved outdoor basketball.
3) How long/what did it take to get things going for PB?
I got my first large grant about a year after starting Project Backboard but it was another year before I did the court with William LaChance in St. Louis that really got a lot of attention and opened the door for Project Backboard to become what it is today.
4) What was the initial reaction? How have reactions changed since you started?
The initial reaction was overwhelmingly positive and that is the reaction I have continued to get. That said, this style of court has become surprisingly common over the past 12-18 months that the reaction now may be a bit more restrained than the early courts. No one had ever seen anything like the William LaChance court when we first painted it.
5) How have you gone about getting funding for these projects?
A lot of the courts are funded either by community or corporate foundations.
6) What is the process like from the original idea for a court to the final execution?
The painting process is different for each court depending on what the artist has in mind for the court artwork. Sometimes its a lot of measuring and straight lines or curves and other times we create a grid across the entire surface of the court and drawing the artwork box by box.
7) PB has teamed up with some big name companies. How have those relationships come about?
People reach out and I respond! I am always open to collaborating but the successful projects have been ones were the brands are able to be a little less “corporate” in their approach and allow the artist the freedom to create and lead the project vision.
8) What is the overall goal for PB?
For every community to have a safe and inviting basketball court. I love outdoor basketball and want to share that with others but, from my perspective, the way that will happen is when individual community members step up to help care for public spaces and hold those charged with maintaining those spaces accountable.
9) Any big projects in the works that you want to share?
Yes! Looking forward to a few courts in the Bay Area and a court in Puerto Rico along with a handful of others.
10) Random thoughts on PB......
I appreciate all the support and, as I said, always open to collaborating and helping others follow my example so don't hesitate to reach out!
Venice Beach Basketball Courts
Venice Beach Basketball Basketball Court designed by Project Backboard
Basketball Court -Crete Academy - Crenshaw
Nipsey Hussle Basketball Court -Crete Academy - Crenshaw
Watts Oasis Basketball Court
Watts Oasis Basketball Court
White Men Can't Jump - Basketball Court
Watts Oasis Basketball Court
Watts Oasis Basketball Court
Dan Peterson - Project Backboard
InBasketballTagsBasketball, sport, hoop dreams, Project Backboard, basketball court, basketball hoop, refurbished basketball court, color, Los Angeles, Watts, Crenshaw, Nipsey Hussle, Nipsey Hussle Basketball Court, Watts oasis, the basketball hoops project, Nikon, LA, basketball never stops, NBA, NCAA, streetball, Venice Beach, VBL, Venice Basketball League, California, White Men Can't Jump, Billy Hoyle
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Photographs of New York City Basketball Courts
Rob HammerNovember 13, 2018
NYC Basketball Court Photography
There’s nothing like a basketball court in New York City. From the iconic cages of West 4th Street to a beat-up hoop in the Bronx tucked between apartment buildings, they are some of the most iconic courts with a character all their own. They’re loud, gritty, full of energy, and they carry decades of stories in their cracked paint and chain nets.
The NYC courts in these photographs aren’t so famous, but each one says something about the rhythm of the city and the people who ball in it. Whether you're working on a creative project that calls for authentic New York visuals or you’re looking for wall art that captures the soul of the city’s basketball culture, this series was made for that. The photographs are available as high-resolution files for licensing and as fine art prints for collectors and fans of the game.
New York basketball has always had its own style—fast, creative, tough. These courts are where that style lives, even when no one’s playing.
Looking to license or purchase NYC court photos?
Get in touch for usage details, print sizes, or a curated selection based on your project - rob@robhammerphotography.com
Click here to shop prints of basketball hoops across America
Flatbush, Brooklyn - Basketball Court
Basketball Court - Brooklyn
Basketball court in Coney Island, Brooklyn
Photograph of a Brooklyn basketball court - NYC
InDocumentary Photography, BasketballTagsBrooklyn, Coney Island, Flathbush, NYC, New York City, NYC Basketball, basketball court, The Basketball Hoops Project, Basketball, Hoop Dreams
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The Basketball Hoops Project
Rob HammerDecember 18, 2017
Just a reminder that I'm having a show at Fathom Gallery tonight after the Kobe jersey retirement ceremony at Staples Center. I'll have a bunch of limited edition prints on display, along with some 1/1 signed Kobe game jerseys by a group of really talented street artists. Hope to see you all there!
HOOPS GALLERY
Fathom Gallery: 110 East 9th St. Suite CL002, Los Angeles, CA 90079
TagsThe Basketball Hoops Project, basketball, basketball hoop, basketball court, hoop dreams, kobe bryant, kobe bryant jersey retirement, staples center, lakers, los angeles, los angeles lakers, fine art, fine art prints
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The Basketball Hoops Project
Rob HammerDecember 4, 2017
On 12/18/17 the Los Angeles Lakers will be retiring Kobe Bryant's jersey at the Staples Center. If you're around for it, or just live in LA, come by Fathom Gallery afterward. I'll be showing some signed limited edition hoops prints. Alongside my prints will be a bunch of signed one of a kind Kobe jerseys that have been made in art pieces by a number of extremely talented street artists. Hope to see you there!!
Fathom Gallery.12/18/17. 9pm-12am. -110 E 9th St, Suite CL002, Los Angeles, CA 90079
These images were made a few weeks ago during a trip to Indonesia, and won't be in the show, but wanted to post some updated images anyway. For more, check out my HOOPS gallery.
Jakarta, Indonesia
Jakarta, Indonesia
TagsBasketball, basketball hoop, basketball court, sport, The Basketball Hoops Project, kobe, kobe bryant, the black mamba, kobe bryant jersey retirement, staples center, hoop dreams, nikon, jakarta, indonesia
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7945 | http://hyperphysics.phy-astr.gsu.edu/hbase/orbv3.html | Earth Orbits
Earth Orbit Velocity ==================== The velocity of a satellite in circular orbit around the Earth depends upon the radius of the orbit and the acceleration of gravity at the orbit. Above the earth's surface at a height of h =m = x 10 6 m, which corresponds to a radius r = x earth radius, the acceleration of gravity is g =m/s 2 = x g on the earth's surface. At the specified orbit radius, the required orbit velocity is v =m/s=km/h=mi/h. The period of the orbit is minutes = hours.Index Gravity concepts
HyperPhysics Mechanics RotationR NaveGo Back
Syncom Satellites ================= Communication satellites are most valuable when they stay above the same point on the earth, in what are called "geostationary orbits". This occurs when the orbital period is 24 hours. From an orbit velocity calculation, it can be seen that a satellite at a radius 6.62 times the Earth's radius will have a period of 24 hours. These satellites are some 35,900 kilometers or about 22,300 miles above the Earth's surface. The minimum transit time to a syncom satellite and back from the surface of the Earth is about a quarter of a second. Any kind of electromagnetic wave travels at the speed of light. So in satellite-linked communication, there will be a minimum of a quarter of a second delay.Index Gravity concepts
HyperPhysics Mechanics RotationR NaveGo Back |
7946 | https://www.youtube.com/watch?v=oMzTZOBa7_E | The Chain Rule for ln(x) and ln|x|
MathDoctorBob
66300 subscribers
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Posted: 18 Jun 2011
Calculus: We consider derivatives for functions of the form ln(g(x)) and ln|g(x)|. As preparation, we note some details concerning the domain of ln(x). Example include y(x) = ln|x^2-4x| and y(x) = ln|ln(x)|.
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Transcript:
for the next phase of natural logarithm study derivatives and anti derivatives first we need to make sure we understand how to use natural log and compositions so if i have natural log of x the domain is just x square than zero so we add this to our list of items we need to consider or find the domain of a function so items we have first you can take the square root of a negative number so five square root of box box is greater than or equal to zero can't divide by zero so if I have one over box box has to be nonzero we have our trig functions like tangent and secant now we have natural log of box which requires the Box is greater than zero now you understand how to find a domain for something of type square root of box pretty much understand what you need for natural log of box so let's take a look at an example so I want to find the domain y equal to natural log of x squared minus 4x take x squared minus 4x that's what's in our box it's going to take that set it greater than zero since x squared minus 4x is a polynomial it's continuous everywhere so to find the regions where it's positive and negative I just find the zeros check one point in each region so 0 0 and 4 for a point each region will check -1 1 & 5 so when you work those out we'll have positive negative and positive that means our domain it's going to be X strictly less than zero X strictly greater than 4 and we note because we're using strictly greater than here we're to leave the endpoints out now I want a roll we're taking the derivative of a composition with natural log so we'll need the chain rule will call this says I take F composer G take the derivative evaluate at X the rule is take the derivative of F times the derivative G just have to evaluate at the right points so we'll have F prime evaluated G of x times G prime of X now we want to apply this rule we have F equal to natural log so if I have F a box equal to natural log of box then F prime of box is equal to 1 over box I take the derivative of natural log of G of X then G of X is in the box so for the derivative we're just going to take 1 over G of X multiplied by the derivative of G then we need to be careful ok so we'll have to have that restriction that G of X is greater than 0 otherwise our original function is not defined at X let's take a look at an example so we use y equal to natural log of x squared minus 4x we know we've seen the domain is X less than 0 X square then for I take the derivative what's in the box is x squared minus 4 X so we'll have box prime over box so we have x squared minus 4 in the denominator the derivative which is 2x minus 4 goes into the numerator so that's our derivative note we have to have X less than zero X square then for for instance this makes perfect sense as a function if I put a 1 in there but note we can't put a 1 into our original function here we have a special case that we'll need for anti derivatives so I have y equal to natural log of absolute value of x how do we make sense of this absolute value of X is a piecewise defined function so when X is greater than or equal to zero I just return an X when X is negative we will take away the minus sign we could do that by multiplying by minus 1 so that returns minus ax now that means our function with X is greater than zero we get natural log of X when X is less than zero we'd have natural log of minus X because absolute value of X is then minus X and if X is equal to zero we get natural log of zero but that's undefined so our domain is going to be X not equal to zero the way we should think of this by taking the absolute value we're extending the domain of natural log as much as we can now we take the derivative how do we take the derivative of a piecewise defined function well we're just going to take the derivative on each piece then we have to worry about what's happening at the endpoints now for the first piece derivative of natural log is just 1 over X so we record that for our second piece I'm going to need the chain rule so derivative of natural log of minus X we have minus X in the box the rule is box prime / box so we'd have minus 1 over minus X or 1 over X again so we have 1 over X 1 or positive 1 over X or negative so I put all that together the derivative of natural log of absolute value of X is 1 over X and note the domain here is just going to be X not equal to 0 now if we compare pictures if I just considered natural log of X that's just going to be everything to the right of the y-axis so here's our graph the derivative is going to be 1 over X with X greater than zero since this is increasing everywhere that means our derivative is going to be greater than or equal to 0 and that pans out when I take the absolute value of x put it in natural log we're to add another piece everything to the left of the y-axis is going to be natural log of minus X now if we put minus X into a function what we're doing is flipping over the y-axis so our graph goes like this to this we know the derivative is still going to be 1 over X here we're always decreasing on this piece so we're expecting the derivative is always less than or equal to zero and that's going to pan out for 1 over X now if we rework our derivative with the absolute value we'll have the derivative the natural log of the absolute value of G of X is equal to G prime of X over G of X with the restriction that G of X is nonzero so note the expression is the same the difference is we're now allowing negative values for G of X so example will now have y equals natural log the absolute value of x squared minus 4x I could put whatever I want here as long as what comes out is not equal to zero so our domain is going to be x squared minus 4 X not equal to 0 or X not equal to 0 or 4 when I take the derivative the rule is just take what's in the box take its derivative divided by box so we'll get the same expression as before 2x minus 4 over x squared minus 4x the difference we're now going to allow x squared minus 4x to take on negative values so our only restriction is that X not be equal to 0 or 4 now note I don't need to write this down because that'll be handled implicitly by the denominator so look at another example so I'll have y equals natural log natural log of X no absolute value now for the domain we just take what's on the inside so that's natural log of X we put it in the box set it greater than zero so we want to know for what x do we have natural log of x greater than zero we draw the picture this is just asking what Y values are positive so we know that'll happen when X is greater than one so the graph hits the x-axis at the point one comma zero so everything to the right is going to be x greater than one we take the derivative we're just going to take what's in the box flip it over then multiply by its derivative so I've 1 over natural log of x times 1 over X or 1 over x times natural log of X with the restriction that X be greater than 1 we redo it with the absolute value so y equal to natural log the absolute value of natural log of X now for what's in the box we only want it to be nonzero so I want natural log of X not equal to 0 that's just going to mean we use the usual domain for natural log of X which is X greater than zero and I just throw away 1 so we put a 1 in we get a 0 so this is going to be our new domain now but what the derivative I don't need to do any new work I'm allowed to just use this expression here 1 over X natural log of X but the restriction I use is that X be greater than 0 and X not be equal to 1 |
7947 | https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/8th-slope/v/slope-intuition-example | Slope & direction of a line (video) | Slope | Khan Academy
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8th grade math
Course: 8th grade math>Unit 3
Lesson 4: Slope
Intro to slope
Intro to slope
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Slope & direction of a line
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Slope
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Slope & direction of a line
CCSS.Math: 8.F.B.4
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About About this video Transcript
A line that's flat has a slope of 0. If a line goes up as you move to the right, it has a positive slope. The steeper it is, the bigger the slope. If a line goes down as you move to the right, it has a negative slope. The steeper it is, the more negative the slope.Created by Sal Khan.
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Tanisha 10 years ago Posted 10 years ago. Direct link to Tanisha's post “What is an undefined slop...” more What is an undefined slope?
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kvandusen 9 years ago Posted 9 years ago. Direct link to kvandusen's post “My math teacher taught it...” more My math teacher taught it to me as a story, if you go down the ski hill that goes straight up and down, when your body is found, you'll be undefinable.
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jungtay000 a year ago Posted a year ago. Direct link to jungtay000's post “I thought the question sa...” more I thought the question said that it had to be negative and greater than the other slope. How come you made it less? Maybe I read the question wrong.
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TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “Good question. He means "...” more Good question. He means "less negative" as negative but less magnitude.
Eg. -2 is less negative than -5.
Remember that the greater number is more to the right on the number line. With negative numbers, the greater number would be "less negative".
Eg. -2 > -5.
Hope that helped.
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pinky 2 years ago Posted 2 years ago. Direct link to pinky's post “but how do u know its neg...” more but how do u know its negative?
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Paige 4 months ago Posted 4 months ago. Direct link to Paige's post “You know if it is negativ...” more You know if it is negative if the line goes down. When you do the equation for slope, either the x or y will be negative causing the whole answer to be negative. It also makes sense because the line is going down and that makes more sense than if the line going up means negative (which it doesn't).
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superdupershaheer 2 years ago Posted 2 years ago. Direct link to superdupershaheer's post “Man this unit is reallY h...” more Man this unit is reallY hard..
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Clarkaboom 2 years ago Posted 2 years ago. Direct link to Clarkaboom's post “_it is pretty hard but th...” more it is pretty hard but the videos make it much easier just take notes and study them
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Ilina 9 years ago Posted 9 years ago. Direct link to Ilina's post “If you had a line that wa...” more If you had a line that was vertical, would it have a positive slope?
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Steven Donald 9 years ago Posted 9 years ago. Direct link to Steven Donald's post “No, if the line is vertic...” more No, if the line is vertical then the slope is undefined. The slope is the measure of the rise (or change in the y-axis) over the run (or change in the x-axis). Since a vertical line has no change in x-axis, or rather, all points share the same x-value, it would have 0 for denominator, which makes the fraction, and thus the slope, undefined.
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Sathwika Sundaramurthy 2 months ago Posted 2 months ago. Direct link to Sathwika Sundaramurthy 's post “What website does he use ...” more What website does he use in the video to get access to the following question in shown by the math guy.?
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TitaniumGodFire 🤖 🧊 2 months ago Posted 2 months ago. Direct link to TitaniumGodFire 🤖 🧊's post “To answer your question: ...” more To answer your question: Sal Khan is the creator of this video and Khan Academy. He created this video before the website was updated, so he was technically using the same exercise questions you still get on this website.
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dudeman907 4 months ago Posted 4 months ago. Direct link to dudeman907's post “Why is this so hard!?!?!?...” more Why is this so hard!?!?!?!?!?!?!?!??!?!?!
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TitaniumGodFire 🤖 🧊 2 months ago Posted 2 months ago. Direct link to TitaniumGodFire 🤖 🧊's post “Let me try to explain the...” more Let me try to explain the point of the video. Sal need a line that is negative. So, as the "x" value increase, the "y" value decreases. He also requires a greater slope than the example line given. "Greater" in this case means "less steeper" because the smaller the m in y=mx+b, the less steeper the line is. THE PREVIOUS STATEMENT ONLY APPLIES TO NEGATIVE LINES. So, Sal creates a line that is less steeper than the negative line shown, but makes it still negative.
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WarmSausageTea ⛮ 4 years ago Posted 4 years ago. Direct link to WarmSausageTea ⛮'s post “Would a vertical line be ...” more Would a vertical line be infinite slope?
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David Severin 4 years ago Posted 4 years ago. Direct link to David Severin's post “It is usually referred to...” more It is usually referred to as undefined slope rather than infinite slope. Undefined is better math vocabulary because we say dividing anything by 0 is undefined including vertical lines.
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annanora20 11 years ago Posted 11 years ago. Direct link to annanora20's post “i really didn't understan...” more i really didn't understand a thing, were the two yellow dots supposed to mean something?
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IzStar 🫶🏼🌊 5 months ago Posted 5 months ago. Direct link to IzStar 🫶🏼🌊's post “Wait, so what defines a n...” more Wait, so what defines a negative slope? One that's tilted to the left?
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TitaniumGodFire 🤖 🧊 4 months ago Posted 4 months ago. Direct link to TitaniumGodFire 🤖 🧊's post “A negative slope is defin...” more A negative slope is defined to show that the "y" value decreases as the "x" value increases. Hope this helps!
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Video transcript
Graph a line that has a slope that is negative and greater than the slope of the blue line. So let's think for a second about what slope means. So if you use the word slope in your everyday life, you're really talking about how inclined something is, like a ski slope. So for example, this orange line isn't inclined at all. It's flat, so this one actually has a slope of 0. Another way of thinking about it is as x increases, what is happening to y? And you see here that y isn't changing at all, so this orange line right now has a slope of 0. If the orange line looked like this, it now has a positive slope. Notice, when x is negative, your y value-- or say when x is negative 5, your y value is here, and then when x is positive 5, your y value has increased. As x increases, y is increasing, so this has a positive slope. This has an even more positive slope, an even more positive slope. This has an even more positive slope. As x is increasing, the y value is increasing really fast. This line is going up really fast as we move towards the right, so it's a very positive slope. This is less positive, less positive. This is a 0 slope, and then this is a negative slope. Notice, as x is increasing, the line is going down. Your y value is decreasing. When x is negative 5, your y is 7. Well, when x is 5, your y is 5. So x is increased, but y has gone down, so this is a negative slope. So they say graph a line that has a slope that is negative and greater than the slope of the blue line. So the blue line also has a negative slope. As x is increasing, your blue line is going down. Over here, when x has negative values, your y value is quite high. And here, when x has positive values, your y value has gone all the way down. So this has a negative slope, but we want to have a slope greater than this one. So we still want to have a-- they say graph a line that has a slope that is negative-- so my orange line currently has a negative slope-- and greater than the slope of the blue line. So it has a negative slope, but it is less negative than this blue line right over here. If I wanted to be more negative than the blue line, I'd have to do something like that. But I want to be negative but less negative than the blue line, so that would be like that. If we wanted a 0 slope, once again, something like this. If we wanted a positive slope, something like this. So once again, negative slope, less negative than the blue line. Check our answer.
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7948 | https://firmfunda.com/maths/wholenumbers/whole-numbers-fundamentals/whole-numbers-approximation-estimation | Whole Numbers : Approximation and Estimation
maths>wholenumbers
Approximation and Estimation
what you'll learn...
overview
In this page, approximation of a number to the nearest "degree of approximation" is explained with a practical example.
In many applications, the value of measurement or count need not be accurate. It can be approximated or estimated to the nearest value to some degree.
lesson outline
Accurate and Approximate
The word "accurate" means: correct in details; exact.
The word "approximate" means: close to the actual; very near in value.
The accurate length of the pencil in the figure is 73 73 mm.
Let us consider a scale that does not have graduated marks for millimeters. Using the scale, the accurate length cannot be measured. The approximate length of the pencil in the figure is 7 7 centimeter. Note that the value 7 7 is closer to the accurate length.
Estimation
The accurate length of the pencil in the figure is 78 78 mm.
Considering the scale with only centimeter marks, the approximate length of the pencil is measured as 8 8 centimeter. Note that the value 8 8 is closer to the accurate length.
In many applications, the value of measurement or count need not be accurate. It can be approximated to the nearest value to some degree.
The degree of approximation is explained as follows
•3842 approximated to nearest 10 is 3840. The degree of approximation is to the nearest 10.
•3842 approximated to nearest 100 is 3800. The degree of approximation is to the nearest 100.
•3842 approximated to nearest 1000 is 4000. The degree of approximation is to the nearest 1000.
The word "degree" means amount or level to which something is present. Degree of approximation is the level to which something is approximated.
The word "estimation" means: approximation and rough calculation to the nearest value. The estimation of a count or measure is the approximate value of the count or measure.
Approximation of numbers : A number may be approximated or rounded to the nearest 10, or 100, or 1000, etc. If the number is halfway or above, then it is approximated to the higher value. If not, then it is approximated to the lower value.
Examples
The approximate value of 3428 to the nearest 100 is 3400.
Estimate the value of 428 to the nearest 10 : The value of 428 to the nearest 10 is 430.
The approximate value of 3450 to the nearest 100 is 3500.
Summary
Approximation of numbers : A number may be approximated or rounded to the nearest 10, or 100, or 1000, etc. If the number is halfway or above, then it is approximated to the higher value. If not, then it is approximated to the lower value.
The degree of approximation is explained as follows
•3842 approximated to nearest 10 is 3840. The degree of approximation is to the nearest 10.
•3848 approximated to nearest 10 is 3850. The degree of approximation is to the nearest 10.
•3842 approximated to nearest 100 is 3800. The degree of approximation is to the nearest 100.
•3842 approximated to nearest 1000 is 4000. The degree of approximation is to the nearest 1000.
next
Outline
The outline of material to learn whole numbers is as follows
Note: click here for detailed outline of Whole numbers
• Introduction
→ Numbers
→ Large Numbers
→ Expanded form
→ Face and place values
→ Approximation and Estimation
• Comparison
→ Comparing two numbers
→ Number line
→ Predecessor & Successor
→ Largest & Smallest
→ Ascending & Descending
• Addition Subtraction
→ Addtion: First Principles
→ Addition: Simplified Procedure
→ Subtraction: First Principles
→ Subtraction: Simplified Procedure
• Multiplication Division
→ Multiplication: First Principles
→ Multiplication: Simplified Procedure
→ Division: First Principles
→ Division: Simplified Procedure
• Numerical Expression
→ Introducing Numerical Expressions
→ Precedence
→ Sequence
→ Brackets
next |
7949 | https://www.e-education.psu.edu/egee102/node/1942 | | |
| EGEE 102 Energy Conservation for Environmental Protection |
HOME
LESSONS
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The Carnot Efficiency
Print
A general expression for the efficiency of a heat engine can be written as:
We know that all the energy that is put into the engine has to come out either as work or waste heat. So work is equal to Heat at High temperature minus Heat rejected at Low temperature. Therefore, this expression becomes:
Where, QHot = Heat input at high temperature and QCold= Heat rejected at low temperature. The symbol (Greek letter eta) is often used for efficiency this expression can be rewritten as:
The above equation is multiplied by 100 to express the efficiency as percent.
French Engineer Sadi Carnot showed that the ratio of QHighT to QLowT must be the same as the ratio of temperatures of high temperature heat and the rejected low temperature heat. So this equation, also called Carnot Efficiency, can be simplified as:
Note: Unlike the earlier equations, the positions of Tcold and Thot are reversed.
The Carnot Efficiency is the theoretical maximum efficiency one can get when the heat engine is operating between two temperatures:
The temperature at which the high temperature reservoir operates ( THot ).
The temperature at which the low temperature reservoir operates ( TCold ).
In the case of an automobile, the two temperatures are:
The temperature of the combustion gases inside the engine ( THot ).
The temperature at which the gases are exhausted from the engine ( TCold ).
It's like taxes. The more money you earn (heat), the more money is taxed (cold), leaving you with less money to take home (efficiency). However, if you could earn more money (heat) and find a way to have less taxes taken out (better engine material), you would have more money to take home (efficiency).
Below is a table showing two temperature scales. The scale labeled "HOT," shows the range of temperatures for the combustion of gases in a car engine. The scale labeled "COLD," shows the range of temperatures at which gases are exhausted from the car engine.
Car Engine Temperatures: red indicates combustion temperatures and blue indicates exhausted temperatures.
Credit: © Penn State is licensed under CC BY-NC-SA 4.0
Instructions: Look carefully at the efficiency numbers in the body of the table. How do the Hot and Cold temperatures' effect on the efficiency.
Car Engine Efficiency
| Hot 500°C | Hot 600°C | Hot 700°C | Hot 800°C | Hot 900°C | Hot 1,000°C | Hot 1,500°C | Hot 2,000°C |
| Cold 150°C | 45 | 52 | 57 | 61 | 64 | 67 | 76 | 81 |
| Cold 125°C | 49 | 54 | 59 | 63 | 66 | 69 | 78 | 82 |
| Cold 100°C | 52 | 57 | 62 | 65 | 68 | 71 | 79 | 84 |
| Cold 75°C | 55 | 60 | 64 | 68 | 70 | 73 | 80 | 85 |
| Cold 50°C | 58 | 63 | 67 | 70 | 72 | 75 | 82 | 86 |
| Cold 25°C | 61 | 66 | 69 | 72 | 75 | 77 | 83 | 87 |
Answer the following questions based on the information in the Car Engine Efficiency table above.
Example
For a coal-fired utility boiler, the temperature of high pressure steam (Thot)would be about 540°C and Tcold, the cooling tower water temperature, would be about 20°C. Calculate the Carnot efficiency of the power plant:
Solution:
Carnot efficiency depends on high temperature and low temperatures between which the heat engine operates. We are given both temperatures. However, the temperatures need to be converted to Kelvin:
Practice
For a coal fired utility boiler, the temperature of high pressure steam would be about 540 degrees C and Tcold, the cooling tower water temperature, would be about 20 degrees C. Calculate the Carnot efficiency of the power plant.
Step 1 Convert the high and low temperatures from Celsius to Kelvin:
Step 2
Determine the efficiency using the Carnot efficiency formula:
From the Carnot Efficiency formula, it can be inferred that a maximum of 64% of the fuel energy can go to generation. To make the Carnot efficiency as high as possible, either Thot should be increased or Tcold (temperature of heat rejection) should be decreased.
Practice Problem
Use the following link to generate a random practice problem.
‹ Heat Engines up Entropy and Quality of Energy › |
7950 | https://en.wikipedia.org/wiki/Densities_of_the_elements_(data_page) | Jump to content
Densities of the elements (data page)
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From Wikipedia, the free encyclopedia
Chemical data page
Density, solid phase
[edit]
In the following table, the use row is the value recommended for use in other Wikipedia pages in order to maintain consistency across content.
| 2 He helium-4 | | |
---
| Hoffer et al. | 0.19085 g/cm3 (from 20.9730 cm3/mole; hcp crystal melting to He-II superfluid at 0 K, 25.00 atm) |
| 0.19083 g/cm3 (from 20.9749 cm3/mole; at local min. density, hcp melting to He-II: 0.884 K, 25.00 atm) |
| 0.19142 g/cm3 (from 20.910 cm3/mole; hcp at triple point hcp−bcc−He-II: 1.463 K, 26.036 atm) |
| 0.18971 g/cm3 (from 21.098 cm3/mole; bcc at triple point hcp−bcc−He-II: 1.463 K, 26.036 atm) |
| 0.19406 g/cm3 (from 20.626 cm3/mole; hcp at triple point hcp−bcc−He-I: 1.772 K, 30.016 atm) |
| 0.19208 g/cm3 (from 20.8381 cm3/mole; bcc at triple point hcp−bcc−He-I: 1.772 K, 30.016 atm) |
| 3 Li lithium | |
| use | 0.534 g/cm3 |
| WEL | (near r.t.) 535 kg/m3 |
| LNG | (at 20 °C) 0.534 g/cm3 |
| CRC | (near r.t.) 0.534 g/cm3 |
| 4 Be beryllium | |
| use | 1.85 g/cm3 |
| WEL | (near r.t.) 1848 kg/m3 |
| LNG | (at 20 °C) 1.8477 g/cm3 |
| CRC | (near r.t.) 1.85 g/cm3 |
| 5 B boron | |
| use | 2.34 g/cm3 |
| WEL | (near r.t.) 2460 kg/m3 |
| LNG | (at r.t.) 2.34 g/cm3 |
| CRC | (near r.t.) 2.34 g/cm3 |
| 6 C carbon (graphite) | |
| use | 2.267 g/cm3 |
| WEL | (near r.t.) 2267 kg/m3 |
| LNG | (at r.t.) 2.267 g/cm3 |
| CRC | (near r.t.) 2.2 g/cm3 |
| 6 C carbon (diamond) | |
| use | 3.513 g/cm3 |
| LNG | (at r.t.) 3.513 g/cm3 |
| CRC | (near r.t.) 3.513 g/cm3 |
| 11 Na sodium | |
| use | 0.968 g/cm3 |
| WEL | (near r.t.) 968 kg/m3 |
| LNG | (at 20 °C) 0.968 g/cm3 |
| CRC | (near r.t.) 0.97 g/cm3 |
| 12 Mg magnesium | |
| use | 1.738 g/cm3 |
| WEL | (near r.t.) 1738 kg/m3 |
| LNG | (at 20 °C) 1.738 g/cm3 |
| CRC | (near r.t.) 1.74 g/cm3 |
| 13 Al aluminium | |
| use | 2.70 g/cm3 |
| WEL | (near r.t.) 2700 kg/m3 |
| LNG | (at r.t.) 2.70 g/cm3 |
| CRC | (near r.t.) 2.70 g/cm3 |
| 14 Si silicon | |
| use | 2.33 g/cm3 |
| WEL | (near r.t.) 2330 kg/m3 |
| LNG | (at r.t.) 2.33 g/cm3 |
| CRC | (near r.t.) 2.3290 g/cm3 |
| 15 P phosphorus (white) | |
| use | 1.823 g/cm3 |
| WEL | (near r.t.) 1823 kg/m3 |
| LNG | (at 25 °C) 1.823 g/cm3 |
| CRC | (near r.t.) 1.823 g/cm3 |
| 15 P phosphorus (red) | |
| use | 2.34 g/cm3 |
| LNG | (at r.t.) 2.34 g/cm3 |
| CRC | (near r.t.) 2.16 g/cm3 |
| 15 P phosphorus (black) | |
| use | 2.69 g/cm3 |
| CRC | (near r.t.) 2.69 g/cm3 |
| 16 S sulfur (orthorhombic, alpha) | |
| use | 2.08 g/cm3 |
| WEL | ? (1960 kg/m3) |
| LNG | (at 20 °C) 2.08 g/cm3 |
| CRC | (near r.t.) 2.07 g/cm3 |
| 16 S sulfur (monoclinic, beta) | |
| use | 1.96 g/cm3 |
| WEL | ? (1960 kg/m3) |
| LNG | (at r.t.) 1.96 g/cm3 |
| CRC | ? (near r.t.) 2.07 g/cm3 |
| 16 S sulfur (gamma) | |
| use | 1.92 g/cm3 |
| LNG | (at r.t.) 1.92 g/cm3 |
| 19 K potassium | |
| use | 0.89 g/cm3 |
| WEL | (near r.t.) 856 kg/m3 |
| LNG | (at r.t.) 0.89 g/cm3 |
| CRC | (near r.t.) 0.89 g/cm3 |
| 20 Ca calcium | |
| use | 1.55 g/cm3 |
| WEL | (near r.t.) 1550 kg/m3 |
| LNG | (at r.t.) 1.55 g/cm3 |
| CRC | (near r.t.) 1.54 g/cm3 |
| 21 Sc scandium (hexagonal ?) | |
| use | 2.985 g/cm3 |
| WEL | (near r.t.) 2985 kg/m3 |
| LNG | (at r.t.) (hexagonal) 2.985 g/cm3 |
| CRC | (near r.t.) 2.99 g/cm3 |
| 22 Ti titanium (hexagonal ?) | |
| use | 4.506 g/cm3 |
| WEL | (near r.t.) 4507 kg/m3 |
| LNG | (at r.t.) (hexagonal) 4.506 g/cm3 |
| CRC | (near r.t.) 4.506 g/cm3 |
| 23 V vanadium | |
| use | 6.11 g/cm3 |
| WEL | (near r.t.) 6110 kg/m3 |
| LNG | (at 19 °C) 6.11 g/cm3 |
| CRC | (near r.t.) 6.0 g/cm3 |
| 24 Cr chromium | |
| use | 7.15 g/cm3 |
| WEL | (near r.t.) 7140 kg/m3 |
| LNG | (at r.t.) 7.15 g/cm3 |
| CRC | (near r.t.) 7.15 g/cm3 |
| 25 Mn manganese | |
| use | 7.21 g/cm3 |
| WEL | (near r.t.) 7470 kg/m3 |
| LNG | (at 20 °C) 7.21 g/cm3 |
| CRC | (near r.t.) 7.3 g/cm3 |
| 26 Fe iron | |
| use | 7.86 g/cm3 |
| WEL | (near r.t.) 7874 kg/m3 |
| LNG | (at r.t.) 7.86 g/cm3 |
| CRC | (near r.t.) 7.87 g/cm3 |
| 27 Co cobalt | |
| use | 8.90 g/cm3 |
| WEL | (near r.t.) 8900 kg/m3 |
| LNG | (at r.t.) 8.90 g/cm3 |
| CRC | (near r.t.) 8.86 g/cm3 |
| 28 Ni nickel | |
| use | 8.908 g/cm3 |
| WEL | (near r.t.) 8908 kg/m3 |
| LNG | (at 20 °C) 8.908 g/cm3 |
| CRC | (near r.t.) 8.90 g/cm3 |
| 29 Cu copper | |
| use | 8.96 g/cm3 |
| WEL | (near r.t.) 8920 kg/m3 |
| LNG | (at 20 °C) 8.96 g/cm3 |
| CRC | (near r.t.) 8.96 g/cm3 |
| 30 Zn zinc | |
| use | 7.14 g/cm3 |
| WEL | (near r.t.) 7140 kg/m3 |
| LNG | (at r.t.) 7.14 g/cm3 |
| CRC | (near r.t.) 7.14 g/cm3 |
| 31 Ga gallium | |
| use | 5.91 g/cm3 |
| WEL | (near r.t.) 5904 kg/m3 |
| LNG | (at 29.6 °C) 5.904 g/cm3 |
| CRC | (near r.t.) 5.91 g/cm3 |
| 32 Ge germanium | |
| use | 5.323 g/cm3 |
| WEL | (near r.t.) 5323 kg/m3 |
| LNG | (at r.t.) 5.323 g/cm3 |
| CRC | (near r.t.) 5.3234 g/cm3 |
| 33 As arsenic | |
| use | 5.727 g/cm3 |
| WEL | (near r.t.) 5727 kg/m3 |
| LNG | (at 25 °C) (5.727 rel. to water at 4 °C) |
| CRC | (near r.t.) 5.75 g/cm3 |
| 34 Se selenium (hexagonal, gray) | |
| use | 4.81 g/cm3 |
| WEL | (near r.t.) 4819 kg/m3 |
| LNG | (at 20 °C) (4.81 rel. to water at 4 °C) |
| CRC | (near r.t.) 4.81 g/cm3 |
| 34 Se selenium (alpha) | |
| use | 4.39 g/cm3 |
| WEL | ? (4819 kg/m3) |
| CRC | (near r.t.) 4.39 g/cm3 |
| 34 Se selenium (vitreous) | |
| use | 4.28 g/cm3 |
| WEL | ? (4819 kg/m3) |
| CRC | (near r.t.) 4.28 g/cm3 |
| 37 Rb rubidium | |
| use | 1.532 g/cm3 |
| WEL | (near r.t.) 1532 kg/m3 |
| LNG | (at r.t.) 1.532 g/cm3 |
| CRC | (near r.t.) 1.53 g/cm3 |
| 38 Sr strontium | |
| use | 2.64 g/cm3 |
| WEL | (near r.t.) 2630 kg/m3 |
| LNG | (at r.t.) 2.64 g/cm3 |
| CRC | (near r.t.) 2.64 g/cm3 |
| 39 Y yttrium | |
| use | 4.472 g/cm3 |
| WEL | (near r.t.) 4472 kg/m3 |
| LNG | (at r.t.) 4.472 g/cm3 |
| CRC | (near r.t.) 4.47 g/cm3 |
| 40 Zr zirconium | |
| use | 6.52 g/cm3 |
| WEL | (near r.t.) 6511 kg/m3 |
| LNG | (at r.t.) 6.52 g/cm3 |
| CRC | (near r.t.) 6.52 g/cm3 |
| 41 Nb niobium | |
| use | 8.57 g/cm3 |
| WEL | (near r.t.) 8570 kg/m3 |
| LNG | (at 20 °C) 8.57 g/cm3 |
| CRC | (near r.t.) 8.57 g/cm3 |
| 42 Mo molybdenum | |
| use | 10.28 g/cm3 |
| WEL | (near r.t.) 10280 kg/m3 |
| LNG | (at r.t.) 10.28 g/cm3 |
| CRC | (near r.t.) 10.2 g/cm3 |
| 43 Tc technetium (Tc-98 ?) | |
| use | 11 g/cm3 |
| WEL | (near r.t.) 11500 kg/m3 |
| LNG | (at r.t.) (Tc-98) 11 g/cm3 |
| CRC | (near r.t.) 11 g/cm3 |
| 44 Ru ruthenium | |
| use | 12.45 g/cm3 |
| WEL | (near r.t.) 12370 kg/m3 |
| LNG | (at 20 °C) (12.45 rel. to water at 4 °C) |
| CRC | (near r.t.) 12.1 g/cm3 |
| 45 Rh rhodium | |
| use | 12.41 g/cm3 |
| WEL | (near r.t.) 12450 kg/m3 |
| LNG | (at 20 °C) 12.41 g/cm3 |
| CRC | (near r.t.) 12.4 g/cm3 |
| 46 Pd palladium | |
| use | 12.023 g/cm3 |
| WEL | (near r.t.) 12023 kg/m3 |
| LNG | (at 20 °C) 12.023 g/cm3 |
| CRC | (near r.t.) 12.0 g/cm3 |
| 47 Ag silver | |
| use | 10.49 g/cm3 |
| WEL | (near r.t.) 10490 kg/m3 |
| LNG | (at r.t.) 10.49 g/cm3 |
| CRC | (near r.t.) 10.5 g/cm3 |
| 48 Cd cadmium | |
| use | 8.65 g/cm3 |
| WEL | (near r.t.) 8650 kg/m3 |
| LNG | (at 25 °C) 8.65 g/cm3 |
| CRC | (near r.t.) 8.69 g/cm3 |
| 49 In indium | |
| use | 7.31 g/cm3 |
| WEL | (near r.t.) 7310 kg/m3 |
| LNG | (at r.t.) 7.31 g/cm3 |
| CRC | (near r.t.) 7.31 g/cm3 |
| 50 Sn tin (white) | |
| use | 7.265 g/cm3 |
| WEL | (near r.t.) 7310 kg/m3 |
| LNG | (at r.t.) 7.265 g/cm3 |
| CRC | (near r.t.) 7.265 g/cm3 |
| 50 Sn tin (gray) | |
| use | 5.769 g/cm3 |
| CRC | (near r.t.) 5.769 g/cm3 |
| 51 Sb antimony | |
| use | 6.697 g/cm3 |
| WEL | (near r.t.) 6697 kg/m3 |
| LNG | (at 25 °C) 6.697 g/cm3 |
| CRC | (near r.t.) 6.68 g/cm3 |
| 52 Te tellurium | |
| use | 6.24 g/cm3 |
| WEL | (near r.t.) 6240 kg/m3 |
| LNG | (at r.t.) 6.24 g/cm3 |
| CRC | (near r.t.) 6.24 g/cm3 |
| 53 I iodine (I2) | |
| use | 4.933 g/cm3 |
| WEL | (near r.t.) 4940 kg/m3 |
| LNG | (at 25 °C) 4.63 g/cm3 |
| CRC | (near r.t.) 4.933 g/cm3 |
| 55 Cs caesium | |
| use | 1.93 g/cm3 |
| WEL | (near r.t.) 1879 kg/m3 |
| LNG | (at 15 °C) 1.8785 g/cm3 |
| CRC | (near r.t.) 1.93 g/cm3 |
| 56 Ba barium | |
| use | 3.51 g/cm3 |
| WEL | (near r.t.) 3510 kg/m3 |
| LNG | (at 20 °C) 3.51 g/cm3 |
| CRC | (near r.t.) 3.62 g/cm3 |
| 57 La lanthanum | |
| use | 6.162 g/cm3 |
| WEL | (near r.t.) 6146 kg/m3 |
| LNG | (at r.t.) 6.162 g/cm3 |
| CRC | (near r.t.) 6.15 g/cm3 |
| 58 Ce cerium | |
| use | 6.770 g/cm3 |
| WEL | (near r.t.) 6689 kg/m3 |
| LNG | (at r.t.) 6.773 g/cm3 |
| CRC | (near r.t.) 6.770 g/cm3 |
| 59 Pr praseodymium (alpha ?) | |
| use | 6.77 g/cm3 |
| WEL | (near r.t.) 6640 kg/m3 |
| LNG | (at r.t.) (alpha) 6.475 g/cm3 |
| CRC | (near r.t.) 6.77 g/cm3 |
| 60 Nd neodymium | |
| use | 7.01 g/cm3 |
| WEL | (near r.t.) 6800 kg/m3 |
| LNG | (at r.t.) 7.01 g/cm3 |
| CRC | (near r.t.) 7.01 g/cm3 |
| 61 Pm promethium (Pm-147 ?) | |
| use | 7.26 g/cm3 |
| WEL | (near r.t.) 7264 kg/m3 |
| LNG | (at r.t.) (Pm-147) 7.22 g/cm3 |
| CRC | (near r.t.) 7.26 g/cm3 |
| 62 Sm samarium | |
| use | 7.52 g/cm3 |
| WEL | (near r.t.) 7353 kg/m3 |
| LNG | (at r.t.) 7.52 g/cm3 |
| CRC | (near r.t.) 7.52 g/cm3 |
| 63 Eu europium | |
| use | 5.244 g/cm3 |
| WEL | (near r.t.) 5244 kg/m3 |
| LNG | (at r.t.) 5.244 g/cm3 |
| CRC | (near r.t.) 5.24 g/cm3 |
| 64 Gd gadolinium | |
| use | 7.90 g/cm3 |
| WEL | (near r.t.) 7901 kg/m3 |
| LNG | (at r.t.) 7.90 g/cm3 |
| CRC | (near r.t.) 7.90 g/cm3 |
| 65 Tb terbium | |
| use | 8.23 g/cm3 |
| WEL | (near r.t.) 8219 kg/m3 |
| LNG | (at r.t.) 8.23 g/cm3 |
| CRC | (near r.t.) 8.23 g/cm3 |
| 66 Dy dysprosium | |
| use | 8.540 g/cm3 |
| WEL | (near r.t.) 8551 kg/m3 |
| LNG | (at 25 °C) 8.540 g/cm3 |
| CRC | (near r.t.) 8.55 g/cm3 |
| 67 Ho holmium | |
| use | 8.79 g/cm3 |
| WEL | (near r.t.) 8795 kg/m3 |
| LNG | (at r.t.) 8.79 g/cm3 |
| CRC | (near r.t.) 8.80 g/cm3 |
| 68 Er erbium | |
| use | 9.066 g/cm3 |
| WEL | (near r.t.) 9066 kg/m3 |
| LNG | (at r.t.) 9.066 g/cm3 |
| CRC | (near r.t.) 9.07 g/cm3 |
| 69 Tm thulium | |
| use | 9.32 g/cm3 |
| WEL | (near r.t.) 9321 kg/m3 |
| LNG | (at r.t.) 9.32 g/cm3 |
| CRC | (near r.t.) 9.32 g/cm3 |
| 70 Yb ytterbium | |
| use | 6.90 g/cm3 |
| WEL | (near r.t.) 6570 kg/m3 |
| LNG | (at r.t.) 6.90 g/cm3 |
| CRC | (near r.t.) 6.90 g/cm3 |
| 71 Lu lutetium | |
| use | 9.841 g/cm3 |
| WEL | (near r.t.) 9841 kg/m3 |
| LNG | (at r.t.) 9.841 g/cm3 |
| CRC | (near r.t.) 9.84 g/cm3 |
| 72 Hf hafnium | |
| use | 13.31 g/cm3 |
| WEL | (near r.t.) 13310 kg/m3 |
| LNG | (at r.t.) 13.31 g/cm3 |
| CRC | (near r.t.) 13.3 g/cm3 |
| 73 Ta tantalum | |
| use | 16.69 g/cm3 |
| WEL | (near r.t.) 16650 kg/m3 |
| LNG | (at r.t.) 16.69 g/cm3 |
| CRC | (near r.t.) 16.4 g/cm3 |
| 74 W tungsten | |
| use | 19.25 g/cm3 |
| WEL | (near r.t.) 19250 kg/m3 |
| LNG | (at r.t.) 19.25 g/cm3 |
| CRC | (near r.t.) 19.3 g/cm3 |
| 75 Re rhenium | |
| use | 21.02 g/cm3 |
| WEL | (near r.t.) 21020 kg/m3 |
| LNG | (at r.t.) 21.02 g/cm3 |
| CRC | (near r.t.) 20.8 g/cm3 |
| 76 Os osmium | |
| use | 22.59 g/cm3 |
| WEL | (near r.t.) 22610 kg/m3 |
| LNG | (at 20 °C) 22.61 g/cm3 |
| CRC | (near r.t.) 22.59 g/cm3 |
| 77 Ir iridium | |
| use | 22.56 g/cm3 |
| WEL | (near r.t.) 22650 kg/m3 |
| LNG | (at 20 °C) (22.65 rel. to water at 4 °C) |
| CRC | (near r.t.) 22.5 g/cm3 |
| 78 Pt platinum | |
| use | 21.45 g/cm3 |
| WEL | (near r.t.) 21090 kg/m3 |
| LNG | (at 20 °C) 21.09 g/cm3 |
| CRC | (near r.t.) 21.5 g/cm3 |
| References according to 21.45 g/cm3 — Zumdahl, Steven S., Zumdahl, Susan L., & Decoste, Donald J. World of Chemistry. Boston: Houghton Mifflin Company, 2002: 141. 21.45 × 103 kg/m3 — Grigoriev, Igor S. & Meilikhov, Evgenii Z. Handbook of Physical Quantities. Boca Raton: CRC Press, 1997: 116. 21.450 g/cm3 — Savitskii, E.M. Physical Metallurgy of Platinum Metals. New York: Pergamon Press, 1978: 31. (20 °C) 21.45 g/cm3 — Vines, R.F. The Platinum Metals and their Alloys. New York: The International Nickel Company, Inc., 1941: 16. — "Values ranging from 21.3 to 21.5 gm/cm3 at 20 °C have been reported for the density of annealed platinum; the best value being about 21.45 gm/cm3 at 20 °C." 21.46 g/cm3 — Rose, T. Kirke. The Precious Metals, Comprising Gold, Silver and Platinum. New York: D. Van Nostrand Company, 1909: 255. — "Pure platinum, according to G. Matthey has a density of 21.46." | |
| 79 Au gold | |
| use | 19.3 g/cm3 |
| WEL | (near r.t.) 19300 kg/m3 |
| LNG | (at r.t.) 19.3 g/cm3 |
| CRC | (near r.t.) 19.3 g/cm3 |
| 81 Tl thallium | |
| use | 11.85 g/cm3 |
| WEL | (near r.t.) 11850 kg/m3 |
| LNG | (at r.t.) 11.85 g/cm3 |
| CRC | (near r.t.) 11.8 g/cm3 |
| 82 Pb lead | |
| use | 11.34 g/cm3 |
| WEL | (near r.t.) 11340 kg/m3 |
| LNG | (at 20 °C) (face-centered cubic) (11.34 rel. to water at 4 °C) |
| CRC | (near r.t.) 11.3 g/cm3 |
| 83 Bi bismuth | |
| use | 9.78 g/cm3 |
| WEL | (near r.t.) 9780 kg/m3 |
| LNG | (at r.t.) 9.78 g/cm3 |
| CRC | (near r.t.) 9.79 g/cm3 |
| 84 Po polonium (alpha) | |
| use | 9.196 g/cm3 |
| WEL | (near r.t.) 9196 kg/m3 |
| LNG | (at r.t.) 9.196 g/cm3 |
| CRC | (near r.t.) 9.20 g/cm3 |
| 84 Po polonium (beta) | |
| use | 9.398 g/cm3 |
| LNG | (at r.t.) 9.398 g/cm3 |
| 85 At astatine | |
| use | ? 7 g/cm3 |
| 87 Fr francium | |
| use | ? 2.48 g/cm3 |
| 88 Ra radium | |
| use | 5.5 g/cm3 |
| WEL | (near r.t.) 5000 kg/m3 |
| LNG | (at r.t.) 5.5 g/cm3 |
| CRC | (near r.t.) 5 g/cm3 |
| 89 Ac actinium (Ac-227 ?) | |
| use | 10 g/cm3 |
| WEL | (near r.t.) 10070 kg/m3 |
| LNG | (at r.t.) (Ac-227) 10.07 g/cm3 |
| CRC | (near r.t.) 10 g/cm3 |
| 90 Th thorium | |
| use | 11.7 g/cm3 |
| WEL | (near r.t.) 11724 kg/m3 |
| LNG | (at r.t.) 11.7 g/cm3 |
| CRC | (near r.t.) 11.7 g/cm3 |
| 91 Pa protactinium | |
| use | 15.37 g/cm3 |
| WEL | (near r.t.) 15370 kg/m3 |
| CRC | (near r.t.) 15.4 g/cm3 |
| 92 U uranium | |
| use | 19.1 g/cm3 |
| WEL | (near r.t.) 19050 kg/m3 |
| LNG | (at r.t.) 19.1 g/cm3 |
| CRC | (near r.t.) 19.1 g/cm3 |
| 93 Np neptunium | |
| use | 20.2 g/cm3 |
| WEL | (near r.t.) 20450 kg/m3 |
| LNG | (at r.t.) 20.2 g/cm3 |
| CRC | (near r.t.) 20.2 g/cm3 |
| 94 Pu plutonium | |
| use | 19.816 g/cm3 |
| WEL | (near r.t.) 19816 kg/m3 |
| LNG | (at 20 °C) (19.816 rel. to water at 4 °C) |
| CRC | (near r.t.) 19.7 g/cm3 |
| 95 Am americium | |
| use | 12 g/cm3 |
| LNG | (at r.t.) 12 g/cm3 |
| CRC | (near r.t.) 12 g/cm3 |
| 96 Cm curium (Cm-244 ?) | |
| use | 13.51 g/cm3 |
| WEL | (near r.t.) 13510 kg/m3 |
| LNG | (at r.t.) (Cm-244) 13.51 g/cm3 |
| CRC | (near r.t.) 13.51 g/cm3 |
| 97 Bk berkelium (alpha) | |
| use | 14.78 g/cm3 |
| WEL | (near r.t.) 14780 kg/m3 |
| LNG | (at r.t.) 14.78 g/cm3 |
| CRC | (near r.t.) 14.78 g/cm3 |
| 97 Bk berkelium (beta) | |
| use | 13.25 g/cm3 |
| LNG | (at r.t.) 13.25 g/cm3 |
| CRC | (near r.t.) 13.25 g/cm3 |
| 98 Cf californium | |
| use | 15.1 g/cm3 |
| WEL | (near r.t.) 15100 kg/m3 |
| CRC | (near r.t.) 15.1 g/cm3 |
| 99 Es einsteinium | |
| use | 8.84 g/cm3 |
| LNG | (at r.t.) 8.84 g/cm3 |
Density, liquid phase
[edit]
| 2 He helium-4 | |
--- |
| Donnelly et al. | 0.1249772 g/cm3 (He-I at boiling point: 4.222 K) |
| 0.1461087 g/cm3 (at lambda transition He-I/He-II: 2.1768 K, saturated vapor pressure) |
| 0.1451397 g/cm3 (He-II superfluid at 0 K, saturated vapor pressure) |
| Hoffer et al. | 0.17309 g/cm3 (from 23.125 cm3/mole; He-II from hcp melt at 0 K, 25.00 atm) |
| 0.17308 g/cm3 (from 23.1256 cm3/mole; at local min. density, from hcp melt at 0.699 K, 24.993 atm) |
| 0.17443 g/cm3 (from 22.947 cm3/mole; He-II at triple point hcp−bcc−He-II: 1.463 K, 26.036 atm) |
| 0.1807 g/cm3 (from 22.150 cm3/mole; He-I at triple point hcp−bcc−He-I: 1.772 K, 30.016 atm) |
| 3 Li lithium | |
| use | 0.512 g/cm3 |
| CR2 | (at m.p.) 0.512 g/cm3 |
| 4 Be beryllium | |
| use | 1.690 g/cm3 |
| CR2 | (at m.p.) 1.690 g/cm3 |
| 5 B boron | |
| use | 2.08 g/cm3 |
| CR2 | (at m.p.) 2.08 g/cm3 |
| 11 Na sodium | |
| use | 0.927 g/cm3 |
| CR2 | (at m.p.) 0.927 g/cm3 |
| 12 Mg magnesium | |
| use | 1.584 g/cm3 |
| CR2 | (at m.p.) 1.584 g/cm3 |
| 13 Al aluminium | |
| use | 2.375 g/cm3 |
| CR2 | (at m.p.) 2.375 g/cm3 |
| 14 Si silicon | |
| use | 2.57 g/cm3 |
| CR2 | (at m.p.) 2.57 g/cm3 |
| 16 S sulfur | |
| use | 1.819 g/cm3 |
| CR2 | (at m.p.) 1.819 g/cm3 |
| 17 Cl chlorine | |
| use | (at −35 °C) 1.5649 g/cm3 |
| LNG | (at −35 °C) 1.5649 g/cm3 |
| 19 K potassium | |
| use | 0.828 g/cm3 |
| CR2 | (at m.p.) 0.828 g/cm3 |
| 20 Ca calcium | |
| use | 1.378 g/cm3 |
| CR2 | (at m.p.) 1.378 g/cm3 |
| 21 Sc scandium | |
| use | 2.80 g/cm3 |
| CR2 | (at m.p.) 2.80 g/cm3 |
| 22 Ti titanium | |
| use | 4.11 g/cm3 |
| CR2 | (at m.p.) 4.11 g/cm3 |
| 23 V vanadium | |
| use | 5.5 g/cm3 |
| CR2 | (at m.p.) 5.5 g/cm3 |
| 24 Cr chromium | |
| use | 6.3 g/cm3 |
| CR2 | (at m.p.) 6.3 g/cm3 |
| 25 Mn manganese | |
| use | 5.95 g/cm3 |
| CR2 | (at m.p.) 5.95 g/cm3 |
| 26 Fe iron | |
| use | 6.98 g/cm3 |
| CR2 | (at m.p.) 6.98 g/cm3 |
| 27 Co cobalt | |
| use | 7.75 g/cm3 |
| CR2 | (at m.p.) 7.75 g/cm3 |
| 28 Ni nickel | |
| use | 7.81 g/cm3 |
| CR2 | (at m.p.) 7.81 g/cm3 |
| 29 Cu copper | |
| use | 8.02 g/cm3 |
| CR2 | (at m.p.) 8.02 g/cm3 |
| 30 Zn zinc | |
| use | 6.57 g/cm3 |
| CR2 | (at m.p.) 6.57 g/cm3 |
| 31 Ga gallium | |
| use | 6.095 g/cm3 |
| LNG | (at 29.8 °C, m.p. is 29.7646 °C) 6.095 g/cm3 |
| CR2 | (at m.p.) 6.08 g/cm3 |
| 32 Ge germanium | |
| use | 5.60 g/cm3 |
| CR2 | (at m.p.) 5.60 g/cm3 |
| 33 As arsenic | |
| use | 5.22 g/cm3 |
| CR2 | (at m.p.) 5.22 g/cm3 |
| 34 Se selenium | |
| use | 3.99 g/cm3 |
| CR2 | (at m.p.) 3.99 g/cm3 |
| 35 Br bromine (Br2) | |
| use | (near r.t.) 3.1028 g/cm3 |
| LNG | (at 25 °C) (3.1023 rel. to water at 4 °C) |
| CRC | (near r.t.) 3.1028 g/cm3 |
| 37 Rb rubidium | |
| use | 1.46 g/cm3 |
| CR2 | (at m.p.) 1.46 g/cm3 |
| 38 Sr strontium | |
| use | 6.980 g/cm3 |
| CR2 | (at m.p.) 6.980 g/cm3 |
| 39 Y yttrium | |
| use | 4.24 g/cm3 |
| CR2 | (at m.p.) 4.24 g/cm3 |
| 40 Zr zirconium | |
| use | 5.8 g/cm3 |
| CR2 | (at m.p.) 5.8 g/cm3 |
| 42 Mo molybdenum | |
| use | 9.33 g/cm3 |
| CR2 | (at m.p.) 9.33 g/cm3 |
| 44 Ru ruthenium | |
| use | 10.65 g/cm3 |
| CR2 | (at m.p.) 10.65 g/cm3 |
| 45 Rh rhodium | |
| use | 10.7 g/cm3 |
| CR2 | (at m.p.) 10.7 g/cm3 |
| 46 Pd palladium | |
| use | 10.38 g/cm3 |
| CR2 | (at m.p.) 10.38 g/cm3 |
| 47 Ag silver | |
| use | 9.320 g/cm3 |
| CR2 | (at m.p.) 9.320 g/cm3 |
| 48 Cd cadmium | |
| use | 7.996 g/cm3 |
| CR2 | (at m.p.) 7.996 g/cm3 |
| 49 In indium | |
| use | 7.02 g/cm3 |
| CR2 | (at m.p.) 7.02 g/cm3 |
| 50 Sn tin | |
| use | 6.99 g/cm3 |
| CR2 | (at m.p.) 6.99 g/cm3 |
| 51 Sb antimony | |
| use | 6.53 g/cm3 |
| CR2 | (at m.p.) 6.53 g/cm3 |
| 52 Te tellurium | |
| use | 5.70 g/cm3 |
| CR2 | (at m.p.) 5.70 g/cm3 |
| 55 Cs caesium | |
| use | 1.843 g/cm3 |
| CR2 | (at m.p. (28.44 °C, near r.t.)) 1.843 g/cm3 |
| 56 Ba barium | |
| use | 3.338 g/cm3 |
| CR2 | (at m.p.) 3.338 g/cm3 |
| 57 La lanthanum | |
| use | 5.94 g/cm3 |
| CR2 | (at m.p.) 5.94 g/cm3 |
| 58 Ce cerium | |
| use | 6.55 g/cm3 |
| CR2 | (at m.p.) 6.55 g/cm3 |
| 59 Pr praseodymium | |
| use | 6.50 g/cm3 |
| CR2 | (at m.p.) 6.50 g/cm3 |
| 60 Nd neodymium | |
| use | 6.89 g/cm3 |
| CR2 | (at m.p.) 6.89 g/cm3 |
| 62 Sm samarium | |
| use | 7.16 g/cm3 |
| CR2 | (at m.p.) 7.16 g/cm3 |
| 63 Eu europium | |
| use | 5.13 g/cm3 |
| CR2 | (at m.p.) 5.13 g/cm3 |
| 64 Gd gadolinium | |
| use | 7.4 g/cm3 |
| CR2 | (at m.p.) 7.4 g/cm3 |
| 65 Tb terbium | |
| use | 7.65 g/cm3 |
| CR2 | (at m.p.) 7.65 g/cm3 |
| 66 Dy dysprosium | |
| use | 8.37 g/cm3 |
| CR2 | (at m.p.) 8.37 g/cm3 |
| 67 Ho holmium | |
| use | 8.34 g/cm3 |
| CR2 | (at m.p.) 8.34 g/cm3 |
| 68 Er erbium | |
| use | 8.86 g/cm3 |
| CR2 | (at m.p.) 8.86 g/cm3 |
| 69 Tm thulium | |
| use | 8.56 g/cm3 |
| CR2 | (at m.p.) 8.56 g/cm3 |
| 70 Yb ytterbium | |
| use | 6.21 g/cm3 |
| CR2 | (at m.p.) 6.21 g/cm3 |
| 71 Lu lutetium | |
| use | 9.3 g/cm3 |
| CR2 | (at m.p.) 9.3 g/cm3 |
| 72 Hf hafnium | |
| use | 12 g/cm3 |
| CR2 | (at m.p.) 12 g/cm3 |
| 73 Ta tantalum | |
| use | 15 g/cm3 |
| CR2 | (at m.p.) 15 g/cm3 |
| 74 W tungsten | |
| use | 17.6 g/cm3 |
| CR2 | (at m.p.) 17.6 g/cm3 |
| 75 Re rhenium | |
| use | 18.9 g/cm3 |
| CR2 | (at m.p.) 18.9 g/cm3 |
| 76 Os osmium | |
| use | 20 g/cm3 |
| CR2 | (at m.p.) 20 g/cm3 |
| 77 Ir iridium | |
| use | 19 g/cm3 |
| CR2 | (at m.p.) 19 g/cm3 |
| 78 Pt platinum | |
| use | 19.77 g/cm3 |
| CR2 | (at m.p.) 19.77 g/cm3 |
| 79 Au gold | |
| use | 17.31 g/cm3 |
| CR2 | (at m.p.) 17.31 g/cm3 |
| 80 Hg mercury | |
| use | (at r.t.) 13.534 g/cm3 |
| LNG | (at r.t.) 13.534 g/cm3 |
| CRC | (near r.t.) 13.5336 g/cm3 |
| 81 Tl thallium | |
| use | 11.22 g/cm3 |
| CR2 | (at m.p.) 11.22 g/cm3 |
| 82 Pb lead | |
| use | 10.66 g/cm3 |
| CR2 | (at m.p.) 10.66 g/cm3 |
| 83 Bi bismuth | |
| use | 10.05 g/cm3 |
| CR2 | (at m.p.) 10.05 g/cm3 |
| 92 U uranium | |
| use | 17.3 g/cm3 |
| CR2 | (at m.p.) 17.3 g/cm3 |
| 94 Pu plutonium | |
| use | 16.63 g/cm3 |
| CR2 | (at m.p.) 16.63 g/cm3 |
Density, gas phase
[edit]
| | value | conditions |
---
| 1 H hydrogen (H2) | | |
| use | 0.08988 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 0.082 g/L | 25 °C, 101.325 kPa |
| KCH | 0.08988 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 0.08988 g/L | 0 °C, 101.325 kPa |
| (lit. source) 0.08988 g/L | |
| 2 He helium | | |
| use | 0.1786 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 0.164 g/L | 25 °C, 101.325 kPa |
| LNG | 0.176 g/L | room temperature |
| KCH | 0.1785 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 0.1786 g/L | 0 °C, 101.325 kPa |
| (lit. source) 0.1785 g/L | 0 °C, 1 atm |
| 7 N nitrogen (N2) | | |
| use | 1.251 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 1.145 g/L | 25 °C, 101.325 kPa |
| LNG | 1.165 g/L | 20 °C |
| KCH | 1.2505 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 1.251 g/L | 0 °C, 101.325 kPa |
| (lit. source) 1.2506 g/L | |
| 8 O oxygen (O2) | | |
| use | 1.429 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 1.308 g/L | 25 °C, 101.325 kPa |
| LNG | 1.331 g/L | 20 °C |
| KCH | 1.42895 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 1.429 g/L | 0 °C, 101.325 kPa |
| (lit. source) 1.429 g/L | 0 °C |
| 9 F fluorine (F2) | | |
| use | 1.7 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 1.553 g/L | 25 °C, 101.325 kPa |
| VDW | (lit. source) 1.696 g/L | |
| | 1.6074 kg/m3 | 15 °C, 1.013 bar |
| Archived 2007-09-28 at the Wayback Machine | 0.0983 lb/ft3 | 70 °F (21 °C), 1 atm |
| 10 Ne neon | | |
| use | 0.9002 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 0.825 g/L | 25 °C, 101.325 kPa |
| LNG | 0.8999 g/L | 0 °C |
| KCH | 0.9002 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 0.9002 g/L | 0 °C, 101.325 kPa |
| (lit. source) 0.89990 g/L | 0 °C, 1 atm |
| 17 Cl chlorine (Cl2) | | |
| use | 3.2 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 2.898 g/L | 25 °C, 101.325 kPa |
| LNG | 2.98 g/L | 20 °C |
| KCH | 3.214 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 3.116 g/L | 0 °C, 101.325 kPa |
| (lit. source) 3.214 g/L | |
| 18 Ar argon | | |
| use | 1.784 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 1.633 g/L | 25 °C, 101.325 kPa |
| LNG | 1.7824 g/L | 0 °C |
| KCH | 1.784 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 1.784 g/L | 0 °C, 101.325 kPa |
| (lit. source) 1.7837 g/L | |
| 36 Kr krypton | | |
| use | 3.749 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 3.425 g/L | 25 °C, 101.325 kPa |
| LNG | 3.7493 g/L | room temperature |
| KCH | 3.744 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 3.749 g/L | 0 °C, 101.325 kPa |
| (lit. source) 3.733 g/L | 0 °C |
| 54 Xe xenon | | |
| use | 5.894 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 5.366 g/L | 25 °C, 101.325 kPa |
| LNG | 5.761 g/L | room temperature |
| KCH | 5.897 kg/m3 | 0 °C, 101.3 kPa |
| VDW | 5.894 g/L | 0 °C, 101.325 kPa |
| (lit. source) 5.88 g/L | |
| 86 Rn radon (Radon-222 ?) | | |
| use | 9.73 g/L | 0 °C, 101.325 kPa |
| CRC | (calc. ideal gas) 9.074 g/L | 25 °C, 101.325 kPa |
| LNG | 9.73 g/L | room temperature |
| VDW | (lit. source) 9.73 g/L | |
| ICT.a | 9.73 g/L | 0 °C, 1 An (=101.325 kPa) formula weight = 222 |
Notes
[edit]
The suggested values for solid densities refer to "near room temperature (r.t.)" by default.
The suggested values for liquid densities refer to "at the melting point (m.p.)" by default.
See also
[edit]
Hardnesses of the elements (data page)
References
[edit]
WEL
[edit]
As quoted at from these sources:
A.M. James and M.P. Lord in Macmillan's Chemical and Physical Data, Macmillan, London, UK, 1992.
D.R. Lide, (ed.) in Chemical Rubber Company handbook of chemistry and physics, CRC Press, Boca Raton, Florida, USA, 77th edition, 1996.
J.A. Dean (ed) in Lange's Handbook of Chemistry, McGraw-Hill, New York, USA, 14th edition, 1992.
G.W.C. Kaye and T.H. Laby in Tables of physical and chemical constants, Longman, London, UK, 15th edition, 1993.
CRC
[edit]
As quoted from various sources in an online version of:
David R. Lide (ed), CRC Handbook of Chemistry and Physics, 84th Edition. CRC Press. Boca Raton, Florida, 2003; Section 4, Properties of the Elements and Inorganic Compounds; Physical Constants of Inorganic Compounds
CR2
[edit]
David R. Lide (ed), CRC Handbook of Chemistry and Physics, 84th Edition. CRC Press. Boca Raton, Florida, 2003; Section 4, Properties of the Elements and Inorganic Compounds; Density of Molten Elements and Representative Salts
LNG
[edit]
As quoted from an online version of:
J.A. Dean (ed), Lange's Handbook of Chemistry (15th Edition), McGraw-Hill, 1999; Section 3; Table 3.2 Physical Constants of Inorganic Compounds
VDW
[edit]
The following molar volumes and densities for the majority of the gaseous elements were calculated from the van der Waals equation of state, using the quoted values of the van der Waals constants. The source for the van der Waals constants and for the literature densities was: R. C. Weast (Ed.), Handbook of Chemistry and Physics (53rd Edn.), Cleveland:Chemical Rubber Co., 1972.
Donnelly et al.
[edit]
Donnelly, Russell J.; Barenghi, Carlo F. (1998). "The Observed Properties of Liquid Helium at the Saturated Vapor Pressure". Journal of Physical and Chemical Reference Data. 27 (6): 1217–74. Bibcode:1998JPCRD..27.1217D. doi:10.1063/1.556028.
Hoffer et al.
[edit]
Hoffer, J. K.; Gardner, W. R.; Waterfield, C. G.; Phillips, N. E. (April 1976). "Thermodynamic properties of 4He. II. The bcc phase and the P-T and VT phase diagrams below 2 K". Journal of Low Temperature Physics. 23 (1): 63–102. Bibcode:1976JLTP...23...63H. doi:10.1007/BF00117245. S2CID 120473493.
| | | | | | | | |
--- --- --- --- |
| | van der Waals constants | | 25 °C, 100.0 kPa | | 0 °C, 1 atm | | Lit. |
| | a (L2⋅bar/mol2) | b (L/mol) | Vm (L) | d (g/L) | Vm (L) | d (g/L) | d (g/L) |
| Argon | 1.363 | 0.03219 | 24.77 | 1.613 | 22.39 | 1.784 | 1.7837 |
| Chlorine | 6.579 | 0.05622 | 25.11 | 2.824 | 22.75 | 3.116 | 3.214 |
| Fluorine | not available | | | | | | 1.696 |
| Helium | 0.03457 | 0.02370 | 24.81 | 0.1613 | 22.44 | 0.1786 | 0.1785 (0 °C, 1 atm) |
| Hydrogen | 0.2476 | 0.02661 | 24.81 | 0.08127 | 22.43 | 0.08988 | 0.08988 |
| Krypton | 2.349 | 0.03978 | 24.73 | 3.388 | 22.35 | 3.749 | 3.733 (0 °C) |
| Neon | 0.2135 | 0.01709 | 24.80 | 0.8139 | 22.42 | 0.9002 | 0.89990 (0 °C, 1 atm) |
| Nitrogen | 1.408 | 0.03913 | 24.77 | 1.131 | 22.39 | 1.251 | 1.2506 |
| Oxygen | 1.378 | 0.03183 | 24.77 | 1.292 | 22.39 | 1.429 | 1.429 (0 °C) |
| Radon | not available | | | | | | 9.73 |
| Xenon | 4.250 | 0.05105 | 24.69 | 5.323 | 22.28 | 5.894 | 5.88 |
Other
[edit]
KCH: Kuchling, Horst, Taschenbuch der Physik, 13. Auflage, Verlag Harri Deutsch, Thun und Frankfurt/Main, German edition, 1991. ISBN 3-8171-1020-0
: (a) Gray and Ramsay, Proceedings of the Royal Society (London). A. Mathematical and Physical Sciences. 84: 536; (1911)
| v t e Chemical elements data | |
--- |
| Elements | List of chemical elements—atomic mass, atomic number, symbol, name Periodic table |
| Data | Abundance of the chemical elements in Earth's crust, sea water, Sun and Solar System + data page Atomic radius empirical, calculated, van der Waals radius, covalent radius + data page Boiling point + data page Critical point + data page Density solid, liquid, gas + data page Elastic properties of the elements: Young's modulus, Poisson's ratio, bulk modulus, shear modulus + data page Electrical resistivity + data page Electron affinity + data page Electron configuration + data page Electronegativity (Pauling, Allen scale) + data page Hardness: Mohs hardness, Vickers hardness, Brinell hardness + data page Heat capacity + data page Heat of fusion + data page Heat of vaporization + data page Ionization energy (in eV) and molar ionization energies (in kJ/mol) + data page Melting point + data page Molar ionization energy Oxidation state + data table Speed of sound + data page Standard atomic weight Thermal conductivity + data page Specific heat capacity Thermal expansion + data page Vapor pressure + data page |
| v t e Periodic table | |
--- |
| Periodic table forms | Alternatives Extended periodic table |
| Sets of elements | | | | | | | | | | --- --- --- --- | | By periodic table structure | | | | --- | | Groups | 1 (Hydrogen and alkali metals) 2 (Alkaline earth metals) 3 4 5 6 7 8 9 10 11 12 13 (Triels) 14 (Tetrels) 15 (Pnictogens) 16 (Chalcogens) 17 (Halogens) 18 (Noble gases) | | Periods | 1 2 3 4 5 6 7 8+ + Aufbau + Fricke + Pyykkö | | Blocks | Aufbau principle | | | By metallicity | | | | --- | | Metals | Lanthanides Actinides Transition metals Post-transition metals | | Metalloids | Lists of metalloids by source Dividing line | | Nonmetals | Noble gases | | | Other sets | Platinum-group metals (PGM) Rare-earth elements Refractory metals Precious metals Coinage metals Noble metals Heavy metals Native metals Transuranium elements Superheavy elements Major actinides Minor actinides | |
| Elements | | | | --- | | Lists | By: Abundance (in humans) Atomic properties Nuclear stability Symbol | | Properties | Aqueous chemistry Crystal structure Electron configuration Electronegativity Goldschmidt classification Term symbol | | Data pages | Abundance Atomic radius Boiling point Critical point Density Elasticity Electrical resistivity Electron affinity Electron configuration Electronegativity Hardness Heat capacity Heat of fusion Heat of vaporization Ionization energy Melting point Oxidation state Speed of sound Thermal conductivity Thermal expansion coefficient Vapor pressure | |
| History | Element discoveries + Dmitri Mendeleev + 1871 table + 1869 predictions Naming + etymology + controversies + for places + for people + in East Asian languages |
| See also | IUPAC + nomenclature + systematic element name Trivial name Dmitri Mendeleev |
| Category WikiProject | |
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7951 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10060259/ | Multiple myeloma with acute light chain cast nephropathy - PMC
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. 2023 Mar 29;13(1):46. doi: 10.1038/s41408-023-00806-w
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Multiple myeloma with acute light chain cast nephropathy
Nelson Leung
Nelson Leung
1 Division of Hematology, Mayo Clinic, Rochester, MN USA
2 Division of Nephrology and Hypertension, Mayo Clinic, Rochester, MN USA
Find articles by Nelson Leung
1,2,✉, S Vincent Rajkumar
S Vincent Rajkumar
1 Division of Hematology, Mayo Clinic, Rochester, MN USA
Find articles by S Vincent Rajkumar
1
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1 Division of Hematology, Mayo Clinic, Rochester, MN USA
2 Division of Nephrology and Hypertension, Mayo Clinic, Rochester, MN USA
✉
Corresponding author.
Received 2022 Nov 18; Revised 2023 Feb 21; Accepted 2023 Feb 27; Collection date 2023 Dec.
© The Author(s) 2023
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons license, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons license and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this license, visit
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PMCID: PMC10060259 PMID: 36990996
Abstract
Light chain cast nephropathy (LCCN) is a leading cause of acute kidney injury (AKI) in patients with multiple myeloma (MM) and is now defined as a myeloma defining event. While the long-term prognosis has improved with novel agents, short-term mortality remains significantly higher in patients with LCCN especially if the renal failure is not reversed. Recovery of renal function requires a rapid and significant reduction of the involved serum free light chain. Therefore, proper treatment of these patients is of the utmost importance. In this paper, we provide an algorithm for treatment of MM patients who present with biopsy-proven LCCN or in those where other causes of AKI have been ruled out. The algorithm is based on data from randomized trial whenever possible. When trial data is not available, our recommendations is based on non-randomized data and expert opinions on best practices. We recommend that all patients should enroll in a clinical trial if available prior to resorting to the treatment algorithm we outlined.
Subject terms: Myeloma, Therapeutics
Introduction
Acute kidney injury (AKI) caused by light chain cast nephropathy (LCCN) is one of the major complications from multiple myeloma (MM). It is most commonly seen at initial MM diagnosis but can also develop later in the course of disease during relapse. The incidence AKI at diagnosis is 16–31% when measured by serum creatinine (Scr) concentration >1.4 mg/dl, and 16–22% when defined using Scr >2 mg/dl [1, 2]. By estimated glomerular filtration rate (eGFR), 36–45% have an eGFR < 60 ml/min/1.73 m 2, while 12–17% have an eGFR <30 ml/min/1.73 m 2 [2–4]. A large French study of 1038 patients with MM found that ~25% met the 2014 International Myeloma Working Group (IMWG) renal impairment criterion (Scr > 2 mg/dl or eGFR < 40 ml/min/1.73 m 2), with 12.9% requiring dialysis . Other have reported dialysis dependence in 6–8% of patients during the clinical course of MM [4, 6].
Of the four myeloma defining events (MDEs) (hyperCalcemia, Renal impairment, Anemia and Bone lytic lesions), renal impairment imposes the greatest impact on overall survival (OS) even after adjusting for other cofactors and comorbidities [2–6]. This was especially evident in the alkylator era [6, 7]. Although this effect is diminished by novel agents particularly bortezomib in clinical trial settings [8, 9], real world data still reflect that AKI at the time of myeloma diagnosis imposes a negative impact on mortality particularly in the first 6 months of therapy [4, 5, 10]. Fortunately, recovery of kidney function reverses the negative impact on OS [5–7, 10]. But while long-term survival has improved, short-term mortality remains higher in patients without kidney recovery regardless of the treatment regimen .
Mechanisms of renal impairment
In 2014, the IMWG clarified that only AKI secondary to LCCN qualifies as a MDE . LCCN occurs when the overproduced monoclonal free light chain (FLC) by the myeloma cells interacts with the Tamm Horsfall protein in the loop of Henle to form light chain casts . The light chain casts obstruct the tubules causing rupture which induces an immune response further injuring the tubules . Injury is also mediated by hydrogen peroxide produced by the FLC and FLC activation of the Nuclear factor kappa-light-chain-enhancer of activated B cells (NF-κB) and Apoptosis signal-regulating kinase 1 (ASK1) or Janus kinases, signal transducer and activator of transcription proteins (STATs) pathway [13–15]. These pathways induce apoptosis, inflammation and promote tubulointerstitial fibrosis that further the damage beyond tubular objection.
Serum FLC concentrations are predictive of the development of AKI and its recovery. AKI is rare when serum FLC concentration is <50 mg/dl but increases significantly when the concentration exceeds 80–200 mg/dl [16, 17]. A high serum FLC concentration alone may not be sufficient as a high urinary FLC excretion appears to be necessary for AKI to occur [7, 18, 19]. Conversely, a rapid reduction of serum FLC concentration is the key to reversing the kidney injury. A minimum reduction of 50–60% of serum FLC has been found to be associated with renal recovery in LCCN [20, 21]. One study finds that fewer patients recovered kidney function with the same degree of FLC reduction achieved at day 21 as compared to day 12 . Finally, a serum FLC concentration of <50 mg/dl at the end of cycle 1 of chemotherapy is associated with better renal recovery in a recent clinical trial using a high cutoff (HCO) dialyzer .
Diagnosis
Renal impairment in MM is currently defined by an eGFR of <40 ml/min/1.73 m 2 or a Scr >2 mg/dl . For diagnostic and management purposes, the etiology of AKI in MM must be established. Only LCCN qualifies as a MDE, and its management is different from the management needs for AKI due to other causes. Thus, in patients with MM presenting with AKI, other causes of renal failure such as dehydration, hypercalcemia, drug-induced nephritis, unrelated diabetes or hypertension, and other monoclonal gammopathy related renal pathology must all be excluded. There are a number of renal disorders than are caused by monoclonal proteins besides LCCN that are collectively incorporated in the diagnostic umbrella now termed monoclonal gammopathy of renal significance (MGRS) . Kidney biopsy is the gold standard for differentiating between LCCN, MGRS lesions, and other unrelated causes of AKI . Unfortunately, a kidney biopsy is not always feasible for various reasons, especially in the acute setting. Once dehydration, hypercalcemia, and other causes of AKI are excluded based on clinical presentation, the main differential is between LCCN and MGRS renal lesions. Since most MGRS kidney lesions affect the glomeruli, they cause a high degree of albuminuria. In contrast, LCCN is associated with mainly Bence Jones proteinuria often with <10% albuminuria . Therefore, in patients with >1 g/d proteinuria with <10% albuminuria, and a serum FLC concentration >150 mg/dl, the probability of LCCN is high enough that a kidney biopsy can be omitted [7, 26, 27]. On the other hand, patients with high degree of albuminuria or lower serum FLC, or where there is any uncertainty about the etiology of AKI should undergo a kidney biopsy.
Current frontline treatment options
Proteasome inhibitors (Table 1)
Table 1.
Multiple myeloma trials with renal impaired patients.
| Trial | Regimen | Patients | Inclusion criteria | Hematologic and Renal Outcomes | Toxicity |
--- --- --- |
| Phase III VISTA | MP vs VMP | NDMM 450 with CrCl >50 ml/min 227 CrCl ≤ 50 ml/min | Scr <2 mg/dl | HR in patients with CrCl <50 ml/min ORR: MP = 46% vs VMP = 68% CR: MP = 5% vs VMP = 31% Renal response Overall: MP = 34% vs VMP = 44% CrCl < 30 ml/min: MP = 7% vs VMP = 37% CrCl 30 - < 50 ml/min: MP = 39% vs VMP = 46% | Grade 4 hematologic AEs (42%) and SAEs (63%) were higher in patients with CrCl ≤ 30 ml/min but discontinuation rates were similar. Patients who achieved reversal of RI had less grade 5 AEs (8% vs 15%), SAEs (43% vs 60%) and discontinuation (6% vs 24%) than patients without reversible RI. |
| Phase II study Ludwig et al. | BDD | 68 NDMM and RMM | eGFR < 50 ml/min | Hematologic response Renal response CR/nCR = 38% CR renal = 31% VGPR = 15% PR renal = 7% PR = 13% MR renal = 24% MR = 6% | Grade 3/4 AEs Anemia - 50% Neutropenia - 14.7% Thrombocytopenia -14.7% Infections: Grade 4 - 19.1%, Grade 3 - 48.5% |
| Phase III Hovon-65/ HMMG-HD4 | PAD/ASCT/Bortezomib maintenance vs VAD/ASCT/thalidomide maintenance | NDMM 81 with Scr ≥ 2 mg/dl 746 with <2 mg/dl | No exclusion for kidney function | HR in patients with Scr > 2 mg/dl ORR: PAD 75% vs VAD 36% >VGPR: PAD 33% vs VAD 9% Renal response ORR: PAD = 81% vs VAD = 63% CR renal: PAD = 58% vs VAD = 43% PR renal: PAD = 0% vs VAD = 3% MR renal: PAD = 23% vs VAD = 17% | Grade 4 AEs Anemia: VAD (7%) vs PAD (8%) Neutropenia: VAD (1%) vs PAD (3%) Thrombocytopenia: VAD (5%) vs PAD (10%) Infections: VAD (21%) vs PAD (26%) Nonhematologic AE GI symptoms: VAD (7%) vs PAD (11%)# PN: VAD (10%) vs PAD (24%)# = p< 0.01, # = p< 0.001 |
| Phase III MYRE | BD: bortezomib 1.3 mg/m 2, dex 20 mg Days (1,2,4,5, 8,9,11,12) cycle 1, cycle 2 is 28 days for age >70 years vs CBD: BD + iv CTX 750 mg/m 2 Day 1 | 184 NDMM with LCCN | Scr > 1.92 mg/dl or eGFR < 40 ml/min/1.73m2 | Hematologic response at 3 months ORR: BD = 78.3% vs CBD = 77.2% CR/VGPR: BD 39.1% vs CBD = 51.1% Renal response at 3 months ORR: BD = 44.6% vs CBD = 51.1% | Serious AEs BD – 32.6% vs CBD – 43.5% Grade ≥ 3 Cytopenia: BD (5.4%) vs CBD (9.8%) Sepsis/pneumonia: BD (2.2%) vs CBD (5.4%) None were statistically significant |
| Tosi et al. | T or TD Thalidomide 100–400 mg daily Dex- 40 mg days 1–4 | 20 RRMM | Scr > 1.5 mg/dl or CrCl < 60 ml/min | Hematologic response at 3 months ORR: 75% PR: 45% Renal response (Scr < 1.5 mg/dl) ORR: 80% | Constipation – 25% Lethargy – 25% Leukopenia – 5% 50% of patients on 400 mg dose could not tolerate the dose due to Grade ≥ 2 toxicity. |
| Analysis of the phase III MM-009 and MM-010 Dimopoulos et al. | Lenalidomide 25 mg days 1–21 Dex 40 mg Days 1–4, 9–12, 17–21 | 341 RRMM | Scr ≥ 2.0 mg/dl | Hematologic response CR: mild RI 16%, mod RI 16%. sRI 6% VGPR: mild RI 19%, mod RI 11%, sRI 31% PR: mild RI 30%, mod RI 29%, sRI 13% Improvement of renal function by 1 level was experienced by 72% | Neutropenia: 32–48%. Thrombocytopenia: 38% severe RI, 22% moderate RI and 9% mild/no RI. Anemia: 44% severe RI, 21% moderate RI, 5% mild/no RI Pneumonia: 25% severe RI, 9% moderate RI, 7% mild/no RI. Dehydration: 13% severe RI, 2% moderate RI and 0.8% mild/no RI |
| Phase II study Ludwig et al. | Lenalidomide dosed based on CrCl Dex 40 mg Days 1–4, 9–12, 17–21 cycle 1 and weekly afterwards | 35 NDMM RRMM LCCN with 15 biopsy confirmed | Scr > 2 mg/dl or CrCl < 50 ml/min | Hematologic and Renal response CR: 20% VGPR: 8% PR: 40% CR renal: 14.2% PR renal: 11.4% MR renal: 20% | Grade 3/4 toxicities Anemia: 43% Thrombocytopenia: 23% Leukopenia: 12% Neutropenia: 15% 3 patients withdrew due to AE and 4 deaths during the first 2 cycles |
| Phase II study Bridoux et al. | Lenalidomide (dose based on kidney function) Dex 40 mg weekly | 38 RRMM | No Scr cutoff but renal function needed to be stable | Hematologic response ORR: 76% CR: 0% VGPR: 26% PR: 50% ≥50 ml/min: 75% vs <50 ml/min: 78% | Nonserious AEs Thrombocytopenia: normal (40%) vs sRI/HD (63.6%) Leukopenia: normal (20%) vs sRI/HD (36.4%) Anemia: normal (0%) vs sRI/HD (27.3%) Fatigue: normal (20%) vs sRI/HD (45.5%) |
| PrE1003 | Lenalidomide dose escalation Dex 40 mg weekly | 63 RRMM | | Hematologic response at 3 months CrCl 30–60 ml/min: 60% CrCl < 30 ml/min not on dialysis: 60% CrCl < 30 ml/min on dialysis: 20% | Grade 3/4 AEs 45% Most common AEs were anemia, decreased appetite, muscle weakness/fatigue Pneumonia: 19.3% |
| Phase II MM-013 | Pomalidomide 4 mg days 1–21 Dex 40 mg weekly if <75 years of age or 20 mg weekly if >75 years of age | 81 RRMM | eGFR < 45 ml/min/1.73 m 2 | HR ORR/VGPR or better A: 39.4%/18.2% B: 32.4%/8/8% C: 14.3%/7.1% Renal response A: 18.2%; complete RR in 18.2$% B: 35.3% C: 7.1% | Grade 3/4 AEs (A/B/C) Neutropenia - 60.6%/44.1%/57.1% Anemia - 27.3%/35.3%/57.1% Thrombocytopenia – 27.3%/17.6%/50% Infections – 39.4%/26.5%/28.6% Pneumonia – 12.1%/5.9%/7.1% Hyperkalemia – 0%/8.8%/14.3% |
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AE adverse event, BDD and PAD bortezomib doxorubicin dexamethasone, CBD cyclophosphamide bortezomib dexamethasone, CR complete response, CrCl – creatinine clearance by Cockcroft Gault method, dex dexamethasone, eGFR estimated glomerular filtration rate by Modification of Diet in Renal Disease (MDRD) method, Group A eGFR 30–<45 ml/min/1.73 m 2, Group B eGFR < 30 ml/min/1.73 m 2 not dialysis dependent, Group C dialysis dependent, HR hematologic response, LCCN light chain cast nephropathy, MP melphalan prednisone, nCR – near complete response, NDMM newly diagnosed multiple myeloma, ORR overall response rate, PR partial response, RI renal impairment (mild = CrCl ≥ 60 ml/min, moderate = ≥30 to <60 ml/min, severe =<30 ml/min), RR renal response, RRMM relapse or refractory multiple myeloma, SAE severe adverse event, Scr serum creatinine, sRI severe renal impairment with CrCl <30 ml/min, VGPR very good partial response, VMP bortezomib melphalan prednisone.
Bortezomib is a reversible proteasome inhibitor that is not renally cleared nor nephrotoxic . In the VISTA trial, the addition of bortezomib (V) to melphalan and prednisone (MP) significantly increased the overall response rate (ORR) from 46 to 68%, and the complete response (CR) rate from 5 to 31%, respectively in patients with eGFR < 50 ml/min/1.73 m 2 . Renal recovery rate in those with an eGFR < 30 ml/min/1.73 m 2 was 37% in the VMP arm vs 7% in the MP treated patients. Bortezomib was combined with doxorubicin and dexamethasone (BDD) in a phase II study in both newly diagnosed (NDMM) and relapsed myeloma patients with a median eGFR of 20.5 (3.7–49.9) ml/min. BDD produced an ORR of 72%, with a very good partial response (VGPR) or better rate of 52% . Improvement in eGFR correlated with the depth of hematologic response (HR) with the median posttreatment eGFR of 59.6 ml/min, 38.9 ml/min and 16.8 ml/min in patients who achieved >VGPR, partial response (PR) or minimal response (MR), and stable disease or less, respectively. Median progression free survival (PFS) of this study was 12.1 months and OS had not been reached after a median follow-up of 22.4 months. In the phase III Hovon-65/HMMG-HD4 trial, BDD was compared to vincristine Adriamycin and dexamethasone (VAD) induction followed by autologous stem cell transplant (ASCT) followed by thalidomide maintenance in the VAD group and bortezomib maintenance in the BDD group . Despite a significantly higher prevalence of high risk cytogenetics [del17p and t(4;14)] in the patient with Scr >2 mg/dl, the OS of patients treated with BDD was similar to those with baseline creatinine ≤2 mg/dl . BDD achieved a significantly higher ORR and >VGPR rate (75% and 33%, respectively) in the patients with a baseline creatinine >2 mg/dl as compared to VAD (36% and 9%, respectively). Patients with a baseline creatinine ≤2 mg/dl had similar OS regardless of treatment but those with Scr >2 mg/dl treated with VAD had an inferior OS. Overall renal response was 81% (58% renal CR) in the BDD treated patients vs 63% (43% renal CR) in the VAD treated patients. A randomized trial comparing bortezomib plus dexamethasone (BD) to bortezomib, cyclophosphamide, dexamethasone (CBD) found similar HR (78.3% in BD vs 77.2% in CBD, p = 1.00) and >VGPR rate (39.1% in BD and 51.1% in CBD, p = 0.14) in NDMM with renal impairment not requiring dialysis at 3 months . There was no difference in the overall renal response (44.6% in BD vs 51.1% in CBD, p = 0.46) but recovery of patients with AKI stage 3 (creatinine increased >3 times of baseline) trended toward CBD therapy (23.2% in BD vs 46.7% in CBD, p = 0.07).
Limited data exist in the frontline therapy of severe renally impaired MM patients treated with the other proteasome inhibitors. A phase I/Ib study was conducted with ixazomib in severe renally impaired patients including 7 patients on dialysis . A dose of 3 mg on Days 1, 8, 15 on a 28-days cycle was established for safety and pharmacokinetics standpoint but no efficacy data was provided. A study of carfilzomib in severe renally impaired patients including 10 patients on dialysis recommended the doses of 27 and 56 mg/m 2 . In a phase II trial of carfilzomib dexamethasone in renally impaired relapsed refractory (RRMM) patients using the 15 and 27 mg/m 2 dosing, HR was similar in patients with varying degree of renal impairment including patients on dialysis . They achieved an ORR of 25.5%, with all being PR or less. Adverse effects were also similar except for AKI in the 3 of 10 patients with moderate renal impairment (creatinine clearance between 30 and 49 ml/min) . Unfortunately, the risk of renal toxicity including thrombotic microangiopathy makes carfilzomib an unattractive choice in MM patients with AKI [35, 36]. Currently, neither ixazomib nor carfilzomib is approved for frontline treatment of MM in the United States.
Immunomodulators (Table 1)
Immunomodulators have been used in MM patients with renal impairment. Despite a similar mechanism of actions, they possess different pharmacologic properties that give each a unique pharmacokinetics and different side effects. Thalidomide is hydrolyzed by all body fluids and probably has limited renal clearance. Pharmacokinetic studies however showed clearance with dialysis, thus it should be given after dialysis . Lenalidomide is renally cleared and dialyzable. It requires dose adjustment according to GFR and dialysis status . Pomalidomide is renally secreted, but it is metabolized by the liver resulting in only mild extension of its half-life even in severe renal impairment making dosage adjustment for renal function unnecessary but should be given post dialysis .
There are only two small studies of thalidomide in renal failure patients with MM. The first is a 7 patient study where thalidomide was dosed between 100 mg and 400 mg daily in patients with creatinine clearance (CrCl) between 47 ml/min and dialysis dependence . Three patients achieved a CR that lasted 5–8 months and 1 patient achieved a PR for 1 year. Adverse effects included constipation, peripheral neuropathy and hyperkalemia were noted in the dialysis dependent patient. In a separate study of 20 RRMM patients with renal impairment treated with thalidomide or thalidomide dexamethasone, 75% achieved a HR with 45% achieving a PR or better after stem cell transplant . Dose reduction to 200 mg daily was required in 50% of patients treated with the 400 mg daily dose due to adverse effects.
Several phase II trials had been performed with lenalidomide dexamethasone (Rd) in renally impaired MM patients. The dosing schedule used was: 10 mg daily for CrCl >30 but ≤50 ml/min, 15 mg q48 hours for CrCl <30 ml/min but not on dialysis, and 5 mg daily after dialysis in patients on dialysis for 21 days on a 28-days cycle in one study . In a small study of 35 NDMM and relapsed patients with high dose dexamethasone, 4 died during the first 2 cycles and 5 withdrew from the study, 3 were for adverse events. On an intention to treat analysis, 20% had a CR and 8% had a VGPR. Sixteen (45.7%) of the 35 patients had a renal response, with 14.2% achieving a renal CR, 11.4% with renal PR and 20% in renal MR. Another phase II trial of Rd using the same lenalidomide dosing but weekly dexamethasone achieved an ORR of 76% with 50% PR and 26% VGPR . The ORR was similar between patients with normal kidney function and mild renal impairment vs patients with moderate renal impairment to ESRD; however, only 1 patient with severe renal impairment (eGFR < 30 ml/min/1.73 m 2 or ESRD) achieved a VGPR. A dose escalation trial in renally impaired RRMM patients used a similar dosing schedule for dialysis independent patients dosed lenalidomide at 15 mg 3 times a week after dialysis for dialysis dependent patients . The highest dosing achieved in the study was 25 mg daily for patients with CrCl 30–59 ml/min, 15 mg daily for those with CrCl < 30 ml/min regardless of dialysis status. The ORR was 54.3% with the poorest response seen in the dialysis group (20%). Twenty-eight patients experienced grade 3/4 adverse events with 1 death in a dialysis dependent patient from lung infection, sepsis and multiorgan failure attributed to therapy while 5 others died of unrelated deaths due to cirrhosis, intra-abdominal hemorrhage, sudden death and 2 ESRD. Median PFS was 12.6 months and OS was 20.0 months.
A phase II trial was conducted with pomalidomide and low dose dexamethasone in RRMM patients . All 81 patients had an eGFR <45 ml/min/1.73 m 2 with 14 dialysis dependent patients. Pomalidomide was dosed at 4 mg days 1–21 on a 28-days cycle. ORR was 39.4% in group A (patients with eGFR between 30 and 45 ml/min/1.73 m 2), 32.4% in group B (patients with <30 ml/min/1.73 m 2 not on dialysis) and 14.3% in group C (patients on dialysis). No CR was recorded but VGPR was noted in 18.2% of group A, 8.8% of group B and 7.1% of group C patients. Median PFS correlated with baseline kidney function [6.5 m (A) vs 4.2 m (B) vs 2.4 m (C)] and OS [16.4 m (A) vs 11.8 m (B) vs 5.2 m (C)] was the shortest among dialysis dependent patients. Leukopenia was more than twice as common in dialysis dependent patients but infection rates were similar. Thrombocytopenia was also most severe in the dialysis dependent patients. Dose reduction and discontinuation rates were similar amongst the 3 groups. Pomalidomide is currently approved for relapsed MM with 2 prior therapies including lenalidomide and a proteasome inhibitor.
Anti-CD38 monoclonal antibodies
Currently, 3 randomized trials have been conducted with daratumumab in the upfront setting in MM. The addition of daratumumab to the backbone of Rd was compared in MAIA, bortezomib thalidomide dexamethasone (VTD) in CASSIOPEIA and bortezomib lenalidomide dexamethasone (VRD) in the GRIFFIN trial [46–48]. Unfortunately, all 3 trials excluded patients with an eGFR <30 ml/min/1.73 m 2 so data on severe renal impairment or dialysis are limited to case reports for daratumumab . A study comparing isatuximab, carfilzomib, dexamethasone vs carfilzomib plus dexamethasone has been conducted in RRMM with renal impairment (IKEMA trial) . Unfortunately, in this study, only 2.7% and 2.4% of the patients had severe renal impairment.
Extracorporeal therapies
Plasmapheresis (PLEX) for the treatment of AKI in MM was first reported in 1976 . Since then, three randomized trials had been performed with differing outcomes. Zucchelli et al. randomized 15 patients to daily PLEX with hemodialysis vs 14 patients to peritoneal dialysis (PD) . Patients who underwent PLEX had greater reduction of Bence Jones proteinuria than PD (p< 0.001). Eleven of 13 dialysis dependent patients treated with PLEX became dialysis independent but only 2 of 11 PD patients recovered kidney function. Although positive, this study was criticized for high early mortality (35.7%) in the PD group vs 6.7% in the PLEX treated patients. Johnson et al. randomized 21 patients to hemodialysis vs hemodialysis plus thrice weekly PLEX. Dialysis was required for 7 of 11 PLEX and 5 of 10 hemodialysis patients . Kidney function improved in 63.6% of the PLEX patients vs 50% of the hemodialysis patients (p = NS), but of the dialysis dependent patients, all 3 who recovered had received PLEX. Clark et al. randomized 58 patients to 5–7 PLEX over the first 10 days and 39 patients to the control group. Hemodialysis was required for 25.9% of the PLEX patients vs 36% of the controls . At the end of the study, dialysis requirement and death were noted in 17.9% and 33.3% of the controls and 8.6% and 32.8% of the PLEX patients respectively, (p = NS). The primary endpoint (a composite of death, dialysis dependence and eGFR < 29 ml/min/1.73 m 2) was documented in 57.9% of PLEX patients vs 69.2% of the controls, p = 0.36.
Although some felt that PLEX was ineffective based on the Clark study, it is important to point out significant differences among these studies. LCCN was confirmed by kidney biopsy in majority of the patients in the Zucchelli and Johnson studies, but few biopsies were performed in the Clark study [52–54]. Dialysis requirement was the threefold higher in positive Zucchelli study than the negative Clark study. Investigators in the Johnson study also felt that PLEX was more beneficial in patients with the more severe AKI. Finally, patients in the Zucchelli study received daily PLEX while the Johnson and Clark studies were limited to every other day [52–54].
Other extracorporeal devices used in the treatment of LCCN includes the high cut-off (HCO) dialyzers. HCO dialyzers are dialyzers with pore size up to 50 kd (vs 5 kd in normal dialyzers) which can reduce serum FLC levels by >70% . Two randomized trials have been performed to date. MYRE was the first trial to report enrolled 98 dialysis dependent patients in France from 2011 to 2016 . Patient were treated with BD on a 21-day cycle and cyclophosphamide could be added after cycle 3 if response was insufficient. Eight 5 h HCO dialysis were performed within the first 10 days while the control group received the same with a regular dialyzer. HCO dialyzer treatment produced a higher ORR (78.3% vs 60.4%, p 0.06) and significantly higher >VGPR rate (69.6% vs 47.9% respectively, OR - 2.37, p = 0.03) as compared to control. HCO hemodialysis resulted in a significantly higher renal recovery at 6 months (secondary endpoint, 56.4% vs 35.4% respectively, odds ratio – 2.37, p = 0.04) and 12 months (60.9% of the HCO vs 37.5% of control, OR 2.59, p = 0.02) but did not reach significance at 3 months (primary endpoint, 41.3% vs 33.3% respectively, p = 0.42). There was no difference in OS.
The EuLITE study enrolled 90 dialysis dependent patients mainly in the United Kingdom from 2008 to 2013 used BDD as the chemotherapy . The HCO dialysis was scheduled for 6 h the first day followed by another seven 8 h sessions over 10 days. Control patients received 4 h dialysis thrice weekly. Despite similar single session FLC reductions (-77% for κ, -72% for λ) as MYRE, patients in EuLITE treated with the HCO dialyzer had a lower CR (14% vs 30%) and VGPR (23% vs 32%) at 6 months as compared to control patients [23, 56]. Moreover, the HCO group had an inferior age adjusted OS (hazard ratio – 2.63, p = 0.03) as compared to control despite similar renal recovery rates at 90 days (56% HCO vs 51% control, p - 0.81) . Higher infectious complications especially pulmonary infection in the HCO group vs control (31% vs 9% respectively) may have played a role.
Recommendations
Given the importance of renal recovery and the requirement for rapid serum FLC reduction, an aggressive therapeutic approach is justified in patients MM patients with AKI. Regimens used should have a high and rapid response rate; the drugs used should not require modifications for renal function and should be readily available for immediate administration. The goal should be to reduce circulating serum FLCs as quickly as possible, including the use of PLEX (Fig. 1). In NDMM, based on the activity of various drugs and treatment regimens used as initial therapy so far, we prefer daratumumab combined with VCD (bortezomib cyclophosphamide dexamethasone) or VD as initial therapy. We have found high activity with VCD/VD in the past , but in our clinical experience and trial data from Andromeda , the addition of daratumumab hastens the response and limits the number of days extracorporeal light chain removal that is needed. In countries where IMiDs can be easily obtained for hospitalized patients, daratumumab plus VTD such as the regimen used in CASSIOPEIA could be an option instead of daratumumab plus VCD . Lenalidomide should probably be avoid in the upfront setting due to requirement for dose adjustment.
Fig. 1. Treatment algorithm for newly diagnosed multiple myeloma patients with acute kidney injury.
Open in a new tab
AKI acute kidney injury, FLC free light chains, M monoclonal, LCCN light chain cast nephropathy, VCD bortezomib, cyclophosphamide, dexamethasone, VTD bortezomib, thalidomide, dexamethasone, PLEX plasma exchange.
Extracorporeal light chain removal with PLEX should started as soon as possible to help reduce the serum FLC concentration more rapidly. Even though data on PLEX is controversial, the procedure has low risk, and in our opinion provides patients with the best chance at renal recovery [20, 57]. PLEX should be performed daily until the involved FLC is below 150 mg/dl or >60% reduction from baseline, if possible. Since PLEX could potentially remove daratumumab, daratumumab should be given after PLEX. In countries where HCO dialyzers are available, these could be an option instead of PLEX, although the 5 h sessions used in the MYRE trial should be performed.
The situation is more complicated with RRMM patients since many of the relapsed medications have not been tested in severe renal impaired patients. In these patients, renal recovery is equally important since eligibility of CAR-T cell therapy and clinical trials require an eGFR >40–50 ml/min/1.73 m 2. If the patient did not relapse on an anti-CD38 antibody drug, daratumumab or isatuximab could be used. Combination therapy based on bortezomib, dexamethasone, thalidomide and a 4-day continuous infusion of cisplatin, doxorubicin, cyclophosphamide, and etoposide (VDT-PACE) can be used in renally impaired patients. In these patients, cisplatin is usually omitted and cyclophosphamide should be dose adjusted for kidney function. PLEX should be offered especially in patients with options of potential future therapies.
Acknowledgements
Supported in part by grants CA 168762 and CA186781 from the National Cancer Institute, Rockville, MD, USA, and the Marvin Family Grant.
Author contributions
All of the authors collectively conceived of the paper, researched the literature, and wrote the paper.
Data availability
This is a current treatment algorithm. There are no new data generated for this paper and data sharing is not applicable.
Competing interests
Dr NL reports Stock and Other Ownership Interests at Senseonics, AbbVie, Verrica, and research funding from Omeros, all outside the submitted work. Dr SVR reports grants from NIH, outside the submitted work.
Footnotes
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Data Availability Statement
This is a current treatment algorithm. There are no new data generated for this paper and data sharing is not applicable.
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7952 | http://dslavsk.sites.luc.edu/courses/phys111/homework/phys111-2016hw9s.pdf | PHYS 111 HOMEWORK #9 Due : 10 November 2016 1. A geosynchronous satellite orbits the Earth in a period of 24 hours, thereby remaining over the same spot on the surface of the Earth. Determine the radius of this orbit (measured from the center of the Earth).
Solution : For this problem we make use of Kepler' s third law (derived in class) : M P2 = 4 π2 G a3 where M is the mass of the central object (in this case the Earth), P is the period of orbit (1 day = 86400s), G is the Newtonian Gravitational constant, and a is the radius of the orbit (actually, the semi-major axis but if we assume circular orbits, this is the same as the radius). Solving for the radius of the orbit gives us: a = G M P2 4 π2 1/3 = 6.67 × 10-11 m3 kg-1 s-26 × 1024 kg(86400 s)2 4 π2 1/3 = 4.2 × 107 m = 42, 000 km ≈26, 000 miles from the center of the Earth 2. Determine the period of a satellite in low Earth orbit (about 200 km above the surface of the Earth).
Solution : We start with the same equation and solve for period : P = 4 π2 a3 G M = 4 π2 6.4 × 106 m 3 6.67 × 10-11 m3 kg-1 s-26 × 1024 kg = 5082 s = 1.41 hours = 85 mins And this is the approximate orbital period of a satellite in low Earth orbit.
3. Problem 51, p. 178.
Solution : Let' s refer to the diagram below : 1m 1m 1.25m 1.25m r 4kg We will begin our analysis by considering the forces acting on the 4 kg block Since the block is not accelerating in the y direction, we know the sum of forces in the y direction is zero, or : ΣFy = Tu,y - Tl,y - m g = 0 where Tu,y is the y component of the tension in the upper string and Tu.l is the y component of the tension in the lower string. We can write this as: ΣFy = Tu cos θ - Tl cos θ - m g = 0 where we define θ to be the angle between the vertical rod and the string. We can use the data provided in the diagram to see that cos θ = 1/1.25 ⇒ θ = 37o. Knowing that the tension in the upper string is 80N, we can solve for the tension in the lower string: (Tu - Tl) cos θ = m g ⇒Tl = Tu - m g cos θ = 80 N -4 kg · 9.8 m s2 0.8 = 31 N We are then asked to find the speed of the block in its rotation. We find this by considering the horizontal forces. The horizontal force is generated by the horizontal component of tension in the two strings, and this horizontal force generates the centripetal force: 2 phys111-2016hw9s.nb ΣFx = Tu,x + Tl,x = m v2 r The horizontal components of tension are: Tu sin θ + Tl sin θ = m v2 r ⇒v = r sin θ (Tu + Tl) m = Before we can compute a speed, we need to determine the radius of the orbit. We can find this from the Pythoagorean theorem: r = (1.25 m)2 - (1 m)2 = 0.75 m Putting all this together, we get: v = 0.75 m sin 37o (80 N + 31 N) / 4 kg = 3.54 m / s 4. A mass starts at rest from position A on the sphere as shown below. The mass is set in motion and slides without friction to point B.
a) What is the vertical difference between points A and B?
Solution : We first compute the vertical drop from A to B so we can use the conservation of energy to compute the speed at B. The distance from the center of the circle to A is the radius, R, of the circle. The "height" of B above the center is R cos θ, so the vertical drop from A to B is R - R cos θ = R (1 - cos θ) b) What is the speed of the mass at B (assuming it started at rest at A)? Express your answer in terms of R, g and θ. Solution : We make use of the conservation of energy. In the absence of friction, we know that the total energy at A equals the total energy at B. The total energy is the sum of kinetic and potential, so we can write : KA + UA = KB + UB At A, K = 0 and U = m g R; at B K = 1/2 m v2 and U = m g R cos θ. Thus we have: m g R = m g R cos θ + 1 2 m v2 ⇒m g R (1 - cos θ) = 1 2 m v2 Another way to interpret this equation is that the loss in potential energy of m g R (1-cos θ) equals the gain in kinetic energy. This yields for the speed at B: : v = 2 g R (1 - cos θ) phys111-2016hw9s.nb 3 A B R R θ 5. Problem 4, p. 213 Solution : We will use the equation W = F cos θ s for the various parts of this problem. In the first case, we are told a constant force of 8.5N drags a box through a displacement of 17.4m. The work done is quite simply W = 8.5 N · 17.4 m = 148 J In the second case, we are told the force is directed at angle with respect to the floor, and does a total of 65J of work. We find the angle between the force and displacement via: W = 65 J = 8.5 N (17.4 m) cos θ ⇒θ = cos-1 65 J (8.5 · 17.4 J = 64o 6. Problem 8, p. 214. Answer parts a) - d) and also : If the package is at rest at the top of the chute, what is its speed at the bottom? (Five points each part).
Solution : The box travels 2 m down the ramp; the forces acting on the box are gravity and friction. The component of gravity down the plane is m g sin θ and the component of gravity perpendicular to the plane is m g cos θ. Friction acts along the plane with a force equal to μk m g cos θ. This allows us to compute: a) Work done by gravity = (m g sin θ) s = 8 kg 9.8 m s2sin 53 (2 m) = 125 J b) work done by friction = (μk m g cos θ ) s = 0.4 · 8 kg · 9.8 m s2 · cos 53 × 2 m = 37.8 J c) The normal force does no work since it is perpendicular to the displacement. d) The total work is 125 J - 37.8 J = 87.2 J e) Let' s solve this problem three different ways. First, we can use either the work - energy theorem to find the speed at the bottom. We know that the work done on the mass will equal the change in 4 phys111-2016hw9s.nb kinetic energy. Since the initial kinetic energy is zero, the work done will equal the final kinetic energy, so we have : Work done = 87.2 J = 1 2 m v2 ⇒v = 2 (87.2 J) / 8 kg = 4.67 m / s Second, we can obtain the same answer by using methods of energy conservation: Ki + Ui + Kf + Uf + Wother where the subscripts i, f refer respectively to initial and final states, and Wother is work done by dissipative forces. For the situation presented here, we have: 0 + m g h = 1 2 m v2 + Wother At the top of the ramp, the box is 2 sin 53 = 1.6m above the ground, and we know the work done by friction is 37.8J, this gives us: 8 kg · 9.8 m s2 · 1.6 m = 1 2 8 kg v2 + 37.8 J ⇒v = 4.68 m / s Finally, we can solve this via kinematic methods. We can Use vf 2 = v0 2 + 2 a s to compute the final speed once we compute the acceleration. We find the acceleration from apply-ing Newton’s second law to forces acting along the plane. Considering these forces we get: ΣF|| = m g sin θ - μk m g cos θ = m a ⇒a = g (sin θ - μk cos θ) = 5.47 m s2 Then, we get : vf 2 = 0 + 2 5.47 m s2(2 m) ⇒vf = 4.68 m / s which gives the same result for finding the speed at the bottom of the ramp.
7. Problem 18, p. 214.
Solution : The statement of the problem (in part b) suggests we might want to try to do som compar-ative reasoning, so let' s try to do everything in symbols, using numbers only at the end. A skier of mass m has an initial speed of v and then coasts for a distance d on level ground. We are asked to use the work energy theorem to find the coefficient of kinetic friction. The work energy theorem relates the amount of work done to the change in kinetic energy. In this case we have : ΔKE = KEf - KEi = 0 - 1 2 m v2 The force doing the work is friction; the force of friction between the skier and the snow is μk m g, so the total work done is: phys111-2016hw9s.nb 5 Wf = μk m g d cos 180 = - μk m g d (the factor of cos 180 arises because the force acts in the direction opposite displacement). Equating these expressions gives us : 1 2 m v2 = μk m g d ⇒μk = v2 2 g d = (12 m / s)2 2 · 9.8 m s2 · 184 m = 0.04 Now, having determined an expression for coefficient of friction as a function of initial speed and stopping distance, we can see that the mass of the skier is irrelevant and plays no role in determining the coefficient of friction. If the skier' s initial speed is doubled, and the coasting distance remains the same, we can see from our expression that μk varies as the square of the initial speed. If the initial speed is squared, μk is quadrupled, and in this case would equal 0.16.
8. Problem 37, p. 215 Solution : Conservation of energy allows us to relate the final kinetic energy to the initial potential energy of the hailstone : Kaloft + Ualoft = Kground + Uground Using the ground as our reference level for potential energy, Ualoft = m g h where h is 500m, The potential on the ground is then zero, as is the kinetic at the start aloft. This gives us: 0 + m g h = 1 2 m v2 + 0 ⇒v = 2 g h = 2 · 9.8 m s2 · 500 m = 99 m / s = 220 mi / hr.
Hailstones at this speed would fracture our skulls; fortunately, most the initial potential energy is converted to work done against the atmosphere rather than to final kinetic energy.
9. Problem 44, p. 216.
Solution : This is a conservation of energy problem, in which the initial elastic potential energy is converted into either kinetic or elastic potential energy. Originally, the mass m compresses a spring of constant k1by an amount (Δx1). This means the spring has elastic potential energy of Uel1 = 1 2 k1 (Δx1)2 If the spring then travels without losing any energy, it will compress the second spring (of constant k2 by an amount (Δx2), meaning the second spring acquires elastic potential energy equal to Uel2 = 1 2 k2 (Δx2)2 Since there is no loss of energy, the initial potential energy must equal the final potential energy, or 1 2 k1 (Δx1)2 = 1 2 k2 (Δx2)2 6 phys111-2016hw9s.nb The first part of the problem asks us to find the compression of the second spring: Δx2 = Δx1 k1 k2 = 0.04 m 32 N / cm 16 N / cm = 0.04 m 2 = 0.057 m Once the mass has left contact with the first spring, we know all of the initial elastic potential energy was converted to kinetic energy, or 1 2 k1 (Δx1)2 = 1 2 m v2 ⇒v = Δx1 k1 m = 0.04 m 3200 N / m 1.5 kg = 1.8 m / s 10. Problem 52, p. 216. Use the diagram provided in the text for reference, but solve the problems using only symbols. In other words, the mass is m, the spring of constant k is compressed by an amount Δx, it travels for a distance L. Use this information to determine an expression for the coefficient of friction between the block and the table.
Solution : A mass m compress a spring of constant k by an amount Δx. When released, the mass moves a distance d before coming to rest. We are asked to find the coefficient of friction μk. In this case, we know all of the initial elastic potential energy goes into doing work against friction. So we can write: Uel = Wf or 1 2 k (Δx)2 = fk L Now, since the force of friction is given by : fk = μk N = μk m g we have : 1 2 k (Δx)2 = μk m g L ⇒μk = k (Δx)2 2 m g L phys111-2016hw9s.nb 7 |
7953 | https://www.cut-the-knot.org/Curriculum/Geometry/OrthoCircumOrtho.shtml | Nagel's Theorem from Interactive Mathematics Miscellany and Puzzles
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This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at download and install Java VM and enjoy the applet.
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Explanation
|Activities||Contact||Front page||Contents||Geometry|
Copyright © 1996-2018 Alexander Bogomolny
The applet attempts to suggest Nagel's theorem:
Let AD and BE be two altitudes in ΔABC and O its circumcenter. Then DE and CO are perpendicular.
Since DE is a side of the orthic triangle, the statement of Nagel's theorem is equivalent to the assertion that the sides of the orthic triangle are perpendicular to the radius-vectors of the circumcircle drawn to the corresponding vertices of the reference triangle.
This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at download and install Java VM and enjoy the applet.
What if applet does not run?
In any triangle, circumcenter and orthocenter are isogonal conjugate. In other words, the altitudes and suitable circum-radius-vectors are reflections of each other in the corresponding angle bisectors. Also, the sides of the orthic triangle are antiparallels relative to the opposite sides of ΔABC, which also means that the directions of the sides of the orthic and reference triangles are reflections in the suitable angle bisectors. Since, the altitudes are perpendicular to the sides of the reference triangle, it follows that the circum-radius-vectors are perpendicular to the sides of the orthic triangle.
(There is another elementary proof.)
References
R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 2007, p. 172
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A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus i) R and ii) the current i. It is found that when R = [4\Omega ], the current is 1A when R is increased to 9$\Omega $, the current reduces to 0.5A. Find the values of the emf E and internal resistance r.
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The nrow Function in R (4 Examples)
Basic R Syntax:
The nrow R function returns the number of rows that are present in a data frame or matrix. Above, you can find the R code for the usage of nrow in R.
You want to know more details? In this article, I’m going to provide you with several reproducible examples of typical applications of the nrow function in R.
Example 1: Count the Number of Rows of a Data Frame
For the following example, I’m going to use the iris data set. Load the data set:
Table 1: Iris Data Frame as Example for the Application of nrow in R.
After loading the data frame in R, we can apply the nrow function as follows:
The number of lines of the iris database is 150.
Example 2: Using nrow in R with Condition
Let’s assume we want to count the rows of the iris data set where the variable Sepal.Length is larger than 5. With the following R code, we can examine this condition:
118 rows (i.e. observations) have a Sepal.Length larger than 5.
Example 3: nrow with NA or NULL
Often, your data will have missing values (usually labeled as NA or NULL). You want to know, how many complete rows are available in your data? The complete.cases function can help:
Example 4: For Loop Using nrow in R
In the next example, I’m showing you how to use the nrow R function as condition within a for loop – Probably the situation where I use nrow the most often.
Assume that we want to calculate the cumulative sum of the column Petal.Width:
Note: The cumulative sum could be calculated much easier with the cumsum R function. The for loop above is just for illustration.
Video Examples: nrow and Related Functions in Practice
In case you want to see more examples for the application of nrow in R, you may have a look at the following video of my YouTube channel:
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7956 | https://mathspace.co/textbooks/syllabuses/Syllabus-943/topics/Topic-19781/subtopics/Subtopic-262812/ | 12.04 Inscribed angles and polygons | Math II Math | Utah Math 2 - 2020 Edition | Mathspace
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Math II
12.04 Inscribed angles and polygons
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An inscribed angle is an angle with a vertex is on the circumference of a circle and legs that contain the chords of a circle. The arc that lies in the interior of the inscribed angle is called the intercepted arc.
Exploration
In the applet below $$∠D F E is the inscribed angle and arc $$D E is its intercepted arc. Move points $$D, $$E, and $$F around the circle and try to answer the following questions.
What relationship exists between the measure of the inscribed angle and its intercepted arc?
What has to happen for $$∠F to be a right angle?
Created with Geogebra
From the applet, we can see that the inscribed angle is always half the measure of its intercepted arc. Since we can prove this is true, it's known as the inscribed angle theorem.
Inscribed angle theorem
If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc.
In the diagram below, $$a=1 2b.
The measure of an angle is half the measure of the intercepted arc, so $$a=1 2b.
In the applet above, we saw that if we move the point on the circumference of the circle, while keeping the arc the same, the measure of the inscribed angle remains the same. This is a corollary to the inscribed angle theorem.
Corollary
If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent.
In the diagram below, $$∠A D B and $$∠B C A intercept the same arc. Therefore, $$∠A D B≅∠B C A.
Angles intercepting the same arc are congruent.
Inscribed polygons
We say that a polygon is inscribed in a circle if all of its vertices are on the circumference of the circle. The polygon is sometimes referred to as an inscribed polygon.
Triangles and quadrilaterals that are inscribed in circles have special properties. First, let's think about the relationship between a semicircle and an inscribed right angle.
Exploration
Suppose that an inscribed angle has a measure of $$9 0°. Then it follows from the inscribed angle theorem that its intercepted arc is $$2×9 0°=1 8 0°. An arc measuring $$1 8 0°is a semicircle. Therefore, an inscribed right angle intercepts a semicircle and the two chords form a diameter of the circle. We can prove the converse in a similar way.
Corollary
An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle.
For the diagram below, $$m∠A C B=9 0°if and only if $$A B is a diameter of the circle.
Since $$A B is a diameter of circle $$O, $$∠C is a right angle.
The following theorem also follows from the inscribed angle theorem. As an exercise, see if you can prove its truth in a two-column or paragraph proof.
Theorem
If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.
For example, in quadrilateral $$A B C D below, we know that $$m∠A+m∠C=1 8 0° and$$m∠B+m∠D=1 8 0°.
In quadrilateral $$A B C D, opposite angles are supplementary.
Practice questions
Question 1
Solve for $$x.
An inscribed angle is formed by the diameter and a shorter chord of the circle. The inscribed angle measures $$3 8° and is marked with an orange-colored arc. The inscribed angle also intercepts an orange-colored arc on the circle labeled $$x degrees.
Reveal Solution Watch video
Question 2
Find the value of $$x.
A circle with an inscribed angle is depicted. The inscribed angle intercepts an arc that has the same end point as the diameter of the circle. The inscribed angle is denoted as (2 x + 8) degrees.
Reveal Solution Watch video
Question 3
In the diagram, $$O is the center of the circle.
A quadrilateral is inscribed in a circle, with its top side being the circle's diameter. A diagonal is drawn from the top-left vertex to the bottom-right vertex of the quadrilateral, forming two triangles, one on top of the other. In the top triangle, the sides are the diameter of the circle, the diagonal of the quadrilateral and the right side of the quadrilateral. The interior angles of the top triangle are labeled. The top-left angle, the angle between the diameter and the diagonal, is labeled as $$2 7°. The top-right angle, the angle between the diameter and the right side, is labeled $$q degrees. The bottom-right angle, the angle between the diagonal and the right side, is labeled as $$p degrees. The $$p-degree angle is inscribed in a semi-circle. The bottom-left angle of the quadrilateral is labeled as $$r degrees. The $$r-degree and the $$q-degree angles are opposite angles of the inscribed quadrilateral.
Solve for $$p.
Solve for $$q.
Solve for $$r.
Reveal Solution Watch video
Outcomes
II.G.C.2
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.
II.G.C.3
Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle.
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7957 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9248856/ | PERSPECTIVE: TEMPERATURE-DEPENDENT DENSITY AND THERMAL EXPANSION OF CRYOPROTECTIVE AGENTS - PMC
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Cryo Letters
. Author manuscript; available in PMC: 2023 Jan 1.
Published in final edited form as: Cryo Letters. 2022 Jan-Feb;43(1):1–9.
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PERSPECTIVE: TEMPERATURE-DEPENDENT DENSITY AND THERMAL EXPANSION OF CRYOPROTECTIVE AGENTS
Prem K Solanki
Prem K Solanki
1 Biothermal Technology Laboratory, Department of Mechanical Engineering, Carnegie Mellon University, Pittsburgh, PA 15213, USA.
Find articles by Prem K Solanki
1, Yoed Rabin
Yoed Rabin
1 Biothermal Technology Laboratory, Department of Mechanical Engineering, Carnegie Mellon University, Pittsburgh, PA 15213, USA.
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1,
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1 Biothermal Technology Laboratory, Department of Mechanical Engineering, Carnegie Mellon University, Pittsburgh, PA 15213, USA.
Corresponding author’s: rabin@cmu.edu
PMC Copyright notice
PMCID: PMC9248856 NIHMSID: NIHMS1808687 PMID: 35315864
Abstract
The density is a key thermophysical property, affecting the response of the material to temperature changes in different ways, consistent with the phase of state. In fluids, temperature variation across the domain leads to colder areas being heavier than warmer areas, where buoyancy effects drive fluid flow and thereby increase heat transfer. This phenomenon is known as natural heat convection, which in general is a more efficient heat transfer mechanism than heat conduction in the absence of flow. In solids, where the material is locked in place, colder areas tend to contract while warmer areas tend to expand, leading the material to deform. When this deformation is constraint by the geometry of the domain and/or its container, mechanical stresses develop. This phenomenon is known as thermomechanical stress (or thermal stress), which can lead to structural damage such as fractures. The picture becomes even more complex during vitrification (or glass formation), where the material gradually changes from liquid to an amorphous solid over a significant temperature range. There, due to temperature variation across the domain, fluid mechanics and solid mechanics effects may coexist. It follows that characterization of the density as a function of temperature is crucial for the analyses of thermal, fluid, and mechanical effects during cryopreservation, with the goals of protocol planning, optimization, and preserving structural integrity. For this purpose, the current study focuses on the density of the material and its companion property of thermal expansion. Specifically, this paper reviews literature data on thermal expansion of cryoprotective agents (CPAs), discusses the mathematical relationship between thermal expansion and density, and presents new calculated density data. This study focuses on the CPA cocktails DP6, VS55, M22, and their key ingredients at various concentrations, including DMSO, Propylene Glycol, and Formamide. Data for DP6 combined with a selection of synthetic ice modulators (SIMs) are further presented.
Keywords: Density, Thermal Expansion, Vitrification, Synthetic Ice Modulators, Thermo-Mechanics, Thermo-Fluids
INTRODUCTION
The long-term preservation of organs and tissues will not only help tens of thousands of patients on the official transplant waiting list, but also millions of others suffering from end-stage organ disease (1). A promising means for long-term preservation of large-size specimens is cryopreservation via vitrification (vitreous means glassy in Latin) (2). Vitrification involves relatively high cooling and rewarming rates of cryoprotective agents (CPAs), which are loaded into the tissue in order to suppress crystallization and, thereby, avoid its associated devastating effects on mammalian cells and tissues (2,3).
Cooling and rewarming rate limits imposed by the sheer size of organs, the underlying principles of heat transfer, and the kinetics of crystallization often require additional means to promote vitrification, with the addition of synthetic ice modulators (SIMs) as a promising alternative (4). In general, while CPAs promote vitrification by exponential increase in viscosity with decreasing temperature, and thereby suppress molecular motion, SIMs are compounds which inhibit ice nucleation and growth by various mechanisms not directly related to viscosity (4). For example, compounds such as 1,3-cyclohexanediol limits the growth of ice crystals owing to its special chemical structure (5), while synthetic compounds like the very-low-molecular-weight polyvinyl alcohol (PVA) copolymer inhibit heterogeneous nucleation of ice (6).
An additional and potentially devastating effect on the recovery of preserved tissues and organs from cryogenic storage is the thermomechanical stress, which is driven by differential expansion of the cryopreserved material due to thermal effects. Such differential thermal expansion may result from temperature gradients in the specimen and the tendency of the material to change its volume with temperature, thereby leading to thermal expansion gradients (7,8). Another source of variable thermal expansion is partial ice formation, where pure water expands by about 9% upon freezing in standard conditions (9). Yet another source of thermomechanical stress is the thermal expansion mismatch between the specimen and the container (10,11), between the specimen and its surrounding solution (12), and even between the different constituents of the tissue (12). Either way, excessive thermomechanical stress can lead to structural damage, with fracture formation and plastic deformation as possible outcomes (10,11,13–15).
Successful vitrification is the result of a multiphysics process optimization associated with the coupled effects of CPA toxicity, thermal effects, kinetics of crystallization, and thermomechanical stresses (16). Given the virtually endless combination of key parameters involved, physical modeling (17–19) and computer simulations (20–24) appear to be critical and cost-effective tools in seeking for improved cryopreservation protocols. The current study summarizes experimental data relating to a key thermophysical property of the material – the density. In heat transfer analysis, the heat capacity of the material is proportional to its density (that is the product of the intrinsic properties of density and the specific heat), in fluid mechanics analysis it is associated with the buoyancy effects and convection, while in the solid mechanics analysis, the intrinsic property of thermal expansion—the driving mechanism of thermomechanical stress—is in fact inversely proportional to the density. For this purpose, the current study presents the mathematical relationship between the thermal expansion coefficient and the density of the material. In addition, this study presents compiled experimental data from literature for the benefit of physical modeling and computational cryobiology.
THE RLEATIONSHIP BETWEEN THERMAL EXPANSION AND DENSITY
The linear thermal expansion coefficient is a relative property describing how much a unit length of the material changes with temperature compared with its known original length at a reference temperature. For this purpose, a small isotropic cube of mass m is assumed:
where its density is ρ and its typical dimension is l. Assuming mass conservation, changes in density and dimensions is given by:
where the subscript 0 refers to a reference value at temperature T 0.
Using the linear thermal expansion coefficient, β, the length of the cube as a function of temperature can be calculated as:
The integral of the thermal expansion coefficient over temperature is defined as the thermal strain, ε, in solid mechanics analyses. Since the integral in Eq. is at least one order of magnitude smaller than 1, the corresponding expansion of the unit volume can be approximated as:
while ignoring higher order terms of the cubic expansion. In practice, the approximation presented in Eq. leads up to 2% error in the calculation of the change in volume, which is less than the uncertainty in the experimental measurement of the thermal expansion coefficient β (25).
Finally, by substituting the volume in Eq. into Eq. , the density can be calculated as:
For the purpose of this study, a reference temperature of 25°C is selected, where the thermal expansion coefficient is extrapolated to the reference temperature using polynomial approximation listed in Table 1. When not readily available, the reference density was calculated by assuming the CPA to be an ideal solution (26). Since the total volume of an ideal solution is equal to the sum of the volumes of the components (26), the reference density can be derived as:
where n c is the number of components, and w i and ρ i are the mass fraction and density of the i th component, respectively. The uncertainty of this assumption is addressed in the discussions section below.
Table 1:
Coefficients of best-fit polynomial approximation for thermal strain, ε = a 2 T 2+a 1 T+a 0, and the coefficient of thermal expansion, β=2a 2 T+a 1, °C−1.
| Solution | Material | Temperature Range, °C | a 0 × 10 3 | a 1 × 10 4 | a 2 × 10 7 | Ref. |
:---: :---: :---:
| DP6 | Pure Solution | −41.9…25 | −3.621 | 2.142 | 9.504 | (34) |
| DP6+UCV | −87.8…25 | −7.442 | 2.279 | 6.839 |
| DP6+UCV+1% X1000 | −84.4…25 | −7.326 | 2.321 | 6.571 |
| DP6+UCV+1% Z1000 | −84.3…25 | −7.374 | 2.487 | 6.548 |
| DP6+UCV+1% X1000+1%Z1000 | −84.4…25 | −5.748 | 2.432 | 6.723 |
| DP6+UCV+0.5M Sucrose | −81.7…25 | −5.069 | 2.274 | 5.157 |
| M22 | −91.1…25 | −4.709 | 2.520 | 6.235 |
| 2.2 M Propylene Glycol | −4.1…25 | −1.923 | 0.583 | 16.77 | (33) |
| 3 M Propylene Glycol | −6.6…25 | −2.346 | 0.847 | 14.53 |
| 3.1 M Formamide | −3.4…25 | −2.0 | 0.849 | 13.33 |
| 3.1 M DMSO | −7.7…25 | −1.983 | 0.956 | 9.324 |
| 6 M DMSO | −36.2…25 | −4.609 | 1.981 | 2.296 |
| 8.4 M DMSO | −95.6…25 | −5.205 | 2.383 | −1.014 |
| DP6 + EC | Pure Solution | −55…25 | 0.247 | 1.992 | 2.705 | (4) |
| Bovine Muscles | −45…25 | −0.428 | 1.957 | 11.50 |
| Goat Artery | −40.5…25 | −4.447 | 1.653 | 10.41 |
| −167.2…25 | −4.321 | 1.564 | 3.825 | (31) |
| DP6 + 12% PEG400 | Pure Solution | −95…25 | −0.496 | 2.313 | −0.219 | (4) |
| Bovine Muscles | −85…25 | 0.982 | 2.118 | 1.697 |
| Goat Artery | −80…25 | −0.487 | 2.392 | 1.777 |
| DP6 + 6% 1,3 - CHD | Pure Solution | −80…25 | 2.699×10−2 | 1.957 | −0.627 |
| Bovine Muscles | −80…25 | −1.366 | 2.175 | 4.411 |
| Goat Artery | −80…25 | −0.548 | 2.232 | 1.054 |
| DP6 + 6% 2,3 - BD | Pure Solution | −90…25 | 0.452 | 2.080 | 0.631 |
| Bovine Muscles | −85…25 | −1.117 | 2.094 | 1.431 |
| Goat Artery | −90…25 | −1.104 | 1.912 | 0.738 |
| VS55 | Pure Solution | −77.1…25 | −3.838 | 1.841 | 2.012 | (32) |
| Bovine Muscles | −47.8…25 | −2.655 | 1.852 | 4.118 |
| Goat Artery | −63.6…25 | −3.711 | 2.081 | 9.425 |
| −165.3…25 | −3.652 | 1.911 | 5.312 | (31) |
| 7.05M DMSO | Pure Solution | −95.8…25 | −4.462 | 2.084 | 0.563 | (32) |
| Bovine Muscles | −58.3…25 | −3.373 | 2.215 | 7.578 |
| Goat Artery | −93.1…25 | −4.895 | 1.995 | 6.127 |
| −167.7…25 | −4.939 | 1.890 | 4.796 | (31) |
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A previous study on density in freshly isolated tissues (27) demonstrated that the fat-free and water-free density of fresh tissues is in a close range for different tissues at about 1390 ± 102 kg/m 3. The water content is found to be 76.5% for bovine muscles (27) and 72.1% for carotid arteries (28). For the case of tissue permeated with CPA, the permeation time is assumed to be sufficiently long such that all the water in the specimen is replaced by CPA. The reference density for the CPA permeated tissue systems is given by Eq. .
Finally, uncertainty in density calculations is calculated by applying the technique presented previously (29) on Eq. :
where Δ represent the uncertainty range.
CALCULATED DENSITY DATA
Literature data on the thermal expansion coefficient (4,30–34) is used to calculate the temperature dependent density. The materials analyzed include DMSO, VS55, M22 and DP6 in combination with various SIMs, in presence and absence of tissues, Table 1. There, EC refers to EuroCollins carrier solution (35), UCV is Unisol cryoprotectant vehicle solution (35), PEG is polyethylene glycol, 1,3-CHD is 1,3 cyclohexanediol and 2,3-BD is 2,3 butanediol. Distilled water is used instead of a vehicle solution wherever EC or UCV are not listed.
Tables 1 and 2 list the coefficients of polynomial approximations for the literature data on thermal strain and the currently compiled data on density, respectively. Note that, due to the dramatic change of twelve orders of magnitude in CPA viscosity during vitrification, two different techniques have been employed to measure the thermal expansion coefficient: on a volumetric basis at higher temperatures when the material is free to flow (4,32–34), and on a linear basis at lower temperatures when the material is highly viscous (30,31,36), where the cutoff is roughly around −90°C. Further note that due to the experimentation techniques used, thermal expansion of CPAs in the presence and absence of tissues were measured in the upper part of the cryogenic temperature range, while only CPA-loaded tissues were measured in lower temperatures.
Table 2:
Coefficients of best-fit polynomial approximation for density, ρ = b 2 T 2+b 1 T+b 0, kg/m 3.
| Solution | Material | Temperature Range, °C | b 0 × 10−3 | b 1 × 10 | b 2 × 10 3 | Ref. for β |
:---: :---: :---:
| DP6 | Pure Solution | −41.9…25 | 1.048 | −6.839 | −2.683 | (34) |
| DP6+UCV | −87.8…25 | 1.062 | −7.425 | −1.946 |
| DP6+UCV+1% X1000 | −84.4…25 | 1.064 | −7.570 | −1.825 |
| DP6+UCV+1% Z1000 | −84.3…25 | 1.066 | −8.137 | −1.757 |
| DP6+UCV+1% X1000+1%Z1000 | −84.4…25 | 1.067 | −7.963 | −1.844 |
| DP6+UCV+0.5M Sucrose | −81.7…25 | 1.124 | −7.822 | −1.402 |
| M22 | −91.1…25 | 1.092 | −8.452 | −1.692 |
| 2.2 M Propylene Glycol | −4.1…25 | 1.013 | −1.791 | −5.027 | (33) |
| 3 M Propylene Glycol | −6.6…25 | 1.017 | −2.611 | −4.334 |
| 3.1 M Formamide | −3.4…25 | 1.025 | −2.641 | −3.989 |
| 3.1 M DMSO | −7.7…25 | 1.039 | −3.005 | −2.798 |
| 6 M DMSO | −36.2…25 | 1.078 | −6.502 | −0.373 |
| 8.4 M DMSO | −95.6…25 | 1.104 | −8.019 | 1.037 |
| DP6 + EC | Pure Solution | −55…25 | 1.056 | −6.405 | −0.518 | (4) |
| Bovine Muscles | −45…25 | 1.124 | −6.691 | −3.659 |
| Goat Artery | −40.5…25 | 1.134 | −5.691 | −3.378 |
| −167.2…25 | 1.132 | −5.463 | −1.264 | (31) |
| DP6 + 12% PEG400 | Pure Solution | −95…25 | 1.069 | −7.543 | 0.659 | (4) |
| Bovine Muscles | −85…25 | 1.133 | −7.322 | −0.156 |
| Goat Artery | −80…25 | 1.149 | −8.399 | −5.054×10−2 |
| DP6 + 6% 1,3 - CHD | Pure Solution | −80…25 | 1.042 | −6.204 | 0.609 |
| Bovine Muscles | −80…25 | 1.113 | −7.398 | −1.142 |
| Goat Artery | −80…25 | 1.127 | −7.674 | 0.148 |
| DP6 + 6% 2,3 - BD | Pure Solution | −90…25 | 1.054 | −6.682 | 0.222 |
| Bovine Muscles | −85…25 | 1.121 | −7.160 | −6.372×10−2 |
| Goat Artery | −90…25 | 1.133 | −6.595 | 0.121 |
| VS55 | Pure Solution | −77.1…25 | 1.068 | −5.985 | −0.362 | (32) |
| Bovine Muscles | −47.8…25 | 1.133 | −6.383 | −1.108 |
| Goat Artery | −63.6…25 | 1.149 | −7.297 | −3.046 |
| −165.3…25 | 1.147 | −6.811 | −1.805 | (31) |
| 7.05M DMSO | Pure Solution | −95.8…25 | 1.090 | −6.922 | 0.257 | (32) |
| Bovine Muscles | −58.3…25 | 1.154 | −7.804 | −2.298 |
| Goat Artery | −93.1…25 | 1.164 | −7.134 | −1.991 |
| −167.7…25 | 1.162 | −7.147 | −1.780 | (31) |
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Figure 1 displays the temperature-dependent density of selected combinations of CPAs and SIMs in higher cryogenic temperatures. Figure 2 displays the temperature-dependent density for tissues loaded with representative CPAs in higher cryogenic temperatures. Figure 3 displays the combined density curves over the entire cryogenic temperature range of interest, where the combination includes different measurement techniques in higher (32) and lower temperatures (31).
Figure 1:
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Density of CPA+SIM cocktails in higher cryogenic temperatures, where the coefficients for polynomial approximations are listed in Table 1.
Figure 2:
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Density of CPA cocktails in presence and absence of tissues in higher cryogenic temperatures, where the coefficients for polynomial approximations are listed in Table 1.
Figure 3:
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Comparison of selected temperature-dependent density curves of CPAs in presence of tissues calculated from two sources of measurements in different temperature ranges (4,31,32).
Figure 4 displays a summary of thermal properties compiled for 7.05M DMSO, which is a reference solution shown useful for thermomechanical stress analysis of cryopreservation by vitrification of highly concentrated CPA solutions (24,37).
Figure 4:
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Thermophysical properties for 7.05 M DMSO, where the density and thermal diffusivity are compiled in the current study, while the thermal conductivity (7) and specific heat (42) are taken from literature data. The uncertainty in thermal diffusivity estimation is displayed by the gray region.
DISCUSSION
In broad terms, in the absence of phase change during vitrification, the material contracts continuously (negative thermal strain) while its density increases monotonically, although at a variable rate. The tendency to contract is somewhat higher at higher temperatures. For example, the density of DP6 increases by 4.11% when the temperature is reduced from 25°C to −41°C, Fig. 1.
The addition of the reported SIMs results in increased density as can be observed from Fig. 1. For example, the addition of 0.5M Sucrose results in the largest density increase compared to the density of DP6+UCV, amounting to 6.35% at 25°C. The density of DP6+UCV+0.5M Sucrose increases by 6.76% as the temperature decreases from 25°C to −82°C.
While the above density changes are significant for fluid mechanics analysis when calculating free convection effects during vitrification (13,38), and for solid mechanics analysis when calculating thermomechanical stress (20,22,24), it is important to compare it with the uncertainty in measurement. For example, the compiled density of M22 increases by 7.43% as the temperature decreases from 25°C to −91°C. The uncertainty in the measured thermal strain within this range has been estimated as 4% (34), while the uncertainty in the compiled density value is 2.5% based on Eq. (Fig. 1). Note that the experimental techniques and corresponding literature data used as the basis for the current study were designed to filter out random errors in measurements (25), and the uncertainty analysis presented here refers essentially to systematic (bias) errors.
A different source of uncertainty in the data compilation in the current study is associated with the calculation of the reference density by means of Eq. . For DMSO, this uncertainty is estimated as 2.5%, based on comparing the outcome of Eq. and literature data for aqueous DMSO solutions (39). Recall that Eq. assumes DMSO solution to follow an ideal mixture law, where the ingredients do not chemically interact with one another, while the DMSO solution is in fact known to be a complex solution (39). Nonetheless, the ideal solution simplification serves well for the compilation of the density, when compared with other sources of uncertainty. Furthermore, the uncertainty associated with this assumption is expected to lead a bias in the density profile, while maintaining its trend, and not to a random uncertainty as would be expected from a discrete measurement. For this reason, the effect of this assumption further diminishes when the data is used to calculate fluid mechanics or solid mechanics effects.
The presence of tissues is demonstrated to increase the density of the specimen At any specific temperature, Figs. 2 and 3. For example, the density of VS55 is 6.05% and 7.25% higher in the presence of bovine muscle and goat artery, respectively, compared to the pure solution at the reference temperature of 25°C. However, the rate of change of density with temperature is not significantly affected by the presence of tissues. For example, as the temperature decreases from 25°C to −45°C, the density of pure solution VS55 increases by 3.93%, while the density of VS55 in presence of bovine muscle and goat artery increases by 3.86% and 4%, respectively. For DP6+EC, as the temperature decreases from 25°C to −45°C, the density for pure solution, solution with bovine muscle and solution with goat artery increases by 4.24%, 3.77% and 3.26%, respectively. Similarly for 7.05 M DMSO, as the temperature decreases from 25°C to −45°C, the density for pure solution, solution with bovine muscle and solution with goat artery increases by 4.55%, 4.53% and 4.16%, respectively.
Recall that the reference densities for tissues permeated with CPA is calculated using Eq. which assumes that total volume of the specimen is the linear sum of the tissue and constituents of the CPA solution. The validity of this assumption remains untested. Possible deviations from this assumption might occur if the tissue swells or shrinks when permeated with CPA (40,41).
Of particular interest for thermal analyses of cryopreservation is the physical property of thermal diffusivity, which represents the ratio of thermal conductivity to the product of density and specific heat . While the thermal conductivity relates to the ability of the material to conduct heat, the thermal diffusivity relates to the rate at which this tends to happen. Figure 4 displays the dependency of all those thermophysical properties for 7.05M DMSO, which has been used repeatedly as a reference solution for thermo-fluids and thermo-mechanics analyses. Interestingly, while the density of 7.05M DMSO increases almost linearly with the decreasing temperature, the specific heat and thermal conductivity display an opposing trend, and the thermal diffusivity is only moderately and non-monotonically changing with temperature. For this solution, the thermal diffusivity varies from a minimum value of 1.15×10−7 m 2/s at −30°C to a maximum value of 1.34×10−7 m 2/s at −96°C, which represents a range of 16.5 %. At the same time, the uncertainty in thermal diffusivity compilation is estimated within the range of 14.1%. The uncertainty in the compiled thermal diffusivity value is calculated using the propagation of uncertainty in the basic thermophysical properties following Eq. (29). The uncertainties in the thermophysical properties was taken as 2.5% for calculation of the reference density based on Eq., 10% for the calculation of the specific heat (42), ±0.3 W/m°C for the measurement of the thermal conductivity (7) and 3% for the measurement of the thermal strain (32). While the uncertainty in thermal diffusivity is significant, as with other thermophysical properties, the trend of property change is noteworthy, even if it may be shifted.
SUMMARY
All materials change volume with temperature, where the thermal expansion coefficient is a physical property that describes the corresponding rate of change. Those volume changes are inversely proportional to the density of the material. In general, direct density measurements are more convenient in fluids, while linear thermal expansion measurements are more convenient in solids. However, from a practical perspective in the extreme conditions of cryogenic temperatures, thermal expansion of CPAs is more commonly available in the literature.
In fluids, thermal expansion may intensify heat transfer by the mechanism of natural convection. In solid, thermal expansion may drive the development of thermomechanical stresses, potentially causing structural damage. During vitrification, fluid mechanics and solid mechanics effects may coexist, leading to complex effects such as macro-scale deformations. In this context, vitrification is widely considered as the only alternative for long-term preservation of organ and large-size tissues. Furthermore, the anomalous expansion of water upon freezing may lead to unexpected thermal stresses, an effect which should be considered when only partial vitrification is achieved during vitrification attempts.
In order to account for the above effects during modeling and computer-assisted analyses of experiments, the current study reviews literature data on thermal expansion, discusses the mathematical relationship between thermal expansion and density, and presents calculated density data relevant to cryopreservation by vitrification.
Specifically, the current study derives density data for selected CPAs and CPAs+SIMs in the presence and absence of tissues. The reported materials include M22 which has shown promising results for kidney cryopreservation (43), VS55 which demonstrated high vitrification tendencies, and DP6 mixed with SIMs, such as sucrose which has drawn significant attention in recent years (44).
The broader impact of the current study on the overall interpretation of thermomechanical effects is presented using the example of thermal diffusivity of 7.05 M DMSO.
In conclusion, this study addresses only one but a very significant aspect of the unmet need for temperature-dependent material properties for improved theoretical investigations and computer modeling of cryopreservation.
Acknowledgements:
Research reported in this publication was supported by the National Heart Lung and Blood Institute (NHLBI) of the National Institutes of Health under award number R01HL127618. The content is solely the responsibility of the authors and does not necessarily represent the official views of the National Institutes of Health.
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Abstract
INTRODUCTION
THE RLEATIONSHIP BETWEEN THERMAL EXPANSION AND DENSITY
CALCULATED DENSITY DATA
DISCUSSION
SUMMARY
Acknowledgements:
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Zeros and poles
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From Wikipedia, the free encyclopedia
Concept in complex analysis
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In complex analysis (a branch of mathematics), a pole is a certain type of singularity of a complex-valued function of a complex variable. It is the simplest type of non-removable singularity of such a function (see essential singularity). Technically, a point z0 is a pole of a function f if it is a zero of the function 1/f and 1/f is holomorphic (i.e. complex differentiable) in some neighbourhood of z0.
A function f is meromorphic in an open set U if for every point z of U there is a neighborhood of z in which at least one of f and 1/f is holomorphic.
If f is meromorphic in U, then a zero of f is a pole of 1/f, and a pole of f is a zero of 1/f. This induces a duality between zeros and poles, that is fundamental for the study of meromorphic functions. For example, if a function is meromorphic on the whole complex plane plus the point at infinity, then the sum of the multiplicities of its poles equals the sum of the multiplicities of its zeros.
Definitions
[edit]
A function of a complex variable z is holomorphic in an open domain U if it is differentiable with respect to z at every point of U. Equivalently, it is holomorphic if it is analytic, that is, if its Taylor series exists at every point of U, and converges to the function in some neighbourhood of the point. A function is meromorphic in U if every point of U has a neighbourhood such that at least one of f and 1/f is holomorphic in it.
A zero of a meromorphic function f is a complex number z such that f(z) = 0. A pole of f is a zero of 1/f.
If f is a function that is meromorphic in a neighbourhood of a point of the complex plane, then there exists an integer n such that
is holomorphic and nonzero in a neighbourhood of (this is a consequence of the analytic property).
If n > 0, then is a pole of order (or multiplicity) n of f. If n < 0, then is a zero of order of f. Simple zero and simple pole are terms used for zeroes and poles of order Degree is sometimes used synonymously to order.
This characterization of zeros and poles implies that zeros and poles are isolated, that is, every zero or pole has a neighbourhood that does not contain any other zero and pole.
Because of the order of zeros and poles being defined as a non-negative number n and the symmetry between them, it is often useful to consider a pole of order n as a zero of order −n and a zero of order n as a pole of order −n. In this case a point that is neither a pole nor a zero is viewed as a pole (or zero) of order 0.
A meromorphic function may have infinitely many zeros and poles. This is the case for the gamma function (see the image in the infobox), which is meromorphic in the whole complex plane, and has a simple pole at every non-positive integer. The Riemann zeta function is also meromorphic in the whole complex plane, with a single pole of order 1 at z = 1. Its zeros in the left halfplane are all the negative even integers, and the Riemann hypothesis is the conjecture that all other zeros are along Re(z) = 1/2.
In a neighbourhood of a point a nonzero meromorphic function f is the sum of a Laurent series with at most finite principal part (the terms with negative index values):
where n is an integer, and Again, if n > 0 (the sum starts with , the principal part has n terms), one has a pole of order n, and if n ≤ 0 (the sum starts with , there is no principal part), one has a zero of order .
At infinity
[edit]
A function is meromorphic at infinity if it is meromorphic in some neighbourhood of infinity (that is outside some disk), and there is an integer n such that
exists and is a nonzero complex number.
In this case, the point at infinity is a pole of order n if n > 0, and a zero of order if n < 0.
For example, a polynomial of degree n has a pole of degree n at infinity.
The complex plane extended by a point at infinity is called the Riemann sphere.
If f is a function that is meromorphic on the whole Riemann sphere, then it has a finite number of zeros and poles, and the sum of the orders of its poles equals the sum of the orders of its zeros.
Every rational function is meromorphic on the whole Riemann sphere, and, in this case, the sum of orders of the zeros or of the poles is the maximum of the degrees of the numerator and the denominator.
Examples
[edit]
The function
: is meromorphic on the whole Riemann sphere. It has a pole of order 1 or simple pole at and a simple zero at infinity.
The function
: is meromorphic on the whole Riemann sphere. It has a pole of order 2 at and a pole of order 3 at . It has a simple zero at and a quadruple zero at infinity.
The function
: is meromorphic in the whole complex plane, but not at infinity. It has poles of order 1 at . This can be seen by writing the Taylor series of around the origin.
The function
: has a single pole at infinity of order 1, and a single zero at the origin.
All above examples except for the third are rational functions. For a general discussion of zeros and poles of such functions, see Pole–zero plot § Continuous-time systems.
Function on a curve
[edit]
The concept of zeros and poles extends naturally to functions on a complex curve, that is complex analytic manifold of dimension one (over the complex numbers). The simplest examples of such curves are the complex plane and the Riemann surface. This extension is done by transferring structures and properties through charts, which are analytic isomorphisms.
More precisely, let f be a function from a complex curve M to the complex numbers. This function is holomorphic (resp. meromorphic) in a neighbourhood of a point z of M if there is a chart such that is holomorphic (resp. meromorphic) in a neighbourhood of Then, z is a pole or a zero of order n if the same is true for
If the curve is compact, and the function f is meromorphic on the whole curve, then the number of zeros and poles is finite, and the sum of the orders of the poles equals the sum of the orders of the zeros. This is one of the basic facts that are involved in Riemann–Roch theorem.
See also
[edit]
Argument principle
Control theory § Stability
Filter design
Filter (signal processing)
Gauss–Lucas theorem
Hurwitz's theorem (complex analysis)
Marden's theorem
Nyquist stability criterion
Pole–zero plot
Residue (complex analysis)
Rouché's theorem
Sendov's conjecture
References
[edit]
Conway, John B. (1986). Functions of One Complex Variable I. Springer. ISBN 0-387-90328-3.
Conway, John B. (1995). Functions of One Complex Variable II. Springer. ISBN 0-387-94460-5.
Henrici, Peter (1974). Applied and Computational Complex Analysis 1. John Wiley & Sons.
External links
[edit]
Weisstein, Eric W. "Pole". MathWorld.
Retrieved from "
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7959 | https://www.quora.com/What-is-the-reason-for-the-function-f-x-sinx-x-being-strictly-increasing | What is the reason for the function f(x) = sinx + x being strictly increasing? - Quora
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What is the reason for the function f(x) = sinx + x being strictly increasing?
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Ishaan Gupta
Studied at Allen Career Institute, Chandigarh
·10mo
The function f(x) = sinx + x is strictly increasing because the derivative of the function is cosx+1 which is a non negative function
This means that the function is positive for all real values of x except when x is an odd multiple of pi
When x is an odd mutiple of pi derivative and double derivative (-sin x in this case) both are 0
This means that these are points of inflection i.e. the curative of the function switches from concave to convex or vice versa
This is a common misconeption that if derivative is 0 then the point has to be a maxima or minima
But in this case the derivative is 0 but it
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The function f(x) = sinx + x is strictly increasing because the derivative of the function is cosx+1 which is a non negative function
This means that the function is positive for all real values of x except when x is an odd multiple of pi
When x is an odd mutiple of pi derivative and double derivative (-sin x in this case) both are 0
This means that these are points of inflection i.e. the curative of the function switches from concave to convex or vice versa
This is a common misconeption that if derivative is 0 then the point has to be a maxima or minima
But in this case the derivative is 0 but it is still an always increasing function because the derivative is 0 only at discrete points
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What is the definition of a function f(x) such that f(f(x)) =f(x)?
Is f(x) =1/x a function?
Is there a differentiable, strictly increasing function that satisfies the equation f(f(x)) =x for all real x?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
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How can it be proven that the function f(x) = (1/x) √(x+1) is strictly increasing?
The logical idea is that if we can show the gradient is always positive then the function must al...
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David Joyce
Professor Emeritus of Mathematics at Clark University · Upvoted by
Siddhant Grover
, MSc in Statistics & BSc Mathematics, Hindu College, University of Delhi (2022) · Author has 9.9K answers and 68.4M answer views
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Can we call sinx/ (sinx + cosx) to be a strictly increasing function? The derivative 1/ (sinx + cosx) ^2 is positive but it does not fit into the definition of strictly increasing function (i.e., f(π/2) > f(π) although π/2 < π).
The function is not defined when the denominator is 0, and that occurs when x=3 4 π,x=3 4 π, so the derivative is not defined there.
If you know a function is differentiable on an interval and its derivative is positive on that interval, then the function will be increasing. So this function is increasing on any interval in which it’s defined.
The graph of this function has a vertical asymptote at x=3 4 π x=3 4 π. It’s periodic with period 2 π 2 π, so it has vertical asymptotes at x=3 4 π x=3 4 π plus multiples of 2 π 2 π.
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The function is not defined when the denominator is 0, and that occurs when x=3 4 π,x=3 4 π, so the derivative is not defined there.
If you know a function is differentiable on an interval and its derivative is positive on that interval, then the function will be increasing. So this function is increasing on any interval in which it’s defined.
The graph of this function has a vertical asymptote at x=3 4 π x=3 4 π. It’s periodic with period 2 π 2 π, so it has vertical asymptotes at x=3 4 π x=3 4 π plus multiples of 2 π 2 π.
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Sohel Zibara
Studied at Doctor of Philosophy Degrees (Graduated 2000) · Author has 5.1K answers and 2.6M answer views
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What are the intervals on which the function f(x) = sin x + cos x is increasing or decreasing?
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Francesco Amato
Studied at University of Bari (Graduated 1999) · Author has 4.5K answers and 1M answer views
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If f(x) =x+sinx, what is the local maximum and minimum value?
The stationary points of the function
f(x)=x+sin x f(x)=x+sinx
are those at which
f′(x)=1+cos x=0 f′(x)=1+cosx=0→→cos x=−1 cosx=−1→→x n=−π±2 n π x n=−π±2 n π.
To establish their kind the second derivative
f′′(x)=−sin x f″(x)=−sinx
must be evaluated at the x n.x n.
Since f′′(x=x n=−π±2 n π)=−sin x|x n=0 f″(x=x n=−π±2 n π)=−sinx|x n=0
the x n x n are inflection points, neither maxima nor minima; the function increases everywhere due to the everywhere positivity of f’(x) (except at the flectional points)
−1≤cos x≤+1−1≤cosx≤+1
0≤1+cos x≤2 0≤1+cosx≤2
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The stationary points of the function
f(x)=x+sin x f(x)=x+sinx
are those at which
f′(x)=1+cos x=0 f′(x)=1+cosx=0→→cos x=−1 cosx=−1→→x n=−π±2 n π x n=−π±2 n π.
To establish their kind the second derivative
f′′(x)=−sin x f″(x)=−sinx
must be evaluated at the x n.x n.
Since f′′(x=x n=−π±2 n π)=−sin x|x n=0 f″(x=x n=−π±2 n π)=−sinx|x n=0
the x n x n are inflection points, neither maxima nor minima; the function increases everywhere due to the everywhere positivity of f’(x) (except at the flectional points)
−1≤cos x≤+1−1≤cosx≤+1
0≤1+cos x≤2 0≤1+cosx≤2
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Sohel Zibara
Studied at Doctor of Philosophy Degrees (Graduated 2000) · Author has 5.1K answers and 2.6M answer views
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If f(x) =x+sinx, what is the local maximum and minimum value?
[Math Processing Error]\displaystyle{\textbf{}\\textbf{Since}\,f'\left(x\right)\,=\,1\,+\,\cos\,x\,\ge\,0\,\,\forall\,x\,\in\,\R,\,\textbf{then}\,f\left(x\right)\,\textbf{is strictly increasing}\\textbf{on}\,\R\,\textbf{and therefore posesses neither maximum nor minimum.However:}\\textbf{}\f\left(x\right)\,\textbf{has}\,\begin{cases}\textbf{global minima and maxima at}\,x\,=\,a\,\textbf{respectively}\x\,=\,b\,\textbf{on any closed interval}\,\left[a\,,\,\right]\\textbf{}\\textbf{global minimum at}\,x\,=\,a\,\textbf{on any interval}\,\left[a\,,\,b\right)\\textbf{}\\textbf{global maximu
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Since f′(x)=1+cos x≥0∀x∈R,then f(x)is strictly increasing on R and therefore posesses neither maximum nor minimum. However :f(x)has⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩global minima and maxima at x=a respectively x=b on any closed interval[a,]global minimum at x=a on any interval[a,b)global maximum at x=b on any interval(a,b]Since f′(x)=1+cos x≥0∀x∈R,then f(x)is strictly increasing on R and therefore posesses neither maximum nor minimum. However :f(x)has{global minima and maxima at x=a respectively x=b on any closed interval[a,]global minimum at x=a on any interval[a,b)global maximum at x=b on any interval(a,b]
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Enrico Gregorio
Associate professor in Algebra · Author has 18.4K answers and 16M answer views
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Why is min (x, x²) an increasing function?
We have x 2>x x 2>x if and only if x(x−1)>0 x(x−1)>0, that is, for x<0 x<0 or x>1 x>1.
Thus the function can be rewritten as
f(x)=min{x,x 2}=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x x<0 0 x=0 x 2 0<x<1 1 x=1 x x>1 f(x)=min{x,x 2}={x x<0 0 x=0 x 2 0<x<1 1 x=1 x x>1
Suppose x<y x<y.
If x<y<0 x<y<0, then f(x)=x<f(y)=y f(x)=x<f(y)=y.
If x<0<y x<0<y, then f(x)<0 f(x)<0 and f(y)>0 f(y)>0, so f(x)<f(y)f(x)<f(y).
If 0<x<y<1 0<x<y<1, then f(x)=x 2<f(y)=y 2 f(x)=x 2<f(y)=y 2.
If 0<x<1<y 0<x<1<y, then f(x)=x 2<1 f(x)=x 2<1 and f(y)=y>1 f(y)=y>1.
If 1<x<y 1<x<y, then f(x)=x<f(y)=y f(x)=x<f(y)=y.
You can discuss the limiting cases, when x∈{0,1}x∈{0,1} or y∈{0,1}y∈{0,1}.
You could also use derivatives: we have
f′(x)=⎧⎪ ⎪⎨⎪ ⎪⎩1 x<0 2 x 0<x<1 1 x>1 f′(x)={1 x<0 2 x 0<x<1 1 x>1
so th
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We have x 2>x x 2>x if and only if x(x−1)>0 x(x−1)>0, that is, for x<0 x<0 or x>1 x>1.
Thus the function can be rewritten as
f(x)=min{x,x 2}=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩x x<0 0 x=0 x 2 0<x<1 1 x=1 x x>1 f(x)=min{x,x 2}={x x<0 0 x=0 x 2 0<x<1 1 x=1 x x>1
Suppose x<y x<y.
If x<y<0 x<y<0, then f(x)=x<f(y)=y f(x)=x<f(y)=y.
If x<0<y x<0<y, then f(x)<0 f(x)<0 and f(y)>0 f(y)>0, so f(x)<f(y)f(x)<f(y).
If 0<x<y<1 0<x<y<1, then f(x)=x 2<f(y)=y 2 f(x)=x 2<f(y)=y 2.
If 0<x<1<y 0<x<1<y, then f(x)=x 2<1 f(x)=x 2<1 and f(y)=y>1 f(y)=y>1.
If 1<x<y 1<x<y, then f(x)=x<f(y)=y f(x)=x<f(y)=y.
You can discuss the limiting cases, when x∈{0,1}x∈{0,1} or y∈{0,1}y∈{0,1}.
You could also use derivatives: we have
f′(x)=⎧⎪ ⎪⎨⎪ ⎪⎩1 x<0 2 x 0<x<1 1 x>1 f′(x)={1 x<0 2 x 0<x<1 1 x>1
so the function is increasing in the intervals (−∞,0)(−∞,0), (0,1)(0,1) and (1,∞)(1,∞). As it is everywhere continuous, we can conclude that the function is increasing over (−∞,0](−∞,0], [0,1][0,1] and [1,∞)[1,∞). Thus the conclusion is obtained by observing that if x 1<0 x 1<0, 0<x 2<1 0<x 2<1 and x 3∈(1,∞)x 3∈(1,∞), we have
f(x 1)<f(0)=0<f(x 2)<f(1)=1<f(x 3)f(x 1)<f(0)=0<f(x 2)<f(1)=1<f(x 3)
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Dave Buchfuhrer
B.S. in Math · Upvoted by
Ali Dursen
, M.S. Mathematics, Sabanci University (2013) · Author has 922 answers and 2.7M answer views
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Is there a strictly increasing function f:R→R f:R→R such that f′(x)=f(f(x))f′(x)=f(f(x)) for all x x?
No such function exists, and we can prove it with fairly simple tools. First, note that f′(x)≥0 f′(x)≥0 because f f is increasing. Because it’s strictly increasing, f′(x)f′(x) can’t always be zero. So there’s some z z such that f(f(z))=f′(z)=ϵ>0 f(f(z))=f′(z)=ϵ>0. Because f(x)f(x) is strictly increasing, so is f′(x)=f(f(x))f′(x)=f(f(x)). So if f′(x)=ϵ f′(x)=ϵ, this means that f(f(z)+y/ϵ)≥f(f(z))+y.f(f(z)+y/ϵ)≥f(f(z))+y. So f(x)f(x) can get arbitrarily large. This also means that f′(x)=f(f(x))f′(x)=f(f(x)) can be arbitrarily large. In particular, there’s some x x such that f′(x)>1 f′(x)>1. Since f′(x)f′(x) is strictly increasing and f f is growing faster than x x, e
Continue Reading
No such function exists, and we can prove it with fairly simple tools. First, note that f′(x)≥0 f′(x)≥0 because f f is increasing. Because it’s strictly increasing, f′(x)f′(x) can’t always be zero. So there’s some z z such that f(f(z))=f′(z)=ϵ>0 f(f(z))=f′(z)=ϵ>0. Because f(x)f(x) is strictly increasing, so is f′(x)=f(f(x))f′(x)=f(f(x)). So if f′(x)=ϵ f′(x)=ϵ, this means that f(f(z)+y/ϵ)≥f(f(z))+y.f(f(z)+y/ϵ)≥f(f(z))+y. So f(x)f(x) can get arbitrarily large. This also means that f′(x)=f(f(x))f′(x)=f(f(x)) can be arbitrarily large. In particular, there’s some x x such that f′(x)>1 f′(x)>1. Since f′(x)f′(x) is strictly increasing and f f is growing faster than x x, eventually we’ll have f(x)>x+1 f(x)>x+1 for all sufficiently large x x.
Now that we know that f(x)>x+1>0 f(x)>x+1>0 for large enough x x, we can combine that with the fact that f′(x)f′(x) is strictly increasing to get that f(f(x))>f(x+1)>f(x)+f′(x)>f′(x)=f(f(x))f(f(x))>f(x+1)>f(x)+f′(x)>f′(x)=f(f(x)). This is a contradiction, as you can’t have f(f(x))>f(f(x))f(f(x))>f(f(x)). The contradiction demonstrates that no such f f exists.
Upvote ·
99 14
9 2
Max Wong
read many books about calculus
·7y
Related
What is the maximum value of the function f(x) =6sinx+8cosx?
Upvote ·
9 9
9 1
Vishal Chandratreya
Former Senior Engineer at Samsung Semiconductor India (2019–2021) · Author has 931 answers and 2.9M answer views
·6y
Related
Why is min (x, x²) an increasing function?
Why is min{x,x 2}min{x,x 2} an increasing function?
Because it is continuous and its components are strictly increasing wherever relevant.
Choosing the lower function at each point, you end up with the following graph.
min{x,x 2}=⎧⎨⎩x x∈(−∞,0)x 2 x∈[0,1)x x∈[1,∞)min{x,x 2}={x x∈(−∞,0)x 2 x∈[0,1)x x∈[1,∞)
Individually, x x and x 2 x 2 are both increasing functions in the intervals above. At the end points of the intervals, there is no discontinuity. Consequently, the function formed by taking their minimum is also an increasing function.
Continue Reading
Why is min{x,x 2}min{x,x 2} an increasing function?
Because it is continuous and its components are strictly increasing wherever relevant.
Choosing the lower function at each point, you end up with the following graph.
min{x,x 2}=⎧⎨⎩x x∈(−∞,0)x 2 x∈[0,1)x x∈[1,∞)min{x,x 2}={x x∈(−∞,0)x 2 x∈[0,1)x x∈[1,∞)
Individually, x x and x 2 x 2 are both increasing functions in the intervals above. At the end points of the intervals, there is no discontinuity. Consequently, the function formed by taking their minimum is also an increasing function.
Upvote ·
A. Skodras
Studied Mathematics&Applied Mathematics · Upvoted by
John Hubbard
, PhD Mathematics, Cornell Univewrsity (1973) · Author has 731 answers and 2.5M answer views
·5y
Related
How do you prove that the function f(x) =3sin(x)-3x is "strictly decreasing" and not just "decreasing"?
The derivative of the function is
f′(x)=3 cos x−3 f′(x)=3 cosx−3
For zeros of f′(x)=0 f′(x)=0 we get
3 cos x−3=0⟺3 cosx−3=0⟺
cos x=1⟺cosx=1⟺
x=2 k π,k∈Z x=2 k π,k∈Z
and f′(x)<0,∀x∈R/{2 k π}′(x)<0,∀x∈R/{2 k π}
So according to the theory f f is strictly decreasing because left and right of every zero the derivative is negative.
Now it is understandable that what bothers many is what happens at the point that the derivative is equal to 0
Just to make it more clear lets consider x∈(0,4 π)x∈(0,4 π)and try to understand the behavior of the function at 2 π 2 π
According to the findings above f f is decreasing in the interval (0,2
Continue Reading
The derivative of the function is
f′(x)=3 cos x−3 f′(x)=3 cosx−3
For zeros of f′(x)=0 f′(x)=0 we get
3 cos x−3=0⟺3 cosx−3=0⟺
cos x=1⟺cosx=1⟺
x=2 k π,k∈Z x=2 k π,k∈Z
and f′(x)<0,∀x∈R/{2 k π}′(x)<0,∀x∈R/{2 k π}
So according to the theory f f is strictly decreasing because left and right of every zero the derivative is negative.
Now it is understandable that what bothers many is what happens at the point that the derivative is equal to 0
Just to make it more clear lets consider x∈(0,4 π)x∈(0,4 π)and try to understand the behavior of the function at 2 π 2 π
According to the findings above f f is decreasing in the interval (0,2π] and we can agree that for any x 1,x 2∈(0,2 π]x 1,x 2∈(0,2 π]with x 1f(x 2)f(x 1)>f(x 2)
The same happens in the interval 2 π,4 π)[2 π,4 π) , for any x 1,x 2∈[2 π,4 π)x 1,x 2∈[2 π,4 π)with x 1f(x 2)f(x 1)>f(x 2)
Merging the above 2 we get that
for any x 1,x 2∈(0,4 π)x 1,x 2∈(0,4 π)with x 1f(x 2)f(x 1)>f(x 2)
So f f is strictly decreasing.
A similar case is for the function f(x)=x 3 f(x)=x 3 considering the point x=0 x=0
f f is increasing in the intervals (−∞,0[0,+∞)so f f is strictly increasing everywhere.
Upvote ·
9 6
9 1
Ron Davis
I earn my living with mathematics. · Author has 6.7K answers and 18.8M answer views
·4y
Related
How do you prove that the function f:(0,∞)→R f:(0,∞)→R, f(x)=a x+1 a x f(x)=a x+1 a x, a>0,a≠1 a>0,a≠1 is a strictly increasing function?
a x+1 a x=2 cosh(x log(a)).a x+1 a x=2 cosh(x log(a)).
∴d(a x+1 a x)d x=2 d cosh(x log(a))d x∴d(a x+1 a x)d x=2 d cosh(x log(a))d x
∴d(a x+1 a x)d x=2 log(a)sinh(x log(a)).∴d(a x+1 a x)d x=2 log(a)sinh(x log(a)).
Upvote ·
9 1
David Joyce
Dave's Short Course in Trig, · Author has 9.9K answers and 68.4M answer views
·1y
Related
What is the range of function f(x) = sinx?
I imagine what you’re asking about is if x x is a real number, then what are the possible values of sin x.sinx. In other words, treating sine as a function R→R,R→R, what is the image of that function? (Sometimes images are referred to as ranges.)
Here’s the graph, y=sin x.y=sinx.
The sine function takes all values from –1 to +1, inclusive. So the image is the closed interval [−1,1].[−1,1].
Continue Reading
I imagine what you’re asking about is if x x is a real number, then what are the possible values of sin x.sinx. In other words, treating sine as a function R→R,R→R, what is the image of that function? (Sometimes images are referred to as ranges.)
Here’s the graph, y=sin x.y=sinx.
The sine function takes all values from –1 to +1, inclusive. So the image is the closed interval [−1,1].[−1,1].
Upvote ·
9 2
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Linear Diophantine Equations
Last Updated : 21 Jan, 2025
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A Diophantine equation is a polynomial equation, usually in two or more unknowns, such that only the integral solutions are required. Linear Diophantine equations are a class of equations of the form:
ax + by = c
where:
a, b, and c are integers,
x and y are variables that are also integers.
Note:A Linear Diophantine equation has integer solutions if and only if the greatest common divisor (gcd) of a and b divides c i.e., for the equation ax + by = c to have a solution, c must be divisible by gcd(a, b),
c = k ⋅ gcd(a, b)
Where k is an integer.
The goal is to determine whether there exist integer solutions for x and y that satisfy the equation.
Examples:
Input : a = 3, b = 6, c = 9
Output: Possible
Explanation: The Equation turns out to be, 3x + 6y = 9 one integral solution would be x = 1 , y = 1
Input : a = 3, b = 6, c = 8
Output:Not Possible
Explanation: o integral values of x and y exists that can satisfy the equation 3x + 6y = 8
Input : a = 2, b = 5, c = 1
Output : Possible
Explanation : Various integral solutions possible are, (-2,1) , (3,-1) etc.
Solution:
For linear Diophantine equation equations, integral solutions exist if and only if, the GCD of coefficients of the two variables divides the constant term perfectly. In other words, the integral solution exists if, GCD(a ,b) divides c.
Thus the algorithm to determine if an equation has integral solution is pretty straightforward.
Find GCD of a and b
Check if c % GCD(a ,b) ==0
If yes then print Possible
Else print Not Possible
Below is the implementation of the above approach.
Try it on GfG Practice C++
```cpp
// C++ program to check for solutions of diophantine
// equations
include
using namespace std;
//utility function to find the GCD of two numbers
int gcd(int a, int b)
{
return (a%b == 0)? abs(b) : gcd(b,a%b);
}
// This function checks if integral solutions are
// possible
bool isPossible(int a, int b, int c)
{
return (c%gcd(a,b) == 0);
}
//driver function
int main()
{
// First example
int a = 3, b = 6, c = 9;
isPossible(a, b, c)? cout << "Possible\n" :
cout << "Not Possible\n";
// Second example
a = 3, b = 6, c = 8;
isPossible(a, b, c)? cout << "Possible\n" :
cout << "Not Possible\n";
// Third example
a = 2, b = 5, c = 1;
isPossible(a, b, c)? cout << "Possible\n" :
cout << "Not Possible\n";
return 0;
}
```
// C++ program to check for solutions of diophantine
// equations#include using namespace std;//utility function to find the GCD of two numbers int gcd(int a, int b){ return (a%b == 0)? abs(b) : gcd(b,a%b);}// This function checks if integral solutions are// possible bool isPossible(int a, int b, int c){ return (c%gcd(a,b) == 0);}//driver function int main(){ // First example int a = 3, b = 6, c = 9; isPossible(a, b, c)? cout << "Possible\n" : cout << "Not Possible\n"; // Second example a = 3, b = 6, c = 8; isPossible(a, b, c)? cout << "Possible\n" : cout << "Not Possible\n"; // Third example a = 2, b = 5, c = 1; isPossible(a, b, c)? cout << "Possible\n" : cout << "Not Possible\n"; return 0;}
Java
```java
// Java program to check for solutions of
// diophantine equations
import java.io.;
class GFG {
// Utility function to find the GCD
// of two numbers
static int gcd(int a, int b)
{
return (a % b == 0) ?
Math.abs(b) : gcd(b,a%b);
}
// This function checks if integral
// solutions are possible
static boolean isPossible(int a,
int b, int c)
{
return (c % gcd(a, b) == 0);
}
// Driver function
public static void main (String[] args)
{
// First example
int a = 3, b = 6, c = 9;
if(isPossible(a, b, c))
System.out.println( "Possible" );
else
System.out.println( "Not Possible");
// Second example
a = 3; b = 6; c = 8;
if(isPossible(a, b, c))
System.out.println( "Possible") ;
else
System.out.println( "Not Possible");
// Third example
a = 2; b = 5; c = 1;
if(isPossible(a, b, c))
System.out.println( "Possible" );
else
System.out.println( "Not Possible");
}
}
// This code is contributed by anuj_67.
Pythonpython3
Python 3 program to check for solutions
of diophantine equations
from math import gcd
This function checks if integral
solutions are possible
def isPossible(a, b, c):
return (c % gcd(a, b) == 0)
Driver Code
if name == 'main':
# First example
a = 3
b = 6
c = 9
if (isPossible(a, b, c)):
print("Possible")
else:
print("Not Possible")
# Second example
a = 3
b = 6
c = 8
if (isPossible(a, b, c)):
print("Possible")
else:
print("Not Possible")
# Third example
a = 2
b = 5
c = 1
if (isPossible(a, b, c)):
print("Possible")
else:
print("Not Possible")
This code is contributed by
Surendra_Gangwar
C#csharp
// C# program to check for
// solutions of diophantine
// equations
using System;
class GFG
{
// Utility function to find
// the GCD of two numbers
static int gcd(int a, int b)
{
return (a % b == 0) ?
Math.Abs(b) :
gcd(b, a % b);
}
// This function checks
// if integral solutions
// are possible
static bool isPossible(int a,
int b,
int c)
{
return (c % gcd(a, b) == 0);
}
// Driver Code
public static void Main ()
{
// First example
int a = 3, b = 6, c = 9;
if(isPossible(a, b, c))
Console.WriteLine("Possible");
else
Console.WriteLine("Not Possible");
// Second example
a = 3; b = 6; c = 8;
if(isPossible(a, b, c))
Console.WriteLine("Possible") ;
else
Console.WriteLine("Not Possible");
// Third example
a = 2; b = 5; c = 1;
if(isPossible(a, b, c))
Console.WriteLine("Possible");
else
Console.WriteLine("Not Possible");
}
}
// This code is contributed by anuj_67.
JavaScriptjavascript
// Javascript program to check for solutions of
// diophantine equations
// Utility function to find the GCD
// of two numbers
function gcd(a, b)
{
return (a % b == 0) ?
Math.abs(b) : gcd(b,a%b);
}
// This function checks if integral
// solutions are possible
function isPossible(a,
b, c)
{
return (c % gcd(a, b) == 0);
}
// Driver Code
// First example
let a = 3, b = 6, c = 9;
if(isPossible(a, b, c))
document.write( "Possible" + "<br/>" );
else
document.write( "Not Possible" + "<br/>" );
// Second example
a = 3; b = 6; c = 8;
if(isPossible(a, b, c))
document.write( "Possible" + "<br/>" ) ;
else
document.write( "Not Possible" + "<br/>" );
// Third example
a = 2; b = 5; c = 1;
if(isPossible(a, b, c))
document.write( "Possible" + "<br/>" );
else
document.write( "Not Possible" + "<br/>" );
PHPphp
php
// PHP program to check for solutions of
// diophantine equations
// utility function to find the
// GCD of two numbers
function gcd($a, $b)
{
return ($a % $b == 0) ?
abs($b) : gcd($b, $a % $b);
}
// This function checks if integral
// solutions are possible
function isPossible($a, $b, $c)
{
return ($c % gcd($a, $b) == 0);
}
// Driver Code
// First example
$a = 3;
$b = 6;
$c = 9;
if(isPossible($a, $b, $c) == true)
echo "Possible\n" ;
else
echo "Not Possible\n";
// Second example
$a = 3;
$b = 6;
$c = 8;
if(isPossible($a, $b, $c) == true)
echo "Possible\n" ;
else
echo "Not Possible\n";
// Third example
$a = 2;
$b = 5;
$c = 1;
if(isPossible($a, $b, $c) == true)
echo "Possible\n" ;
else
echo "Not Possible\n";
// This code is contributed by ajit..
?
```
Output :
Possible
Not Possible
Possible
Time Complexity: O(min(a,b))
Auxiliary Space: O(1)
How does this work?Let GCD of 'a' and 'b' be 'g'. g divides a and b. This implies g also divides (ax + by) (if x and y are integers). This implies gcd also divides 'c' using the relation that ax + by = c. Refer this wiki link for more details.
New Approach:-
Find GCD of a and b using Euclidean algorithm:
Divide the larger number by the smaller number and find the remainder.
Repeat the process with the divisor (smaller number) and the remainder.
Continue this process until the remainder becomes zero.
The GCD will be the last non-zero remainder.
Check if c is divisible by GCD(a, b).
If c is divisible by GCD(a, b), then there exist integer solutions.
Otherwise, there are no integer solutions.
Below is the implementation of the above approach:-
C++
```cpp
include
using namespace std;
// Function to find the GCD of two numbers
int gcd(int a, int b)
{
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
// Function to check if integral solutions are possible
bool isPossible(int a, int b, int c)
{
int gcd_val = gcd(a, b);
return (c % gcd_val == 0);
}
// Driver function
int main()
{
int a = 3, b = 6, c = 9;
if (isPossible(a, b, c)) {
cout << "Possible" << endl;
}
else {
cout << "Not Possible" << endl;
}
return 0;
}
```
include
using namespace std;// Function to find the GCD of two numbers int gcd(int a, int b){ if (a == 0) { return b; } return gcd(b % a, a);}// Function to check if integral solutions are possible bool isPossible(int a, int b, int c){ int gcd_val = gcd(a, b); return (c % gcd_val == 0);}// Driver function int main(){ int a = 3, b = 6, c = 9; if (isPossible(a, b, c)) { cout << "Possible" << endl; } else { cout << "Not Possible" << endl; } return 0;}
Java
```java
import java.util.;
class LinearDiophantineEquations {
// Function to find the GCD of two numbers
public static int gcd(int a, int b) {
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
// Function to check if integral solutions are possible
public static boolean isPossible(int a, int b, int c) {
int gcd = gcd(a, b);
return (c % gcd == 0);
}
// Driver function
public static void main(String[] args) {
int a = 3, b = 6, c = 9;
if (isPossible(a, b, c)) {
System.out.println("Possible");
} else {
System.out.println("Not Possible");
}
}
}
Pythonpython3
class LinearDiophantineEquations:
# Function to find the GCD of two numbers
def gcd(self, a, b):
if a == 0:
return b
return self.gcd(b % a, a)
# Function to check if integral solutions are possible
def isPossible(self, a, b, c):
gcd = self.gcd(a, b)
return c % gcd == 0
# Driver function
def main(self):
a, b, c = 3, 6, 9
if self.isPossible(a, b, c):
print("Possible")
else:
print("Not Possible")
Create an object of the class and call the main function
ld = LinearDiophantineEquations()
ld.main()
C#csharp
using System;
class GFG
{
// Function to find the GCD of two numbers
static int GCD(int a, int b)
{
if (a == 0)
{
return b;
}
return GCD(b % a, a);
}
// Function to check if integral solutions are possible
static bool IsPossible(int a, int b, int c)
{
int gcdVal = GCD(a, b);
return (c % gcdVal == 0);
}
// Driver function
static void Main(string[] args)
{
int a = 3, b = 6, c = 9;
if (IsPossible(a, b, c))
{
Console.WriteLine("Possible");
}
else
{
Console.WriteLine("Not Possible");
}
}
}
JavaScriptjavascript
// Function to find the GCD of two numbers
function gcd(a, b) {
if (a === 0) {
return b;
}
return gcd(b % a, a);
}
// Function to check if integral solutions are possible
function isPossible(a, b, c) {
let gcdValue = gcd(a, b);
return (c % gcdValue === 0);
}
// Driver function
let a = 3,
b = 6,
c = 9;
if (isPossible(a, b, c)) {
console.log("Possible");
} else {
console.log("Not Possible");
}
```
Output:-
Possible
Time Complexity:-The time complexity of this program is dominated by the gcd function, which uses the Euclidean algorithm to compute the GCD of two numbers. The time complexity of the Euclidean algorithm is O(log min(a, b)), so the time complexity of the gcd function is O(log min(a, b)). Since the isPossible function calls the gcd function once, its time complexity is also O(log min(a, b)).
Auxiliary Space:-The space complexity of this program is O(1), since we are only using a few integer variables (a, b, c, and gcd) and a few boolean and string variables (isPossible and the output string). None of these variables grow with the input size, so the space complexity is constant.
Therefore, the time complexity of this program is O(log min(a, b)) and the space complexity is O(1).
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7961 | https://www.vedantu.com/maths/cos-45-degrees | Published Time: 2022-03-08, 9:01:10,+5:30
Cos 45 Degrees Value, Formula, Derivation & Uses Explained
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Maths
Cos 45 Degrees: Value, Derivation, Formula & Examples
Cos 45 Degrees: Value, Derivation, Formula & Examples
Reviewed by:
Rama Sharma
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What is the value of cos 45 degrees?
The concept of cos 45 degrees is a core topic in trigonometry, often tested in maths exams and used in STEM fields. Understanding and remembering its value helps students solve a wide range of geometry and algebra problems efficiently.
What Is Cos 45 Degrees?
Cos 45 degrees is the cosine of a 45° angle. It is a trigonometric ratio that relates the length of the adjacent side to the hypotenuse in a right-angled triangle. You'll encounter this value in geometry, the unit circle, and various real-life applications like physics and engineering. In trigonometry, cos 45° is considered a "standard angle" because its value is easy to learn, use, and apply to different problems.
Key Formula for Cos 45 Degrees
Here’s the standard formula: cos45∘=Adjacent side Hypotenuse
Value of Cos 45 Degrees (Exact, Fraction, Decimal, Radian)
| Angle | Fraction (Surd) | Decimal | Radian Equivalent |
--- --- |
| 45° | 2 2 | 0.7071 | π 4 |
How to Derive the Value of Cos 45 Degrees
Let's derive cos 45 degrees using an isosceles right triangle and the unit circle for a clear understanding.
Draw a right triangle where both non-right angles are 45° (isosceles right triangle).
Let the two shorter sides be 1 unit each. By Pythagoras’ theorem, hypotenuse = 1 2+1 2=2
Cos 45° = Adjacent/Hypotenuse = 1/2
Rationalize: 1/2=2/2 (exact value)
On the unit circle, the x-coordinate at 45° is also 2/2.
Cos 45 Degrees in Trigonometric Tables
The value of cos 45 degrees is a key entry in any trigonometric table. It is regularly used in competitive and board exams for solving MCQs, geometry, and trigonometric identities. Being able to recall cos 45° quickly allows you to solve questions faster and score better.
| Angle | Cosine Value |
--- |
| 0° | 1 |
| 30° | 3 2 |
| 45° | 2 2 |
| 60° | 1 2 |
| 90° | 0 |
Cos 45 Formulae and Related Identities
Here are some important identities involving cos 45 degrees:
cos 245∘+sin 245∘=1
cos45∘=sin45∘
cos(90∘−45∘)=sin45∘
Trigonometric identities often include cos45∘ in compound angles and sum/difference formulas.
Solved Examples Using Cos 45 Degrees
Here are a few sample problems to help you master cos 45 degrees value in real scenarios.
Q1. Find the value of: 2 sin60∘−4 cos45∘
Substitute known values: sin60∘=3/2, cos45∘=2/2
Calculate: 2×(3/2)−4×(2/2)
Simplify: 3−2 2
Final Answer: 3−2 2
Q2. What is cos45∘+cos90∘?
Substitute: cos45∘=2/2, cos90∘=0
Add: 2/2+0=2/2
Final Answer: 2/2
Q3. Evaluate: 3 cos45∘+2 sin60∘
Known values: cos45∘=2/2; sin60∘=3/2
Multiply: 3×(2/2)+2×(3/2)
Simplify: (3 2+2 3)/2
Final Answer: (3 2+2 3)/2
Speed Trick to Remember Cos 45 Degrees
A simple trick for remembering the cosine values of standard angles (0°, 30°, 45°, 60°, and 90°) is to use the square root pattern:
cosθ=n/2 where n = 4, 3, 2, 1, 0 for 0°, 30°, 45°, 60°, 90°, respectively.
Thus, for cos 45°, n = 2: cos45∘=2/2.
Relation to Other Trigonometric Concepts
Learning cos 45 degrees helps you connect with important topics like sin 45 degrees (which equals cos 45°), the trignometric table, and trigonometric ratios for other standard angles.
Wrapping It All Up
We explored cos 45 degrees—its definition, derivation, conversion to radian, decimal and fraction, usage in tables and examples, along with memory tricks. Keep practicing and refer to Vedantu's resources for doubt clearance and more interactive learning on trigonometry.
Try These Yourself
Calculate cos 45 degrees as a decimal without a calculator.
Prove cos 45 = sin 45 using a right triangle.
Find the area of an isosceles right triangle with legs of length 5 using cos 45°.
Solve: If cos θ = cos 45°, what is θ between 0° and 360°?
Internal Links for Further Learning
Sin 45 Degrees: Understand why sin 45° = cos 45°.
Trigonometry Table: Look up all major angles and their sine, cosine, and tangent values.
Unit Circle: Visual guide to trig ratios and their coordinates.
Trigonometric Functions: Dive deeper into trigonometric graphs, domains, and inverses.
Degrees to Radians: Convert 45° to radians and vice versa for advanced applications.
Continue your maths journey with Vedantu to master cos 45 degrees and all other important trigonometric values for exams and real life!
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7962 | https://blogs.sas.com/content/iml/2011/06/01/the-trapezoidal-rule-of-integration.html | # Blogs
The trapezoidal rule of integration
17
By Rick Wicklin on The DO Loop
In a previous article I discussed the situation where you have a sequence of (x,y) points and you want to find the area under the curve that is defined by those points. I pointed out that usually you need to use statistical modeling before it makes sense to compute the area.
However, there is a numerical technique that is very useful for a wide range of numerical integration scenarios, and that is the trapezoidal rule.
The following graph illustrates the trapezoidal rule. Given a set of points (x1, y1), (x2, y2), ..., (xn, yn), with x1 ≤ x2 ≤ ... ≤ xn, the trapezoidal rule computes the area of the piecewise linear curve that passes through the points. You can compute the area under the piecewise linear segments by summing the area of the trapezoids A1, A2, A3, and A4. (Sometimes a trapezoid is degenerate and is actually a rectangle or a triangle.)
Implementing the Trapezoidal Rule in SAS/IML Software
It is easy to use SAS/IML software (or the SAS DATA step) to implement the trapezoidal rule. The area of a trapezoid defined by (xi, yi) and (xi+1, yi+1) is
( xi+1 – xi ) ( yi + yi+1 ) / 2
The first term is just the width of the subinterval [xi, xi+1] and the second term is the average of the heights at each end of the subinterval.
The following user-defined function computes the area under the piecewise linear segments that connect the points. The function does not check that the x values are sorted in nondecreasing order.
| |
| proc iml; /Given two vectors x,y where y=f(x), this module approximates the definite integral int_a^b f(x) dx by the trapezoid rule. The vector x is assumed to be in numerically increasing order so that a=x and b=x[nrow(x)]. The module does not assume equally spaced intervals. The formula is Integral = Sum( (x[i+1] - x[i]) (y[i] + y[i+1])/2 ) / start TrapIntegral(x,y); N = nrow(x); dx = x[2:N] - x[1:N-1]; meanY = ( y[2:N] + y[1:N-1] )/2; return( dx` meanY ); finish; / test it / x = {0.0, 0.2, 0.4, 0.8, 1.0}; y = {0.5, 0.8, 0.9, 1.0, 1.0}; area = TrapIntegral(x,y); print area; |
Notice that the implementation does not require equally spaced points. The width of all subintervals are computed in a single statement and assigned to the vector dx. Similarly, the average of the heights are computed in a single statement and assigned to the vector meanY. The summation of all the areas is then computed by using a dot product of vectors. (Equivalently, the module could also return the quantity sum(dx # meanY), but the dot product is the more efficient computation.)
The simplicity of the trapezoidal rule makes it an ideal for many numerical integration tasks. Also, the trapezoidal rule is exact for piecewise linear curves such as an ROC curve. Also, as John D. Cook points out, there are other situations in which the trapezoidal rule performs more accurately than other, fancier, integration techniques.
The trapezoidal rule is not as accurate as Simpson's Rule when the underlying function is smooth, because Simpson's rule uses quadratic approximations instead of linear approximations. The formula is usually given in the case of an odd number of equally spaced points. Leave a comment to discuss the relative advantages and disadvantages of Simpson's rule as compared to the trapezoidal rule.
In a future blog post, I will use the TrapIntegral function to integrate some functions that arise in statistical data analysis.
Tags Numerical Analysis
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About Author
Rick WicklinDistinguished Researcher in Computational Statistics
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Rick Wicklin, PhD, is a distinguished researcher in computational statistics at SAS and is a principal developer of SAS/IML software. His areas of expertise include computational statistics, simulation, statistical graphics, and modern methods in statistical data analysis. Rick is author of the books Statistical Programming with SAS/IML Software and Simulating Data with SAS.
17 Comments
Charlie Huang on
Fantastic work! I applied it in my daily routine immediately. Thanks a lot.
Reply
2. Pingback: A statistical application of numerical integration: The area under an ROC curve - The DO Loop
3. Pingback: The area under a density estimate curve: Nonparametric estimates - The DO Loop
4. Steven on
If you could state the strengths and limitations of both simpson's rule and trapezodial rule that would be greatly appreciated. However, the work already posted is great.
Reply
Rick Wicklin on
The strength of the trapezoidal rule is that it is fast and it is exact for piecewise linear functions. The strength of Simpson's rule is that it is usually more accurate: it has a smaller error when integrating smooth functions. The limitations for both algorithms is that they are designed for finite intervals. They are also require that you choose the number of subdivisions in advance, which is why I use the QUAD subroutine in SAS/IML software for serious work. The QUAD subroutine implements an adaptive Romberg-type algorithm that is is more accurate than Simposon's Rule and can work on infinite domains.
Reply
5. willard rupia on
I need some help on the application of trapezoidal rule or simpson rule in mining metallurgy, where exactly can we apply these two rules?
Reply
Rick Wicklin on
These rules enable you to numerically approximate an integral. Integrals appear in many areas of material science, including loads, shears, stresses, torsion, etc. See any engineering or calculus textbook.
Reply
6. Pingback: An easy way to approximate a cumulative distribution function - The DO Loop
7. laura on
what about missing data points? what is the best way to estimate area under the curve if there is a missing data point?
Reply
Rick Wicklin on
Use the trapezoid rule applied to the nonmissing data, which is equivalent to linear interpolation of missing Y values.
Reply
Greg on
So if I have two groups (placebo vs active) and 5 timepoints (i.e., 4 trapezoids) for a measurement scale. Several of the subjects are missing the last 1-2 timepoints. If I sum the values of the trapezoids to a final AUC for each subject, can I simply compare mean AUC's with a t-test (assuming normally distributed)?.
If so, some of the total AUCs will be much smaller because of missing values--how can I handle this post hoc?
Reply
Rick Wicklin on
If you want to compare two ROC curves, use the ROC and ROCCONTRAST statements in PROC LOGISTIC. If you have further questions, you can post them to the SAS Support Communities.
Reply
- Christian Badillo on
One of the objectives of pharmacokinetic studies is the determination of the area under the curve (AUC) parameter. The area under the curve represents the proportion of the drug absorbed into the systemic circulation and the time required to do so.
This pharmacokinetic parameter is used to determine the total exposure of a person to the drug over a period of time and is required by regulatory authorities to determine bioavailability and conclude bioequivalence or the absence of pharmacokinetic interactions.
The area under the curve can be obtained through two methodologies: the first involves nonlinear regression model analysis, or through non-compartmental analysis (NCA).
NCA is usually the preferred methodology as it requires fewer assumptions than model-based approaches. There are four methods to determine the AUC through non-compartmental analysis: Linear Log Trapezoidal, Linear Trapezoidal Interpolation, Linear Up Log Down, and Linear/Logarithmic Trapezoidal Interpolation.
The linear trapezoidal method uses linear interpolation between data points to calculate the AUC. This method is required by the OGD and the FDA, and it is the standard for bioequivalence trials.
The logarithmic trapezoidal method uses logarithmic interpolation between data points to calculate the AUC. This method is more accurate when concentrations are decreasing because drug elimination is exponential (making it linear on a logarithmic scale).
Below you will find a slide about your issue.
Reply
8. sanaz on
Hello,Thanks for your explanation. How can I implement this code in to the Malab script?
Reply
9. Karrar on
I want to make a report in Trapezoid rule and simpson rule.What the advantage and disadvantage for Trapezoid rule and simpson ruleWhat the theory of each one and their applications?Thank you
Reply
Rick Wicklin on
Theoretically, each method is based on evenly dividing an interval into subintervals. The trapezoidal rule uses a linear approximation to the function on each interval, whereas Simpson's rule uses a quadratic approximation. The area under the (approximate) curve is computed for each subinterval, and the areas are summed to approximate the integral on the full interval. Because Simpson's rule uses a quadratic approximation on each subinterval, Simpson's rule is more accurate when each method uses the same number of subintervals. The advantage of the trapezoidal rule is that it is very fast and it is exact for piecewise linear functions.
Reply
10. arkam on
can you please discuss the limitation of trapezoidal method
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7963 | https://static1.squarespace.com/static/5ebed9b237eead74e35f143d/t/5ed5edd3a8c2082cb64d8a7c/1591078356036/Chapter+3+HW.pdf | Geometry 13 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Lesson 3.1 Practice A For use with pages 129–134 3.1 LESSON NAME ________ DATE __ Think of each segment in the diagram as part of a line. Fill in the blank with parallel, skew, or perpendicular.
1.
are 2.
3.
4.
Think of each segment in the diagram as part of a line. There may be more than one correct answer.
5. Name a line parallel to 6. Name a line perpendicular to 7. Name a line skew to 8. Name a plane parallel to plane RPL.
Complete the statement with corresponding, alternate interior, alternate exterior, or consecutive interior.
9.
and are angles.
10.
and are angles.
11.
and are angles.
12.
and are angles.
13.
and are angles.
14.
and are angles.
Answer true or false.
15. The hands of a clock are perpendicular at 3:00 and 9:00.
16. If two lines do not intersect, then they are parallel.
17. The perpendicular postulate states that for a point on a line, there is exact-ly one line through the point perpendicular to the line.
18. The parallel postulate states that for a point not on a line, there is exactly one line through the point parallel to the line.
Use the diagram to answer the question.
19. Name all pairs of vertical angles.
20. Name all pairs of corresponding angles.
21. Name all pairs of alternate interior angles.
1 8 2 7 3 6 4 5 ? 7 5 ? 5 9 ? 6 8 ? 8 5 ? 10 4 ? 7 3 4 3 5 6 8 7 9 10 ↔ SN.
↔ PR.
↔ MN.
Q R S P N M L ↔ AB and ↔ FG are ? .
↔ BF and ↔ FG are ? .
↔ AB and↔ BC are ? . ? .
↔ AB and ↔ DC A D C B G F E 14 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Lesson 3.1 Practice B For use with pages 129–134 3.1 LESSON NAME ________ DATE ___ Think of each segment in the diagram as part of a line. Fill in the blank with parallel, skew, or perpendicular.
1.
are 2.
are 3.
are 4. plane VWT and plane RSX are 5.
are Think of each segment in the diagram as part of a line. There may be more than one correct answer.
6. Name a line parallel to 7. Name a line perpendicular to 8. Name a line skew to 9. Name a plane parallel to plane GHJ.
10. Name a line perpendicular to Complete the statement with corresponding, alternate interior, alternate exterior, or consecutive interior.
11.
and are angles.
12.
and are angles.
13.
and are angles.
14.
and are angles.
15.
and are angles.
16.
and are angles.
Use the diagram of the Ferris wheel to decide whether the statement is true or false.
17. At any position around the wheel, the line containing the crossbar, of each cart is parallel to the ground.
18. For any cart of the Ferris wheel, the line containing the back support, and the line containing the crossbar, are skew lines.
19. At any position around the wheel, the line containing the back support, is perpendicular to the ground. (Assume the carts are hanging as shown in the diagram.) ↔ DC, ↔ AB, ↔ CD, ↔ AB, C D A B ? 10 8 ? 11 5 ? 8 12 ? 9 8 ? 9 7 ? 10 6 6 5 8 7 10 9 11 12 ↔ JH.
↔ GH.
↔ LM.
↔ HJ.
K G H J N M L ? .
↔ TW and ↔ WX ? . ? .
↔ TU and ↔ WX ? .
↔ RS and ↔ VW ? .
↔ UT and ↔ WT U V W T X S R Geometry 15 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice C For use with pages 129–134 3.1 LESSON NAME ________ DATE __ Lesson 3.1 Lesson 3.1 Think of each segment in the diagram as part of a line. Fill in the blank with parallel, skew, or perpendicular.
1.
and are 2.
and are 3.
and are 4.
and are 5. plane CGF and plane ABH are Think of each segment in the diagram as part of a line.
There may be more than one correct answer.
6. Name a line parallel to 7. Name a line perpendicular to 8. Name a line skew to 9. Name a plane parallel to plane PNM.
10. Name a line perpendicular to Complete the statement with corresponding, alternate interior, alternate exterior, or consecutive interior.
11.
and are angles.
12.
and are angles.
13.
and are angles.
14.
and are angles.
15.
and are angles.
16.
and are angles.
Use the diagram of the ski lift to decide whether the statement is true or false.
17. At any position around the lift, the line containing the crossbar, of each chair is parallel to the ground.
18. For any chair of the lift, the line containing the back support, and the line containing the crossbar, are skew lines.
19. At any position around the lift, the line containing the back support, is perpendicular to the ground. (Assume the carts are hanging as shown in the diagram.) ↔ DC, ↔ AB, ↔ CD, ↔ AB, C D A B ? 1 12 ? 9 4 ? 5 10 ? 2 8 ? 11 7 ? 10 1 2 4 10 9 11 12 6 5 7 8 1 3 ↔ NM.
↔ PN.
↔ PN.
↔ PN.
L N M K J I P H ? . ? .
↔ HG ↔ AD ? .
↔ BH ↔ CD ? .
↔ GH ↔ FG ? .
↔ AB ↔ DC A B C D E F G H 26 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice A For use with pages 136–141 3.2 LESSON NAME ________ DATE _ Lesson 3.2 State the reason for the conclusion.
1. Given: Conclusion: 2. Given: Conclusion: is a right angle 3. Given: and are vertical angles Conclusion: 4. Given: and are complementary angles Conclusion: 5. Given: and are supplementary angles Conclusion: Find the value of x.
6.
7.
8.
9. Complete the flow proof of Theorem 3.2.
Given: Prove: and are complementary.
C D E 1 2 2 1 → CD→ CE DCE is a right .
b.
DCE = 90° c.
m DCE = m 1 + m 2 d.
m 1 + m 2 = 90° m e.
2 are complementary.
1 and f.
a.
CD CE 29° x 47° x x m8 m9 180 9 8 m6 m7 90 7 6 4 5 5 4 3 n m1 m2 1 2 1 2 3 n 4 5 6 7 8 9 Geometry 27 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice B For use with pages 136–141 3.2 LESSON NAME ________ DATE __ Lesson 3.2 State the reason for the conclusion.
1. Given: Conclusion: 2. Given: and are a linear pair.
Conclusion: and are supplementary.
3. Given: Conclusion: 4. Given: X is the midpoint of Conclusion: 5. Given: bisects Conclusion; Find the value of x.
6.
7.
8.
9. Complete the two-column proof of Theorem 3.2.
Given: Prove: and are complementary.
10. Complete the flow proof of a portion of Theorem 3.3.
Given: is a right angle.
Prove: is a right angle.
1 3 3 m 1 = 90° e.
m 3 = 90° f.
1 ≅ 3 b.
m 1 = m 3 c.
d.
1 is a right .
3 is a right .
g.
3 are vertical s.
1 and a.
1 C D E 1 2 2 1 → CD→ CE 3x x (x 12)° 49° (x 38) BAD DAC BAC.
→ AD MX NX MN.
6 5 5 6 4 3 4 3 1 2 m1 m2 1 2 4 3 5 6 M X N A B C D Statements Reasons 1.
1.
2.
is a right 2.
3.
3. Def. of right 4.
4.
5.
5. Substitution 6.
and are complementary.
6.
2 1 mDCE m1 m2 .
DCE → CD→ CE 28 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice C For use with pages 136–141 3.2 LESSON NAME ________ DATE __ Lesson 3.2 What can you conclude from the given information? State the reason for your conclusion.
1.
2.
3.
Find the value of x.
4.
5.
6.
7. Complete the two-column proof of Theorem 3.1.
Given: and are a linear pair Prove: 8. Write a paragraph proof of Theorem 3.2.
Given: Prove: and are complementary 2 1 → FG→ FH 2 1 F G H 2 1 n n 2 1 2, 1 2x° 3x° (2x 9)° 27° (2x 18)° 1 4 2 3 p q 1 2 A C B 1 2 m pq → AB→ AC 1 2 Statements Reasons 1.
1.
2.
2.
3.
and are a linear pair 3.
4.
and are supplementary 4.
5.
5.
6.
6.
7.
7.
8.
8.
9.
is a right 9.
10.
10.
n 1 m1 90 2m1 180 m1 m1 180 m1 m2 180 2 1 2 1 m1 m2 1 2 40 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice A For use with pages 143–149 3.3 LESSON NAME ________ DATE ___ Lesson 3.3 Name the relationship between the pair of angles.
1.
and 2.
and 3.
and 4.
and 5.
and 6.
and State the postulate or theorem that justifies the statement.
7.
8.
9.
10.
Find the values of x and y.
11.
12.
13.
14.
15.
16.
Find the value of x.
17.
18.
19.
135° (3x 15)° 120° 3x (2x 10)° 80° 130° x y 120° x y 100° x y x y 115° x y 70° x y m4 m6 180 2 7 3 6 3 7 2 4 1 3 6 8 5 7 4 8 6 4 5 8 6 3 7 2 5 1 2 4 1 3 6 8 5 7 Geometry 41 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice B For use with pages 143–149 3.3 LESSON NAME ________ DATE __ Lesson 3.3 Find and Explain your reasoning.
1.
2.
3.
Find the values of x and y.
4.
5.
6.
Find the value of x.
7.
8.
9.
10. Complete the flow proof of the Alternate Exterior Angles Theorem.
Given: Prove: 1 2 3 m 1 2 1 ≅ 3 b.
1 ≅ 2 d.
3 ≅ 2 c.
a.
m m (4x 9) 75 3(x 9)° 129° (5x 15) 80 x y 98 x y 81 x y 127° 1 2 72° 1 2 118° 1 2 m2.
m1 42 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice C For use with pages 143–149 3.3 LESSON NAME ________ DATE _ Lesson 3.3 Find the measure of all labeled angles in the diagram.
1.
2.
Find the value of x and y.
3.
4.
5.
Find the value of x.
6.
7.
8.
(3x 9) (5x 25) 84 7(x 19) 148 (6x 32) (x 25) (y 18) 98 (x 9) 2y 83 (y 13) x 53° 1 3 2 10 8 9 4 5 6 7 42° 1 2 3 10 8 9 4 5 6 7 9. Write a proof of the Alternate Exterior Angles Theorem.
Given: Prove: 10. Write a proof.
Given: Prove: 2 1 m q p 1 2 m, p q 2 1 m 1 2 m Geometry 57 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Lesson 3.4 Practice A For use with pages 150–156 3.4 LESSON NAME ________ DATE __ Is it possible to prove that lines p and q are parallel? If so, explain how.
1.
2.
3.
4.
5.
6.
Find the value of x that makes 7.
8.
9.
Use the diagram and the given information to determine which lines are parallel.
10.
11.
12.
13.
14. Complete the two-column proof of the Alternate Exterior Angles Converse Theorem.
Given: Prove: m 1 2 7 9 16 2 4 8 13 11 1 2 4 3 5 6 8 7 13 14 16 15 9 10 12 11 m n q q p 80 (2x 20) q p 120 (2x 10) q p 45 3x p q.
q p q p q p x° x° 115° 75° q p 70° 70° q p 50° 130° q p 1 2 3 m Statements Reasons 1.
1.
2.
2.
3.
3.
4.
4.
m 2 3 1 3 1 2 58 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Lesson 3.4 Practice B For use with pages 150–156 3.4 LESSON NAME ________ DATE __ Is it possible to prove that lines p and q are parallel? If so, explain how.
1.
2.
3.
Find the value of x that makes 4.
5.
6.
Give the choice or choices that make the statement true.
7. If two lines are cut by a transversal so that alternate interior angles are (congruent, supplementary, complementary), then the lines are parallel.
8. If two lines are cut by a transversal so that consecutive interior angles are (congruent, supplementary, complementary), then the lines are parallel.
9. If two lines are cut by a transversal so that (alternate interior, alternate exterior, corresponding) angles are congruent, then the lines are parallel.
10. Complete the two-column proof.
Given: Prove: 11. Write a two-column proof.
Given: Prove: a b a b m 1 3 2 m, 1 2 a b a b m 1 3 2 m, 1 2 (5x 10) (2x 50) p q x (3x 24) p q 4x (3x 30) p q p q.
53° 79° 48° p q 132° 74° 58° p q 124° 56° p q Statements Reasons 1.
1.
2.
2.
3.
3.
4.
4.
5.
5.
a b 2 3 1 2 1 3 m Geometry 59 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Lesson 3.4 Practice C For use with pages 150–156 3.4 LESSON NAME ________ DATE ___ Is it possible to prove that lines p and q are parallel? If so, explain how.
1.
2.
3.
Find the value of x that makes 4.
5.
6.
7. Write a two-column proof.
Given: Prove: 8. Write a two-column proof.
Given: Prove: 9. Write a flow proof.
Given: are supplementary Prove: a b m, 1 and 2 a b m 2 1 3 a b m, 1 2 a b m 2 1 3 a b m, 1 2 a b m 2 1 3 q p 2x 4x 7x (3x 10) q p (4x 22) (3x 17) q p p q.
24° 27° 128° q p 53° 37° q p 38° 105° 143° q p 70 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice A For use with pages 157–164 3.5 LESSON NAME ________ DATE __ Lesson 3.5 State the postulate or theorem that allows you to conclude that 1. Given: 2. Given: 3. Given: Explain how you would show that State any theorems or postulates that you would use.
4.
5.
6.
7. Construct a line parallel to through point P.
8. Complete the flow proof of Theorem 3.12.
Given: Prove: m n m, n P 70 110 c d 80 80 c d 110 110 c d c d.
a b 1 2 a b c a c b 1 2 ac, bc a c, b c a b.
m n 1 2 a.
m b.
1 is a right .
c.
n f.
n m d.
2 is a right .
1 ≅ 2 e.
Geometry 71 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice B For use with pages 157–164 3.5 LESSON NAME ________ DATE _ Lesson 3.5 State the postulate or theorem that allows you to conclude that 1. Given: 2. Given: 3. Given: Explain how you would show that State any postulates or thoerems that you would use.
4.
5.
6.
7. Construct a line parallel to through point P.
8. Complete the two-column proof of Theorem 3.12.
Given: Prove: m n m, n P c a b 42 143 37 42 b c a 1 2 c a b a b.
b c a a b 1 2 a b c a c, b c 1 2 ac, bc a b.
m n 1 2 Statements Reasons 1.
1.
2.
is a rt. 2.
3.
is a rt. 3.
4.
4.
5.
5.
m n 1 2 .
2 .
1 m, n 72 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice C For use with pages 157–164 3.5 LESSON NAME ________ DATE __ Lesson 3.5 Explain how you would show that 1.
2.
3.
Determine which lines, if any, must be parallel. Explain your reasoning.
4.
5.
6. Draw an obtuse angle. Construct an angle congruent to it.
7. Draw a horizontal line. Construct a line parallel to it through a point not on the line.
8. Proof: Write a two-column proof of Theorem 3.12.
Given: Prove: 9. Proof: Write a two-column proof.
Given: Prove: 10. Proof: Write a two-column proof.
Given: Prove: m 1 2, 3 4 m n 1 2 3 4 AB CD 1 2, 1 3 A B E C D 1 3 2 m n m, n m n m n p m n p 113 67 66 (2x) (2x) (180 2x) a b x x a b (5x) (21 2x) (18x 19) a b a b.
Geometry 85 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Lesson 3.6 Practice A For use with pages 165–171 3.6 LESSON NAME ________ DATE __ Calculate the slope of the line shown.
1.
2.
3.
Calculate the slope of the line that passes through the labeled points on the graph.
4.
5.
6.
Find the slope of each line. Are the lines parallel?
7.
8.
9.
Write an equation of the line.
10. slope 11. slope 12. slope y-intercept y-intercept y-intercept 13. parallel to 14. parallel to 15. parallel to y-intercept y-intercept y-intercept 3 4 0 1 3 y x 1 y 3x 7 y 2x 5 4 3 5 1 2 2 4 1 1 y x A B C D 1 1 y x A B C D 1 1 y x A B C D 1 1 y x (2, 2) (3, 2) 1 1 y x (1, 2) (1, 4) 1 1 y x (2, 1) (3, 3) 1 1 y x 1 2 2 2 y x 3 2 1 1 3 1 x y 86 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Lesson 3.6 Practice B For use with pages 165–171 3.6 LESSON NAME ________ DATE ___ Calculate the slope of the line that passes through the labeled points on the graph.
1.
2.
3.
Find the slope of each line. Are the lines parallel?
4.
5.
6.
Write an equation of the line.
7. slope 8. parallel to 9. parallel to y-intercept y-intercept y-intercept Write an equation of the line that passes through the given point P and has the given slope.
10.
slope 11.
slope 12.
slope Use the following information.
A parallelogram is a four-sided figure whose opposite sides are parallel. Given and 13. Plot and label the points. Connect the points with a segment to form quadrilateral ABCD.
14. Determine the slopes of and 15. Is quadrilateral ABCD a parallelogram? Explain.
DA.
AB, BC, CD, D2, 5.
C3, 4, B1, 6, A2, 3, 1 P4, 2, 4 5 P5, 6, 2 P0, 5, 6 1 3 3 y 1 2x 3 y 3x 2 1 1 y x A B C D 3 3 y x A B C D 3 3 y x A B C D 1 1 y x (2, 2) (2, 3) 1 1 y x (3, 2) (2, 1) 1 1 y x (2, 2) (3, 1) Geometry 87 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Lesson 3.6 Practice C For use with pages 165–171 3.6 LESSON NAME ________ DATE __ Calculate the slope of the line that passes through the labeled points on the graph.
1.
2.
3.
Find the slope of each line. Are the lines parallel?
4.
5.
6.
Write an equation of the line.
7. slope 8. parallel to 9. parallel to y-intercept y-intercept y-intercept Write an equation of the line that passes through the given point P and has the given slope.
10.
slope 11.
slope 12.
slope Use the following information.
A parallelogram is a four-sided figure whose opposite sides are parallel. Given and 13. Plot and label the three points.
14. Determine the coordinates of point D so that the points are the vertices of a parallelogram. Hint: There is more than one location.
15. If one pair of opposite sides of a parallelogram have positive slopes, will the other pair of sides have negative slopes? Explain.
C2, 5.
B1, 6, A2, 3, 3 P3, 3, 2 3 P2, 4, 5 P0, 2, 0 3 5 2 y 8 y 4x 4 2 5 1 1 y x A B C D 1 1 y x A B C D 1 1 y x A B C D 2 2 y x (6, 6) (3, 6) 2 2 y x (4, 3) (8, 0) 1 1 y x (3, 1) (2, 1) 100 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice A For use with pages 172–178 3.7 LESSON NAME ________ DATE _ Lesson 3.7 Find the slope of and Decide whether is perpendicular to 1.
2.
3.
The slopes of two lines are given. Are the lines perpendicular?
4.
5.
6.
7.
8.
9.
Lines a and b are perpendicular. The slope of line a is given. What is the slope of line b?
10. 3 11.
12.
13.
14.
15.
16. 1 17.
Decide whether lines and are perpendicular.
18. line 19. line line line 20. line 21. line line line Line j is perpendicular to the line with the given equation and line j passes through P. Write an equation of line j.
22.
23.
24.
25.
Write an equation parallel to the given line. Write an equation perpendicular to the given line.
26.
27.
28. y 1 3x 2 y x 5 y 2x 4 y 2 3x 4, P2, 0 y 4 5x 4, P1, 1 y 3x 4, P0, 2 y 1 3x 4, P0, 5 p2: 4x 2y 10 p2: 2x 4y 6 p1: x 8y 4 p1: 4x 2y 6 p2: 5x 3y 18 p2: y 1 3x 5 p1: 3x 5y 12 y 3x 5 p1: p2 p1 6 7 2 5 1 2 5 2 2 3 4 m1 1, m2 1 m1 3 4, m2 4 3 m1 2 3, m2 3 2 m1 4, m2 1 4 m1 1 2, m2 2 m1 2, m2 1 2 3 1 y x B(4, 1) C(2, 1) D(1, 2) A(1, 2) 2 1 y x B(2, 0) A(4, 1) C(2, 2) D(4, 3) 4 1 y x A(2, 2) B(2, 3) C(0, 2) D(2, 1) ↔ BD.
↔ AC ↔ BD.
↔ AC Geometry 101 Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice B For use with pages 172–178 3.7 LESSON NAME ________ DATE __ Lesson 3.7 Find the slope of and Decide whether is perpendicular to 1.
2.
3.
The slopes of two lines are given. Are the lines perpendicular?
4.
5.
6.
7.
8.
9.
Decide whether lines and are perpendicular.
10. line 11. line line line 12. line 13. line line line Determine if the intersection of and forms a right angle.
Explain your reasoning.
14.
15.
16.
17.
Line j is perpendicular to the line with the given equation and line j passes through P. Write an equation of line j.
18.
19.
20.
21.
Write an equation parallel to the given line. Write an equation perpendicular to the given line.
22.
23.
24. 2x 4y 3 y 1 3x 1 y 5x y 2 7x 4, P2, 3 y 5 6x 4, P10, 12 y 4x 7, P1, 2 y 1 3x 5, P6, 2 A4, 5, B1, 3, C2, 4, D4, 1 A5, 3, B3, 2, C0, 2, D1, 6 A3, 6, B1, 4, C4, 0, D0, 8 A9, 2, B0, 1, C1, 8, D2, 1 ↔ CD ↔ AB p2: 6x 3y 8 p2: 7x 9y 5 p1: x 2y 4 p1: 9x 7y 6 p2: 6x 8y 18 p2: y 1 2x 5 p1: 6x 8y 12 y 2x 5 p1: p2 p1 m1 3, m2 3 m1 31 2, m2 2 7 m1 2 5, m2 5 2 m1 2, m2 1 2 m1 4 3, m2 4 3 m1 3, m2 1 3 1 1 y x C(2, 2) B(1, 4) D(3, 0) A(2, 1) 1 1 y x B(2, 3) D(4, 3) C(3, 0) A(2, 2) 4 y x B(1, 3) C(1, 1) D(0, 1) A(3, 0) 1 ↔ BD.
↔ AC ↔ BD.
↔ AC 102 Geometry Chapter 3 Resource Book Copyright © McDougal Littell Inc. All rights reserved.
Practice C For use with pages 172–178 3.7 LESSON NAME _________ DATE _____ Lesson 3.7 Find the slope of and Decide whether is perpendicular to 1.
2.
3.
Decide whether lines and are perpendicular.
4. line 5. line line line 6. line 7. line line line Determine if the intersection of and forms a right angle. Explain your reasoning.
8.
9.
10.
11.
Line j is perpendicular to the line with the given equation and line j passes through P. Write an equation of line j.
12.
13.
14.
15.
Decide whether the lines with the given equations are parallel, perpendicular, or neither.
16.
17.
18.
19.
20.
21.
3x 6y 10 y 6 5x 4 5x 2y 2 x 2y 12 y 5 6x 8 2x 5y 8 4x 8y 7 y 3x 2 y 5x 2 2x 4y 3 y 1 3x 1 y 5x 2 y 2 3x 4, P6, 2 y 5 2x 1, P5, 6 y 0.1x 7, P1, 2 y 1 6x 5, P3, 1 A1, 2, B2, 6, C1, 5, D5, 2 A4, 4, B4, 3, C2, 4, D1, 4 A5, 8, B1, 6, C1, 3, D3, 5 A7, 0, B2, 1, C3, 6, D4, 3 ↔ CD ↔ AB p2: 6x 2y 8 7x 9y 5 p2: x 3y 4 p1: 9x 4 7y p1: p2: 2x 7y 5 p2: y 1 3x 5 p1: 7x 2y 5 y 3x 5 p1: p2 p1 1 y x D(1, 3) C(1, 2) A(3, 3) B(4, 1) 1 1 y x D(3, 1) C(1, 3) A(2, 3) B(0, 3) 1 1 y x A(2, 2) D(2, 2) B(1, 1) C(0, 0) 1 ↔ BD.
↔ AC ↔ BD.
↔ AC |
7964 | https://www.youtube.com/watch?v=3I7kiaKSoKc | Calculus - What's a Sequence of Partial Sums?
The Infinite Looper
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Posted: 1 Oct 2015
Continuing the previous video, this video explains what a sequence of partial sums is.
11 comments
Transcript:
okay in the last video we talked about sequences and series and uh specifically we talked about the differences between them what's a sequence versus what's a series and we mentioned that series have uh two sequences associated with them there's that sequence a subn that actually list all the terms that are being added and then there's also the sequence of partial sums and in this video we're going to talk about the sequence of partial sums so I did want to keep them separate just so that things aren't too confusing um because sequence of partial sums might be something that's relatively new to you uh if first starting with Cal 2 but sequence is a series you may have seen before in earlier courses uh way back when so anyway with the sequence of partial sums so if we have a series some from Nal 1 to positive Infinity of a subn so we know that that's a sub 1 plus a sub 2 plus a sub3 plus dot dot dot there's actually an Associated sequence of partial sums and the sequence is capital S Sub 1 comma capital S Sub 2 comma capital S sub3 comma dot dot dot so remember the sequence is a list so this here is a sequence of partial sums and because it's a sequence we're writing it like this okay or we could use the more compact notation like this so we could also say this let me get rid of that uh we could also say s subn from Nal 1 to positive Infinity okay we could also write the sequence of partial sums like that now what exactly is the sequence what are these sub subn well Sub sub one is just the first term uh a sub 1 s sub2 so that's not very interesting but S Sub 2 is a sub 1 plus a sub 2 okay or if we want to use the uh Sigma notation here that's going to be the sum from Nal 1 to 2 of the a subn so s sub 1 is just a sub 1 or n = 1 to 1 of a sub one but writing that's kind of ridiculous we would never really write that we would just write a sub one or sorry this would be a subn sorry about that um but this whole thing right here just equals a sub one so we would never write uh we would never write this that's silly but it we could write it if we wanted to and it would be correct um it's just kind of Overkill so S Sub 1 is a sub 1 s sub2 is a sub 1 plus a sub 2 which we can write more compactly like this and what do you think s sub3 would be so I'm kind of running out of room over here so let me extend a little bit so give some space over here so s sub3 you may guess is a sub 1 plus a sub2 plus a sub3 okay and that's we can write that more compactly as the sum from n = 1 to 3 of a subn okay so remember we call this a sequence of partial sums because each elements in this sequence is a partial sum of the series okay so the series here is really the sum from nals 1 all the way up to positive Infinity so we're adding up all of these infinitely many values here but if we want to take a partial sum we would just add up and stop at some U one of the finite values here so we would just add up a finite number of them so S Sub one is just a sub One S sub2 is a sub 1 plus a sub 2 uh s sub3 is a sub 1 plus a sub 2 plus a sub3 and so on and so forth so what we could say in general and I'm totally out of room so I'm going to come over here and get rid of this so let's write down a general formula for that so what we saw was um S Sub 2 equals the sum from Nal 1 to 2 of a subn and we saw that s sub3 equals the sum from Nal 1 to 3 of a subn right so what do you think s sub4 would be well s sub4 would be the sum from n equal 1 to 4 of a subn which would be a sub 1 plus a sub 2 plus a sub3 plus a sub 4 okay and this pattern is going to continue so we could have five here five and then plus a sub five and so on and then this is getting really hard to read so let's uh stop with that so if we get rid of all this here um now in general what Would S subk be so this is the kith partial sum okay what we're going to do right here is give a for or just a write down a little formula for the K partial sum well for the K partial sum we would have the sum from n = 1 to K of a subn okay so this is always going to be a subn here because this Index right here has to correspond with this Index right here okay now you might be wondering um well first of all let's analyze this just real quick so s subk is basically the sum of all the first K terms okay so S Sub remember Sub sub 2 is A1 plus A2 so if we have S Sub 2 then our K is going to be two so this will be two this will be two up here okay which is what we have uh here here's two and two over here okay so um if we were to write this out what would that be so let's switch colors here just to make this a little bit easier to read so this would be a sub 1 plus a sub 2 plus a sub 3 plus dot dot dot plus a subk so this s subk represents the k partial sum which is the sum of the first K elements in the series okay so S Sub 2 is the sum of the first two elements uh in the series um and then s or sorry the first two terms I should say s sub2 is the sum of the first two terms in the series A1 and A2 are the first two terms uh S Sub one is the sum of the first one term which is kind of a silly thing to say but it's still true okay just A1 by itself uh s sub3 is the sum of the first three terms A1 Plus A2 2 plus A3 and S subk is the sum of the first K terms A1 Plus A2 plus A3 plus dot dot dot plus a k okay so um now just real quick you might be wondering why don't we say s subn well we can't say s subn because n is already being used here okay we can't use that index to represent something else this index n and this index K represent two different things okay so we can't use them we can't use the same variable for both of these CU these are two separate things here so we can't present them with the same thing so just uh be careful of that so anyway uh that's all about sequences of partial sums and just to recap real quick we have the concept of a sequence that we talked about in the previous video and remember a sequence is just a list okay and we also have the concept of a series and remember a series is a sum of the elements in the sequence okay so a series is a sum and the series has two sequences associated with it there's the sequence that actually tells you what the terms are and then there's the sequence of partial sums which is what we talked about in this video here okay so that about wraps it up for sequences a series and what their differences are uh what a sequence is what a series is and what a sequence of partial sums is |
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Tri-Triangles
In the problem Tri-Triangles, students use algebraic thinking to solve problems involving patterns, sequences, generalizations, and linear and non-linear functions. The mathematical topics that underlie this problem are finding and extending patterns, creating generalizations, finding functions, developing inverse processes, exploring non-linear functions, and justifying solutions. In each level, students must make sense of the problem and persevere in solving it (MP.1). Each problem is divided into five levels of difficulty, Level A through Level E, to allow access and scaffolding for students into different aspects of the problem and to stretch students to go deeper into mathematical complexity.
PRE-K
In this task, students are shown a triangle pattern and use toothpicks to extend the pattern and identify the number of toothpicks that would be needed in subsequent patterns.
LEVEL A
In this level, students view a pattern of triangles composed of toothpicks. Their task is to determine the number of toothpicks that make up each pattern.
In this level, students use multiplication and division within 100 to solve word problems in situations involving equal groups (3.OA.A.3).
LEVEL B
In this level, students examine a linear pattern that involves a constant. The task involves a set of triangular tables that are arranged adjacently in a row. The task asks students to determine the relationship between the number of tables and the number of people who can sit around the tables. Students need to extend the pattern. They also find the inverse relationship, i.e., find the pattern number when given the total number of people seated.
In this level, students are given a shape pattern and asked to extend the pattern and informally describe the rule of the pattern (4.OA.C.5).
LEVEL C
In this level, students determine how a pattern grows. Students need to see that the pattern grows by square numbers. They identify the relationship and then explain a valid process for finding these values.
In this level, students determine how a pattern of triangles grows. Students use their understanding of expressions and exponents (6.EE.A.1) to identify the relationship between the pattern and number of triangles and then explain a valid process for finding these values (6.EE.B.6, 6.EE.C.9).
LEVEL D
In this level, students are asked to generalize a rule for finding a value in the triangular number sequence. They are also asked to explain the process for finding an inverse value for the triangular number sequence by finding the term when given the total.
Students explore a pattern of triangular numbers and write an explicit expression describing the pattern (F-BF.A.1a). Students use the expression to write a quadratic equation to find the pattern number given the triangular number (A-CED.A.1). The quadratic equation can be solved with reasoning, factoring, or the quadratic formula (A-REI.B.4b). Students explain their reasoning (SMP.3) and use visual models to explore the pattern (SMP.5).
LEVEL E
In this level, students generate a closed expression for a sequence that grows as the sum of two exponential functions. In addition, the students must justify their findings. This level builds on the work students do in level D.
In this level, students build on their experience with triangular numbers from level D to generate an explicit expression to describe a pattern (F-BF.A.1a). As they build the explicit expression, they interpret parts of the expression independently and then combine the parts to arrive at the final expression (A-SSE.A.2). The rule that is developed may use exponential or polynomial functions. This level requires students to look for and make use of structure in the expressions they build (SMP.7).
PROBLEM OF THE MONTH
Download the complete packet of Tri-Triangles Levels A-E here.
You can learn more about how to implement these problems in a school-wide Problem of the Month initiative in “Jumpstarting a Schoolwide Culture of Mathematical Thinking: Problems of the Month,” a practitioner’s guide. Download the guide as iBook with embedded videos or Download as PDF without embedded videos.
SOLUTIONS
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7966 | https://www.sciencedirect.com/science/article/pii/S2772416622000286 | Interaction of microplastics with metal(oid)s in aquatic environments: What is done so far? - ScienceDirect
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Abstract
Graphical abstract
Keywords
1. Introduction
2. Sorption of metal(oid)s on microplastics in aquatic environment
3. Release of metal(oid)s additives from plastics
4. Studies focused on the metal(oid)s-microplastics interaction in aquatic environments
5. Ecological repercussions from interaction between microplastics and metal(oid)s
6. Contribution and challenges of laboratory studies to ecological risk assessment
7. Final remarks
Declaration of Competing Interest
Acknowledgments
References
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Journal of Hazardous Materials Advances
Volume 6, May 2022, 100072
Interaction of microplastics with metal(oid)s in aquatic environments: What is done so far?
Author links open overlay panel Joana Patrício Rodrigues a, Armando C.Duarte a, Juan Santos-Echeandía b
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Highlights
•Environmental exposure of MPs strongly facilitates metal(oid)s adsorption from water.
•Polymer surface changes create active zones enhancing binding capacity to metal(oid)s.
•PVC polymer with structural polar groups allows the strongest bond to metal(oid)s.
•Electrostatic interaction and complexation lead bonding of metal(oid)s on polymers.
•MPs alter bioavailability of hazardous elements to aquatic organisms.
Abstract
Microplastics (MPs) are being recognized as an emergent route of contaminants to aquatic environments, which initially attracted the research interest on their interactions with organic pollutants. Lately, a turning point of attention is evident, with more published studies reporting the presence of metal(oid)s in plastics. This review assembles the mechanisms occurring on microplastics surfaces that enhance sorption of hazardous elements (i.e., metals and metalloids) over environmental exposure. Reported findings of experimental studies are of major importance to understand the factors controlling the sorption/desorption of metal(oid)s to/from microplastics as much as determination of metal(oid)s in environmental plastics. Existence or formation of oxygen-containing functional groups and complexes from surface coatings strongly allow bond of metal(oid)s on reactive surfaces while sorption dynamics are strongly controlled by water chemistry parameters. Moreover, the present work evidences the potential impacts caused by metal(oid)s-MPs interactions to aquatic organisms, prioritizing the need of environmental realistic parameters to test. Bioaccumulation of metal(oid)s desorbed from ingested MPs prove the significant influence of these plastic particles in the bioavailability of pollutants to aquatic biota. In this way, this is a comprehensive manuscript committed to the estimation of the potential ecological risk of MPs to aquatic environments due to their association with metal(oid)s.
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Keywords
Polymer surface reactivity
Microplastic
Sorption
Ecotoxicology
Bioaccumulation
1. Introduction
Land- based plastic debris are considered a problem of marine pollution, composing about 80% of the marine litter (GESAMP,2015, GESAMP,2019; UNEP,2016; Plastics Europe,2017). In addition to land sources, over than 11,000 tons of ALDFG (Abandoned, Lost and Discarded Fishing Gear) and marine accidents that lose containers transporting plastic pellets are contributing to annual numbers of plastic waste in Europe (GESAMP,2019). Plastics are synthetic polymers produced from organic materials such as fossil fuels, or by bio-based feedstocks, which are widely applied in routine industries as food packaging or personal care products. According to their size, microplastics (MPs) are plastics ranging from 100 nm to 5 mm. They can be divided into primary (originally produced with this size for several purposes) or secondary microplastics which represent fragmented plastics arising from the effect of environmental conditions (e.g., photo-oxidation, wave action) that promote their degradation (Andrady,2017). Despite the non-establishment of an acknowledged quantity of MPs present at regional or global scale, due to the lack of data or studies giving reliable concentrations, coastal environments usually host a higher amount of plastic debris (estimated values of 1–100,000 particles per m 3) than open ocean waters (Norén and Naustvoll,2010; Wright et al., 2013). As distinctively ubiquitous and small, MPs can enter on biogeochemical cycles and food webs, spreading from surface layer of coastal waters to the sediment compartment of the open ocean (Rogers et al., 2020).
The role of metal(oid)s as harmful to human health is of global knowledge. Being potential toxic elements, metal(oid)s can cause carcinogenic, teratogenic and allergenic effects (Cooksey,2012; Yang et al., 2015), as well as to induce alterations in behavior and at neurological level (Tchounwou et al., 2014; Koller and Saleh,2018). Some metals as chromium (Cr), nickel (Ni), cadmium (Cd), cobalt (Co) and lead (Pb) and a very toxic metalloid arsenic (As) represent potential carcinogenic elements that are commonly found in drinking water (Fakhri et al., 2018). The persistence, bioaccumulation and high toxicity of this hazardous substances make them a serious threat to environments. These elements can be found in aquatic environments as a result of leaching from metal-based materials, farming or mining, introduced through wastewater effluents (Förstner,1980; Deheyn and Latz,2006; Almeida and Sousa,2007; Brennecke et al., 2016; Koller and Saleh,2018; Barletta et al., 2019). On its ionic form, metal(oid)s can have a great affinity by dissolved organic matter (Santos-Echeandia et al., 2008b) strongly present in media, or different anions (e.g., carbonates, chlorides, hydroxides) (Millero,2001).
The increasing awareness about the presence of MPs in the aquatic environment has highlighted to warning consequences concerning ecosystem's health (Brennecke et al., 2016). Physical impact is inflicted on marine species of different trophic levels as a result of their direct contact with these small pieces in their natural environments (Wright et al., 2013). However, this is not the only implication caused by the spread of MPs in nature, since they contain or can interact with other pollutants (i.e., metals, metalloids), resulting in a chemical impact beyond the physical damage (Roman et al., 2020). The transfer of contaminants from MPs to the tissues of organisms has been already reported, potentially causing increased concentrations and toxicological effects on aquatic biota (GESAMP,2015; Luís et al., 2015; Barletta et al., 2019; Rivera-Hernández et al., 2019). Plastics are produced with the addition of organic and inorganic components, acting as plasticizers, stabilizers, surface modifiers, flame retardants, or pigments. Alternatively, these contaminants can also be later associated with plastics by their sorption from the surrounding water once in the environment (Town et al., 2018; Santos-Echeandía et al., 2020). The essential features that guide the processes of sorption and desorption occurring between a microplastic particle and a pollutant are mediated by their chemical and physical characteristics, with abiotic parameters having a key-role in this process (Rodrigues et al., 2019). To understand this phenomenon, studies frequently focus on the comparison of the amount of elements accumulated on MPs represented by virgin plastic particles and the ones collected from the environment, often resorting to pellets (Ashton et al., 2010; Holmes et al., 2014; Rochman et al., 2014; Fisner et al., 2017; Vedolin et al., 2017; Munier and Bendell,2018; Gao et al., 2019) or small fragments (Brennecke et al., 2016; Santos-Echeandía et al., 2020; Wang et al., 2020). Accordingly, metal(oid)s can be transported to and between aquatic environments through their interaction with MPs (Akhbarizadeh et al., 2016; Wang et al., 2016; Vedolin et al., 2017). As representants of potential hazardous substances, the reduction of metal concentrations in the environment is the aim of environmental protection agencies and European Directives such as the WFD (2000/60/CE) and the MSFD (2008/56/CE). Wherefore, the emergent route of metal(oid)s to aquatic environments through MPs as vehicles deserves global attention. Nonetheless, to date, organic compounds have been greater explored regarding the impact of their interaction with MPs to environmental contamination, probably due to the hydrophobic trait in common (Antunes et al., 2013; Ferretto et al., 2014; Pittura et al., 2018; Endo and Koelmans,2016; Rodrigues et al., 2019). A turning point has been observed in the last years, as more studies investigating interactions with metal(oid)s have been published (Akhbarizadeh et al., 2016; Brennecke et al., 2016; Wang et al., 2016; Vedolin et al., 2017; Dobaradaran et al., 2018; Lu et al., 2018; Maršić-Lučić et al., 2018; Munier and Bendell,2018; Town et al., 2018; Acosta-coley et al., 2019; Gao et al., 2019; Mohsen et al., 2019; Richard et al., 2019). However, overall mechanisms and factors involved in the metal(oid)s-MPs interactions have not been well enlightened to date.
The present manuscript is a critical evaluation of the current state of investigation concerning the interaction of microplastics with metal(oid)s in the aquatic environment. Despite a few reviews have been published regarding the interactions of MPs with metals, this the first one including both metals and metalloids, specifically. Available published research, essentially from the last 6 years, was reunited and well detailed here, focused on the mechanisms and circumstances allowing sorption/desorption of these metallic pollutants on/from polymers with emphasis on the modifications occurring in the MPs surface, particularly over environmental exposure. More than that, we enlighten all the possible scenarios reported with organisms to estimate ecological impacts of metal(oid)s-MPs interactions, from bioaccumulation to toxicity effects as a result of their coexposure or estimation through physiologically simulated extraction tests. In this way, this review will provide an insight on the impact of MPs to marine pollution and potential ecological risk due to their interaction with metal(oid)s.
2. Sorption of metal(oid)s on microplastics in aquatic environment
2.1. Surface modification promoting adsorption
Studies carried out since the 1960s demonstrated the tendency of metals to adsorb on plastic surfaces (Robertson,1968; Struempler,1973; Good and Schroder,1984; Grossmann et al., 1990; Giusti et al., 1994; Weijuan et al., 2001; Cobelo-Garcia et al., 2007; Fischer et al., 2007). Even so, the research interest in order to understand their association has been recently globally triggered. The early lack of investigation on metal(oid)s, regarding the interactions of polymers with contaminants could be explained by the scepticism to look on the possibility of metal(oid)s adsorption onto plastic from water. Overall, plastic particles were considered biochemically inert in relation to aqueous metal ions (European Commission,2019), due to neutral charge of polymers. However, this prior believe was demystified, since their interaction can be feasible (Turner and Holmes,2015; Rivera-Hernández et al., 2019; Dong et al., 2020; Fernández et al., 2020; Wang et al., 2021) when the particles are suspended in water, allowing them to acquire charge. The MPs-metal(oid)s binding can even be enhanced as plastic properties are modified, mainly driven by environmental processes (weathering) (Santos-Echeandía et al., 2020), or artificially (oxidized MPs), conferring a new more reactive surface to polymers (Wang et al., 2020). Environmental action promotes surface degradation through mechanical, chemical, and biological activities which are fundamentally responsible for structural changes on the polymer (Fig.1). First, mechanical action (sand abrasion and washing over wave transportation) induce fragmentation of plastic debris introducing surface modification. The decrease of size leads to increase of specific surface area, and more cavities are created, enhancing polymer porosity and conferring a higher roughness ideal to uptake of metal(oid)s (van Leeuwen et al., 2013; Brennecke et al., 2016; Town et al., 2018). Apart from this, the formation of coatings of hydrogenous and biogenic nature (Brennecke et al., 2016; Richard et al., 2019) on the surface of polymers allow adhesion of organic matter (Turner and Holmes,2015); agglomerates of mineral nature, as silt and clay suspended particles (Johansen et al., 2018); and microbial colonization forming biofilms (Tien and Chen,2013). The formation of superficial complexes would act as an enhancer to allow interaction with hydrous oxides (which interfere on cation-exchange) and metal/metalloid ions. Such hydrogenous-biogenic coatings occur through formation of metallic oxides on the surface of polymers which promote co-precipitation of metal(oid)s with iron (Fe) and manganese (Mn) oxo-hydroxides (Leiser et al., 2020). Additionally, coatings (i.e., biofilms) also have the capacity to sink MPs, finally reaching the sediments, preventing low density polymers from being restricted to the water surface and dragging them to bottom waters. This trip through the water column allows a greater interaction of the polymers with the metal(oid)s present in the water and on the sedimentary matrix. Alternatively, biofilm growing and organic blooms over buoyant polymers may block the UV oxidation reactions (Weinstein et al., 2016; Oberbeckmann et al., 2018; Acosta-coley et al., 2019), also influencing sorption performances. Chemical degradation (mainly hydrolysis) induces modification with creation of new functional groups on the surface of polymers. This occur through breakdown of structural linkages, setting up instable groups able to react with aqueous metal(oid)s (Yu et al., 2019). This polar functional groups usually consist in esters, keto groups and quinone-related structures (Yu et al., 2019; Santos-Echeandía et al., 2020). Besides this, oxygen-based functional groups (OH, CO and COH) created by UV degradation are also essential to metal(oid)ds-polymer bond (Liu et al., 2019; Yu et al., 2019). Biological degradation, as a result of microbial communities fouling over MPs can also modify functional groups. In fact, polarity and electronegativity of polymers are determined by the diversity of functional groups whom create anionic binding sites suitable for the uptake of cations from water (Holmes et al., 2012, 2014; Santos-Echeandía et al., 2020). Consequently, as active sites environmentally created can be limited, metal/metalloid species present at the surrounding environment can compete for binding sites (Yang et al., 2019). In the end, once a metal/metalloid is bound to a functional group, they can increase their stability in the particle surface (Kochanke et al., 2019).
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Fig. 1. Summary of surface changes of polymers as a result from weathering that allow adsorption of metal(oid)s on microplastics.
As each polymer type has its own structure, this feature is also important to define adsorption capacities towards metal(oid)s. Despite polymers are essentially apolar, Polyvinyl chloride (PVC) and Chlorinated polyethylene (CPE) have chlorine on their structure which corresponds to one of the most electronegative atoms, so chlorine based functional groups provide polar regions to form a strong bond with metal(oid)s, before or after environmental exposure (Zou et al., 2020). For example, under influence of electrostatic attraction (i.e., high water pH), new polystyrene (PS) MPs exhibit negative charge which promotes the bond of positively charged ions, even before surface alterations through weathering (Mao et al., 2020). Even more, when a polymer is surface modified during the manufacturing process (i.e., oxidized MPs) its sorption capacity can be strengthened, even before suffering from environmental action (Wang et al., 2021). This is allowed by the changes in the superficial functional groups, enhancing the bond with different chemical species.
MPs color appears to be also important since different colors have been associated with different species of metal(oid)s, according to degradation-color relation of environmental polymers (Filella and Turner,2018; Acosta-Coley et al., 2019). Wang et al.(2016) reported high concentration of metals in dark colors of plastic (bags and screw caps), over hundred times higher than in lighter ones. As matter of fact, darker pellets collected from environment have been associated to MPs that have suffered significant weathering (Fisner et al., 2017; Acosta-Coley et al., 2019; Santos-Echeandía et al., 2020). However, the association of colored polymers with certain metal(oid)s could be due to their use as additives, demonstrating a pigmentation-color relation as well, later discussed on this paper (Section 3).
In brief, polymers have determined properties which can be modified through biotic and abiotic degradation (e.g., functional groups). The match of their unique traits will determine the level of interaction with metal(oid)s and the strength of bindings that will occur in aquatic environments (Min et al., 2020).
2.2. Dynamics of adsorption/desorption influenced by environmental parameters
To investigate interactions of MPs with metal and metalloid ions, studies perform analysis of sorption kinetics, with estimation of sorption efficiencies and mechanisms involved in the process. Primarily, an efficient increase of adsorption of metal(oid)s onto surfaces is frequently observed until equilibrium time is normally reached over 24 (Lin et al., 2021; Wang et al., 2021) or 48 h (Mao et al., 2020; Xue et al., 2021) up to 96 h (Wang et al., 2020) on an aqueous solution (Turner and Holmes,2015; Fernández et al., 2020; Mao et al., 2020; Tang et al., 2020; Zou et al., 2020). This behavior follows stabilization, which depends on the conditions involved. Regarding desorption, the diffusion of metal(oid)s from polymers towards water can stabilize within 10 h (Tang et al., 2020). Metal ions diffuse into or out of a polymer saturated with water, being partially hydrated, and their rate of release is equal to their diffusion rate in the aqueous media (Reuvers et al., 2015; Town et al., 2018). Town et al., 2018 described a simple model, based on spherical geometry, to explain the description of release kinetics of metal species from nano- and micro-plastic particles, highlighting that is not comparable the velocity and effusion of metal ions in water to metals within polymers. For micro(nano)plastics and metals, the researchers acknowledged the size of the particles and the timescale of the exposure as the main responsible parameters for the dynamics of their relation (Town et al., 2018). Despite equilibrium time in the laboratory studies is normally relatively low (48 to 100 h) (Holmes et al., 2012; 2015), for field studies this scenario can change. In the study of Rochman et al.(2014) it took more than 12 months to achieve equilibrium between metals and microplastic pellets. It is expectable that equilibrium be faster in the controlled conditions of a laboratory experiment comparing to an environment much more dynamic and even unpredictable. Essentially, water chemistry parameters and other conditions from surroundings will dictate the dynamics of metal(oid)s-MPs interactions.
In an aquatic environment, pH and ionic strength significantly influence the process of adsorption/desorption (Tang et al., 2020; Zou et al., 2020), as these parameters determine the reactivity and charge state of polymer surfaces. Overall, the increase of pH solution lead to higher adsorption efficiencies, whereas decrease of pH enhances substantial desorption (Kalčíková et al., 2020; Tang et al., 2020; Lin et al., 2021). At a low water pH, more H+ ions surround the surface of a polymer, what consequently promotes repulsion of metal cations, resulting in the decrease of adsorption interactions and likely higher desorption from surfaces (Tang et al., 2020). In turn, due to low solubility of metal(oid)s at a basic pH medium (>7), the precipitation of metal(oid)s onto microplastic surfaces is enhanced (Kalčíková et al., 2020; Lin et al., 2021). However, this scenario can be different if anions are involved, as for As ions. Dong et al.(2020) observed decrease of As adsorption onto MPs with increase of pH, reversely to what was acknowledged for cations. Acidic pH promoted adsorption here since observed low concentration of OH− avoid competition for binding sites with the metalloid anions (Meng et al., 2002; Dong et al., 2020). Furthermore, some polymers (i.e., PS) have a tendency to become negatively charged as the pH increase, resulting in inhibition of the bond with anions through electrostatic repulsion on the surface (Dong et al., 2020), and affinity for cations is enhanced. As matter of fact, positively charge surface attract anions and repulse cations, and vice versa in the case of negatively charged ones. Since water pH establish metal and metalloid ion species, and as bonds with polymer surfaces involve charged sites, the behavior of metal(oid)s towards MPs can be in constant change concerning sorption, over a dynamic pH surroundings. This also supports the main role of electrostatic interaction in the process. Nonetheless, adsorption performance in seawater can fluctuate, as, for the majority of metals, adsorption usually increases under a basic pH (characteristic of seawater) but decreases with high salt content (specific of seawater). In line with this, different behaviors amongst the metal/metalloid ionic forms subject to the chemical properties of the water can be observed. Turner and Holmes(2015) found adsorption of Cd, Co, Ni and Pb to be efficient in freshwater (and high pH), conversely to lower adsorption of Cr, which was found to be higher with respect to seawater (Holmes et al., 2012). Differences concerning water environments suggested that a higher ionic strength promotes complexation and ion pair with seawater anions and also competition with the divalent seawater cations, decreasing the boding of metal cations (i.e., Cd 2+, Co 2+, Ni 2+, Pb 2+, Ag 2) and available binding sites on MPs surfaces (Holmes et al., 2012; Turner and Holmes,2015). The previous example highlights the complexity of the process which needs to consider all intervening variables. Increase of ionic strength entail high levels of sodium (Na+) or potassium (K+) ions what can suppress adsorption of metal(oid)s on surfaces since the presence of more ions encourage competition for adsorption sites with other metal(oid) cations (Holmes et al., 2014; Johansen et al., 2018; Tang et al., 2020). Organic matter also has a role on sorption dynamics, since high concentrations (e.g., fluvic acid) behave as competitor to microplastic particles regarding the uptake of metal(oid)s (Tang et al., 2020). In this way, the higher presence of organic ligands in freshwaters (Santos-Echeandia et al., 2008a) will reduce metal adsorption to MPs that could increase when reaching seawater with a lower ligand content (Santos-Echeandia et al., 2008b). High temperature can also break hydrogen bonds between surface and metal(oid)s, decreasing adsorption (Dong et al., 2020). Additionally, biological activity (e.g., bacterial, worms) potentially foster metal(oid)s removal, as metal and metalloid ions are weakly linked within a biofilm on polymers (Richard et al., 2019). Accordingly, once on environment, MPs can either acquire and/or release associated hazardous substances (i.e., metals and metalloids) enhanced by natural factors and depending on local environmental conditions.
3. Release of metal(oid)s additives from plastics
The presence of metal(oid)s on MPs can also be due to their previous and intended addition. Virgin or recently manufactured plastics already contain chemicals themselves used as catalysts, fillers and plasticizers, applied to keep their properties and resistance (GESAMP,2015; Castro et al., 2018). Those inherent metal(oid)s can be found in the matrix of polymers and are usually linked by physical and not chemical forces (Turner et al., 2020). Metal(oid)s used as additives in plastics can contribute essentially as fillers, pigments and stabilizers (heat and flame retardants) or they can confer antimicrobial properties (Turner and Filella,2021). In addition, As is a known toxic metalloid used as a biocide in plastic production. Cadmium, zinc (Zn) and lead are examples of metals commonly used as stabilizers to reinforce the durability of polymers (about 3% composition), and also applied in color pigments of plastics along with Cr and mercury (Hg) (Imhof et al., 2016; Hahladakis et al., 2018; Munier and Bendell,2018; Town et al., 2018). Highly toxic Hg and Pb are strongly related to production of chlorine groups of PVC polymer (Salvaggio et al., 2019). The exposure to metal(oid)s additives from polymers in the environment is real, as they can be decoupled from the polymer to surrounding water following assimilation on external tissues of organisms (Oliviero et al., 2019) or dissociated inside their bodies to accumulate on internal organs. Turner et al.(2020) suggested the release of desorbed metal(oid)s from the surface of polymers occurring through desorption process. In turn, the element additives are released from the polymer matrix to surrounding water by diffusion (Turner et al., 2020). Regardless, both processes take place in water environments. Even though, in line with predictions of Nakashima et al.(2016), and despite the researchers have quantified the leaching of Pb from marine plastic debris, the hazardous additives of a plastic material are not released in relevant amounts to environment. From the 5100 µg/g of maximum Pb quantified in marine plastic floats, just 0.1% was predicted to be lost to water (Nakashima et al., 2016). In the mentioned study was underlined that is more likely that additive-Pb present at high concentrations in plastic debris is stocked and transported to long-distances than released (leached) from polymers to water during their migration on environment. Accordingly, diffusion of hazardous elements to water occurs at low rates, being potentially no relevant to soon threat ecosystem's health. A reasonable justification for this finding is the fact that the additives were added to the polymer bulk to be an inherent component of the particle itself, being more stable and unlikely to dissociate (Town et al., 2018) unless high degradation of the polymer occurs (e.g., plastic particles washed ashore suffer from more physical degradation over deposition on beaches) (Nakashima et al., 2016). Furthermore, Turner et al.(2020) found high levels of Pb as additive in beached microplastic (PVC, PE, PP) as well, comparing to its presence as adsorbed Pb (PE beached MPs).
Potential transfer of hazardous metals from leaching of plastic toys has been spotlighted as a concern issue to health ware. To minimize the damages associated to the hazardous role of heavy metals, their application in plastic manufacturing became restricted in Europe (Hahladakis et al., 2018). Cd and Pb have been quantified in mostly PVC-based former plastic toys (Miller and Harris,2015) at levels exceeding the actual EU permissible limits for their production (17 and 23 µg/g, respectively) (Turner,2018b). This scenario could be worst if plastics are ingested following desorption and accumulation of toxicants in the human body, as ecotoxicological effects have been observed in aquatic organisms (Oliviero et al., 2019) as a result of exposure to plastic toys. Complementarily, no detectable levels for heavy metals were detected in most-recently produced plastic toys (Miller and Harris,2015), in line with their exclusion in production of these materials. Thus, the high load of heavy metals (e.g., Hg and Pb) used as additives and that have been restricted or banned, is rather found in “old” sampled plastics and reflect the age and residence time of the plastic debris (Filella and Turner,2018). Apart from the metal(oid)s additives used to strengthen raw plastic material, it is possible to find other sources of their relationship. The industry of electric and electronic equipment (EEE) is a great contributor to plastic pollution, since a large proportion of their resultant waste (WEEE) is composed by plastic, amongst 300 different types (Wäger et al., 2012; Peisino et al., 2019). Although plastic from WEEE is highly potential renewable, it is also composed by metal(oid)s residues (Wäger et al., 2012; Peisino et al., 2019) or reused to make black plastic (Turner,2018a; Shaw and Turner,2019), which often end up in aquatic systems. Therefore, the use of Pb, Hg, Cr and Cd as additives with concentrations higher than 1000 mg/L cannot also be applied in the plastic production and recycling of EEE, according to the restriction of hazardous substances (RoHS) directive (Hahladakis et al., 2018). However, until recently, heavy metals were detected in plastic environmental samples exhibiting values above the RoHS directive (Turner,2018c). Even in recycled plastics as polyethylene terephthalate (PET) bottles, the leaching of hazardous elements was found (Cheng et al., 2010). Despite renewed plastics can have metal(oid)s as a result from reaction of recycled residues, they can also adsorb metal(oid)s once on environment. At last, the transferring of metal(oid)s additives from MPs is causing worldwide preoccupation since they are strongly present in our life with potential consequences.
High color diversity attributed to MPs is frequently associated with different species of metal(oid)s pigments or with different level of degradation of polymers, as mentioned before. Cu can be found in blue and green-colored plastics, while Pb and Cd are also associated to green pigmentation of plastics (Carbery et al., 2020). Reddish-brown plastics are often attributed to Hg and Cd addition. Black and grey colored plastic debris are normally deriving from WEEE, whom contain an expected high metal content (Haarman and Gasser,2016; Shaw and Turner,2019). Nonetheless, dark and yellow-colored MPs (e.g., pellets) are also associated to have a high degradation degree promoted by strong weathering, thus with strong capacity to concentrate metal(oid)s from environment. This could explain why colored MPs sampled from environment usually have high metal(oid)s content, as they have higher sorptive capacity and/or high inherent additive content, to further release to environment. In this way, colored MPs, especially the darker ones, can represent a higher risk considering their role as vectors of higher amounts of hazardous substances to aquatic environment. To clarify this and other variables and mechanisms intervening in this process, it is of great relevance to study the polymer-potentially toxic element interactions in controlled laboratory conditions, upon determination of leachable metal(oid)s in plastics with environmental past.
4. Studies focused on the metal(oid)s-microplastics interaction in aquatic environments
4.1. Determination of elemental content in environmental plastics
In the last six years, an increasing number of studies have been dedicated to investigate the interactions between plastic and metal(oid)s in the aquatic environment (Turner,2016, 2018c; Turner and Lau,2016; Turner and Solman,2016; Massos and Turner,2017; Filella and Turner,2018; Carbery et al., 2020; Turner et al., 2020). Among these studies, large part of them were carried out with plastic material samples collected in the field (Fig.2) in order to evaluate the environmental behaviors and real contamination concerning MPs and metal(oid)s interactions, while some laboratory experiments explained and reinforce those observations. The regions near Plymouth (UK), mainly beaches as representants of marine environments with high tendency to accumulate plastics, both of land and sea origin, have been commonly chosen as study area in the several reunited studies. In this nearby port region, high amounts of heavy metals (i.e., Pb) were quantified in plastics with total concentrations exceeding the 10,000 μg/g on PE (Turner and Solman,2016; Massos and Turner,2017; Turner et al., 2020), polyurethane (PUR) (Turner,2016; Turner and Lau,2016), PP and PVC (Turner,2016; Turner and Solman,2016; Turner et al., 2020) and lower levels (> 1000 µg/g) also identified on PE and PP samples (Massos and Turner,2017). Cd was found with maximum concentrations up to 1000 µg/g (Turner and Solman,2016) on PVC (Filella and Turner,2018; Turner,2018c) and Pb even exceeding the 16,000 µg/g on PU (Turner and Lau,2016; Turner,2016), plus substantial levels of other heavy metals as cooper (Cu) (> 1000 µg/g) and Cr (> 1000 µg/g) (Turner,2016). In addition, Zn (> 1000 µg/g on PUR), iron (Fe) (> 1000 µg/g on PE and PUR) (Turner and Lau,2016), (titanium (Ti) and barium (Ba) (> 10,000 µg/g) (Turner and Solman,2016) are examples of reported metals in local plastic debris. Metalloids As and antimony (Sb) were also found in environmental plastics but in different orders of magnitude, as 21 µg/g and up to 10,000 µg/g maximum concentrations, respectively (Turner,2016). In the study of Turner(2016) the presence of Pb was highly associated to prior introduction of lead chromate as additive into plastics, since Cr was tightly quantified together with Pb on the same beach samples. Besides Pb identification in all the beached plastics collected, this potentially toxic element was also estimated as highly bioavailable to local marine birds, according to Turner and Lau(2016) work. Carbery et al.(2020) quantified higher content of metal(oid)s (Mn, Cr, Cu, As, Zn and Pb) in MPs from coastal areas related with industrial regions (e.g., Pb > 8 mg/kg; As > 0.6 mg/kg) comparing to MPs from other associated uses (urban, residential or rural) (e.g., Pb > 1 mg/kg; As > 0.1 mg/kg), evidencing the capacity of MPs to storage (potentially adsorbed) metal(doi)s and transport them over long-distance ranges. Moreover, a color dependency was suggested, as the Cu levels found were likely due to the highest number of blue-colored plastic items sampled there. In this way, as Cu is typically used as metal pigment to confer blue color to polymers, maybe its presence on MPs was more due to its introduction as additives than adsorption from environment. The elemental content found in plastics pieces can reveal the residence time of plastics in environment or demonstrate long-range distances travelled. For instance, all three of the most hazardous elements as Pb (>10,000 μg/g), Cd (>1000 μg/g) and Hg (> 100 μg/g) were found on PVC plastics with levels nowadays restricted, as mentioned in the previous section (Filella and Turner,2018). Also reported before, was the high amounts of the potential highly toxic Hg being strongly associated to aged darker plastics (blank and reddish-brown) (Acosta-Coley et al., 2019; Santos-Echeandía et al., 2020). Santos-Echeandía et al.(2020) reported initial evidence of the transfer of Hg to biogeochemical cycles through plastics. They quantified Hg (1 to 1600 μg/kg) in all the virgin and sampled beached plastics (mostly PUR and PVC, but also on PE and PP), with higher levels at coastline comparing to upper regions in the beach. Accordingly, additive and adsorbed Hg on MPs flows in ocean, transported until deposition on beaches, where volatilization to atmosphere occur through UV radiation. From this point of view, plastics could remove hazardous elements from water and concentrate them, thus reducing the concentration of dissolved metals on aquatic environments (Santos-Echeandía et al., 2020).
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Fig. 2. Field studies all over the world focusing on the determination of metals in plastics from aquatic environments, considering late years (2015–2021). The numbers on the map represent the study and the site where it was performed. (1) Guadeloupe, France (Catrouillet et al., 2021); (2) Guadeloupe, France (El Hadri et al., 2020) (3) Australia (Carbery et al., 2020); (4) Murcia, Spain (Santos-Echeandía et al., 2020); (5) Plymouth, UK (Turner et al., 2020); (6) Cartagena, Colombia (Acosta-coley et al., 2019); (7) Bandar Abbas, Iran (Yazdani Foshtomi et al., 2019); (8) China (Mohsen et al., 2019); (9) Burrard Inlet, Vancouver, Canada (Munier and Bendell,2018); (10) Croatia (Maršić-Lučić et al., 2018); (11) Lake Geneva, Switzerland (Filella and Turner,2018); (12) Persian Gulf (Dobaradaran et al., 2018); (13) Whitsand beach, Plymouth (Massos and Turner,2017); (14) San Paulo, Brazil (Vedolin et al., 2017); (15) Whitsand Bay, Plymouth, UK (Turner and Lau,2016); (16) Plymouth, UK (Turner,2016); (17) Plymouth, UK (Turner and Solman,2016); (18) Beijiang River, China (Wang et al., 2016); (19) Persian Gulf, Iran (Akhbarizadeh et al., 2016); (20) Lake Garda, Italy (Imhof et al., 2016). Moreover, the colors of the bubbles classify the studies by year range, according to the published year.
In MPs-metal(oid)s studies, identifying the source of interactions of these pollutants is frequently a struggle that some researchers aim to clear up. Turner et al.(2020) have quantified Pb in marine MPs, which was considerably higher (maximum concentration) as an additive (40,000 µg/g on PVC, PE, PP) than the amounts found to be adsorbed from water (0.1 µg/g on PE). Furthermore, percentage levels of Pb estimated to be released through desorption (i.e., adsorbed Pb) from MPs to a simulated gastric avian solution had a higher proportion (70%) than the levels released from diffusion (i.e., additive Pb) (16%) considering initial concentrations of Pb identified as adsorbed or additive, respectively. However, with respect to absolute concentration of Pb released, the levels were higher (two orders of magnitude) through release of additives. Additives exhibit physical interactions within a polymeric matrix, whereas adsorbed metals link to surface of polymers on active sites forming a chemical bond (Turner et al., 2020). For this reason, reactivity of adsorbed Pb on MPs was higher. Considering that PVC is highly associated with Pb present as additive, perhaps if only the other polymers were used to comparison, as PE which was found not to be an efficient adsorbent of metals (Yu et al., 2019), the disparity between the levels of Pb adsorbed and quantified as additives would not be as significant as was verified in the Turner et al.(2020) work. Another two recent studies were engaged in to develop methods for determination of elements as additives or adsorbed on beached MPs (pellets and fragments) (El Hadri et al., 2020; Catrouillet et al., 2021). El Hadri et al.(2020) identified Cd (< 8 mg/kg) concentrated on PE, and Pb (> 1000 mg/kg), antimony (Sb) (< 25 mg/kg) and Zn (20 mg/kg) on PP as additives. Regardless, Pb and Sb were also found to be adsorbed onto PE, based on generated profiles (qualitative and quantitative criteria) concerning their distribution on MPs subsurface. On the other hand, Catrouillet et al.(2021) applied different release processes to quantify additives (i.e., acid leaching) and bioavailable (i.e., acid total digestion) metal(oid)s from polymers. They found lower metal(oid)s release through leaching (additives) than through digestion (potentially adsorbed elements from environment). However, the authors retained that they method did not estimate additive-sources for leached elements (environmental vs industrial). In this way, more similar comparative studies should be performed resorting to other polymers and metal(oi)s in order to obtain conclusions about differences in the leaching of metal(oid)s added during the manufacturing process or adsorbed once in the environment.
A few researchers compared their results from quantification of hazardous elements in environmental plastics with data of metal contamination in local water (Maršić-Lučić et al., 2018) or sediment (Dobaradaran et al., 2018; Mohsen et al., 2019). Maršić-Lučić et al.(2018) found higher (two orders of magnitude) mean concentration of metals quantified as Cd (2.9 µg/g), Cu (0.21 µg/g) and Ni (0.14 µg/g) in the pellets (unidentified polymers) collected from beach sediment, comparing with their dissolved levels reported on the local seawater (8.5; 338 and 423 ng/L, respectively), and even three orders of magnitude higher on pellets for Pb (0.26 µg/g) and Zn (2.08 µg/g) comparing to their dissolved levels (182 and 722 ng/L, respectively). This results support occurrence of adsorption of metals from the aqueous media to remain on polymers. Mohsen et al.(2019) determined metal(oid)s in MPs accumulated on sediment, and in the sediment itself, from a marine environment in a sea cucumber farm. Among the elements identified, they found higher mean concentration of Cd, Pb, and Zn (0.058–0.99; 2.56–40.8; 16.44–1190 mg/kg, respectively) on the MPs than on the sediment. On the other hand, Dobaradaran et al.(2018) found no significant differences in metal content (i.e., Al, Mn, Cd, Cr, Ni, Pb, Cu) between the two solid phases sampled, except for Fe, which was higher in the sediment (3050 µg/g) than in the MPs (531 µg/g) mean concentration.
As previously indicated, MPs are transported by wind, waves and marine currents and are finally accumulated on the beaches. Once there, the adsorbed metal(oid)s can escape from MPs particles due to the high temperatures reached in the sand of beaches, with respect to the mean temperature of seawater, by means of volatilization. This fact has been already verified for Pb in PE and PE-LD bags (Alam et al., 2018) or Hg volatilized from several types of polymers found on the beach (Santos-Echeandía et al., 2020). Furthermore, chlorine was found to induce volatilization of metals in PVC waste with temperature rise (Osada et al., 2009), namely Pb, Ni and Cu (Yu et al., 2017). This mechanism will act as a source of metal(oid)s accumulated from seawater onto MPs back to the atmosphere and lithosphere. At last, something important to point out in some of the field studies is the lack of polymer identification, in part due to the different binding behavior of metal(oid)s and different polymers. As matter of fact, polymer characterization should be performed in parallel to metal(oid)s determination, since it is an important criteria to scientifically validate studies (Hermsen et al., 2018) and compare the results obtained in laboratory.
4.2. Complementary findings from laboratorial studies
Interaction between MPs and metal(oid)s is complex and has a high variety of variables involved (i.e., polymer type, degradation status, element, presence of organic matter, biogenic coatings, salinity, temperature). The control of this great number of variables in the field is difficult and limits the understanding of the weight that each variable has in the kinetics or potential binding between metal(oid)s and MPs. In order to learn more about this process that could allow to draw up reliable conclusions, several researchers have performed laboratory-controlled experiments testing most part of these variables (Table 1).
Table 1. Characterization of experimental studies investigating sorption mechanisms/capacity of MPs towards metal(oid)s and other hazardous elements with influencing variables.
| Polymer features | Element | Tested medium | Analytical method | Element initial conc. tested | Time of exposure | Env. factors tested | Ref. |
--- --- --- --- |
| Virgin and beached pellets (PE) | Cd Co Cr Cu Ni Pb | Estuarine conditions | Inductively Coupled Plasma Mass Spectrometry (ICP-MS) | 0–20 μg/L | 24 h | pH | Holmes et al.(2014) |
| Beached and virgin pellets (PE, PP) Virgin pellets (PE) | Ag Cd Co Cr Cu Hg Ni Pb Zn | Beach | ICP-MS | 5 μg/L | 168 h | pH | Turner and Holmes(2015) |
| Aged fragments (PVC) Virgin beads (PS) | Cu Zn | Seawater | Flame Atomic Absorption Spectrometry (FAAS) | – | 14 days | – | Brennecke et al.(2016) |
| Virgin and aged (PE-HD) | Cs Sr | Freshwater Estuary Seawater | X-ray Fluorescence Microscopy | 26–35 mg/L | 142 days | Biofilms | Johansen et al.(2018) |
| Virgin pellets (PE, PP, PA, PVC, POM) | Pb Cu Cd | Metallic solutions Seawater | ICP-MS | 5 mg/L | 11 days 9 months | – | Gao et al.(2019) |
| Virgin and aged fragments (PET) | Cu Zn | Aqueous solution | ICP-MS | 5 mg/L | 144–500 h | UV light | Wang et al.(2020) |
| PVC, PE-LD, PE-HD, CPE | Cu Cd Pb | Metallic solutions | FAAS | 0.1–50 mg/L | 24 h | pH ionic strength | Zou et al.(2020) |
| Virgin and aged | Pb | Aqueous solution | FAAS | 8 mg/L | 48 h | Fulvic acid pH Ionic strength | Tang et al.(2020) |
| Virgin pellets (PS) | As | As(III) solution | Atomic fluorescence spectroscopy | 10–50 mg/L | 32 h | pH Ionic strength | Dong et al.(2020) |
| Aged pellets and virgin films (LD-PE) | Pb | Tap water | ICP-MS | 15 μg/L | 48 h | UV light | Ahamed et al.(2020) |
| Virgin and aged (PS) | Pb Cu Cd Ni Zn | Pure water Seawater | FAAS | 50 mg/L | 48 h | UV light | Mao et al.(2020) |
| Virgin and aged (PS, PVC) | B | Boric acid solution (dissolved in water) | Inductively Coupled Plasma Emission Spectrometry (ICP-OES) | 1 g/L | 30 h | Humic acid Ionic strength pH | Wang et al.(2021) |
| Virgin and aged thermoplastic polyurethane (TPU) | Cu | Deionized Water | FAAS | 2.5–50 mg/L | 72 h | – | Xue et al.(2021) |
| Virgin (PE) and pellets (PP) | Cu Cd Pb Zn | Aqueous solution | Atomic Absorption Spectrometry (AAS) | 1 mg/L | 9 days | pH Salinity | Ahechti et al.(2020) |
| Virgin (PVC, PE, PS) | Pb | Distilled water | Dithizone method | 25 mg/L | 24 h | Temperature pH Ionic strength | Lin et al.(2021) |
For the characterization of the studies, were considered the following parameters: Polymers features, Element, Tested medium, Analytical method to quantification of elements (Analytical method), Elemental initial concentration tested (Element initial conc. tested), Time of exposure, Environmental factors tested (Env. Factors tested), and reference (Ref.).
Brennecke et al.(2016) observed high adsorption of Cu and Zn onto MPs (virgin PS and aged PVC fragments) previously leached from an antifouling paint to water. PVC obtained the highest adsorption efficiency (850 partition coefficient between polymers and water) with respect to Cu, highlighting the strong adsorption capacity of plastics towards metals and comparing to their dissolved levels on water. Both aged and pristine particles adsorbed the available metals from water, evidencing the reactivity of MPs towards metal ions, even before their properties are modified after environmental exposure. The adsorption capacity of virgin MPs to metal(oid)s was also demonstrated in other studies (Turner and Holmes,2015; Rivera-Hernández et al., 2019; Dong et al., 2020; Fernández et al., 2020; Mao et al., 2020), although, for the largest part, higher adsorption capacities are attributed to the aged MPs, as reviewed here. Gao et al.(2019) investigated sorption through performance of both laboratory and field tests, considering different times of exposure (9 days and 9 months, respectively) of new commercial plastic particles (PP, PE, PVC, polyamide (PA) and polyoxymethylene (POM)) with different sizes (2–5 mm). Adsorption capacity increased with decrease of particle size and the equilibrium between the heavy metals Pb, Cu and Cd and the MPs was achieved within the 9 days of the laboratorial suspension. The highest adsorbed levels were quantified for Pb, ranging from 0.627 μg/g in PP to 1.318 μg/g in PVC, while the lowest amounts corresponded to Cd (0.002 μg/g) in POM, finding different adsorption trends for the metals tested depending on the polymer type. Considering the field exposure, the highest metal levels (i.e., Pb) were found to be adsorbed on PP (0.109 - 0.144 μg/g) rather than on PVC (0.028–0.107 μg/g), conversely to what was observed from the laboratory experiment. The fluctuation observed in the quantified metals on MPs from the environmental suspension between the two sampling times of 6 and 9 months (first decreased to following increase), suggests a dynamic behavior of metal adsorption onto polymers on nature. That is to say that ions are in constant exchange between the solid and liquid interface, until they reach equilibrium, guided by environmental parameters (Gao et al., 2019). Earlier, Turner and Holmes(2015) performed reaction kinetics between trace metals (Ag, Cd, Co, Cr, Cu, Hg, Ni, Pb, Zn) and (beached or new) pellets on a freshwater solution. Pb was identified with a maximum of distribution coefficient in both beached (10 2 mL/g) and virgin pellets (6 mL/g) from water, suggesting about two orders of magnitude higher adsorption capacity for the aged MPs. In turn, the water parameters influence sorption dynamics between metals. Increase of river water pH facilitated adsorption of Ag, Cd, Co, Ni, Pb and Zn, but not for Cr. Surface complexes formed with cations, oxyanions and organic complexes on the natural environment was considered to be the main responsible for the higher reactivity of aged MPs.
A range of several environmental conditions are tested as they influence sorption behaviors between polymer surfaces and metal(oid)s. From a very complete study, Tang et al.(2020) found adsorption of Pb onto nylon MPs increasing with pH rise but decreasing with increase of ionic strength and fulvic acid concentration, representing significant key factors for sorption dynamics. High concentrations of fulvic acid inhibited Pb uptake onto MPs, as this dissolved form of organic matter has more affinity for metals than MPs have, through formation of complexes, and thus, illustrating competition for metal bond. Pb bond was higher with aged MPs and due to electrostatic interaction, with carboxylate groups representing the most important functional group in this process, along with surface complexation (Tang et al., 2020). Surface complexation and electrostatic interactions were reported for many studies to be the main responsible to bond metal(oid)s (e.g., Pb, B, Cd, Cu) to polymer surfaces (Tang et al., 2020; Zou et al., 2020; Wang et al., 2021). Zou et al.(2020) confirmed the mentioned mechanisms as main responsible for sorption of Cd and Cu, but stronger electrostatic interaction for Pb, on commercial polymers. Here, higher adsorption of heavy metals on the chlorinated polymers tested CPE and PVC was identified (adsorption capacities followed CPE > PVC > PE-HD > PE-LD) with higher adsorption efficiency for Pb, followed by Cu and finally Cd (Zou et al., 2020). Electronegativity of CPE polymer dictated its strong bond with cations. Also, the oxidation of PE-HD with formation of CeO/C=O groups allowed higher adsorption for this polymer, in relation to PE-LD. Despite high pH allowed strong adsorption, ionic strength had not a significant influence on the process, conversely to the findings from other studies. In turn, adsorption of Pb demonstrating the influence of different salinities and depending on polymer type of MPs was reported by Lin et al.(2021). Here, the maximum adsorption amount for Pb between polymers was as follows: PE (531/459 μg/g), PVC (476.9/328 μg/g) and PS (246.4/227.1 μg/g), respectively for lower and higher ionic strength. Holmes et al.(2014) early related high salinity with decrease of adsorption capacity towards metals (i.e., Pb) promoted by the competition between cations for available surfaces to settle from water. Furthermore, the effect of ageing mechanism on MPs concerning sorption has been investigated by Mao et al.(2020) and Wang et al.(2020). Mao et al.(2020) induced photo-aging on PS and their sorption behavior towards a few metals (Pb, Cu, Cd, Ni, Zn), while Wang et al.(2020) tested influence of UV light on PET and their affinity for Cu and Zn. From both studies was possible to acknowledge that exposure to UV irradiation is a determinant factor enhancing the sorption capacity of wasted plastics towards metals on environment. After UV induced-aging, an increase of cracking, pore size, roughness, and creation of new oxygen-containing functional groups to reaction with metals on water was observed. This consists in the major alterations on the polymer surface as photolysis enable the break of chemical bounds (e.g., C-H), which originate new functional groups by reaction with free radicals with oxygen, thus forming C-O and C=O bounds. Additionally, the influence of external factors was also explored by Wang et al.(2020) in the same study, who found enhancement of adsorption of metals onto surfaces when exposed to high temperature (with maximum sorption capacity of 268.4 μg/g for Cu at 318 K) and high pH (Cu achieved 282.5 μg/g on MPs at pH 7).
Additionally, Richard et al.(2019) identified biofilm (adhered to plastic surfaces after exposure in estuarine waters) as the most significant variable promoting uptake of metals on polylactic acid (PLA) and low-density polyethylene (PE-LD) new pellets. Biofilm coverage was identified on the same polymer (PE-LD) for Ahamed et al.(2020), subject to potable water plumbing pipes, to strongly facilitate adsorption of Pb (1602 μg/m 2) comparing to their absence (124 μg/m 2). In terms of sorption equilibrium, Pb onto MPs with no biofilm was achieved over 6 h while the 5 days of exposure for biofilm-covered MPs were not enough to achieve equilibrium. This observation could reflect the more available sites for metal(oid)s created by complexes with biofilms on MPs surface. Furthermore, microbial communities contain natural polymers which have functional groups with negative charges (e.g., OH−, SO 4 2−) likely favoring the bond with metal cations (i.e., Pb 2+). The competition of Pb 2+ with other cations for limited available active sites and the ion pair with anions (forming more soluble complexes) decreases its activity towards polymers (Ahamed et al., 2020). Accordingly, the presence of chlorine residuals on water (characteristic of tap water tested as background solution here) delayed the equilibrium achievement of Pb onto pellets (135 μg/m 2 and 95 μg/m 2 in their absence and presence, respectively). Formation of chloro‑complexes with Pb and oxidation to precipitation of Pb ions is suggested to inhibited metals-MPs adsorption. Overall, the uptake of metal(oid)s essentially occurs on freshwater environments (biofilm growth) to be further transported until marine environments to be potentially released there.
Among the revised studies, a higher affinity of polymers for Pb is evident, as its cations have a strong electronegative interaction with polymers allowing their bond (Zou et al., 2020). The high toxicity and high potential bioaccumulation, adding up their wide use as plastic additive makes Pb (Lavers and Bond,2016a; Nakashima et al., 2016; Tang et al., 2020; Turner et al., 2020) and also Cu (Holmes et al., 2014; Turner and Holmes,2015; Brennecke et al., 2016; Gao et al., 2019; Xue et al., 2021) as the main elements used as metal models in field and laboratory studies. Additionally, Pb was demonstrated to be highly adsorbed onto polymers, especially PVC (Brennecke et al., 2016; Gao et al., 2019; Tang et al., 2020; Zou et al., 2020), and with a high desorption potential (Turner et al., 2020). The high adsorption capacity of PVC is common for several studies, especially for Cu, Pb and Cd, which is attributed to its higher surface and polarity (Brennecke et al., 2016; Nakashima et al., 2016; Filella and Turner,2018; Munier and Bendell,2018; Gao et al., 2019; Santos-Echeandía et al., 2020). Wang et al.(2021) supported high adsorption of PVC towards boron (B) metalloid compared to PS polymer, which were as follows: aged PVC (0.91 mg/g) > aged PS (0.197 mg/g) > virgin PVC (0.1 mg/g) > virgin PS (0.005 mg/g). Accordingly, aged was an enhancer of adsorption through the link of B(OH) 3 species with functional groups with oxygen formed, while PVC stand out likely due to C/Cl bond with B. Moreover, from other parameters tested, the presence of humic acid and high pH restrained the bond of B to MPs. Although, polymer type, ion type (represented here by Mg 2+, Ca 2+, Cu 2+, and Al 3+) and ion concentration were defined as main drivers in sorption process of boron. As an example, at pH 7 (with domain of B(OH)3 species), adsorption of polymers (aged and virgin PVC, aged PS) decreased under low concentration of Ca 2+ (0.04 – 0.4 g/L), suggesting competition of other cations with B(OH)3. However, cations can also behave as bridges for boron ([B(OH)4]− species) for negatively charged MPs, as PS. Despite increasing of ionic strength (with more Na+) enhanced adsorption of B onto aged PVC, inhibition caused by ion competition (i.e., Mg 2+, Ca 2+) suggests annulation of salinity effect. At last, desorption of previously adsorbed B onto aged PVC under a simulating digestive solution of warm-blooded animals resulted in high desorption rates (35.9%). Another study with a metalloid (i.e., As) was performed in order to understand the mechanisms towards sorption of MPs (Dong et al., 2020). Lower sizes of particles (higher surface area) and lower temperatures increased adsorption. As matter of fact, with temperature increase, hydrogen principal bounds between As(III) and carboxyl functional groups of PS were broken. Moreover, pH rise and also NO 3 and PO 4 3− also inhibited adsorption (resulted from competition of arsenate ions with OH−), supporting the role of electrostatic interaction occurring between As and PS particles here (exhibiting a positive electrostatic potential on surface). Despite the tests conducted in laboratory considering one or a few metal(oid)s to test, on a real aquatic environment it is needed to take in account competition with other ions for binding sites, thus behaviors can be countered there.
From the several types of plastics tested, pellets are the main representatives of primary MPs used in both field and laboratory studies assembled here (Turner and Holmes,2015; Vedolin et al., 2017; Maršić-Lučić et al., 2018; Richard et al., 2019; Holmes et al., 2020). In environmental studies, these MPs are chosen to be good targets of biofouling and proper indicators of environmental pollution from their quick spread. The findings of studies confirmed pellets as efficient adsorbents of metal(oid)s, with potential repercussion to further transport them from their original source to long distances to be available to a high diversity if organisms.
5. Ecological repercussions from interaction between microplastics and metal(oid)s
5.1. Role of microplastics on the bioavailability of metal(oid)s
Aquatic organisms are daily exposed to a high diversity of plastic debris in their environment from nano- to micro- and macro- plastics. Because of their small size, MPs can easily enter the bodies of organisms either by direct ingestion or through filtration on epithelial organs, following accumulation on their tissues (Wang et al., 2018). The high surface-volume ratio of MPs enables the formation of several bonds with different elements present in the surrounding water. This fact makes these microparticles being responsible for the transfer of a potentially toxic mix of metal(oid)s (adsorbed or as additives) to the circulatory systems of aquatic biota. The acidified environment in the gut of an organism is what promotes the undirect assimilation of chemicals, since desorption and release of metal(oid)s from polymers can be much higher in acidic conditions (30 times faster) than in fresh and seawater (Bakir et al., 2014; Holmes et al., 2020; Tang et al., 2020). Later, after assimilation and upon bioaccumulation, metal(oid)s can turn back to the aqueous media through faeces (Rivera-Hernández et al., 2019) or metabolic mechanisms to elimination of the toxicants in a more soluble and available form to be adsorbed again (Fig.3). The retention time of MPs in the stomach and digestive tract of aquatic animals (e.g., fishes, bivalves) normally reflect short-time periods (hours to a few days). Although, this time is enough to allow desorption of metal(oid)s from MPs and uptake on tissues. Subsequently, organisms tend to excrete the microparticles, as they have no nutritional content (Gonçalves et al., 2019). In this case, if the bond of metal(oid)s to the MPs is strong, the microparticles can reduce or prevent potential effects from metal(oid)s in contact with organisms, even if they are ingested. In this way, MPs can be vectors of chemicals to aquatic species in two ways (Rodrigues et al., 2019), since the contact with these particles can contribute to increase or reduction on the assimilation of coexisting hazardous elements and toxicological effects (Khan et al., 2015; Lu et al., 2018; Rivera-Hernández et al., 2019; Rodrigues et al., 2019).
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Fig. 3. Process of vectorization of metal(oid)s through microplastics resulted from exposure of aquatic organisms to those related pollutants that result in an ecological impact.
5.2. Bioaccumulation, ecotoxicity and reported scenario effects
It is relevant to estimate the potential of MPs-associate metal(oid)s to be desorbed upon (or before) ingestion by aquatic organisms and subsequent effects (i.e., bioaccumulation and toxicity). Studies reporting the impact of MPs on aquatic organisms are extensive. Yet, a few years ago, just a few of them focus on toxicity concerning exposure to metals associated to MPs (de Sá et al., 2018). Characterization of several bioassays with single, coexposure and exposure to MPs and (preadsorbed) metal(oid)s with reported effects derived from their interaction are presented in the Table 2. Khan et al.(2015) tested the influence of MPs in the uptake and path-route of silver (Ag) on Danio rerio. The treatments with MPs differed between (1) exposure to previously bounded Ag to MPs and (2) coexposure of Ag and MPs initially dissociated in the aqueous solution. Exposure of bonded Ag to MPs resulted in the reduction of metal uptake by the organisms, comparing with the levels accumulated from Ag alone, and different accumulation between the organs of the fish, which was higher in their gut. However, the mixture of MPs and Ag (dissociated) had no influence in the bioaccumulation and localization of Ag. These observations demonstrate that the differences in the way of exposure of organisms to chemicals can result in different observations (as they are previously bounded before exposure or not), concerning the bioaccumulation of metals through MPs and potential toxic effects. According to the findings of Rivera-Hernández et al.(2019), Hg ingested via its association with MPs did not affect substantially the levels of Hg accumulated in Mytilus galloprovincialis in comparison to the Hg associated to microalgae or to dissolved Hg. The tested mussels ingested the microparticles to further excrete them through faeces over the depuration time and even the Hg levels in their bodies was higher via water than through MPs (Rivera-Hernández et al., 2019). Additionally, the levels and target tissues for this metal were different between microparticles (microalgae and MPs) and via water, with higher Hg accumulation via MPs or microalgae in the digestive gland whereas via water higher levels of Hg were quantified in gills. Accumulation of Hg is readily facilitated through Hg-microalgae than for MPs, since dissociation of Hg from microalgae is easier to followed absorption on tissues, despite they allowed similar Hg accumulation on organisms, as consolidated by Fernández et al.(2020). These observations suggest that ingestion through MPs facilitates elimination of an hazardous element due to the strong bond of Hg to polymers (Rivera-Hernández et al., 2019). These studies provide evidence towards the role of MPs on the bioavailability of metal(oid)s, even tough, they promoted the reduction on the assimilation of associated hazardous elements and potentially preventing toxicity effects, pointing to a positive action of MPs concerning other toxicants. Bioaccumulation of Hg was also verified to be reduced in the presence of MPs on the clam Corbicula fluminea (Oliveira et al., 2018), compared to the heavy metal exposed alone. Ecotoxicological effects were estimated here, with observation of antagonism between the two pollutants in the coexposure of Hg and MPs, since the effects were minimized in relation to their single exposure effects and highlighting existence of interaction and prevention of potential toxic effects from Hg to clams. However, exposure to MPs alone induced neurotoxicity on the organisms. Besides these results support MPs as favorable to prevention of toxicity from the associated metal, MPs themselves are toxic to organisms, perhaps due to inherent metal(oid)s additives they have. One thing in common in the previous cited studies (Oliveira et al., 2018; Rivera-Hernández et al., 2019) is that the tested organisms are bivalves so they easily ingest (or filter) the MPs as they represent particle filter-feeding organisms so they have efficient strategies/mechanisms associated to their digestive system to elimination of non-nutritive particles (and associated toxicants). This could be an advantage over other classes of organisms, dictating different results of bioaccumulation and toxic damage from exposure to metal(oid)s via MPs. For instance, Barboza et al.(2018) reported high toxicity in fish (Dicentrarchus labrax) from the exposure to MPs with Hg, as neurotoxicity, oxidative stress and damage. Besides the interference of MPs in the levels of bioavailable Hg, the single exposure of MPs also induced high toxicity on organisms. Luís et al.(2015) estimated toxicity effects on fish juveniles (Pomatoschistus microps), as neurotoxicity with inhibition of Acetylcholinesterase (AchE) enzyme activity and decline of predatory performance, inflicted by coexposure to MPs and Cr; and also by MPs alone. However, single exposure of Cr caused mortality, which points to MPs suppressing mortality of the organisms here. Jinhui et al.(2019) observed low body growth and decrease on survival of Hippocampus kuda exposed to MPs carrying Cu, Cd and Pb. Despite bioaccumulation of the heavy metals was observed in the sea horses, it was not determined to be as a consequence of MPs vectorization. This work suggested that the toxicity inflicted is more due to the metals than to the MPs themselves. In line with the previous study, Roda et al.(2020) did not observe a vector effect from MPs in the levels of Cu absorbed on Prochilodus lineatus tissues, although, genotoxicity in fishes was identified, but with no aggravation caused by the interference of MPs. Conversely to the previous findings (Jinhui et al., 2019; Roda et al., 2020), MPs prompted bioaccumulation and toxicity of Cd, such as oxidative damage and inflammation on Danio rerio ( adult fish), resulted from the long-term coexposure of Cd and MPs (Lu et al., 2018). Other studies with Danio rerio were conducted in order to investigate effects from co-exposure of MPs and Cu in fish early life stages were conducted (Santos et al., 2020; Santos et al., 2021b, 2021a). In these studies, toxicity (neurotoxicity, oxidative damage, avoidance behavior disruption) and even mortality (Santos et al., 2021a, 2021b) for all the single and mixture exposures were reported. Neurotoxicity is a commonly reported toxicity effect on fish resulting from coexposure to MPs and several metals (Luís et al., 2015; Oliveira et al., 2018; Roda et al., 2020; Santos et al., 2020, 2021b, 2021a). Even so, antagonism over 96 h (Santos et al., 2020) or synergism after 14 days (Santos et al., 2021a) were observed between Cu and MPs, regarding biomarkers response, reflecting the relevant impact of MPs over toxicity. Moreover, bioaccumulation of Cu through MPs was observed after 10 and 14 days of exposure (Santos et al., 2021a). Those findings also highlight the influence of the time of exposure concerning the impacts of MPs on toxicity of hazardous substances.
Table 2. Experimental studies of impact of microplastics on bioaccumulation and toxicity effects of metals on aquatic organisms resulted from the simultaneous exposure of microplastics and metals.
| Exp. | Elem. conc. (µg/L) | MPs conc. (mg/L) | Organism | Bioac. of metals by MPs | Analytical method | Effects | Time exp. | Ref. |
--- --- --- ---
| Ag bounded to MPs and coexposure of Ag and MPs | 1 (ER) | 10–1000 (PE) | Danio rerio | No (Decrease of Ag) | Radiotracer | No effect from co-exposure Decrease of Ag accumulation (Ag-bounded MP) | 24 h | Khan et al.(2015) |
| Coexposure of Cr and MPs | 5600–28,400 (NER) | 10–1000 | Pomatoschistus microps | – | – | Oxidative damage Neurotoxicity Decrease of predatory performance MPs suppressed mortality from Cr | 96 h | Luís et al.(2015) |
| Ni and MP mixture (bonded and dissociated) | 1000–5000 (NER) | 1–30 (PS, PS-COOH) | Daphnia magna | Yes | ICP-MS | Immobilization from MPs copresence (higher with PS-COOH) Higher accumulation of Ni through MPs | 48 h | Kim et al.(2017) |
| Cd combined with MPs | 10 (ER) | 0.02–0.2 (PS) | Danio rerio | Yes | ICP-MS | MPs increased Cd toxicity Oxidative damage and inflammation | 3 weeks | Lu et al.(2018) |
| Hg and MPs mixture | 30 (ER) | 0.13 | Corbicula fluminea | No (decrease of Hg) | Silicon UV diode detector in an automatic Mercury analyzer | Antagonism in FR, ChE activity, GST activity and LPO levels | 14 days | Oliveira et al.(2018) |
| Hg and MPs mixture | 10 – 16 (ER) | 0.26- 0.69 | Dicentrarchus labrax | Yes | AAS with silicon UV diode detector | Oxidative damage | 96 h | Barboza et al.(2018) |
| Cu Cd Pb alone and MPs alone | 10 - 50 (ER) | 0.33 (PE-HD) | Hippocampus kuda | No | ICP-MS | Growth reduction | 45 days | Jinhui et al.(2019) |
| Cu and MPs mixture | 10 (ER) | 0.02 (PE) | Prochilodus lineatus | No | AAS | Genotoxicity Neurotoxicity Physiological effects | 24–96 h | Roda et al.(2020) |
| Ag bounded to MPs | 10,000 (NER) | 10–100 (PE) | Lemna minor Daphnia magna | – | – | Decrease of specific growth Immobility | 48 h | Kalčíková et al.(2020) |
| Hg bounded to MPs | 2.5 (ER) | 2 (PE-HD) | Mytillus galloprovincialis | Yes | ICP-MS | Higher accumulation of Hg on gut from MPs Faster elimination of Hg by MPs | 7 days | Rivera-Hernández et al.(2019) |
| Cu and MPs mixture | 500 (NER) | 10 (PVC) | Chlorella vulgaris | – | – | Growth inhibition (at low MPs conc. single exposure but not at high levels of MPs) Growth increase from coexposure of MPs and Cu | 10 days | Fu et al.(2019) |
| Pb bounded toMPs | 1500–2500 (NER) | 5–20 (PS-COOH) | Ceriodaphnia dubia | Yes | Graphite furnace atomic absorbence spectrometry (GFAA) | MPs enhanced Pb toxicity and accumulation Algae reduced toxicity of MPs-Pb Accumulation of MPs and fast elimination | 24 h and 7 days | Liu and Wang(2020) |
| Cu and MPs coexposure | 15 – 125 (ER) | 2 | Danio rerio (larvae) | – | – | Neurotoxicity Avoidance behavior disruption Oxidative stress Antagonism | 96 h | Santos et al.(2020) |
| Cu and MPs coexposure | 60-125 (ER) | 2 | Danio rerio (larvae) | Yes | Electrothermal atomic absorption spectrometry | Synergism Neurotoxicity Oxidative stress Mortality | 14 days | Santos et al. (2021a) |
| Cu and MPs coexposure | 60-125 (ER) | 2 | Danio rerio (larvae) | Yes | Electrothermal atomic absorption spectrometry | Neurotoxicity Avoidance behavior Mortality | 14 days | Santos et al. (2021b) |
| Cu-MPs and Cd-MPs co exposure | 1000-2 000 (NER) | 100 (PS) | Triticum aestivum | No (decrease of Cd and Cd) | AAS | MPs promote decrease of toxic effects | 2 days | Zong et al.(2021) |
For the characterization of the studies, were considered the following parameters: Exposure way (Exp.), Elemental concentration tested (Elem. conc.), Microplastics polymer (MPs polymer), Microplastics concentration tested and polymer (MPs conc.), Tested organism (Organism), Bioaccumulation of metals through MPs (Bioac. of metals by MPs), Analytical method, Effects, Time of exposure (Time exp.), and Reference (Ref.). ER identifies Environmentally Relevant tested concentrations and NER correspond to Non-Environmentally Relevant tested concentrations.
Environmental or environmentally dependent characteristics (e.g., food content, residence time and degradation status of particles, exposure time) are important to influence response of organisms towards exposure of MPs and associated chemicals. One would expect that a long-term exposure of metal(oid)s through MPs cause higher bioaccumulation and toxic damage to organisms than acute toxicity from a brief exposure (Lu et al., 2018). Since metal(oid)s globally exist in trace concentrations in the marine environments, the amounts potentially transferred there in a short period of time could not be high enough to cause immediate effects or accumulation. Even so, a long-term contact with contaminated MPs, as they are ubiquitous and potential sink of metal(oid)s, can magnify the bioaccumulation of toxicants through gradual uptake (Brennecke et al., 2016). The modifications that MPs went through on environment (e.g., formation of new functional groups, biofouling) conferring them a higher frequency bond with metal(oid)s obtained from water, potentially represent higher risk for aquatic biota or preventing them to suffer from toxicity effects, due to the strong bond between the pollutants. Kalčíková et al.(2020) observed higher toxicity on Daphnia magna and Lemna minor exposed to virgin MPs and also simulated aged MPs containing biofilms both with adsorbed silver than the toxicity observed from pristine and aged MPs alone. Virgin MPs and aged MPs both with adsorbed Ag induced decrease of specific growth rates (at the highest concentration of MPs) on L. minor and immobility on D. magna, compared to their single exposure without Ag. However, no difference in the magnitude of toxicity between aged and pristine MPs (with Ag) was demonstrated. Moreover, biofilm was found to also contribute to higher sorption capacity to MPs carrying metal(oid)s, since MPs with a biofilm adsorbed about 44% more Ag than pristine ones, and with higher released amounts of Ag as well, concerning desorption to an acidic medium. Kim et al.(2017) exposed Daphnia magna to Ni and MPs (PS and PS-COOH) and observed synergistic effects and bioaccumulation increase from their mixture. Immobilization of the crustaceans was higher through PS-COOH, however, the functional group or lack of it did not influence on the bioaccumulation level of Ni. It is important to say that, despite bioaccumulation overall increased in a dose-dependence perspective over the mixture treatments, it was observed a reversal concerning the highest tested concentration of Ni (5 mg/L). This finding is important to highlight the importance of chemical tested concentrations, as the results can change considering different circumstances. Liu and Wang(2020) introduced algae to test the influence of a natural food pathway on the impact of MPs to Pb accumulation and toxicity on Ceriodaphnia dubia. Bioaccumulation and toxicity of Pb to the crustaceans was higher when the metal was bond to MPs, compared with their single exposure. However, in the presence of algae, the Pb toxicity caused through MP significantly decreased. It's important to retain that the algae also allowed assimilation of Pb, but reduced toxicity. Synergism with aggravation of the effects from coexposure of polymers and metal(oid)s, compared with their single exposure, being size- and dose- dependent was demonstrated in a study with nanoplastics (Lee et al., 2019). Bioaccumulation and induced inflammatory responses on Danio rerio from nano PS and gold (Au) exposure were higher considering the lowest size of nanoplastics (50 µm) with accumulation in the whole body through the penetration of membranes and particularly accumulation in the lipidic areas. The smaller the plastic particles (i.e., microplastic and nanoplastics) enter inside organisms, the richer in hazardous elements they will be exposed and bioaccumulate.
At last, a few studies investigated the impact of leaching additives from MPs (or adsorbed afterwards) to different classes of aquatic organisms. Oliviero et al.(2019) exposed marine planktonic organisms (Paracentrotus lividus) to commercial plastic material toys (micronized PVC at 0.3–30 mg/L) and also to their hazardous elemental additives (mainly Zn with maximum concentration of 1446.1 µg/L) previously leached to water. A decline and delay of urchin embryo and larvae development, and even its blockage (from the highest concentration), was observed from the MPs exposure. From the released hazardous substances exposure, an early delay on development and deformities was also reported. These results evidence the toxicity on organisms caused by MPs and their metal(oid)s components. Furthermore, the researchers also detected a color dependence for the toxicity observed, mainly explained by the different mix metallic pigments added to confer it. Despite the toxicity observed from microplastic leachates, no toxicity was observed when the organisms were exposed directly to PVC virgin ones or to their leachates (tested as control). In other respect, Karami et al.(2017) used low concentrations of MPs (LDPE) fragments (5, 50 and 500 µg/L) in their bioassay with Danio rerio to investigate the toxicity effects of potentially leached pollutants (concentration of heavy metals and other chemicals were below the LOD) and find that environmental realistic concentrations of MPs just cause minimal ecological impact. Nobre et al.(2015) obtained higher toxicity in sea urchins (Lytechinus variegatus) exposed to virgin MPs (PE) than from those beach sampled, suggesting higher toxicity attributed to hazardous additives than those adsorbed on environment. Although, they did not determine or quantify the substances present on the polymers. Nonetheless, those findings point to leachate metal(oid)s additives from virgin MPs being more hazardous to aquatic species, as they are typically associated to high concentrations of heavy metals, especially the old-produced ones. These findings are in accordance with the study of Turner et al.(2020) who used simulated gastric conditions of seabirds to estimate bioaccessibile Pb from MPs, referred in a previous topic. Here, the adsorbed metal was more biaccessible (70% of the quantified adsorbed Pb versus 16% of quantified additive Pb). However, metal additive content released was much higher (two orders of magnitude) due to higher content quantified on the MPs, so it was considered to represent a higher ecological risk (Turner et al., 2020). Despite metal(oid)s adsorbed from the environment are more easily desorbed to water from polymers, the metal(oid)s added during manufacturing have a higher impact to aquatic biota and ecosystems. This is justified if ingested by organisms, as MPs and metal(oid)s are dissociated in acid conditions (i.e., gastrointestinal tract) in a larger scale to follow assimilation of toxicants in biological tissues. Other studies used the previous referred method with adaptations to estimation of the bioaccessible levels of hazardous elements (additive or adsorbed) released from MPs to digestion solutions. Avian physiologically-based extraction test (PBET) (Turner,2018c; Holmes et al., 2020) or dietary-adapted with fish oil (DA-PBET) (Smith and Turner,2020) were applied to the method. All the quantified elements (i.e., Pb, Br, Cd, Cr, Co, Fe, Mn, Se, Ba, Sb, Hg) tested were bioaccessible, among the different studies. Although, Pb, known as one of the most toxic metals, was found to be totally bioaccessible (100%), with maximum release from beached polyurethane exposed to the DA-PBET (Smith and Turner,2020), and also high levels bioaccessible (80%) from beached PE (Holmes et al., 2020) to PBET. Although, bioaccessible levels for Cd were found to substantially vary (65%) from consumer polycarbonate-acrylonitrile (Smith and Turner,2020) to less than 1% from PE and PP marine MPs (Turner,2018c) through PBET digestion. This fact demonstrates that the release and bioavailability of elements from MPs vary and is dependent on nature and source of polymers, type of additive/residue, extraction solution and model. Therefore, high levels of metal(oid)s can be released in high amount from both pristine and weathered MPs.
Susceptibility of seabirds on their own environment to exposure to either adsorbed or inherent metals through local plastic debris has been investigated. Lavers et al.(2014) related the trace elements amounts found in the feathers of seabirds (Puffinus carneipes) with the ones determined in sampled plastics. The high levels of Cr found were likely associated to the plastic load, as this metal is known as an additive for the main polymers found inside the organisms and likely responsible for the observed reduction of body condition, caused by vectorization through MPs. These observations support plastics as responsible for transferring of hazardous elements on environment and demonstrate that aquatic organisms (e.g., seabirds, bivalves) can monitor the local load of metal(oid)s not just accumulating them from water (in their dissolved form), but also through MPs ingestion; such as the MPs themselves.
One of the factors or plastic properties that was mentioned here to potentially have an influence on metal type or concentration is the color. Additionally, color of polymers can influence the uptake of MPs by organisms, as they ingest selectively a plastic piece by color for resemblance with the natural color of their prey (Lavers and Bond,2016b). Consequently, polymer color can determine the assimilated elements (due to leaching of remaining pigments) on their bodies with corresponding toxicities. Since MPs are easily mistaken by any prey and are present in endless numbers on aquatic environments and numerous colors (especially pellets used to make plastic products), the potential impact by metal(oid) additives/adsorbed transferring is worrying.
Regarding plastic size, MPs represent higher risk to aquatic organisms than larger plastics, as well as they can reach almost all levels of the trophic chain, from planktonic organisms (Kalčíková et al., 2020) through bivalves (Rivera-Hernández et al., 2019), fishes (Khan et al., 2015; Luís et al., 2015; Lu et al., 2018) and birds (Weitzel et al., 2021) up to cetaceans (Moore et al., 2020), directly by ingestion/filtration or through bioaccumulation on their bodies (Wang et al., 2018). One would expect nanoplastics being even more alarming than other sized plastics, as they have an even larger surface area, which is directly proportional to a higher sorptive capacity, inducing higher reactivity to metal(oid)s species (van Leeuwen et al., 2013; Town et al., 2018). Furthermore, nanoplastic particles are so small that can go through chorion membranes (by penetration of the pores) of early life stages of fish (i.e., zebrafish) just by their contact on water to further achievement of all organs to accumulation and undergo potential toxicity of related hazardous elements (Lee et al., 2019). Thus, plastic particles have a size-dependent accumulation pattern on organisms, exemplified here by the chorion holes acting as a barrier to the entry of foreign larger particles, which is also correlated to the extent of induced accumulation/toxicity of associated toxicants on body organisms. It is not clarified whether organisms are or not selective in relation to polymer type regarding MPs uptake (Digka et al., 2018; Town et al., 2018). However, as charged chemical species can be related to a specific polymer type available to bind, regarding its structure and functional groups, a microparticle of a specific polymer ingested could dictate the potential toxicological effects inflicted on an organism.
It is worth to say that the majority of the studies mentioned employed pristine MPs, instead of expose organisms to aged ones. However, aquatic biota are much more likely to be exposed to environmental and degraded plastics, with higher sorptive capacity to concentrate metal(oid)s, reducing bioavailability (Nobre et al., 2015), or enhancing it through transference of the metal(oid)s upon ingestion of MPs by the organisms. Furthermore, benthic and demersal organisms are presumably more exposed to high density polymers, such as PVC (non-buoyant), whereas pelagic ones should be more exposed to low density ones, just as PE and PP, that also correspond to the most produced and wasted plastics. PE was the most tested polymer with respect to bioassays, thus, desorption studies to estimate the potential effects of associated metals should be conducted in the rest of existing polymers, mainly in the most produced ones like PP, PS or PVC, and according to their susceptibility to each polymer in order to guarantee the environmentally reliability of the study. Nonetheless, low density MPs can be scavenged from the water column and reach the sediments following the incorporation of silt and clay materials or biofilms, as described on previous sections.
Which is common for the most part of the studies reported here is that MPs alter the bioavailability of associated metal(oid)s with inflicted toxic effects on organisms just by themselves which are aggravated or not by their association to metal(oid)s. Micronized size allows ingestion of plastics for the majority of organisms (independent feeding stages), making MPs more dangerous than larger ones. At last, concentration of MPs can manipulate impact results.
6. Contribution and challenges of laboratory studies to ecological risk assessment
To date, literature has been inconsistent in respect to estimation of ecotoxicological impact caused by MPs and associated contaminants spread on aquatic environments (Rodrigues et al., 2019). Field studies are more representative of what is factually occurring on nature, although exposure effect and all variables involved are difficult to manage (Anbumani and Kakkar,2018). Apart from this, availability of controlled experimental studies investigating dynamics and effects is massive and reveals some heterogeneity on methodologies. Hermsen et al.(2018) reunited ten essential criteria to quality assessment of ecological studies focused on uptake of MPs by marine biota. Among them, it is highlighted the sample size, negative and positive controls, target component and polymer identification. Provide all this information on a scientific paper is mandatory in order to assure reliability and reproducibility of a study and enable comparison between obtained findings and other similar studies. The researchers have also classified quantitatively field studies on the subject and obtained an average of negative results regarding accumulative reliability scores (as they failed in at least one criterion), highlighting the lack of standardized methodology. As matter of fact, one key element often revealing the discrepancy between studies is the concentrations used of both MPs and chemicals in test. Reports of Scientific Advice Mechanisms (SAM, 2019) and Scientific Advice for Policy by European Academies (SAPEA,2019) are specially critical with the concentrations used in laboratory works. Reliable concentrations should be the main element to take into consideration when a study is focused in the real impact of MPs to environmental toxicology, as evaluation of ecosystem's health represents the scientific relevance prompted to perform such work. However, it is possible to observe a mismatch on the tested concentrations and environmental realistic ones on published data, caused by the existence of a gap between theoretical methods to estimate the number of plastics and their impact on food web transfer SAPEA(2019). For instance, concentrations of pollutants in sea and open ocean are lower than in coastal areas which should be considered when designing the experiments. Since impact of MPs, considering their relation to contaminants, is a novel research field, the focus to soon publish data frequently overlaps the optimization of reliable methodologies. Nevertheless, some studies concern in to use environmental or low concentrations to obtain reliable answers on toxicity of MPs and associated contaminants (i.e., metal(oid)s) to aquatic biota (Jinhui et al., 2019; Roda et al., 2020; Santos et al., 2020, 2021a, 2021b). This is the example of Roda et al.(2020), who tested the coexposure effect of Cu and MPs using extremely low concentrations of MPs, and also tested environmentally realistic concentrations for Cu, and yet obtained ecotoxicological effects. On the other hand, for Jinhui et al.(2019), the low tested concentrations resulted in observations of little effects in the organisms and no vectorization of the metals (Cd, Cu, Pb) through MPs exposure. The studies reporting no effects, especially the ones using reliable concentrations are of extremely importance and should not be ignored, as they reflect the actual ecological risk of MPs (European Commission,2019). Notwithstanding the seriousness to use low concentrations of MPs, the use of deflected amounts of metal(oid)s deserves also critical remark from scientific community. Since a few metal(oid)s exist at trace levels in seawater, real concentrations should be primarily applied, to endorse the relevance of a study aimed to address ecological risk assessment. Ultimately, SAM strongly recommend stiffening the publication criteria of scientific papers of MPs, assuring their quality, pertinence, and comparability, through prioritization of accurate studies.
7. Final remarks
The increasing release of microplastics into aquatic environment allows dissolved metal(oid)s to interact with this new particulate fraction, altering their biogeochemical cycle. Understanding the mechanisms of sorption between both contaminants can help to take measures in order to revert the environmental impact of this phenomenon. From the information collected, considering impulse mechanisms for this process, it was accomplished that weathering from environmental exposure wields mechanical, chemical, and biological action over a polymer surface promoting its reactivity. This process primarily occurs through creation of new functional groups (i.e., containing oxygen) which confer active sites allowing adsorption of metal(oid)s. Coatings (i.e., biogenic nature, organic matter, biofilms) create complexes on polymer surfaces that also allow the bond of metal(oid)s. Furthermore, the biotic and abiotic forces can cause leaching on MPs resulting in introduction of the adsorbed metal(oid)s or inherent metal(oid)s additives included during manufacturing process of polymers to surrounding water. This suggests the intervention of MPs on bioavailability of contaminants as they can contribute to transportation towards release of hazardous substances on aquatic environments and potential toxicity effects. The small size of MPs makes them reachable to all levels of the trophic chain, favoring their ingestion/contact from early life stages of organisms with higher potential toxicological impact to ecosystems. In most of the studies reported here, MPs alter the bioavailability of associated metal(oid)s either by increasing or decreasing bioaccumulation and associated effects on aquatic organisms, also caused by the MPs themselves. Time-scale exposure and concentration of the exposed chemicals state the magnitude of effects caused in biota, with damage reported here. For this reason, chronic studies, with gradual exposure and accumulation of environmentally relevant concentrations of the toxicants will provide better information of what is factually occurring on aquatic environments, compared with acute ones. Color of polymers intervenes both in metal(oid)s content (due to be related with degree of weathering influencing sorption efficiencies and to be associated with specific additive metal(oid)s responsible to confer color to polymers) and in the uptake of MPs by species with associated effects. Old-produced plastics can be more dangerous to organisms as they have concentrated legacy contents of highly toxic heavy metals (i.e., Pb, Cd, Hg) and are highly bioaccessible through ingestion of plastic pieces, exhibiting potential higher ecological risk to aquatic environments. However, organisms were essentially exposed to virgin MPs in the majority of studies reported, what is inconsistent to what happens in natural environments, with exposure to aged ones closely representing environmental MPs, with their initial properties modified, and consequently, with adsorption capacity strengthened. Moreover, bond of polymers with metal(oid)s can differ between the elements involved, thus, a strong bond between MPs and a metal or metalloid could reduce or prevent accumulation and toxicity effects on exposed organisms. Undoubtedly, PE and PVC were the most mentioned or sampled polymers in consulted literature explored as model plastics. Due to high residual metal content of old PVC and affinity for metal(oid)s uptake, particularly Pb, PVC revealed to be the most relevant polymer regarding interactions with metal(oid)s. This polymer reunites features, as non-buoyancy and chlorine structural electronegative polar groups, that makes it a strong adsorbent of metal(oid)s, even before environmental modification on surface (although adsorption is still enhanced by weathering). Consequently, stricter rules for both use and production levels of this polymer may be required. In addition, electrostatic interaction and complexation occurring on surface of MPs were reported as most responsible to bonding metal(oid)s. In a dynamic aquatic medium, occurrence of either adsorption or desorption from/onto polymers can fluctuate, driven by water parameters (i.e., pH, ionic strength, temperature) and environmental contents (i.e., organic matter, biofilms) of water.
Trace metal(oid)s present on aquatic environments are usually highly toxic even at low concentrations to aquatic organisms, more than organic compounds (which are hydrophobic, thus, existing at low dissolved concentrations on water). Even more, surface modified polymers with environmental background either have a strong capacity to bond with metal(oid)s. Therefore, MPs could not be potentially dissociated from having metal(oid)s on their surfaces carrying them through aquatic systems to marine ones. For those reasons, they can enlarge the bioavailability of hazardous substances to aquatic biota with risk to induce ecotoxicological effects, especially if they are ingested. To clarify the ecological risk of MPs to aquic environments, more experimental studies with organisms using environmental plastic pieces at realistic concentrations, containing coatings and modified surfaces, are of great relevance.
Declaration of Competing Interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgments
The authors thank to CESAM for the financial support (UIDP/50017/2020+UIDB/50017/2020 + LA/P/0094/2020), through national funds of FCT/MCTES. Thanks are also due to financial support from the Portuguese Science Foundation (FCT) through one doctoral scholarship (PD/BD/143086/2018 attributed to Joana Patrício Rodrigues) under POCH funds, co-financed by the European Social Fund and Portuguese National Funds from MEC. This work was carried out within the SEEME project (PID2019-109355RA-I00, MICIU/AEI/FEDER, EU) supported by the Spanish State Research Agency (AEI) and the Ministry of Science, Innovation and Universities (MICIU).
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Why is the index of refraction different for different wavelengths? [duplicate]
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The index of refraction can be written as
n=λvλm
where λv is the wavelength in a vacuum and λm is the wavelength in the medium. I’ve been told that since wavelength appears in the definition of an index of refraction, an index of refraction varies with wavelength. However, why would that be the case? The index of refraction is a ratio; if a wavelength of one wave is different from that of another wave passing through the same medium, the index of refraction should not be different for each wave, since they would have had different wavelengths in a vacuum too. So why is the index of refraction dependent on the wavelength?
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asked Jul 30, 2015 at 9:40
lightweaverlightweaver
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possible duplicate of Why do prisms work (why is refraction frequency dependent)?
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You are mixing up two different things.
The refractive index is usually defined in terms of the velocity of light:
n=cv
where v is the velocity in the medium. However the velocity is related to the frequency and wavelength by:
v=λf
so:
n=λ0f0λf
The frequency of the light, f, doesn't change as the light enters the medium, so f0=f and the fs cancel to give:
n=λ0λ
which is the equation you cite. The equation is not based upon any assumptions about the variation of n with the frequency/wavelength of the light.
However the refractive index does change with frequency. This effect is called optical dispersion. The cause is the way the interaction of the light and the electrons in the medium change with frequency. See for example the questions Why do prisms work (why is refraction frequency dependent)? and Why does the refractive index depend on wavelength?.
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edited Apr 13, 2017 at 12:39
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answered Jul 30, 2015 at 10:22
John RennieJohn Rennie
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I think you will have an easier time viewing the index of refraction from a speed-point of view.
Consider the following: The energy of a given photon is determined by its frequency (color):
E=hν
(h being the Planck constant)
Assuming the photon does not lose energy when entering the material, its frequency must be conserved.
However, as light is an electromagnetic wave, its propagation in a material differs fundamentally between the vacuum and a crystalline material. In the material, the electric field of the photon will be harder to 'produce', since the crystal's electrons will react to it. The higher the frequency, the stronger this effect is - the light gets increasingly slower. The refractive index for photons with a certain wavelength is
n(ν)=c0cm(ν) with c0 being the speed of light in vacuum (equal for all wavelengths as far as we know) and cm(ν) the speed of a photon of a certain frequency ν in the material.
Using c=λν (and assuming constant frequency) you will arrive at the expression you were using.
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answered Jul 30, 2015 at 10:32
Patrick KrausPatrick Kraus
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Not correct: there are materials with negative slopes to their dispersion curve; further most materials have a nonlinear relationship between input wavelength and index of refraction.
Carl Witthoft
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Carl Witthoft
2015-07-30 12:03:29 +00:00
Commented
Jul 30, 2015 at 12:03
To be fair this is true for most optically transparent materials at visible wavelengths. The refractive index becomes anomalous when the light energy reaches the band gap energy, and for most transparent materials the band gap energy is in the uv.
John Rennie
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John Rennie
2015-07-30 12:08:15 +00:00
Commented
Jul 30, 2015 at 12:08
True - I was refering to 'classical', linear materials. No no nonlinear effects or effects found in metamaterials are accounted for. But if we would go along this path I would start with the more realistic complex valued refractive index - which would also mean the energy- and thus the frequency would not be constant for the photon alone.
Patrick Kraus
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Patrick Kraus
2015-07-30 12:08:18 +00:00
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Jul 30, 2015 at 12:08
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7968 | https://byjus.com/biology/what-does-bacterial-plasmid-contains/ | An Overview
Bacteria are the most abundant living organisms on the Earth and one of the earliest life to evolve, i.e. it was over two to three billion years before eukaryotes appeared.
A bacterial plasmid is an extrachromosomal circular DNA found naturally in bacteria, archaea, and eukaryotes.
Explore more: Plasmids
Structure and Functions of Bacterial Plasmid
These plasmids are small, circular, double-stranded DNA structures naturally found in all bacterial cells. These plasmids are much smaller than the primary chromosomal DNA. The size of plasmids vary from 1 to 1000 kbp, and they usually contain 5 to 100 genes.
All bacteria have plasmids that occur naturally, and they play an essential role as vectors and help in the survival of the bacteria under stressful conditions. Every bacterial cell has its plasmids involved in food digestion, facilitating replication, etc.
pBR322, pUC18, Ti plasmid are a few examples of bacterial plasmids.
Explore more:Do All Bacteria Have Plasmids?
Components of a Bacterial Plasmid
The bacterial plasmid contains DNA as genetic material, replicating independently and comprising antibiotic-resistant genes.
Many antibiotic-resistant genes in bacteria are present in plasmids.
The three necessary functional regions present within the plasmid are:
It is a sequence of DNA that initiates replication on a plasmid.
There are several antibiotic resistance genes present on the same plasmid.
Resistance A, B, and C are a few examples of the antibiotic resistance gene. These genes are transferred from cell to cell by conjugation, transformation, or transduction.
Explore more:Bacterial Genetics
This is also called a multiple cloning site where the restriction enzymes cleave.
It is an essential component required for expression vectors. These vectors help determine which cell types the gene is expressed in and the amount of recombinant protein obtained. Some plasmids also have selectable markers for use in other cell types.
Explore more:What Are Selectable Markers?
Stay tuned to BYJU’S Biology to know more about the plasmids, bacterial plasmids, structure, functions, and characteristics.
Frequently Asked Questions on Bacterial Plasmid Contain
What are the applications of plasmids?
Plasmids are used to prepare recombinant DNA with the desired gene to transfer genes from one organism to another. This is known as genetic engineering.
Are plasmids found in all bacteria?
Yes. Plasmids naturally exist in all bacterial cells.
These plasmids are found within the bacterial cell’s cytoplasm, and they are usually found separated from the chromosomes.
What is the role of plasmids in bacteria?
In bacterial cells, the plasmids play a vital role in:
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7969 | https://www.purplemath.com/modules/exponent4.htm | Home
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What is engineering notation?
Engineering notation is similar to scientific notation, in that numbers are converted to (a number) times (10 raised to some power). But the powers in engineering notation will always be multiples of 3.
Because the powers are always multiples of three, the resulting numbers correspond to our common names for numbers, such as millions, billions, and trillions (and millionths, billionths, and trillionths).
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Scientific Notation
For instance, 13,460,972 is thirteen million and some. In the newspaper, this number would probably be abbreviated as "13.5 million". In engineering notation, you would move the decimal point six places to the left to get 13.460972 × 106. Once you get used to this notation, you recognize that 106 means "millions", so you would see right away that this is around 13.5 million.
Any time you see a reference to some number of millions or billions or trillions, rather than a complete enumeration of the entire number with all its digits, the writer is, in effect, using engineering notation.
Is engineering notation between 1 and 1,000?
Since engineering notation converts numbers to their units (millions, say, or millionths), the coefficient will be between 1 and 999. If the coefficient were 1,000, then the number would not have been converted properly, since this would be, say, "a thousand million" when it should have been converted, in this example, to "one billion".
In other words, in engineering notation, there should be between one and three digits to the left of (that is, before) the decimal point. And if there is just one digit before the decimal point, that digit must be an "interesting" number, so that digit cannot be zero.
What are some examples of engineering notation?
The following is an example of converting a "regular" number to engineering notation.
Express 472,690,128,340 in engineering notation.
This is a twelve-digit number. I need to move the decimal point from the end of the number toward the beginning of the number, but I must move it in steps of three decimal places.
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In this case, I must move the decimal point to between the 2 and the 6 (that is, to the location in the original number of the first comma), because this will leave nine digits (and nine is a multiple of 3) after the decimal point, and no more than three digits before the decimal point.
(Yes, 12 is a multiple of 3, but if I move the decimal point twelve places to the left, I'll have no non-zero digits to the left of the decimal point. This would be wrong; it would be like saying "about 0.4 quadrillion", which is just silly. I have to have non-zero stuff to the left of the dot so, in this case, I have to stop at nine decimal places.)
This is a large number and I moved the decimal point nine places, so the power on 10 will be a positive 9. Then the answer is:
472.690128340 × 109
...or about 472.7 billion.
It may be simpler to think in terms of commas when converting larger numbers to engineering notation, because those commas are placed specifically in "regular" numbers in order to demarcate hundreds from thousands, thousands from millions, millions from billions, and so forth. So just look for the comma furthest to the left, and move the decimal point to that location.
Express 83,201 in engineering notation.
I need to move the decimal point over to the left in sets of three digits (that is, in comma-delimited groups of digits). I can't move the decimal point any further than to the left of the 2, which is three places, so the answer is:
83.201 × 103
...or 83.201 thousand.
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When working with small numbers (that is, with numbers whose interesting digits are all the right of the decimal point), we don't have commas, but we can still think in terms of sets of three.
Express 0.000 063 8 in engineering notation.
I need to move the decimal point over in sets of three. If I move the decimal point to the right three places, I'll be left with "0.0638", which won't do because it'll leave me with just zero to the left of the decimal point. If I move the decimal point to the right nine places, I'll get "63800", which is too many digits to the left of the decimal point. So I need to move the decimal point six places.
Since this started out as a small number, the power on 10 will be negative; since I moved the decimal point six places, the power will be a negative 6. Then the answer is:
63.8 × 10−6
...or 63.8 millionths.
Express 0.397 53 in engineering notation.
I need to move the decimal point to the right three places. Since this started as a small number, the power on 10 will be negative:
397.53 × 10−3
...or 397.53 thousandths.
You should notice that, in engineering notation, it is perfectly okay to have more than one digit to the left of the decimal point; in fact, you should expect to have something other than always only one digit. Just make sure that the power on 10 is a multiple of three.
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7970 | https://www.cuemath.com/geometry/slope-of-parallel-lines/ | LearnPracticeDownload
Slope of Parallel Lines
Slope of parallel lines are equal. The parallel lines are equally inclined with the positive x-axis and hence the slope of parallel lines are equal. If the slopes of two parallel lines are represented as m1, m2 then we have m1 = m2.
Let us learn more about the slope of parallel lines, their derivation, with the help of examples, FAQs.
| | |
--- |
| 1. | What Is the Slope of Parallel Lines? |
| 2. | Derivation of Slope of Parallel Lines |
| 3. | Examples on Slope of Parallel Lines |
| 4. | Practice Questions |
| 5. | FAQs on Slope of Parallel Lines |
What is the Slope of Parallel Lines?
Slopes of parallel lines are equal. The slope of a line is computed with respect to the positive x-axis and the parallel lines are equally inclined with respect to the positive x-axis. If the slope of one line is m1 and the slope of another line is m2 and if it is given that both the lines are parallel, then we have m1 = m2.
The equations representing parallel lines have equal coefficients for x and y. The line parallel to ax + by + c1 = 0, is ax + by + c2 = 0. On observation, we find that the coefficients of x ad y in both the equations are equal.
Derivation of Slope of Parallel Lines
The condition for slope of the parallel lines can be derived from the formula of the angle between two lines. The angle between two parallel lines is 0º or 180º. For two lines having slopes m1 and m2, the angle between the two lines can be calculated using Tanθ.
(Tanθ = \dfrac{m_1 - m_2}{1 + m_1.m_2})
The angle between two parallel lines is 0º, and we have Tan0º = Tan180º = 0.
(Tan0º = \dfrac{m_1 - m_2}{1 + m_1.m_2})
(0 = \dfrac{m_1 - m_2}{1 + m_1.m_2})
(0 = m_1 - m_2)
( m_1 - m_2 = 0)
( m_1 = m_2)
Thus the slope of two parallel lines is equal in magnitude.
☛Related Topics
Slope Intercept Form
Point Slope Form
Coordinate Geometry
Cartesian Coordinate System
Examples on Slope of Parallel Lines
Example 1: What is the slope of a line parallel to y = 2x + 3, and is passing through (-1, 2)?
Solution:
The given equation of a line is y = 2x + 3. Comparing this with the slope-intercept form of the equation of line y = mx + c, we have m = 2
The required slope of the parallel line is equal to the slope of this given line and is equal to 2. Also, the given point (-1, 2) is not required to find the slope of the parallel line.
Therefore the slope of the parallel line is m = 2.
2. Example 2: Find the equation of a line passing through (-1, 2) and the slope of a parallel line is 2/3.
Solution:
The given slope of the parallel line is m = 2/3, and hence the slope of this required line is also m = 2/3.
The given point is ((x_1, y_1)) = (-1, 2)
The equation of a line can be calculated using point slope form of equation of a line.
((y - y_1) = m(x - x_1))
((y - 2) = \frac{2}{3} (x - (-1)))
(3(y - 2) = 2(x + 1))
3y -6 = 2x + 2
2x - 3y + 2 + 6 = 0
2x -3y + 8 = 0
Therefore the required equation of the line is 2x - 3y + 8 = 0.
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Practice Questions on Slope of Parallel Lines
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FAQs on Slope of Parallel Lines
What Is Slope of Parallel Lines in Maths?
The slope of parallel lines are equal. The parallel lines are equally inclined with respect to the positive x-axis and hence the slope of parallel lines are equal. If m1, m2 are the slopes of parallel lines then we have m1 = m2.
Where Do We Use the Slope of Parallel Lines?
The slope of parallel lines is useful to find the equation of the other parallel line. Also, the slope of parallel lines is helpful to find if both the lines are equally inclined with the positive x-axis.
What Are The Formulas For Slope of Parallel Lines?
The slopes of parallel lines are equal and the formula for the slope of parallel lines is m1 = m2. The parallel lines are equally inclined with respect to the positive x-axis and hence the slope of parallel lines are equal.
How to Find Equation Of A Line From Slope Of Parallel Lines?
The equation of a line from the slope of parallel lines can be calculated using the point-slope form or the slope-intercept form. The slope of a line is equal to the slope of parallel lines, and we have m1 = m2. And the equation of a parallel line can be calculated using the formulas ((y - y_1) = m(x - x_1)), and y = mx + c.
What Is The Difference Between Slope of Parallel Lines And Slope of Perpendicular Lines?
The slope of parallel lines are equal in magnitude. The slope of perpendicular lines is such that the slope of one line is the negative inverse of the slope of another line. For two lines having slopes m1 and m2 the condition for the slope of parallel lines is m1 = m2 and the condition for the slope of perpendicular lines is m1.m2 = -1.
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7971 | https://www.youtube.com/watch?v=vtE-wWgAdOY | To find the diagram power and diagram efficiency of impulse steam turbine| Engr. Arshad Ali Khan
Dr. Arshad Ali Khan Official
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Description
4279 views
Posted: 19 Jun 2021
Book: Applied Thermodynamics by T.D Eastop & McConkey,
Chapter #11: Rotodynamic Machinery,
Problem 11.3:
The nozzles of the impulse stage of a turbine receive steam at 15 bar and 300 C and discharge it at 10 bar. The nozzle efficiency is 95% and the nozzle angle is 20. The blade speed is that required for maximum work, and the inlet angle of the blades is that required from entry of the steam without shock. The blade exit angle is 5 less than the inlet angle. The blade velocity coefficient is 0.9. Calculate for a steam flow of 1350 kg/h:
i. The diagram power
ii. The diagram efficiency
6 comments
Transcript:
hello students welcome to my channel engineering education please subscribe my channel thank you this lecture is about the solution of problem 11.3 chapter 11 proto-dynamic machinery book applied thermodynamics by tds top and monkey the statement of the problem is the nozzles of the impulse stage of a turbine receive steam at 15 bar and 300 degree centigrade and discharge is at 8 10 bar the nozzle efficiency is 95 and the nozzle angle is 20 degrees the blade speed is that required for maximum work and the inlet angle of the blades is that required for entry of the steam without shock the blade exit angle is 5 degree less than the inlet angle the blade velocity coefficient is 0.9 calculate for a steam flow of 1350 kilogram per hour the diagram power in the diagram efficiency so first of all we will extract the given information in the statement so we have given the initial condition of the steam that is p1 15 bar this is the initial condition of the steam at inlet and temperature t1 is 300 degree centigrade while at state 2 when it discharges the pressure becomes 10 bar the nozzle efficiency is 95 so the nozzle efficiency is 95 percent this is the nozzle efficiency while the nozzle angle alpha i is 20 degrees the blade speed is is that required for entry of the steam without shock blade exit angle is 5 degree less than the inlet angle so beta e is equal to beta i minus 5 db blade co velocity coefficient k is 0.9 while the mass flow rate is 350 kilogram per hour we have to find the diagram power so two things are required in this problem first to find the diagram power and second we have to find the diagram efficiency efficiency represented by eta d so these two things are required in this problem so before starting the solution let's discuss the velocity triangle for this specific problem this is the absolute velocity at inlet and this is the relative velocity with respect to the blade at inlet this is blade speed or blade velocity cre is the relative velocity at exit with respect to the blade and cae is the absolute velocity at exit and the angles alpha i is the nozzle angle at the inlet beta e is the blade angle at exit and beta i is the blade angle at inlet from the this velocity triangle the angle oab if we take the sine o a b it is equal to 180 degree minus beta i and it we if we expand it we get that sine o av is equal to sine beta i this will be used later on in the solution of this problem so it is highlighted over here in the start similarly nozzle efficiency is equal to the actual in enthalpy drop which is h1 minus h2 and to the enthalpy isentropic enzyme enthalpy drop h1 minus h2s so nozzle efficiency expression is h1 minus h2 which is the actual implant that we drop and h1 minus h2s which is the isentropic enthalpy drop so coming to the solution of this problem so first of all from the condition of the steam at inlet at p1 15 bar and t1 300 degree c if we go to the steam tables first we visit the saturated steam tables from there we see that the saturation temperature is less than 300 degree c so it means that the steam is in superheated region so going to the superheated tables from there f 15 bar and 300 degrees c we will note the values of enthalpy so the value of enthalpy at 15 bar and 300 degree c is 3 0 3 9 kilo joule per kilogram similarly we will note the value of specific entropy and that specific entropy s1 is 6.919 joule per kilogram per kilogram similarly to find the value of h2s which we need here for that again visiting the saturated steam table from there we see that the specific entropy value 6.919 is greater than the value of the sg meaning that s1 is greater than sg from the saturated steam tables so at 10 bar and at this specific entropy value again we predict that the steam is in superheated form so again referring to the superheated tables from there we calculate the value of h2s and the value of h2s from the table is two nine four zero point five kilo joule per kilogram now taking the difference of these two values which is h1 minus h2s so h 1 minus h 2 s is equal to h 1 is 3 0 3 9 while h 2 s is 2 9 4 0 so subtracting it we get the value uh 40 and we get the value of by subtraction 99 kilo joule per kilogram now to find the actual enthalpy drop so for that we will use the nozzle efficiency expression so actually that we drop will be h1 minus h2 and it is equal to nozzle efficiency n2 h1 minus h2s so nozzle efficiency is 95 percent which is 0.95 and the value of h1 minus h2 is is 99 kilo joule per kilogram so from this we get the value which is three point six into ten wrist power three joule per kilogram or it is ninety three point six kilo joule per kilogram now to find the value of c r c a i which is the absolute velocity at inlet it is equal to from the nozzle analysis study we know that it is equal to 2 into h1 minus h2 putting the values square root of 2 n2 h1 minus h2 is basically 93.6 into 10 raised to power 3 so the value of c a i calculated from here is 432.7 meter per second so this is the value of cai which is the absolute velocity at n later later on this value will be used to find some other missing parameters are turns now for maximum work we have to find the blade velocity and the blade velocity for maximum work is equal to c a i cos alpha by 2 so putting the values cai is 432.7 cos of alpha i is 20 divided by 2 so simplifying this we get the value which is 203.3 meter per second so this is the blade velocity our blade speed now also to find another velocity there is relative velocity at inlet for that we will use the cosine law so from velocity triangles cosine is used c r i square is equal to c a i square plus cb square minus 2 cai cb cos alpha i we have used this expression again and again in other problem solution in the first two problems problem 11.1 and 11.2 now putting the values cai is 430 2.7 square plus 203 point 3 square minus 2 into 432.7 into 203.3 cos of 20 degree so simplifying this and solving it for the relative velocity at inlet c r i is equal to [Music] 251.5 meter per second also to find the blade angle for that we have that sine beta i is equal to c c a i by c r i sine alpha i actually here we implemented the cosine sine sorry the sine so from this we will find the angle beta i which is the blade angle at inlet so sine beta i is equal to cai is 2 0 4 4 3 2 0.7 divided by cri is calculated here 251.5 and sine alpha i is 20 degree so simplifying this and calculating it for beta i the value of beta i is equal to 36 point degree so now to find the relative velocity at exit for that we have the expression that c r e is equal to k c r i so k is known to us which is 0.9 and the value of cri is 251.5 meter per second so simplifying this we get 226.4 meter per second similarly the blade angle at exit is equal to blade angle at like my minus 5 degree is given in the problem so beta i is 36.05 degree minus 5 degree so the blade angle it exit comes out 31.05 degree similarly to find the change in world velocity which is delta cw we have that cri cos beta i plus c r e cos beta e this is expression for delta cw so putting the value cri is 2 51.5 meter per second cos beta i beta i is 36.05 degree plus cre is 2 2 6.4 meter per second and beta e is 31.05 degree so simplifying this and solving it for delta cw we get the value 397.3 397.3 meter per second now to find the diagram power now we are able to find the missing are the required parameters our terms so the diagram power is equal to m dot delta cw into cb so m dot is 1350 kilogram per hour to convert it into kilogram per second we will divide it by 3600 seconds also the value of delta cw is 397.3 meter per second and the value of blade velocity is calculated already let's check from the above value 203.3 meter per second so putting that value over here 203.3 meter per second so simplifying it we get the value of diagram power which is equal to 26 30.3 kilowatt 30.3 into 10 power 3 watt are 30.3 kilowatt so this is the value of diagram power required similarly to find another from there is the diagram efficiency so for that we have the expression that the diagram efficiency is equal to 2 delta c w into cb by c a i square so putting the values 2 n2 delta cw is 397.3 per second and cb is 203.3 divided by cai cai is 432.7 square so simplifying this we get the value which is 86.3 0.863 in percentage the diagram efficiency is equal to eighty six point three percent so this is the solution of problem eleven point three thanks for watching [Music] please like and comment on my videos also subscribe my channel to get new videos |
7972 | https://www.wardandsmith.com/articles/issues-with-a-piece-rate-compensation-system-for-construction-businesses | A Piece-Rate Compensation System and Overtime for Construction Businesses
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CAUTION! CONSTRUCTION ZONE: A Piece-Rate Compensation System Can Be An Effective Tool For Construction Businesses, But What About Overtime?
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CAUTION! CONSTRUCTION ZONE: A Piece-Rate…
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October 21, 2013
If your business involves construction, then you may be familiar with the concept of piece‑rate compensation. Under this system, employees are paid based on the number of units, or pieces, they complete, rather than on the number of hours they work. For example, a plumbing subcontractor may pay employees based on the number of sink faucets installed in an office building, rather than on the number of hours it took to install the faucets during a particular workweek. Although this is an entirely permissible and legal way to pay employees, and is quite common in the construction industry, it is not a cure‑all. Employees paid on a piece-rate basis are not exempt from the various requirements of the Fair Labor Standards Act ("Act"), including minimum wage, overtime, and record‑keeping obligations.
The Problems with a Piece‑Rate Compensation System
This unfortunate reality may come as a surprise to employers who have successfully used a piece-rate compensation system for years. After all, the entire concept behind piece‑rate compensation is that the time it takes an employee to complete a task is not relevant. This is generally a true statement so long as:
The employee does not work more than 40 hours in a workweek; and,
The employee's total compensation for the week averages at least the applicable minimum wage.
However, even assuming the above standards are satisfied, it still is essential for employers to maintain accurate daily and weekly time records for employees paid on a piece‑rate basis. Otherwise, without such records, the employer has no way to prove that the minimum wage and overtime pay requirements have been properly satisfied if a complaint is filed or government investigators come looking. To make matters worse, the method for properly calculating overtime compensation for employees compensated on a piece‑rate basis can become a mathematical nightmare for some employers. As you can imagine, a piece‑rate compensation system is a breeding ground for wage and hour problems, which can result in substantial penalties for violations.
Why Have a Piece‑Rate Compensation System?
At this stage, you may be asking yourself: "Why would an employer ever choose to pay piece‑rate?" Paying on a piece‑rate basis can be advantageous for both employees and employers. Employers find it attractive because the compensation scheme motivates employees to work more efficiently which, in turn, helps the employer's bottom line. In theory, piece‑rate compensation provides an employee the ability to earn more in less time than the employee would have earned with regular hourly pay. At the same time, the employer creates a work atmosphere that promotes and rewards productivity.
There are downsides, of course, one being that employees can be overly "incentivized" to work too quickly, which could jeopardize the quality of the job performed. And, for the employer, piece‑rate compensation does not come without a few administrative headaches, as discussed below.
Piece‑rate is not a new or novel compensation technique; however, in recent years, the U.S. Department of Labor has dedicated substantial resources to investigating, penalizing, and prosecuting employers who misuse the piece‑rate system. This undesired scrutiny has caused many employers to revisit how to properly and legally implement a piece‑rate payment program.
How to Make the System Work
So what must an employer do if it wants to pay its employees by the piece rather than by the hour? The Act contains three basic requirements to which an employer must adhere regardless of whether its employees are paid hourly or by the piece:
Minimum wage;
Overtime compensation; and,
Record‑keeping.
These basic requirements are summarized below.
Minimum Wage on Piece-Rate
Even under a piece‑rate compensation scheme, employers must pay employees at least the applicable minimum wage for each hour of work (currently, $7.25 per hour). This does not mean that employers are required to pay a certain rate per piece; rather, employers must ensure that at the conclusion of each workweek, all nonexempt employees have received at least $7.25 for each hour of work performed during the workweek.
As an example of the mathematics involved, suppose a piece‑rate employee was paid $10.00 for each faucet installed. In a workweek, the employee installed 38 faucets and, thus, received $380.00. It took the employee 40 hours to install the 38 faucets. By dividing the employee's total compensation for the week ($380.00) by the number of hours worked (40), the employee's effective hourly rate for the week was $9.50. Therefore, the employee received at least minimum wage for that particular workweek.
On the other hand, if the employee had worked 40 hours, but managed to install only 28 faucets and received $280.00 for that particular workweek, then the employee's effective rate for the week was only $7.00 per hour ($280.00 divided by 40). In that situation, the employer must make up the difference to ensure that the employee receives the minimum wage. Therefore, the employer must pay the employee an extra $10.00 to bring the employee's total compensation for the week to $290.00, which represents 40 hours of work at minimum wage.
Overtime Compensation on Piece‑Rate
Now, let's suppose that the nonexempt employee worked 45 hours during a particular workweek. Under the Act, irrespective of whether the employee is paid hourly or on a piece‑rate basis, the employer must pay the employee one and one‑half times the employee's regular rate of pay for all hours worked in excess of 40 hours in a workweek. Unlike regular hourly pay (where an employer simply multiplies the hourly rate by one and one-half), calculating overtime compensation for piece‑rate work can be somewhat tricky. Under a piece‑rate system, an employer must perform the following calculation for each workweek that the employee works over 40 hours.
First, similar to the minimum wage determination discussed above, the employer must determine the employee's effective hourly rate for the workweek based on the number of hours worked. It is important to understand that the piece‑rate is not the regular rate of pay; rather, for a piece‑rate employee, the regular rate of pay is calculated by taking the total compensation for the week (including piece‑rate compensation, as well as any bonuses or commissions) and dividing it by the total number of hours worked. In our above example, if the employee worked 45 hours during a workweek, installed 45 faucets, and earned $450.00, then the employee's effective regular rate of pay for that workweek was $10.00 per hour. This is sometimes referred to as the "straight time" pay.
Second, after calculating the regular rate of pay for the workweek, or straight time pay, the employer must determine the additional overtime compensation due. The key here is to remember that, in our example, the employee already received straight time pay of $10.00 per hour for each hour worked, including the hours of overtime. Thus, the employee is entitled to receive only an additional one‑half of the employee's regular rate of pay (not the full one and one‑half) for all hours worked in excess of 40.
In our example, the employee worked 45 hours and, thus, had five hours of overtime. Since the employee already received $10.00 per hour for all 45 hours, the employee is entitled to only an additional $5.00 (i.e., half of the employee's regular rate of pay) for each of the five hours of overtime. Therefore, for the week, the employee should receive overtime compensation of $25.00 on top of the $450.00 received as straight time compensation. In total, the employee is paid $475.00 for that particular week.
There is another, perhaps less complicated, way to pay overtime compensation when using a piece‑rate system; however, the employee must agree to this method before any work is performed. If the employer and employee agree beforehand, the employer may pay the employee one and one-half times the employee's regular piece‑rate for all pieces completed during overtime hours. Again, however, the minimum wage requirements are still present, meaning the employee must receive at least one and one‑half times the minimum wage ($7.25 + 50% ($3.63) = $10.88), or $10.88 per hour for each hour of overtime.
Record-Keeping Requirements
As you have probably gathered at this point, it is imperative for an employer to maintain an accurate record of the hours the employee actually works, regardless of how the employee is paid. These records are vitally important in ensuring that employees are properly paid under the Act. If an employer fails to keep accurate records, or worse, fails to keep records at all, then the above calculations would be impossible to perform.
Even more damaging, failing to create and retain accurate time and pay records will place an employer in the crosshairs of the U.S. Department of Labor. Without proper time records, an employer will be significantly handicapped in its efforts to defend itself against alleged violations of the Act. Although keeping records may be time‑consuming and tedious (and, in many instances, difficult, as a practical matter), the penalties for violating the Act are far too serious to ignore these responsibilities.
Conclusion
Don't become the Department of Labor's next victim. If you are paying your employees on a piece‑rate basis, make sure you continue to fulfill your obligations under the Act. If you are unsure of your obligations, consult legal counsel well‑versed on what the Act requires and how to comply.
--
© 2025 Ward and Smith, P.A. For further information regarding the issues described above, please contact Devon D. Williams.
This article is not intended to give, and should not be relied upon for, legal advice in any particular circumstance or fact situation. No action should be taken in reliance upon the information contained in this article without obtaining the advice of an attorney.
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7973 | https://stats.stackexchange.com/questions/312646/simulating-data-correlation-vs-causation | simulation - Simulating data - correlation vs causation - Cross Validated
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Simulating data - correlation vs causation
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I'm just dipping my toe into correlation vs causation, so forgive me if I butcher some of the concepts here.
To get a better understanding of these concepts from data, I would like to simulate data under a correlational assumption and a causational assumption. I'm not sure where to start, but here's where I'm headed.
For correlation, this is fairly easy.
Generate a sequence of normal random variables to serve as a basis.
Duplicate that sequence but add random normal noise to the variable - essentially produce a new sequence using aX + b, with a being a single random value, and b being a new sequence of small random normal values to introduce noise. The magnitude of a and b will determine the final correlation.
If I do this multiple times, I'll have a set of variables that are all correlated with the original sequence and will likely have correlation structure among each other. If I remove the original sequence, the remaining sequences will have some degree of pairwise correlation - the covariance matrix will be non-diagonal - but there will be no causation between them.
For causation, my thoughts were:
Choose a distribution (Normal, Gamma, Beta, ...)
For the parameters of that distribution, draw a random variable from another distribution - standard normal is probably fine.
Generate a sequence of random variables from the first distribution using the determined parameters determined from the second.
In my mind, this creates a causal relationship, but I'm sure I'm wrong. Regardless, the result is a data matrix that has some correlation - different variables from the same first distribution with parameters drawn from the second distribution may have a correlation but they are not related to each other causally.
I appreciate any comments/suggestions regarding how to approach this.
correlation
simulation
causality
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edited Nov 8, 2017 at 14:52
Tim
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asked Nov 8, 2017 at 14:47
KirkD_COKirkD_CO
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My understanding about this issue is fundamental but from what I read during my learning of directed acyclic graph and causal influence, when simulating causally related variables, we only also rely on correlation. In fact, data alone are insufficient, to support causality one will have to take study design and plausible mechanism into account as well.Penguin_Knight –Penguin_Knight 2017-11-08 14:54:00 +00:00 Commented Nov 8, 2017 at 14:54
To echo Penguin_Knight, the data itself are not causal or correlational on the basis of correlation alone. The real context in which the data were generated matters.RickyB –RickyB 2017-11-08 15:09:14 +00:00 Commented Nov 8, 2017 at 15:09
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All this is easier with a theory of causality. For example, let's use here the Structural Causal Models (which includes the Potential Outcomes) approach.
A Structural Causal Model (SCM) is triplet M=⟨V,U,F⟩M=⟨V,U,F⟩ where U U is a set of exogeneous variables, V V a set of endogenous variables and F F is a set of structural equations that determines the values of each endogenous variable. The structural equations are in the sense of assignments not equalities. For example, consider the simple structural equation Y=X 2 Y=X 2. This is meant to be read Y←X 2 Y←X 2, in the sense that if I experimentally set X=2 X=2 then this causally determines the value of Y=4 Y=4 but experimentally setting Y=4 Y=4 does nothing to X X. The asymmetry is important/fundamental in causality: rain causes the floor to be wet, but making the floor wet does not cause rain.
So our causal model can be thought as functional relationships among variables and we are considering these relationships as autonomous. You can think of it as an idealized representation of the real world, where the variables V V, the endogenous variables, are what we choose to model, and the variables U U are the aspects we chose to ignore. Since we chose not to model the U U, what we usually do is to represent our ignorance about U U with a probability distribution P(U)P(U) over the domain of U U, giving us a probabilistic SCM which is pair ⟨M,P(U)⟩⟨M,P(U)⟩. Notice this means that causal relationships are ultimately functional relationships, therefore causal relationships may or may not translate to specific probabilistic dependencies. Finally, every causal model can be associated with a directed (acyclic) graph G(M)G(M).
Hence, one way to simulate from a probabilistic causal model is by specifying: (i) the endogenous variables V V you are going to model; (ii) the exogenous variables U U which are usually the "disturbances", along with their joint probability distribution; and, (iii) the (causal) structural relationships among the variables. It might be easier to start this process qualitatively by first drawing the causal DAG with the main features that you want to illustrate and then add the details of the simulation (functional forms) later.
To see how this can be easily done in practice, let's simulate a simple causal model that illustrates simpson's paradox (for more see Pearl). Suppose our model M M is given by the following causal DAG, where the variables in parenthesis are "unobserved" and each variable has an associated exogenous disturbance U U which is omitted for convenience:
More specifically we will assume the following structural equations F F:
W 1 W 2 Z X Y=U W 1=U W 2=W 1+W 2+U Z=W 1+U x=X+10 W 2+U y W 1=U W 1 W 2=U W 2 Z=W 1+W 2+U Z X=W 1+U x Y=X+10 W 2+U y
Finally, assume all disturbances in U U are independent standard normal random variables. Now it's easy to simulate from our causal model. In R for instance:
rm(list = ls())
set.seed(1)
n <- 1e5
w1 <- rnorm(n)
w2 <- rnorm(n)
z <- w1 + w2 + rnorm(n)
x <- w1 + rnorm(n)
y <- x + 10w2 + rnorm(n)
This example is interesting because if you run the regression Y∼X Y∼X you get 1 1:
```
lm(y ~ x)
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
0.01036 1.00081
```
But if you further "control" for Z Z, which is a pre-treatment variable correlated with both X X and Y Y --- and some people still erroneously would say it's a confounder --- you will actually see a sign reversal of the estimate and get −1−1:
```
lm(y ~ x + z)
Call:
lm(formula = y ~ x + z)
Coefficients:
(Intercept) x z
0.00845 -1.01127 4.00041
```
In this example, since we simulated the data, we know the true causal effect is 1 1 which is captured by the first regression. But you can only know that if you know the true causal structure. There's nothing in the data itself that tells you which one is the correct answer. Hence if you simulate this and give to a researcher only the variables x x, y y and z z he can't tell the right answer just from looking at the correlations. If you want further play/simulate causal models with multi-stage Simpson's paradox reversals, you can check it here.
Simulating correlations/dependencies
To simulate correlations/dependencies you can take a similar approach. You can simply create a causal model that gives you the correlations/dependencies you want (adding latent variables if needed), simulate as above and the resulting data will have the desired correlations/dependencies. To make it easier, you can start by drawing the causal DAG (bayesian network) and read from the graph if the desired conditional dependencies/independencies are implied by your model. After that you might think of specific functional forms to get other quantitative aspects that you want. Notice that several models with different causal interpretations can give you the same correlations.
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edited Nov 9, 2017 at 17:36
answered Nov 9, 2017 at 3:42
Carlos CinelliCarlos Cinelli
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Thanks for the detailed post! I'll need to read it thoroughly to understand...KirkD_CO –KirkD_CO 2017-11-09 03:49:44 +00:00 Commented Nov 9, 2017 at 3:49
+1: Terrific answer, as always. One clarification question. Is it the case that when we do not condition on the collider y i=β 0+β 1 x i+ϵ i y i=β 0+β 1 x i+ϵ i, the assumption E(ϵ i|x i)=0 E(ϵ i|x i)=0 holds and we get the correct causal effect, but when we erroneously condition on the collider y i=β 0+β 1 x i+β 2 z i+ϵ i y i=β 0+β 1 x i+β 2 z i+ϵ i, we get the incorrect causal effect of x on y and therefore we can conclude that it must be that E(ϵ i|x i)≠0 E(ϵ i|x i)≠0? How can I see that E(ϵ i|x i)≠0 E(ϵ i|x i)≠0 in the second regression?ColorStatistics –ColorStatistics 2021-03-01 17:44:35 +00:00 Commented Mar 1, 2021 at 17:44
And more broadly, I am asking myself if every time regression does not correctly estimate the causal effect, it must be that E(ϵ i|x i)≠0 E(ϵ i|x i)≠0?ColorStatistics –ColorStatistics 2021-03-01 18:00:10 +00:00 Commented Mar 1, 2021 at 18:00
1 @ColorStatistics that's correct, in this particular model, not conditioning in anything gives the correct answer. Regarding knowing that the second regression is wrong, unfortunatelly this is impossible with the data alone. There's nothing on the data that tells us the second regression is wrong and the first regression is right.Carlos Cinelli –Carlos Cinelli 2021-03-05 20:31:59 +00:00 Commented Mar 5, 2021 at 20:31
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First, causality cannot be observed in the output. Only correlation.
You can imagine two functions that output exactly the same (x 1,x 2)(x 1,x 2) pairs (given the same random seed). Yet you could interpret one function as creating causally related pairs, the other not. It depends on how the code is written (or how you interpret it), not what it actually computes as a final result.
Causality as defined with DAGs requires the possibility of intervention on one of the two variables at least as a thought experiment. Imagine it as a debugger: you interrupt your program just after the first variable was computed, reset it to a certain value, and then restore execution. Will this impact the second variable?
In the way you explain the process (second case) call:`
A A: choice for the distribution
B B: choice for its parameters
(X 1,X 2)(X 1,X 2): final output
The DAG is :
There is no causal relationship between X 1 X 1 and X 2 X 2.
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answered Nov 8, 2017 at 17:39
Benoit SanchezBenoit Sanchez
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exactly. X1 and X2 are likely correlated, but not causally related. However, A and B are causally related to X1 and X2, correct? Your graphical model looks like a Bayesian Net idea, which is consistent with my question.KirkD_CO –KirkD_CO 2017-11-08 21:07:20 +00:00 Commented Nov 8, 2017 at 21:07
1 Yes. The graph here is a "direct acyclic graph." (DAG). You can read more about it here: stats.stackexchange.com/questions/45999/… or in Carlos answer. SCM and DAG are equivalent ways to define causality (as far as I know). Actually those theories are rather modern and still presented in several different ways but importantly they show how causality can't be defined by probabilities alone but can be studied together with probabilities. The graph + the joint distribution is the full causal model.Benoit Sanchez –Benoit Sanchez 2017-11-09 11:05:46 +00:00 Commented Nov 9, 2017 at 11:05
I'm disappointed that I cannot vote both of these as answers. Both have been truly helpful.KirkD_CO –KirkD_CO 2017-11-11 22:18:49 +00:00 Commented Nov 11, 2017 at 22:18
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Hardy-Weinberg Equilibrium Calculator
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This genetics calculator determines allele frequencies and expected genotype frequencies under Hardy-Weinberg equilibrium. It also calculates the chi-square statistic to test whether a population is in genetic equilibrium.
Hardy-Weinberg Frequency and Chi-Square Test
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Hardy-Weinberg Equilibrium and Chi-Square Formula
Where:
$$CH$$ = number of homozygous dominant individuals
$$H$$ = number of heterozygous individuals
$$RH$$ = number of homozygous recessive individuals
$$N$$ = total individuals = CH + H + RH
$$p$$, $$q$$ = allele frequencies
$$E_{CH}$$, $$E_H$$, $$E_{RH}$$ = expected genotype counts
$$\chi^2$$ = chi-square value for deviation from equilibrium
The Hardy-Weinberg principle predicts genotype frequencies in a population under ideal conditions (no mutation, migration, selection, or genetic drift). This calculator allows you to input observed counts of dominant, heterozygous, and recessive individuals to determine allele frequencies (p and q), expected genotype frequencies, and whether the population deviates from equilibrium using a chi-square test. It is useful in population genetics, evolutionary biology, and academic research.
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10 thoughts on “Hardy-Weinberg Equilibrium Calculator”
Asmaa says:
16.06.2025 at 08:25
how can I know if there is deviation from HWE or not ?
Reply
2. Yurany Granada says:
09.06.2021 at 23:19
Buenas tardes, quiero saber como debo citarlos. Gracias
Reply
3. Adriana Mena says:
07.06.2021 at 16:16
How are the df calculated?
Reply
Monica says:
04.11.2022 at 14:15
df = number of classes compared – no. of parameters estimated – 1.
4. Oladejo Aliu Abiola says:
03.04.2021 at 15:10
How do you calculate the expected value for the alleles
Reply
5. Daniel says:
28.07.2020 at 18:51
Si una enfermedad es recesiva y la frecuencia del alelo que la genera es de0,08, ¿cómo serán las frecuencias genotípicas?
Reply
6. Nada says:
25.07.2020 at 17:21
I have one question please, the equation was as follows;ECH = pfreq pfreq (CH + H + RH)yet you have used the frequency of the rare allele in the following example ,Step 3 : Expected CH
Expected CH =
= 0.78 0.2121 (21 + 10 + 2)
can you explain why, please?Plus,what do you mean exactly by the following;chi2 = chi1 + chi2 + chi3
Reply
7. Aaron Castle says:
02.06.2020 at 18:56
I love you, thank you for saving my grade in bio and letting me keep an A. Thank you!
Reply
8. hamzah H kzar says:
24.03.2020 at 21:28
thank you so much to helping me by your this website
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9. kamna srivastava srivastava says:
24.09.2019 at 10:30
very useful. with you examples anybody can learn easily and population genetics.
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Home/Blog - Carpathen/Signs of overwatering plants - how to identify and fix overwatered plants
Signs of overwatering plants - how to identify and fix overwatered plants
Veronica Sticlaru on 04/03/2025
Overwatering is a prevalent issue that can severely impact plant health, affecting both indoor and outdoor plants. It occurs when plants receive more water than they can effectively use or drain, leading to waterlogged soil and oxygen-deprived roots. This problem is particularly common among novice gardeners who may mistakenly believe that more water equates to better care. However, excessive watering can be just as harmful as underwatering, if not more so.
The primary causes of overwatering include poor drainage, frequent watering without allowing the soil to dry between sessions, and using containers without adequate drainage holes. Environmental factors such as high humidity, low light, and cool temperatures can also contribute to overwatering by reducing the plant's water uptake and evaporation rates. These conditions are often present in indoor environments, making houseplants particularly susceptible to overwatering. Recognizing the signs of overwatering and understanding how to address this issue is essential for any gardener or plant enthusiast seeking to cultivate healthy, vibrant plants.
In this article, we will discuss the key signs of overwatering, their impact on plant health, and effective strategies to prevent and remedy this common issue.
Table of Contents
Understanding the effects of overwatering and its impact on plants
Common signs of overwatering plants
Yellowing or wilting leaves
Root rot and soil issues
How to tell if a plant is overwatered or underwatered?
Steps to save an overwatered plant
Understanding the effects of overwatering and its impact on plants
When plants receive excessive water, a cascade of negative effects compromises their health and vitality. The primary damage occurs at the root level, where waterlogged soil displaces oxygen from the spaces between soil particles. This lack of oxygen creates an anaerobic environment that prevents roots from respiring properly, leading to suffocation and eventual decay. Known as hypoxia, this condition can rapidly deteriorate the entire root system.
As root function declines, the plant struggles to absorb essential nutrients, resulting in visible symptoms such as yellowing leaves (chlorosis), stunted growth, and premature leaf drop. The weakened state of overwatered plants also makes them more vulnerable to pests and diseases. Moist conditions encourage the growth of pathogens, especially soil-borne fungi that cause root rot.
Surprisingly, overwatering can mimic the symptoms of underwatering. Damaged roots lose their ability to transport moisture effectively, causing leaves to wilt despite the presence of excess water in the soil. This phenomenon, known as physiological drought, can also manifest as leaf edema (blisters or lesions) and necrosis at the leaf tips and margins, appearing as brown or yellow discoloration.
Prolonged exposure to excessive moisture can have lasting effects on plant development. Growth may slow, stems may become weak, and flowering or fruiting can be significantly reduced. Additionally, essential nutrients may leach from the soil, exacerbating deficiencies and further weakening the plant.
Common signs of overwatering plants
Recognizing the indicators of overwatering is crucial for maintaining healthy plants. Here are some key signs that your plants may be receiving too much water:
Yellowing leaves, particularly on the lower parts of the plant, are one of the most common signs of overwatering plants. This discoloration occurs due to nutrient deficiencies caused by root damage and is often referred to as chlorosis.
Paradoxically, overwatered plants may also wilt, despite having plenty of moisture. The leaves often feel soft and limp to the touch, unlike the crisp texture of underwatered plants. This wilting is a result of the plant's inability to transport water effectively due to root damage.
Root rot is a serious condition that can be fatal to plants. Signs include mushy, dark-colored roots, a foul odor emanating from the soil, and stunted growth or sudden decline in plant health. The roots may appear brown or black instead of the healthy white or tan color.
Edema, where cells in the leaves burst due to excess water absorption, creates raised, watery blisters on leaves or stems. These blisters may eventually turn into corky, brown spots.
Consistently wet soil creates an ideal environment for fungal growth. You may notice white, fuzzy mold on the soil surface, green algae growth, or the presence of fungus gnats. These pests are attracted to the moist conditions and can further damage plant roots. Overwatered plants may also shed both old and new leaves at an accelerated rate as the plant attempts to reduce water uptake and transpiration. This leaf drop is a stress response to the excess water.
Brown leaf tips and edges, while often associated with underwatering, can also indicate overwatering, especially when combined with other symptoms like soft, yellowing leaves. This browning occurs due to the accumulation of salts and minerals in the leaf tips, which cannot be properly transported through the damaged root system. Plants that are consistently overwatered may exhibit slow or stunted growth as their root systems struggle to function properly in waterlogged soil. This reduced growth is a result of the plant diverting energy to survival rather than new growth.
Yellowing or wilting leaves
As we explained in a concise form before, yellowing or wilting leaves are prominent indicators of overwatering. While it may seem counterintuitive, an overwatered plant can display symptoms similar to underwatering. The key difference lies in the texture and appearance of the affected foliage.
When a plant receives excessive water, its leaves often turn yellow, particularly on the lower parts. This discoloration occurs because the waterlogged roots cannot effectively absorb nutrients, leading to deficiencies that manifest as yellowing. The process, known as chlorosis, begins with the older leaves at the bottom of the plant and progresses upward. For instance, the leaves may appear pale or have a washed-out look, starting from the interveinal areas and spreading to the entire leaf.
Wilting is another telltale sign of overwatering. Despite the abundance of moisture in the soil, the plant's roots struggle to take up water efficiently due to oxygen deprivation. As a result, the leaves may droop and look limp, even when the soil is visibly wet. This wilting can be particularly confusing for gardeners, as it mimics the appearance of a plant in need of water. The phenomenon, known as physiological drought, occurs because the damaged roots cannot supply water to the upper parts of the plant effectively.
To distinguish between overwatering and underwatering, pay attention to the leaf texture. Overwatered plants have leaves that feel soft and limp to the touch, while underwatered plants have leaves that feel crisp and dry. If you notice yellowing or wilting leaves accompanied by consistently moist soil, it's likely that your plant is suffering from overwatering. Additionally, overwatered plants may develop water-soaked spots or blisters on their leaves, a condition known as edema. These blisters occur when cells take up more water than they can process, causing them to rupture and form corky lesions as they heal.
Root rot and soil issues
Root rot is one of the most serious consequences of overwatering plants and can be fatal if left untreated. It occurs when roots are constantly saturated, depriving them of oxygen and creating an ideal environment for fungal pathogens to thrive. Key indicators of root rot include mushy, dark-colored roots that may have a foul odor, stunted growth or sudden decline in plant health, and wilting despite moist soil. The roots affected by rot will appear brown or black, in contrast to healthy roots which are typically white or light-colored.
In addition to root rot, overwatering can lead to other soil-related issues. These include compacted, waterlogged soil that lacks proper aeration, growth of mold or algae on the soil surface, and the presence of fungus gnats, which thrive in consistently wet conditions. Also, the soil may develop a sour or musty smell, indicating anaerobic decomposition processes. This anaerobic environment not only harms the roots but also alters the soil microbiome, potentially leading to long-term soil health issues.
To identify these problems, carefully remove the plant from its pot and examine the root system and soil. Healthy roots should be firm and light-colored, while the soil should be moist but not saturated. If you notice any signs of overwatered plants, such as soggy soil or discolored roots, immediate action is necessary to save the plant. The soil should have a balanced moisture content, neither waterlogged nor completely dry, and should not emit any unpleasant odors.
Addressing root rot and soil issues often involves trimming away affected roots with clean, sharp tools, repotting the plant in fresh, well-draining soil, and adjusting watering practices to prevent future overwatering. It's crucial to improve soil drainage by adding materials like perlite or coarse sand to the potting mix. Using nursery pots with proper drainage holes can also help prevent waterlogging. When repotting, ensure that the new container is not significantly larger than the previous one, as excess soil can retain more moisture than the plant needs.
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By recognizing these overwatered plant problems early and taking appropriate action, you can often rescue affected plants and prevent further damage to your garden or indoor plant collection.
How to tell if a plant is overwatered or underwatered?
Determining whether a plant is overwatered or underwatered can be challenging, as both conditions can produce similar symptoms. However, there are key differences to look out for:
Signs of overwatering can be traced by identifying yellowing leaves, especially on the lower parts of the plant, wilting despite moist soil, soft, limp leaves that feel mushy, brown spots or edges surrounded by yellow halos on leaves, mold or algae growth on soil surface, and fungus gnats flying around the plant. The soil may feel consistently wet or waterlogged, and there might be a musty odor emanating from the pot. Overwatered plants may also exhibit slow growth or leaf drop as the plant struggles to cope with excess moisture.
On the other hand, signs of underwatering include crisp, dry leaves, particularly at the edges, wilting with dry soil, slow growth or leaf drop, brown, brittle leaf tips, and compact, hard soil. Underwatered plants may also have leaves that curl inward or develop a dull, lackluster appearance. The soil will feel dry and may pull away from the sides of the pot. Underwatered plants often have a more uniform pattern of wilting compared to overwatered plants.
To differentiate between an overwatered plant and an underwatered one, check soil moisture by inserting your finger about an inch into the soil. If it feels soggy or waterlogged, it's likely overwatered. If it's bone dry, it's underwatered. Assess leaf texture: overwatered leaves feel soft and limp, while underwatered leaves are crispy and dry. Observe root health by carefully removing the plant from its pot. Overwatered roots appear dark and mushy, while underwatered roots are dry and brittle.
Consider recent care and reflect on your watering habits and environmental conditions to determine which scenario is more likely. Factors such as pot size, soil type, and plant species can all influence water requirements. For example, using planters with proper drainage can help prevent overwatering issues. Consequently, pay attention to the plant's overall appearance and growth rate, as overwatered plants often show signs of stress throughout the entire plant, while underwatered plants may have more localized symptoms.
By carefully observing these signs and considering the plant's recent care, you can accurately diagnose whether your plant is suffering from too much or too little water and take appropriate action to restore its health. Remember that consistent monitoring and adjusting your care routine based on your plant's specific needs is key to maintaining optimal plant health.
Steps to save an overwatered plant
If you've identified signs of overwatering in your plants, there are several steps you can take to save them:
First, stop watering immediately. This is the most crucial step. Allow the soil to dry out significantly before considering any further watering. This may take several days to a week, depending on the severity of overwatering and environmental conditions. During this time, monitor the plant closely for any changes in its condition.
Next, improve drainage. Ensure your plant has proper drainage. If it's in a pot, check that there are drainage holes at the bottom. You may need to repot the plant in a container with better drainage or add materials like perlite or coarse sand to improve soil aeration. For outdoor plants, consider using raised garden beds to enhance drainage and soil quality. Raised beds allow for better control over soil composition and can significantly reduce the risk of waterlogging.
Photo source: Carpathen.com
Carefully remove the plant from wet soil. Gently extract the plant from its current pot or soil. Shake off excess soil and inspect the roots thoroughly. Healthy roots should be firm and white, while overwatered roots may appear brown, black, or mushy. This step allows you to assess the extent of root damage and take appropriate action. Be gentle during this process to avoid further stress to the plant.
Trim damaged roots using clean, sharp scissors or pruning shears. Remove any rotted or damaged roots. This helps prevent further decay and allows the plant to focus energy on healthy root growth. Be sure to sterilize your tools before and after use to prevent the spread of pathogens. However, get rid only of the clearly damaged portions, as excessive pruning can shock the plant.
Repot in fresh, well-draining soil. Choose a pot with adequate drainage holes and fill it with a fresh, well-draining potting mix. Avoid reusing old soil, as it may contain harmful pathogens. When repotting, ensure the plant sits at the same depth as it was previously. Consider adding a layer of gravel or small stones at the bottom of the pot to further improve drainage. Select a pot that is appropriate for the plant's size, as an oversized pot can retain excess moisture.
Prune affected foliage to help the plant recover. Remove any yellowed, wilted, or damaged leaves. This allows the plant to direct its energy towards recovery rather than maintaining compromised foliage. Use clean, sharp pruning tools to make clean cuts and avoid tearing the plant tissue. Be conservative in your pruning, removing only the most affected parts to maintain as much healthy foliage as possible.
Adjust environmental conditions to support recovery. Place the plant in a location with bright, indirect light and good air circulation. Avoid direct sunlight, which can stress an already weakened plant. Maintain a consistent temperature and avoid drafts. If possible, increase humidity around the plant to support recovery, but avoid misting the leaves directly. Consider using a humidity tray or a small humidifier if the air is particularly dry.
Photo source: Carpathen.com
Monitor and adjust watering habits as your plant recovers. Be vigilant about watering. Only water when the top inch of soil feels dry to the touch. Use the finger test or a moisture meter to assess soil moisture in the future. For outdoor plants, consider implementing adrip irrigation system to ensure consistent and controlled watering. Adjust watering frequency based on factors such as light exposure, temperature, and humidity. Observe the plant's response to your new watering routine and make adjustments as necessary.
In conclusion, recovery from overwatering requires patience and consistent care. Regularly monitor your plants for signs of stress and adjust watering habits accordingly to create a healthier growing environment. Ensuring proper drainage, maintaining balanced humidity levels, and tailoring care to each plant’s specific needs will help prevent future issues. With the right attention, many overwatered plants can recover and regain their vitality, allowing them to thrive once again.
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7976 | https://ocw.mit.edu/courses/18-s096-matrix-calculus-for-machine-learning-and-beyond-january-iap-2023/ocw_18s096_lecture05-part1-new_2023jan27_transcript.pdf | Page 1/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player [SQUEAKING]¬ 1 [RUSTLING]¬ 2 [CLICKING]¬ 3 STEVEN G. JOHNSON:¬ 4 OK, so last time¬ 5 I talked about how in¬ 6 order to define a gradient,¬ 7 you need an inner product.¬ 8 So that way, if you have a¬ 9 scalar function of a vector,¬ 10 the gradient is defined--¬ 11 basically the¬ 12 derivative has to be¬ 13 a linear function that takes¬ 14 a vector in and gives you¬ 15 a scalar out.¬ 16 So it turns out this has to¬ 17 be-- if you have a dot product,¬ 18 this has to be a dot product¬ 19 of something with the x.¬ 20 And we call the gradient.¬ 21 So the gradient is the¬ 22 thing with the same shape¬ 23 as x that we take the¬ 24 dot product with the x¬ 25 to get the f.¬ 26 So what I didn't mention is¬ 27 that, in fact, not only did¬ 28 we need a dot product¬ 29 to define a gradient,¬ 30 actually we swept something¬ 31 under the rug earlier.¬ 32 We actually need a norm in order¬ 33 to even define a derivative¬ 34 in the first place.¬ 35 All right.¬ 36 If you have a¬ 37 vector space, a norm¬ 38 is some measure of the¬ 39 length of the vector¬ 40 or a measure of distance.¬ 41 A norm takes in a vector v¬ 42 and gives you out a scalar.¬ 43 And technically, to¬ 44 qualify as a norm,¬ 45 this map has to be non-negative.¬ 46 It can't be negative.¬ 47 It can only be 0 if v is 0.¬ 48 If you multiply¬ 49 the vector by 2, it¬ 50 Page 2/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player has to multiply the length by 2.¬ 51 Or if you multiply the¬ 52 length by negative 2,¬ 53 it has to multiply the length by¬ 54 2, basically the absolute value¬ 55 of any scalar.¬ 56 It has to satisfy something¬ 57 called a triangle inequality.¬ 58 So usually, most commonly¬ 59 we're going to get a norm just¬ 60 from an inner product.¬ 61 So once you define¬ 62 an inner product--¬ 63 we talked about those last time.¬ 64 You can even define inner¬ 65 products of matrices--¬ 66 you get a norm for free.¬ 67 You can take the norm¬ 68 is just a square root¬ 69 of the inner¬ 70 product with itself.¬ 71 And this satisfies¬ 72 all these properties¬ 73 for any inner product.¬ 74 So the reason I mention¬ 75 this-- oh, and, by the way,¬ 76 just cultural note.¬ 77 So if you have a¬ 78 continuous vector space¬ 79 with an inner product, we¬ 80 call that a Hilbert space.¬ 81 If you have a continuous¬ 82 vector space with a norm,¬ 83 that's called a Banach space.¬ 84 So it's a fancy-sounding¬ 85 name, but it just¬ 86 means you have a norm.¬ 87 ALAN EDELMAN: You can¬ 88 impress your friends¬ 89 with your fancy mathematics,¬ 90 but that's all it is.¬ 91 STEVEN G. JOHNSON: Yes.¬ 92 So the reason I¬ 93 wanted to mention this¬ 94 is that really the definition¬ 95 of the derivative that we¬ 96 used earlier implicitly¬ 97 requires us to have a norm.¬ 98 So it actually is both¬ 99 the input and the output.¬ 100 Page 3/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player So it really only¬ 101 applies to Banach spaces.¬ 102 So the reason for that¬ 103 is remember I define¬ 104 the derivative to start with.¬ 105 If you look at the¬ 106 change in the output,¬ 107 f of x plus delta¬ 108 x minus f of x,¬ 109 for not an infinitesimal but¬ 110 a finite delta x that may be¬ 111 small, remember that we defined¬ 112 the derivative as the linear¬ 113 part, as the linear operator¬ 114 that gives the change to first¬ 115 order, which means we¬ 116 dropped any term that's--¬ 117 we called it little o of¬ 118 delta x-- any term that¬ 119 goes to 0 faster than delta x.¬ 120 So any term that's¬ 121 small compared to this.¬ 122 But in order to define¬ 123 what it means to be small,¬ 124 you need a norm.¬ 125 If I have two vectors,¬ 126 a column vector,¬ 127 and I want to say is this column¬ 128 vector smaller than that column¬ 129 vector, how do I check it?¬ 130 I check the length.¬ 131 You need to map it¬ 132 to a real number¬ 133 to get a distance¬ 134 or a smallness.¬ 135 So formally, the definition¬ 136 of this little o dx¬ 137 is basically any function¬ 138 such that the norm¬ 139 of this over the norm¬ 140 of delta x goes to 0,¬ 141 as delta x goes to 0.¬ 142 And in fact, even¬ 143 to define a limit,¬ 144 you need a norm of delta¬ 145 x because if you've taken¬ 146 [INAUDIBLE],, there's this¬ 147 epsilon delta meaning¬ 148 of a norm--¬ 149 of a limit, sorry.¬ 150 Page 4/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player You can make this¬ 151 arbitrarily small.¬ 152 You can make this less¬ 153 than or equal to epsilon¬ 154 for all epsilon greater than 0.¬ 155 And I'm not going to go¬ 156 through the definition.¬ 157 If you've seen the¬ 158 definition of a limit,¬ 159 there's some absolute values¬ 160 in there that for vector spaces¬ 161 have to turn into norms.¬ 162 But basically it's just--¬ 163 ALAN EDELMAN: My¬ 164 experience is everyone's¬ 165 seen the definition of¬ 166 delta and epsilon limits,¬ 167 and no one really¬ 168 understands it.¬ 169 STEVEN G. JOHNSON: Yeah.¬ 170 ALAN EDELMAN: Is that fair?¬ 171 Maybe some of you guys really¬ 172 do, but most of us don't.¬ 173 STEVEN G. JOHNSON: Yeah,¬ 174 I mean, to be fair,¬ 175 it took people 2,000¬ 176 years to figure it out.¬ 177 The concept of a limit¬ 178 and an infinitesimal¬ 179 was a big struggle¬ 180 in mathematics¬ 181 going back to the ancient¬ 182 Greeks, Zeno's paradoxes¬ 183 and so forth.¬ 184 So it really took a¬ 185 long time for people¬ 186 to nail down what this meant.¬ 187 But yeah, you need¬ 188 to be able to have¬ 189 a length, a norm of the¬ 190 output, because this has¬ 191 the same shape as the output.¬ 192 These are the same shape as f.¬ 193 To say that these terms¬ 194 are small compared to delta¬ 195 x, which you also need¬ 196 a norm of delta x.¬ 197 So just implicitly,¬ 198 you always need¬ 199 a norm of all of these¬ 200 Page 5/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player things to define it.¬ 201 And usually, we're¬ 202 going to get it¬ 203 because we're going to define--¬ 204 in most cases, we'll¬ 205 define an inner product,¬ 206 as we'll want that¬ 207 anyway because if we want¬ 208 to take derivatives¬ 209 of scalar functions,¬ 210 we want to be able to¬ 211 write down gradients.¬ 212 But this is what¬ 213 you really need.¬ 214 So anyway, so I just wanted to--¬ 215 this is something we swept¬ 216 under the rug in the beginning.¬ 217 But since we defined¬ 218 Hilbert spaces,¬ 219 so I thought I should¬ 220 define a Banach space.¬ 221 I mean, I'm still sweeping¬ 222 some things under the rug.¬ 223 I'm sweeping what¬ 224 does it mean for it¬ 225 to be continuous under the rug?¬ 226 But yeah, I wanted to¬ 227 throw that out there.¬ 228 That's all I wanted to say.¬ 229 ALAN EDELMAN: That's it?¬ 230 STEVEN G. JOHNSON: Yeah.¬ 231 Any questions about that?¬ 232 ALAN EDELMAN: Questions?¬ 233 By all means.¬ 234 OK, good.¬ 235 All right.¬ 236 So this is just a¬ 237 little notebook.¬ 238 And if we really need--¬ 239 this isn't the live version,¬ 240 so I can't really do anything.¬ 241 But I have a feeling that¬ 242 this will be good enough.¬ 243 But if we need the¬ 244 live version, we¬ 245 can just press a few buttons.¬ 246 So if I understood¬ 247 correctly, last time¬ 248 you got the answer for¬ 249 what is the gradient¬ 250 Page 6/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player of the determinant.¬ 251 Is that right?¬ 252 Did you derive this formula?¬ 253 STEVEN G. JOHNSON:¬ 254 I did not derive it.¬ 255 I just gave the answer.¬ 256 ALAN EDELMAN: You¬ 257 gave the answer.¬ 258 And there's a few¬ 259 different formats.¬ 260 Did you--¬ 261 STEVEN G. JOHNSON: I¬ 262 give it the first one.¬ 263 Determinant A--¬ 264 ALAN EDELMAN: Is the cofactor.¬ 265 STEVEN G. JOHNSON:¬ 266 --inverse transpose, yeah.¬ 267 ALAN EDELMAN: Oh,¬ 268 the gradient is¬ 269 the determinant of A times--¬ 270 that's the second one, right?¬ 271 STEVEN G. JOHNSON: Yeah.¬ 272 ALAN EDELMAN: So the¬ 273 first one is the cofactor¬ 274 of A, which is one of¬ 275 those linear algebra terms¬ 276 that you may or¬ 277 may not remember.¬ 278 This is the one.¬ 279 And just another¬ 280 term is the adjugate¬ 281 of A transpose, which is the--¬ 282 the adjugate of a¬ 283 matrix is the inverse¬ 284 of the matrix divided¬ 285 by the determinant.¬ 286 STEVEN G. JOHNSON: Multiplied¬ 287 by the determinant.¬ 288 ALAN EDELMAN: Multiplied¬ 289 by the determinant, right.¬ 290 Let's see.¬ 291 Of course, if you have¬ 292 the gradient, then--¬ 293 did you write down this¬ 294 version as well last time?¬ 295 The d of the determinant.¬ 296 STEVEN G. JOHNSON: Well, I did d¬ 297 of any matrix function, so yes.¬ 298 I defined the dot product¬ 299 and matrix dot product.¬ 300 Page 7/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player ALAN EDELMAN: Right, right.¬ 301 I see.¬ 302 So d, the determinant, will be¬ 303 the trace of whatever formula¬ 304 you have over here,¬ 305 this formula times dA.¬ 306 dA.¬ 307 STEVEN G. JOHNSON: Transposed.¬ 308 Transposed times it, yeah.¬ 309 ALAN EDELMAN: Yes, right, sorry.¬ 310 STEVEN G. JOHNSON: [INAUDIBLE]¬ 311 ALAN EDELMAN: Let¬ 312 me clear my head.¬ 313 Yes, the trace of this¬ 314 thing transposed times dA,¬ 315 which is the element-wise--¬ 316 it's just like the dot product.¬ 317 You all get that.¬ 318 It's just like the¬ 319 vector dot product.¬ 320 You multiply corresponding¬ 321 elements, and you take the sum.¬ 322 Whenever you have the¬ 323 trace of A transpose B,¬ 324 as Steven is writing very nicely¬ 325 over here, that's the A dot B.¬ 326 So if we know the gradient, then¬ 327 the d has to be this formula.¬ 328 I'm just defining the¬ 329 adjugate right here so¬ 330 that I can have it handy.¬ 331 As Steven was saying,¬ 332 it's just the determinant¬ 333 times the inverse.¬ 334 This is just a definition.¬ 335 And then there's the¬ 336 cofactor matrix, which¬ 337 is the adjugate of A transpose.¬ 338 Once I've defined this¬ 339 one, to define this one,¬ 340 I just get to do the equality.¬ 341 So this defines these functions.¬ 342 And here I've sort of¬ 343 written it every which way.¬ 344 The inverse in terms of the¬ 345 adjugate and the cofactor,¬ 346 the adjugate in terms¬ 347 of the determinant,¬ 348 the inverse and the cofactor.¬ 349 You get it.¬ 350 Page 8/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player All three possibilities¬ 351 are written here.¬ 352 So for 2-by-2 matrices,¬ 353 here's the 2-by-2 matrix,¬ 354 and here's the cofactor matrix.¬ 355 Some of you will¬ 356 recognize that when¬ 357 you form the inverse¬ 358 of a 2-by-2 matrix,¬ 359 the determinant goes¬ 360 in the denominator.¬ 361 And the thing that goes¬ 362 in the numerator-- right,¬ 363 you're all good at¬ 364 2-by-2 inverses.¬ 365 Do you know that by heart?¬ 366 Would you be able to¬ 367 do it in your sleep?¬ 368 You switch the a and the d,¬ 369 and you negate the b and the c.¬ 370 Well, let's see.¬ 371 You negate the b and the c, but¬ 372 I'm also doing the transpose.¬ 373 So you negate the b¬ 374 and c and transpose¬ 375 because it's the cofactor.¬ 376 For the adjugate, you¬ 377 just take the minus.¬ 378 And anyway, these¬ 379 are all the formulas.¬ 380 Here's the inverse.¬ 381 So the inverse is the adjugate¬ 382 divided by the determinant.¬ 383 Doing all this¬ 384 numerically just for fun.¬ 385 So numerically, here's¬ 386 a random matrix,¬ 387 and here's a random¬ 388 perturbation.¬ 389 What we're going to do is¬ 390 look at the determinant¬ 391 of the perturbed A minus¬ 392 the determinant of A.¬ 393 So there's the numerical value.¬ 394 And by the way, I know¬ 395 Steven has recommended always¬ 396 using things on the order¬ 397 of square root of epsilon¬ 398 to make the perturbations¬ 399 10 to the minus eighth.¬ 400 Page 9/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player And he's right.¬ 401 I never do that, but you¬ 402 should listen to him.¬ 403 I just start typing three¬ 404 or four 0's and a 1.¬ 405 And actually, it's been good¬ 406 enough for my purposes just¬ 407 to check things.¬ 408 I mean, Steven's is more--¬ 409 it's the best possible one,¬ 410 a square root of epsilon.¬ 411 But with a quick¬ 412 and dirty test, I¬ 413 don't have the time to type¬ 414 all those 0's, and I never¬ 415 remember to type 1e minus 8.¬ 416 So I just type these four¬ 417 or five 0's or three, four,¬ 418 or five 0's.¬ 419 But in any event, here's¬ 420 what the finite difference.¬ 421 Here's the trace of¬ 422 the adjugate times.¬ 423 We see that they're correct to¬ 424 enough digits to believe it.¬ 425 STEVEN G. JOHNSON: How¬ 426 come the adjugate is not¬ 427 transposed there?¬ 428 There's something missing here.¬ 429 Oh, no, it's the¬ 430 adjugate-- yeah, OK, right.¬ 431 The determinant is the¬ 432 transpose of the adjugate.¬ 433 Never mind.¬ 434 Never mind.¬ 435 Adjugate is the great--¬ 436 ALAN EDELMAN: I have to go¬ 437 back and look at these formulas¬ 438 to answer your--¬ 439 STEVEN G. JOHNSON:¬ 440 It's the transpose¬ 441 of the gradient, so yes.¬ 442 ALAN EDELMAN: Let me¬ 443 say, yes, what you just¬ 444 said, that the adjugate¬ 445 of the transpose¬ 446 is the thing that you want.¬ 447 And so the trace needs¬ 448 to transpose it twice,¬ 449 so it's left non-transposed.¬ 450 Page 10/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player Yeah.¬ 451 You got it.¬ 452 The gradient is¬ 453 the one transposed,¬ 454 and this has to be the¬ 455 transpose of the gradient.¬ 456 STEVEN G. JOHNSON: Yeah.¬ 457 This is gradient of determinant.¬ 458 ALAN EDELMAN: Right,¬ 459 like a double negative¬ 460 makes a positive, a double¬ 461 transpose makes for a no op.¬ 462 STEVEN G. JOHNSON: This¬ 463 is our dot product.¬ 464 ALAN EDELMAN: Yep.¬ 465 That's right.¬ 466 OK.¬ 467 So to actually see¬ 468 the gradient, we¬ 469 can rely on Julia's¬ 470 internal forward difference¬ 471 mode, for example, which is--¬ 472 forward differentiation.¬ 473 It's not forward difference.¬ 474 It's automatic differentiation.¬ 475 It's different from¬ 476 forward differencing.¬ 477 It's the forward¬ 478 mode automatic--¬ 479 I see this, and I think forward¬ 480 differences, but it's not.¬ 481 I think in this¬ 482 lecture, if I get¬ 483 a chance in just¬ 484 a little bit, I'll¬ 485 tell you about how this¬ 486 forward mode works.¬ 487 Steven kind of gave¬ 488 you one view of it.¬ 489 I'll give you¬ 490 another view today.¬ 491 But you just say, give me the¬ 492 derivative or the gradient¬ 493 of the determinant function,¬ 494 and Julia will happily do it.¬ 495 And of course, I¬ 496 can compare that¬ 497 with the adjugate¬ 498 of A transpose.¬ 499 And you guys know me by now.¬ 500 Page 11/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player When I see these¬ 501 things matching,¬ 502 it looks like to all the¬ 503 digits, it makes me happy.¬ 504 It makes me think, wow, this¬ 505 formula for the derivative¬ 506 is correct.¬ 507 Right, so this, for sure, is¬ 508 the derivative of the gradient.¬ 509 OK.¬ 510 I don't know.¬ 511 Maybe I tried to¬ 512 say this before,¬ 513 but I'm just going to repeat¬ 514 it, if you've heard me say it.¬ 515 But just philosophically,¬ 516 I find it remarkable¬ 517 that you could think of a¬ 518 limit of a finite difference,¬ 519 and the great mathematical¬ 520 gods let us have a formula.¬ 521 I mean, you've all¬ 522 done it in calculus,¬ 523 like the difference of a-- you¬ 524 take a sine and a little bit¬ 525 more sine, you get a cosine.¬ 526 Or the log, you get 1 over x.¬ 527 Or x squared, you get 2x.¬ 528 But I don't know.¬ 529 Could you guys¬ 530 imagine a universe¬ 531 where the mathematical¬ 532 gods weren't kind enough?¬ 533 Not every integral could¬ 534 be written as a formula.¬ 535 I mean, as you know,¬ 536 lots of integrals¬ 537 can't be written in terms¬ 538 of elementary functions.¬ 539 But derivatives, you¬ 540 could always do it.¬ 541 And you could do it¬ 542 for scalar calculus.¬ 543 That's why calculus¬ 544 is so easy to teach¬ 545 and is a beginning subject.¬ 546 You could do it for¬ 547 vector calculus,¬ 548 and you could do it for these¬ 549 complicated matrix functions.¬ 550 Page 12/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player I don't know.¬ 551 Do you ever stop and think¬ 552 about that being remarkable,¬ 553 or you just take it as a given¬ 554 and move on with your lives?¬ 555 I think it's amazing¬ 556 that we could have¬ 557 a formula for this difference.¬ 558 I just do.¬ 559 And a simple formula, in effect.¬ 560 But maybe you guys just¬ 561 take it as a granted given,¬ 562 but I don't know.¬ 563 I think formulas are¬ 564 gifts from the gods,¬ 565 and I don't take¬ 566 them for granted.¬ 567 All right.¬ 568 So this is really¬ 569 just to show you¬ 570 how to do reverse mode in Julia.¬ 571 So it's not much different.¬ 572 I'm just calling¬ 573 Zygote, which is¬ 574 one of the big, popular¬ 575 packages in Julia¬ 576 to do reverse mode autodiff.¬ 577 And you see it's¬ 578 not much different.¬ 579 This is the ForwardDiff¬ 580 package dot gradient,¬ 581 and this is the¬ 582 Zygote dot gradient.¬ 583 Under the hood, it's¬ 584 getting the answer¬ 585 in a completely different¬ 586 way, by going reverse mode.¬ 587 But it actually gives¬ 588 the same answer.¬ 589 Though I wouldn't be surprised--¬ 590 maybe Steven knows better--¬ 591 I wouldn't be surprised if¬ 592 this is just built-in formulas¬ 593 in both cases.¬ 594 I don't know.¬ 595 Let's see.¬ 596 We could do it symbolically,¬ 597 but let's get to the proof.¬ 598 So there are a couple of ways¬ 599 to prove this mathematically.¬ 600 Page 13/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player So one relatively¬ 601 simple proof is¬ 602 to remember the Laplace¬ 603 expansion of determinants.¬ 604 So I suspect you all¬ 605 remember that if you¬ 606 want to calculate the¬ 607 determinant of a big matrix,¬ 608 usually people take¬ 609 maybe the first row.¬ 610 But in fact, you¬ 611 could take any row.¬ 612 But do you remember?¬ 613 You take the¬ 614 top-left entry times¬ 615 the determinant of what¬ 616 happens if you cross out¬ 617 the first row and first column.¬ 618 Then the second entry, and you¬ 619 cross out that row and column¬ 620 with a minus sign.¬ 621 Plus, minus, plus, minus.¬ 622 You remember that rule?¬ 623 So that's the Laplace expansion.¬ 624 And the key fact is that if--¬ 625 for example, if I'm¬ 626 working with Ai1.¬ 627 Let's say I'm starting¬ 628 with the i-th row.¬ 629 Then Ai1 is not inside¬ 630 any of these Cs.¬ 631 Ai1 only appears here.¬ 632 Everything else you see¬ 633 depends on other elements¬ 634 of the matrix, but it¬ 635 doesn't depend on Ai1.¬ 636 Similarly, if you look at¬ 637 Ai2, Ci2, and every other term¬ 638 only depends on Ai2.¬ 639 To make that point very¬ 640 clear, here what I did was I¬ 641 took this matrix, and¬ 642 I made this matrix--¬ 643 this 3-by-3 matrix¬ 644 you'll see, it's¬ 645 almost completely¬ 646 numerical, but I put¬ 647 one symbol in the bottom right.¬ 648 And if you take the determinant,¬ 649 you see that this is an--¬ 650 Page 14/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player I don't know whether¬ 651 to call this a linear¬ 652 or an affine function.¬ 653 STEVEN G. JOHNSON:¬ 654 It would be affine.¬ 655 ALAN EDELMAN: Affine¬ 656 for this class.¬ 657 Some people would¬ 658 actually say it's¬ 659 linear in the sense of¬ 660 linear, quadratic, cubic.¬ 661 It's first-degree polynomial.¬ 662 But let's call it affine for¬ 663 the purposes of this class¬ 664 because it's not 13a plus 0.¬ 665 But whatever it is, it's¬ 666 a first-degree polynomial¬ 667 is what it is.¬ 668 And the fact of the¬ 669 matter is the coefficient¬ 670 of a is exactly¬ 671 this determinant.¬ 672 It's 4 times 4 minus 3.¬ 673 It's 16 minus 3.¬ 674 It's 13.¬ 675 And so the¬ 676 coefficient of every--¬ 677 if you make any¬ 678 element a symbol,¬ 679 the coefficient that's in¬ 680 front of it is just this minor.¬ 681 And so taking derivatives of¬ 682 first-degree functions is easy.¬ 683 It's just the slope.¬ 684 The derivative of¬ 685 this determinant¬ 686 with respect to this element¬ 687 is clearly the number 13.¬ 688 And so the way to¬ 689 say this, in general,¬ 690 is if I want to take the¬ 691 derivative of determinant¬ 692 with respect to any¬ 693 Aij element, the slope¬ 694 is the thing that multiplies it.¬ 695 So it's Cij.¬ 696 And so that is one¬ 697 immediate way to conclude¬ 698 that the gradient¬ 699 of the determinant¬ 700 Page 15/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player is the cofactor matrix.¬ 701 It's that simple.¬ 702 There's another proof¬ 703 that is sort of--¬ 704 I mean, this proof¬ 705 is pretty simple.¬ 706 I think it's easy to agree that¬ 707 this is a nice, simple proof.¬ 708 There's another proof that¬ 709 might seem a slightly harder,¬ 710 in one way, but in a¬ 711 way, it's sort of--¬ 712 mathematicians like¬ 713 this kind of proof.¬ 714 And so you get to take¬ 715 your pick which one you¬ 716 like best, but let me just¬ 717 show you an alternative proof.¬ 718 So in this alternative¬ 719 proof, what we're going to do¬ 720 is we're going to figure out the¬ 721 right answer near the identity.¬ 722 And then we're going to--¬ 723 and then we're going to use¬ 724 that to bootstrap ourselves¬ 725 to any other matrix.¬ 726 You know how to expand--¬ 727 you're all familiar,¬ 728 if I ask you¬ 729 to compute the characteristic¬ 730 polynomial of a matrix,¬ 731 let's call it M. If¬ 732 I need to do the--¬ 733 here, I'll do what Steven¬ 734 does, and I'll do it like this.¬ 735 So if anybody asked¬ 736 you to calculate¬ 737 the characteristic polynomial--¬ 738 and I'm using the mouse, which¬ 739 means it's really sloppy.¬ 740 Not that my handwriting is so¬ 741 great, but it's not this bad.¬ 742 All right.¬ 743 So the characteristic¬ 744 polynomial of any matrix M¬ 745 is usually written like this.¬ 746 And there's lots of factors.¬ 747 There's lambda to the¬ 748 n and all the way down¬ 749 to the determinant, plus or¬ 750 Page 16/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player minus the determinant of M.¬ 751 So you remember that.¬ 752 And if you want, you can--¬ 753 if you want, you can¬ 754 make this a plus sign,¬ 755 and then you get plus signs¬ 756 in this whole formula.¬ 757 And so this is not¬ 758 much different.¬ 759 Here if lambda was 1, if you¬ 760 just took lambda equals 1,¬ 761 you'd have determinant¬ 762 of I-- well,¬ 763 let's just see it this way.¬ 764 Determinant of I plus a¬ 765 matrix would be 1 plus.¬ 766 And then there would be the¬ 767 terms that you would get.¬ 768 There would be the next terms.¬ 769 This thing here,¬ 770 as you all know,¬ 771 is the trace of the matrix.¬ 772 So maybe I should¬ 773 have put that in.¬ 774 You get lambda to¬ 775 the n plus lambda¬ 776 to the n minus 1 times¬ 777 the trace of the matrix.¬ 778 So if you make a tiny,¬ 779 little perturbation,¬ 780 the determinant of I plus dA--¬ 781 I guess I should have made¬ 782 this 1 plus the trace of dA,¬ 783 to be honest.¬ 784 Let's fix that right now.¬ 785 So the determinant of¬ 786 I plus dA would be 1.¬ 787 That would be 1 to the n.¬ 788 Plus 1 to the n minus 1 times¬ 789 the trace of the matrix.¬ 790 And then there's the¬ 791 lower-order terms.¬ 792 So that's one way¬ 793 to think of it.¬ 794 And so there we immediately get¬ 795 the answer around the identity.¬ 796 And now if we want¬ 797 to get this anywhere,¬ 798 all we have to do is¬ 799 recognize that if we want¬ 800 Page 17/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player to go to the determinant¬ 801 of A plus dA,¬ 802 then we just go A times¬ 803 A inverse over here,¬ 804 and that's just the identity.¬ 805 But then we can use the¬ 806 properties of determinants¬ 807 to pull out the¬ 808 determinant of A.¬ 809 And you just get I¬ 810 plus A inverse dA.¬ 811 And this basically here, we¬ 812 just think of this A inverse dA¬ 813 as the trace formula.¬ 814 And therefore, we¬ 815 get our answer,¬ 816 the very answer¬ 817 we're looking for.¬ 818 In a way, this is¬ 819 more complicated,¬ 820 but mathematicians like¬ 821 this one better than--¬ 822 I don't know why.¬ 823 They're both valid.¬ 824 You get to take your pick.¬ 825 There's something I like¬ 826 about this, though it is¬ 827 a little bit more complicated.¬ 828 But in any event, we¬ 829 get the same answer.¬ 830 So what do we have here?¬ 831 So application to¬ 832 the derivative of¬ 833 the characteristic polynomial.¬ 834 So once again, there's¬ 835 the simple proof.¬ 836 The characteristic¬ 837 polynomial of a matrix¬ 838 is the product of x¬ 839 minus the eigenvalues.¬ 840 Probably a different sign¬ 841 from what I have here.¬ 842 You take the derivative¬ 843 of this product.¬ 844 You get the sum of these¬ 845 products, n minus 1¬ 846 at a time, which you¬ 847 could rewrite like this.¬ 848 But you can also directly¬ 849 do-- with our technology,¬ 850 Page 18/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player you can do this and get¬ 851 basically the same answer¬ 852 as the direct proof.¬ 853 And then I have some¬ 854 numerical checks.¬ 855 Let's see.¬ 856 And the derivative of¬ 857 the log determinant.¬ 858 Log determinant comes up a lot.¬ 859 Logs have lots of¬ 860 functions come up a lot.¬ 861 For example, Steven,¬ 862 I don't know,¬ 863 a few lectures ago talked¬ 864 about this f over f prime.¬ 865 It's what shows up whenever you¬ 866 do anybody's Newton's method.¬ 867 And of course, this¬ 868 could be written as 1¬ 869 over the log f prime.¬ 870 So basically, the logarithmic¬ 871 derivative and its reciprocal¬ 872 come up all over mathematics.¬ 873 So the derivative of the¬ 874 log of the determinant¬ 875 is simply the trace of¬ 876 the inverse times the dA.¬ 877 This you've seen, A inverse.¬ 878 And that's it.¬ 879 Any questions?¬ 880 That basically covers the¬ 881 gradient of the determinant.¬ 882 Any questions about that?¬ 883 So maybe a few words¬ 884 about determinant.¬ 885 Interestingly¬ 886 enough, people often¬ 887 tell you that you should¬ 888 never compute a determinant.¬ 889 Or hardly ever might¬ 890 be a fair term.¬ 891 So determinants are real.¬ 892 It's a real favorite of¬ 893 elementary linear algebra¬ 894 classes.¬ 895 Determinants are great for¬ 896 telling you in exact arithmetic¬ 897 whether a matrix¬ 898 is singular or not.¬ 899 So a matrix has determinant¬ 900 Page 19/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player 0, it's singular.¬ 901 If the determinant¬ 902 is not 0, it's not.¬ 903 And that sounds like¬ 904 a really good idea,¬ 905 to have something like that.¬ 906 But it turns out that when¬ 907 you're doing computations¬ 908 in finite precision, if¬ 909 you're doing it on a computer,¬ 910 the determinant turns out¬ 911 to be not so meaningful.¬ 912 It gets to be hard to¬ 913 compute accurately.¬ 914 There are a lot of issues with¬ 915 calculating the determinant.¬ 916 It turns out that while the¬ 917 pure mathematicians live¬ 918 in a binary world where a matrix¬ 919 is singular or non-singular,¬ 920 the truth of the matter is¬ 921 is that the world of matrices¬ 922 is not so binary.¬ 923 It's a bit more of¬ 924 a spectrum where¬ 925 matrices are singular or nearly¬ 926 singular or a little bit bad¬ 927 or not at all bad.¬ 928 And probably you've all heard¬ 929 the word that I'm referring to.¬ 930 The word that we use¬ 931 in numerical analysis¬ 932 is conditioning.¬ 933 So ill conditioned means a¬ 934 matrix is nearly singular.¬ 935 And well conditioned means¬ 936 that it's very non-singular.¬ 937 Too many double¬ 938 negatives there, but it's¬ 939 sort of the good¬ 940 side of singular¬ 941 when we say it's¬ 942 well conditioned.¬ 943 And so the determinant¬ 944 doesn't really give--¬ 945 the determinant it's¬ 946 not a really good--¬ 947 give a good job of talking¬ 948 about how nearly singular¬ 949 matrices are.¬ 950 Page 20/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player The condition number, which is¬ 951 related to singular values--¬ 952 I'm not going to talk¬ 953 about that today--¬ 954 is a much better way of¬ 955 talking about matrices¬ 956 being singular or not.¬ 957 So you learned it¬ 958 all in a course,¬ 959 like 18.06 or elementary¬ 960 linear algebra.¬ 961 You learned about determinants.¬ 962 And then later on,¬ 963 when you compute,¬ 964 people tell you to forget¬ 965 about determinants mostly.¬ 966 There are times, but mostly.¬ 967 And the other thing¬ 968 we tell people to do¬ 969 is forget about the¬ 970 characteristic polynomial¬ 971 as well.¬ 972 That's not how we calculate¬ 973 eigenvalues either.¬ 974 We don't take roots¬ 975 of polynomials.¬ 976 Anybody happen to know¬ 977 how we compute eigenvalues¬ 978 in the real world?¬ 979 We don't do characteristic¬ 980 polynomials.¬ 981 Anybody know the¬ 982 magic two letters¬ 983 that happen when you¬ 984 type eigenvalues?¬ 985 How many of you¬ 986 just thought it was¬ 987 the characteristic polynomial?¬ 988 You take the roots.¬ 989 How many of you had any¬ 990 idea how roots got taken--¬ 991 eigenvalues got taken¬ 992 on the computer?¬ 993 So do you have any idea what¬ 994 the algorithm is being used?¬ 995 AUDIENCE: [INAUDIBLE]¬ 996 ALAN EDELMAN: The power method.¬ 997 AUDIENCE: Yeah.¬ 998 ALAN EDELMAN: OK,¬ 999 so you probably¬ 1000 Page 21/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player didn't hear the student¬ 1001 saying that, well, in 18.06,¬ 1002 I learned about something like¬ 1003 the power method, which gives¬ 1004 you the dominant eigenvalue.¬ 1005 Yeah.¬ 1006 And then nobody else in¬ 1007 the room has any idea¬ 1008 how eigenvalues get calculated?¬ 1009 Just a little bit¬ 1010 of culture here.¬ 1011 So people don't know.¬ 1012 I see.¬ 1013 I kind of feel like I¬ 1014 ruined the question then.¬ 1015 I should have just asked how¬ 1016 are eigenvalues computed?¬ 1017 Because I imagine many¬ 1018 of you would have said,¬ 1019 isn't it the¬ 1020 characteristic polynomial?¬ 1021 You get the roots.¬ 1022 Because every one¬ 1023 of you have formed¬ 1024 the characteristic polynomials¬ 1025 of 2-by-2 matrices.¬ 1026 I know you have.¬ 1027 You got that quadratic equation,¬ 1028 and you solve for the roots.¬ 1029 You remember?¬ 1030 Who remembers doing that?¬ 1031 Quadratics, you get the roots.¬ 1032 If you had a mean teacher, maybe¬ 1033 they forced you to do a cubic,¬ 1034 but I bet they didn't.¬ 1035 Anyone ever do it for a cubic?¬ 1036 Maybe it was rigged¬ 1037 to be easy though.¬ 1038 So right, so none of that¬ 1039 happens on the computer.¬ 1040 I'm not going to tell you¬ 1041 in detail how it's done,¬ 1042 but I will mention just¬ 1043 the fact that it's not¬ 1044 the characteristic¬ 1045 polynomial is half¬ 1046 of what I want you to know.¬ 1047 And the other half is there's¬ 1048 something called the QR¬ 1049 algorithm for eigenvalue.¬ 1050 Page 22/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player So in general, a QR for¬ 1051 a matrix factors a matrix¬ 1052 to orthogonal times¬ 1053 upper triangular.¬ 1054 And a funny thing happens.¬ 1055 If you factor a matrix into QR¬ 1056 and then reverse it and get RQ,¬ 1057 and if you do that again,¬ 1058 factor that new matrix into QR¬ 1059 and reverse it to RQ,¬ 1060 and you keep doing that,¬ 1061 essentially the eigenvalues¬ 1062 magically appear.¬ 1063 And there are some details.¬ 1064 If the matrix is¬ 1065 symmetric, the matrix¬ 1066 will actually become more¬ 1067 and more diagonal as you go.¬ 1068 If it's not symmetric, but¬ 1069 it has real eigenvalues,¬ 1070 it will become triangular.¬ 1071 And you'll see the eigenvalues¬ 1072 on the diagonal eventually.¬ 1073 And if it's complex, you'll¬ 1074 get these little 2-by-2 pieces¬ 1075 which are easy to get¬ 1076 the eigenvalues from.¬ 1077 So there are a bunch of¬ 1078 tricks to accelerate all this,¬ 1079 but the basic idea¬ 1080 is QR, RQ, QR, RQ.¬ 1081 You might actually try it in¬ 1082 Julia or Python or whatever.¬ 1083 MATLAB, whatever you¬ 1084 like to do one day.¬ 1085 You'll see it just works.¬ 1086 STEVEN G. JOHNSON:¬ 1087 But it is related¬ 1088 to the power method under¬ 1089 the hood, if you dig deep.¬ 1090 ALAN EDELMAN: Oh,¬ 1091 dig really deep.¬ 1092 That's right.¬ 1093 It's sort of like a block power¬ 1094 method in multiple dimensions¬ 1095 all at once.¬ 1096 It's crazy.¬ 1097 Yeah.¬ 1098 OK.¬ 1099 All right.¬ 1100 Page 23/23 /Users/jcplayer/Desktop/18.S096/OCW_18.S096_Lecture05-Part1-New_2023jan27.txt Saved: 11/28/23, 4:15:58 PM Printed for: Jason Player So that's that one.¬ 1101 1102 |
7977 | https://users.highland.edu/~jsullivan/genchem/s09_molar_mass.html | This is "Unit 9", section 9.2 from the book General
Chemistry (v. 1.0).
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9.2 Molar Mass
Learning Objective
To calculate the molecular mass of a covalent compound and the formula
mass of an ionic compound and to calculate the number of atoms, molecules,
or formula units in a sample of a substance.
As you learned in
Chapter 1
"Introduction to Chemistry", the mass
number is the sum of the numbers of protons and neutrons present in
the nucleus of an atom. The mass number is an integer that is approximately
equal to the numerical value of the atomic mass. Although the mass number is
unitless, it is assigned units called atomic mass
units (amu). Because a molecule or a polyatomic ion is an assembly of
atoms whose identities are given in its molecular or ionic formula, we can
calculate the average atomic mass of any molecule or polyatomic ion from its
composition by adding together the masses of the constituent atoms. The average
mass of a monatomic ion is the same as the average mass of an atom of the
element because the mass of electrons is so small that it is insignificant in
most calculations.
Molecular and Formula Masses
The molecular massThe sum of the average masses of the atoms in one molecule of
a substance, each multiplied by its subscript. of a substance is
the sum of the average masses of the atoms in one molecule of a substance. It
is calculated by adding together the atomic masses of the elements in the
substance, each multiplied by its subscript (written or implied) in the
molecular formula. Because the units of atomic mass are atomic mass units, the
units of molecular mass are also atomic mass units. The procedure for
calculating molecular masses is illustrated in Example 9.2-1.
Example 9.2-1
Calculate the molecular mass
of ethanol, whose condensed structural formula is CH3CH2OH. Among its many
uses, ethanol is a fuel for internal combustion engines.
Given: molecule
Asked for: molecular mass
A Determine the number of atoms of each element
in the molecule.
B Obtain the atomic masses of each element from
the periodic table and multiply the atomic mass of each element by the number
of atoms of that element.
C Add together the masses to give the molecular
mass.
Solution:
A The molecular formula of ethanol may be
written in three different ways: CH3CH2OH (which illustrates the presence of an ethyl group,
CH3CH2−, and an
−OH group), C2H5OH, and C2H6O; all show that ethanol has two carbon atoms, six
hydrogen atoms, and one oxygen atom.
B Taking the atomic masses from the periodic
table, we obtain
2×atomic mass of carbon=2atoms(12.011 amuatom)=24.022amu6×atomic mass of hydrogen=6atoms(1.0079 amuatom)=6.0474amu1×atomic mass of oxygen=1atom(15.9994 amuatom)=15.9994amu
C Adding together the masses gives the molecular
mass:
24.022 amu + 6.0474 amu
+ 15.9994 amu = 46.069 amu
Alternatively, we could have
used unit conversions to reach the result in one step:
[2a C(12.011amu1atom C)]+[6a H(1.0079amu1atom H)]+[1a O(15.9994amu1atom O)]=46.069amu
The same calculation can also
be done in a tabular format, which is especially helpful for more complex
molecules:
2C(2 atoms)(12.011 amu/atom)=24.022 amu6H(6 atoms)(1.0079 amu/atom)=6.0474 amu+1O(1 atom)(15.9994 amu/atom)=15.9994 amuC2H6Omolecular mass of ethanol=46.069 amu
Exercise
Calculate the molecular mass
of trichlorofluoromethane, also known as Freon-11, whose condensed structural
formula is CCl3F. Until recently, it was used as a
refrigerant. The structure of a molecule of Freon-11 is as follows:
Answer: 137.368 amu
Unlike
molecules, which have covalent bonds, ionic compounds do not have a readily
identifiable molecular unit. So for ionic compounds we use the formula mass (also called the empirical formula massAnother name for formula mass.) of the compound
rather than the molecular mass. The formula massThe sum of the atomic
masses of all the elements in the empirical formula, each multiplied by its
subscript. is the sum of the atomic masses of all the elements in
the empirical formula, each multiplied by its subscript (written or implied).
It is directly analogous to the molecular mass of a covalent compound. Once
again, the units are atomic mass units.
Note the Pattern
Atomic mass, molecular mass,
and formula mass all have the same units: atomic mass units.
Example 9.2-2
Calculate the formula mass of
Ca3(PO4)2, commonly called calcium phosphate. This compound is
the principal source of calcium found in bovine milk.
Given: ionic compound
Asked for: formula mass
Strategy:
A Determine the number of atoms of each element
in the empirical formula.
B Obtain the atomic masses of each element from
the periodic table and multiply the atomic mass of each element by the number
of atoms of that element.
C Add together the masses to give the formula
mass.
Solution:
A The empirical formula—Ca3(PO4)2—indicates that the simplest electrically
neutral unit of calcium phosphate contains three Ca2+ ions and two PO43− ions. The formula mass of this molecular
unit is calculated by adding together the atomic masses of three calcium atoms,
two phosphorus atoms, and eight oxygen atoms.
B Taking atomic masses from the periodic table,
we obtain
3×atomic mass of calcium=3atoms(40.078 amuatom)=120.234amu2×atomic mass of phosphorus=2atoms(30.973761 amuatom)=61.947522amu8×atomic mass of oxygen=8atoms(15.9994 amuatom)=127.9952amu
C Adding together the masses gives the formula
mass of Ca3(PO4)2:
120.234 amu + 61.947522
amu + 127.9952 amu = 310.177 amu
We could also find the
formula mass of Ca3(PO4)2 in one step by using
unit conversions or a tabular format:
[3atoms Ca(40.078 amu1atom Ca)]+[2atoms P(30.973761 amu1atom P)]+[8atoms O(15.9994 amu1atom O)]=310.177amu 3Ca(3 atoms)(40.078 amu/atom)=120.234 amu2P(2 atoms)(30.973761 amu/atom)=61.947522 amu+8O(8 atoms)(15.9994 amu/atom)=127.9952 amuCa3P2O8formula mass of Ca3(PO4)2=310.177 amu
Exercise
Calculate the formula mass of
Si3N4, commonly
called silicon nitride. It is an extremely hard and inert material that is used
to make cutting tools for machining hard metal alloys.
Answer: 140.29 amu
Molar Mass
The concept of
the mole allows us to count a specific number of individual atoms and molecules
by weighing measurable quantities of elements and compounds. To obtain 1 mol of
carbon-12 atoms, we would weigh out 12 g of isotopically pure carbon-12.
Because each element has a different atomic mass, however, a mole of each
element has a different mass, even though it contains the same number of atoms
(6.022 × 1023). This is analogous to the
fact that a dozen extra large eggs weighs more than a dozen small eggs, or that
the total weight of 50 adult humans is greater than the total weight of 50
children. Because of the way in which the mole is defined, for every element
the number of grams in a mole is the same as the number of atomic mass units in
the atomic mass of the element. For example, the mass of 1 mol of magnesium
(atomic mass = 24.305 amu) is 24.305 g. Because the atomic mass of magnesium
(24.305 amu) is slightly more than twice that of a carbon-12 atom (12 amu), the
mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of
1 mol of carbon-12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass
= 4.002602 amu) is 4.002602 g, which is about one-third that of 1 mol of
carbon-12.
The molar massThe mass in grams of 1 mol of a substance. of a
substance is defined as the mass in grams of 1 mol of that substance. One mole
of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar
mass is the mass of 1 mol of atoms of that element; for a covalent molecular
compound, it is the mass of 1 mol of molecules of that compound; for an ionic
compound, it is the mass of 1 mol of formula units. That is, the molar mass of
a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that
substance. In each case, the number of grams in 1 mol is the same as the number
of atomic mass units that describe the atomic mass, the molecular mass, or the
formula mass, respectively.
Note the Pattern
The molar mass of any
substance is its atomic mass, molecular mass, or formula mass in grams per
mole.
The periodic
table lists the atomic mass of carbon as 12.011 amu; the average molar mass of
carbon—the mass of 6.022 × 1023
carbon atoms—is therefore 12.011 g/mol:
| Substance (formula) | Atomic, Molecular, or Formula Mass (amu) | Molar Mass (g/mol) |
---
| carbon (C) | 12.011 (atomic mass) | 12.011 |
| ethanol (C2H5OH) | 46.069 (molecular mass) | 46.069 |
| calcium phosphate [Ca3(PO4)2] | 310.177 (formula mass) | 310.177 |
The molar mass
of naturally occurring carbon is different from that of carbon-12 and is not an
integer because carbon occurs as a mixture of carbon-12, carbon-13, and
carbon-14. One mole of carbon still has 6.022 × 1023 carbon atoms, but 98.89% of those atoms are
carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 1012) are carbon-14. Similarly, the molar mass of uranium is
238.03 g/mol, and the molar mass of iodine is 126.90 g/mol.
The molar mass
of ethanol is the mass of ethanol (C2H5OH) that contains 6.022 × 1023 ethanol molecules. As you calculated in Example 1,
the molecular mass of ethanol is 46.069 amu. Because 1 mol of ethanol contains
2 mol of carbon atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 ×
1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994 g), its molar mass is
46.069 g/mol. Similarly, the formula mass of calcium phosphate [Ca3(PO4)2] is 310.177 amu, so its molar mass is 310.177 g/mol.
This is the mass of calcium phosphate that contains 6.022 × 1023 formula units. Figure 9.2(a) shows samples that contain precisely one molar mass of
several common substances.
Figure 9.2(a) Samples of 1 Mol
of Some Common Substances
Image Credit: Highland Community College, Freeport, IL
The mole is
the basis of quantitative chemistry. It provides chemists with a way to convert
easily between the mass of a substance and the number of individual atoms,
molecules, or formula units of that substance. Conversely, it enables chemists
to calculate the mass of a substance needed to obtain a desired number of
atoms, molecules, or formula units. For example, to convert moles of a
substance to mass, we use the relationship
Equation 9.2(eq1)
(moles)(molar mass) → mass
or, more
specifically,
moles(gramsmole)=grams
Conversely, to
convert the mass of a substance to moles, we use
Equation 9.2(eq2)
(massmolar mass)→moles(gramsgrams/mole)=grams(molegrams)=moles
Be sure to pay
attention to the units when converting between mass and moles.
Figure 9.2(b) is a flowchart for converting between mass;
the number of moles; and the number of atoms, molecules, or formula units. The
use of these conversions is illustrated in Example 9.2-3 and Example 9.2-4.
Figure 9.2(b) A Flowchart for
Converting between Mass; the Number of Moles; and the Number of Atoms,
Molecules, or Formula Units
Example 9.2-3
For 35.00 g of ethylene
glycol (HOCH2CH2OH),
which is used in inks for ballpoint pens, calculate the number of
moles.
molecules.
Given: mass and molecular formula
Asked for: number of moles and number of
molecules
Strategy:
A Use the molecular formula of the compound to
calculate its molecular mass in grams per mole.
B Convert from mass to moles by dividing the
mass given by the compound’s molar mass.
C Convert from moles to molecules by multiplying
the number of moles by Avogadro’s number.
Solution:
A The molecular
mass of ethylene glycol can be calculated from its molecular formula using
the method illustrated in Example 1:
2C(2 atoms)(12.011 amu/atom)=24.022 amu6H(6 atoms)(1.0079 amu/atom)=6.0474 amu+2O(2 atoms)(15.9994 amu/atom)=31.9988 amuC2H6O2molecular mass of ethylene glycol=62.068 amu
The molar mass of
ethylene glycol is 62.068 g/mol.
B The number of moles of ethylene glycol
present in 35.00 g can be calculated by dividing the mass (in grams) by the
molar mass (in grams per mole):
mass of ethylene glycol (g)molar mass (g/mol)=moles ethylene glycol (mol)
So
35.00g ethylene glycol(1 mol ethylene glycol62.068g ethylene glycol)=0.5639mol ethylene glycol
It
is always a good idea to estimate the answer before you do the
actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be
less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1
mol, so we have probably not made a major error in the calculations.
2. C To calculate the
number of molecules in the sample, we multiply the number of moles by
Avogadro’s number:
molecules of ethylene glycol=0.5639mol(6.022×1023molecules1mol)=3.396×1023molecules
Because we are dealing
with slightly more than 0.5 mol of ethylene glycol, we expect the number of
molecules present to be slightly more than one-half of Avogadro’s
number, or slightly more than 3 × 1023
molecules, which is indeed the case.
Exercise
For 75.0 g of CCl3F (Freon-11), calculate the number of
moles.
molecules.
Answer:
0.546 mol
3.29 × 1023 molecules
Example 9.2-4
Calculate the mass of 1.75
mol of each compound.
S2Cl2 (common
name: sulfur monochloride; systematic name: disulfur dichloride)
Ca(ClO)2 (calcium hypochlorite)
Given: number of moles and molecular or
empirical formula
Asked for: mass
Strategy:
A Calculate the molecular mass of the compound
in grams from its molecular formula (if covalent) or empirical formula (if
ionic).
B Convert from moles to mass by multiplying the
moles of the compound given by its molar mass.
Solution:
We begin by calculating the
molecular mass of S2Cl2 and the formula mass of Ca(ClO)2.
A The molar mass
of S2Cl2 is
obtained from its molecular mass as follows:
2S(2 atoms)(32.065 amu/atom)=64.130 amu+2Cl(2 atoms)(35.453 amu/atom)=70.906 amuS2Cl2molecular mass of S2Cl2=135.036amu
The molar mass of S2Cl2 is 135.036
g/mol.
B The mass of 1.75 mol of S2Cl2 is calculated as
follows:
moles S2Cl2[molar mass(gmol)]→mass of S2Cl2(g)1.75mol S2Cl2(135.036 g S2Cl21mol S2Cl2)=236 g S2Cl2
2. A The formula mass
of Ca(ClO)2 is obtained as follows:
1Ca(1 atom)(40.078 amu/atom)=40.078 amu2Cl(2 atoms)(35.453 amu/atom)=70.906 amu+2O(2 atoms)(15.9994 amu/atom)=31.9988 amuCa(ClO)2formula mass of Ca(ClO)2=142.983amu
The molar mass of
Ca(ClO)2 142.983 g/mol.
B The mass of 1.75 mol of Ca(ClO)2 is calculated as follows:
moles Ca(ClO)2[molar mass Ca(ClO)21 mol Ca(ClO)2]=mass Ca(ClO)21.75mol Ca(ClO)2[142.983 g Ca(ClO)21mol Ca(ClO)2]=250 g Ca(ClO)2
Because 1.75 mol is less
than 2 mol, the final quantity in grams in both cases should be less than
twice the molar mass, which it is.
Exercise
Calculate the mass of 0.0122
mol of each compound.
Si3N4 (silicon
nitride), used as bearings and rollers
(CH3)3N
(trimethylamine), a corrosion inhibitor
Answer:
1.71 g
0.721 g
Example 9.2-5
What is the mass of 3.56 mol of HgCl2? The molar mass of HgCl2 is 271.49 g/mol.
Solution
Use the molar mass as a conversion factor between moles and grams. Because we want to cancel the mole unit and introduce the gram unit, we can use the molar mass as given:
(3.56mol HgCl2)(271.49g HgCl2mol HgCl2)=967g HgCl2
Test Yourself
What is the mass of 33.7 mol of H2O?
Answer
607 g
Example 9.2-6
How many moles of H2O are present in 240.0 g of water (about the mass of a cup of water)?
Solution
Use the molar mass of H2O as a conversion factor from mass to moles. The molar mass of water is (1.0079 + 1.0079 + 15.999) = 18.015 g/mol. However, because we want to cancel the gram unit and introduce moles, we need to take the reciprocal of this quantity, or 1 mol/18.015 g:
(240.0g H2O)(1 mol H2O18.015g H2O)=13.32mol H2O
Test Yourself
How many moles are present in 35.6 g of H2SO4 (molar mass = 98.08 g/mol)?
Answer
0.363 mol
Other conversion factors, for example, density, can be combined with the definition of mole.
Example 9.2-7
The density of ethanol is 0.789 g/mL. How many moles are in 100.0 mL of ethanol? The molar mass of ethanol is 46.08 g/mol.
Solution
Here, we use density to convert from volume to mass and then use the molar mass to determine the number of moles.
(100.0mLethanol)(0.789gmL)(1mol46.08g)=1.71mol ethanol
Test Yourself
If the density of benzene, C6H6, is 0.879 g/mL, how many moles are present in 17.9 mL of benzene?
Answer
0.201 mol
Summary
The molecular mass and the formula mass of a compound are obtained by
adding together the atomic masses of the atoms present in the molecular formula
or empirical formula, respectively; the units of both are atomic mass units
(amu). The molar mass of a substance
is defined as the mass of 1 mol of that substance, expressed in grams per mole,
and is equal to the mass of 6.022 × 1023
atoms, molecules, or formula units of that substance.
Key Takeaway
To analyze chemical transformations, it is essential to use a
standardized unit of measure called the mole.
Numerical Problems
Derive an expression that relates the number of molecules
in a sample of a substance to its mass and molecular mass.
Calculate the molecular mass or formula mass of each
compound.
KCl (potassium chloride)
NaCN (sodium cyanide)
H2S (hydrogen sulfide)
NaN3 (sodium azide)
H2CO3
(carbonic acid)
K2O (potassium oxide)
Al(NO3)3
(aluminum nitrate)
Cu(ClO4)2
[copper(II) perchlorate]
Calculate the molecular mass or formula mass of each
compound.
V2O4
(vanadium(IV) oxide)
CaSiO3 (calcium silicate)
BiOCl (bismuth oxychloride)
CH3COOH (acetic acid)
Ag2SO4
(silver sulfate)
Na2CO3
(sodium carbonate)
(CH3)2CHOH
(isopropyl alcohol)
Calculate the molar mass of each compound.
blue = Si, green = Cl
yellow = S, red = O
blue = N, red = O
gray = C, white = H
gray = C, white = H, red = O
Calculate the number of moles in 5.00 × 102 g of each substance. How many molecules or
formula units are present in each sample?
CaO (lime)
CaCO3 (chalk)
C12H22O11 [sucrose (cane sugar)]
NaOCl (bleach)
CO2 (dry ice)
Calculate the mass in grams of each sample.
0.520 mol of N2O4
1.63 mol of C6H4Br2
4.62 mol of (NH4)2SO3
Solutions of iodine are used as antiseptics and
disinfectants. How many iodine atoms correspond to 11.0 g of molecular
iodine (I2)?
What is the total number of atoms in each sample?
2.48 g of HBr
4.77 g of CS2
1.89 g of NaOH
1.46 g of SrC2O4
Decide whether each statement is true or false and explain
your reasoning.
There are more molecules in 0.5 mol of Cl2 than in 0.5 mol of H2.
One mole of H2 has 6.022 × 1023 hydrogen atoms.
The molecular mass of H2O is 18.0
amu.
The formula mass of benzene (C6H12) is 78 amu.
Complete the following table.
| Substance | Mass (g) | Number of Moles | Number of Molecules or Formula Units | Number of Atoms or Ions |
--- ---
| MgCl2 | 37.62 | | | |
| AgNO3 | | 2.84 | | |
| BH4Cl | | | 8.93 × 1025 | |
| K2S | | | | 7.69 × 1026 |
| H2SO4 | | 1.29 | | |
| C6H14 | 11.84 | | | |
| HClO3 | | | 2.45 × 1026 | |
11. Give the formula mass or the molecular mass of each
substance.
PbClF
Cu2P2O7
BiONO3
Tl2SeO4
Give the formula mass or the molecular mass of each
substance.
MoCl5
B2O3
UO2CO3
NH4UO2AsO4
Determine the molar mass of each substance.
Si
SiH4
K2O
Determine the molar mass of each substance.
Cl2
SeCl2
Ca(C2H3O2)2
Determine the molar mass of each substance.
Al
Al2O3
CoCl3
Determine the molar mass of each substance.
O3
NaI
C12H22O11
What is the mass of 4.44 mol of Rb?
What is the mass of 0.311 mol of Xe?
What is the mass of 12.34 mol of Al2(SO4)3?
What is the mass of 0.0656 mol of PbCl2?
How many moles are present in 45.6 g of CO?
How many moles are present in 0.00339 g of LiF?
How many moles are present in 1.223 g of SF6?
How many moles are present in 48.8 g of BaCO3?
How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?
How many moles are present in 56.83 mL of O2 if the density of O2 is 0.00133 g/mL?
Answers
(mass/molar mass) × 6.02 × 1023= number of molecules
165.9 amu
116.2 amu
260.4 amu
60.03 amu
311.8 amu
106.0 amu
60.03 amu
8.92 mol, 5.37 × 1024 formula units
5.00 mol, 3.01 × 1024 formula units
1.46 mol, 8.80 × 1023 molecules
6.71 mol, 4.04 × 1024 formula units
11.4 mol, 6.84 × 1024 molecules
5.22 × 1022 atoms
False, see the answer to question 5 in the numerical exercises for section 9.1.
False, see the answer to question 5 in the numerical exercises for section 9.1.
True, H2O is a molecular substance. Each H contributes 1 amu to the mass and oxygen contributes 16 amu to the mass.
True, though formula mass is often used for salts, and perhaps molecular mass would be more appropriate, benzene certainly has a formula, C6H12, and can therefore have a formula mass. Each carbon contributes 12 amu and each hydrogen contributes 1 amu.
261.7 amu
301 amu
287 amu
551.8 amu
28.086 g
32.118 g
94.195 g
26.981 g
101.959 g
165.292 g
379 g
4222 g
1.63 mol
0.008374 mol
3.72 mol
Previous Section
Table of Contents
Next Section
The sum of the average masses of the atoms in one molecule of
a substance, each multiplied by its subscript.
Another name for formula mass.
The sum of the atomic
masses of all the elements in the empirical formula, each multiplied by its
subscript.
The mass in grams of 1 mol of a substance. |
7978 | https://math.stackexchange.com/questions/2968643/maximinzing-and-minimizing-a-circle-in-an-ellipse-using-lagrange-multipliers | Skip to main content
Maximinzing and minimizing a circle in an ellipse using lagrange multipliers
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Consider the following question:
The equation 73x2 + 72xy + 52y2=100 defines an ellipse which is
centered at the origin, but has been rotated about it. Find the
semiaxes of this ellipse by maximizing and minimizing f(x,y)=x2+y2 on it.
This problem seems very complicated to me. I'm self-studying multivariable calculus and I cannot figure out this problem. Let g(x,y,z)=73x2+72xy+52y2−100 and f(x,y,z)=x2+y2
First, I computed
gx=146x+72y
gy=104y+72x
gz=0
fx=2x
fy=2y.
fz=0.
Then, I have
2x=λ(146x+72y)
2y=λ(104y+72x)
73x2+72xy+52y2=100
I don't know how to proceed since this is a system of three variables. Also, this is my first lagrange multiplier attempt so I'm not confident about what to do next either. I've been using Paul's Online Notes to try and understand. My ultimate goal is to be able to apply this on inequalities because I've heard lagrange multipliers are helpful in math olympiad. This seemed like a good example.
multivariable-calculus
lagrange-multiplier
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edited Oct 24, 2018 at 4:22
asked Oct 24, 2018 at 3:50
user400359user400359
6
The problem says that you should maximize x and y on the ellipse. So the last equation should be that of the ellipse. Then you will have 3 variables and 3 equations
– Amphiaraos
Commented
Oct 24, 2018 at 3:58
Sorry, I don't understand what you mean by this? So like have another equation doing 73x2+72xy+52y2−100= what?
– user400359
Commented
Oct 24, 2018 at 4:00
The first two equations are those with the Lagrange multipliers. You need another equation to be able to solve the system. That last equation should be 73x2+72xy+52y2=100 which means that you are taking x, y on the ellipse
– Amphiaraos
Commented
Oct 24, 2018 at 4:03
Why did you introduce the variable z?
– amd
Commented
Oct 24, 2018 at 4:03
I don't know, I was sort of just outlining an example I saw earlier because this is the first problem I am trying with lagrange multipliers.. I guess I don't need it here.
– user400359
Commented
Oct 24, 2018 at 4:06
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This is a homogeneous problem and can be handled easily. Making y=μx we have.
{x2+y2=x2(1+μ2)73x2+72xy+52y2=x2(73+72μ+52μ2)=100
so the problem is equivalent to
minμf(μ)=100(1+μ2)73+72μ+52μ2
now
f′(μ)=0≡(1+μ)(16μ−37)=0
and thus we have μ=−1 and μ=1637 as stationary points etc.
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answered Oct 24, 2018 at 9:19
CesareoCesareo
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The problem we have is:
minimizex,y x2+y2s.t. 73x2+72xy+52y2=100
We first find the gradients for f and g:
∇f=(2x, 2y)∇g=(146x+72y, 72x+104y)
.
The Lagrange multipliers method tells us that ∇f=λ∇g. This gives us the following two equations:
2x=λ(146x+72y) (1)2y=λ(72x+104y) (2)
This is a system of 2 equations with 3 variables. In order to solve it, we need a third equation. We know that the solution must lie on the ellipse, this gives us the last equation:
73x2+72xy+52y2=100 (3)
The problem is now to solve this system of equations.
Finding λ in (1) and replacing in (2), and simplifying we get the following:
12x2−7xy−12y2=0
This equation has two solutions, which are straight lines:
x=7y±25y24
and those are the semiaxes of the ellipse.
The points where these lines intercept the ellipse are the solutions to the optimisation problem. The points are (0.8,0.6) and (−0.8,−0.6) for the minimization problem and (−1.2,1.6), (1.2,−1.6) for the maximization one.
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answered Oct 24, 2018 at 4:42
AmphiaraosAmphiaraos
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Lagrange multiplier, in essence, says the gradient of a function f(x1,x2..xn) being maximized or minimized subjected to a constraint g(y1,y2...ym) will be such that, at the critical points ▽f(x1,x2..xn)=λ▽g(y1,y2...yn).
You have the equation set up wrong, In your question the constraint is 73x2+72xy+52y2=100=g(x,y) and the function you are maximizing is x2+y2=f(x,y).
So your equation then becomes ▽(x2+y2)=λ▽(73x2+72xy+52y2=100). after solving it you will have two equations and 3 unknowns; however, your third equation will be 73x2+72xy+52y2=100.
fx=λgx
fy=λgy
73x2+72xy+52y2=100
Solve the equations for x,y, and z. That is the point where your function will reach its maximum and minimum. In this particular case, though, I believe you will get more than one max and min point. I am not sure, you will have to solve it.
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answered Oct 24, 2018 at 4:22
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
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do i solve the system for λ? or x and y? what do i do with λ?
– user400359
Commented
Oct 24, 2018 at 4:27
@stackofhay42 To be honest, it depends on the kind of equations you get. If it is easier to solve for lλ first, then solve for lambda. However, If you can solve for x and y directly, then that is the best. It will make the problem easier. That said, generally the questions are such that you will have to solve for λ, but there are times where you don't even care about λ but those are rare.
– Bertrand Wittgenstein's Ghost
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Oct 24, 2018 at 4:30
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7979 | https://www.intmath.com/blog/mathematics/calculating-weight-using-different-gravity-loads-12519 | Interactive Mathematics
Calculating Weight Using Different Gravity Loads
By Kathleen Cantor, 15 Oct 2020
One can define gravity as a universal force that acts between two objects. It tends to pull objects towards the center of the earth. Each body in the universe possesses a particular amount of matter. This is known as mass, which is defined as the amount of matter contained in a substance. Anything that occupies space and has weight is known as matter. The SI unit for mass is the kilograms (KG), although mass can also be expressed in grams(g), milligrams (mg), and even tones (t) and is measured using a beam balance.
How to calculate weight using different gravity loads
When gravity acts on a body, it experiences a pull effect. This pull effect of gravity is known as weight. Weight is measured in Newton and changes depending on the place. The SI unit for weight is Newton (N). The difference between mass and weight is brought about by the presence of gravity in the latter. Weight is a vector quantity, and it has both magnitude and direction.
An astronomer's weight on earth slightly differs from the same astronomer's weight on a different planet, but the mass remains constant. Many people think it's fine to refer to mass and weight as the same thing, but a science student needs to understand the difference between them. Weight is directly proportional to the gravitational force available; thus, mathematically, this can be expressed as:
Weight = mass gravity
W= mg
Lifting a 1-kilogram bag requires less physical effort compared to a 2-kilogram bag. This is because a 1-kilogram bag will weigh more than a 2-kilogram bag. The 1 Kg bag has a force of about 10 Newton acting on it, and the 2 Kg bag has a force of about 20 Newton acting on it; thus, the higher the mass, the larger the force required to carry it.
How to calculate weight on the moon
A man on the moon weighs slightly lighter than he does on earth. The man's mass is constant on both the moon and earth; what brings about the difference in weight then? This is brought about by the different gravitational forces on both the earth and the moon's surface. The gravity at the surface of the moon is about 1/6th of the acceleration on earth. Thus, for a 2-kilogram bag of maize, a force of about 20 newton acts on it on earth, which translates to about 4 newton on the moon. This also happens because the moon is less heavy than the earth.
Even on the earth's surface, there are local variations in the gravitational force due to altitude, latitude, and geological structures. When a body is at rest, it experiences some acceleration due to its own weight and contents. To calculate the gravity load, one needs to get the product between the object's mass, the earth's gravitational acceleration, and the height above the ground in meters.
i.e. mg which is (9.8 m/s2) h
=mgh
Weight is the pull effect felt on a body due to gravity. The formulae for calculating weight as stated earlier is w = m g ¦.. (i)
Where 'w' is the weight of the object, 'm' is the mass of the body measured in kilograms (Kg), and 'g' represents gravitational acceleration, which is 9.8 m/s2 when expressed in meters and 32.2 f/s2 when expressed in term of feet. Weight is a force; thus, the equation (i) above can also be written as F=mg.. .(ii)¦this is expressed in Newton.
How to calculate the magnitude of weight
When calculating weight, the mass of the object in question should always be converted to kilograms. You shouldn't mix up units of measurement; also, during the calculation, you should only use the scientific units available and convert them accordingly if need be. The next step is to figure out the gravitational acceleration depending on where gravity is acting from as the gravitation forces vary.
For example, the gravitation acceleration on the sun is about 274.0 m/s2. This is about 28 times the acceleration on earth. This means, for instance, if a man weighs 20 kilograms, he would have a weight force of 200 Newton on earth and 560 Newton on the sun. This is proof that the wrong estimated value of gravitational acceleration could lead to finding fault results; therefore, enough research must be done concerning this before proceeding with the math.
Lastly, once the mass and gravitational force with the right units have been noted, the values are then substituted in the equation (ii), i.e., F=mg accurately and the final weight force calculated.
Example:
Let's say an object has a mass of 90 Kg; what is its weight on earth?
I) identify the mass and convert it to kilograms
In this case, the mass is already in kilograms, and therefore it remains m=90 Kg
II) Identify the gravitational acceleration
In our case above, the space in question is the earth, which ah a gravitational acceleration of 9.8 m/s2
III) Finally, after identifying the 'g' and 'm,' substitute these values to equation [ii] above
F=mg
=9.8 90 =882N
For this particular object, the weight is 882 Newton
A body has a mass of 100,000grams; what is its weight on the moon?.
In this above example, we have to convert the mass from grams to kilograms by diving by 1000
The mass, in this case, is 100 Kilograms. The gravitational acceleration on the moon is 1.6 m/s2
Therefore, the body's final weight after calculation is; 160 Newton after finding the product of the mass and the gravitational acceleration.
It should be noted that the gravity force in the moon and the earth are different.
Let's take into consideration the weight of a body in a lift. At rest, the weight of the body is W=mg. During upward movement of the lift with an acceleration 'a,' then the weight becomes; W=m (a + g). and during downward movement of the lift with a deceleration '-a' the weight becomes W= m ( g a). This explains the elevator phenomenon where one feels lighter than usual when accelerating down and heavier than usual when accelerating upwards. In the upward movement, one feels heavy because of the force of moving upwards and the gravity acting on it. For the case of a freely falling lift, the weight is zero since there is no pressure on the feet or floor .i.e there is no support force.
In weight calculations, sometimes, one may be required to determine the gravity load of different structures. This load mainly comprises the weight of the structure and its occupants. The gravity load is majorly done for objects at rest. It is like determining the potential energy of a body, and it is measured in joules. Gravity load is determined by finding the products of mass, acceleration, and height. For a particular body, let's say a cuboid-shaped structure, made out of wood, you can determine the mass of this structure from the volume and density of the wood used. With the value of the cuboid's gravitational force and height, you can determine the gravitational load and weight.
Conclusion
We can conclude that the calculations involving weight are not hard or tiresome; only two factors that is mass, and gravitational acceleration, are put into consideration. It is necessary, though, that one stays keen with the scientific units.
Be the first to comment below.
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7980 | https://www.math.emory.edu/~rg/P117Y06.pdf | Digital Object Identifier (DOI) 10.1007/s00373-006-0651-6 Graphs and Combinatorics (2006) 22:217–224 Graphs and Combinatorics © Springer-Verlag 2006 On H-Linked Graphs Michael Ferrara1, Ronald Gould2, Gerard Tansey3 and Thor Whalen4 1 Dept. Math, University of Colorado at Denver, Denver, CO 80217, USA.
2 Dept. Math and CS, Emory University, Atlanta, GA 30322, USA.
3 Dept. Math, Agnes Scott College, Atlanta, GA 30030, USA.
4 Metron Inc. Reston, VA 20190, USA.
Abstract. For a fixed multigraph H, possibly containing loops, with V (H) = {h1, . . . , hk}, we say a graph G is H-linked if for every choice of k vertices v1, . . . , vk in G, there exists a subdivision of H in G such that vi represents hi (for all i). This notion clearly generalizes the concept of k-linked graphs (as well as other properties). In this paper we determine, for a connected multigraph H and for any sufficiently large graph G, a sharp lower bound on δ(G) (depending upon H) such that G is H-linked.
1. Introduction For terms not defined here, see . Let H be a multigraph, possibly containing loops.
For any graph G, let P(G) denote the set of paths in G. An H-subdivision in G is a pair of mappings f1 : V (H) →V (G) and f2 : E(H) →P(G) such that: (i) f1 is injective; (ii) for every edge xy ∈E(H), f2(xy) is an f1(x) −f1(y) path in G and distinct edges of H map to internally disjoint paths in G.
A graph G is H-linked if every injective map f1 : V (H) →V (G) can be extended to an H-subdivision. The vertices in f1(V (H)) are called the ground vertices. Thus, we can say that G has a subdivision of H whose ground vertices are prescribed. This idea originated with Jung , but had not been considered for arbitrary H until recently, when the concept was considered in and .
A graph is k-linked if for every sequence of 2k vertices, v1, . . . , vk, w1, . . . , wk, there are internally disjoint paths P1, . . . , Pk such that Pi joins vi and wi. Clearly, the concept of graphs being H-linked generalizes that of being k-linked. In fact, if H = kK2, then G is k-linked if, and only if, G is H-linked. Further, a graph G is said to be k-ordered if for every sequence of k vertices v1, . . . , vk there is a cycle in G that encounters the vertices v1, . . . , vk in the specified order. If H is a k-cycle, then G is H-linked if and only if G is k-ordered. Thus, the property of being H-linked also generalizes the property of being k-ordered. Other such connections are explored in .
A common question dealing with k-linked graphs is to find the minimum con-nectivity f (k) such that every f (k)-connected graph is k-linked. At this time the 218 M. Ferrara et al.
best know result is that every 10k-connected graph is k-linked . The literature also contains numerous results pertaining to the minimum number of edges needed to assure a graph G contains a subdivision of some graph H. For a survey of results of this type, see .
The purpose of this paper is to provide a sharp minimum degree condition such that any such sufficiently large graph G will be H-linked. This bound will depend on the multigraph H. In order to present our result, we first need some notation.
All multigraphs in this paper will be assumed to have labeled vertices. An edge is proper if it is not a loop. We denote by dH(w), the degree of w in H, which is the number of proper edges incident to w, plus twice the number of loops at w.
Additionally, for a given subgraph H ′ of G, let NH ′(x) denote the set of vertices in H ′ that are adjacent to x in G. This set is the neighborhood of x in H ′. If F1 and F2 are two subgraphs of H, then EH(F1, F2) will represent the set of edges having one end-vertex in F1 and the other end vertex in F2 and eH(F1, F2) = |EH(F1, F2)|.
Now consider a connected multigraph H, possibly with loops. Let η(H) = maxX⊂V (G)e(X, V (G) −X) denote the maximum size of a bipartite subgraph in H, or in other words, the size of a maximum edge cut in H.
Our main result will be: Given a connected multigraph H, possibly with loops, if G is sufficiently large and δ(G) ≥n+η(H)−2 2 , then G is H-linked. To see that this minimum degree is needed, suppose that the multigraph H has η(H) as the maxi-mum size of a bipartite subgraph. Also suppose that this cut determines a partition of V (H) into sets X and Y. Now suppose that G is formed from two complete graphs G1 and G2, each of order m, that intersect on η(H) −1 vertices. If the set S chosen as the image of V (H) under f1 is such that the vertices of X lie in G1 −G2 and the vertices of Y lie in G2 −G1, then clearly G1 ∩G2 is not large enough to allow an embedding of H. Further, δ(G) = m −1. Since |V (G)| = 2m −η(H) + 1, we see that δ(G) = n+η(H)−3 2 . Thus, the minimum degree condition is necessary.
2. Main Result For convenience we let η = η(H). We now state our main result.
Theorem 2.1. Let H be a connected multigraph on k vertices. If G is a graph of suffi-ciently large order n and δ(G) ≥n+η−2 2 , then G is H-linked. Furthermore, on any path between ground vertices in G, there will be at most two intermediate vertices.
Proof. Let G be as above and let S ⊂V (G) be the image of f1, that is, the ground vertices in G. For convenience, until the end of the proof, we will remove any loops from H. Our goal is to show that we can construct the necessary subdivision of H in G, using paths with at most two intermediate vertices between ground vertices.
After this, we will build the paths corresponding to loops. These will also contain at most three intermediate vertices.
On H-Linked Graphs 219 Clearly, by the minimum degree condition, any two nonadjacent vertices x, y ∈ G satisfy |NG(x) ∩NG(y)| ≥η. We first note that for any two vertices x, y ∈S such that f −1 1 (x)f −1 1 (y) ∈E(H) and x and y are adjacent in G, then x and y already have the desired path between them.
We now define the auxiliary graph L as follows: Let V (L) = S and let E(L) consist of all edges xy where x, y ∈S are such that f −1 1 (x)f −1 1 (y) ∈E(H), x and y are not adjacent in G and |NG−S(x) ∩NG−S(y)| ≤3|E(H)| −1.
(1) Our goal will be to link in G these pairs from L with at most two intermediate vertices. We will then show we can link the remaining nonadjacent pairs with single vertices.
□ Claim 2.1.1. The graph L is bipartite.
Proof. Suppose not and let C be the shortest odd cycle in L, and assume that for some integer t > 1 the vertices of C are, in order, x1, y1, . . . , xt−1, yt−1, xt.
By our construction of L, the common neighborhood in G of y1 with either x1 or x2 is at most 3|E(H)| < 3k2. This, combined with the fact that δ(G) > n 2 implies that |NG(x1) ∩NG(x2)| = (1 −o(1))n 2.
(2) Since there are at most k(= o(n)) vertices in C, the reader can verify that the neighborhood intersections of x1, . . . , xt and y1, . . . , yt−1 both have order n 2. How-ever, this contradicts the fact that x1xt is an edge in L, implying that no such odd cycle C exists.
□ Let X and Y be the partite sets of L. We will impose this partition on S and H when necessary. Let ZX = {x ∈X : dL(x) = 0} and ZY = {y ∈Y : dL(y) = 0}.
Further, let UX = {u ∈ x∈X−ZX NG−S(x) : u / ∈( y∈Y−ZY NG−S(y))}, that is, UX is the set of common neighbors in G −S of all vertices in X −ZX, that are also not neighbors of any vertex in Y −ZY . Similarly, let VY = {v ∈ y∈Y−ZY NG−S(y) : v / ∈( x∈X−ZX NG−S(x))}.
Lemma 2.1. |UX| = (1 −o(1)) n 2(k/2)+1 and |UY | = (1 −o(1)) n 2(k/2)+1 .
220 M. Ferrara et al.
Proof. Let C1 = X1 ∪Y1 and (if it exists) C2 = X2 ∪Y2 be nontrivial components of L. Let NX i := x∈Xi NG−S(x), and let NY i := y∈Yi NG−S(y). As in the proof of Claim 2.1.1, |NX 1 | ≥(1 −o(1)) n 2, and |NY 1 | ≥(1 −o(1)) n 2. Also, note that NX 1 and NY 1 are almost disjoint. Now choose vertices x ∈X2 and y ∈Y2 such that xy ∈E(L). The vertices of X2 within G have a large common neighborhood in G −S, and the same is true for the vertices of Y2 within G. Thus, either (|N(x) ∩NX 1 | ≥(1 −o(1))|NX 1 | 2 and |N(y) ∩NY 1 | ≥(1 −o(1))|NY 1 | 2 ); or, (|N(x) ∩NY 1 | ≥(1 −o(1))|NY 1 | 2 and |N(y) ∩NX 1 | ≥(1 −o(1))|NX 1 | 2 ).
If the latter case is true, reverse the labels of X2 and Y2. These observations, combined with (2), imply that all of the vertices in X1 ∪X2 have a common neigh-borhood within G −S of order (1 −o(1)) n 4, and all of the vertices in Y1 ∪Y2 have a common neighborhood within G −S of order (1 −o(1)) n 4. Inductively, it is easy to see that if C1, . . . , Cr are the nontrivial components of L, then | r i=1 UG−S(Xi)| ≥(1 −o(1)) n 2r+1 ≥(1 −o(1)) n 2(k/2)+1 and | r i=1 UG−S(Yi)| ≥(1 −o(1)) n 2r+1 ≥(1 −o(1)) n 2(k/2)+1 , where UG−S(Xi) represents the common neighborhood in G −S of the vertices in Xi, and UG−S(Yi) represents the common neighborhood in G −S of the vertices in Yi.
Let NX and NY denote the common neighborhoods of the vertices in X −ZX and Y −ZY respectively. Note that NX −S = ∩i=1UG−s(X −i). We wish to show almost none of the vertices in NX are adjacent to any vertex in Y −ZY and, similarly, almost none of the vertices in NY are adjacent to any vertex in X −ZX.
We can accomplish this by showing that for any x in X −ZX and all but at most O(k4) vertices v in NG−S(x), v is not adjacent to any vertex in Y −ZY that lies in the same component of L as x. Consider any edge xy in some component C of L.
By the definition of L, we know that in G, x and y have at most 3|E(H)| < 3k2 vertices in their common neighborhood. As above, the vertices of C∩X have at least (1 −o(1)) n 2 other vertices in their common intersection. As each vertex in C ∩X is adjacent in L to at least one vertex from Y, and possibly all of them, there are at least (1 −o(1))n 2 −3k2|X||Y| ≥(1 −o(1))n 2 −3k4 = (1 −o(1))n 2 (3) vertices in the common neighborhood of C ∩X that are not adjacent to any of the vertices in C ∩Y.
On H-Linked Graphs 221 Thus, for any non-trivial component Ci of L, nearly all of the vertices in NX i are not adjacent to Ci ∩Y. Therefore, at least (1 −o(1))|NX| ≥(1 −o(1)) n 2(k/2)+1 (4) vertices in NX are not adjacent to any vertex in Y −ZY and hence are in UX. The proof is similar for VY .
□ Lemma 2.2. Let u ∈UX, v ∈VY , x ∈X and y ∈Y and let z be the number of vertices of degree 0 in L. Then d(u, NG−S(y)) + d(v, NG−S(x)) ≥η −z.
Proof. Let u be a vertex of UX. Let zX and zY be the number of vertices of degree 0 in L lying in the sets X and Y, respectively. Then d(u, NG−S(y)) = d(u) −d(u, S) −d(u, G −S −NG−S(y)) ≥n + η −2 2 −d(u, X) −d(u, Y) −(n −k −1 −dG−S(y)).
However by our definitions of UX and VY , u is not adjacent to any vertices of Y with nonzero degree in L. Thus, d(u, NG−S(y)) ≥n + η −2 2 −|X| −zY −n + k + 1 + dG−S(y) ≥n + η −2 2 −|X| −zY −n + k + 1 + d(y) −d(y, X) −d(y, Y) ≥n + η −2 −|X| −zY −n + k + 1 −d(y, X) −|Y| + 1 ≥η + k −|X| −|Y| −zY −d(y, X) = η −d(y, X) −zY .
Similarly, d(v, NG−S(y)) ≥η −d(x, Y) −zX. Therefore, d(u, NG−S(x)) + d(v, NG−S(y)) ≥2η −d(x, Y) −d(y, X) −zX −zY ≥η −z.
□ Lemma 2.3. Let M be a connected graph and let W, Z be a vertex partition of V (M).
Then for any F ⊆V (M), e(W −F, Z −F) ≤η(M) −|F|.
We proceed by induction. First, assume that F = {v}. Without loss of generality, we may assume that v is in W. Let W ′ denote W −{v}.
If the result fails, then e(W ′, Z) = η(M). As M is connected, either W ′ or Z must contain some x that is adjacent to v in M, say W ′. Then, W ′, Z ∪{v} is a partition of the vertices of M, but e(W ′, Z ∪{v}) > e(W ′, Z) = η(M), a contradiction.
222 M. Ferrara et al.
Now, assume that |F| = p and assume that there is some partition of V (M) into sets W and Z such that e(W −F, Z−F) ≥η(G)−p+1. Since M is connected, there is some vertex v in F such that v is adjacent to, without loss of generality, some w in W. If we denote F −{v} by F ′, the induction hypothesis implies that there is no par-tition of V (M−F ′) into sets W ′ and Z′ such that e(W ′−F ′, Z′−F ′) ≥η(M)−p+2.
However, this is clearly contradicted by choosing W ′ = W and Z′ = Z ∪{v}. Hence the lemma holds.
□ This implies that if there are exactly j isolated vertices in L, then L has at most η(H) −j edges. Now we can proceed to patch the edges of H represented in L.
Assume that we have constructed as many paths in G representing edges from L as possible using two intermediate vertices, and then constructed as many remaining paths as we could using one intermediate vertex. Let xy be an edge in L that has not yet been mapped, and let u be in UX and v be in VY . Applying Lemma 2.3 to the degree sum in Lemma 2.2 we have that |NG−S(u) ∩N(y)| + |NG−S(v) ∩N(x)| ≥|E(L)|.
(5) Let x′y′ be some edge in L with x′ in X and y′ in Y that has already been mapped.
We consider two cases.
Case 1. The path representing x′y′ has been constructed with one intermediate ver-tex.
Let w be the intermediate vertex used and assume that both u and v are adjacent to w in G. Recall that by our choice of u and the definition of UX, ux′ is an edge in G. Then, x′uwy′ would patch the edge x′y′ and increase the number of patched edges using 2 intermediate vertices, a contradiction. Thus, there is at most one edge from u or v to w.
Case 2. The path representing x′y′ has been constructed with two intermediate ver-tices.
Let these intermediate vertices be u′ and v′. One may assume that they lie in UX and VY respectively. We wish to show here, as in Case 1, that there can be at most one edge from u or v to u′ and v′ that is accounted for in (5).
Note that (5) counts only edges from u to the neighborhood of y and from v to the neighborhood of x. Since u′ is not adjacent to any y in Y and v′ is not adjacent to any x in X, the only possible way to account for two edge from (5) is if u′v and v′u were both edges in G. However, then we could use the paths xu′vy and x′uv′y to construct an additional path, contradicting our maximality assumption.
Thus, we have used at most one edge from the degree count in (5) for each path already constructed. As there are at least η(H) −j such edges, and at most η(H) −j −1 paths already constructed, either u has an unused common neighbor with y or v has an unused common neighbor with x, allowing us to construct the path representing xy. Hence, by our maximality assumption, we can construct all of the paths for edges represented in L.
Now that we have joined all of the pairs of vertices that correspond to edges in L, it remains to show that we can join those pairs of vertices that correspond to the On H-Linked Graphs 223 remaining edges. Let T ⊂E(H) represent those edges already linked in G such that E(L) ⊂T . As above, assume that we have used at most 2 intermediate vertices to link each pair.
Let x and y be vertices in S corresponding to an edge in E(H) −T . As xy is not in L, (1) implies that there are at least 3|E(H)| vertices in G adjacent to both x and y. At this point, we have linked at most |E(H)| −1 pairs of vertices in T with at most 2 intermediate vertices each. Together with the k −2 other vertices in S, there are at most 2(|E(H)| −1) + k −2 < 3|E(H)| (6) vertices accounted for thus far. Hence, there is some vertex w adjacent to both x and y that is not in S and is not being used to link any pair of vertices in S that correspond to an edge in H. We may therefore link x and y in G with the path xwy.
At the start of the proof, we removed all loops from H. Thus far, we have man-aged to embed this subgraph of H. Now, assume that H had loops l1, . . . , lm where m may be arbitrarily large with respect to k. We will patch these loops using at most three intermediate vertices. Consider some loop li+1 adjacent to some vertex x in H, and assume that we have already dealt with l1, . . . , li. There are at most 3k2 + k + 3i = o(n) vertices from the neighborhood of x already in use in our embedding. Call this set of vertices V (x). If there is a pair of adjacent vertices in N(x) \ V (x), we can easily embed li+1. If there is not, however, N(x)\V (x) is an independent set in G of order at least (1 −o(1)) n 2. In this case, our minimum degree condition implies that for any pair of vertices a, b in N(x) \ V (x) there is some vertex p in V (G) \ N(x) that is not being utilized in our embedding and is adjacent to both a and b. If so, we will construct the path representing li+1 with the path apb.
□ 3. Remarks As stated, this result holds for n sufficiently large. If H does not have exceedingly many loops relative to k, then we require the order of n to exceed 2(k/2)+1 as in Lemma 2.1. It is possible, however, that H may have arbitrarily many loops, in which case, our bound for n would be based on |E(H)|. A. Kostochka and G. Yu have provided a linear bound on n for the case that H is a connected loopless multigraph with minimum degree at least 2.
Once a graph G is known to be H-linked, it then becomes interesting to know under what conditions an H-subdivision can be extended to a spanning H-subdi-vision. This question is addressed in and provides a generalization of a number of well-known results.
Acknowledgements. The authors would like to thank the referees for their helpful comments.
224 M. Ferrara et al.
References 1. Chartrand, G., Lesniak, L.: Graphs & Digraphs, 3rd edn. Chapman & Hall, London, 1996 2. Jung, H.A.: Eine verallgemeinerung des n-fachen zusammenhangs f¨ ur graphen. Math Ann 187, 95–103 (1970) 3. Gould, R.J., Whalen, T.: Subdivision extendability. Graph and Combinatorics (to appear) 4. Kostochka, A., Yu, G.: An extremal problem for H-linked graphs. J. Graph Theory (to appear) 5. Mader, W .: Topological minors in graphs of minimum degree n. Contemporary trends in discrete mathematics (ˇ Stiˇ rin Castle, 1997), 199–211, DIMACS Ser. Discrete Math. Theo-ret. Comput. Sci., 49, Amer. Math. Soc., Providence, RI, 1999 6. Thomas, R., Wollen, P.: An improved bound for graph linkage. European J. Combinatorics 26, 309–324 (2005) 7. Whalen, T.: Degree conditions and relations to distance, extendability, and levels of con-nectivity in graphs. Ph.D. Thesis, Emory University, August, 2003 Received: June 24, 2004 Final version received: December 5, 2005 |
7981 | https://fiveable.me/key-terms/combinatorics/cycle-elimination-lemma | Cycle Elimination Lemma - (Combinatorics) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Cycle Elimination Lemma
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key term - Cycle Elimination Lemma
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Definition
The Cycle Elimination Lemma states that in a weighted graph, if a cycle is formed by adding an edge to a spanning tree, that edge can be removed without affecting the minimum spanning tree properties of the graph. This lemma is crucial for proving the correctness of algorithms used to find minimum spanning trees, as it helps identify unnecessary edges that can be eliminated while maintaining optimality.
5 Must Know Facts For Your Next Test
The Cycle Elimination Lemma helps streamline the process of constructing minimum spanning trees by identifying redundant edges.
This lemma emphasizes that if an edge creates a cycle in a spanning tree, it can be safely discarded as it won't contribute to minimizing the total edge weight.
In practical applications, using the Cycle Elimination Lemma can significantly reduce the number of edges to consider when implementing algorithms like Kruskal's and Prim's.
Understanding this lemma is key for proving that both Kruskal's and Prim's algorithms yield valid minimum spanning trees.
The lemma underlines the relationship between cycles and trees in graph theory, showcasing how cycles can indicate potential inefficiencies in graph connectivity.
Review Questions
How does the Cycle Elimination Lemma contribute to finding a minimum spanning tree in a weighted graph?
The Cycle Elimination Lemma simplifies the process of finding a minimum spanning tree by allowing us to identify and remove edges that form cycles. When constructing a spanning tree, if adding an edge creates a cycle, this lemma tells us that we can eliminate that edge without losing the minimum spanning tree properties. This leads to more efficient algorithms since we can focus only on relevant edges that help minimize the total weight.
Discuss how Kruskal's Algorithm utilizes the Cycle Elimination Lemma in its approach to constructing minimum spanning trees.
Kruskal's Algorithm relies heavily on the Cycle Elimination Lemma by processing edges in increasing order of weight. As edges are added to the growing forest, the algorithm checks for cycles using union-find data structures. If adding an edge creates a cycle, according to the lemma, it is discarded since it won't contribute to forming a minimum spanning tree. This ensures that only edges leading to optimal connections remain in the final result.
Evaluate the importance of the Cycle Elimination Lemma within the broader context of graph theory and algorithm design.
The Cycle Elimination Lemma holds significant importance within graph theory and algorithm design as it provides foundational principles for understanding how cycles affect connectivity and efficiency in graphs. By applying this lemma, algorithm designers can create more effective solutions for problems involving network design and resource optimization. Furthermore, its application in well-known algorithms like Kruskal's and Prim's reinforces its relevance, showcasing how eliminating unnecessary complexity leads to faster computations and more reliable results in real-world scenarios.
Related terms
Spanning Tree:A spanning tree is a subgraph that includes all the vertices of the original graph, connected in such a way that there are no cycles and the total edge weight is minimized.
Kruskal's Algorithm:
A greedy algorithm that finds a minimum spanning tree by sorting edges by weight and adding them one by one, while ensuring no cycles are formed.
Prim's Algorithm:
A greedy algorithm that builds a minimum spanning tree by starting from an arbitrary vertex and continuously adding the cheapest edge that expands the tree until all vertices are included.
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7982 | https://www.mheducation.com/unitas/school/explore/sites/reveal-math/reveal-math-algebra-teacher-edition.pdf | Module 4 Linear and Nonlinear Functions Module 4 • Linear and Nonlinear Functions 207a Module Goals • Students graph linear, piecewise-defined, step, and absolute value functions.
• Students find and interpret the rate of change and slope of lines.
• Students identify the effects of transformations on the graphs of linear and absolute value functions.
Focus Domain: Functions Standards for Mathematical Content: F.IF.7a Graph linear and quadratic functions and show intercepts, maxima, and minima.
F.IF.7b Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. F.BF.3 Identify the effect on the graph of replacing f (x) by f (x) + k, k f (x), f (kx), and f (x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology.
Also addresses A.CED.2, A.REI.10, F.IF.4, F.IF.6, F.BF.1a, F.BF.2, F.LE.1a, F.LE.2, and F.LE.5.
Standards for Mathematical Practice: All Standards for Mathematical Practice will be addressed in this module. Be Sure to Cover To completely cover F.LE.1a, go online to assign the following activity: • Linear Growth Patterns (Expand 4-3) Coherence Vertical Alignment Next Students will create linear equations and analyze data to make predictions. A.CED.2 Now Students write and graph linear and nonlinear equations. F.IF.7a, F.IF.7b, F.BF.3 Previous Students graphed functions and interpreted key features in graphs of functions. F.IF.1, F.IF.4 Rigor The Three Pillars of Rigor To help students meet standards, they need to illustrate their ability to use the three pillars of rigor. Students gain conceptual understanding as they move from the Explore to Learn sections within a lesson. Once they understand the concept, they practice procedural skills and fluency and apply their mathematical knowledge as they go through the Examples and Practice.
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION EXAMPLE & PRACTICE LEARN EXPLORE Suggested Pacing Lessons Standards 45-min classes 90-min classes Module Pretest and Launch the Module Video 1 0.5 4-1 Graphing Linear Functions A.REI.10, F.IF.7a, F.LE.5 1 0.5 4-2 Rate of Change and Slope F.IF.6, F.LE.5 1 0.5 4-3 Slope-Intercept Form A.CED.2, F.IF.7a, F.LE.5 2 1 4-3 Expand Linear Growth Patterns F.LE.1a 1 0.5 4-4 Transformations of Linear Functions F.IF.7a, F.BF.3 2 1 4-5 Arithmetic Sequences F.BF.1a, F.BF.2, F.LE.2 1 0.5 4-6 Piecewise and Step Functions F.IF.4, F.IF.7b 1 0.5 4-7 Absolute Value Functions F.IF.7b, F.BF.3 2 1 Put It All Together: Lessons 4-6 through 4-7 1 0.5 Module Review 1 0.5 Module Assessment 1 0.5 Total Days 15 7.5 Analyze the Probe Review the probe prior to assigning it to your students.
In this probe, students will determine which graph matches the correct function and explain their choices.
Targeted Concepts Certain modifications to the parent function of an absolute value function will result in predictable transformations of the graph.
Targeted Misconceptions • Students may not recognize a horizontal transformation and/or predict an incorrect direction of a horizontal transformation.
• Students may not recognize a vertical transformation and/or predict an incorrect direction of a vertical transformation.
Use the Probe after Lesson 4-7.
Collect and Assess Student Answers If the student selects these responses… Then the student likely...
1. D 3. B recognizes the horizontal shift but fails to use the opposite value of the number associated with x to determine the direction of the shift. Example: For Item 1, the student recognizes that positive 4 is associated with the x-value (horizontal shift) but moves the graph to the right.
2. A 4. C recognizes the vertical shift but fails to use the same value of the number associated with y to determine the direction of the shift. Example: For Item 2, the student recognizes that positive 4 is associated with the y-value (vertical shift) but moves the graph down.
5. A recognizes the vertical shift but is confused with the direction of the shift when the number is placed on the same side as y. Example: For Item 5, the student recognizes that negative 4 is associated with the y-value (vertical shift) but does not solve for y before using the “rules” of transformation and moves the graph down.
1. A 2. D 3. C 4. B confuses a horizontal shift with a vertical shift Example: For Item 3, the student incorrectly moves the graph up 4 units instead of to the right 4 units.
Take Action After the Probe Design a plan to address any possible misconceptions. You may wish to assign the following resources.
• Absolute Value Functions • Lesson 4-7, all Learns, all Examples Revisit the probe at the end of the module to be sure that your students no longer carry these misconceptions.
Correct Answers: 1. B 2. C 3. D 4. A 5. C Module Resource MATH PROBES CHERYL TOBEY Formative Assessment Math Probe Absolute Value Functions 207b Module 4 • Linear and Nonlinear Functions Copyright © McGraw-Hill Education Essential Question What can a function tell you about the relationship that it represents?
Module 4 Linear and Nonlinear Functions What Will You Learn?
Place a check mark (✓) in each row that corresponds with how much you already know about each topic before starting this module.
KEY — I don’t know. — I’ve heard of it. — I know it!
Before After graph linear equations by using a table graph linear equations by using intercepts find rates of change determine slopes of linear equations write linear equations in slope-intercept form graph linear functions in slope-intercept form translate, dilate, and reflect linear functions identify and find missing terms in arithmetic sequences write arithmetic sequences as linear functions model and use piecewise functions, step functions, and absolute value functions translate absolute value functions Foldables Make this Foldable to help you organize your notes about functions. Begin with five sheets of grid paper.
1. Fold five sheets of grid paper in half from top to bottom.
2. Cut along fold. Staple the eight half-sheets together to form a booklet.
3. Cut tabs into margin. The top tab is 4 lines wide, the next tab is 8 lines wide, and so on. When you reach the bottom of a sheet, start the next tab at the top of the page.
4. Label each tab with a lesson number. Use the extra pages for vocabulary.
4-4 4-1-4-3 4-5 4-6-4-7 1 2 3 4 Module 4 • Linear and Nonlinear Functions 207 Program: Reveal Math Component: Module Intro COM4_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 207_208_HSM_NA_S_A1M04_CO_662599.indd Page 207 14/01/19 5:05 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Interactive Presentation Module 4 • Linear and Nonlinear Functions 207 The Ignite! activities, created by Dr. Raj Shah, cultivate curiosity and engage and challenge students. Use these open-ended, collaborative activities, located online in the module Launch section, to encourage your students to develop a growth mindset towards mathematics and problem solving. Use the teacher notes for implementation suggestions and support for encouraging productive struggle. Essential Question At the end of this module, students should be able to answer the Essential Question.
What can a function tell you about the relationship that it represents? Sample answer: It can tell you about the rate of change, whether the relationship is positive or negative, the locations of the x- and y-intercepts, and what points fall on the graph.
What Will You Learn?
Prior to beginning this module, have your students rate their knowledge of each item listed. Then, at the end of the module, you will be reminded to have your students return to these pages to rate their knowledge again. They should see that their knowledge and skills have increased.
Focus As students read and study this module, they should show examples and write notes about linear functions and relations.
Teach Have students make and label their Foldables as illustrated. Students should label the front of each half page with the lesson title. On the back of each of these pages, they can record concepts and notes from that particular lesson. When to Use It Encourage students to add to their Foldables as they work through the module and to use them to review for the module test.
Launch the Module For this module, the Launch the Module video uses real-world scenarios to illustrate how functions and their graphs can be used to model both linear and nonlinear relationships. Students learn about using graphs to model the change in altitude of an airplane and the change in strength of a Wi-Fi signal.
What Vocabulary Will You Learn? ELL As you proceed through the module, introduce the key vocabulary by using the following routine.
Define The slope of a line is the rate of change in the y-coordinates (rise) for the corresponding change in the x-coordinates (run) for points on the line.
Example A line passes through the points (1, 4) and (3, 8).
Ask What is the slope of the line? 2 Are You Ready?
Students may need to review the following prerequisite skills to succeed in this module.
• identifying domain and range • identifying slopes • translating and reflecting geometric figures • finding the next terms in patterns • graphing linear functions • evaluating absolute value expressions ALEKS is an adaptive, personalized learning environment that identifies precisely what each student knows and is ready to learn, ensuring student success at all levels.
You can use the ALEKS pie report to see which students know the topics in the Functions and Lines module—who is ready to learn these topics and who isn’t quite ready to learn them yet—in order to adjust your instruction as appropriate. Mindset Matters Collaborative Risk Taking Some students may be averse to taking risks during math class, like sharing an idea, strategy, or solution. They may worry about their grades or scores on tests, or some might feel less confident solving math problems, especially in front of their peers.
How Can I Apply It?
Assign the Practice problems of each lesson and encourage students to take risks as they solve problems, try new paths, and discuss their strategies with their partner or group. Copyright © McGraw-Hill Education What Vocabulary Will You Learn?
Check the box next to each vocabulary term that you may already know.
☐ absolute value function ☐ arithmetic sequence ☐ common difference ☐ constant function ☐ dilation ☐ family of graphs ☐ greatest integer function ☐ identity function ☐ interval ☐ nth term of an arithmetic sequence ☐ parameter ☐ parent function ☐ piecewise-defined function ☐ piecewise-linear function ☐ rate of change ☐ reflection ☐ sequence ☐ slope ☐ step function ☐ term of a sequence ☐ transformation ☐ translation ☐ vertex Are You Ready?
Complete the Quick Review to see if you are ready to start this module. Then complete the Quick Check.
Quick Review Example 1 Graph A(3, –2) on a coordinate grid.
Start at the origin. Since the x-coordinate is positive, move 3 units to the right. Then move 2 units down since the y-coordinate is negative. Draw a dot and label it A.
Example 2 Solve x − 2y = 8 for y. x − 2y = 8 Original expression x − x − 2y = 8 − x Subtract x from each side. −2y = 8 − x Simplify. − 2y _ − 2 = 8 − x − 2 Divide each side by −2. y = 1 __ 2 x − 4 Simplify. Quick Check Graph and label each point on the coordinate plane.
1. B(−3, 3) 2. C(−2, 1) 3. D(3, 0) 4. E(−5, −4) 5. F(0, −3) 6. G(2, −1) Solve each equation for y. 7. 3x + y = 1 8. 8 − y = x 9. 5x − 2y = 12 10. 3x + 4y = 10 11. 3 − 1 __ 2 y = 5x 12. y + 1 _ 3 = x + 2 How did you do?
Which exercises did you answer correctly in the Quick Check? Shade those exercise numbers below. 1 2 3 4 5 6 7 8 9 10 11 12 y x O A 3 2 208 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Module Intro COM4A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 1-6. See margin. 7. y = −3x + 1 8. y = −x + 8 9. y = 5 __ 2 x − 6 10. y = − 3 __ 4 x + 5 __ 2 11. y = −10x + 6 12. y = 3x + 5 207_208_HSM_NA_S_A1M04_CO_662599.indd Page 208 5/9/19 3:25 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
y x O C D E F G B Answer 1-6.
208 Module 4 • Linear and Nonlinear Functions LESSON GOAL Students graph linear functions by using tables and intercepts.
1 LAUNCH Launch the lesson with a Warm Up and an introduction.
2 EXPLORE AND DEVELOP Explore: Points on a Line Develop: Graphing Linear Functions by Using Tables • Graph by Making a Table • Choose Appropriate Domain Values • Graph y = a • Graph x = a Explore: Lines Through Two Points Develop: Graphing Linear Functions by Using the Intercepts • Graph by Using Intercepts • Use Intercepts You may want your students to complete the Checks online.
3 REFLECT AND PRACTICE Exit Ticket Practice DIFFERENTIATE View reports of student progress on the Checks after each example.
Resources AL OL BL ELL Remediation: Proportional Relationships and Slope ● ● ● Extension: Graphing Equations in Three Dimensions ● ● ● Language Development Handbook Assign page 20 of the Language Development Handbook to help your students build mathematical language related to graphing linear functions.
ELL You can use the tips and suggestions on page T20 of the handbook to support students who are building English proficiency. Lesson 4-1 Graphing Linear Functions Lesson 4-1 • Graphing Linear Functions 209a A.REI.10, F.IF.7a, F.LE.5 Suggested Pacing 90 min 0.5 day 45 min 1 day Focus Domain: Algebra, Functions Standards for Mathematical Content: A.REI.10 Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).
F.IF.7a Graph linear and quadratic functions and show intercepts, maxima, and minima.
F.LE.5 Interpret the parameters in a linear or exponential function in terms of a context.
Standards for Mathematical Practice: 1 Make sense of problems and persevere in solving them.
5 Use appropriate tools strategically.
Coherence Vertical Alignment Rigor The Three Pillars of Rigor 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Conceptual Bridge In this lesson, students expand on their understanding of and fluency with linear functions (first studied in Grade 8) to graphing linear functions by using a table and by using intercepts. They apply their understanding of linear functions by solving real-world problems.
Next Students will investigate rate of change and slope. F.IF.6, F.LE.5 Now Students graph linear functions using tables and intercepts. A.REI.10, F.IF.7a, F.LE.5 Previous Students sketched graphs and compared graphs of functions.
F.IF.4, F.IF.9 Interactive Presentation 1 LAUNCH Warm Up Prerequisite Skills The Warm Up exercises address the following prerequisite skill for this lesson: • identifying domain and range Answers: 1. D: {2, 3, 4, 7}, R: {2, 3, 4}; yes 2. D: {0.9, 1.4, 3.2}, R: {0.8, 1.4}; yes 3. D: {–5, –1, 3, 4}, R: {–4, –1, 3, 4}; yes 4. D: {Ohio, Texas}, R: {Cleveland, Columbus, Dallas, Houston}; no 5. D: {dog, fish, cat, other, bird, rabbit, hamster, horse, snake}, R: {5, 6, 20, 21, 22, 24, 42, 60, 71}; yes Launch the Lesson Launch the Lesson MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain how the verbal description of the relationship between the device’s strength and the distance from the router can be modeled by a function, which can be used to create a table of values and a graph. Go Online to find additional teaching notes and questions to promote classroom discourse.
Today’s Standards Tell students that they will address these content and practice standards in this lesson. You may wish to have a student volunteer read aloud How can I meet these standards? and How can I use these practices?, and connect these to the standards.
See the Interactive Presentation for I Can statements that align with the standards covered in this lesson.
Mathematical Background The graph of a linear function is a line. The coordinates of the points on the line are the solutions of the related linear equation. If you know at least two solutions of the equation, you can use them to graph the line. You can also use the x- and y-intercepts to graph the line. The intercepts can be found by alternately replacing x and y with 0. The line that connects the intercepts is the graph of the linear equation.
Warm Up 209b Module 4 • Linear and Nonlinear Functions A.REI.10, F.IF.7a, F.LE.5 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Points on a Line Objective Students explore the relationship between graphs of linear equations and their solutions.
MP MP Teaching the Mathematical Practices 7 Look for a Pattern Help students to see the pattern in this Explore.
Ideas for Use Recommended Use Present the Inquiry Question, or have a student volunteer read it aloud. Have students work in pairs to complete the Explore activity on their devices. Pairs should discuss each of the questions. Monitor student progress during the activity. Upon completion of the Explore activity, have student volunteers share their responses to the Inquiry Question. What if my students don’t have devices? You may choose to project the activity on a whiteboard. A printable worksheet for each Explore is available online. You may choose to print the worksheet so that individuals or pairs of students can use it to record their observations.
Summary of the Activity Students will complete guiding exercises throughout the Explore activity. Students will be presented with a linear equation and its graph. Several points on the coordinate plane are marked and labeled, some on the graph, and some not on the graph. Students will record the coordinates of the marked points, and determine whether each pair of coordinates makes the equation true. Then, students will answer the Inquiry Question.
(continued on the next page) Explore Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students tap on each point to explore the relationship between points on a graph and solutions of an equation.
TAP Students complete a table and answer questions about the points that make an equation true.
TYPE a Lesson 4-1 • Graphing Linear Functions 209c A.REI.10 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Points on a Line (continued) Questions Have students complete the Explore activity.
Ask: • Why is it important to know if coordinates make an equation true? Sample answer: It is important to know when the substituted values make both sides of the equation equal. The coordinates that make the equation true are solutions of the equation.
• Given a graph of a linear function, how could you find a solution of the related equation? Sample answer: I could look for coordinates on the line because any point on the line is a solution of the related equation. Inquiry How is the graph of a linear equation related to its solutions? Sample answer: The graph of a line is all of the solutions of its equation plotted on a coordinate plane. Go Online to find additional teaching notes and sample answers for the guiding exercises.
Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students respond to the Inquiry Question and can view a sample answer.
TYPE a 209d Module 4 • Linear and Nonlinear Functions A.REI.10 2 EXPLORE AND DEVELOP Explore Lines Through Two Points Objective Students use a sketch to explore the number of lines that pass through two points.
MP MP Teaching the Mathematical Practices 5 Use Mathematical Tools Point out that to solve the problem in this Explore, students will need to use a sketch. Work with students to explore and deepen their understanding of lines through two points.
Ideas for Use Recommended Use Present the Inquiry Question, or have a student volunteer read it aloud. Have students work in pairs to complete the Explore activity on their devices. Pairs should discuss each of the questions. Monitor student progress during the activity. Upon completion of the Explore activity, have student volunteers share their responses to the Inquiry Question. What if my students don’t have devices? You may choose to project the activity on a whiteboard. A printable worksheet for each Explore is available online. You may choose to print the worksheet so that individuals or pairs of students can use it to record their observations.
Summary of the Activity Students will complete guiding exercises throughout the Explore activity. Students will use a sketch to explore the number of lines that can be drawn through a single point. They will then explore the number of lines that can be drawn through two points. Then, students will answer the Inquiry Question.
(continued on the next page) Explore Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students answer questions about the graphed functions.
TYPE a Students use a sketch to explore the graphs of linear functions.
WEB SKETCHPAD Interactive Presentation Lesson 4-1 • Graphing Linear Functions 209e HEADER Interactive Presentation Explore Lines Through Two Points (continued) Questions Have students complete the Explore activity.
Ask: • Can you graph a function from a table that has only two points? Sample answer: As long as you know that the function is linear, it is okay for the table to only list two points.
• When graphing, do you think it would be better to use two points close together or farther apart? Sample answer: Farther apart would help you get a better idea of where the line should be drawn. If the points are too close together, you might not have your ruler or line tool lined up correctly. Inquiry How many lines can be formed with two given points? Sample answer: There is only one line that can be formed with two given points. Go Online to find additional teaching notes and sample answers for the guiding exercises.
Explore Students respond to the Inquiry Question and can view a sample answer.
TYPE a 2 EXPLORE AND DEVELOP 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION 209f Module 4 • Linear and Nonlinear Functions Interactive Presentation Learn Graphing Linear Functions by Using Tables Objective Students graph linear functions by making a table of values.
MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the table, coordinates, equation, and graph of a linear function.
What Students Are Learning Students come to understand that although a table of values can be used to construct the graph of a linear function, the graph represents all of the solutions of the equation. They learn that every point on the graph represents a pair of coordinates that is a solution of the equation.
Example 1 Graph by Making a Table MP MP Teaching the Mathematical Practices 3 Construct Arguments In this example, students will use stated assumptions, definitions, and previously established results to construct an argument.
Questions for Mathematical Discourse AL What values are in the domain of the function? all real numbers OL Why is it helpful to choose both positive and negative values? Sample answer: Choosing positive and negative values gives you a better idea of what the graph will look like and will show you where the graph crosses the y-axis.
BL What should you do if one of the points you graph is not on the same line as the others? Sample answer: Check your work to see if you miscalculated the y-value. Go Online • Find additional teaching notes.
• View performance reports of the Checks.
• Assign or present an Extra Example.
Copyright © McGraw-Hill Education Lesson 4-1 Graphing Linear Functions Learn Graphing Linear Functions by Using Tables A table of values can be used to graph a linear function. Every ordered pair that makes the equation true represents a point on its graph. So, a graph represents all the solutions of an equation.
Linear functions can be represented by equations in two variables.
Example 1 Graph by Making a Table Graph -2x - 3 = y by making a table. Step 1 Choose any values of x from the domain and make a table. Step 2 Substitute each x-value into the equation to find the corresponding y-value. Then, write the x- and y-values as an ordered pair.
Step 3 Graph the ordered pairs in the table and connect them with a line. y x O −4 −2 −8−6 8 6 4 2 4 6 2 8 −4 −6 −8 Talk About It!
What values of x might be easiest to use when graphing a linear equation when the x-coefficient is a whole number? Justify your argument. Go Online You can complete an Extra Example online.
x -2x - 3 y (x, y) -4 -2 0 1 3 Study Tip Exactness Although only two points are needed to graph a linear function, choosing three to five x-values that are spaced out can verify that your graph is correct. Explore Points on a Line Online Activity Use an interactive tool to complete an Explore. INQUIRY How is the graph of a linear equation related to its solutions?
Today’s Goals ● Graph linear functions by making tables of values.
● Graph linear functions by using the x- and y-intercepts.
Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-1 • Graphing Linear Functions 209 Sample answer: -2, -1, 0, 1, and 2 would be easiest to use because they are small and therefore easy to substitute into the equation to find the y-values.
-2(-4) -3 5 (-4, 5) -2(-2) -3 1 (-2, 1) -2(0) -3 -3 (0, -3) -2(1) -3 -5 (1, -5) -2(3) -3 -9 (3, -9) 209_218_HSM_NA_S_A1M04_L01_662599.indd Page 209 5/17/19 3:26 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 1 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students discuss the x-values that would be easiest to use when graphing a linear equation if the coefficient of x is an integer.
TYPE Students move through the steps to see how to make a table of values for a line.
TAP a Lesson 4-1 • Graphing Linear Functions 209 A.REI.10, F.IF.7a, F.LE.5 Interactive Presentation 2 EXPLORE AND DEVELOP Example 2 Choose Appropriate Domain Values MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the equation, table, and graph in this example.
Questions for Mathematical Discourse AL What values are in the domain? all real numbers OL Why were the values selected for x in the table -8, -4, 0, 4, and 8? Sample answer: They were all multiples of 4 and since the coefficient is 1 __ 4 this makes multiplication easier.
BL What would happen if you used multiples of 2 for x in the table? Sample answer: Multiples of 2 that are also multiples of 4 would cancel out the denominator, but others would reduce to have a denominator of 2.
Common Error Some students may make calculation errors when working with a coefficient that is a fraction. Help them avoid this by suggesting that they write the integer that they are substituting for x as a fraction with a denominator of 1.
Copyright © McGraw-Hill Education Check Graph y = 2x + 5 by using a table.
x y -5 -3 -1 0 2 Example 2 Choose Appropriate Domain Values Graph y = 1 __ 4 x + 3 by making a table. Step 1 Make a table.
Step 2 Find the y-values. Step 3 Graph the ordered pairs in the table and connect them with a line. Check Graph y = 3 __ 5 x - 2 by making a table. x y -10 -5 0 5 10 Go Online You can complete an Extra Example online. Think About It!
What are some values of x that you might choose in order to graph y = 1 _ 7 x - 12?
Watch Out!
Equivalent Equations Sometimes, the variables are on the same side of the equal sign. Rewrite these equations by solving for y to make it easier to find values for y.
x 1 __ 4 x + 3 y (x, y) -8 -4 0 4 8 y x O −4 −6 −2 −8 8 4 6 2 4 6 2 8 −4 −6 −8 Your Notes y x O −4 −8 −2 −6 8 4 6 2 4 8 2 6 −4 −8 −6 −8 −4−2 −6 8 4 6 2 8 4 6 2 −8 −4 −6 y x O 210 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: {-14, -7, 0, 7, 14} 1 __ 4 (-8) + 3 1 (-8, 1) 1 __ 4 (-4) + 3 2 (-4, 2) 1 __ 4 (0) + 3 3 (0, 3) 1 __ 4 (4) + 3 4 (4, 4) 1 __ 4 (8) + 3 5 (8, 5) -5 -1 3 5 9 -8 -5 -2 1 4 209_218_HSM_NA_S_A1M04_L01_662599.indd Page 210 14/01/19 5:05 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 2 Students give the possible domain values for a given linear equation with a rational slope.
TYPE 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION a Students use a sketch to graph the ordered pairs from the table of values.
WEB SKETCHPAD 210 Module 4 • Linear and Nonlinear Functions A.REI.10, F.IF.7a, F.LE.5 Interactive Presentation Copyright © McGraw-Hill Education Example 3 Graph y = a Graph y = 5 by making a table. Step 1 Rewrite the equation. y = 0x + 5 Step 2 Make a table. Step 3 Graph the line.
The graph of y = 5 is a horizontal line through (x, 5) for all values of x in the domain.
Example 4 Graph x = a Graph x = -2. You learned in the previous example that equations of the form y = a have graphs that are horizontal lines. Equations of the form x = a have graphs that are vertical lines.
The graph of x = −2 is a vertical line through (−2, y ) for all real values of y. Graph ordered pairs that have x-coordinates of -2 and connect them with a vertical line.
Check Graph x = 6. y x O Think About It!
In general, what does the graph of an equation of the form y = a, where a is any real number, look like? Think About It!
Is x = a a function? Why or why not? Go Online You can complete an Extra Example online.
x 0x + 5 y (x, y) -2 -1 0 1 2 y x O y x O −4 −8 −2 −6 8 4 6 2 4 8 2 6 −4 −8 −6 Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-1 • Graphing Linear Functions 211 Sample answer: a horizontal line through (x, a) for all values of x in the domain No; sample answer: The element a in the domain is paired with more than one element of the range.
0(-2) + 5 5 (-2, 5) 0(-1) + 5 5 (-1, 5) 0(0) + 5 5 (0, 5) 0(1) + 5 5 (1, 5) 0(2) + 5 5 (2, 5) 209_218_HSM_NA_S_A1M04_L01_662599.indd Page 211 5/10/19 3:00 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 3 Students explain what the graph of an equation in the form y = a will look like.
TYPE a 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 3 Graph y = a MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Questions for Mathematical Discourse AL How is this equation different from other linear equations that you have worked with? Sample answer: The coefficient of x is zero.
OL What would the table look like for other values of x? Sample answer: The y-values would all be 5.
BL Is the graph a function? Explain Yes; sample answer: This is a function because it passes the vertical line test.
Common Error Some students may interpret an equation such as y = 5 as a point, not a line. Help them to see that although the equation specifies that y = 5, x could be infinitely many values. Use a table to show how this leads to the graph of y = 5 consisting of more than one point.
Example 4 Graph x = a MP MP Teaching the Mathematical Practices 5 Use Mathematical Tools Point out that to solve the problem in this example, students will need to use a sketch. Work with students to explore and deepen their understanding of graphs of horizontal lines.
Questions for Mathematical Discourse AL How is this equation different from other linear equations that you have worked with? Sample answer: There is only one variable, x.
OL What is the x-intercept for the graph of an equation of the form x = a? (a, 0) BL Why does every point of the form (-2, y) satisfy the equation? Sample answer: Because the equation has no y-variable, substituting any point (-2, y) into the equation will result in the true statement -2 = -2.
DIFFERENTIATE Language Development Activity AL BL ELL IF students are having difficulty remembering which equations represent horizontal lines and which represent vertical lines, THEN have them use the acronyms HOY and VUX to remember which is which. HOY stands for “Horizontal, 0 slope, y =,” and VUX stands for “Vertical, Undefined slope, x = .” Students complete the Check online to determine whether they are ready to move on.
CHECK Students move through the steps to graph a line in the form y = a.
TAP Lesson 4-1 • Graphing Linear Functions 211 A.REI.10, F.IF.7a, F.LE.5 Interactive Presentation 2 EXPLORE AND DEVELOP Learn Graphing Linear Functions by Using the Intercepts Objective Students graph linear functions by using the x- and y-intercepts.
MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Common Misconception Some students may think that the x- and y-intercepts are the coefficients of x and y. Use an example such as 3x + 2y = 12 to review the process of finding intercepts and show that neither coefficient is an intercept.
Copyright © McGraw-Hill Education Go Online You can complete an Extra Example online. Think About It!
Why are the x- and y-intercepts easy to find? Think About It!
What does a line that only has an x-intercept look like? a line that only has a y-intercept?
Learn Graphing Linear Functions by Using the Intercepts You can graph a linear function given only two points on the line. Using the x- and y-intercepts is common because they are easy to find. The intercepts provide the ordered pairs of two points through which the graph of the linear function passes.
Example 5 Graph by Using Intercepts Graph -x + 2y = 8 by using the x- and y-intercepts.
To find the x-intercept, let . Original equation Replace y with 0. Simplify. Divide.
This means that the graph intersects the x-axis at . To find the y-intercept, let . Original equation Replace x with 0. Simplify. Divide.
This means that the graph intersects the y-axis at . Graph the equation.
Step 1 Graph the x-intercept.
Step 2 Graph the y-intercept.
Step 3 Draw a line through the points.
y x O −6 −8 −9 −5−4−3−2 −1 1 −7 4 5 2 3 1 −2 −4 −5 −3 Study Tip Tools When drawing lines by hand, it is helpful to use a straightedge or a ruler.
Explore Lines Through Two Points Online Activity Use graphing technology to complete an Explore. INQUIRY How many lines can be formed with two given points? Go Online You can watch a video to see how to graph linear functions.
212 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: Because either the x- or y-value of an intercept is 0.
Sample answer: A line that only has an x-intercept is a vertical line. A line that only has a y-intercept is a horizontal line.
y = 0 (-8, 0) (0, 4) x = 0 -x + 2y = 8 -x + 2(0) = 8 -x = 8 x = -8 -x + 2y = 8 -0 + 2y = 8 2y = 8 y = 4 209_218_HSM_NA_S_A1M04_L01_662599.indd Page 212 5/9/19 3:26 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 5 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students describe what a line looks like that only has an x– or y– intercept.
TYPE a Students use a sketch to graph a linear function.
WEB SKETCHPAD Example 5 Graph by Using Intercepts MP MP Teaching the Mathematical Practices 5 Decide When to Use Tools Mathematically proficient students can make sound decisions about when to use mathematical tools such as a straightedge. Help them see why using these tools will help to solve problems and what the limitations are of using the tool.
Questions for Mathematical Discourse AL What are the intercepts of the graph of a linear function? the points where the line crosses the x- and y-axes OL How does finding the x- and y-intercepts help you to graph the function? Sample answer: Two points make a line, so a line can be drawn using the intercepts as the two points.
BL When finding the x-intercept, why do you substitute 0 for y in the equation? Sample answer: The y-coordinate of any point on the x-axis is 0, so substituting 0 for y in the equation tells you the value of x when y = 0, which is the x-intercept of the graph of the function.
212 Module 4 • Linear and Nonlinear Functions A.REI.10, F.IF.7a, F.LE.5 Interactive Presentation Example 6 Use Intercepts MP MP Teaching the Mathematical Practices 4 Apply Mathematics In this example, students apply what they have learned about graphing linear functions to solving a real-world problem.
Questions for Mathematical Discourse AL What information is given in the problem? Angelina starts with 60 cups of dog food and feeds her dog 5 __ 2 cups per day.
OL What does each variable represent, and what does this tell you about the intercepts? Sample answer: x represents days, and y represents cups of food remaining. So the x-intercept represents the number of days when there are 0 cups of food left, and the y-intercept represents the amount of food when 0 days have passed.
BL Explain what the intercepts mean in the context of the problem. Sample answer: At 24 days, there is no food left. The bag started with 60 cups of food and after 24 days, the bag was empty.
Common Error Some students may interchange the intercepts, thinking that when they let x = 0, they are finding the x-intercept or vice versa. Help students avoid this error by having them write the ordered pairs with the zeros in place before they solve algebraically. Then have them fill in the values they find, and plot the points from the ordered pairs. Essential Question Follow-Up Students have used a variety of methods to graph linear equations.
Ask: Why is it helpful to have different ways to graph linear functions? Sample answer: Some methods of graphing are easier in different contexts. For instance, graphing by finding the x- and y-intercepts might be obvious from inspecting the particular equation. For a function that represents a real-world situation, it might be easier to create a table of values for the situation.
DIFFERENTIATE Enrichment Activity BL Have students work in pairs to create a poster about graphing linear equations. Have them include information about tables of values, intercepts, and the solutions of the equations in their display.
Copyright © McGraw-Hill Education (continued on the next page) Check Graph 4y = -12x + 36 by using the x- and y- intercepts. x-intercept: y-Intercept: Example 6 Use Intercepts PETS Angelina bought a 15-pound bag of food for her dog. The bag contains about 60 cups of food, and she feeds her dog 2 1 __ 2 or 5 __ 2 cups of food per day. The function y + 5 __ 2 x = 60 represents the amount of food left in the bag y after x days. Graph the amount of dog food left in the bag as a function of time. Part A Find the x- and y-intercepts and interpret their meaning in the context of the situation.
To find the x-intercept, let . Original equation Replace y with 0. Simplify. Multiply each side by 2 __ 5 .
The x-intercept is 24. This means that the graph intersects the x-axis at . So, after 24 days, there is no dog food left in the bag.
To find the y-intercept, let . Original equation Replace x with 0. Simplify.
The y-intercept is 60. This means that the graph intersects the y-axis at . So, after 0 days, there are 60 cups of food in the bag.
−4 −2−1 −3 4 2 3 1 8 4 6 2 −8 −4 −6 y x O Think About It!
Find another point on the graph. What does it mean in the context of the problem? Go Online You can watch a video to see how to use a graphing calculator with this example.
Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-1 • Graphing Linear Functions 213 y = 0 x = 0 (24, 0) (0, 60) 3 9 y + 5 __ 2 x = 60 0 + 5 __ 2 x = 60 5 __ 2 x = 60 x = 24 y + 5 __ 2 x = 60 y + 5 __ 2 (0) = 60 y = 60 Sample answer: (10, 35); After 10 days, there are 35 cups of dog food left in the bag.
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Example 6 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students explain the meaning of another point in context and identify the assumptions made.
TYPE a Students use a sketch to plot the intercepts and graph the line.
WEB SKETCHPAD Students can watch a video to review how to graph a linear function using a graphing calculator.
WATCH Lesson 4-1 • Graphing Linear Functions 213 A.REI.10, F.IF.7a, F.LE.5 2 EXPLORE AND DEVELOP Copyright © McGraw-Hill Education Part B Graph the equation by using the intercepts.
y x O 10 5 10 15 20 25 30 35 45 40 20 30 40 50 60 70 80 90 Check PEANUTS A farm produces about 4362 pounds of peanuts per acre. One cup of peanut butter requires about 2 __ 3 pound of peanuts. If one acre of peanuts is harvested to make peanut butter, the function y = - 2 __ 3 x + 4362 represents the pounds of peanuts remaining y after x cups of peanut butter are made.
x-intercept: y-intercept: Which graph uses the x- and y-intercepts to correctly graph the equation? A. 2000 0 4000 6000 Amount of Peanuts (lb) 6000 4000 2000 0 Amount of Peanut Butter (c) Peanut Farming B.
2000 0 4000 6000 Amount of Peanuts (lb) 6000 4000 2000 0 Amount of Peanut Butter (c) Peanut Farming C.
2000 0 4000 6000 Amount of Peanuts (lb) 6000 4000 2000 0 Amount of Peanut Butter (c) Peanut Farming D.
2000 0 4000 6000 Amount of Peanuts (lb) 6000 4000 2000 0 Amount of Peanut Butter (c) Peanut Farming Go Online You can complete an Extra Example online. Think About It!
What assumptions did you make about the amount of food Angelina feeds her dog each day? Go Online You can watch a video to see how to graph a linear function using a graphing calculator.
214 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 6543 4362 C Sample answer: I assumed that the bag of food contains exactly 60 cups and that Angelina feeds her dog the exact same amount each day.
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1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Exit Ticket Recommended Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond using a separate piece of paper. Have students hand you their responses as they leave the room.
Alternate Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond verbally or by using a mini-whiteboard. Have students hold up their whiteboards so that you can see all student responses. Tap to reveal the answer when most or all students have completed the Exit Ticket.
Students complete the Check online to determine whether they are ready to move on.
CHECK Interactive Presentation 214 Module 4 • Linear and Nonlinear Functions A.REI.10, F.IF.7a, F.LE.5 Interactive Presentation Practice and Homework Suggested Assignments Use the table below to select appropriate exercises.
DOK Topic Exercises 1, 2 exercises that mirror the examples 1–16 2 exercises that use a variety of skills from this lesson 17–25 2 exercises that extend concepts learned in this lesson to new contexts 26–29 3 exercises that emphasize higher-order and critical thinking skills 30–37 Copyright © McGraw-Hill Education Name Period Date Practice Go Online You can complete your homework online.
Examples 1 through 4 Graph each equation by making a table.
1. x = -2 2. y = −4 x y y x O x y y x O 3. y = -8x 4. 3x = y x y −4 −2−1 −3 4 2 3 1 8 4 6 2 −8 −4 −6 y x O x y y x O 5. y - 8 = -x 6. x = 10 − y x y −4−2 8 10 12 4 6 8 10 12 4 6 2 −4 2 y x O x y −4−2 8 10 12 4 6 2 8 10 12 4 6 2 −4 y x O 7. y = 1 __ 2 x + 1 8. y + 2 = 1 _ 4 x x y −4 −2−1 −3 4 2 3 1 4 2 3 1 −4 −2 −3 y x O x y −8 −4−2 −6 8 4 6 2 4 2 3 1 −4 −2 −3 y x O Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-1 • Graphing Linear Functions 215 −2 −1 0 0 1 1 2 −4 0 10 −2 1 −4 3 9 −1 2 −4 6 8 0 3 2 2 2 8 4 1 1 1 4 2 0 0 0 0 0 0 8 8 0 7 −8 6 −2 1 −2 2 209_218_HSM_NA_S_A1M04_L01_662599.indd Page 215 5/9/19 3:26 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education Example 5 Graph each equation by using the x-and y-intercepts. 9. y = 4 + 2x 10. 5 - y = -3x 11. x = 5y + 5 y x O y x O y x O 12. x + y = 4 13. x - y = -3 14. y = 8 - 6x y x O y x O y x O Example 6 15. SCHOOL LUNCH Amanda has $210 in her school lunch account. She spends $35 each week on school lunches. The equation y = 210 - 35x represents the total amount in Amanda’s school lunch account y for x weeks of purchasing lunches. a. Find the x- and y-intercepts and interpret their meaning in the context of the situation. b. Graph the equation by using the intercepts. 16. SHIPPING The OOCL Shenzhen, one of the world’s largest container ships, carries 8063 TEUs (1280-cubic-feet containers). Workers can unload a ship at a rate of 1 TEU every minute. The equation y = 8063 - 60x represents the number of TEUs on the ship y after x hours of the workers unloading the containers from the Shenzhen. a. Find the x- and y-intercepts and interpret their meaning in the context of the situation. b. Graph the equation by using the intercepts.
216 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 15a. The x-intercept is 6. This means that after 6 weeks, Amanda will have $0 in her school lunch account. The y-intercept is 210. This means that there was initially $210 in Amanda’s school lunch account.
16a. The x-intercept is about 134. This means that after about 134 hours, there are 0 TEUs on the ship. The y-intercept is 8063. This means that there were initially 8063 TEUs on the ship.
2 4 6 8 1 3 5 7 9 0 40 80 120 160 220 20 60 100 140 180 200 Amount in Account ($) Weeks 40 80 120 160 0 2 4 6 8 1 3 5 7 9 10 TEUs on Ship (thousands) Time (hours) 209_218_HSM_NA_S_A1M04_L01_662599.indd Page 216 5/10/19 3:04 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
3 REFLECT AND PRACTICE ASSESS AND DIFFERENTIATE Use the data from the Checks to determine whether to provide resources for extension, remediation, or intervention.
IF students score 90% or more on the Checks, THEN assign: BL • Practice, Exercises 1–25 odd, 30–37 • Extension: Graphing Equations in Three Dimensions • Ordered Pairs; Graphing Lines IF students score 66%–89% on the Checks, THEN assign: OL • Practice, Exercises 1–37 odd • Remediation, Review Resources: Proportional Relationships and Slope • Personal Tutors • Extra Examples 1–6 • Proportional Relationships; Slope IF students score 65% or less on the Checks, THEN assign: AL • Practice, Exercises 1–15 odd • Remediation, Review Resources: Proportional Relationships and Slope • Quick Review Math Handbook: Linear Functions • ArriveMATH Take Another Look • Proportional Relationships; Slope 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Lesson 4-1 • Graphing Linear Functions 215-216 A.REI.10, F.IF.7a, F.LE.5 Interactive Presentation Answers 26c. 120 80 40 −10 −20 10 20 y x O 27. y = 1.7x + 40; The y-intercept is 40. This means that it would cost $40 to hook up the car. 28. 3x + 2y = 18; The x-intercept is 6. The x-intercept represents how many pounds of peanuts can be bought if no pretzels are bought. The y-intercept is 9. The y-intercept represents how many pounds of pretzels can be bought if no peanuts are bought. 30c. Time (min) x Height (ft) y 0 28 1 21 2 14 4 0 3 REFLECT AND PRACTICE Copyright © McGraw-Hill Education Name Period Date Mixed Exercises Graph each equation. 17. 1.25x + 7.5 = y 18. 2x - 3 = 4y + 6 19. 3y - 7 = 4x + 1 y x O y x O −8 −4−2 −6 8 4 6 2 8 10 4 6 2 −4 −6 y x O Find the x-intercept and y-intercept of the graph of each equation. 20. 5x + 3y = 15 21. 2x − 7y = 14 22. 2x − 3y = 5 23. 6x + 2y = 8 24. y = 1 _ 4 x − 3 25. y = 2 __ 3 x + 1 26. HEIGHT The height of a woman can be predicted by the equation h = 81.2 + 3.34r, where h is her height in centimeters and r is the length of her radius bone in centimeters.
a. Is this a linear function? Explain.
b. What are the r- and h-intercepts of the equation? Do they make sense in the situation? Explain.
c. Graph the equation by using the intercepts.
d. Use the graph to find the approximate height of a woman whose radius bone is 25 centimeters long. 27. TOWING Pick-M-Up Towing Company charges $40 to hook a car and $1.70 for each mile that it is towed. Write an equation that represents the total cost y for x miles towed. Graph the equation. Find the y-intercept, and interpret its meaning in the context of the situation. 28. USE A MODEL Elias has $18 to spend on peanuts and pretzels for a party. Peanuts cost $3 per pound and pretzels cost $2 per pound. Write an equation that relates the number of pounds of pretzels y and the number of pounds of peanuts x. Graph the equation. Find the x- and y-intercepts. What does each intercept represent in terms of context?
Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-1 • Graphing Linear Functions 217 See margin.
See margin.
x-int: 3; y-int: 5 x-int: 7; y-int: –2 x-int: 2 1 __ 2 ; y-int: −1 2 __ 3 x-int: 1 1 __ 3 ; y-int: 4 x-int: 12; y-int: −3 x-int: −1 1 __ 2 ; y-int: 1 Yes; the equation can be written in standard form where A = 3.34, B = −1, and C = −81.2.
h-int: 81.2; r-int: about −24.3; no, we would expect a woman 81.2-cm tall to have a radius bone of some length, and a negative radius bone length has no real meaning.
See margin.
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Copyright © McGraw-Hill Education 30. WRITE Consider real-world situations that can be modeled by linear functions. a. Write a real-world situation that can be modeled by a linear function. b. Write an equation to model your real-world situation. Be sure to define variables. Then find the x- and y-intercepts. What does each intercept represent in your context? c. Graph your equation by making a table. Include a title for the graph as well as labels and titles for each axis. Explain how you labeled the x- and y-axes. State a reasonable domain for this situation. What does the domain represent? 31. FIND THE ERROR Geroy claims that every line has both an x- and a y-intercept. Is he correct? Explain your reasoning. 32. WHICH ONE DOESN’T BELONG? Which equation does not belong with the other equations? Justify your conclusion. y = 2 – 3x 5x = y – 4 y = 2x + 5 y – 4 = 0 33. ANALYZE Robert sketched a graph of a linear equation 2x + y = 4. What are the x- and y-intercepts of the graph? Explain how Robert could have graphed this equation using the x- and y-intercepts. 34. ANALYZE Compare and contrast the graph of y = 2x + 1 with the domain {1, 2, 3, 4} and y = 2x + 1 with the domain all real numbers. CREATE Give an example of a linear equation in the form Ax + By = C for each condition. Then describe the graph of the equation. 35. A = 0 36. B = 0 37. C = 0 y x O 29. REASONING One football season, a football team won 4 more games than they lost. The function y = x + 4 represents the number of games won y and the number of games lost x. Find the x- and y-intercepts. Are the x- and y-intercepts reasonable in this situation? Explain.
218 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-1_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE y − 4 = 0; When graphed, all of the other equations have x- and y-intercepts, but y − 4 = 0 only has a y-intercept.
Sample answer: y = 8; horizontal line Sample answer: x = 5; vertical line Sample answer: x − y = 0; line through (0, 0) Sample answer: In the equation, let y = 0 to find the x-intercept; 2x + 0 = 4. So the x-intercept is 2. In the equation, let x = 0 to find the y-intercept; 2(0) + y = 4. So the y-intercept is 4. Robert graphed points at (2, 0) and (0, 4) and connected the points with a line.
Sample answer: The first graph is a set of points that are not connected. The second graph is of a line. The points of the first graph are points on the line in the second graph.
See margin.
Sample answer: Keith is climbing down a 28-foot cliff. Keith descends 7 feet per minute. Sample answer: y = 28 – 7x, where y is Keith’s height in feet after x minutes. The x-intercept, 4, represents the number of minutes it takes Keith to reach the bottom of the cliff. The y-intercept, 28, represents Keith’s initial height above the bottom of the cliff. No; sample answer: A horizontal line only has a y-intercept and a vertical line only has an x-intercept. Higher-Order Thinking Skills Sample answer: The x-intercept is −4. The x-intercept is not reasonable because the football team cannot lose −4 games. The y-intercept is 4. The y-intercept is reasonable because the y-intercept means that if the football team won 4 games, they lost 0 games.
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1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION 2 4 6 8 9 1 3 5 7 10 0 10 20 30 40 60 5 15 25 35 45 50 55 Total Cost ($) Miles 2 1 4 3 6 5 8 7 10 9 0 1 2 3 4 5 6 7 8 9 10 Pretzels (lb) Peanuts (lb) Party Snack Purchases 2 1 4 3 5 0 4 8 12 16 20 24 28 32 36 40 Height Above Bottom (ft) Time (min) Keith’s Clif Climb 217-218 Module 4 • Linear and Nonlinear Functions A.REI.10, F.IF.7a, F.LE.5 LESSON GOAL Students find and interpret the rate of change and slopes of lines.
1 LAUNCH Launch the lesson with a Warm Up and an introduction.
2 EXPLORE AND DEVELOP Develop: Rate of Change of a Linear Function • Find the Rate of Change • Compare Rates of Change • Constant Rate of Change • Rate of Change Explore: Investigating Slope Develop: Slope of a Line • Positive Slope • Negative Slope • Slopes of Horizontal Lines • Slopes of Vertical Lines • Find Coordinates Given the Slope • Use Slope You may want your students to complete the Checks online.
3 REFLECT AND PRACTICE Exit Ticket Practice DIFFERENTIATE View reports of student progress on the Checks after each example.
Resources AL OL BL ELL Remediation: Order of Integer Operations ● ● ● Extension: Treasure Hunt with Slopes ● ● ● Language Development Handbook Assign page 21 of the Language Development Handbook to help your students build mathematical language related to rates of change and slopes.
ELL You can use the tips and suggestions on page T21 of the handbook to support students who are building English proficiency.
Lesson 4-2 Rate of Change and Slope Lesson 4-2 • Rate of Change and Slope 219a F.IF.6, F.LE.5 Suggested Pacing 90 min 0.5 day 45 min 1 day Focus Domain: Functions Standards for Mathematical Content: F.IF.6 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.
F.LE.5 Interpret the parameters in a linear or exponential function in terms of a context.
Standards for Mathematical Practice: 2 Reason abstractly and quantitatively.
4 Model with mathematics.
Coherence Vertical Alignment Rigor The Three Pillars of Rigor 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Conceptual Bridge In this lesson, students expand on their understanding of and fluency with slope and rate of change (first studied in Grade 8). They apply their understanding of slope and rate of change by solving real-world problems.
Mathematical Background Rate of change is a ratio that describes, on average, how one quantity changes with respect to a change in another quantity. Slope can be used to describe rate of change. The slope of a line is the ratio of the vertical change (the rise) to the horizontal change (the run). The slope formula, m = y2 - y1 _ x2 - x1 , where ( x1, y1) and ( x2, y2) are two points that lie on the line, can be used to find the slope of a line without graphing.
Next Students will graph equations in slope-intercept form. A.CED.2, F.IF.7a, F.LE.5 Now Students find and interpret the rate of change and slopes of lines.
F.IF.6, F.LE.5 Previous Students graphed linear functions using tables and intercepts.
A.REI.10, F.IF.7a, F.LE.5 Interactive Presentation 1 LAUNCH Warm Up Prerequisite Skills The Warm Up exercises address the following prerequisite skill for this lesson: • subtracting integers in fractions Answers: 1. 1 2. – 5 __ 2 3. undefined 4. – 3 5. 13,100 people Launch the Lesson MP MP Teaching the Mathematical Practices 2 Make Sense of Quantities Mathematically proficient students need to be able to make sense of quantities, such as slope and rate of change, and their relationships. Go Online to find additional teaching notes and questions to promote classroom discourse.
Today’s Standards Tell students that they will address these content and practice standards in this lesson. You may wish to have a student volunteer read aloud How can I meet these standards? and How can I use these practices?, and connect these to the standards.
See the Interactive Presentation for I Can statements that align with the standards covered in this lesson.
Today’s Vocabulary Tell students that they will be using these vocabulary terms in this lesson. You can expand each row if you wish to share the definitions. Then discuss the questions below with the class.
Warm Up Launch the Lesson Today’s Vocabulary 219b Module 4 • Linear and Nonlinear Functions F.IF.6, F.LE.5 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Investigating Slope Objective Students use a sketch to explore how the slope of a line affects its graph.
MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Ideas for Use Recommended Use Present the Inquiry Question, or have a student volunteer read it aloud. Have students work in pairs to complete the Explore activity on their devices. Pairs should discuss each of the questions. Monitor student progress during the activity. Upon completion of the Explore activity, have student volunteers share their responses to the Inquiry Question. What if my students don’t have devices? You may choose to project the activity on a whiteboard. A printable worksheet for each Explore is available online. You may choose to print the worksheet so that individuals or pairs of students can use it to record their observations.
Summary of the Activity Students will complete guiding exercises throughout the Explore activity. Students will use the sketch to see how the slope of a line changes as the line is rotated. They will observe how the rise and the run are affected as the line is rotated, and how that affects the calculation of the slope. They will explore lines with positive slopes and negative slopes and will investigate the slopes of horizontal and vertical lines. Then, students will answer the Inquiry Question.
(continued on the next page) Explore Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students answer questions about the slope of a line.
TYPE a Students use a sketch to investigate the slope of a line.
WEB SKETCHPAD Lesson 4-2 • Rate of Change and Slope 219c F.IF.6 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Investigating Slope (continued) Questions Have students complete the Explore activity.
Ask: • How can the words “rise” and “run” remind you whether to look for a change in y or a change in x? Sample answer: You can think of rise as something going up or down, which goes along with a change in vertical distance along the y-axis. You can think of “run” as something you do on the ground, which is horizontal or along the x-axis.
• What does a slope of -3 tell you about the line? Sample answer: The negative sign tells me that the line will be decreasing as it moves from left to right. I also know that the line will go down three units for every one unit to the right. Inquiry How does slope help to describe a line? Sample answer: The slope of a line can tell you whether the graph of the line will slope up or down from left to right or if it will be a horizontal or vertical line. Go Online to find additional teaching notes and sample answers for the guiding exercises.
Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students respond to the Inquiry Question and can view a sample answer.
TYPE a 219d Module 4 • Linear and Nonlinear Functions F.IF.6 Interactive Presentation Learn Rate of Change of a Linear Function Objective Students calculate and interpret rate of change by identifying the change in the independent and dependent variables.
MP MP Teaching the Mathematical Practices 2 Make Sense of Quantities In this Learn, help students to notice the relationship between the variables when calculating rate of change. Example 1 Find the Rate of Change MP MP Teaching the Mathematical Practices 2 Attend to Quantities Point out that it is important to note the meaning of the quantities used in this problem.
Questions for Mathematical Discourse AL What are the quantities being compared in the table? the number of pancakes and the number of cups of flour OL How do you know which is the independent variable and which is the dependent variable? Sample answer: The pancakes depend on the flour because the number of pancakes you can make depends on how much flour you use. So the number of pancakes is the dependent variable, and the amount of flour is the independent variable.
BL Would the ratio be different if you used the first and last pairs of values from the table to calculate the rate of change? Explain. No; sample answer: 36 - 12 _ 6 - 2 = 24 ___ 4 or 6. Go Online • Find additional teaching notes.
• View performance reports of the Checks.
• Assign or present an Extra Example.
Copyright © McGraw-Hill Education Today’s Goals ● Calculate and interpret rate of change.
● Calculate and interpret slope.
Rate of Change and Slope Lesson 4-2 Today’s Vocabulary rate of change slope Learn Rate of Change of a Linear Function The rate of change is how a quantity is changing with respect to a change in another quantity.
If x is the independent variable and y is the dependent variable, then rate of change = change in y _ change in x . Example 1 Find the Rate of Change COOKING Find the rate of change of the function by using two points from the table. Amount of Flour x (cups) Pancakes y 2 12 4 24 6 36 rate of change = change in y __ change in x = change in pancakes __ change in flour = 24 - 12 _ 4 - 2 = or The rate is or . This means that you could make pancakes for each cup of flour.
Check Find the rate of change. dollars __ gallons Amount of Gasoline Purchased (Gallons) Cost (Dollars) 4.75 15.77 6 19.92 7.25 24.07 8.5 28.22 Study Tip Placement Be sure that the dependent variable is in the numerator and the independent variable is in the denominator. In this example, the number of pancakes you can make depends on the amount of flour you can use. Think About It!
Suppose you found a new recipe that makes 6 pancakes when using 2 cups of flour, 12 pancakes when using 4 cups of flour, and 18 pancakes when using 6 cups of flour. How does this change the rate you found for the original recipe? Go Online You can complete an Extra Example online. 6 __ 1 3.32 6 6 Sample answer: The new recipe makes half as many pancakes for the same amount of flour. So, the rate of the new recipe is half the rate of the original recipe and I could make only 3 pancakes per cup of flour. 12 __ 2 6 __ 1 Lesson 4-2 • Rate of Change and Slope 219 Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 219_228_HSM_NA_S_A1M04_L02_662599.indd Page 219 14/01/19 5:04 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Students explain how the rate of change can be used to find the number of pancakes for a given number of cups of flour.
Example 1 TYPE a 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Lesson 4-2 • Rate of Change and Slope 219 F.IF.6, F.LE.5 Interactive Presentation 2 EXPLORE AND DEVELOP Example 2 Compare Rates of Change MP MP Teaching the Mathematical Practices 4 Make Assumptions In the Study Tip, have students point out where an assumption or approximation was made in the solution.
Questions for Mathematical Discourse AL What are you trying to determine? the rate of change for two time periods: 2000–2005 and 2010–2015 OL What do x and y in the formula for the rate of change represent? y represents dollars, and x represents years BL Can you find the rate of change by simply subtracting the numbers that are called out on the graph? Why or why not? No; sample answer: Subtracting those numbers will not produce a rate. Each listed point is at an interval of 5 years, so you need to divide by 5 to determine the rate of change per year.
Common Error When interpreting a solution, some students may ignore the sign of a rate that is negative. Explain that in any real-world problem, the sign of a quantity has meaning. Help them to see that in this example, the negative rate of change means that the budget was reduced over that time period.
Copyright © McGraw-Hill Education Example 2 Compare Rates of Change STUDENT COUNCIL The Jackson High School Student Council budget varies based on the fundraising of the previous year. Part A Find the rate of change for 2000–2005 and describe its meaning in the context of the situation. change in budget __ change in time = 1675 - 1350 _ 2005 - 2000 = , or This means that the student council’s budget increased by $ over the -year period, with a rate of change of $ per year. Part B Find the rate of change for 2010–2015 and describe its meaning in the context of the situation. change in budget __ change in time = 1325 - 1550 _ 2015 - 2010 = , or This means that the student council’s budget was reduced by $ over the -year period, with a rate of change of -$ per year.
Check TICKETS The graph shows the average ticket prices for the Miami Dolphins football team.
Part A Find the rate of change in ticket prices between 2009–2010. dollars _ year Part B The ticket prices have the greatest rate of change between .
Part C Between and , the rate of change is negative. Think About It!
How is a greater increase or decrease of funds represented graphically? Go Online You can complete an Extra Example online.
Study Tip Assumptions In this example, we assumed that the rate of change for the budget was constant between each 5-year period. Although the budget might have varied from year to year, analyzing in larger periods of time allows us to see trends within data.
2000 0 2005 2010 2015 0 500 1000 1500 2000 Dollars Years 1350 1675 1550 1325 Jackson High School Student Council Budget 2008 2009 2010 2011 2012 2013 2014 0 64 Cost (Dollars) Year 66 68 70 66.11 66.74 70.32 71.14 70.54 65.16 71.14 Miami Dolphins Average Ticket Prices Your Notes 220 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: The line between points is steeper when there is a greater change in funds.
65 65 -45 5 325 225 5 45 3.80 2013–2014 2010–2011 2013–2014 -225 5 325 ___ 5 219228_HSM_NA_S_A1M04_L02_662599.indd Page 220 04/10/19 11:21 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 2 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students answer a question about how a greater increase or decrease of funds is represented graphically.
TYPE a Students complete the rate of change formula by dragging the values to the correct bins.
DRAG & DROP 220 Module 4 • Linear and Nonlinear Functions F.IF.6, F.LE.5 Interactive Presentation Copyright © McGraw-Hill Education Example 3 Constant Rate of Change Determine whether the function is linear. If it is, state the rate of change.
Find the changes in the x-values and the changes in the y-values.
Notice that the rate of change for each pair of points shown is The rates of change are , so the function is . The rate of change is .
Example 4 Rate of Change Determine whether the function is linear. If it is, state the rate of change.
Find the changes in the x-values and the changes in the y-values.
The rates of change are . Between some pairs of points the rate of change is , and between the other pairs it is . Therefore, this is .
Check Complete the table so that the function is linear.
x y -2.25 1 11 10.5 7.5 10 10.75 9.5 Go Online You can complete an Extra Example online.
x y 11 -5 8 -3 5 -1 2 1 -1 3 x y 22 -4 29 -1 36 1 43 4 50 6 Study Tip Linear Versus Not Linear Remember that the word linear means that the graph of the function is a straight line. For the graph of a function to be a line, it has to be increasing or decreasing at a constant rate.
constant not constant not a linear function linear − 2 __ 3 − 2 __ 3 3 __ 7 12 11.5 4.25 14 2 __ 7 Lesson 4-2 • Rate of Change and Slope 221 Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 219_228_HSM_NA_S_A1M04_L02_662599.indd Page 221 14/01/19 5:04 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 3 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 3 Constant Rate of Change MP MP Teaching the Mathematical Practices 8 Look for a Pattern Help students to see the pattern in this example.
Questions for Mathematical Discourse AL What do you need to know in order to determine whether the function is linear? whether there is a constant rate of change OL How does finding the differences between successive values in the table help you determine whether a function is linear? Sample answer: If the differences are the same, then I know that the rate of change is constant, and therefore the function is linear.
BL What is another way you can use the table to determine if the rate of change is constant? Sample answer: Because consecutive x-values decrease by 3, I can check to see if consecutive y-values increase or decrease by the same number. Because consecutive y-values increase by 2, I know there is a constant rate of change.
Students tap on the card to see the changes in the x- and y-values.
TAP Students complete the Check online to determine whether they are ready to move on.
CHECK Students select the correct word to complete the sentence.
MULTI-SELECT Example 4 Rate of Change MP MP Teaching the Mathematical Practices 1 Explain Correspondences Use the Study Tip to encourage students to explain the relationship between the graph and rate of change of a linear function.
Questions for Mathematical Discourse AL How will you determine whether the function is linear? Sample answer: I will find the changes in the x-values and the changes in the y-values, and see if those changes are constant.
OL Is it necessary to calculate the rate of change between every pair of points to determine linearity? Explain. No; sample answer: Once you have found two pairs that have different rates of change, you have shown that the function is not linear.
BL If you graphed the points from the table, would they lie on a straight line? How do you know? No; sample answer: Because the rates of change are not constant, the function is not linear, and therefore the graph of the points will not lie on a line.
Common Error Some students may observe the pattern in the differences between the y–values (3, 2, 3, 2) and think that this regularity indicates that the function is linear. Correct this reasoning, and reinforce that when the differences in the x-values are the same, the differences in the y-values must also be the same for the function to be linear.
Lesson 4-2 • Rate of Change and Slope 221 F.IF.6, F.LE.5 Interactive Presentation 2 EXPLORE AND DEVELOP Copyright © McGraw-Hill Education Go Online You can watch a video to see how to find the slope of a nonvertical line. Think About It!
If the point (1, 3) is on a line, what other point could be on the line to make the slope positive? negative? zero? undefined? Think About It!
How would lines with slopes of m = 1 __ 8 and m = 80 compare on the same coordinate plane? Think About It!
Can a line that passes through two specific points, such as the origin and (2, 4), have more than one slope? Explain your reasoning. Go Online You can complete an Extra Example online.
Explore Investigating Slope Online Activity Use graphing technology to complete an Explore. INQUIRY How does slope help to describe a line?
Learn Slope of a Line The slope of a line is the rate of change in the y-coordinates (rise) for the corresponding change in the x-coordinates (run) for points on the line.
Key Concept • Slope Words The slope of a nonvertical line is the ratio of the rise to the run.
Symbols The slope m of a nonvertical line through any two points (x1, y1) and (x2, y2) can be found as follows. m = y2 - y1 x2 - x1 Example y x O x2 - x1 (x2, y2) (x1, y1) y2 - y1 The slope of a line can show how a quantity changes over time. When finding the slope of a line that represents a real-world situation, it is often referred to as the rate of change.
Example 5 Positive Slope Find the slope of a line that passes through (−3, 4) and (1, 7).
y x O (1, 7) (-3, 4) m = y2 - y1 x2 - x1 = 7 - 4 __ 1 - (-3) = Check Determine the slope of a line passing through the given points. If the slope is undefined, write undefined. Enter your answer as a decimal if necessary.
(-1, 8) and (7, 10) 222 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 3 __ 4 Sample answer: Positive: (3, 6) Negative: (2, 0) Zero: (5, 3) Undefined: (1, 6) Sample answer: The line with the smaller slope would be more horizontal with only a slight positive slope. The line with the greater slope would look nearly vertical with a very steep positive slope.
No; sample answer: A line is defined by two points, and slope is determined by any two points on the line.
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Example 5 Students describe and correct the error made when finding the slope of the line.
TYPE a 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Learn Slope of a Line Objective Students calculate and interpret slope by using the Slope Formula.
MP MP Teaching the Mathematical Practices 7 Use Structure Help students to explore the structure of slopes of lines in this Learn.
Example 5 Positive Slope Questions for Mathematical Discourse AL Finding the slope is the same as finding what other measure? the rate of change OL Is the slope of this line positive, negative, or zero? How can you tell by looking at the graph? Positive; sample answer: The line slopes upward from left to right.
BL Does it matter which coordinates you use as x2 and y2? Explain. No; sample answer: You can use either of the x-coordinates as x2, but the value for y2 must then be the y-coordinate that corresponds with x2.
DIFFERENTIATE Enrichment Activity AL BL ELL IF students automatically assume that the left-most point has to be (x1, y1 ) and the point farther right is (x2, y2), THEN explain that the designation of (x1, y1) and (x2, y2) is arbitrary. Write pairs of points on index cards. Give one card to each student. Have them find the slope both ways. Then ask which way made the subtraction easier.
Students can watch a video to see how find the slope of a nonvertical line.
WATCH DIFFERENTIATE Language Development Activity ELL Intermediate Instruct a small group of students to write a paragraph describing what is happening in the illustration of slope in the Key Concept box. Their paragraphs should describe all parts of the diagram in their own words. Ask for volunteers to read their paragraphs. Have students ask for clarification as needed. Then, have students revise their paragraphs based on the feedback and questions from the group.
222 Module 4 • Linear and Nonlinear Functions F.IF.6, F.LE.5 Interactive Presentation Copyright © McGraw-Hill Education Example 6 Negative Slope Find the slope of a line that passes through (-1, 3) and (4, 1).
y x O (4, 1) (-1, 3) m = y2 - y1 x2 - x1 = 1 - 3 4 - (-1) = Check Determine the slope of a line passing through the given points. If the slope is undefined, write undefined. Enter your answer as a decimal if necessary.
a. (5, -4) and (0, 1) Example 7 Slopes of Horizontal Lines Find the slope of a line that passes through (-2, −5) and (4, -5).
y x O (-2, -5) (4, -5) m = y2 - y1 x2 - x1 = -5 - (-5) _ 4 - (-2) = or Example 8 Slopes of Vertical Lines Find the slope of a line that passes through (-3, 4) and (-3, -2).
y O (-3, 4) (-3, -2) x m = y2 - y1 _ x2 - x1 = -2 - 4 _ -3 - (-3) = or Talk About It!
Why is the slope for vertical lines always undefined? Justify your argument. Go Online You can complete an Extra Example online.
Study Tip Positive and Negative Slope To know whether a line has a positive or negative slope, read the graph of the line just like you would read a sentence, from left to right. If the line “goes uphill,” then the slope is positive. If the line “goes downhill,” then the slope is negative.
-1 0 __ 6 - 6 __ 0 0 undefined Sample answer: The x-values are always the same in vertical lines, so the difference in the denominator of the Slope Formula will always be zero. Dividing any number by zero is undefined. − 2 __ 5 Lesson 4-2 • Rate of Change and Slope 223 Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 219_228_HSM_NA_S_A1M04_L02_662599.indd Page 223 14/01/19 5:04 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 7 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 6 Negative Slope MP MP Teaching the Mathematical Practices 8 Use Slope Help students to pay attention to the calculation of the slope of the line.
Questions for Mathematical Discourse AL If x1 = -1, what is the value of y1? 3 OL Is the slope of this line positive, negative, or zero? How can you tell by looking at the graph? Negative; sample answer: The line slopes downward from left to right.
BL What would the value of the slope be if you used (4, 1) for (x1, y1) and (-1, 3) for (x2, y2)? It would still be - 2 __ 5 .
Example 7 Slopes of Horizontal Lines MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationship between the graph, points, and slope in this example.
Questions for Mathematical Discourse AL How would you describe what is meant by the slope of a line? Sample answer: It is the steepness of the line.
OL Is the slope positive, negative, or zero? zero BL Why is the slope zero? Sample answer: The slope is zero because there is no change in y-values, so the numerator will be zero and zero divided by any number is zero.
Example 8 Slopes of Vertical Lines MP MP Teaching the Mathematical Practices 8 Use Slope Help students to pay attention to the calculation of slope for a vertical line.
Questions for Mathematical Discourse AL Which values are the same? x-values: -3 OL Why is the slope undefined instead of zero? It is not possible to divide by 0. So, the slope of a vertical line is undefined.
BL Does the graph of a line with an undefined slope represent a function? Why or why not? No; sample answer: In a function, every x-value is paired with exactly one y-value. In a relation that is represented by a vertical line, there is one x-value paired with infinitely many y-values.
Lesson 4-2 • Rate of Change and Slope 223 F.IF.6, F.LE.5 Interactive Presentation 2 EXPLORE AND DEVELOP Copyright © McGraw-Hill Education Example 9 Find Coordinates Given the Slope Find the value of r so that the line passing through (-4, 5) and (4, r) has a slope of 3 __ 4 . m = y2 - y1 x2 - x1 3 _ 4 = r - 5 4 - (-4) 3 _ 4 = r - 5 8 = = = = Use the Slope Formula.
(−4, 5) = (x1, y1) and (4, r) = (x2, y2) Subtract.
Multiply each side by 8.
Simplify.
Add 5 to each side.
Simplify.
Check Find the value of r so that the line passing through (−3, r) and (7, −6) has a slope of 2 2 __ 5 .
r = Example 10 Use Slope OCEANS What is the slope of the continental slope at Cape Hatteras?
m = y2 - y1 _ x2 - x1 = -2700 - (-65) ___ 125 - 75 = or Use the Slope Formula (75, -65) = (x1, y1) and (125, - 2700) = (x2, y2) Simplify.
The continental slope at Cape Hatteras has a slope of . Think About It!
If a crab is walking along the ocean floor 112 meters away from the shoreline to 114 meters away from the shoreline, how far does it descend? Go Online You can complete an Extra Example online.
Study Tip Converting Slope When solving for an unknown coordinate, like the previous example, converting a slope from a decimal or mixed number to an improper fraction might make the problem easier to solve. For example, a slope of 1. ¯ 333 can be rewritten as 4 _ 3 .
Continental Shelf (75, –65) 75 m from the shoreline 65m below the ocean surface (125, –2700) 125 m from the shoreline 2700 m deep Continental Slope Oceanic Crust 224 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: 105.4 meters 11 r -30 8 ( 3 __ 4 ) 6 6 + 5 r - 5 + 5 8(r - 5) _ 8 r - 5 -2635 _ 50 -52.7 -52.7 219_228_HSM_NA_S_A1M04_L02_662599.indd Page 224 5/9/19 3:26 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 10 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 9 Find Coordinates Given the Slope MP MP Teaching the Mathematical Practices 1 Check Answers Mathematically proficient students continually ask themselves, “Does this make sense?” In this example, encourage students to check their answer.
Questions for Mathematical Discourse AL For what variable in the equation do you substitute 3 __ 4 ? m OL How could a graph help determine the missing coordinate? Sample answer: I can plot the given point and then use the slope to move to the next point. I can continue using the slope until I get to the point with the x-coordinate of 4.
BL Name another point on the same line. Sample answers: (0, 8), (8, 14) Example 10 Use Slope MP MP Teaching the Mathematical Practices 4 Interpret Mathematical Results In this example, point out that to solve the problem, students should interpret their mathematical results in the context of the problem.
Questions for Mathematical Discourse AL What are the two ordered pairs you can use to find the slope? (75, -65) and (125, -2700) OL Interpret the value of the slope in the context of the problem. Sample answer: The slope means that the water gets 52.7 meters deeper for every meter you move farther from shore.
BL Do you think the continental slope is constant? Sample answer: No, there are probably places where the drop is less steep and places where it is more steep.
Exit Ticket Recommended Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond using a separate piece of paper. Have students hand you their responses as they leave the room.
Alternate Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond verbally or by using a mini-whiteboard. Have students hold up their whiteboards so that you can see all student responses. Tap to reveal the answer when most or all students have completed the Exit Ticket.
Students tap on each marker to find the points to use in the Slope Formula.
TAP Students complete the Check online to determine whether they are ready to move on.
CHECK Students use the slope to answer a question about another point along the continental slope.
TYPE a 224 Module 4 • Linear and Nonlinear Functions F.IF.6, F.LE.5 Copyright © McGraw-Hill Education Name Period Date Practice Go Online You can complete your homework online.
Example 1 Find the rate of change of the function by using two points from the table.
1. x y 5 2 10 3 15 4 20 5 2. x y 1 15 2 9 3 3 4 -3 3. POPULATION DENSITY The table shows the population density for the state of Texas in various years. Find the average annual rate of change in the population density from 2000 to 2009. 4. BAND In 2012, there were approximately 275 students in the Delaware High School band. In 2018, that number increased to 305. Find the annual rate of change in the number of students in the band. Example 2 5. TEMPERATURE The graph shows the temperature in a city during different hours of one day.
a. Find the rate of change in temperature between 6 a.m. and 7 a.m. and describe its meaning in the context of the situation. b. Find the rate of change in temperature from 1 p.m. and 2 p.m. and describe its meaning in the context of the situation. 6. COAL EXPORTS The graph shows the annual coal exports from U.S. mines in millions of short tons.
a. Find the rate of change in coal exports between 2000 and 2002 and describe its meaning in the context of the situation. b. Find the rate of change in coal exports between 2005 and 2006 and describe its meaning in the context of the situation. 6 A.M.
8 A.M.
10 A.M.
12 P.M.
2 P.M.
10 0 20 30 40 50 60 70 80 90 100 Temperature (°F) 2000 30 0 40 50 60 70 80 90 100 Million Short Tons Source: Energy Information Association 2001 2002 2003 2004 2005 2006 Total Exports Population Density Year People Per Square Mile 1930 22.1 1960 36.4 1980 54.3 2000 79.6 2009 96.7 Source: Bureau of the Census, U.S. Dept. of Commerce Lesson 4-2 • Rate of Change and Slope 225 Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 1 __ 5 -6 increased about 1.9 people per square mile increased by 5/yr -5; This means the temperature decreased 5°F per hour from 6 a.m. to 7 a.m.
-5; This means the temperature decreased 5°F per hour from 1 p.m. to 2 p.m.
-10; This means the coal exports decreased 10 million tons per year between 2000 and 2002.
0; This means the coal exports did not change between 2005 and 2006.
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Copyright © McGraw-Hill Education Examples 3 and 4 Determine whether the function is linear. If it is, state the rate of change.
7. x 4 2 0 -2 -4 y -1 1 3 5 7 8. x -7 -5 -3 -1 0 y 11 14 17 20 23 9. x -0.2 0 0.2 0.4 0.6 y 0.7 0.4 0.1 0.3 0.6 10. x 1 __ 2 3 __ 2 5 __ 2 7 __ 2 9 __ 2 y 1 __ 2 1 3 __ 2 2 5 __ 2 Examples 5 through 8 Find the slope of the line that passes through each pair of points. 11. (4, 3), (-1, 6) 12. (8, -2), (1, 1) 13. (2, 2), (-2, -2) 14. (6, -10), (6, 14) 15. (5, -4), (9, -4) 16. (11, 7), (-6, 2) 17. (-3, 5), (3, 6) 18. (-3, 2), (7, 2) 19. (8, 10), (-4, -6) 20. (-12, 15), (18, -13) 21. (-8, 6), (-8, 4) 22. (-8, -15), (-2, 5) 23. (2, 5), (3, 6) 24. (6, 1), (-6, 1) 25. (4, 6), (4, 8) 26. (-5, -8), (-8, 1) 27. (2, 5), (-3, -5) 28. (9, 8), (7, -8) 29. (5, 2), (5, -2) 30. (10, 0), (-2, 4) 31. (17, 18), (18, 17) 32. (-6, -4), (4, 1) 33. (-3, 10), (-3, 7) 34. (2, -1), (-8, -2) 35. (5, -9), (3, -2) 36. (12, 6), (3, -5) 37. (-4, 5), (-8, -5) Example 9 Find the value of r so the line that passes through each pair of points has the given slope. 38. (12, 10), (-2, r), m = -4 39. (r, -5), (3, 13), m = 8 40. (3, 5), (-3, r), m = 3 _ 4 41. (-2, 8), (r, 4), m = - 1 __ 2 42. (r, 3), (5, 9), m = 2 43. (5, 9), (r, -3), m = -4 44. (r, 2), (6, 3), m = 1 __ 2 45. (r, 4), (7, 1), m = 3 _ 4 linear; − 1 _ 1 or -1 not linear not linear linear; 1 __ 2 - 3 __ 5 1 __ 2 - 1 __ 3 1 __ 10 -3 - 7 __ 2 11 __ 9 5 __ 2 -1 - 14 __ 15 10 __ 3 undefined undefined undefined undefined 1 66 1 __ 2 6 8 2 4 11 3 __ 4 0 2 undefined 8 - 3 __ 7 0 1 5 __ 17 1 __ 6 0 4 __ 3 226 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 219_228_HSM_NA_S_A1M04_L02_662599.indd Page 226 14/01/19 5:04 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
3 REFLECT AND PRACTICE ASSESS AND DIFFERENTIATE Use the data from the Checks to determine whether to provide resources for extension, remediation, or intervention.
IF students score 90% or more on the Checks, THEN assign: BL • Practice, Exercises 1–61 odd, 63–68 • Extension: Treasure Hunt with Slopes • Equations of Lines IF students score 66%–89% on the Checks, THEN assign: OL • Practice, Exercises 1–67 odd • Remediation, Review Resources: Order of Integer Operations • Personal Tutors • Extra Examples 1–10 • Multiplication and Division with Integers IF students score 65% or less on the Checks, THEN assign: AL • Practice, Exercises 1–49 odd • Remediation, Review Resources: Order of Integer Operations • Quick Review Math Handbook: Rate of Change and Slope • ArriveMATH Take Another Look • Multiplication and Division with Integers 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Practice and Homework Suggested Assignments Use the table below to select appropriate exercises.
DOK Topic Exercises 1, 2 exercises that mirror the examples 1–49 2 exercises that use a variety of skills from this lesson 50–58 2 exercises that extend concepts learned in this lesson to new contexts 59–62 3 exercises that emphasize higher-order and critical thinking skills 63–68 Lesson 4-2 • Rate of Change and Slope 225-226 F.IF.6, F.LE.5 3 REFLECT AND PRACTICE Copyright © McGraw-Hill Education Name Period Date Example 10 46. ROAD SIGNS Roadway signs such as the one shown are used to warn drivers of an upcoming steep down grade. What is the grade, or slope, of the hill described on the sign? 47. HOME MAINTENANCE Grading the soil around the foundation of a house can reduce interior home damage from water runoff. For every 6 inches in height, the soil should extend 10 feet from the foundation. What is the slope of the soil grade? 48. USE A SOURCE Research the Americans with Disabilities Act (ADA) regulation for the slope of a wheelchair ramp. What is the maximum slope of an ADA regulation ramp? Use the slope to determine the length and height of an ADA regulation ramp. 49. DIVERS A boat is located at sea level. A scuba diver is 80 feet along the surface of the water from the boat and 30 feet below the water surface. A fish is 20 feet along the horizontal plane from the scuba diver and 10 feet below the scuba diver. What is the slope between the scuba diver and fish? 80 ft 30 ft 20 ft 10 ft (80, –30) Mixed Exercises STRUCTURE Find the slope of the line that passes through each pair of points. 50. 51. 52. y x O (0, 1) (2, 5) y x O (0, 0) (3, 1) y x O (0, 1) (1, −2) 53. (6, -7), (4, -8) 54. (0, 5), (5, 5) 55. (-2, 6), (-5, 9) 56. (5, 8), (-4, 6) 57. (9, 4), (5, -3) 58. (1, 4), (3, -1) 8% Lesson 4-2 • Rate of Change and Slope 227 Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 1 __ 12 ; Sample answer: length: 60 inches, height: 5 inches 2 ___ 25 1 ___ 20 - 1 __ 2 2 1 __ 3 1 __ 2 2 __ 9 7 __ 4 - 5 __ 2 0 -1 -3 219_228_HSM_NA_S_A1M04_L02_662599.indd Page 227 04/10/19 10:22 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education 59. REASONING Find the value of r that gives the line passing through (3, 2) and (r, -4) a slope that is undefined. 60. REASONING Find the value of r that gives the line passing through (-5, 2) and (3, r) a slope of 0. 61. CREATE Draw a line on a coordinate plane so that you can determine at least two points on the graph. Describe how you would determine the slope of the graph and justify the slope you found. 62. ARGUMENTS The graph shows median prices for small cottages on a lake since 2005. A real estate agent says that since 2005, the rate of change for house prices is $10,000 each year. Do you agree? Use the graph to justify your answer. 63. CREATE Use what you know about rate of change to describe the function represented by the table. 64. WRITE Explain how the rate of change and slope are related and how to find the slope of a line. 65. FIND THE ERROR Fern is finding the slope of the line that passes through (-2, 8) and (4, 6). Determine in which step she made an error. Explain your reasoning.
66. PERSEVERE Find the value of d so that the line that passes through (a, b) and (c, d) has a slope of 1 __ 2 . 67. ANALYZE Why is the slope undefined for vertical lines? Explain.
68. WRITE Tarak wants to find the value of a so that the line that passes through (10, a) and (-2, 8) has a slope of 1 __ 4 . Explain how Tarak can find the value of a.
= 6 – 8 –2 –4 –2 –6 1 3 = = m Step 1 Step 2 Step 3 Time (wk) Height of Plant (in.) 4 9.0 6 13.5 8 18.0 5 10 0 50 100 Price ($ thousand) Years since 2005 Cottage Prices Since 2005 228 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-2_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 3 2 Step 1; she reversed the order of the x-coordinates in the formula. c - a + 2b _ 2 No; The graph appears to show an increase in price of about $10,000 over 5 years or about $2000 per year.
The difference in the x- values is always 0, and division by 0 is undefined.
Use the slope formula. Substitute (10, a) for (x1, y1), (-2, 8) for (x2, y2), and 1 __ 4 for m. Cross multiply and then solve the equation to find that a = 11.
The rate of change is 2 1 __ 4 inches of growth per week.
After drawing a graph, use the two points on the graph to determine the slope. This can be done by counting squares for the rise and run of the line or by using the coordinates of the points in the slope formula.
See margin. Higher-Order Thinking Skills 219_228_HSM_NA_S_A1M04_L02_662599.indd Page 228 04/10/19 10:23 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Answer 64. Sample answer: Slope can be used to describe a rate of change. Rate of change is a ratio that describes how much one quantity changes with respect to a change in another quantity. The slope of a line is also a ratio and it is the ratio of the change in the y-coordinates to the change in the x-coordinates.
227-228 Module 4 • Linear and Nonlinear Functions F.IF.6, F.LE.5 LESSON GOAL Students graph equations in slope-intercept form.
1 LAUNCH Launch the lesson with a Warm Up and an introduction.
2 EXPLORE AND DEVELOP Develop: Writing Linear Equations in Slope-Intercept Form • Write Linear Equations in Slope-Intercept Form • Rewrite Linear Equations in Slope-Intercept Form • Write Linear Equations Explore: Graphing Linear Functions by Using the Slope-Intercept Form Develop: Graphing Linear Functions in Slope-Intercept Form • Graph Linear Functions in Slope-Intercept Form • Graph Linear Functions • Graph Constant Functions • Use Graphs of Linear Functions You may want your students to complete the Checks online.
3 REFLECT AND PRACTICE Exit Ticket Practice DIFFERENTIATE View reports of student progress on the Checks after each example.
Resources AL OL BL ELL Remediation: Slope of a Line ● ● ● Extension: Pencils of Lines ● ● ● Language Development Handbook Assign page 22 of the Language Development Handbook to help your students build mathematical language related to equations in slope-intercept form.
ELL You can use the tips and suggestions on page T22 of the handbook to support students who are building English proficiency.
Lesson 4-3 Slope-Intercept Form Lesson 4-3 • Slope-Intercept Form 229a A.CED.2, F.IF.7a, F.LE.5 Suggested Pacing 90 min 1 day 45 min 2 days Focus Domain: Algebra, Functions Standards for Mathematical Content: A.CED.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
F.IF.7a Graph linear and quadratic functions and show intercepts, maxima, and minima.
F.LE.5 Interpret the parameters in a linear or exponential function in terms of a context.
Standards for Mathematical Practice: 1 Make sense of problems and persevere in solving them.
4 Model with mathematics.
5 Use appropriate tools strategically.
Coherence Vertical Alignment Rigor The Three Pillars of Rigor 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Conceptual Bridge In this lesson, students extend their understanding of slope. They build fluency by rewriting equations in slope-intercept form to find the slope and y-intercept. They apply their understanding by solving real-world problems involving slope and y-intercept.
Next Students will Identify the effects of transformations of the graphs of linear functions.
F.IF.7a, F.BF.3 Now Students graph equations in slope-intercept form.
A.CED.2, F.IF.7a, F.LE.5 Previous Students found and interpreted the rate of change and slopes of lines.
F.IF.6, F.LE.5 Interactive Presentation 1 LAUNCH Warm Up Warm Up Prerequisite Skills The Warm Up exercises address the following prerequisite skill for this lesson: • identifying slopes Launch the Lesson MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the verbal description and graphs representing the minimum salary for a Major League Baseball player. Go Online to find additional teaching notes and questions to promote classroom discourse.
Today’s Standards Tell students that they will address these content and practice standards in this lesson. You may wish to have a student volunteer read aloud How can I meet these standards? and How can I use these practices?, and connect these to the standards.
See the Interactive Presentation for I Can statements that align with the standards covered in this lesson.
Launch the Lesson Today’s Vocabulary Today’s Vocabulary Tell students that they will use these vocabulary terms in this lesson. You can expand each row if you wish to share the definitions. Then discuss the questions below with the class.
Mathematical Background The slope-intercept form of a linear equation is y = mx + b, where m is the slope, and b is the y-intercept. Writing a linear equation in this form is helpful when you want to graph the function. There are two methods that can be used. The first is to select two values of x, substitute those values into the equation to calculate the corresponding values of y, plot the resulting ordered pairs, and draw the line that passes through the points. The second method is to plot the y-intercept, use it as a starting point, and then use the slope to determine another point on the line. The line can then be drawn through the two points.
Answers: 1. undefined 2. -1 3. - 2 __ 3 4. 1 __ 2 5. -2 229b Module 4 • Linear and Nonlinear Functions A.CED.2, F.IF.7a, F.LE.5 Interactive Presentation 2 EXPLORE AND DEVELOP 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Explore Graphing Linear Functions by Using the Slope-Intercept Form Objective Students use a sketch to explore how changing the slope and y-intercept changes the graph of the line.
MP MP Teaching the Mathematical Practices 5 Use Mathematical Tools Point out that to solve the problem in the Explore, students will need to use a sketch. Work with students to explore and deepen their understanding of slope-intercept form of a linear equation.
Ideas for Use Recommended Use Present the Inquiry Question, or have a student volunteer read it aloud. Have students work in pairs to complete the Explore activity on their devices. Pairs should discuss each of the questions. Monitor student progress during the activity. Upon completion of the Explore activity, have student volunteers share their responses to the Inquiry Question. What if my students don’t have devices? You may choose to project the activity on a whiteboard. A printable worksheet for each Explore is available online. You may choose to print the worksheet so that individuals or pairs of students can use it to record their observations.
Summary of the Activity Students will complete guiding exercises throughout the Explore activity. Students will use the sketch to explore how changing the value of m and b in the equation of a line affects the graph of the function. They will use sliders and animations to change the values of m and/or b in a linear equation, and observe the change in orientation of the related line. Then, students will answer the Inquiry Question.
(continued on the next page) Explore Explore Students answer questions about changing the parameters in the slope-intercept form of a line.
TYPE a Students use a sketch to graph a line by changing the slope and y-intercept.
WEB SKETCHPAD Lesson 4-3 • Slope-Intercept Form 229c F.IF.7a Interactive Presentation 2 EXPLORE AND DEVELOP 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Explore Graphing Linear Functions by Using the Slope-Intercept Form (continued ) Questions Have students complete the Explore activity.
Ask: • Describe the graph when 0 < m < 1 . Sample answer: When the slope is a fraction between 0 and 1, the run is greater than the rise. This means that the slant of the line is more gradual.
• What are the slope and y-intercept of y = 2 __ 3 x − 4? The slope is 2 __ 3 and the y-intercept is -4. Inquiry How do the quantities m and b affect the graph of a linear function in slope-intercept form? Sample answer: Changing the slope affects the steepness of the graph. Changing the y-intercept determines the distance and direction that the graph is shifted from the origin. Go Online to find additional teaching notes and sample answers for the guiding exercises.
Explore Students respond to the Inquiry Question and can view a sample answer.
TYPE a 229d Module 4 • Linear and Nonlinear Functions F.IF.7a Interactive Presentation Learn Writing Linear Equations in Slope-Intercept Form Objective Students rewrite equations in slope-intercept form by applying the properties of equality.
MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Example 1 Write Linear Equations in Slope-Intercept Form MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the verbal description and equation in this example.
Questions for Mathematical Discourse AL What is the slope of the line? 4 __ 7 OL Which variable represents the slope in y = mx + b? m BL How would this equation have changed if the slope had been - 4 __ 7 ? It would have been y = - 4 __ 7 x + 5. Go Online • Find additional teaching notes. • View performance reports of the Checks.
• Assign or present an Extra Example.
Copyright © McGraw-Hill Education Lesson 4-3 Slope-Intercept Form Learn Writing Linear Equations in Slope-Intercept Form An equation of the form y = mx + b, where m is the slope and b is the y-intercept, is written in slope-intercept form. When an equation is not in slope-intercept form, it might be easier to rewrite it before graphing. An equation can be rewritten in slope-intercept form by using the properties of equality.
Key Concept • Slope-Intercept Form Words The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.
Example y = mx + b y = -2x + 7 Example 1 Write Linear Equations in Slope-Intercept Form Write an equation in slope-intercept form for the line with a slope of 4 __ 7 and a y-intercept of 5.
Write the equation in slope-intercept form. y = mx + b Slope-intercept form. y = ( ) x + 5 m = 4 _ 7 , b = 5 y = Simplify.
Check Write an equation for the line with a slope of −5 and a y-intercept of 12. Think About It!
Explain why the y-intercept of a linear equation can be written as (0, b), where b is the y-intercept.
Today’s Vocabulary parameter constant function Go Online You can complete an Extra Example online.
Today’s Goals ● Rewrite linear equations in slope-intercept form.
● Graph and interpret linear functions.
Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-3 • Slope-Intercept Form 229 Sample answer: The y-intercept is the y-coordinate of a point where a graph crosses the y-axis. The point where the graph crosses the y-axis will always have an x-coordinate of 0. 4 __ 7 4 __ 7 x + 5 y = −5x + 12 229_238_HSM_NA_S_A1M04_L03_662599.indd Page 229 04/10/19 11:11 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Students explain how the equation would change if the y-intercept was negative.
Example 1 TYPE a 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION A.CED.2, F.IF.7a, F.LE.5 Lesson 4-3 • Slope-Intercept Form 229 Interactive Presentation 2 EXPLORE AND DEVELOP Copyright © McGraw-Hill Education Example 2 Rewrite Linear Equations in Slope-Intercept Form Write −22x + 8y = 4 in slope-intercept form. - 22x + 8y = 4 Original equation Add 22x to each side. Simplify. Divide each side by 8. Simplify. Check What is the slope intercept form of −16x − 4y = −56? Example 3 Write Linear Equations JOBS The number of job openings in the United States during a recent year increased by an average of 0.06 million per month since May. In May, there were about 4.61 million job openings in the United States. Write an equation in slope-intercept form to represent the number of job openings in the United States in the months since May.
Use the given information to write an equation in slope-intercept form. • You are given that there were million job openings in May. • Let and . • Because the number of job openings is 4.61 million when , , and because the number of job openings has increased by million each month, . • So, the equation represents the number of job openings in the United States since May.
Check SOCIAL MEDIA In the first quarter of 2012, there were 183 million users of a popular social media site in North America. The number of users increased by an average of 9 million per year since 2012. Write an equation that represents the number of users in millions of the social media site in North America after 2012. Go Online You can complete an Extra Example online. Think About It!
When x = 2, describe the meaning of the equation in the context of the situation. Think About It!
Can x = 5 be rewritten in slope-intercept form? Justify your argument.
Your Notes 230 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: When x = 2, the equation represents the number of job openings in July, or two months after May.
No; sample answer: x = 5 is a vertical line, and vertical lines have undefined slopes. So, x = 5 cannot be rewritten in slope-intercept form.
4.61 x = the number of months since May number of job openings in millions y = the x = 0 b = 4.61 0.06 m = 0.06 y = 0.06x + 4.61 y = 9x + 183 −22x + 8y + 22x = 4 + 22x 8y = 22x + 4 8y ___ 8 = 22x + 4 _ 8 y = 2.75x + 0.5 y = −4x + 14 229_238_HSM_NA_S_A1M04_L03_662599.indd Page 230 5/16/19 11:16 PM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 3 Students complete the Check online to determine whether they are ready to move on.
CHECK Students explain the meaning of a certain x-value in context of the situation.
TYPE 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION a Example 2 Rewrite Linear Equations in Slope-Intercept Form MP MP Teaching the Mathematical Practices 3 Justify Conclusions Mathematically proficient students can explain the conclusions drawn when solving a problem. The Think About It! feature asks students to justify their conclusions.
Questions for Mathematical Discourse AL Is this equation in slope-intercept form? Why? No; sample answer: Slope-intercept form is y = mx + b, and in this equation, the y-variable is not isolated. OL How do you know if a linear equation is in slope-intercept form? Sample answer: The y-variable is isolated and it is in the form y = mx + b BL How would this problem be different if the original equation had been -22x - 8y = 4? The last step would have involved dividing by –8 instead of 8, resulting in y = -2.75x - 0.5. Example 3 Write Linear Equations MP MP Teaching the Mathematical Practices 2 Attend to Quantities Point out that it is important to note the meaning of the quantities used in this problem.
Questions for Mathematical Discourse AL Which number is the y-intercept? the slope? 4.61, 0.06 OL What do the slope and y-intercept represent in the context of this situation? the increase in the number of millions of job openings per month since May; 4.61 million job openings in May BL What would it mean if the rate of change was -0.06 in the context of the situation? Sample answer: It would mean a decrease of 0.06 million job openings per month.
A.CED.2, F.IF.7a, F.LE.5 230 Module 4 • Linear and Nonlinear Functions Interactive Presentation Copyright © McGraw-Hill Education Think About It!
Use the slope to find another point on the graph. Explain how you found the point.
Study Tip Negative Slope When counting rise and run, a negative sign may be associated with the value in the numerator or denominator. In this case, we associated the negative sign with the numerator. If we had associated it with the denominator, we would have moved up 3 and left 2 to the point (-2, 7). Notice that this point is also on the line. The resulting line will be the same whether the negative sign is associated with the numerator or denominator.
Explore Graphing Linear Functions by Using the Slope-Intercept Form Online Activity Use graphing technology to complete an Explore. INQUIRY How do the quantities m and b affect the graph of a linear function in slope-intercept form?
Learn Graphing Linear Functions in Slope-Intercept Form The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept. The variables m and b are called parameters of the equation because changing either value changes the graph of the function.
A constant function is a linear function of the form y = b. Constant functions where b ≠ 0 do not cross the x-axis. The graphs of constant functions have a slope of 0. The domain of a constant function is all real numbers, and the range is b.
Example 4 Graph Linear Functions in Slope-Intercept Form Graph a linear function with a slope of − 3 __ 2 and a y-intercept of 4.
Write the equation in slope-intercept form and graph the function.
y = mx + b y = ( ) x y = y x O rise = –3 run = 2 Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-3 • Slope-Intercept Form 231 Sample answer: (4, -2); I moved down 3 units and right 2 units from the point (2, 1) - 3 __ 2 + 4 - 3 __ 2 x + 4 229_238_HSM_NA_S_A1M04_L03_662599.indd Page 231 5/16/19 11:21 PM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Learn Graphing Linear Functions in Slope-Intercept Form Objective Students graph and interpret linear functions by writing them in slope-intercept form.
MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between linear functions in slope-intercept form and their graphs.
Common Misconception Some students may think that when the slope is negative, they should count down for the rise and left for the run to find additional points. Show students that this would lead to a line that is rising from left to right, not falling, as would be the orientation for a line with a negative slope. Tell them to count up and to the right for positive slopes, and down and to the right for negative slopes.
Example 4 Graph Linear Functions in Slope-Intercept Form MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Questions for Mathematical Discourse AL In slope-intercept form, which variable represents the slope? m the y-intercept? b OL When graphing a line in slope-intercept form, why is b graphed first? Sample answer: In order to use the slope, you have to have a starting point.
BL Why do you find the next point by counting down 3 and to the right 2? Sample answer: The slope is negative, so instead of counting up and to the right, you count down and to the right.
Example 4 Students explain how to find another point on the graph by using the slope.
TYPE a Students use a sketch to graph a linear function in slope-intercept form.
WEB SKETCHPAD A.CED.2, F.IF.7a, F.LE.5 Lesson 4-3 • Slope-Intercept Form 231 Interactive Presentation 2 EXPLORE AND DEVELOP Example 5 Graph Linear Functions MP MP Teaching the Mathematical Practices 1 Seek Information Mathematically proficient students must be able to transform algebraic expressions to reach solutions. Point out that gaining fluency in this skill is as important as learning their math facts was in the elementary grades.
Questions for Mathematical Discourse AL What variable must you solve for in order to write the equation in slope-intercept form? y OL What are the slope and the y-intercept of the line? The slope is 4. The y-intercept is -6.
BL How can the intercepts of the line be used to check your answer? Sample answer: Using the given form of the line, I know the x-intercept will be (1.5, 0) and the y-intercept will be (0, -6). My graph crosses at those points, so the graph is correct.
Common Error For an equation such as y = 4x - 6, some students may state that b = 6. Review the general form of the slope-intercept form of a linear equation ( y = mx + b), and highlight the plus sign. Help students to see that y = 4x - 6 is equivalent to y = 4x + (-6), so b = -6. Therefore, the y-intercept is -6.
DIFFERENTIATE Reteaching Activity AL ELL IF students have difficulty distinguishing between the variables and the parameters in the equation, THEN write several different equations on the board, each in slope-intercept form. Point out that in each case, the equation contains numbers where m and b, which are fixed values, would be the parameters while the variables x and y, which vary in value, represent the coordinates of the solutions of the equation. Examining several equations side by side helps to strengthen understanding of the concept.
Copyright © McGraw-Hill Education Check Graph a linear function with a slope of −2 and a y-intercept of 7.
y x O 6 7 5 4 3 2 1 −1 3 2 1 4 5 6 7 Example 5 Graph Linear Functions Graph 12x – 3y = 18.
Rewrite the equation in slope-intercept form. 12x - 3y = 18 Original equation Subtract 12x from each side. Simplify. Divide each side by −3. Simplify. Graph the function.
Plot the y-intercept (0, −6).
The slope is rise ___ run = 4. From (0, −6), move up 4 units and right 1 unit. Plot the point (1, −2).
Draw a line through the points (0, -6) and (1, −2). y x O −4 −8 8 4 4 8 −4 −8 (1, −2) (0, −6) Talk About It!
Why is it useful to write an equation in slope-intercept form before graphing the function? Go Online You can complete an Extra Example online.
232 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: When an equation is in slope-intercept form, you can easily determine the slope and y-intercept and use them to create the graph.
12x - 3y − 12x = 18 − 12x -3y = −12x + 18 -3y _ - 3 = −12x + 18 __ -3 y = 4x − 6 229_238_HSM_NA_S_A1M04_L03_662599.indd Page 232 5/16/19 11:24 PM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 5 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students explain why it is useful to write an equation in slope-intercept form before graphing.
TYPE a Students move through the steps to graph a linear function.
TAP A.CED.2, F.IF.7a, F.LE.5 232 Module 4 • Linear and Nonlinear Functions Interactive Presentation DIFFERENTIATE Enrichment Activity BL Write 3x + 2y = 8 and -3x + 2y = 8 on the board. Ask students to tell how the equations are alike and how they are different. Then, ask students to tell how the graphs of the two functions are alike and how they are different without graphing them. Finally, have them graph the functions and check their answers.
Example 6 Graph Constant Functions MP MP Teaching the Mathematical Practices 5 Use Mathematical Tools Point out that to solve the problem in this example, students will need to use a sketch. Work with students to explore and deepen their understanding of slope-intercept form.
Questions for Mathematical Discourse AL What is the y-intercept? x-intercept? 2; There is no x-intercept.
OL Why is the graph a horizontal line? Sample answer: Because the slope is 0, the graph will not rise, but can run left to right any amount.
BL What is the domain of this function? the range? D = all real numbers; R = 2 Common Error Some students may think that the slope is 2, since, when an equation is written in slope-intercept form, the slope is the number after the equal sign. Point out that if the slope were 2, the equation would be y = 2x. Since there is no x term, the slope is 0, and the equation is y = 0x + 2. Essential Question Follow-Up Students have explored the relationship between the parameters of a linear function and its graph.
Ask: What can you learn about the graph of a linear function by analyzing its equation? Sample answer: If the equation is in slope-intercept form, I can tell where the graph intersects the y-axis and what the slope of the line is.
Copyright © McGraw-Hill Education Example 6 Graph Constant Functions Graph y = 2.
Step 1 Plot (0, 2). Step 2 The slope of y = 2 is 0.
Step 3 Draw a line through all the points that have a y-coordinate of 2.
y x 2 3 1 −2 −3 O −2 −3 −1 2 3 1 Check Graph y = 1. y x 2 1 3 −2 −3 O −2 −3 −1 2 3 1 Match each graph with its function. y = 8 3x + 7y = − 28 y = 3 _ 7 x − 4 y = −4 y = − 3x + 8 3x − y = 8 A.
y x 8 4 −8 −4 O 2 4 6 B.
y x 8 4 −8 −4 O −2 −4 2 4 C. y x 4 2 −4 −2 O −2 −4 2 4 D.
y x 8 4 −8 −4 O −2 −4 2 4 E.
y x 8 4 −8 −4 O −4 −8 4 8 F.
y x 8 4 −8 −4 O −4 −8 4 8 Go Online You can complete an Extra Example online. Think About It!
How do you know that the graph of y = 2 has a slope of 0?
Watch Out!
Slope A line with zero slope is not the same as a line with no slope. A line with zero slope is horizontal, and a line with no slope is vertical.
Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-3 • Slope-Intercept Form 233 D F E C B A Sample answer: The equation y = 2 can be rewritten as y = 0x + 2. Then I can see that m = 0 and b = 2.
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Example 6 Conceptual Understanding Fluency Application 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students move through the steps of graphing a constant function.
TAP Students explain why the graph has a slope of 0.
TYPE a Students use a sketch to graph a constant function.
WEB SKETCHPAD A.CED.2, F.IF.7a, F.LE.5 Lesson 4-3 • Slope-Intercept Form 233 Interactive Presentation Interactive Presentation 2 EXPLORE AND DEVELOP Apply Example 7 Use Graphs of Linear Functions MP MP Teaching the Mathematical Practices 1 Make Sense of Problems and Persevere in Solving Them, 4 Model with Mathematics Students will be presented with a task. They will first seek to understand the task, and then determine possible entry points to solving it. As students come up with their own strategies, they may propose mathematical models to aid them. As they work to solve the problem, encourage them to evaluate their model and/or progress, and change direction, if necessary.
Recommended Use Have students work in pairs or small groups. You may wish to present the task, or have a volunteer read it aloud. Then allow students the time to make sure they understand the task, think of possible strategies, and work to solve the problem.
Encourage Productive Struggle As students work, monitor their progress. Instead of instructing them on a particular strategy, encourage them to use their own strategies to solve the problem and to evaluate their progress along the way. They may or may not find that they need to change direction or try out several strategies.
Signs of Non-Productive Struggle If students show signs of non-productive struggle, such as feeling overwhelmed, frustrated, or disengaged, intervene to encourage them to think of alternate approaches to the problem. Some sample questions are shown.
• How can you determine the domain?
• How can you use the graph to estimate how many people will be shopping online in 2020? Write About It!
Have students share their responses with another pair/group of students or the entire class. Have them clearly state or describe the mathematical reasoning they can use to defend their solution.
Exit Ticket Recommended Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond using a separate piece of paper. Have students hand you their responses as they leave the room.
Alternate Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond verbally or by using a mini-whiteboard. Have students hold up their whiteboards so that you can see all student responses. Tap to reveal the answer when most or all students have completed the Exit Ticket.
Copyright © McGraw-Hill Education Think About It!
Estimate the year when the number of online shoppers in the United States will reach 271 million. Go Online to learn about intervals in linear growth patterns in Expand 4-3. Go Online You can complete an Extra Example online. Apply Example 7 Use Graphs of Linear Functions SHOPPING The number of online shoppers in the United States can be modeled by the equation −5.88x + y = 172.3, where y represents the number of millions of online shoppers in the United States x years after 2010. Estimate the number of people shopping online in 2020. 1. What is the task?
Describe the task in your own words. Then list any questions that you may have. How can you find answers to your questions?
2. How will you approach the task? What have you learned that you can use to help you complete the task?
3. What is your solution?
Use your strategy to solve the problem. Graph the function.
In 2020, there will be approximately online shoppers in the United States.
4. How can you know that your solution is reasonable? Write About It! Write an argument that can be used to defend your solution.
2 4 6 8 10 Number of Shoppers (millions) 250 200 150 100 50 0 Years After 2010 Online Shoppers in the United States 234 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: 17 years after 2010 or 2027 Sample answer: I need to find the number of people who shop online in 2020.
Sample answer: I will graph the given function. Then I can figure out from the graph how many people will be shopping online in 2020.
230 million Sample answer: Rewriting the equation in slope-intercept form shows that b = 172.3 and m = 5.88. This means that there were 172.3 million online shoppers in 2010. The number of online shoppers is increasing at a rate of 5.88 million per year. The graph of this line shows that in 2020 the number of online shoppers is more than 225 million but less than 250 million. From the graph, there will be approximately 230 million online shoppers in 2020.
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1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Apply Example 7 Students estimate the year when the online shoppers will reach a certain number.
TYPE a Students complete the Check online to determine whether they are ready to move on.
CHECK A.CED.2, F.IF.7a, F.LE.5 234 Module 4 • Linear and Nonlinear Functions 3 REFLECT AND PRACTICE Practice and Homework Suggested Assignments Use the table below to select appropriate exercises.
DOK Topic Exercises 1, 2 exercises that mirror the examples 1–30 2 exercises that use a variety of skills from this lesson 31–39 2 exercises that extend concepts learned in this lesson to new contexts 40–43 3 exercises that emphasize higher-order and critical thinking skills 44–47 ASSESS AND DIFFERENTIATE Use the data from the Checks to determine whether to provide resources for extension, remediation, or intervention.
IF students score 90% or more on the Checks, THEN assign: BL • Practice, Exercises 1–43 odd, 44–47 • Extension: Pencils of Lines • Equations of Lines IF students score 66%–89% on the Checks, THEN assign: OL • Practice, Exercises 1–47 odd • Remediation, Review Resources: Slope of a Line • BrainPOP Video: Slope and Intercepts • Extra Examples 1–7 • Slope IF students score 65% or less on the Checks, THEN assign: AL • Practice, Exercises 1–29 odd • Remediation, Review Resources: Slope of a Line • Quick Review Math Handbook: Writing Equations in Slope-Intercept Form • ArriveMATH Take Another Look • Slope Copyright © McGraw-Hill Education Name Period Date Practice Go Online You can complete your homework online.
Example 1 Write an equation of a line in slope-intercept form with the given slope and y-intercept. 1. slope: 5, y-intercept: -3 2. slope: -2, y-intercept: 7 3. slope: -6, y-intercept: -2 4. slope: 7, y-intercept: 1 5. slope: 3, y-intercept: 2 6. slope: -4, y-intercept: -9 7. slope: 1, y-intercept: -12 8. slope: 0, y-intercept: 8 Example 2 Write each equation in slope-intercept form. 9. -10x + 2y = 12 10. 4y + 12x = 16 11. -5x + 15y = -30 12. 6x - 3y = -18 13. –2x – 8y = 24 14. -4x - 10y = -7 Example 3 15. SAVINGS Wade’s grandmother gave him $100 for his birthday. Wade wants to save his money to buy a portable game console. Each month, he adds $25 to his savings. Write an equation in slope-intercept form to represent Wade’s savings y after x months. 16. FITNESS CLASSES Toshelle wants to take strength training classes at the community center. She has to pay a one-time enrollment fee of $25 to join the community center, and then $45 for each class she wants to take. Write an equation in slope-intercept form for the cost of taking x classes. 17. EARNINGS Macario works part time at a clothing store in the mall. He is paid $9 per hour plus 12% commission on the items he sells in the store. Write an equation in slope-intercept form to represent Macario’s hourly wage y. 18. ENERGY From 2002 to 2005, U.S. consumption of renewable energy increased an average of 0.17 quadrillion BTUs per year. About 6.07 quadrillion BTUs of renewable power were produced in the year 2002. Write an equation in slope-intercept form to find the amount of renewable power P in quadrillion BTUs produced in year y between 2002 and 2005. Example 4 Graph a linear function with the given slope and y-intercept. 19. slope: 5, y-intercept: 8 20. slope: 3, y-intercept: 10 21. slope: −4, y-intercept: 6 22. slope: −2, y-intercept: 8 Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-3 • Slope-Intercept Form 235 y = 5x − 3 y = −2x + 7 y = −6x − 2 y = 7x + 1 y = 3x + 2 y = −4x − 9 y = x − 12 y = 8 y = 5x + 6 y = −3x + 4 y = 1 __ 3 x − 2 y = 2x + 6 y = −0.25x − 3 y = −0.4x + 0.7 y = 25x + 100 y = 45x + 25 y = 0.12x + 9 P = 0.17y + 6.07 19–22. See margin.
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Copyright © McGraw-Hill Education Examples 5 and 6 Graph each function. 23. 5x + 2y = 8 24. 4x + 9y = 27 25. y = 7 26. y = - 2 __ 3 27. 21 = 7y 28. 3y − 6 = 2x Example 7 29. STREAMING An online company charges $13 per month for the basic plan. They offer premium channels for an additional $8 per month. a. Write an equation in slope-intercept form for the total cost c of the basic plan with p premium channels in one month.
b. Graph the function.
c. What would the monthly cost be for a basic plan plus 3 premium channels? 30. CAR CARE Suppose regular gasoline costs $2.76 per gallon. You can purchase a car wash at the gas station for $3.
a. Write an equation in slope-intercept form for the total cost y of purchasing a car wash and x gallons of gasoline. b. Graph the function. c. Find the cost of purchasing a car wash and 8 gallons of gasoline. Mixed Exercises Write an equation of a line in slope-intercept form with the given slope and y-intercept. 31. slope: 1 __ 2 , y-intercept: -3 32. slope: 2 __ 3 , y-intercept: -5 Graph a function with the given slope and y-intercept. 33. slope: 3, y-intercept: -4 34. slope: 4, y-intercept: -6 Graph each function. 35. −3x + y = 6 36. −5x + y = 1 Write an equation in slope-intercept form for each graph shown. 37. (2, 1) (0, -3) y x O 38.
(0, 2) (2, -4) y x O 39.
(2, -3) (0, -1) y x O 236 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE y = 1 __ 2 x − 3 y = 2 __ 3 x − 5 c = 13 + 8p $25.08 $37 y = 2.76x + 3 2 3 1 4 5 6 7 8 9 10 Cost of Gas and Car Wash ($) 24 0 2 4 6 8 10 12 14 16 18 20 22 Gasoline (gal) (0, 3) (0, 3) (2, 8.52) (4, 14.04) y = 2x − 3 y = −3x + 2 y = −x − 1 33–36. See Mod. 4 Answer Appendix.
23–28. See margin.
2 4 6 8 10 1 3 5 7 9 5 0 10 15 20 25 30 35 40 45 50 Total Cost ($) Premium Channels Streaming Television Plan 229_238_HSM_NA_S_A1M04_L03_662599.indd Page 236 05/10/19 7:51 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION A.CED.2, F.IF.7a, F.LE.5 Lesson 4-3 • Slope-Intercept Form 235–236 Interactive Presentation 3 REFLECT AND PRACTICE Copyright © McGraw-Hill Education Name Period Date 40. MOVIES MovieMania, an online movie rental Web site charges a one-time fee of $6.85 and $2.99 per movie rental. Let m represent the number of movies you watch and let C represent the total cost to watch the movies.
a. Write an equation that relates the total cost to the number of movies you watch from MovieMania. b. Graph the function. c. Explain how to use the graph to estimate the cost of watching 13 movies at MovieMania. d. SuperFlix has no sign-up fee, just a flat rate per movie. If renting 13 movies at MovieMania costs the same as renting 9 movies at SuperFlix, what does SuperFlix charge per movie? Explain your reasoning. e. Write an equation that relates the total cost to the number of movies you watch from SuperFlix. Round to the nearest whole number. 41. FACTORY A factory uses a heater in part of its manufacturing process. The product cannot be heated too quickly, nor can it be cooled too quickly after the heating portion of the process is complete.
a. The heater is digitally controlled to raise the temperature inside the chamber by 10°F each minute until it reaches the set temperature. Write an equation to represent the temperature, T, inside the chamber after x minutes if the starting temperature is 80°F. b. Graph the function. c. The heating process takes 22 minutes. Use your graph to find the temperature in the chamber at this point. d. After the heater reaches the temperature determined in part c, the temperature is kept constant for 20 minutes before cooling begins. Fans within the heater control the cooling so that the temperature inside the chamber decreases by 5°F each minute. Write an equation to represent the temperature, T, inside the chamber x minutes after the cooling begins. 42. SAVINGS When Santo was born, his uncle started saving money to help pay for a car when Santo became a teenager. Santo’s uncle initially saved $2000. Each year, his uncle saved an additional $200. a. Write an equation that represents the amount, in dollars, Santo’s uncle saved y after x years.
b. Graph the function. c. Santo starts shopping for a car when he turns 16. The car he wants to buy costs $6000. Does he have enough money in the account to buy the car? Explain.
Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-3 • Slope-Intercept Form 237 C = 2.99m + 6.85 See Mod. 4 Answer Appendix.
See Mod. 4 Answer Appendix.
See Mod. 4 Answer Appendix.
C = 5m T = 10x + 80 See Mod. 4 Answer Appendix.
300°F T = -5x + 300 See Mod. 4 Answer Appendix.
No; Santo only has $5200 in his account. He needs to save an additional $800 to buy the car he wants.
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Copyright © McGraw-Hill Education 43. STRUCTURE Jazmin is participating in a 25.5-kilometer charity walk. She walks at a rate of 4.25 km per hour. Jazmin walks at the same pace for the entire event. a. Write an equation in slope-intercept form for the remaining distance y in kilometers of walking for x hours. b. Graph the function. c. What do the x- and y-intercepts represent in this situation?
d. After Jazmin has walked 17 kilometers, how much longer will it take her to complete the walk? Explain how you can use your graph to answer the question. For Exercises 44 and 45, refer to the equation y = − 4 __ 5 x + 2 __ 5 where -2 ≤ x ≤ 5. 44. ANALYZE Complete the table to help you graph the function y = − 4 __ 5 x + 2 __ 5 over the interval. Identify any values of x where maximum or minimum values of y occur. x − 4 __ 5 x + 2 __ 5 y (x, y) −2 0 5 45. WRITE A student says you can find the solution to - 4 __ 5 x + 2 __ 5 = 0 using the graph. Do you agree? Explain your reasoning. Include the solution to the equation in your response. 46. PERSEVERE Consider three points that lie on the same line, (3, 7), (-6, 1), and (9, p). Find the value of p and explain your reasoning. 47. CREATE Linear equations are useful in predicting future events. Create a linear equation that models a real-world situation. Make a prediction from your equation.
238 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-3_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE y = −4.25x + 25.5 The x-intercept (6) represents the number of hours it will take Jazmin to complete the walk. The y-intercept (25.5) represents the length of the walk. Using the graph, I can determine the value of x when y equals −17 + 25.5 or 8.5 km, and use the value of the x-intercept. The value of x is 4 when y = 8.5 and the x-intercept is 6. Therefore, Jazmin has 6 − 4 or 2 hours more to walk.
Yes; you can find the value of x on the graph when y = 0; x = 1 __ 2 .
11; Use the first two points to find the equation of the line, then replace x and y with 9 and p, respectively, to solve for p.
Sample answer: y = 25x + 200; I have $200 in savings and will save $25 per week until I have enough money to buy a new phone. I can predict how much money I’ll have after x number of weeks. Maximum value of y occurs when x = −2; Minimum value of y occurs when x = 5 2 4 6 8 10 12 Remaining Distance (km) 26 0 2 4 6 8 10 12 14 16 18 20 22 24 Hours (0, 25.5) (6, 0) Charity Walk − 4 __ 5 (−2) + 2 __ 5 2 (−2, 2) − 4 __ 5 (0) + 2 __ 5 2 __ 5 (0, 2 __ 5 ) − 4 __ 5 (5) + 2 __ 5 − 18 __ 5 (5, − 18 __ 5 ) y x 4 2 −4 −2 O −2 2 4 Higher-Order Thinking Skills 229_238_HSM_NA_S_A1M04_L03_662599.indd Page 238 05/10/19 7:53 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Answers 19. y x 8 4 −8 −4 O −4 −8 4 8 20. y x 12 8 4 −4 O −4 −8 4 8 21. y x 8 4 −8 −4 O −4 −8 4 8 22. y x 8 4 −8 −4 O −4 −8 4 8 23. y x O 24. y x O 25. y x O y = 7 26. y x O y = − 2 – – 3 27. y x O 21 = 7y 28. y x O A.CED.2, F.IF.7a, F.LE.5 237–238 Module 4 • Linear and Nonlinear Functions LESSON GOAL Students identify the effects of transformations of the graphs of linear functions.
1 LAUNCH Launch the lesson with a Warm Up and an introduction.
2 EXPLORE AND DEVELOP Explore: Transforming Linear Functions Develop: Translations of Linear Functions • Vertical Translations of Linear Functions • Horizontal Translations of Linear Functions • Multiple Translations of Linear Functions • Translations of Linear Functions Dilations of Linear Functions • Vertical Dilations of Linear Functions • Horizontal Dilations of Linear Functions Reflections of Linear Functions • Reflections of Linear Functions Across the x-Axis • Reflections of Linear Functions Across the y-Axis You may want your students to complete the Checks online.
3 REFLECT AND PRACTICE Exit Ticket Practice DIFFERENTIATE View reports of student progress on the Checks after each example.
Resources AL OL BL ELL Remediation: Reflections ● ● ● Extension: Transformations of Other Families of Functions ● ● ● Language Development Handbook Assign page 23 of the Language Development Handbook to help your students build mathematical language related to transformations of the graphs of linear functions.
ELL You can use the tips and suggestions on page T23 of the handbook to support students who are building English proficiency.
Lesson 4-4 Transformations of Linear Functions Lesson 4-4 • Transformations of Linear Functions 239a F.IF.7a, F.BF.3 Suggested Pacing 90 min 1 day 45 min 2 days Focus Domain: Functions Standards for Mathematical Content: F.IF.7a Graph linear and quadratic functions and show intercepts, maxima, and minima.
F.BF.3 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology.
Standards for Mathematical Practice: 1 Make sense of problems and persevere in solving them.
5 Use appropriate tools strategically.
7 Look for and make use of structure.
Coherence Vertical Alignment Next Students will write and graph equations of arithmetic sequences. F.BF.1a, F.BF.2, F.LE.2 Now Students Identify the effects of transformations of the graphs of linear functions.
F.IF.7a, F.BF.3 Previous Students graphed equations in slope-intercept form.
A.CED.2, F.IF.7a, F.LE.5 Rigor The Three Pillars of Rigor 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Conceptual Bridge In this lesson, students develop understanding of transformations of functions by examining the family of linear functions. They build fluency by describing transformations and identifying transformed functions. They apply their understanding by solving real-world problems.
Interactive Presentation 1 LAUNCH Warm Up Prerequisite Skills The Warm Up exercises address the following prerequisite skill for this lesson: • translating and reflecting geometric figures Launch the Lesson Launch the Lesson MP MP Teaching the Mathematical Practices 7 Use Structure Help students to use the structure of a linear function to identify the effect on the graph when replacing f(x) with f(x) + k, k · f(x), f(kx), and f(x + k) for specific values of k. Go Online to find additional teaching notes and questions to promote classroom discourse.
Today’s Standards Tell students that they will address these content and practice standards in this lesson. You may wish to have a student volunteer read aloud How can I meet these standards? and How can I use these practices?, and connect these to the standards.
See the Interactive Presentation for I Can statements that align with the standards covered in this lesson.
Today’s Vocabulary Today’s Vocabulary Tell students that they will use these vocabulary terms in this lesson. You can expand each row if you wish to share the definitions. Then discuss the questions below with the class.
Mathematical Background The parent function of the family of linear functions is f (x) = x.
Transformations of the parent graph occur when a constant is added to or subtracted from the function or the argument, or when the function or the argument is multiplied by a number. These transformations alter the graph, translating it in a particular direction, dilating it, or reflecting it. Recognizing the effect produced by each type of transformation allows for the new graph to be easily obtained from the graph of the parent function.
Answers: 1. rotation 4. dilation 2. translation 5. translation 3. reflection Warm Up 239b Module 4 • Linear and Nonlinear Functions F.IF.7a, F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Transforming Linear Functions Objective Students use a sketch to explore how changing the parameters changes the graphs of linear functions.
MP MP Teaching the Mathematical Practices 3 Construct Arguments In this Explore, students will use stated assumptions, definitions, and previously established results to construct arguments.
Ideas for Use Recommended Use Present the Inquiry Question, or have a student volunteer read it aloud. Have students work in pairs to complete the Explore activity on their devices. Pairs should discuss each of the questions. Monitor student progress during the activity. Upon completion of the Explore activity, have student volunteers share their responses to the Inquiry Question. What if my students don’t have devices? You may choose to project the activity on a whiteboard. A printable worksheet for each Explore is available online. You may choose to print the worksheet so that individuals or pairs of students can use it to record their observations.
Summary of the Activity Students complete guiding exercises throughout the Explore activity. Students use a sketch to explore how the graph of a function is affected when a number is added to the function, when a number is subtracted from the argument of the function, or when the function is multiplied by a number. They enter various values for the number and view the resulting graph. Then, students answer the Inquiry Question.
(continued on the next page) Explore Explore Students use a sketch to explore the effects of addition and multiplication on a function.
Students answer questions about transformations of linear functions.
WEB SKETCHPAD TYPE 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION a Lesson 4-4 • Transformations of Linear Functions 239c F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Transforming Linear Functions (continued ) Questions Have students complete the Explore activity.
Ask: • Does adding or subtracting a value to a function change the slope or y-intercept? Sample answer: The line moves up/down or left/right when you add or subtract values to the function. This means that the y-intercept is changing, but not the slope.
• Why does multiplying a function by a value make it more or less steep? Sample answer: If we multiply every value in a function, then we are changing the value of y for every x-value. If we multiply by a value greater than one, then the difference between the y-values will be greater, resulting in a greater slope and a steeper line. Inquiry How does performing an operation on a linear function change its graph? Sample answer: Adding a value to the function moves the graph up or down. Subtracting a value from x moves the graph left or right.
Multiplying the function by a value makes the graph more steep or less steep. Go Online to find additional teaching notes and sample answers for the guiding exercises.
Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students respond to the Inquiry Question and can view a sample answer.
TYPE a 239d Module 4 • Linear and Nonlinear Functions F.BF.3 Interactive Presentation Learn Translations of Linear Functions Objective Students identify the effects on the graphs of linear functions by replacing f(x) with f(x) + k and f(x - h) for positive and negative values.
MP MP Teaching the Mathematical Practices 7 Use Structure Help students to explore the structure of translations in this Learn.
What Students Are Learning The parent function of the family of linear functions is f(x) = x. Its graph is the line that passes through the origin and has a slope of 1. The graph of every other linear function is a transformation of this function. The first type of transformation students will learn about is translations. Under a translation, the graph of a line is slid to a new location.
Common Misconception Students may believe that a translation will change the orientation of the figure. Help them to see that this is not the case. When a figure is slid in its entirety up, down, left, or right, its orientation remains the same. In the case of a line, its slope is not affected, so the new image has the same slope as the original graph.
Vertical Translations MP MP Teaching the Mathematical Practices 7 Use Structure Help students to explore the structure of vertical translations in this Learn.
About the Key Concept When k is added to the function f(x) = x, the graph of the function is translated vertically. This is because adding k to the function increases the y-value that is associated with each x-value by k units. When k is negative, each y-value decreases, which translates the graph down |k| units.
Common Misconception Some students may think that adding k to a function increases (or decreases) the x-value in each ordered pair. Remind students that the notation f(x) represents the y-value that is paired with x. Thus, f(x) + k represents an increase (or decrease) in y-values, resulting in a vertical translation.
Copyright © McGraw-Hill Education Lesson 4-4 Transformations of Linear Functions Study Tip Slope When translating a linear function, the graph of the function moves from one location to another, but the slope remains the same. Today’s Vocabulary family of graphs parent function identity function transformation translation dilation reflection Watch Out!
Translations of f (x) When a translation is the only transformation performed on the identity function, adding a constant before or after evaluating the function has the same effect on the graph. However, when more than one type of transformation is applied, this will not be the case.
Explore Transforming Linear Functions Online Activity Use graphing technology to complete an Explore. INQUIRY How does performing an operation on a linear function change its graph?
Learn Translations of Linear Functions A family of graphs includes graphs and equations of graphs that have at least one characteristic in common. The parent function is the simplest of the functions in a family.
The family of linear functions includes all lines, with the parent function f (x) = x, also called the identity function. A transformation moves the graph on the coordinate plane, which can create new linear functions.
One type of transformation is a translation. A translation is a transformation in which a figure is slid from one position to another without being turned. A linear function can be slid up, down, left, right, or in two directions. Vertical Translations When a constant k is added to a linear function f (x), the result is a vertical translation. The y-intercept of f (x) is translated up or down.
Key Concept • Vertical Translations of Linear Functions The graph of g(x) = x + k is the graph of f (x) = x translated vertically.
If k > 0, the graph of f (x) is translated k units up.
y x O k > 0 k = 0 Every point on the graph of f(x) moves k units up.
If k < 0, the graph of f (x) is translated |k| units down.
y x O k = 0 k < 0 Every point on the graph of f(x) moves |k| units down.
Today’s Goals ● Apply translations to linear functions.
● Apply dilations to linear functions.
● Apply reflections to linear functions.
Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 Lesson 4-4 • Transformations of Linear Functions 239 239_250_HSM_NA_S_A1M04_L04_662599.indd Page 239 04/10/19 11:48 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Learn 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students tap each flash card to learn more about vertical translations.
TAP Lesson 4-4 • Transformations of Linear Functions 239 F.IF.7a, F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Example 1 Vertical Translations of Linear Functions MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others. Questions for Mathematical Discourse AL Looking at only the equation, how do you know the type of transformation? 2 is being subtracted from the parent function so this is a vertical translation.
OL How is the y-value of each ordered pair in the parent function affected? Each y-value decreases by 2.
BL How would you write this function as a vertical translation of the parent graph up 2 units? g(x) = f(x) + 2 or g(x) = x + 2 Horizontal Translations MP MP Teaching the Mathematical Practices 3 Analyze Cases Guide students to examine the cases of different translations. Encourage students to familiarize themselves with all of the cases.
Common Misconception Some students may think that the graph of f(x + h), where h is a positive number, is a translation of the parent graph h units to the right. Point out that f(x + h) = f(x - (-h)), so the number being subtracted is a negative number. Thus, the shift is to the left, not to the right.
Copyright © McGraw-Hill Education Example 1 Vertical Translations of Linear Functions Describe the translation in g(x) = x - 2 as it relates to the graph of the parent function.
Graph the parent graph for linear functions.
Because f(x) = x, g(x) = f(x) + k where . g(x) = x - 2 → The constant k is not grouped with x, so k affects the , or . The value of k is less than 0, so the graph of f(x) = x is translated units down, or 2 units down.
g(x) = x - 2 is the translation of the graph of the parent function 2 units down.
Check Describe the translation in g(x) = x - 1 as it relates to the graph of the parent function.
The graph of g(x) = x - 1 is a translation of the graph of the parent function 1 unit . Horizontal Translations When a constant h is subtracted from the x-value before the function f (x) is performed, the result is a horizontal translation. The x-intercept of f (x) is translated right or left.
Key Concept • Horizontal Translations of Linear Functions The graph of g(x) = (x − h) is the graph of f (x) = x translated horizontally.
If h > 0, the graph of f (x) is translated h units right.
y x O h = 0 h > 0 Every point on the graph of f (x) moves h units right.
If h < 0, the graph of f (x) is translated |h| units left.
y x O h < 0 h = 0 Every point on the graph of f (x) moves |h| units left. Go Online You can complete an Extra Example online. Think About It!
What do you notice about the y-intercepts of vertically translated functions compared to the y-intercept of the parent function? x f(x) f(x) - 2 (x, g(x)) -2 -2 -4 (-2, -4) 0 0 -2 (0, -2) 1 1 -1 (1, -1) y x O g(x) f(x) Go Online You can watch a video to see how to describe translations of functions.
Your Notes Go Online You may want to complete the Concept Check to check your understanding.
240 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 Sample answer: The y-intercepts move up k units or down |k| units from the y-intercept of the parent function.
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1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 1 Students describe how the y-intercept of the translated function compares to the parent function.
TYPE a Students move through the steps to graph a vertical translation.
TAP 240 Module 4 • Linear and Nonlinear Functions F.IF.7a, F.BF.3 Interactive Presentation Copyright © McGraw-Hill Education Example 2 Horizontal Translations of Linear Functions Describe the translation in g(x) = (x + 5) as it relates to the graph of the parent function.
Graph the parent graph for linear functions.
Because f(x) = x, where h = -5. g(x) = (x + 5) → The constant h is grouped with x, so k affects the , or . The value of h is less than 0, so the graph of f(x) = x is translated units left, or 5 units left. g(x) = (x + 5) is the translation of the graph of the parent function 5 units left.
Check Describe the translation in g(x) = (x + 12) as it relates to the graph of the parent function.
The graph of g(x) = (x + 12) is a translation of the graph of the parent function 12 units . Example 3 Multiple Translations of Linear Functions Describe the translation in g(x) = (x - 6) + 3 as it relates to the graph of the parent function.
Graph the parent graph for linear functions. Because f(x) = x, where and . g(x) = (x - 6) + 3 → The value of h is grouped with x and is greater than 0, so the graph of f(x) = x is translated . The value of k is not grouped with x and is greater than 0, so the graph of f(x) = x is translated . g(x) = (x - 6) + 3 is the translation of the graph of the parent function 6 units right and 3 units up. Think About It!
What do you notice about the x-intercepts of horizontally translated functions compared to the x-intercept of the parent function? Go Online You can complete an Extra Example online.
x x + 5 f(x + 5) (x, g(x)) -2 3 3 (-2, 3) 0 5 5 (0, 5) 1 6 6 (1, 6) y x O f(x) g(x) x x - 6 f(x - 6) f(x - 6) + 3 (x, g(x)) -2 -8 -8 -5 (-2, -5) 0 -6 -6 -3 (0, -3) 1 -5 -5 -2 (1, -2) y x O g(x) f(x) Think About It!
Eleni described the graph of g(x) = (x − 6) + 3 as the graph of the parent function translated down 3 units. Is she correct? Explain your reasoning. Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 Lesson 4-4 • Transformations of Linear Functions 241 Sample answer: The x-intercepts move right h units or left |h| units from the x-intercept of the parent function.
input x-values |-5| g(x) = f(x - h) g(x) = f(x - (-5)) left g(x) = f(x - h) + k g(x) = f(x - 6) + 3 h = 6 k = 3 6 units right 3 units up Yes; sample answer: g(x) = (x − 6) + 3 can be simplified to g(x) = x − 3, which is the same as g(x) = f(x) − 3.
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Example 2 Students describe how the x-intercept of the translated function compares to the parent.
TYPE a 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 2 Horizontal Translations of Linear Functions MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the graph of the translated function and the graph of the parent function used in this example.
Questions for Mathematical Discourse AL Looking at only the equation, how do you know the type of transformation? The +5 is grouped with the x in the parentheses so this is a horizontal translation.
OL What are the coordinates when g(x) = 0? (-5, 0) BL Write the function that shows a horizontal translation of the parent function 3 units right. f(x - 3) 7 units left f(x + 7) Example 3 Multiple Translations of Linear Functions MP MP Teaching the Mathematical Practices 7 Use Structure Help students to use the structure of the transformed function to identify the translations in the function.
Questions for Mathematical Discourse AL Looking at only the equation, how many translations are there? 2 OL Looking at only the equation, how do you know that the horizontal translation is to the right? because the number being subtracted from x is positive 6 BL Write a function that represents a translation 6 units left and 3 units down. f(x) = (x + 6) - 3 Students move through the steps to graph a horizontal translation.
TAP Lesson 4-4 • Transformations of Linear Functions 241 F.IF.7a, F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Example 4 Translations of Linear Functions MP MP Teaching the Mathematical Practices 4 Apply Mathematics In this example, students apply what they have learned about translations of linear functions to solving a real-world problem.
Questions for Mathematical Discourse AL What does the 12 in the function represent? the cost per ticket What does the t in the function represent? the number of tickets OL What does the parent function represent in the context of the situation? cost of tickets without the online service fee BL What would the function be if, in addition to the service fee, there was also a $5 charge for tax? g(t) = 12t + 4 + 5 or g(t) = 12t + 9 Common Error Some students may try to work the 12 into the translation. Remind these students that translations occur when numbers are added or subtracted, not multiplied. Essential Question Follow-Up Students have observed how a function that models a real-world situation can be a transformation of another function.
Ask: Why is it important to understand how the structure of a function models a situation? Sample answer: The structure helps you understand how the different quantities in the situation affect the function.
Copyright © McGraw-Hill Education Go Online You can complete an Extra Example online. Example 4 Translations of Linear Functions TICKETS A Web site sells tickets to concerts and sporting events. The total price of the tickets to a certain game can be modeled by f(t) = 12t, where t represents the number of tickets purchased. The Web site then charges a standard service fee of $4 per order. The total price of an order can be modeled by g(t) = 12t + 4. Describe the translation of g(t) as it relates to f(t).
Complete the steps to describe the translation of g(t) as it relates to f(t). Because f(t) = 12t, g(t) = f(t) + k, where k = 4. g(t) = 12t + 4 → f(t) + The constant k is added to f(t) after the total price of the tickets has been evaluated and is greater than 0, so the graph will be shifted 4 units up. g(t) = 12t + 4 is the translation of the graph of f(t) units .
Graph the parent function and the translated function. y x O 1 2 1 −1 3 4 −1 −2 f(t) g(t) Check RETAIL Jerome is buying paint for a mural. The total cost of the paint can be modeled by the function f(p) = 6.99p. He has a coupon for $5.95 off his purchase at the art supply store, so the final cost of his purchase can be modeled by g(p) = 6.99p - 5.95. Describe the translation in g(p) as it relates to f(p).
242 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 4 The graph of g(p) = 6.99p - 5.95 is the translation of the graph of f (p) 5.95 units down.
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Example 4 Students complete the Check online to determine whether they are ready to move on.
CHECK 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students use a sketch to graph the translated function.
WEB SKETCHPAD Students complete the statements to identify the value for the translation.
TYPE a DIFFERENTIATE Enrichment Activity AL BL ELL IF students are having difficulty determining the direction of a translation, THEN have them create four examples of functions that represent each type of translation, and write each one on an index card. Have them sketch the transformation on a coordinate plane on the back of the card, and write the description. Then have them use the flash cards (in both directions) to practice what they have learned.
242 Module 4 • Linear and Nonlinear Functions F.IF.7a, F.BF.3 Interactive Presentation Learn Dilations of Linear Functions Objective Students identify the effects on the graphs of linear functions by replacing f(x) with af(x) and by replacing f(x) with f(ax).
MP MP Teaching the Mathematical Practices 7 Use Structure Help students to explore the structure of vertical and horizontal dilations in this Learn.
About the Key Concept When the function f(x) = x is multiplied by a number a, the graph of the function is dilated vertically. This is because multiplying the function by a number affects the y-value that is associated with each x-value. When |a| > 1, the graph is stretched vertically, making it steeper. When |a| < 1, the graph is compressed vertically, making it less steep.
Common Misconception Some students may think that when a is positive, the dilation stretches the graph, and when it is negative, the dilation compresses the graph. Use a table of values for several functions to show students the error in this reasoning. Sample functions: f(x) = x, g(x) = 2f(x), g(x) = −2f(x), g(x) = 0.5f(x), g(x) = −0.5f(x) Copyright © McGraw-Hill Education Go Online You can complete an Extra Example online.
Learn Dilations of Linear Functions A dilation stretches or compresses the graph of a function.
When a linear function f (x) is multiplied by a positive constant a, the result a∙f (x) is a vertical dilation. Key Concept • Vertical Dilations of Linear Functions The graph of g(x) = ax is the graph of f (x) = x stretched or compressed vertically.
If |a| > 1, the graph of f (x) is stretched vertically away from the x-axis.
y x O |a| > 1 a = 1 (x, y) (x, ay) The slope of the graph of a·f (x) is steeper than that of the graph of f (x).
If 0 < |a| < 1, the graph of f (x) is compressed vertically toward the x-axis.
y x O 0 < |a| < 1 a = 1 (x, y) (x, ay) The slope of the graph of a·f (x) is less steep than that of the graph of f (x).
When x is multiplied by a positive constant a before a linear function f (a ∙ x) is evaluated, the result is a horizontal dilation. Key Concept • Horizontal Dilations of Linear Functions The graph of g(x) = (a · x) is the graph of f(x) = x stretched or compressed horizontally.
If |a| > 1, the graph of f (x) is compressed horizontally toward the y-axis.
y x O |a| > 1 a = 1 (x, y) x, y 1 a ( ) The slope of the graph of f(a · x) is steeper than that of the graph of f(x).
If 0 < |a| < 1, the graph of f (x) is stretched horizontally away from the y-axis.
y x O 0 < |a| < 1 a = 1 (x, y) x, y 1 a ( ) The slope of the graph of f(a · x) is less steep than that of the graph of f(x).
Watch Out!
Dilations of f(x) = x When a dilation is the only transformation performed on the identity function, multiplying by a constant before or after evaluating the function has the same effect on the graph. However, when more than one type of transformation is applied, this will not be the case. Go Online You can watch a video to see how to describe dilations of functions.
Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 Lesson 4-4 • Transformations of Linear Functions 243 239_250_HSM_NA_S_A1M04_L04_662599.indd Page 243 14/01/19 5:02 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Learn Students drag the correct term to identify vertical and horizontal dilations.
DRAG & DROP Conceptual Understanding Fluency Application 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students explain the effect of dilations on the intercepts of the parent function.
TYPE a Students tap on each flash card to learn more about vertical and horizontal dilations.
TAP About the Key Concept When the argument of the function f(x) = x is multiplied by a number a, the graph of the function is dilated horizontally. This is because multiplying the argument by a number affects the x-value that is associated with each y-value. When |a| > 1, the graph is compressed horizontally, making it steeper. When |a| < 1, the graph is stretched horizontally, making it less steep.
Lesson 4-4 • Transformations of Linear Functions 243 F.IF.7a, F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Example 5 Vertical Dilations of Linear Functions MP MP Teaching the Mathematical Practices 7 Look for a Pattern Help students to see the pattern in calculating the coordinates for g(x) in this example.
Questions for Mathematical Discourse AL Will the placement of the 2 cause a change to the x-value or to the y-value of each ordered pair of the parent function? y-value; Sample answer: Because the 2 is not grouped with the x-variable, it will change the y-value.
OL Looking at only the equation, what kind of dilation is this? a vertical stretch by a factor of 2 BL How would the transformation be different if the function had been g(x) = 1 __ 2 f(x)? There would be a vertical compression instead of a vertical stretch.
Example 6 Horizontal Dilations of Linear Functions MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the graphs and equations of the functions in this example.
Questions for Mathematical Discourse AL What value is grouped with the x? 1 __ 4 Will this cause a change to the x-value or to the y-value of each ordered pair of the parent function? the x-value OL Looking at only the equation, what kind of dilation is this? a horizontal stretch by a factor of 4 BL How is a horizontal stretch by a factor of 4 related to a vertical compression by a factor of 4? They result in the same line.
Copyright © McGraw-Hill Education Think About It!
What do you notice about the slope of the horizontal dilation g(x) compared to the slope of f(x)?
How does this relate to the constant a in the horizontal dilation?
Example 5 Vertical Dilations of Linear Functions Describe the dilation in g(x) = 2(x) as it relates to the graph of the parent function.
Graph the parent graph for linear functions. Since f(x) = x, where . g(x) = 2(x) → The positive constant a is not grouped with x, and |a| is greater than 1, so the graph of f(x) = x is by a factor of a, or . g(x) = 2(x) is a vertical stretch of the graph of the parent function. The slope of the graph of g(x) is steeper than that of f(x). Check Describe the transformation in g(x) = 6(x) as it relates to the graph of the parent function.
The graph of g(x) = 6(x) is a of the graph of the parent function. The slope of the graph g(x) is than that of the parent function.
Example 6 Horizontal Dilations of Linear Functions Describe the dilation in g(x) = ( 1 __ 4 x) as it relates to the graph of the parent function.
Graph the parent graph for linear functions. Since f(x) = x, where . g(x) = ( 1 _ 4 x) → The positive constant a is grouped with x, and |a| is between 0 and 1, so the graph of f(x) = x is by a factor of 1 __ a , or 4. g(x) = ( 1 _ 4 x) is a horizontal stretch of the graph of the parent function. The slope of the graph of g(x) is less steep than that of f(x). Think About It!
What do you notice about the slope of the vertical dilation g(x) compared to the slope of f(x)?
x f(x) 2f(x) (x, g(x)) -2 -2 -4 (-2,-4) 0 0 0 (0, 0) 1 1 2 (1, 2) y x O f(x) g(x) Go Online You can complete an Extra Example online.
How does this relate to the constant a in the vertical dilation?
x 1 __ 4 x f ( 1 __ 4 x) (x, g(x)) -4 -1 -1 (-4, -1) 0 0 0 (0, 0) 4 1 1 (4, 1) y x O g(x) f(x) 244 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 Sample answer: The slope of g(x) is one fourth the slope of f(x).
Sample answer: a = 1 __ 4 , so, the slope of g(x) is the slope of f(x) multiplied by a.
Sample answer: The slope of g(x) is twice the slope of f (x).
Sample answer: a = 2, so the slope of g(x) is the slope of f (x) multiplied by a.
g(x) = a ∙ f(x) g(x) = 2f(x) stretched vertically a = 2 2 vertical stretch steeper g(x) = f(a ∙ x) a = 1 __ 4 g(x) = f ( 1 __ 4 x) stretched horizontally 239_250_HSM_NA_S_A1M04_L04_662599.indd Page 244 14/01/19 5:02 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 5 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students complete the Check online to determine whether they are ready to move on.
CHECK Students compare the slope and y-intercept of the vertical dilation to the parent graph.
TYPE a Students move through the steps to graph a vertical dilation.
TAP 244 Module 4 • Linear and Nonlinear Functions F.IF.7a, F.BF.3 Interactive Presentation Copyright © McGraw-Hill Education Learn Reflections of Linear Functions A reflection is a transformation in which a figure, line, or curve, is flipped across a line. When a linear function f(x) is multiplied by -1 before or after the function has been evaluated, the result is a reflection across the x- or y-axis. Every x- or y-coordinate of f(x) is multiplied by −1. Key Concept • Reflections of Linear Functions The graph of −f (x) is the reflection of the graph of f (x) = x across the x-axis.
y x O (x, y) (x, -y) -f(x) f(x) Every y-coordinate of −f (x) is the corresponding y-coordinate of f (x) multiplied by −1.
The graph of f(−x) is the reflection of the graph of f (x) = x across the y-axis.
y x O (x, y) (-x, y) f(-x) f(x) Every x-coordinate of f (−x) is the corresponding x-coordinate of f (x) multiplied by −1.
Example 7 Reflections of Linear Functions Across the x-Axis Describe how the graph of g(x) = - 1 __ 2 (x) is related to the graph of the parent function.
Graph the parent graph for linear functions.
Since f(x) = x, where . g(x) = - 1 __ 2 (x) → The constant a is not grouped with x, and |a| is less than 1, so the graph of f(x) = x is . The negative is not grouped with x, so the graph is also reflected across the . The graph of g(x) = - 1 __ 2 (x) is the graph of the parent function vertically compressed and reflected across the x-axis. x f(x) - 1 __ 2 f(x) (x, g(x)) -2 -2 1 (-2, 1) 0 0 0 (0, 0) 4 4 -2 (4, -2) y x O g(x) f(x) Go Online You can complete an Extra Example online. Go Online You can watch a video to see how to describe reflections of functions. Talk About It!
In the example, the slope of g(x) is negative. Will this always be the case when multiplying a linear function by -1? Justify your argument. Watch Out!
Reflections of f(x) = x When a reflection is the only transformation performed on the identity function, multiplying by -1 before or after evaluating the function appears to have the same effect on the graph. However, when more than one type of transformation is applied, this will not be the case. Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 Lesson 4-4 • Transformations of Linear Functions 245 g(x) = -1 ∙ a ∙ f(x) g(x) = - 1 __ 2 f(x) a = 1 __ 2 x-axis vertically compressed No; sample answer: If you multiply a linear function that is already negative by -1, the resulting function will have a positive slope.
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Learn 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Learn Reflections of Linear Functions Objective Students identify the effects on the graphs of linear functions by replacing f(x) with -af(x) and f(-ax).
MP MP Teaching the Mathematical Practices 1 Explain Correspondence Encourage students to explain the relationships between the coordinates, equations, and graphs of reflected functions and the parent function.
Students drag the correct term to identify if a reflection is over the x- or y-axis.
DRAG & DROP Students write a function that is a dilation and reflection of a parent function.
TYPE a Students tap on each flash card to learn more about reflections of functions.
TAP Example 7 Reflections of Linear Functions Across the x-Axis MP MP Teaching the Mathematical Practices 3 Construct Arguments In this example, students will use stated assumptions, definitions, and previously established results to construct an argument in the Talk About It! feature.
Questions for Mathematical Discourse AL Looking at only the equation, is this function a reflection? yes What other type of transformation is it? a dilation OL How do you know - 1 __ 2 is not grouped with x? Sample answer: It is not inside the parentheses with x.
BL The point (2, 2) lies on the graph of the parent function. To what point does this correspond on the graph of g(x)? (2, -1) Lesson 4-4 • Transformations of Linear Functions 245 F.IF.7a, F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Example 8 Reflections of Linear Functions Across the y-Axis MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the graph of the reflected function and the graph of the parent function used in this example.
Questions for Mathematical Discourse AL How do you know the negative sign is grouped with the x? Sample answer: Because the negative sign is inside the parentheses with x.
OL How do you know if the parent function will be reflected over the y-axis? Sample answer: If the negative is inside the parentheses with x, the reflection will be over the y-axis.
BL How would the function have been written if the reflection was across the x-axis? g(x) = -3f(x) Common Error Students may have difficulty seeing how the graph of g(x) is related to the graph of f(x). For these students, you may want to show the transformation in two different steps, first dilating the graph by a factor of 3, and then reflecting the resulting graph across the y-axis.
Example 8 Students complete the Check online to determine whether they are ready to move on.
CHECK 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students move through the steps to graph a reflection of a linear function across the y-axis.
TAP Copyright © McGraw-Hill Education Go Online You can complete an Extra Example online.
Example 8 Reflections of Linear Functions Across the y-Axis Describe how the graph of g(x) = (-3x) is related to the graph of the parent function.
Graph the parent graph for linear functions. Since f (x) = x, where . g(x) = -3x → The constant a is grouped with x, and |a| is greater than 1, so the graph of f(x) = x is . The negative is grouped with x, so the graph is also reflected across the . The graph of g(x) = (-3x) is the graph of the parent function horizontally compressed and reflected across the y-axis.
Check Describe how the graph of g(x) = (-10x) is related to the graph of the parent function.
The graph of g(x) = (-10x) is the graph of the parent function compressed horizontally and reflected across the .
x -3x f (-3x) (x, g(x)) -1 3 3 (-1, 3) 0 0 0 (0, 0) 1 -3 -3 (1, -3) y x O f(x) g(x) Go Online You can watch a video to see how to graph transformations of a linear function using a graphing calculator. Check How can you tell whether multiplying −1 by the parent function will result in a reflection across the x-axis? A. If the constant is not grouped with x, the result will be a reflection across the x-axis. B. If the constant is grouped with x, the result will be a reflection across the x-axis.
C. If the constant is greater than 0, the result will be a reflection across the x-axis.
D. If the constant is less than 0, the result will be a reflection across the x-axis.
246 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 a = 3 y-axis y-axis g(x) = f (-3x) horizontally compressed g(x) = f (-1 ∙ a ∙ x) A 239_250_HSM_NA_S_A1M04_L04_662599.indd Page 246 11/10/19 2:49 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
246 Module 4 • Linear and Nonlinear Functions F.IF.7a, F.BF.3 Exit Ticket Recommended Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond using a separate piece of paper. Have students hand you their responses as they leave the room.
Alternate Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond verbally or by using a mini-whiteboard. Have students hold up their whiteboards so that you can see all student responses. Tap to reveal the answer when most or all students have completed the Exit Ticket.
Interactive Presentation Practice and Homework Suggested Assignments Use the table below to select appropriate exercises.
DOK Topic Exercises 1, 2 exercises that mirror the examples 1–21 2 exercises that use a variety of skills from this lesson 22–29 2 exercises that extend concepts learned in this lesson to new contexts 30–33 3 exercises that emphasize higher-order and critical thinking skills 34–36 Copyright © McGraw-Hill Education Name Period Date Practice Go Online You can complete your homework online.
Examples 1 through 3 Describe the translation in each function as it relates to the graph of the parent function. 1. g(x) = x + 11 2. g(x) = x - 8 3. g(x) = (x - 7) y x O 12 6 9 3 −12 −6 −9 −3 −12 −6 −9 6 9 3 12 y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 4. g(x) = (x + 12) 5. g(x) = (x + 10) - 1 6. g(x) = (x - 9) + 5 y x O 12 6 9 3 −12 −6 −9 −3 −12 −6 −9 6 9 3 12 y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 Example 4 7. BOWLING The cost for Nobu to go bowling is $4 per game plus an additional flat fee of $3.50 for the rental of bowling shoes. The cost can be modeled by the function f(x) = 4x + 3.5, where x represents the number of games bowled. Describe the graph of g(x) as it relates to f(x) if Nobu does not rent bowling shoes. 8. SAVINGS Natalie has $250 in her savings account, into which she deposits $10 of her allowance each week. The balance of her savings account can be modeled by the function f(w) = 250 + 10w, where w represents the number of weeks. Write a function g(w) to represent the balance of Natalie’s savings account if she withdraws $40 to purchase a new pair of shoes. Describe the translation of f(w) that results in g(w). 9. BOAT RENTAL The cost to rent a paddle boat at the county park is $8 per hour plus a nonrefundable deposit of $10. The cost can be modeled by the function f(h) = 8h + 10, where h represents the number of hours the boat is rented. Describe the graph of g(h) as it relates to f(h) if the nonrefundable deposit increases to $15. Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 Lesson 4-4 • Transformations of Linear Functions 247 g(x) is a translation of the parent function 11 units up g(x) is a translation of the parent function 12 units left g(x) = 4x; g(x) is the translation of f(x) 3.5 units down.
g(w) = 210 + 10w ; g(w) is the translation of f(w) 40 units down.
g(h) = 8h + 15; g(h) is the translation f(h) of 5 units up.
g(x) is a translation of the parent function 8 units down g(x) is a translation of the parent function 10 units left and 1 unit down g(x) is a translation of the parent function 7 units right g(x) is a translation of the parent function 9 units right and 5 units up 239_250_HSM_NA_S_A1M04_L04_662599.indd Page 247 5/10/19 3:13 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education Examples 5 and 6 Describe the dilation in each function as it relates to the graph of the parent function. 10. g(x) = 5(x) 11. g(x) = 1 _ 3 (x) 12. g(x) = 1.5(x) y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 13. g(x) = (3x) 14. g(x) = ( 3 _ 4 x) 15. g(x) = (0.4x) y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 Example 7 Describe how the graph of each function is related to the graph of the parent function. 16. g(x) = -4(x) 17. g(x) = -8(x) 18. g(x) = - 2 __ 3 (x) y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 248 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 g(x) is a vertical stretch of the parent function by a factor of 5 g(x) is a horizontal compression of the parent function by a factor of 1 __ 3 g(x) is a vertical stretch of the parent function by a factor of 4 and a reflection across the x-axis g(x) is a vertical compression of the parent function by a factor of 1 __ 3 g(x) is a horizontal stretch of the parent function by a factor of 4 __ 3 g(x) is a vertical stretch of the parent function by a factor of 8 and a reflection across the x-axis g(x) is a vertical stretch of the parent function by a factor of 1.5 g(x) is a horizontal stretch of the parent function by a factor of 2.5 g(x) is a vertical compression of the parent function by a factor of 2 __ 3 and a reflection across the x-axis 239_250_HSM_NA_S_A1M04_L04_662599.indd Page 248 11/10/19 2:49 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
3 REFLECT AND PRACTICE ASSESS AND DIFFERENTIATE Use the data from the Checks to determine whether to provide resources for extension, remediation, or intervention.
IF students score 90% or more on the Checks, THEN assign: BL • Practice, Exercises 1–33 odd, 34–36 • Extension: Transformations of Other Families of Functions • Equations of Lines IF students score 66%–89% on the Checks, THEN assign: OL • Practice, Exercises 1–35 odd • Remediation, Review Resources: Reflections • Personal Tutors • Extra Examples 1–8 • Reflections IF students score 65% or less on the Checks, THEN assign: AL • Practice, Exercises 1–21 odd • Remediation, Review Resources: Reflections • Quick Review Math Handbook: Transformations of Linear Functions • ArriveMATH Take Another Look • Reflections 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Lesson 4-4 • Transformations of Linear Functions 247-248 F.IF.7a, F.BF.3 Interactive Presentation Answer 34. No; This is true only if the slope is 1. Consider any constant function. Shifting the graph of a constant function to the right or left doesn’t result in any vertical shift of the same graph. If the slope m of the line described by f(x) is something other than -1, 0, or 1, then a horizontal shift of k units is the same as a vertical shift of -mk units. For example, if f(x) = 3x, then f(x + 5) = 3(x + 5) or 3x + 15. f(x + 5) is shifted 5 units left of f(x) or 15 units up from f(x). Sample graphs shown.
3 REFLECT AND PRACTICE Copyright © McGraw-Hill Education Name Period Date Example 8 Describe how the graph of each function is related to the graph of the parent function. 19. g(x) = (- 4 __ 5 x) 20. g(x) = (-6x) 21. g(x) = (-1.5x) y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 Mixed Exercises Describe the transformation in each function as it relates to the graph of the parent function. 22. g(x) = x + 4 23. g(x) = (x - 2) - 8 24. g(x) = (- 5 __ 8 x) y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 25. g(x) = 1 __ 5 (x) 26. g(x) = -3(x) 27. g(x) = (2.5x) y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 y x O 8 4 −8 −4 −8 −4 4 8 Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 Lesson 4-4 • Transformations of Linear Functions 249 g(x) is a horizontal stretch of the parent function by a factor of 5 __ 4 and a reflection across the y-axis g(x) is a translation of the parent function 4 units up g(x) is a vertical compression of the parent function by a factor of 1 __ 5 g(x) is a horizontal compression of the parent function by a factor of 1 __ 6 and a reflection across the y-axis g(x) is a translation of the parent function 2 units right and 8 units down g(x) is a vertical stretch of the parent function by a factor of 3 and a reflection across the x-axis g(x) is a horizontal compression of the parent function by a factor of 2 __ 3 and a reflection across the y-axis g(x) is a horizontal stretch of the parent function by a factor of 8 __ 5 and a reflection across the y-axis g(x) is a horizontal compression of the parent function by a factor of 0.4 239_250_HSM_NA_S_A1M04_L04_662599.indd Page 249 11/10/19 2:49 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education REASONING Write a function g(x) to represent the translated graph. 28. f(x) = 3x + 7 translated 4 units up. 29. f(x) = x - 5 translated 2 units down. 30. PERIMETER The function f(s) = 4s represents the perimeter of a square with side length s. Write a function g(s) to represent the perimeter of a square with side lengths that are twice as great. Describe the graph of g(s) compared to f(s). 31. GAMES The function f(x) = 0.50x gives the average cost in dollars for x cell phone game downloads that cost an average of $0.50 each. Write a function g(x) to represent the cost in dollars for x cell phone game downloads that cost $1.50 each. Describe the graph of g(x) compared to f(x). 32. TRAINER The function f(x) = 90x gives the cost of working out with a personal trainer, where $90 is the trainer’s hourly rate, and x represents the number of hours spent working out with the trainer. Describe the dilation, g(x) of the function f(x), if the trainer increases her hourly rate to $100. 33. DOWNLOADS Hannah wants to download songs. She researches the price to download songs from Site F. Hannah wrote the function f(x) = x, which represents the cost in dollars for x songs downloaded that cost $1.00 each. a. Hannah researches the price to download songs from Site G. Write a function g(x) to represent the cost in dollars for x songs downloaded that cost $1.29 each. b. Describe the graph of g(x) compared to the graph of f(x). 34. PERSEVERE For any linear function, replacing f(x) with f(x + k) results in the graph of f(x) being shifted k units to the right for k < 0 and shifted k units to the left for k > 0. Does shifting the graph horizontally k units have the same effect as shifting the graph vertically −k units? Justify your answer. Include graphs in your response. 35. CREATE Write an equation that is a vertical compression by a factor of a of the parent function y = x. What can you say about the horizontal dilation of the function? 36. WHICH ONE DOESN’T BELONG Consider the four functions. Which one does not belong in this group? Justify your conclusion. f(x) = 2(x + 1) - 3 f(x) = 1 _ 2 x - 4 f(x) = -3x + 10 f(x) = 5(x - 7) + 3 250 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson_4 PDF Pass Vendor: Aptara Grade: A1-M04 g(x) = 1.29x g(s) = 8s; The graph of g(s) = 8s is the graph of f(s) = 4s stretched vertically by a factor of 2. g(x) = 1.5x; The graph of g(x) = 1.5x is the graph of f(x) = 0.50x stretched vertically by a factor of 3. g(x) = 100x; The graph of g(x) = 100x is the graph of f(x) = 90x stretched vertically by a factor of 10 __ 9 . The graph of g(x) = 1.29x is the graph of f(x) = x stretched vertically by a factor of 1.29.
y = 1 __ a x; The function is horizontally stretched by a factor of a.
See margin. f(x) = -3x + 10; This function is a reflection and translation of the parent graph. The other three functions are dilations and translations of the parent graph. Higher-Order Thinking Skills g(x) = 3x + 11 g(x) = x - 7 239_250_HSM_NA_S_A1M04_L04_662599.indd Page 250 5/9/19 3:27 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION y O x -4 4 8 -8 4 8 −4 −8 y O x -8 8 16 -16 8 16 −8 −16 249-250 Module 4 • Linear and Nonlinear Functions F.IF.7a, F.BF.3 LESSON GOAL Students write and graph equations of arithmetic sequences.
1 LAUNCH Launch the lesson with a Warm Up and an introduction.
2 EXPLORE AND DEVELOP Develop: Arithmetic Sequences • Identify Arithmetic Sequences • Find the Next Term Explore: Common Differences Develop: Arithmetic Sequences as Linear Functions • Find the nth Term • Apply Arithmetic Sequences as Linear Functions You may want your students to complete the Checks online.
3 REFLECT AND PRACTICE Exit Ticket Practice DIFFERENTIATE View reports of student progress on the Checks after each example.
Resources AL OL BL ELL Remediation: Add Integers ● ● ● Extension: Arithmetic Series ● ● ● Language Development Handbook Assign page 24 of the Language Development Handbook to help your students build mathematical language related to arithmetic sequences.
ELL You can use the tips and suggestions on page T24 of the handbook to support students who are building English proficiency.
Lesson 4-5 Arithmetic Sequences Lesson 4-5 • Arithmetic Sequences 251a F.BF.1a, F.BF.2, F.LE.2 Suggested Pacing 90 min 0.5 day 45 min 1 day Focus Domain: Functions Standards for Mathematical Content: F.BF.1a Determine an explicit expression, a recursive process, or steps for calculation from a context.
F.BF.2 Write arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms.
F.LE.2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a description of a relationship, or two input-output pairs (include reading these from a table).
Standards for Mathematical Practice: 3 Construct viable arguments and critique the reasoning of others.
8 Look for and express regularity in repeated reasoning.
Coherence Vertical Alignment Rigor The Three Pillars of Rigor 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Conceptual Bridge In this lesson, students expand on their understanding of and build fluency with sequences (first studied in Grade 4) by writing formulas for arithmetic sequences and relating them to linear functions. They apply their understanding by solving real-world problems related to arithmetic sequences.
Now Students will graph piecewise-defined and step functions. F.IF.4, F.IF.7b Now Students write and graph equations of arithmetic sequences. F.BF.1a, F.BF.2, F.LE.2 Previous Students identified the effects of transformations of the graphs of linear functions.
F.IF.7a, F.BF.3 Interactive Presentation 1 LAUNCH Warm Up Warm Up Prerequisite Skills The Warm Up exercises address the following prerequisite skill for this lesson: • finding the next terms in patterns Launch the Lesson Launch the Lesson MP MP Teaching the Mathematical Practices 8 Look for a Pattern Help students to see the pattern in the triangle structures that compose The Nima Sand Museum and in the Pyramid of Oranges. Go Online to find additional teaching notes and questions to promote classroom discourse.
Today’s Standards Tell students that they will address these content and practice standards in this lesson. You may wish to have a student volunteer read aloud How can I meet these standards? and How can I use these practices?, and connect these to the standards.
See the Interactive Presentation for I Can statements that align with the standards covered in this lesson.
Today’s Vocabulary Today’s Vocabulary Tell students that they will use these vocabulary terms in this lesson. You can expand each row if you wish to share the definitions. Then discuss the questions below with the class.
Mathematical Background A sequence is a set of numbers in a specific order. The numbers in a sequence are called terms. If the terms of a sequence increase or decrease at a constant rate, the sequence is called an arithmetic sequence. The difference between successive terms of an arithmetic sequence is called the common difference. Any term of an arithmetic sequence can be found by adding the common difference to the preceding term. The formula for finding a specific term in an arithmetic sequence is an = a1 + (n - 1)d, where an is the nth term, a1 is the first term, and d is the common difference.
Answers: 1. 3, 6, 5 2. 21, 28, 36 3. a + 16, a + 25, a + 36 4. 6d - 4, 7d - 5, 8d - 6 5. 12, 18, 24; 9 wk 251b Module 4 • Linear and Nonlinear Functions F.BF.1a, F.BF.2, F.LE.2 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Common Differences Objective Students use a sketch to explore the relationship between arithmetic sequences and linear functions.
MP MP Teaching the Mathematical Practices 4 Use Tools Point out that to solve the problem in this Explore, students will need to use the table and sketch.
Ideas for Use Recommended Use Present the Inquiry Question, or have a student volunteer read it aloud. Have students work in pairs to complete the Explore activity on their devices. Pairs should discuss each of the questions. Monitor student progress during the activity. Upon completion of the Explore activity, have student volunteers share their responses to the Inquiry Question. What if my students don’t have devices? You may choose to project the activity on a whiteboard. A printable worksheet for each Explore is available online. You may choose to print the worksheet so that individuals or pairs of students can use it to record their observations.
Summary of the Activity Students will complete guiding exercises throughout the Explore activity. Students will use the sketch to graph a linear function to solve a real-world problem. They will observe as the data points are plotted, and then answer questions related to the resulting graph. Then, students will answer the Inquiry Question. (continued on the next page) Explore Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students answer questions about the pattern in the data.
Students use a sketch to explore the amount of food remaining as cats are fed.
WEB SKETCHPAD TYPE a Lesson 4-5 • Arithmetic Sequences 251c Interactive Presentation 2 EXPLORE AND DEVELOP Explore Common Differences (continued) Questions Have students complete the Explore activity.
Ask: • Why is the amount of food decreasing with each cat? Sample answer: Each cat is being fed a certain amount of food, so there will be less after each cat is fed. • How does the amount of food each cat is fed relate to the slope of the linear function that models the situation? Sample answer: The amount of food each cat is fed represents the change in the amount of food, which is the slope of the function. As long as there is a constant change, or constant slope, then you have a linear function. Inquiry How can you tell if a set of numbers models a linear function? Sample answer: The points are on the same line and have a constant slope. Go Online to find additional teaching notes and sample answers for the guiding exercises.
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Explore Students respond to the Inquiry Question and can view a sample answer.
TYPE a 251d Module 4 • Linear and Nonlinear Functions Interactive Presentation Learn Arithmetic Sequences Objective Students construct arithmetic sequences by using the common difference.
MP MP Teaching the Mathematical Practices 8 Look for a Pattern Help students see the pattern in this Learn.
Common Misconception Students may think that the terms of all arithmetic sequences must increase. They may believe this because the definition of an arithmetic sequence refers to the use of addition to find successive terms. Point out that when the number being added is negative, the terms will decrease.
Example 1 Identify Arithmetic Sequences MP MP Teaching the Mathematical Practices 3 Reason Inductively In this example, students will use inductive reasoning to make plausible arguments.
Questions for Mathematical Discourse AL How are the terms of an arithmetic sequence found? The same number is added to each term to find the next term.
OL What requirement must be met for the sequence to represent an arithmetic sequence? The difference between the terms must be constant. BL Does the sequence follow a pattern? Explain. Yes; sample answer: The difference in the numbers repeats itself, so the next difference would be -3. Go Online • Find additional teaching notes.
• View performance reports of the Checks.
• Assign or present an Extra Example.
Copyright © McGraw-Hill Education Arithmetic Sequences Lesson 4-5 Learn Arithmetic Sequences A sequence is a set of numbers that are ordered in a specific way. Each number within a sequence is called a term of a sequence. In an arithmetic sequence, each term after the first is found by adding a constant, the common difference d, to the previous term.
Words An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference.
Examples 22 The common diference is -5.
The common diference is 8.
17 12 7 2, ...
-5 -5 -5 -5 -13 -5 3 11 19, ...
+8 +8 +8 +8 Example 1 Identify Arithmetic Sequences Determine whether the sequence is an arithmetic sequence. Justify your reasoning.
17, 14, 10, 7, 3 17 14 10 7 3 -3 -4 -3 -4 This sequence a common difference between its terms. This an arithmetic sequence.
Check Determine whether the sequence is an arithmetic sequence. Justify your reasoning.
82, 73, 64, 55, . . .
Check the difference between terms.
Today’s Vocabulary sequence term of sequence arithmetic sequence common difference nth term of an arithmetic function Think About It!
How are arithmetic sequences and number patterns alike and different? Go Online You can complete an Extra Example online.
Today’s Goals ● Construct arithmetic sequences.
● Apply the arithmetic sequence formula.
Lesson 4-5 • Arithmetic Sequences 251 Program: Reveal Math Component: Lesson 4-5_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE does not have yes, there is a common difference of −9 is not Sample answer: An arithmetic sequence is a series of numbers that has a common difference between two consecutive terms, which is used to find other terms in the sequence. A number pattern is a series of numbers that follows a rule but might not have a common difference between terms.
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Students enter the number of chorus members in the 6th row.
Learn TYPE a 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Lesson 4-5 • Arithmetic Sequences 251 F.BF.1a, F.BF.2, F.LE.2 Interactive Presentation 2 EXPLORE AND DEVELOP Example 2 Find the Next Term MP MP Teaching the Mathematical Practices 1 Monitor and Evaluate Point out that in this example, students must stop and evaluate their progress when determining the next terms in the sequence.
Questions for Mathematical Discourse AL What is the relationship between the terms? Each term is 4 less than the previous term.
OL If this sequence continues on, will all of the subsequent terms be negative, or will they go back to being positive? Explain. Sample answer: They will stay negative because each term is less than the one before.
BL What is the tenth term in the arithmetic sequence? -25 Learn Arithmetic Sequences as Linear Functions Objective Students apply the arithmetic sequence formula by examining the common differences in arithmetic sequences.
MP MP Teaching the Mathematical Practices 3 Construct Arguments In this Learn, students will use stated assumptions, definitions, and previously established results to construct an argument.
8 Look for a Pattern Help students to see the pattern in the formula for the nth term of an arithmetic sequence.
Important to Know In the context of a function, the term numbers represent the input values, and the terms of the sequence represent the output values. The common difference is a constant that represents the slope. The function rule is linear, defining how each term is determined by its term number, n.
Common Misconception Some students may think that the function rule contains more than one variable. Use several examples to show students that for any particular sequence, a1 and d are known constants, and only n is variable.
Copyright © McGraw-Hill Education Example 2 Find the Next Term Determine the next three terms in the sequence.
11, 7, 3, –1 Find the common difference between terms. Add the common difference to the last term of the sequence to find the next terms.
-1 + ( ) = + ( ) = + ( ) = Check Determine the next three terms in the sequence.
31, 18, 5, , , Talk About It!
Why would it be useful to develop a rule to find terms of a sequence? Explain. Go Online You can complete an Extra Example online. Go Online You can complete an Extra Example online.
Explore Common Differences Online Activity Use a real-world situation to complete the Explore.
Learn Arithmetic Sequences as Linear Functions Each term of an arithmetic sequence can be expressed in terms of the first term a1 and the common difference d. Key Concept • nth Term of an Arithmetic Sequence The nth term of an arithmetic sequence with the first term a1 and common difference d is given by an = a1 + (n - 1)d, where n is a positive integer.
The graph of an arithmetic sequence includes points that lie along a line. Because there is a constant difference between each pair of points, the function is linear. For the equation of an arithmetic sequence, an = a1 + (n - 1)d • n is the independent variable, • an is the dependent variable, and • d is the slope.
The function of an arithmetic sequence is written as f (n) = a1 + (n - 1)d, where n is a counting number.
Watch Out!
Subscripts Subscripts are used to indicate a specific term. For example, a8 means the 8th term of the sequence. It does not mean a × 8. Think About It!
Why is the domain of a sequence counting numbers instead of all real numbers?
Your Notes INQUIRY How can you tell if a set of numbers models a linear function?
252 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-5_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: By developing a rule, you could find large terms, such as the 100th term, more easily than finding each of the sequence’s previous terms.
Sample answer: The independent variable for an arithmetic sequence is the term number. Because there cannot be negative or fractional term numbers, the domain has to be only counting numbers.
-4 -4 -8 -21 -34 -5 -5 -4 -9 -9 -4 -13 251_258_HSM_NA_S_A1M04_L05_662599.indd Page 252 04/10/19 12:25 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 2 Students complete the Check online to determine whether they are ready to move on.
CHECK 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students enter the consecutive members of the arithmetic sequences.
Students explain how developing a rule or equation would be useful in finding terms of a sequence.
TYPE TYPE a a 252 Module 4 • Linear and Nonlinear Functions F.BF.1a, F.BF.2, F.LE.2 Interactive Presentation Copyright © McGraw-Hill Education Example 3 Find the nth Term Use the arithmetic sequence -4, -1, 2, 5, . . . to complete the following.
Part A Write an equation.
Part B Find the 16th term of the sequence.
Use the equation from Part A to find the 16th term in the arithmetic sequence. Equation from Part A Substitute 16 for n. Multiply. Simplify.
Check RUNNING Randi has been training for a marathon, and it is important for her to keep a constant pace. She recorded her time each mile for the first several miles that she ran. • At 1 mile, her time was 10 minutes and 30 seconds. • At 2 miles, her time was 21 minutes. • At 3 miles, her time was 31 minutes and 30 seconds. • At 4 miles, her time was 42 minutes.
Part A Write a function to represent her sequence of data. Use n as the variable.
Part B How long will it take her to run a whole marathon? Round your answer to the nearest thousandth if necessary. (Hint: a marathon is 26.2 miles.) Example 4 Apply Arithmetic Sequences as Linear Functions MONEY Laniqua opened a savings account to save for a trip to Spain. With the cost of plane tickets, food, hotel, and other expenses, she needs to save $1600. She opened the account with $525. Every month, she adds the same amount to her account using the money she earns at her after school job. From her bank statement, Laniqua can write a function that represents the balance of her savings account. Go Online You can complete an Extra Example online.
Lesson 4-5 • Arithmetic Sequences 253 Program: Reveal Math Component: Lesson 4-5_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE an = 3n – 7 an = 3n - 7 a16 = 3(16) - 7 a16 = 48 - 7 a16 = 41 f(n) = 10.5n 4.585 hours (continued on the next page) 251_258_HSM_NA_S_A1M04_L05_662599.indd Page 253 5/9/19 3:28 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 3 Students can tap to see the steps to writing and using an equation for arithmetic sequences.
EXPAND 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 3 Find the nth Term MP MP Teaching the Mathematical Practices 3 Reason Inductively In this example, students will use inductive reasoning to make plausible arguments.
Questions for Mathematical Discourse AL What values do you need to know in order to write the equation? You need to know the first term, a1, and the difference, d.
OL How are the values substituted to find the equation? Sample answer: -4 + 3(n - 1) = -4 + 3n - 3 = 3n - 7 BL Will n always be a positive number? Explain. Yes; sample answer: Since n refers to the number of the term, like the 1st term or the 15th term, it will always be a positive whole number.
DIFFERENTIATE Reteaching Activity AL ELL IF students have difficulty following the progression of steps that lead to the building of the equation, THEN have them cycle through the steps again, using a simpler sequence, such as 1, 4, 7, 10, … DIFFERENTIATE Enrichment Activity BL Have students work with a partner. Tell them that you know of an arithmetic sequence in which the 4th term is 27 and the 8th term is 59. Ask them to find the first term and the common difference. Have pairs share how they solved the problem, and describe how they checked that their solution is correct. a1 = 3, d = 8 Lesson 4-5 • Arithmetic Sequences 253 F.BF.1a, F.BF.2, F.LE.2 Interactive Presentation 2 EXPLORE AND DEVELOP Example 4 Apply Arithmetic Sequences as Linear Functions MP MP Teaching the Mathematical Practices 5 Use a Source Guide students to find external information to answer the questions posed in the Use a Source feature.
Questions for Mathematical Discourse AL Why does the list of balances represent an arithmetic sequence? because there is a common difference between the balances OL What does the common difference mean in the context of the problem? Laniqua is saving $55 each month, so her account is increasing by $55 per month.
BL Is the function discrete or continuous? Explain. Discrete; sample answer: The domain is the counting numbers, so the graph would consist of points, not a line.
DIFFERENTIATE Enrichment Activity BL Arithmetic sequences can be programmed into graphing calculators with results displayed in lists. Have advanced learners locate a set of directions for programming a sequence and develop a lesson for their classmates on analyzing sequences using the calculator. Exit Ticket Recommended Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond using a separate piece of paper. Have students hand you their responses as they leave the room.
Alternate Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond verbally or by using a mini-whiteboard. Have students hold up their whiteboards so that you can see all student responses. Tap to reveal the answer when most or all students have completed the Exit Ticket.
Copyright © McGraw-Hill Education Laniqua Jones Current Balance as of 03/01/2019 Balance as of 02/01/2019 Balance as of 01/01/2019 Starting Balance as of 12/01/2018 $ 690 $ 635 $ 580 $ 525 Part A Create a function to represent the sequence.
First, find the .
525 580 635 690 +55 +55 +55 The common difference is .
The balance after 1 month is $580, so let a1 = 580. Notice that the starting balance is $525. You can think of this starting point as a0 = 580. Formula for the nth term a1 = 580 and d = 55 Simplify. Part B Graph the function and determine its domain.
n f(n) 0 1 2 3 4 5 6 The domain is the number of months since Laniqua opened her savings account. The domain is {0, 1, 2, 3, 4, 5, …}. Go Online You can complete an Extra Example online.
Use a Source Find the cost of a flight from the airport closest to you to Madrid, the capital of Spain. How many months would Laniqua need to save to afford the ticket?
Study Tip Graphing You might not need to create a table of the sequence first. However, it might serve as a reminder that an arithmetic sequence is a series of points, not a line.
2 3 1 4 5 n 100 0 200 300 400 500 600 700 800 f(n) 254 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-5_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: The cost to fly from the closest airport to Madrid is $1416. To afford the ticket, Laniqua would have to save for 17 months.
common difference 55 855 635 580 525 690 745 800 f(n) = a1 + (n - 1)d = 580 + (n - 1)(55) = 580 + 55n - 55 = 55n + 525 251_258_HSM_NA_S_A1M04_L05_662599.indd Page 254 05/10/19 7:57 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 4 Students complete the Check online to determine whether they are ready to move on.
CHECK 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students can watch an animation that describes the balance of an account over time.
WATCH Students research costs of flights to Spain and determine the number of months they would need to save.
TYPE a 254 Module 4 • Linear and Nonlinear Functions F.BF.1a, F.BF.2, F.LE.2 Interactive Presentation Practice and Homework Suggested Assignments Use the table below to select appropriate exercises. DOK Topic Exercises 1, 2 exercises that mirror the examples 1–26 2 exercises that use a variety of skills from this lesson 27–34 2 exercises that extend concepts learned in this lesson to new contexts 35–37 3 exercises that emphasize higher-order and critical thinking skills 38–46 Copyright © McGraw-Hill Education Name Period Date Practice Go Online You can complete your homework online.
Example 1 ARGUMENTS Determine whether each sequence is an arithmetic sequence. Justify your reasoning. 1. -3, 1, 5, 9, … 2. 1 __ 2 , 3 _ 4 , 5 __ 8, 7 __ 16 , … 3. -10, -7, -4, 1, … 4. -12.3, -9.7, -7.1, -4.5, … 5. 4, 7, 9, 12, ... 6. 15, 13, 11, 9, ... 7. 7, 10, 13, 16, ... 8. -6, -5, -3, -1, ...
Example 2 Find the common difference of each arithmetic sequence. Then find the next three terms. 9. 0.02, 1.08, 2.14, 3.2, ... 10. 6, 12, 18, 24, ... 11. 21, 19, 17, 15, ... 12. - 1 __ 2 , 0, 1 __ 2 , 1, ... 13. 2 1 _ 3 , 2 2 __ 3 , 3, 3 1 _ 3 , ... 14. 7 __ 12 , 1 1 _ 3 , 2 1 __ 12 , 2 5 __ 6 , ... 15. 3, 7, 11, 15, ... 16. 22, 19.5, 17, 14.5, ... 17. -13, -11, -9, -7, ... 18. -2, -5, -8, -11, ...
Example 3 Use the given arithmetic sequence to write an equation and then find the 7th term of the sequence. 19. –3, –8, –13, –18, … 20. –2, 3, 8, 13, … 21. –11, –15, –19, –23, … 22. -0.75, –0.5, –0.25, 0, … Lesson 4-5 • Arithmetic Sequences 255 Program: Reveal Math Component: Lesson 4-5_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE This sequence has a common difference of 4 between its terms. This is an arithmetic sequence.
1.06; 4.26, 5.32, 6.38 –2; 13, 11, 9 1 __ 3 ; 3 2 __ 3 , 4, 4 1 __ 3 4; 19, 23, 27 2; –5, –3, –1 This sequence does not have a common difference between its terms. This is not an arithmetic sequence.
This sequence does not have a common difference between its terms. This is not an arithmetic sequence.
This sequence has a common difference of 3 between its terms. This is an arithmetic sequence.
This sequence does not have a common difference between its terms. This is not an arithmetic sequence.
6; 30, 36, 42 1 __ 2 ; 1 1 __ 2 , 2, 2 1 __ 2 3 __ 4 ; 3 7 __ 12 , 4 1 __ 3 , 5 1 __ 12 –2.5; 12, 9.5, 7 –3; –14, –17, –20 This sequence has a common difference of 2.6 between its terms. This is an arithmetic sequence.
This sequence has a common difference of –2 between its terms. This is an arithmetic sequence.
This sequence does not have a common difference between its terms. This is not an arithmetic sequence.
an = –5n + 2; –33 an = –4n – 7; –35 an = 5n – 7; 28 an = 0.25n – 1; 0.75 251_258_HSM_NA_S_A1M04_L05_662599.indd Page 255 5/9/19 3:28 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education Example 4 23. SPORTS Wanda is the manager for the soccer team. One of her duties is to hand out cups of water at practice. Each cup of water is 4 ounces. She begins practice with a 128-ounce cooler of water. a. Create a function to represent the arithmetic sequence. b. Graph the function. c. How much water is remaining after Wanda hands out the 14th cup? 24. THEATER A theater has 20 seats in the first row, 22 in the second row, 24 in the third row, and so on for 25 rows. a. Create a function to represent the arithmetic sequence. b. Graph the function. c. How many seats are in the last row? 25. POSTAGE The price to send a large envelope first class mail is 88 cents for the first ounce and 17 cents for each additional ounce. The table shows the cost for weights up to 5 ounces.
Weight (ounces) 1 2 3 4 5 Postage (dollars) 0.88 1.05 1.22 1.39 1.56 Source: United States Postal Service a. Create a function to represent the arithmetic sequence. b. Graph the function. c. How much did a large envelope weigh that cost $2.07 to send? 26. VIDEO DOWNLOADING Brian is downloading episodes of his favorite TV show to play on his personal media device. The cost to download 1 episode is $1.99. The cost to download 2 episodes is $3.98. The cost to download 3 episodes is $5.97. a. Create a function to represent the arithmetic sequence. b. Graph the function. c. What is the cost to download 9 episodes? 256 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-5_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE f(n) = –4n + 128 68 seats 72 ounces 8 ounces f(n) = 2n + 18 f(n) = 0.17n + 0.71 2 3 1 4 5 6 7 8 9 10 12 0 24 36 48 60 72 84 96 108 120 132 Ounces Remaining Cups 2 3 1 4 5 6 7 8 9 10 4 0 8 12 16 20 24 28 32 36 40 Seats Rows 2 3 1 4 5 6 7 8 9 10 0.16 0 0.32 0.48 0.64 0.8 0.96 1.12 1.28 1.44 1.6 Cost (S) Ounces 2 3 1 4 5 6 7 8 9 10 1 0 2 3 4 5 6 7 8 9 10 Cost ($) Episodes f(n) = 1.99n $17.91 251_258_HSM_NA_S_A1M04_L05_662599.indd Page 256 5/10/19 3:15 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
3 REFLECT AND PRACTICE ASSESS AND DIFFERENTIATE Use the data from the Checks to determine whether to provide resources for extension, remediation, or intervention.
IF students score 90% or more on the Checks, THEN assign: BL • Practice, Exercises 1–37 odd, 38–46 • Extension: Arithmetic Series • Arithmetic Sequences IF students score 66%–89% on the Checks, THEN assign: OL • Practice, Exercises 1–45 odd • Remediation, Review Resources: Add Integers • Personal Tutors • Extra Examples 1–4 • Addition and Subtraction with Integers IF students score 65% or less on the Checks, THEN assign: AL • Practice, Exercises 1–25 odd • Remediation, Review Resources: Add Integers • Quick Review Math Handbook: Arithmetic Sequences as Linear Functions • ArriveMATH Take Another Look • Addition and Subtraction with Integers 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Lesson 4-5 • Arithmetic Sequences 255-256 F.BF.1a, F.BF.2, F.LE.2 Interactive Presentation Answers 32. an = 6n + 1; 33. an =-4n + 34; 34. an = 3n - 10; an n O 20 30 10 2 4 6 an n O 20 30 10 2 4 6 an n 4 2 4 6 O −4 −8 3 REFLECT AND PRACTICE Copyright © McGraw-Hill Education Name Period Date 27. USE A MODEL Chapa is beginning an exercise program that calls for 30 push-ups each day for the first week. Each week thereafter, she has to increase her push-ups by 2. a. Write a function to represent the arithmetic sequence. b. Graph the function. c. Which week of her program will be the first one in which she will do at least 50 push-ups a day? Mixed Exercises CONSTRUCT ARGUMENTS Determine whether each sequence is an arithmetic sequence. Justify your argument. 28. -9, -12, -15, -18, ... 29. 10, 15, 25, 40, ... 30. -10, -5, 0, 5, ... 31. -5, -3, -1, 1, ...
Write an equation for the nth term of each arithmetic sequence. Then graph the first five terms of the sequence. 32. 7, 13, 19, 25, … 33. 30, 26, 22, 18, … 34. -7, -4, -1, 2, … 35. SAVINGS Fabiana decides to save the money she’s earning from her after-school job for college. She makes an initial contribution of $3000 and each month deposits an additional $500. After one month, she will have contributed $3500. a. Write an equation for the nth term of the sequence. b. How much money will Fabiana have contributed after 24 months?
36. NUMBER THEORY One of the most famous sequences in mathematics is the Fibonacci sequence. It is named after Leonardo de Pisa (1170–1250) or Filius Bonacci, alias Leonardo Fibonacci. The first several numbers in the Fibonacci sequence are shown. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . . Does this represent an arithmetic sequence? Why or why not? 37. STRUCTURE Use the arithmetic sequence 2, 5, 8, 11, ... a. Write an equation for the nth term of the sequence. b. What is the 20th term in the sequence? Lesson 4-5 • Arithmetic Sequences 257 Program: Reveal Math Component: Lesson 4-5_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 2 3 1 4 5 6 7 8 9 10 4 0 8 12 16 20 24 28 32 36 40 Push Ups Week f(n) = 2n + 28 11th week This sequence has a common difference of –3 between its terms. This is an arithmetic sequence.
This sequence has a common difference of 5 between its terms. This is an arithmetic sequence.
This sequence does not have a common difference between its terms. This is not an arithmetic sequence.
This sequence has a common difference of 2 between its terms. This is an arithmetic sequence.
See margin.
See margin.
See margin.
an = 3000 + 500n $15,000 No, because the difference between terms is not constant.
an = 3n – 1 59 251_258_HSM_NA_S_A1M04_L05_662599.indd Page 257 04/10/19 1:30 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education 38. CREATE Write a sequence that is an arithmetic sequence. State the common difference, and find a6. 39. CREATE Write a sequence that is not an arithmetic sequence. Determine whether the sequence has a pattern, and if so describe the pattern. 40. REASONING Determine if the sequence 1, 1, 1, 1, . . . is an arithmetic sequence. Explain your reasoning. 41. CREATE Create an arithmetic sequence with a common difference of -10.
42. PERSEVERE Find the value of x that makes x + 8, 4x + 6, and 3x the first three terms of an arithmetic sequence.
43. CREATE For each arithmetic sequence described, write a formula for the nth term of a sequence that satisfies the description. a. first term is negative, common difference is negative b. second term is –5, common difference is 7 c. a2 = 8, a3 = 6 Andre and Sam are both reading the same novel. Andre reads 30 pages each day. Sam created the table at the right. Refer to this information for Exercises 44–46.
44. ANALYZE Write arithmetic sequences to represent each boy’s daily progress. Then write the function for the nth term of each sequence. 45. PERSEVERE Enter both functions from Exercise 44 into your calculator. Use the table to determine if there is a day when the number of pages Andre has read is equal to the number of pages Sam has left to read. If so, which day is it? Explain how you used the table feature to help you solve the problem. 46. ANALYZE Graph both functions on your calculator, then sketch the graph in the coordinate plane at the right. How can you use the graph to answer the equation from Exercise 45? Sam’s Reading Progress Day Pages Left to Read 1 430 2 410 3 390 4 370 5 10 270 540 y x O 258 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-5_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: 2, 5, 8, 11, …; The common difference is 3; a6 = 17 Sample answer: 2, -8, -18, -28, ...
-1 Sample answer: 5, 3, 8, 6, 11, 9, 14, …; The pattern is to subtract 2 from the first term to find the second term, then add 5 to the second term to find the third term. The sequence 1, 1, 1, 1, … is a set of numbers whose difference between successive terms is the constant number 0. Thus, this sequence is an arithmetic sequence by the definition.
On day 9, Andre has read 270 pages while Sam has 270 pages left to read. The table shows that both functions have a value of 270 when x = 9.
The intersection of the lines, which appears to be at (9, 270), represents the day (Day 9) when Andre has read the same number of pages that Sam has left to read (270 pages).
Sample answer: an = –2 – 3n an = –19 + 7n an = 12 – 2n Andre: 30, 60, 90, 120, …; A(n) = 30n; Sam: 430, 410, 390, 370, …; S(n) = 450 – 20n Higher-Order Thinking Skills 251_258_HSM_NA_S_A1M04_L05_662599.indd Page 258 5/9/19 3:28 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION 257-258 Module 4 • Linear and Nonlinear Functions F.BF.1a, F.BF.2, F.LE.2 LESSON GOAL Students graph piecewise-defined and step functions.
1 LAUNCH Launch the lesson with a Warm Up and an introduction.
2 EXPLORE AND DEVELOP Develop: Graphing Piecewise-Defined Functions • Graph a Piecewise-Defined Function Explore: Age as a Function Develop: Graphing Step Functions • Graph a Greatest Integer Function • Graph a Step Function You may want your students to complete the Checks online.
3 REFLECT AND PRACTICE Exit Ticket Practice DIFFERENTIATE View reports of student progress on the Checks after each example.
Resources AL OL BL ELL Remediation: Construct Linear Functions ● ● ● Extension: Taxicab Graphs ● ● ● Language Development Handbook Assign page 25 of the Language Development Handbook to help your students build mathematical language related to piecewise-defined and step functions.
ELL You can use the tips and suggestions on page T25 of the handbook to support students who are building English proficiency.
Lesson 4-6 Piecewise and Step Functions Lesson 4-6 • Piecewise and Step Functions 259a F.IF.4, F.IF.7b Suggested Pacing 90 min 0.5 day 45 min 1 day Focus Domain: Functions Standards for Mathematical Content: F.IF.4 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship.
F.IF.7b Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions.
Standards for Mathematical Practice: 4 Model with mathematics.
6 Attend to precision.
Coherence Vertical Alignment Rigor The Three Pillars of Rigor 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Conceptual Bridge In this lesson, students extend their understanding of linear functions to piecewise-defined and step functions. They build fluency by graphing both types of functions, and they apply their understanding by solving real-world problems related to piecewise-defined and step functions.
Mathematical Background Piecewise-defined functions are functions that are defined by two or more functions, each with its own domain. The graph consists of the graph of each piece over its domain. A step function is a function whose graph consists of segments that look like a set of steps. The graph of the greatest integer function is an example of a step function.
Next Students will identify the effects of transformations of the graphs of absolute value functions. F.IF.7b, F.BF.3 Now Students graph piecewise-defined and step functions.
F.IF.4, F.IF.7b Previous Students wrote and graphed equations of arithmetic sequences.
F.BF.1a, F.BF.2, F.LE.2 Interactive Presentation 1 LAUNCH Warm Up Warm Up Prerequisite Skills The Warm Up exercises address the following prerequisite skill for this lesson: • graphing linear functions Answers: 1. e 2. a 3. d 4. b 5. c Launch the Lesson Launch the Lesson MP MP Teaching the Mathematical Practices 4 Apply Mathematics In this Launch, students learn how to apply what they have learned about special functions to a real-world situation about the discounts offered at a store. Go Online to find additional teaching notes and questions to promote classroom discourse.
Today’s Standards Tell students that they will address these content and practice standards in this lesson. You may wish to have a student volunteer read aloud How can I meet these standards? and How can I use these practices?, and connect these to the standards.
See the Interactive Presentation for I Can statements that align with the standards covered in this lesson.
Today’s Vocabulary Today’s Vocabulary Tell students that they will use these vocabulary terms in this lesson. You can expand each row if you wish to share the definitions. Then discuss the questions below with the class.
259b Module 4 • Linear and Nonlinear Functions F.IF.4, F.IF.7b Interactive Presentation 2 EXPLORE AND DEVELOP Explore Age as a Function Objective Students collect data to explore how real-world data can be represented by a step function.
MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Ideas for Use Recommended Use Present the Inquiry Question, or have a student volunteer read it aloud. Have students work in pairs to complete the Explore activity on their devices. Pairs should discuss each of the questions. Monitor student progress during the activity. Upon completion of the Explore activity, have student volunteers share their responses to the Inquiry Question. What if my students don’t have devices? You may choose to project the activity on a whiteboard. A printable worksheet for each Explore is available online. You may choose to print the worksheet so that individuals or pairs of students can use it to record their observations.
Summary of the Activity Students complete guiding exercises throughout the Explore activity. Students will explore how data in a real-world scenario involving age groups can be modeled by a step function. They will use their own age to create a table that shows the group in which they would be placed after various periods of time and answer questions regarding the data in their table. They will then explore how the graph of a step function represents this type of data. Then, students answer the Inquiry Question. (continued on the next page) Explore Explore Students answer the questions and complete a table based on age.
TYPE 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Lesson 4-6 • Piecewise and Step Functions 259c F.IF.4 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Age as a Function (continued) Questions Have students complete the Explore activity.
Ask: • In which age group would you place someone who will be 13 next week? Why? 11-12; Sample answer: According to the rules, the person would be in the 11-12 group because he or she is still 12.
• What other situations could be modeled by a step function? Sample answer: Movie ticket prices that depend on age could be modeled by a step function. Inquiry When can real-world data be described using a step function? Sample answer: When domain values in intervals have the same range value, real-world data can be described using a step function. Go Online to find additional teaching notes and sample answers for the guiding exercises.
Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students respond to the Inquiry Question and can view a sample answer.
TYPE 259d Module 4 • Linear and Nonlinear Functions F.IF.4 Interactive Presentation Learn Graphing Piecewise-Defined Functions Objective Students graph piecewise-defined functions and identify their domain and range by determining the intervals where each part of the function should be graphed.
MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between equations and graphs of piecewise-defined and piecewise-linear functions. Example 1 Graph a Piecewise-Defined Function MP MP Teaching the Mathematical Practices 2 Different Properties Mathematically proficient students looks for different ways to solve problem. Encourage them to consider an alternate method in the Think About It! feature.
Questions for Mathematical Discourse AL Why do you think this is called a piecewise-defined function? Sample answer: The function has different rules for different “pieces” of the graph.
OL Why is (1, 6) included in the graph, but (1, 2) is not? Sample answer: The first domain includes 1 because it states that x ≤ 1 while the second domain does not include 1. So the y-value that corresponds with x = 1 is 2(1) + 4, or 6.
BL Why is the range not the set of real numbers? There are no values of x that are paired with numbers greater than 6. Go Online • Find additional teaching notes.
• View performance reports of the Checks.
• Assign or present an Extra Example.
Copyright © McGraw-Hill Education Piecewise and Step Functions Lesson 4-6 Learn Graphing Piecewise-Defined Functions Some functions cannot be described by a single expression because they are defined differently depending on the interval of x. These functions are piecewise-defined functions. A piecewise-linear function has a graph that is composed of some number of linear pieces.
Example 1 Graph a Piecewise-Defined Function To graph a piecewise-defined function, graph each “piece” separately. Graph f(x) = { 2x + 4 if x ≤ 1 -x + 3 if x > 1 . State the domain and range.
First, graph f(x) = 2x + 4 if x ≤ 1.
• Create a table for f(x) = 2x + 4 using values of x > 1.
• Because x is less than or equal to 1, place a at to indicate that the endpoint is included in the graph.
• Then, plot the points and draw the graph beginning at (1, 6).
x y 1 0 −1 −2 −3 Next, graph f(x) = −x + 3 if x > 1.
• Create a table for f(x) = −x + 3 using values of x > 1.
• Because x is greater than but not equal to 1, place a at to indicate that the endpoint is not included in the graph.
• Then, plot the points and draw the graph beginning at (1, 2).
x y 1 2 3 4 5 The domain is all real numbers. The range is y ≤ .
Today’s Vocabulary piecewise-defined function piecewise-linear function step function greatest integer function Think About It!
What would be an advantage of graphing the entire expression and removing the portion that is not in the interval?
y x O Go Online You can complete an Extra Example online.
y x O Go Online An alternate method is available for this example.
Today’s Goals ● Identify and graph piecewise-defined functions.
● Identify and graph step functions.
Program: Reveal Math Component: Lesson 4-6_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-6 • Piecewise and Step Functions 259 Sample answer: You can quickly graph the expressions using slope-intercept form or the intercepts and avoid having to make a table.
dot circle (1, 6) (1, 2) 6 6 4 2 0 −2 2 1 0 −1 −2 259_266_HSM_NA_S_A1M04_L06_662599.indd Page 259 05/10/19 8:00 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 1 Students explain the advantage of graphing the entire expression and then removing the portion not in the interval.
TYPE 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students select a method for graphing a piecewise-defined function.
TAP Lesson 4-6 • Piecewise and Step Functions 259 F.IF.4, F.IF.7b Interactive Presentation 2 EXPLORE AND DEVELOP Learn Graphing Step Functions Objective Students graph step functions by making a table of values.
MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the x- and y-values of each horizontal line segment used in this Learn. About the Key Concept The graph of a greatest integer function always consists of infinitely many “steps,” each with one closed endpoint and one open endpoint. The parameters of the function determine the length of the steps. The greatest integer function is a type of piecewise-defined linear function, as the function is equal to a different constant for different intervals in the domain.
Common Misconception Some students may think that the steps on the graph of a greatest integer function are always 1 unit long. Explain that while this is true of the graph of the parent greatest integer function, other greatest integer functions will contain parameters that may affect the length of each step.
Copyright © McGraw-Hill Education Study Tip Piecewise-Defined Functions When graphing piecewise-defined functions, there should be a dot or line that contains each member of the domain.
Check Part A Graph f(x) = ⎧ ⎨ ⎩ -x + 1 if x ≤ -2 -3x - 2 if x > -2 .
y x O Part B Find the domain and range of the function. Go Online You can complete an Extra Example online.
Learn Graphing Step Functions A step function is a type of piecewise-linear function with a graph that is a series of horizontal line segments. One example of a step function is the greatest integer function, written as f(x) = 〚 x 〛 in which f(x) is the greatest integer less than or equal to x.
Key Concept • Greatest Integer Function Type of graph: disjointed line segments The graph of a step function is a series of disconnected horizontal line segments.
Domain: all real numbers; Because the dots and circles overlap, the domain is all real numbers.
Range: all integers; Because the function represents the greatest integer less than or equal to x, the range is all integers.
Parent function: f(x) = 〚 x 〛 y x O f(x) = [x] Watch Out!
Circles and Dots Do not forget to examine the endpoint(s) of each piece to determine whether there should be a circle or a dot. > and < mean that a circle should be used, while ≥ and ≤ mean that a dot should be used. Go Online You can watch a video to see how to graph a piecewise-defined function on a graphing calculator Your Notes Explore Age as a Function Online Activity Use a real-world situation to complete an Explore. INQUIRY When can real-world data be described using a step function?
260 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-6_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE D = all real numbers; R = all real numbers 259_266_HSM_NA_S_A1M04_L06_662599.indd Page 260 10/05/19 8:25 PM f-0204 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
DIFFERENTIATE Enrichment Activity AL BL ELL IF students have difficulty understanding the nature of the graph of the greatest integer function, THEN have them create a table of values for the function. Instruct them to include decimals and fractions in their tables. Then have them describe how they determined the y-values for the x-values that they chose.
Learn 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students tap on each marker to learn about graphing step functions.
TAP Students explain the relationship between the x- and y-values of each horizontal line segment.
TYPE a Students complete the Check online to determine whether they are ready to move on.
CHECK 260 Module 4 • Linear and Nonlinear Functions F.IF.4, F.IF.7b Interactive Presentation Copyright © McGraw-Hill Education Go Online You can complete an Extra Example online.
Example 2 Graph a Greatest Integer Function Graph f(x) = 〚 x + 1〛. State the domain and range.
First, make a table. Select a few values that are between integers.
x x + 1 〚x + 1〛 −2 −1 −1 –1, –0.75, and –0.25 are greater than or equal to –1 but less than 0. So, –1 is the greatest integer that is not greater than –1, –0.75, or –0.25.
−1.75 −0.75 −1 −1.25 −0.25 −1 −1 0 0 0, 0.5, and 0.75 are greater than or equal to 0 but less than 1. So, 0 is the greatest integer that is not greater than 0, 0.5, or 0.75.
−0.5 0.5 0 −0.25 0.75 0 0 1 1 1, 1.25, and 1.5 are greater than or equal to 1 but less than 2. So, 1 is the greatest integer that is not greater than 1, 1.25, or 1.5.
0.25 1.25 1 0.5 1.5 1 1 2 2 2, 2.25, and 2.75 are greater than or equal to 2 but less than 3. So, 2 is the greatest integer that is not greater than 2, 2.25, or 2.75.
1.25 2.25 2 1.75 2.75 2 On the graph, dots represent included points. Circles represent points that are not included. The domain is all real numbers. The range is all integers. Note that this is the graph of f(x) = 〚 x 〛 shifted 1 unit to the left.
Check Graph f(x) = 〚 x - 2〛 by making a table.
x x - 2 〚x - 2〛 −1 −3 −0.75 −2.75 −0.25 −2.25 −3 0 −2 −2 0.25 −1.75 −2 0.5 −1.5 1 −1 1.25 −0.75 −1 1.5 −0.5 2 0 2.25 0.25 Talk About It!
What do you notice about the symmetry, extrema, and end behavior of the function?
Watch Out!
Greatest Integer Function When finding the value of a greatest integer function, do not round to the nearest integer. Instead, always round nonintegers down to the greatest integer that is not greater than the number.
y x O y x O Program: Reveal Math Component: Lesson 4-6_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-6 • Piecewise and Step Functions 261 Sample answer: The function has no symmetry and no minimum or maximum values. As x increases, f(x) increases, and as x decreases, f(x) decreases.
-3 -3 -2 -1 -1 0 0 259_266_HSM_NA_S_A1M04_L06_662599.indd Page 261 04/10/19 1:41 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 2 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students move through the slides to learn more about the table and graph of a greatest integer function.
Students describe what they noticed about the symmetry, extrema, and end behavior of the function.
TYPE TAP a Example 2 Graph a Greatest Integer Function MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Questions for Mathematical Discourse AL Why do you think this is called a step function? The graph looks like steps on a staircase.
OL Why is (0, 1) included in the graph but (1, 1) is not? 〚0 + 1〛 = 1 and 〚1 + 1〛 = 2 BL The greatest integer function is sometimes called the floor function. Why do you think that is? Sample answer: The value truncates to the integer portion of the value, like standing on a chair on the 2nd floor still means you are on the 2nd floor.
Common Error Some students may take the greatest integer of the x-value before adding 1. Explain that the greatest integer symbols act as grouping symbols, requiring that the operation inside the symbols be performed first, before finding the greatest integer of the resulting value.
DIFFERENTIATE Enrichment Activity BL Have students work with a partner. Ask them to create a story problem that can be modeled using the function f(x) = 15〚 x 〛. Have students construct a graph for the model, and share their problems with the class.
Lesson 4-6 • Piecewise and Step Functions 261 F.IF.4, F.IF.7b 2 EXPLORE AND DEVELOP Copyright © McGraw-Hill Education 2 1 4 5 3 6 35 0 70 105 140 175 210 Cost ($) Days Dog Boarding Think About It!
How would the graph change if 1 certified lifeguard could watch up to 59 swimmers? For example, if there are greater than or equal to 60, but fewer than 120 swimmers, there must be 2 lifeguards on duty. Go Online You can complete an Extra Example online. Example 3 Graph a Step Function SAFETY A state requires a ratio of 1 lifeguard to 60 swimmers in a swimming pool. This means that 1 lifeguard can watch up to and including 60 swimmers. Make a table and draw a graph that shows the number of lifeguards that must be on duty f(x) based on the number of swimmers in the pool x.
The number of lifeguards that must be on duty can be represented by a step function.
• If the number of swimmers is greater than 0 but fewer than or equal to 60, only 1 lifeguard must be on duty.
• If the number of swimmers is greater than 60 but fewer than or equal to 120, there must be 2 lifeguards on duty.
• If the number of swimmers is greater than 180 but fewer than or equal to 240, there must be 4 lifeguards on duty.
The circles mean that when there are more than a multiple of 60 swimmers, .
The dots represent the number of swimmers that can be in the pool for that particular number of on duty.
Check PETS At Luciana’s pet boarding facility, it costs $35 per day to board a dog. Every fraction of a day is rounded up to the next day. Graph the function representing this situation by making a table. x f(x) 0 < x ≤ 60 1 60 < x ≤ 120 2 120 < x ≤ 180 180 < x ≤ 240 4 240 < x ≤ 300 300 < x ≤ 360 360 < x ≤ 420 Days Cost ($) 0 < x ≤ 1 1 < x ≤ 2 2 < x ≤ 3 3 < x ≤ 4 4 < x ≤ 5 5 < x ≤ 6 120 240 360 1 0 2 3 4 5 6 7 Number of Lifeguards Number of Swimmers Lifeguard Requirements Math History Minute Oliver Heaviside (1850–1925) was a self-taught electrical engineer, mathematician, and physicist who laid much of the groundwork for telecommunications in the 21st century. Heaviside invented the Heaviside step function, f(x) = ⎧ ⎜ ⎨ ⎜ ⎩ 0 if x < 0 1 __ 2 if x = 0, 1 if x > 0 which he used to model the current in an electric circuit.
262 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-6_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: Every dot would become a circle, and every circle would become a dot.
3 5 6 7 another lifeguard is required maximum lifeguards 35 70 105 140 175 210 259_266_HSM_NA_S_A1M04_L06_662599.indd Page 262 04/10/19 1:42 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 3 Graph a Step Function MP MP Teaching the Mathematical Practices 4 Apply Mathematics In this example, students apply what they have learned about step functions to solving a real-world problem.
Questions for Mathematical Discourse AL How many lifeguards are needed for 59 swimmers? 1 For 61 swimmers? 2 OL Why is this situation represented by a step function? Sample answer: Every x-value in each interval of 60 is paired with the same y-value, forming a graph that consists of steps.
BL How would the graph of the function change if the number of lifeguards required for the number of swimmers is cut in half? Sample answer: The graph would be stretched horizontally because more swimmers could be watched by each lifeguard. Essential Question Follow-Up Students have analyzed and graphed step functions.
Ask: If you know that a function is a step function, what do you know about how the elements of the domain are paired with the elements of the range? Sample answer: The domain is grouped into intervals, and every number in the interval is paired with the same number in the range.
Exit Ticket Recommended Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond using a separate piece of paper. Have students hand you their responses as they leave the room.
Alternate Use At the end of class, go online to display the Exit Ticket prompt and ask students to respond verbally or by using a mini-whiteboard. Have students hold up their whiteboards so that you can see all student responses. Tap to reveal the answer when most or all students have completed the Exit Ticket.
Example 3 Students tap on each marker to see the difference between the circles and dots.
TAP Interactive Presentation Students complete the Check online to determine whether they are ready to move on.
CHECK Students drag the correct values to complete the table.
DRAG & DROP 262 Module 4 • Linear and Nonlinear Functions F.IF.4, F.IF.7b Practice and Homework Suggested Assignments Use the table below to select appropriate exercises.
DOK Topic Exercises 1, 2 exercises that mirror the examples 1–14 2 exercises that use a variety of skills from this lesson 15–20 2 exercises that extend concepts learned in this lesson to new contexts 21–22 3 exercises that emphasize higher-order and critical thinking skills 23–31 ASSESS AND DIFFERENTIATE Use the data from the Checks to determine whether to provide resources for extension, remediation, or intervention.
IF students score 90% or more on the Checks, THEN assign: BL • Practice, Exercises 1–21 odd, 23–31 • Extension: Taxicab Graphs IF students score 66%–89% on the Checks, THEN assign: OL • Practice, Exercises 1–31 odd • Remediation, Review Resources: Construct Linear Functions • Personal Tutors • Extra Examples 1–3 • Tables and Graphs of Lines IF students score 65% or less on the Checks, THEN assign: AL • Practice, Exercises 1–13 odd • Remediation, Review Resources: Construct Linear Functions • Quick Review Math Handbook: Special Functions • ArriveMATH Take Another Look • Tables and Graphs of Lines 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION 3 REFLECT AND PRACTICE Copyright © McGraw-Hill Education Weight (pounds) Rate (dollars) 1 16.20 2 19.30 3 22.40 4 25.50 5 28.60 4 6 2 8 10 24 0 48 72 96 120 Daily Earnings ($) Hours Worked x y (8, 64) 10. g(x) = 〚x〛 + 3 11. h(x) = 〚x〛 - 1 12. h(x) = 1 __ 2 〚x〛 + 1 Example 3 13. BABYSITTING Ariel charges $8 per hour as a babysitter. She rounds every fraction of an hour up to the next half-hour. Draw a graph to represent Ariel’s total earnings y after x hours.
1 2 3 5 0 10 15 20 25 Ariel’s Earnings ($) Hours Babysitting y x 14 FUNDRAISING Students are selling boxes of cookies at a fundraiser. The boxes of cookies can only be ordered by the case, with 12 boxes per case. Draw a graph to represent the number of cases needed y when x boxes of cookies are sold.
x y 24 36 12 48 60 72 1 0 2 3 4 5 6 Cases Needed Boxes Sold Mixed Exercises 15. PRECISION A package delivery service determines rates for express shipping by the weight of a package, with every fraction of a pound rounded up to the next pound. The table shows the cost of express shipping packages that weigh no more than 5 pounds. Write a piecewise-linear function representing the cost to ship a package that weighs no more than 5 pounds. State the domain and range. 16. EARNINGS Kelly works in a hospital as a medical assistant. She earns $8 per hour the first 8 hours she works in a day and $11.50 per hour each hour thereafter. a. Organize the information into a table. Include a row for hours worked x, and a row for daily earnings f(x). b. Write the piecewise equation describing Kelly’s daily earnings f(x) for x hours. c. Draw a graph to represent Kelly’s daily earnings.
y x O y x O y x O 264 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-6_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE a–c. See margin.
See margin for function. D = {x | 0 < x ≤ 5}; R = {16.20, 19.30, 22.40, 25.50, 28.60} D = all real numbers, R = all integers D = all real numbers, R = all multiples of 1 __ 2 D = all real numbers, R = all integers 259_266_HSM_NA_S_A1M04_L06_662599.indd Page 264 11/10/19 2:50 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education Name Period Date Practice Go Online You can complete your homework online.
Example 1 Graph each function. State the domain and range. 1. f(x) = ⎧ ⎨ ⎩ 2. f(x) = ⎧ ⎨ ⎩ 3. f(x) = ⎧ ⎨ ⎩ 4. f(x) = ⎧ ⎨ ⎩ 3x + 4 if x ≥ 1 x + 3 if x < 1 5. f(x) = ⎧ ⎨ ⎩ 3x + 2 if x > -1 - 1 __ 2 x - 3 if x ≤ -1 6. f(x) = ⎧ ⎨ ⎩ Example 2 Graph each function. State the domain and range. 7. f(x) = 3 〚x〛 8. f(x) = 〚-x〛 9. g(x) = -2 〚x〛 1 __ 2 x - 1 if x > 3 -2x + 3 if x ≤ 3 2x + 3 if x ≥ -3 - 1 __ 3 x + 1 if x < -3 2x - 5 if x > 1 4x - 3 if x ≤ 1 y x O 2x + 1 if x < -2 -3x - 1 if x ≥ -2 y x O y x O y x O y x O y x O f(x) x O f(x) x O y x O Program: Reveal Math Component: Lesson 4-6_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-6 • Piecewise and Step Functions 263 D = all real numbers, R = f(x) ≥ -3 D = all real numbers, R = f(x) < 4 or f(x) ≥ 7 D = all real numbers, R = all integer multiples of 3 D = all real numbers, R = f(x) ≥ -3 D = all real numbers, R = f(x) ≤ 5 D = all real numbers, R = all even integers D = all real numbers, R = all real numbers D = all real numbers, R = f(x) ≥ -2.5 D = all real numbers, R = all integers 259_266_HSM_NA_S_A1M04_L06_662599.indd Page 263 14/01/19 5:02 PM f-0204 /115/GO02573_R1/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Answers 16.20 if 0 < x ≤ 1 19.30 if 1 < x ≤ 2 15. f(x) = 22.40 if 2 < x ≤ 3 25.50 if 3 < x ≤ 4 28.60 if 4 < x ≤ 5 16a. 16b. f(x) = 8x if x ≤ 8 64 + 11.5(x - 8) if x ≥ 8 ⎧ ⎨ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x 0 4 8 12 16 f(x) 0 32 64 110 156 Lesson 4-6 • Piecewise and Step Functions 263-264 F.IF.4, F.IF.7b 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION 3 REFLECT AND PRACTICE 3.50 0 < x ≤ 1 7.00 1 < x ≤ 2 20a. C(p) = 10.50 2 < x ≤ 3 14.00 3 < x ≤ 4 17.50 4 < x ≤ 5 1 __ 2 x - 3 x 6 27. f(x) = - 1 __ 2 x + 3 x ≤ 6 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎨ ⎪ ⎩ Copyright © McGraw-Hill Education 4 8 12 −4 4 8 12 x O f(x) −4 −8 8 4 4 8 −4 −8 y x O 22. INVENTORY Malik owns a bakery. Every week he orders chocolate chips from a supplier. The supplier’s pricing is shown in the table.
a. Write a function to represent the cost of chocolate chips. b. Malik’s budget for chocolate chips for the week is $25. How many whole pounds of chocolate chips can he order? 23. CREATE Write a piecewise-defined function with three linear pieces. Then graph the function. 24. FIND THE ERROR Amy graphed a function that gives the height of a car on a roller coaster as a function of time. She said her graph is the graph of a step function. Is this possible? Explain your reasoning. 25. WRITE What is the difference between a step function and a piecewise-defined function? 26. ANALYZE Does the piecewise relation y = ⎧ ⎨ ⎩ represent a function? Justify your argument.
ANALYZE Refer to the graph for Exercises 27–31. 27. Write a piecewise function to represent the graph. 28. What is the domain? 29. What is the range? 30. Find f(8.5). 31. Find f(1.2). -2x + 4 if x ≥ 2 - 1 __ 2 x - 1 if x ≤ 4 Chocolate Chip Pricing $4 per pound Up to 3 pounds $1.50 per pound For each pound over 3 pounds 266 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-6_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE No; the pieces of the graph overlap vertically, so the graph fails the vertical line test.
No; the height of the car must change continuously, so it cannot “jump” from one constant height to another constant height as it would if the graph were a step function.
A step function has different constants over different intervals of its domain. A piecewise-defined function can have different algebraic rules over different intervals of its domain.
See margin.
D = all real numbers R = f(x) ≥ 0 1.25 2.4 Higher-Order Thinking Skills 11 lbs ⎧ ⎪ ⎨ ⎪ ⎩ Sample answer: y = -x if x < - 4 2x if - 4 ≤ x ≤ 2 x - 2 if x > 2 ⎧ ⎨ ⎩ f(x) = 4x if 0 < x ≤ 3 12 + 1.5(x - 3) if x > 3 259_266_HSM_NA_S_A1M04_L06_662599.indd Page 266 11/10/19 2:50 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education Name Period Date 17. REASONING Write a piecewise function that represents the graph. 18. STRUCTURE Suppose f(x) = 2⟦x - 1⟧.
a. Find f(1.5). b. Find f(2.2).
c. Find f(9.7). d. Find f(-1.25). 19. RENTAL CARS Mr. Aronsohn wants to rent a car on vacation. The rate the car rental company charges is $19 per day. If any fraction of a day is counted as a whole day, how much would it cost for Mr. Aronsohn to rent a car for 6.4 days?
g(x) x O 20. USE A MODEL A roadside fruit and vegetable stand determines rates for selling produce, with every fraction of a pound rounded up to the next pound. The table shows the cost of tomatoes by weight in pounds.
a. Write a piecewise-linear function representing the cost of purchasing 0 to 5 pounds of tomatoes, where C is the cost in dollars and p is the number of pounds. Weight (pounds) Rate (dollars) 1 3.50 2 7.00 3 10.50 4 14.00 5 17.50 b. Graph the function. c. State the domain and range. d. What would be the cost of purchasing 4.3 pounds of tomatoes at the roadside stand? 21. ELECTRONIC REPAIRS Tech Repairs charges $25 for an electronic device repair that takes up to one hour. For each additional hour of labor, there is a charge of $50. The repair shop charges for the next full hour for any part of an hour.
a. Complete the table to organize the information. Include a row for hours of repair x, and a row for total cost f(x).
x 0 2 4 6 8 f(x) b. Write a step function to represent the total cost for every hour x of repair. c. Graph the function. d. Devesh was charged $125 to repair his tablet. How long did the repair take to complete?
p C 2 3 1 4 5 6 7 2 0 4 6 8 10 12 14 16 18 Cost ($) Number of Pounds f(x) x 2 3 1 4 5 6 7 8 50 0 100 150 200 250 300 350 400 450 Cost ($) Number of Hours Program: Reveal Math Component: Lesson 4-6_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-6 • Piecewise and Step Functions 265 0 $133.00 2 < x ≤ 3 0 75 175 275 375 f(x) = 25 + 50〚x〛 See margin.
$17.50 D = {x | 0 < x ≤ 5}; R = {3.50, 7.00, 10.50, 14.00, 17.50} 16 2 -6 2x + 1 if x ≤ 2 x - 2 if x > 2 g(x) = ⎧ ⎨ ⎩ 259_266_HSM_NA_S_A1M04_L06_662599.indd Page 265 11/10/19 2:50 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
265-266 Module 4 • Linear and Nonlinear Functions F.IF.4, F.IF.7b LESSON GOAL Students identify the effects of transformations of the graphs of absolute value functions.
1 LAUNCH Launch the lesson with a Warm Up and an introduction.
2 EXPLORE AND DEVELOP Explore: Parameters of an Absolute Value Function Develop: Graphing Absolute Value Functions; Translations of Absolute Value Functions • Vertical Translations of Absolute Value Functions • Horizontal Translations of Absolute Value Functions • Multiple Translations of Absolute Value Functions • Identify Absolute Value Functions from Graphs • Identify Absolute Value Functions from Graphs (Multiple Translations) Dilations of Absolute Value Functions • Dilations of Form a|x| When x > 1 • Dilations of the Form |ax| • Dilations When 0 < a < 1 Reflections of Absolute Value Functions • Graphs of Reflections with Transformations • Graphs of y = –a|x| • Graphs of y = |–ax| Transformations of Absolute Value Functions • Graph an Absolute Value Function with Multiple Translations • Graph an Absolute Value Function with Translations and Dilation • Graph an Absolute Value Function with Translations and Reflection • Apply Graphs of Absolute Value Functions You may want your students to complete the Checks online.
3 REFLECT AND PRACTICE Exit Ticket Practice MATH PROBES CHERYL TOBEY Formative Assessment Math Probe DIFFERENTIATE View reports of student progress on the Checks after each example.
Resources AL OL BL ELL Remediation: Integers: Opposites and Absolute Value ● ● ● Extension: Parametric Equations ● ● ● Lesson 4-7 Absolute Value Functions Lesson 4-7 • Absolute Value Functions 267a F.IF.7b, F.BF.3 Suggested Pacing 90 min 1 day 45 min 2 days Focus Domain: Functions Standards for Mathematical Content: F.IF.7b Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions.
F.BF.3 Identify the effect on the graph of replacing f (x) by f (x) + k, k f (x), f (kx), and f (x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Standards for Mathematical Practice: 1 Make sense of problems and persevere in solving them.
5 Use appropriate tools strategically.
7 Look for and make use of structure.
Coherence Next Students will create linear equations in slope-intercept form.
A.CED.2, S.ID.7 Now Students identify the effects of transformations of the graphs of absolute value functions.
F.IF.7b, F.BF.3 Previous Students graphed piecewise-defined and step functions.
F.IF.4, F.IF.7b Rigor The Three Pillars of Rigor 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Conceptual Bridge In this lesson, students extend their understanding of absolute value to absolute value functions. They build fluency by graphing absolute value functions, and they apply their understanding by solving real-world problems related to absolute value functions.
Mathematical Background The graph of the absolute value parent function is V-shaped, with the vertex at the origin. The right side of the V is the graph of y = x; the left side is the graph of y = –x. Translations, dilations, and reflections of the graph of the absolute value parent function, f (x) = |x|, result in shifts, stretches or compressions, and flips (respectively), of the V-shaped graph.
Interactive Presentation 1 LAUNCH Warm Up Warm Up Prerequisite Skills The Warm Up exercises address the following prerequisite skill for this lesson: • evaluating absolute value expressions Answers: 1. > 2. = 3. < 4. = 5. > Launch the Lesson Launch the Lesson MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationship between the shape of the Palace of Peace and Reconciliation and the graph of an absolute value function. Go Online to find additional teaching notes and questions to promote classroom discourse.
Today’s Standards Tell students that they will address these content and practice standards in this lesson. You may wish to have a student volunteer read aloud How can I meet these standards? and How can I use these practices?, and connect these to the standards.
See the Interactive Presentation for I Can statements that align with the standards covered in this lesson.
Today’s Vocabulary Today’s Vocabulary Tell students that they will use this vocabulary term in this lesson. You can expand the row if you wish to share the definition. Then discuss the questions below with the class.
Language Development Handbook Assign page 26 of the Language Development Handbook to help your students build mathematical language related to transformations of the graphs of absolute value functions.
ELL You can use the tips and suggestions on page T26 of the handbook to support students who are building English proficiency.
267b Module 4 • Linear and Nonlinear Functions F.IF.7b, F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Parameters of an Absolute Value Function Objective Students use a sketch to explore how changing the parameters changes the graphs of absolute value functions.
MP MP Teaching the Mathematical Practices 5 Use Mathematical Tools Point out that to complete this Explore activity, students will need to use the sketch. Work with students to explore and deepen their understanding of absolute value functions.
Ideas for Use Recommended Use Present the Inquiry Question, or have a student volunteer read it aloud. Have students work in pairs to complete the Explore activity on their devices. Pairs should discuss each of the questions. Monitor student progress during the activity. Upon completion of the Explore activity, have student volunteers share their responses to the Inquiry Question. What if my students don’t have devices? You may choose to project the activity on a whiteboard. A printable worksheet for each Explore is available online. You may choose to print the worksheet so that individuals or pairs of students can use it to record their observations.
Summary of the Activity Students will complete guiding exercises throughout the Explore activity. Students will use a sketch to explore how changing the parameters of an absolute value function affects its graph. Students explore the graphs on their own and through an animation. They will answer questions and form generalizations based on their observations. Then, students will answer the Inquiry Question. (continued on the next page) Students use a sketch to explore transformations of absolute value functions.
WEB SKETCHPAD 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Explore Explore Students answer questions about the transformations of absolute value functions.
TYPE a Lesson 4-7 • Absolute Value Functions 267c F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Explore Parameters of an Absolute Value Function (continued) Questions Have students complete the Explore activity.
Ask: • How is changing the value of a for the absolute value graph similar to a linear function? Sample answer: The graphs get steeper as the value of a increases and less steep as a decreases. • How can looking at point V help you determine the transformations in the function? Sample answer: Point V is moved up, down, left or right depending on how values were added or subtracted to the function. Inquiry How does performing an operation on an absolute value function change its graph? Sample answer: Adding a value to the function moves the graph up or down. Subtracting a value from x moves the graph left or right. Multiplying the function by a value makes the graph wider or narrower or flips it over the x-axis. Go Online to find additional teaching notes and sample answers for the guiding exercises.
Explore 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students respond to the Inquiry Question and can view a sample answer.
TYPE a 267d Module 4 • Linear and Nonlinear Functions F.BF.3 Interactive Presentation Copyright © McGraw-Hill Education Absolute Value Functions Lesson 4-7 Today’s Vocabulary absolute value function vertex Learn Graphing Absolute Value Functions The absolute value function is a type of piecewise-linear function. An absolute value function is written as f(x) = a |x - h| + k, where a, h, and k are constants and f(x) ≥ 0 for all values of x.
The vertex is either the lowest point or the highest point of a function. For the parent function, y = |x|, the vertex is at the origin.
Key Concept • Absolute Value Function Parent Function f(x) = |x|, defined as f(x) = { x if x ≥ 0 - x if x < 0 Type of Graph V-Shaped Domain: all real numbers Range: all nonnegative real numbers Learn Translations of Absolute Value Functions Key Concept • Vertical Translations of Absolute Value Functions If k > 0, the graph of f(x) = |x| is translated k units up.
If k < 0, the graph of f(x) = |x| is translated |k| units down.
Key Concept • Horizontal Translations of Linear Functions If h > 0, the graph of f(x) = |x| is translated h units right.
If h < 0, the graph of f(x) = |x| is translated h units left.
Example 1 Vertical Translations of Absolute Value Functions Describe the translation in g(x) = |x| - 3 as it relates to the graph of the parent function.
Graph the parent function, f(x) = |x|, for absolute values.
The constant, k, is outside the absolute value signs, so k affects the y-values. The graph will be a vertical translation. Think About It!
Why does adding a positive value of k shift the graph k units up? Study Tip Horizontal Shifts Remember that the general form of an absolute value function is y = a |x - h| + k. So, y = |x + 7| is actually y = |x − (−7)| in the function’s general form. Go Online You can watch a video to see how to describe translations of functions.
Explore Parameters of an Absolute Value Function Online Activity Use graphing technology to complete the Explore. INQUIRY How does performing an operation on an absolute value function change its graph?
(continued on the next page) y x O f(x) g(x) Today’s Goals ● Graph absolute value functions.
● Apply translations to absolute value functions.
● Apply dilations to absolute value functions.
● Apply reflections to absolute value functions.
● Interpret constants within equations of absolute value functions.
Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-7 • Absolute Value Functions 267 Sample answer: This adds k to every value of the output of the absolute value. Since the output corresponds to the y-values, the graph moves up k units.
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Learn 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students tap on each card to see how vertical transformations affect the graph.
TAP Students explain why adding a positive value shifts the graph the same number of units up.
TYPE a Lesson 4-7 • Absolute Value Functions 267 F.IF.7b, F.BF.3 Learn Graphing Absolute Value Functions Common Misconception Some students may think that the graph of any absolute value function will lie completely above the x-axis. Explain that just as with other functions, transformations of the function will relocate the graph, and the resulting graph may, in fact, contain points that lie below the x-axis.
Learn Translations of Absolute Value Functions Objective Students identify the effect on the graph of an absolute value function by replacing f(x) with f(x) + k or f(x - h) for positive and negative values.
MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Example 1 Vertical Translations of Absolute Value Functions MP MP Teaching the Mathematical Practices 7 Use Structure Help students to use the structure of the transformed function to identify the translation in the function.
Questions for Mathematical Discourse AL What type of transformation occurs in g(x)? a vertical translation How do you know? 3 is being subtracted from the parent function.
OL How is the y-value of each ordered pair in the parent function affected? Each y-value decreases by 3 units.
BL How would the graph of f(x) = |x| + 3 compare to this graph? Sample answer: It would be shifted up 3 instead of down 3. Go Online • Find additional teaching notes.
• View performance reports of the Checks.
• Assign or present an Extra Example.
Interactive Presentation 2 EXPLORE AND DEVELOP Example 2 Horizontal Translations of Absolute Value Functions MP MP Teaching the Mathematical Practices 7 Use Structure Help students to determine the structure of the translated absolute value function in this example.
Questions for Mathematical Discourse AL What type of transformation occurs in j(x)? a horizontal translation How do you know? The -4 is inside the absolute value symbols.
OL How would the graph of f (x) = |x + 4| compare to this graph? The graph of the parent function would be shifted 4 units to the left instead of to the right.
BL How would the graph of f (x) = |x| - 4 compare to this graph? The graph of the parent function would be shifted 4 units down instead of to the right.
Example 3 Multiple Translations of Absolute Value Functions Questions for Mathematical Discourse AL Looking at only the equation, which value shifts the graph vertically? +3 OL Looking at only the equation, how do you know that the horizontal translation is to the right? Sample answer: If you use the form f(x - h) for the translation, then |x - 2| means that h = 2. This represents a translation to the right 2 units. BL What would the function be if it was a horizontal translation of 2 units left and 3 units down? g(x) = |x + 2| - 3 Common Error As the function becomes more complex, some students may have difficulty seeing the relationship to the parent function. Encourage them to rewrite functions like the one in this example using f(x). For example, for the function in this problem, students would write f(x – 2) + 3. In this way, they can see that 2 is being subtracted from x, and 3 is being added to the function values (i.e., the y-values).
Copyright © McGraw-Hill Education Since f(x) = |x|, g(x) = f(x) + k where k = -3.
g(x) = |x| - 3 ⟶ g(x) = f(x) + (-3) The value of k is less than 0, so the graph will be translated |k| units down, or 3 units down.
g(x) = |x| - 3 is a translation of the graph of the parent function 3 units down.
Example 2 Horizontal Translations of Absolute Value Functions Describe the translation in j(x) = |x - 4| as it relates to the parent function.
Graph the parent function, f(x) = |x|, for absolute values.
The constant, h, is inside the absolute value signs, so h affects the input or, x-values. The graph will be a horizontal translation.
Since f(x) = |x|, j(x) = f(x - h), where h = 4.
j(x) = |x - 4| ⟶ j(x) = f(x - 4) The value of h is greater than 0, so the graph will be translated h units right, or 4 units right.
j(x) = |x - 4| is the translation of the graph of the parent function 4 units right.
Example 3 Multiple Translations of Absolute Value Functions Describe the translation in g(x) = |x - 2| + 3 as it relates to the graph of the parent function.
The equation has both h and k values. The input and output will be affected by the constants. The graph of f(x) = |x| is vertically and horizontally translated. Since f(x) = |x|, g(x) = f(x - h) + k where h = 2 and k = 3. Because and , the graph is translated 2 units and 3 units . g(x) = |x - 2| + 3 is the translation of the graph of the parent function 2 units right and 3 units . Think About It!
Since the vertex of the parent function is at the origin, what is a quick way to determine where the vertex is of q(x) = |x − h| + k?
Emilio says that the graph of g(x) = |x + 1| − 1 is the same graph as f(x) = |x|. Is he correct? Why or why not? Go Online You can complete an Extra Example online.
Your Notes y x O j(x) f(x) y x O g(x) f(x) 268 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE No; sample answer: the graph of g(x) = |x + 1| − 1 is a graph that has been translated 1 unit left and 1 unit down from the parent function, f(x) = |x|.
Sample answer: The vertex will be at (h, k).
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Example 3 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students answer questions about the graphs of translated functions.
TYPE a DIFFERENTIATE Enrichment Activity AL ELL BL IF students are having difficulty determining the direction of a translation, THEN have them create four examples of absolute value functions that represent each type of translation, and write each one on an index card. Have them sketch the transformation on a coordinate plane on the back of the card, and write the description. Then have them use the flash cards (in both directions) to practice what they have learned. Students move through the steps to see how the graph of the function relates to the parent function.
TAP 268 Module 4 • Linear and Nonlinear Functions F.IF.7b, F.BF.3 Interactive Presentation Copyright © McGraw-Hill Education Example 4 Identify Absolute Value Functions from Graphs Use the graph of the function to write its equation.
The graph is the translation of the parent graph 1 unit to the right.
g(x) = |x - h| General equation for a horizontal translation g(x) = |x - 1| The vertex is 1 unit to the right of the origin.
Example 5 Identify Absolute Value Functions from Graphs (Multiple Translations) Use the graph of the function to write its equation.
The graph is a translation of the parent graph 2 units to the left and 5 units down.
g(x) = |x - h| + k General equation for translations g(x) = |x - (-2)| + k The vertex is 2 units left of the origin.
g(x) = |x - (-2)| + (-5) The vertex is 5 units down from the origin.
g(x) = |x + 2| - 5 Simplify. Go Online You can complete an Extra Example online.
y x O g(x) f(x) y x O g(x) f(x) Learn Dilations of Absolute Value Functions Multiplying by a constant a after evaluating an absolute value function creates a vertical change, either a stretch or compression.
Key Concept • Vertical Dilations of Absolute Value Functions If |a| > 1, the graph of f(x) = |x| is stretched vertically.
If 0 < |a| < 1, the graph of f(x) = |x| is compressed vertically.
When an input is multiplied by a constant a before for the absolute value is evaluated, a horizontal change occurs.
Key Concept • Horizontal Dilations of Absolute Value Functions If |a| > 1, the graph of f(x) = |x| is compressed horizontally.
If 0 < |a| < 1, the graph of f(x) = |x| is stretched horizontally. Talk About It!
How is the value of a in an absolute value function related to slope? Explain.
Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-7 • Absolute Value Functions 269 Sample answer: The value of a determines the slope of each part of the graph. The function y = a |x| can also be written as f(x) = { ax it x ≥ 0 -ax if x < 0 where a and -a are the slopes of the rays.
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Example 4 Example 4 Identify Absolute Value Functions from Graphs MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the graph and its equation used in this example.
Questions for Mathematical Discourse AL What translation is shown on the graph? a horizontal shift of 1 to the right OL Does this indicate that the value being added or subtracted should go inside or outside the absolute value symbols? inside BL A classmate argues that the function should be f(x) = |x + 1| because the shift is in the positive direction. Explain why this is incorrect. Sample answer: Translations are written in the form f(x) = |x – h| + k, so f(x) = |x + 1| would be f(x) = |x – (–1)|, which would be a horizontal shift to the left.
Common Error Some students may write the equation using a plus sign instead of a minus sign. Remind them that once they determine how many units and in what direction the graph is translated, they need to subtract that number from x.
Example 5 Identify Absolute Value Functions from Graphs (Multiple Translations) Questions for Mathematical Discourse AL How do you know that this graph represents a function with more than one transformation? Sample answer: The vertex is not on an axis.
OL How many transformations are there, and what type are they? 2; Sample answer: a horizontal translation of 2 units to the left and a vertical translation of 5 units down BL What are the coordinates of the vertex? (–2, –5) How does identifying the coordinates help you solve the problem? Sample answer: I can use the x-coordinate for h and the y-coordinate for k in the equation g(x) = |x – h| + k.
Learn Dilations of Absolute Value Functions Objective Students identify the effect on the graph of an absolute value function by replacing f (x) with af (x) or f(ax).
MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Students tap on the graph to see the parent function.
TAP Students complete the Check online to determine whether they are ready to move on.
CHECK 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Lesson 4-7 • Absolute Value Functions 269 F.IF.7b, F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Copyright © McGraw-Hill Education Example 6 Dilations of the Form a |x| When a > 1 Describe the dilation in g(x) = 5 __ 2 |x| as it relates to the graph of the parent function.
Since f(x) = |x|, g(x) = a • f(x), where a = 5 __ 2 .
g(x) = 5 __ 2 |x| ⟶ g(x) = 5 __ 2 • f(x) g(x) = 5 __ 2 |x| is a vertical stretch of the graph of the parent graph.
x |x| 5 __ 2 |x| (x, g(x)) -4 |-4| = 4 10 (-4, 10) -2 |-2| = 2 5 (-2, 5) 0 |0| = 0 0 (0, 0) 2 |2| = 2 5 (2, 5) 4 |4| = 4 10 (4, 10) Example 7 Dilations of the Form |ax| Describe the dilation in p(x) = |2x| as it relates to the graph of the parent function.
For p(x) = |2x| , . Since a is inside the absolute value symbols, the input is first multiplied by a. Then, the absolute value of ax is evaluated. Plot the points from the table. Since f(x) =|x|, where . p(x) = |2x| → p(x) = |2x| is a of the graph of the parent graph. Check Match each description of the dilation with its equation. stretched vertically j(x) = | 4 _ 3 x| compressed vertically q(x) = | 1 __ 5 x| stretched horizontally p(x) = 6|x| compressed horizontally g(x) = 5 __ 7 |x| Think About It!
How are a |x| and |ax| evaluated differently? Go Online You can complete an Extra Example online. Go Online You can watch a video to see how to describe dilations of functions.
Watch Out!
Differences in Dilations Although a|x| and |ax| appear to have the same effect on a function, they are evaluated differently and that difference is more apparent when a function is dilated and translated horizontally. For a function with multiple transformations, it is best to first create a table.
x |2x| p(x) (x, p(x)) -4 |2(-4)| = |-8| 8 (-4, 8) -2 |2(-2)| = |-4| 4 (-2, 4) 0 |2(0)| = |0| 0 (0, 0) 2 |2(2)| = |4| 4 (2, 4) 4 |2(4)| = |8| 8 (4, 8) y x O p(x) f(x) y x O f(x) g(x) 270 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: In a|x|, the absolute value of the input is evaluated before multiplying by a. In |ax|, the input is first multiplied by a and then the absolute value is evaluated.
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Example 7 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students will move through the slides to see how to graph a dilation of an absolute value function.
TAP 270 Module 4 • Linear and Nonlinear Functions F.IF.7b, F.BF.3 Example 6 Dilations of the Form a|x| When a > 1 MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the graphs and equations of the dilated function and the parent function in this example.
Questions for Mathematical Discourse AL Is |a| greater than 1 or between 0 and 1? Why? Sample answer: a is greater than 1 because a = 5 __ 2 , and 5 __ 2 > 1.
OL What kind of dilation does this represent? Explain. It is a vertical stretch by a factor of 5 __ 2 . Sample answer: The 5 __ 2 is outside of the absolute value symbols and it is greater than 1.
BL How would the function be different if it was a horizontal compression where a = 5 __ 2 ? Sample answer: The function would be g(x) = | 5 __ 2 x|.
Example 7 Dilations of the Form |ax| MP MP Teaching the Mathematical Practices 3 Construct Arguments In this example, students will use stated assumptions, definitions, and previously established results to construct an argument.
Questions for Mathematical Discourse AL When the absolute value function is in the form f (x) = |ax|, what will be the effect of a? Sample answer: The graph will be horizontally stretched or compressed. OL How would the transformation have changed if the function was p (x) = | 1 __ 2 x| ? Sample answer: It would be a horizontal stretch instead of a compression.
BL What would be an equivalent vertical dilation? Sample answer: a vertical stretch, p(x) = 2|x| Interactive Presentation Copyright © McGraw-Hill Education Example 8 Dilations When 0 < a < 1 Describe the dilation in j(x) = 1 __ 3 |x| as it relates to the graph of the parent function.
For j(x) = 1 __ 3 |x|, . Because a is outside the absolute value signs, the absolute value of the input is evaluated first. Then, the function is multiplied by a. Plot the points from the table. Because f(x) = |x| , where . j(x) = 1 __ 3 |x|→ j(x) = 1 __ 3 |x| is a of the graph of the parent function.
Check Write an equation for each graph shown. Learn Reflections of Absolute Value Functions The graph of -a |x| appears to be flipped upside down compared to a |x| , and they are symmetric about the x-axis. Key Concept • Reflections of Absolute Value Functions Across the x-axis The graph of -af(x) is the reflection of the graph of af(x) = a|x| across the x-axis. When the only transformation occurring is a reflection or a dilation and reflection, the graphs of f(ax) and f(-ax) appear the same.
Key Concept • Reflections of Absolute Value Functions Across the y-axis The graph of f(-ax) is the reflection of the graph of f(ax) = |ax| across the y-axis. x |x| 1 __ 3 |x| (x, j(x)) -6 |-6| = 6 2 (-6, 2) -3 |-3| = 3 1 (-3, 1) 0 |0| = 0 0 (0, 0) 3 |3| = 3 1 (3, 1) 6 |6| = 6 2 (6, 2) y x O j(x) f(x) Go Online You can complete an Extra Example online. Go Online You can watch a video to see how to describe reflections of functions. Think About It!
Why would g(x) = |-2x| and j(x) = |2x| appear to be the same graphs?
y x O y x O Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-7 • Absolute Value Functions 271 j(x) = a · f( x) vertical compression j(x) = 1 __ 3 · f( x) a = 1 __ 3 a = 1 __ 3 Sample answer: Because the absolute values of -2x and 2x are the same nonnegative numbers, the graphs are the same. j(x) = 0.125|x| p(x) = |6x| 267_280_HSM_NA_S_A1M04_L07_662599.indd Page 271 11/10/19 2:51 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 8 Students move through the steps to see how the given function relates to the parent function.
TAP 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Students complete the Check online to determine whether they are ready to move on.
CHECK Lesson 4-7 • Absolute Value Functions 271 F.IF.7b, F.BF.3 Example 8 Dilations When 0 < a < 1 MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the graph and its equation used in this example.
Questions for Mathematical Discourse AL Looking at only the equation, how do you know this is a vertical dilation and not a horizontal dilation? Sample answer: The 1 __ 3 is being multiplied on the outside of the function, not with x.
OL How would the dilation change if the function were j (x) = 3|x|? Sample answer: It would be a vertical stretch by a factor of 3.
BL How would this function change if it was a horizontal stretch where a = 1 __ 3 ? The function would be j (x) = | 1 __ 3 x|.
Learn Reflections of Absolute Value Functions Objective Students identify the effect on the graph of an absolute value function by replacing f (x) with –af (x) or f(-ax).
MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
2 EXPLORE AND DEVELOP Copyright © McGraw-Hill Education Example 9 Graphs of Reflections with Transformations Describe how the graph of j(x) = − |x + 3| + 5 is related to the graph of the parent function. x |x + 3x + 3x + 3| + 5 (x, j(x)) -5 |-5 + 3| = |-2| = 2 -2 -2 + 5 = 3 (-5, 3) -4 |-4 + 3| = |-1| = 1 -1 -1 + 5 = 4 (-4, 4) -3 |-3 + 3| = |0| = 0 0 0 + 5 = 5 (-3, 5) -2 |-2 + 3| = |1| = 1 -1 -1 + 5 = 4 (-2, 4) -1 |-1 + 3| = |2| = 2 -2 -2 + 5 = 3 (-1, 3) First, the absolute value of x + 3 is evaluated. Then, the function is multiplied by -1 · a. Finally, 5 is added to the function. Plot the points from the table. Because f(x) = |x|, where . j(x) = - |x + 3| + 5→ j(x) = - |x + 3| + 5 is the graph of the parent function reflected across the and translated 3 units and 5 units . Go Online You can complete an Extra Example online.
y x O f(x) j(x) −4 −2 −8−6 8 6 4 2 4 6 2 8 −4 −6 −8 Example 10 Graphs of y = −a |x| Describe how the graph of q(x) = - 3 __ 4 |x| is related to the graph of the parent function. First, the absolute value of x is evaluated. Then, the function is multiplied by -1 · a. Plot the points from the table. Because f(x) = |x| , where .
q(x) = - 3 _ 4 |x| → q(x) = - 3 _ 4 |x| is the graph of the parent function reflected across the and . x |x| - 3 __ 4 |x| (x, q(x)) -8 |-8| = 8 -6 (-8, -6) -4 |-4| = 4 -3 (-4, -3) 0 |0| = 0 0 (0, 0) 4 |4| = 4 3 (4, 3) 8 |8| = 8 6 (8, 6) y x O q(x) f(x) 272 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE j(x) = −1 · a · f(x) a = 1, h = −3, and k = 5 j(x) = -1 · f(x + 3) + 5 up left x-axis x-axis q(x) = -1 · a · f(x) vertically compressed q(x) = - 3 __ 4 f(x) a =- 3 __ 4 267_280_HSM_NA_S_A1M04_L07_662599.indd Page 272 11/10/19 2:51 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 9 Graphs of Reflections with Transformations MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the graph of the reflected function and the graph of the parent function used in this example.
Questions for Mathematical Discourse AL How do you know whether there is a horizontal translation? There is a 3 being added to x inside the absolute value symbols. OL What is the effect of the negative in front of the absolute value symbols? It reflects the graph across the x-axis.
BL Why do you need to add 3 and take the absolute value before multiplying by -1? Sample answer: When evaluating to find the coordinates, you have to use the order of operations. In this case, you add 3 first because it is the operation inside the parentheses or grouping symbols.
Common Error Remind students that the order in which they perform the operations when evaluating the function is important. Tell students that when creating the table, they must first add 3, then take the absolute value, then multiply by –1, then add 5.
Example 10 Graphs of y = –a|x| Questions for Mathematical Discourse AL How does the rule for q(x) compare to the rule for the parent function? The rule for q(x) is the rule for the parent function multiplied by - 3 __ 4 . OL How do you expect the vertex of q(x) to compare to the vertex of the parent function? Explain. Sample answer: They will be the same because q(x) has not been translated.
BL The point (12, 12) lies on the graph of the parent function. To what point does this map to on the graph of q(x)? (12, -9) Common Error Students may have difficulty seeing how the graph of q(x) is related to the graph of the parent function. For these students, you may want to show the transformation in two different steps, first dilating the graph by a factor of 3 __ 4 , and then reflecting the resulting graph across the x–axis.
Interactive Presentation Students move through the slides to see how a given function is related to the parent function.
TAP Example 10 DIFFERENTIATE Enrichment Activity BL Give students the function f (x) = -|x - 4| - 2. Have students create a step-by-step list of instructions for how to graph this function. Then have them graph the function.
272 Module 4 • Linear and Nonlinear Functions F.IF.7b, F.BF.3 Interactive Presentation Example 11 Graphs of y = |-ax| MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Questions for Mathematical Discourse AL What is the coefficient of x? -4 OL Looking at only the equation, what type of transformations does this function represent? a horizontal compression and a reflection across the y-axis BL How would this function be different if it was a vertical stretch where a = 4 and a reflection across the x-axis? The function would be f (x ) = -4|x |.
Common Error Some students may think that this function is equivalent to f(x) = -|4x|. Have them create a table of values for both functions so that they can see that the two functions produce different sets of ordered pairs.
Learn Transformations of Absolute Value Functions Objective Students graph absolute value functions by interpreting constants within the equation or by making a table of values.
MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the graph of the transformed functions and the graph of the parent function used in this Learn.
Common Misconception Some students may think that translations should be applied before dilations and reflections. Use an example, such as f (x) = -2|x - 3| + 4, to show students that if they apply the vertical translation before the dilation and reflection, the resulting graph is not the same as when the transformations are applied in the correct order, with the vertical translation as the last transformation.
1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 11 Students move through the slides to see how a given function is related to the parent function.
TAP Students complete the Check online to determine whether they are ready to move on.
CHECK DIFFERENTIATE Enrichment Activity BL Have students work with a partner to create a poster showing examples of graphs that represent dilations of the graph of the parent function, including vertical and horizontal compressions and stretches, and have them use arrows to illustrate the stretch and compression. Have them also provide a description of each transformation.
Copyright © McGraw-Hill Education Go Online You can complete an Extra Example online.
Example 11 Graphs of y = |-ax| Describe how the graph of g(x) = |−4x| is related to the graph of the parent function.
First, the input is multiplied by - 1 · a. Then the absolute value of - ax is evaluated. Because f(x) = |x|, where . g(x) = |-4x| → g(x) = |-4x| is the graph of the parent function reflected across the and . y x O g(x) f(x) Think About It!
Describe how the graph of y = |-ax| is related to the parent function.
Why does there appear to be no reflection for the graph of y = |−ax| ?
Learn Transformations of Absolute Value Functions You can use the equation of a function to understand the behavior of the function. Because the constants a, h, and k affect the function in different ways, they can help develop an accurate graph of the function. Concept Summary Transformations of Graphs of Absolute Value Functions g(x) = a|x - h| + k Horizontal Translation, h If , the graph of f(x) = |x| is translated units .
If h < 0, the graph of f(x) = |x| is translated |h| units left.
y x O h > 0 h < 0 Vertical Translation, k If , the graph of f(x) = |x| is translated units .
If k < 0, the graph of f(x) = |x| is translated |k| units down.
y x O k > 0 k < 0 Reflection, a If a > 0, the graph opens up.
If , the graph opens down.
y x O a > 0 a < 0 Dilation, a If , the graph of f(x) = |x| is stretched vertically.
If 0 < |a| < 1, the graph is compressed vertically.
y x O a > 1 0 < a < 1 Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-7 • Absolute Value Functions 273 a = 4 g(x) = f(−1 · a · x) g(x) = f(−1 · 4 · x) y-axis horizontally compressed Sample answer: The graph is being reflected across the y-axis as a result of multiplying x by -a. There is a reflection occurring, but the final graph appears the same as y = |ax| .
Sample answer: Because x is multiplied by -a before the absolute value is evaluated, the graph would be stretched or compressed, depending on the value of a, but not reflected across the x-axis.
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Lesson 4-7 • Absolute Value Functions 273 F.IF.7b, F.BF.3 Interactive Presentation 2 EXPLORE AND DEVELOP Copyright © McGraw-Hill Education Example 12 Graph an Absolute Value Function with Multiple Translations Graph g(x) = |x + 1| − 4. State the domain and range.
a = 1 The graph is not reflected or dilated in relation to the parent function.
y x O h = -1 The graph is 1 unit left from the parent function.
k = -4 The graph is translated units down from the parent function.
y x O The graph of g(x) = |x + 1| - 4 is the graph of the parent function translated 1 unit left and 4 units down without dilation or reflection. The domain is . The range is . Go Online You can complete an Extra Example online.
y x O g(x) f(x) Go Online You can watch a video to see how to graph a transformed absolute value function.
Watch Out!
Dilations and Translations Don’t assume that j(x) = 2 |x − 5| + 1 and p(x) = |2x - 5| + 1 are the same graph. Functions are evaluated differently depending on whether a is inside or outside the absolute value symbols. It might be best to create a table to generate an accurate graph.
Choose the phrase that best describes how each parameter affects the graph of g(x) = -5|x - 2| + 3 in relation to the parent function.
-5 2 3 Translates right Translates up Stretches vertically only Compresses vertically only Translates left Translates down Reflects and compresses vertically Reflects and stretches vertically 274 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE translated 4 all real numbers all real numbers greater than or equal to -4 Reflects and stretches vertically Translates right Translates up 267_280_HSM_NA_S_A1M04_L07_662599.indd Page 274 5/9/19 3:29 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Students tap on each marker to analyze the parameters in the function.
TAP 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Example 12 274 Module 4 • Linear and Nonlinear Functions F.IF.7b, F.BF.3 Example 12 Graph an Absolute Value Function with Multiple Translations MP MP Teaching the Mathematical Practices 7 Use Structure Helps students to use the structure of the transformed function to identify the transformations in g(x) and graph g(x).
Questions for Mathematical Discourse AL Looking at only the equation, what transformations occur in g(x)? a horizontal translation 1 unit to the left and a vertical translation 4 units down OL How do you know that there is no reflection in this transformation? Sample answer: There are no negative coefficients in the function.
BL How would the function be different if it also represented a reflection over the x-axis? Sample answer: The function would be f (x ) = -|x + 1| - 4.
Interactive Presentation Example 13 Graph an Absolute Value Function with Translations and Dilation MP MP Teaching the Mathematical Practices 1 Explain Correspondences Encourage students to explain the relationships between the equations and graphs of the transformed function and the parent function.
Questions for Mathematical Discourse AL What types of transformations occur in j(x)? a horizontal compression and a horizontal shift OL What is the vertex of the graph of j(x)? (2, 0) BL How could the Distributive Property help explain the horizontal shift 2 units to the right? Sample answer: If we apply the Distributive Property to factor the expression inside the absolute value function, we get |3(x - 2)|. This shows that we would first perform a translation of 2 units to the right, then a horizontal compression of 3. Example 14 Graph an Absolute Value Function with Translations and Reflection MP MP Teaching the Mathematical Practices 6 Communicate Precisely Encourage students to routinely write or explain their solution methods. Point out that they should use clear definitions when they discuss their solutions with others.
Questions for Mathematical Discourse AL Will the graph open up or down? How do you know? Down; sample answer: There is a negative sign in front of the absolute value symbols.
OL What types of transformations occur in p(x)? a horizontal translation 3 units to the right, a reflection across the x-axis, and a vertical translation 5 units up BL How would the function be different if the graph had been translated 3 units to the right and then reflected over the y-axis instead of over the x-axis? The function would be f(x) = |-x - 3| + 5.
Example 13 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Copyright © McGraw-Hill Education Think About It!
How is the vertical translation k of an absolute value function related to its range? Go Online You can complete an Extra Example online.
Example 13 Graph an Absolute Value Function with Translations and Dilation Graph j(x) = |3x - 6|. State the domain and range.
Because a is inside the absolute value symbols, the effect of h on the translation changes.
Evaluate the function for several values of x to find points on the graph.
x (x, j(x)) 0 (0, 6) 1 (1, 3) 2 (2, 0) 3 (3, 3) 4 (4, 6) The graph of j(x) = |3x - 6| is the graph of the parent function compressed horizontally and translated 2 units right.
The domain is . The range is .
Example 14 Graph an Absolute Value Function with Translations and Reflection Graph p(x) = -|x - 3| + 5. State the domain and range.
In p(x) = -|x - 3| + 5, the parent function is reflected across the x-axis because the absolute value is being multiplied by -1. The function is then translated 3 units right.
Finally, the function is translated 5 units up.
p(x) = -|x - 3| + 5 is the graph of the parent function translated 3 units and 5 units and reflected across the . The domain is . The range is .
y x O f(x) j(x) x O f(x) p(x) y Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-7 • Absolute Value Functions 275 all real numbers all real numbers greater than or equal to 0 all real numbers less than or equal to 5 all real numbers right up x-axis Sample answer: Since a vertical translation affects the location of the minimum or maximum point, the range will be greater than or equal to or less than or equal to the value of k, depending upon whether the function has been reflected.
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Students move through the steps to graph the function.
TAP Lesson 4-7 • Absolute Value Functions 275 F.IF.7b, F.BF.3 Interactive Presentation Copyright © McGraw-Hill Education Example 15 Apply Graphs of Absolute Value Functions BUILDINGS Determine an absolute value function that models the shape of The Palace of Peace and Reconciliation.
To write the equation for the absolute value function, we must determine the values of a, h, and k in f(x) = a |x - h| + k from the graph. If we consider the absolute value as a piecewise function, we can find the slope of one side of the graph to determine the value of a.
Because this function opens downward, the graph is a reflection of the parent graph across the x-axis. So we know that the a-value in the equation should be negative. m = y2 - y1 x2 - x1 The Slope Formula = (0, 62) = (x1, y1) and (31, 0) = (x2, y2) = or Next, notice that the vertex is not located at the origin. It has been translated. The absolute value function is not shifted left or right, but has been translated 62 units up from the origin. y = a = −2, h = 0, k = 2 y = Simplify.
So, y = models the shape of The Palace of Peace and Reconciliation.
Check GLASS PRODUCTION Certain types of glass heat and cool at a nearly constant rate when they are melted to create new glass products. Use the graph to determine the equation that represents this process.
y = |x − | + Go Online to practice what you’ve learned about graphing special functions in the Put It All Together over Lessons 4–6 through 4–7.
y x O (0, 62) (−31, 0) (31, 0) 4 8 12 16 20 0 200 400 600 800 1000 1200 Temperature (°C) Time (hours) (0, 29) (18, 29) (9, 1100) Go Online You can complete an Extra Example online.
Dmitry Chulov/123RF 276 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-7A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 0 - 62 _ 31 - 0 - 2 -2 |x| + 62 −119 9 1100 -2 |x - 0| + 62 -2 |x| + 62 - 62 ___ 31 267280_HSM_NA_S_A1M04_L07_662599.indd Page 276 04/10/19 4:00 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Example 15 Example 15 Apply Graphs of Absolute Value Functions MP MP Teaching the Mathematical Practices 4 Apply Mathematics Students will explore how to use an absolute value function to model the shape of a building. They will learn how to use the physical attributes of the building to calculate the parameters of the function.
Questions for Mathematical Discourse AL How do you know that the value of a will be a negative number? Sample answer: The shape of the building is a V that opens down.
OL How do you know that the value of k will be 62? Sample answer: The vertex of the building is 62 units above the origin.
BL Why is it important to find the slope of the sides of the building? Sample answer: The slope tells you if there is a vertical or horizontal stretch or compression.
Common Error After studying the photo, some students may try to incorporate a parameter representing a horizontal translation of 31 units. Help students to see that the diagram shows that the building is symmetric with respect to the y-axis, so there is no horizontal translation. Explain that the purpose of the marked points on the x-axis is for determining the dilation.
Exit Ticket Recommended Use At the end of class, have students respond to the Exit Ticket prompt using a separate piece of paper. Have students hand you their responses as they leave the room.
Alternate Use At the end of class, have students respond to the Exit Ticket prompt verbally or by using a mini-whiteboard. Have students hold up their whiteboards so that you can see all student responses. Tap to reveal the answer when most or all students have completed the Exit Ticket. 2 EXPLORE AND DEVELOP Students enter the values to find the slope of one side of the function.
TYPE a Students complete the Check online to determine whether they are ready to move on.
CHECK 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION 276 Module 4 • Linear and Nonlinear Functions F.IF.7b, F.BF.3 Practice and Homework Suggested Assignments Use the table below to select appropriate exercises.
DOK Topic Exercises 1, 2 exercises that mirror the examples 1–35 2 exercises that use a variety of skills from this lesson 26–42 2 exercises that extend concepts learned in this lesson to new contexts 43–48 3 exercises that emphasize higher-order and critical thinking skills 49–52 ASSESS AND DIFFERENTIATE Use the data from the Checks to determine whether to provide resources for extension, remediation, or intervention.
IF students score 90% or more on the Checks, THEN assign: BL • Practice, Exercises 1–47 odd, 49–52 • Extension: Parametric Equations • Absolute Value Functions IF students score 66%–89% on the Checks, THEN assign: OL • Practice, Exercises 1–51 odd • Remediation, Review Resources: Absolute Value and Distance • Personal Tutors • Extra Examples 1–15 • Plotting and Comparing Signed Numbers IF students score 65% or less on the Checks, THEN assign: AL • Practice, Exercises 1–35 odd • Remediation, Review Resources: Absolute Value and Distance • Quick Review Math Handbook: Special Functions • ArriveMATH Take Another Look • Plotting and Comparing Signed Numbers Answers 19. The graph of g(x) is a reflection of the parent function across the x-axis and a vertical stretch. 20. The graph of g(x) is a reflection of the parent function across the x-axis and translated 2 units down. 21. The graph of g(x) is a reflection of the parent function across the y-axis and a horizontal stretch. 22. The graph of g(x) is a reflection of the parent function across the x-axis and translated 7 units right and 3 units up. 23. The graph of g(x) is a reflection of the parent function across the y-axis and a horizontal compression. 24. The graph of g(x) is a reflection of the parent function across the x-axis and a vertical compression.
Copyright © McGraw-Hill Education x O y x O y x O y y x O x O y x O y Name Period Date Practice Go Online You can complete your homework online.
Examples 1 through 3 Describe the translation in g(x) as it relates to the graph of the parent function. 1. g(x) = |x| - 5 2. g(x) = |x + 6| 3. g(x) = |x - 2| + 7 4. g(x) = |x + 1| - 3 5. g(x) = |x| + 1 6. g(x) = |x - 8| Examples 4 and 5 Use the graph of the function to write its equation. 7. 8. 9. 10. 11. 12. Examples 6 through 8 Describe the dilation in g(x) as it relates to the graph of the parent function. 13. g(x) = 2 __ 5 |x| 14. g(x) = |0.7x| 15. g(x) = 1.3 |x| 16. g(x) = |3x| 17. g(x) = | 1 __ 6 x| 18. g(x) = 5 __ 4 |x| Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-7 • Absolute Value Functions 277 The graph of g(x) is the The graph of g(x) is the The graph of g(x) is the parent function translated parent function translated parent function translated 5 units down. 6 units left. 2 units right and 7 units up. The graph of g(x) is a The graph of g(x) is a The graph of g(x) is a vertical compression horizontal stretch vertical stretch of the parent function. of the parent function. of the parent function. f(x) = |x + 5| f(x) = |x| + 1 f(x) = |x - 4| - 2 f(x) = |x + 2| f(x) = |x + 1| - 2 f(x) = |x| – 3 The graph of g(x) is the The graph of g(x) is the The graph of g(x) is the parent function translated parent function translated parent function translated 1 unit left and 3 units down. 1 unit up. 8 units right. The graph of g(x) is a The graph of g(x) is a The graph of g(x) is a horizontal compression horizontal stretch vertical stretch of the parent function. of the parent function. of the parent function 267_280_HSM_NA_S_A1M04_L07_662599.indd Page 277 11/10/19 3:25 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
3 REFLECT AND PRACTICE Copyright © McGraw-Hill Education Examples 9 through 11 Describe how the graph of g(x) is related to the graph of the parent function. 19. g(x) = -3 |x| 20. g(x) = - |x| - 2 21. g(x) = |- 1 _ 4 x| 22. g(x) = - |x - 7| + 3 23. g(x) = |-2x| 24. g(x) = - 2 __ 3 |x| Examples 12 through 14 USE TOOLS Graph each function. State the domain and range. 25. g(x) = |x + 2| + 3 26. g(x) = |2x - 2| + 1 27. f(x) = | 1 __ 2 x - 2| 28. f(x) = |2x - 1| 29. f(x) = 1 __ 2 |x| + 2 30. h(x) = -2 |x - 3| + 2 31. f(x) = -4 |x + 2| - 3 32. g(x) = - 2 __ 3 |x + 6| - 1 33. h(x) = - 3 _ 4 |x - 8| + 1 Example 15 Determine an absolute value function that models each situation. 34. ESCALATORS An escalator travels at a constant speed. The graph models the escalator’s distance, in floors, from the second floor x seconds after leaving the ground floor. −40 −20 40 20 4 6 2 −4 −6 −2 y x O 35. TRAVEL The graph models the distance, in miles, a car traveling from Chicago, Illinois is from Annapolis, Maryland, where x is the number of hours since the car departed from Chicago, Illinois. 4 8 12 100 200 150 50 −50 −100 y x O Mixed Exercises MODELING Graph each function. State the domain and range. Describe how each graph is related to its parent graph. 36. f(x) = -4 |x - 2| + 3 37. f(x) = |2x| 38. f(x) = |2x + 5| 278 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 19–24. See margin.
25–33. See margin.
y = 0.1|x + 20| y = 65|10 − x| 36–38. See Mod 4 Answer Appendix.
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1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Lesson 4-7 • Absolute Value Functions 277-278 F.IF.7b, F.BF.3 3 REFLECT AND PRACTICE 1 CONCEPTUAL UNDERSTANDING 2 FLUENCY 3 APPLICATION Answers 25. y x O 26. y x O D = all real numbers, D = all real numbers, R = g(x) ≥ 3 R = g(x) ≥ 1 27. y x O 28. y x O D = all real numbers, D = all real numbers, R = f(x) ≥ 0 R = f(x) ≥ 0 29. y x O 30. y x O D = all real numbers, D = all real numbers, R = f(x) ≥ 2 R = h(x) ≤ 2 31. y x O 32. y x O −4 −6 2 4 2 −4 −6 −8 −10 −12 −8 −10 −12 −14 D = all real numbers, D = all real numbers, R = f(x) ≤ -3 R = g(x) ≤ -1 33. h(x) x O 2 4 6 8 2 4 6 8 10 12 14 −4 −6 −8 D = all real numbers, R = h(x) ≤ 1 Copyright © McGraw-Hill Education Name Period Date Use the graph of the function to write its equation. 39. 40. 41. 42. 43. SUNFLOWER SEEDS A company produces and sells bags of sunflower seeds. A medium-sized bag of sunflower seeds must contain 16 ounces of seeds. If the amount of sunflower seeds s in the medium-sized bag differs from the desired 16 ounces by more than x, the bag cannot be delivered to companies to be sold. Write an equation that can be used to find the highest and lowest amounts of sunflower seeds in a medium-sized bag. 44. REASONING The function y = 5 __ 4 |x - 5| models a car’s distance in miles from a parking lot after x minutes. Graph the function. After how many minutes will the car reach the parking lot? 45. STATE YOUR ASSUMPTION A track coach set up an agility drill for members of the track team. According to the coach, 21.7 seconds is the target time to complete the agility drill. If the time differs from the desired 21.7 seconds by more than x, the track coach may require members of the track team to change their training. Write an equation that can be used to find the fastest and slowest times members of the track team can complete the agility drill so that their training does not have to change. If x = 3.2, what can you assume about the range of times the coach wants the members of the track team to complete the agility drill? Solve your equation for x = 3.2 and use the results to justify your assumption. 46. SCUBA DIVING The function y = 3 |x - 12| - 36 models a scuba diver’s elevation in feet compared to sea level after x minutes. Graph the function. How far below sea level is the scuba diver at the deepest point in their dive? y x O y x O −12 −8 −4 4 −4 4 8 y x O y x O Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Lesson 4-7 • Absolute Value Functions 279 x = |s - 16| f(x) = | -x - 3| f(x) = | -3x - 5| f(x) = | 1 __ 3 x + 2| f(x) = 1 __ 3 |x| + 4 x = |t - 21.7|; The range of times is twice the value of x, 3.2(2) = 6.4 s; The solution to the equation is 24.9 and 18.5, which has a range of 24.9 - 18.5 = 6.4 s.
36 feet below sea level 10 20 30 40 50 0 10 20 30 40 50 Distance from Parking Lot (mi) Time (min) O Time (min) Elevation (ft) 6 12 18 24 30 3 6 9 12 15 18 21 24 27 30 −18 −12 −24 −30 5 minutes 267_280_HSM_NA_S_A1M04_L07_662599.indd Page 279 11/10/19 3:26 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Copyright © McGraw-Hill Education 47. MANUFACTURING A manufacturing company produces boxes of cereal. A small box of cereal must have 12 ounces. If the amount of cereal b in a small box differs from the desired 12 ounces by more than x, the box cannot be shipped for selling. Write an equation that can be used to find the highest and lowest amounts of cereal in a small box. 48. STRUCTURE Amelia is competing in a bicycle race. The race is along a circular path. She is 6 miles from the start line. She is approaching the start line at a speed of 0.2 mile per minute. After Amelia reaches the start line, she continues at the same speed, taking another lap around the track.
a. Organize the information into a table. Include a row for time in minutes x, and a row for distance from start line f(x). b. Draw a graph to represent Amelia’s distance from the start line. 49. WRITE Use transformations to describe how the graph of h(x) = - |x + 2| – 3 is related to the graph of the parent absolute value function. 50. ANALYZE On a straight highway, the town of Garvey is located at mile marker 200. A car is located at mile marker x and is traveling at an average speed of 50 miles per hour.
a. Write a function T(x) that gives the time, in hours, it will take the car to reach Garvey. Then graph the function on the coordinate plane. b. Does the graph have a maximum or minimum? If so, name it and describe what it represents in the context of the problem. 51. PERSEVERE Write the equation y = |x - 3| + 2 as a piecewise-defined function. Then graph the piecewise function. 52. CREATE Write an absolute value function, f(x), that has a domain of all real numbers and a range that is greater than or equal to 4. Be sure your function also includes a dilation of the parent function. Describe how your function relates to the parent absolute value graph. Then graph your function. 280 Module 4 • Linear and Nonlinear Functions Program: Reveal Math Component: Lesson 4-7_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 10 20 30 40 50 1 0 2 3 4 5 6 7 8 9 10 Distance from Start Line (mi) Time (min) x 0 10 20 30 40 50 f(x) 6 4 2 0 2 4 To get the graph of h(x), the parent absolute value function is reflected in the x-axis, then translated 2 units left and 3 units down. T(x) = 1 ___ 50 |x - 200| ; See Mod 4. Answer Appendix for graph. Minimum at (200, 0); If the car is at mile marker 200, the time to reach Garvey is 0 hours.
f(x) = { -x + 5 if x < 3 x - 1 if x ≥ 3 ; See Mod 4. Answer Appendix for graph.
Sample answer: f(x) = 1 __ 3 |x| + 4; The graph of f(x) is the parent function translated 4 units up, and vertically compressed by a factor of 3. See Mod 4. Answer Appendix for graph. Higher-Order Thinking Skills x = |b - 12| 267_280_HSM_NA_S_A1M04_L07_662599.indd Page 280 11/10/19 2:51 AM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
279-280 Module 4 • Linear and Nonlinear Functions F.IF.7b, F.BF.3 Module 4 • Linear and Nonlinear Functions 281 Module 4 • Linear and Nonlinear Functions Review Rate Yourself! Have students return to the Module Opener to rate their understanding of the concepts presented in this module. They should see that their knowledge and skills have increased. After completing the chart, have them respond to the prompts in their Interactive Student Edition and share their responses with a partner. Answering the Essential Question Before answering the Essential Question, have students review their answers to the Essential Question Follow-Up questions found throughout the module.
• Why is it helpful to have different ways to graph linear functions?
• What can you learn about the graph of a linear function by analyzing its equation?
• Why is it important to understand how the structure of a function models a situation?
• If you know that a function is a step function, what do you know about how the elements of the domain are paired with the elements of the range?
Then have them write their answer to the Essential Question in the space provided.
ELL A completed Foldable for this module should include the key concepts related to linear and nonlinear functions. LearnSmart Use LearnSmart as part of your test preparation plan to measure student topic retention. You can create a student assignment in LearnSmart for additional practice on these topics for Linear and Exponential Relationships, Descriptive Statistics, and Quadratic Functions and Modeling. • Interpret Expressions for Functions • Build Linear and Exponential Function Models • Interpret Linear Models • Construct and Compare Linear, Quadratic, and Exponential Models and Solve Problems Copyright © McGraw-Hill Education Module 4 • Linear and Nonlinear Functions Review Essential Question What can a function tell you about the relationship that it represents?
Module Summary Lessons 4-1 through 4-3 Graphing Linear Functions, Rate of Change, and Slope • The graph of an equation represents all of its solutions.
• The x-value of the y-intercept is 0. The y-value of the x-intercept is 0.
• The rate of change is how a quantity is changing with respect to a change in another quantity. If x is the independent variable and y is the dependent variable, then rate of change = change in y __ change in x .
• The slope m of a nonvertical line through any two points can be found using m = y2 - y1 ___ x2 - x1 .
• A line with positive slope slopes upward from left to right. A line with negative slope slopes downward from left to right. A horizontal line has a slope of 0. The slope of a vertical line is undefined.
Lesson 4-4 Transformations of Linear Functions • When a constant k is added to a linear function f(x), the result is a vertical translation.
• When a linear function f(x) is multiplied by a constant a, the result a·f(x) is a vertical dilation.
• When a linear function f(x) is multiplied by -1 before or after the function has been evaluated, the result is a reflection across the x- or y-axis.
Lesson 4-5 Arithmetic Sequences • An arithmetic sequence is a numerical pattern that increases or decreases at a constant rate called the common difference.
• The nth term of an arithmetic sequence with the first term a1 and common difference d is given by an = a1 + (n - 1)d, where n is a positive integer.
Lessons 4-6, 4-7 Special Functions • A piecewise-linear function has a graph that is composed of a number of linear pieces.
• A step function is a type of piecewise-linear function with a graph that is a series of horizontal line segments. • An absolute value function is V-shaped.
Study Organizer Foldables Use your Foldable to review this module. Working with a partner can be helpful. Ask for clarification of concepts as needed.
4-4 4-1-4-3 4-5 4-6-4-7 Module 4 Review • Linear and Nonlinear Functions 281 Program: Reveal Math Component: Module 4 Review_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE Sample answer: It can tell you about the rate of change, whether the relationship is positive or negative, the locations of the x- and y-intercepts, and what points fall on the graph.
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282 Module 4 • Linear and Nonlinear Functions Review and Assessment Options The following online review and assessment resources are available for you to assign to your students. These resources include technology-enhanced questions that are auto-scored, as well as essay questions.
Review Resources Put It All Together: Lessons 4-1 through 4-7 Vocabulary Activity Module Review Assessment Resources Vocabulary Test AL Module Test Form B OL Module Test Form A BL Module Test Form C Performance Task The module-level performance task is available online as a printable document. A scoring rubric is included.
Copyright © McGraw-Hill Education Name Period Date 1. GRAPH Jalyn made a table of how much money she will earn from babysitting. (Lesson 4-1) Hours Babysitting Money Earned 1 5 2 10 3 15 4 20 Use the table to graph the function in the coordinate grid.
2 4 6 0 10 20 Money Earned ($) Hours Babysitting 2. TABLE ITEM What are the missing values in the table that show the points on the graph of f(x) = 2x - 4? (Lesson 4-1) x −2 0 2 4 6 f (x) −8 −4 3. OPEN RESPONSE Mr. Hernandez is draining his pool to have it cleaned. At 8:00 A.M., it had 2000 gallons of water and at 11:00 A.M. it had 500 gallons left to drain. What is the rate of change in the amount of water in the pool? (Lesson 4-2) 4. MULTIPLE CHOICE Find the slope of the graphed line. (Lesson 4-2) f(x) x O (−4, 0) (0, 3) A − 4 _ 3 B − 3 _ 4 C 3 _ 4 D 4 _ 3 5. MULTIPLE CHOICE Determine the slope of the line that passes through the points (4, 10) and (2, 10). (Lesson 4-2) A −1 B 0 C 1 D undefined 6. GRAPH Graph the equation of a line with a slope of −3 and a y-intercept of 2. (Lesson 4-3) 7. MULTIPLE CHOICE What is the slope of the line that passes through (3, 4) and (−7, 4)? (Lesson 4-3) A 0 B undefined C –2 D –10 Test Practice y x O 282 Module 4 Review • Linear and Nonlinear Functions Program: Reveal Math Component: Module 4 Review_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 0 4 8 - 500 gallons/hr 281_284_HSM_NA_S_A1M04_Review_662599.indd Page 282 04/10/19 4:16 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Module 4 • Linear and Nonlinear Functions 283 Question Type Description Exercise(s) Multiple Choice Students select one correct answer.
4, 5, 7, 8, 11, 17 Multi-Select Multiple answers may be correct. Students must select all correct answers.
18 Table Item Students complete a table by entering in the correct values.
2, 15 Graph Students create a graph on an online coordinate plane.
1, 6, 16 Open Response Students construct their own response in the area provided.
3, 9, 10, 12, 13, 14, 19, 20, 21 Copyright © McGraw-Hill Education 8. MULTIPLE CHOICE A teacher buys 100 pencils to keep in her classroom at the beginning of the school year. She allows the students to borrow pencils, but they are not always returned. On average, she loses about 8 pencils a month. Write an equation in slope-intercept form that represents the number of pencils she has left y after a number of x months. (Lesson 4-3) A y = −8x − 100 B y = −8x + 100 C y = 8x + 100 D y = 8x − 100 9. OPEN RESPONSE Name the transformation that changes the slope, or the steepness of a graph. (Lesson 4-4) 10. OPEN RESPONSE Describe the dilation of g(x) = 1 __ 2 (x) as it relates to the graph of the parent function, f(x) = x. (Lesson 4-4) 11. MULTIPLE CHOICE Arjun begins the calendar year with $40 in his bank account. Each week he receives an allowance of $20, half of which he deposits into his bank account. The situation describes an arithmetic sequence. Which function represents the amount in Arjun’s account after n weeks? (Lesson 4-5) A f(n) = 20n + 40 B f(n) = 40n + 20 C f(n) = 40 + 10n D f(n) = 10 + 40n 12. OPEN RESPONSE What number can be used to complete the equation below that describes the nth term of the arithmetic sequence -2, -1.5, -1, 0, 0.5, …? (Lesson 4-5) an = 0.5n - ___ 13. OPEN RESPONSE Write and graph a function to represent the sequence 1, 10, 19, 28, … (Lesson 4-5) an n 5 0 1 2 3 4 5 6 7 8 10 15 20 25 30 35 40 14. OPEN RESPONSE Christa has a box of chocolate candies. The number of chocolates in each row forms an arithmetic sequence, as shown in the table. (Lesson 4-5) Row 1 2 3 4 Number of Chocolates 3 6 9 12 Write an arithmetic function that can be used to find the number of chocolates in each row.
Name Period Date Module 4 Review • Linear and Nonlinear Functions 283 Program: Reveal Math Component: Module 4 Review_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE dilation g(x) is a vertical compression of the parent function by a factor of 1 __ 2 .
2.5 f(n) = 9n − 8 an = 3n 281_284_HSM_NA_S_A1M04_Review_662599.indd Page 283 05/10/19 8:18 PM t /137/GO02573_R3/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
Practice You can use these pages to help your students review module content and prepare for online assessments. Exercises 1–21 mirror the types of questions your students will see on online assessments.
To ensure that students understand the standards, check students’ success on individual exercises.
Standard Lesson(s) Exercise(s) A.CED.2 4-3 8 A.REI.10 4-1 2 F.IF.2 4-6 15 F.IF.6 4-2 3, 4, 5, 7 F.IF.7 4-1, 4-3, 4-4, 4-6, 4-7 1, 6, 16, 21 F.BF.1a 4-5 12 F.BF.3 4-4, 4-7 9, 10, 17, 18, 19, 20 F.LE.2 4-5 11, 13, 14 284 Module 4 • Linear and Nonlinear Functions Copyright © McGraw-Hill Education 15. TABLE ITEM Daniel earns $9 per hour at his job for the first 40 hours he works each week. However, his pay rate increases to $13.50 per hour thereafter. This situation can be represented with the function f(x) = ⎧ ⎨ ⎩ 9x, if x ≤ 40 360 + 13.5(x - 40), if x > 40 Use this function to complete the table with the correct values. (Lesson 4-6) Hours Worked, x Money Earned, f(x) 30 35 315 40 45 427.5 50 16. GRAPH Graph the function f(x) = 2. (Lesson 4-6) y x O 17. MULTIPLE CHOICE Which of the following describes the effect a dilation has upon the graph of the absolute value parent function? (Lesson 4-7) A Flipped across axis B Stretch or compression C Rotated about the origin D Shifted horizontally or vertically 18. MULTI-SELECT Describe the transformation(s) of the function graphed below in relation to the absolute value parent function. Select all that apply. (Lesson 4-7) x O y A Reflected across x-axis B Vertical stretch C Vertical compression D Reflected across y-axis E Translated right 3 F Translated up 3 19. OPEN RESPONSE Describe the graph of g(x) = |x| + 5 in relation to the graph of the absolute value parent function. (Lesson 4-7) 20. OPEN RESPONSE Across which axis is the graph of h(x) = -5|x| reflected? (Lesson 4-7) 21. OPEN RESPONSE Use the graph of the function to write its equation. (Lesson 4-7) y x O Name Period Date 284 Module 4 Review • Linear and Nonlinear Functions Program: Reveal Math Component: Module 4 Review_A1 PDF Pass Vendor: Aptara Grade: 9-12, SE 270 Sample answer: It is translated 5 units up.
x-axis f(x) = -|x − 4| + 3 360 495 281_284_HSM_NA_S_A1M04_Review_662599.indd Page 284 5/10/19 4:37 AM gg-57 /112/GO02573_R2/Reveal_Algebra_Interactive/NA/SE/Vol_1/007_662599_0_P1/Applicatio ...
MODULE 4 ANSWER APPENDIX Lesson 4-3 33. y x O 34. y x 8 4 −8 −4 O −4 −8 4 8 35. y x 8 4 −8 −4 O −4 −8 4 8 36. y x O 40b. 2 8 16 24 32 40 48 4 6 10 12 8 m c 0 40c. Sample answer: Find 13 along the horizontal axis. Move up to the line. The corresponding value along the vertical axis is about 45. So, the cost of watching 13 movies from MovieMania is about $45. 40d. Sample answer: The cost of watching 13 movies from MovieMania is about $45, so divide $45 by 9 to get $5. So, the cost of watch a movie from SuperFlix is about $5. 41b. 4 50 100 150 200 250 300 8 12 20 24 16 x T 0 42b 2 0 1000 2000 3000 4000 5000 6000 4 6 10 12 14 16 8 x y Lesson 4-7 36. f(x) x O D = all real numbers, R = f(x) ≤ 3 The graph of f(x) is a reflection of the parent function across the x-axis, vertically stretched by a factor of 4, and translated 2 units right and 3 units up. 37. f(x) x O 38. f(x) x O D = all real numbers, D = all real numbers, R = f(x) ≥ 0 R = f(x) ≥ 0 The graph of f(x) is the The graph of f(x) is the parent function horizontally parent function horizontally compressed by a factor of 1 2. compressed by a factor of 1 2, and translated 2.5 units left.
Module 4 • Answer Appendix 284a MODULE 4 ANSWER APPENDIX 50a. 100 200 300 400 500 1 0 2 3 4 5 6 7 8 9 10 Time (h) Mile Marker 51. y x O 52. y x O 284b Module 4 • Answer Appendix |
7983 | http://astro.vaporia.com/start/columndensity.html | column density
Astrophysics (Index)About
column density
(measure of the matter through which EMR is passing)
Column density is a measure of the amount of matter along some particular span, such as that along a line-of-sight toward a star. Specifically, it is the amount of matter per unit of area, the area being the cross-section of a tube-shape along the distance. Column mass density is mass per unit area, e.g., g/cm 2 and column number density is the molecule count (specifically, any particles present that the photons could react with) per unit area, e.g., cm-2. These may also be used for some specific particle type or types, e.g., electron column density, for the column number density specifically of free electrons (i.e., not counting other particles).
Column density is equivalent to surface density, each being a measure of the density through one dimension of a volume. The term surface density is generally used for measures through relatively flat volumes whereas column density is generally used for such a measurement through a volume that must traverse some significant distance. The term column density is likely to be used for the case of particles along the line-of-sight to star, whereas surface density, of particles over the course of the thickness of a disk.
(measure,EMR)Further reading:
Referenced by pages:
dispersion measure (DM)
line broadening
Lockman Hole
particle number (N)
pulsar (PSR)
surface density (Σ)
Thomson optical depth (τ T)
tomographyNotes to myself regarding this page
Index |
7984 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3222735/ | Nasopharyngeal cancer mimicking otitic barotrauma in a resource-challenged center: a case report - PMC
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J Med Case Rep
. 2011 Oct 31;5:532. doi: 10.1186/1752-1947-5-532
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Nasopharyngeal cancer mimicking otitic barotrauma in a resource-challenged center: a case report
Adekunle Daniel
Adekunle Daniel
1 Department of Otorhinolaryngology, College of Medicine and University College Hospital, PMB 5116, Queen Elizabeth Road, Ibadan, Oyo-State, Nigeria
Find articles by Adekunle Daniel
1,✉, Ayotunde James Fasunla
Ayotunde James Fasunla
1 Department of Otorhinolaryngology, College of Medicine and University College Hospital, PMB 5116, Queen Elizabeth Road, Ibadan, Oyo-State, Nigeria
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1
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1 Department of Otorhinolaryngology, College of Medicine and University College Hospital, PMB 5116, Queen Elizabeth Road, Ibadan, Oyo-State, Nigeria
✉
Corresponding author.
Received 2011 Jun 28; Accepted 2011 Oct 31; Collection date 2011.
Copyright ©2011 Daniel and Fasunla; licensee BioMed Central Ltd.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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PMCID: PMC3222735 PMID: 22040358
Abstract
Introduction
Nasopharyngeal cancer commonly manifests with cervical lymphadenopathy, recurrent epistaxis and progressive nasal obstruction. Neuro-ophthalmic and otologic manifestations can also occur. Isolated otologic presentations of nasopharyngeal cancer are rare and the diagnosis of nasopharyngeal cancer may not be foremost in the list of differentials.
Case presentation
We present the case of a 29-year-old Nigerian woman with bilateral conductive hearing loss and tinnitus after air travel. There were no other symptoms. The persistence of the symptoms after adequate treatment for otitic barotrauma necessitated re-evaluation, which led to a diagnosis of nasopharyngeal cancer.
Conclusion
Isolated otologic manifestations of nasopharyngeal cancer are rare in regions with low incidence of the disease. There is a need for it to be considered as a possible differential in patients presenting with bilateral serous otitis media.
Introduction
The clinical presentations of nasopharyngeal cancer may sometimes be insidious and nonspecific. They are usually related to the local, regional and distant spread or metastasis of the lesion. They may include cervical lymphadenopathy, nasal blockage, epistaxis, hyponasal speech and otologic and neuro-ophthalmic manifestations . The clinical morphology of the lesion may be infiltrative, ulcerative or exophytic.
The otological manifestations of this disease entity are commonly unilateral Eustachian tube dysfunction, fluid accumulation within the middle ear, conductive hearing loss, otalgia and tinnitus . However, these presentations are not pathognomonic of nasopharyngeal cancer. It is quite uncommon for nasopharyngeal cancer patients to present with only isolated otologic symptoms, especially in regions where the incidence of this disease is low. When they do occur, other more common benign ear diseases that present with similar symptoms are usually considered. A high index of suspicion is required to evaluate the patient for nasopharyngeal cancer as a differential diagnosis. Hence, we report an unexpected presentation of nasopharyngeal cancer, with isolated otologic symptoms, which was initially managed as otitic barotrauma.
Case presentation
A 29-year-old Nigerian woman, who frequently travels by air, presented with a six-month history of persistent bilateral hearing impairment following a flight. She erstwhile had experienced repeated episodes of this symptom, which occurred each time she flew, but there was always complete resolution after a few days following treatment from an outside health facility. There was associated tinnitus but no otalgia, no ear discharge and no sensation of disequilibrium or vertiginous spells. She did not have any nasal blockage, nasal discharge, epistaxis or postnasal drip. There were no throat or neuro-ophthalmic symptoms. She did not complain of neck swelling. There was no history suggestive of exposure to carcinogens.
She had received treatment at peripheral hospitals for barotrauma before presenting to our hospital due to persistence of the symptoms.
Examination revealed a young woman with a nevus on the lobule of her right pinna. Both tympanic membranes were dull with a loss of light reflex. The tuning fork test showed evidence of bilateral conductive hearing loss. No evidence of spontaneous nystagmus was noted. A nasal and oropharyngeal examination revealed essentially normal findings. Indirect laryngoscopy findings appeared normal. Her cranial nerves and both eyes were grossly normal. Examination of her other systems did not reveal any abnormalities.
A pure tone audiogram confirmed the bilateral conductive hearing loss (Figure 1A). Impedance audiometry showed type B curves bilaterally.
Figure 1.
Open in a new tab
Pure tone audiogram. (A)Audiogram of our patient at presentation with evidence of bilateral conductive hearing loss. (B)Audiogram shows improvement in hearing thresholds after commencement of treatment.
A diagnosis of bilateral otitic barotrauma was made. She was treated with nasal decongestants, prophylactic antibiotics and asked to perform the Vasalva maneuver frequently. However, her symptoms still persisted after two weeks. This necessitated a re-evaluation; during examination her tympanic membranes were now hyperemic and bulging. A computerized tomographic (CT) scan of her paranasal sinuses was done, which revealed isodense lesions in both fossae of Rosenmüller with complete occlusion of the openings of the Eustachian tubes bilaterally (Figure 2). Nasopharyngoscopy, which would have been pivotal in reaching a diagnosis, was not done before the CT scan because nasopharyngeal cancer had not been in our list of differentials. She underwent examination of the nasopharynx under general anesthesia and a biopsy of the lesion was performed. The histology revealed an undifferentiated carcinoma of the nasopharynx (World Health Organization type III). She was referred to the clinical oncologist and radiotherapist in our center for treatment. The hearing loss improved after commencement of chemoradiation; a pure tone audiogram thereafter showed socially adequate hearing thresholds in most frequencies (Figure 1B).
Figure 2.
Open in a new tab
Computerized tomography scan of our patient shows an isodense lesion in her nasopharynx.
Discussion
This present study clearly demonstrates a case of bilateral serous otitis media which was the only clinical finding in a patient who was initially thought to have otitic barotrauma. Thorough evaluation after the failure of initial treatment led to a diagnosis of nasopharyngeal cancer. The otologic manifestations of nasopharyngeal cancer are usually unilateral. Bilateral presentation is quite uncommon . Bilateral serous otitis media or Eustachian tube dysfunction as the only clinical manifestation of nasopharyngeal cancer is uncommon and rarely reported in the literature. A high index of suspicion is therefore needed to evaluate patients with bilateral serous otitis media or Eustachian tube dysfunction for possible nasopharyngeal cancer.
The otologic manifestations of nasopharyngeal cancer occur as a result of the sheer tumor bulk within the nasopharynx and paranasopharyngeal space extension [4,5]. These manifestations may include Eustachian tube dysfunction, fluid accumulation within the middle ear (otitis media with effusion), conductive hearing loss, tinnitus and otalgia . These symptoms are usually unilateral and are more common in regions with a high incidence of the disease . It has been postulated that the altered Eustachian tube compliance in these patients is a result of cartilage erosion by the tumor and not necessarily the destruction of the tensor veli palatinus . Bilateral Eustachian tube dysfunction in nasopharyngeal cancer is rarely reported in the literature. It can occur if the tumor grows to obstruct the openings of the Eustachian tubes in the nasopharynx, especially in the exophytic or infiltrative morphological type. In that instance, the otologic presentation will initially be unilateral. In our patient, both ears were simultaneously affected after air travel. Usually, mild conductive hearing loss accompanies otitis media with effusion. However in this patient, the severe bilateral conductive hearing loss may be due to the summative effects of both the sheer bulk of the tumor in the nasopharynx and the otitic barotrauma on the Eustachian tube.
The hidden nature of the nasopharyngeal space poses diagnostic and therapeutic challenges, thus allowing significant spread of the disease before diagnosis . The inclusion of nasopharyngoscopy in the clinical setting has greatly increased early diagnosis of nasopharyngeal cancer with consequently improved prognosis of the disease . This was not done in our patient because nasopharyngeal cancer was not in our list of differentials. In a study by Grandawa et al. of 40 patients with nasopharyngeal carcinoma in north-eastern Nigeria, otologic symptoms were not noted. The clinical profile reported in these patients included cervical lymphadenopathy (72.5%), rhinorrhea (55%) and epistaxis (45%) . However, a study by Iseh et al. of 30 patients in north-western Nigeria reported clinical presentations of deafness and otalgia in 36.3% and 30% of patients, respectively. Other clinical presentations included cervical lymphadenopathy (93.3%), epistaxis (83.3%), nasal obstruction (66.7%), palatal swelling (26.7%), cranial nerve involvement (23.3%) and visual impairment (20%) . A study by Sham et al. of 237 Chinese patients with nasopharyngeal cancer showed that 41% of them had unilateral serous otitis media . This value is quite high and may be related to the fact that nasopharyngeal cancer is seen more commonly among Asians . The true incidence of this disease in Africa, however, is largely unknown: Nwaorgu et al. reported a steady increase in the disease occurrence over the last two decades in Nigeria . Inner ear symptoms, such as vertigo, in nasopharyngeal cancer are rare . In our patient, bilateral hearing impairment and tinnitus were the only presenting symptoms. Nasopharyngeal cancer is unlikely to be easily thought of as a possible diagnosis, especially when the symptoms occur after air travel. Our patient was initially treated for barotitis and only when the symptoms did not improve was she re-evaluated and a diagnosis of nasopharyngeal cancer confirmed.
Otitic barotrauma (barotitis) is a traumatic inflammation of the middle ear occurring as a result of pressure difference between the air in the middle ear and the external atmosphere, developing after ascent or, more usually, descent during air travel. It occurs because of the failure of the Eustachian tube to equilibrate middle ear and atmospheric pressure. It is quite common and presents with ear fullness, otalgia and deafness . Severe cases may result in tympanic membrane perforation and even round window perforation . It is an uncommon differential diagnosis of nasopharyngeal cancer . The treatment of nasopharyngeal carcinoma is chemoradiation. This was the treatment administered to our patient and she has shown remarkable improvement in her clinical condition to date. The observed significant improvement in hearing thresholds in the repeat pure tone audiogram may be a result of the combined effect of both the gross tumor excision during the biopsy and chemoradiation therapy, which might have relieved the Eustachian tube obstruction.
Conclusion
In this case report, it is suggested that isolated bilateral otologic symptoms can be the only or initial manifestation of nasopharyngeal cancer even in regions of low disease incidence. It is therefore recommended that, in cases of bilateral serous otitis media or Eustachian tube dysfunction in an adult, nasopharyngeal cancer should be considered.
Consent
Written informed consent was obtained from the patient for the publication of this case report and any accompanying images. A copy of this consent is available for review by the Editor- in-Chief of this journal.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
DA was the principal investigator, performed the literature search and wrote the manuscript. FAJ assisted in preparing and proofreading the manuscript for intellectual content and gave final approval for the publication. DA and FAJ read and approved the final manuscript and take responsibility for its publication.
Contributor Information
Adekunle Daniel, Email: kunle_d2002@yahoo.com.
Ayotunde James Fasunla, Email: ayofasunla@yahoo.com.
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7985 | https://uomus.edu.iq/img/lectures21/MUCLecture_2024_12842322.pdf | ]ً[اكتب نصا Boolean Expression for a Logic Circuit • To derive the Boolean expression for a given combinational logic circuit, begin at the left-most inputs and work toward the final output, writing the expression for each gate. • Example: Determine the Boolean expression for the following logic circuit • Solution: A(B + CD) 1 2 اسن المادة : تقنيات رقمية : اسن التدريسي م.م علياء دمحم جواد : المرحلة الثانية : السنة الدراسية2023 _ 2024 Simplifying logic circuit: عنوان المحاضرة ]ً[اكتب نصا Constructing a Truth Table for a Logic Circuit • Step2: Putting the Results in Truth Table Format • First, list the sixteen input variable combinations of 1s and 0s in a binary sequence. Next, place a 1 in the output column for each combination of input variables that was determined in the evaluation. Finally, place a 0 in the output column for all other combinations of input variables. These results are shown in the truth table below 3 4 Constructing a Truth Table for a Logic Circuit • Once the Boolean expression for a given logic circuit has been determined, a truth table that shows the output for all possible values of the input variables can be developed. Step1: Evaluating the Expression • For example, to evaluate the expression A(B + CD), find the values of the variables that make the expression equal to 1 using the rules for Boolean addition and multiplication. • Thus, the expression A(B + CD) equals 1 only if A = 1 and B + CD = 1. • Now determine when the B + CD term equals 1. The term B + CD = 1 if either B = 1 or CD = 1 or if both B and CD equal 1. • The term CD = 1 only if C = 1 and D = 1. • Therefore, the expression A(B + CD) = 1 when A = 1 and B = 1 regardless of the values of C and D or when A = 1 and C = 1 and D = 1 regardless of the value of B. And, the expression A(B + CD) = 0 for all other value combinations of the variables. ]ً[اكتب نصا Simplification reduces gates for the same function • The figure below shows that five gates are required to implement the expression AB + A(B + C) + B(B + C) in its original form; however, only two gates are needed for the simplified expression (B + AC ). 5 6 Logic Simplification Using Boolean Algebra Example: Using Boolean algebra techniques, simplify this expression: AB + A(B + C) + B(B + C) Solution: Note: The following is not necessarily the only approach. AB + A(B + C) + B(B + C)= AB + AB + AC + BB + BC = AB + AB + AC + B + BC =AB + AC + B + BC = AB + AC + B = B + AC ]ً[اكتب نصا Simplification reduces gates for the same function Example: Simplify the following Boolean expression: Solution: = = = = = = = (using rule 11 ) 7 8 Simplification reduces gates for the same function Example: Simplify the following Boolean expression Solution: = = = = = = = = = ]ً[اكتب نصا = : Standard Forms of Boolean Expressions • All Boolean expressions can be converted into either of two standard forms: the sum-of-products form or the product-of-sums form. • Standardization makes the evaluation, simplification, and implementation of Boolean expressions much more systematic and easier. 9 10 Simplification reduces gates for the same function Example: Simplify the following Boolean expression Solution: = (By applying DeMorgan’s theorem ) = (By applying DeMorgan’s theorem to each term in the parentheses) = = = ]ً[اكتب نصا . AND/OR Implementation of an SOP Expression • Implementing an SOP expression simply requires ORing the outputs of two or more AND gates. • A product term is produced by an AND operation, and the sum (addition) of two or more product terms is produced by an OR operation. • Therefore, an SOP expression can be implemented by AND-OR logic in which the outputs of a number (equal to the number of product terms in the expression) of AND gates connect to the inputs of an OR gate, as shown in the figure below for the expression 11 12 The Sum-of-Products (SOP) Form • When two or more product terms are summed by Boolean addition, the resulting expression is a sum-of-products (SOP). Some examples are: • Also, an SOP expression can contain a single-variable term, as in Note: • In an SOP expression, a single overbar cannot extend over more than one variable; however, more than one variable in a term can have an overbar. • For example, an SOP expression can have the term but not ]ً[اكتب نصا Conversion of a General Expression to SOP Form • Any logic expression can be changed into SOP form by applying Boolean algebra techniques. • For example, the expression A(B + CD) can be converted to SOP form by applying the distributive law: • Example: Convert each of the following Boolean expressions to SOP form: 13 14 NAND/NAND Implementation of an SOP Expression • NAND gates can be used to implement an SOP expression. • By using only NAND gates, an AND/OR function can be accomplished, as illustrated in figure (a) below to implement . (a) (b) • The first level of NAND gates feed into a NAND gate that acts as a negative-OR gate. The NAND and negative-OR inversions cancel and the result is effectively an AND/OR circuit. ]ً[اكتب نصا Converting Product Terms to Standard SOP • Each product term in an SOP expression that does not contain all the variables in the domain can be expanded to standard form to include all variables in the domain and their complements. • As stated in the following steps, a nonstandard SOP expression is converted into standard form using Boolean algebra rule 6 (A + A = 1). • Step 1: Multiply each nonstandard product term by a term made up of the sum of a missing variable and its complement (you can multiply anything by 1 without changing its value). • Step 2: Repeat Step 1 until all resulting product terms contain all variables in the domain in either complemented or uncomplemented form. • In converting a product term to standard form, the number of product terms is doubled for each missing variable. 15 16 The Standard SOP Form • A standard SOP expression is one in which all the variables in the domain appear in each product term in the expression. • For example, the expression is not in the standard SOP while is in a standard SOP expression. • Note: has a domain made up of the variables A, B, C, and D. However, the complete set of variables in the domain is not represented in the first two terms of the expression; that is, D or D is missing from the first term and C or C is missing from the second term. ]ً[اكتب نصا The Product-of-Sums (POS) Form • When two or more sum terms are multiplied, the resulting expression is a product-of-sums (POS). Some examples are • A POS expression can contain a single-variable term, as in In a POS expression, a single overbar cannot extend over more than one variable; however, more than one variable in a term can have an overbar. • For example, a POS expression can have the term but not 17 18 Converting Product Terms to Standard SOP • Example: Convert the following Boolean expression into standard SOP form: • Solution: = = = = + + + + + + + + ]ً[اكتب نصا • The Standard POS Form • A standard POS expression is one in which all the variables in the domain appear in each sum term in the expression. For example, is a standard POS expression. • Converting a Sum Term to Standard POS • A nonstandard POS expression is converted into standard form by using the following steps: • Step1: Add to each nonstandard product term a term made up of the product of the missing variable and its complement . • Step2: Apply rule 12: A + BC = (A + B)(A + C) • Step3: Repeat Step 1 until all resulting sum terms contain all variables in the domain in either complemented or uncomplemented form. 19 20 Implementation of a POS Expression • Implementing a POS expression simply requires ANDing the outputs of two or more OR gates. A sum term is produced by an OR operation, and the product of two or more sum terms is produced by an AND operation. • For example, the implementation of the POS expression (A + B)(B + C + D)(A + C) is shown in the figure below: ]ً[اكتب نصا Converting Standard SOP to Standard POS • Step 1: Evaluate each product term in the SOP expression. That is, determine the binary numbers that represent the product terms. • Step 2: Determine all of the binary numbers not included in the evaluation in Step 1. • Step 3: Write the equivalent sum term for each binary number from Step 2 and express in POS form. Note: Using a similar procedure, you can go from POS to SOP. 21 22 Converting a Sum Term to Standard POS • Example: Convert the following Boolean expression into standard POS form: • Solution ]ً[اكتب نصا Boolean Expressions and Truth Tables • A truth table is simply a list of the possible combinations of input variable values and the corresponding output values (1 or 0). • The truth table is a common way of presenting the logical operation of a circuit. • Standard SOP or POS expressions can be determined from a truth table. Converting SOP Expressions to Truth Table Format Step1: construct a truth table by listing all possible combinations of binary values of the variables in the expression. Step2: convert the SOP expression to standard form if it is not already. Step3: place a 1 in the output column (X) for each binary value that makes the standard SOP expression a 1 and place a 0 for all the remaining binary values. • Note: an SOP expression is equal to 1 only if at least one of the product terms is equal to 1 23 24 Converting Standard SOP to Standard POS • Example: Convert the following SOP expression to an equivalent POS expression: • Solution: The evaluation is as follows: 000 + 010 + 011 + 101 + 111 • Since there are three variables in the domain of this expression, there are a total of eight (23) possible combinations. The SOP expression contains five of these combinations, so the POS must contain the other three which are 001, 100, and 110. • Remember, these are the binary values that make the sum term 0. The equivalent POS expression is ]ً[اكتب نصا Boolean Expressions and Truth Tables Converting POS Expressions to Truth Table Format Step1: list all the possible combinations of binary values of the variables just as was done for the SOP expression. Step2: convert the POS expression to standard form if it is not already. Step3: place a 0 in the output column (X) for each binary value that makes the expression a 0 and place a 1 for all the remaining binary values. • Note: a POS expression is equal to 0 only if at least one of the sum terms is equal to 0. 25 26 Converting SOP Expressions to Truth Table Format Example: Develop a truth table for the standard SOP expression Solution: ]ً[اكتب نصا Determining Standard Expressions from a Truth Table Determining the standard SOP expression represented by a truth table • Step1: list the binary values of the input variables for which the output is 1. • Step2: Convert each binary value to the corresponding product term by replacing each 1 with the corresponding variable and each 0 with the corresponding variable complement. For example, the binary value 1010 is converted to a product term as follows: 27 28 Converting POS Expressions to Truth Table Format • Example: Determine the truth table for the following standard POS expression: • Solution ]ً[اكتب نصا Determining Standard Expressions from a Truth Table • Example: From the truth table below, determine the standard SOP expression and the equivalent standard POS expression. Solution: 29 30 Determining Standard Expressions from a Truth Table • Determining the standard POS expression represented by a truth table Step1: list the binary values for which the output is 0. Step2: Convert each binary value to the corresponding sum term by replacing each 1 with the corresponding variable complement and each 0 with the corresponding variable. For example, the binary value 1001 is converted to a sum term as follows: ]ً[اكتب نصا Karnaugh Map • The 3-variable Karnaugh map is an array of eight cells, as shown in the figure below: • The 4-variable Karnaugh map is an array of sixteen cells, as shown in the figure In “wrap-around” adjacency, the cells in the top row are adjacent to the corresponding cells in the bottom row and the cells in the outer left column are adjacent to the corresponding cells in the outer right column. 010 cell is adjacent to the 000 cell, the 011 cell, and the 110 cell 31 32 Karnaugh map • A Karnaugh map provides a systematic method for simplifying Boolean expressions and, if properly used, will produce the simplest SOP or POS expression possible, known as the minimum expression. • A Karnaugh map is similar to a truth table because it presents all of the possible values of input variables and the resulting output for each value. • The Karnaugh map is an array of cells in which each cell represents a binary value of the input variables. • The cells are arranged in a way so that simplification of a given expression is simply a matter of properly grouping the cells. ]ً[اكتب نصا Karnaugh Map SOP Minimization • Example: Map the following standard SOP expression on a Karnaugh map: • Solution: 33 34 Karnaugh Map SOP Minimization Mapping a Standard SOP Expression • For an SOP expression in standard form, a 1 is placed on the Karnaugh map for each product term in the expression. • Each 1 is placed in a cell corresponding to the value of a product term. For example, for the product term ABC, a 1 goes in the 101 cell on a 3-variable map. • When an SOP expression is completely mapped, there will be a number of 1s on the Karnaugh map equal to the number of product terms in the standard SOP expression as shown in the figure below. ]ً[اكتب نصا Mapping a Nonstandard SOP Expression • Example: Map the following SOP expression on a Karnaugh map: Solution: • Notice that some of the values in the expanded expression are redundant. 35 36 Mapping a Nonstandard SOP Expression • Note: A Boolean expression must first be in standard form before you use a Karnaugh map. If an expression is not in standard form, then it must be converted to standard form. • Numerical Expansion of a Nonstandard Product Term • Assume that one of the product terms in a 3-variable expression is B, then it can be expanded numerically to standard form as follow: Write the binary value of the variable; then attach all possible values for the missing variables A and C as follows: ]ً[اكتب نصا Karnaugh Map Simplification of SOP Expressions • Grouping the 1s • You can group 1s on the Karnaugh map according to the following rules by enclosing those adjacent cells containing 1s. The goal is to maximize the size of the groups and to minimize the number of groups. 1.
A group must contain either 1, 2, 4, 8, or 16 cells, which are all powers of two. In the case of a 3-variable map, 23 = 8 cells is the maximum group. 2.
Each cell in a group must be adjacent to one or more cells in that same group, but all cells in the group do not have to be adjacent to each other. 3.
Always include the largest possible number of 1s in a group in accordance with rule 1. 4.
Each 1 on the map must be included in at least one group. The 1s already in a group can be included in another group as long as the overlapping groups include noncommon 1s. 37 38 Karnaugh Map Simplification of SOP Expressions • The process that results in an expression containing the fewest possible terms with the fewest possible variables is called minimization. • After an SOP expression has been mapped, a minimum SOP expression is obtained by grouping the 1s and determining the minimum SOP expression from the map. ]ً[اكتب نصا Determining the Minimum SOP Expression from the Map 1. Group the cells that have 1s. Each group of cells containing 1s creates one product term composed of all variables that occur in only one form (either uncomplemented or complemented) within the group. Variables that occur both uncomplemented and complemented within the group are eliminated. These are called contradictory variables. 2. Determine the minimum product term for each group. 3. When all the minimum product terms are derived from the Karnaugh map, they are summed to form the minimum SOP expression. 39 40 Grouping 1s on the Karnaugh map • Example: Group the 1s in each of the Karnaugh maps • Solution ]ً[اكتب نصا Determining the Minimum SOP Expression from the Map • Example: Determine the product terms for each of the Karnaugh maps in the figures below and write the resulting minimum SOP expression. • Solution: The minimum SOP expressions for each of the Karnaugh maps in the figures are: 41 42 Determining the Minimum SOP Expression from the Map Example: Determine the product terms for the Karnaugh map in the figure below and write the resulting minimum SOP expression. Solution: • The resulting minimum SOP expression is the sum of these product terms: ]ً[اكتب نصا Karnaugh map • Example: Use a Karnaugh map to minimize the following SOP expression: • Solution: • The first term must be expanded into and to get the standard SOP expression, which is then mapped; the cells are grouped as shown below: • The resulting minimum SOP expression is 43 44 Karnaugh map • Example: Use a Karnaugh map to minimize the following standard SOP expression: • Solution: • The binary values of the expression are 101 + 011 + 001 + 000 + 100 • Map the standard SOP expression and group the cells as shown in the figure below: • Then, the resulting minimum SOP expression is |
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Fundamentals
Supplemental Modules (Organic Chemistry)
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The Use of Curly Arrows
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Contents
1. Home
2. Bookshelves
3. Organic Chemistry
4. Supplemental Modules (Organic Chemistry)
5. Fundamentals
6. The Use of Curly Arrows
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Homolytic C-H Bond Dissociation Energies of Organic Molecules
How to Draw Organic Molecules
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Intermolecular Forces
Structure & Bonding
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Isomerism in Organic Compounds
Lewis Structures
Nomenclature
Organic Acids and Bases
Oxidation States of Organic Molecules
Reactive Intermediates
Resonance Forms
Rotation in Substituted Ethanes
Solubility - What dissolves in What?
Structure of Organic Molecules
The Golden Rules of Organic Chemistry
The Use of Curly Arrows
What is the pKa of water?
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The Use of Curly Arrows
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The most important question in chemistry is "Where are the electrons?"
What is the pKa of water?
Page ID 3648
Jim Clark
Truro School in Cornwall
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Using curly arrows to show the movement of electron pairs
2. Using curly arrows to show the movement of single electrons
3. Contributors
This page explains the use of curly arrows to show the movement both of electron pairs and of single electrons during organic reaction mechanisms.
Using curly arrows to show the movement of electron pairs
Curly arrows (and that's exactly what they are called!) are used in mechanisms to show the various electron pairs moving around.
The arrow tail is where the electron pair starts from. That's always fairly obvious, but you must show the electron pair either as a bond or, if it is a lone pair, as a pair of dots. Remember that a lone pair is a pair of electrons at the bonding level which isn't currently being used to join on to anything else.
The arrow head is where you want the electron pair to end up.
For example, in the reaction between ethene and hydrogen bromide, one of the two bonds between the two carbon atoms breaks. That bond is simply a pair of electrons.
Those electrons move to form a new bond with the hydrogen from the HBr. At the same time the pair of electrons in the hydrogen-bromine bond moves down on to the bromine atom.
There's no need to draw the pairs of electrons in the bonds as two dots. Drawing the bond as a line is enough, but you could put two dots in as well if you wanted to.
Notice that the arrow head points between the C and H because that's where the electron pair ends up. Notice also that the electron movement between the H and Br is shown as a curly arrow even though the electron pair moves straight down. You have to show electron pair movements as curly arrows - not as straight ones.
The second stage of this reaction nicely illustrates how you use a curly arrow if a lone pair of electrons is involved.
The first stage leaves you with a positive charge on the right hand carbon atom and a negative bromide ion. You can think of the electrons shown on the bromide ion as being the ones which originally made up the hydrogen-bromine bond.
The lone pair on the bromide ion moves to form a new bond between the bromine and the right hand carbon atom. That movement is again shown by a curly arrow. Notice again, that the curly arrow points between the carbon and the bromine because that's where the electron pair ends up.
That leaves you with the product of this reaction, bromoethane:
Using curly arrows to show the movement of single electrons
The most common use of "curly arrows" is to show the movement of pairs of electrons. You can also use similar arrows to show the movement of single electrons - except that the heads of these arrows only have a single line rather than two lines.
shows the movement of an electron pair
shows the movement of a single electron
The first stage of the polymerization of ethene, for example, could be shown as:
You should draw the dots showing the interesting electrons. The half arrows show where they go. This is very much a "belt-and-braces" job, and the arrows don't add much. Whether you choose to use these half arrows to show the movement of a single electron should be governed by what your syllabus says. If your syllabus encourages the use of these arrows, then it makes sense to use them. If not - if the syllabus says that they "may" be used, or just ignores them altogether - then they are as well avoided.
There is some danger of confusing them with the arrows showing electron pair movements, which you will use all the time. If, by mistake, you use an ordinary full arrow to show the movement of a single electron you run the risk of losing marks.
Contributors
Jim Clark (Chemguide.co.uk)
This page titled The Use of Curly Arrows is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jim Clark.
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Matlab - product-like simplification of trigonometric function
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Asked 9 years, 3 months ago
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Consider the following:
matlab
syms T fi t real
fun = symfun(sin(T+fi)+cos(T+fi),[T fi]);
fun = expand(fun);
which yields:
matlab
cos(T)cos(fi) - sin(T)sin(fi) + cos(T)sin(fi) + sin(T)cos(fi)
Now if I use either simplify or combine I get:
matlab
2^(1/2)sin(pi/4 + T + fi)
Could you please tell me which function will let me obtain product form, i.e.:
matlab
cos(fi)(cos(T)+sin(T)) + sin(fi)(cos(T)-sin(T))
matlab
trigonometry
symbolic-math
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edited Jun 22, 2016 at 22:05
horchler
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asked Jun 22, 2016 at 20:53
MaverickMaverick
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I think I may have found a way to do this using collect – it works in R2016a:
```matlab
syms T fi t real
fun = symfun(sin(T+fi)+cos(T+fi),[T fi]);
fun = expand(fun);
fun2 = collect(fun,[cos(fi) sin(fi)])
```
which returns (cos(T) + sin(T))cos(fi) + (cos(T) - sin(T))sin(fi).
This usage of collect (collecting functions of a variable) isn't really documented. I tried this out after reading through the examples for MuPAD's collect upon which collect is likely based on or related to.
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edited Jun 22, 2016 at 22:08
answered Jun 22, 2016 at 21:58
horchlerhorchler
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Andras Deak -- Слава Україні
Andras Deak -- Слава УкраїніOver a year ago
Maybe calling children on that and separately simplifying, then simplifying the rejoined version will help;)
2016-06-22T22:01:28.637Z+00:00
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In R2012b simplify gives me
matlab
cos(T + fi) + sin(T + fi)
so you probably have a newer version. Still: it's hard to come up with a specific simplification of a symbolic expression, especially if trigonometric functions are involved.
If you're specifically looking for an expression in terms of sines, you could try rewrite:
```matlab
rewrite(fun,'sin')
ans(T, fi) =
sin(T + fi) - 2sin(T/2 + fi/2)^2 + 1
```
The above output is again from R2012b, it's highly likely that your newer version will do better.
I originally suggested that you try simple, which by default (with 0 output variables) will try a bunch of various simplification attempts, and tell you the result. However, as @horchler pointed out, this function has been deprecated and no longer available after 2015a. Anyway, here's the output from this now-gone function from R2012b, which might give you hints regarding more low-level functions to try:
```matlab
syms T fi t real
fun = symfun(sin(T+fi)+cos(T+fi),[T fi]);
simple(fun)
simplify:
cos(T + fi) + sin(T + fi)
radsimp:
cos(T + fi) + sin(T + fi)
simplify(100):
cos(T + fi) + sin(T + fi)
combine(sincos):
cos(T + fi) + sin(T + fi)
combine(sinhcosh):
cos(T + fi) + sin(T + fi)
combine(ln):
cos(T + fi) + sin(T + fi)
factor:
cos(T + fi) + sin(T + fi)
expand:
cos(T)cos(fi) - sin(T)sin(fi) + cos(T)sin(fi) + sin(T)cos(fi)
combine:
cos(T + fi) + sin(T + fi)
rewrite(exp):
exp(- Ti - fii)(1/2 + i/2) + exp(Ti + fii)(1/2 - i/2)
rewrite(sincos):
cos(T + fi) + sin(T + fi)
rewrite(sinhcosh):
cosh(Ti + fii) - sinh(Ti + fii)i
rewrite(tan):
(2tan(T/2 + fi/2))/(tan(T/2 + fi/2)^2 + 1) - (tan(T/2 + fi/2)^2 - 1)/(tan(T/2 + fi/2)^2 + 1)
mwcos2sin:
sin(T + fi) - 2sin(T/2 + fi/2)^2 + 1
collect(T):
cos(T + fi) + sin(T + fi)
ans(T, fi) =
cos(T + fi) + sin(T + fi)
```
where the final line is the shortest representation found from the previous list, that would be your single return value. You can take your pick from the list. Your version will probably produce a more diverse set of outputs.
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edited May 23, 2017 at 12:08
CommunityBot
1 1 1 silver badge
answered Jun 22, 2016 at 21:19
Andras Deak -- Слава УкраїніAndras Deak -- Слава Україні
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Andras Deak -- Слава Україні
Andras Deak -- Слава УкраїніOver a year ago
@Maverick well...does it help?:) Does your version find this expression?
2016-06-22T21:29:16.277Z+00:00
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horchler
horchlerOver a year ago
I think it's fine. I especially like "it's hard to come up with a specific simplification of a symbolic expression, especially if trigonometric functions are involved." There's no way to write a perfect function that can guess at what rearrangement is desired. The only way to solve the OP's request would be to write a bunch of custom code that probably would assume a lot about the format of the input.
2016-06-22T21:39:49.777Z+00:00
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Andras Deak -- Слава Україні
Andras Deak -- Слава УкраїніOver a year ago
@horchler thank you for the feedback. I moved the deprecated bit to the end to have the currently available rewrite-based attempt on top.
2016-06-22T21:42:27.63Z+00:00
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horchler
horchlerOver a year ago
Unfortunately the output of rewrite(fun,'sin') in R2016a is not as nice: (2sin(T/2)^2 - 1)(2sin(fi/2)^2 - 1) - sin(T)(2sin(fi/2)^2 - 1) - sin(fi)(2sin(T/2)^2 - 1) - sin(T)sin(fi). If you don't call expand first (third line of OP's code) then it's the same.
2016-06-22T21:44:50.497Z+00:00
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Maverick
MaverickOver a year ago
@Andras Deak, I'm affraid I wasn't able to obtain this particular form, but as horchler put it: it's tricky to explain Matlab what we really want to aquire.
2016-06-22T21:48:21.797Z+00:00
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Probabilities of random geometric objects having certain properties (enclosing the origin, having an acute angle,...); expected counts, areas, ... of random geometric objects. For questions about the geometric distribution, use the tag [probability-distributions] instead.
Geometric probability is a tool to deal with the problem of infinite outcomes by measuring the number of outcomes geometrically, in terms of length, area, or volume. In basic probability, we usually encounter problems that are "discrete" (e.g. the outcome of a dice roll; see probability by outcomes for more). However, some of the most interesting problems involve "continuous" variables (e.g., the arrival time of your bus).
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7989 | https://www.physicsclassroom.com/Physics-Interactives/Momentum-and-Collisions/Collision-Carts/Collision-Carts-Interactive | Skip to Content Go to sign in Skip to Primary Navigation Skip to Secondary Navigation Skip to Page Navigation Skip to Header Navigation Skip to Footer Navigation Read more about accessability options and our navigation
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7990 | https://www.wyzant.com/resources/answers/908770/find-the-exact-solutions-to-sin-x-cos-x-in-the-interval-0-2-in-radians-if-t | Find the exact solutions to sin ( x ) = cos ( x ) in the interval [ 0 , 2 π ) in radians. If the equation has no solutions, respond with DNE. | Wyzant Ask An Expert
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Monica C.
asked • 10/14/22
Find the exact solutions to sin ( x ) = cos ( x ) in the interval [ 0 , 2 π ) in radians. If the equation has no solutions, respond with DNE.
Please show the work thoroughly so I can understand.
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Do you have a unit circle?
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If sinx = cosx, then dividing both sides by cosx gives tanx = 1
So, in the interval [0 , 2π], the solutions are π/4 and 5π/4.
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7991 | https://en.wikipedia.org/wiki/Equioscillation_theorem | Jump to content
Equioscillation theorem
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From Wikipedia, the free encyclopedia
Theorem
In mathematics, the equioscillation theorem concerns the approximation of continuous functions using polynomials when the merit function is the maximum difference (uniform norm). Its discovery is attributed to Chebyshev.
Statement
[edit]
Let be a continuous function from to . Among all the polynomials of degree , the polynomial minimizes the uniform norm of the difference if and only if there are points such that where is either -1 or +1.
That is, the polynomial oscillates above and below at the interpolation points, and does so to the same degree.
Proof
[edit]
Let us define the equioscillation condition as the condition in the theorem statement, that is, the condition that there exists ordered points in the interval such that the difference alternates in sign and is equal in magnitude to the uniform-norm of .
We need to prove that this condition is 'sufficient' for the polynomial being the best uniform approximation to , and we need to prove that this condition is 'necessary' for a polynomial to be the best uniform approximation.
Sufficiency
[edit]
Assume by contradiction that a polynomial of degree less than or equal to existed that provides a uniformly better approximation to , which means that . Then the polynomial
is also of degree less than or equal to . However, for every of the points , we know that because and (since is a better approximation than ).
Therefore, will have the same sign as (because the second term has a smaller magnitude than the first). Thus, will also alternate sign on these points, and thus have at least roots. However, since is a 'polynomial' of at most degree , it should only have at most roots. This is a contradiction.
Necessity
[edit]
Given a polynomial , let us define .
We will call a point an upper point if and a lower point if it equals instead.
Define an alternating set (given polynomial and function ) to be a set of ordered points in such that for every point in the alternating set, we have , where equals or as before.
Define a sectioned alternating set to be an alternating set along with nonempty closed intervals called sections such that
1. the sections partition (meaning that the union of the sections is the whole interval, and the intersection of any two sections is either empty or a single common endpoint)
2. For every , the th alternating point is in the th section
3. If is an upper point, then contains no lower points. Likewise, if is a lower point, then contains no upper points.
Given an approximating polynomial that does not satisfy the equioscillation condition, it is possible to show that the polynomial will have a two point alternating set. This alternating set can then be expanded to a sectioned alternating set. We can then use this sectioned alternating set to improve the approximation, unless the sectioned alternating set has more than points in which case our improvement cannot be guaranteed to still be a polynomial of degree at most [clarification needed]
Variants
[edit]
The equioscillation theorem is also valid when polynomials are replaced by rational functions: among all rational functions whose numerator has degree and denominator has degree , the rational function , with and being relatively prime polynomials of degree and , minimizes the uniform norm of the difference if and only if there are points such that where is either -1 or +1.
Algorithms
[edit]
Several minimax approximation algorithms are available, the most common being the Remez algorithm.
References
[edit]
^ Jump up to: a b c Golomb, Michael (1962). Lectures on Theory of Approximation.
^ "Notes on how to prove Chebyshev's equioscillation theorem" (PDF). Archived from the original (PDF) on 2 July 2011. Retrieved 2022-04-22.
External links
[edit]
Notes on how to prove Chebyshev’s equioscillation theorem at the Wayback Machine (archived July 2, 2011)
The Chebyshev Equioscillation Theorem by Robert Mayans
The de la Vallée-Poussin alternation theorem at the Encyclopedia of Mathematics
Approximation theory by Remco Bloemen
Retrieved from "
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7992 | https://www.fs.usda.gov/rm/pubs_journals/2020/rmrs_2020_mccormick_f001.pdf | from the Latin word esox meaning Pike, which came origi-nally from the Greek isox or possibly the Gaelic eog, ehawe (= salmon) (Boschung & Mayden 2004).
The three members of the Umbridae (genus Umbra) are small, secretive species living primarily in vegetated wet-lands, sloughs, and ditches; slow- moving creeks and river margins; and off- channel habitats such as oxbows. Occa-sionally used as baitfish or sporadically popu lar in the aquar ium trade, they are rarely seen except by fishery biolo-gists and ichthyologists sampling such habitats. They can breathe atmospheric oxygen using a modified swim blad-der, which allows them to survive in hypoxic (low dissolved oxygen) conditions in the lowland habitats they usually oc-cupy. They also feed and digest food under relatively cold winter temperatures in their northern haunts. The family name is derived from the Latin word for shade or shadow, umbra, apparently in reference to the propensity of the group to inhabit darkly stained waters with abundant cover (Scott & Crossman 1973; Etnier & Starnes 1993).
DIVERSITY AND DISTRIBUTION The number of living species of Esociformes is relatively small ( Table 19.1). Worldwide, 14 species are recognized; eight of these are native to North Amer i ca. All fossil and extant species assigned to this order are restricted to continents in the Northern Hemi sphere. Extant esoci-form species are assigned to four genera: Esox, Novumbra, Dallia, and Umbra. All are represented in North Amer i ca. These four genera are recognized either as one family, the Esocidae (Page & Burr 2011; Campbell et al. 2013; Page et al. 2013) or as two families, Esocidae (including By blue lake marge, upon whose breast The water- lilies love to rest, Lurking beneath those leaves of green The fierce pike seeks his covert screen, And thence with sudden plunge and leap, Swift as a shaft through the air may sweep, He seizes, rends, and bears away To hidden lair his struggling prey.
— Fishing in American Waters by Genio C. Scott I saw, dimly, Once a big pike rush And small fish fly like splinters . . .
— Fish by D. H. Lawrence The order Esociformes (Pikes and Mudminnows) comprises two families, Esocidae (Pikes) and Umbridae (Mudmin-nows). The Pikes are a small Holarctic (Northern Hemi sphere) family, that includes large, elongate predators with duckbill- like snouts full of sharp teeth. Popu lar with sport fishers, the largest Pikes fight fiercely on hook and line. As piscivorous, voracious, ambush predators, the Pikes play an impor tant functional role in the trophic ecol ogy and fish assemblage structure of many aquatic systems, especially in northern lakes. Other esocids, such as the Olympic Mud-minnow, Novumbra hubbsi, and Blackfishes, genus Dallia, are in ter est ing because of their tolerance of low dissolved oxygen and pH. The Alaska Blackfish, Dallia pectoralis, and the Northern Pike, Esox lucius, can also withstand the ex-tremely cold conditions of the Arctic and subarctic waters of Canada, Alaska, and Siberia. The name Esocidae is de-rived from Linnaeus’s (1758) generic name for Pike, Esox, Chapter 19 Esociformes: Esocidae, Pikes, and Umbridae (Mudminnows) Frank H. McCormick, Terry Grande, Cheryl Theile, Melvin L. Warren, Jr., J. Andrés López, Mark V. H. Wilson, Roger A. Tabor, Julian D. Olden, and Lauren M. Kuehne © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 194 FRESHWATER FISHES OF NORTH AMERICA Esox, Novumbra, and Dallia) and Umbridae (including Umbra). We adopt the latter approach herein (see phyloge ne tic relationships section) (López et al. 2000; Grande et al. 2004, 2013; Betancur- R. et al. 2013ab, 2017; Nelson et al. 2016).
Genus Esox Figure 19.1. Geographic range of Esox in North Amer i ca.
Figure 19.2. Northern Pike, Esox lucius, this group in the Yellowknife River, Northwest Territories, has the largest geographic range of any esociform in North Amer i ca (photo graph in May 2012 by and used with permission of © Paul Vecsei / Engbretson Underwater Photography).
Figure 19.3. The Chain Pickerel, Esox niger, one of two species in the subgenus Kenoza, the other being the Grass Pickerel, Esox americanus. Kenoza contains the smallest North American Pikes (photo graphs taken in Nov 2013 by and used with permission of Lance Merry). Table 19.1. Classification of the Mudminnows (Umbridae) and Pikes (Esocidae) based primarily on ge ne tic evidence (López et al. 2000; Grande et al. 2004; Near et al. 2012b, 2013; Betancur- R. et al. 2013ab; Grande et al. 2013). After each genus name the number of extant, recognized species is given in parentheses.
Classification Order Esociformes Family Umbridae (Mudminnows) Genus Umbra (3) Family Esocidae (Pikes) Subfamily Dallinae (Blackfishes) Genus Dallia (3) Subfamily Esocinae (Pikes) Genus Novumbra (1) Genus Esox (7) Subgenus Esox (5) Subgenus Kenoza (2) Genus Esox The genus Esox, the Pikes (and pickerels), is the most di-verse and has the widest geographic distribution among esociform genera. Seven living species are currently rec-© 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 195 France), and Aquitanian Pike (endemic to southwestern France) are the only extant species in the genus that are not native to North Amer i ca (Bianco & Delmastro 2011; Lucentini et al. 2011; Denys et al. 2014).
The Northern Pike exhibits the widest distribution of any esociform species, occurring at temperate latitudes in Asia, Eu rope, and North Amer i ca (Crossman 1996; Berra 2007; see Bianco & Delmastro 2011; Denys et al. 2014). In North Amer i ca, the native range of the species includes most of Canada except the Maritime Provinces, most of western and southern British Columbia, and the Arctic Archipelago. In the United States, the native range of the Northern Pike includes the northern half of the country (as far south as Missouri) be-tween the Rocky and Appalachian Mountains, a range cover-ing 23 states (Page & Burr 2011). Generally, the Northern Pike is restricted to fresh water, but some Eu ro pean populations regularly enter brackish water and spawn in coastal areas of the Baltic Sea (Scott & Crossman 1973). The pre sent distribu-tion of the species and apparent ge ne tic homogeneity throughout its range prob ably resulted from a rapid range ex-pansion of some small number of populations, which took place after receding glaciers made habitat available; however, ognized in Esox; three are endemic to North Amer i ca, one has a Holarctic distribution ( Table 19.1; Fig. 19.1), one is native to Asia, and two are endemic to southern Eu rope.
The species of Esox are divided between the subgenera Esox and Kenoza. The subgenus Esox comprises of the Northern Pike (Esox lucius), Holarctic distribution (Fig. 19.2); Muskellunge (or musky), Esox masquinongy, endemic to North Amer i ca; Amur Pike, Esox reichertii, of Asia; and Southern Pike, Esox cisaplinus = E. flaviae and Aquitanian Pike, Esox aquitanicus, of southern Eu rope (Launey et al. 2006; Berra 2007; Bianco & Delmastro 2011; Lucentini et al. 2011; Bianco 2014; Denys et al. 2014; Gandolfi et al. 2015; Nelson et al. 2016). The subge-nus Kenoza comprises the Chain Pickerel, Esox niger (Fig. 19.3), and Grass Pickerel, Esox americanus (Fig. 19.4), both of which are endemic to North Amer i ca. The Grass Pickerel has two subspecies, the Redfin Pickerel, E. a. americanus, and the Grass Pickerel, E. a. vermiculatus (Page & Burr 2011; Fig. 19.4). The eastern Asian endemic, Amur Pike, and the Eu ro pean endemics, Southern Pike (native to Padano- Veneto, Tuscany and Lazio, Italy, and apparently historically to Lake Geneva, Switzerland, and Figure 19.4. Grass Pickerel, Esox americanus vermiculatus (adult, upper, and juvenile, lower), another member of the subgenus Kenoza (photo graphs taken in Kankakee County, Illinois, upper, and Newton County, Indiana, lower, in April 2008 by and used with permission of Uland Thomas).
Plate 19.1. Muskellunge, Esox masquinongy.
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NOT FOR SALE OR DISTRIBUTION 196 FRESHWATER FISHES OF NORTH AMERICA (Fig. 19.4) includes Atlantic and Gulf Coast drainages as far north as the U.S.– Canadian border (Page & Burr 2011). The two species are morphologically distinct (e.g., E. america-nus, unlike E. niger, exhibits an elongation of the maxillae beyond the midpoint of the eye, and reduction in the num-ber of branchiostegal rays, pelvic- fin rays, and lateral- line scales) (Grande et al. 2004).
Populations of the Grass Pickerel in the Atlantic Coast drainages correspond to the subspecies E. a. americanus, Redfin Pickerel, which gradually intergrades with popula-tions of E. a. vermiculatus, Grass Pickerel, in the southern part of its range between western Florida and western Mississippi (Crossman 1978). The Mississippi drainage populations consist mostly of Redfin Pickerel. Individuals of both subspecies occur in Coastal Plain drainages of Al-abama and Georgia. Morphological analyses (i.e., mor-phometrics, cardioid scale frequencies) confirm distinc-tiveness of the two subspecies. In the intergrade zone fish show intermediacy in some characters and tendency to overlap with one subspecies or the other in other charac-ters (Crossman 1966; Reist & Crossman 1987).
Mitochondrial DNA barcoding, using partial sequences of cytochrome c oxidase subunit I among geo graph i cally dispersed individuals of E. americanus revealed two un-confirmed candidate species with ge ne tic divergences of >2% (April et al. 2011). A similar analy sis of Canadian populations detected ge ne tic differentiation between E. a. americanus from the St. Lawrence River to the east and E. a. vermiculatus from the Laurentian Great Lakes farther west (Hubert et al. 2008). DNA barcoding did not detect any unconfirmed candidate species in the Chain Pickerel, but the species did share DNA barcodes with the Grass Pickerel (April et al. 2011; see ge ne tics section); other ge ne tic analyses of these taxa produced similar results (Grande et al. 2004; Hubert et al. 2008).
debate is ongoing over the number and location of the popula-tions of the Northern Pike that survived the glacial maxima and gave rise to living populations (see ge ne tics section).
The other three North American species of Esox are more limited in native distribution than the Northern Pike. They are restricted to the Great Lakes, Mississippi River, and Atlantic Coast drainages of central and eastern North Amer i ca. The Muskellunge (from the Ojibwa word maash-kinoozhe) is native to the Great Lakes and the upper Missis-sippi River system from southern Canada to Tennessee. The native range of the Muskellunge is sometimes divided into three geographic regions reflecting variation in color pat-tern among populations, which may be indicative of incipi-ent taxonomic differentiation. The three regions are Wis-consin, Minnesota, southwestern Ontario, and southeastern Manitoba; Great Lakes and St. Lawrence River; and the Ohio River system (Scott & Crossman 1973) with possibly a fourth variant in the Tennessee and Cum-berland Rivers (Etnier & Starnes 1993). The taxonomic sta-tus of the color variants prevalent in these regions is the subject of debate (e.g., Hourston 1955; Crossman 1978; Trautman 1981; Etnier & Starnes 1993). Dramatic popula-tion declines in much of the Ohio River system (e.g., Cum-berland and Tennessee River populations) and introduc-tions and artificial hybridization confound understanding of these variants (Casselman et al. 1986; Etnier & Starnes 1993; Jenkins & Burkhead 1994).
The native range of the two species in the subgenus Kenoza (Chain Pickerel and Grass Pickerel) extends south into central Florida and includes the southernmost lati-tudes naturally inhabited by any esociform. The Chain Pickerel (Fig. 19.3) inhabits Atlantic Coastal Plain drain-ages from southern Canada to Florida, in the lower reaches of the Mississippi River basin, and Gulf Coast drainages (Page & Burr 2011). The native range of the Grass Pickerel Figure 19.5. The Olympic Mudminnow, Novumbra hubbsi, is the smallest and most range restricted esocid, reaching only about 80 cm TL, and being restricted to the Olympic Peninsula, Washington (photo graph taken by and used with permission of Tom Baugh).
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NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 197 Fossil locations of Novumbra suggest bogs, swamps, and freshwater marshes were the preferred habitat since the Oligocene (33.9–23.03 mya) (Schultz 1929, 1930).
Genus Dallia The three currently recognized species of Dallia are native to western Alaska and the northeastern Siberian Chuckot, although controversy exists over the validity of the two forms restricted to Asia (Mecklenburg et al. 2002; Campbell & López 2014). The Alaska Blackfish, (can’giiq in the Yup’ik language) (Fig. 19.7), is the only member of the genus occur-ring in North Amer i ca (Fig. 19.8). The complete geographic range of D. pectoralis covers the northeastern portion of the Chukotsky (Chukchi) Peninsula in Asia to the coastal areas of the Bristol Gulf in the far northeast of Rus sia, the coastal plains of the Arctic and Bering Sea drainages of western Alaska, and St. Lawrence, St. Matthew, and Nunivak Islands in the Bering Sea. The two other species in the genus, the Pilkhykay Blackfish, Dallia delicatissima, and the Amguema Blackfish, Dallia admirabilis, are endemic to the northern coastal drainages of the Chukotka Peninsula, Amguyema River drainage, in far northeastern Siberia (Chereshnev & Balushkin 1981; Gudkov 1998).
From a zoogeographic perspective, the genus is unique among strictly freshwater fishes in being restricted to the Beringia Ice Age refugium. The distribution of the Alaska Blackfish is mysteriously circumscribed without evident bar-riers to its post- glacial dispersal along Arctic Coastal low-lands ( toward the Mackenzie or Kolyna Rivers) or upstream in the Yukon River beyond its pre sent limit near Fairbanks (Lindsey & McPhail 1986). Differences in chromosome numbers between populations of the Alaska Blackfish from Genus Novumbra The monotypic genus Novumbra includes the Olympic Mudminnow, Novumbra hubbsi, (Fig. 19.5), but one fossil species, †Novumbra oregonensis, is also assigned to the ge-nus (Cavender 1969). The Olympic Mudminnow is con spic u ous as the smallest, most range- restricted species of esocid. It occurs in the Chehalis River drainage, Washing-ton, in direct tributaries of southern Puget Sound, and in lowland habitats of the Olympic Peninsula from North Bay of Grays Harbor to Ozette Lake (Harris 1974). The Olympic Mudminnow distribution reflects the glacial re-fugia that existed at the margins of the Vashon Glacier during the Pleistocene (1.8–0.01 mya) (Fig. 19.6). Mor-phological differences among populations suggest limited dispersal since that time (Meldrim 1968). Ge ne tic evi-dence indicates that Olympic Mudminnow populations in eastern Puget Sound represent undocumented introduc-tions of the species from an unknown location on the southern Olympic Coast (DeHaan et al. 2014). As with other Mudminnow species, the Olympic Mudminnow is strongly associated with shallow, sluggish water bodies, dense vegetation, and fine substrates (Meldrim 1968). Novumbra hubbsi Figure 19.6. Geographic range of the Olympic Mudminnow, Novumbra hubbsi.
Figure 19.7. Male Alaska Blackfish, Dallia pectoralis, about 203 mm TL, captured from ditches near the vicinity of Anchorage, Alaska, by K. Stoops in 1982 (courtesy of John Brill).
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NOT FOR SALE OR DISTRIBUTION 198 FRESHWATER FISHES OF NORTH AMERICA Dallia pectoralis Figure 19.8. Geographic range of the Alaska Blackfish, Dallia pectoralis, in North Amer i ca, including an introduced population near Anchorage, Alaska.
Genus Umbra Figure 19.9. Geographic range of Umbra in North Amer i ca.
Figure 19.10. Male (upper) and female (lower) Central Mudminnows, Umbra limi, during the breeding season (photo graphs taken April 2008, Kankakee County, Illinois, by and used with permission of Uland Thomas).
the Yukon and Colville River drainages and ge ne tic struc-ture revealed by mitochondrial DNA analyses suggest the presence of undescribed taxonomic units (see ge ne tics sec-tion) (Crossman & Ráb 1996; Campbell & Lopez 2014).
Genus Umbra The genus Umbra includes three extant species of Mud-minnows; of these, two are endemic to North Amer i ca (Fig. 19.9) and one to Eu rope (Eu ro pean Mudminnow, Umbra krameri). The Central Mudminnow, Umbra limi (Fig. 19.10), inhabits the Great Lakes region, Hudson Bay, the upper and middle Mississippi River drainage, mostly northern tributaries of the Ohio River, and sparsely occurs in the Missouri River drainage, Iowa, South Da-kota, and Missouri. The Eastern Mudminnow, Umbra pyg-maea (Fig. 19.11), occurs in Atlantic Coast drainages from New York to South Carolina (Page & Burr 2011). Problem-atic specimens of Umbra pygmaea from south of Okefeno-kee Swamp in Baker County, Florida, have higher dorsal- and anal- fin ray counts than either U. pygmaea or U. limi (Laerm & Freeman 2008).
Esociforms as Non- natives The native distribution of species of Esox has been altered as a result of introduction and extirpation. The Northern Pike and Muskellunge are introduced extensively outside of their natu ral ranges because of their value to recre-© 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 199 fishes, Salmonidae), as the top predator in a subarctic lake in northern Sweden and produced strong cascading ef-fects on both prey abundance and trophic structure (Bys-tröm et al. 2007).
State and federal programs have undertaken vigorous control programs for the Northern Pike, which include penalties for illegal introduction, monitoring, and active re-moval to protect native and recreationally and commer-cially impor tant fishes. The Upper Colorado River Endan-gered Fish Recovery Program includes an eradication program for Northern Pike in an attempt to reduce their impact on native minnows and Suckers (Catostomidae) (USFWS 2008; Zelasko et al. 2016). The state of Washing-ton similarly regards the Northern Pike as an invasive spe-cies in the Pend Oreille and Spokane Rivers, Colombia River drainage, and of grave concern to fisheries manage-ment agencies throughout the Pacific Northwest. Predation by non- native Northern Pike in that region has contributed to reduced recruitment in adfluvial Westslope Cutthroat Trout, Oncorhynchus clarkia lewisi, in Coeur d’Alene Lake, Idaho, and declines in Cutthroat Trout populations across much of the Pacific Northwest (Muhlfeld et al. 2008; Wal-rath et al. 2015). The Alaska Department of Fish and Game identifies Northern Pike as an invasive species in south cen-tral Alaska. In that region the species is decimating native ational anglers. Government- funded stocking programs maintain populations of these two species in water bodies where they did not occur naturally or from which native populations are extirpated or are in decline (e.g., Simon-son 2003). In the United States, Northern Pike are intro-duced in 20 states and Muskellunge in 16 states outside of their natu ral distributions. The combined result of legal and illegal introductions is the presence of self- sustaining populations of all four North American species of Esox outside of their natu ral range (Fuller et al. 1999). The Asian- native Amur Pike was stocked in central Pennsylva-nia between 1968 and 1971, but its status is uncertain (Fuller et al. 1999). Cooper (1983) reported Amur Pike as established; however, Robins et al. (1991) considered them as not established.
Introductions of top piscivores, such as the Northern Pike and Muskellunge, affect native fish assemblages and trophic relationships in aquatic ecosystems (Persson et al. 1996; Findlay et al. 2005; Eby et al. 2006). For example, introduced Northern Pike dramatically reduced native minnow (Cyprinidae, Carps and Minnows) richness in Adirondack lakes (Findlay et al. 2000) and reduced the abundance of native salmonids in Flathead Lake, Mon-tana (Muhlfeld et al. 2008). Northern Pike replaced Arc-tic Char, Salvelinus alpinus (Trouts, Salmons, and White-Figure 19.11. Adult female Eastern Mudminnow, Umbra pygmaea (60 mm SL), from Sussex County, Coppahaunk Swamp, Virginia, July 1984 (courtesy of N. Burkhead and R. Jenkins and courtesy of Virginia Division of Game and Inland Fisheries).
Plate 19.2. Central Mudminnow, Umbra limi.
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NOT FOR SALE OR DISTRIBUTION 200 FRESHWATER FISHES OF NORTH AMERICA are also documented in Montana, Oklahoma, and Texas (Fuller et al. 1999). The Central Mudminnow has dis-persed into Atlantic Slope drainages, New York, presum-ably via the old Erie Canal and has expanded its range along the Mohawk River watershed from Fort Hunter to the Hudson River (Smith 1985; Daniels 2001; Waldman et al. 2006) where it appears to be hybridizing with the Eastern Mudminnow (Schmidt & Daniels 2006). It has re-cently been reported from the Saint John River near Ed-mundston, New Brunswick, presumably as a result of a bait bucket transfer by fishermen (Curry et al. 2018).
PHYLOGENETIC RELATIONSHIPS Early Views of Esociform Relationships The position of esociforms among the classification of fishes has under gone many changes. At the onset of for-mal classification schemes, Linnaeus accepted Peter Arte-di’s work on fishes in which the only esociform recog-nized at the time, Esox lucius, was placed among the malacoptergyians (fishes with soft fins). Much later, the genera Esox and Umbra were grouped in the order Hap-lomi, which was thought to have close affinities to the orders Isospondyli and Cyprinodontes of the superorder Teleostei. In the original description of Dallia pectoralis, Bean (1880) provisionally placed the species with Umbra in the family Umbridae of haplomous fishes. Distinct fea-tures of the anatomy of the pectoral girdle and maxillary skeleton in Dallia led Gill (1885) to place the genus in its own order, Xenomi. Schultz (1929) in the description of Novumbra hubbsi, erected a new family, Novumbridae, as part of the Haplomi related to Umbridae and Dalliidae. Gill’s evidence in support of Xenomi was later reinter-preted and eventually Dallia, Novumbra, and Umbra were grouped in the family Umbridae (Berg 1931). Chapman (1934) proposed an alternative arrangement of haplomous fishes in which Novumbra, Umbra, and Dallia were each placed in its own family. Gradually, the order Haplomi, which also included Esox, was replaced by either the sub-order Esocoidei or the order Esociformes without changes in membership. Berg (1948) placed the Esocoidei in the order Clupeiformes, which he recognized as an artificial assemblage. In his classification, Esocoidei contained the families Dalliidae (Dallia), Umbridae (Novumbra and Umbra), and Esocidae (Esox). In mid-20th century classifica-tions, esociforms were recognized as a basal lineage of the higher Division III teleosts beginning with the seminal work of Greenwood et al. (1966), who placed esociforms in populations of Pacific salmon, Oncorhynchus spp., poses an economic threat to recreational and subsistence fishing, has resulted in the extirpation of Coho Salmon, Oncorhynchus kisutch, Rainbow Trout, O. mykiss, and Arctic Grayling, Thy-mallus arcticus (Salmonidae), from lakes in the Susitna River drainage, and has caused declines in Suckers (Catos-tomidae) and whitefishes (Coregonus spp., Salmonidae) (ADFG 2002; Southcentral Alaska Northern Pike Control Committee 2007; Sepulveda et al. 2013, 2015).
Several western states have begun introducing the hybrid tiger muskellunge (Northern Pike, Esox lucius × Muskellunge, E. masquinongy) into aquatic systems to ostensibly control perceived nuisance species (e.g., Northern Pikeminnow, Pty-chocheilus oregonensis, in some reservoirs in western Wash-ington) and overabundant nongame fishes to aid in estab-lishing a recreational fishery. The use of the tiger muskellunge as biological control agents to benefit sport fish populations, however, may be counterproductive to manage-ment goals. Stable isotope analyses in Colorado reservoirs where tiger muskellunge were stocked indicated that it pri-marily consumed stocked salmonids (53–84% by mass) (Lepak et al. 2014).
Other species of umbrids and esocids also occur outside their native ranges. Olympic Mudminnow populations in several drainages east of Puget Sound are apparently the result of introductions (Mongillo & Hallock 1999; De-Haan et al. 2014). Introduced populations of the Alaska Blackfish thrive in lakes near Anchorage, Alaska, where it creates a serious prob lem in the management of the Rain-bow Trout, Oncorhynchus mykiss, fishery, and on St. Paul Island in the Pribilofs (Morrow 1980; Lindsey & McPhail 1986; Fuller et al. 1999).
Several populations of the Eastern Mudminnow are es-tablished in Eu rope, likely resulting from the release of aquar ium specimens (Fuller et al. 1999; Verreycken et al. 2007). Although introduced in the early 1900s, the East-ern Mudminnow is currently found in only six Eu ro pean countries: Belgium, Denmark, France, Germany, Poland, and The Netherlands. Assessments of invasive potential vary, but Verreycken et al. (2010) ascribed low to medium risk of invasiveness to Eastern Mudminnows. However, Copp et al. (2009) applied the Fish Invasiveness Scoring Kit (FISK) analy sis to 67 fish species including the Eastern Mudminnow and determined it was a high- risk species for the United Kingdom. In North Amer i ca, acci-dental release of laboratory specimens gave rise to a self- sustaining population of the Central Mudminnow in the Connecticut River system, Connecticut and Mas sa chu setts (Fuller et al. 1999; Hartel et al. 2002). Introductions © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 201 combined mitochondrial and nuclear data: López et al. 2004; Burridge et al. 2012), some of which used broad taxon sampling and numbers of nuclear genes (Near et al. 2012b, 2013; Betancur- R. et al. 2013ab, 2017). However, unlike the strong support for a salmoniform- esociform clade, the posi-tion of this clade among other groups of teleosts is less cer-tain. The salmoniform- esociform clade is resolved with the Argentiniformes (Near et al. 2012b, 2013) or with poor sup-port closest to the Galaxiidae (Galaxids) (minus Lepidogal-axias) with the argentiniforms basal to these two clades (Betancur- R. et al. 2013ab, 2017). Complete mitogenomic analyses yielded a fossil- calibrated divergence time between esociforms and salmoniforms in the Early Cretaceous (113.02 mya, 96.34–134.11 mya highest probability density) (Campbell et al. 2013), although these dates should be in-terpreted cautiously because no fossils of either group are available before the early Campanian (Late Cretaceous; about 84 mya) (Wilson et al. 1992; Grande 1999).
Esociform Intraordinal Relationships In the past, two esociform families were recognized: Esocidae containing only Esox, and Umbridae containing Dallia, Novumbra, and Umbra. This classification neatly divided the larger, piscivorous esocids from the smaller umbrids and seemed to explain the evolution of their dif fer ent body forms parsimoniously. Cavender (1969) exam-ined umbrid osteology, finding evidence to support a close relationship between Dallia and Novumbra (Fig. 19.12A) (e.g., an elongation of the subopercle, presence of a single row of well- developed incurved teeth on the dentary, presence of a crescent ridge of bone on the posterior part of each frontal forming the anterior limit of the posttem-poral fossa) but also found characters to unite Umbra and the Protacanthopterygii with groups such as argentinids (Herring Smelts), salmonids, and osmerids (Smelts).
Esociform Interordinal Relationships Into the late 20th century the relationships of the order continued to be debated (summaries by Nelson 2006; Nelson et al. 2016). Some considered esociforms to be a primitive sister- group to osmeriforms (Freshwater Smelts) + salmoniforms (Trouts, Salmons, and Whitefishes). Others viewed them as the primitive sister- group to the eu-teleosts (Fink & Weitzman 1982; Lauder & Liem 1983; Fink 1984). A morphological study of lower euteleostean rela-tionships supported placement of esociforms as the closest relatives of the Neoteleostei based on four synapomorphic (shared, derived) characters, including Type 4 tooth attach-ment, acellular skeleton, loss of the third uroneural (a bone in the caudal skeleton), and scaling of the cheek and oper-culum (Johnson & Patterson 1996). In contrast, Williams (1987) considered osmeroids and argentinoids (Marine Smelts) as the primitive sister- group of the neoteleosts and suggested salmonids and esocoids were closest relatives (see also Arratia 1997, 1999; Wilson & Williams 2010) and more primitive than osmeroids and other higher forms. Reaching agreement on the lineage among basal euteleosts closest to Esociformes proved to be difficult. Morphologically based phyloge ne tic studies of relationships among the basal lin-eages of the Euteleostei did not lead to a consensus view on the placement of esociforms perhaps because of a scarcity of shared, derived morphological traits between this group and other basal euteleost lineages (e.g., Rosen 1974).
Sister- group to the Esociforms DNA sequence- based studies that included representative esociforms and salmonids agree among themselves in placing these two groups as closest relatives. The sister re-lationship of esociforms and salmoniforms was morphologi-cally supported earlier by evidence from the suspensorium (cartilage and bone supporting lower jaw) and associated musculature, including presence of an anteroventral wing of the hyomandibular bone that overlaps the metapterygoid and by loss of the supramaxillary ligament in both groups (Williams 1987). In sequence- based studies, the salmonids and esociforms are resolved consistently as a strongly sup-ported sister pair by multiple lines of evidence (summarized by Zaragüeta- Bagils et al. 2002; mitochondrial genome: Ishiguro et al. 2003; Li et al. 2010; Campbell et al. 2013; nu-clear sequence data: López et al. 2004; Santini et al. 2009; A Novumbra Dallia Umbra Esox B Dallia Umbra Novumbra Esox Figure 19.12. Historically proposed hypotheses of relationships among genera within Pikes (Esocidae) and Mudminnows (Umbridae): (A) Cavender (1969); (B) Nelson (1972) (redrawn from López et al. 2000).
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NOT FOR SALE OR DISTRIBUTION 202 FRESHWATER FISHES OF NORTH AMERICA the parasphenoid length, number of abdominal vertebrae 39–48, caudal vertebrae 17–21, and the presence of three epurals. The subgenus Kenoza was diagnosed by 12 de-rived character states, including the presence of four man-dibular canal pores, a discontinuous infraorbital canal, a vomerine tooth patch <50% of the vomer length and con-sisting of small teeth and with only a few teeth along the neck of the vomer, a close connection between epicentral and epineural intermuscular bones, an expansion of the second neural arch in the transverse plane, two epurals in the caudal skeleton, and the presence of cardioid scales between the pelvic fins. Denys et al. (2014), using partial mtDNA sequences, investigated the relationships of the two recently named Eu ro pean species of Pikes. They con-cluded that Esox cisalpinus and E. flaviae are more closely related to E. lucius than to any other species of Esox.
Molecular evidence also strongly supports the subgenera Esox and Kenoza as monophyletic (Grande et al. 2004; López et al. 2004) in agreement with the morphological evidence of Grande et al. (2004) cited above. Evidence from gene sequences also supports a sister relationship between Novumbra and a monophyletic Esox (López et al. 2000, 2004; Campbell et al. 2013; Near et al. 2012b, 2013; see Be-tancur- R. et al. 2013ab, 2017; Fig. 19.13). Complete mitoge-nomic analyses yielded a fossil- calibrated divergence time between the two genera at the Paleocene- Eocene boundary (56.31 mya, 48.48–64.44 mya highest probability density) (Campbell et al. 2013); however, this is after the first ap-pearance of Esox in the fossil rec ord 66–59 mya (Wilson 1980, 1984). The clade Esox + Novumbra is resolved as sister Novumbra (e.g., presence of an inframandibular bone, pres-ence of a transverse pro cess on vertebra one with a liga-mentous connection to the cleithrum). In a cladistic study based on characters of the cephalic (head) sensory sys-tem, Nelson (1972) cited evidence that supported uniting Novumbra, Dallia, and Umbra, which were at that time in Umbridae (e.g., reduction or loss of the mandibular, jaw, sensory canal), placing Novumbra as the most basal of the three umbrid genera based on loss of the infraorbital and extrascapular (dorsal posterior head) canals in Dallia and Umbra (Fig. 19.12B) and supporting division of Esox into the subgenera Esox for the Pikes (including the Muskellunge) and Kenoza for the pickerels. In Kenoza, for example, the infraorbital canal is discontinuous and the extrascapular canal is lost. In a broad study of the salmoniforms, Rosen (1974) discovered characters of the axial skeleton (skull and vertebral column) (e.g., reduced height of the fourth epibranchial and loss of the fifth epibranchial, epurals re-duced to one, a full neural spine on preural centrum one, and an anteriorly foreshortened uroneural) supporting a Dallia + Novumbra clade as sister to Esox with this group being sister to Umbra. Further evidence from osteological traits (Wilson & Veilleux 1982) supported Nelson’s (1972) hypothesis of umbrid relationships. López et al. (2000), however, later critiqued the morphological evidence as be-ing mostly based on skeletal reduction, a pro cess that could happen convergently or as expressed in a mosaic fashion in the taxa in question. Their mitochondrial DNA sequence evidence suggested the following relationships: Umbra (Dallia (Novumbra + Esox)). López et al. (2004) pro-vided additional molecular support for that view. A mor-phological and molecular cladistic analy sis of the species of Esox using Novumbra and Umbra as outgroups, found 38 synapomorphies, which supported the following set of in-terrelationships: (Esox masquinongy (E. lucius + E. reicher-tii) + (E. niger + E. americanus)) (Grande et al. 2004). Monophyly of the genus Esox was supported by the pres-ence of a posttemporal canal; the anterior part of the pal-atine articulating with the premaxilla forming a toothed biting surface; the mandibular length being >50% of the head length; a pro cess on the anterolateral margin of the palatine that articulates with the maxilla; the presence of depressible teeth on the dentary, vomer, and palatine; the presence of toothplates on basibranchials one and two; ex-pansion of the anterior supraneural; and the presence of notched or cardioid (heart- shaped) scales along the lateral line. The monophyly of the subgenus Esox was supported by eight derived character states, including a continuous and complete infraorbital canal, a vomer length <50% of Esox lucius E. reicherti E. masquinongy E. americanus E. niger Novumbra Dallia Umbra limi U. pygmaea U. krameri Salmonidae Figure 19.13. Phylogeny of species of the Pikes (Esocidae: Esox, Novumbra, Dallia,) and Mudminnows (Umbridae: Umbra) including the sister family to these two clades, the Trouts and Salmons (Salmonidae), as resolved from extensive taxon sampling and analyses of multiple nuclear and mitochondrial gene sequences (modified from López et al. 2000, 2004; Grande et al. 2004; Near et al. 2012b, 2013; Betancur- R. et al. 2013ab). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 203 these display many of the characteristics that identify ex-tant species of Esox, including the duckbill- like snout, de-pressible teeth and fixed canine teeth, the elongate body with posteriorly situated dorsal and anal fins, and the dis-tinctive, tri- lobed body scales (Wilson 1980). Another noteworthy extinct species assigned to this genus is †Esox kronneri, which was described from a single fossil in good condition from the Green River Formation and dating to the early Eocene (about 52–50 mya) (Grande 1999; Fig. 19.14). The significance of †E. kronneri is due in part to its hypothesized close relationship to the subgenus Ke-noza. Characters placing †E. kronneri in Kenoza include a smaller number of vertebrae and having only two epurals in the caudal skeleton (Grande 1999). If it indeed belongs in Kenoza, the fossil is evidence for an ancient divergence between the two subgenera of Esox (>50 mya).
Abundant fossils of Esox are known from the Oligocene to the Pleistocene occurring throughout the Holarctic (re-viewed by Grande 1999). Notably, a fossil of a Pike from the Miocene of Oregon (23.0–5.33 mya) (Cavender et al. 1970) had fixed canine teeth as well as depressible teeth on its palatine bones, a characteristic of the extant Mus-kellunge. Another fossil relative of the Muskellunge is †E. columbianus from the Pliocene (5.3–2.6 mya) of southern Washington, which also had fixed canines on its palatines and vomer (Smith et al. 2000). This rich fossil rec ord has the potential to inform our understanding of the biogeo graph i cal factors influencing the distribution of the living members of the genus once the relationships among the fossil and living forms are better understood.
to Dallia, leaving a monophyletic Umbra as the most basal extant esociform lineage and the only member of its family. The mean divergence estimate between Umbra and the other two genera is 88.61 mya (Late Cretaceous). We follow Nelson et al. (2016) using the best available evidence and recognize Umbridae as containing only the genus Umbra and place the remaining genera (Dallia, Novumbra, and Esox) in the family Esocidae ( Table 19.1; Fig. 19.13).
FOSSIL REC ORD The oldest fossils assigned to the Esociformes lived in the Late Cretaceous in Western North Amer i ca beginning with fossils from the Campanian (84–72 mya) (Wilson et al. 1992; see also the review of fossil Esocidae by Grande 1999). The Cretaceous fossils consist of palatine and dentary bones with distinctive C- shaped tooth bases on their tooth- bearing surfaces. The shape and orientation of the tooth bases are diagnostic of a par tic u lar type of de-pressible teeth (teeth that fold back toward the throat) seen, among extant fishes, only in the modern genus Esox (Cavender et al. 1970; Wilson et al. 1992). All of the Cre-taceous esociform fossil remains come from small fishes about the size of living species of Mudminnows. Wilson et al. (1992) erected two genera for the pre- Cenozoic (>66 mya) esociform fossils: †Estesox and †Oldmanesox. †Estesox is the more abundant and more ancient of the two with fossils as old as early Campanian (about 83.6– 72.1 mya); its jaws and palatines had only depressible teeth. †Oldmanesox had jaws with depressible and fixed canine teeth as in the extant and fossil members of Esox. As in all species of Esox, the depressible teeth on the lower jaw are located near the front of the mouth, and the fixed canine teeth are farther back, closer to the angle of the jaws (Wilson et al. 1992). Extant Umbra, Novumbra, and Dallia have neither depressible teeth nor fixed canine teeth similar to those of Esox (Wilson & Veilleux 1982). Thus, the fixed teeth of Esox possibly evolved from de-pressible teeth by secondary loss of the depression mecha-nism coupled with enlargement of the teeth. If this is cor-rect, the Cretaceous †Oldmanesox may well be more closely related to Esox than it is to Novumbra, Dallia, or Umbra. The earliest fossils assigned to an extant esociform genus were discovered in deposits of mid- Paleocene age (66–59 mya) in Alberta, Canada. From these deposits Wilson (1980, 1984) described †Esox tiemani, based on one well- preserved and nearly complete fossilized skeleton and about a dozen partial skele tons of dif fer ent sizes; Figure 19.14. Holotype of †Esox kronneri (118 mm TL; FMNH PF14918) from the Fossil Butte Member of the Green River Formation (early Eocene, about 52 mya), the only known specimen of an esociform from the fish- fossil rich Green River Formation. Although small, the specimen is one of the best- preserved fossil esociforms ever found (from Grande, Lance. 2013. The Lost World of Fossil Lake: Snapshots from Deep Time. University of Chicago Press, 425 pp.) (courtesy Lance Grande.) © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 204 FRESHWATER FISHES OF NORTH AMERICA †Proumbra is another extinct esociform genus displaying combinations of traits not observed in living members (Sytchevskaya 1968; Gaudant 2012). The genus was de-scribed from fossils dating to the late Oligocene of western Siberia, and its affinities are the subject of debate. The ge-nus was proposed as a close ally of the lineage leading to Dallia or of the lineage leading to Umbra, although Gaudant (2012) suggested †Proumbra is intermediate between †Pal-aeoesox and the most primitive species of Umbra.
MORPHOLOGY External Morphology All esociform fishes are characterized by the placement of the dorsal fin far back on the body opposite the anal fin (Figs. 19.2–19.5, 19.7, 19.10, 19.11, and 19.14). The bodies of esociforms are elongate with a round to oval cross sec-tion. The caudal peduncle is robust, and the caudal fin ranges in shape from rounded (i.e., Dallia, Umbra; Figs. 19.7, 19.10, and 19.11; Table 19.2) to truncate (i.e., Novum-bra; Fig. 19.5) to forked (i.e., Esox; Figs. 19.2–19.4). The pelvic fins are abdominal in position, but poorly devel-oped in Alaska Blackfish (Fig. 19.7). The pectoral fins originate ventrally immediately behind the opercular opening. A well- developed fleshy base supports large, rounded pectoral fins in the Alaska Blackfish (Fig. 19.7). In all esociforms, both paired and median fins are with-out spines. Members of Esox stand out by their flattened, elongated snouts that resemble a duck’s bill (Figs. 19.2– 19.4 and 19.15). Other esociform species have robust, rounded to square heads, and terminal mouths (Figs. 19.5, 19.7, 19.10, and 19.11). Esox spp. possess a complete, straight lateral line (weakly developed in the Muskel-lunge); the lateral line is inconspicuous in Dallia, incom-plete or lacking in the Eastern Mudminnow, and lacking in Novumbra and the Central Mudminnow (Page & Burr 2011; Table 19.2). Although lacking a lateral line per se, the Central Mudminnow has an elaborate development of superficial neuromasts on the head and body. Notably, groups of 3–18 vertically and horizontally alternating neu-romasts occur along the sides in the position of the lateral line and elsewhere on the trunk (Fig. 19.16) and head (see sensory morphology and biology subsection, this section). The body and head of esociforms are covered with cycloid scales. The cycloid scales of Dallia and Novumbra resemble those of Esox in size; the scales in Umbra are large (i.e., relatively low scale counts) ( Table 19.2). Species of Esox have distinctive tri- lobed body scales (Casselman et al. A new species from the early Pleistocene of the Ukraine, †Esox nogaicus, has a massive dentary with deep symphysis (fibrocartilaginous fusion between two bones) and the pos si ble presence of a pair of fixed canine- like teeth near the anterior end of the vomer. Such canine teeth are also seen in a few species known only from North Amer i ca, the extant Muskellunge, the fossil species †E. columbianus, and an unnamed Miocene form (Grande et al. 2017). Fixed canines also occur anteriorly on the palatines in the three North American species.
In contrast to Esox, the other extant esociform genera are poorly represented in the fossil rec ord. †Novumbra orego-nensis from middle Oligocene deposits (about 31.1 mya) of the John Day Formation of Oregon is the sole fossil species presently assigned to Novumbra (Cavender 1969, 1986). Similarly, a single fossil consisting of a partial skeleton from late Miocene deposits (about 7.2 mya) in the Kenai Penin-sula, Alaska, is assigned to the genus Dallia. Lastly, otoliths assigned to Umbra from Oligocene (33.9–23 mya) and Plio-cene (5.3–2.6 mya) deposits in Eu rope form part of the fossil rec ord of this genus. Additional Paleocene through Oligocene (66–23 mya) esociform fossils from Eu rope (e.g., Sytchevskaya 1976; Gaudant 2012) are potentially impor tant for future understanding of the origins and radiation of this order. Fossils with combinations of derived and primi-tive traits not seen in living forms are potentially informa-tive in the development of evolutionary hypotheses. Apart from Cretaceous material (Wilson et al. 1992), several ex-tinct esociform genera are thought to represent transitional forms or intermediate lineages between extant esociform genera. Gaudant (2012) reviewed Eu ro pean fossil umbrids and included †Boltyschia from the early late Paleocene (58 mya) of Ukraine, †Palaeoesox from the Eocene (56.0– 33.9 mya) and Oligocene (33.9–23.0 mya) of Germany and Switzerland, and †Umbra prochazkai from the Oligocene of the Czech Republic. These were all placed within the Umb-ridae as currently conceived (i.e., closer to Umbra than to other esociforms). †Boltyschia has a combination of charac-ters seen in umbrids (e.g., several rows of small teeth on the dentary and premaxillary) and esocids (e.g., a narrower frontal, a face bone); even so, Gaudant (2012) suggested that it is a primitive umbrid.
†Palaeoesox fritzchei, described from middle Eocene de-posits (about 45 mya) of Geiseltal, Germany, has proven particularly difficult to classify. Until the description of more ancient fossils of Esox, †P. fritzchei was thought to represent a close relative of the stem lineage of Esox. Others placed it among the umbrids (Cavender 1969; Nel-son 1972; Gaudant 2012).
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NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 205 highest number of notched scales between the pelvic fins (Fig. 19.17), possibly being correlated with the numerous notched scales on the body. The higher number of notched scales on the body and the higher number of lateral- line scales of the Redfin Pickerel differentiates this subspecies from the Grass Pickerel (Crossman 1966; Grande at al. 2004).
Pigmentation The pigmentation patterns of esociforms are varied and include mottling (Alaska Blackfish; Fig. 19.7), vertical bars (Olympic Mudminnow; Fig. 19.5), horizontal bars (Eastern Mudminnow; Fig. 19.11), spots (Northern Pike; 1986) and modified (cardiform or cardioid) scales with notched posterior margins (Fig. 19.17). Notched scales are randomly scattered and intermixed with tri- lobed cycloid scales along the lateral line in all Esox species. In the Grass Pickerel, the notched scales are more numerous and far outnumber the typical lateral- line scales. Notched scales positioned between the pelvic fins occur in all spe-cies of Kenoza, but are absent in the subgenus Esox and Umbra (Grande et al. 2004). The Redfin Pickerel has the Figure 19.15. Close-up of the head of the Grass Pickerel, Esox americanus vermiculatus, emphasizing the duckbill- like snout and relatively large mouth that characterizes all members of Esox. The individual was in Lake Tomahawk, Oneida County, Wisconsin (photo graph in July by and used with permission of © Eric Engbretson / Engbretson Underwater Photography). h v v h A B Figure 19.16. (A) Distribution of groups of neuromasts along the sides and back of the Central Mudminnow, Umbra limi. Although the lateral line canal is lacking, note the groups (about 23–26) of 3–18 superficial neuromasts extending from the upper angle of the gill cleft to the caudal fin and then extending in two rows onto the caudal fin (one short and one long extending to the end of the fin). (B) Close-up of groups of oval- shaped neuromasts along the sides with groups aligned vertically (v) and horizontally (h). In laboratory experiments, enucleated individuals (eyes removed) conditioned to food responded to water currents ≥1.6 mm/s, including orientation to surface waves, apparently using at least in part the elaborate system of neuromasts to detect the currents (Schwartz & Hasler 1966) (redrawn from Schwartz & Hasler 1966).
A B C Figure 19.17. Some of the cycloid scales of Pikes (Esocidae; genus Esox) are (A) singly or doubly notched at the posterior margin (cardioid scales) or are tri- lobed (not shown). The relative numbers of cardioid scales (B) between the pelvic fins and (C) along the sides (dorsal to anal fin origin) are used to separate the Redfin Pickerel, Esox americanus americanus (more smaller scales, left), and Grass Pickerel, E. a. vermiculatus (fewer larger scales, right) (redrawn from Crossman 1966).
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NOT FOR SALE OR DISTRIBUTION Table 19.2. Morphological comparison of Pikes, Esocidae, and Mudminnows, Umbridae (Esociformes) (from Crossman 1996).
Dallia pectoralis Novumbra hubbsi Esox lucius Esox masquinongy Esox a. americanus/ a. vermiculatus Esox niger Umbra limi Umbra pygmaea Dorsal- fin spines; soft rays 0; 10–14 0; 12–15 0; 17–25 0; 15–19 0; 13–21 0; 14–15 0; 13–17 0; 14–15 0; 14–17 Anal- fin spines; soft rays 0; 12–16 0; 10–13 0; 10–22 0; 14–16 0; 13–18 0; 11–13 0; 7–10 0; 8–11 0; 13–15 Caudal fin Rounded Truncated Forked Forked Forked Forked Rounded Rounded Vertebrae 40–42 37–38 57–65 64–66 42–51 52–54 35–37 42–47 Lateral series scales 76–100 52–58 105–148 132–167 93–124 117–135 30–37 28–35 97–118 Pored lateral- line scales Pre sent but inconspicuous Absent 55–65 Weakly developed Pre sent Pre sent Absent 0–7 Pectoral spines; soft rays 0; 32–36 0; 18–25 0; 14–17 0; 14–19 0; 8–18 0; 12–15 0; 11–16 0; 12–16 0; 14–15 Pelvic spines; soft rays 0; 2–3 0; 6–7 0; 9–11 0; 11–12 0; 8–11 0; 9–10 0; 6 0; 9–10 Branchiostegal rays 14–18 8 14–15 16–19 14–16 14–17 4–5 0–14 Submandibular pores 5–6 5 (6–10) 4 (3–5) 3–5 4 (3–5) © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 207 stripe), which extends anteriorly to the snout (Figs. 19.10 and 19.11). Eye stripes and bars are lacking in U. krameri, Dallia, and Novumbra.
Size and Longevity: Esox In general, Muskellunge live longer than Northern Pike and achieve larger size (Scott & Crossman 1973; Trautman 1981; Becker 1983; Burdi & Grande 2010; Fig. 19.19). The Muskellunge normally lives 5–15 years but can reach 30 years of age and achieves an average maximum size of about 130 cm TL and weight of about 30 kg. The normal life expectancy of the Northern Pike is about seven years but may reach 24–26 years, reaching an average maximum length of about 120 cm TL and weight of about 20 kg. Fish tales being what they are, the North American angling rec ord for the Northern Pike is 21 kg, 133.3 cm TL and 31.8 kg, 163.8 cm TL for the Muskellunge (International Game Fish Association 2015; NYDEC 2016). Ultimate size in the Northern Pike and Muskellunge differs between sexes (females are generally larger than males) and varies regionally (Mann 1976; Casselman et al. 1999). A general trend is evident within species, such as the Northern Pike for the more northern populations to grow more slowly, achieve larger sizes, and live longer than southern popula-tions of the same species (Scott & Crossman 1973).
Species of pickerels also show differential sizes and ages. The Chain Pickerel lives ≤9 years and can reach 99 cm TL and about 4 kg in weight (Scott & Crossman 1973; Trautman 1981; Becker 1983; Page & Burr 2011). The smaller Grass Pickerel reaches about 38 cm TL and 0.40 kg in weight (Trautman 1981; Scott & Crossman 1973; Rohde et al. 1994; Page & Burr 2011). Both subspe-Fig. 19.2), and netlike pigmentation (Chain Pickerel; Figs. 19.3 and 19.18). Three geographic color variants of the Muskellunge occur throughout its range (see diversity and distribution section): populations in the Great Lakes basin have spots on the body arranged in oblique rows (Scott & Crossman 1973); Muskellunge in the Ohio River system, Chautauqua Lake, Lake Ontario, and the St. Lawrence River tend to have a barred or striped pattern (Trautman 1981); and populations in the inland lakes of Wisconsin, Minnesota, northwestern Ontario, and southeastern Manitoba lack distinct markings (Becker 1983).
Species of Esox tend to have brighter or more iridescent body color than other esociforms, which are mostly drab olive- green to brown in color. Overall coloration in Umbra is dark olive- green to brown- black with vertical dark brown bars along the lateral surfaces (Figs. 19.10 and 19.11). Cen-tral Mudminnows and Eastern Mudminnows have a promi-nent, vertical, dusky bar at the caudal base. Notable excep-tions to the subdued pigmentation patterns are the irides-cent blue- green highlights on the anal and dorsal fins of Central Mudminnows and the breeding colors of male Olympic Mudminnows (see reproduction section).
All species of Esox have a suborbital (below the eye) bar or tear drop, which varies among species in prominence and length. For example, the suborbital bar in Kenoza is complete and extends from the ventral margin of the eye to the ventral margin of the head (Figs. 19.3, 19.4, 19.15, and 19.18). The suborbital bar in the Northern Pike ex-tends from the ventral margin of the orbit to two- thirds the distance to the ventral margin of the head, but in the Muskellunge, the bar extends only halfway to the ventral margin of the head. The North American species of Um-bra have an indistinct horizontal eye stripe (preorbital Figure 19.18. A Chain Pickerel, Esox niger, with typical chain- like lateral coloration, swims over an algae bed in Ponce de Leon Springs, Holmes County, Florida, in October. This species inhabits vegetated lakes, swamps, and backwaters and can also occur in quiet pools of creeks and small rivers (Page & Burr 2011) (courtesy of © Isaac Szabo / Engbretson Underwater Photography).
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NOT FOR SALE OR DISTRIBUTION 208 FRESHWATER FISHES OF NORTH AMERICA 2010) and a maximum of 140 mm SL (Schilling et al. 2006). Eastern Mudminnow adult lengths are variable, ranging from 91 mm TL ( Virginia; Jenkins & Burkhead 1994) to 149 mm TL (North Carolina; Pardue & Huish 1981) to 152 mm TL (Mary land tributaries to Chesapeake Bay; Mansueti & Hardy 1967). Longevity in these species is typically ≤4 years (Applegate 1943; Becker 1983).
In a study contrasting growth characteristics of New York State lake and stream populations of the Central Mudminnow, individuals in the lake population were larger and achieved a larger maximum size, especially at older ages, than individuals from the stream population of the same age (Robinson et al. 2010). Growth rate was also greater in the lake population. Females were larger at age than males in both populations; however, all individu-als 200 mm TL and 30–35 g (Morrow 1980); however, maximum size varies across the range of the species. Individuals from near Point Barrow, Alaska, averaged about 71 mm and 4 g; the largest indi-vidual reported was 116 mm TL (Ostdiek & Nardone 1959). A 165 mm Alaska Blackfish in the 3+ age group was the longest fish collected from Big Eldorado Creek (Black-ett 1962). Fish in the Yukon- Kuskokwim delta commonly reach 255 mm TL. Introduced populations near Anchor-age, Alaska, yield larger specimens on average, and fish ≤304 mm and ≤366 g are recorded (Trent & Kubik 1974), including one specimen of about 330 mm (Morrow 1980). Alaska Blackfish live ≤8 years (Aspinwall 1965).
In Novumbra, the maximum size is seldom >90 mm TL (McPhail 1969) with most specimens being <77 mm TL (Mongillo & Hallock 1999). The average size of individu-als from several populations was about 52 mm TL with fe-males averaging slightly larger than males (Hagen et al. 1972; Mongillo & Hallock 1999). Olympic Mudminnows may live 4–5 years (A. Mearns, pers. comm.).
Size and Longevity: Umbra The Central and Eastern Mudminnows appear to have similar growth rates, average sizes, and life spans, al-though data are rather scant. The average adult size of the Central Mudminnow is 60–80 mm TL with a range of 50–115 mm (Peckham & Dineen 1957; Robinson et al. Figure 19.19. The Muskellunge, Esox masquinongy, here cruising near large logs in Sparkling Lake, Wisconsin, is a voracious, top predator. The fast- growing species is the largest member of the family and among North Amer i ca’s largest freshwater fishes (photo graph in May 2006 by and used with permission of © Engbretson Underwater Photography). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Table 19.3. Life- history characteristics of Pikes, Esocidae, and Mudminnows, Umbridae (Esociformes) from references in the text. IUCN, International Union for Conservation of Nature; AFS, Jelks et al. 2008.
Life- history attribute Dallia pectoralis Novumbra hubbsi Esox lucius Esox masquinongy Esox a. americanus/ a. vermiculatus Esox niger Umbra limi Umbra pygmaea Strictly fresh water (ppt) Yes Yes Transient in brackish water Yes Yes, infrequent in brackish water Yes Yes Yes, rarely in brackish water Maximum length (cm TL) 33 (usually <20) 8.9 (usually ≤5.3) 124 138 38 99 15 9–15 Maximum weight 366 g Unknown 28.4 kg 31.8 kg 1.0 kg 4.25 kg Unknown Unknown Maximum age (years) 8 4–5 24–26 30 7–8 8–9 4–7 4–5 Length at maturity (cm TL, unless noted) 8.0 5.4 (range, 4.1–6.5) 35–56 59–80 Male, 15.0; female 14.0 17.0 5–10 SL Unknown Age at maturity (years) 2–3 Unknown 1–6 3–5 1–4 1–4 2 1–2 Iteroparous Yes Yes Yes Yes Yes Yes Yes Yes Fecundity estimates range (average) 40–300 Unknown 7,600–595,000 (32,000) 6,000–265,000 (120,000) 186–800 (3,700– 15,000 eggs total; see text for explanation) 300 to 8,000 (see text for explanation) 220–2,286 (425–450) 31–2,566 (342) Mature egg dia meter (mm) 2.0 1.9 2.5–3.0 2.5–3.0 2.8 2–2.5 1.6 1.65 (1.4–2.2) (fertilized) Egg deposition sites Attached to vegetation Attached to vegetation Scattered on vegetation or debris in shallow water Scattered on vegetation or debris in shallow water Scattered on vegetation or debris in shallow water Scattered on vegetation or debris in shallow water Attached to vegetation Attached to vegetation Range of spawning dates and temperatures (°C) May- August; 9.2–16.8 10.0–17.9 Late March- April; 4.4–11 Mid- April- May; 9–15 March- April; 10 Early spring; 6–11 Mid- March- April; 12.8–15.6 March- April or later; 9.0–15.0 Habitat of spawning sites; water depth (cm) Vegetation at the bottom of shallow ponds and quiet streams Shallow, flooded stream margins Shallow flooded marshes, 15–25 Shallow bays, 15–75 Vegetated stream margins and floodplains Vegetated stream margins and floodplains Vegetated margins of ponds, lakes, and stream floodplains; 12–20 Vegetated margins of ponds, lakes, and stream floodplains; <10 (continued) © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Table 19.3, continued Life- history attribute Dallia pectoralis Novumbra hubbsi Esox lucius Esox masquinongy Esox a. americanus/ a. vermiculatus Esox niger Umbra limi Umbra pygmaea Incubation period 9 days at 12°C; 14 days at 10°C 9 days at 15–17°C; 13 days at 10.1°C 12–14 days at 7.2–15.6°C 8–14 days at 12.2–16.7°C 10–14 days at 10°C 11 days at 10°C 6 days; temperature not reported Unknown Mean size at hatching (mm TL) 6.0 About 5.0 6.0–10.0 7.9–9.2 5.0–6.2 4.0–8.0 5 Unknown Parental care Unknown No parental care; male defends territory after first spawning No parental care No parental care No parental care No parental care Guarding by male Courtship; nest- building, guarding, fanning by female Minimum dissolved oxygen (ppm) 2.3 2.5 1.0 3.2 Unknown Unknown 1.6 2.0 Minimum pH 6.8 3.8 5.0 Unknown Unknown 3.8 6.0 4.0 Major dietary items Aquatic invertebrates and fishes Aquatic invertebrates Fishes and other small vertebrates Fishes and other small vertebrates Aquatic insects, other invertebrates, small fishes Aquatic insects, small fishes Aquatic invertebrates Aquatic invertebrates General year- round habitat Heavi ly vegetated lowland swamps, ponds, and medium to large rivers and lakes Vegetated portions of slow streams, sloughs, marshes, and off- channel habitats Vegetated portions of lakes, ponds, slow rivers Vegetated portions of lakes, pools, and slow rivers Vegetated portions of slow streams, swamps, and lakes Vegetated portions of slow streams, swamps, and lakes Vegetated portions of slow streams, swamps, ponds, and sloughs Vegetated portions of slow streams, swamps, ponds, and sloughs Migratory or diadromous Yes, in fresh water Yes, in fresh water Yes, in fresh water Yes, in fresh water Yes, in fresh water Yes, in fresh water Yes, in fresh water Yes, in fresh water Imperilment status IUCN/AFS Least concern/ no status Least concern/ vulnerable Least concern/ no status Least concern/ no status Least concern/no status Least concern/ no status Least concern/no status Least concern/no status © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 211 Umbra spp. have four supraneurals positioned above neu-ral spines 1–4 and Novumbra and Dallia have a single su-praneural above neural spine 4 (Wilson & Veilleux 1982; Grande et al. 2004; Fig. 19.21).
Jaw (dentary and palatine) teeth in Esox spp. are com-pletely mineralized at the point of attachment of the tooth base to the associated bone (Fink 1981). The large, fang- like teeth of Pikes (and pickerels) (Fig. 19.22) are typically at-tached by full ankyloses (rigid fusion of bones) and are re-lated to a piscivorous (or at least carnivorous) diet (Type 1). Type 1 attachment also occurs in Umbra. In Dallia, the at-tachment is similar, but a narrow ring of unmineralized collagen lies between the dentine and the bone (Type 2). Pharyngeal (throat) teeth in Esox have a depressible, poste-riorly hinged tooth attachment (Type 4) with the axis of ro-tation posterior to the tooth attachment site. Pharyngeal tooth attachment is more fixed in Umbra (Type 1) and Dal-lia (Type 2).
Esociforms share several characteristics in the caudal skeleton, including one pair of uroneurals, two ural centra, and the presence of an autogenous (unfused) first preural centrum (Rosen 1974; Wilson & Veilleux 1982; Grande et al. 2004; Burdi & Grande 2010). In the caudal fin skele-ton (Fig. 19.20), the subgenus Esox has three epurals and six hypurals, but the subgenus Kenoza and Umbra have two epurals (Grande et al. 2004). The adult caudal skele tons of Alaska Blackfish have one epural, Olympic Mudminnows have five or six hypurals, and Umbra spp. have five hypurals (Rosen 1974; Wilson & Veilleux 1982).
Esox is also distinguished from all other genera by an expansion of the anterior supraneural (Fig. 19.21). The po-sition of the anterior supraneural relative to the corre-sponding neural arches is variable within Esox (Grande et al. 2004). In Kenoza the supraneural sits directly above neural arches two and three (Fig. 19.21); in the Northern Pike the supraneural is expanded and positioned above neural arches one through three; and in Muskellunge the supraneural is positioned above arches one through four. A u1 pu2 pu1 hy1 hy6 u2 un un B u1 pu2 pu1 hy1 hy6 u2 Figure 19.20. Caudal fin skeleton of the (A) Muskellunge, Esox masquinongy (133 mm SL, CU 9118) representing the subgenus Esox and (B) Grass Pickerel, Esox americanus vermiculatus (128 mm SL, FMNH 7187) representing the subgenus Kenoza (redrawn from Grande et al. 2004). Key: Epurals drawn in black; hy1 = hypural 1; hy6 = hypural 6; pu1 = preural centrum 1; pu2 = preural centrum 2; u1 = ural centrum 1; u2 = ural centrum 2; un = uroneural. Anterior to the left.
A B C sn ns4 ana ns4 ns4 sn sn sn sn sn Figure 19.21. Anterior vertebral region of the (A) Chain Pickerel, Esox niger (105 mm SL, FMNH 21811), (B) Central Mudminnow, Umbra limi (74 mm SL, FMNH 99738), and (C) Olympic Mudminnow, Novumbra hubbsi (48 mm SL, composite of UMMZ 187427 and UAMZ 3714) (A by T. Grande, B and C redrawn from Grande et al. 2004). Key: ana = accessory neural arch; ns4 = neural spine 4; sn = supraneural. Anterior to the left.
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NOT FOR SALE OR DISTRIBUTION 212 FRESHWATER FISHES OF NORTH AMERICA Mineralization of the hyomandibula and dentary occur first, followed by the maxilla, symplectic, and opercle. The last bones to mineralize are the supraorbitals and der-mosphenotics (Jollie 1975; Burdi & Grande 2010). Al-though a consistent pattern of bone development occurs, the Northern Pike grows faster than the Muskellunge, yet bone formation and mineralization at a given size is Larval Adaptations Regardless of individual variation, the relative sequence in development of the axial skeleton bones is consistent between the Northern Pike and Muskellunge (Burdi & Grande 2010). The hyomandibular, dentary, and quadrate bones form first in cartilage, followed by the hypurals. pmx pal ect ent mtg sym hy op iop sop pop ret q aa smx den mx A pmx pal ect ent mtg sym hy op iop sop pop ret q aa int den B pmx pal ect ent mtg sym hy op iop sop pop ret q aa smx den mx C D pal ect ent mtg sym hy op iop sop pop ret q aa den Figure 19.22. As top predators, the Pikes (Esocidae; genus Esox) characteristically bear large, fang- like teeth on the jaws and other bones as illustrated here on the dentary (lower jaw) and palatine (upper jaw) bones of the suspensorium (series of bones suspending lower jaw from skull) of the Northern Pike, Esox lucius, and Muskellunge, Esox masquinongy. Suspensorium of Esox lucius, lateral (A) and medial (B) view, LUD.F 241811, 78.4 mm SL; 41811; E.masquinongy, lateral (C) and medial (D) view, LUD.F 241812, 81.0 mm SL.
Anterior to left (redrawn from Burdi & Grande 2010). Key: aa = anguloarticular; den = dentary; ect = ectopterygoid; ent = entopterygoid; hy = hyomandibula; int = interhyal; iop = interopercle; mx = maxilla; mtg = metapterygoid; op = opercle; pal = palatine; pmx = premaxilla; pop = preopercle; q = quadrate; ret = retroarticular; smx = supramaxilla; sop = subopercle; sym = symplectic.
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NOT FOR SALE OR DISTRIBUTION 213 © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 214 FRESHWATER FISHES OF NORTH AMERICA Figure 19.23. First appearance or mineralization of individual bones within the axial skeleton of (A) cranial and (C) post- cranial bone formation in Northern Pike, Esox lucius, and (B) cranial and (D) post- cranial bone formation in the Muskellunge, Esox masquinongy, relative to SL (mm) and development time (h). Note, for example, the greater number of mineralized cranial bones in Muskellunge relative to Northern Pike at similar lengths (e.g., at 20 mm SL 18 skull bones have begun to form in Muskellunge versus 11 in Northern Pike) and the faster early growth of Northern Pike versus Muskellunge (e.g., SL at 1,000 h). A similar pattern is apparent in post- cranial bone formation and mineralization. Cranial and post- cranial bones are divided by species and arranged in order from the last to form (top) to the earliest. Blue denotes cartilage, red mineralization. Specific ages in hours (h) are notated on the x- axis for comparisons (redrawn from Burdi & Grande 2010).
0 5 10 15 20 25 30 35 40 45 50 55 1000 hrs 700 hrs SL (mm) Preural Centrum 2 Preural Centrum 1 Ural Centrum 1 Ural Centrum 2 Caudal Centra Uroneural Abdominal Centra Anterior Centra Supraneural Hypural 6 Epural 3 Hypural 5 Hypural 4 Epural 2 Hypural 3 Epural 1 Parhypural Hypural 2 Hypural 1 D 700 hrs SL (mm) 0 5 10 15 20 25 30 35 40 45 50 55 Preural Centrum 2 Preural Centrum 1 Ural Centrum 1 Ural Centrum 2 Caudal Centra Uroneural Abdominal Centra Anterior Centra Supraneural Hypural 6 Epural 3 Hypural 5 Hypural 4 Epural 2 Hypural 3 Epural 1 Parhypural Hypural 2 Hypural 1 C slower than in the Muskellunge (Figs. 19.23–19.25). At 25 mm SL, the axial skeleton of the Muskellunge is 55% mineralized but that of the Northern Pike is only 25% mineralized; this suggests early skeletal development in the Muskellunge is favored over increased size. This dif-ference may reflect differences in early foraging strategy between the two species. Juveniles of the Muskellunge ab-sorb their yolk sac earlier than those of the Northern Pike and become active predators just a few days after hatch-ing. A well- developed axial skeleton with fully mineral-ized teeth is likely necessary for early predacious be hav ior (Burdi & Grande 2010).
Adhesive organs, often seen in larvae, are described for a variety of teleosts (Braum et al. 1996; Grande & Young 1997; Echelle & Grande 2014; Kuhajda 2014). These struc-tures enable the larvae to adhere to rocks or surfaces while © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 215 are tall, prismatic- type granular cells with basal nuclei. Secretory activity of the organs begins shortly before hatching (Braum et al. 1996). After hatching, larvae begin to swim upward, stopping once they hit an object. Secre-tions from the adhesive organs then hold the larvae in place during the remainder of the yolk- sac stage (Kot-lyarevskaya 1969). By adhering to objects higher in the water, column larvae can respire in areas of higher dis-solved oxygen than is generally available near the substrate (Braum et al. 1996). Northern Pike and Muskellunge have active adhesive organs at hatching. Northern Pike young, however, may have a greater propensity to display adhesive be hav ior (Cooper et al. 2008). Adhesive organs also de-velop in Kenoza (Chain Pickerel, Underhill 1949; Grass Pickerel, Scott & Crossman 1973) and are used for larval attachment (Scott & Crossman 1973). Olympic Mudmin-now larvae have two pores on the side of the head that ex-trude a sticky mucus allowing them to attach to vegetation (Meldrim 1968). Neither adhesive organs nor attaching be hav ior are documented in larvae of Dallia or Umbra.
Adaptations for Air Breathing in Dallia and Umbra Anatomical, physiological, and behavioral aspects of aer-ial respiration are described in Umbra spp. and the Alaska Blackfish (Crawford 1974; Gee 1980, 1981; Krout & Dun-son 1985; Currie et al. 2010; Nelson & Dehn 2010; Lefevre Figure 19.24. Skull development and mineralization in the Muskellunge, Esox masquinongy. (A) 10.6 mm NL (notochord length), LUD.F 241800; (B) 12.1 mm SL, LUD.F 241801; (C) 13.8 mm SL, LUD.F 241802; (D) 19.9 mm SL, LUD.F 241805; (E) 21.1 mm SL, LUD.F 241806; (F) 42.4 mm SL, LUD.F 241810. Cartilage is stained blue; bone is stained red (from Burdi & Grande 2010) (used with permission of TG).
2000 1500 1000 500 Age (h) D 90 80 70 60 50 40 30 SL (mm) C 24 23 22 21 20 SL (mm) E. masquinongy E. lucius E. lucius E. masquinongy E. masquinongy E. lucius E. lucius E. masquinongy A 1100 1000 900 800 700 600 Age (h) B Figure 19.25. (A) Mean SL of the Northern Pike, Esox lucius, and Muskellunge, E. masquinongy, at the 3 cartilaginous hypural mark. (B) Mean ages (hours) of the Northern Pike and Muskellunge at the 3 hypural mark. (C) Mean SL of the Northern Pike and Muskellunge at the 20 vertebrae mark.
(D) Mean ages of the Northern Pike and Muskellunge at the 20 vertebrae mark.
Error bars denote 1 SE (redrawn from Burdi & Grande 2010).
living in fast- moving waters. The adhesive organ of the yolk- sac larvae of the Northern Pike consists of two convex epidermal patches of folds and grooves located on the lat-eral aspect of the head (Fig. 19.26). The cells of the organs © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 216 FRESHWATER FISHES OF NORTH AMERICA esophagus of Dallia was proposed based on these histolog-ical and anatomical characteristics (Crawford 1974), but functional studies of this respiratory organ are unavail-able. The esophageal skeletal musculature may be in-volved in expelling gas from the esophagus (Crawford 1971). Dallia also possesses a highly vascularized swim bladder similar in gross anatomy to that of Umbra; how-ever, capillaries do not penetrate the inner epithelial lin-ing of the swim bladder and are separated from the epi-thelium by a layer of loose connective tissue and smooth muscle (Crawford 1974). The vascularization and its con-nection to the esophagus by a pneumatic duct makes it a good candidate for gas secretion (e.g., for buoyancy regu-lation) but a poor one for gas absorption.
The internal surface of the swim bladder of the Eu ro pean Mudminnow has the histological characteristics of respiratory epithelium (Rauther 1914). Blood reaches the swim bladder of the species via the dorsal aorta (Fig. 19.27). The aorta branches into smaller blood vessels, which give rise to a dense network of capillaries that sur-round the swim bladder and cover about 80% of its sur-face (Jasiński 1965). Blood from this capillary network flows to a vein that empties into the posterior vena cava. In contrast, blood leaving the swim bladder of fishes that do not use this organ to obtain oxygen typically flows to the hepatic portal vein as would be appropriate for a part of the digestive system. The vascularization and blood cir-culation of the swim bladder of the Central Mudminnow is comparable to that described in the Eu ro pean Mudmin-now, and it presumably functions similarly (Black 1945). The Eastern Mudminnow also uses atmospheric air for respiration and prob ably also uses a modified swim blad-der for oxygen absorption (Jasiński 1965).
Vascularization of the swim bladder of the Northern Pike does not suggest a significant role in oxygen absorption et al. 2014; Nelson 2014). The morphology of these struc-tures is described for Alaska Blackfish (esophageal wall; Crawford 1974) and Eu ro pean Mudminnows (swim blad-der; Jasiński 1965). The swim bladder may have a similar function in the Central Mudminnow (Crawford 1974).
The epithelium on the internal esophageal wall of the Alaska Blackfish is profusely vascularized and densely populated by glandular cells. Posteriorly, the esophagus is separated from the stomach by a sphincter. Blood flows to the esophagus from the gastrointestinal artery, a branch of the celiac artery. Blood leaving the esophagus flows to the posterior cardinal vein. An air- breathing role for the A B eye adhesive gland nasal opening mouth Figure 19.26. The adhesive gland of the Northern Pike, Esox lucius, from scanning electron micrographs. Lower micrograph (B) is an enlargement of the right adhesive gland shown in the upper micrograph (A). The specimen is about 10 days post- fertilization and <24 h post- hatching (9.16 mm TL, 9.04 mm NL, notochord length) (photomicrograph by CT).
VN AM DP AN 0.5 CM Figure 19.27. Vascularized swim bladder of the Eurasian Mudminnow, Umbra krameri, used for oxygen absorption. Key: AM = median dorsal aorta; AN = arterial vesicae natatoriae; DP = ductus pneumaticus; VN = venous vesicae natatoriae (redrawn from Jasiński 1965).
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NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 217 in the Central Mudminnow and Eastern Mudminnow (Nelson 1972; Figs. 19.30 and 19.31). The preopercular canal is pre sent in all esociforms. Six pores are pre sent in Esox, five in Novumbra, and four in Dallia and Umbra. Esoci-forms also have a complete supraorbital canal; six or seven pores are pre sent in the esocids, and four pores occur in the umbrids. In all species of Esox plus Novumbra, three pores occur in the temporal canal, but Dallia and Umbra have two. Fi nally, a posttemporal canal occurs in all species of Esox but not in Umbra and Novumbra (Nelson 1972; Grande et al. 2004; Figs. 19.28 and 19.30).
Vision As might be expected in shallow- water visual predators, the eyes of esociforms are moderately large and well de-veloped (Fig. 19.15), possessing morphological adaptations for both acute vision in low light and the ability to distin-guish color (Braekevelt 1975, 1980). Photoreceptor type and arrangement as well as photopigments in Esox are typical of crepuscular predators. Three photoreceptor types occur in the Northern Pike: rods, single cones, and twin cones (Braekevelt 1975). Rod photoreceptors pre-dominate and perhaps reflect the crepuscular feeding be hav ior of the species. Single and twin cones are arranged in a well- defined, repeating mosaic pattern and possess a more elaborate and extensive synaptic region than that of the rods. The Northern Pike and Muskellunge color vision is most sensitive in the red- green spectrum between 500– 600 nm wavelengths. Muskellunge also have a yellow cor-nea and lens that filters out short wavelengths (Bridges 1969). Vision in Esox spp. may be more attuned to detec-tion and spatial orientation than to pro cessing visual cues about size and shape attributes of prey. Acute vision of the Muskellunge is also used in the initial acquisition of the prey, and vision and the lateral- line system are impor tant in determining the initiation of the fast start. Interest-ingly, suppression of the lateral line altered the approach of the Muskellunge to more distant prey, but bilaterally blinded fish did not stalk prey at all and only lunged at close prey (New et al. 2001). The pineal organ of the Northern Pike is photosensitive and as with the ret ina may play a role in circadian rhythm (Falcón & Meissl 1981; Falcón & Collin 1991). Although apparently little visual research is available on most non- Esox esociforms, early experimental evidence suggested that Central Mud-minnows have acute color vision allowing them to dis-criminate between edible prey items and inedible mate-rial (Hineline 1927).
for respiration (Jasiński 1965). A small gas gland occurs in the anterior ventral aspect of the swim bladder of the Northern Pike. Elsewhere on the swim bladder surface, bundles of arteries from the median aorta and the con-necting veins form a rete mirabile. Through comparison of swim- bladder circulation in dif fer ent fishes, Jasiński (1965) concluded the vascularization of the swim bladder of the Northern Pike may indicate an accessory role in buoyancy control.
Body Form and Function The body form of Esox is characteristic of an ambush predator. The body is elongate and slender, the snout, par-ticularly anterior to the eye, is elongate resulting in elon-gated jaws, and the median fins are positioned posteriorly (Maxwell & Wilson 2013). In Esox, the elongation of the body, characteristic of transient- propulsion specialists, minimizes drag and maximizes thrust (Webb 1984). This elongated body- type is hypothesized to have evolved mul-tiple times among Ray- finned Fishes (Actinopterygii) and in the case of Esox, is associated with an increase in num-bers of abdominal vertebrae added at the boundary be-tween the abdominal and caudal vertebral regions. This modularity of abdominal vertebrae and caudal vertebrae results from dif fer ent Hox gene expression and develop-mental mechanisms in each region (Bird & Mabee 2003; Ward & Brainerd 2007).
Sensory Morphology and Biology Within esociforms, the cephalic sensory canal system ranges in pattern from a complete set of canals as in the Northern Pike, Muskellunge, and Amur Pike to a reduction of sensory canals and an increase and elaboration of pit-lines in Kenoza, Novumbra, Dallia, and Umbra (Figs. 19.28– 19.32). The infraorbital canal in the Northern Pike (7–8 pores) and Muskellunge (7–9 pores) is closed and continu-ous. In the Redfin, Grass, and Chain Pickerel (8–9 pores), the canal is incomplete and interrupted, but the number of pores is not reduced (Fig. 19.28). In the other genera, how-ever, the canal is greatly reduced and replaced by a pitline, and the number of pores is reduced (0–3) (Figs. 19.30 and 19.31). The mandibular canal, which runs the ventral length of the dentary, is continuous in Esox. Four (occasion-ally ≥5) pores occur in Kenoza, five in Northern Pike, and 8–9 in Muskellunge (Nelson 1972; Grande et al. 2004; Fig. 19.28). The canal is reduced in size (two pores) in the Olympic Mudminnow and Eu ro pean Mudminnow and lost © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 218 FRESHWATER FISHES OF NORTH AMERICA and Muskellunge rely primarily on vision and lateral sen-sory systems for stalking prey (New et al. 2001). Northern Pike are attracted by the release of alarm pheromone of Fathead Minnow, Pimephales promelas (Mathis et al. 1995; Chivers et al. 1996; Chivers & Smith 1998) and exhibit distinct foraging responses to hypoxanthin-3(N)- oxide, a component of fish alarm pheromones (Mathis et al. 1995). Olfaction may play a role in predator avoidance in esoci-forms, although the species examined (Northern Pike, Alaska Blackfish, and Central Mudminnow) apparently lack specialized epithelial cells that secrete alarm phero-mones (Pfeiffer 1977; Wisenden et al. 2008). Northern Pike do not exhibit any alarm responses to conspecific skin extracts, but larvae decreased their feeding fre-quency on zooplankton and showed other anti- predator responses to chemical cues of Eurasian Perch, Perca fluvia-tilis (Lehtiniemi 2005; Lehtiniemi et al. 2005). In con-trast, Central Mudminnows exhibit distinct alarm re-sponses to conspecific skin extracts (Wisenden & Chivers 2006; Wisenden et al. 2008), which may reflect their small size and resulting vulnerability to predators.
GENETICS Karyology The chromosomal organ ization of the genomes of esoci-form fishes is a particularly well- studied aspect of their bi-ology. The genome size, karyotype, and location of the nu-PTC EC TC ML SC AL SC SNL EHL MC MPPL CL IC PPC OL A PTC EC TC ML SC AL SC SNL EHL MC MPPL CL IC PPC OL B PTC EL TC ML SC AL SC SNL EHL MC MPPL CL ICP PPC OL ICM ICA ILA ILP C PTC EL TC ML SC AL SC SNL MC MPPL CL ICP PPC OL ICM ICA ILA ANL ILP D E PTC EL TC ML SC AL SC SNL MC MPPL CL ICP PPC OL ICM ICA ILA ILP Figure 19.28. Head sensory canals, pores, and pitlines (lateral view) in the subgenus Esox (A) Northern Pike, Esox lucius; and (B) Muskellunge, E. masquinongy Esox; and subgenus Kenoza (C) Redfin Pickerel, E. americanus vermiculatus; (D) Grass Pickerel, E. americanus americanus; and (E) Chain Pickerel, E. niger (redrawn from Nelson 1972). (A) AMNH 20268, 40 mm SL; (B) AMNH, uncat., 88 mm SL; (C) AMNH, uncat., 56 mm SL; (D) ROM 25878, 126 mm SL; (E) AMNH 23714, 47 mm SL. Key: AL = antorbital pitline; ANL = anterior pitline; CL = cheek line; EC = extrascapular canal; EHL = ethmoidal pitline; EL = extrascapular pitline; IC = infraorbital canal; ICA = anterior infraorbital canal; ICM = middle infraorbital canal; ICP = posterior infraorbital canal; IL = infraorbital pitline; ILA = anterior infraorbital pitline; ILP = posterior infraorbital pitline; MC = mandibular canal; ML = middle pitline; MLI = mandibular pitline; MML = mentomandibular pitline; MPPL = mandibulopreopercular pitline; OL = opercular pitline; POL = postocular pitlines; PPC = preopercular canal; PC = posttemporal canal; SC = supraorbital canal; SL = supraorbital pitline; SNL = subnasal pitline; TC = temporal canal.
Olfaction Structurally, the olfactory system of the Northern Pike is similar to many fish species. Olfactory receptors are found on radially arranged lamellae in a bi- narial olfactory sac (Kasumyan 2004). Behavioral experiments conducted in captivity suggest a relatively minor role for olfaction in the sensory biology of esocids, at least in feeding be hav ior. Chemical cues from foraging Northern Pike elicit minor responses from conspecifics (Nilsson & Brönmark 1999), © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 219 1987; Ráb et al. 2002). Together, these mea sures of ge-nome organ ization revealed that the relatively low species diversity of living esociforms belies a rich diversity of ge-nomic arrangements. Each of the four esociform genera is characterized by a distinct genomic organ ization with diploid chromosome numbers ranging from 22 in the North American Mudminnows to 71–79 in Dallia spp. Dallia and Umbra exhibit intrageneric variation in chro-mosome number. Chromosomes may be exclusively acro-centric (centromere near one end) (Esox spp.), exclusively metacentric (centromere in the middle) (both North American species of Umbra), or complicated and distinct (Novumbra and Dallia) (Crossman & Ráb 2001). The ex-tent of this variation fits well with the ancient origin of the esociform lineages inferred from fossil evidence. The information on karyological characteristics of the ge-nomes of esociforms suggested karyotypic evolution pro-ceeded along three distinct lineages represented today by Esox, Dallia + Novumbra, and Umbra, relationships com-patible with those derived from ge ne tic and other mor-phological evidence (Crossman & Ráb 1996; Ráb et al. 2002) (see phyloge ne tic relationships section).
Karyotypic Stability in Esox Esox has a conserved genomic organ ization. The karyo-types of the five extant species consist of 25 pairs of acro-centric chromosomes (2n = 50) of gradually differing sizes (Beamish et al. 1971). The NORs in all North American species of Esox are located near the centromere of an in-termediately sized chromosome pair (Ráb & Crossman 1994). Notably, the techniques used to determine NOR localization did not produce information regarding the homology of chromosomes from dif fer ent species. Hence, the fact that NORs in the dif fer ent species of Esox are located on similarly sized chromosomes does not rule out the possibility that these actually are non- homologous ele ments. The genome size of esocids range from 2.2 to 2.7 pg/cell (Beamish et al. 1971).
A remarkable aspect of the cytological observations is that if the fossil E. kronneri is assigned correctly to the subgenus Kenoza, then the gross genomic organ ization of this genus (i.e., 2n = 50, all acrocentrics and pericentric NORs on chromosome pair 11 by size) has remained un-changed for >50 million years despite observed changes in morphology and genes among extant taxa (see phyloge ne tic relationships and morphology sections). This remark-able genomic stability may partly explain the viability of hybrid offspring from intra- subgeneric crosses of species PTC EC TC ML SC IC AL SNL EHL A PTC EC TC ML SC IC AL SNL EHL B PTC ICA TC ML SC ICP AL SNL EHL EL C PTC ICA TC ML SC ICP AL SNL EHL EL ANL ICM D E PC ICA TC ML SC ICP AL SNL EHL EL Figure 19.29. Head sensory canals, pores, and pitlines (dorsal view) in subgenus Esox, (A) Northern Pike, Esox lucius, and (B) Muskellunge, E. masquinongy, and subgenus Kenoza, (C) Redfin Pickerel, E. americanus vermiculatus, (D) Grass Pickerel, E. a. americanus, and (E) Chain Pickerel, E. niger (redrawn from Nelson 1972). Abbreviations as in Fig. 19.28.
cleolar organ izing region (NOR) is known for most species (Beamish et al. 1971). In addition, vari ous chromosome- banding techniques were applied to the karyotypes of most esociform species (e.g., Ráb & Mayr © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 220 FRESHWATER FISHES OF NORTH AMERICA EHL SC ANL ML EC TC PPC AL ICP ICA IL SNL MC MPPL CL OL A POL SC ANL ML EL TC PPC AL IC IL SNL MLI MPPL CL OL B SC ANL ML EL TC PPC AL IC IL SNL MLI MPPL CL OL C SC ANL ML EL TC PPC AL IC IL SNL MML MPPL CL OL MC D Figure 19.30. Head sensory canals, pores, and pitlines (lateral view) in the (A) Olympic Mudminnow, Novumbra hubbsi; (B) Alaska Blackfish, Dallia pectoralis; (C) Eastern Mudminnow, Umbra pygmaea; and (D) Eu ro pean Mudminow, U. krameri (redrawn from Nelson 1972).
Abbreviations as in Fig. 19.28. (A) UMMZ187427, 18 mm SL; (B) AMNH 1526, 34 mm SL; (C) AMNH 22745, 25 mm SL; (D) UMMZ 185076, 13 mm SL.
EHL SC ANL ML EC TC AL ICP ICA A EHL SC ANL ML EL TC AL PPC SNL B EHL SC ANL ML EL TC AL PPC SNL SL IL IC C EHL SC ANL ML EL TC AL SL IC D Figure 19.31. Head sensory canals, pores, and pitlines (dorsal view) in the (A) Olympic Mudminnow, Novumbra hubbsi; (B) Alaska Blackfish, Dallia pectoralis; (C) Eastern Mudminnow, Umbra pygmaea; and (D) Eu ro pean Mudminow, U. krameri (redrawn from Nelson 1972). Abbreviations as in Fig. 19.28.
EHL SL SC ANL TC MLI EL PPC AL IL SNL MPPL CL OL A MLI PPC MPPL CL OL B EHL SL SC ANL TC MLI EL PPC AL SNL IC C Figure 19.32. Head sensory canals, pores, and pitlines in the Central Mudminnow, Umbra limi. (A) Lateral view; (B) ventral view; (C) dorsal view (redrawn from Nelson 1972). Abbreviations as in Fig. 19.28.
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NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 221 tremely small acrocentric chromosomes with 2n = 78 (Crossman & Ráb 1996; Beamish et al. 1971), but 2n var-ies within individuals and among populations. Interest-ingly, dif fer ent modal diploid chromosome numbers char-acterize populations of Alaska Blackfish in the Yukon and Colville river drainages. Fish from the Yukon River system exhibited an unstable karyotype with a modal 2n number of 78 (range 71–79). In contrast, Colville River fish had an invariable, fixed karyotype of 2n = 74 (Crossman & Ráb 1996). This difference may be grounds for taxonomic sep-aration, but to date separation is not formally proposed.
Genome Size and Karyotype Changes in Umbra Gross genomic organ ization in Umbra shows significant intraordinal and interspecific variation. The estimated DNA content of Central Mudminnow and Eastern Mud-minnow cells is about two times that of all Esox, Novum-bra, and Dallia, leading Beamish et al. (1971) to conclude that the tendency at that time to include Umbra, Novum-bra, and Dallia in the same family might be ill advised. Although the North American species of Umbra have genomes >2.5 times larger than the Eu ro pean Mudmin-now (Crossman & Ráb 1996), they possess half as many chromosomes (2n = 22, metacentric and submetacentric) as their Eu ro pean congener (2n = 44, acrocentric; Beamish et al. 1971; Ráb 1981). The genome size and chromosome number of Eu ro pean Mudminnows more closely resemble those of other esociforms and other basal fishes. As such the evolution of the genome of the North American species of Umbra must have involved significant amplification of DNA sequences accompanied by the pair-wise fusion of all chromosome ele ments in the ancestral karyotype (Gornung 2013). Whether the two events are related is unknown, but the coinciding karyotypic and genome- size changes evident among species of Umbra are possibly the result of pro cesses with genome- wide conse-quences, which may play an impor tant role in molecular and organismal evolution (e.g., repetitive ele ment amplifi-cation and repetitive sequence- mediated chromosomal rearrangements).
The relatively simple gross genomic organ ization in the North American species of Umbra (22 large, metacentric and submetacentric chromosomes) made this karyotype suitable for the study of factors that affect genomic stabil-ity, such as the mutagenic effects of radiation and envi-ronmental pollutants (Kligerman et al. 1975). At least two laboratory cell lines derived from fin epithelium of the Central Mudminnow were developed, and the effects of of Esox and suggests in ter est ing ave nues of research into the evolution and maintenance of genomic synteny.
Sex Determination in Esox lucius and E. masquinongy Eggs of the Northern Pike experimentally manipulated into developing without ge ne tic input from sperm gener-ate exclusively female offspring. This indicates females of this species are homogametic, and the species has an XX:XY sex- determination system (Luczynski et al. 1997). Similar experiments on the Muskellunge produced more ambiguous results, but females of this species apparently are heterogametic because a few gynoge ne tically (no ge ne tic contribution by sperm, only initiates ovum develop-ment) derived males were observed, suggesting the action of a WZ:ZZ sex- determination system (Rinchard et al. 2002). Although no readily identifiable sex chromosomes are in the karyotypes of these Esox, incipient heterochro-matic differentiation between homologues was correlated with sex in a Eu ro pean population of the Northern Pike in which only male karyotypes possess a chromosome with a distinct telomeric C- band.
Karyotype of Novumbra The Olympic Mudminnow karyotype has 48 chromosomes (Beamish et al. 1971; Crossman & Ráb 1996) consisting of two pairs of metacentric, five pairs of submetacentric (cen-tromere closer to the middle), seven pairs of subtelocentric (centromere closer to one end), and 10 pairs of acrocentric chromosomes. Crossman & Ráb (1996) consider the meta-centric and submetacentric chromosomes as biarmed result-ing in an FN = 62 (number of major chromosome arms). This suggests that the karyotype of the Olympic Mudmin-now may represent a linkage between a hy po thet i cal esocoid ancestor (with 2n = 48, FN = 48) with the unusual karyo-type of the Alaska Blackfish by pericentromeric inversions (chromosome inversion that includes the centromere) and amplifications of NOR sites (Crossman & Ráb 2001). The ge-nome size in the Olympic Mudminnow is 2.08 pg/cell (Beamish et al. 1971).
Intraspecific Karyotypic Variation in Dallia pectoralis Among species of Dallia, the karyotypic characteristics are described only for the Alaska Blackfish. The species has a mixture of metacentric, submetacentric, and ex-© 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 222 FRESHWATER FISHES OF NORTH AMERICA ne tic basis for the morphometric variables, but hybrid in-dividuals tended to resemble the male parent. Interest-ingly, the intergrade individuals most closely resembled the Grass Pickerel, a result consistent with an earlier study (Crossman 1966), although they did show variability in multivariate discriminant space. The multivariate cen-troid for the intergrades was near the centroid of the Grass Pickerel (Fig. 19.34). The authors offered five expla-nations for the non- intermediacy of the intergrades. First, the variation is genet ically based, and the fish examined were intergrades, but proximity to pure Grass Pickerel populations caused the skewed placement toward that subspecies. Second, the variation is ge ne tic but distrib-uted clinally to cause the gradation. Third, the specimens thought to be intergrades were actually Grass Pickerel, implying a narrower zone of intergradation than Cross-man (1966) originally delineated. Fourth, the variation is nonge ne tic and caused by clinal responses to the environ-ment. Fi nally, the specimens were indeed intergrades, but equal ge ne tic contribution by the parental types is untrue for fish; that is, the overlap of F1 hybrid crosses with the parental centroid indicated in some crosses the offspring resemble the male parent. They concluded the correct ex-planation is likely some combination of these. Despite the solid morphological evidence for differentiation in the two subspecies and the well- defined morphological inter-grade zone based on two scale counts and one mea sure ment (Crossman 1966; Reist & Crossman 1987), ge ne tic investigations of the population structure or historical de-mographics across the range of these two wide- ranging and clearly differentiated forms or their intergrade zone have limited geographic sampling (Grande et al. 2004; see April et al. 2011, 2013; see phyloge ne tic relationships sec-tion). DNA barcoding provided strong support (2–2.8%) radiation and mutagenic substances on their karyotypes investigated (Suyama & Etoh 1988; Park et al. 1989).
Hybridization and Introgression Hybridization among species of Esox is common in nature and is documented experimentally (Crossman & Buss 1965; Crossman & Meade 1977; Reist & Crossman 1987). Fertile hybrids of the Chain and Grass Pickerel occur in the wild where their ranges overlap. Similarly, hybrids between the Northern Pike and Muskellunge (the tiger muskellunge) are observed frequently in the wild and some anglers favor them. Female tiger muskellunge are sometimes fertile, but males are always infertile. Infertile hybrids between the Northern Pike and members of Kenoza were produced in the lab and documented in the wild (Crossman & Buss 1965; Herke et al. 1990). In light of the fossil evidence suggesting that the two subgenera of Esox diverged >50 mya (see fossil rec ord section), it is remarkable that hybrids between spe-cies of these two lineages are viable even if sterile.
To document the ge ne tic basis of morphometric vari-ables, artificial hybrids between the Redfin Pickerel and Grass Pickerel were produced and reared in a common laboratory environment (Reist & Crossman 1987). Ten morphometric variables were mea sured on the F1 and F2 hybrids (parentage known), parental samples across the range of each subspecies, and specimens from the inter-grade zone (unknown but likely mixed parentage) (Figs. 19.33 and 19.34). Univariate and multivariate analyses sta-tistically confirmed the distinctiveness of the two subspe-cies. The multivariate centroid for the F1 and F2 hybrids was intermediate between Redfin Pickerel and Grass Pickerel samples in discriminant space (Fig. 19.34). The intermediacy of the hybrids indicated at least a partial ge Figure 19.33. Intergrade of the Redfin Pickerel × Grass Pickerel, Esox americanus americanus × E. a. vermiculatus, in McBride Slough Spring, Wakulla County, Florida, in October (courtesy of © Isaac Szabo / Engbretson Underwater Photography).
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NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 223 sistent ge ne tic differences among populations sampled from Eu rope, Asia, and North Amer i ca. In addition, vari-ability in mtDNA amplified fragment length polymor-phism supported the distinctiveness of the Southern Pike (Esox cisalpinus) (Lucentini et al. 2011). Overall, ge ne tic variability in extant populations of the Northern Pike is consistent with the idea this species survived the latest gla-cial maxima in a limited number of refugia and the sizes of the populations that survived in those refugia to eventually expand into its pre sent range were effectively small.
Seventeen pairs of primers were developed and micro-satellite markers amplified on 30 Northern Pike from the St. Lawrence River and tested on all five species of Esox (Ouellet- Cauchon et al. 2014). Average ge ne tic variation was moderate but highly variable (mean number of al-leles = 6.88, range = 2–23; mean expected heterozygos-ity = 0.49, range = 0.033–0.950; mean observed heterozy-gosity = 0.51, range = 0.033–0.967). All loci were successfully amplified in the Northern Pike from North Amer i ca and Eu rope and Amur Pike. Among other spe-cies the number of successfully amplified loci varied from eight to 11 and the quality of amplification was lower than for Northern Pike and Amur Pike.
Microsatellite marker variation also was used to test natal site fidelity in the Northern Pike. Spawning individ-uals of the species may return to the same spawning grounds year after year (see reproduction section), but whether these individuals spawn in the same area where they hatched is unclear (see Engstedt et al. 2014). A small but significant amount of ge ne tic differentiation as mea for divergence between Redfin Pickerel and Grass Pick-erel (April et al. 2013) and provided evidence for large- scale hybridization between the Redfin Pickerel and Chain Pickerel (Grande et al. 2004; Hubert et al. 2008; April et al. 2011).
Interspecific Ge ne tic Variability: Esox Few reports on the population ge ne tics of esociform spe-cies are available, and most focus on the eco nom ically impor tant Northern Pike and Muskellunge (see genome size and karyotype changes in Umbra subsection, hybrid-ization and introgression subsection, and gene expression levels and tissue specificity subsection, this section). Allo-zyme and microsatellite allele variation were used to ex-amine the ge ne tic diversity of and relationships between dif fer ent populations of the two species with some in ter est ing results, particularly concerning the ge ne tic vari-ability of the Northern Pike. In addition, sequencing of in-trons on the growth hormone across all Esox revealed unique minisatellites.
Studies of allozyme loci and microsatellite variation in populations of the Northern Pike indicated relatively low ge ne tic diversity. Only a small percentage of the loci ex-amined were polymorphic (e.g., 3–10%) (Seeb et al. 1987). The microsatellite ge ne tic diversity of this species is re-portedly lower than that of other animals with notori-ously genet ically homogeneous populations (e.g., cheetah, social wasps) (Senanan & Kapuscinski 2000). Even so, microsatellite- based studies revealed detectable and con-6 6 6 6 5 5 5 5 5 5 6 4 2 = Choctawhatchee R.
= Escambia R.
(A) Intergrades 1 3 3 1 4 2 3 3 4 2 2 2 4 4 3 4 1 2 2 2 2 2 2 4 2 (B) Hybrids 1=EAA♂ x EAV♀ 2=EAV♂ x EAA♀ 3=(EAV♂ x EAA♀)2 4=(EAA♂ x EAV♀)2 4 2 -4 -3 -2 -1 0 1 2 3 4 8 6 (C) “Pure” Forms subspecies of specimens =Misclassified =EAV =EAA A A A A V V V AV V V A A Discriminant Score Number of Individuals Figure 19.34. Histograms of discriminant function scores for the Grass Pickerel (Esox americanus vermiculatus), Redfin Pickerel (Esox a. americanus), their hybrids, and intergrades. Arrows indicate respective centroids; EAA or A, E. a. americanus; EAV or V, E. a.
vermiculatus. (A) Scores of intergrades. (B) Scores of hybrids. Symbols 1 and 2 are F1 hybrid crosses as indicated on the figure, 3 is the F2 cross (EAV♂ × EAA♀) × (EAV♂ × EAA♀), and 4 is the F2 cross (EAA♂ × EAV♀) × (EAA♂ × EAV♀). (C) Distribution of scores of reference parental samples. The open boxes indicate misclassified individuals (redrawn from Reist & Crossman 1987).
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NOT FOR SALE OR DISTRIBUTION 224 FRESHWATER FISHES OF NORTH AMERICA site fidelity (Miller et al. 2001) and reproductive homing (Crossman 1990). Although Muskellunge can travel much greater distances (LaPan et al. 1996; Weeks & Hansen 2009), sites 20 km apart were genet ically distinct, and di-vergence increased with distance between spawning sites (Wilson et al. 2016). Muskellunge in the North Channel and Georgian Bay showed significant spatial ge ne tic structure despite the lack of habitat barriers to movement, and greater interpopulation ge ne tic differences relative to other Great Lakes populations (Kapuscinski et al. 2013; Scribner et al. 2015) may result from a dif fer ent native an-cestry and a lack of introgression from stocking (Wilson et al. 2016). Clearly, further studies are needed to identify ge ne tic structuring across the range of the species and to determine if the three geo graph i cal variants recognized in this species have a ge ne tic basis (see genus Esox subsec-tion, diversity and distribution section).
Seven polymorphic microsatellite loci with 2–11 alleles and heterozygosities between 0.190–0.917 were devel-oped in Muskellunge and amplified across two other spe-cies of Esox (Reading et al. 2003). The seven microsatellite loci developed in Muskellunge amplified monomorphic products in Northern Pike and three of the seven ampli-fied monomorphic products in Grass Pickerel. Northern Pike exhibited alleles at five and Grass Pickerel at one of the loci that was diagnostically dif fer ent in size from the range of alleles found in Muskellunge. The amplification of products highly dif fer ent in size, even though mono-morphic, may prove useful as species diagnostic markers (e.g., identification of hybrids) (Reading et al. 2003).
Ge ne tic studies to date in Kenoza used allozymes and DNA bar codes. A study of allozyme variation in popula-tions of Grass Pickerel occupying dif fer ent areas along a short stream reach in Pennsylvania revealed some ge ne tic diversity. This variability, however, did not differentiate among spatially defined populations, and it remained sta-ble over the time period of the study (1971–1973) (Eckroat 1975). Because the study included only one locus (Lactate Dehydrogenase- A), diversity mea sures could not be calcu-lated. Redfin and Chain Pickerel in two regions in Canada had high levels of ge ne tic divergence in the cytochrome oxidase c I gene (>3%) with E. a. americanus from the St. Laurence River differentiated from E. a. vermiculatus from the Laurentian Great Lakes (Hubert et al. 2008). More ge ne tic divergence was observed between the two subspecies than with Esox niger. Samples of Chain Pick-erel from Quebec and New Brunswick were more closely aligned with Quebec samples of Redfin Pickerel than Redfin Pickerel was with Grass Pickerel.
sured by microsatellite allele frequencies occurred be-tween sympatric populations of the Northern Pike that spawn in dif fer ent areas of the same lake (Miller et al. 2001). These results are consistent with natal site fidelity because in its absence, spawning site fidelity alone is not expected to give rise to ge ne tic differences between popu-lations spawning in close proximity. However, the low level of ge ne tic variation observed in the study makes in-terpretation of the results problematic. Low ge ne tic varia-tion and the lack of structure among populations in the northcentral United States may have resulted from recolo-nization from a common glacial refugium, but stocking programs may also have obscured past structure (Miller & Senanan 2003).
Ge ne tic analyses have helped determine the rapidity and extent of post- glacial dispersal from non- glaciated refugia. Ancient DNA was extracted from the remains of a Northern Pike (E. cf. lucius) in interior Alaska dated to 8,820 years be-fore pre sent. The DNA of the specimen was identical to modern Northern Pike and supports a biogeographic sce-nario involving rapid dispersal of this species from glacial re-fugia in the northern hemi sphere after the last glaciation (Wooler et al. 2015). In another study, three well- defined mtDNA lineages of the Northern Pike indicated population expansion following isolation and dispersal from discrete glacial refugia. One lineage has a modern Holarctic distribu-tion suggesting transcontinental dispersal from a single Eur-asian or Beringian refugium (Skog et al. 2014).
The limited information available for populations of Muskellunge indicates that ge ne tic variability, as mea sured with allozymes, mitochondrial DNA haplotypes, and randomly amplified polymorphic DNA (RAPDs), is considerably higher than that mea sured in populations of the Northern Pike (summarized by Senanan & Kapuscin-ski 2000; Kapuscinski et al. 2013). An early study of allo-zyme variation (57 loci, 9 populations) of the Muskel-lunge, including some hatchery fish, revealed polymorphism at 10 loci (Koppelman & Philipp 1986). In-terestingly, the distribution of this variability among pop-ulations suggested some degree of ge ne tic distinctiveness. Because information on the ge ne tic structure was lacking, Muskellunge are typically managed as single stocks and have been stocked across drainage bound aries, even where evidence for stock structure exists (Crossman 1990; Farrell et al. 2007; Miller et al. 2009, 2012; Kapus-cinski et al. 2013; Scribner et al. 2015). Microsatellite ge-notyping of the Muskellunge in Lake Huron and Georgian Bay indicated small, genet ically discrete populations (Wil-son et al. 2016), providing evidence for both high natal © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 225 graphic areas. The study examined the cytochrome c oxi-dase I gene and the control region and included 188 individ-uals from 23 localities broadly representing the range of the species across mainland Alaska, St. Lawrence Island in the Bering Sea, and three sites in eastern Siberia (Chukotka Peninsula), Rus sia (Campbell & López 2014). Phyloge ne tic and molecular variance analyses supported delineation of four distinct phylogeographic units: Interior Alaska (sample locations 1–2, 4–6); Norton Sound coastal drainages (13– 14); Arctic Coastal Plain (samples north of the Brooks Range) (16–20); and Bering Coast, including populations from areas on or surrounding the Bering land bridge (east-ern Siberia, islands on the Bering Sea, and western Coastal Alaska) (3, 7–12, 15, 21–23) (Fig. 19.35). Interestingly and perhaps of taxonomic relevance, populations north and south of the Brooks Range shared no haplotypes, which is consistent with chromosome differences noted between populations to the north and south of the range (see karyo-typic stability in Esox subsection, this section). This geo-graphic barrier apparently promoted development of distinct gene pools of Alaska Blackfish. Notably, no evidence was in-dicated for a distinct mitochondrial lineage at the three lo-calities sampled in eastern Asia. Subsequent work including two nuclear gene introns, however, indicated an eastern Asian cluster (the West Beringia population) (Fig. 19.35) re-lated to but distinct from the Alaskan coastal population (Campbell et al. 2015). The mitochondrial and nuclear di-versity and spatial distribution patterns are consistent with per sis tence of lineages in multiple refugia through the last glacial maximum (Campbell & Lopez 2014; Campbell et al. 2015). Ge ne tic structure within Alaskan populations was greater than that across the Bering Sea (Campbell et al. 2015). Interestingly, estimated pairwise migration rates be-tween the populations was highest (2.42 individuals/genera-tion) from West Beringia to Coastal Alaska populations fol-lowed by migration from Interior Alaska to Coastal Alaska populations (1.57). Other migration estimates ranged from 0.15 to 0.52 immigrants/generation.
Intraspecific Ge ne tics of Umbra Only scant ge ne tic information is available for the North American Mudminnows. Kopp et al. (1992, 1994) mea sured heterozygosity (allozyme alleles at 21 loci) in popu-lations of Central Mudminnows to determine the effect of stressful environmental variables (e.g., low pH, high alu-minum concentration) on the ge ne tic variability of individu-als and the population. Control populations exhibited het-erozygosity values ranging from 0.130 to 0.142. The growth hormone gene was examined in five species of Esox by cloning the fourth intron of the gene (Barnett et al. 2007). A 33- nucleotide minisatellite in the fourth in-tron is pre sent in copy numbers ranging from seven to 16 among the species with no observed intraspecific variation in copy number (7 repeats in Chain and Grass Pickerels; 12 in Northern Pike; 13 in Muskellunge; and 16 in Amur Pike) (Barnett et al. 2007). This apparently unique minis-atellite is pre sent as a single copy ele ment in all salmonids, indicating a recent expansion in the species of Esox since their divergence from a common ancestor. Point mutations and deletions in the minisatellites suggest a model for the evolution of this ge ne tic ele ment and corroborate molecu-lar phylogenies (e.g., López et al. 2004) for the five mem-bers of the genus. Alignment of introns among the five species based on repeat number and shared polymor-phisms places them into their hypothesized phyloge ne tic relationships. The discovery of the novel structure of the growth hormone gene across Esox raises the question of whether it is pre sent in other esociforms.
Phylogeography: Novumbra Range- wide ge ne tic analy sis from 21 whole- genome micro-satellites of the Olympic Mudminnow (DeHaan et al. 2014) supported an earlier hypothesis that its biogeogra-phy was structured by glacial refugia, which existed during the Pleistocene, with limited subsequent range expansions (Meldrim 1968). High population ge ne tic structure among sampled populations (FST = 0.273) and a correspondingly high level of ge ne tic isolation in populations along the Olympic Coast (FST = 0.350) indicate historically limited gene flow (DeHaan et al. 2014). Populations of the Olym-pic Mudminnow in eastern Puget Sound (Cherry Creek drainage, Snohomish River drainage, and Issaquah Creek, Lake Washington drainage) had an allelic structure more similar to populations on the southern coast of the Olym-pic Peninsula than the nearby Chehalis River drainage. The lack of unique mtDNA alleles (Pickens 2003), which would reflect glacial relicts if pre sent, suggests Olympic Mudminnow populations in eastern Puget Sound drain-ages are the result of intentional introductions.
Phylogeography: Dallia A phylogeographic study of the Alaska Blackfish revealed a significant level of interspecific variability largely corre-sponding to historical and con temporary barriers with little to no mixing of mitochondrial haplotypes among geo-© 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 226 FRESHWATER FISHES OF NORTH AMERICA 1994). Two Canadian populations of Central Mudmin-nows exhibited >2.7% ge ne tic divergence of cytochrome c oxidase I (COI) (Hubert et al. 2008).
Gene Expression Levels and Tissue Specificity Studies on gene expression illustrate another aspect of es-ociform ge ne tics beyond karyotypes and intraspecific ge ne tic variation. Ge ne tic regulation of white muscle cy-tochrome c oxidase (COX) activity in the Northern Pike and Central Mudminnow indicated COX activity was sig-nificantly higher in winter in Central Mudminnows (3.5fold) but did not increase in the Northern Pike (Bremer & Moyes 2011). Laboratory acclimation to winter temperatures did not alter COX activity in Central Mud-minnows, suggesting a role for nonthermal cues.
Exposure to mutagens affects tissue- specific gene ex-pression, produces DNA adducts (DNA bound to a cancer- causing substance), and induces chromosomal and nu-clear anomalies in Pikes and Mudminnows. Northern Pike exposure to perflourinated compounds in municipal waste water resulted in tissue- specific expression of eight gene products in blood (metallothionein and vitellogenin), muscle and gill (metallothionein, glutathion- S- transferase, superoxide dismutase, catalase and cytochrome P450 1A1), and liver (vitellogenin, estrogen receptor beta-2, and cytochrome P450 3A27) tissues (Houde et al. 2013). Juve-nile Northern Pike subjected to a single exposure (intra-peritoneal injection or food) of a mutagenic mixture of compounds exhibited rapid, dose- and time- dependent ex-pression of DNA adducts in liver and intestine. Experi-mentally induced chromosomal mutations ( sister chroma-tid exchanges) and nuclear anomalies (micronuclei) are described for Umbra spp. by several authors (Kligerman & Bloom 1976; Kligerman 1979; Metcalfe 1988; Alink et al. 2007). These abnormalities were induced by direct expo-sure to or injection of known mutagens, such as ethyl methanesulphonate and benzo(a)pyrene (Hooftman & Vink 1981; Hooftman & de Raat 1982).
Patterns of gene expression in Novumbra, Dallia, and Umbra revealed significant changes in the tissue specific-ity and the level of gene expression at 27 dif fer ent loci (Kettler & Whitt 1986; Kettler et al. 1986). Two related, but dif fer ent, mea sures of ge ne tic distance from patterns of gene expression revealed a greater similarity between Dallia and Novumbra than between Umbra and Dallia or Novumbra, reflecting current phyloge ne tic relationships. Unfortunately, no comparable expression data are avail-able for any of the species of Esox.
0 250 500 1000 km 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 3 2 1 5 4 Interior Alaska Arctic Coastal Plain Coastal Alaska West Beringia 0.15 2.42 0.12 0.52 1.57 0.08 Figure 19.35. (upper) Map of the major phylogeographic regions revealed in a mtDNA analy sis of 188 specimens of the Alaska Blackfish, Dallia pectoralis, sampled across the Alaska mainland and St. Lawrence Island (sample locations 1–20) and Chukotka, Rus sia (sample locations 21–23). (lower) Migration rates among Beringian Dallia populations. Migration rates in individuals/ generation since the time of divergence (t) between populations of Dallia estimated by Isolation with Migration analyses. Approximate geographic distribution of populations defined by clustering algorithms are outlined with black dashed lines (Norton sound samples excluded). Arrows reflect the magnitude of migration rate. Current coastlines and international borders (yellow lines) are overlaid onto a map available from the U.S. National Geophysical Data Center of -110 m sea level (dark green layer) corresponding to the last glacial maximum (redrawn from Campbell & Lopéz 2014, upper, and Campbell et al. 2015, lower).
Heterozygosity values were significantly lower in stressed populations. In these experiments, the more genet ically variable (i.e., heterozygous) individuals were significantly more tolerant of acutely toxic conditions (Kopp et al. © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 227 feeding at temperatures of 32.0–35.3°C and mortality ap-peared to result from starvation (Scott 1964). Muskellunge ranging in age from 1–51 days post- hatching had a CTM of 27–36°C (acclimation temperature range, 7–22°C) (re-viewed by Beitinger et al. 2000). Northern Pike and Mus-kellunge showed comparable results based on experimen-tal determinations of growth optima and preferred and lethal temperatures (reviewed by Jobling 1981). Optimum temperature for growth is slightly higher for young- of- the- year (22–23°C), than for juveniles and subadults (Cassel-man 1996). Muskellunge larvae held at ambient tempera-tures of about 20°C had an average CTM of 32.8°C; CTM increased from 29.9°C on day 4 (at the onset of swim-up) to 35.6°C on day 31 (Bonin & Spotila 1978). Laboratory ex-periments and predictions from bioenergetics models sug-gested that Northern Pike should grow faster than Muskel-lunge at cool temperatures, but slower than Muskellunge at warm temperatures (>25°C), although no significant dif-ferences in metabolic rates were observed (Bevelhimer et al. 1985). Preferred temperatures of Chain Pickerel was 24°C and Grass Pickerel was 26°C; Chain Pickerel avoided temperatures >27°C and <20°C (Ferguson 1958; Coutant 1977; Wismer & Christie 1987).
Temperatures affect abundance, growth, metabolic rate, and habitats occupied by esociforms. Canadian fish-eries surveys report an increase in Northern Pike abun-dance in eastern Lake Ontario over seven de cades, corre-lated with a significant increase in water temperature (Casselman & Dietrich 2003). Growth rate increases rap-idly at >10°C and is highest between 19°C (for biomass) and 21°C (for length); upper lethal temperature is 29.4°C, and the species can tolerate temperatures ≥0.1°C as lakes approached freeze-up (Casselman 1978). In laboratory studies, the final preferred temperature of the Northern Pike was slightly higher (2–3°C) than the optimum for growth (McCauley & Casselman 1981). Use of temperature- sensitive radio transmitters on adult North-ern Pike in two southern Ohio impoundments indicated that, when water temperature was >20°C, fish sought a cooler average temperature, but they sought significantly cooler temperatures only when the surface was >25°C. Once the surface temperature reached 25°C, fish sought the coolest available water with >3 mg/l dissolved oxygen. Northern Pike maintained body temperatures ≤25°C when surface water temperatures were >26°C. Adult fish lost weight during this period of habitat constriction (Headrick & Carline 1993).
The ability of many esociforms to survive extremely low temperatures is poorly understood despite long- standing PHYSIOLOGY Although factors affecting growth and reproduction of the large esocids are best known (see ecol ogy and reproduc-tion sections, respectively), esociforms exhibit several impor tant, and extreme, physiological adaptations. Nota-ble among these are cold tolerance in Dallia and Umbra, adaptations for air breathing in hypoxic conditions, and tolerance of low pH and brackish waters. Here we focus on aspects of esociform tolerances to temperature, hy-poxia, pH and alkalinity, salinity, and swimming ability. Sensory adaptations are treated in the morphology section.
Temperature Tolerance The temperature tolerances of esociform species cover a broad range. Latitudinal ranges and the mean annual tem-peratures characteristic of those latitudes differ among species (Newbrey et al. 2008). For example, Esox lucius in-habits latitudes from 37 to 71 degrees N latitude in areas with mean annual temperatures ranging from −7° to 15°C. Its southern extent appears to be temperature- limited; Northern Pike are absent from the Mississippi and Ohio Rivers south of the 31–32°C isotherm and summer die- offs occur in midwestern lakes and reservoirs when surface temperatures reach 32°C. Dallia range from 58 to 71 de-grees N latitude with mean annual temperatures of 0–3°C. At the other extreme, the ranges of Esox niger and E. ameri-canus extend from 27 to about 46 degrees N latitude in re-gions with mean annual temperatures of 2–23°C. The spe-cies of Umbra, Novumbra, and both Esox reichertii and E. masquinongy inhabit regions with intermediate mean an-nual temperatures. The oldest esociforms likely inhabited much warmer climate zones between the Cretaceous and Early Eocene. Large body sizes in the subgenus Esox and to some extent, Dallia, may represent adaptations to climatic cooling during the latter part of the Cenozoic (Newbrey et al. 2008). Muskellunge occur throughout the Ohio River basin (including the Tennessee- Cumberland River systems) south of the 31°C isotherm. Novumbra individuals tolerate temperatures of 0–27°C but preferred tempera-tures are 11–14°C as inferred from a habitat occupancy study (Meldrim 1968).
Some information is also available for large Esox spp. on critical thermal maxima (CTM) and preferred temperature versus optimal temperature for growth. Northern Pike ac-climated to temperatures of 25.0, 27.5, and 30.0°C ceased © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 228 FRESHWATER FISHES OF NORTH AMERICA condition. Northern Pike exhibited 100% mortality dur-ing 1 h exposures to oxygen concentrations of 0.5– 2.0 mg/l in highly eutrophic lake water. As in most fishes, critical oxygen concentration depends on temperature (re-viewed by Casselman 1978, 1996; Casselman & Lewis 1996). In Northern Pike acclimated at 15 and 19°C, dis-solved oxygen concentrations of 4.5 and 5.0 mg/l had no effect on survival, larval development, and onset of feed-ing. At concentrations <2.5 mg/l, hatching and survival were significantly impaired (Siefert et al. 1973). Under ex-perimentally induced hypoxia (dissolved oxygen <3 ppm), Northern Pike growth rate and food consumption were reduced (Adelman & Smith 1970). Greater relative sur-vival of smaller individuals suggested that small Northern Pike were more resistant to hypoxia- related winterkill (Casselman & Harvey 1975).
Low dissolved oxygen responses affect feeding, vertical and horizontal distribution, and survival in Pikes and Mudminnows. Stress avoidance (seeking more oxygenated conditions) in the Northern Pike begins at concentra-tions <4 mg/l and feeding ceases when oxygen concentra-tion is <2 mg/l. The vertical and horizontal fish distribu-tion under ice was a function of low dissolved oxygen and high carbon dioxide (Magnuson et al. 1985). Northern Pike can avoid low dissolved oxygen in late winter by oc-cupying deeper habitats (Casselman 1978), and juveniles are attracted to traps with an aerated discharge under the ice (Johnson & Moyle 1969). Eutrophic lakes experience winterkill, particularly when surface ice is cloudy or snow- covered. Resident Northern Pike and Grass Pickerel appeared to actively seek areas of higher dissolved oxygen in a stream inlet. In lakes where no inlet or outflow oc-curred, survival of Northern Pike was greater at dissolved oxygen concentrations >0.5 mg/l with severe winterkill at 0.2–0.0 mg/l; Grass Pickerel survived at lower concentra-tions (dissolved oxygen between 0.1 and 0.3 mg/l) (Coo-per & Washburn 1949). Sudden onset of near- anoxic con-ditions (0.01–2.0 mg/l) following Hurricanes Fran, Bonnie, and Floyd resulted in mortality of Chain Pickerel and other fishes (Mallin et al. 2002).
Although it lacks morphological adaptations for air breathing, individuals of Novumbra can tolerate oxygen levels <0.2 mg/l (Meldrim 1968). Olympic Mudminnows occurred in wetland habitats as dissolved oxygen concen-trations decreased from 10 to 0 mg/l (Henning et al. 2007) and occupied riverine habitats with dissolved oxy-gen concentrations of 1.1–11.5 mg/l (Kuehne & Olden 2016). Olympic Mudminnows may actively seek areas with low dissolved oxygen, low temperatures, and high interest in this aspect of esociform physiology, which origi-nated in anecdotal accounts from the earliest encounters of populations of Alaska Blackfish by western explorers. Ac-cording to these accounts individual Alaska Blackfish can survive complete freezing of their tissues for prolonged pe-riods of time. Barton Evermann (in a letter to the editors of Forest and Stream, 18 January 1913) recounted the endur-ing story (often repeated but largely undocumented) of na-tive Alaskans capturing Alaska Blackfish in the autumn and retaining them frozen in baskets to be fed to their sled dogs during winter. The fish, allegedly thawed by the warmth of the dog’s stomachs, revived sufficiently to cause the dogs to regurgitate them alive (Turner 1886:101; “This I have seen, but have heard some even more wonderful sto-ries of this fish.”). Scholander et al. (1953) investigated the claim and found that Alaska Blackfish can survive partial freezing, if only for a few days, but not being frozen solid. Extensive tissue damage eventually led to death. Over a temperature range of 0–15°C, Alaska Blackfish had the low-est metabolic rate of any Arctic fish examined.
Other anecdotes confer similar tolerance to the Eu ro pean and Central Mudminnows. For example, Berg (1948) related an account of individuals of the Eu ro pean Mudminnow sur-viving >2 days of winter temperatures without water. The fact that some populations of Alaska Blackfish and the Cen-tral Mudminnow occur in shallow water bodies that un-dergo extensive winter freezing nurtured the expectations raised by those early accounts (e.g., Gudkov 1998). So far, however, careful observation indicates that, although well suited to survive cold- water temperatures, these species can-not survive complete freezing and partial freezing usually results in necrosis of the affected tissue (e.g., Scholander et al. 1953). Whether physiological (e.g., seasonal production of freeze- resistant substances) or behavioral (e.g., local movement to suitable microhabitat) adaptations allow these species to inhabit areas with apparent complete winter freezing (e.g., shallow Siberian lakes) is unstudied. Regard-less, most esociforms are clearly well adapted to low water temperatures and remain active during the winter even under ice cover (Scott & Crossman 1973). For example, 14 of 66 Central Mudminnows examined had fed (food in stom-ach or intestine) in Fish Lake, Ontario, under the ice at water temperatures of 4°C (Keast 1968).
Hypoxia Tolerance: Esox and Novumbra Northern Pike tolerate depressed oxygen concentrations ≤0.3 mg/l in shallow lakes and reportedly survive levels ≤0.04 mg/l although survival may depend on lake trophic © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 229 servations of winterkill may have resulted from methane discharge that is typical of the thermokarst lakes where Alaska Blackfish are frequently found on the Chukchi Peninsula (Gudkov 1998) and in Interior Alaska (Blackett 1962). Methane production in these lakes can be ex-tremely rapid and average 73–88% by gas volume (Walter et al. 2008, 2010). When trapped under ice, methane can be oxidized aerobically (producing carbon dioxide) or an-aerobically (producing hydrogen sulfide) in lethal concen-trations (Scidmore 1957; Magnuson et al. 1985; Reiffen-stein et al. 1992). If Alaska Blackfish exhibit the same be hav ior as described in Mudminnows, methane dis-charge trapped under ice would reduce the concentration of oxygen in trapped bubbles, leaving insufficient oxygen for survival. Methane bubbling, however, may be large enough to maintain open leads in the ice and contribute to the long- term survival of Alaska Blackfish. Subsistence fishermen frequently exploit the be hav ior as fish aggre-gate at the holes (Andersen et al. 2004) and refer to them as blackfish holes (Campbell et al. 2014).
The physiology of air breathing and some of its environ-mental and behavioral correlates were studied in the Cen-tral Mudminnow (Gee 1980), which is a facultative air breather that continuously breathes air when it is accessi-ble. The fish appears to efficiently use both air and water- derived oxygen and to switch between the two modes of breathing depending on many variables, including water temperature, dissolved oxygen concentration, presence of conspecifics, availability of vegetation cover, and predator risk. Because fish swimming to the surface to gulp air conductivity associated with groundwater contributions from springs (Meldrim 1968; Kuehne & Olden 2016).
Hypoxia and Air Breathing: Dallia and Umbra The ability to breathe atmospheric oxygen is documented in Dallia and Umbra. In both, modified vascularization of areas of the gut or swim bladder surface serve in gas ex-change (see morphology section). Facultative air breath-ing enables Alaska Blackfish to survive under hypoxic conditions that occur in warm shallow wetlands when summertime dissolved oxygen is <2.3 mg/l (Ostdiek & Nardone 1959; Crawford 1974; Morrow 1980) or by using trapped air bubbles in ice covered habitats during winter (Haynes et al. 2014). Individual Alaska Blackfish main-tained normal metabolic rates at <2 mg/l dissolved oxygen by supplementing aquatic respiration with air breathing (Crawford 1971), which appears to be influenced by cen-tral carbon dioxide chemosensitivity (Hoffman et al. 2009). Alaska Blackfish could maintain standard oxygen uptake (ṀO2) in hypoxic conditions by increased aerial respiration (Fig. 19.36); when air was denied, standard ṀO2 was reduced by about 30–50%. Additional mecha-nisms must exist for Alaska Blackfish to survive hypoxic submergence during the winter, such as hypoxia- induced enhancement in the capacities for carry ing and binding blood oxygen, behavioural avoidance of hypoxia, and sup-pression of metabolic rate (Lafevre et al. 2014).
Despite these adaptations, winter mortality in Alaska Blackfish may be significant (Campbell et al. 2014). Ob-180 150 120 90 60 30 0 180 150 120 90 60 30 0 0 5 10 15 20 25 0 5 10 15 20 25 B Time (h) D C M · O2 (mg O2 kg–1 h–1) Water (6) (5) Total (6) (5) Air A Figure 19.36. Bimodal oxygen uptake over 24 h in the Alaska Blackfish, Dallia pectoralis. Oxygen uptake (ṀO2) from air, water, and in total recorded over 24 h for 5°C (A, B) and 15°C (C, D) acclimated fish in normoxia (A, C: 19.8 kPa) and hypoxia (B, D: 2.5 kPa). Values are mean ± SE. Dif fer ent individuals were used in the four experimental treatment groups; sample sizes are given in parentheses (redrawn from Lafevre et al. 2014). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 230 FRESHWATER FISHES OF NORTH AMERICA ratory simulations of winterkill conditions (1.6–3.9°C water temperature, water surface covered), Central Mud-minnows also frequently engulfed gaseous bubbles (Magnuson et al. 1983; Fig. 19.38). The use of bubbles was unrelated to methane or nitrogen content (0–80%) if all bubbles contained 20% oxygen. If oxygen content was varied (0–20%), fish visited bubbles randomly but re-mained longer and engulfed gas at lower frequencies at increases their predation risk, individuals of the Central Mudminnow switch from random to synchronized air breathing in the presence of simulated predators. An in ter est ing aspect of the physiology of air breathing in the species is the relationship between buoyancy and the changes in swim bladder volume caused by air breathing and subsequent gas absorption. In the absence of simu-lated predators, buoyancy and air breathing are tightly co-ordinated to minimize the energy cost (increased swim-ming effort) incurred by changes in buoyancy caused by air breathing and gas absorption. When predator presence is simulated, this coordination is disrupted, which results in the fish spending more energy swimming against their changing or increasing buoyancy (Gee 1980).
Under ice- covered habitats, Central Mudminnows use bubbles composed of gas mixtures other than air for respi-ration (Klinger et al. 1982), likely an adaptation to the oxygen- depleted, carbon- dioxide rich water of northern winterkill lakes (Magnuson et al. 1983). Central Mudmin-nows held in field enclosures with air bubbles in a winter-kill lake (dissolved oxygen near 0.0 mg/l) survived longer than fish in enclosures lacking air bubbles (Fig. 19.37). The oxygen content of the naturally occurring bubbles was 3% (range, 0–11%) at dissolved oxygen levels of 0.5 mg/l; the bubbles also contained 1–75% nitrogen and 23–98% methane (Klinger et al. 1982). In a series of labo-with air bubbles without air bubbles 0 25 50 75 100 0 3 6 9 12 15 18 21 24 ELAPSED TIME (h) SURVIVAL (% alive) Figure 19.37. Survival of Central Mudminnows, Umbra limi, held in enclosures with air bubbles (closed circles) and without (open circles) during winter ice over in Mystery Lake, Wisconsin (redrawn from Klinger et al. 1982).
Low dissolved O2, Naive fish High dissolved O2 Low dissolved O2, experienced fish Major gases vary O2 (20%), CH4 varies O2 varies A B C D E F 0 20 40 60 0 20 Time (%) Time (%) Air bubble Open Plugged Ice Bottom Hole { Air bubble Open Plugged Ice Bottom Hole { Air bubble Open Plugged Ice Bottom Hole { Air N2 CH4 0 20 40 60 80 0 2.5 5 10 20 CH4 (%) O2 (%) Figure 19.38. (upper) Time (mean ± 95% CI) spent by Central Mudminnows at air bubbles, open holes, plugged holes, simulated ice at surface, and the bottom under three experimental conditions to simulate winterkill conditions in northern lakes: (A) low dissolved oxygen- high carbon dioxide (naïve fish); (B) high dissolved oxygen- low carbon dioxide; and (C) low dissolved oxygen- high carbon dioxide (experienced fish). At open holes fish could ventilate at the holes and pull in oxygenated water but could not breathe air. (lower) Time spent by Central Mudminnows at bubbles with vari ous gas mixtures: (D) air, nitrogen, and methane bubbles; (E) bubbles varying in nitrogen and methane content; and (F) bubbles varying in oxygen content. Values that do not share a common underline are significantly dif fer ent (P<0.05) from each other.
Water temperatures were 1.6–3.9°C in all experiments (redrawn from Magnuson et al. 1983).
© 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 231 rence of Redfin Pickerel increased gradually as pH in-creased from about 4.5 to 8.5 (median, about 7.2), al-though sample sizes of that species were somewhat low (Graham 1993). Habitats of the Olympic Mudminnow generally have pH levels of 5.9–8.2, although the fish tol-erates pH ≤3.8 (Washington Department of Ecol ogy 2012; Kuehne & Olden 2014). Those for Alaska Blackfish range from pH 6.8–8.0 (Ostdiek & Nardone 1959).
As would be expected from their preferred habitats, Um-bra spp. are highly acid tolerant occurring at a pH that would stress or even kill many fishes. The species may pos-sess physiological and cellular adaptations, which allow it to compensate for sodium loss at low pH (Krout & Dunson 1985; Flik et al. 1987). Across 85 New Jersey lakes, Eastern Mudminnows were captured at pH of 4.0–8.2 (median, about 5.6); their probability of occurrence was near 0.8 at pH = 4, but decreased with increasing pH, and approached a probability of occurrence of 0.1 at pH = 8 (Graham 1993). In The Netherlands, Eastern Mudminnows occurred at pH of 3.5–4.0 (alkalinity <0.1 meq/l) (Dederen et al. 1986). In experimental settings, Eastern Mudminnows showed no mortality at a pH of 2.8 (den Hartog & Wendelaar Bonga 1990). Notably, growth rates in the Eastern Mudminnow acclimated to acid waters surpassed those held at a pH of 6.5–7.5 (Flik et al. 1987). In another laboratory trial (Ra-hel & Magnuson 1983), Central Mudminnows were ranked 10 out of 12 species tested for ability to tolerate low pH. Of individuals tested, 41% survived a pH of 3.05 for a median of 240 h. In 43 northern Wisconsin lakes, Central Mud-minnows occurred at the lowest pH (4.0) of 31 species cap-tured (Rahel & Magnusson 1983). They occurred most of-ten in lakes with median pH of 5.1–5.3 (overall range, 4.3–6.9) and least often in lakes with median pH of 6.5 (range, 6.1–7.4) (Rahel 1984).
Salinity Tolerances Tolerances of salinity are best documented in Esox spp., but research in that arena is confined primarily to the Northern Pike. Although most populations of Northern Pike are confined to fresh waters in lakes and rivers, the species can also tolerate brackish water. In a review of es-tuarine salt marsh fishes in eastern North Amer i ca, the Northern Pike was categorized as a freshwater transient species (Nordlie 2003) based on its occurrence in the brackish waters of five estuaries in eastern James and Hud-son Bays (Morin et al. 1980). Northern Pike are also occa-sionally caught in commercial coastal fisheries (Lee 1999; Southcentral Alaska Northern Pike Control Committee bubbles with higher oxygen content. Although fish occa-sionally visited open holes in the surface (used to simulate cracks in the ice), they were not recorded there during ob-servation periods, indicating that the Central Mudmin-nows preferred to use bubbles rather than to pump oxy-genated water down through the holes. Whether the bubbles come from sediments, from gases extruded in water when it freezes, or from gases exhaled or escaping from the fur of aquatic mammals (e.g., muskrats, beavers), they are a critical winter resource for Central Mudmin-nows, allowing them to survive in environments that are lethal to other fishes (Magnuson et al. 1983). Central Mudminnows also actively seek areas of higher dissolved oxygen in a stream inlet in ice covered lakes.
The ability of Central Mudminnows to use their lung- like gas bladder to serve as an oxygen reservoir also has foraging implications in stratified lakes with hypoxic bot-tom waters. Using field observations and laboratory simu-lations, Central Mudminnows in a stratified northern Wisconsin lake were documented foraging during the day in hypoxic bottom sediments for phantom midge larvae. The Central Mudminnows ingested air bubbles at the sur-face to enhance the time they could forage on the bottom in the other wise lethal hypoxic conditions of the hypolim-nion (Rahel & Nutzman 1994).
Between 5–20°C the Central Mudminnow could com-pensate all of the systems involved in energy metabolism so that its metabolic rate and oxygen consumption were nearly constant (Hanson & Stanley 1969, 1970; Currie et al. 2010). This allows the fish to sustain a high degree of activity at both cold and warm temperatures.
Alkalinity and Acidity Esociforms show varying levels of tolerance to low and high pH levels. Northern Pike tolerate a wide range of al-kalinity. In Wisconsin lakes they occur at pH of 6.1–8.6 (Margenau et al. 1998). In highly alkaline Nebraska lakes, a population survived about 14 months at pH 9.5 until to-tal alkalinity was >1200 mg/l and carbonate alkalinity was >600 mg/l. In another Nebraska lake a population survived and recruited, albeit sporadically, for de cades at pH 9.3 (total alkalinity range, 460–950 mg/l; carbonate alkalinity mean = 130 mg/l) (McCarraher 1971). At the other extreme, a Northern Pike population in Saskatche-wan reportedly sustained itself at pH 5.0 (Inskip 1982). Chain Pickerel had a probability of occurrence of >0.8 across a pH range of about 4.0–9.0 (median, about 7.0) in 85 New Jersey lakes. In contrast, the probability of occur-© 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 232 FRESHWATER FISHES OF NORTH AMERICA 1984). The sustained swimming effort of Northern Pike over a 10 min period was related to size but was among the lowest of 12 species tested (Jones et al. 1974). For exam-ple, 20 cm FL individuals were fatigued in 10 min at a water velocity of only 10 cm/s and 50 cm FL individuals fa-tigued at about 25 cm/s. In comparison, Walleye, Sander vit-reus, at the same sizes fatigued at much higher flow rates of 45 and 80 cm/s. Swimming activity of the Northern Pike in the laboratory is highly correlated with temperature with an activity optimum at 19.5°C (Casselman 1978).
Similarly, swimming endurance studies comparing a continuous swimmer, the Creek Chub, Semotilus atromac-ulatus, a cyprinid (Carps and Minnows), with the esoci-form, the Central Mudminnow, revealed the esociform was best suited for short bursts of activity. The metabo-lism of the Central Mudminnow is adapted to meet the anaerobic energy demands associated with fast strikes (Goolish 1991). From these results and the common mor-phological and ecological features shared by all esoci-forms, all of the species can be presumed to fare poorly under sustained strong flow.
Other Aspects of Esociform Physiology The Northern Pike was used as a model organism in the study of a number of disparate aspects of physiology, such as circadian rhythm (Gaildrat & Falcón 2000 and refer-ences therein), and cadmium tolerance (Norey et al. 1990). Considerable effort was directed at understanding factors related to the physiological variables affecting pro-duction in the Northern Pike, and to a lesser extent, the Muskellunge. These efforts were motivated in part by the need to improve management programs. Diana (1996) presented an in- depth review of the lit er a ture on the en-ergy demands and bud get of the Northern Pike with allu-sions to relevant information for the Muskellunge.
BE HAV IOR Movement and Activity A behavioral feature shared by all esociform species is the tendency to be sedentary. Esociforms may remain at the same location either resting on the bottom or hovering among aquatic vegetation for extended periods of time (e.g., Cook & Bergersen 1988; Jepsen et al. 2001). In Dal-lia, Novumbra, and Umbra foraging or air breathing peri-odically interrupts these sedentary periods. Local move-ment and holding in place are typically accomplished by 2007). Such reports raise the possibility of the species seeding new freshwater habitat through coastal infiltra-tion. Its importance as a commercial species in the Baltic Sea has prompted several studies of salinity tolerance (Mann 1996). Microsatellite analy sis of Northern Pike populations in the southern Baltic Sea showed that stock-ing programs aimed at supplementing the Baltic Sea com-mercial harvest failed to increase recruitment when freshwater- reared fingerlings were used. This suggests that wild- spawned fish are adapted to the slightly saline water of the southern Baltic Sea (around 6 ppt) (Larsen et al. 2005). In a study of salinity tolerances of embryos and lar-vae of the Northern Pike, brackish- water populations ex-hibited greater salinity tolerance (LC50 = 13.2 ppt at 15°C) than those of freshwater origin (LC50 = 12 ppt at 14°C), al-though differences were relatively slight (Jacobsen et al. 2007; Jørgensen et al. 2010). Salinity tolerance of the Northern Pike varies during dif fer ent life stages, and high salinity affects both fertilization and development (Raat 1988; Jacobsen et al. 2008). Growth was also reduced at high salinities, suggesting metabolic energy that other wise might support growth is diverted to osmoregulation (Engström- Öst et al. 2005; Jacobsen et al. 2008). Tempera-ture may be a factor affecting salinity tolerance. Larvae reared at 3, 14, and 18°C showed the least re sis tance to os-motic stress in a 90 min exposure in saline solution (2% sodium chloride in hatchery water) with optimum temper-atures for fertilization, hatching rates, and quality of larvae of 6–10°C (Bondarenko et al. 2015). Chain Pickerel are re-ported from brackish water (Scott & Crossman 1973; Jen-kins & Burkhead 1994). The species occurred at 3 ppt sa-linity and Redfin Pickerel at 10 ppt (Keup & Bayless 1964).
Little is known definitively about salinity tolerances in other esociform genera. The Eastern Mudminnow occurs in brackish (salinity <5 ppt) habitats in Delaware (Wang & Kernehan 1979) and survived salinities of 10 ppt in labora-tory experiments (Hoese 1963). Olympic Mudminnows can tolerate salinities of 11.5 ppt but prefer salinities <0.5 ppt (Meldrim 1968). The apparent absence of any reports of capture in brackish waters or laboratory exposure to salinity suggests that Olympic Mudminnows, Alaska Blackfish, and Central Mudminnows are obligate freshwater species. An analy sis of otolith Sr:Ca ratios in Alaska Blackfish supports that observation for the species (Brown & Severin 2009).
Aspects of Esociform Swimming Ability Although esociforms excel at fast strikes, they are poorly suited for maneuvering and sustained swimming (Webb © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 233 water (Miller & Menzel 1986) with forays into open water being rare (Cook & Bergersen 1988). In Alaska, movement into and out of lakes occurs during the ice- out (May– June) then declines between June and July. Northern Pike may aggregate in groups shortly before movement to overwin-tering habitats.
Northern Pike home range and frequency and distance of movement are extensively studied but do not provide a clear picture of the characteristics that govern their be hav iors. Habitat types (e.g., lakes, reservoirs, rivers; clear and turbid systems) affect be hav ior, habitat use, and the location and size of home ranges (Andersen et al. 2008; Kobler et al. 2008b, 2009). Populations and individuals within populations are described as highly mobile or resi-dent (Mann 1980; Chapman & Mackay 1984ab; Minns 1995; Jepsen et al. 2001). Individual use and size of home range varies seasonally, and the effect of fish size on home range variability is contradictory (Minns 1995; Cook & Bergersen 1998; Knight et al. 2008; Kobler et al. 2008a, 2009). Fish with larger winter and spring home ranges may compete more successfully with conspecifics for ac-cess to larger areas, prey and access to spawning grounds and mates (Billard 1996). Large Northern Pike moved at a greater rate than small individuals and movement differed significantly between seasons (Vehanen et al. 2006).
Capture site fidelity for the Muskellunge can vary greatly (Miller & Menzel 1986). Tagged Muskellunge moved between lakes from year to year (Weeks & Hansen 2009) or showed strong homing and site affinities, return-ing to the same lake after spawning in multiple years or to their original capture locations after ≥1 year at large (Crossman 1956; Miles 1978; Diana et al. 2015). Trans-planted Muskellunge returned to a specific locality and fish return to previous year’s activity areas after extensive spawning movements. After spawning, males moved downstream and immediately established home range ar-eas (Crossman 1977; Minor & Crossman 1978).
Following spawning, Muskellunge generally remain in spawning areas until water temperature approaches 16°C before moving to deeper water to establish summer home ranges (Miller & Menzel 1986). Some Wisconsin popula-tions exhibited home range tendencies throughout the year (Dombeck 1979, 1986). The duration of the post- spawning occupancy of spawning areas before movement to summer home ranges varied considerably among the fish (Miller & Menzel 1986). Males establish home ranges earlier in sum-mer than females (Minor & Crossman 1978).
Both the Northern Pike and Muskellunge become more active with the higher water temperatures of the summer the sculling undulation of median or pectoral fins. An impor tant determinant of behavioral changes in these and all other esociform species is the onset of the spawning season when relatively short migrations to the spawning grounds are undertaken and other behavioral changes as-sociated with reproduction occur (e.g., territoriality in No-vumbra and Umbra) (see reproduction section). Short mi-grations not associated with reproduction are known in some species (e.g., Alaska Blackfish, Laske et al. 2016; Olympic Mudminnow, Henning et al. 2007; Kuehne & Olden 2014; Northern Pike, Cook & Bergersen 1988; Mus-kellunge, Weeks & Hansen 2009).
Radio tagging and mark- recapture studies aimed at un-derstanding the daily and seasonal patterns of activity and movement of the Northern Pike and Muskellunge indi-cate many factors interact in complex ways to influence be hav ior of individuals in a given population. Noting that Northern Pike exhibited a much greater versatility in the range of habitats they used than was previously believed, most studies confirmed that the Northern Pike primarily occupied littoral areas with dense aquatic vegetation but display little site fidelity (Bregazzi & Kennedy 1980; Jep-sen et al. 2001; Koed et al. 2006).
This complexity in activity and movement results in con-siderable behavioral variation among individuals in a popu-lation and between dif fer ent populations. Some factors apparently influencing activity in these species are water turbidity and temperature, availability of prey and vegeta-tion cover, current velocity, water- level changes, and weather conditions; even so aquatic vegetation is the key factor in habitat se lection (Cook & Bergersen 1988; Chap-man & Mackay 1984ab; Jepsen et al. 2001; Rosell & MacOscar 2002).
Adult Northern Pike and Muskellunge are often recap-tured in the vicinity of the location of original capture. Thus, migration between populations is generally assumed to be rare. Further, natal site fidelity might help maintain genet ically distinct co- occurring populations (Westin & Limburg 2002; Rohtla et al. 2012). Reproductively mature Northern Pike exhibit relatively high spawning site fidelity, returning from year to year or in alternating years (Roach 1998). Earlier telemetry- based studies failed to demonstrate that Northern Pike established well- defined home ranges or detect differences in the extent of movement or habitat se lection between summer and winter (Diana et al. 1977). On a smaller scale, two types of movement are reported in the Northern Pike: localized, directionless roving parallel to shore along the edge of submerged vegetation, and me-dium range (100–2,000 m) directed swimming across open © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 234 FRESHWATER FISHES OF NORTH AMERICA spawning. When water levels were dropped for the win-ter, the Muskellunge moved to deeper areas to overwinter and displayed restricted movements (Gillis et al. 2010). Increased discharge rate, however, causes Muskellunge to move to shallower waters during the summer and winter and occupy slack- water habitats along riverbanks during the winter (Brenden et al. 2006). Potentially, these Mus-kellunge could be forced into habitats with higher ener-getic demands than the habitats that would have been se-lected other wise, a consideration for the health of populations in canals or near dams.
Canals and water management are implicated in the range expansion of Chain and Redfin Pickerels into areas of southern Florida, including Big Cypress National Preserve and Everglades National Park, but the specific mechanism for their occupancy and dispersal is unidentified (Loftus & Kushlan 1987; Gandy et al. 2012; Kline et al. 2014; Zokan et al. 2015). Canals may break down dispersal barriers such as shallow- vegetated wetlands that experience seasonal dry- down or brackish coastal marshes.
Studies of diel activity in Esox have produced somewhat mixed results. The diel rhythm of the Northern Pike in a lake and reservoir in Denmark changed significantly over the course of a year, but no differences occurred in the de-gree of movement between dif fer ent times of the day (e.g., Jepsen et al. 2001). Others found Northern Pike to be most active at lower light intensities at dawn and dusk when peak feeding seemed to occur and inactive during night (Dobler 1977; Diana 1980; Chapman & Mackay 1984a, 1990; Casselman 1996). Northern Pike were active about 37% of the time with crepuscular activity peaks during summer and a diurnal peak during winter. In a Colorado reservoir, Northern Pike were most active during April and May and least active during October (Cook & Bergersen 1988). Muskellunge, on the other hand, display an increase in movements during nocturnal periods in comparison to crepuscular periods but only during the summer (Cross-man 1977; Wagner & Wahl 2011). Chain Pickerel move into shallow water at dusk and appear to show some site fidelity (Scott & Crossman 1973; Helfman 1981).
Activity levels and movement in the smaller esocids and umbrids typify the sedentary be hav ior of the esociforms. Olympic Mudminnows generally avoid swift currents and prefer sloughs, wetlands, oxbows, and other habitats with little or no flow (Meldrim 1968). In the mainstem habitats of the Chehalis River, Washington, Olympic Mudminnows occupy stream margins, backwater and off- channel habi-tats, frequently where groundwater reaches the surface (Kuehne & Olden 2014). They occur in large numbers in and early autumn (Younk et al. 1996). During this period feeding is most intensive. As water temperatures decline in the late autumn and winter, both species reduce swimming activity and tend to occupy deeper waters, but they con-tinue feeding at a lower rate. Seasonal differences in site oc-cupancy and movements of Northern Pike range from in-creased frequency of movement over shorter distances in winter and increased activity when winter water tempera-tures increased (Diana et al. 1977; Jepsen et al. 2001; Koed et al. 2006). Seasonal differences in site occupancy and movement are similarly masked by individual differences rather than a species, habitat, or sex- related characteristic.
Northern Pike behaviorally thermoregulate in response to changes in both internal and external conditions (Wahl & Stein 1991). Where habitats including thermal refugia are available, they may avoid temperatures above their op-timum due to the high metabolic costs, allowing meta-bolic energy to be allocated for activity, somatic growth and maintenance, and gonad development (Wieser & Medgyesy 1991). Evidence for a strong seasonal effect of temperature on Northern Pike movement is mixed. Al-though extremes of low and high- water temperatures may inhibit activity and movements (Vehanen et al. 2006), many field studies found no difference in the level of movement with temperature (Diana et al. 1977; Rogers & Bergersen 1995; Jepsen et al. 2001).
Similarly, Muskellunge display seasonal changes in movement with the greatest movement in spring and de-creasing in summer (Gillis et al. 2010; Wagner & Wahl 2011). Summer home ranges are generally large (mean = 612 ha) but can vary greatly (range 17–1,309 ha) (Younk et al. 1996; Diana et al. 2015). Muskellunge may undertake large movements over the course of a year, but focus on limited areas during spawning when they tend to move to shallow embayments and marginal river wetlands for spawning (Crossman 1977; Weeks & Hansen 2009). Movement by radio- tagged Muskellunge in the Mississippi River varied seasonally but was also influenced by habitat characteristics (free- flowing versus regulated river). Winter habitat use (defined by home range) was greater in the reg-ulated section of the river than in the free- flowing section. Muskellunge established winter ranges that were distinct from and smaller than summer ranges (Younk et al. 1996).
Other factors may influence Muskellunge habitat se lection and use, including water level and discharge rate in human altered lotic systems. In one canal, Muskellunge showed expected seasonal movements; increased water levels in the spring still corresponded with increased home ranges and increased movement, most likely tied to © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 235 and its diet (Brown et al. 1995; see summary by Gidmark & Simons 2014).
Anti- predator Be hav ior: Umbra Central Mudminnows exhibit several anti- predator be hav iors. Anti- predator behavioral responses to chemical cues in conspecific skin extract were tested in field avoidance experiments (Wisenden et al. 2008). Treated field traps caught fewer individuals than controls, and in laboratory trials, activity was reduced, and fish were displaced to the bottom of the tank (Fig. 19.39); both components of an anti- predator response. In the wild, Central Mudminnows escape predators by concealing themselves in the mud, their brownish coloration helping to camouflage them (Peckham & Dineen 1957; Becker 1983; Tonn & Pasz-kowski 1986). Central Mudminnows also exhibit shoaling be hav ior (fish grouped together), another anti- predator strategy, but show no preference for a large over a small shoal of conspecifics (Jenkins & Miller 2007).
REPRODUCTION Esociforms share some common features in their repro-ductive biology, including timing of spawning, pre- spawning movements, preferred spawning habitats, and, except in Eastern Mudminnow, the absence of nesting be hav iors. The lit er a ture pre sents conflicting reports about differential growth between males and females in the both regulated and natu ral wetlands and may move into and out of those habitats when they are inundated (Hen-ning et al. 2007; Kuehne & Olden 2014), but longer- range movements have not been reported. Dietary studies and be-havioral observations in aquaria indicate that they are visual predators, responding strongly to prey movements (Meldrim 1968; Tabor et al. 2014), which would suggest that they are diurnal predators. The dispersal be hav ior of Alaska Blackfish has not been observed, although their ab-sence from ephemerally connected waterbodies (Ostdiek & Nardone 1959; Haynes et al. 2014), divergent life histories among populations (Ostdiek & Nardone 1959; Blackett 1962; Aspinwall 1965), and limited gene flow (Campbell et al. 2015) suggest limited dispersal capabilities.
In aquar ium studies, Central Mudminnows rested on the bottom in shelters or artificial vegetation during day-light hours, swimming slowly along the bottom along the edge of the tank or darting rapidly across the open bottom. Feeding attempts decreased in the presence of a predator (Tonn et al. 1986). Central Mudminnows exhibited more crepuscular and nocturnal activity and overall greater ac-tivity in autumn than summer (Spencer 1939). Diurnal rhythms changed with seasons as activity during the day increased in autumn. Perhaps to exploit the cover afforded by darkness, fish actively sought darkened areas in response to light. In situ enclosure experiments confirmed crepuscu-lar peaks in Central Mudminnow activity (Tonn & Pasz-cowcksi 1987). Central Mudminnows were more active off-shore and during the day in winter but activity shifted beginning in the spring when fish were concentrated in-shore near the bottom. Winter distributions appeared to be responses to abiotic conditions (ice cover and low oxygen availability); overall patterns during open- water periods ap-peared to be linked to prey emergence and availability.
Alarm Pheromones: Esox Prey species significantly alter their be hav ior (activity) when Northern Pike are pre sent (Ranåker et al. 2012). Minnows exhibited a fright response to a stimulus of Northern Pike feces if the Northern Pike had eaten min-nows. Apparently, the minnow alarm pheromone is in the feces; the minnows exhibit the same reaction to alarm pheromone in water without feces. In what may indicate a behavioral adaptation to avoid detection, captive North-ern Pike spent significantly more time in the area where they were fed but deposited feces in the opposite end of the tank. Such chemosensory cues would potentially re-veal the presence of a predator in hiding, its activity level, Water Stimulus injection period. No data recorded.
Skin Extract Time (min) Depth Distribution Score 3 4 5 6 7 8 9 10 11 12 0 2 4 6 8 10 12 16 18 14 Figure 19.39. Mean (± SE) vertical distribution of Central Mudminnows, Umbra limi, in test tanks (15 trials, 3 fish/trial) before and after addition of water or conspecific skin extract. Minimum distribution score is 3 (at surface); maximum score is 12 (on bottom). Mann- Whitney U- test and P <0.001 (redrawn from Wisenden et al. 2008). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 236 FRESHWATER FISHES OF NORTH AMERICA period of 7–10 days at temperatures in the middle of that range (Armbruster 1959; Scott & Crossman 1973; Jenkins & Burkhead 1994). Late summer and autumn spawning also occurs (Lagler & Hubbs 1943; Underhill 1949; Arm-bruster 1959; Miller 1962; Kleinert & Mraz 1966; Scott & Crossman 1973), and in support of these observations, large fingerlings of Chain Pickerel sometimes are cap-tured in the spring (Kleinert & Mraz 1966).
Spawning Time and Temperatures: Dallia and Novumbra When water temperatures increase 5–10°C and ice off oc-curs, Dallia migrate to off- channel areas where spawning occurs from May to July in the interior of Alaska but appar-ently occurs only in late July in the Bristol Bay area (Black-ett 1962; Aspinwall 1965). Most spawning in the Olympic Mudminnow occurs between March and mid- June with the greatest activity in April and May at water temperatures of 10–18°C (Meldrim 1968; Hagen et al. 1972). Some addi-tional spawning takes place in late autumn, subsides in the winter, and resumes in spring (Mongillo & Hallock 1999).
Spawning Time and Temperatures: Umbra In Umbra spp., spawning is most likely prompted by warm-ing water temperatures and flooding in spring (Peckham & Dineen 1957; Becker 1983). Central Mudminnows spawn in early spring, usually in March or April. Preferred water tem-perature for spawning is 12.8°C but can occur at ≤15.3°C (Becker 1983). In the Eastern Mudminnow, courtship re-portedly begins in March with peak spawning occurring in mid- April at temperatures of 9–12°C, and all females are spent by late April (13–15°C) (Panek & Weis 2013).
Reproductive Cycle in Esox Reviews of reproductive biology in the Northern Pike (Billard 1996) and Muskellunge (Parsons 1959; Scott & Crossman 1973; Becker 1983; Hall 1986) provide a gener-alized depiction of reproduction in Esox. Even so, the Northern Pike differs from other esociforms in producing a single synchronous group of maturing oocytes / spawn-ing season. In contrast, the oocytes of other esociforms mature in two distinct groups and presumably result in fractional spawning (Lebeau 1991).
In Northern Pike, gonad growth begins in July and pro-ceeds at a similar rate between sexes through late sum-mer. Testicular growth ceases in September, but ovarian Esociformes, and size at maturity appears to vary within and among species as a function of locality, spawning, and rearing conditions ( Table 19.3). This appears to vary as much by locality and conditions as by species. Pri-mary spawning is in the early spring, starting in February to as late as May (although Dallia spawns until August), and spawning preceded by movements (see be hav ior sec-tion) in some species. Where winter conditions result in freezing of surface waters, initial spawning movements coincide with ice off, which provides access to smaller tributary streams, ditches, off- channel marshes, and inun-dated shorelines. In most esociforms, the preferred spawning habitat is densely vegetated areas of moderate to shallow depths and mud or sand bottoms ( Table 19.3). Many of the differences in spawning timing among esoci-form species result from variation in the prevailing envi-ronmental conditions impor tant for reproduction that prevail in the areas they inhabit (e.g., seasonal fluctuation of water levels and temperature). The timing of spawning for any given population is directly correlated with the latitude where it is located because latitude dictates sea-sonal water temperature changes and the onset of optimal spawning temperatures. Day length does not appear to be a primary factor in the onset of spawning but is one of several non- temperature variables that stimulate spawn-ing activity (Svardson 1949; Fabricius 1950; Fabricius & Gustafson 1958; Franklin & Smith 1963). Windy or cloudy days seem to reduce spawning activities and cool nights appear to delay spawning activities the following morning (Clark 1950a). In most esociform species, spring spawning usually takes place over a period of 2–4 weeks.
Spawning Time and Temperatures: Esox Peak spawning in the spring is triggered by rising water temperatures, but for wide- ranging species, identifying spawning temperatures depends as much on local condi-tions as on latitude. Northern Pike typically spawn earlier and at cooler water temperatures (about 4.4–11°C) than Muskellunge (9.4–17.2°C). As a result, spawning for either species at the southern limits of their distribution may be-gin as early as late February or early March and be com-pleted in April, but in the north spawning may not begin until May. Where the Northern Pike and Muskellunge co- occur, the spawning period of the two species may par-tially overlap at water temperatures of 13–18°C (Farrell et al. 1996). Spawning in species of Kenoza typically oc-curs in late winter through early spring at water tempera-tures from 2–22°C depending on latitude, peaking over a © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 237 no relationship between reproductive age and latitude. Age at maturation depends on temperature, food avail-ability, growth rate, and stock exploitation (Billard 1996; Diana 1996). Heavy fishing pressure and associated mor-tality appears to induce earlier maturity and increase en-ergy allocation to reproduction in some populations (Di-ana 1983). Muskellunge reportedly reach sexual maturity at age 3–5 with males maturing earlier than females (Scott & Crossman 1973; Cook & Solomon 1987). River populations grow and mature more rapidly than fish from lake populations (Harrison & Hadley 1979). Pickerels typ-ically mature at 3–4 years but some populations report-edly were mature as early as year 1 (Underhill 1949).
Reproductive Cycle in Novumbra Olympic Mudminnows are sexually mature at 41–65 mm TL (Mongillo & Hallock 1999). Water temperatures dur-ing the breeding season ranged from 10–18°C (Meldrim 1968). Eggs are adhesive and are usually deposited one or two at a time on vegetation or algal mats near the bottom where they are fertilized during courtship and left unat-tended. In aquaria, eggs hatched in nine days at 15–17°C. The sedentary larvae adhere strongly to vegetation.
Reproductive cycle in Dallia Like the Muskellunge and pickerels, the Alaska Blackfish appears to be a fractional spawner with two distinct size groups of eggs in the ovaries during spawning (Blackett 1962; Aspinwall 1965; Morrow 1980). As in Esox, inter- populational differences exist. By August only one rather uniform size class of eggs remains in the ovaries; those are assumed to be recruitment stock to be developed by the next spawning season (Blackett 1962). Females de-posit about 40–300 eggs; the number increases with in-creased size of the female (Morrow 1980). Partially spawned females are captured during most of the spawn-ing season, suggesting only a few eggs are extruded with each spawning act. Sexual maturity is reached at age 2+ or 3+ (about 80 mm TL, 5 g for females) so some fe-males likely spawn in multiple seasons over their lifetime.
Reproductive Cycle in Umbra Gonadal development in Umbra begins in autumn or win-ter (September– January) and peaks in the spring (Dederen et al.1986). Average fecundity in Umbra limi is 425–450 eggs/female; fecundity increases with age to nearly 1,500 growth continues through autumn and winter until April (Medford & Mackay 1978; Lenhardt & Cakić 2002). Sperm maturation is completed over a short period be-tween July and September preceding the spawning season in the Northern Pike. In females, oocyte maturation be-gins in the summer with the ovaries continuing to gain weight until spawning. The somatic body weight of fe-males does not decline during the period of gamete devel-opment and gonad growth. Somatic body weight de-creases in males and females during spawning (Medford & Mackay 1978; Diana & Mackay 1979).
Fecundity of the Northern Pike increases with size of the female; Inskip (1982) cites an average fecundity of 32,000 eggs/female, although much larger numbers are reported. Fe-cundity for several populations of the species in the British Isles varied between 10–24 eggs/g (Trea surer 1990). Further, the relationship between population fecundity and recruit-ment is weak (see ecol ogy section). Although bigger females produce bigger eggs, the average size of the month- old larva is the same regardless of mature female size; ge ne tic and en-vironmental factors, primarily water level and temperature, appear to affect growth and survival more than egg and lar-val size (Craig 1996). Even though the estimates cited above serve as rough estimates of fecundity, fecundity of the North-ern Pike is a difficult- to- measure, highly variable characteris-tic, which depends on many factors such as female health, available energy, temperature, and even population biomass (Billard 1996). As with the Northern Pike, fecundity in the Muskellunge increases with female size, ranging from about 6,000–265,000 eggs. Average fecundity is about 120,000 (Scott and Crossman 1973; Cook & Solomon 1987). Relative to the largest Esox spp., females of species of the subgenus Ke-noza produce similar sized but fewer eggs/season ( Table 19.3), both in absolute numbers and in proportion to body weight (e.g., from a few hundred in Redfin and Grass Pickerel to an average of 8,000 in Chain Pickerel) (Scott & Crossman 1973; Becker 1984; Jenkins & Burkhead 1994). Fecundity estimates in pickerel are complicated by the fact that ovaries may con-tain three dif fer ent stages of eggs with mature, ready for spawning eggs usually the fewest in number. For example, Grass Pickerel may have 800 ripe eggs and 4,000 and 11,000 eggs in two developmental classes (Scott & Crossman 1973).
Northern Pike can reach reproductive maturity in a year (females at 30 cm TL and males at 19 cm TL), but age at maturation appears to be highly variable (Raat 1988). In Scotland, the age at first spawning is variously reported as two years for males and three years for females. First spawning of more northernly populations may occur at a later age (Trea surer 1990); however, Diana (1983) found © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 238 FRESHWATER FISHES OF NORTH AMERICA man 1973). One to three males may attend one female with many matings over a period of several days (Arm-bruster 1959; Clark 1950a; Ming 1968; Scott & Crossman 1973; Billard 1996). The behavioral cues causing the fe-male to be receptive to a given male and triggering gam-ete release are unknown. After successful courtship, a spawning group swims slowly side- by- side with their vents aligned, paused frequently to complete the spawning act, then continue swimming until spawning is complete. Eggs (<100) and milt are broadcast over the vegetation. On gamete release, the spawning fish direct a strong, quick body undulation against each other, helping scatter the fertilized eggs, which sink and stick to vegetation or the bottom, hatching about two weeks later without pa-rental care (Billard 1996). Hatching success, at least in the Northern Pike in one study, was higher in eggs depos-ited on sand or silt compared with those laid on aquatic plants (Wright & Shoesmith 1988).
Muskellunge in inland lakes commonly spawn in shal-low bays (<1 m deep) with detritus, vegetation, submerged logs, and gravel, sand, silt, or muck substrates (Dombeck et al. 1984; Zorn et al. 1998; Rust et al. 2002; Nohner & Diana 2015). In larger systems (e.g., Great Lakes), Muskel-lunge spawn at sites with greater depths (1–3 m) and with vegetation coverage of 0–100% and substrates of gravel, sand, and silt (Haas 1978; Farrell 2001; Crane et al. 2014).
Spawning Be hav ior in Novumbra Among esociforms, the Olympic Mudminnow is unique because the males exhibit spawning site territoriality and aggressively repel rival males and intruders of other spe-cies (Hagen et al. 1972). During the spawning period, males acquire their distinct breeding coloration, which consists of a dark choco late brown to almost black body with a series of about 15 thin iridescent- green, white, or blue vertical bars along the entire length of the fish. The dorsal and anal fins, used in display, are conspicuously banded on their edges with a sky- blue coloration (some-times appearing white or yellow). In contrast, females are a dull olive- brown becoming lighter ventrally with a few faint vertical bars along the sides. The outer margin of the branchiostegal membrane of males also develops a wide black band along the edges of the membranes during the spawning season. These can be greatly expanded in a striking display that greatly enlarges the apparent size of the head. Gular flaring is more common in aggressive in-teractions between resident males with abutting territo-ries when they meet at their territorial bound aries. The for age-5 individuals (81–94 mm TL) (Evermann & Clark 1920; Peckham & Dineen 1957; Martin- Bergmann & Gee 1985; Table 19.3). Within individual ovaries all eggs were the same size before spawning season and ripen at the same time, suggesting a single, short spawning season (Peckham & Dineen 1957). Age- specific fecundity esti-mates for Umbra pygmaea ranged from 250 eggs/female at age-1 to 2,168 at age-5 (Panek & Weis 2012). Umbra, par-ticularly males, may reach sexual maturity at age-1, but maturity is usually delayed in females until age-2.
Reproductive Be hav ior Esociforms undertake migrations of varying lengths to reach their preferred spawning areas. These may entail short migrations from streams into lakes (or vice versa) or moving inshore from deep water or onto spring- flooded ar-eas. Many references note spring migration of esociforms upstream into flooded or marshy areas shortly after the ice dis appears (e.g., Umbra, Abbott 1870a; Dallia, Blackett 1962; Morrow 1980; Esox, Clark 1950a; McNamara 1937; Becker 1983; Tomelleri & Eberle 2011). Northern Pike may migrate to distant localities during spring, despite suitable spawning areas near where they resided most of the year (Koed et al. 2006). Mark- recapture, ge ne tic, and otolith mi-crochemistry studies uncovered evidence of annual returns to preferred spawning areas and natal site fidelity in the Northern Pike and Muskellunge (Crossman 1990; LaPan et al. 1996; Miller et al. 2001; Engstedt et al. 2014).
Aside from this directed migration, spawning also is ac-companied by a significant increase in swimming activity in the normally sedentary Northern Pike. Changes in ac-tivity levels associated with spawning are dif fer ent for males and females. This disparity may be related to sex differences in spawning be hav ior, but information on this aspect of esocid biology is limited. At the spawning grounds, males outnumber females, which arrive later and persist longer. In most populations of esocids, females grow faster and live longer, thus the females are usually larger than the courting males. Spawning aggregations can be impressive; in the 1950s, a spawning run of 6,000 Northern Pike was tabulated in a creek in Saskatchewan (Scott & Crossman 1973 citing Schultz 1955).
Spawning Be hav ior in Esox Spawning be hav ior is similar in the Northern Pike, Mus-kellunge, and pickerels. The large esocids exhibit no terri-toriality or nesting be hav ior (Ming 1968; Scott & Cross-© 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 239 Spawning Be hav ior in Dallia Nothing so elaborate as observed for the Olympic Mudmin-now is described in the spawning be hav iors of the Alaska Blackfish, but little is published on the species. During the spawning season, males develop a reddish fringe along bor-ders of the medial fins (Blackett 1962; Morrow 1980). No direct observations of spawning exist but massive upstream movement to shallower waters occurs following ice break up when temperatures warm 6–8°C and flows decrease (Blacket 1962). The eggs are prob ably deposited in vegeta-tion at the bottom of shallow ponds and quiet streams (Nel-son 1884; Morrow 1980). The capture of partially spawned females through the season suggests that spawning pro-ceeds over a period of several days with only a few eggs be-ing extruded at each spawning act.
Spawning Be hav ior in Umbra Detailed accounts of reproductive be hav ior in North American Umbra spp. are lacking. As with other esoci-forms, Central Mudminnows move toward margins of streams or lakes following ice off before moving into channels connected to marshes or inundated shorelines during periods of spring flooding and heavy rains (Peck-ham & Dineen 1957; Jones 1973). Adams & Hankinson (1926) noted marked migratory movements to streams for breeding; females preceded males. Sexual dimorphism oc-curs in the Central Mudminnow in the form of an elonga-tion of the anal fin and an intensification of bluish- green coloration of the anal and pelvic fins in males during the breeding season. In Eastern Mudminnows, adults during the breeding season can have ragged fins suggestive of ag-gressive be hav ior during courtship. Males flare their pec-toral fins and approach gravid females and nudge them to induce egg release. Females affix eggs individually di-rectly to vegetation (Peckham & Dineen 1957), but nest-ing and attaching eggs to the underside of loose rocks is also described (Breder & Rosen 1966).
The only detailed accounts of reproductive be hav ior of any species in Umbra are from laboratory observations in the Eu ro pean Mudminnow (Bohlen 1995; Kovác 1997). Preceding spawning, a period of courting lasts 1–4 days. During courting, males pursue females and attempt to nudge the vent of the female with their mouths. Eventu-ally, females become receptive and pick a suitable spawn-ing site. Females reportedly often excavate shallow nests in the substrate before spawning (Kovác 1997); however, Bohlen (1995) never observed this be hav ior. The preferred gular- flaring posture may be held ≤10 s and may be fol-lowed by circular fighting in which the two males circle one another closely and attempt to strike one another with the tail (Hagen et al. 1972).
Male Novumbra select a suitable patch of vegetation and establish a territory, which they patrol and defend against intruders. Territories are large (about 1.0 × 0.5 m) relative to the size of the male and are maintained for the entire length of the spawning season (>7 weeks; Hagen et al. 1972). Residents patrol the margin of their territory with a stylized swimming movement in which the tail is slowly swept from side to side in a wide arc. The male abruptly halts, hovering for a few seconds, and then patrols again for a short distance. If intruding fishes ( whether conspecifics or not) approach a territory, the resident male rapidly turns to face the intruder and hovers. Holding the body rigid, the male fully expands the colorful dorsal, anal fins, and pelvic fins and rapidly vibrates them. Per sis tent intruders may be attacked with a forceful blow with the head or the side of the body with an impact sufficient to turn the opponent up-side down. In the wild, strikes against conspecific intruders were rare and were more commonly directed against in-truders of other species (Hagen et al. 1972). Coloration of breeding males and patrolling be hav ior intensifies once eggs are deposited on their territory. Colors are also more con spic u ous during courtship or fighting. Submissive males or those defeated in territorial fighting become a light brown and lose the colored bands on the fins.
The most elaborate display during Olympic Mudminnow courtship is the wigwag dance of the male. The body is strongly oscillated from side to side in a wide arc along the direction of swimming with the dorsal and anal fins maxi-mally expanded. During each dance, males move only about 7 cm with bouts of dancing alternating with freezing and fins displayed. Display is usually parallel to the female, and males then sweep in an arc round her head while dancing.
If the display is successful, the receptive female enters the male’s territory, where the male repeatedly swims along the female and tries to make contact. After an inter-val of courtship lasting 5–20 min, the female swims into the vegetation and allows the male to swim along her side and make body contact. The female apparently selects the exact site for depositing eggs and moves into the vegeta-tion, and rather than moving away, responds to the male by mutually pushing against his side while vibrating. The pair maintains contact while their bodies vibrate until they fi nally push off each other with a sudden thrust at which point eggs and milt are released into the vegeta-tion. The pair may spawn repeatedly for >1 h.
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NOT FOR SALE OR DISTRIBUTION 240 FRESHWATER FISHES OF NORTH AMERICA Northern Pike by Bry 1996). The substrate at spawning sites is typically covered by abundant loose organic matter or vegetation. Water velocity is low or negligible. Under these conditions, dissolved oxygen levels may vary widely because of microbial respiration in the substrate and lack of mixing currents. Because an adequate oxygen supply is critical for embryo development and oxygen levels may be temporarily low in the spawning habitat, embryo asphyxi-ation from hypoxia can cause catastrophic mortality in the Muskellunge (Zorn et al. 1998). The tendency of lar-vae to attach to standing vegetation via adhesive glands located on the head may help avoid anoxic or toxic condi-tions directly above the substrate.
Eggs and Larvae Mature esociform eggs are adhesive and demersal, pale yellow to orange in color and range in dia meter from 1.83 to 3.29 mm in Northern Pike (Raat 1988) and 1.6 to about 2.0 mm in Dallia, Novumbra, and Umbra. The time re-quired for embryos to hatch is temperature dependent, but hatching usually occurs in six days to two weeks at 10–13°C (Underhill 1949; Aspinwall 1965; Hokanson et al. 1973; Morrow 1980; Kendall & Mearns 1996; Panek & Weis 2012). In Novumbra, embryos hatch in about nine days at 15–17°C (Meldrim 1968).
Development of Esox Embryos of Esox spp. develop on the substrate or attached to debris and vegetation where they are subject to preda-tion, fungal infestation, and other environmental condi-tions contributing to high mortality during the first sum-mer of life (Farrell 2001). Yolk- sac larvae are 6.5–10.0 mm TL at hatching (Franklin & Smith 1963; Kotlyarevskaya 1969; Fuiman 1982) and can swim soon after hatching (Fig. 19.40). Muskellunge yolk- sac larvae are 7.9–9.2 mm TL at hatching. Embryo and larval development in Kenoza species is generally similar to that of the large esocids ex-cept that yolk- sac larvae are slightly smaller at hatching (4–8 mm TL) (Jones et al. 1978; Fuiman 1982). Typically, newly hatched larvae of Esox spp. swim upward facilitated by the heterocercal shape of the tail and attach vertically to available substrates by means of a nasal adhesive gland (Bry 1996). They remain relatively inactive if undisturbed until the yolk sac is absorbed, at which point they begin to swim and feed. Depending on water temperature, yolk ab-sorption occurs during the next 5 days at 19°C and 16 days at 10°C (Underhill 1949; Raat 1988). Oxygen exchange is spawning substrate is dense, soft vegetation, but spawning can proceed even in the absence of the preferred substrate (Bohlen 1995). Receptive females swim into the spawning site followed by one to five males. One male pairs with the female and swims up to her side. Their bodies vibrate, and the gametes are released and fertilized. The pair leaves the site until the female is ready for another spawning bout. As spawning proceeds the female becomes progres-sively more aggressive toward the males until all males are repelled from the nest area. Similar aggressive display be hav ior, although seemingly unrelated to spawning, was observed in female Olympic Mudminnows and in male and female Central Mudminnows and Eastern Mudmin-nows kept in aquaria (Peckham & Dineen 1957).
Parental Care in Esox, Novumbra, and Dallia Parental care is lacking completely or minimal across all esocids. Evidence of parental care to the embryos or lar-vae is lacking in any species of Esox or in Alaska Blackfish. Males of the Olympic Mudminnow may still defend their territories for a period of about seven weeks but do not tend to the developing eggs or larvae (Meldrim 1968; Ha-gen et al. 1972). When spawning is completed the adults leave the spawning grounds; the fertilized eggs develop unattended. In the laboratory, larvae dispersed from the egg deposition site after seven days (Hagen et al. 1972).
Parental Care in Umbra In Umbra spp., females exhibit some egg- guarding be hav ior and weak fanning of the embryos, but other parental be hav iors are unknown (Peckham & Dineen 1957; Becker 1983). Females reportedly consume undeveloped eggs. The lit er a ture differs regarding the quality and extent of parental care in Eu ro pean Mudminnows. In one aquarium- based study, females fanned the nest over a period of two weeks, al-though the females seemed unaware of the developing em-bryos and occasionally ate them (Bohlen 1995). In contrast, Kovác (1997) reported that aquarium- held females guarded the nest after hatching and could discriminate offspring from food. Further, one female carried fertilized eggs in her mouth from their place of deposition to a presumably better site (i.e., more secluded).
Spawning Sites and Dissolved Oxygen Spawning habitats eventually provide shelter to the larvae (see review of the role of vegetation in the life cycle of the © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 241 and live longer than populations in Interior Alaska (Mor-row 1980).
Development of Umbra In Umbra, embryos hatch in about six days (Ryder 1866), hatching time being dependent on water temperature. The transparent larvae hatch at about 5 mm TL becom-ing pigmented and dark on their sixteenth day after hatching (Peckham & Dineen 1957; Fig. 19.43). In East-ern Mudminnows, the yolk sac is absorbed in ≤2 weeks. Until young reach 25 mm TL they have a notochord lobe above the developing caudal fin (Breder 1933; Becker 1983). Young Central Mudminnows migrated from breeding areas to the main stream at about 30 mm TL (Peckham & Dineen 1957; Wallus 1990). The young grow rapidly and may reach 55 mm by October (Becker 1983).
The yolk sac of developing Eastern Mudminnow larvae is covered by a circulatory rete formed by a sub- intestinal vitelline vein, which extends over the posterior two- thirds of its surface. This rete may function as adaptation to the unpredictable dissolved oxygen concentrations associated with esociform spawning habitat as is hypothesized for Eu ro pean Mudminnows (Kovác 1995).
Embryo, Larval, and Juvenile Survival Artificial fertilization in hatchery conditions affords much higher embryo survival than under natu ral condi-through the body surface until gills develop (at 10.5– 12.0 mm TL in Northern Pike; Franklin & Smith 1963), so some water movement is necessary (Braum et al. 1996). Lin et al. (1997) gave a detailed description of the post- hatching (14–250 mm TL) development of the gonads and sex differences in young of the Muskellunge.
Development of Novumbra Newly hatched larvae of the Olympic Mudminnow (5 mm TL) remain attached to surrounding vegetation by adhe-sive glands located on the head (Meldrim 1968) until yolk- sac absorption is completed and they become mobile (7 mm TL under laboratory conditions; Hagen et al. 1972). Kendall & Mearns (1996) detailed embryo and larval de-velopment of Olympic Mudminnows (Fig. 19.41).
Development of Dallia Newly hatched larvae of the Alaska Blackfish are about 6 mm TL at hatching and have a large yolk sac (Fig. 19.42). They transition from yolk- sac to post yolk- sac larvae at about day 10 post hatching at 12–13°C in the laboratory. By day 22 the fish are about 12 mm TL and are beginning to resemble adults. By day 44 at about 20–21 mm TL, the transition to juvenile is complete (Aspinwall 1965; Mor-row 1980). Alaska Blackfish growth is rapid in the first year; annual incremental growth declines thereafter (Blackett 1962; Aspinwall 1965), but growth is continu-ous. Blackfish in Bristol Bay, Alaska, are slower growing A B C D 6.5 mm TL 8 mm TL 9.1 mm TL 10 mm TL E F G H 14 mm TL 19.5 mm TL 24.3 mm TL 50 mm TL Figure 19.40. Larvae and juveniles of the Chain Pickerel, Esox niger. (A– D) Yolk- sac larvae; (E– H) Post yolk- sac larvae and juveniles (redrawn from Yeager 1990).
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NOT FOR SALE OR DISTRIBUTION 242 FRESHWATER FISHES OF NORTH AMERICA submerged vegetation, which serves as shelter and sup-ports food for juveniles (Kipling 1983; Werner & Gil-liam 1984; le Cren 1987). Maintaining ecological flows in regulated rivers can benefit Northern Pike reproduc-tion and recruitment by maintaining access to littoral habitats for spawning and rearing (Mingelbier et al. 2008). Variation in Northern Pike population dynamics among spawning and nursery habitats and between years highlights the role of temperature in influencing the timing of spawning and subsequent growth (Craig & Kipling 1983; Casselman & Lewis 1996; Minns et al. 1996; Farrell et al. 2006). High, stable water levels through larval development produce larger year classes (Johnson 1957) as do higher summer temperatures (Craig 1996). Access to shallow nearshore areas with abundant aquatic vegetation for cover, abundance of prey species inhabiting nearshore areas, and low shore-line development contribute to increased recruitment success in young- of- the- year Muskellunge (Murry & Farrell 2007).
tions. Billard (1996) reviewed the lit er a ture on incuba-tion and hatching of Northern Pike embryos under con-trolled conditions, providing many useful insights into the tolerance of developing embryos to temperature, ox-ygen concentration, and physical shock. Embryo mor-tality is generally high and determined by many vari-ables (e.g., oxygen levels, water temperature fluctuations, nursery habitat conditions) (Casselman & Lewis 1996; Zorn et al. 1998; Farrell 2001). Two impor tant factors determining year- class strength and mortal-ity rates in juvenile Northern Pike are temperature and water level (Kipling 1983; le Cren 1987; Casselman 2002). Temperature dictates how quickly the juveniles reach a size rendering them less vulnerable to cannibal-ism, and water level affects the extent of habitat with A B C D E F Figure 19.41. Larvae and juvenile of the Olympic Mudminnow, Novumbra hubbsi. (A) Yolk- sac larva, 5.2 mm SL; (B) pre- flexion larva, 7.7 mm SL; (C) early flexion larva, 12.2 mm SL; (D) late flexion larva, 14.5 mm SL; (E) post- flexion larva, 15.7 mm SL; (F) juvenile, 22.0 mm SL (redrawn from Kendall & Mearns 1966).
A B C D Figure 19.42. Larvae (A– C) and juvenile (D) Alaska Blackfish, Dallia pectoralis, reared in the laboratory (mean water temperature 12–12.5°C). Parental stock was obtained in Lake Aleknagik, Bristol Bay, Alaska (redrawn from Aspinwall 1965). (A) 5.7 mm TL, hatchling, x35; (B) 9.0 mm TL, 10.5 days post- hatch, x22; (C) 12 mm TL, 21.75 days post- hatch, x16.7; and (D) 19.5 mm TL, 44 days post hatch, x10.1.
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NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 243 man 1973; Becker 1983; Rohde et al. 1994; Page & Burr 2011; Figs. 19.18, 19.33, and 19.44). Aquatic vegetation is an impor tant factor for esocid reproduction and recruit-ment, and vegetated habitats provide cover for hunting as well as refuge from other predators (Grimm 1983, 1989; summarized in Raat 1988). Within lotic systems, Grass Pickerel avoid riffle habitats and prefer to be associated with submerged vegetation or underwater woody debris (Cain et al. 2008). This environment is ideal for waiting and ambushing prey, a tactic shared by all members of Esox. Habitat use, however, may not remain static. For ex-ample, seasonal changes in habitat use by Muskellunge in West Okoboji Lake, Iowa, was related to spawning and foraging (Miller & Menzel 1986). In regulated rivers, Muskellunge occurred in deep water habitats but moved into shallower areas closer to shore when discharge in-creased (Brenden et al. 2004), and Muskellunge also showed seasonal movements in the Mississippi River (Younk et al. 1996). Seasonal changes in temperature may also alter the habitat se lection of esocids (Chapman & Mackay 1984a). For example, Chain Pickerel inhabit deep, cool waters during mid- summer, but move into the shal-low weedy areas in the autumn when the water cools (Raney 1942; Armbruster 1959). Habitat partitioning be-tween juvenile and adult esocids is not apparent; they ap-pear to equally exploit all habitats frequented.
Habitat of Other Esociforms Among the other esociforms, little differentiates the types of habitats most occupied, other than perhaps the extremely harsh winter conditions imposed on the habitat of the Alaska Blackfish. Tolerant of a wide range of environmental conditions, Alaska Blackfish and Olympic, Central, and Eastern Mudminnows are sedentary inhabitants of heavi ly vegetated lowland marshes, swamps, ponds, and sloughs and occasionally exploit medium to large rivers with slack flows at channel margins, braided and backwater habitats, and lakes with abundant aquatic macrophytes (Peckham & Dineen 1957; Scott & Crossman 1973; Harris 1974; Smith 1985; Mongillo & Hallock 1999; Currie et al. 2010; Page & Burr 2011; Tabor et al. 2014; Haynes et al. 2014). Most spe-cies occur in habitats with thick layers of organic material at the bottom in which they occasionally bury themselves.
Olympic Mudminnows usually occur in heavi ly vege-tated lowland swamps and ponds (Fig. 19.45), but also persist in disturbed areas such as sloughs and roadside ditches, tolerating low dissolved oxygen levels and high water temperatures that might other wise be stressful for ECOLOGY From an ecological perspective the esociforms are broadly divisible into two groups. Ecologically, the traditional taxo-nomic grouping of Dallia, Novumbra, and Umbra is easy to understand. These generally diminutive fish have similar habitat requirements, tolerate similar conditions, and have similar diets in the majority of their life histories. Like-wise, the habits and habitats of their larger Esox relatives overlap markedly. The main difference in diet between young and old Esox and among the large esocids and small esociforms is in the sizes of their prey. Like many other fishes, diets shift from microcrustaceans and smaller in-vertebrates to larger food items as fish size increases.
Habitat of Esox Species of Esox generally inhabit slow moving, vegetated waters in lakes, swamps, and large rivers (Scott & Cross-A B C D E 7.6 mm TL 9 mm TL 12 mm TL 15.5 mm TL 19 mm TL Figure 19.43. Development of young Central Mudminnows, Umbra limi. (A) Late yolk- sac larvae; (B– D) post yolk- sac larvae; (E) early juvenile. A– C, drawings of wild- caught specimens from Big Sandy River floodplain, Tennessee; D– E wild- caught from same locality and reared in the laboratory (redrawn from Wallus 1990). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 244 FRESHWATER FISHES OF NORTH AMERICA other native fishes of western Washington (Mongillo & Halleck 1999; summary by Wydoski & Whitney 2003).
The Alaska Blackfish is most abundant in tundra re-gions but does occur in forested areas (Fig. 19.46). They are tolerant of harsh conditions and exploit late succes-sional beaver ponds over other habitat types presumably because these habitats afford greater respite from stream flow and provide off- channel refugia from predators (Mal-ison et al. 2014). Occupancy models showed that Alaska Blackfish are more likely to occur in lakes with more deep- water refugia, prob ably because they overwinter in these lakes (Haynes et al. 2014). The species does not ap-pear to exploit ephemeral connections between waterbod-ies for dispersal (Laske et al. 2016).
During periods of high flows, Umbra species move from their normal habitats and venture into the flooded areas along the banks to avoid strong water currents. They can breathe air and thus can survive for short periods of time under extreme hypoxic conditions (Peckham & Dineen 1957; Klinger et al. 1982; Becker 1983; Chilton et al. 1984; Rahel & Nutzman 1994; Schilling et al. 2006). Central Mudminnows form shoals of several individuals, perhaps to avoid predators or find food (Jenkins & Miller 2007). When threatened, they flee into soft sediment and ooze (Applegate 1943; Peckham & Dineen 1957).
Feeding Esociforms feed almost exclusively on animals, although small amounts of algae, plant material, and detritus are of-ten found in the stomachs of Dallia and Umbra (Ostdiek & Figure 19.44. A solitary Northern Pike, Esox lucius, cruising along the edge of vegetation, its usual habitat, in Baker Creek, a tributary to Great Slave Lake, Northwest Territories, in June (courtesy of © Paul Vecsei / Engbretson Underwater Photography). Figure 19.45. Habitat of the Olympic Mudminnow, Novumbra hubbsi, in Conner Creek, Washington (courtesy of RT). Figure 19.46. Habitat of the Alaska Blackfish, Dallia pectoralis, in Big Eldorado Creek, Tanana River drainage, Alaska (photography by AL).
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NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 245 Mudminnows in aquaria suggest they respond strongly to prey movements (Meldrim 1968; McPhail 1969). Typically, they remain motionless and then stalk and fi nally strike at passing prey (Meldrim 1968). They use their large pectoral fins to paddle slowly about the vegetation in search of food. Once a prey organism is spotted, they capture it with a quick dart, much like Esox. Olympic Mudminnows are diurnal foragers (Tabor et al. 2014).
Species of Umbra appear to use their vision to detect and target prey. In laboratory observations, these fishes hover in the water using slow, continuous undulations of their fins or slowly stalk a prey item to within striking distance and then strike with a quick lunge similar to Esox spp. (Keast & Webb 1966). In laboratory experiments, the Eastern Mud-minnow exhibited no preference for prey location and cap-tured items with equal success from the substrate, mid- water, surface, or on submerged plants (Paszkowski 1984, 1985). Aquar ium observations of escape be hav ior of Central Mudminnows when startled revealed the fish froze and as-sumed a posture similar to the C- or S- start pose described in esocids (Jones 1973). Apparently, information on feeding be hav ior of Alaska Blackfish has not been published.
Prey Se lection in Esox Laboratory and field experiments indicated the Northern Pike, Muskellunge, and tiger muskellunge practiced prey se lection, grew more slowly in pond systems where Sun-fish (Centrarchidae) prey were pre sent, and strongly pre-ferred Gizzard Shad (Dorosoma cepedianum) over Bluegills (Lepomis macrochirus) when both species were available (Wahl & Stein 1988). Gape limits on prey size are linearly related to Northern Pike body length; a similar relation-ship is seen in the Muskellunge (Fig. 19.48). Northern Pike occasionally select prey smaller than would be pre-dicted; the increased handling time for larger prey makes Nardone 1959; Martin- Bergmann & Gee 1985; Panek & Weis 2013). They are primarily diurnal or crepuscular for-agers as befits visual predators with well- developed eyes (except for Dallia).
The mainly piscivorous esocids strike at potential prey with a power ful lunge once they have visually identified the target. In Esox, feeding strikes are usually launched from ambush at the edge of submerged vegetation after a slow stalking propelled by sculling the pectoral and cau-dal fins (Diana et al. 1977; Webb & Skadsen 1980). The posterior placement of the dorsal fin characteristic of all esociforms and a well- developed anal fin facilitate the fast, power ful strikes used by larger esociforms to capture prey (Higham 2007). In the species studied (i.e., Esox), vision is the most impor tant sensory input in the earliest phases of prey capture, then, as the distance to the target is reduced, the cephalic and lateral- line sensor systems provide all the information necessary for successful prey capture (New & Kang 2000; New et al. 2001; New 2002). Large esocids may switch from strict ambush to pursuit as water clarity declines. The increased encounter rate may functionally offset the added metabolic costs (Turesson & Brönmark 2007; Carey & Wahl 2010; Pintor et al. 2014).
Esox exhibits fast- start swimming be hav iors in feeding strikes (Webb & Skadsen 1980; Frith & Blake 1995; Do-menici & Blake 1997; Hale 2002; Schriefer & Hale 2004). Muskellunge feeding be hav ior consists of a slow stalk of its prey (Fig. 19.47) followed by a C- or S- start lunge from a sta-tionary position (Domenici & Blake 1997; New et al. 2001). Fast- start per for mance (mean maximum acceleration) for the Northern Pike is estimated at 119.2 ± 19.0 m/s2 with mean velocities approaching 4 m/s2 (Harper & Blake 1990). S- start strikes usually occur over shorter distances and have higher success rates than C- start strikes (95 versus 58%).
Feeding be hav ior among Novumbra and Umbra is similar and shows some parallels to Esox. Observations of Olympic Figure 19.47. Muskellunge, Esox masquinongy, in search of prey in Spread Ea gle Chain of Lakes, Florence County, Wisconsin, in May (courtesy of © Eric Engbretson / Engbretson Underwater Photography). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 246 FRESHWATER FISHES OF NORTH AMERICA the predator and may be acquired at less metabolic cost (Eklöv & Hamrin 1989). In Esox, large fusiform prey may be impor tant for fast growth (Beyerle & Williams 1968; Diana 1979; Gillen et al. 1981; Jacobson 1992).
Seasonality of Feeding Northern Pike, and prob ably most Esox spp., feed through-out the year even though rates of feeding may change. Feeding appears to decline significantly in August, al-though it is not clear if the decrease is related to water temperature (Chapman & Mackay 1990). Winter preda-tion by the Northern Pike is well documented (Casselman 1978; Diana & Mackay 1979); Northern Pike gained en-ergy reserves from winter feeding and this may serve to prepare fish for spring spawning.
Alaska Blackfish feed actively in winter and exhibit sig-nificant differences among seasons. Prey shifted from gas-tropods (summer) to gastropods and ostracods (autumn). In winter, ostracods and dipterans were the primary prey item but fish were the dominant food item by biomass (Eidam them more susceptible to predation by conspecifics and kleptoparasitism (i.e., one fish taking food from another fish) (Nilsson & Brönmark 1999). Northern Pike diets var-ied considerably with season and habitats and included higher than expected frequencies of invertebrate prey, es-pecially in small (<600 mm TL) individuals, despite the role of the Northern Pike as a top piscivore in most envi-ronments (Lawler 1965; Mann 1976; Chapman et al. 1989). Fishes are an impor tant food throughout the year, and Northern Pike exploit small cyprinids throughout the year, however, they shift to large- bodied prey when avail-able (Frost 1954; Lawler 1965; Diana 1979). Small fishes serve as a primary component in the diets of small North-ern Pike (200–500 mm TL) but decline in importance in the diets of the largest individuals (>600 mm TL) (Frost 1954; Lawler 1965; Bregazzi & Kennedy 1980; Mann 1982; Sammons et al. 1994). Similarly, small Grass Pick-erel (≤95 mm TL) have a diet dominated by fishes com-pared to one dominated by crayfishes in large Grass Pick-erel (>150 mm TL) (Weinman & Lauer 2007). Soft- bodied fishes without hard spines require less manipulation by 0 10 20 30 40 50 Prey TL (cm) 20 40 60 80 100 Muskellunge TL (cm) Figure 19.48. Relationship between TL of 427 prey fishes and TL of Muskellunge, Esox masquinongy (R2 = 0.340, p<0.001). Muskellunge ate prey fishes that ranged from 6 to 47% of their own TL. Muskellunge were captured from 34 Wisconsin water bodies from spring through autumn. Unidentifiable fish and non- fish prey items were excluded (redrawn from Bozek et al. 1999).
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NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 247 mayflies in the ≤4°C sub- ice waters (Martin- Bergmann & Gee 1985). Two populations of Eastern Mudminnows fed continuously through the winter and showed no signifi-cant difference in diet (Panek & Weis 2013).
Diet of Esox Post yolk- sac larval Esox feed on planktonic organisms, then as their size increases they switch to macroinverte-brates and soon become primarily piscivores, and depend-ing on gape size they occasionally eat crayfishes, amphibi-ans, water fowl, and small mammals (Figs. 19.49 and 19.50). The timing of their diet shifts varies among species but shifts to piscivory among Esox occurs as young- of- the- year (Mittelbach & Persson 1998). Muskellunge shift to a mainly piscivorous diet in ≤5 days after exogenous feeding has begun or at 3 cm SL, but Northern Pike begin feeding et al. 2016). Prey se lection also differed among stream, lake, and wetland habitats. Olympic Mudminnows also feed regularly in winter. The winter diet can show high overlap with the diet in spring and summer, although this can vary with size, class, and habitat (Tabor et al. 2014).
Umbra spp. feed actively during winter and digest food relatively rapidly at cold temperatures (Chilton et al. 1984). Of six Great Lakes fish species examined, Central Mudminnows exhibited the greatest tendency for winter feeding (4°C water temperature) with 50–65% of individ-uals age 0–2 containing food in their gut (Keast 1968); >85% of Central Mudminnows (40–94 mm TL) in a small Indiana stream contained food in their stomach in winter (Peckham & Dineen 1957). In small streams in southern Manitoba in winter, the percentage of full or distended stomachs of the species was as high as early summer; fish were primarily consuming small fishes and Spring (N=131) Summer (N=107) Autumn (N=344) 0 10 20 30 40 50 0 10 20 30 40 50 0 10 20 30 40 50 Catostomidae Cyprinidae Yellow Perch Lepomis Pomoxis Rock bass Micropterus Walleye Cisco Sculpin Stickleback Mudminnow Darter Esocidae Bullhead Trout-perch Carp Unidentified fish Crayfish Aquatic insect Frog Mudpuppy Tadpole Mouse Relative Importance Figure 19.49. Relative importance of prey consumed by Muskellunge, Esox masquinongy, captured from 34 Wisconsin water bodies during spring through autumn. Of the 1,092 Muskellunge (22.6–118.0 cm TL) examined, 375 contained food items; N is the number of items identified (total N = 582) (redrawn from Bozek et al. 1999). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 248 FRESHWATER FISHES OF NORTH AMERICA be consistent and may be masked by local conditions. Stom-ach samples from Olympic Mudminnows ≥40 mm TL re-vealed their diet comprised primarily aquatic invertebrates (oligochaetes, chironomid larvae, and copepods) (Fig. 19.51). Though larger fish displayed more dietary spe-cialization, no consistent pattern of ontoge ne tic shifts in diet occurred. Similarly, Alaska Blackfish diets in summer showed a high occurrence of aquatic insect larvae, espe-cially dipteran larvae, as well as microcrustacea (Fig. 19.52). Small fishes are also eaten by Alaska Blackfish (Ostdiek & Nardone 1959; Chlupach 1975).
Terrestrial insects occur in the diet of Umbra as do juve-nile conspecifics, but stomach contents of both Central and Eastern Mudminnows suggest they are primarily benthic feeders (i.e., high volumes of detritus and benthic organ-isms) (Peckham & Dineen 1957; Panek & Weis 2013; Table 19.4). Juvenile winter diet of Umbra spp. during their first year included small crustaceans and crustacean on fishes at a later point in development at ≥5 cm SL (see morphology section) (Scott & Crossman 1973; Becker 1983). In communities with low prey fish abundance, Northern Pike may rely heavi ly on macroinvertebrates (Be-audoin et al. 1999). The preferred prey species of Northern Pike varies by location, but in North Amer i ca prey gener-ally includes Yellow Perch (Perca flavescens), Suckers, Core-gonus spp. (ciscoes and whitefishes), and cyprinids (Lawler 1965). Similarly, Yellow Perch and Suckers are typically the most impor tant components of the diet of the Muskellunge, but the species consumes a diverse array of prey (Fig. 19.49). Whether this diet reflects a selective preference on the part of the predator or simply the relative abundance of those prey species and their habitat overlap with the predator is unclear. Fishes with deep bodies and spiny fins are rarely eaten by either species, but other environmental factors may change this preference. In short, esocids can be consid-ered opportunistic predators, which will take any suitable prey. The Northern Pike and Muskellunge are considered voracious predators and as such are thought to be impor tant competitors of other piscivores and to potentially affect biomass of prey species and indirectly all lower trophic lev-els (Søndergaard et al. 1997).
Post- yolk sac pickerel larvae begin feeding almost immedi-ately on copepods and other micro- crustaceans (Raney 1942). Larvae switch to increasingly larger prey after feeding on in-vertebrates at <140 mm TL; yearlings and adults are piscivo-rous but opportunistically exploit aquatic insects, crayfish, and other invertebrates (Fig. 19.50). Fishes and crayfishes are about equally impor tant food for pickerels (Raney 1942; Becker 1983; Weinman & Lauer 2007). Cyprinids and cen-trarchids are usually numerically dominant prey items, but catostomids predominate by volume (McIlwain 1970; Lewis 1974). Chain Pickerel diets differed between populations from freshwater and brackish water habitats; fishes and crusta-ceans dominated the diet in brackish environments, and the use of aquatic insects decreased (Meyers & Muncy 1962).
Diets of Novumbra, Dallia, and Umbra The diets of Novumbra, Dallia, and Umbra are similar (Figs. 19.51 and 19.52; Table 19.4). These species are carnivorous, generalist predators whose diet varies among sites, seasons, and individual fish sizes. They consume a wide variety of benthic aquatic invertebrates but also occasionally fishes and terrestrial invertebrates (Peckham & Dineen 1957; Ost-diek & Nardone 1959; Keast 1978; Martin- Bergmann & Gee 1985; Gudkov 1998; Panek & Weis 2013; Tabor et al. 2014; Eidam et al. 2016). Dietary shifts with size do not appear to Mean Percentage Volume Relative Importance Index 57-95 mm 96-150 mm >150 mm Fish Size (TL) Insects Fishes Crayfishes Other Prey Category Figure 19.50. Stomach analyses for Grass Pickerel, Esox americanus, in three size classes and four prey categories (insects, fishes, crayfishes, and other) showing mean percentage volume and relative importance index value for individuals captured in northern and east central Indiana. Relative importance was not calculated for the other category, which included annelids, isopods, frogs, and unidentified (data from Weinman & Lauer 2007). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 249 and ≤546 mm TL by the end of the first year depending on latitude (Scott & Crossman 1973; Becker 1983). Growth rate remains high for several years before slowing once adults reach maturity. Females in the larger esocids are generally both larger and heavier than males of similar age. Northern populations of Northern Pike grow more slowly but live longer than southern populations (Scott & Crossman 1973). As with the larger esocids, growth of young- of- the- year pickerels is rapid, reaching >120 mm FL for Redfin and Grass Pickerel and >170 mm FL for Chain Pickerel (Scott & Crossman 1973; Trautman 1981; Becker 1983). Growth in Canadian populations of Chain, Grass, and Redfin Pickerel appears to be continuous and females grow more rapidly than males, achieving larger size (Scott and insect larvae (Peckham & Dineen 1957; Becker 1983; Chilton et al. 1984; Rahel & Nutzman 1994). In addition to aquatic invertebrates, females of the Central Mudmin-now will feed on small fishes (e.g., cyprinids), especially in the winter at the time of egg development when energy demands are expected to be particularly high. Only adult female Central Mudminnows were piscivorous during winter, and age 2+ females were nearly entirely piscivo-rous during this season (Chilton et al. 1984).
Age and Growth of Esociformes Growth of Northern Pike and Muskellunge is rapid, reaching 150–250 mm TL by the end of the first summer ND 0 20 40 60 80 100 0 20 40 60 80 100 0 20 40 60 80 100 Winter Spring Summer Green Cove wetland Hopkins Ditch Woodard Creek pond Percent Percent Percent 40-49 50-59 >60 40-49 50-59 >60 40-49 50-59 >60 Size Class (mm TL) Other Ostracods Cladocerans Copepods Mollusks Oligochaetes Other aquatic insects Aquatic dipterans Figure 19.51. Diet composition (percentage by weight) of the Olympic Mudminnow, Novumbra hubbsi, at three sites in Thurston County, Washington, February to August. Size class of fish sampled given below bottom bars. ND = no data. The other category includes plant material, detritus, amphipods, water mites, and terrestrial invertebrates (redrawn from Tabor et al. 2014). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 250 FRESHWATER FISHES OF NORTH AMERICA tion rather than flesh feeding (Renaud 2002). Large Muskel-lunge (>122 cm TL) were preferred over small individuals.
Cannibalism and Interspecific Predation A leading cause for mortality in larval Northern Pike was identified as cannibalism between the larvae near Hough-ton Lake, Michigan (Hunt & Carbine 1951) and Pleasant Lake Marsh, Wisconsin (Fago 1977). Cannibalism also plays a role in older pike as well with some variation in the rate of cannibalism across the seasons. Cannibalism by the Northern Pike also increases with seasonal changes in availability of prey species (e.g., Coregonus spp.) associated with spawning be hav ior and habitat prefer-ences of preferred prey (Lawler 1965). The frequency of cannibalism in the Northern Pike varies depending on the size differences between individuals and on prey abundance (Nursall 1973; Eklöv & VanKooten 2001). Because populations of Muskellunge spawn after those of co- occurring Northern Pike, predation of Muskellunge larvae by slightly older, larger juveniles of Northern Pike might contribute to early mortality and the apparent com-petitive disadvantage of the Muskellunge relative to sym-patric Northern Pike (Scott & Crossman 1973).
Laboratory experiments with Northern Pike larvae showed that cannibalism appears as early as four weeks of age and accounted for 54–96% of daily mortality in the experimental tanks (Giles et al. 1986). The onset of canni-balism altered larval be hav ior; larvae established even & Crossman 1973). Alaska Blackfish growth is rapid in the first year; annual incremental growth declines thereafter (Blackett 1962; Aspinwall 1965), but growth is continu-ous. Bristol Bay Alaska Blackfish are slower growing and live longer than Interior Alaska populations (Morrow 1980). Information on growth of young- of- the- year Olym-pic Mudminnows is lacking. Growth of young- of- the- year umbrids is difficult to determine because scale annuli do not accurately reflect age. Central Mudminnows reach 37.5 mm SL by the end of the first summer and 90 mm SL at year 1 (Scott & Crossman 1973; Trautman 1981; Becker 1983).
Predators In general, esociforms are subject to predation by a wide va-riety of piscivorous wildlife, including wading and diving birds, mammals, and reptiles (Peckham & Dineen 1957; McEwan & Hirth 1980; Becker 1983) and large predatory fish, such as the Largemouth Bass, Micropterus salmoides, and large esociforms, including conspecifics (Wahl & Stein 1989; Craig 2008). Larval and juvenile esociforms are preyed upon by aquatic invertebrates (e.g., diving beetles) and fishes (Lewis 1974; Bry 1996; Le Louarn & Cloarec 1997; Panek & Weis 2013; Eidam et al. 2016). Predation by Silver Lamprey (Ichthyomyzon unicuspis) on Muskellunge was documented in the lower Ottawa River, Ontario and Que-bec. Sites of attachment, shallowness of the wounds, and evidence of buccal gland secretions suggested blood preda-20 40 60 80 100 PERCENT NEMATODA ACANTHOCE-PHALA CLADOCERA COPEPODA OSTRACODA TRICHOPTERA LARVAE DIPTERA LARVAE DIPTERA ADULT HYDRACARINA PHYSIDAE PLANORBIDAE FRY TETRASPORA OEDOGONIUM ZYGNEMA NOSTOC OTHER ALGAE Figure 19.52. Percentage of occurrence of classes of organisms in the stomachs of Alaska Blackfish, Dallia pectoralis, sampled in July (4 stations; n = 36) and August (same 4 stations plus an additional station; n = 41) in a shallow tundra lake and adjacent tundra marshes, near Point Barrow, Alaska. The dark bars represent July samples (4 stations), the cross- hatched bars (same 4 stations) August samples, and the dotted bars August samples from the additional station (redrawn from Ostdiek & Nardone 1959). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 251 Table 19.4. Percentage composition by number (volume) of major food groups in the diet of the Eastern Mudminnow, Umbra pygmaea (22–89 mm TL), from the Great Swamp National Wildlife Refuge (GS), New Jersey, and Croatan National Forest (CNF), North Carolina (data from Panek & Weiss 2013).
Winter January- March Spring April- June Summer July- September Autumn October- December GS CNF GS CNF GS CNF GS CNF Food Group (n = 68) (n = 38) (n = 32) (n = 85) (n = 30) (n = 50) (n = 17) (n = 17) Detritus — — — — — — — — (24.2) (25.8) (33.4) (47.2) (38.3) (35.0) (36.9) (37.8) Amphipoda 0.1 0.9 0.1 2.1 0.1 3.0 1.4 4.7 (4.2) (10.4) (3.1) (4.2) (3.0) (5.7) (2.6) (16.8) Cladocera — 79.6 4.8 26.4 — 20.7 1.8 0.9 (6.2) (9.3) (1.1) (4.2) — (0.5) (0.1) (0.4) Copepoda 0.9 0.7 34.2 4.6 — 6.1 21.2 14.2 (1.0) (1.8) (3.8) (2.2) — (1.1) (1.6) (2.4) Ostracoda 24.7 0.5 50.4 3.4 97.4 9.7 53.8 4.7 (22.0) (0.6) (28.2) (0.3) (50.6) (4.0) (19.5) (2.5) Diptera 73.9 15.3 5.5 44.1 — 37.1 10.1 62.0 (31.2) (23.5) (9.0) (13.2) — (9.1) (4.5) (10.0) Coleoptera 0.1 0.8 0.8 5.9 0.1 6.7 5.3 5.7 (10.6) (15.7) (6.4) (17.8) (3.0) (15.8) (17.9) (20.0) Odonata 0.1 — 0.4 0.4 — 3.0 1.6 — (2.3) — (5.3) (0.9) — (6.2) (3.5) — Trichoptera — — 3.2 3.8 — — — 0.9 (9.4) — (3.4) (2.9) — — — (5.7) Lepidoptera — — 0.8 — — — — (0.6) — — Ephemeroptera — — — 3.0 — — — — (6.2) — Collembola — 0.1 — 0.7 — — (0.1) — (1.0) — Heteroptera — 1.0 3.4 1.5 — — (7.1) (4.8) (2.3) Hydrachnidae — 0.2 0.4 — 4.7 — (1.0) (0.1) (1.2) Formicidae — — — 0.8 — 2.3 — — — — — (0.3) — (1.3) — — Nematoda — — — — — 1.5 — — — — — — — (0.2) — — Oligochaeta — 0.8 0.4 1.7 0.1 0.8 — — (3.1) (3.8) (3.5) (0.4) (1.7) (0.1) — — Pelecypoda 0.1 — 0.1 — — 0.8 2.4 — (0.9) — (1.3) — — (1.8) (6.3) — Gastropoda 0.1 — 0.1 — — 1.5 2.4 — (2.2) — (2.5) — — (5.9) (6.3) — U. pygmaea — — — — 0.8 0.8 — — — — — — (3.2) (3.2) — — Duckweed — — — — — — — — (1.2) — — — — — (1.0) — Filamentous algae — — — — — — — — (0.2) — — — — — (0.6) — Unidentified — 0.1 — 0.3 — — — 2.2 — (0.9) — (1.5) — — — (3.3) Percentage empty stomachs 30.9 10.5 40.7 26.5 37.5 38.0 28.3 17.6 © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 252 FRESHWATER FISHES OF NORTH AMERICA Stunting is more common in small lakes where availability of littoral habitat, water temperature, dissolved oxygen, water visibility and productivity, and amount and type of food resources are limiting (Margenau et al. 1998). Stunt-ing is less common in larger waters where Northern Pike density is lower and a greater diversity of prey exists (Mar-genau et al. 1998).
The importance of the functional role of the Northern Pike in ecosystems is the subject of debate. Because mem-bers of this species are voracious predators and because they can consume relatively large prey, they may be a sig-nificant factor in shaping fish communities, which in turn may influence entire ecosystems. In Denmark, Northern Pike are stocked in eutrophic lakes with the goal of reduc-ing zooplanktivorous fishes and in turn increasing zoo-plankton levels so that water turbidity is decreased (Søn-dergaard et al. 1997). The results of this program are mixed. Whether the ambiguous results are caused by high mortality in the stocking pro cess or the inability of North-ern Pike to have a significant effect on prey populations is unclear. In the United States, a similar biomanipulation program was deemed successful (Lathrop et al. 2002); however, this program involved the introduction of the Northern Pike and Walleye so the specific contribution of the Northern Pike to the success of the experiment is un-clear. Evidence from modeling suggested certain charac-teristics of the predatory be hav ior of the Northern Pike, such as cannibalism and kleptoparasitism, may make their effect on prey abundance and community structure short lived and relatively minor when compared with the effect other predators may have (Nilsson 2001).
In contrast, some populations of Northern Pike can in-deed consume enough prey biomass to cause significant changes in prey abundance and potentially shape the structure of the community (Grimm 1983; Paukert et al. 2003). Craig (1996) reviewed the role of the Northern Pike in shaping the distribution of prey species, their growth rate, and their species composition. In the pres-ence of the Northern Pike, species with deeper bodies and spiny fins, which are less vulnerable to predation by Northern Pike, increased in abundance, but this effect was somewhat dependent on habitat type (Eklöv & Ham-rin 1989). Where vegetation is scarce, the Northern Pike preferentially takes the soft- rayed cyprinid Scardinius erythrophthalmus (Rudd), but in abundant submerged veg-etation the preferred prey is the spiny Yellow Perch. This suggests Northern Pike are opportunistic, and the preva-lence of Yellow Perch in their diet is the result of habitat overlap rather than prey se lection. Seasonal change is an-spacing in the tanks and remained still for long periods, particularly following the onset of cannibalism. Cannibals were attracted by larval movements which often resulted in attacks. Non- cannibals reduced feeding rates, and while stationary, oriented to cannibals.
Abundance, Population Dynamics, and Ecological Role of Esox An extensive lit er a ture exists on population dynamics and ecological role of the Northern Pike (reviewed Craig 1996, 2008). In addition, Hall (1986) edited a compilation of studies on the Muskellunge propagation, ecol ogy, and pop-ulation biology. Mittelbach & Persson (1998) compared the ontogeny of piscivory among several families. Pike and pickerel generally tend to spawn at lower temperatures, have larger egg dia meters, and hatch at larger sizes than other piscivorous genera except for some salmonids.
Estimates of population density and biomass of the Northern Pike vary extensively among populations and between years within a population. Densities of 2.5–80.0 individuals/ha and biomass of 1.5–31.3 kg/ha are known (Bregazzi & Kennedy 1980). Part of this variability is caused by wide fluctuations in recruitment success from year to year. A suitable prey base and the water tempera-ture regime also impact population density. The size of the parental cohort and the size of the resulting offspring cohort are not correlated, pointing to the importance of mortality rates of eggs, larvae, and juveniles in determin-ing recruitment success. Broad climatological patterns like the strength and direction of the North Atlantic Oscillation displacement (corresponding to dif fer ent cli-matic conditions in autumn and winter) may be impor tant explanatory variables for recruitment and year- class success (Paxton et al. 2009). Female recruitment was also influenced by young- of- the- year winter temperature.
Management strategies aimed at improving the quality of recreational fisheries of Northern Pike and Muskellunge generally include use of length and creel limits, critical hab-itat protection (i.e., spawning, rearing, and feeding), and stocking programs to supplement natu ral reproduction. The results of these programs are mixed as the effects of limits may be masked by recruitment failures and mortality in the population (Margenau & AveLallement 2000; Mar-genau et al. 2008). Stunting (poor growth) in the Northern Pike may be associated with a variety of biotic and abiotic factors, including high population abundance, competition for food, overstocking, and factors associated with lake size (Snow 1974; Goeman et al. 1993; Margenau et al. 1998). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 253 ral wetlands with a peak in abundance in April and May when young- of- the- year emerge and access is increased due to inundation of off- channel wetland and oxbow habi-tats (Henning 2004). Electrofishing at 16 sites historically supporting the species produced 0–144 individuals (CPUE 0–1.9 individuals/s) (Mongillo & Hallock 1999). The oc-currence of Olympic Mudminnows in wetland sites may be related more to habitat preference (Henning et al. 2007) than competitive exclusion or predation by non- native species (Beecher & Fernau 1983). An intensive sur-vey of habitats in the Chehalis River system showed strong affinities for lower water temperatures, suggesting that Olympic Mudminnows per sis tence in marginal habi-tats may depend on cool- water refugia associated with groundwater springs (Kuehne & Olden 2016). Ordination and modeling of quadrat- scale environmental variables showed that detection and abundance of Olympic Mud-minnows were strongly related to lower temperatures fol-lowed by shallow depths (associated with shorelines), low dissolved oxygen, and greater extent of macrophyte cover.
Although quantitative data are lacking, the Alaska Blackfish often occurs in vast numbers in tiny, weed- choked pools and shallow lakes in the tundra (Nelson 1884; Turner 1886; Aspinwall 1965; Haynes et al. 2014). In a multimethod survey in summer of 86 lakes across the interior of the Arctic Coastal Plain in Alaska, detection probabilities for Alaska Blackfish were highest in minnow traps set in the deepest portions of the shallow lakes rela-tive to minnow traps set along the shorelines, pelagic and benthic gill nets, and fyke nets (Haynes et al. 2013). The probability of occupancy of Alaska Blackfish decreased as lakes became less densely distributed at the regional level, suggesting connectivity is an impor tant life- history factor. Occupancy also increased as a greater percentage of the lake remained unfrozen during winter, prob ably because they overwinter in lakes with deep, unfrozen waters (Haynes et al. 2014). Alaska Blackfish had an observed oc-cupancy of 0.74 and an estimated occupancy 0.76 (Haynes et al. 2014).
In suitable habits Umbra spp. can be quite common, al-though quantitative estimates of population sizes are few. For example, in a multimethod survey of fishes in the Great Swamp National Wildlife Refuge, New Jersey, the Eastern Mudminnow accounted for 73.8% (248 individu-als) of the seven fishes captured over multiple months (17–68 individuals/season) (Panek & Weiss 2012). Simi-larly, in central Wisconsin, >3,000 Central Mudminnows were captured from undercut banks along a 200 m reach of an old ditch downstream of a marsh (Becker 1983).
other source of variation in the composition of the diet of the Northern Pike (Sammons et al. 1994; Paukert et al. 2003). Brenden et al. (2006) estimated Muskellunge pre-dation on other species in the New River, West Virginia, where they prey primarily on cyprinids and catostomids (0.31 kg ha/year and 0.63 kg/ha/year, respectively for 100 age-1 Muskellunge).
An in ter est ing effect of the Northern Pike on prey spe-cies was a documented increase in body depth of resident Crucian Carp, Carassius carassius, after the introduction of the Northern Pike (Brönmark & Miner 1992). This effect is mediated by chemical cues detected by Crucian Carp in the odor of Northern Pike (Brönmark & Pettersson 1994). In another prey- related effect, larger gapes were observed in populations of Northern Pike living in conditions of low prey abundance (Magnhagen & Heibo 2001). The effect was attributed to either phenotypic plasticity or ge ne tic dif-ferences driven by se lection for those individuals likely to capitalize on the infrequent encounters with prey under conditions of low resource abundance. The emergence of a novel ecotype of Chain Pickerel (deeper body, more slender head, and shorter upper jaw) was documented in im-pounded coastal lakes with landlocked populations of Ale-wife (Alosa pseudoharengus). Morphological changes corre-sponded with differences in stable isotopes, dietary observations, and elevated lipid content and suggested that pelagic foraging by Chain Pickerel conferred a fitness ad-vantage (Brodersen et al. 2015).
Due to the potential effect of Northern Pike introduc-tions on native species, special attention is given to the re-lationship between introduced Northern Pike and salmo-nids, which are eco nom ically valuable in commercial and recreational fisheries. In this re spect, the illegal introduc-tion of Northern Pike to Pacific Coast drainages in Alaska has generated some alarm (Dalton 2002). Declines in fish-eries stocks of Northern Pike around the Baltic Sea are attributed to eutrophication, climate change, and over-fishing and have produced a trophic cascade resulting in a shift from a predator- dominated to a prey- dominated food web (Lehtonen et al. 2009; Ljunggren et al. 2010).
Abundance and Occurrence of Novumbra, Dallia, and Umbra Olympic Mudminnows occur in greatest abundance in shallow, emergent wetlands with muddy substrate, dense vegetation, and little or no current, regardless of whether non- native species are pre sent (Henning et al. 2007). They occur in large numbers in both regulated and natu © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 254 FRESHWATER FISHES OF NORTH AMERICA the U.S. Atlantic Coastal Plain (Bunnell 2006; Panek & Weis 2012). This suggests the species is ecologically impor tant to the functioning of these systems. Like the Central Mudminnow, the Eastern Mudminnow is hardy and often lives in marginal habitats and withstands ex-treme environmental temperatures, periods of hypoxia, and low pH. Evidence suggests abundance of Umbra spp. is inversely related to the presence of predatory species (Dederen et al. 1986) and to the number of fish species pre sent (Panek 1981).
Dallia, Novumbra, and Umbra all have some characteris-tics (small fish, inhabit shallow waters, locally abundant, associated with dense aquatic macrophytes, tolerant of a wide range of environmental conditions) of fishes that are used to control nuisance insects such as mosquitoes (Ahmed et al. 1988; Walton 2007; Pyke 2008). Because Dallia, Novumbra, and Umbra occupy marginal habitats and may be the only species pre sent, they may be a useful control agent for reducing mosquito populations, either directly through predation or indirectly by affecting adult insect choice in oviposition sites (Lombardi 2009; Walton et al. 2009; Tabor et al. 2014).
Parasites Common parasites in esociforms include myxozoans, ciliates, trematodes, cestodes, helminths, and acantho-cephalans (Amin 1980; Amin & Myer 1982; Muzzall & Buckner 1982; Muzzall 1984; Hoffman 1999; Basson & Van As 2006). The diversity of parasites that occur in any one species can be quite high. The Northern Pike, one of the best studied species, for example, is documented as harboring parasitic fungi (2 taxa), protozoans (about 24), monogeneans (7), trematodes (about 50), cestodes (about 20), nematodes (about 27), acanthocephalans (about 21), hirudineans (6), molluscans (prob ably several), and crus-taceans (about 13) (review by Hoffman 1999). Although most parasites of esociforms occur in a variety of fishes, some parasites apparently occur only in a par tic u lar esoci-form, such as the spiny headed worm, Neoechinorhynchus limi, in the Central Mudminnow (Muzzall 1984), the ces-tode, Diphyllobothrium dalliae, in the Alaska Blackfish (Rausch 1956), the monogeneans, Gyrodactylus lucii, in the Northern Pike (Cone & Dechtiar 1986), G. fryi, in the Muskellunge (Cone & Dechtiar 1984), and G. neili, in the Chain Pickerel (LeBlanc et al. 2006).
Parasitic infestations can be heavy in some species and predominate in larval and small fishes. For example, D. dalliae occurs free in the body cavity of Alaska Blackfish Other Ecological Aspects of Non- Esox Esociforms Little else is known about the ecological role played by other esociform species, most notably for Novumbra and Dallia. Generally, these species form part of fish communi-ties that include only a few species (<4), and they appear to compete poorly with other fishes. The Olympic Mudmin-now can be locally abundant and may be the only fish spe-cies pre sent (Meldrim 1968; Tabor et al. 2014). As such, they may have impor tant effects on the aquatic community, but their par tic u lar role in regulating aquatic communities is uncertain. Catch / unit effort of Olympic Mudminnows declines as fish species richness increases, suggesting they fare poorly when in competition with or potentially preyed on by other species (Mongillo & Hallock 1999).
Populations of Alaska Blackfish most often coexist with Gasterosteus aculeatus (Threespine Stickleback), Cottus spp. (freshwater Sculpins), and juveniles of vari ous species of Pacific Salmon (Oncorhynchus spp.), but they occur at greatest densities in the absence of these or any other spe-cies (Reynolds 1997). Similarly, individuals of Alaska Blackfish in Siberia exhibit faster growth and attain larger sizes in the absence of other fish species (Gudkov 1998).
Although species of Umbra frequently form part of more diverse fish communities, they too appear to be poor competitors and to be more abundant in the absence of other species (Tonn & Paszkowski 1987). This may be at-tributed to the tolerance of these species to extreme envi-ronmental conditions that are stressful or lethal to other fishes. The Central Mudminnow was described as a re-source generalist based on the breadth and variability of their diet and as a specialist based on their specific habitat requirements (Martin- Bergmann & Gee 1985).
In acidic systems, Central Mudminnows and Eastern Mudminnows may be impor tant predators (Rahel & Magnusson 1983; Dederen et al. 1986). At high densities, Central Mudminnows may affect the structure of littoral macroinvertebrate communities (Brown & DeVries 1985), however their effect on littoral fish communities is un-known. Central Mudminnows may also serve as impor tant prey items in some systems (Tonn 1985; Ward et al. 2008) but not in others (Weidel et al. 2000, 2007). As female Cen-tral Mudminnows apparently shift to piscivory during win-ter (Chilton et al. 1984), the size and age structure of a large adult female population could strongly influence the ability of other competing fish species to become estab-lished or maintain populations under some conditions.
Eastern Mudminnows often represent a significant component of the fish community in flooded marshes on © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 255 transformation is unconfirmed; salmonids are the primary host fishes for Freshwater Pearl Mussels (Haag 2012).
CONSERVATION Among esociform species, members of Esox and Dallia are considered of least conservation concern by the Interna-tional Union for Conservation of Nature and Natu ral Re-sources (IUCN 2016; NatureServe 2018). These species are generally wide- ranging and abundant. As impor tant sport fish, the Northern Pike and Muskellunge are also widely stocked to support recreational and commercial fisheries. Stocking of Muskellunge may be obscuring the taxonomic status of the color variants, and management strategies to conserve ge ne tic diversity in Muskellunge populations are proposed (Casselman et al. 1986; Jen-nings et al. 2010).
Owing to its rarity and limited distribution, the Olym-pic Mudminnow is categorized as Vulnerable by the American Fisheries Society (Jelks et al. 2008), and as a Sensitive protected species in Washington because of its restricted range and highly vulnerable habitat (Mongillo & Hallock 1999; WDFW 2012). It was considered for pro-tection under the U.S. Endangered Species Act until the rule change governing the status of candidate species. The current IUCN assessment lists it as Least Concern (IUCN 2016). Fishing for or collecting this species is pro-hibited even though it is not commonly fished or used for bait, and the main threat to its survival remains the de-struction of suitable wetland habitat (Mongillo & Hallock 1999). A proposed impoundment on the upper Chehalis River would inundate significant sections of habitat of flu-vial populations (Scott et al. 2013; Kuehne & Olden 2016).
Other umbrids are at risk in portions of their home ranges, but no range- wide threats are currently identified (Jelks et al. 2008; NatureServe 2018). In states where Cen-tral and Eastern Mudminnows are species of conservation concern, the species have limited distributions or occur at the periphery of their range (Page & Burr 2011). The Eu ro pean Mudminnow is classified as Vulnerable (IUCN 2016).
Climate Change Effects on Distribution Climate change associated effects in freshwater ecosystems (including rising stream temperatures and sea levels) pose potential threats to esociform species (Ficke et al. 2007). Under the Canadian Climate Center general circulation model, Eaton & Scheller (1996) predicted a <50% loss of throughout its range in Alaska (Hilliard 1959); infections transferred to dogs and humans may cause pernicious anemia (loss of stomach cells cause anemia) (Rausch et al. 1967). In an Indiana population, 10% of Central Mudmin-nows contained heavy infections of yellow grub (Trema-toda: Clinostomum sp.) that hindered swimming move-ments and appeared to afflict small fish the greatest (Peckham & Dineen 1957). Gyrodactylus limi was common on the skin and gills of larvae (10–15 mm TL) and ap-peared to pass readily from one fish to another. A com-mon parasite of the urinary bladder was a trematode Phyl-lodistomum brevicecum. In Central Mudminnows, seasonal patterns of infection of two intestinal acanthocephalan parasites could be related to movements of the species, changes in water temperature, and possibly to dietary changes (Muzzall 1984). Eutrophication of aquatic sys-tems linked to watershed disturbance in the acidic streams of the New Jersey Pine Barrens is correlated with high susceptibility and infection rates by helminth para-sites in the Eastern Mudminnow (Acanthocephala, 2 taxa; Digenea, 2 taxa; Nematoda, 3 taxa; 8–75% prevalence) and Chain Pickerel (Acanthocephala, 2 taxa; Digenea, 2 taxa; Nematoda, 3 taxa; Cestoda, 1 taxon; 14–100% preva-lence) (Hernandez et al. 2007). Schaufler et al. (2014) highlighted an unintended consequence of stocking pro-grams of Northern Pike, reporting a subsequent epizootic outbreak of cestodes in resident char.
Unlike many other North American fishes, the esoci-forms apparently do not serve as specialized hosts for any North American freshwater mussel taxa (Haag 2012). Glo-chidia of the Creeper, Strophitus undulatus, a species that uses an array of fishes as host, transformed on the Central Mudminnow in laboratory trials. Laboratory infection of Chain Pickerel with glochidia of Fatmucket (Lampsilis sili-quoidea) apparently caused the death of the host fish (Keller & Ruessler 1997). Some fishes in the wild, like the Redfin Pickerel, can be infected with glochidia (larvae) of generalist mussel species, like the Paper Pondshell, Utter-backia imbecilis, but transformation is unconfirmed. The occurrence of Yellow Lampmussel (Lampsilis cariosa) glo-chidia in a juvenile Chain Pickerel may indicate it is a potential host (Kneeland & Rhymer 2008). Likewise, Northern Pike in the wild carry glochidia of the Three-ridge, Amblema plicata, another generalist species, but transformation is unconfirmed (Parmalee & Bogan 1998; Williams et al. 2008, 2014). Glochidia of Eu ro pean unionid mussels (Unio sp.) and Freshwater Pearl Mussel (Margaritifera margaritifera) occur naturally in the North-ern Pike (Blažek & Gelnar 2006; Geist et al. 2006), but © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 256 FRESHWATER FISHES OF NORTH AMERICA the Alaska Blackfish (see ge ne tics section) is maintained by the strictly freshwater nature and low dispersal ability of this genus. These ecological characteristics make the di-verse populations vulnerable to rising temperatures and rising sea levels associated with climate change.
Habitat Alteration Other than Olympic Mudminnows, all esociform species have relatively wide distributions that presently include many healthy, self- sustaining populations. Therefore, as species they are not considered at risk. However, because all esociforms require slow moving, heavi ly vegetated, shallow waters, their suitable habitat is usually adversely affected by land development practices that require water- level alteration and drainage of shallow wetlands. In fact, habitat alteration, not harvesting, likely is responsible for declines or extirpation of some populations of esociforms from their historical range (e.g., Umbra, Kenoza, Trautman & Gartman 1974; Esox spp., Axon & Kornman 1986; Cook & Solomon 1987; Minns et al. 1996; Rust et al. 2002; Coo-per et al. 2008; Novumbra, Mongillo & Hallock 1999; Glasgow & Hallock 2009). Management programs are in place to prevent further losses of populations and de-creases in population size of Northern Pike and Muskel-lunge, both of which are highly regarded targets for recre-ational anglers (see commercial importance section).
Pollution As top piscivores, Esox spp. have the potential to accumu-late contaminants in their tissues in environments ex-posed to human pollution, including mercury (Gariboldi et al. 1998; Stafford & Haines 2001; Jewett et al. 2003; Tang et al. 2013), other heavy metals (Harrison & Klaverkamp 1990), and organic pollutants (Lazorchak et al. 2003; Burreau et al. 2004; Sharma et al. 2009). Tis-sue concentrations of these pollutants tend to be strongly and positively correlated with size and age (Jewett et al. 2003; Kamman et al. 2005; Jewett & Duffy 2007) and with the length and complexity of lake food webs (Stem-berger & Chen 1998) and catchment area (Peterson et al. 2007). In regional- scale assessments of mercury contami-nation, the highest mean mercury concentrations oc-curred in the Muskellunge (0.98 ng/g, fillets) and North-ern Pike (0.64 ng/g, fillets; 0.56 ng/g, whole- body) (Kamman et al. 2005). Chain Pickerel had tissue mercury concentrations between 0.58–1.22 ng/g body weight (mean = 0.88 ng/g), the highest of nine species sampled the Northern Pike from gaged stations in the conterminous United States. Similar declines are predicted for the North-ern Pike and Central Mudminnow in Wisconsin under multiple climate scenarios (Lyons et al. 2010). Though the successful invasion of subarctic lake ecosystems by the Northern Pike could not be directly associated with in-creased temperatures, the dramatic changes observed in abundance of prey species and trophic relationships in the lakes typified the pos si ble effects of future Northern Pike invasions that could be enhanced by climate change (Minns & Moore 1992; Byström et al. 2007).
Climate change may affect food availability and pro-ductivity in freshwater systems (Linnansaari & Cunjak 2012) and in turn affect Esox spp. as well as other esoci-forms. Although reduced prey would suppress Northern Pike populations, increased nutrient bud gets could in-crease Northern Pike production in systems with low pro-ductivity (Reist et al. 2006). A shifting thermal regime, however, may affect condition indices as metabolic de-mands on resident fishes increase (Spotila et al. 1979; Clapp & Wahl 1996). Rising temperatures may disrupt growth, reproduction, and recruitment as well as the phe-nology of prey availability to larvae and juveniles. For ex-ample, increases in summer and early winter water tem-peratures impaired year- class strength for Northern Pike (Casselman 2002). Rising temperatures can also trigger a cascade of effects, including increased susceptibility to epizootic diseases, magnification of pollutants, oxygen de-pletion, and degraded water quality (Jeppesen et al. 2010). As a geo graph i cally restricted, stenothermic cold- water fish, the Alaska Blackfish is among the most vulnerable species to climate change (Lehtonen 1996; Reist et al. 2006). Thawing permafrost may increase methane emis-sions; methane trapped under ice is linked with winterkill in Alaska Blackfish (Campbell et al. 2014).
Climate change can also impose habitat constraints because changes in timing of spring floods may affect availability and access to spawning habitats. Lower water levels from drought could also impose an added energy burden as fish seek to avoid temperatures >25°C. North-ern Pike lose weight when their habitats are restricted as when thermoclines separating warm surface waters from cooler water approach hypoxic depths (Reist et al. 2006).
Rising sea levels pre sent pos si ble threats to populations of Olympic Mudminnows, Alaska Blackfish, and Eastern Mudminnows in coastal sloughs and marshes. Intrusion of saltwater into freshwater and brackish marshes could raise salinities above the maximum tolerance thresholds for those species. The high degree of ge ne tic structuring of © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 257 et al. 1992). Laboratory toxicity tests revealed the most tolerant fish had greater ge ne tic variation, suggesting ge ne tic diversity conferred differential survival to individu-als under acute physiological stress. Because of their occurrence in sloughs, small ponds, ditches, and backwa-ter habitats, small esocids and umbrids may be dispropor-tionately affected by exposure to pesticides used in mos-quito control (Bender & Westman 1976).
Diseases of Esociforms Diseases associated with pathogens and parasites vary across the native and introduced range of esociforms (Dick & Choudhury 1996; Harshbarger & Clark 1990; Roberts 2012). Owing to their commercial and recre-ational importance and their propagation in aquaculture, more attention has been devoted to the characterization of diseases in the Northern Pike and Muskellunge than in other esociforms. Northern Pike, particularly the larvae, in aquaculture are susceptible to a variety of diseases that have high infection rates and mortality (Bootsma 1971; Bootsma & van Vorstenbosch 1973; Graham et al. 2004; reviewed by Coffee et al. 2013).
Viruses can cause significant pathology in esocids. Viral haemorrhagic septicaemia virus (VHSV) occurs in the Muskellunge causing damage to internal organs and thickening and formation of fluid- filled vesicles on the in-ner wall of the swim bladder (Elsayed et al. 2006). An outbreak in the Upper St. Lawrence River caused size- and- age specific mortality of about 50%; the culling effect resulted in a growth- rate change (Casselman 2011). Red Disease (or Pike Fry Rhabdovirus Disease), which pro-duces hemorrhagic areas in the trunk in the Northern Pike, is caused by a rhabdovirus and can be transmitted through water (De Kinkelin et al. 1973; Bootsma et al. 1975; Jørgensen et al. 1993).
Blue Spot Disease (caused by Esocid Herpesvirus-1, EsHV-1) occurs in the Northern Pike and Muskellunge and is associated with small, pale bluish- white, granular skin lesions that occur mostly over the dorsal skin and fins (Yamamoto et al. 1984; Margenau et al. 1995). Its prevalence in Wisconsin differed among lakes, years, and seasons, and was generally higher for female Northern Pike. Infection rates averaged 11% (200 ng/g wet weight, an estimated tissue threshold associated with sublethal and reproductive effects in fish (Rolfhus et al. 2015), and exceeding wildlife exposure values for piscivorous mam-mals (e.g., river otter, Lutra canadensis and mink, Mustela vison) and birds (e.g., belted kingfisher, Ceryle alcyon) (Lazorchak et al. 2003).
Environmental contaminants may also affect recruit-ment and year class success. Larval deformities character-istic of selenium exposure (i.e., skeletal curvatures, cra-niofacial deformities, fin deformities, and edema) were 20% more frequent than background levels in eggs ob-tained from female Northern Pike exposed to metal min-ing effluent (Muscatello et al. 2006). Pulp mill effluent in-duced severe skeletal deformities in the jaw and skull (Lindesjöö & Thulin 1992).
Esociforms may be negatively affected by other sources of pollution. For example, degraded water clarity associ-ated with increased turbidity and eutrophication can de-crease visual range and reaction distance in visual preda-tors like esocids resulting in lower predation success (Turesson & Brönmark 2004, 2007; Nilsson et al. 2009; Ranåker et al. 2012). Increasing eutrophication depressed larval condition of the Northern Pike in some parts of the Baltic Sea and has reduced population size (Salonen et al. 2009). Similarly in a small esociform, Central Mudmin-nows exhibited a threshold response of increased growth as agricultural and urban land use increased (Filgueira et al. 2016). Central Mudminnow populations from Ad-irondack Mountain sites exposed to acid deposition had lower ge ne tic heterozygosity than at control sites (Kopp © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 258 FRESHWATER FISHES OF NORTH AMERICA resource to the indigenous peoples of Alaska (Brown et al. 2005; Jewett & Duffy 2007; Fig. 19.53), although the im-portance of that subsistence fishery is much less today than in the past.
The Northern Pike and Muskellunge, both of which are impor tant sportfish, are the only esociforms that have a significant economic impact. A substantial recreational fishery targeting these two species thrives in the United States and Canada (Fig. 19.54). The commercial value of have observable lesions. The pathogenesis of Blue Spot Disease is unknown. Clinical signs dissipated when water temperatures reach 14°C. The infection rates found by Margenau et al. (1995) were generally higher on average than those reported by Yamamoto et al. (1984) (<1% to 7%) in Canada.
Esocid lymphosarcoma, a malignancy caused by a Type C retrovirus that produces sores and tumors on the skin, muscles, and internal organs of fishes, occurs widely in the Northern Pike and Muskellunge adults in North Amer i ca and Eu rope (Papas et al. 1976; Quackenbush et al. 2010). The tumor has a low prevalence in the summer, but its oc-currence increases in the autumn, winter, and spring. The frequency in Northern Pike (19.9%) is the highest fre-quency of a malignant neoplasm in any known free- living vertebrate (Papas et al. 1976). Pike epidermal proliferation, another Type C retroviral disease, causes smooth, either transparent or translucent, plaque- like lesions with occa-sional hemorrhaging (Bowser & Casey 1993).
In experimental water- borne challenges, Northern Pike larvae showed significant infection rates and mortality from several Ranavirus isolates. Mortality rates following infection were higher at warmer temperatures (22 versus 12°C) (Jensen et al. 2009). Horizontal transmission of es-ocid lymphosarcoma may occur by contact during spawn-ing (Coffee et al. 2013). Viral diseases can be transmitted to the Northern Pike from prey fishes (Ahne 1985).
COMMERCIAL IMPORTANCE Subsistence and commercial fishing are of modest scale for some esociform species where regulations allow it. In the past, these fisheries had a significantly greater com-mercial monetary value. The Great Lakes historically sup-ported a Northern Pike commercial harvest of 1.6 million kg in the early 1900s that declined to <0.05 million kg by the 1960s (Casselman & Lewis 1996). The decline coin-cided with the timing and extent of Eu ro pean settlement and shoreline development. The state of New York re-ported a commercial harvest of 40,370 kg of Muskellunge in 1894 and 46,779 kg in 1895, for a combined value at that time of $29,270 (about $857,000 in 2016). In the United States, however, the commercial fishing for Mus-kellunge was largely limited to the 19th century and was prohibited at the turn of the century (Crossman 1986). Other esociforms are unimportant in commercial harvest and recreational angling, but Alaska Blackfish historically represented an impor tant, almost essential, subsistence Figure 19.53. Into the mid-20th century, native peoples often used wooden, funnel- shaped fish traps known as taluyat (Yup’ik language) to catch Alaska Blackfish, Dallia pectoralis. These traps varied in size and shape and were set in holes in the ice or placed to block narrow, shallow stream channels. The funnel- shaped top (on the ground to the right) was inverted over the mouth of the trap. The narrow end of the funnel had a small orifice through which Alaska Blackfish entered. Many traps were made from split spruce, which was bound with spruce root. The tendency of the fish to congregate at holes in the ice to gulp air (Fig. 19.36) and then sink back to the bottom made them particularly vulnerable to these traps. The Alaska Blackfish was an extremely impor tant staple in the diet of Alaskan natives, providing food for the people as well as their sled dogs at times of the year when other food was scarce. One native name for the fish roughly translates as the ones you survive on (Andersen et al. 2004). In the 1880s, ≥140,600 kg (or perhaps double that amount) of the fish were being harvested by Alaskan natives each year (Nelson 1884; Turner 1886; Morrow 1980). The photo graph of the fisher about to reset his trap was taken in February 1940 near Nunapitchuk, Alaska, by G. Dale and E. Butler. (photo graph P306-0581 Alaska State Library Butler / Dale Photo Collection, used with permission). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION Esociformes: Esocidae, Pikes, and Umbridae, Mudminnows 259 LIT ER A TURE GUIDE The first written description of reproductive be hav ior in the Northern Pike was prob ably that of Walton (1653), who noted the basic pattern of external fertilization in shallow waters into which the fish migrated to spawn. The Northern Pike has the distinction of being studied by Frost and Kipling (Frost & Kipling 1967; Kipling & Frost 1969, 1970; Kipling 1983). Their work, and that of their collaborators, is significant for the long- term studies of Northern Pike biology in Lake Windemere.
As autecological studies have fallen from favor, many of the best sources for information on the diet, ecol ogy, and habitats of species that lack importance as commercial or fisheries or are at conservation risk are older studies from the middle of the 20th century. The digitization of myriad older publications has restored the availability of this work. Information about the sporting species of interest to anglers is now widely available in agency lit er a ture because much of it is available on the Internet. Indeed, the lit er a ture is rife with population and even single lake studies of growth, reproduction, recruitment, and popula-tion dynamics of the Muskellunge and Northern Pike, but few broad- scale syntheses of these diverse findings exist. Crossman & Goodchild (1978) compiled an extensive bib-liography on Muskellunge biology (Fig. 19.55), and Hall (1986) edited a treatise on their biology and propagation. Craig (2008) summarized advances in Northern Pike ecol ogy in the de cade since the volume he edited was pub-lished (authors cited herein). Harvey (2009) summarized the biology of the Northern Pike. Particularly succinct, this recreational fishery extends well beyond the transac-tions immediately associated with fishing, such as sales of permits and fishing supplies. In 2011, freshwater anglers spent about $2.1 billion on tackle, trips, and other recre-ational items fishing for the Northern Pike, Muskellunge, and other esocids (USFWS 2014). In the Great Lakes re-gion, the allure of landing a trophy Northern Pike or bet-ter yet, a Muskellunge, helps fuel a strong recreational fishing industry with the associated benefits to tourism and related operations. Great Lakes fishing alone had trip and equipment expenditures in 2011 that totaled $1.9 bil-lion, and 200,000 of the 1.7 million anglers who fished the Great Lakes reported fishing specifically for the Northern Pike and other esocids (USFWS 2014). Early in this century, an estimated 360,000 anglers targeted Mus-kellunge (Simonson 2003).
Another consideration when gauging the economic value of these species is the level of investment by govern-mental institutions through management programs. These programs often involve substantial stocking opera-tions to supplement natu ral reproduction in declining populations or to maintain introduced populations to sup-port active recreational fishing. Fishing for either species is under strict local government regulations with mini-mum size and bag limits in most areas. Some states apply trophy regulations (e.g., large minimum size limits, low bag limits) in the Muskellunge fishery with the goal of in-creasing the abundance of large fish (Kerr 2011). Further, thanks to angler- led catch and release initiatives, even regulation- sized Muskellunge are released at a high rate (Isermann et al. 2011).
Figure 19.54. An angler displays a large female Muskellunge, Esox masquinongy (1.4 m TL), caught as it leaped out of the water to engulf a lure being retrieved into the boat. The fish was caught (and released) in the St. Lawrence River, Montreal, in July. The species is a highly sought after gamefish in a specialized sport fishery. Finding, hooking, and catching fish of this size can take an enormous amount of effort (100s of hours) (courtesy of Marc Thorpe / Marc Thorpe Guiding Ser vice). © 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION 260 FRESHWATER FISHES OF NORTH AMERICA the life history of the Olympic Mudminnow, and Scott & Crossman (1973) and Morrow (1980) synthesized nearly all that was known at the time on the biology of the Alaska Blackfish. Unfortunately little has been added since. Wanzenböck (1995) summarized the knowledge up to that time of the Eu ro pean Mudminnow; some refer-ences therein are relevant to the study of North American Umbridae and esociforms in general.
An abundant lit er a ture is available on management strategies for improving recruitment of the Northern Pike and Muskellunge (e.g., Hanson et al. 1986; Farrell et al. 2006, 2007, 2014), but many are focused on specific reaches of individual waterbodies. Management of large esocids may be improved if stream– lake (and wetland) network characteristics are incorporated into manage-ment plans (Weeks & Hansen 2009; Diana et al. 2015). Such networks support greater habitat and biological di-versity (and productivity) than systems that lack lentic habitats. These considerations could improve Muskel-lunge management in regulated rivers like the Ohio and Tennessee Rivers.
Acknowl edgments We thank Gayle Henderson for painstakingly redrawing numerous figures. Mickey Bland patiently worked with li-brary ser vices to locate masses of often obscure lit er a ture. Ken Sterling and Gordon McWhirter were essential in for-matting a large lit er a ture cited and proofing text, tables, and figures. Carolyn McCormick assisted with proofing the chapter and the lit er a ture cited.
but complete accounts summarizing the biology and ecol ogy of the Central Mudminnow and Esox spp. are avail-able in Scott & Crossman (1973) and Becker (1983). Wydoski & Whitney (2003) provide a thorough account of Figure 19.55. Edwin J. Crossman (1929–2003), long an employee of the Royal Ontario Museum, was a world- renowned expert on Pikes (Esocidae). His major works on the esocids are cited in this chapter, including his classic study on geographic variation in the Grass Pickerel, Esox americanus. In the massive volume on Freshwater Fishes of Canada (Scott & Crossman 1973) much of the earlier lit er a ture on esocids and umbrids (Mudminnows) was reviewed (courtesy of ichthyologist photo graph collection of Brooks M. Burr).
© 2019 The Johns Hopkins University Press UNCORRECTED PROOF Do not quote for publication until verified with finished book All rights reserved. No portion of this may be reproduced or distributed without permission.
NOT FOR SALE OR DISTRIBUTION |
7993 | https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf?srsltid=AfmBOor2GmYg_NCfgnVX1rupb-RtYdNTwk1-ZYQfdNgKq5SAq74UXYCB | Olympiad Inequalities
Thomas J. Mildorf December 22, 2005
It is the purpose of this document to familiarize the reader with a wide range of theorems and techniques that can be used to solve inequalities of the variety typically appearing on mathematical olympiads or other elementary proof contests. The Standard Dozen is an exhibition of twelve famous inequalities which can be cited and applied without proof in a solution. It is expected that most problems will fall entirely within the span of these inequalities. The Examples section provides numerous complete solutions as well as remarks on inequality-solving intuition, all intended to increase the reader’s aptitude for the material covered here. It is organized in rough order of difficulty. Finally, the Problems section contains exercises without solutions, ranging from straightforward to quite difficult, for the purpose of practicing techniques contained in this document. I have compiled much of this from posts by my peers in a number of mathematical communities, particularly the Mathlinks-Art of Problem Solving forums, 1 as well as from various MOP lectures, 2 Kiran Kedlaya’s inequalities packet, 3 and John Scholes’ site. 4 I have tried to take note of original sources where possible. This work in progress is distributed for personal educational use only. In particular, any publication of all or part of this manuscript without prior consent of the author, as well as any original sources noted herein, is strictly prohibited. Please send comments - suggestions, corrections, missing information, 5 or other interesting problems - to the author at tmildorf@mit.edu .Without further delay...
1
and respectively, though they have merged into a single, very large and robust group. The forums there are also host to a considerable wealth of additional material outside of inequalities.
2
Math Olympiad Program. Although some people would try to convince me it is the Math Olympiad Summer Program and therefore is due the acronym MOSP, those who know acknowledge that the traditional “MOP” is the preferred appellation.
3
The particularly diligent student of inequalities would be interested in this document, which is available online at Further ma-terial is also available in the books Andreescu-Cartoaje-Dospinescu-Lascu, Old and New Inequalities , GIL Publishing House, and Hardy-Littlewood-P´ olya, Inequalities , Cambridge University Press. (The former is elementary and geared towards contests, the latter is more technical.)
4
where a seemingly inexhaustible supply of Olympiads is available.
5
Such as the source of the last problem in this document.
11 The Standard Dozen
Throughout this lecture, we refer to convex and concave functions. Write I and I′ for the intervals [ a, b ] and ( a, b ) respectively. A function f is said to be convex on I if and only if
λf (x) + (1 − λ)f (y) ≥ f (λx + (1 − λ)y) for all x, y ∈ I and 0 ≤ λ ≤ 1. Conversely, if the inequality always holds in the opposite direction, the function is said to be concave on the interval. A function f that is continuous on I and twice differentiable on I′ is convex on I
if and only if f ′′ (x) ≥ 0 for all x ∈ I (Concave if the inequality is flipped.) Let x1 ≥ x2 ≥ · · · ≥ xn; y1 ≥ y2 ≥ · · · ≥ yn be two sequences of real numbers. If
x1 + · · · + xk ≥ y1 + · · · + yk for k = 1 , 2, . . . , n with equality where k = n, then the sequence
{xi} is said to majorize the sequence {yi}. An equivalent criterion is that for all real numbers
t,
|t − x1| + |t − x2| + · · · + |t − xn| ≥ | t − y1| + |t − y2| + · · · + |t − yn|
We use these definitions to introduce some famous inequalities.
Theorem 1 (Jensen) Let f : I → R be a convex function. Then for any x1, . . . , x n ∈ I
and any nonnegative reals ω1, . . . , ω n,
ω1f (x1) + · · · + ωnf (xn) ≥ (ω1 + · · · + ωn) f
(ω1x1 + · · · + ωnxn
ω1 + · · · + ωn
)
If f is concave, then the inequality is flipped.
Theorem 2 (Weighted Power Mean) If x1, . . . , x n are nonnegative reals and ω1, . . . , ω n
are nonnegative reals with a postive sum, then
f (r) :=
(ω1xr
1
· · · + ωnxrn
ω1 + · · · + ωn
)1
r
is a non-decreasing function of r, with the convention that r = 0 is the weighted geometric mean. f is strictly increasing unless all the xi are equal except possibly for r ∈ (−∞ , 0] ,where if some xi is zero f is identically 0. In particular, f (1) ≥ f (0) ≥ f (−1) gives the AM-GM-HM inequality.
Theorem 3 (H¨ older) Let a1, . . . , a n; b1, . . . , b n; · · · ; z1, . . . , z n be sequences of nonnegative real numbers, and let λa, λ b, . . . , λ z positive reals which sum to 1. Then
(a1 + · · · + an)λa (b1 + · · · + bn)λb · · · (z1 + · · · + zn)λz ≥ aλa
1
bλb
1
· · · zλz
1
· · · + aλz
n
bλb
n
· · · zλz
n
This theorem is customarily identified as Cauchy when there are just two sequences.
Theorem 4 (Rearrangement) Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be two nondecreasing sequences of real numbers. Then, for any permutation π of {1, 2, . . . , n }, we have
a1b1 + a2b2 + · · · + anbn ≥ a1bπ(1) + a2bπ(2) + · · · + anbπ(n) ≥ a1bn + a2bn−1 + · · · + anb1
with equality on the left and right holding if and only if the sequence π(1) , . . . , π (n) is de-creasing and increasing respectively.
2Theorem 5 (Chebyshev) Let a1 ≤ a2 ≤ · · · ≤ an; b1 ≤ b2 ≤ · · · ≤ bn be two nondecreas-ing sequences of real numbers. Then
a1b1 + a2b2 + · · · + anbn
n ≥ a1 + a2 + · · · + an
n ·b1 + b2 + · · · + bn
n ≥ a1bn + a2bn−1 + · · · + anb1
n
Theorem 6 (Schur) Let a, b, c be nonnegative reals and r > 0. Then
ar(a − b)( a − c) + br(b − c)( b − a) + cr(c − a)( c − b) ≥ 0
with equality if and only if a = b = c or some two of a, b, c are equal and the other is 0.
Remark - This can be improved considerably. (See the problems section.) However, they are not as well known (as of now) as this form of Schur, and so should be proven whenever used on a contest.
Theorem 7 (Newton) Let x1, . . . , x n be nonnegative real numbers. Define the symmetric polynomials s0, s 1, . . . , s n by (x + x1)( x + x2) · · · (x + xn) = snxn + · · · + s1x + s0, and define the symmetric averages by di = si/(ni
). Then
d2
i
≥ di+1 di−1
Theorem 8 (Maclaurin) Let di be defined as above. Then
d1 ≥ √d2 ≥ 3
√d3 ≥ · · · ≥ n
√dn
Theorem 9 (Majorization) Let f : I → R be a convex on I and suppose that the sequence
x1, . . . , x n majorizes the sequence y1, . . . , y n, where xi, y i ∈ I. Then
f (x1) + · · · + f (xn) ≥ f (y1) + · · · + f (yn)
Theorem 10 (Popoviciu) Let f : I → R be convex on I, and let x, y, z ∈ I. Then for any positive reals p, q, r ,
pf (x) + qf (y) + rf (z) + (p + q + r)f
(px + qy + rz p + q + r
)
≥ (p + q)f
(px + qy p + q
)
( q + r)f
(qy + rz q + r
)
( r + p)f
(rz + px r + p
)
Theorem 11 (Bernoulli) For all r ≥ 1 and x ≥ − 1,
(1 + x)r ≥ 1 + xr
3Theorem 12 (Muirhead) Suppose the sequence a1, . . . , a n majorizes the sequence b1, . . . , b n.Then for any positive reals x1, . . . , x n,
∑
sym
xa1
1
xa2
2
· · · xan
n
≥ ∑
sym
xb1
1
xb2
2
· · · xbn
n
where the sums are taken over all permutations of n variables.
Remark - Although Muirhead’s theorem is a named theorem, it is generally not favor-ably regarded as part of a formal olympiad solution. Essentially, the majorization criterion guarantees that Muirhead’s inequality can be deduced from a suitable application of AM-GM. Hence, whenever possible, you should use Muirhead’s inequality only to deduce the correct relationship and then explicitly write all of the necessary applications of AM-GM. For a particular case this is a simple matter.
We now present an array of problems and solutions based primarily on these inequalities and ideas.
2 Examples
When solving any kind of problem, we should always look for a comparatively easy solu-tion first, and only later try medium or hard approaches. Although what constitutes this notoriously indeterminate “difficulty” varies widely from person to person, I usually con-sider “Dumbassing,” AM-GM (Power Mean), Cauchy, Chebyshev (Rearrangement), Jensen, H¨ older, in that order before moving to more clever techniques. (The first technique is de-scribed in remarks after example 1.) Weak inequalities will fall to AM-GM, which blatantly pins a sum to its smallest term. Weighted Jensen and H¨ older are “smarter” in that the effect of widely unequal terms does not cost a large degree of sharpness 6 - observe what happens when a weight of 0 appears. Especially sharp inequalities may be assailable only through clever algebra. Anyway, I have arranged the following with that in mind. 1. Show that for positive reals a, b, c
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥ 9a2b2c2
Solution 1. Simply use AM-GM on the terms within each factor, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥
(
3 3
√a3b3c3
) (
3 3
√a3b3c3
)
= 9 a2b2c2
6The sharpness of an inequality generally refers to the extent to which the two sides mimic each other, particularly near equality cases.
4Solution 2. Rearrange the terms of each factor and apply Cauchy,
(a2b + b2c + c2a) ( bc 2 + ca 2 + ab 2) ≥
(√a3b3c3 + √a3b3c3 + √a3b3c3
)2
= 9 a2b2c2
Solution 3. Expand the left hand side, then apply AM-GM, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) = a3b3 + a2b2c2 + a4bc
ab 4c + b3c3 + a2b2c2
a2b2c2 + abc 4 + a3c3
≥ 9 9
√a18 b18 c18 = 9 a2b2c2
We knew this solution existed by Muirhead, since (4 , 1, 1) , (3 , 3, 0), and (2 , 2, 2) all majorize (2 , 2, 2). The strategy of multiplying out all polynomial expressions and ap-plying AM-GM in conjunction with Schur is generally knowing as dumbassing because it requires only the calculational fortitude to compute polynomial products and no real ingenuity. As we shall see, dumbassing is a valuable technique. We also remark that the AM-GM combining all of the terms together was a particularly weak inequality, but the desired was a multiple of a2b2c2’s, the smallest 6th degree symmetric polynomial of three variables; such a singular AM-GM may not always suffice. 2. Let a, b, c be positive reals such that abc = 1. Prove that
a + b + c ≤ a2 + b2 + c2
Solution. First, we homogenize the inequality. that is, apply the constraint so as to make all terms of the same degree. Once an inequality is homogenous in degree
d, we may scale all of the variables by an arbitrary factor of k, causing both sides of the inequality to scale by the factor kd. This is valid in that it does not change the correctness of an inequality for any positive k, and if d is even, for any nonzero
k. Hence, we need consider a nonhomogenous constraint no futher. In this case, we multiply the left hand side by 3
√abc , obtaining
a43 b13 c13 + a13 b43 c13 + a13 b13 c43 ≤ a2 + b2 + c2
As abc = 1 is not homogenous, the above inequality must be true for all nonnegative
a, b, c . As (2 , 0, 0) majorizes (4 /3, 1/3, 1/3), we know it is true, and the necessary AM-GM is 2a2
3 + b2
6 + c2
6 = a2 + a2 + a2 + a2 + b2 + c2
6 ≥ 6
√a8b2c2 = a43 b13 c13
Let P (x) be a polynomial with positive coefficients. Prove that if
P
( 1
x
)
≥ 1
P (x)5holds for x = 1, then it holds for all x > 0.
Solution. Let P (x) = anxn + an−1xn−1 + · · · + a1x + a0. The first thing we notice is that the given is P (1) ≥ 1. Hence, the natural strategy is to combine P (x) and P ( 1
x
)
into P (1) in some fashion. The best way to accomplish this is Cauchy, which gives
P (x)P
( 1
x
)
= (anxn + · · · + a1x + a0)
(
an
1
xn + · · · + a1
1
x + a0
)
≥ (an + · · · + a1 + a0)2 = P (1) 2 ≥ 1as desired. This illustrates a useful means of eliminating denominators - by introducing similar factors weighted by reciprocals and applying Cauchy / H¨ older. 4. (USAMO 78/1) a, b, c, d, e are real numbers such that
a + b + c + d + e = 8
a2 + b2 + c2 + d2 + e2 = 16 What is the largest possible value of e?
Solution. Observe that the givens can be effectively combined by considering squares: (a − r)2 + ( b − r)2 + ( c − r)2 + ( d − r)2 + ( e − r)2 = (a2 + b2 + c2 + d2 + e2)
− 2r(a + b + c + d + e) + 5 r2
= 16 − 16 r + 5 r2
Since these squares are nonnegative, e ≤ √5r2 − 16 r + 16 + r = f (r) for all r. Since equality e = f (r) can be achieved when a = b = c = d = r, we need only compute the smallest value f (r). Since f grows large at either infinity, the minimum occurs when
f ′(r) = 1 + 10 r−16 2√5r2−16 r+16 = 0. The resultant quadratic is easily solved for r = 65 and
r = 2, with the latter being an extraneous root introduced by squaring. The largest possible e and greatest lower bound of f (r) is then f (6 /5) = 16 /5, which occurs when
a = b = c = d = 6 /5 and e = 16 /5. Alternatively, proceed as before except write
a = b = c = d = 8−e
4
since the maximum e must occur when the other four variables are equal. The second condition becomes a quadratic, and the largest solution is seen to be e = 16 5 .The notion of equating a, b, c, d is closely related to the idea of smoothing and Jensen’s inequality. If we are working with S1 = f (x1) + · · · + f (xn) under the constraint of a fixed sum x1 + · · · + xn, we can decrease S1 by moving several xi in the same interval
I together (that is, replacing xi1 < x i2 with x′
i1
= xi1 + ≤ < x i2 − ≤ = x′
i2
for any sufficiently small ≤) for any I where f is convex. S1 can also be decreased by spreading
xi in the same interval where f is concave. When seeking the maximum of S1, we proceed in the opposite fashion, pushing xi on the concave intervals of f together and moving xi on the convex intervals apart. 65. Show that for all positive reals a, b, c, d ,1
a + 1
b + 4
c + 16
d ≥ 64
a + b + c + d
Solution. Upon noticing that the numerators are all squares with √1 + √1 + √4 + √16 = √64, Cauchy should seem a natural choice. Indeed, multiplying through by
a + b + c + d and applying Cauchy, we have (a + b + c + d)
(12
a + 12
b + 22
c + 42
d
)
≥ (1 + 1 + 2 + 4) 2 = 64 as desired. 6. (USAMO 80/5) Show that for all non-negative reals a, b, c ≤ 1,
ab + c + 1 + bc + a + 1 + ca + b + 1 + (1 − a)(1 − b)(1 − c) ≤ 1
Solution. Let f (a, b, c ) denote the left hand side of the inequality. Since ∂2
∂a 2
f =
2b
(c+a+1) 3
2c
(a+b+1) 3
≥ 0, we have that f is convex in each of the three variables; hence, the maximum must occur where a, b, c ∈ { 0, 1}. Since f is 1 at each of these 8 points, the inequality follows. Second derivative testing for convexity/concavity is one of the few places where the use of Calculus is not seriously loathed by olympiad graders. It is one of the standard techniques in inequalities and deserves to be in any mental checklist of inequality solving. In this instance, it led to an easy solution. 7. (USAMO 77/5) If a, b, c, d, e are positive reals bounded by p and q with 0 < p ≤ q,prove that (a + b + c + d + e)
( 1
a + 1
b + 1
c + 1
d + 1
e
)
≤ 25 + 6
(√ pq −
√ qp
)2
and determine when equality holds.
Solution. As a function f of five variables, the left hand side is convex in each of
a, b, c, d, e ; hence, its maximum must occur when a, b, c, d, e ∈ { p, q }. When all five variables are p or all five are q, f is 25. If one is p and the other four are q, or vice versa, f becomes 17 + 4( pq + qp ), and when three are of one value and two of the other,
f = 13 + 6( pq + qp ). pq + qp ≥ 2, with equality if and only if p = q. Clearly, equality holds where p = q. Otherwise, the largest value assumed by f is 13 + 6( pq + qp ), which is obtained only when two of a, b, c, d, e are p and the other three are q, or vice versa. In such instances, f is identically the right hand side. This is a particular case of the Schweitzer inequality, which, in its weighted form, is sometimes known as the Kantorovich inequality. 78. a, b, c, are non-negative reals such that a + b + c = 1. Prove that
a3 + b3 + c3 + 6 abc ≥ 14
Solution. Multiplying by 4 and homogenizing, we seek 4a3 + 4 b3 + 4 c3 + 24 abc ≥ (a + b + c)3
= a3 + b3 + c3 + 3 (a2(b + c) + b2(c + a) + c2(a + b)) + 6 abc
⇐⇒ a3 + b3 + c3 + 6 abc ≥ a2(b + c) + b2(c + a) + c2(a + b)Recalling that Schur’s inequality gives a3 +b3 +c3 +3 abc ≥ a2(b+c)+ b2(c+a)+ c2(a+b), the inequality follows. In particular, equality necessitates that the extra 3 abc on the left is 0. Combined with the equality condition of Schur, we have equality where two of a, b, c are 12 and the third is 0. This is a typical dumbass solution.
Solution 2. Without loss of generality, take a ≥ b ≥ c. As a+b+c = 1, we have c ≤ 13
or 1 −3c ≥ 0. Write the left hand side as ( a+b)3 −3ab (a+b−2c) = ( a+b)3 −3ab (1 −3c). This is minimized for a fixed sum a + b where ab is made as large as possible. As by AM-GM ( a + b)2 ≥ 4ab , this minimum occurs if and only if a = b. Hence, we need only consider the one variable inequality 2 (1−c
2
)3 + c3 + 6 (1−c
2
)2 c = 14 · (9 c3 − 9c2 + 3 c + 1). Since c ≤ 13 , 3 c ≥ 9c2. Dropping this term and 9 c3, the inequality follows. Particularly, 9c3 = 0 if and only if c = 0, and the equality cases are when two variables are 12 and the third is 0. 9. (IMO 74/5) If a, b, c, d are positive reals, then determine the possible values of
aa + b + d + bb + c + a + cb + c + d + da + c + d
Solution. We can obtain any real value in (1 , 2). The lower bound is approached by
a → ∞ , b = d = √a, and c = 1. The upper bound is approached by a = c → ∞ ,
b = d = 1. As the expression is a continuous function of the variables, we can obtain all of the values in between these bounds. Finally, these bounds are strict because
aa + b + d + bb + c + a + cb + c + d + da + c + d >aa + b + c + d + ba + b + c + d + ca + b + c + d + da + b + c + d = 1 and
aa + b + d + bb + c + a + cb + c + d + da + c + d <aa + b + ba + b + cc + d + dc + d = 2 Whenever extrema occur for unusual parameterizations, we should expect the need for non-classical inequalities such as those of this problem where terms were completely dropped. 810. (IMO 95/2) a, b, c are positive reals with abc = 1. Prove that 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) ≥ 32
Solution 1. Let x = 1
a
, y = 1
b
, and z = 1
c
. We perform this substitution to move terms out of the denominator. Since abc = xyz = 1, we have 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) = x2
y + z + y2
x + z + z2
x + y
Now, multiplying through by ( x + y)( y + z)( z + x), we seek
x4 + y4 + z4 + x3y + x3z + y3z + xy 3 + xz 3 + yz 3 + x2yz + xy 2z + xyz 2 ≥
3
√xyz ·
(
3xyz + 32 · (x2y + x2z + y2x + xy 2 + xz 2 + yz 2))
which follows immediately by AM-GM, since x2yz +xy 2z+xyz 2 ≥ 3 3
√x4y4z4, x3y+xy 3+x3z
3
≥
3
√x7y4z and 7x4+4 y4+z4
12
≥ 3
√x7y4z - as guaranteed by Muirhead’s inequality.
Solution 2. Substitute x, y, z as before. Now, consider the convex function f (x) = x−1
for x > 0. ( f (x) = xc is convex for c < 0 and c ≥ 1, and concave for 0 < c ≤ 1, verify this with the second derivative test.) Now, by Jensen,
x2
y + z + y2
z + x + z2
x + y = xf
(y + zx
)
yf
(z + xy
)
zf
(x + yz
)
≥ (x + y + z)f
((y + z) + ( z + x) + ( x + y)
x + y + z
)
= x + y + z
2But x + y + z ≥ 3 3
√xyz = 3, as desired.
Solution 3. Perform the same substitution. Now, multiplying by ( x + y + z) and applying Cauchy, we have 12 (( y + z) + ( z + x) + ( x + y))
( x2
y + z + y2
z + x + z2
x + y
)
≥ 12(x + y + z)2
Upon recalling that x+y +z ≥ 3 we are done. Incidentally, the progress of this solution with Cauchy is very similar to the weighted Jensen solution shown above. This is no coincidence, it happens for many convex f (x) = xc.
Solution 4. Apply the same substitution, and put x ≥ y ≥ z. Simultaneously,
xy+z
≥ yz+x ≥ zx+y . Hence, by Chebyshev,
x ·
( xy + z
)
y ·
( yz + x
)
z ·
( zx + y
)
≥ x + y + z
3
( xy + z + yx + z + zx + y
)
Again, x + y + z ≥ 3. But now we have Nesbitt’s inequality, xy+z + yx+z + zx+y ≥ 32 . This follows immediately from AM-HM upon adding 1 to each term. 911. Let a, b, c be positive reals such that abc = 1. Show that 2(a + 1) 2 + b2 + 1 + 2(b + 1) 2 + c2 + 1 + 2(c + 1) 2 + a2 + 1 ≤ 1
Solution. The previous problem showed the substitution offers a way to rewrite an inequality in a more convenient form. Substitution can also be used to implicity use a given. First, expand the denominators and apply AM-GM, obtaining 2(a + 1) 2 + b2 + 1 = 2
a2 + b2 + 2 a + 2 ≤ 1
ab + a + 1 Now, write a = xy , b = yz , c = zx . We have 1
ab +a+1
= 1
xz+xy+1
= yz xy +yz +zx . It is now evident that the sum of the new fractions is 1. 12. (USAMO 98/3) Let a0, . . . , a n real numbers in the interval (0 , π
2
) such that tan
(
a0 − π
4
)
tan
(
a1 − π
4
)
· · · + tan
(
an − π
4
)
≥ n − 1Prove that tan( a0) tan( a1) · · · tan( an) ≥ nn+1
Solution 1. Let yi = tan (x − π
4
). We have tan( xi) = tan ((xi − π
4
) + π
4
) = yi+1 1−yi .Hence, given s = y0 + · · · + yn ≥ n − 1 we seek to prove ∏ni=0 1+ yi
1−yi
≥ nn+1 . Observe that for any a > b and fixed sum a + b, the expression
( 1 + a
1 − a
)
·
( 1 + b
1 − b
)
= 1 + 2( a + b)(1 − a)(1 − b)can be decreased by moving a and b any amount closer together. Hence, for any sequence y0, . . . , y n, we can replace any yi > sn+1 and yj < sn+1 with y′
i
= sn+1 and
y′
j
= yi + yj − sn+1 , decreasing the product. Now we have
n
∏
i=0
1 + yi
1 − yi
≥
( 1 + sn+1
1 − sn+1
)n+1
≥
( 2nn+1 2
n+1
)n+1
= nn+1
Where the last inequality follows from the fact that 1+ x
1−x
is an increasing function of x.
Solution 2. Perform the same substitution. The given can be written as 1 + yi ≥
∑
j6=i
(1 − yj ), which by AM-GM gives 1+ yn
n
≥ ∏
j6=i
(1 − yj ) 1
n
. Now we have
n
∏
i=0
1 + yi
n ≥
n
∏
i=0
∏
j6=i
(1 − yj ) 1
n
=
n
∏
i=0
(1 − yi)as desired. 10 13. Let a, b, c be positive reals. Prove that 1
a(1 + b) + 1
b(1 + c) + 1
c(1 + a) ≥ 31 + abc
with equality if and only if a = b = c = 1.
Solution. Multiply through by 1+ abc and add three to each side, on the left obtaining 1 + a + ab + abc a(1 + b) + 1 + b + bc + abc b(1 + c) + 1 + c + ac + abc c(1 + a)= (1 + a) + ab (1 + c)
a(1 + b) + (1 + b) + bc (1 + a)
b(1 + c) + (1 + c) + ac (1 + b)
c(1 + a)which is at least 6 by AM-GM, as desired. In particular, this AM-GM asserts the equivalence of (1+ a)
a(1+ b)
and a(1+ b)1+ a , or that they are both one. Likewise, all of the other terms must be 1. Now, (1 + a)2 = a2(1 + b)2 = a2b2(1 + c)2 = a2b2c2(1 + a)2, so the product abc = 1. Hence, 1+ aa(1+ b) = bc (1+ a)1+ b = bc (1+ a)
b(1+ c)
so that 1 + b = b + bc = b + 1
a
. It is now easy to see that equality holds if and only if a = b = c = 1. 14. (Romanian TST) Let a, b, x, y, z be positive reals. Show that
xay + bz + yaz + bx + zax + by ≥ 3
a + b
Solution. Note that ( a + b)( xy + yz + xz ) = ( x(ay + bz ) + y(az + bx ) + z(ax + by )). We introduce this factor in the inequality, obtaining (x(ay + bz ) + y(az + bx ) + z(ax + by ))
( xay + bz + yaz + bx + zax + by
)
≥
(x + y + z)2 ≥ 3( xy + yz + xz )Where the last inequality is simple AM-GM. The desired follows by simple algebra. Again we have used the idea of introducing a convenient factor to clear denominators with Cauchy. 15. The numbers x1, x 2, . . . , x n obey −1 ≤ x1, x 2, . . . , x n ≤ 1 and x 31 + x 32 + · · · + x 3
n
= 0. Prove that
x1 + x2 + · · · + xn ≤ n
3
Solution 1. Substitute yi = x3
i
so that y1 + · · · + yn = 0. In maximizing 3
√y1 +
· · · + 3
√yn, we note that f (y) = y 13 is concave over [0 , 1] and convex over [ −1, 0], with
|f ′(y1)| ≥ | f ′(y2)| ⇐⇒ 0 < |y1| ≤ | y2|. Hence, we may put y1 = · · · = yk = −1;
−1 ≤ yk+1 < 0, and yk+2 = · · · = yn = k−yk+1
n−k−1
. We first show that yk+1 leads to a maximal sum of 3
√yi if it is -1 or can be made positive. If |yk+1 | < |yk+2 |, we set 11 y′
k+1
= y′
k+2
= yk+1 +yk+2
2
, increasing the sum while making yk+1 positive. Otherwise, set y′
k+1
= −1 and y′
k+2
= 1 − yk+1 − yk+2 , again increasing the sum of the 3
√yi. Now we may apply Jensen to equate all positive variables, so that we need only show
k 3
√−1 + ( n − k) 3
√ kn − k ≤ n
3But we have (n + 3 k)3 − 27( n − k)2k = n3 − 18 n2k + 81 nk 2 = n(n − 9k)2 ≥ 0as desired. Particularly, as k is an integer, equality can hold only if 9 |n and then if and only if one ninth of the variables yi are -1 and the rest are 1/8.
Solution 2. Let xi = sin( αi), and write 0 = x31 + · · · + x3
n
= sin 3(α1) + · · · + sin 3(αn) =
14
((3 sin( α1) − sin(3 α1)) + · · · + (3 sin( αn) − sin(3 αn))). It follows that x1 + · · · + xn =sin( α1) + · · · + sin( αn) = sin(3 α1)+ ··· +sin(3 αn)3 ≤ n
3
. The only values of sin( α) which lead to sin(3 α) = 1 are 12 and -1. The condition for equality follows. 16. (Turkey) Let n ≥ 2 be an integer, and x1, x 2, . . . , x n positive reals such that x21 + x22 +
· · · + x2
n
= 1. Determine the smallest possible value of
x51
x2 + x3 + · · · + xn
x52
x3 + · · · + xn + x1
· · · + x5
n
x1 + · · · + xn−1
Solution. Observe that ∑ni=1 xi
∑
j6=i
xj ≤ n − 1, so that
( n∑
i=1
xi
(∑
j6=i
xj
)) ( n∑
i=1
x5
i
∑
j6=i
xi
)
≥ (x31 + · · · + x3
n
)2
= n2
(x31 + · · · + x3
n
n
)2
≥ n2
(x21 + · · · + x2
n
n
)3
= 1
n
Leads to n∑
i=1
x5
i
∑
j6=i
xi
≥ 1
n(n − 1) with equality if and only if x1 = · · · = xn = 1√n .17. (Poland 95) Let n be a positive integer. Compute the minimum value of the sum
x1 + x22
2 + x33
3 + · · · + xnn
n
12 where x1, x 2, . . . , x n are positive reals such that 1
x1
1
x2
· · · + 1
xn
= n
Solution. The given is that the harmonic mean of x1, . . . , x n is 1, which implies that the product x1x2 · · · xn is at least 1. Now, we apply weighted AM-GM
x1 + x22
2 + x33
3 + · · · + xnn
n ≥
(
1 + 12 + 13 + · · · + 1
n
)
1+ 12 +··· + 1
n
√x1x2 · · · xn
= 1 + 12 + 13 + · · · + 1
n
Prove that for all positive reals a, b, c, d ,
a4b + b4c + c4d + d4a ≥ abcd (a + b + c + d)
Solution. By AM-GM, 23 a4b + 7 b4c + 11 c4d + 10 ad 4
51 ≥ 51
√a102 b51 c51 d51 = a2bcd
from which the desired follows easily. Indeed, the most difficult part of this problem is determining suitable weights for the AM-GM. One way is to suppose arbitrary weights
x1, x 2, x 3, x 4 for a4b, b 4c, c 4d, ad 4 respectively, and solve the system
x1 + x2 + x3 + x4 = 14x1 + x2 = 24x2 + x3 = 14x3 + x4 = 119. (USAMO 01/3) Let a, b, c be nonnegative reals such that
a2 + b2 + c2 + abc = 4 Prove that 0 ≤ ab + bc + ca − abc ≤ 2
Solution [by Tony Zhang.] For the left hand side, note that we cannot have a, b, c >
Suppose WLOG that c ≤ 1. Then ab +bc +ca −abc ≥ ab +bc +ca −ab = c(a+b) ≥ 0. For the right, 4 = a2 + b2 + c2 + abc ≥ 4( abc )43 =⇒ abc ≤ 1. Since by the pigeon hole principle, among three numbers either two exceed 1 or two are at most 1. Hence, we assume WLOG that ( a − 1)( b − 1) ≥ 0, which gives ab + 1 ≥ a + b ⇐⇒ abc + c ≥ ac +
bc ⇐⇒ c ≥ ac + bc − abc . Now, we have ab + bc + ca − abc ≤ ab + c. Either we are done or ab +c > 2. But in the latter case, 4 = ( a2 +b2)+ c(c+2 ab ) > 2ab +2 c = 2( ab +c) > 4, a contradiction. 13 20. (Vietnam 98) Let x1, . . . , x n be positive reals such that 1
x1 + 1998 + 1
x2 + 1998 + · · · + 1
xn + 1998 = 11998 Prove that n
√x1x2 · · · xn
n − 1 ≥ 1998
Solution. Let yi = 1
xi+1998
so that y1 + · · · + yn = 11998 and xi = 1
yi
− 1998. Now
n
∏
i=1
xi =
n
∏
i=1
( 1
yi
− 1998
)
= e
Pni=1 ln
“1
yi−1998
”
Hence, to minimize the product of the xi, we equivalently minimize the sum of ln
( 1
yi
− 1998
)
.In particular,
ddy
(
ln
( 1
y − 1998
))
= 1
( 1
y
− 1998
)2 · −1
y2
= −1
y − 1998 y2
d2
dy 2
(
ln
( 1
y − 1998
))
= 1 − 3996 y
(y − 1998 y2)2
So ln
( 1
y
− 1998
)
is convex on [0 , 1/3996]. If we had 0 < y i ≤ 1/3996 for all i we could apply Jensen. Since yi + yj ≤ 1/1998 for all i, j , we consider
( 1
a − 1998
) ( 1
b − 1998
)
≥
( 2
a + b − 1998
)2
⇐⇒ 1
ab − 1998
( 1
a + 1
b
)
≥ 4(a + b)2 − 4 · 1998
a + b
⇐⇒ (a + b)2 − 1998( a + b)3 ≥ 4ab − 4ab (a + b) · 1998
⇐⇒ (a − b)2 ≥ 1998( a + b)( a − b)2
which incidentally holds for any a + b ≤ 11998 . Hence, any two yi and yj may be set to their average while decreasing the sum in question; hence, we may assume yi ∈ (0 , 13996 ]. Now Jensen’s inequality shows that the minimum occurs when yi = 11998 n for all i, or when xi = 1998( n − 1) for all i. It is easy to see that this yields equality. 21. (Romania 99) Show that for all positive reals x1, . . . , x n with x1x2 · · · xn = 1, we have 1
n − 1 + x1
· · · + 1
n − 1 + xn
≤ 114 Solution. First, we prove a lemma: the maximum of the sum occurs when n − 1 of the xi are equal. Consider f (y) = 1
k+ey
for an arbitrary nonnegative constant k. We have f ′(y) = −ey
(k+ey)2
and f ′′ (y) = ey (ey −k)(k+ey )3 . Evidently f ′′ (y) ≥ 0 ⇐⇒ ey ≥ k. Hence,
f (y) has a single inflexion point where y = ln( k), where f (y) is convex over the interval ((ln( k), ∞). Now, we employ the substitution yi = ln( xi) so that y1 + · · · + yn = 0 and
n
∑
i=1
1
n − 1 + xi
=
n
∑
i=1
f (yi)We take k = n − 1 and write k0 = ln( n − 1). Suppose that y1 ≥ · · · ≥ ym ≥ k0 ≥
ym+1 ≥ · · · xn for some positive m. Then by, Majorization,
f (y1) + · · · + f (ym) ≤ (m − 1) f (k0) + f (y1 + · · · + ym − (m − 1) k0)But then, also by Majorization, (m − 1) f (k0) + f (ym+1 ) + · · · + f (yn) ≤ (n − 1) f
((m − 1) k0 + ym+1 + · · · + yn
n − 1
)
Otherwise, all of the yi are less than k0. In that case we may directly apply Majorization to equate n − 1 of the yi whilst increasing the sum in question. Hence, the lemma is valid. 7 N
Applying the lemma, it would suffice to show
kk + x + 1
k + 1
xk
≤ 1Clearing the denominators,
(
k2 + kxk
)
( k + x) ≤ k2 + k
(
x + 1
xk
)
x1−k
−xk + x + k ≤ x1−k
But now this is evident. We have Bernoulli’s inequality, since x1−k = (1 + ( x − 1)) 1−k ≥
1 + ( x − 1)(1 − k) = x + k − xk . Equality holds only where x = 1 or n = 2. 22. (Darij Grinberg) Show that for all positive reals a, b, c ,
√b + ca +
√c + ab +
√a + bc ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
7This n−1 equal value principle is particularly useful. If a differentiable function has a single inflexion point and is evaluated at narbitrary reals with a fixed sum, any minimum or maximum must occur where some n−1 variables are equal.
15 Solution 1. By Cauchy, we have √(a + b)( a + c) ≥ a + √bc . Now,
∑
cyc
√b + ca ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
⇐⇒ ∑
cyc
b + ca
√(a + b)( a + c) ≥ 4( a + b + c)Substituting our result from Cauchy, it would suffice to show
∑
cyc
(b + c)
√bc a ≥ 2( a + b + c)WLOG a ≥ b ≥ c, implying b + c ≤ c + a ≤ a + b and √bc a ≤ √ca b ≤ √ab c . Hence, by Chebyshev and AM-GM,
∑
cyc
(b + c)
√bc a ≥
(2( a + b + c))
(√bc a + √ca b + √ab c
)
3
≥ 2( a + b + c)as desired.
Solution 2. Let x = √b + c, y = √c + a, z = √a + b. Then x, y, z are the sides of acute triangle XY Z (in the typical manner), since x2 + y2 = a + b + 2 c > a + b = z2.The inequality is equivalent to
∑
cyc
xy2 + z2 − x2 ≥ x2 + y2 + z2
xyz
Recalling that y2 + z2 − x2 = 2 yz cos( X), we reduce this to the equivalent
∑
cyc
x2
cos( X) ≥ 2( x2 + y2 + z2)WLOG, we have x ≥ y ≥ z, implying 1cos( X) ≥ 1cos( Y ) ≥ 1cos( Z) , so that applying Chebyshev to the left reduces the desired to proving that the sum of the reciprocals of the cosines is at least 6. By AM-HM, 1cos( X) + 1cos( Y ) + 1cos( Z) ≥ 9cos( X) + cos( Y ) + cos( Z)But recall from triangle geometry that cos( X) + cos( Y ) + cos( Z) = 1 + rR and R ≥ 2r.The desired is now evident. 16 23. Show that for all positive numbers x1, . . . , x n,
x31
x21 + x1x2 + x22
x32
x22 + x2x3 + x23
· · · + x3
n
x2
n
xnx1 + x21
≥ x1 + · · · + xn
3
Solution. Observe that 0 = ( x1 −x2)+( x2 −x3)+ · · · +( xn −x1) = ∑ni=1
x3
i−x3
i+1
x2
i+xixi+1 +x2
i+1
.Hence, (where xn+1 = x1)
n
∑
i=1
x3
i
x2
i
xixi+1 x2
i+1
= 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
But now a3 + b3 ≥ 13 a3 + 23 a2b + 23 ab 2 + 13 b3 = 13 (a + b)( a2 + ab + b2). Hence, 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
≥ 12
n
∑
i=1
xi + xi+3
3 = 13
n
∑
i=1
xi
as desired. 24. Let a, b, c be positive reals such that a + b ≥ c; b + c ≥ a; and c + a ≥ b, we have 2a2(b + c) + 2 b2(c + a) + 2 c2(a + b) ≥ a3 + b3 + c3 + 9 abc
Solution. After checking that equality holds for ( a, b, c ) = ( t, t, t ) and (2 t, t, t ), it is apparent that more than straight AM-GM will be required. To handle the condition, put a = y + z, b = z + x, c = x + y with x, y, z ≥ 0. Now, the left hand side becomes 4x3 + 4 y3 + 4 z3 + 10 x2(y + z) + 10 y2(z + x) + 10 z2(x + y) + 24 xyz while the right hand side becomes 2 x3 + 2 y3 + 2 z3 + 12 x2(y + z) + 12 y2(z + x) + 12 z2(x + y) + 18 xyz .The desired is seen to be equivalent to x3 + y3 + z3 + 3 xyz ≥ x2(y + z) + y2(z +
x) + z2(x + y), which is Schur’s inequality. Equality holds where x = y = z, which gives ( a, b, c ) = ( t, t, t ), or when two of x, y, z are equal and the third is 0, which gives (a, b, c ) ∈ { (2 t, t, t ), (t, 2t, t ), (t, t, 2t)}.25. Let a, b, c be the lengths of the sides of a triangle. Prove that
a
√2b2 + 2 c2 − a2 + b
√2c2 + 2 a2 − b2 + c
√2a2 + 2 b2 − c2 ≥ √3
Solution 1. Again write a = y + z, b = z + x, and c = x + y, noting that x, y, z are positive. (Triangles are generally taken to be non-degenerate when used in inequalities.) We have
∑
cyc
a
√2b2 + 2 c2 − a2 = ∑
cyc
y + z
√4x2 + 4 xy + 4 xz + y2 + z2 − 2yz
17 Consider the convex function f (x) = 1√x . (As we shall see, Jensen almost always provides a tractable means of eliminating radicals from inequalities.) Put x+y +z = 1. We have
∑
cyc
(y + z)f (4x2 + 4 xy + 4 xz + y2 + z2 − 2yz ) ≥
(( y + z) + ( z + x) + ( x + y)) f
(∑
cyc
(y + z) (4 x2 + 4 xy + 4 xz + y2 + z2 − 2yz )(y + z) + ( z + x) + ( x + y)
)
= 2√2
√∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2
Noting that ∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2 =
∑
cyc
2x3 + 7 x2(y + z) + 8 xyz ,8( x + y + z)3 ≥ 3 ∑
cyc
2x3 + 7 x2(y + z) + 8 xyz
⇐⇒ ∑
sym
4x3 + 24 x2y + 8 xyz ≥ ∑
sym
3x3 + 21 x2y + 12 xyz
⇐⇒ 2x3 + 2 y3 + 2 z3 + 3 (x2(y + z) + y2(z + x) + z2(x + y)) ≥ 24 xyz
which follows by AM-GM. As a follow up on an earlier mentioned connection, oberserve the similarity of the above application of Jensen and the following inequality (which follows by H¨ older’s inequality)
(∑
i
αiβi
) ( ∑
i
αi
1
√βi
)2
≥
(∑
i
αi
)3
Solution 2 [by Darij Grinberg.] Let ABC be a triangle of side lengths a, b, c
in the usual order. Denote by ma, m b, m c the lengths of the medians from A, B, C
respectively. Recall from triangle goemetry that ma = 12
√2b2 + 2 c2 − a2, so that we need only show ama + bmb + cmc ≥ 2√3. But a triangle with side lengths ma, m b, m c, in turn, has medians of length 3a
4
, 3b
4
, and 3c
4
. The desired inequality is therefore equivalent to 43 ma
a
43 mb
b
43 mc
c
≥ 2√3 where we refer to the new triangle ABC . Recalling that
23
ma = AG , where G is the centroid, the desired is seen to be equivalent to the geometric inequality AG a + BG b + CG c ≥ √3. But we are done as we recall from triangle geometry that AM a + BM b + CM c ≥ √3 holds for any point inside triangle ABC .8
8For a complete proof of this last inequality, see post #14.
18 26. (IMO 99/2) For n ≥ 2 a fixed positive integer, find the smallest constant C such that for all nonnegative reals x1, . . . , x n,
∑
1≤i<j ≤n
xixj (x2
i
x2
j
) ≤ C
( n∑
i=1
xi
)4
Solution. The answer is C = 18 , which is obtained when any two xi are non-zero and equal and the rest are 0. Observe that by AM-GM, (x1 + · · · + xn)4 =
( n∑
i=1
x2
i
2 ∑
1≤i<j ≤n
xixj
)2
≥ 4
( n∑
i=1
x2
i
) (
2 ∑
1≤i<j ≤n
xixj
)
= 8 ∑
1≤i<j ≤n
xixjn∑
k=1
x2
k
But x21 + · · · + x2
n
≥ x2
i
x2
j
with equality iff xk = 0 for all k 6 = i, j . It follows that (x1 + · · · + xn)4 ≥ 8 ∑
1≤i<j ≤n
xixj
(x2
i
x2
j
)
as desired. 27. Show that for nonnegative reals a, b, c ,2a6 + 2 b6 + 2 c6 + 16 a3b3 + 16 b3c3 + 16 c3a3 ≥ 9a4(b2 + c2) + 9 b4(c2 + a2) + 9 c4(a2 + b2)
Solution 1. Consider
∑
cyc
(a − b)6 = ∑
cyc
a6 − 6a5b + 15 a4b2 − 20 a3b3 + 15 a2b4 − 6ab 5 + b6 ≥ 0and ∑
cyc
ab (a − b)4 = ∑
cyc
a5b − 4a4b2 + 6 a3b3 − 4a2b4 + ab 5 ≥ 0Adding six times the latter to the former yields the desired result.
Solution 2. We shall prove a6 − 9a4b2 + 16 a3b3 − 9a2b4 + b6 ≥ 0. We have
a6 − 2a3b3 + b6 = (a3 − b3)2
= ((a − b)( a2 + ab + b2))2
≥ (a − b)2(3 ab )2 = 9 a4b2 − 18 a3b3 + 9 a2b4
19 As desired. The result now follows from adding this lemma cyclicly. The main difficulty with this problem is the absence of a5b terms on the right and also the presence of
a4b2 terms on the right - contrary to where Schur’s inequality would generate them. Evidently AM-GM is too weak to be applied directly, since a6 + 2 a3b3 ≥ 3a4b2 cannot be added symmetrically to deduce the problem. By introducing the factor ( a − b)2,however, we weight the AM-GM by a factor which we “know” will be zero at equality, thereby increasing its sharpness. 28. Let 0 ≤ a, b, c ≤ 12 be real numbers with a + b + c = 1. Show that
a3 + b3 + c3 + 4 abc ≤ 932
Solution. Let f (a, b, c ) = a3 + b3 + c3 + 4 abc and g(a, b, c ) = a + b + c = 1. Because
f and g are polynomials, they have continuous first partial derivatives. Moreover, the gradient of g is never zero. Hence, by the theorem of Lagrange Multipliers ,any extrema occur on the boundary or where ∇f = λ∇g for suitable scalars λ. As
∇f =< 3a2 + 4 bc, 3b2 + 4 ca, 3c2 + 4 ab > and ∇g =< 1, 1, 1 >, we have
λ = 3a2 + 4 bc
= 3b2 + 4 ca
= 3c2 + 4 ab g(a, b, c ) = a + b + c = 1 We have 3 a2 + 4 bc = 3 b2 + 4 ca or ( a − b)(3 a + 3 b − 4c) = ( a − b)(3 − 7c) = 0 for any permutation of a, b, c . Hence, without loss of generality, a = b. Now, 3 a2 + 4 ac =3c2 + 4 a2 and a2 − 4ac + 3 c2 = ( a − c)( a − 3c) = 0. The interior local extrema therefore occur when a = b = c or when two of {a, b, c } are three times as large as the third. Checking, we have f (13 , 13 , 13 ) = 7 /27 < 13 /49 = f (17 , 37 , 37 ). Recalling that f (a, b, c ) is symmetric in a, b, c , the only boundary check we need is f (12 , t, 12 −t) ≤ 932 for 0 ≤ t ≤ 12 .We solve
h(t) = f
(12, t, 12 − t
)
= 18 + t3 +
(12 − t
)3
2 t
(12 − t
)
= 14 + t
4 − t2
2
h(t) is 14 at either endpoint. Its derivative h′(t) = 14 − t is zero only at t = 14 . Checking,
h(14 ) = f (12 , 14 , 14 ) = 932 . Since h(t) has a continuous derivative, we are done. (As a further check, we could observe that h′′ (t) = −1 < 0, which guarantees that h(14 ) is a local minimum.) 20 Usage Note. The use of Lagrange Multipliers in any solution will almost certainly draw hostile review, in the sense that the tiniest of errors will be grounds for null marks. If you consider multipliers on Olympiads, be diligent and provide explicit, kosher remarks about the continuous first partial derivatives of both f (x1, . . . , x n) and the constraint g(x1, . . . , x n) = k, as well as ∇g 6 = 0, before proceeding to solve the system ∇f = λ∇g. The main reason this approach is so severely detested is that, given sufficient computational fortitude (if you are able to sort through the relevant algebra and Calculus), it can and will produce a complete solution. The example provided here is included for completeness of instruction; typical multipliers solutions will not be as clean or painless. 9
(Vascile Cartoaje) Let p ≥ 2 be a real number. Show that for all nonnegative reals
a, b, c ,
3
√
a3 + pabc
1 + p + 3
√
b3 + pabc
1 + p + 3
√
c3 + pabc
1 + p ≤ a + b + c
Solution. By H¨ older,
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤
(∑
cyc
11 + p
) ( ∑
cyc
a
) ( ∑
cyc
a2 + pbc
)
But a2 + b2 + c2 ≥ ab + bc + ca (proven by AM-GM, factoring, or a number of other methods) implies that
∑
cyc
a2 + pbc ≤ (p + 1) ∑
cyc
a2 + 2 bc
3 = p + 1 3 (a + b + c)2
From which we conclude
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤ (a + b + c)3
as desired. 30. Let a, b, c be real numbers such that abc = −1. Show that
a4 + b4 + c4 + 3( a + b + c) ≥ a2
b + a2
c + b2
c + b2
a + c2
a + c2
b
Solution. First we homogenize, obtaining a4 + b4 + c4 + a3(b + c) + b3(c + a) + c3(a +
b) − 3abc (a + b + c) ≥ 0. As this is homogenous in the fourth degree, we can scale a, b, c
9Just how painful can the calculations get? Most multipliers solutions will tend to look more like than this solution.
21 by any real k and hence may now ignore abc = −1. Equality holds at a = b = c = 1, but also at a = b = 1 , c = −2, a = 1 , b = 0 , c = −1, and a number of unusual locations with the commonality that a + b + c = 0. Indeed, c = −a − b is a parametric solution, and we discover the factorization ( a + b + c)2(a2 + b2 + c2 − ab − bc − ca ) ≥ 0. (We are motivated to work with factorizations because there are essentially no other inequalities with a + b + c = 0 as an equality condition.) 31. (MOP 2003) Show that for all nonnegative reals a, b, c ,
a4(b2 + c2) + b4(c2 + a2) + c4(a2 + b2) +2abc (a2b + a2c + b2c + b2a + c2a + c2b − a3 − b3 − c3 − 3abc )
≥ 2a3b3 + 2 b3c3 + 2 c3a3
Solution. As was suggested by the previous problem, checking for equality cases is important when deciding how to solve a problem. We see that setting a = b produces equality. As the expression is symmetric, this certainly implies that b = c and c = a are equality cases. Hence, if P (a, b, c ) is the difference LHS - RHS, then ( a − b)( b − c)( c −
a)|P (a, b, c ). Obviously, if the problem is going to be true, ( a−b) must be a double root of P , and accordingly we discover the factorization P (a, b, c ) = ( a − b)2(b − c)2(c − a)2.The result illustrated above was no accident. If ( x−y) divides a symmetric polynomial
P (x, y, z ), then ( x − y)2 divides the same polynomial. If we write P (x, y, z ) = ( x −
y)Q(x, y, z ), then ( x − y)Q(x, y, z ) = P (x, y, z ) = P (y, x, z ) = ( y − x)Q(y, x, z ), which gives Q(x, y, z ) = −Q(y, x, z ). Hence Q(x, x, z ) = 0, and ( x − y) also divides Q(x, y, z ). 32. (Cezar Lupu) Let a, b, c be positive reals such that a + b + c + abc = 4. Prove that
a
√b + c + b
√c + a + c
√a + b ≥√22 · (a + b + c)
Solution. By Cauchy
(∑
cyc
a√b + c
) ( ∑
cyc
a
√b + c
)
≥ (a + b + c)2
But, also by Cauchy,
√(a + b + c) ( a(b + c) + b(c + a) + c(a + b)) ≥ ∑
cyc
a√b + c
Hence, ∑
cyc
a
√b + c ≥√22 · (a + b + c) ·
√ a + b + cab + bc + ca
22 And we need only show a + b + c ≥ ab + bc + ca . Schur’s inequality for r = 1 can be expressed as 9abc a+b+c ≥ 4( ab + bc + ca ) − (a + b + c)2. Now, we suppose that
ab + bc + ca > a + b + c, and have 9abc a + b + c ≥ 4( ab + bc + ca ) − (a + b + c)2
(a + b + c) (4 − (a + b + c)) = abc (a + b + c)Hence, a + b + c < 3. But then abc < 1, which implies 4 = a + b + c + abc < 4. Contradiction, as desired. 33. (Iran 1996) Show that for all positive real numbers a, b, c ,(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
≥ 94
Solution. Fearless courage is the foundation of all success. 10 When everything else fails, return to the sure-fire strategy of clearing all denominators. In this case, we obtain 4( a + b)2(b + c)2(c + a)2(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
=
∑
sym
4a5b + 8 a4b2 + 10 a4bc + 6 a3b3 + 52 a3b2c + 16 a2b2c2
on the left, and on the right, 9( a + b)2(b + c)2(c + a)2 =
∑
sym
9a4b2 + 9 a4bc + 9 a3b3 + 54 a3b2c + 15 a2b2c2
Canceling like terms, we seek
∑
sym
4a5b − a4b2 + a4bc − 3a3b3 − 2a3b2c + a2b2c2
Sure enough, this is true, since 3a5b+ab 5
4
≥ a4b2 and a4b2+a2b4
2
≥ a3b3 by AM-GM, and
abc (a3 + b3 + c3 − a2(b + c) + b2(c + a) + c2(a + b) + 3 abc ) ≥ 0 by Schur. 34. (Japan 1997) Show that for all positive reals a, b, c ,(a + b − c)2
(a + b)2 + c2 + (b + c − a)2
(b + c)2 + a2 + (c + a − b)2
(c + a)2 + b2 ≥ 35
10 Found on a fortune cookie by Po-Ru Loh while grading an inequality on 2005 Mock IMO Day 2 that was solved by brutal force.
23 Solution. Put a + b + c = 3 so that equality will hold at a = b = c = 1 and suppose that there exists some k for which (b + c − a)2
(b + c)2 + a2 = (3 − 2a)2
(3 − a)2 + a2 ≥ 15 + ka − k
for all positive a, b, c ; such an inequality would allow us to add cyclicly to deduce the desired inequality. As the inequality is parametrically contrived to yield equality where
a = 1, we need to find k such that a = 1 is a double root. At a = 1, the derivative on the left is (2(3 −2a)·− 2)((3 −a)2+a2)−((3 −2a)2)(2(3 −a)·− 1+2 a)((3 −a)2+a2)2 = −18 25 . The derivative on the right is k, so we set k = −18 25 . But for this k we find (3 − 2a)2 −
(15 + ka − k
) ((3 − a)2 + a2) = 18 25 − 54 a2
25 + 36 a3
25 = 18 25 (a − 1) 2(2 a + 1) ≥ 0as desired. Alternatively, we could have used AM-GM to show a3 + a3 + 1 ≥ 3a2. As hinted at by a previous problem, inequalities are closely linked to polynomials with roots of even multiplicity. The isolated manipulation idea used in this solution offers a completely different approach to the inequalities which work with every term. 35. (MOP 02) Let a, b, c be positive reals. Prove that
( 2ab + c
)23
+
( 2bc + a
)23
+
( 2ca + b
)23
≥ 3
Solution. Suppose that there exists some r such that
( 2ab + c
)23
≥ 3ar
ar + br + cr
We could sum the inequality cyclicly to deduce what we want. Since equality holds at
a = b = c = 1, we use derivatives to find a suitable r. At the said equality case, on the left, the partial derivative with respect to a is 23 , while the same derivative on the right is 23 r. Equating the two we have r = 1. (This is necessary since otherwise the inequality will not hold for either a = 1 + ≤ or a = 1 − ≤.) 11 Now, 3aa + b + c ≤ 3a
3 3
√
a · (b+c
2
)2
11 Actually, even this is a special case of the general sense that the convexity of one side must exceed the convexity of the other. More precisely, we have the following result: Let fand gfunctions over the domain Dwith continuous partial derivatives. If f(ν)≥g(ν) for all ν∈D, then at every equality case ν0,
∇(f−g)( ν0) = 0and every component of ∇2(f−g) ( ν0) is nonnegative.
24 = a23
(b+c
2
)23
=
( 2ab + c
)23
by AM-GM, as desired. 36. (Mildorf) Let n ≥ 2 be an integer. Prove that for all reals a1, a 2, . . . , a n > 0 and reals
p, k ≥ 1, ( a1 + a2 + · · · + an
ap
1
ap
2
· · · + apn
)k
≥ ak
1
ak
2
· · · + akn
apk
1
apk
2
· · · + apk n
where inequality holds iff p = 1 or k = 1 or a1 = a2 = · · · = an, flips if instead 0 < p < 1, and flips (possibly again) if instead 0 < k < 1.
Solution. Taking the kth root of both sides, we see that the inequality is equivalent to n∑
i=1
k
√
aki
ak
1
ak
2
· · · + akn
≥
n
∑
i=1
k
√
apk i
apk
1
apk
2
· · · apk n
WLOG, suppose that a1 ≥ a2 ≥ · · · ≥ an. We prove a lemma. Let Si = api
ap
1+··· +apn
and
Ti = aqi
aq
1+··· +aqn
for i = 1 , 2, . . . , n where 0 < q < p . Then the sequence S1, S 2, . . . , S n
majorizes the sequence T1, T 2, . . . , T n.To prove the claim, we note that S1 ≥ · · · ≥ Sn and T1 ≥ · · · ≥ Tn and have, for
m ≤ n,
m
∑
i=1
Si ≥
m
∑
i=1
Ti
⇐⇒ (ap
1
· · · + apm) ( aq
1
· · · + aqn) ≥ (aq
1
· · · + aqm) ( ap
1
· · · + apn)
⇐⇒ (ap
1
· · · + apm) (aqm+1 + · · · + aqn
) ≥ (aq
1
· · · + aqm) (apm+1 + · · · + apn
)
⇐⇒ ∑
(i,j )| { 1≤i≤m<j ≤n}
api aqj − aqi apj ≥ 0Which is obvious. In particular, m = n is the equality case, and the claim is established. But now the desired is a direct consequence of the Majorization inequality applied to the sequences in question and the function f (x) = k
√x.37. (Vascile Cartoaje) Show that for all real numbers a, b, c,(a2 + b2 + c2)2 ≥ 3 (a3b + b3c + c3a)
25 Solution. We will be content to give the identity (a2 + b2 + c2)2 − 3( a3b + b3c + c3a) = 12
∑
cyc
(a2 − 2ab + bc − c2 + ca )2
Any Olympiad partipant should be comfortable constructing various inequalities through well-chosen squares. Here, we could certainly have figured we were summing the square of a quadratic that is 0 when a = b = c such that no term a2bc is left uncancelled. A good exercise is to show that equality actually holds iff a = b = c or, for some cyclic permutation, a : b : c ≡ sin 2 (4π
7
) : sin 2 (2π
7
) : sin 2 (π
7
).38. (Anh-Cuong) Show that for all nonnegative reals a, b, c ,
a3 + b3 + c3 + 3 abc ≥ ab √2a2 + 2 b2 + bc √2b2 + 2 c2 + ca √2c2 + 2 a2
Solution. Upon observing that this inequality is stronger than Schur’s inequality for
r = 1, we are inspired to prove a sharp lemma to eliminate the radical. Knowing that √2x2 + 2 y2 ≥ x + y ≥ 2xy x+y , we seek a combination of the latter two that exceeds the former. We find 3x2 + 2 xy + 3 y2
2( x + y) ≥ √2x2 + 2 y2
This follows from algebra, since (3 x2 + 2 xy + 3 y2)2 = 9 x4 + 12 x3y + 22 x2y2 + 12 xy 3 +9y4 ≥ 8x4 + 16 x3y + 16 x2y2 + 16 xy 3 + 8 y4 = 4( x + y)2(2 x2 + 2 y2), so that (3 x2 + 2 xy +3y2)2 − 4( x + y)2(2 x2 + 2 y2) = x4 − 4x3y + 6 x2y2 − 4xy 3 + y4 = ( x − y)4 ≥ 0. Now,
∑
cyc
ab √2a2 + 2 b2 ≤ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b)So it would suffice to show
∑
cyc
a(a − b)( a − c) = ∑
cyc
(a3 + abc − ab (a + b))
≥ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b) − ab (a + b)= ∑
cyc
3a3b + 2 a2b2 + 3 ab 3 − 2a3b − 4a2c2 − 2ab 3
2( a + b)= ∑
cyc
ab (a − b)2
2( a + b)But ∑
cyc
(b + c − a)( b − c)2 = 2 ∑
cyc
a(a − b)( a − c)26 so that the desired is
∑
cyc
(
b + c − a − bc b + c
)
(b − c)2 ≥ 0which is evident, since without loss of generality we may assume a ≥ b ≥ c and find
(
a + b − c − ab a + b
)
(a − b)2 ≥ 0
(
c + a − b − ac a + c
) ((a − c)2 − (b − c)2) ≥ 0
(
b + c − a − bc b + c
)
(b − c)2 +
(
c + a − b − ac a + c
)
(b − c)2 ≥ 0The key to this solution was the sharp upper bound on the root-mean-square. At first glance our lemma seems rather arbitrary and contrived. Actually, it is a special case of a very sharp bound on the two variable power mean that I have conjectured and proved.
Mildorf’s Lemma 1 Let k ≥ − 1 be an integer. Then for all positive reals a and b,
(1 + k)( a − b)2 + 8 ab
4( a + b) ≥ k
√ak + bk
2
with equality if and only if a = b or k = ±1, where the power mean k = 0 is interpreted to be the geometric mean √ab . Moreover, if k < −1, then the inequality holds in the reverse direction, with equality if and only if a = b.
Usage Note. As of early November 2005, I have proven an extension of this lemma to additional values of k.12 Thus, you may rest assured that the result stated above is true. I was unable to get this result published, so I have instead posted the proof here as “ASharpBound.pdf.” However, the proof is rather difficult (or at least so I think, being as though it took me nearly half a year) and the lemma is far from mainstream. Thus, should you require it on an Olympiad, you should prove it for whatever particular value of k you are invoking. This is not terribly difficult if k is a small integer. One simply takes the kth power of both sides and factors the difference of the two sides as
(a − b)4 · P (a, b ), etc.
For x ≥ y ≥ 1, prove that
x
√x + y + y
√y + 1 + 1
√x + 1 ≥ y
√x + y + x
√x + 1 + 1
√y + 1
12 In particular, the inequality holds for all kin ( −∞ ,−1) ,{− 1,0,1},(1 ,3/2] ,[2 ,∞) with the signs ≤,≥,≤
,≥respectively, with equality iff a=bor k=±1.
27 Solution. By observation, equality holds when y = 1 and when x = y. Combining this with the restriction, it makes sense to write x = y + a and y = 1 + b where a, b ≥ 0. Now we can write
x − y
√x + y + y − 1
√y + 1 + 1 − x
√1 + x ≥ 0
⇐⇒ a
√2 + a + 2 b + b
√2 + b ≥ a + b
√2 + a + b
But this is evident by Jensen’s inequality applied to the convex function f (x) = 1√x ,since
af (2 + a + 2 b) + bf (2 + b) ≥ (a + b)f
(a(2 + a + 2 b) + b(2 + b)
a + b
)
= (a + b)f
((a + b)2 + 2( a + b)
a + b
)
= a + b
√2 + a + b
as desired. 40. (MOP) For n ≥ 2 a fixed positive integer, let x1, . . . , x n be positive reals such that
x1 + x2 + · · · + xn = 1
x1
1
x2
· · · + 1
xn
Prove that 1
n − 1 + x1
1
n − 1 + x2
· · · + 1
n − 1 + xn
≤ 1
Solution. We will prove the contrapositive. (We are motivated to do this for two good reasons: 1) it is usually difficult the show that the sum of some reciprocals is bounded above, and 2) the given relation in its current form is an abomination.) Take
yi = 1
n−1+ xi
, and for the sake of contradiction assume y1 + · · · + yn > 1. Since the yi
are too large, the xi are too small and we shall prove 1
x1
· · · + 1
xn
x 1 + · · · + xn.Since xiyi = 1 − (n − 1) yi, we have (n − 1) yi > (n − 1)
(
yi + 1 −
n
∑
j=1
yj
)
= (n − 1) yi − 1 +
n
∑
j=1
(1 − (n − 1) yj )= −xiyi +
n
∑
j=1
xj yj (∗)=⇒ n − 1
xi
−1 +
n
∑
j=1
xj yj
xiyi
(∗∗ )28 Summing () over i,(n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(( n∑
j=1
1
xj yj
)
− 1
xiyi
)
But by Cauchy and (), we have
( n∑
j=1
1
xj yj
)
− 1
xiyi
≥ (n − 1) 2
(∑nj=1 xj yj
)
− xiyi
(n − 1) 2
(n − 1) yi
= n − 1
yi
Hence, (n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(n − 1
yi
)
= ( n − 1)( x1 + · · · + xn)as desired. 41. (Vascile Cartoaje) Show that for positive reals a, b, c ,14a2 − ab + 4 b2 + 14b2 − bc + 4 c2 + 14c2 − ca + 4 a2 ≥ 97( a2 + b2 + c2)
Solution. Upon expansion, we see that it is equivalent to
∑
sym
56 a6 − 28 a5b + 128 a4b2 + 44 a3b3 + 95 2 a4bc + 31 a3b2c − 45 2 a2b2c2 ≥ 0We conjure up the following inequalities:
∑
sym
a6 − 2a5b + a4bc ≥ 0 (1)
∑
sym
a5b − 4a4b2 + 3 a3b3 ≥ 0 (2)
∑
sym
a4b2 − a4bc − a3b3 + 2 a3b2c − a2b2c2 ≥ 0 (3)
∑
sym
a4bc − 2a3b2c + a2b2c2 ≥ 0 (4) (1) and (4) follow from Schur’s inequality for r = 4 and r = 1 (multiplied by abc )respectively. (2) is the result of expanding ∑
cyc
ab (a − b)4 ≥ 0, and (3) is the expanded form of the famous ( a − b)2(b − c)2(c − a)2 ≥ 0. The desired now follows by subtracting 56 times (1), 84 times (2), 208 times (3), 399 2 times (4), and then simple AM-GM to clear the remaining a2b2c2.29 This is about as difficult as a dumbass solution can get. A good general strategy is to work with the sharpest inequalities you can find until you reduce a problem to something obvious, starting with the most powerful (most bunched, in this case ∑
sym
a6) term and work your way down to the weak terms while keeping the most powerful term’s coefficient positive. My solution to this problem starts with (1), Schur with r = 4 (Schur is stronger for larger r), which is almost certainly sharper than the inequality in question. Next, inequality (2) is a sharp cyclic sum to use the a5b terms. In particular, it relates terms involving only two of the three variables. Most of the time, the only inequality that can “pull up” symmetric sums involving three variables to stronger ones involving just two is Schur, although it does so at the expense of a very strong term with only one variable. Hence, we made a logical choice. Inequality (3) is extremely sharp, and allowed us to obtain more a4bc and a3b3 terms simultaneously. In particular, it was necessary to cancel the a3b3 terms. I’ll note that this inequality is peculiar to sixth degree symmetry in three variables - it does not belong to a family of similar, nice inequalities. Finally, inequality (4), which is a handy corollary to (3), is another Schur. Every inequality we have used so far is quite sharp, and so it is no surprise that the leftovers are the comparatively weak AM-GM. 42. (Reid Barton, IMO Shortlist 03/A6.) Let n ≥ 2 be a positive integer and x1, x 2, . . . , x n,y1, y 2, . . . , y n a sequence of 2 n positive reals. Suppose z2, z 3, . . . , z 2n is such that z2
i+j
≥
xiyj for all i, j ∈ { 1, . . . , n }. Let M = max {z2, z 3, . . . , z 2n}. Prove that
(M + z2 + z3 + · · · + z2n
2n
)2
≥
(x1 + · · · + xn
n
) ( y1 + · · · + yn
n
)
Reid’s official solution. Let max( x1, . . . , x n) = max( y1, . . . , y n) = 1. (We can do this by factoring X from every xi, Y from every yj , and √XY from every zi+j
without changing the sign of the inequality.) We will prove M + z2 + · · · + z2n ≥
x1 + x2 + · · · + xn + y1 + y2 + · · · + yn, after which the desired follows by AM-GM. We will show that the number of terms on the left which are greater than r is at least as large as the number of terms on the right which are greater than r, for all r ≥ 0. For r ≥ 1, the claim is obvious, since all terms on the right are at most 1. Now take
r < 1. Let A and B denote the set of i for which xi > r and the set of j for which
yj > r respectively, and write a = |A|, b = |B|. Evidently, from our scaling, a, b ≥ 1. Now, xi > r and yj > r implies zi+j ≥ √xiyj ≥ r. Hence, if C is the set of k for which zk > r , we have |C| ≥ | A + B|, where the set addition is defined by the set of possible values if we take an element of A and add it to an element of B. How-ever, |A + B| ≥ | A| + |B| − 1, since if A and B consist of the values p1 < · · · < p a and
q1 < · · · < q b respectively we have all of the values p1 +q1 < . . . < p a +q1 < · · · < p a +qb
in A + B. Hence, |C| ≥ a + b − 1. Since |C| ≥ 1, there is some zk > r , and hence,
M > r . Therefore, the left side of the inequality in question has at least a + b terms which exceed r, as desired. •
30 The preponderance of difficulty here stemmed from dealing with the superabundance of givens, especially the mysterious M . Scaling allowed us to introduce some degree of control and, with marked audacity, a profoundly clever idea. As it turned out, the in-equality was no sharper than simple AM-GM! It is my opinion that it is highly unlikely that a problem as staggeringly pernicious as this one will appear on an Olympiad - at least in the foreseeable future. Nevertheless, I have included it here for the purpose of illustrating just how unusual and creative a solution can be.
3 Problems
(MOP 04) Show that for all positive reals a, b, c ,
( a + 2 ba + 2 c
)3
+
( b + 2 cb + 2 a
)3
+
(c + 2 ac + 2 b
)3
≥ 32. (MOP) Show that if k is a positive integer and a1, a 2, . . . , a n are positive reals which sum to 1, then n∏
i=1
1 − aki
aki
≥ (nk − 1)n
Let a1, a 2, . . . , a n be nonnegative reals with a sum of 1. Prove that
a1a2 + a2a3 + · · · + an−1an ≤ 144. (Ukraine 01) Let a, b, c, x, y, z be nonnegative reals such that x + y + z = 1. Show that
ax + by + cz + 2 √(ab + bc + ca )( xy + yz + zx ) ≤ a + b + c
Let n > 1 be a positive integer and a1, a 2, . . . , a n positive reals such that a1a2 . . . a n = 1. Show that 11 + a1
· · · + 11 + an
≤ a1 + · · · + an + n
(Aaron Pixton) Let a, b, c be positive reals with product 1. Show that 5 + ab + bc + ca ≥ (1 + a)(1 + b)(1 + c)7. (Valentin Vornicu 13 ) Let a, b, c, x, y, z be arbitrary reals such that a ≥ b ≥ c and either
x ≥ y ≥ z or x ≤ y ≤ z. Let f : R → R+0 be either monotonic or convex, and let k be a positive integer. Prove that
f (x)( a − b)k(a − c)k + f (y)( b − c)k(b − a)k + f (z)( c − a)k(c − b)k ≥ 0
13 This improvement is more widely known than the other one in this packet, and is published in his book,
Olimpiada de Matematica... de la provocare la experienta , GIL Publishing House, Zalau, Romania. (In English, “The Math Olympiad... from challenge to experience.”)
31 8. (IMO 01/2) Let a, b, c be positive reals. Prove that
a
√a2 + 8 bc + b
√b2 + 8 ca + c
√c2 + 8 ab ≥ 19. (USAMO 04/5) Let a, b, c be positive reals. Prove that
(a5 − a2 + 3 ) ( b5 − b2 + 3 ) ( c5 − c2 + 3 ) ≥ (a + b + c)3
(Titu Andreescu) Show that for all nonzero reals a, b, c ,
a2
b2 + b2
c2 + c2
a2 ≥ ac + cb + ba
(IMO 96 Shortlist) Let a, b, c be positive reals with abc = 1. Show that
ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤ 112. Let a, b, c be positive reals such that a + b + c = 1. Prove that
√ab + c + √bc + a + √ca + b ≥ 1 + √ab + √bc + √ca
(APMO 2005/2) Let a, b, c be positive reals with abc = 8. Prove that
a2
√(a3 + 1) ( b3 + 1) + b2
√(b3 + 1) ( c3 + 1) + c2
√(c3 + 1) ( a3 + 1) ≥ 4314. Show that for all positive reals a, b, c ,
a3
b2 − bc + c2 + b3
c2 − ca + a2 + c3
a2 − ab + b2 ≥ a + b + c
(USAMO 97/5) Prove that for all positive reals a, b, c ,1
a3 + b3 + abc + 1
b3 + c3 + abc + 1
c3 + a3 + abc ≤ 1
abc
(Mathlinks Lore) Show that for all positive reals a, b, c, d with abcd = 1, and k ≥ 2, 1(1 + a)k + 1(1 + b)k + 1(1 + c)k + 1(1 + d)k ≥ 22−k
(IMO 05/3) Prove that for all positive a, b, c with product at least 1,
a5 − a2
a5 + b2 + c2 + b5 − b2
b5 + c2 + a2 + c5 − c2
c5 + a2 + b2 ≥ 032 18. (Mildorf) Let a, b, c, k be positive reals. Determine a simple, necessary and sufficient condition for the following inequality to hold: (a + b + c)k (akbk + bkck + ckak) ≤ (ab + bc + ca )k(ak + bk + ck)19. Let a, b, c be reals with a + b + c = 1 and a, b, c ≥ − 34 . Prove that
aa2 + 1 + bb2 + 1 + cc2 + 1 ≤ 910 20. (Mildorf) Show that for all positive reals a, b, c ,
3
√4a3 + 4 b3 + 3
√4b3 + 4 c3 + 3
√4c3 + 4 a3 ≤ 4a2
a + b + 4b2
b + c + 4c2
c + a
Let a, b, c, x, y, z be real numbers such that (a + b + c)( x + y + z) = 3 , (a2 + b2 + c2)( x2 + y2 + z2) = 4 Prove that
ax + by + cz ≥ 022. (Po-Ru Loh) Let a, b, c be reals with a, b, c > 1 such that 1
a2 − 1 + 1
b2 − 1 + 1
c2 − 1 = 1 Prove that 1
a + 1 + 1
b + 1 + 1
c + 1 ≤ 123. (Weighao Wu) Prove that (sin x)sin x < (cos x)cos x
for all real numbers 0 < x < π
4
.24. (Mock IMO 05/2) Let a, b, c be positive reals. Show that 1 < a
√a2 + b2 + b
√b2 + c2 + c
√c2 + a2 ≤ 3√2225. (Gabriel Dospinescu) Let n ≥ 2 be a positive integer. Show that for all positive reals
a1, a 2, . . . , a n with a1a2 . . . a n = 1,
√a21 + 1 2 + · · · +
√a2
n
1 2 ≤ a1 + · · · + an
33 26. Let n ≥ 2 be a positive integer, and let k ≥ n−1
n
be a real number. Show that for all positive reals a1, a 2, . . . , a n,
( (n − 1) a1
a2 + · · · + an
)k
+
( (n − 1) a2
a3 + · · · + an + a1
)k
· · · +
( (n − 1) an
a1 + · · · + an−1
)k
≥ n
(Mildorf) Let a, b, c be arbitrary reals such that a ≥ b ≥ c, and let x, y, z be nonnegative reals with x + z ≥ y. Prove that
x2(a − b)( a − c) + y2(b − c)( b − a) + z2(c − a)( c − b) ≥ 0and determine where equality holds. 28. (USAMO 00/6) Let n ≥ 2 be an integer and S = {1, 2, . . . , n }. Show that for all nonnegative reals a1, a 2, . . . , a n, b 1, b 2, . . . , b n,
∑
i,j ∈S
min {aiaj , b ibj } ≤ ∑
i,j ∈S
min {aibj , a j bi}
(Kiran Kedlaya) Show that for all nonnegative a1, a 2, . . . , a n,
a1 + √a1a2 + · · · + n
√a1 · · · an
n ≤ n
√
a1 · a1 + a2
2 · · · a1 + · · · + an
n
(Vascile Cartoaje) Prove that for all positive reals a, b, c such that a + b + c = 3,
aab + 1 + bbc + 1 + cca + 1 ≥ 3231. (Gabriel Dospinescu) Prove that ∀a, b, c, x, y, z ∈ R+| xy + yz + zx = 3,
a(y + z)
b + c + b(z + x)
c + a + c(x + y)
a + b ≥ 332. (Mildorf) Let a, b, c be non-negative reals. Show that for all real k,
∑
cyc
max( ak, b k)( a − b)2
2 ≥ ∑
cyc
ak(a − b)( a − c) ≥ ∑
cyc
min( ak, b k)( a − b)2
2(where a, b, c 6 = 0 if k ≤ 0) and determine where equality holds for k > 0, k = 0, and
k < 0 respectively. 33. (Vascile Cartoaje) Let a, b, c, k be positive reals. Prove that
ab + ( k − 3) bc + ca
(b − c)2 + kbc + bc + ( k − 3) ca + ab
(c − a)2 + kca + ca + ( k − 3) ab + bc
(a − b)2 + kab ≥ 3( k − 1)
k
(Taiwan? 02) Show that for all positive a, b, c, d ≤ k, we have
abcd
(2 k − a)(2 k − b)(2 k − c)(2 k − d) ≤ a4 + b4 + c4 + d4
(2 k − a)4 + (2 k − b)4 + (2 k − c)4 + (2 k − d)4
34 |
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Binomial Theorem
The Binomial Theorem states that for real or complex, , and non-negativeinteger,
where is a binomial coefficient. In other words, the coefficients when is expanded and like terms are collected are the same as the entries in the th row of Pascal's Triangle.
For example, , with coefficients , , , etc.
Contents
[hide]
1 Proof
1.1 Proof via Induction
1.2 Proof using calculus
2 Generalizations
2.1 Proof
3 Usage
4 See also
Proof
There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof:
We can write . Repeatedly using the distributive property, we see that for a term , we must choose of the terms to contribute an to the term, and then each of the other terms of the product must contribute a . Thus, the coefficient of is the number of ways to choose objects from a set of size , or . Extending this to all possible values of from to , we see that , as claimed.
Similarly, the coefficients of will be the entries of the row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS].
Proof via Induction
Given the constants are all natural numbers, it's clear to see that . Assuming that , Therefore, if the theorem holds under , it must be valid. (Note that for )
Proof using calculus
The Taylor series for is for all .
Since , and power series for the same function are termwise equal, the series at is the convolution of the series at and . Examining the degree- term of each, which simplifies to for all natural numbers.
Generalizations
The Binomial Theorem was generalized by Isaac Newton, who used an infiniteseries to allow for complex exponents: For any real or complex, , and ,
.
Proof
Consider the function for constants . It is easy to see that . Then, we have . So, the Taylor series for centered at is
Usage
Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial , one could factor it as such: . It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.
See also
Combinatorics
Multinomial Theorem
Retrieved from "
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Theorems
Combinatorics
Algebra
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7995 | https://resident360.amboss.com/adult-medicine/surgery/fluid-and-electrolyte-management/fluid-and-electrolyte-management.html | Resident 360 Study Plans on AMBOSS
Find all Resident 360 study plans on AMBOSS
Fast Facts
A brief refresher with useful tables, figures, and research summaries
Fluid and Electrolyte Management
Fluid and electrolyte management is key to all phases of care in surgical patients, but assessment of fluid and electrolyte status can present challenges in surgical patients different from those found in other types of hospitalized patients. For example, patients who undergo prolonged abdominal surgeries can experience significant and unaccounted insensible fluid losses and fluid shifts. Insensible fluid losses refer to the body’s loss of water through nonvisible and unmeasurable means (e.g., evaporation from the skin and respiratory tract). These losses can have a significant affect on fluid and electrolyte balance, particularly in settings where accurate assessment and management of fluid status are crucial. Similarly, patients recovering from bariatric surgery or small-bowel resection can experience electrolyte abnormalities that are not common in other surgical scenarios.
This rotation guide focuses on general principles of fluid and electrolyte management, including preoperative optimization, intraoperative considerations, and postoperative management.
The topics in this rotation guide are organized as follows:
Fluid Resuscitation
Common Intravenous Fluid Solutions Used in Hospitalized Patients
Fluid Choice
Fluid Administration
Electrolyte Abnormalities
More information on fluid and electrolyte management can be found in the following rotation guides:
Resuscitation Fluids and Blood Transfusion (Adult Critical Care)
Fluids, Electrolytes, and Nutrition (Neonatal Care)
Sodium Homeostasis Disorders (Adult Nephrology)
Calcium-Based Disorders (Adult Endocrinology)
Fluid Resuscitation
Fluid selection is an essential component of medical care. Both the initial decision to give fluid and the choice of fluids requires a careful history and physical exam, and consideration of clinical adjuncts such as laboratory tests and imaging.
Common Intravenous Fluid Solutions Used in Hospitalized Patients
Crystalloid solutions (e.g., lactated Ringer solution, normal saline) contain ions that are movable across membranes and confer a degree of osmolarity. Most surgeons prefer balanced crystalloid solutions (e.g., lactated Ringer) over unbalanced fluids (e.g., normal saline) due to modestly better outcomes in surgical patients.
Lactated Ringer (LR) solution
Chloride concentration composition more closely approximates human plasma than normal saline (NS).
LR is hypotonic to plasma and preferred for patients with hypernatremia.
LR contains a physiologic concentration of potassium and limits the tendency to hypokalemia that may occur with infusion of a large volume of potassium-free solutions.
Maintenance LR has been associated with modest improvement in renal outcomes as compared with NS in patients who are not in an intensive care unit (ICU) and may be associated with a small mortality benefit in patients in an ICU.
Normal saline (NS)
similar osmolarity as plasma water
hyperchloremic as compared to plasma
large volume can cause mild metabolic acidosis
less likely to lead to hyponatremia (e.g., preferred in patients with significant gastric losses and in patients with traumatic brain injury)
recommended over LR when mixed with some medications to avoid precipitation
Plasmalyte
available as a fluid with similar ion concentrations as plasma but less commonly used due to its high cost
Colloid solutions (e.g., albumin) contain proteins or other large molecules that maintain intravascular oncotic pressure better than crystalloid solutions. Colloids do not easily pass outside of vasculature and therefore expand intravascular volumes better than crystalloids do (via oncotic pressure), without adding high total-body-fluid volume.
Albumin is the most commonly used colloid solution in surgical patients.
Other types of colloids can be used in certain practice settings:
Hydroxyethyl starches (HES) are synthetic colloids that can be used for volume expansion during surgery or trauma, or when patients have significant edema. However, their use has declined due to concerns about kidney function impairment and coagulopathy.
Dextrans are glucose polymers that theoretically improve microcirculatory blood flow. Dextrans are sometimes used in vascular surgery to prevent venous thrombosis, but they can also interfere with blood typing and crossmatching as well as promote bleeding.
Gelatin-based solutions are derived from bovine collagen and can be used as plasma volume expanders. They are rapidly metabolized and have short duration of action, making them suitable for short-term use (e.g., in an operative setting where rapid fluid turnover is expected).
Artificial colloids are new synthetic colloid preparations similar to those based on modified fluid gelatin or newer starches, designed to minimize adverse effects.
See the Critical Care rotation guide for more information on choosing a crystalloid, colloid, and blood product for resuscitation.
Fluid Choice
Fluid choice influences the volume administered and affects outcomes. The composition of varying solutes can lead to metabolic changes. The following table compares the composition of body electrolytes and common fluids used for resuscitation and maintenance. This information can be used by physicians to compare the electrolyte composition of various intravenous (IV) solutions against normal plasma levels, aiding in the selection of the appropriate fluid therapy to correct or maintain a patient’s electrolyte balance.
Composition of Body Electrolytes and Common Fluids Used for Resuscitation and Maintenance
| Electrolyte | Plasma | 0.9% Normal Saline | Lactated Ringer | Albumin |
--- ---
| Sodium (mEq/L) | 134-144 | 154 | 131 | 135-160 |
| Potassium (mEq/L) | 3.5-5 | | 5.4 | 2 |
| Chloride (mEq/L) | 96-106 | 154 | 111 | |
| Calcium (mEq/L) | 2.2 | | 2 | |
| Magnesium (mmol/L) | 2 | | 1 | |
| Bicarbonate (mmol/L) | 24 | | | |
| pH | 7.4 | 5.4 | 6.5 | 7.4 |
The plasma water concentration of sodium is 145-156 mEq/L. (References: Electrolytes. StatPearls 2023; Serum Chloride. Clinical Methods: The History, Physical, and Laboratory Examinations, 3rd edition 1990; Choice of Fluid Therapy in the Initial Management of Sepsis, Severe Sepsis, and Septic Shock. Shock 2016; Selection of Intravenous Fluids. Am J Kidney Dis 2018.)
Fluid Administration
Maintenance: Maintenance fluids address the patient’s physiological needs, encompassing sensible and insensible fluid losses. Sensible losses pertain to conventional forms of excretion, such as urination and defecation. Insensible losses pertain to less obvious fluid expenditure, including sweating and respiratory evaporation.
The 4-2-1 rule is typically used to calculate maintenance rate for crystalloid fluids.
4 mL/kg/hr for first 10 kg of patient weight
2 mL/kg/hr for second 10 kg of patient weight
1 mL/kg/hr for every kg above 20 kg of patient weight
Example: A 70-kg patient requiring maintenance fluids should receive IV fluids at a rate of 110 mL/hr [(4 × 10) + (2 × 10) + (1 × 50)] = 110 mL/hr
Maintenance fluids are generally only used in patients who have reduced fluid intake (e.g., nothing by mouth [NPO] status, altered mental status leading to poor intake) or increased/dysregulated outputs (e.g., large wounds, burns).
NS should be used in patients on certain medications (e.g., ceftriaxone) to avoid calcium precipitation from calcium-containing LR.
Resuscitation/replacement of fuid losses: Vital-sign changes and physical exam findings provide early clues of volume depletion in surgical patients. Read more on the administration of different types of fluids for resuscitation. The following special considerations should be kept in mind during fluid replacement:
Special considerations:
In patients with oliguria, the fractional excretion of sodium (FENa) can be used to determine whether the renin-angiotensin-aldosterone system is active and promoting sodium retention (assuming the patient is not receiving diuretics).
The fractional excretion of urea (FEUrea) is a similar calculation that can be used for patients on diuretics.
In burn patients, LR should be used for fluid resuscitation due to its physiologic electrolyte concentrations and to prevent acidosis. Read more on fluid resuscitation in burn patients.
The Parkland formula (4 mL × %TBSA [total body surface area burned] × body weight [kg]) can be used to calculate the amount of LR. The Parkland formula calculates the total volume of fluid to be administered during the first 24 hours following presentation. Half the amount is given within the first 8 hours and the other half is given during the remaining 16 hours.
Adequate resuscitation is measured with urine output (0.5-1 cc/kg/hr in adults, 1-2 cc/kg/hr in pediatric populations), examination of acid-base status, and vital signs.
Neonatal and pediatric patients, and those on dialysis, are important to identify early in the preoperative period because of the following special considerations during fluid administration:
Neonatal and pediatric patients have increased surface area-to-body volume ratio, resulting in higher rates of insensible losses. Read more on administration of fluids in neonates.
Patients with kidney disease may not adequately excrete fluid by urination and are particularly susceptible to developing volume overload.
Fluids during surgery that involves acute blood loss (e.g., trauma or transplantation) necessitate administration of blood products over crystalloid or colloid solutions to avoid dilution of coagulation factors.
Insensible (incalculable) losses (not urine, blood, or stool) are usually 8-12 mL/kg/day but can increase to one liter per hour in open thoracic or abdominal surgery due to evaporation from the exposed pleural or peritoneal cavity.
Electrolyte Abnormalities
Many electrolyte derangements in surgical patients, particularly in the postoperative period, are preventable, treatable, or both when recognized promptly. Most deficiencies can be attributed to NPO status of patients before and after surgery, as well as changes in absorptive capacities due to resections of the alimentary tract. Eectrolyte derangements can be similar in surgical patients and nonsurgical patients, but the etiology and treatment in the acute and chronic settings may differ.
Sodium
Sodium regulation and homeostasis are important factors in many surgical patients. Read more on the normal regulation of sodium and signs and symptoms of sodium dysregulation in the Nephrology rotation guide.
Hypernatremia: Hypernatremia is defined as a sodium concentration >145 mmol/liter and usually reflects a water deficit.
Causes: In surgical patients, causes include diuretic medications, diabetes insipidus, trauma, emesis/diarrhea, and even paraneoplastic syndromes. Hypernatremia can also result from excess sodium intake from sodium-containing medications without sufficient water intake in patients who have diminished thirst or ability to drink (e.g., intubated patients).
Complications: Neurologic complications include severe hypernatremia, which can result in shrinkage of the brain and cause venous bleeding
Treatment: Slow correction is not required in adults but physicians must have a low threshold to consult nephrology for help with titration.
A table summarizing the calculation for infusions to correct hypernatremia can be found here.
Hyponatremia: Hyponatremia is defined as a sodium concentration <135 mmol/liter and reflects water in excess of sodium in the blood. See a helpful algorithm for determining the cause of hyponatremia.
Causes: In the surgical setting, hyponatremia most commonly is associated with hypovolemia coincident with intake of dilute fluids. Although the entirety of causes are beyond the scope of this guide, patients with common surgical issues are predisposed to developing hyponatremia from the following:
Gastrointestinal (GI) and blood losses: Patients with hypovolemia due to vomiting, diarrhea, or hemorrhage become hyponatremic with intake of dilute fluids (oral or IV) due to appropriate relase of vasopressin and renal retention of water.
Syndrome of inappropriate antidiuretic hormone secretion (SIADH): Major surgery and anesthesia are risk factors for SIADH, which leads to retention of excess water.
Complications: Cerebral edema can lead to seizures, altered mental status, and even brain-stem herniation/death in extreme cases.
Treatment: Management is directed at the etiology.
In the setting of volume depletion, colloid or isotonic crystalloid should be given. Once volume depletion is treated, the stimulus for vasopressin is removed and the kidneys can excrete excess water. The serum sodium in this setting may increase rapidly and may require intervention to slow the rate of correction.
In the setting of SIADH, hypertonic saline can be infused slowly to provide solute to help excrete excess water.
Calculate as a “sodium deficit” using the serum sodium level and estimated total body water (TBW, 0.6 × body weight for male patients, and 0.55 × body weight for femal patients). The formula is as follows:
Correction of hyponatremia should be slow, with a goal of 6 mEq/day or 0.25 mEq/hour.
Potassium
Hyperkalemia: Hyperkalemia is defined as a serum potassium level >5.5 mEq/liter, although the laboratory value for the upper limit of normal may be slightly lower. Hyperkalemia is common in surgical patients both intraoperatively and postoperatively. Severe hyperkalemia can be life-threatening.
Causes: Many surgical patients are at risk for hyperkalemia. The causes are extensive and include the following:
release of potassium from cell lysis, as with rhabdomyolysis or ischemia-reperfusion injury
pancreatic insufficiency or diabetes mellitus (due to lack of or resistance to insulin), which put patients at risk since insulin helps shift potassium into muscle cells
iatrogenic causes due to medication administration
acidosis contributing to hyperkalemia, as potassium shifts into serum in exchange for intracellular movement of hydrogen ions
volume depletion, acute kidney injury, and chronic kidney disease, which all limit renal excretion of potassium
high potassium values occurring in the setting of red blood cell (RBC) lysis related to phlebotomy or laboratory handling
Complications: Hyperkalemia can lead to symptoms of muscle weakness and life-threatening arrhythmias.
Treatment: Urgent electrocardiogram (ECG) is essential to the workup of hyperkalemia, as changes in cardiac electrical activity (e.g., peaked T waves) confirm that hyperkalemia is not a laboratory artifact.
Acute treatment of hyperkalemia with ECG changes focuses on shifting potassium intracellularly, as well as preventing the cardiotoxic effects of hyperkalemia.
initial treatment: 1-2 grams of calcium gluconate administered over 3 minutes to stabilize cardiac membranes and allow for time to correct the cause of hyperkalemia
subsequent acute treatment: 10 units of regular insulin to move potassium intracellularly followed by one ampule of 50% dextrose (D50) to limit the risk of hypoglycemia
sodium bicarbonate infusion: can be used in the presence of evidence of acidemia
a beta-agonist (e.g., albuterol)
acute treatments: do not decrease the total amount of potassium in the body and should be followed by more-definitive treatments (e.g., oral sodium polysterene and loop diuretics)
Definitive treatment increases renal and GI excretion of potassium using furosemide and a potassium binder.
The older potassium binder, sodium polystyrene sulfonate (SPS), has been associated with bowel necrosis and has fallen out of favor.
Sodium zirconium cyclosilicate and patiromer are approved by the U.S. Food and Drug Administration for chronic hyperkalemia, but these agents are now commonly used in the acute setting given the concerns related to SPS.
Hypokalemia: Hypokalemia is defined as serum potassium level below 3.5 mEq/liter.
Causes: Surgical patients often become hypokalemic because of renal or GI losses.
Patients with high-volume emesis or nasogastric tube output can develop hypokalemic, hypochloremic metabolic alkalosis if they are not receiving a proton pump inhibitor (PPI). Renal excretion of excess bicarbonate leads to renal wasting of potassium.
Patients with diarrheal losses and excess laxative use can develop hypokalemia.
Infusion of large volumes of IV fluids such as NS without potassium leads to renal potassium losses.
Complications:
Similar to hyperkalemia, hypokalemia affects cardiac electrical activity and can cause dangerous arrhythmias if not treated promptly.
ECG can reveal flattened T waves or the presence of U waves, which can be indicative of impending severe arrhythmias.
Patients undergoing GI surgery may be predisposed to decreased oral intake, gut dysmotility and ileus, or malabsorption after intestinal resection. Therefore, the different causes for hypokalemia may require alternative management solutions than in nonsurgical patients.
For example, although oral potassium has excellent bioavailability, patients who are NPO prior to or after surgery often require IV potassium for potassium repletion.
Bariatric surgical patients are particularly susceptible to hypokalemia and may be prophylactically repleted following surgery.
Patients are also often deficient in other electrolytes or vitamins, with hypomagnesemia being a common codeficiency.
Magnesium is a cofactor in many potassium transporters and must be repleted first so that potassium repletion is effectively absorbed.
Treatment:
For both oral and IV potassium repletion, 10 mEq of potassium is expected to increase serum potassium concentration by 0.1 mEq provided there are no ongoing losses.
For example, in a patient with a potassium level of 3.3, 20 mEq of oral or IV potassium should improve serum potassium levels to 3.5.
Unlike hyperkalemia, calcium gluconate administration usually does not have a role in the acute treatment of hypokalemia.
Calcium
Hypercalcemia: Hypercalcemia is defined as total serum calcium levels >10.4 mg/dL, or ionized calcium concentration >5.6 mg/dL (>1.31 mmol/liter). Hypercalcemia can be acute or chronic, and signs and symptoms may present at much higher calcium levels in chronic hypercalcemia.
Causes: Hypercalcemia is often caused by pathology that can be treated surgically, such as primary hyperparathyroidism due to parathyroid adenomas.
Parathyroid hormone (PTH), along with vitamin D, is a primary driver of increasing calcium levels in serum.
Hypercalcemia can be caused by malignancy due to bone metastases or dysregulation of calcium hormones including PTH, vitamin D, and calcitonin.
Additional causes include, but are not limited to, medication administration (e.g., hydrochlorothiazide), sarcoidosis, familial hypocalciuric hypercalcemia, and excess intake of calcium or vitamins D and A.
Complications: Hypercalcemia can lead to altered mental status, nephrolithiasis, nausea/vomiting/constipation, arthralgias and osteoporosis, volume depletion, and polyuria.
Treatment:
Initial treatment of hypercalcemia is administration of IV fluids (usually at least 200 mL/hour of normal saline) to promote excretion because it often leads to volume depletion. Loop diuretics can be administered in the setting of volume overload.
IV fluids can be followed by additional medications if the hypercalcemia is severe (>14 mg/dL).
Calcitonin or glucocorticoids can be administered to lower serum calcium levels.
Bisphosphonates such as zoledronic acid can also be used to treat hypercalcemia, particularly in hypercalcemia of malignancy. However, these medications often require at least 48 hours to take effect.
Hypocalcemia: Hypocalcemia is defined as a serum total calcium level <8.5 mg/dL (ionized calcium concentration <4.65 mg/dL or <1.16 mmol/liter), can be classified as acute or chronic, and as disorders associated with high and low parathyroid hormone.
Causes: Hypocalcemia with low PTH is commonly caused by hypomagnesemia or acquired hypoparathyroidism (often secondary to surgical removal of parathyroid glands or disruption of parathyroid blood supply during thyroidectomy).
Complications: Watch for signs of neuromuscular abnormalities, such as perioral numbness/tingling, paresthesia, sensorineurium alteration, and even ECG changes in severe hypocalcemia.
Treatment: Immediately following parathyroidectomy, patients must be closely monitored for signs of hypocalcemia (including paresthesias, Trousseau sign, and Chvostek sign).
Patients may receive prophylactic oral calcium postoperatively, titrated based on the patient’s intraoperative drop in PTH and postoperative calcium levels.
Patients with calcium levels >7.5 mg/dL should be treated with oral calcium and vitamin D to increase the absorption of oral calcium.
Patients with calcium levels <7.5 mg/dL should be given IV calcium gluconate and closely monitored with serial labwork and ECGs.
Magnesium
Hypermagnesemia: Hypermagnesemia is defined as serum magnesium levels >2.3 mg/dL; however, symptoms typically only appear at levels higher than 4.0 mg/dL.
Causes: Hypermagnesemia is rare and typically accompanied by renal failure (because the kidneys can generally handle a large magnesium load).
Hypermagnesemia can occur in patients with trauma or burns, in the peripartum period (associated with magnesium prophylaxis for eclampsia), and in patients who are on hemodialysis.
Complications:
The most common presenting symptom is altered mental status, specifically diminished attentiveness.
Severe hypermagnesemia can cause diarrhea, muscle weakness and diminished deep tendon reflexes, arrhythmias, and seizures.
Treatment:
As with hypercalcemia treatment, treatment includes administration of normal saline at a rate of at least 200 mL/hour, as well as diuresis.
Severe hypermagnesemia may also require treatment with calcium gluconate to stabilize cardiac membranes.
Hypomagnesemia: Hypomagnesemia is defined as serum magnesium levels <1.4 mg/dL; however, symptoms are not usually seen until magnesium level is below 1.0 mg/dL.
Causes: Surgical patients often become hypomagnesemic during massive resuscitation or in the setting of GI losses.
Complications: The most dangerous complication of hypomagnesemia is torsades de pointes, a dangerous ventricular arrhythmia.
Treatment:
Torsades de pointes requires magnesium infusion emergently.
Milder symptoms can be treated with oral magnesium. However, oral magnesium can cause significant diarrhea, and IV magnesium may be better tolerated (although rapid infusion of magnesium can lead to rapid renal excretion).
Phosphate
Hyperphosphatemia is defined as serum phosphate levels > 5.0 mg/dL.
Causes: Alterations in phosphate homeostasis occur in postoperative patients due to the role of phosphate in energy metabolism and calcium regulation.
Hyperphosphatemia is frequently caused by renal failure and contributes to secondary hyperparathyroidism in this setting.
Postoperative hyperphosphatemia can also occur after parathyroid surgery and will correct itself without intervention if the parathyroidectomy was successful.
Treatment: Begin treatment with administration of fluids and phosphate binders.
Aluminum hydroxide should only be used for severe hyperphosphatemia (>7 mg/dL) and should be discontinued in favor of other binders (e.g., calcium carbonate, calcium acetate, or non-calcium-based binders, such as sevelamer).
Hemodialysis may be necessary for refractory hyperphosphatemia.
Hypophosphatemia is defined as serum phosphate levels below 2.5 mg/dL.
Causes: Postoperative patients can develop hypophosphatemia after major surgery due to increased energy expenditure (adenosine triphosphate [ATP] usage) during healing.
Hypophosphatemia is common after hepatic resections due to the significant metabolic demand of liver-tissue regeneration.
Other causes in surgical patients include malabsorption or poor intake.
Complications: Symptoms range from weakness to respiratory failure and leukocyte, erythrocyte, and platelet dysfunction.
Severe hypophosphatemia can develop in the setting of refeeding syndrome and can be life-threatening and result in cardiopulmonary failure due to the sudden increase in serum insulin levels and generation of ATP.
Treatment: Begin empiric phosphate repletion postoperatively after major surgery (e.g., hepatic resection).
Electrolyte Derangements and Associated Findings
| Electrolyte Imbalance | Definition | Associated Findings |
---
| Hyponatremia | Na <135.0 mEq/liter | Altered mental status, dry skin, seizures in severe cases Avoid rapid correction; can lead to central pontine myelinolysis |
| Hypernatremia | Na >145.0 mEq/liter | Thirst, altered mental status Seizures in severe cases |
| Hypokalemia | K <3.5 mEq/liter | Weakness, ileus, arrhythmias U wave on ECG |
| Hyperkalemia | K >5.0 mEq/liter | Weakness Peaked T waves, prolonged QRS on ECG |
| Hypocalcemia | Ca <8.6 mg/dL | Chvostek/Trousseau sign, perioral numbness/tingling |
| Hypercalcemia | Ca >10.2 mg/dL | Symptoms summarized as stones, bones, abdominal groans, thrones, and psychiatric overtones |
| Hypomagnesemia | Mg <1.4 mg/dL | Altered mental status, weakness, seizures in severe cases Torsades de pointes on ECG |
| Hypermagnesemia | Mg >2.3 mg/dL | Altered mental status (diminished attentiveness), diarrhea, seizures |
| Hypophosphatemia | Phos <2.5 mg/dL | Mental status changes, weakness |
| Hyperphosphatemia | Phos >4.5 mg/dL | Changes in calcium metabolism, usually asymptomatic |
Stones, kidney stones; bones, bone pain; groans, abdominal pain; thrones, polyuria and constipation; psychiatric overtones, altered mental status.
Research
Landmark clinical trials and other important studies
Effect of Intravenous Fluid Treatment with a Balanced Solution vs 0.9% Saline Solution on Mortality in Critically Ill Patients: The BaSICS Randomized Clinical Trial
Zampieri FG et al. JAMA 2021.
Use of a balanced solution compared with 0.9% saline solution did not significantly reduce 90-day mortality in critically ill patients requiring fluid challenges.
Balanced Crystalloids versus Saline in Noncritically Ill Adults
Self WH et al. N Engl J Med 2018.
Among noncritically ill adults treated with intravenous fluids in the emergency department, there was no difference in hospital-free days between treatment with balanced crystalloids and treatment with saline.
Read the NEJM Journal Watch Summary
Balanced Crystalloids versus Saline in the Intensive Care Unit: Study Protocol for a Cluster-Randomized, Multiple-Crossover Trial
Semler MW et al. Trials 2017.
This pragmatic trial is the largest and most comprehensive comparison to date of clinical outcomes with saline versus balanced crystalloids among critically ill adults.
A Comparison of Balanced and Unbalanced Crystalloid Solutions in Surgery Patient Outcomes
Kuca T et al. Anaesth Crit Care Pain Med 2017.
Reviews
The best overviews of the literature on this topic
Treatment of Hypercalcemia of Malignancy with Bisphosphonates
Berenson JR. Semin Oncol 2002.
Hypernatremia
Adrogué HJ and Madias NE. N Engl J Med 2000.
Serum Sodium
Ackerman GL. Clinical Methods: The History, Physical, and Laboratory Examinations 1990.
Guidelines
The current guidelines from the major specialty associations in the field
Intravenous Fluid Therapy in Adults in Hospital
National Institute for Health and Care Excellence (UK) 2017.
Part 8: Advanced Challenges in Resuscitation. Section 1: Life-Threatening Electrolyte Abnormalities
European Resuscitation Council. Resuscitation 2000. |
7996 | https://www.youtube.com/watch?v=GDUdUumkv0o | Common Free-Fall Pitfalls
Flipping Physics
174000 subscribers
471 likes
Description
50905 views
Posted: 5 Dec 2013
Yes, there are mistakes that many people make when it comes to free-fall acceleration problems. I dispel many misconceptions and explain both why people think they are true and why they actually aren't. Oh, and there are some special effects too!
0:00 Intro
0:14 Review of the Basics of Free-Fall
1:04 1st Misconception - The acceleration on the way up is positive
2:09 2nd Misconception - The initial velocity going upward is zero
2:45 3rd Misconception - A thrown ball will accelerate faster than a dropped ball
4:00 Reminder - Velocity at the top is zero
4:29 4th Misconception - The acceleration at the top is zero
6:36 Review
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42 comments
Transcript:
Intro Mr. P: Good morning. Today we are going to
understand the reality behind some common
mistakes or misconceptions students have about objects in free fall. Here we go! Bo: Hey guys. Billy: Hey Bo! Bobby: H-hi Bo. ♫ (lyrics) Flipping Physics ♫ Review of the Basics of Free-Fall Mr. P: Let's start with a quick review to make sure we're all on the same page. Bo, could you please remind me, how do we identify an object in free fall? Bo: An object in free fall
is one that isn't touching anything else and there can't be any air. Mr. P: True. I like to call it being in the
vacuum that you can breathe. And Billy, what do we know about an object that's in free fall on Earth? Billy: Little g, or the
acceleration due to gravity on Earth, is positive 9.81
meters per second squared. And the acceleration in the y direction is equal to negative g,
so the acceleration is negative 9.81 meters per second squared. Bobby: Or 9.81 meters
per second squared down. Mr. P: Great. And remind me, class. Is the acceleration due to gravity, little g, positive or negative? All: Positive. Mr. P: And this makes the acceleration in the y direction for an object in free fall positive or negative? All: Negative. 1st Misconception - The acceleration on the way up is positive Mr. P: Great. The first misconception is that if I throw this ball upward,
then when it is moving upward, because the velocity is
positive the acceleration would also be positive
and therefore it would be a positive 9.81 meters per second squared. Bobby, please explain
why this is incorrect. Bobby: Uh, well, we know
the ball will slow down as it moves upward,
therefore the direction of the acceleration needs to be opposite to the direction of the velocity. Therefore, because the velocity
is upward and positive, the acceleration must be
downward and negative. Mr. P: Exactly, Bobby. The ball slows down on the way up. So the acceleration is opposite to the direction of the velocity, therefore the acceleration
is down and negative. Who can explain to me what would happen if the acceleration were
positive on the way up? Bobby: Uh, the ball would
shoot upward like a rocket. Right? Mr. P: Yep.
(suspenseful sound effects) If the acceleration were positive, the ball would accelerate
upward like a rocket. Just like this. (explosion sound effect) (laughs) Billy: Wow! Bo: Woah. Bobby: Cool. Billy: Yeah, I don't
think that's possible. 2nd Misconception - The initial velocity going upward is zero Mr. P: Next. Sometimes students tell me
that the initial velocity for an object being thrown upward is zero. I love this, because then I get to show an object being thrown upward with an initial velocity of zero. Watch. (anticipatory sound effects) (Mr. P grunts in effort) (laughs) See, it won't go anywhere. Bobby: Yeah, that's not
going to go up very far. Mr. P. Right. An initial velocity of zero will not cause an object to move upward. In fact, the initial
velocity in the y direction must be positive in order
for an object to move upward. 3rd Misconception - A thrown ball will accelerate faster than a dropped ball Another misconception is
that if I drop a ball, and if I take a ball
and I throw it downward, that the ball that I threw downward will have an acceleration downward that is larger than if
I just drop the ball. Billy, please help dispel
this misconception. Billy: Well, we know the dropped ball and the ball thrown downward
will both have accelerations of negative 9.81 meters
per second squared, because they are both
objects in free fall. So I don't really see why people would think the thrown
one would accelerate more. Bo: Uh, Billy, can I give it a try? Billy: Sure, Bo, go ahead. Bo: I think it's because people often confuse velocity and acceleration. The ball that is thrown is moving faster, however, its velocity
is not changing faster. So it must be because people
see the ball moving faster they assume it must be
accelerating more quickly. Right, Mr. P? Mr. P: Yes, Bo, very nice. The thrown ball is moving faster, therefore people think it
is accelerating faster. However, it is not. As Billy pointed out, both objects are objects in free
fall, and therefore both have an acceleration in the y direction of negative 9.81 meters
per second squared. Reminder - Velocity at the top is zero OK, remind me, Billy, what is the velocity in the y direction for an object in free fall at the very top of its path. Right there. Billy: The velocity at the top
in the y direction is zero. Mr. P: And Bobby, please explain to me one reason we know why. Bobby: Um, on the way up
the velocity is positive, and on the way down the
velocity is negative, so at the top the velocity must be zero. Mr. P: Which leads us to
our last misconception. Class, what is the
acceleration in the y direction 4th Misconception - The acceleration at the top is zero of the ball at the very top of its path? Again, right there. All: Uh... Billy: Zero! Bobby: Negative 9.81
meters per second squared. Mr. P: Right, which is why
we need to talk about this. Class, when the ball is at the very top of its path, is it touching anything? All: No. Mr. P: Is it in the vacuum
that you can breathe? All: Yes. Mr. P: Then is it in free fall? All: Yes. Mr. P: Then its acceleration is... All: Negative 9.81 meters
per second squared. Billy: But it's velocity is zero, so its acceleration must also be zero. Mr. P: It is very common to assume that because the velocity at the top of the path is equal zero, then also the acceleration must be equal to zero. This is not true. To help us to understand why, let's start with the equation
definition for acceleration. Please, Billy? Billy: Acceleration equals change in velocity over change in time. Mr. P: I think it's safe to assume that time doesn't stop,
therefore the change in time is not going to be equal to zero. Therefore if the acceleration at the top is equal to zero, then
the change in velocity at the top must also be equal to zero. Therefore, the velocity would not change at the top if the acceleration at the top were equal to zero. So, Billy, explain to me what would happen if the acceleration at the top in the y direction were equal to zero. Billy: Oh, it would stop
in mid-air, and just float! Because if the velocity is zero at the top and the velocity isn't changing, then the velocity would continue to stay zero and the ball
would just float there! (dramatic sound effect) Mr. P: Right. It would look like this. (eerie sound effects) Eh? I know. Bo: Awesome. Bobby: Mr. P, that's cool. Billy: Wait, I thought
it wouldn't do that! Mr. P: Therefore, the
acceleration clearly isn't zero... Sorry. (claps hands) And it is an object in free fall, therefore the acceleration
in the y direction is negative 9.81 meters
per second squared. Any questions? Bobby: Nope, I'm good. Bo: No, I got it. Billy: I think I'm OK. Review Mr. P: So ends our common
free fall pitfalls lesson. Let's do a quick review. First off, for an object going up, the acceleration is still negative. For an object thrown
upward, the initial velocity in the y direction must
be greater than zero. For an object thrown downward, again, it's still in free
fall, so the acceleration is negative 9.81 meters
per second squared. Yes, it's moving faster, but
it's not accelerating more. The velocity at the top in the
y direction is equal to zero. And lastly, the acceleration at the top is still negative 9.81
meters per second squared, because it's an object in free
fall, regardless of the fact that the velocity at the
top is equal to zero. Thank you very much for
learning with me today, I enjoyed learning with you. |
7997 | https://www.youtube.com/watch?v=HXNyr2N3euc | Fleury's algorithm
OCLPhase2
4750 subscribers
613 likes
Description
81451 views
Posted: 27 Apr 2012
Video to accompany the open textbook Math in Society ( Part of the Washington Open Course Library Math&107 course.
24 comments
Transcript:
(male narrator)
So let's find an Euler circuit
on this graph. We can tell
from all the even degrees that it's gonna have
an Euler circuit. Now most of the time,
it's gonna be easiest to just guess and check
at a, uh...at an Euler circuit, but if we want a more
sort of formulaic approach here, we use something called
"Fleury's algorithm." The way it works is we pick
sort of a-a-a starting point. Uh...so we pick any old vertex
that we wanna start at. Uh...and...and-and we
choose one of the edges that is...that is leaving
our-our current vertex. And we can pick
any edge we want as long as it doesn't separate
or disconnect the graph. So let's say we pick A
as our starting point here, and the edge
that we choose is A to D. So A to D is not going
to disconnect the graph, so we'll go ahead and add
that in as our first step. Now over here in sort
of our duplicate copy here, we're gonna delete
that edge. So we're gonna delete that edge,
because...well, we feel like it. So...then...from D, we're
gonna look at these two edges and say, you know, which one
do I want to go to next? Now we'd
wanna be careful here. We would not want
to remove, uh... or choose edge D to C, because notice
if we choose D to C, it's gonna disconnect
the graph. A and C will no longer
be connected to the rest of the graph. So instead,
maybe let's choose D to E. We'll choose that
as our second edge, and we'll delete that. So now we're here at E, and we choose an edge
that we wanna follow. In this case, we wanna... because we've deleted
these edges, that's preventing backtracking. Uh...there's only one route
we can take now, and that is to...B. And then from B, there's only one route
we can follow, and that's to D. And then from D, there's only one route
we can follow to C. And from C, there's only one route
we can follow back to A. Uh...and there is
our Euler circuit. Now depending
upon how we had made our original choices
at the beginning, we could...we may have ended up
with different Euler circuits. For most graphs, there is
not just one Euler circuit. Uh...if there is
an Euler circuit, there's usually not just one
but multiple Euler circuits, depending
upon sort of the choice of, uh...what direction
you go first. |
7998 | https://docs.timefold.ai/timefold-solver/latest/optimization-algorithms/local-search | Local search
1. Overview
Local Search starts from an initial solution and evolves that single solution into a mostly better and better solution.
It uses a single search path of solutions, not a search tree.
At each solution in this path it evaluates a number of moves on the solution and applies the most suitable move to take the step to the next solution.
It does that for a high number of iterations until it’s terminated (usually because its time has run out).
Local Search acts a lot like a human planner: it uses a single search path and moves facts around to find a good feasible solution.
Therefore it’s pretty natural to implement.
Local Search needs to start from an initialized solution, therefore it’s usually required to configure a Construction Heuristic phase before it.
2. Local search concepts
2.1. Step by step
A step is the winning Move.
Local Search tries a number of moves on the current solution and picks the best accepted move as the step:
Figure 1. Decide the next step at step 0 (four queens example)
Because the move B0 to B3 has the highest score (-3), it is picked as the next step.
If multiple moves have the same highest score, one is picked randomly, in this case B0 to B3.
Note that C0 to C3 (not shown) could also have been picked because it also has the score -3.
The step is applied on the solution.
From that new solution, Local Search tries every move again, to decide the next step after that.
It continually does this in a loop, and we get something like this:
Figure 2. All steps (four queens example)
Notice that Local Search doesn’t use a search tree, but a search path.
The search path is highlighted by the green arrows.
At each step it tries all selected moves, but unless it’s the step, it doesn’t investigate that solution further.
This is one of the reasons why Local Search is very scalable.
Local Search solves the four queens problem by starting with the starting solution and make the following steps sequentially:
B0 to B3
D0 to D2
A0 to A1
A naive Local Search configuration solves the four queens problem in three steps,
by evaluating only 37 possible solutions (three steps with 12 moves each + one starting solution),
which is only a fraction of all 256 possible solutions.
It solves 16 queens in 31 steps, by evaluating only 7441 out of 18446744073709551616 possible solutions.
By using a Construction Heuristics phase first,
it’s even a lot more efficient.
2.2. Decide the next step
Local Search decides the next step with the aid of three configurable components:
A MoveSelector which selects the possible moves of the current solution.
See move and neighborhood selection.
An Acceptor which filters out unacceptable moves.
A Forager which gathers accepted moves and picks the next step from them.
The solver phase configuration looks like this:
<localSearch>
<unionMoveSelector>
...
</unionMoveSelector>
<acceptor>
...
</acceptor>
<forager>
...
</forager>
</localSearch>
xml
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In the example below, the MoveSelector generated the moves shown with the blue lines,
the Acceptor accepted all of them and the Forager picked the move B0 to B3.
Turn on trace logging to show the decision making in the log.
Because the last solution can degrade (such as in Tabu Search),
the Solver remembers the best solution it has encountered through the entire search path.
Each time the current solution is better than the last best solution,
the current solution is cloned
and referenced as the new best solution.
2.3. Acceptor
Use an Acceptor (together with a Forager) to activate Tabu Search, Simulated Annealing, Late Acceptance, … For each move it checks whether it is accepted or not.
By changing a few lines of configuration, you can easily switch from Tabu Search to Simulated Annealing or Late Acceptance and back.
You can implement your own Acceptor, but the built-in acceptors should suffice for most needs.
You can also combine multiple acceptors.
2.4. Forager
A Forager gathers all accepted moves and picks the move which is the next step.
Normally it picks the accepted move with the highest score.
If several accepted moves have the highest score, one is picked randomly to break the tie.
Breaking ties randomly leads to better results.
| | |
--- |
| | It is possible to disable breaking ties randomly by explicitly setting breakTieRandomly to false, but that’s almost never a good idea: If an earlier move is better than a later move with the same score, the score calculator should add an extra softer score level to score the first move as slightly better. Don’t rely on move selection order to enforce that. Random tie breaking does not affect reproducibility. |
2.4.1. Accepted count limit
When there are many possible moves, it becomes inefficient to evaluate all of them at every step.
To evaluate only a random subset of all the moves, use:
An acceptedCountLimit integer, which specifies how many accepted moves should be evaluated during each step. By default, all accepted moves are evaluated at every step.
<forager>
<acceptedCountLimit>1000</acceptedCountLimit>
</forager>
xml
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Unlike the N-queens problem, real world problems require the use of acceptedCountLimit.
Start from an acceptedCountLimit that takes a step in less than two seconds.
Turn on INFO logging to see the step times.
Use the Benchmarker to tweak the value.
| | |
--- |
| | With a low acceptedCountLimit (so a fast stepping algorithm), it is recommended to avoid using selectionOrder SHUFFLED because the shuffling generates a random number for every element in the selector, taking up a lot of time, but only a few elements are actually selected. |
2.4.2. Pick early type
A forager can pick a move early during a step, ignoring subsequent selected moves.
There are three pick early types for Local Search:
NEVER: A move is never picked early: all accepted moves are evaluated that the selection allows. This is the default.
<forager>
<pickEarlyType>NEVER</pickEarlyType>
</forager>
xml
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FIRST_BEST_SCORE_IMPROVING: Pick the first accepted move that improves the best score. If none improve the best score, it behaves exactly like the pickEarlyType NEVER.
<forager>
<pickEarlyType>FIRST_BEST_SCORE_IMPROVING</pickEarlyType>
</forager>
xml
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FIRST_LAST_STEP_SCORE_IMPROVING: Pick the first accepted move that improves the last step score. If none improve the last step score, it behaves exactly like the pickEarlyType NEVER.
<forager>
<pickEarlyType>FIRST_LAST_STEP_SCORE_IMPROVING</pickEarlyType>
</forager>
xml
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3. Hill climbing (simple local search)
3.1. Algorithm description
Hill Climbing tries all selected moves and then takes the best move, which is the move which leads to the solution with the highest score.
That best move is called the step move.
From that new solution, it again tries all selected moves and takes the best move and continues like that iteratively.
If multiple selected moves tie for the best move, one of them is randomly chosen as the best move.
Notice that once a queen has moved, it can be moved again later.
This is a good thing,
because in an NP-complete problem it’s impossible to predict what will be the optimal final value for a planning variable.
3.2. Stuck in local optima
Hill climbing always takes improving moves.
This may seem like a good thing, but it’s not: Hill Climbing can easily get stuck in a local optimum. This happens when it reaches a solution for which all the moves deteriorate the score.
Even if it picks one of those moves, the next step might go back to the original solution and which case chasing its own tail:
Improvements upon Hill Climbing (such as Tabu Search, Simulated Annealing and Late Acceptance) address the problem of being stuck in local optima.
Therefore, it’s recommended to never use Hill Climbing, unless you’re absolutely sure there are no local optima in your planning problem.
3.3. Configuration
Simplest configuration:
<localSearch>
<localSearchType>HILL_CLIMBING</localSearchType>
</localSearch>
xml
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Advanced configuration:
<localSearch>
...
<acceptor>
<acceptorType>HILL_CLIMBING</acceptorType>
</acceptor>
<forager>
<acceptedCountLimit>1</acceptedCountLimit>
</forager>
</localSearch>
xml
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4. Tabu search
4.1. Algorithm description
Tabu Search is a Local Search that maintains a tabu list to avoid getting stuck in local optima.
The tabu list holds recently used objects that are taboo to use for now.
Moves that involve an object in the tabu list, are not accepted.
The tabu list objects can be anything related to the move, such as the planning entity, planning value, move, solution, …
See example with entity tabu for four queens, so the queens are put in the tabu list:
4.2. Configuration
Simplest configuration:
<localSearch>
<localSearchType>TABU_SEARCH</localSearchType>
</localSearch>
xml
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When Tabu Search takes steps it creates one or more tabus.
For a number of steps, it does not accept a move if that move breaks tabu.
That number of steps is the tabu size.
Advanced configuration:
<localSearch>
...
<acceptor>
<entityTabuSize>7</entityTabuSize>
</acceptor>
<forager>
<acceptedCountLimit>1000</acceptedCountLimit>
</forager>
</localSearch>
xml
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| | |
--- |
| | A Tabu Search acceptor should be combined with a high acceptedCountLimit, such as 1000. |
Timefold Solver implements several tabu types:
Planning entity tabu (recommended) makes the planning entities of recent steps tabu.
For example, for school timetabling it makes the recently moved lessons tabu.
It’s recommended to start with this tabu type.
<acceptor>
<entityTabuSize>7</entityTabuSize>
</acceptor>
xml
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To avoid hard coding the tabu size, configure a tabu ratio, relative to the number of entities, for example 2%:
<acceptor>
<entityTabuRatio>0.02</entityTabuRatio>
</acceptor>
xml
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Planning value tabu makes the planning values of recent steps tabu.
For example, for school timetablig it makes the recently assigned timeslots tabu.
<acceptor>
<valueTabuSize>7</valueTabuSize>
</acceptor>
xml
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To avoid hard coding the tabu size, configure a tabu ratio, relative to the number of values, for example 2%:
<acceptor>
<valueTabuRatio>0.02</valueTabuRatio>
</acceptor>
xml
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Move tabu makes recent steps tabu. It does not accept a move equal to one of those steps.
<acceptor>
<moveTabuSize>7</moveTabuSize>
</acceptor>
xml
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| | |
--- |
| | When using move tabu with custom moves, make sure that the planning entities do not include planning variables in their hashCode methods. Failure to do so results in runtime exceptions being thrown due to the hashCode not being constant, as the entities have their values changed by the local search algorithm. |
Sometimes it’s useful to combine tabu types:
<acceptor>
<entityTabuSize>7</entityTabuSize>
<valueTabuSize>3</valueTabuSize>
</acceptor>
xml
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If the tabu size is too small, the solver can still get stuck in a local optimum.
On the other hand, if the tabu size is too large, the solver can be inefficient by bouncing off the walls.
Use the Benchmarker to fine tweak your configuration.
5. Simulated annealing
5.1. Algorithm description
Simulated Annealing evaluates only a few moves per step, so it steps quickly.
In the classic implementation, the first accepted move is the winning step.
A move is accepted if it doesn’t decrease the score or - in case it does decrease the score - it passes a random check.
The chance that a decreasing move passes the random check decreases relative to the size of the score decrement and the time the phase has been running (which is represented as the temperature).
Simulated Annealing does not always pick the move with the highest score, neither does it evaluate many moves per step.
At least at first.
Instead, it gives non improving moves also a chance to be picked, depending on its score and the time gradient of the Termination.
In the end, it gradually turns into Hill Climbing, only accepting improving moves.
5.2. Configuration
Start with a simulatedAnnealingStartingTemperature set to the maximum score delta a single move can cause.
Use the Benchmarker to tweak the value.
Advanced configuration:
<localSearch>
...
<acceptor>
<simulatedAnnealingStartingTemperature>2hard/100soft</simulatedAnnealingStartingTemperature>
</acceptor>
<forager>
<acceptedCountLimit>1</acceptedCountLimit>
</forager>
</localSearch>
xml
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Simulated Annealing should use a low acceptedCountLimit.
The classic algorithm uses an acceptedCountLimit of 1, but often 4 performs better.
Simulated Annealing can be combined with a tabu acceptor at the same time.
That gives Simulated Annealing salted with a bit of Tabu.
Use a lower tabu size than in a pure Tabu Search configuration.
<localSearch>
...
<acceptor>
<entityTabuSize>5</entityTabuSize>
<simulatedAnnealingStartingTemperature>2hard/100soft</simulatedAnnealingStartingTemperature>
</acceptor>
<forager>
<acceptedCountLimit>1</acceptedCountLimit>
</forager>
</localSearch>
xml
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6. Late acceptance
6.1. Algorithm description
Late Acceptance (also known as Late Acceptance Hill Climbing) also evaluates only a few moves per step.
A move is accepted if it does not decrease the score, or if it leads to a score that is at least the late score (which is the winning score of a fixed number of steps ago).
Scientific paper: The Late Acceptance Hill-Climbing Heuristic by Edmund K. Burke, Yuri Bykov (2012)
6.2. Configuration
Simplest configuration:
<localSearch>
<localSearchType>LATE_ACCEPTANCE</localSearchType>
</localSearch>
xml
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Late Acceptance accepts any move that has a score which is higher than the best score of a number of steps ago.
That number of steps is the lateAcceptanceSize.
Advanced configuration:
<localSearch>
...
<acceptor>
<lateAcceptanceSize>400</lateAcceptanceSize>
</acceptor>
<forager>
<acceptedCountLimit>1</acceptedCountLimit>
</forager>
</localSearch>
xml
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Late Acceptance should use a low acceptedCountLimit.
Late Acceptance can be combined with a tabu acceptor at the same time.
That gives Late Acceptance salted with a bit of Tabu.
Use a lower tabu size than in a pure Tabu Search configuration.
<localSearch>
...
<acceptor>
<entityTabuSize>5</entityTabuSize>
<lateAcceptanceSize>400</lateAcceptanceSize>
</acceptor>
<forager>
<acceptedCountLimit>1</acceptedCountLimit>
</forager>
</localSearch>
xml
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7. Diversified Late acceptance
7.1. Algorithm description
Diversified Late Acceptance is similar to Late Acceptance,
but it offers different acceptance and replacement strategies.
A move is accepted if its score matches the current solution score
or is better than the late score (which is the winning score of a fixed number of steps ago).
Diversified Late Acceptance was first proposed in
Diversified Late Acceptance Search by M. Namazi, C. Sanderson, M. A. H. Newton, M. M. A. Polash, and A. Sattar
7.2. Configuration
Simplest configuration:
<solver xmlns="
<enablePreviewFeature>DIVERSIFIED_LATE_ACCEPTANCE</enablePreviewFeature>
...
<localSearch>
<localSearchType>DIVERSIFIED_LATE_ACCEPTANCE</localSearchType>
</localSearch>
</solver>
xml
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The late elements list is updated as follows:
The current solution score is worse than the late score.
The current solution score is better than the late score and different from the previous one.
The size of the late elements list is typically smaller.
Advanced configuration:
...
<localSearch>
...
<acceptor>
<lateAcceptanceSize>5</lateAcceptanceSize>
</acceptor>
<forager>
<acceptedCountLimit>1</acceptedCountLimit>
</forager>
</localSearch>
xml
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| | |
--- |
| | The new acceptor is available as a preview feature It may be subject to change and must be enabled in the solver configuration by setting: <enablePreviewFeature>DIVERSIFIED_LATE_ACCEPTANCE</enablePreviewFeature> |
8. Great Deluge
8.1. Algorithm description
Great Deluge algorithm is similar to the Simulated Annealing algorithm, it evaluates only a few moves per steps,
so it steps quickly. The first accepted move is the winning step. A move is accepted only if it is not lower than
the score value (water level) that we are working with. It means Great Deluge is deterministic and opposite
of Simulated Annealing has no randomization in it. The water level is increased after every step either about the fixed value
or by percentual value.
A gradual increase in water level gives Great Deluge more time to escape from local maxima.
8.2. Configuration
Simplest configuration:
<localSearch>
<localSearchType>GREAT_DELUGE</localSearchType>
</localSearch>
xml
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Great Deluge takes as starting water level best score from construction heuristic and uses default rain speed ratio. Advanced configuration:
<localSearch>
...
<acceptor>
<greatDelugeWaterLevelIncrementRatio>0.00000005</greatDelugeWaterLevelIncrementRatio>
</acceptor>
<forager>
<acceptedCountLimit>1</acceptedCountLimit>
</forager>
</localSearch>
xml
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Timefold Solver implements two water level increment options:
If greatDelugeWaterLevelIncrementScore is set, the water level is increased by a constant value.
<acceptor>
<greatDelugeWaterLevelIncrementScore>10</greatDelugeWaterLevelIncrementScore>
</acceptor>
xml
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To avoid hard coding the water level increment, configure a greatDelugeWaterLevelIncrementRatio (recommended) when the water level is increased by percentual value, so there is no need to know the size of the problem or value of a scoring function.
<acceptor>
<greatDelugeWaterLevelIncrementRatio>0.00000005</greatDelugeWaterLevelIncrementRatio>
</acceptor>
xml
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The algorithm takes as starting value the best score from the construction heuristic.
Use the Benchmarker to fine-tune tweak your configuration.
9. Step counting hill climbing
9.1. Algorithm description
Step Counting Hill Climbing also evaluates only a few moves per step.
For a number of steps, it keeps the step score as a threshold.
A move is accepted if it does not decrease the score, or if it leads to a score that is at least the threshold score.
Scientific paper: An initial study of a novel Step Counting Hill Climbing heuristic applied to timetabling problems by Yuri Bykov, Sanja Petrovic (2013)
9.2. Configuration
Step Counting Hill Climbing accepts any move that has a score which is higher than a threshold score.
Every number of steps (specified by stepCountingHillClimbingSize), the threshold score is set to the step score.
<localSearch>
...
<acceptor>
<stepCountingHillClimbingSize>400</stepCountingHillClimbingSize>
</acceptor>
<forager>
<acceptedCountLimit>1</acceptedCountLimit>
</forager>
</localSearch>
xml
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Step Counting Hill Climbing should use a low acceptedCountLimit.
Step Counting Hill Climbing can be combined with a tabu acceptor at the same time, similar as shown in the Late Acceptance section.
10. Strategic oscillation
10.1. Algorithm description
Strategic Oscillation is an add-on, which works especially well with Tabu Search.
Instead of picking the accepted move with the highest score, it employs a different mechanism: If there’s an improving move, it picks it.
If there’s no improving move however, it prefers moves which improve a softer score level, over moves which break a harder score level less.
10.2. Configuration
Configure a finalistPodiumType, such as in a Tabu Search configuration:
<localSearch>
...
<acceptor>
<entityTabuSize>7</entityTabuSize>
</acceptor>
<forager>
<acceptedCountLimit>1000</acceptedCountLimit>
<finalistPodiumType>STRATEGIC_OSCILLATION</finalistPodiumType>
</forager>
</localSearch>
xml
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The following finalistPodiumTypes are supported:
HIGHEST_SCORE (default): Pick the accepted move with the highest score.
STRATEGIC_OSCILLATION: Alias for the default strategic oscillation variant.
STRATEGIC_OSCILLATION_BY_LEVEL: If there is an accepted improving move, pick it. If no such move exists, prefer an accepted move which improves a softer score level over one that doesn’t (even if it has a better harder score level). A move is improving if it’s better than the last completed step score.
STRATEGIC_OSCILLATION_BY_LEVEL_ON_BEST_SCORE: Like STRATEGIC_OSCILLATION_BY_LEVEL, but define improving as better than the best score (instead of the last completed step score).
11. Variable neighborhood descent
11.1. Algorithm description
Variable Neighborhood Descent iteratively tries multiple move selectors
in original order (depleting each selector entirely before trying the next one),
picking the first improving move (which also resets the iterator back to the first move selector).
| | |
--- |
| | Despite that VND has a name that ends with descent (from the research papers), the implementation will ascend to a higher score (which is a better score). |
11.2. Configuration
Simplest configuration:
<localSearch>
<localSearchType>VARIABLE_NEIGHBORHOOD_DESCENT</localSearchType>
</localSearch>
xml
CopyCopied!
Advanced configuration:
<localSearch>
<unionMoveSelector>
<selectionOrder>ORIGINAL</selectionOrder>
<changeMoveSelector/>
<swapMoveSelector/>
...
</unionMoveSelector>
<acceptor>
<acceptorType>HILL_CLIMBING</acceptorType>
</acceptor>
<forager>
<pickEarlyType>FIRST_LAST_STEP_SCORE_IMPROVING</pickEarlyType>
</forager>
</localSearch>
xml
CopyCopied!
Variable Neighborhood Descent doesn’t scale well,
but it is useful in some use cases with a very erratic score landscape. |
7999 | https://math.libretexts.org/Bookshelves/Applied_Mathematics/Developmental_Math_(NROC)/02%3A_Fractions_and_Mixed_Numbers/2.01%3A_Introduction_to_Fractions_and_Mixed_Numbers/2.1.03%3A_Factors_and_Primes | 2.1.3: Factors and Primes - Mathematics LibreTexts
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2.1: Introduction to Fractions and Mixed Numbers
2: Fractions and Mixed Numbers
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Sun, 05 Sep 2021 07:00:08 GMT
2.1.3: Factors and Primes
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Contents
1. Home
2. Bookshelves
3. Applied Mathematics
4. Developmental Math (NROC)
5. 2: Fractions and Mixed Numbers
6. 2.1: Introduction to Fractions and Mixed Numbers
7. 2.1.3: Factors and Primes
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Developmental Math (NROC)
Front Matter
1: Whole Numbers
2: Fractions and Mixed Numbers
3: Decimals
4: Ratios, Rates, and Proportions
5: Percents
6: Measurement
7: Geometry
8: Concepts in Statistics
9: Real Numbers
10: Solving Equations and Inequalities
11: Exponents and Polynomials
12: Factoring
13: Graphing
14: Systems of Equations and Inequalities
15: Rational Expressions
16: Radical Expressions and Quadratic Equations
17: Functions
18: Exponential and Logarithmic Functions
19: Trigonometry
Back Matter
2.1.3: Factors and Primes
Last updated Sep 5, 2021
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2.1.2: Proper and Improper Fractions
2.1.4: Simplifying Fractions
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Table of contents
1. Learning Objectives
2. Introduction
3. Tests of Divisibility
1. Other useful divisibility tests:
2. Exercise
Factoring Numbers
Exercise
Prime Factorization
Exercise
Summary
Learning Objectives
Recognize (by using the divisibility rule) if a number is divisible by 2, 3, 4, 5, 6, 9, or 10.
Find the factors of a number.
Determine whether a number is prime, composite, or neither.
Find the prime factorization of a number.
Introduction
Natural numbers, also called
counting numbers
(1, 2, 3, and so on), can be expressed as a
product
of their factors. When working with a fraction, you often need to make the fraction as simple as possible. This means that the numerator and the denominator have no common
factors
other than . It will help to find
factors
, so that later you can simplify and compare
fractions
.
Tests of Divisibility
When a natural number is expressed as a
product
of two other
natural numbers
, those other numbers are
factors
of the original number. For example, two
factors
of 12 are 3 and 4, because 3⋅4=12.
When one number can be divided by another number with no remainder, we say the first number is divisible by the other number. For example, 20 is
divisible
by 4(20÷4=5). If a number is
divisible
by another number, it is also a multiple of that number. For example, 20 is
divisible
by 4, so 20 is a
multiple
of 4.
Divisibility tests are rules that let you quickly tell if one number is
divisible
by another. There are many
divisibility tests
. Here are some of the most useful and easy to remember:
A number is divisible by 2 if the last (ones) digit is divisible by 2. That is, the last digit is 0, 2, 4, 6, or 8. (We then say the number is an even number.) For example, in the number 236, the last digit is 6. Since 6 is divisible by 2(6÷2=3), 236 is divisible by 2.
A number is divisible by 3 if the sum of all the digits is divisible by 3. For example, the sum of the digits of 411 is 4+1+1=6. Since 6 is divisible by 3(6÷3=2), 411 is divisible by 3.
A number is divisible by 5 if the last digit is 0 or 5. For example, 275 and 1,340 are divisible by 5 because the last digits are 5 and 0.
A number is divisible by 10 if the last digit is 0. For example, 520 is divisible by 10 (last digit is 0).
Other useful
divisibility tests
:
: A number is divisible by 4 if the last two digits are divisible by 4.
: A number is divisible by 6 if it is divisible by both 2 and 3.
: A number is divisible by 9 if the sum of its digits is divisible by 9
Here is a summary of the most commonly used divisibility rules.
A number is
divisible
byExample
2 if the last
digit
is even (0, 2, 4, 6, 8).426 yes
273 no
3 if the
sum
of the digits is
divisible
by 3.642 yes
(6+4+2=12, 12 is
divisible
by 3)
721 no
(7+2+1=10, 10 is not
divisible
by 3)
4 if the last two digits form a number that is
divisible
by 4.164 yes
(64 is
divisible
by 4)
135 no
(35 is not
divisible
by 4)
5 if the last
digit
is 0 or 5.685 yes
432 no
6 if the number is
divisible
by 2 and 3.324 yes
(it is even and 3+2+4=9)
411 no
(although
divisible
by 3, it is not even)
9 if the
sum
of the digits is
divisible
by 9.279 yes
(2+7+9=18)
512 no
(5+1+2=8)
10 if the last
digit
is a 0.620 yes.
238 no
If you need to check for divisibility of a number without a rule, divide (either using a calculator or by hand). If the result is a number without any fractional part or remainder, then the number is
divisible
by the divisor. If you forget a rule, you can also use this strategy.
Exercise
Determine whether 522 is
divisible
by 2, 3, 4, 5, 6, 9, or 10.
1. 2 and 3 only
2. 4 only
3. 2, 3, 6, and 9 only
4. 4, 5, and 10 only
Answer
1. Incorrect. Although 522 is divisible by 2 (the last digit is even) and 3 (5+2+2=9, which is a multiple of 3), it is also divisible by 6 and 9. The correct answer is 2, 3, 6, and 9 only.
2. Incorrect. The last two digits (22) are not divisible by 4, so 522 is not divisible by 4. 522 has a last digit divisible by 2, so 522 is divisible by 2. The sum of the digits is divisible by 3 (5+2+2=9) and by 9, so 522 is divisible by 3 and 9. Since 522 is divisible by 2 and 3, it is divisible by 6. Since the last digit is not 0 or 5, 522 is not divisible by 5 or 10. The correct answer is 2, 3, 6, and 9 only.
3. Correct. 522 is divisible by 2 (the last digit is even) and 3 (5+2+2=9, which is a multiple of 3). Since it is divisible by 2 and 3, it is also divisible by 6. Also, the sum of the digits is divisible by 9, so 522 is divisible by 9. Since the last digit is not 0 or 5, 522 is not divisible by 5 or 10. The number formed by the last two digits, 22, is not divisible by 4, so 522 is not divisible by 4.
4. Incorrect. The last two digits are not divisible by 4, so 522 is not divisible by 4. The last digit is not 0 or 5 so 522 is not divisible by 5. The last digit is not 0, so 522 is not divisible by 10. However, 522 is divisible by 2 (the last digit is even) and 3 (5+2+2=9, which is a multiple of 3). Since it is divisible by 2 and 3, it is also divisible by 6. Also, the sum of the digits is divisible by 9, so 522 is divisible by 9. The correct answer is 2, 3, 6, and 9 only.
Factoring
Numbers
To find all the
factors
of a number, you need to find all numbers that can divide into the original number without a remainder. The divisibility rules from above will be extremely useful!
Suppose you need to find the
factors
of 30. Since 30 is a number you are familiar with, and small enough, you should know many of the
factors
without applying any rules. You can start by listing the
factors
as they come to mind:
2⋅15
3⋅10
5⋅6
Is that it? Not quite. All
natural numbers
except 1 also have 1 and the number itself as
factors
:
1⋅30
The
factors
of 30 are 1,2,3,5,6,10,15, and 30.
When you find one
factor
of a number, you can easily find another
factor
: it is the
quotient
using that first
factor
as the
divisor
. For example, once you know 2 is a
factor
of 30, then 30÷2 is another
factor
. A pair of
factors
whose
product
is a given number is a factor
pair of the original number. So, 2 and 15 are a
factor
pair for 30.
What do you do if you need to
factor
a greater number and you can’t easily see its
factors
? That’s where the divisibility rules will come in quite handy. Here is a general
set
of steps that you may follow:
1. Begin with and check the numbers sequentially, using divisibility rules or division.
2. When you find a factor , find the other number in the factor pair.
3. Keep checking sequentially, until you reach the second number in the last factor pair you found, or until the result of dividing gives a number less than the divisor .
Note that you can stop checking when the result of dividing is less than the number you’re checking. This means that you have already found all
factor
pairs, and continuing the process would find pairs that have been previously found.
If a number has exactly two
factors
, 1 and itself, the number is a prime number. A number that has more
factors
than itself and 1 is called a composite number. The number 1 is considered neither prime nor composite, as its only
factor
is 1. To determine whether a number is prime, composite, or neither, check
factors
. Here are some examples.
NumberComposite, Prime, or Neither?Explanation
1 Neither 1 does not have two different
factors
, so it is not prime.
2 Prime 2 has only the
factors
2 and 1.
3 Prime 3 has only the
factors
3 and 1.
4 Composite 4 has more than two
factors
: 1, 2, and 4, so it is composite.
5, 7, 11, 13 Prime Each number has only two
factors
: 1 and itself.
6, 8, 9, 10, 50, 63 Composite Each number has more than two
factors
.
Exercise
Find all the
factors
of 48.Answer
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Prime Factorization
A
composite number
written as a
product
of only prime numbers is called the prime factorization of the number. One way to find the
prime factorization
of a number is to begin with the prime numbers 2, 3, 5, 7, 11 and so on, and determine whether the number is
divisible
by the primes.
For example, if you want to find the
prime factorization
of 20, start by checking if 20 is
divisible
by 2. Yes, 2⋅10=20.
Then
factor
10, which is also
divisible
by 2(2⋅5=10).
Both of those
factors
are prime, so you can stop. The
prime factorization
of 20 is 2⋅2⋅5, which you can write using exponential notation as 2 2⋅5.
One way to find the
prime factorization
of a number is to use successive divisions.
10 2\longdiv 20 Divide 20 by 2 to get 10. 2 is being used because it is a
prime number
and a
factor
of 20. You could also have started with 5.
5 2\longdiv 10 2\longdiv 20 Then divide 10 by 2 to get 5.
2⋅2⋅5 Multiplying these divisors forms the
prime factorization
of 20.
To help you organize the
factoring
process, you can create a factor tree. This is a diagram that shows a
factor
pair for a
composite number
. Then, each
factor
that isn’t prime is also shown as a
factor
pair. You can continue showing
factor
pairs for composite
factors
, until you have only prime
factors
. When a
prime number
is found as a
factor
, circle it so you can find it more easily later.
Written using
exponential notation
, the
prime factorization
of 20 is again 2 2⋅5.
Notice that you don’t have to start checking the number using divisibility of prime numbers. You can
factor
20 to 4⋅5, and then
factor
4 to 2⋅2, giving the same
prime factorization
: 2⋅2⋅5.
Now look at a more complicated factorization.
Notice that there are two different trees, but they both produce the same result: five 2s and one 3. Every number will only have one, unique
prime factorization
. You can use any
sets
of
factor
pairs you wish, as long as you keep
factoring
composite numbers.
When you rewrite the
prime factorization
of 96(2⋅2⋅2⋅2⋅2⋅3) in
exponential notation
, the five 2s can be written as 2 5. So, 96=2 5⋅3.
Exercise
When finding the
prime factorization
of 72, Marie began a
tree diagram
using the two
factors
9 and 8. Which of the following statements are true?
1. Marie started the diagram incorrectly and should have started the tree diagram using the factors 2 and 36.
2. Marie’s next set of factor pairs could be 3, 3 and 2, 4.
3. Marie’s next set of factor pairs could be 3, 3 and 9, 8.
4. Marie didn’t have to use a tree diagram .
1 only
2 only
3 and 4 only
2 and 4 only
Answer
1. Incorrect. Marie could have started her tree diagram with the factors 2 and 36, but she does not have to start with those factors . Starting with 9 and 8 is fine. The correct answer is D.
2. Incorrect. Yes, Marie’s next set of factor pairs could contain 3, 3, and 2, 4, but statement 4 is also correct. The correct answer is D.
3. Incorrect. Statement 4 is correct: Marie does not have to use a tree diagram , but 3 is not true. When creating a prime factorization , the factors are not combined with the previous composite. The correct answer is D.
4. Correct. Marie’s next set of factor pairs could read 3, 3, and 2, 4, as 3⋅3 is a factorization of 9 and 2⋅4 is a factorization of 8. Marie could also find the prime factorization by using successive divisions.
Summary
Finding the
factors
of a natural number means that you find all the possible numbers that will divide into the given number without a remainder. There are many rules of divisibility to help you to find
factors
more quickly. A
prime number
is a number that has exactly two
factors
. A
composite number
is a number that has more than two
factors
. The
prime factorization
of a number is the
product
of the number’s prime
factors
.
This page titled 2.1.3: Factors and Primes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by The NROC Project via source content that was edited to the style and standards of the LibreTexts platform.
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